Application of Structural System in Building Design by Archineer

SA BAH SH AWK AT RICHA RD SCHLESIN G ER

Application of Structural System in Building Design

Sabah Shawkat © S ABAH S H AWKAT R I C H AR D S C H L E SI N G E R

Sabah Shawkat ©

Application of Structural System in Building Design

Sabah Shawkat I Richard Schlesinger

Reviewer: Cover Design: Editor: Software Support: Publisher: Printed and Bound:

Prof. Dipl. Ing. Ján Hudák, PhD. Peter Nosáľ Sabah Shawkat I Richard Schlesinger asc. Applied Software Consultants, s.r.o., Bratislava, Slovakia Tribun EU, s.r.o., Brno, Czech Republic Tribun EU, s.r.o., Brno, Czech Republic

Sabah Shawkat © All rights reserved. No part of this book may be reprinted, or reproduced or utilized in any form or by any electronic, mechanical or other means, including photocopying, without permission in writing from the author.

Application of Structural System in Building Design ©

Sabah Shawkat I Richard Schlesinger 1. Edition, Tribun EU, s.r.o. Brno, Czech republic 2020 ISBN 978-80-263-1561-2

Sabah Shawkat © Sabah Shawkat

Richard Schlesinger

Sabah Shawkat is a structural designer, specializing in lightweight structures, such as tensile structures, tensile integrity structures, grid shells and reciprocal frames. He focuses on transforming these beautiful structures into design components such as chairs, tables, illuminated lamps or hammocks for interiors, gardens or public spaces.

Richard Schlesinger works as an assistant professor at the Engineering Room of the Academy of Fine Arts and Design in Bratislava. He defended his dissertation on design and technological procedures used in monument protection at the Department of Architecture, Faculty of Civil Engineering of the Slovak University of Technology.

Sabah Shawkat is also a passionate expert in traditional fibre reinforced and prestressed structures. He has published numerous articles in professional journals and has written several books.

Richard Schlesinger teaches students of architecture several structural engineering subjects such as Introduction to Structural Engineering and Construction in Architecture. Moreover, he regularly organizes workshops and exhibitions of student projects and construction models. He is also actively involved in projecting, building construction and reconstructions as well as modernisations of buildings.

He teaches students of architecture and building engineering. Moreover, he regularly organizes workshops for students and exhibitions of his own as well as student projects and construction models. He is also actively involved in projecting and building constructions as well as reconstructions and modernizations of buildings.

R C - F R C - H P C I R E I N F O R C E D C O N C R E T E , F I B RE REINFORCED C O N C R E T E A N D H I G H P E R F O R M A N C E C O N C RETE

TRADITIO N A L C O N C R E T E

CONTENTS

01 02 03 04 05 06 07 08

TABLE OF

INTRODU C T I O N

Sabah Shawkat © PRESTRE S S C O N C R E T E PRECAST C O N C R E T E

INVESTIG AT I O N O F R E I N F O R C E D C O N C R E T E

R E V I TA L I Z AT I O N A M O D E R N I S AT I O N O F B U I LDINGS – ” P E N TA G O N ” ( B R AT I S L A V A ) R E N O V AT I O N A N D E X T E N S I O N O F FA M I LY H OUSE

REALISAT I O N P R O J E C T S

1-40

R C - F R C - H P C I R E I N F O R C E D C O N C R E T E , FIBRE REINFORCED C O N C R E T E A N D H I G H P E R F O R M A N C E C O NCRETE

Author : Sabah Shawkat

41-212

AUTHORS

Author : Sabah Shawkat

TABLE OF

INTRO D U C T I O N

TRAD I T I O N A L C O N C R E T E Author : Sabah Shawkat

Sabah Shawkat ©

213-318

PRES T R E S S C O N C R E T E

319-364

PREC A S T C O N C R E T E

365-402

INVES T I G AT I O N O F R E I N F O R C E D C O N C RETE

403-420

R E V I TA L I Z AT I O N A M O D E R N I S AT I O N O F B UILDINGS– ” P E N TA G O N ” ( B R AT I S L A V A )

Author : Sabah Shawkat Author : Sabah Shawkat

Author : Sabah Shawkat

Head of Project : Sabah Shawkat I Assistant : Richard Schlesinger Students : Lea Debnárová I Silvia Galová I Petra Garajová I Rúth Sýkorová

421-462 463-498

R E N OV AT I O N A N D E X T E N S I O N O F FA M I LY HOUSE Head of Project : Sabah Shawkat I Assistant : Richard Schlesinger Students : Patrik Olejňák I Rebeka Hauskrechtová I Silvia Galová I Vanesa Rybárová I Eva Kvaššayová Chris Varga I Karin Brániková I Viliam Jankovič I Lea Debnárová I Petra Garajová I Peter Galdík Jana Arnoulová I Monika Studničná

REAL I S AT I O N P R O J E C T S

Author : Sabah Shawkat, Richard Schlesinger

Sabah Shawkat ©

Introduction This textbook is designed to teach students but it can serve as a reference for practising engineering or researchers as well. The text offers a simple, comprehensive, and methodical presentation of the basic concepts in the analysis of reinforced concrete, fibre reinforced concrete and high performance concrete. Many worked examples are included and design aids in the form of tables are presented. The major quantities which are considered in the design calculations, namely the loads, dimensions, and material properties, are subject to varying degrees of uncertainty and randomness. Further, there are idealizations and simplifying assumptions used in the theories of structural analysis and design. There are also several other variable and often unforeseen factors that influence the prediction of ultimate strength and performance of a structure, such as construction methods, workmanship and quality control, probable service life of the structure, possible future change of use, frequency of loadings, etc.

Primary considerations in structural design are safety and serviceability.Safety requires that the structure remains without damage under the normal expected loads. Serviceability requires that under the expected loads the structure performs satisfactorily with regard to its intended use, without discomfort to the user due to excessive deflection, permanent deformations, cracking etc.This textbook contains design aides such methods as tables and diagrams, developed by authors which enable very simple and rapid determination of reinforcement in a given variable concrete cross-section subjected to bending moment or bending moment with axial force.Design rules of this textbook also are based on the concept of Concrete Structures EuroDesign.

Sabah Shawkat © Reinforced concrete structures (RC, FRC, HPFRC), and structural components are designed to have sufficient strength and stability to withstand the effects of factored loads, thereby satisfying the safety requirements. The design for serviceability limits is made at specified (service) loads (generally, the design is first made for strength and the serviceability limits are checked).

Kindly I ask readers of this textbook who have questions, suggestions for improvements, or who find errors, to write to me. I thank you in advance for taking the time and interest to do so. My sincere thanks to my friend Jan Hudak who is the rewire of this book. Finally, I am grateful to my wife Ivana for her help, constructive criticism, patience and encouragement that have made this project possible. The authors dedicate this book to Czech academic city in Erbil. Sabah Shawkat

Bratislava 02/2020

Sabah Shawkat © 01. RC- FRC -HPC I reinforced concrete, fibre reinforced concrete and high performance concrete

In this chapter we will deal with the characteristics and properties of concrete historically, where we will gradually talk about improving the strength of concrete over time. We start with ordinary concrete such as RC and then fiber concrete and high strength concrete.

In some parts of China and Japan, clay walls build around 2000 BC were reinforced by thin strips of bamboos. Thin sheet cement products reinforced with asbestos fibres have been in production by the Hatcheck process since the early 1900s.

The maximum design strength of reinforced concrete has changed its value over the years. In the 1950’s concrete with a maximum compressive strength of 35 MPa was obtained without the use of any chemical and mineral admixtures. In the 1960’s commercial usage of 50 MPa concrete was achieved. It was connected with developing superplasticizers – effective chemical admixtures to decrease the water content in concrete while maintaining the workability of the concrete. In the 1970’s, the combined use of superplasticizers and ultra-fine materials such as silica fume, finely ground blast furnace slag or anhydrous gypsum based additives were studied and has been applied to concrete structures. Currently 100 to 120 MPa is considered as the upper limit for concrete strength used in the construction industry.

Concrete reinforced with steel wires and fibres can be traced back to the patents of Joseph Lambot in 1847 and A. Berard in 1874. Since then, an increasing variety of fibres, including steel, polymeric (polypropylene, polyethylene, polyvinyl alcohol, aramid, acrylic, nylon, etc.) and natural (asbestos, cellulose, sisal, etc.), as well as glass and carbon fibres have been applied to concrete, mortar or cement reinforcement in continuous and discontinuous forms, employing a variety of processes. Most of these Fibre Reinforced Concrete (FRC) are used in non-structural applications or in secondary elements, meaning that the fibres are employed to serve function such as minimizing shrinkage cracking and limit crack widths due to mechanical loading. Structural use of FRC is scarce, and attempts to use fibre reinforcement as structural reinforcement has so far been concentrated on replacement of shear reinforcing stirrups in structural members such as beams as well as replacement of complicated reinforcement arrangements in areas where concentrated loading is applied to the structure.

Sabah Shawkat ©

So it has increased the quality of concrete throughout history, but the problem of cracking and their number on the constructions was still a big problem, it was necessary to check the condition well and honestly and therefore research in the world was concerned with the production of steel fibers and their types to it was possible to analyze in detail the formation of cracks and their opening. The use of fibre reinforcement is not a particularly recent idea. Historically, Egyptians around 1500 BC employed straws to reinforced mud bricks which otherwise would become very brittle upon drying.

Concrete is a mixture of cement, aggregate and water. The addition of water to the

The tensile strength of a bar, is the stress, calculated as force divided by the original

concrete mix starts a chemical reaction called hydration that results in the binding of the sand

area of the bar, which causes fracture of the bar in tension. The yield stress, is the stress at the

and aggregate to produce a sandstone conglomerate. Concrete is placed in forms prior to

yield point, the yield point is the point at which plastic flow of the material begins to occur.

hardening. As concrete dries and cures, fine cracks are produced from thermal effects caused

Below the yield point the material is elastic and is able to recover its original shape when the

from shrinkage. Concrete reaches about ninety percent of its full strength in 28 days. Other

load is removed. Above this point the material is said to be plastic and undergoes

materials called admixtures are added to concrete to improve strength, workability and frost

a permanent deformation, which remains after the load is removed.

resistance, and to reduce shrinking, cracking and permeability.

The tensile strength of concrete is neglected for the strength of reinforced and

Admixtures are ingredients added to the concrete mix that alter and/or improve its’ properties

prestressed concrete structures, in general, it is an important characteristic for the

Cement and water factor.

development of cracking and therefore, for the prediction of deformations and the durability

Cement and Water factors for Concrete Concrete: Strength Non-Air-Entrained Max. Water Min. 28 Day Comp. Min. Cement Cement Ratio Str. kg/m3 MPa 35 375 0.45 30 325 0.55 25 280 0.65 25 300 *

of concrete. Other characteristics such as bond and development length of reinforcement and Air-Entrained Min. Cement Max. Water kg/m3 Cement Ratio 385 340 290 310

0.40 0.50 0.55 *

the concrete contribution to the shear and torsion capacity are closely related to the tensile strength of concrete. Bond Strength is the measure of effective grip between the concrete and the

Sabah Shawkat ©

embedded-steel bar. The design theory of reinforced-concrete beam is based on the assumption that a bond develops between the reinforcement and the concrete that prevents relative movement between them as the load is applied. How much bond strength develops depends largely on the area of contact between the two materials. Because of their superior

The most important mechanical properties of bulding materials are their weight,

tensile strength, yield strength, shear strength, compressive strength, ductility, toughness,

bonding value, bars having a very rough surface (deformed bars or rebars) have replaced plain bars as steel reinforcement

impact resistance, fatigue resistance, elasticity and creep under load. Other important

The bond behavior of a reinforcing bar and the surrounding concrete has a decisive

mechanical properties are their strain capacity and their resistance to rain and moisture

importance regarding the bearing capacity and the serviceability of reinforced concrete

penetration. Some properties, for example the compressive strength of concrete, depend on

members. This knowledge is an indispensable requirement to give design rules for anchorage

the way the test is done and may reflect other, more fundamental properties.

and lap lengths of reinforcing bars, for the calculation of deflections taking into account the

The concrete's shear strength is about one-third the unit compressive strength, whereas, tensile strength is less than one-half the shear strength. The failure of a concrete slab subjected to a downward concentrated load is due to diagonal tension. However, web reinforcement can prevent beams from failing in diagonal tension.

tension-stiffening effect, for the control of crack width, and, thus, the necessary minimum amount of reinforcement. In practice many materials exhibit plastic behaviour, in which permanent deformations occur, as well as elastic behaviour. The modulus of elasticity of a material is the

The concrete's tensile strength is such a small percentage of the compressive strength

ratio of stress to strain and is low if the material has a large stretch under load. For a linear-

that it is ignored in calculations for reinforced-concrete beam. Instead, horizontal steel bars

elastic material, the modulus of elasticity is constant up to a point just below the yield point

well embedded in the tension area provide tensile resistance.

but, above this, large deformations or strains may occur with only small or no increases of stress. The modulus of elasticity is needed to calculate deflection, although a slightly

Properties of Materials

different value of the modulus may be used to calculate deflection due to bending to that used to calculate the extension of a bar in tension, for example. Concrete has a high ratio of compressive strength to bulk cost, so that it is particularly suitable for use in walls and columns. Unreinforced concrete is able to accept modest amounts of tension, up to aproximately a tenth of its compressive strength. This property is

Tensile strength for various ND concrete grades [MPa] Concrete grade C12 C20 C30 C40 C50 12 20 30 40 50 fck 1,6 2,2 2,9 3,5 4,1 fctm 1,1 1,5 2,0 2,5 2,9 fck,min 2,0 2,9 3,8 4,6 5,3 fck,max

C60 60 4,6 3,2 6,0

C70 70 5,1 3,6 6,6

C80 80 5,6 3,9 7,3

crucial, enabling concrete to accept some shear stress without cracking. Compressive Strength According to CEB/FIB Model Code (1993) the characteristic strength fck is determined by uniaxial compressive test of the concretes at an age of 28 days on cylindersf 150x300 mm. The highest grade is 80 MPa for ND concrete. The mean strength fcm is estimated from the characteristic strength fck. f cm

f ck  f CEB/FIB MC 90 - The examples of the stress-strain curve in compression.

Sabah Shawkat ©

where f = 8 MPa

Modulus of Elasticity Values of modulus of elasticity for ND concrete can be estimated from the

Tensile Strength

characteristic strength.

The tensile strength is determined in accordance with RILEM CPC 7. In case of the

absence more precise data, the values of the upper and the lower characteristic tensile strength can be obtained from characteristic compressive strength fck as follows:

 f ck    f cko 

f ctkomin 

f ctkmax

Eci

 f cm 

Eco 

  f cmo 



10000 f cm

3

measurement of Eci are recommended. Where only an elastic analysis of a concrete structure is carried out, a reduced modulus of elasticity Ec should be used in order to account for the

For the mean axial tensile strength fctm the following equation is recommended: 2

 f ck    f cko 

 fck  f 3

To take full account of differences in aggregate stiffness or modulus, direct

Where fcko = 10 MPa, fctko,min = 0.95 MPa, fctko,max = 1.85 MPa

f ctm

10000

 f ck  f ctkomax    f cko 

If the actual compressive strength of concrete at an age of 28 days fcm is known, Eci can be estimated from

f ctkmin

Eci

 f ck  f  Eco    f cmo 

initial plastic strain

f ctkom 

0.85Eci

The values of tangent elastic modulus Eci and a reduced modulus of elasticity Ec for various ND concrete grades are presented in.

Where fctko,m = 1.40 Mpa The values of the tensile strength for various ND concrete grades are indicated in Table

Properties of Materials

ACI 318-95 (1995)

Numerical values of Ec, Ec1 and cu for various ND concrete grades [MPa]. C20 30 26 13 -4,2

C30 34 29 18 -3,7

C40 36 31 22 -3,3

C50 39 33 27 -3,0

C60 41 35 31 -2,8

C70 43 36 36 -2,6

C80 44 38 40 -2,4

STRESS

C12 27 23 9 -5,0

Combinated Bending

 co

for |c| < |c,lim|

c

 cu

 1  cu

E ci  c . E c1  c1 1

 

f c'

 1 f c'

Concrete grade Eci Ec Ec1 c,lim

c

E ci

E c1



STRAIN

 c1 .f cm  . c

The compressive strength of concrete fc' versus the ultimate strain cu. The maximum compressive strain, the ACI Committe 363 recommends the values of the

 c1

If the strain |c| > |c,lim| , then the descending part of the stress-strain curve is defined as follows:

maximum compressive strain for HSC equal 0,003.

Sabah Shawkat © 1

c

 c.lim

.

 c1

 c.lim

c

 c1

 c.lim

 .

c

 c1

Tensile Strength

Modulus of rupture fr is determined on plain beams (152 x 152 x 762mm) loaded in

 c1

flexure at the third points of a (610mm) span. The tensile strength can be determined by splitting test on cylinders (152 x 304mm). In case of the absence of experimental results, the

Where

 c.lim E ci . E c1  c1 2



 c.lim E ci . E c1  c1

 c.lim

E ci

 c1

E c1

modulus of rupture can be estimated from characteristic compressive strength fc ' as follows:

0.94 f c

7.5

[MPa ] for 21 MPa < fc'< 83 MPa

Commitee ACI 363 provides the following relationship:

The Figure below shows the example of the sress-strain curve for uniaxial compression. ft

0.94

MPa for 21 MPa < fc'< 83 MPa fc The flexural tensile strength or modulus of rupture fr, from a modulus of rupture test is

calculated using the following equation, assuming the concrete is linearly elastic. In the splitting cylinder test, an element on the vertical diameter of the specimen is stressed in biaxial tension and compression, as shown in figure below In the elastic homogeneous cylinder this loading procedures a nearly uniform tensile stress across the loaded plane. The splitting tensile strength fct may be computed as: f ct

2 P

 d L

Properties of Materials

Cylinder splitting test for tensile strength The modulus of elasticity is defined as the secant modulus corresponding to a strain 0,5.fc'. The values of the modulus of elasticity are determined according to ASTM C469. For normal weight concrete, Ec may be taken as MPa E c 4700. f c The empirical relationship have been further developed and the ACI Commitee 363 proposes:

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3320.

6900

1.5

for 21 MPa < f c'< 83 MPa

2300

Where w is unit weight of concrete; [kg/m3]

The Design Compressive strength f cd

f cn m

0.56f ck

 28

1.4

Concrete strength properties GRADE Char.compr. strength of cube (100) fck Char.compr. strength of cylinders fcck In situ compressive strength fcn Char. tensile strength ftk In situ tensile strength ftn

LC15

C25 C35 C45 C55 C65 C75 C85 LC25 LC35 LC45 LC55 LC65 LC75 LC85

C95

C105

11.2

16.8

22.4

28.0

33.6

39.2

44.8

50.4

56.0

61.6

1.55

2.10

2.55

2.95

3.30

3.65

4.00

4.30

4.60

4.90

1.0

1.40

1.70

2.00

2.25

2.50

2.70

Properties of Materials

Sabah Shawkat ©

Geometry of FRC and Creating Hinge Model at the Middle of the Span for the Crack Mouth Opening Displacement (CMOD)

Fibre Reinforced Concrete (FRC) Can be seen as a composite material consisting of concrete mass and fibres, which can be dispersed diffrent way. The main reason why such a composite cementinous material was invented and has been using for practical application is the improvement of some mechanical properties. By adding different types of fibres can be increase the toughness of the concrete. There ara three diffrent types of fibre orientation in the hardened concrete mass. Most common are randomly oriented low fibre volume (Fig.a) or randomly oriented high fibre volume (Fig.b). Less common are uni-directional fibre composites consisting of non-metal fibres (Fig.c)

Sabah Shawkat © HSC is a brittle material, and as the concrete strenght increases the post-peak portion of the stress-strain diagram almost vanishes or descends steeply. The increase in concrete strenght reduces its ductility- the higher the strenght of concrete, the lower is its ductility. This inverse relation between strenght and ductility is a serious drawback for the use of HPC and a compromise between these two characteristics of concrete can be obtained by adding discontinuous fibres. The concept of using fibres to improve the characteristics of construction materials is very old. Addition of fibers to concrete makes it more homogeneous and isotropic and transforms it from a brittle to a more ductile material. When concrete cracks, the randomly oriented fibers arrest a microcracking mechanism and limit crac propagation, thus improving strength and ducility.

High Performance Fibre Concrete (HPFC)

9 (1) They may increase the strength of the composite over that of the matrix, by providing a means of transferring stresses and loads across cracks.

Simplified description of the multiple cracking process and the resulting

In FRC composites, the major role played by the fibres occurs in the post-cracking zone, in which the fibres bridge across the cracked matrix. In a well-designed composite the fibres can serve two functions in the post-cracking zone:

(2) More importantly, they increase the toughness of the composite by providing energy absorption mechanics, related to the debonding and pull-out processes of the fibres bridging the cracks.The sequence of events following first cracking in the composite determines whether these strengthening and toughening effects will occur. As cracking occurs in the brittle matrix, the load is transferred to the fibres. If failure is to be prevented at this stage, the load bearing capacity of the fibres, sfu Vf in the case of aligned and continuous fibres, should be greater than the load on the composite at first crack, which can be calculated on the basis of the elastic stresses at the cracking strain of the matrix in the composite, emu.

Sabah Shawkat © Schematic description of the stress-strain curve, based on the ACK model

Thus, the overall mechanical behaviour of the FRC composite can usually be described in terms of the three stages of the tensile stress versus strain curve of: (1) Elastic range, up to the point of first crack: the matrix and the fibres are both in their linear, elastic range. (2) Multiple cracking range, in which the composite strain has exceeded the ultimate strain of the matrix. (3) Post-multiple cracking stage, during which the fibres are being stretched or the first crack to occur in the composite will not lead to catastrophic failure, but will result in redistribution of the load between the matrix and the fibres. That is , the load carried by the matrix in the cracked zone will be imposed on the bridging fibres and matrix in the cracked zone will be imposed on the bridging fibres and the matrix at the edges of the crack will become stress free. Additional loading will result in additional cracks, until the matrix is divided into a number of segments, separated by cracks. This proces is known as multiple cracking. It occurs at an approximately constant stress, which is equal to the first crack stress emu . Ec, where Ec is the modulus of elasticity of the composite.

Mechanics of Fibre Reinforced Cementious Composites

The steel used for making steel fibres are generally carbon steels or alloy steels; the latter are used primarily for corrosion-resistant fibres, in refractory applications and marine structures. Steel fibres may be produced in a number of ways. Round fibres are produced by cutting or chopping wires, with diameters typically in the range of 0.25 mm to 0.75 mm. Flat fibres may be produced either by shearing sheets or flattening wire; cross-sectional dimensions are in the range of 0.15 to 0.41 mm thick, 0.25-0.90 mm wide. Crimped and deformed fibres of a number of different shapes have also been produced. The deformations may extend along the full length of the fibre or be restricted to the end portions. In order to make handling and mixing easier, some fibres are collated into bundles of 10-30 fibres using a water-soluble glue, which dissolves during the mixing process. In addition, fibres are sometimes produced by the hot melt extraction process, in which a rotating wheel is brought in contact with the molten steel surface, lifting off some liquid metal which then freezes and is thrown off in the form of fibres. Depending on the type of steel and the type of production process, steel fibres may have tensile strengths in the range of about 345-2100 MPa and ultimate elongations of 0.5% to 35%.

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The basic problem is to introduce a sufficient volume of uniformly dispersed fibres to achieve the desired improvements in mechanical behaviour, while retaining sufficient workability in the fresh mix to permit proper mixing, placing and finishing. The performance of the hardened concrete is enhanced more by fibres with a higher aspect ratio, since this improves the fibre-matrix bond. On the other hand, a high aspect ratio adversely affects the workability of the fresh mix. In general, the problems of both workability and uniform distribution increase with increasing fibre length and volume. It is these contradictory requirements that have led to the development of deformed fibres, with which bonding is achieved largely through mechanical anchoring, which is more efficient than the frictional shear bond stress mechanism associated with straight fibres. For istance, mangat and Azari have shown that the apparent coefficient of friction is about 0.09 for hooked fibres, compared to only about 0.04 for straight fibres.

Technologies for Producing Steel Fibers

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Developement of Concrete Strenght – NSC vs HSC

Silica fume Silica fume (SF), is a by-product of the melting process used to produce silicon metal and ferrosilicon alloys. The main characteristics of SF are its high content of amorphous SiO2 ranging from 85 to 98%, mean particle size of 0.1 - 0.2 micron and its spherical shape. SF acts both as filler and as a pozzolan.

influenced by the matrix strength (cement paste) surrounding the aggregate particles. For given conditions of curing and age, concrete compressive strength depends primarily on the w/c ratio, the type and quality of cement. Any marked increase or decrease in strength is usually attributed to the cement.

The use of SF as a replacement of a part of the cement gives a considerable strength gain. SF is 2-4 times more efficient than Portland Cement as far as long-term strength is concerned for NSC. This efficiency factor rises with increasing strength. The use of SF is therefore of particular interest in HPC, where the amount of cement should not be too high. The amount of 10 –15 % of SF of total amount of cement content can be used. For most binder combinations, the use of SF is one of the most usual way of producing concrete of normal workability with a strength level exceeding 80 N/mm2. To ensure a proper dispersion of the ultra-fine SF particles, plasticizers should be used in these mixtures. SF can be used by mixing in different forms as powder, granulates or slurry.

In HPC, however, the paste and the transition zone are generally much denser and stronger than in NSC, thus providing good stress transfer between aggregate and paste. In another words, the aggregates participate more actively in the concrete’s mechanical responses. Researchers and practitioners alike are coming to recognize the vitally important role that aggregates play when it comes to making HPC. In this respect, two points that assume some significance are the aggregate strength and the strength of the interface, i.e.., the bond between the aggregate and matrix. The strength of the bond between HCP and aggregate is a function of the amount of Ca(OH)2 crystal formation and degree to which voids or other defects are present at the interface. Above mentioned calcium hydroxide (Ca(OH)2) forms during the hydration of the cement, does not contribute to the strength development and reduces quality of the transition zone.

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In theory, when concrete is subjected to compression, cracks first begin to develop in the transition zone. Microscopic examination of hydrated cement paste (HCP) in NSC reveals a rather weak zone around the aggregate, where stress transfer between the HCP and aggregate is poor; actually the aggregates share only a very small part of these stresses. With NSC (w/c ratio in the range of 0.50-0.70), the aggregate is normally assumed to be an inert material, which, in very broad terms, is used in the mix for reasons of economy and volumetric stability of the finished product. Thus, the concrete strength is evidently strongly

By introducing mineral admixtures such as natural pozzolans, silica fume and fly ash, the calcium hydroxide is transformed during the secondary reaction to the calcium silicate hydrates (CSH) which contribute to the strength of cement and concrete.

Silica Fume

Cement Strength development and strength potential in HPC depend on the choice of cement. The clinker composition and the fineness are factors that influence both early and final strength. The clinker minerals C3S, C2S and C3A have the greatest influence on the strength development in cement paste. C3S contributes both to a rapid early age strength development and a high final strength. C2S hydrates somewhat slower, but can contribute significantly to the final strength. C3A has particular influence on the early strength. The hydratation of the clinker minerals may be influenced by the fineness of the cement. A high specific surface leads to a rapid reaction. A high degree of grinding fineness may, however, reduce the strength development after 28 days of curing.

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Other components in the cement may also influence the development of strength and heat. Especially, a high content of alkalies will result in an increased early strength and reduced final strength potential.

Cement

The main differences between the stress-strain curves of NSC and HPC are: • a more linear stress-strain relationship to a higher % of the maximum stress • a slightly higher strain at maximum stress • a steeper shape of the descending part of the curve • reduced ultimate strain may be observed for medium HPC These changes in the load-response are a consequence of improved aggregate-paste bond for HPC. The more linear stress-strain relationship reflects the reduced amount of microcracking at lower levels of loading for these concretes. Typical examples of the stress-strain behaviour of HPC is shown in adjacent figure. Behaviour of very HPC is represented with a linear ascending part of the stress-strain curve, a relatively high strain at maximum stress ( e u= 4%) and a sudden drop after the ultimate stress is reached.

Sabah Shawkat © The difference in rigidity between cement paste and aggregates is far less in HPC than in NSC. Consequently, the internal stress-distribution is more homogeneous. As the tendency of early microcracking is reduced, the stress-strain curve is more linear. A less developed microcrack pattern also results in a more sudden failure, because the ability of redistributing stress is reduced.drawback for the use of HPC and a compromise between these two characteristics of concrete can be obtained by adding discontinuous fibres.

Behaviour of Concrete in Compression

Sabah Shawkat © The strain in tension zone is limited to 5%o. The flexural strenght bBZ of the concrete without fibres is used for the calculation up to the first crack (state I), after which a plastic hinge is introduced (state II,III). At this point the stress decreases and the values (bIfb,BZm bIIfb, BZ) can be determined by the testing or calculation.

The values of equivalent flexural strenght fct,eq 150 and fct,eq,300 are determined by testing or calculation as well. Determination by the test is performed by deformation-conrolled bending tests according to Belgian standart NBN B15-238. The load-deflection diagram is drawn and the energy B absorbed during the test is measured from the start of loading to a deflection of l/300 and l/150, i.e. a deflection of 1,5mm and 3mm in case of span of beam 450mm.

German and Belgium Diagram of FRC

Correspondence between the softening curve of the cohesive crack model (a), and the stress-strain curve of the crack band model

(a) Homogenous bar. (b) Elastic-softening stress strain curve. (c) Stress-fracturing strain curve. (d) Bar with a softening band of length h.

Consider a quasi-static process in which the bar is monotonically stretched. Up to the peak, the strain is uniform, equal to the elastic strain. At peak, just as seen before, a bifurcation can occur so that a portion of the bar, of length h < L, continues stretching, while the rest of the bar unloads elastically (d).

Sabah Shawkat ©

Softening curve (a), and resulting stress-elongation curve when the bulk material behaviour is assumed to be linear elastic (b).

To simplify the computations while retaining the essentials of the model, Hillerborg further assumed that the inelastic strain in the loadingunloading path was negligible, i.e., the behaviour of the bulk material was linear elastic, so that, given the softening curve in a, the loadelongation curve is constructed as b shows

Basic Concept in Crack Band Model (CBM) and Smeared Cracking

Sabah Shawkat © Consider a uni-axial tension test on a concrete bar. If no special care is exerted, a nearly linear stress-elongation curve will be recorded up to the peak stress (arc OP). At the peak load, a cohesive crack normal to the axis of the bar appears somewhere in the specimen (at the weakest cross section).

After the peak, the crack develops a finite opening w while still transferring stress, and at the same time the remainder of the specimen unloads and its strain decreases uniformly along the arc PB. Thus, the total elongation at point A is the addition of a uniform strain corresponding to point B and the crack opening w

Basic Concept in Crack Band Model (CBM) and Smeared Cracking

18 Although the tensile strength of concrete is neglected for the strength of reinforced and prestressed concrete structures, in general, it is an important characteristic for the development of cracking and therefore, for the prediction of deformations and the durability of concrete. Other characteristics such as bond and development length of reinforcement and the concrete contribution to the shear and torsion capacity are closely related to the tensile strength of concrete. With SF it is possible to improve the microstructure in the surface of intersection between cement stone and aggregate. This leads to a totally different surface of fracture. In HPC the surface of fracture passes right through the aggregates – a relatively smooth surface of fracture is obtained. Brittle behaviour of the concrete is the result.The fracture energy increases asymptotically as w/c ratio decreases. Among other parameters, the creep behaviour of concrete is strongly affected by internal factors such as the material properties of the phases of concrete and its composition. In particular microstructural changes of the hardened cement paste matrix have a significant effect on the creep characteristics. The tests on HPC published in the literature show mainly two basic differences in comparison with NSC: • The magnitude of the total creep strains, i.e. the sum of the basic creep component is considerably lower for HPC. • While the drying creep component is substantially lower for HPC, the magnitude of the basic creep component is less reduced. The ratio of the drying creep component to the basic creep component decreases as the strength of concrete increases.

Sabah Shawkat ©

The first difference results from the significantly higher strength and stiffness, and the lower porosity of the hardened cement paste matrix of HPC in comparison with that of NSC. As a consequence, all effects on the creep of concrete resulting from interactions between the components of a composite material, such as the effects of the aggregate content and the stiffness of the aggregate as well as microcracking, should be reduced for HPC being made with ordinary aggregates.

Durability of Concrete

Composite Materials Approach The composite material approach is usually based on the rule of mixtures. In the precracked zone (stage I), we consider the composite material consisting of two individual components, which are linear elastic and the perfect bond is between them. Model of this component is shown in adjacent figure. where sf is the stress in the fibre at the first crack strain and smu is the tensile strength of the matrix in the absence of fibres.In the post-cracked zone (stage III), the contribution of the matrix is small or even negligible, because of its multiple cracking. Once first cracking has taken place in the brittle matrix, the fibres serve to inhibit unstable crack propagation. At this stage, the cracking patterns are complex, with discontinuous microcracks present ahead of the principal crack. This can be deduced from the various analytical models and has also been observed microscopically, assuming different mechanisms of stress transformation across the crack.Wacharatana & Shah’s Model (COD Crack Opening Displacement) COD is used to characterize fracture behaviour in the vicinity of a sharp notch by considering the opening of the notch face (which can be measured using clip gauges). Three distinct zones can be identified: (1) Traction free zone. (2) Fibre bridging zone, in which stress is transferred by frictional slip of the fibres. (3) Matrix process zone, containing microcracks, but with enough continuity and aggregate interlock to transfer some stress in the matrix itself..

Sabah Shawkat ©

- fracture energy Gf - uniaxial strength ft - width of the crack band wc This approach lends itself particularly to computer-based finite element modelling of the cracks. For very large structures, this theory becomes equivalent to the LEFM approach.

Crack Stabilization

20 The efficiency of fibre reinforcement can be judged on the basis of two criteria:the enhancement in strength and the enhancement in toughness of the composite, compared with the brittle matrix. These effects depend upon the fibre length, the orientation, fibre-matrix shear bond strength and ultimate strength of fibres (functions of the critical length lc). Toughening mechanisms can be explained consequently. OA. The cracks initiate at a very low load level and are few and widely dispersed, no more than 1 mm in length. A. Cracks begin to localize near the point where the matrix contribution is maximum; this point is referred to as the bend-over-point (BOP) OA. The cracks initiate at a very low load level and are few and widely dispersed, no more than 1 mm in length. A. Cracks begin to localize near the point where the matrix contribution is maximum; this point is referred to as the bend-over-point (BOP) AB. With further straining, the localized bands seem to move closer together. B. Above mentioned process leads to the homogenisation of microcracking.

Sabah Shawkat ©

Homogenous distribution of cracking observed at very high strain levels (point B) may explain the apparently high level of matrix contribution observed at this level (multiple cracking).When the matrix is divided by parallel cracks, any additional tensile load will cause stretching or pullout of the fibres. In case of long fibres (l > lc), fibres will be stretching under load and the slope in this range is Ef Vf and failure will occur when the fibres reach their load bearing capacity, at a composite stress of sfu Vf. This type of composite is termed as composite exhibiting strain hardening.

Crack Stabilization

A fracture mechanics approach has also been applied to the problem of fibre debonding and pull-out, as an alternative to the treatment based on the analysis of elastic and frictional shear stress. The object is to develop material parameters to account for debonding which are more reliable and easier to evaluate experimentally than the interfacial shear bond strength values. In this treatment it is assumed that the debonded region is stress free (t f = 0) and this zone is treated as an interfacial crack of length b. Using the classical Griffith theory (of LEFM) the conditions leading to the propagation of this crack and to spontaneous debonding, can be calculated.

Sabah Shawkat ©

It should be remembered that, when examining the effects of fibre additions, changes in SFRC properties are always expressed in terms of average fibre contents. It is implicitly assumed that the fibres are uniformly distributed throughout the matrix and moreover, that they are randomly oriented. Unfortunately, neither assumption is likely to be correct after the SFRC has been placed and compacted by vibration and this leads not only to a considerable amount of scatter in the data, but also to a considerable variation in measured values due to the direction of loading. When using table vibration, fibres tend to align themselves in planes perpendicular to the direction of casting (or gravity)

Fibre-Matrix Debonding

Sabah Shawkat © The role of fibres in the concrete is to transfer the load from the matrix. There are two cases, which can occur and the stress transfer effects between the fibre and matrix are different in: - the pre-cracking case - the post-cracking case Assumptions applied in this theory: • the matrix and the fibre are both elastic materials • no slip between the fibre and the matrix at the interface • the interface is infinitesimally thin • the properties of the matrix in the vicinity of the fibre are the same as those of the bulk matrix • there is no effect of the stress field around one fibre on neighbouring fibres • the tensile strain in the matrix, e m, at a distance R from the fibre, is equal to the tensile strain of the composite, e c

Pullout Testing

- Equivalent flexural strength eqv.b BZ3R, is defined for a deflection up to 3 mm where DfBZ3 is the energy absorbed by fibres considered as the product of an average load eqvF3 causing this deflection.

Sabah Shawkat © Distribution of stresses, when the ultimate load is reached. The values of equivalent flexural strength fct,eq,150 and fct,eq,300 are determined by testing or calculation as well. Determination by the test is performed by deformation-controlled bending tests according to Belgian standard NBN B15-238. The load-deflection diagram is drawn, and the energy B absorbed during the test is measured from the start of loading to a deflection of l/300 and l/150, i.e. a deflection of 1.5 mm and 3 mm in case of a span of beam 450 mm.

Load vs Deflectiom

OA - Intact interface Force PA is required to start breaking the adhesion. AB – Gradual debonding BC – Debonding completed Force PB is generated by the resistance to slip which is at this stage is effective over the embedded length of fibre. C - Stick-Slip behaviour Frictional resistance decreases, while embedded length continuously decreases The efficiency of fibre reinforcement can be judged on the basis of two criteria: the enhancement in strength and the enhancement in toughness of the composite, compared with the brittle matrix..These effects depend upon the fibre length, the orientation of the fibres and the fibre-matrix shear bond strength. These three factors are not independent, since the effects of both fibre length and orientation are highly sensitive to the bond.A critical length parameter lc, can be defined as the minimum fibre length required for the build-up of a stress (or load) in the fibre which is equal to its strength (or failure load)

Sabah Shawkat © - equivalent flexural strength eqv.b BZ2R. Is defined for a slight deflection up to 0.5 mmwhere DfBZ2 is the energy absorbed by fibres considered as the product of an average load eqvF2 causing this deflection.

Pullout Testing

The composite tensile strength scu can be then obtained as follows (for aligned fibers): The stress scu is initial stress, when the crack starts to develop (the crack opening u is nearly equal to zero). With increasing crack opening u, more and more fibers are fully out and , for u = lf / 2, the stress s(u) drops to zero. Because the embedded length L(u) and the number of bounded fibers Vf (u) decreases linearly with increasing u, for 0<u<lf/2

Sabah Shawkat ©

The Developement of Cracks Propagation

Due to the microcracking the stresses in front of a crack tip may have a typical distribution according to adjacent Fig.. In this case we have no well defined crack tip, but rather a fracture zone, within which the cracking increases and the stresses decrease as the deformation increases. We cannot identify any well defined crack tip, but some points of interest may be noted: 1. The point where the stress has its maximum. On further deformation the stress decreases due to increasing microcracking. This may be looked upon as the first sign of fracture and this point may serve as a limit for the fracture zone. 2. The point where the crack becomes visible with or without a microscope.

Sabah Shawkat ©

3. The point where the stress transfer ends. This point may serve as a limit between the fracture zone and what may be termed the “real crack” i e that part of the crack, over which no stresses are transferred. From the above it is evident that no well-defined crack tip exists for concrete (like it does for metals), as there is a gradual distribution with increasing deformation and cracking within the fracture zone.

The Developement of Cracks Propagation

The stress scu is initial stress, when the crack starts to develop (the crack opening u is nearly equal to zero). With increasing crack opening u, more and more fibers are fully out and , for u = lf / 2, the stress s(u) drops to zero.

Sabah Shawkat ©

The flexural hinge in a member of depth h shown inadjacent Fig. is characterized by the compression zone z and the hinge rotation of q Assuming a compressive stress of 0.85 fc over 80 % of the depth z, introducing the crack opening parameter x

Flexure General Analysis

Flexural testing is usually obtained in four-point loading and various specifications are available, recommending the specimen size and the test span, depending on the type of components. Tensile testing is seldom specified in standards or recommended practices and is usually carried out only in research. One of the major problems is to provide a gripping arrangement which will not lead to cracking at the grips. various gripping methods, similar to those applied to unreinforced concrete have been used successfully. This problem is not as critical in fibre reinforced concrete because of their higher toughness. In order to get the complete load-deflection curves, the testing system must be equipped with strain or deflection measurement gauges. In practice, to obtain the full curve, a servo-mechanical (or servo-hydraulic) system must be used. Because of the importance of obtaining a reliable curve in the post-cracking zone, a rigid testing machine, or a closed loop testing arrangement are necessary, in particular for the tensile test.

Sabah Shawkat © Schematic load-deflection curves and derived toughness parameters, according to (a) ACI method; (b) JCI method; (c) Barr; (d) ASTM C1018. (After Johnston.)

The static mechanical tests of particular interest for FRC composites are their behaviour under tensile and flexural loading. To characterize the tensile and flexural behaviour it is necessary to measure the stress-strain (or load-deflection) curves, which reflect the effect of the fibres on the toughness of the composite and its crack control potential. The compressive properties are not much different from those of the matrix and the test methods for this property are essentially the same as those for ordinary hardened concrete.

There are a number of other ways of defining and measuring the toughness of FRC. A comparison of these various methods of calculating the toughness index. ACI Committee 544 has recommended that the area under the load versus mid-point deflection curve, out to deflection 1.9 mm (measured on a 100 mm x 100 mm x 356 mm beam tested in third-point loading on a 300 mm span), divided by the area under the curve up to first cracking, would provide a suitable ‘toughness index’. The Japan Concrete Institute recommends the measurement of the area under the load-deflection curve out to a centre-point deflection of (1/150) span. Barr proposed still another toughness index, again in terms of the areas under the load versus deflection curves

Static Testing of Tensile and Flexural Properties

Sabah Shawkat © The redistribution of stress in a body due to introducing a crack or notch may be begun by methods of linear-elastic stress analysis. Of course the greatest attention should be paid to the high elevation of stresses at or surrounding the crack tip, which will usually be accompanied by other non-linear effects.

Principles of Linear Elastic Fracture Mechanics (LEFM)

Four phases of stress distribution in elastic layer of hinge: Phase 0 = state of stress prior to cracking; Phase I – III = states of stresses during crack propagation.

Sabah Shawkat ©

In this phase the sectional stresses are in interval from ft to sw(w1) – stress at the end of the first slope of bilinear stress-crack opening diagram. Nc + Nt + Nf = Next => a Nc.ec + Nt.et + Nf.ef = Mext => m

Here the constant c has been introduced as c = (1 – b2). (1 - b1) / (b2 - b1). In terms of q the point of transition from Phase I to Phase II, qI-II, may be found from the condition that y1 = h and the point of transition from Phase II to Phase III, qII-III, may similarly be found from y2 = h. These transition points, together with the point of transition between Phase 0 and Phase I, q0-I, are given by

Cracks Propagation in FRC I, II, III

Sabah Shawkat ©

Cracks Propagation in FRC I, II, III

32 The solution for the mid-point deflection u can be expressed in non-dimensional form by introducing the non-dimensional mid-point deflection U as a function of q. where ue is the elastic deflection of the beam, which after a few modifications become depending on mq), where un is the deflection due to presence of notch according to Tada depending on v2, function of the ratio between the length of initial notch a0 and distance from the top to bottom of specimen H

Sabah Shawkat ©

The normalized deformation due to the crack Ucb (q) is found by subtracting the normalized elastic deformation qe from the total deformation of the hinge q. The normalized elastic deformation of the hinge is given by qe = m(q) Since the normalized deflection due to hinge deformation is equal to the normalized hinge deformation q

The Mid Span Deflections

The whole aim of inverse analysis is to start forward analysis on basis of some initial parameters of the stress-crack opening curve and to receive the P-CMOD, respectively P-u (deflection) curves, called in this section predictions. Afterward, observations and predictions are compared and by means of optimalization techniques to modify the parameters of the initial stress-crack opening curve to achieve the smallest differences between observations and predictions.

Where N0max and NImax represents the last observation made belonging to phase 0 and I respectively. These numbers are functions of the values of ft, a1, a2 and b2, since the phase change values of q,q0-I, qI-II are functions thereof. Thus, the number of observations in a single phase is dependent on parameters not included in the optimalization of that phase. This means that the initial guess on the parameters must be close to the true values in order to predict the approximately correct number of observations for which the optimalization must be performed, or that the optimalization procedure must be performed more than once in order to approach the correct number. This problem must be taken into account when performing inverse analysis using the above described method.

Sabah Shawkat ©

On basis of this model can be calculated the entire P-CMOD, respectively P-u (deflection) curve for a given geometry and mass of the specimen and a given stress-crack opening curve. In case of searching the P-CMOD, respectively P-u (deflection) response from a fully specified softening law, the issue is called forward analysis.

Forward vs Inverse Analysis

Different forms of steel fibres Characteristics of concrete according to Euro-code 2 Strength class C12/15 C16/20 C20/25 C25/30 C30/37 C35/45 C40/50 C45/55 C50/60 fckcyl [MPa]

fckcube [MPa] Ecm [MPa] fct,m [MPa] fctk,0.05 [MPa] fctk,0.95 [MPa] fcm [MPa] cu . 103 Rd [MPa]

26000 1.6 1.1 2.0 20 -3.6 0.18

27500 1.6 1.3 2.5 24 -3.5 0.22

29000 2.2 1.5 2.9 28 -3.4 0.26

30500 2.6 1.8 3.3 33 -3.3 0.30

32000 2.9 2.0 3.8 38 -3.2 0.34

33500 3.2 2.2 4.2 43 -3.1 0.37

35000 3.5 2.5 4.6 48 -3.0 0.41

36000 3.8 2.7 4.9 53 -2.9 0.44

37000 4.1 2.9 5.3 58 -2.8 0.48

Final values of concrete shrinkage cs,inf according to Euro-code 2 Replacement thicknesses Relative hm = 2Ac / u [v mm] location humidity RH [%] < 150 600 drought, inside ~50 -60 . 10-5 -50 . 10-5 damp, outside ~80 -33 . 10-5 -28 . 10-5

Different forms of steel fibres Form

Description production

Dimensions [mm] diameter length

Tensile streng [Mpa]

Strainght, plain, round Svabet kamma Scanovator

0.25-0.50 0.60

25 60

780

Harex, Stahlfaser-technik irregular

0.8 x 2.0

16-32

1630-2100

Melt extracted fibres also stain steel irregular shape Battelle Corp. Ohio USA Trefil-ARBED Straight, plain, round Type 3 Type 2 Trefil-ARBED Waved on the ends

0.3 x 1.0 0.40 0.58

10-50 35 38

0.30 0.38

25 25

0.25-0.50

960

Sabah Shawkat ©

Final values of creep coefficient of concreteinf according to Euro-code 2 Replacement thicknesses The age of the hm = 2 Ac / u [v mm] concrete at the 50 150 600 50 150 600 beginning of the load Placement conditions application (Relative humidity RH = 50%) (Relative humidity RH = 80%) to [days] (inside) (outside) 1 5.5 4.6 3.7 3.6 3.2 2.9 7 3.9 3.1 2.6 2.6 2.3 2.0 28 3.0 2.5 2.0 1.9 1.7 1.5 90 2.4 2.0 1.6 1.5 1.4 1.2 365 1.8 1.5 1.2 1.1 1.0 1.0

Table values of creep coefficient of concrete 70 according to ENV 1992-1 and CEB-FIP MC 90 The age of Replacement thicknesses the concrete 2 Ac / u [v mm] at the 50 150 600 50 150 600 beginning Relative humidity (inside) Relative humidity (outside) of the load application [RH = 50%] [RH = 80%] to [days] 1 5.4 4.4 3.6 3.5 3.0 2.6 7 3.9 3.2 2.5 2.5 2.1 1.9 28 3.2 2.5 2.0 1.9 1.7 1.5 90 2.6 2.1 1.6 1.6 1.4 1.2 365 2.0 1.6 1.2 1.2 1.0 1.0

30 40 50 60

Plain with hooks at the ends, gued together in bundles Dramix ZL Bekaert Wire Corp Belgium Waved, made with plain wire Jonson and Nephew UK

0.35 0.40 0.50 0.60

Straight, plain,square or rectangular cross-section Sumitomo Metal Ind. Japan Metal strips FIBRAFLEX, France

0.80

1-2x30 �m

Straight, intented in two perpendicular planes Douform,National Standard Straight, plain, with enlarged Ends Tibo

0.40-0.60

0.25 x 13 / 25 0.38 x 13 / 25 0.64 x 13 / 25 0.80

20-40

2000

1000

1150

Straight, plain, with enlarged e EE Australian Wire Ind. Straight, plain fibres glued together in bundles Dramix OL Bekaert Belgium

1400

0.25-0.50

6-30

Characteristic of Concrete

Flexure - general analysis f  40

lf  30 mm

d f  0.5 mm

fc  30 MPa

If 

   lf      2  x   1  cosh  1      mm   f ( x)  Ef   m  lf         2    cosh   1 mm    

  lf  1   2  x  2 sinh  1  Gm     mm   ( x)  Ef   m    lf    2 Ef  ln R     r    2  cosh   1 mm  

Shear stress  fc    MPa 

 b  0.6 

Gf 

f  lf   b MPa

 1 

 2 Gm     mm 2 R  Ef  r  ln     r 

lf  40 mm

lf  0.04m

Ef  210 MPa

u  0 mm1 mm

12 d f  7850

lf 2

  0 0.01 1

2 d f  7850

x  0 mm0.01 mm lf

f  lf   b MPa

Initial stress o 

Specific fracture stress

( u )  o  1  2



u

 lf 

z(  )  1



2.04 fc

o 3  3   

 b(  ) 



z(  )   h  z(  )

Sabah Shawkat ©

 m  0.001

R  3 d f

r 

df 2

The ultimate moment  m(  )  0.68 fc  z(  )  0.6 z(  )   

Gm  2 MPa

f ( x) 

-0.21

MPa

-0.209 -0.208 -0.207 -0.206 -0.206 -0.205 -0.204 -0.203 -0.202 -0.201 -0.201

-0.2 ...

Examples of FRC

2    h  z(  )  6  8   3    2  12  12   4   

36 Example:

 2  2  ( u) du  4.4277220185735886985e6 Pa  mm 3.0    3.0 e6�Pa  0  mm



 ( u)

z(  ) 

 0.886

MPa

0.021



 b ( )

0.213

0.771

0.021

4.298·10-4

0.211

0.665

0.02

8.509·10-4

0.209

0.567

0.02

1.264·10-3

0.207

0.476

0.02

1.668·10-3

0.205

0.394

0.02

2.064·10-3

0.203

0.319

0.02

2.451·10-3

0.201

0.252

0.019

2.831·10-3

0.199

0.193

0.019

3.202·10-3

0.197

0.142 0.098 0.063 0.035 0.016 3.936·10-3 0

n  2.035294

17 MPa



fcm



MPa

Eci   3320

k  0.67 



 6900  MPa

 c1 



 clim  1.99  c1

m ( ) 



fcm

n  0.8 

m MPa

fc   c 

c  c1

n n 1

210

k  1.00871

fcm

n  Eci n  1

3

 c1  1.866863 10

 c  0 0.0001  clim



310

fcm 62 MPa

 c     c1 

n k

 fcm

Eci   c 

d  c   d c   c1

 



n n 1

 c     c1 

n k

 fcm

  

 clim fcm

 

fc  c

Sabah Shawkat © 0.019

3.566·10-3

0.195

0.019

3.921·10-3

0.193

0.019

4.269·10-3

0.191

0.018

4.609·10-3

0.189

0.018

4.941·10-3

0.188

0.018

5.266·10-3

...

110

3

210

3

310

3

410

3

c

   clim   c n   fcm   c d c     c1 n k     c  n 1         c1    0      clim  c n     fcm d c  c1 n k    c   n 1       c1    0  f  40

lf  30 mm

d f  0.5 mm

1   

 clim

 

fc  c d c

P max  clim   

 clim

 

f c  c d c

Pmax

fc  30 MPa

            

h  500 mm

if    0 0.01 1 2

 fc    MPa 

 b  0.6 

 b  5.793

Examples of FRC

Gf 

f  lf   b MPa 12 d f  7850

3 1

Gf  4.428  10 m

u  0 mm1 mm

o 

z(  )  1



3



(3) There is no slip between the fibre and the matrix at the interface, i.e. 'perfect 'bond exists

o  0.886MPa

2 d f  7850

z( 1)  7.132 10

2.04 fc

o 3  3       

f  lf   b MPa

 b(  ) 



between the two. (4) The properties of the matrix in the vicinity of the fibre are the same as those of the bulk

z(  )   h  z(  )

matrix

(5) The fibres are arranged in a regular, repeating array.

lf 2



(6) The tensile strain in the matrix, m, at a distance R from the fibre, is equal to the tensile



strain of the composite, c.

( u )  u du  1.6603957569650957619e7  Pa   mm  8.0   3.0   6.0

0 mm

( u )  o  1  2



u

(7) No stress is transmitted through the fibre ends.



(8) There is no effect of the stress field around one fibre on neighbouring fibres.

lf 

110

0.025

810

 ( u)

lf  40 mm

0.02

610

z( )

410

lf  0.04m Ef  210 MPa

r 

Gm  2 MPa

0.01

R  3 d f

Tensile stress

0.015

210

 m  0.001

Sabah Shawkat © 3

510

3

510

0.01

0.015

0.2

0.4

0.6

0.8

0.015

2.510

m(  ) 1.5105

510

110

0.2

0.4

0.6

0.8



510

 2 Gm     mm 2 R  Ef  r  ln     r 

  lf        x  2  1  cosh  1      mm   f ( x)  Ef   m    lf       2    cosh   1   mm  

210

0.01 3

 1 



 b(  )

x  0 mm0.01 mm lf

0.2

0.4

0.6

0.8

110

 1  0.412

   lf    x 2  2 sinh  1  Gm     mm   ( x)  Ef   m    lf    2 Ef  ln R       r  2  cosh   1  mm  1

 0

Stress transfer in the un-cracked composite membre Elastic Stress Transfer

 f ( x)

During the early staged of loading, the interaction between the fibre and the matrix is elastic in

 ( x)

nature. The first analytical model to describe the stress transfer in the elastic zone was developed by Cox. Later models are usually referred to as shear lag theories. They are based on the analysis of the stress field around a discontinuous fibre embedded in an elastic matrix. In calculating the stress field developed due to these deformations, several simplifying assumptions are made: (1) The matrix and the fibre are both elastic materials. (2) The interface is infinitesimally thin.

Examples of FRC

 110

 210

 310

0.01

0.02 x

0.03

0.04

38 Since the crack width w in the fictitious crack model can be converted into a strain (=w/l) by APPLICATION OF FRACTURE MECHANICS, Modelling of the crack

The sequence of crack formation from the initiation of micro cracks in concrete under tension to development into major macroscopic cracks is modelled according to tension softening

means of the length ofwthe the stress-crack relationship be transformed Since the crack width inrod the element fictitiousl, crack model canwidth be converted into acan strain (=w/l) by into a stress-strain relationship. The length l of the rod element is assumed to be unity (l=1) in means of the length of the rod element l, the stress-crack width relationship can be transformed

 crack the analysis. In the relationship. elastic regionThe before the tensile is exceeded (<ftot, be p), no into a stress-strain length l of thestrength rod element is assumed unity (l=1) in

behaviour. The tension softening behaviour means that, once the tensile stress reaches the

is theinitiated. analysis. In the elastic region before the tensile strength is exceeded (<ft, p), no crack

tensile strength of the concrete, the stress decreases as the fictitious crack increases in width.

is initiated.

This behaviour is represented by a tension softening curve. The area enclosed by the tension softening curve corresponds to the fracture energy of the concrete. The fracture energy is defined as the energy required to create a fully cracked unit surface of concrete across which the tensile stress cannot be transferred. ft - tensile strength of concrete

Steel wire fibre reinforced concrete element: structural element having a m

wo - crack width over which stress cannot be

Steel wire reinforced concrete element: structural element having a m fibre content rf offibre 0,0025.

transferred GF

  

Stress-strain relationship

 dw

fibre content rf of 0,0025. Additionally, the spacing s between fibres calculated as Additionally, the spacing s between fibres calculated as

Stress-strain relationship Steel wire fibre reinforced concrete element: structural element having a minimum

fibre content rf of 0,0025.

Sabah Shawkat © Additionally, the spacing s between fibres calculated as S S

  d f  lf

4 f

  df

Tension softening curve and fracture energy

The tension softening curve has been obtained experimentally by various methods (Uchida at

al. (1991)). There are also various proposals for numerical modes of the tension softening curve.

must be smaller than 0,45 lf, where

  d f  lf

 4df  lf f

4 f

Af Af

  df

 4 df

must be smaller than 0,45 lf, where

thanpercentage 0,45 lf, where the volume of fibres; -mustrfbeis smaller --

r f is the volume percentage of fibres; length of fibre lf - distance between the two extremities of the fibre;

length of fibre lf - distance between the two of extremities the fibre; (equivalent) diameter of fibre df - diameter a straight of circular wire with sam

(equivalent) of length fibre dfof- the diameter of a straight circular wire with sam fibre after stretching) and weight as the straightened diameter length ls (=

In this analysis, the one-fourth bilinear model (Figure bellow) is adopted because it is widely

rf is the volume percentage of fibres;

accepted as the standard tension softening model.

fibre. For a fibre with constant cross-section of thea fibre Af iscircular definedcross as: section, the diameter is measur length of fibre lf - distance between the two extremities of- thethe fibre;

of the fibre Af is defined as: - the cross-section (equivalent) diameter of fibre df - diameter of a straight circular wire with same For the steel fibres of length to diameter ratio lf > 60, the minimum fibre content o straightened length ls (= length of the fibre after stretching) and weight as the considered For the determining. steel fibres of length to diameter ratio lf > 60, the minimum fibre content o always fibre. For a fibre with a constant circular cross section, the diameter is measured directly; always determining. the cross-section of the fibre Af is defined as: df diameter or equivalent diameter of steel fibres

length ofcircular the fibre aftersection, stretching) and weight as the straightened length ls (= fibre. For a fibre with a constant cross the diameter is measur

equivalent length oforsteel fibres diameter of steel fibres ldff diameter For the steel fibres of length to diameter ratio lf > 60, the minimum fibre content of 0,0025 is length of steel fibres lfs straightened length of steel fibres always determining. straightened steel fibres sls spacing factorlength of theoffibres df diameter or equivalent diameter of steel fibres lf

The one-fourth model

length of steel fibres

s spacing factor of the fibres rf volume percentage of steel fibres

ls straightened length of steel fibres

rf volume percentage of steel fibres l f length to diameter ratio of steel fibres

s spacing factor of the fibres

lf length to diameter ratio of steel fibres

rf volume percentage of steel fibres lf length to diameter ratio of steel fibres

Examples of FRC

[1] Hillerborg, A. (1980) “Analysis of fracture by means of fictitious crack model,

[10] Design of Concrete Structure, Norwegian Standard NS3473 (1992),

particularly for fibre reinforced concrete.” The Int. J. Cem. Comp. 2(4), 177-184

Norwegian Council for Building Standardization, Oslo, 1992

[2]Jenq, Y. S. and Shah, S. P. (1985) “Two parameter fracture model for

[11] Rossi, P. “Mechanical Behaviour of Metal-fibre Reinforced Concretes.”

concrete.” J.Eng. Mech.-ASCE, 11(10), 1227-1241.

Cement & Concrete Composites 14 , pp. 3-16, 1992

[3] Petersson, P.-E. (1981) “Crack Growth and Development of Fracture Zone

[12] Moranville-Rgourd, M.:“Durability of High Performance Concrete : Alkali-

in Plain Concrete and Similar Materials.” Report No. TVBM-1006, Division of

Aggregate Reaction and Carbonation“, High Performance Concrete: From

Building Materials, Lund Institute of Technology, Lund, Sweden.

material to structure, 1992 E & FN Spon

[4] Hillerborg, A. (1985) “Numerical methods to simulate softening and fracture of

[13] Hassanzadeh, M.; Haghpassand, A.:”Brittleness of Normal and High-

concrete.” Fracture Mechanics of Concrete: Structural Application and Numerical

Strength Concrete”,

Calculation, G. C. Sih and A. DiTomasso, eds., Martinus Nijhoff, Dordrecht,

Symposium in Lillehammer, Norway, June 20-23, 1993

Utilization of High Strength Concrete, Proceedings,

Sabah Shawkat ©

141-170.

[14] Zhou, F. P.; Barr, B. I. G.:” Effect of coarse aggregate on elastic modulus and

[5] Wittmann, F. H., Roelfstra, P. E., Mihashi, H., Huang, Y.-Y. and Zhang, X.-H.

compressive strength of High Performance Concrete”, Cement and Concrete

(1987) “Influence of age of loading, water-cement ration and rate of loading on

Research, Vol. 25, No. 1, 1995 Elsevier Science Ltd.

fracture energy of concrete.” Materials and Structures, 20, 103-110.

[15] Casanova, P. and Rossi, P. (1996) “Analysis of metallic fiber-reinforced

[6] Balaguru, P. & Kendzulak, J.:“ Mechanical properties of Slurry Infiltrated

concrete beams submitted to bending.” Materials and Structures, 29, 354-361.

Fiber Concrete (SIFCON). American concrete Institute, Detroit, 1987

[16] Holzmann, P.: “ High strength concrete C 105 with increased Fire-resistance [7] “ Fracture energy and strain softening of concrete as determined by means

due to polypropylen Fibres”, Utilization of High Strength /High Performance

of compact tension specimens.” Mater. Struct., 21, 21-32.

Concrete, Proceedings, Symposium in Paris, France, May 29-31 1996

[8] Mindes, S. & Bentur, A.:”Fibre Reinforced cementitious Composites”, 1990

[17] Muller, H. S.: “Creep of High Performance Concrete – Characteristics and

Elsevier Science Publishers Ltd

Code-Type prediction Model.” Fourth International Symposium on the Utilization of High Strength / High Performance Concrete, 1996, Paris, France

[9] European pre-standard: ENV 1992-1-1: Eurocode 2: Design of concrete structures – part 1: General rules and rules for buildings.

[18] Casanova, P. and Rossi, P. (1997) “Analysis and design of steel fiber reinforced concrete beams.” ACI Structural J., 94(5), 595-602.

References

[19] Stang, H. and Olesen, J. F. (1998) “Evaluation of crack width in FRC with conventional reinforcement.” Cement & Concrete Comp., 14(2), 143-154.

[20] Stang, H. and Olesen, J. F. (2000) “A fracture mechanics based design approach to FRC.”in P. Rossi and G. Chanvillard (eds.), Fiber-Reinforced Concretes (FRC), BEFIB’ 2000, RILEM Publications S.A.R.L., ENS – 61 Av. Pdt. Wilson, F-94235 Cachan Cedex, France, 315-324, proceedings of the Fifth International RILEM Symposium.

[21] RILEM-Committee-TDF-162 “Test and Design Methods for Steel Fiber Reinforced Concrete. Recommendations for Uni-axial Tension Test.” (2000), Mater. Struct.,.

Sabah Shawkat ©

[22] RILEM-Committee-TDF-162 “Test and Design Methods for Steel Fiber

Reinforced Concrete. Recommendations for Bending Test (2000).” Mater. Struct., 33, 3-5.

[23] RILEM-Committeee-TDF-162 “Test and Design methods for Steel Fiber Reinforced Concrete. Recommendations for

s - e Design Method.” (2000),

Mater. Struct., 33(3), 75-81.

References

Sabah Shawkat © 02. Traditional Reinforced Concrete

Normally or conventional concrete is known as long as concrete strength reaches about 40MPa, and concrete deformation reaches 0.003 or 0.0035.

Sabah Shawkat ©

In this chapter we briefly mention the different types of loads such as the permanent load, which includes the inherent self-weight of the structure and the permanent loads as the layers of the individual materials laid on the structure itself.

Furthermore, the loads are thrown in by the influence of people and furniture in addition, we deal as an example of the effects of wind and temperature. As we know that the effect of the load from temperature induce the stresses and not the cross-sectional forces. We also give examples of ideal loads that are dimensionless, and gradually calculate the reaction and cross-sectional forces, but we do not deal with the earthquake load. The reader will also find a simple way to calculate the bending moments and transverse forces on individual reinforced bearing concrete members such as beams, slabs, columns, foundation and shearing wall. While the reinforcement design of all elements are made according to the method suggested by the author of this book.

43 Loads are forces that act or may act on a structure. For the purpose of predicting the resulting behaviour of the structure, the loads, or external influences, including forces, consequent displacements, and support settlements, are presumed to be known. Loads are typically divided into two general classes: dead load, which is the weight of a structure including all of its permanent components, and live load, which is comprised of all loads other than dead loads.

Structural system consists of the primary load-bearing structure, including its members and connections. An analysis of a structural system consists of determining the reactions, deflections, and sectional forces and corresponding stresses caused by external loads. Methods for determining these depend on both the external loading and the type of structural system that is assumed to resist these loads.

beams must resist bending, diagonal tension, shear and torsion and must be such as to transmit forces through a bond without causing internal cracking. The details must be able to optimize the behaviour of the beams under load. The shapes of the beams can be square, rectangular, flanged or tee (T). Although it is more economical to use concrete in compression, it is not always possible to obtain an adequate sectional area of concrete owing to restrictions imposed on the size of the beam (such as restrictive head room). The flexural capacity of the beam is increased by providing compression reinforcement in the compression zone of the beam which acts with tensile reinforcement. It is then called a doubly reinforced concrete beam. As beams usually support slabs, it is possible to make use of the slab as part of a T-beam. In this case the slab is generally not doubly reinforced.

Sabah Shawkat ©

In a statically determinate system, all reactions and internal member forces can be calculated solely from equations of equilibrium. However, if equations of equilibrium alone do not provide enough information to calculate these forces, the system is statically indeterminate. In this case, adequate information for analysing the system will only be gained by also considering the resulting structural deformations. A member subjected to pure compression, such as a column, can fail under axial load in either of two modes. One is characterized by excessive axial deformation and the second by flexural buckling or excessive lateral deformation.

Reinforced Concrete Beams Beams are structural elements carrying external loads that cause bending moments. Shear forces and torsional moments along their length. The beams can be singly or doubly reinforced and can be simply supported, fixed or continuous. The structural details of such

Where beams are carried over a series of supports, they are called continuous beams. A simple beam bends under a load and a maximum positive bending moment exists at the centre of the beam. The bottom of the beam which is in tension is reinforced. The bars are cut off where bending moments and shear forces allow it. In a continuous beam the sag (deflection) of the centre of the beam is coupled with the hog at the support. An adequate structural detailing is required to cater for these changes. The reinforcement bars and their cut-off must follow the final shape of the final bending moment diagram. Where beams, either straight or curved, are subjected to in-plane loading, they are subjected to torsional moments in addition to flexural bending and shear. The shape of such a moment must be carefully studied prior to detailing of reinforcement. The structural detailing of reinforcing bars must prevent relative movement or slip between them and the concrete.

Structural System of RC

7 kt ce c r2 ct2 1 2,66 c r ct

Action on structures BUILDING PARAMETERS: - width - length - wall height - height - wall frames

B= D= H= Hmax =

65,00 60,00 8,00 10,00

[m] [m] [m] [m]

ar =

7,10

[m]

- roff - chosen rooftype :

max wind pressure:

(Walls no. 2 and 4) (Walls no. 1 and 3)

8,7 [%] duopitch

5,0

3 kt ci c r2 ct2 1 1,78 c r ct

[o]

A.2. ROOF LOAD trapezoidalprofiles thickness glasswool thickness steel roof tile thickness

wall wind 0,18

[ kN/m ]

[ cm ]

0,32

[ kN/m ]

0,75

[ cm ]

0,06

[ kN/m2 ]

0,56

[ kN/m2 ]

characteristic roof load:

gr =

load for wall frames ar = 7,10 [ m ]

characteristic Gr = 3,97 [ kN/m ]

12,8 1,090

qi,max = qb,0*ci(zi) =

0,44

External loading coefficients for vertical walls:

zi = cr = kt*ln(zi/z0) =

reference hight for internal wind load: internal wind pressure:

[ mm ]

0,67

Cpe,10 Cpe,1 Cpe,10 Cpe,1 Cpe,10 Cpe,1 -1,0 -1,5 -0,8 -1,5 0,6 g = gkr * f

f 1,35

grc =

1,35

Grc =

0,75

[ kN/m2 ]

max wind pressure for internal wind loading

A.1. STRUCTURAL STEEL Rm-win - pc program, structural steel calculations

1,25

qmax = qb,0*ce(ze) =

[-]

[-] [m]

[ kN/m2 ]

Cpe,10 Cpe,1 1,5 -0,5 -1,5

Design wind loading for vertical walls:

[ kN/m2 ]

we = qmax*cpe(ze)*cpe Wind on Wall no. 1 and 3. e = 20,0 [ m ] load for wall frames [kN/m] we[kN/m²] ar [ m ] character. f cpe Area bredth hight area 1 20,0 8,0 160 -1,00 -0,67 7,10 -4,72 1,5 2 45,0 8,0 360 -0,80 -0,53 7,10 -3,78 1,5 3 60,0 8,0 480 0,60 0,40 7,10 2,83 1,5 4 60,0 8,0 480 -0,50 -0,33 7,10 -2,36 1,5

calcul. -7,08 -5,67 4,25 -3,54

Wind on Wall no. 2 and 4. e = 20,0 [ m ] load for wall frames [kN/m] we[kN/m²] ar [ m ] character. f cpe Area bredth hight area A 20,0 8,0 160,0 -1,0 -0,67 7,10 -4,72 1,5 B 40,0 8,0 320,0 -0,8 -0,53 7,10 -3,78 1,5 D 65,0 8,0 520,0 0,6 0,40 7,10 2,83 1,5 E 65,0 8,0 520,0 -0,5 -0,33 7,10 -2,36 1,5

calcul. -7,08 -5,67 4,25 -3,54

Sabah Shawkat © calculation 5,36 [ kN/m ]

ENV 1991-2-3:1995

A.3. SNOW LOADS

form factor for snowload on a monopitchroof: ci = 0,80 reference area: A= 1

[-] [-]

Sk =

2,1

[ kN/m ]

0,6

[-]

Ct =

1,0

[-]

S = ci*Ce*Ct*Sk =

1,01

[ kN/m2 ]

characteristic snow load:

load for wall frames ar = 7,10 [ m ] A.4. WIND ACTIONS general Informations

[o]

Ce =

reference snow load:

5,00

characteristic S = 7,16 [ kN/m ]

[ kN/m2 ]

1,5

Sc =

1,50

calculation Sc = 10,74 [ kN/m ]

1,51

roof wind external pressure coefficients for duopitch roofs: zone for wind direction = 0° angle = 5o

ENV 1991-2-4:1995 basic velocity:

vb,0 =

20,0

[ m/s ]

roughness length:

z0 =

0,03

[m]

terrain factor:

kt =

0,18

[-]

minimum hight:

zmin =

3,0

[m]

topografi factor:

ct =

1,0

[-]

direction factor:

Cdir =

1,0

[-]

time factor :

Cårs =

1,0

[-]

air density : reference wind force:

= qb,0 = 0.5* *vb,02 =

1,25 0,25

[ kg/m3 ] [ kN/m2 ]

reference wind velocity:

vb = Cdir*Cårs*vb,0 =

20,0

[ m/s ]

reference hight:

ze = h =

15,0

[m]

roughness factor:

cr = kt*ln(ze/z0) =

1,12

[-]

Cpe,10 Cpe,1 Cpe,10 Cpe,1 Cpe,10 Cpe,1 Cpe,10 Cpe,1 Cpe,10 Cpe,1 5,0° -1,7 -2,5 -1,2 -2,0 -0,50 -1,2 -0,4 -0,4 -0,4 -0,4 15,0° -0,9 -2,0 -0,9 -1,5 -1,075 -0,3 -0,4 -0,4 -1,0 -1,5 o zone for wind direction = 90° angle = 5

Cpe,10 5,0° 15,0°

-1,6 -1,2

Cpe,1

Cpe,10 -2,2 -2,0

-1,2 -1,2

Cpe,1

Cpe,10 -2,0 -2,0

-0,7 -0,7

Cpe,1

Cpe,10 -1,2 -1,2

-0,5 -0,5

Cpe,1 -0,5 -0,5

design wind loading for vertical walls: we = qmax*cpe(ze)*cpe External wind load: = Roof slope: 5,0 ° 5,0° (nearest ref. value ) 15,0° (nearest ref. value ) = 0° Min values Wind loading on duopitch roof, wind on the side of the building

Actions on Structures

X 2,00

CF =

Cpe,10 Cpe,10 Cpe,10 (nearest ref. value) (nearest ref. (found by value) interp.) (m²)

Y Area 5,00 10,00

-1,70

-2,50

2,00

5,00

100,00

-1,20

-0,90

-1,20

-2,00

-1,50

-2,00

5,00 1830,00

-0,50

-1,08

-0,50

-1,20

-0,30

-1,20

CI =

2,00

5,00 1830,00

-0,40

CJ =

2,00

5,00

-0,40

-1,00

-0,40 -0,40 -1,50 load for wall frames [kN/m] 2 we [kN/m ] ar [ m ] character. f

-0,40

CI =

-2,00

2,00

120,00

-1,70

-1,13

7,10

-8,03

1,5

masa suwnicy load for wall frames dynamic calculation reakcja od nacisku kó�

-12,04

CG =

-1,20

-0,80

7,10

-5,67

1,5

-8,50

-0,50

-0,33

7,10

-2,36

1,5

-3,54

CI =

-0,40

-0,27

7,10

-1,89

1,5

-2,83

CJ =

-0,40

-0,27

7,10

-1,89

1,5

-2,83

Wind loading on duopitch roof, wind on the side of the building

Area (m²)

Cpe,10 Cpe,10 Cpe,10 (nearest ref. value) (nearest ref. (found by value) interp.)

Cpe,1 Cpe,1 (nearest (nearest ref. ref. value value

Cpe,1 (found by interp.)

5,00

10,00

-1,60

-1,20

-1,60

-2,20

-2,00

2,00

5,00

55,00

-1,20

-2,00

2,00

5,00

260,00

-0,70

-1,20

2,00

5,00

1625,00

-0,50

-0,50 -0,50 -0,50 load for wall frames [kN/m] we [kN/m ] ar [ m ] character. f

-0,50

1,65

Rmax c =

1,65

G1cra c =

-2,20

calculation 94,05 [ kN ] 18,15

[ kN ]

wd2 =

1,00 = 1,50 characteristic Rmax = 110,10 [ kN ]

cranes brake for beams Hri maxk = 0,12 * Rmax = 6,84 [ kN ] characteristic Hp maxk = k * Nmax = 17,10 [ kN ]

cranes brake for beams

2,00

2 = wd1 * 1

3 = wd2 * 2 1,5

masa suwnicy G1cra = 21,20 [ kN ] 1,5 Rm-win - pc program, structural steel calculations

calcul.

CH =

characteristic Rmax = 57,00 [ kN ] G1cra = 11,00 [ kN ]

= 90° Min values

CH =

-2,50

CG =

Cpe

CG =

-1,70

Cpe,1 (found by interp.)

CH =

CF =

-0,90

Cpe,1 Cpe,1 (nearest (nearest ref. ref. value value

0,3

calculation Rmax c = 165,15 [ kN ] G1cra c = 31,80 [ kN ]

1,50 1,50

10,26 [ kN ] calculation HP max = 25,65 [ kN ]

1,50

HP max =

Hr max =

[-]

characteristic Hp maxk = 8,50 [ kN ]

calculation 12,75 [ kN ]

Sabah Shawkat © 10,00

Cpe

-1,60

-1,06

7,10

-7,56

1,5

CG =

-1,20

-0,80

7,10

-5,67

1,5

-8,50

CH =

-0,70

-0,47

7,10

-3,31

1,5

-4,96

CI =

-0,50

-0,33

7,10

-2,36

1,5

-3,54

A.5. SERVICE LOAD ( ROOF AND FLOOR ) 2 gs = 0,30 [ kN/m ] load for wall frames characteristic ar = 7,10 [ m ] Gs = 2,13 [ kN/m ] A.6. OFFICE FLOOR LOAD internal walls

concrete and steel thickness

[ cm ]

1,00

[ kN/m2 ]

3,00

[ kN/m2 ]

1,6 1,6

characteristic live floor load:

2 gf = 4,00 [ kN/m ] characteristic 2 gf = 3,00 [ kN/m ]

1,35 1,6

-11,34

0,48 [ kN/m ] calculation Gsc = 3,41 [ kN/m ] gsc =

g = gf * f

characteristic floor load:

calcul.

CF =

2 5,40 [ kN/m ] calculation 2 gfc = 4,80 [ kN/m ]

gfc =

A.7. TRAVELLING CRANES LOAD A.7.1

L = 13,5m G = 10t

load for beams dynamic calculation

Rmax =

57,00

[ kN ]

G4cra =

40,00

[ kN ]

wd1 =

1,10 1,50

G1cra = G4 / 4 =

10,00

[ kN ]

Action on Structures

(m)

Region 11

Region 1

extreme normal pressure pressure

extreme pressure

normal pressure

(kg/m ) Region III normal pressure

extreme pressure

(m)

q5 0

.:_ 195 �

�

-=t

190

160

-=-

1s5 �

-185 1

200

t--

/80

150 '-··- 11.0 130 :.:_ 120

t� : �

150 11.5

constant

tľ-- :::

200

constant

500 1.00 350

Sabah Shawkat © " _ f L:t:: : :

80 · � 70 �

�

-f

115

-- 110 ·

r. L

� �-

fr�

Action on Structures

q eqv L a)

Sabah Shawkat ©

Calculation of Equivalent Load

Calculation of equivalent load on simple supported beam subjected to uniform load

The bending moment and the moment of inertia at each point do not have two values. Wm

Sm 1  Mm Mm 1  Sm  Mm 1 Mm    2  2     6 Em  Jm 1 Jm  6 Em 1  Jm Jm 1 

S1  M0 M1   2    6 E1  J0 J1 

Sn  Mn 1 Mn    2  6 En  Jn 1 Jn 

The calculation of ideal loads will be simplified if Sm

1 1

Mpm 1



Em Jpm 1

2 1

Mlm

3 2

Em Jlm

 S m 



2 1

Mpm

3 2

Em 1 Jpm

 S m  1 



1 1

 Sm 1 

3 2

Mlm 1

Em 1 Jlm 1

6 E

S  M0 M1   2    J1   J0

6 E

Mn  S  Mn 1   2  J Jn   n 1

Where we get after editing Wm

Sm 1  Mpm Mlm 1  Sm  Mpm 1 Mlm    2  2     6 Em  Jpm 1 Jlm  6Em 1  Jpm Jlm 1 

For the ideal load calculated at the beginning, respectively. At the end of the beam we can write (m-1= 0, m+1= n) W0

Sabah Shawkat ©

 S m 

3 2

Em 1

Mm Mm 1  S  Mm 1   4   J Jm Jm 1   m 1

S, Em

Sm  1

2 1

Mp0

3 2

E1 Jp0

 S 1 



1 1

Ml1

3 2

E1 Jl1

 S 1 

1 1

Mpn 1

3 2

En Jpn 1

 S n 



Rotation line and deflection line y obtained on a fictitious beam using ideal loads. The overall process of calculating ordinate lines and y lists below The rotation is obtained as fictitious reaction of ideal load W n1  o l

S1  Mp0 Ml1   2    6 E1  Jp0 Jl1 

6 E





 W i r i

n 1 o

i  0

2 1

Mln

3 2

En Jln

 S n 

Sn  Mpn 1 Mln    2  6 En  Jpn 1 Jln 





 Wi ri

i0

The next rotation is obtained by subtracting the partial rotation over a certain interval as follows: 1

o

 

Example - Ideal Load

2

1

 

3

2

 

We will do this after calculating the last value: n

n

 n 1

  n

o





 i

i1

Furthermore, the relations for the calculation of partial rotations within the interval: 1. Linear course:

M J Sudden change of bending moment M at individual points: Sm  Mpm 1 Ml´m     Jl´m   Jpm 1

 m

2 Em

The bending moment at each point does not have two values:

Sabah Shawkat © Sm  Mm 1 Mlm     Jm   Jm 1

 m

2 Em

Applies to all values up

 m 1

  m

We have derived relations for the calculation of partial surfaces-partial rotations:

Or also: 1

 

o

 

o





 

o

 

o





i1

Generally: o





i1

 i

 1 a  n  1

Mn 1 Mn   Mn 2  8  5  12E  Jn 2 Jn 1 Jn  S

 n

 

The deflection line guidance is obtained as a bending moment from ideal loads from one side of the beam.

m

Mm Mm 1   Mm 1  8   J Jm Jm 1  m  1 

 5 

We use the expression a for the calculation

2

12E

 i

i1 3

 m

2

up to  n.:

2. Quadratic course

Generally: m

 1

 i

  0  W 0  S 1

  0  W0  

i1

Example - Ideal Load

Si  W1 S2

  0  W0  

i1 n



yn 1

0

 W0 



Si  W1 



i2 n

Si  W1 

i1



Si  W2 

Si  Wm 1 Sm

i3 n

Si  W2 

i2



Si  Wn 2 Sn 1

l n

  0  W0 m  W1 ( m  1)  W2 ( m  2)  Wm 1 1

l     0  W0 m  n 



i3

 Wi ( m  i)  

m 1



i1

This relationship implies directly: Generally, we can write:

  0  W0  

Si  W1 

i1



i2

Si  W2 



Si  Wm 1 Sm

Example: calculation of rotation and deflection line for uniform load on a simple supported beam, using equivalent load.

Since it is a regular division of the beam into equal parts S = l/n, we can write equations for rotation and deflection ordinates

 W 0 r 0  W 1 r 1  ...........  W n  1 r n  1 0 where l l l r0 n r1 ( n  1) r n 1 1 n n n then l l l  0 l W0  n  W1  ( n  1)  ..........  Wn 1  1 n n n 1

Data 1

Sabah Shawkat ©

 0 l

Because for m = 0, and for m = n, the expression on the right changes to the moment condition to the point n for calculation of rotation

i3

0

g  20kN m J 

b h

b  0.30m

J  0.03328m

h  1.10m

E=constant

l  10.0m

S  1 m

W0 n  W1 ( n  1)  ............  Wn 1 1 n 1

0





Wi ( n  i )

i0 m

  0  W0  

i1

Si  W1 



i2

Si  W2 



Si  Wm 1 Sm

i3

Because

S ym

l n

  0  W0 

l n

l l l m  W1  ( m  1)  W2  ( m  2)  Wm 1  n n n

1 2

gl x  gx 2

Example - Ideal Load

Bending moment

E  30000MPa



Mx E J

     

x

Curvature



Mx E J

 

dx  C 1 

 l3 1 3   l  4  12EJ  4

Rotation

 

Calculation of deflections:

 g  l2 1 3    3 l   l   C1  12EJ  4 4 

x

g l

 3 

24EJ

Then

    Mx 2  d x  C 1 x  C 2    EJ    

   dx x  

gl  x x   1  6  4   2 3 24EJ l l   3

x

g l

24EJ

3  g  l x



12EJ



2 g x

12EJ

After the introduction of dimensionless coordinates:

    Mx 2    d x  C 1 x  C 2  E  J    

x l





g l 2 3  1  6    4  24EJ





g l K1 (  ) 24EJ

Sabah Shawkat ©  1   EJ   

x



 1   EJ  

 1 1 2 gl x  gx dx 2 2   g

12EJ





24EJ

12EJ







 3l x  2x

K1 (  )

g 24EJ

 1 2 3   3 l x  2 x dx 12EJ  



 2 l  x  x





i

x 



24EJ

( 0  0)  C1 0  C2



gl  x x x    2   3 4 24EJ l l l  

g l K2 (  ) 24EJ

y

x1  1.0m x7  7.0m

x2  2.0m x8  8.0m

x3  3.0m x9  9.0m

x4  4 m x5  5.0m x10  10.0m

xi l

K1 i  1  6  i  4  i 2

gl K1 (  ) 24EJ



 2   i     i 

x0  0 m x6  6.0m



g gl 3 4  2 l x  x  x 24EJ 24EJ

K2 i

i

3 4  2l x  x   C1 x  C2

 6  4 



 1 1 2  gl x  gx dx dx 2  2 

 3 l x  2 x  C1

3 4 x g  x   3 l   2   3 4  12EJ 

2 3 x x  1 1 1  gl   g  2 2 3  EJ  2

g l  0.00035 24EJ

S 6 EJ

K2 i   i  2  i    i

 0m

1

kN

g l  0.00347m 24EJ

1

Theoretical values of rotation and deflections

 i

 g l 3  K1  i  104  24EJ 

 

Example - Ideal Load

 g l 4  K2  i  104  24EJ 

y i  

1 2

gl  500m kN

Calculation of rotation line and the deflection line

Equivalent load: 1

gl x  gx



 1 2  x x2   2 gl   2    l l 

M



2 x  gl   1  2 l 

Linear course M:

First we determine the values  i

x  l



2 EJ

 i

and the resulting of the ordinate rotation line:

 M i 1  M i 105

A final ordinate line rotation: K3  i   i  1   i

W 0 

6 EJ

gl K3 i

n

r1 5

dimensionless

 M i 1  4M i  M i 1 105

r0 0

dimensionless







l n

( n  1)

 Wi yi

r n 1

1

o

 

1 2

1

 

3

2

 

Wn 10 

Generally:

 M 10 1  2M 10 105

6 EJ

K1 i 

0.2 0.3 0.4 0.5 0.6



K2 i 

  i

i

 i 1

3

   0 

0.0981

0.09

0.792

0.1856

0.16

0.568

0.2541

0.21

0.296

0.2976

0.24

0.3125

0.25

-0.296

0.2976

0.24

y i 

W i 

M i  45

m kN

3.40625

2.75

6.44444

6.52778

1.97222

8.82292

105

8.61111

1.02778

10.33333

120

9.86111

10.85069

125

10.27778

-1.02778

10.33333

120

9.86111



3.61111

0

 34.37625

4

 10.20959





 i 





 i1  5   5    0   i1 

K3  i 

0.944

3.27778

2

    i  

1

5



   0 

 

 32.50125





 i 

4

 

i1

2



   0 

 

 27.29292



i1

3

 0.00125

 i



 1.875

 i

0

i0

5.20833 7.70833 9.375 10.20833

Reaction 4



i0

W i 

W 5 2



R  34.375

Wn 10  0.625

 i

 1.875

i1

Example - Ideal Load



i1

 i

 7.08333



i1

 i



 i 

 

 19.58459

10.20833

R 

i0

Sabah Shawkat ©

 0.1

 i

( 2 M ) 0  M 1 105 6 EJ 

W 0  0.625

i

W i 

M i 

 14.79167



 i



 24.16667

i1

 i

   

 34.375

i1

m 1

m 1

 M M ds E J

1 Em  J m 

 

 Sm  Mm  1  1

Em  1  J m  1

 

1 2



1 2





 Mm  Mm  1  S m 

 Sm  1  Mm 

1 2



1 2



2

 

3



 Mm  1  M m  S m  1 

1



3

Deflection line. S1  1 m S2  2 m y0  0 Y1    0  W 0 S1

S3  3 m

S4  4 m

S5  5 m

Y 1  33.75125 m

Y2    0  W 0 S2  W 1 S1

Y 2  63.8914 m

Y3    0  W 0 S3  W 1 S2  W 2 S1

Y 3  87.50376 m

Y4    0  W 0 S4  W 1 S3  W 2 S2  W 3 S1

Y 4  102.50502 m

Y5    0  W 0 S5  W 1 S4  W 2 S3  W 3 S2  W 4 S1

Y 5  107.64516 m

Sabah Shawkat ©

Based on the principle of virtual work we can write: 

 y m 1 

 ym 

Sm  1

ym  ym 1

 m 1

m

Mm  Mm  1

Qm  1

 ym 

Sm  1

 ym  1

   

m 1

m 1

m 1    M M  QQ  ds  ds  E J GA  m 1

ym  1  ym Sm  1

m  1

M m  1  Mm Sm  1

m 1

Lets adjust the left and right sides of the equation Left side 

1 Sm

 ym 1 

1 Sm

ym 

1 Sm  1

 ym 

1 Sm  1

ym  1

ym  ym 1 Sm



ym  1  ym Sm  1

 m 1   m

 m

Right side

Example - Ideal Load

   

m 1

m 1

   

m 1

m 1

   

m 1

m 1

   

m 1

m 1

   

m 1

m 1

 M M ds E J

 Mm  1

Sm Em  J m







Sm  1  Mm Mm  1  Mm      6  Em  1  J m  1  2 

Mm  Mm  1  3

Sm  1  3  M m  1  2  Mm  2  Mm  1   3  Mm  Mm  1  Mm     6 6   Em  1  J m  1  



Sm 6  Em  J m





 M m  1  2  Mm 

  QQ ds GA

Gm  A m

  QQ ds GA

Gm  A m

m

 Qm  Sm 

 Qm 

1 Sm



Sm  1 6  Em  1  J m  1

m 1 Gm  1  A m  1



 2  Mm  Mm  1

 Qm  1  Sm  1 

Reactions A

pl

12  E  J

Deflection y1

A

pl



12  E  J 2

pl

48  E  J



1 Sm  1

 Qm  1

Sabah Shawkat ©

After substituting the obtained relations into the original conditional equation we get: Wm

 m

Sm Sm  1  M m  1  2  Mm   2  Mm  Mm  1  6  Em  J m 6  Em  1  J m  1 m 1 m   Qm   Qm  1 Gm  A m Gm  1  A m  1









It is a formula by which we can easily determine ideal loads if we know the course of M and Q on construction and stiffness characteristics.

Three ideal loads: Ideal loads

The deflection is obtained as a bending moment from ideal loads on the respective fictitious beam. Deflection of simple supported beam, which is in the middle of the span loaded by force P:

Ideal load

 6E J  2

0  2

pl 

0  2

One ideal concentrated load

 6E J  4

2  p  l  0 2   4  6E J  4 

l 12  E  J

pl

pl

pl  8

 



6E J

 pl  pl   4   8

12  E  J

Reactions

Example - Ideal Load

24  E  J



2p1  2pl  2pl

6pl

192 E  J

 pl  2 pl   4 2 pl  pl      4  8  6E J  8 6E J  4 4

2

l 24  E  J



p1  4pl  4pl  pl

10  p  l

192 E  J

W2 55 W1  2

pl

192 E  J

 ( 6  10)

Deflections: y1

A

l 4

11  p  l

A

W1 

11  p  l

 W1 

l 4

A

pl

192 E  J

11  p  l

11  p  l l  192 E  J 4

192 E  J

 ( 6  10)

 Deflections: y3 192 E  J 4 768 E  J 2 y1

11  p  l

6  p3 l l

192 E  J

Calculation of rotation and deflection lines on uniform loaded cantilever beam (see figure) by ideal linear loads assuming a linear of bending moment. Input data

768 E  J

11  p  l

Uniform load:

l6  p  l

y 2 A   W 1      768 E  J 2 768 E  J2 2 4 192  E  J 192 4 E J 4

g  30  kN  m

1

Cross-section stiffness: EJ = constant Modulus of elasticity of concrete: E = 32500 MPa Modulus of the cross-section: b  0.4  m

J 

h  0.8  m

1 12

J  0.01707m

bh

Span of reinforced concrete beam:

Sabah Shawkat © l  20.0 m

S  2  m

x 0  0  m

x 1  2  m

x 2  4  m

x 3  6  m

x 4  8  m

x 6  12  m

x 7  14  m

x 8  16  m

x 9  18  m

x 10  20  m

x

1 2 dy dx

gl x 

      

Example - Ideal Load

1 2

gx 

M E J

1 12

gl



 x dx  C1

 

1 12

gl

1 1 1 2 2 gl x  gx   g  l   C1 2 12 2 



x 5  10  m





   dx x  

12  E  J

  C1



2 2  x3 gl x x  2  3  3 2 l  12  E  J  l  l 

y

3 2  x4 x x    2  3 2 24  E  J  4 l l  l

o

gl

2

  C1 x  C2

gl

12  E  J

gl

 K4

24  E  J

 K5

g l



i 

 i 4  2   i 3   i 2  

K5i 

y

 3  3  i2  i

g l

24  E  J

 K5 

 gl  5   K5i  10  24  E  J  4

yi 

Deformation of the beam: 3

 0.36058m

24  E  J



1 2

gl x 



Wi 

0.2

0.096

0.0256

0.04

346.15385

-40

0.3

0.084

0.0441

-0.26

302.88462

260

0.4

0.048

0.0576

-0.44

173.07692

440

0.5

0.0625

-0.5

500

0.6

-0.048

0.0576

-0.44

-173.07692

440

0.7

-0.084

0.0441

-0.26

-302.88462

260

0.8

-0.096

0.0256

0.04

-346.15385

-40

0.9

-0.072

0.0081

0.46

-259.61538

-460

Wi 

292.06731 m

-173.07692

923.07692

-21.63462

1590.14423

86.53846

2076.92308

151.44231

2253.60577

173.07692

2076.92308

151.44231

1 12

1 2 2

S S 6E J



gx 

1 12

gl

-21.63462

292.06731

-173.07692

R 

2  x x   gl 1  6  6   2 l 12 l  



Wi 

K6

1  6  6



 2  M0  M1  10



86.53846

923.07692

W5

dimensionless

R  18.02885

Linear course:

 g  l  K6 

6E J

1590.14423

i  0

 2

K6i  1  6  i  6  i

W0  147.83654



 Mi 1  4  Mi  Mi 1  10

 i 

Ideal loads assuming linear course M over sections: W0 

259.61538

 g  l  1000 m  kN

Bending moment:

M

-460 m  kN

0.46

Reaction

gl

 0.03606

12  E  J

Mi 

 i 

0.0081

Sabah Shawkat ©

 gl  5   K4i  10  12  E  J  3

gl

K6i 

0.072

yi 

K4i  2  i

K5i 

0.1

 K4 

12  E  J

Theoretical values of rotation and deflections:  i 

K4i 

i 

  x g x 2 x   3l   2  l    C1  x  C2  3 4 2  12  E  J  

 2l x  x  l x

24  E  J





 3l x  2x  l x

x

Mi  

 g  l  K6i





 Mi  1  Mi  10

The next rotation is obtained by subtracting the partial rotation over the interval as follows: r 0 

dimensionless

2E J

 0 

l n

n

1 l

dimensionless



n 1





r n 

l n

 ( n  1)

Wi  ri

r n  1 

l n

1

 0  17.30769

r n  18 m

i

 i  1   i

i  1

 1   0   1



Example - Ideal Load



 3    0 

 



i  1



 i

 



 2    0 

 



i  1



 i

 

5   5    0   i  



  i  1  

4   4    0   i  



  i  

 10

-90.14423 39.66346 126.20192

169.47115 169.47115 126.20192 39.66346

 1  245.91346

 2  336.05769

 4  170.19231

 5  0.72115









t2    M1  L M1 1 t1   L 11m t  L 2 1   2 t1  tM 1 2 1  L M1  10   M  10ρ is curvature  M1 L t1 t2M    t  Lh 2  t L  2 1  2  2  10   M1t1L t1  t2 t1  t2  10   Mt2 M 2 h   t  L   t hL   L 2  t1  t2  2  1   1 2   t  L  20  M2 L   10   M1  L  2 1 2   h h h   M1 2 2  L  h 20 h 2 h 1

  

        





   

The deflection line guidance is obtained as a bending moment from ideal loads from one side

ρ is curvature

ρh is curvature



 3  296.39423

 t110 t2    M     t  L  2 

 L



 h M

-90.14423

ρ is curvature

 t1  t2   t h  L   2   t1  t2      L h h  L   2 t1  tt2 h     t  L 2 curvature h10 h   M  ρh is   10  M M  t1  t2   t1  t2  ht1  t2   t L    t L   t  L 2 2    10  1 M  M    L  10 2 M M  M   M1  L M1h 1 L 1 m h 

 i  -263.22115

 M

 10





     

   t

  M1  L M1 1 L 1 m  t1  t2  2  t t   t    1 2  1 2     t  L  t1  t2    20  2   M2  Lt L  2  10  M M h 1  2   10 h M1  L   2 h 1   t1  t2   M  10   M1  L M1 1 L  1  m 30   t  L 2 X1  10 X X 1   11  X 2   12  10 X 3   31  X 3  133   30  2  30   11  303 X333  X1 X 1   11  X 2   12  10 X 3 20 X 3M 2  L 31  33  t t   12 2   h  33  11    t  L  1  M1  L   2  L  2  2  31  M1  M ds Rotation  11  1  10  2   M1  L    Reaction  11  ML1 (EJ L )  2 L  stiffness  EJ hReaction 1  M 2  3  M Rotation  11  11 ds L  ( EJ )   t t EJ     L 2stiffness   1   M    1 t N  L  L   10 3  N  N  2 t  t 1 dX s  X   ds  X 33  1  10X1    33 X3     30 10 11 2 12 3 31  X 3   33 h L  ( EA )  1  20   M2  L  EA  11 M  1h  N3  L  L 2   N N    t ds

Sabah Shawkat ©

of the beam:

Deflection line: S 1  2  m

S 2  4  m

S 3  6  m

S 4  8  m

S5  10  m

y 0  0





Y 1  261.05769m





Y 2  868.26923m





Y 3  1518.75 m





Y4  1996.15385m





Y 1   0  W0  S 1

Y2   0  W0  S2  W1  S1

Y 3   0  W0  S3  W1  S 2  W2  S 1

Y4   0  W0  S4  W1  S3  W2  S 2  W3  S1

Y 5   0  W0  S 5  W1  S4  W2  S3  W3  S 2  W4  S 1

Y5  2170.67308m

 33

 10  1ds   33  h  EA 2  M  L   t1  t2L ( EA ) 2 1 deflection     10   b  M L  L 2  M  h  M 2 1  3 1 1   M  1 11 L  2 t 10 t ds 1  10 2 11  10

rotation ( 0.5  L)

Reaction Rotation   1EJ 2X stiffness30 L ( EJ)  X     10 X X 1   11  X 2 2 12 X   1 3 deflection rotation  ( 0.5  L )  10   30 3 31 3 33  b t  t   t 1  t 2  1 22  33  h N3   b   3011 ( N  L )   b    30    ds  2    2     1   2     10 1 2 10  2 1    N3  L  L t  t    M 1    t ds  M  L   L  N  N 1   1 2  t  t     1 2 1 10 t M1  MN 33 ds2 3 (N L)  b   30 33 Reaction  b      ds   30   3ds h L  ( EA )Rotation  11 EA 2   2 11   stiffness L  ( EJ)    EJ   10  30 

X3   30  33  M deflection  10  1  N Example 33- Ideal 2  10   t  t 1 ds   N Load  332 h ds 1  h  M  L    L L  ( EA )  EA  2 1  1 3  2  10  1Reaction    10 M  M  2 1   Rotation 

X 1   11  X 2   12

 11

 

 10

X 3   31  X 3   33  t1  t2    N3 Lb L 

1   11  M  L 2  

 )  1  M  L  L  ( EJ  t1  t2   ds 2  30 t1 N3t2b  

stiffness  30

 t 1  t 2  

(N L)  b 

rotation (

 1  M1  L   2  L 2  3

 11

M EJ



M EJ



h 

 E 

 12

h L

3 bh 



  0.5  Ec



h L



Lh

  EJ

L

 EJ Lh

Lh



 1   M1  L 



   121  b  h3 

1    Ec  b  h

dimensionless 1 3 bh 24  L

M M2 EJ

M1

 1 kN m

NM3



1  kN M L

L  (hEJ L)

L  ( EA )

L  ( EJ) 10

 

   10 21  10

 10

 30

1  kN

1 m

 L  N3  L

 33

   1  30 



  M  Lt11  Lt2     

2

X2    21   22

 31 X1   32 X2   33 X3

 30

X3    33

   

 11

   

M1 h

  b  t ds

M 1  M´ EJ

 20

   

M2 h



L  ( EJ)

2

 21

1 3

L ( EJM)  M´ 2   

 ds

 10

N L   b    1   2  ds

 33

2

M L

N   t  t  L ds

 

X1    11   12

 21

 20

   

 21

  b  t ds

M ( EJ)

 10

 30 M2  M´ EJ

  

 30

 33

  

N   t  t  L ds

N3   b  t ds



 d    h

 20

X2    21   22

 M2  L    L



1 3

L  ( EJ)

 10

 t 1  t2 

   b  ds 

 10

XExample  Load - Ideal 10 1    11   12

 20

X2 N 21   22   t  L ds

 33

 

t

M  ds W ( EJ)

 20

1 2

 d    h  ds W

W,R,M are dimensionless

 11 X1   12 X2   13 X3  10  t t 1 2  30  33  31 X1   32 X2    33X3  30   Xb3ds  2   X3  20  1  X1   22   20 21  M211 L    XM21  M´  23  M2  M´ ds  21  21 ds 12  EJ EJ 2 h    21 X1   22 X2   23 X3

   M2  L 

N L   b    1   2  ds

 10

 M1  L  

   2 M1h M 3 2 2   b t dsb  30t 1 t N3   b  t ds   b2 t ds  20  1  10  1 2  1   h h  t1  t2   2    b20  ds  1    b  ds 1 2 2 1 2      20  M2  L   M1  L  2 2 h h

M L

 M2  L   L



1

  1b  ds 

 11 X1   12 X2   13 X3

   b  ds 

 t1  t2       ds 1  2  b  M2  L 

X1    11   12

 10

 M1  L  

 t1  t2       ds 1  2  b  M2  L 

X3   33  31 X1  132 X2  233 X3  30  M1  L   L L  N3  L 3 2    11  33  21 L  ( EA ) L  ( EJ) M 1  M´   21  12  11 ds  EJ  t1  t2  

 10

21  ds 22 1  20 LX2  1 m

 kN 3 120

 t1  t2      b  ds L  1  m 2ds  1  1

 20

bh

X1    11   12L  (EA  10 )

 21  23 M X1 M 22 X12 kN m X3N

2



 11 X1   12 X2   13 X3

L  ( EA )

 1  M1h  L   2  L20 N L  b 2   1   2  ds 3

 30

 33

   b  ds 

L 24 ( EJ  L)

1  kN m

 t1  t2       ds 1  2  b  M1  L 

 11

 L  N3  L

 t2 

Lh

 1  M2  L   1  L     0.5 L  E N 2 c 3  L  b  h  12  21   21    Ec b 3 h  33



 EJ ds 1 J

 10

1  kN

L

3

t1  t2  M1,M are dimensionless 3,ds =1   - ds  2  b 

1 m h

 M1  L  

Sabah Shawkat © h

 1  M1 1L   2 3L 2 E   12  b 3h 

 11

  EJ

 20

 10

24  L

mensionless

 21 X1   22 X2   23 X3

24  11 L X1   12 X2   13 X3

1    Ec  b  h

 10

L  ( EJ)

bh

 12

  0.5  Ec

12 t

L  ( EA )

 1  M2  L   1  L 2  3

 t1  t2      b  ds  2  1

 21

N L   b    1   2  ds 1    Ec 30 bh



L  ( EJ)

 10

1 3    0.5  Ec    bh  1  EJ L  EJ  b12 h  J 12

   121  b  h3 

M1,M3,ds =1 - are   EJ L - are EJdimensionless M1,M3,ds=1 J



M EJ

1 3  E  bh   M  12  M h h EJ

M EJ



 L  N3  L

 33

 M1  L  



Conditional equation:  11 X1

kN 2

m 13

  12 X2   13 X3   10 1

 m kN m

m

4 2

 32

 31

 11

 22

 LX1  3 Ec Ic 2

 12

 21

 LX2  3 Ec Ic 2

1 1.210 5 40

(h )

 21 X1

  22 X2   23 X3   20

 31 X1

  32 X2   33 X3   30

 23



 LX2  3 Ec Ic 2  LX1  3 Ec Ic 2

( 0.5h)

1 1.210 5 

 33

40  0

Ec Ic

T

LX3



Sabah Shawkat ©  10

 20

 T

  11   12 X1

L 2

 10

 30

 ( 0.5h) L

 10

  11   12

 33 X3

The calculation of stress on the upper chord and lower chord of the beam h

 X3  X1 h    A  Ic 2



d

 X3   A



 30 X3

  30     33 



X1 h  Ic 2

Actions due to Temperature on Reinforced Concrete Members

Sabah Shawkat ©

Actions due to Temperature on Reinforced Concrete Members

a result of the development of design methodologies and construction techniques. The concern

In the 20th century, concrete bridges are being built with increasing large spans as

Example for calculation of primary deformations, stresses and fictitious internal forces from the effect of nonlinear temperature on a T-shaped reinforced concrete beam

with durability problems and the use of more accurate methods of analysis led, in the preceding

Data: Material characteristic of concrete Eb  30000MPa

decades, to the design of these structures with reduced numbers of joints and bearings. With

t

 0.000012

i  1  7

this tendency for more monolithic bridges, the thermal effects began to play an important role

in the design and construction process. Several accidents due to inaccurate thermal analysis of

T- the nonlinear temperature distribution in the cross-section of bridge superstructures

bridges have been reported (Moorty and Roeder 1990; Thermal Effects in Concrete Bridge

Tm- uniform temperature in a linear gradient for each direction associated

Superstructures 1985).

ΔT- temperature difference and in nonlinear distribution To

The increase of the temperature of steel and concrete in composite steel-concrete elements, leads to a decrease of mechanical properties such as yield stress, Young's modulus, and ultimate compressive strength of concrete. Thus, when a steel or a composite structure is submitted to a fire action, its load bearing resistance decreases. If the duration and the intensity of the fire are large enough, the load bearing resistance can fall to the level of the applied load resulting in the collapse of the structure.

A- cross-section area, I- inertia h- height along the y-axis, respectively and yG= coordinate of the centroid G E- Young modulus of the material and αe coefficient of thermal expansion

Sabah Shawkat © ht  0  mm

ht  300mm

ht  500 mm

h t  650  mm

bt  200 mm

bt  200mm

b t  400 mm

ht  800 mm

ht  800.1 mm

ht  900 mm

bt  500 mm

bt  900mm

b t  800 mm

Actions due to Temperature on Reinforced Concrete Members

ht  1000mm 8

b t  700  mm 8

Temperatures

Static moment

T t  5

T t   5

Tt  0.000000

T t  5

T t  10

 20

Calculation of area of cross-sections: ht

ei 

( i 1) 3



SM  Ai  ei  ht i



i1

S M  0.2436964995 m

zt 

zt  0.654m

The calculation of modulus moment.

 ht bt  2 bt ( i 1) i i  b t  bt i

SM 

Ai 

( i 1)

1 2

bt  bt



r i  zt  ht  ei i

 ht ( i 1)  ht i

( i 1)



Jt  i

1 2

 b t  b t



2  ht ( i 1)  ht i  r i

( i 1)

Ai 

ei 

SM  0.06 0.04 0.045

0.0675 0.00007 0.084915 0.075

0.009 0.016

0.15 0.1 0.0833333333

3 2 b t b t  ht ( i 1)  ht i  i ( i 1 )  Jyt  b t  b t  36 (  1 ) i i i   bt  bt ( i 1 )  i 

Sabah Shawkat © 0.02625

0.049125

0.0000560038

0.072098829

0.0711666667

0.0777777778 0.0000547619

Jt 

ri 

0.504245136

0.0489705882

0.254245136

0.0488888889

Jyt 

0.0152557894 0.0025856236

0.0709118027

-0.0735326417 -0.1458096259 -0.1948254522 -0.2946437528

0.00045

0.0001333333

0.0002262818

0.00008125

0.0003649758

0.0001260417

0.0000014882

5.6746031746·10-14

0.003223115

0.0000705396

0.0065111206

0.0000624074

The modulus moment of cross-section then will be: JT  Jt  Jyt i

i 7

JT 

JT 

0.0157057894 0.0027189569



i1

JT  0.0290919664m

0.0003075318 0.0004910175

The total area will be:

0.0000014882

A 



A  0.372485 m

0.0032936546 0.006573528

i1

Actions due to Temperature on Reinforced Concrete Members

TB1 

-1

44.4444453333

-44.4444452

10.1111112667

22.2222217778

-15.5555551778

0.7222222044

20000.0000000288

-15997.5000000231

-50.1001502003

120.2103003905

-59.6081486892

-100

250

-136

The calculation of integrals:

 Tt ( i 1) ht i  Tt i ht ( i 1)   T2    i   ht  ht i ( i 1 )  

d v1  ht

 Tt ( i 1 ) i T 1  i ht  ht

( i 1 )

 bt i  ht

( i 1 )

 ht

dv2 

 b t ( i 1) ht i  b t i ht ( i 1)      ht ht i ( i 1 )  

 

dv2  ht  ht i i  ( i 1)

dv1 

0.09

0.027

0.2

0.16

0.15

0.1725

0.15

0.2175

0.237375

0.0001

0.00016001

0.000192024

0.0999

0.16983999

0.216807976

0.1

0.19

0.271

B2   

0.098

0.149625

Sabah Shawkat © i

( i 1 )

T1 

B1 

-33.3333333333 0

1

0 0

33.333334

1.3333333333

33.3333326667

0.6666666667

4000.0000000004

50.0500500501

-1.001001001

100

-1

0.2

-5

0.2

 

1 i

B2 

T2 

TB1

dv4 

dv4 

-0.4666666667

0.0081

-0.0333333333

0.0544

-3199.5000000004

0.11600625

-35.045045045

1.7009009009

0.23109375

-80

1.7

0.0002048384 0.2462951616

0.3439

TB12  T1 B2  T2 B1 i

TB12

i

dv3 

TB2

i

-21.666667

TB1



dv3  i

TB12

TB2

dv1

dv2

 0 -0.2

-0.0999999975 0.1750000032 0.00035 0.6327 1.1166666667

TB2  T2 B2 i

dv2 

-21.6666661333

TB1  T1 B1

 

d v4  ht  ht i i  ( i 1)

Actions due to Temperature on Reinforced Concrete Members

 

dv3  ht  ht i i  ( i 1)

dv3 

0.3

( i 1 )

Integration coefficients:

1

-6.6666666667

B 1 

TB2 

TB12 

1 i

 0.015 0.08 0.0556249985

-0.1315625023 -0.0002800192 -0.5407374712 -1.0658333333

t

The total integral: 7

1







t

1





i1

 1.6247166724 m

1



-4478.7211887085

i

i1 1



 Eb      zt  ht   t Tt  i i 

1 i

1069.6220997174

2

kN

2368.5176253347 1542.6892335476

  1.5877883275 m

716.8609137606 717.5103615234 -433.6913234308

Primary stress: 



t

 1



 0.000052342





t

 zt  1   1



  0.0002164826 m

1

n

 



M  zt  h t J t  J yt i



n



7870.4222770209 24616.0788744349

The calculation of fictive internal forces: N  A  Eb

N  584.898kN

2

kN

94762.9983255307

M  JT  Eb

M  188.9371m kN

1633.4729367333 -18504182.845172636

t





 Eb      zt  ht   t Tt  i i 

t

-8366.8150356337



-7063.5155919293

Sabah Shawkat © i

-4478.7211887085 1069.6220997174

2

kN

z i

2368.5176253347

  n   t i i

z i

3391.7010883124

1542.6892335476

25685.7009741523

716.8609137606

2

kN

3176.162170281

-433.6913234308

-18503465.984258875 -7649.3046741103

We should calculate the effective internal forces:

Ms  188.9371m kN

97131.5159508653

717.5103615234

Ms   JT Eb



-7497.2069153601

Ns   A Eb

N s  584.898 kN

These are introduced as a loaded state into the static structure. From the derived values of the

support bonds we calculate the course of the internal forces in the structure, which really express

Derivation of relations for calculation of primary deformations and stresses and fictitious internal forces

the influence of temperature respectively. Cooling of the structure.

Relative deformation The calculation of secondary and resulting stress then will be: For any cross-section in the structure, we can quantify the course of secondary stresses according to the relation.



i represents here as a z-axis coordinate



   zt  ht i The relative deformation from temperature will be i



t

 t Tt



t

Actions due to Temperature on Reinforced Concrete Members

th(y)  αi ΔT(y)

free thermal strains

while the primary effect we consider as self-load then the sum of internal forces should be 0 Nx = 0

By comparing relations above we get tot(y) = th(y)+ mech(y)

Nx = A  T1(y,z) dA

tot(y) = o + y o

Nx = A  (E (oT yT . zzT . y - T .T(y,z))) dA oT A  dA+yT A  z dA +zT A  y dA - T A  T(y,z) dA = 0

Where o is the strain at the centroid of the cross-section and o is the curvature, Positive curvature is defined in the same sense as positive moment. i(y) = Ei mech(y) = Ei (tot(y)- th(y)) = Ei (tot(y)- αi ΔT(y))  t Tt



t

 t Tt



t

oT = (T/A). A  T(y,z) dA For Moment Equilibrium to The Centroid Axis ycg My = 0

My = A  T1(y,z) z.dA

The total strain field is assumed to be linear (linear) in accordance with the Bernoulli

My = A  (E (oT yT . zzT . y - T .T(y,z))) z. dA

hypothesis, so mechanical (i.e. stress-induced) strains, will be introduced

Sabah Shawkat © yT . Iycg - A  T T(y,z) z. dA = 0

αi = coefficient of the thermal expansion of the material i, and ΔT(y) = temperature change at

yT = (T / Iycg ) / A  T(y,z) z.dA

height y relative to temperature at construction. after adjustment  tz i

Moment Equilibrium to The Centroid Axis zcg



E    zt  ht



   t Tti

Mz = 0

Mz = A  T1(y,z) y.dA

Since we do not consider applying another load, the condition equation to the neutral axis must satisfy.

Mz = A  (E (oT yT . zzT . y - T .T(y,z))) y. dA

primary stress T1(y,z) induce from the differ of section strain from temperature effects and

zT . Izcg - T A  T(y,z) y. dA = 0

plane section deformation (Bernoulli hypothesis) zT = (T / Izcg) A  T(y,z) y.dA

Primary Stress T1(y,z) can be Described by: oTyT . zzT . y = T .T(y,z) T1(y,z)  E T1=E (oT yT . zzT . y - T .T(y,z)) Strain and curvatures oT , yT, zT we can define from basic conditions of equilibrium as follows:

Secondary Stress Similarly, as to transfer the prestress to structures, also in non-uniform warming (temperature drop) induce on the statically indeterminate system secondary internal forces, then secondary internal forces induced stresses. Secondary internal forces we can solve by several methods such as force method or deformation

Conditions of Equilibrium:

method

Actions due to Temperature on Reinforced Concrete Members

N  0kN

Force Method

M  0kN m

Calculating the strain and curvatures, then we used them to define the right side system of

After overriding the right condition,

canonical equation (i0)

Axial force equilibrium requires: 0 =   dA =  Ei [tot(y)‐ αi ΔT(y)] dAi

Deformation Method

From the calculating of strain, we define forces load system as follow: N = oT.E.A

My = yT . E. Iycg

Mz = zT E. Izcg

With these forces effects we able to load the statically indeterminate systems in each section,

 E     zt  ht   t Ttz  dA i i  A





   tzi dA A

The bending moment condition from neutral axis

then the calculating reactions we used on certain system as actions and we define the internal

forces (we make continue as a similar way) Total Stress

   ti dA  zt  ht i A





 E     zt  ht   t Tt   zt  ht dA i i i A 









Sabah Shawkat ©

The total stress is the sum of primary stress and secondary stress

Substituting for tot(y) we get the value for the centroid strain:

Finite Element Method (FEM)

The structure should divide to finite elements, which subjected on thermal load. Then

o ={ ( i Ei) /(  Ei Ai)} .  ΔT(y) dAi

obtaining the total stress we able to solve the structure.

Similar logic for moment equilibrium leads to the following strain gradient of curvature:

The application of theoretical arrangement effects of non-uniform change temperature gradient to solve real bridge construction

o = (  i Ei  ΔT(y) dAi ) / ( Ei Ai)

 Case of temperature drop in two directions  In direction y, z

If the temperature file is shown, the variable o and o , which fully define the strain field, can

To solve the real bridge constructions, we use discreet solving model. Then we begin to divide

be computed, the resulting stress induced at any level can then be computed, then the integrals

the cross-section of bridge to several small surfaces, therefore to each small surfaces we assign

can easily be evaluated numerically.

the temperature load. Camber can be obtained by integrating the curvatures. For example, if the curvature is uniform Then the calculation of strain and curvature will be change from the double integral to simple evaluate summarily

over a simple span, the midspan camber is given by

 =( o L2) /8

oT = (T / A).  (Tm Am) To estimate the effects of variations in service temperatures, it is necessary to select a yT = (T / Iycg).  (Tm zm.Am)

representative temperature distribution and to calculate from it the resulting deformations and

zT = (T / Izcg ) /  (Tm ym.Am)

stresses. The temperature distribution in the concrete depends on the interaction of solar

Actions due to Temperature on Reinforced Concrete Members

radiation and re-radiation and of heat conduction and convection. The selection of the appropriate depends on the zone in which the structure is located.

b ( z)

bt bt zi   bt i  bt z( i 1)  ( i 1) ( i 1)   z   i     zi  z( i 1) z( i 1)  zi   zi  z( i 1) z( i 1)  zi 

So if

For average strain (y,z) we drive that

(y,z)  oTyT . zzT . y

( i 1)

 bt

z( i 1)  zi



and

( i 1)

zi  bt z( i 1) i

z( i 1)  zi

Then For average starin which induced by thermal we write that

b ( z)

(y,z)  T .T(y,z)T1(y,z)  E

B1 Z  B2

Analogously for temperature we can write Thus, from the neutral axis holds

i

T( z)

  Tt  zt  ht d.zi btz i i i Jty  i A t





T1Z  B2

where

( i 1)

 Tt

and

z( i 1)  zi

( i 1)

zi  Tt z( i 1) i

z( i 1)  zi

Next we calculate integrals for one surface

Sabah Shawkat ©

An and Jyt are cross-sectional values and t is the coefficient of thermal expansion.

I

The whole problem is focused on calculating integrals

  Ttibtz i dzi A

 ( i 1)   T1z  T2  B1z  B2 dzi zt zt

I

  Ttibzi dzi A

I

  Tti zt  ht i btz i dzi A





I

Since it is necessary to reasonably divide the cross-section by height into individual faces, we take the trapezoid as a universal geometric shape.

 ( i 1) T1B1z2   T1B2  T2B1 z  T2B2 dzi  zt zt

I

Then we can quantify the given integrals as the sum of the partial integrals of the individual trapezoidal surfaces, see Figure.

TB1  T1B1

According to LaGrange we write a linear course of width b (z) and temperature T (z) along the

and

TB12

 T1B2  T2B1

and

T B2  T 2  B 2

Then

height of the surface. zt ( i 1)

cross section width

b ( z)

z  z( i 1) zi  z( i 1)

I

bt ( i ) 

z  zi z( i 1)  zi

bt

  zt

 TB1z

( i 1)

I

TB1 3



 TB12 z  TB2 dy

zt



 

3 3   zti  

( i 1)

TB12 2

zt



 

2 2   zti   TB2zt( i 1)  zti

( i 1)

Actions due to Temperature on Reinforced Concrete Members

After the sum of the individual integrals we get t



We express fictitious internal forces Ms

 Jyt Ec

These

are

n 1





It

i1

Ns introduced

 A b Ec

load

case

into

the

static

structure.

From the derived values of the support bonds we calculate the course of internal forces in the structure, which really express the influence of warming respectively. Cooling of the structure.

Similarly, we assemble to solve the integral Iw

For any cross-section in the structure, we can calculate the course of secondary stresses 0

I

  Ttibt i zt  ht i dz A





according to the relatio

 nz

  zt  Tt bt dz   Tt bt dz i i i i A A



M  zt  h t i J t  J yt i



The resulting values of stresses in the cross-section are calculated according to the relation. z i

I

 

t  ( i 1)  T1B1z2   T1B2  T2B2 z  T2B2 z dz zt I   

 tz   nz i i

Sabah Shawkat © zt

Thermal effect on simple supported reinforced concrete beam

Then if we mark

I 1

zt zt zt ( i 1) ( i 1) ( i 1)    3 2    T1 B1 z dz  TB12 z dz  TB2 z dz zt zt zt i

At

So after the sum of the individual integrals we get

X1 

n 1  t  n 1  zt  It  I 1   i i   Jyt i i1  i1 



M = EI . Curvature

curvature

1 2

M1 

1 L

  t  t  1   d  2

EI 

At A1

d

 tu  to  t ds 



 L

After calculating the deformation of the cross-section, it is possible to determine its stress from non-linear temperature gradient, see the above formula.

 tz



E    zt  ht

reaction

   t Tti

Calculation of secondary and resulting stresses:

Actions due to Temperature on Reinforced Concrete Members

 10  11

0

  y z   z y

 t T

0

 t T t



69 

y

0

i

 i

  y z   z y

 z  

j y z   zy   t Tt

0

y

i

  z 0

 i

0

 E i

 E

 t T t

 y



z



The deformation and secondary a statically loaded by thermal load can be obtained by anystress of ofthe knownindeterminate methods structure of structural mechanics.

load can be obtained by any of the known methods of structural mechanics.

 z y

 j

There isThe a similarity to and the secondary design load dueofto the prestressing of statically cables deformation stress a statically indeterminate structure indeterminate loaded by thermal There is a similarity to the design load due to the prestressing of statically indeterminate cables

in the external links, rise to additional forces, load can be causing obtained reactions by any that of give the known methods of(secondary) structural internal mechanics. in the external links, causing reactions that give rise to additional (secondary) internal forces,

Therestress is a similarity to the design load due to the prestressing of statically indeterminate cables respectively respectively stress  i

0



  yj

0

 0   y z   zy   t Tt

 t T

0

  y z

  0   y z   zy   t Tt E

The deformation and secondary stress of a statically indeterminate structure loaded by thermal

j

z

  0   y z   zy   t Tt E 0



 j

z

 j

0 0

y

in the external links, causing reactions that give rise to additional (secondary) internal forces, respectively stress

  z y

 t dA



  zz y

 z y

y

M y  z y  z

 y z

Sabah Shawkat ©

When solving real constructions, we can usually divide most of the cross-sections into

elementary patterns bounded by lines. We also divide the temperature course so that the When solving real constructions, we can usually divide most of the cross-sections into

When solving real constructions, we can usually divide most of the cross-sections into temperature gradient over the height of the elementary pattern is linear. Then we integrate elementary patterns bounded by by lines. WeWealso the temperature temperaturecourse course so that elementary patterns bounded lines. alsodivide divide the so that the the within elementary patterns, converting the integrals into a sum of partial integrals. temperature gradient over over the height of of thetheelementary linear. Then integrate temperature gradient the height elementary pattern pattern isislinear. Then we we integrate The secondary stresses can be determined, for example, with a two-pole continuous within elementary patterns, converting the integrals into a sum of partial integrals.

within elementary patterns, converting the integrals into a sum of partial integrals. beam with field span

The secondary stresses can be determined, for example, with a two-pole continuous The secondary stresses can be determined, for example, with a two-pole continuous beam X   with   field 0 span 10

1 10 beam with field11span

X1  10  11 X1  M

X1  10  11

101

103

L 2 L

   

L 2

L  2 E LJ 2

2  



11



11

1 L 2 2  1 2 E J 3

1 L 2 2  1

12 EL J 32 2  1 2 E J 3

L

L 2 2  1 2 E  J 3 L

L L  2 1 2 E1 J L 3 1

X1 2  

2 2  1 2 E J 3

3 M on the continuous beam X1is the  Mvalue ofXSecondary     Etorque J 1

M X1 B   12  E m J H  4.5  m E  36000 MPa b1  8  m 2 M is the value of Secondary torque on the continuous beam

b2  0.6  m

 4.5 h1 12 36000 h  0.2  m b5 12  m 0.25  m 4.05  m  mSecondary B torque m Eh2the  MPa b15  8  m M is theHvalue of on continuous beam

A  b  h b21 0.61 m 1

2 2m  12 b2B  h2 12 0.25 b5 hm5E hA2136000 m hA15  4.05 m H  4.5bA52m m  MPa

 mm bhA15 2 0.2 m 82.43

2 A1 4.05  2m b5  h5 h2  m b5  12A2m b2 hh12  A0.25 5  m

2.43 m hA52 0.2 m

A 2  b2  h2

A 5  b5  h5

A1  2 m

b 2.4 AAb m h1 m 152 10.6

A5A 1  2.4 m b1  h1

A2  2.43 m

Actions due to Temperature on Reinforced Concrete Members

A5  2.4 m

Cross-Section: i  1  5 zh  0.25  m

zh  4.10  m

zh  4.30  m

zh  4.05  m

zh  4.5  m

zd  0  m

zd  0.25  m

zd  4.10  m

zd  4.0  m

zd  4.35  m

bh  8.0  m

bh  1.2  m

bh  12.0  m

bd  8.0  m

bd  1.2  m

bd  12.0  m

Temperature: Th  0

Th  0

Th  3.6

Th  4.5

Th  15.4

Td  2

Td  0

Td  3.6

Td  4.5

Additional auxiliary data for quantification of partial integrals

( i)

0.25

0.2

A5 

0.25

A calculated

zt  H  h5  t

Jyp 

zt  H  h5  t



to Yp



 h1

3

zt  2.361 m

Jyt  Jyp  A  t

1.2

72.6667

T1  i

Jyp  61.4757 m

3 3 3  b5  h5  b1   h1  h2   b1  2  b2  h2  3

 zd

Jyt  26.67481 m

( i)

0 -68.4 -311.6

TB12  T1  B2  T2  B1

1

TB2  T2  B2 i

TB2  i

21.6

-88.56

216

-820.8

872

-3739.2

Jyp  61.4757 m

Actions due to Temperature on Reinforced Concrete Members

-73.8

-64

Jyt  26.67481 m

Jyt  Jyp  A  t

 Td

( i)

Moment of Inertia: Jyp 

T2 

1

-8

TB12 

 b5  h5  b1   h1  h2   b1  2  b2  h2  3

t  1.939 m

T1 

TB1 

  h2  

TB1  T1  B1

t  1.939 m

zt  2.361 m

 A1   2Moment of Inertia: 2  2 A  2  A2  1

3.6c

h0.25 h2  h1  5  2  A2   A1    h2 

The centre of gravity is

t 

A  9.26 mc x 3.6 x

 zd

B2 

The centre4.5 of gravity  0.2 is calculated to Yp

t 

A5 

B1 

4.5  0.2

 Th( i)  zdi  Tdi  zh( i)     zh  zd ( i) i  

A  9.26 m

A1  2  A2  A5

A  A1  24.5  A2 x A 5 0.25 0.2 4.5

 bh( i)  zdi  bdi  zh( i)     zh  zd ( i) i  

B2  

T2  

Cross-Section area

Cross-Section area A 

 bd

Sabah Shawkat © B1 

dv1  zh i

( i)

 

dv4  zh   zd i i  ( i) i



1 i

i

TB1

 

TB1 4

 m

 dv2 

TB12

 dv1

 dv3 

5

TB2

1 i

TB2

 

dv2  zh   zd i i  ( i)

Primary stress: i  0  5

zh  0  m

zh  0.25  m

zh  4.1  m

zh  4.3  m

zh  4.35  m

zh  4.5  m

 dv4 

TB12

 dv3 

 

dv3  zh   zd i i  ( i)

 zd



TB12 2

 dv2  5

TB2 1

 dv1

 dv2

T0  2

0.432

t

-1.8288

2.43

T1  0

T2  0

T3  3.6





T4  4.5

T5  15.4

zi 

Ti 

 MPa

2.3614

-0.22

-79.497

zi  zt  zh







 E      zt  zh   t  Ti

-1.236

-9.783

17.91

Calculate the primary stresses at each cross-sectional point



-0.166667

2.118

2.1114 -1.7386

Sabah Shawkat ©

Sum of integrals:





i

1

 22.772 m

1 i

1

 91.27547 m

-1.9886

4.5

-4.292

-1.9386

3.6

0.326

1

0.685

-2.1386

15.4

i 1 5

1





i 1

Calculation of secondary and output stresses:

Let us express fictitious internal forces

The secondary moment M above the inner support pulls the lower filaments to its size

We calculate the relative deformations from the above expressions after consideration 5  t  1.2  10  

 

t

M

3       Jyt  E

 2

Jyt  26.675 m 3

 1

t

Jyt



  zt   1   1



 24301.1815 m  kN

E  36000 MPa

X1      E  Jyt

X1  24301.1815 m  kN

Ns    A  E

Ns  9837.504 kN

 0.0000295102

M



1

 0.0000168707m

1

 0.0000168707m

We introduce these as a load case into the static structure. From the derived values of the support bonds we calculate the course of the internal forces in the structure, which really express the influence of warming respectively. Cooling of the structure.

Calculation of fictitious internal forces

For any cross-section in the structure, we can quantify the course of secondary N  A    E

N  9837.504 kN

M  Jyt    E

M  16200.788 m  kN

stresses according to the relation

Actions due to Temperature on Reinforced Concrete Members

 

n i

Secondary stress:

 M

 zt  zhi Jyt

 11  x1   10



n i

2.15

 MPa

1.92

 11  10

-1.58 -1.77

 10

-1.81 -1.95

The resulting stresses will be the sum of the primary and secondary stresses

z

  n   t i

z

M1 M0

i 

 0.915 1.704 0.534

-1.082

 MPa

0 1 2 3

   M1  M0 dx EI 0 1

X1  L

1 8

gL

M1  X1  M0

V1  X1  V0

Sabah Shawkat © -1.486 -6.24

we evaluate the equivalent loads and then calculate the deflection.

Cross-Section stress analysis from the effects of no-linear temperature change.

N1  X 1  N0

Tgy Tey

1   A 

y   I 

by  Ty dy

Average temperature uniform

by  Ty   y  yG dy

Linear differential temperature at level y

Ty  Tm  Tgy

self-equilibrating temperature at level y

Actions due to Temperature on Reinforced Concrete Members

Where Ty- temperature at distance y between the neutral axis

T

z

N M  A W

This is the formula for the temperature when the external load acts on the structure, but this

    y dA 

   t  E  y dA 

E  Iz

E  Iy

E  Iz

   T  z d A Iz 

   T  y d A Iz 

t

Then we calculate the primary stress

is not our case because the structures do not exert any external load and therefore we have to start from deformations in the cross-section and then calculate the forces and primarily the

E 

i

E    t   y  z   z  y   t  T





stresses E    t   y  z   z  y   t  T t

 t  T

 t E

i

 t  T

E

 i  Ai

Ai  E   t   T

N i

  

 N i  zi

E    t   y  z   z  y   t  T

T

N  T  dA yt ( y)  0T  y1

Fictitious stresses, which then create fictitious forces



 0 ( T) 



 y  yG

yT ( y)

 0 ( T)

  0  ( T)   y  yG

 N i  yi

Sabah Shawkat © My

  

i

dA i

 i  zi

 i  yi

  tE 

yT ( y)

T dA i

dA i

  t E  

dA i

  t E  

 t  T( y)

T zi dAi



 T ( y)

 0 ( T)

  0 ( T)   y  yG

 t  T( y)



 T ( y)

The calculation of primary stress

T yi dAi

 t ( y)

 0 ( T)   0 ( T)   y  yG   t  T( y)  E

We can write the deformation of the free homogeneous element from their effects

t

N EA

  t  E   T dA  E A

  

   T dA A 

 t ( y)

   

t



T( y) dA

 t



  

 0 ( T)

 dA   0 ( T)   

 y  yG dA

   t 

   

T( y) dA  E

The calculation of curvature in y and z directions  0 ( T)

y

    z dA 

   t  E  z dA 

E  Iy

   T  z dA Iy 

t

   T  z dA Iy 

t

  

  A  t

 t ( y)

T( z) dA

dA   y  yG

  ( T)    0   



 y  yG dA   0 (T)    y  yG 2 dA 

Actions due to Temperature on Reinforced Concrete Members

   t 

T( y)   y  yG dA

   

 

  J 

t

 0 ( T)



T( y)   y  yG dA

 t 

1   A 

 0 ( T)  



  

 y  yG 2 dA

T( y)   y  yG dA

 0T

T( y) dA

 T  Tm

If we introduce a temperature constant (load) Tm then  T  T

 0t

h h   J 

T

Sabah Shawkat ©  

E   T  Tm 

 T ( y)

T( y)   y  yG dA

  T  Tm  

 T  T

 

  y  yG   T  T( y)

 

  y  yG   T  T( y)

 

 T  E  T( y)  Tm  T 

 

 T  E  T( y)  Tm  T 

 y  yG  h

 

 y  yG  h

 

Then the stress can be written  T ( y)

E  T0

Actions due to Temperature on Reinforced Concrete Members

Every building, whether it is large or small, must have a structural system capable of carrying all kinds of loads - vertical, horizontal, temperature, etc. In principle, the entire resisting system of the building should be equally active under all types of loading. In other words, the structure resisting horizontal loads should be able to resist vertical loads as well, and many individual elements should be common to both types of systems.A beam may be determinate or indeterminate Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the equations of static equilibrium. If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.

The area of steel provided over supports with little or no end fixity assumed in design, should be at least 25% of the area of steel provided in the span.Where a beam is supported by a beam instead of a wall or column, reinforcement should be provided and designed to resist the mutual reaction. This reinforcement is in addition to that required for other reasons. This rule also applies to a slab not supported at the top of a beam. A continuous beam carries a uniform load over two equal spans as shown in figure 3.1.1-1. In T-beam construction, the flange and web shall be built integrally or otherwise effectively bonded together. The effectively flange width to be used in the design of symmetrical T-beams shall not exceed 0.40 of the span length of a simply supported beam or 0.25 of the span length of a continuous beam, and its overhanging width on either side of the web shall not exceed 12 times the slab thickness, nor one-half of the clear distance of the next web.

Sabah Shawkat ©

The degree of indeterminacy is taken as the difference between the number of reactions to the number of equations in static equilibrium that can be applied.The degree of indeterminacy is taken as the difference between the number of reactions to the number of equations in static equilibrium that can be applied.

Continuous beams are those that rest over three or more supports, thereby having one or more redundant support reactions. Sufficient reinforcement should be provided at all sections to resist the envelope of the acting tensile force, including the effect of inclined cracks in webs and flanges.

Without shear reinforcement the beam would have a catastrophic failure due to shear-web and flexure-shear cracks. These cracks would form due to the shear forces in the beam and cause equivalent tension stresses that would cause failure in the beam since concrete is very weak in tension. There-fore stirrups at a determined spacing are used to provide a source of tensile strength against these shear forces (and equivalent tensile stresses. Concrete frame structures are very common – or perhaps the most common – type of modern building. This type of building consists of a frame or skeleton of concrete. Horizontal members of this frame are called beams and slabs, and vertical members are called columns. The column is the most important, as it is the primary load-carrying element of the building.

Structural System of RC

The structural system of a building is a complex three-dimensional assembly of interconnected discrete or continuous structural elements. The primary function of the structural system is to carry all the loads acting on the building effectively and safely to the foundation. The structural system is therefore expected to: 1. Carry dynamic and static vertical loads. 2. Carry horizontal loads due to wind and earthquake effects. 3. Resist stresses caused by temperature and shrinkage effects. 4. Resist external or internal blast and impact loads. 5. Resist, and help damp vibrations and fatigue effects.

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The design principle of Strong Beam-Column Joints is essential for building structure to resist horizontal load such as wind or earthquakes.

So the structure is actually a connected frame of members, each of which are firmly connected to each other. These connections are called moment connections, which means that the two members are firmly connected to each other. In concrete frame structures have moment connections in perfect fixed. This frame becomes very strong, and must resist the various loads that act on a structure during service load.

Structural System of RC

77 Redistribution of force effects in reinforced concrete beam structures and beam slabs

Example: Calculation of bending moments on continuous beam. The load applied to the beam is shown in Figure F-3.

Continuous beams and beam slabs loaded with uniform load where the span of the fields does not differ by more than 10% of the largest span.

g1 1

Values of bending moments Bending moment

g3 l 3 /4

under service load

1 2 qdL 11

vs 2  gs   L  12.5 10 

1 2 qdL 16

gs vs 2   L  24 12 

l 3 /4

under extreme load

extreme span

between support Fig. F-3 Continuous beam a) method of loading, b) equivalent load Loads acting on the beam:

internal spans

- triangular load g1 = 60 kN/m - equivalent load g3 = 70 kN/m

beams with two spans

- concentrated force acting in the middle of the span l2 P = 80 kN

1

10 L

1

2 qdL

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Equivalent loads acting in individual spans:

g3

48.125kN m





M 1 l1  2 M 2 l1  l2  M 3 l2







M 2 l2  2 M 3 l2  l3  M 4 l3

Fig. F-4

q 1 l1

q 2 l2 4

beams with three and more spans

The solution of the three-moment equation is done by means of a Claperron Method: 1

q 2 l2 4



q 3 l3 4

1

2 qdL

1

2 qdL

1

q d = gd + v d

1 q L2 11 d 2

Mred =M d (1- b p ) L

2 qsL

qd = gd + vd , (qs = gs + vs) - extreme (service) uniform dead load and live load L - the range of the relevant span, in case of different stresses,

M1 = M4 = 0 kNm

for other internal spans

Where

M3 = 144,28 kNm

1

When calculating the support moments, the span of the adjacent span is used

M2 = 34,92 kNm

at first inner support

0,425.L L

0,165.L 0,15.L

0,410.L

1 q L2 16 d

Mred

1 bp

1 q L2 11 d

1 q L2 16 d 2

24kN m

Over supports

1

qdL

37.5kN m

1

g1

P  2 l 2 3

1

qdL

2 qsL

2 1 q L2 16 d

0,35.L

0,15.L

0,35.L

0,15.L

1 q L2 16 d

0,35.L

Static diagram and bending moments

Calculation of Bending Moments and Shearing Forces on RC beams

0,15.L

0,35.L L

0  17  0  004 



m

min



min

The The rectangular rectangular cross-section cross-section with with tension tension and and compression compression reinforcement reinforcement shown shown in in Figure Figure bellow, is is subjected subjected to to bending bending moment. moment. The The task task is is to to verify verify the the stresses stresses in in concrete concrete and and bellow, reinforcement. reinforcement. ff ck 25  MPa Standard cylindrical cylindrical compressive compressive strength strength of of concrete concrete (MPa): (MPa): Standard ck 25 MPa 410 MPa MPa Standard strength strength of of steel steel (MPa): (MPa): Standard ff yk 410

0  25  v d

gd vd

where gd.min - extreme value of continuous load with load factorfg < 1,0 1 - straight line 2 – quadratic parabola

0.4 m m ): Dimensions of of cross-section cross-section (( m bb 0.4 Dimensions m ): Service value value of of the the applied applied bending bending moment: moment: Service Distances of of the the center center of of gravity gravity of of the the reinforcement reinforcement from from the the Distances edges edges of of the the cross-section cross-section (m): (m):

q d = gd + v d 1 q L2 16 d

1 q L2 16 d +

1 q L2 16 d

+ L

1 q L2 16 d

Assestement Urcenie sírky of Compression face tlacenej casti betonu width in Beam.

1 q L2 16 d

2 Area 2 ): Area of of compression compression reinforcement reinforcement (( m m ):

2 Area 2 ): Area of of tension tension reinforcement reinforcement (( m m ): Area of transformed cross-section: +A A22)) -- (A (A11 + +A A22)) Area of transformed cross-section: A A= = b. b. h h+ + n. n. (A (A11 + Es where E where s 15 nn E E 15 Ess,, E Ecc are are the the modulus modulus of of elasticity elasticity of of steel steel and and concrete concrete Eb

1 q L2 16 d

b b=1 m b

positive negative - kladne - záporné momenty momenty moments moments

b=1 m

positive negative - záporné - kladne momenty momenty moments moments

positive - kladne momenty moments

Calculation of of the the Calculation area: area:

negative - záporné momenty moments

2 x2

The shape of bending moments diagram on the affected continuous beam, resp. on a beam slab loaded with a uniform load

1 q L2 10 d

L/3

1 q L2 11 d 0,425.L

0,165.L 0,165.L

0,410.L



n A11 n A1

c c

Fc Fc

Neutral axis Neutrálna os Neutrálna os

h d h d

L/3

F.L 3

L/3

L/3 L

Rc = 0,425 q dL

+ Rc

2 3 F

L/3

II cr cr



8 3 F

0,425.L

Static diagram and calculation of cross-

sectional forces considering redistribution on

sectional forces on two-span continuous

two-span continuous beam, resp. on a beam

beam loaded by concentrated forces in thirds

slab loaded with a uniform load

of its span

b b

0.006 m

 s2 s2

n  M M s  n s

xx  dd11

IIcr cr (d d x x)) ( n n  M M ss  I Icr cr

Obr. Obr. C-3 C-3

x M s  x M s II cr cr

Ms M s 2

b  d d2 b

cc == 5,7 = 15 MPa  5,7 MPa MPa < < 0,6 0,6 .. ffck ck = 15 MPa => the the assumption assumption was was correct correct =>

120.59 120.59 MPa MPa   0.8 0.8  ffyk yk  

´ ´cc  

(b)Transformed Transformovaný (b) Transformovaný prierez section prierez

66.76  MPa MPa 66.76

Using the the graphs, graphs, we we obtain obtain the the Using coefficient: coefficient:

b b

(a)Beam Nosník (a) Nosník

Stress Stress in in reinforcement reinforcement (MPa) (MPa) s1  s1

n A22 n A2

Stress at at the the most most compression compression edge edge level level Stress concrete cross cross section section (MPa): (MPa): concrete c  c

1 9 F.L



A22 A2

4 0.006 m4

d22 d2

3 bb  xx  n  A  ( d  x) 22  n  A  d  x 22 n n x 33   A 22 ( d  )   A 11 d 11  x

A2 A 2  100  100 b  d d b

1.18 1.18

5.27 5.27 MPa MPa

´ ´s2 s2

Fs1 Fs1

x xx x3 3

A11 A 1

0 0

Moment crossMoment of of inertia inertia of of the the transformed transformed crosssection to to the the neutral neutral axis axis (m (m44)) section II cr cr

2 9 F.L

Rb =(0,575 q d L)x2

0,425.L

0,410.L

Ra = 0,425 q d L

L/3 L

1 q L2 11 d



x  nn  A A 2 (( dd   xx))  A 1 xx   dd 1 bb    nn  A 22 2 1 1 xx 0.228 0.228 m m

depth of of compression compression depth

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q d = gd + v d

Mred =M d (1- b p ) L

yk dd 0.55 0.55 m m M M ss 0.15 0.15 MN MN m m 0.05 m m dd 11 0.05 0.05 m m dd 22 0.05 2 A 0.0008m  2 A 11 0.0008m 2  2 A A 22 0.0026m 0.0026m

98.86 98.86 MPa MPa

1.23967 1.23967

Compressive stress stress at at the the upper upper edge edge of of the the concrete concrete cross-section cross-section (MPa): (MPa): Compressive which is less less than than  c ´ ´ bcin  tension 6.53 MPa MPa c   0.6 0.6  ff ck Tension reinforcement (MPa):  which is   6.53 c ck c bc which is less than  s2 ´ s2   122.55MPa   s  0.8  f yk

Calculation of Bending Moments and Shearing Forces on RC beams

z z

s2 s2  n n

Fs2 F s2

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Verification of Stresses – Concrete vs Steel

Analysis of Stresses-Section cracked and elastic Sections designed for strength (Ultimate Limit States) under factored loads must be checked for serviceability (deflection, crack width, etc.) under the specified loads. Under service loads, flexural members are generally in the cracked phase with linear distribution of strains and stresses. The computation of stresses using the straight-line theory (elastic theory) for cracked section is discussed below. RC beam of rectangular section, subjected to a specified load moment, M s . For this beam, the corresponding transformed section, neglecting the concrete in the tension side of the neutral axis. The centroid of the transformed section locates the neutral axis. Knowing the neutral axis location and moment of inertia, the stresses in the concrete (and steel) in this composite section may be computed from the flexure formula bc

Ms 

x Icr

The same results could be obtained more simply and directly considering the static equilibrium of internal forces and external applied moment.

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The internal resultant forces in concrete and steel will then be: Fc

b  x

bc

A 2 s2

The inner lever arm will be: z

d 

x 3

Equating the external applied moment M and the moment of resistance,

F c z

F s z

b x 



z

A2  s z

The moment of inertia of the section, Icr is given by: 3

Icr

b x 3

 n As ( d  x)

from which

 bc

M

x Icr

and

s

M A2 z

A reinforced concrete beam of rectangular section has the cross-section dimensions shown in Figure bellow. Compute the stresses in concrete and steel reinforcement for a given bending moment.

Analysis of Stresses in RC members

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Determination of Reinforcement to the RC elements

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Determination of Reinforcement to the RC Members

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Determination of Reinforcement to the RC Members



M sd  N sd  

2

Design of reinforced concrete members

 d 2 



b  d f cd

206 MPa

h M sd  N sd    d 2  2 

235 MPa

Bending w ith compression

b  d f cd

0,50 0,48



0,45

fyk

Bending w ith tension

245 MPa 300 MPa

M sd

325 MPa

Pure bending

375 MPa

b  d  f cd

0,43

392 MPa

0,40

410 MPa

0,38

412 MPa

0,35

450 MPa

req

190 MPa

0,30

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0,33

490 MPa

0,28

520 MPa

0,25 0,23 0,20 0,18

 N sd

Bending w ith tension

A req

  b  d  f cd  100  

Bending w ith compression

A req

  b  d  f cd  100  

Pure bending

A req

  b  d  f cd  100

 f yk

0,15 0,13 0,10

 N sd  f yk

0,08 0,05 0,03 0,00

0,03

0,06

0,09

0,12

0,15

0,18

0,21

0,24

0,27

0,3

0,33

0,36

0,39

0,42

4

 10 



4

 10 

0,45

h, d, b, d2 fcd, fyd Nsd Msd Areq

in m in MPa in MN in MNm in cm2



0,48



Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel an bi-linear diagram for concrete

Determination of Reinforcement to the RC Members

Design of reinforced concrete members

 h  d  2 2

M sd  N sd  



h  M sd  N sd    d 2  2 



0,50 0,48 0,45

b  d f cd

M sd



b  d  f cd

0,43

fyk

Bending w ith tension

b  d f cd

190 MPa 206 MPa

Bending w ith compression

235 MPa 245 MPa 300 MPa

Pure bending

325 MPa 375 MPa

0,40

410 MPa

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0,38 0,33

392 MPa

0,30

req

450 MPa 490 MPa

0,28

412 MPa

0,35

0,25

520 MPa

0,23 0,20 0,18 0,15

 N sd

Bending w ith tension

A req

  b  d  f cd  100  

Bending w ith compression

A req

  b  d  f cd  100  

Pure bending

A req

  b  d  f cd  100

 f yk

0,13 0,10 0,08 0,05 0,03 0,00 0,00

0,03

0,06

0,09

0,12

0,15

0,18

0,21

0,24

0,27

0,30

 N sd  f yk

0,33

0,36

0,39

4

h, d, b, d2 fcd, fyd Nsd Msd Areq

 10 



4

 10 



0,42

0,45

0,48

in m in MPa in MN in MNm in cm2



Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel an bi-linear diagram for concrete

Determination of Reinforcement to the RC Members

 d 2 



b  d f cd h M sd  N sd    d 2  2  2

b  d f cd

0,50



M sd 2

b  d  f cd

0,45

Design of reinforced concrete members

Bending w ith tension

. fyk 190 MPa

Bending w ith compression

206 MPa 235 MPa 245 MPa

Pure bending

300 MPa 325 MPa

0,40

375 MPa

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392 MPa

0,35 0,30

410 MPa 412 MPa

req

441 MPa



2



M sd  N sd  

0,25

450 MPa 490 MPa 520 MPa

0,20 0,15

 N sd

Bending w ith tension

A req

  b  d  f cd  100  

Bending w ith compression

A req

  b  d  f cd  100  

Pure bending

A req

  b  d  f cd  100

 f yk

0,10 0,05 0,00 0,00

0,03

0,05

0,08

0,10

0,13

0,15

0,18

0,20

0,23

0,25

0,28

0,30

 N sd  f yk

0,33

0,35

0,38

0,40

4

 10 



4

 10 

0,43

h, d, b, d2 fcd, fyd Nsd Msd Areq

in m in MPa in MN in MNm in cm2



0,45

0,48

0,50

Sections without compression reinforcement for pure bending or bending moment with axial force, based on the parabolic-rectangular diagram for steel an bi-linear diagram for concrete

Diagram of reinforcement ratio r versus coefficient m.

Determination of Reinforcement to the RC Members



 h  d  2 2

M sd  N sd   2

b  d f cd h  M sd  N sd    d 2  2  2

b  d f cd

0,43



0,40

M sd 2

b  d  f cd

0,38

Design of reinforced concrete members Bending w ith tension

Bending w ith compression

Pure bending

0,35

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0,30 0,28

req

0,25

0,33

0,23 0,20 0,18 0,15 0,13

 N sd

Bending w ith tension

A req

  b  d  f cd  100  

Bending w ith compression

A req

  b  d  f cd  100  

Pure bending

A req

  b  d  f cd  100

 f yk

0,10 0,08 0,05 0,03 0,00

0,00

0,04

0,08

0,12

0,16

0,20

0,24

0,28

0,32

0,36

0,40

 N sd  f yk

0,44

0,48

0,52

4

 10 



4

 10 



h, d, b, d2 fcd, fyd Nsd Msd Areq

in m in MPa in MN in MNm in cm2

r 0,56

0,60

0,64

0,68

0,72

Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel an parabolic-rectangular diagram for concrete

Determination of Reinforcement to the RC Members

Material data:

The difference between the graphs of RCM for design of tension reinforcement is:

Characteristic value of concrete cylinder compressive strength (MPa): Design value of concrete cylinder compressive strength (MPa): Characteristic yield stress of reinforcement (MPa):

fyk

fcd

fck 0.85 

25  MPa

fck

Diameter:

14.16  MPa

Provided:

0.3  m

Cover of reinforcement (m):

0.5  m

0.02  m

Effective depth of a cross-section (m):

Msd

0.48  m

0.15  MN  m

n   

t

10.17  cm

As  A2

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Msd



b  d  fcd

0.15

The required tension reinforcement A 2 ( cm ) is as follows:

0.04147

  b  d  fcd  100

8.456  cm

2) From the Design graphs of RCM: 

subjected to bending moment Msd. Use the bi - linear diagram for steel ( ´

0.04581 2

  b  d  fcd  100

9.340 cm

0.04179 2

The required tension reinforcement A 2 ( cm ) is as follows:

Assumptions:

3) From the Design graphs of RCM: 

0) and the bi - linear diagram for

concrete.

The required tension reinforcement A 2 ( cm ) is as follows:

Determine the required tension reinforcement of reinforced concrete rectangular beam shown on Figure bellow,

Characteristic value of concrete cylinder compressive strength (MPa):   b  d  fcd  100

8.521 cm

fck

From the Design graphs of RCM we obtain the reinforcement ratio: 

n

1) To apply the Design graphs pf reinforced concrete members the bending moment Msd has to be brought into a dimensionless form:

4  18

Which is greater than the value:

0.02  m

h  d1

Design maximum bending moment ( MN  m):

Number of bars:

18  mm

Cross-section ( m):

t

 0.90535

410  MPa

9.340

fcd

1.5

8.456

Design value of concrete cylinder compressive strength (MPa): Characteristic yield stress of reinforcement (MPa): Design yield stress of reinforcement (MPa):

Examples of RC Beams

fyk

fcd

245  MPa

0.85 

fck 1.5

25  MPa

fcd

14.16  MPa

Cover of reinforcement (m): Cross - section (m):

Figure bellow, subjected to bending moment Msd and tension axial force Nsd.

1.15

0.03  m

0.70  m

Determine the required tension reinforcement of a reinforced concrete rectangular column shown on

f yk

f yd

0.30  m

Effective depth of a cross - section (m):

h  d2

Design maximum bending moment (MN.m):

Msd

d 0.4

0.67  m

MN  m



Msd



b  d  fcd

From the Design GRCM we obtain:

0.21



0.11191

Diameter:  t

30  mm

Number of bars:

Provided tension reinforcement:

Which is greater than the value:

Reinforcing steel (MPa):

fyk

300  MPa

fyd

  b  d  fcd  100

31.85  cm

n   

As  A2

n

t

0.25  m

fck 1.5

fyk 1.15

fcd

14.16  MPa

fyd

260.86  MPa

0.5  m

0.04  m

Effective depth of a cross - section (m):

d2 d

0.06  m

h  d1

Tension axial force (MN):

Nsd

0.300  MN

0.46  m

Bending moment (MN.m):

Msd

0.030  MN  m

35.34  cm

Nsd  0

We obtain the value of eccentricity (m):

The required tension reinforcement

Provided:

3  12

A 1 ( cm ) is as follows:

Diameter:

A s1

The required tension reinforcement

A 2 ( cm ) is as follows:

Examples of RC Beams

0.1  m

 h  d  e  2 2 

d  d1  fyd

nc   

c

0.08  m

 10

Number of bars:

12  mm

c

Nsd

Nsd  

Msd

0.85 

Sabah Shawkat © A2

fcd

The required tension reinforcement A 2 ( cm ) is as follows:

25  MPa

Cover of reinforcement (m):

fck

Concrete (MPa):

Cross-section (m):

Material data:

nc

e

2.46  cm

As1

 Nsd 4   10  A1   f yd  

3.39  cm

9.0  cm

h 6

Provided:

Diameter:

3  20

Number of bars:

20  mm

t

A s2

nt   

t

nt

As2

Bending moment (MN.m):

9.42  cm

h  d  2 2 

Msd  Nsd  

0.01407

Sabah Shawkat © 2

Reinforcing steel (MPa):

Cross-section (m):

fck

25  MPa

fyk

300  MPa

0.25  m

Cover of reinforcement (m): d1

fyd

 10

fcd

fyd

0.85 

fck 1.5

fyk 1.15

0.5  m

0.04  m

0.06  m

fcd

14.16  MPa

fyd

260.86  MPa

Nsd fyd

 10  A2

Tension axial force (MN):

Nsd

9.20  cm

As above

h  d1

0.46  m

2.29  cm

As above

Conclusion:

With the acting tension force Nsd  0 within the lower and the upper reinforcement layers no concrete

compression zone appears. The tension force is carried completely by reinforcement and nothing by the concrete because the tensile strength of concrete is neglected.

Effective depth of a cross-section (m):

Concrete (MPa):

  b  d  fcd  100 

Nsd

The required tension reinforcement A 1 ( cm ) is as follows:

Material data:



The required tension reinforcement A 2 ( cm ) is as follows:

Or use GRCM

MN  m

0.03603



From the Design GRCM we obtain:

Msds b  d  fcd



0.027

form:

Msds

To apply the Design GRCM the bending moment Msds has to be brought into a dimensionless

0.030  MN  m

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement: Msds

M sd

0.300  MN

Examples of RC Beams

We show example, in which we design and checking a traditional reinforced concrete slab.

h 9 cm We suppose: Design of reinforcement for reinforced concrete slab:

RC slab subjected to bending moment as a result value from extreme load M sd

Geometry coefficient: Material Data: u

Material characteristics, concrete and steel: 11.5 MPa

Rbd

b

Rbd   b

Rbr

Rbtd

0.9 MPa

Rsd

375 MPa

s





1

15 kN m

M sd

ast1

ast1 



2.05078



a st

a st1



h  ast1

1    1 







0.27581

 2

0.8621

0.010 m

The required tension reinforcement to the cross-section will be:

Diameter of steel profile: 

 u  0.85

0.85714

 u  b  Rbr

Cross-sectional width: 1 m

u

 50

M sd

Design the depth of reinforced concrete slab

h mm he

Rsd   s

Rsr

1



M sd

 ast

 u   h e Rsr

0.00068m

0.015m

Sabah Shawkat © 2

We suppose:  st



We propose:



0.01

1





0.9   st 

R sr

R br





0.29348

   1 





0.85326

   1 







2





 2



1.99835



10 mm

A st

n   

A st

0.00079m

Distance the length between the steel profiles: 100 cm

0.25041

Assessment:

 lim 1.25 

 lim

Rsd

0.46667

M sd

420 MPa

ast

15 kN m ast1 



0.09m

ast

0.015m

The effective depth of a cross-section: he

h  ast

Effective depth of a cross-section:



M sd b  Rbr

0.07217m

h e  ast1

0.08217m

Examples of RC Slabs

0.075m

0.00068m

0.08m

92 Geometry coefficient: We design and assess the bending reinforcement of a monolithic RC beam of rectangular u

1

h mm

u

 50

cross-section reinforced on one side with a clear span ln. Extreme load of the beam, including

 u  0.85

0.85714

self-weight, is f d. Uniform load:

The depth of compression area of the cross-section: A s  R sr

b  R br

 lim h e

0.02206m

xu   lim h e

0.035m

1

58.5 kN m

span: 6.5 m

Ultimate bending moment capacity of the cross-section:

1.05 ln

6.825m

Material characteristics, concrete and steel Mu



 u  A st  Rsr  h e  0.5 xu



Md  Mu

16.15kNm

Cross-Section Ok To

11.5 MPa

Rbd

b

Rbd   b

Rbr

0.9 MPa

Rbtd

375 MPa

Rsd

s

Apply the design, diagram the bending moment has to be brought into a dimensionless form: Bending moment: d

fcdcyl.

fyk



Sabah Shawkat ©

0.075m



0.8 0.85



410 MPa

M sd 2

fckcube b

M sd

  

fyk

fyd

s



0.29412

15kNm

fcdcyl.

fyd



20 MPa

fckcube

A provided

0.00079m

M sd

340.62082kNm

356.52MPa

cross sectional dimensions of beam:

0.10089

 b  d  fcd  100

f l 8 d

9.07 MPa

b  d  fcd

A required

M sd

A required

6.86052cm 

6.76645cm 

A provided

A st

0.3 m

 st

0.013

Diameter of main reinforcement and concrete cover: 

12 mm



0.9  st 



   1 



as Rsr Rbr





2

25 mm

ast







0.38

0.30

Examples of RC Beams

as 



ast

2 1

   1  





0.031m

1.79



1

 2 1

 lim 1.25 

Rsd 420 MPa

 lim

 2

0.466



0.80

Effective depth of the cross-section:

M sd



b  R br

Geometry coefficient: u

0.56m

The depth of the Cross-section: h

h e  ast

60 cm

Design of beam reinforcement: h  ast

u

 u  0.85

0.969

0.596m xu

20 h  50

Height of the compression area of cross- section:

Design the depth of cross-section: h

1

A s  Rsr

b  Rbr

 lim h e

0.222m

xulim

0.26367m

Bearing bending capacity of the cross-section:

0.569m

Geometry coefficient: u

1

u

h  50

0.96923

 u  0.85



 u  A st  Rsr  h e  0.5 xu



362.66kNm

M sd

Md  Mu

340.62082kNm

cross-section complies. To apply the design graphs, the transformed action Msd, has to be brought into a

Sabah Shawkat © dimensionless form: d



1.78279





    1   2 

M sd

 u  b  Rbr

0.39112



0.80444

fyk

410 MPa

M sd

 b  d  fcd  100

A provided

A st

n    4

A st

0.0022m

Concrete cover of the Longitudinal reinforcement: as 

 2

ast

0.035m

Effective depth of the cross-section: he

h  a st

0.3m



fckcube



b

0.8 0.85

fyk s

0.39229

A required

0.00205m

ast

fyd

M sd

  

9.07MPa

356.52 MPa

fyd



340.62082kNm

fcdcyl.

0.15 2

23.052cm 

fcdcyl.



A required

 u   h e Rsr

20 mm



0.565m

b  d  fcd

Proposal and assessment: 

1



Required amount of reinforcement: As



20 MPa

fckcube

0.565m

Examples of RC Beams

A st

A provided

0.0022m

20.47414cm 

b

1.5

s

1.15

We design and assess the reinforcement of the monolithic beam of a rectangular cross-section reinforced on both sides subjected by the bending moment Msd.

0.25 m

 sc

Dimensions of a cross-section, width, height: b

We suppose that: 10 mm

0.45 m

M sd

M b1

M sd

0.9 MPa



A sc

 A sc  Rscr  zs

u

Bending moment from extreme load:

M b1

1.75089



Rbd Rsr

11.5 MPa

b

Rsd   s

Rscr

Rbr

Rbd   b

0.9 MPa

Rbtd

Rsd

375 MPa

s



Rsr

0.013

A std

12 mm

25 mm



0.79479

A st1d

h  ast





0.031m

A st1d  A sc

A std

0.419m

u

1



n   st  

 st

0.00132m

0.00148m

20 mm

u

 50

0.96

0.0015708m

1 Rbtd   b  h  A st  0.03 b  h 3 Rsr

0.00009m

Assessment, effective section height, geometry factor:

 u  0.85

h  ast

u

0.419m

1

h mm

1.64071

u

 50

0.96

 u  0.85

Percentage amount of reinforcement:

M sd  u  b  Rbr



1    1 





 2

It is necessary to design compression reinforcement because    lim

 lim 1.25 

A st

1 Rbtd  b h 3 Rsr

mm 

A st1d

Minimum percentage of reinforcement to a cross-section:

st sc st 2 Design of reinforcement to the reinforced concrete beam: he

M b1  h e Rsr

Concrete cover of longitudinal reinforcement: as

   lim

0.41042

Sabah Shawkat © 

A st

a st



Required amount of reinforcement:

Proposal dimensions of the RC beam:  st

1

0.00015708m

164.64491kNm

    1   2 

b  Rbr

A sc



M b1

Material characteristics, concrete, steel 10 425:

n   sc  

Rsd

 lim

0.46667

h e  asc

0.388m

A st

 st

b h

h e  ast

s

0.01396

Ast  Asc  b h

s

0.01536

0.388m

The depth of compression area of cross- section will be:

420 MPa M sd

 lim  lim  lim  1   2  

 st

lim

0.35778

A scd

u

 lim b  h e  Rbr zs  Rscr

A scd

0.00004753m

A st  Rsr  A sc  Rscr b  Rbr

Examples of RC Beams

0.1844m

 lim h e

0.19553m

 s  0.04

We propose the dimensions and reinforcement of a slab monolithic beam acting as a simple beam with span lb and axial distance of the beam ls. We consider the floor load and at the

lever arm of internal forces: zb

he 

same time the live load v with the load factor 1,2 and the purpose factor  . n u

0.3268m

The ultimate bending moment of a cross-section:



 u  b  xu  zb  Rbr  A sc  Rscr  zs



Choose preliminary dimensions:

Md  Mu

188.26238kNm

Slab:

To apply the design diagrams, the bending moment has to be brought into a dimensionless form: d1

asc

1.15

fyk

0.25m

410 MPa

M sd



fckcube



b

fcdcyl.  0.8 0.85

fckcube  20 MPa

s

0.419m

fyd

  

 b  1.5

180kNm

0.08 m

Beam: bw

0.25 m

2.25 m

0.50 m

lb  7.0 m

2

2.5 kN m

2

7 kN m

2

fcdcyl.  9066.67kN m

fyk

fyd

s

356.52MPa

Sabah Shawkat © We design in the compression area of the cross-section: 2

A sc  2.26 cm

2V12

M b1



M sd  M sc

M b1 2



M b1

0.37377

M sc

A 2  A sc



M sc

31.26268kNm

148.73732kNm



0.13747

b  d  fcd

A 2required



A sc  fyd  d  d 1

A 2required

The floor:

15.32 cm

 b  d  fcd  100

A sprovided

A st

13.06 cm

g spodlaha

g p  ls   u

g dpodlaha

g spodlaha  1.3

Self-weight of the reinforced concrete slab: gsdosky hsls  u

g ddosky

g sdosky  1.1

Self-weight of the reinforced concrete beam:

g stramu





bw  h  hs     u

g dtramu

g stramu  1.1

Dead load: gs

g spodlaha  g sdosky  g stramu

12.1125kNm

1

Examples of RC T-Beams

gd gd

g dpodlaha  g ddosky  g dtramu 1

14.3925kNm



3

25 kN m

Live load: vs

v n  ls   u

gs  vs

14.9625 kNm

 1

1

27.075kNm

v s  1.2

gd  vd

Rbtd

0.9 MPa

17.955 kNm

 1

1

32.3475kNm

Bending moment: M sd

f l 8 d b

M sd

198.13kNm

Material characteristics, concrete, steel 10 425:

Rbd Rsr

11.5 MPa

b

Rbr

Rsd   s

Rbd   b

Rscr

Rsd

375 MPa

s

375 MPa

Sabah Shawkat ©

Effective width, choose, diameter of longitudinal reinforcement:

as  25 mm

12 mm



ast

as 



ast



0.031m

0.17

we can consider that: h s  0.05h

Effective height: he

h  ast

0.469m

Geometry coefficient:

u

1

20 h mm



 50

1    1  



u

0.96364

 u  0.85



he M sd



3.85831

 u  b  Rbr 

0.0696

 2

Examples of RC T-Beams

Determine the reinforcement to a rectangular tie with the specified dimensions b h . Extreme load causes in rod normal force N sd and bending moment Msd.

 st1

Assumptions:

 st2   sdtmin

Characteristic materials, concrete, steel 10 425: b

0.30 m 11.5 MPa

Rbd

s

0.30 m

375 MPa

Rsd

h  ast

0.27m

600 kN

Nsd

h  ast  asc

M sd

25 kN m

Rsr

375 MPa

0.24m

Reinforcement design: 1 Rbtd  b h 3 Rsd

 stmin

u

0.00007m

1

M sd

Nsd zs

 50

u

0.94286

u  0.85

0.04167m

M tulim



 st1   sdtmin

0.0067 Ntulim zs2



 u  A s2  zs2  A s1  zs1  Rsr M tu



Nsd

M tulim M tu

M tulim  M tu Ntu  Nsd  Ntu  Ntulim Nsd

A s2

 st2

b h

53.31kNm

27.725kNm

0.05772m

25kNm

fckcube

20 MPa

fcdcyl.

0.8 0.85

Nsd b



fckcube



b

600kN

ast

zs1

small eccentricity

s

0.12m

A s1d

Nsd zs2  ed  u zs  Rsr

A s2d

Nsd zs1  ed  u zs  Rsr

zs2

zs1

zs2

0.00055m

M sds

  

fcdcyl.

9.06667MPa 

410 MPa

fyk

A s2d

0.00114m

A 2provided 2

16 mm

A s1

n

1 4



A s1



0.0006m

2

n

2 4



A s2

20 mm

A 1required

600kN

M sd

Ntu

 u  A s1  A s2  Rsr





Ntulim

 u  A s2  Rsr

Ntulim  Nd

25kNm

0.04167m

Ntu

Ntulim

fyd



657.58022kN

444.31096kN

small eccentricity

Examples of RC Beams

 10

M sds

h M sd  Nsd    d 2 2





M sds

47kNm

0.077

A 2required

A 1required

Nsd  A 2provided  fyd

0.00126m

Nsd

A s2

Assessment: Nsd

0.23703

 b  d  fcd  10 

A s2

356.52174MPa 

fyd

s

As2d   stmin

Proposal: 1

fyk

b  d  fcd

A s1d   stmin

A 2required

fyd

0.12m



A s1d

1.15

0.3m

1.5

Sabah Shawkat © 2

ed 

0.12m

0.01396

see graphs, we design the reinforcement to the given cross-section as follows): M sd

20 mm

 st1

b h

M tu 0.03 m

ast

0.9 MPa

Rbtd

0.03 m

asc

A s1

fyd

A 1provided

11.17439cm

0.00043m

A s1

A 1provided

0.0006m

Design the reinforcement of the lower belt of the Vierendel beam with the given dimensions. Extreme load induces normal force and bending moment in the lower beam

the reinforcement A s1 can be fully counted Nsd

Assumptions: 120 kN

Nsd asc

2.4 cm

b

s

ast

1 Rbtd  3 Rsd

 stmin

Rbd

Rsr

Rsd

 lim

Rsd

0.30 m

2.4 cm

Rbr

 stmin

 lim 1.25 

250 kN m

M sd

Rbd Rsd

Rscr

0.55 m

11.5 MPa

u

 b  xu  Rbr  A s1  Rscr A2

Rsr

0.00159m

Rbtd 0.9 MPa Rscd Rsd

375 MPa Rsr

0.0008

 

 lim

 lim  1 

lim

0.46667

 

lim

0.35778

420 MPa

Suggestion: he zs2

u

zs1

Nsd 1

0.502m

zs1

0.251m

Suggestion:

2.08333m

2

ed  zs2

u





114.04834kNm

20 mm

A s2

n

2 4



A s2

0.00188m

120kN

Ntu

 u  A s1  A s2  Rsr



M sd

250kNm



A s1

Ntu

0.00031 m

794.90148kN

A s2

Nsd

0.00188 m

120kN

M

227.46207kNm

Nsd

u  0.85

0.96667

Assessment:

large eccentricity

M ds2

u

M

h  asc  ast

0.251m

h  50

Nsd  ed  zs2

M ds2

0.526m

zs2

M sd

Sabah Shawkat ©

h  ast

M ds2  lim b  h e  Rbr

Ntu  Nd

Ntulim

 u  A s2  Rsr

Ntulim

683.2964kN

Ntulim  Nd

M  0

large eccentricity A s2  Rsr  A s1  Rscr 

Compressive reinforcement is not statically necessary with regard to the structural

 st2

arrangement and shrinkage of concrete, we choose:

0.01142

b  Rbr

Nsd u

0.13544m

1

14 mm

M ds2  A s1  zs  Rscr

 lim h e

0.24547m

A s1

n

1 4



169.5044kNm

xu   lim h e

A s1

 lim h e

0.00031m

he  he 

2 asc

0.048m

2 M b b  Rbr

0.10361m

2 asc

0.24547m 

0.048m

xu   lim h e

xu  2 asc

Examples of RC Beams

the reinforcement As1 can be fully counted





 u  b  xu  0.5 h  xu  Rbr  A s1  Rscr  zs1  A s2  Rsr  zs2  

Nsd

2.44289m



ed  eu

See the graphs, then we determine the reinforcement to the reinforced concrete cross-

M sd

SERIE I

2 b. d

7,000 -with stirrups -with compression reinforcement

section as follows:

M sd

250kNm

b

s

Nsd

0.3m



fckcube



b

0.8 0.85

fcdcyl.

1.5

120kN

  

ast

9.07 MPa

fcdcyl.

fckcube

fyk

20 MPa

410 MPa

6,000

5,000

-with compression reinforcement -without stirrups -without compression reinforcement -without stirrups

Sabah Shawkat © fyd

1.15

M sds

fyk

fyd

s

h M sd  Nsd    d 2 2



M sds



356.52

4,000

gs=1

219.88kN

3,000



M sds





0.29218

0.099

b  d  fcd

A 2required

A 2provided

A 1required

 b  d  fcd  10 

A s2

Nsd fyd

2,000

A 1provided

Nsd  A 2provided  fyd fyd

 10

A 2required

17.52998cm

1,000

A s1

fck,cyl=35 MPa A 1required

0.00155m

0,000 0,0E+00

a tot (deflection) [m] 1,5E-03

The Effect of Stirrups on Capacity of RC Beams

3,0E-03

4,5E-03

100

Sabah Shawkat ©

Bending Moments in RC Continuous Beams

101

Simplified method for the calculation of short-term deflection 0,50 0,40 0,30

k s5

k s4 0,20

Sabah Shawkat © 0,10

1,50

1,00

0,50

0,00 0,00

0,50

1,00

0,10 0,20

k s6

0,30 0,40 0,50

12 MPa



Msd 2

b d fcd

ksi  kbid d

a Deflection d Effective depth of a cross-section (m b Overall width of a cross-section (m) L Span of the beam (m) Msd Bending design moment (kN m)



Simplified method for the calculation of short-term deflection

 Deflections in RC Members 0,50 0,40

1,50

16 MPa 20 MPa 25 MPa 30 MPa 35 MPa 40 MPa

0,30

k s6

0,40 0,50

a Deflection d Effective depth of a cross-section (m b Overall width of a cross-section (m) L Span of the beam (m) Msd Bending design moment (kN m)

30 MPa 35 MPa 40 MPa



102

Simplified method for the calculation of short-term deflection



0,50 0,40 0,30



0,20

k s2

Ms 2

b d fcd

L ksi  kbid d k s1

Sabah Shawkat © 0,10

1,50

1,00

0,50

0,00 0,00

0,50

0,10

1,00

1,50

12 MPa 16 MPa

0,20

k s3

20 MPa 25 MPa

0,30

30 MPa 35 MPa

0,40

kb 0,50

Deflections in RC Members

40 MPa

103

Extremely stressed elements by bending moment and axial force Ms >0 Ns>0, tension, Ns<0 compression, effects from service load Input data, for compressive cross-sections



Cross-section: b

0.40 m



fckcube



1.5

0.8  0.85

fcdcyl.

  

h M s  Ns    d 2 2



9.06667MPa 

fcdcyl.







from graph, based on m we find r

0.29

0.194

b  d  fcd

0.50 m

Bending Moment, Axial Force Ms he

130 kN m h  ast

Ns he

460 kN

asc

2.8 cm

ast

3

3 cm

Rsd

15 MPa 190 MPa

Rbtn Rscd

Ns  10

 10

fyk

0.47m

7.39 cm

A 2required

1.15

Material characteristic Rbn

 b  d  fcd  10 

A 2required

Area and position of reinforcement: 1.4 MPa 190 MPa

Eb Es

27 GPa 210 GPa

t

16 mm

A st

nt   

c

12 mm

A sc

nc   

0.28261m

see graphs, we design the reinforcement to the given cross-section as follows):

t

A st

0.0008m

A sc

0.00023m

c

Sabah Shawkat © ef

Ms Ns

has (-) a sign of normal force

First we decide if cracks are expected if N s  0 ,  bg

1 ef

10.5 ef  h 6 ef  h

1

 bg

2.06

12m



3.53846m

1



1 ef

cracks not expected)

7.77



cracks expected

Calculation of ideal full-acting cross-section characteristics Distance of the centre of gravity of the ideal cross-section from the upper edge:



agi

1 b  h  2 n  A st  h e  A sc  asc  2 b  h  n  A st  A sc





agi

0.254m

agi

Moment of inertia to the axis passing through the centre of gravity of the ideal cross section Ii

fckcube

20 MPa

fyk

206 MPa

ast

0.47m Ai

h 2

b h 





 agi agi  h



b  h  n  A st  A sc

Examples of RC Beams





  n Ast  he  agi2  Asc  agi  asc 2 Ai

0.2080m

0.00455m

104

rt, (rc) The core line corresponding to the drawn (compression) edge, we determine it from the relation: rt





A i h  agi

0.85 Eb  Ii

Bfla

0.0892m

If cracks are expected: first determine the stiffness without cracks- bending:

A i agi

Axial stiffness: Bfla

Baxa

rt, (rc) the core line with respect to the drawn (printed) edge is determined as follows: axial force provided that tension concrete is decisive

4498629.29kN

next, we determine the rigidity of the cross-section with solid by eliminating concrete in tension With compressed cross-section

277.22kN

Nr1

1

Baxa

igi2  ef  0.5 h  agi

Ai (Ii) cross-sectional area (moment of inertia) of ideal cross-section

 bg  Rbtn 

104455.92kNm

0.08589m

Rbtn, (Rbn) is the characteristic tensile strength of concrete

Nr1

Bfla

( Ms  0Ns  0 ) in the cracks cross-section, the compressive area is always formed;

xr 3  a2xr 2  a1xr  ao

where



2 0.5 h  ef



0.06522m

axial force provided that compressed concrete is decisive

Sabah Shawkat © Ai

0.6 Rbn 

Nr2

1

Nr2

 4   n  A  0.5 h  e  h  A  0.5 h  e  a    st  f e sc  f sc  b

436.37kN

if it is Nr1  Nr2, then Nr

Nr1 ,







 n  A st  h e 0.5 h  ef  h e  A sc  asc  0.5 h  ef  asc 



0.03251m

0.01481m

The value x is determined by Newton's numerical method, i. iteration by relation: r

if Ns  Nr cracks are not expected,,

xri

0.5  h

if Ns  Nr cracks are expected:

xrii

 

Determination of stiffness- if cracks are not expected: bending stiffness without cracks

xri 

 

 2  xri3

ao  a1 xri  a2 xri

 

 2

a1  2 a2 xri  3 xri

xrii

0.198m

xrii

where the initial value (i = 1) can be considered x 0.5 h (index i stands for the iteration r i number) and iterations are repeated until the condition is met:

Bfl

0.85 Eb  Ii

Bfl

104455.92kNm

r i 1

igi

Next we determine:

0.147m

- axial stiffness of the cross-section without cracks B axa

B fl

 igi 

P

where P is the number of digits in two consecutive iterations agree. igi

1

r i

Radius of inertia



 e f  0.5  h  a gi

Baxa

0.198m





 he

  asc   1  A sc    1  ef xr xr    

b  xr  2 n  A st  

4498629.29kN



Examples of RC Beams

0.01845m

105

cross-section stiffness with fully eliminated drawn concrete

Calculation of curvature in rectangular cross-section subjected to bending moment: Reinforced Concrete Beam

Bending stiffness: Eb  P xr

Bflb

Input data:

Bflb

49431.0kNm

0.3 m

0.5 m

asc

0.038 m

ast

asc

Axial stiffness: Eb  P

Baxb

Bending moment from service load:

3107654.19kN

Baxb

 agi  2 ef    1  xr 

r

 Nr1    5  1 4  Ns 

M s  1.25

M sd

181.25kNm

the bottom surface is only exposed to the environment

The resulting stiffness: 1

145 kN m

r

r  0

0.50333

= (0,0,1)

h  ast

0.462m

Span, concrete class: l

7.0 m

Ecm

30500 MPa

2.6 MPa

fctm

Bending stiffness:

Sabah Shawkat ©

Bfl

1  r   r B   Bflb  fla 

Characteristic yield strength of reinforcement (determined with a probability of below 0.05 (MPa)):

67266.17kNm

fyk

410 MPa

Axial stiffness:

Reinforcement modulus:

B ax

 r  B  axa



1  r B axb

  

Bax

20861377.5

200000MPa 

Concrete mix consistency: S2 Relative humidity:

This stiffness came out with a negative sign, which means that the sign of axial deformation will not coincide with the sign of axial force:

50 100

Age of concrete at the beginning of loading: t0

60 days

Duration of wet concrete treatment: ttr

b

1.5

fck

30 MPa

s

1.15

fyk

410 MPa

days To apply the design diagram, the bending moment, has to be brought into a dimensionless form:

Examples of RC Beams

fcdcyl.

fyd



fck 



b

0.8 0.85

fyk s

 

fcdcyl.

13.6 MPa

fyd

356.52MPa 

106

M sd





0.20813

Defining flexural stiffness: 0.066

b  d  fcdcyl.  c1  c2  c3  tab1  (   0.5)   tab2  (   1.5)  

d

A required

A sc

 b  d  fcdcyl.  100

n sc   

 sc

A required

 sc

12.44074cm

14 mm

n sc

A st1

The bent member is considered satisfactory from the point of view of serviceability limit state, if the condition:

4 2

 st1

0.00031m

n st1   

 st1

14 mm

n st1

 d

The beam complies with the serviceability limit state. It is not necessary to count the deflection:

2 2

A st1

 st2

20 mm

n st2

Ratio of span to effective section height l

A st2

n st2   

 st2

22.72

A sc

d

A st2

0.00157m

A st

A st1  A st2

A st

0.00157m

15.15

Deflection calculation:

Sabah Shawkat © Calculation of creep coefficient:

Assessment according to the defining flexural stiffness:

Area of the whole concrete cross-section:

Type of construction: 

( 1825)

tab 1

 tab1

 tab2

b h

0.15m

Perimeter of concrete cross-section exposed to the environment:

Reinforcement percentage:

Ast is the area of the tension reinforcement, b the width of the cross-section 

A st b d



 100

kh1

kh2

kh3

kh1  kh3 2  kh2 h

1.13333

Reinforcement area required to transfer moment to failure: A required

12.44 cm

A provided

A st

Reinforcement area designed to cross-section (assumed) A s prov  A s req A provided

0.00157m

Determination of coefficients:

 c1

 c2

7 8

 c3

410 MPa A required    fyk  A  provided  

 c3

1.2627

Table values of equivalent cross-section thickness: ht

( 0.050.150.60.050.150.6)  m

Examples of RC Beams

0.15m

107

( 1 7 2890365)

t i

t0  tt i1 tt i  tt i1

( 0.050.150.60.050.150.6)  m

0.05  h 0  0.6

t i  1

h 0  h t j 1 h t j  h t j 1

0.6m



j 1

A si

0.03441m



Ss0

j 1

A  z 2  s j  s j    



Is0

( 0.6  0.15)  m



Eceff 3

h0  0.15 m

A si

 A s  zs   j j

Ss0

0.00073741m

Is0

0.00033572m

Characteristics of ideal full-acting cross-section ( A a I ) area, the distance of centre gravity g

t0  28

(days)

( 90  28)

from top, moment of inertia to centre of gravity:

0.51613

Area of ideal cross-section:





A c  e A stot

0.18441m

The creep coefficient is then obtained from the relation: Distance of the centre of gravity of the ideal cross-section from its upper edge:





c  

0.6m

 1  d t  d h  d t d h   t i1 j 1  d t d h   t ij 1 1  d h d t   t ij d t d h

b h

t i  1 j  1

Sabah Shawkat © cs

 e Ss0

agi

0.2766m

Moment of inertia of ideal cross-section:

Table values of creep coefficient:

b h

t

 5.4  3.9   3.2  2.6   2.0



2.0

t i  1 j

ccs

  2.5 2.0 1.9 1.7 1.5  2.1 1.6 1.6 1.4 1.2   1.6 1.2 1.2 1.0 1.0 

3.2 2.5 2.5 2.1 1.9



1.6

t i j











1 

A sc

A st1

e

10918.01MPa 

I fctm  h  agi





M cr

52.85165kNm

compressed concrete Scx the static moment of this area to the compression edge, then the Es

e

Eceff

condition from which the height of the compression area is calculated will have the form x : r

18.318

A x  S xc r

1

0.00002017kN

cross-section with the total elimination of concrete in tension. If we designate A cx the area of

0.00188m

Effective modulus of elasticity: Eceff

The height of the compression zone xr and the moment of inertia to the neutral axis Ixr of the

A stot

Ecm

Eceff  I

M cr

1.79

Calculation of the boundary bending stiffness of a fully acting cross-section:

Eceff

0.00454127m

The crack moment Mcr is then



A st  A sc

2.1

t i j  1

1 2.5 1  d t  d h  d t d h  2.0 1  d t  d h  2.1 1  d h  d t  1.6 d t  d h

A stot

 2

 e Is0  A i agi

For suggestion of full cross-section, the stiffness C will be: I

4.4 3.6 3.5 3.0 2.6 





t i  1 j  1



2.5

A st2

( 1  3)

zs 1

asc

zs 2

zs 3



   A e

Examples of RC Beams

s tot

xr  Ss0

108

In general, this equation should be solved numerically (eg by the log method or the falsi

Calculation of boundary curvature since shrinkage:

 r 2 what leads

b x

control method). However, at the rectangular cross-section A cx b  xr , Scx

Calculation of boundary curvature since shrinkage:

to a quadratic equation with the solution: Ss0

ags

A stot



ks   1  

1  2



Scx

 

b  xr

e

A stot

( 1 7 2890365)

 

0.20655m

A cx

b  xr

A cx

ttr  7

( 28  7)

Scx

( 0.050.150.60.050.150.6)  m

0.6m

ttr  tt i1 tt i  tt i1

0.06197m

(days)

0.11471m

0.05  h 0  0.6

ags  ks

0.39252m

0.14286

h0  0.15 m

( 0.6  0.15)  m

t i

t i  1

h 0  h t j 1 h t j  h t j 1 1

0.0064m

The creep coefficient is then obtained from the relation: For the calculation of the moment of inertia to the neutral cross-sectional axis with the elimination of the action of concrete in tension, the following general formula applies: 

 3

Icx



c   cs

 1  d t  d h  d t d h   t i1 j 1  d t d h   t ij 1 1  d h d t   t ij d t d h

t i  1 j  1

Sabah Shawkat ©

b  xr

Icx

0.00088m

Ixr



Icx  Scx xr  e Is0  Ss0  xr



Ixr

0.00292m

Table values of creep coefficient:

The stiffness of the crack-weakened cross-section will then be: 1

C II

CII

E ceff  I xr



1

0.00003138kN

coated rods, long-term load 1

1.0

2

 5.4   3.9  3.2  2.6   2.0

t

0.5

Effect of reduction of ductility of concrete between cracks:



3.2

t i  1 j  1

2.5

t i  1 j

ccs

4.4 3.6 3.5 3.0 2.6 

  2.5 2.0 1.9 1.7 1.5  2.1 1.6 1.6 1.4 1.2   1.6 1.2 1.2 1.0 1.0  3.2 2.5 2.5 2.1 1.9









t i j  1





2.5



t i j

2.0





1 3.2 1  d t  d h  d t d h  2.5 1  d t  d h  2.5 1  d h  d t  2.0 d t d h

2.428

 Mcr   0.93357   Ms  For members subjected to bending, the stress ratio is replaced by the moment ratio: 



Calculation of the boundary bending stiffness of a fully acting cross-section:

1   1  2 

Eceff

8895.83MPa

e

Es Eceff

e

22.48

1 r

m

M s  ( 1  )  CI   CII



Characteristics of ideal full-acting cross-section ( A agI ) area, centre of gravity distance from top, moment of inertia to centre of gravity Area of ideal cross-section:

Resulting curvature from bending moment: m

Ecm 1 

is the stress in the tensile reinforcement calculated on the section weakened by the crack from the

effect of the given load,  is the reinforcement stress at the same cross-section from the crack load.



Eceff



m

1

0.00444168m

0.19224m

Examples of RC Beams

109



Distance of the centre of gravity of the ideal cross-section from its upper edge: b h  ag

h 2



 e A st  d

 ag

ks   1  

1  2



0.27994m

ags  ks

 

 2

b  xr

Scx

0.22025m

A cx

0.06607m

Scx

b  xr

A cx

0.00728m

Moment of inertia of ideal cross-section I

b h

 

 e Is0  A i agi

For the calculation of the moment of inertia to the neutral cross-section axis with the neglect of the action of concrete in tension, the following general formula applies:

0.00534075m

for the full acting of cross-section, the stiffness will be: CI

1 2

Eceff  I

0.00002105kN

 3

b  xr

Icx



0.00107m

Hcx

Icx  Scx xr



Ixr

0.00336228m

Icx  Scx xr  e Is0  Ss0  xr

Ixr

Free shrinkage  cs is determined depending on the replacement thickness h 0 and relative

Icx

shII

 cs  e A stot 

ags  xr

0.00053418m

shII

Ixr

Hcx

1

0.00108203m

Sabah Shawkat ©

humidity r h rh

The stiffness of the crack-weakened cross-section will then be:

100

0.6m

CIIsh

Eceff  Ixr

1 2

0.00003343kN

Final shrinkage  cs  (% o)

Resulting curvature from shrinkage:

if so h  0.15 m, it will succeed in this relationship h 0.15 m 0 0  sh1  0.0006

 cs

 





ccs   sh1   sh2   sh1 

full development of cracks

ags

1 rshI Ss0 A stot

shII

ags

1 rs hII

0.45 m

1 rshI

 

 cs

0.0005

rshII

0.39252m

sh

( 1  )  shI   shII

sh

1

0.00103972m

5 2  l 48 m

0.02267m

Deflection in the middle of the span due to shrinkage: of the fully acting cross-section and curvature after the fsh

the weakened cross-section are then:

Deflection in the middle of the span from the bending effect (uniform load):

h 0  0.15 m

The curvature due to shrinkage

shI



 sh2  0.0005

shI

 cs  e A stot 

e

A stot b

ags  ag I

shI

1

 l 8 sh

fsh

0.00637m

Resulting of the total deflection:

0.00044513m

0.14079m

fm  fsh

Deflection does not fit

Examples of RC Beams

0.02904m

fmedz

l 250

fmedz

0.028m

110 Calculation of Deflections a) Deflection of a simply supported beam of uniform cross – section 1) Simply supported beam AB of uniform cross – section, subjected to

3) Simply supported beam AB of uniform cross - section loaded symmetrically by two forces F (Fig. VIa.3).

uniform distributed load (Fig. VIa.1).



bd fs1 i

fs1

5 48

gl

i

i bd



Ecm 

bd

   b  d2 2 l  1 3  9.6  E  12  b  d 



s i  b i d

l

1  cr i 

8

M l

1 5 12   i   48 Ecm 1  cr i

ks1 i



2





  3 1  cr i   d   Ecm  b   12      b  d2  l2

fs3 i

0.11458 

fs3

0.11458

M l



Ecm  I

  3 P l  0.02864 E  I 

4

Ecm





1  cr i





RDM

d

 

l 4

 12  i

0.11458 

ks3 i

 

 l

 Ecm



    l  1.37496 E  d  

2   bd 3  l l  0.02864 12  4   3 E bd 



1  cr i

i  b  d 12  Ecm 

1 12

bd



1  cr i

l

ks2 i

i Ecm



1  cr i

P

(Fig. VIa.4).

i  b  d  l

fs4 i

P l E

1 12

3 3

bd

0.99984

4M 3 l l 3

Ecm  b  d

9.575  Ecm 

0.99984

 l

Ecm  d



1  cr i

bd

i

ks4 i

9.575  Ecm 

4M

0.02083

unsymmetrically with a triangularly distributed load of maximum intensity w

2 2

fs2 i

4) Simply supported beam AB of uniform cross-section loaded

9.6   i  B el

concentrated force F at the middle of the span (Fig. VIa.2).

Sabah Shawkat © Msd  l

1.25 

2) Simply supported beam AB of uniform cross – section, subjected to a

 a 

P l



76.6  E 

1  cr i

P

l 8

 x 

  2 l

fs4

12 8 l

bd

1  cr i

8M 3 l l 1



76.6  E 

Deflection Examples

1 12

1.25326 d

 2 1 l  d 1  cr i





P l

76.6  E  I

111 5) Simply supported beam AB of uniform cross – section, loaded symmetrically with a triangularly distributed load of maximum intensity w (Fig. VIa.5).

b) Deflection of a free cantilever beam of uniform cross – section 1) Free cantilever beam AB of uniform cross – section, subjected to uniform distributed load (Fig. VIb.1).

2 2

i  b  d  l

fs5 i

10  Ecm 

1 12

P l



10  Ecm 

bd

i

ks5 i



1  cr i

P

l 6

 x 

 

2 2 2

1  cr i

i  b  d  l

fcf1i

4  Ecm 

6 l

1.2 



  Ecm ( 1  cr ) d

1 12



bd

1  cr i

P l

Sabah Shawkat © 60  E  I

60  Ecm 

d

6) Simply supported beam AB of uniform cross - section loaded symmetrically

fcf1

with a triangularly distributed load of maximum intensity w (Fig. VIa.6).

gl

2M 4 l 2 l

8E I

8  Ecm 

bd

 1    l2 1    1 1  cr i   4  Ecm  12  d 

3 

   Ecm  d  1  cr i   l



2) Free cantilever beam AB of uniform cross - section loaded unsymmetrically with a

triangularly distributed load of maximum intensity w (Fig. VIb.2).

2 2

fs6

i b d l

0.10152

Ecm  b 

ks6 i

0.10152

12   i Ecm



1  cr i



1 1  cr i

0.01081 E

1 12

4 3

bd

0.10152

M l

 2

 bd gl

0.10648 w  l

0.01081

0.10648 l E

1 12

l

bd

  2 l 0.01081 0.10648  1 E d  12 

  1.21826  Ecm  d ( 1  cr )     l

2 2

fcf2i

i  b  d  l 5  Ecm 

1 12

bd



1  cr i

Seviceability Limit State of RC Members

P l 3l

P l

15 E  I

112 c) Deflection of a cantilever beam of uniform cross – section

P l

3M 

15  E  I

2 3

l

3 l

15  Ecm 

1) Cantilever beam AB of uniform cross – section, with a uniformly distributed load of

bd

15  Ecm 

 l

1 12

d

5  Ecm 

1 12

 d

maximum intensity w is supported at A (Fig. VI c.1).

1  cr i

3) Free cantilever beam AB loaded unsymmetrically with a triangularly distributed load of maximum intensity w (Fig. VIb.3).

 x 

3 8

 

 l

fc1

M l

0.0768

fc1 i

Ecm  b 

i  l

0.0768

Ecm 



1  cr i

Sabah Shawkat © 2 2

5.5   i  b  d  l 20  Ecm 

pl

fcf3i

9 128



bd

2P l

1  cr i

3l

11 P  l  60 E  I

pl

0.0054

E

0.0054 

bd

128   2 l 9 E

0.0768 

d

 l

E



d

1  cr i

2) Cantilever beam AB of uniform cross – section, loaded unsymmetrically with a

triangularly distributed load of maximum intensity w is supported at A (Fig. VI c.2).

3M l 3

11 P  l  60 E  I

11 60



2l E

1 12

l

3   l

11 3

bd

 E

1 12

d

1 11  3  12   l   60  2 E  d 1  cr i





2 2

fc2 i

i b d l

12.47619 Ecm  b 

fc2

M l



1  cr i

2 3

12.47619 Ecm  b 

Seviceability Limit State of RC Members

P

l 16.8

x  

 

5

113 d) Deflection of a fixed beam of uniform cross – section

M 3 16.8   l l

209.6  E cm 

1 12

16.8  (  )  l 3

209.6  E cm 

bd

1) Fixed beam AB of uniform cross – section, subjected to uniform distributed load of

1 12

maximum intensity w (Fig. VId.1)

d

3) Cantilever beam AB of uniform cross – section, loaded unsymmetrically with a triangularly distributed load of maximum intensity w is supported at B (Fig. VIc.3). ff1i

ff1

fc3 i

0.07199

fc3

0.07199

12   i  b  d  l Ecm  b  d

M l

24    l 3

 x 

 

2

ff1

0.0625

M l

0.00609

E

1 12

0.0846 l

l

bd

384 E 

d

2) Fixed beam AB of uniform cross – section, subjected to a concentrated force F at the middle of the span (Fig. VId.2)

0.0846 P  l

0.671  l)

bd

 bd

bd

1  cr i

384 E 

0.00609

0.0846 E

1 12

l



2 3

bd

0.00609

0.0846 E

1 12

l

d

0.86383

 l

E d



E I



Ecm 

kf1 

24  M 4 l 2 l

pl

Sabah Shawkat ©

2 2

  i  b  d2  l2  1  0.0625   3 1  cr i   d   0.85  Ecm  b   12  

  i  b  d2  l2

  3 1  cr i   d   Ecm  b   12  

ff2i

0.04167 

ff2

kf2 

1  cr i l



pl 192 E 

1 12

 x 

P l 8

8 l 3

bd

Seviceability Limit State of RC Members

192 E 

  2 l

ff2

1 12

ok d

0.04167

M l

E I

114 3) Fixed beam AB of uniform cross – section, loaded symmetrically by two forces F (Fig. VId. 3)

2 2

i  b  d  l

ff3

i  b  d  l

1 4

bd

 l

bd

E d



P  ( 0.25  l)

 ( 3  l  4  a)

 ff3  2 l

2M l

M l

24  EI

12  EI

1     3  l  4   l 4  3 

M l

0.0625 2 

0.0625 l 1

l

ff5 i

24  E 

bd

24  E 

i  b d  l

Ecm 

3840  E 

bd



1  cr i

32  M



bd

1

d

  cr i



6) Fixed beam AB of uniform cross – section, loaded symmetrically with distributed load of maximum intensity w (Fig. VId.6)

P

 x 

l 24

 ff4  2 l

P l

384 EI

0.0625

M l

2 2 2

EI ff6

384  E 

M 3 l l 1 12

24  12    l 3

bd

384  E  d



1  cr i

24 



0.03418

i  b  d  l

384  E 

Ecm  b 

0.05833

  i  b  d 2  l2  1   3  1   cr i    d  Ecm  b   12  

7  32    l

P l

ff5

M l

2 2

24 

  2  l



0.05833

triangularly distributed load of maximum intensity w (Fig. VId.4)

 x 

1  cr i

0.0625

bd

4) Fixed beam AB of uniform cross - section loaded unsymmetrically with a

ff4

p l

Sabah Shawkat ©

 x 

P a 24  E 

P a

M l 0.0625 l

l



3 1  cr i Ecm   b  d  

12  Ecm 

distributed load of maximum intensity w (Fig. VId.5)

5) Fixed beam AB of uniform cross - section loaded symmetrically with a triangularly

0.002321 g  l E

1 12

bd



1  cr i

162

wl

ff6

0.002321 162  E

1 12

d

 11

l



1  cr i

Seviceability Limit State of RC Members

0.03418

M l

115 Deflection calculation Concentrated load in the centre of the beam, EI = constant The greatest bending moment in the centre of the beam is P

a- 0,5. l

0,5. l zoa

A' Pl 4

Fm 2

l /6

Rotation in support A:

If the beam is loaded with a symmetrically acting triangular moment surface Fm, will be a

b

A´

2  EI



1 2



PI

Fig. I-2

Fm 2

Fig. I-1 Simple beam with single loads in the middle

rotation:

0,5. l zob

l /6

0,5. l

ymax

P .a.b l

0,5. l

P l

Mmax



Rotation in support B:

Pl

16  EI

a

l  EI



P a b I



I 2



lb 3



P  a  b  ( l  b)

6  l  EI

Sabah Shawkat © If l  a b , then the deflection will be

Maximum deflection in the middle of the span: y max

P l

Mmax  l

48 EI

Deflection:

y max



Pl

     2 6



max  l

1 P a b l  b  a   6 EI l

P a b

3  EI  l

Mmax

2  max

relationship first

A´

0 determine maxy z and here we calculate a fictitious maximum mmax

moment mmax . Finally, we calculate the maximum deflection yb y max

(I-3)

6 E  h

The maximum bending moment in the centre of the beam will be: P ab

Mmax

Distance of centre of gravity of triangle from support:

z oa

In the same way we get:

z ob

a



Mmax

ql

(I-8)

(I-4)

l l

Uniformly loaded beam is = constant

Concentrated force in any position, EI = constant The greatest bending moment on the beam is:

(I-7)

operate at the same point. So that we can determine y max , we have to get out of the

Deflection y p is not the maximum as seen from the picture because Mmax a mmax do not

(I-2)

Mmax  

12 EI

Stress:

y max

 1 F m 1   A´     2 2 6 EI 

m max

(I-5) (I-6)

P  a  b  ( l  a)

b

(I-1)

F m  z ob

3 l  2 a  l

la

lb 3

Seviceability Limit State of RC Members

ymax

116 Fm

Fm 2

q l2 8

Fm 2

3 l 8 2

Fig. I-3

a

The greatest deflection will be in the center of the beam:

y max

 384 EI

5 Mmax  l  48 EI

From the relationship

y max

b

 q



q l

(I-9)

24  EI

l 3 l  A´      EI  2 8 2 1

1 EI

q

 5

l 16

A´ 2  max

y max

 max 

, we get the equation I-14. In the same

(I-16)

B  z´  MB

To calculate the rotation, first determine the fictitious reactions: (I-10)

Mmax

M A  MB

way we get from the right side: 2

(I-15)

A  z  MA

If we put relation into this equation h A

Uniformly loaded simple beam Fm

ql

0,5. l

rotation:

We get the moment in the cross section from the left

0,5. l

(I-14)

M A  MB    MA  z  l  

(I-11)

A´

a

 1 M l  1 l  1 M l  2 l 2 B  3 2 A  3      l

A´

 Mb  l 

1 3

 MA  l

M B  l  2  MA  l

(I-17)

6 l



 2  MA  MB



b



 2  MB  MA



(I-18)

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Bending moments at both ends, EI = constant

For MA  MB we get reactions in the commanded direction (Fig. I-4) with values

M A  MB

(I-12)

6  EI

The support moment on the analysed support is twice the value. The deflection in the centre

y 0.5l

 M A  M B   l2

(I-19)

16 EI

of the beam will be:

If we assume that, MA y 0.5l

MA MB A'

l /3

Fig. I-4

l /3

maxy

M l

M , then the deflection in the centre of the beam will be:

8  EI

(I-20)

Console, EI = constant

In the Console area, the deflection and rotation are zero, achieving their highest values at the free end. Now if we change the Moment diagram as a dummy load of the unchanged system,

Support moments on simple beam

The moment surface is now trapezoidal. Imagine it consisting of two triangles, the moment at any point in the field is as follows:

6  EI

we get the opposite results: the fictitious transverse force and the fictitious moment have extreme values at fixed area and Zero values at the free end. Thus, it appears that the boundary conditions that apply to the elastic line are not met to determine the fictitious forces

(I-13)

MA  z´  MB  z l

If we decompose the trapezoid into a rectangle and a triangle, then the moment will be

and moments on the actual beam. The solution is easy to find: after the transverse force calculation Q and bending moment M , we move from a real system to a reciprocal or fictitious beam (a console with fixed on the opposite side). Out of load q

Seviceability Limit State of RC Members

M we calculate the

117 fictitious transverse force Q´ and fictitious moment m , thus, as with a simple beam Q´

y´



Just as we applied a positive moment as a downward load on a simple beam, we now apply a negative moment as an upward load (Fig. I-5c). So that we can calculate fictitious

transverse force and fictitious moment anywhere, we assume part of the beam cut

a) Konzola - rovnomerne zatazená

a) Concentrated load at end of console, EI = constant

edge of the imaginary beam (Fig. IM( z)

between the cross-section and any

(I21)

P  ( l  z )

5d). we get:

y b) M - diagram

When perfectly fixed is Mmin

q.l2

P  l . To determine the fictitious transverse force Q´ ( = EI -

multiple rotation) and fictitious moment m (= EI - multiple deflection) it is necessary to load the cantilever with a fixed and the opposite side along the beam with a real moment diagram.

(b) Uniform loads,

Q´

 P  (l  z)  z

 

z

(I-22)

z

 z 2

(I-23)

M.I 3l 4

the console is transmitted on a fictitious beam as a load in the

Fig. I-6

commanded direction (Fig. I-6c):

max m

Cantilever loaded with uniform load.

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1 2

 l 

 z

max

2

2 1 1  P  l  z   z   P  ( l  z)  z   z 3 2 3

z

P l

2  EI

P z

6  EI

maxy

(I-24)

 ( 3 l  z)

P l

(I-25)

Maximum transverse force:

maxQ´

Q´ ( z

Maximum rotation:

maxy´

y´z

Maximum fictional moment:

maxm

Maximum deflection:

maxy

3  EI

max

ql

3  M l  l 3 4

ql

6  EI

ql

1 ql  3 2

 M l



ql

maxm

ql

8  EI

(I-26) (I-27) (I-28) (I-29)

d) Výpocet hodnoty Q a M v mieste z

M(z)=-P(l -z)

cross-sections are rotated relative to each other, with no mutual offset or lateral displacement. The deformer is referred to as an angle deformer (Fig. I-12).

l -z

c) M - diagram ako fiktívne zatazenie fiktívneho nosníka

y'= O

Bending moment: When determining the bending moment influence lines, the two adjacent

P.(l -z)

0,5.P.l .z e) Pootocenie

b) M - diagram

P. l

mz Qz

P. l

0,5.P.(l -z).z

P.l

 Q´ P   l  

f) Priehyb

Fig. I-12

max Q y

P. l y max = 3

Fig. I-5 Console loaded with concentrated load

y max

y´

P l z 

1 3

Q' =

The bending moment diagram on 1

max Q'

c) M - diagram ako fiktívne zatazenie fiktívneho nosníka

EI = constant

klb

M b

Seviceability Limit State of RC Members

r x O=1

118

Transverse force: Two adjacent cross-sections shall be transversely displaced relative to each other, with no twisting or delay (Fig. I-13).

Q Q b Fig. I-13 Normal force: Two adjacent cross-sections span each other. There must be no twisting or transverse displacement of the cross-sections (Fig. I-14).

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b Fig. I-14

Seviceability Limit State of RC Members

119

Determination of the rotation at the end of cantilever beam due to axial force P at the end of beam, E and J are constant, the moment M´ =1 subject at the end of beam. 1 

     M M´ dx   EJ  0  1

The integrals containing Q an N are equal null and, since the virtual load gives N´=0, and Q´=0 M

 P x

M´

kN m m kN 4 m 2 m

    EJ  0

  

l c



M M´ dx

The integration should be carried out here along two sectors, namely at lengths L and c. Part L

1

P c L

Part c

Mx´



M´

1 x L

M´x´

1

x

P x´

 c  P c  1     x x dx   EJ  L  L   x´ 0  l

P x´( 1) dx´

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 1´

    ( P x) ( 1) dx    EJ  0 

 1´

 L2   EJ  2 

  P x dx EJ 0 1

 x2   EJ  2  1

 P 

   

P EJ

P  x  x´  c   2 2 EJ 3 L 3



   





x



dx  

 

P  L  c  c   2 2 EJ 3 L 3



 P 

We ask for the rotation of the right end cross-section of the beam, shown in figure, which is finally loaded by the force P. in the cross-section where the rotation is sought, we apply the torque M=1, as EJ=constant, and the influence of the transverse force neglected.

Seviceability Limit State of RC Members



   x´

x´

P  L c  c    EJ  3  2

dx´

120

Relative rotation of two cross-sections When determining the relative curvature of two cross-sections, we imagine that in their

M´x

x L1

A x 

Mx´

intersections there are two opposite oriented moments M = 1 (1.25), so we obtain the equation of work. l

 ab

 M M´  Q Q´  dx    dx   EJ  GA 0 0

P1 L1

x´ L2



p1 x

 L1 x  x

p2 L2 2



 N N´  dx  EA 0

As an example, we will determine the relative rotation of the end cross-sections of two simple

p2 x´

p2 L2





C x´ 

 ab

L   L1  2 p p1   x x´ 2 2 2   L1 x  x  dx    L2 x´  x´  dx´ 2 2 EJ  L1  L2   0  0 

beams that are jointed and loaded with a uniform load.



x´ 

p2 x´

Mx´



 L2 x´  x´





 ab

L   p  L1 P2  2  1  2 3 2 3   L1 x  x dx  L2 x´  x´ dx´  EJ  2L1  2 L2  0 0  

 ab

 p  L x3 4 3 4  P2  x´ x x´   1  1      L2    4  3 4  EJ  2L1  3 2 L2  









Sabah Shawkat © 

Relative rotation of the beam at a common support point b In the joint we apply a couple of moments M =1. Then the rotation l l  1 2 M M´  dx  EJ 0



 ab



4 4   p   1  4L1  3L1    12  EJ  2  L 1 



p1 P2  1  4 4   L   L2  EJ  2 L1 12 1 2 L2 12 

 

B´

C´ L1 M´ M´  L1 L2

B´



A´

M´ L1

C´



P2   p1 1 4 4   L   L  24EJ  L1 1 L2 2 

The M´ diagram has its highest value 1 in the joint. The value of reactions are

A´

 4 L 4  2 12 2 L 2  P2

M´ L2

Seviceability Limit State of RC Members

 

1 24EJ

3 L 2



   

 p1 L1  p2 L2



121

Simple supported beam



The meaning express 

A displacement of point m in the middle of the beam is required, simple supported beam according to figure below. The beam in m is loaded by the virtual force P =1. then the influence

 M M´  dx  EJ 0 2

q x

q Lx

q x

2  Lx  x  2

kN m m kN 4 m 2 m

result is multiplied by 2

The integral can then be understood as the volume of the body whose view is equal to the will be shown in a few examples. Example 1: M- and M´-diagrams of rectangles V

1 L M  M´

Example 2: M- diagram triangle, M´- diagram rectangle

2 q  2  Lx  x ( x) dx 2EJ 0



In moment diagrams, the moments are represented as lengths. In such a fact, the product

diagram M, the lateral projection to the diagram M´, and all the sections are rectangles. This

Mx A x   M´x A´ x x 2 2 2 2 As the moment diagrams are symmetrical, the integration is in the interval k = 0.2.7. and the

m

as volumes

infinitesimal length dx.

     M M´ dx   EJ  0  1



M  M´ dx

M • M´ • dx can be understood as a quadrate, which has a height M, a depth M´, and an

of the transverse force is neglected and E, I am constant.







2 q  2 3 L x  x dx  2EJ 0





1 2

 L M  M´

Example 3: M- and M´-diagrams triangles, vertices on the same side

Sabah Shawkat © m

m

4

x q  L x    4  2EJ  3 3

4 4 q  L L    64  2 EJ  24

  L3   L   q  8   2EJ  3 4

q L  8  3   2 EJ  192 

 L4      16   4  4

5 q  L

384EJ

1 3

 L M  M´

Example 4: M- and M´-diagrams triangles, vertices at opposite ends V

1 6

 L M  M´

Example 5: M-diagram square parabola, M´- diagram rectangular V

2 3

 L M  M´

Seviceability Limit State of RC Members

122

Assignment from the course Reinforced concrete load-bearing structures

Assignment from the course Reinforced concrete load-bearing structures, 4th year, winter semester Lecturer: Student: ....................................... year: ............ Specified on: .................................

Stropná doska

Trámy

Stužujúce steny

H=n

Celková výška objektu

Stužujúce steny

Stlpy Základová pätka

Sabah Shawkat © L1

Prievlaky

   

Entered data Purpose of construction: apartments, administration, cafes and restaurants, libraries Distance between columns L1 = 4.0, 4.2, 4.6, 4.8, 5.0, 5.2, 5.6, 5.8, 6.0 m L2 = 4.0, 4.5, 5.0, 5.5, 6.0 m Floor construction height: v = 2.8, 3.0, 3.2, 3.6, 4.0, 4.2, 4.8 m Number of floors: n = 4, 5, 6, 7, 8, 9, 10 Used concrete: C20 Used steel: (V) 10425

 

The aim of the assignment is to elaborate Drawing of monolithic reinforced concrete frame structures

      

Preliminary design of dimensions Geometric shape Static scheme Load calculation Calculation of internal forces Design and check of reinforcement Reinforcement scheme

  

Seviceability Limit State of RC Members

- ceiling slab - beam - column (min. 300 x 300 mm) - stairs - foundations

123

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RC Beams and T-Slab

124

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Box Girders Beams in Bridge Design

125

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Preliminary Design of RC Members

126

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RC Roofs

127

Reinforced Concrete Slabs, Determination of Bending Moments z

z x



g h

0 y

z x

x a/2

y

 ga

 

 

g  a4 E  h3

 1ga

 0.0479 0.0553 0.0626 0.0693 0.0753 0.0812 0.0862 0.0908 0.0948 0.0985 0.1017 0.1189 0.1235 0.1246 0.125

 0.0479 0.0494 0.0501 0.0504 0.0506 0.0499 0.0493 0.0486 0.0479 0.0471 0.0464 0.0404 0.0384 0.0375 0.0375

b/2 My

0 Mx

x b/2

 0.0443 0.0530 0.0616 0.0697 0.0770 0.0843 0.0906 0.0964 0.1017 0.1064 0.1106 0.1336 0.1400 0.1416 0.1422

y

 

 

M y1

0 2

a/2

b/a 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 3.0 4.0 5.0



  g  a

g  a4 E  h3

x a/2

a/2

b>a

My1 b/2 My2

M y2

Mx b/2

b/a 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 3.0

 0.0209 0.0274 0.0340 0.0424 0.0502 0.0582 0.0658 0.0730 0.0799 0.0863 0.0987 0.1276 0.1422

 0.070 0.079 0.087 0.094 0.100 0.105 0.109 0.112 0.115 0.117 0.119 0.125 0.125

Sabah Shawkat © 8

a a

 1ga

 2  g  a2

 0.024 0.031 0.038 0.045 0.052 0.059 0.065 0.071 0.077 0.082 0.087 0.114 0.125

 0.033 0.037 0.040 0.043 0.045 0.046 0.047 0.047 0.048 0.048 0.047 0.042 0.038

0.1450 0.1300

0.1500

0.1150

0.1350

  

0.1050 0.0900

0.0850 0.0700 0.0550

0.0750

0.0400

0.0600

0.0250

0.0450 0.0300

   

0.1000

0.1200

0.0100 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 b/a

0.8

1.3

1.8

Determination of Bending Moments in RC Slabs

2.3

2.8

3.3 b/a

128

g h x a/2 My1

a/b b

2.0 1.5 1.4 1.3 1.2 1.1

b/2

y

 

 

  g  b

a a

g  b4

Mx  0.0284 0.0284 0.0270 0.0262 0.0255 0.0243 0.0228

 0.083 0.084 0.083 0.081 0.079 0.077 0.074

E  h3

M y2

b<a

b/2

My2

M y1

a/2

 

z x

 2gb

 0.042 0.042 0.041 0.040 0.039 0.037 0.036



 1  g  b2

 0.001 0.014 0.017 0.019 0.020 0.022 0.023

z x

g x a/2

M y1

a/2

My1 b/2 My2

b/2

b/a 1.0 1.1 1.2 1.3 1.4 1.5 2.0

 

  g  a

  2

g  a4 E  h3

M y2 Mx

 0.0300 0.0380 0.0470 0.0550 0.0630 0.0700 0.1010 0.1420

 0.084 0.092 0.098 0.104 0.109 0.112 0.122 0.125

 2  g  a2  1  g  a2

 0.034 0.041 0.049 0.056 0.063 0.069 0.094 0.125

 0.039 0.042 0.044 0.045 0.047 0.048 0.047 0.037

Sabah Shawkat © a

0.0850

y

b>a

0.1000

0.1300

0.0700

   

0.0550

0.1150 0.1000    

0.0850

0.0400

0.0700 0.0250

0.0550

0.0100 1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0 a/b

0.0400 0.0250 0.0100 0.8

1.0

1.2

1.4

1.6

Determination of Bending Moments in RC Slabs

1.8

2.0

2.2 b/a

129

a 2

y

 

 

a/2

M y1

My1 My2

b/2 x b/2

Mx a a

  g  b

a/b b

2.0 1.5 1.4 1.3 1.2 1.1

 0.0570 0.0530 0.0460 0.0440 0.0410 0.0380 0.0350

E  h3

M y2 Mx

b<a 8

a/2

g  b4

 0.125 0.122 0.111 0.108 0.103 0.098 0.091

 2  g  b2

z x h

a/2

 1  g  b2

 0.019 0.023 0.028 0.030 0.031 0.032 0.033

 0.062 0.060 0.054 0.052 0.050 0.047 0.043

 

M y1

y

 

  g  a

 

My2

3.0 2.0 1.5 1.0

b 2

My1 x

b/a

g  a4 E  h3  2  g  a2

M y2

 1  g  a2

b>a

z x

 

>2h

 0.1660 0.1660 0.1640 0.1540 0.1230

 0.125 0.125 0.125 0.124 0.119

 0.133 0.133 0.131 0.123 0.097

 0.062 0.062 0.061 0.056 0.042

Sabah Shawkat © 0.1300 0.1150

0.1750 0.1600

0.1000

   

0.0850 0.0700 0.0550

0.1450 0.1300

   

0.1150 0.1000 0.0850

0.0400

0.0700

0.0250

0.0550 0.0400

0.0100 1.0 1.1 1.2

1.3 1.4 1.5 1.6 1.7 1.8

1.9 2.0 2.1

a/b

0.0250 0.0100 0.8

1.0

1.2

Determination of Bending Moments in RC Slabs

1.4

1.6

1.8

2.0

2.2

b/a

130

g h

x a/2

a/2

y

  g  b

b<a b

My1 x

 b 

b/a 2/3 1/2 1/3 0

 0.3660 0.6500 1.0300 1.3700

 

0.7100 0.6100

z x

 2  g  b2  1gb

 0.056 0.029 0.008 0.000

0 y

x a/2

M x1

a/2

 0.060 0.073 0.094 0.094

a 2

y

 

 

Mx1

 ga

b/a 0.50 0.67 0.71 0.77 0.83 0.91 1.00 1.10 1.20 1.30 1.40 1.50 2.00 3.00

gb

 0.0775 0.1057 0.1117 0.1192 0.1265 0.1345 0.1404 0.1464 0.1511 0.1547 0.1575 0.1596 0.1646 0.1660 0.1662

M x2

 0.060 0.083 0.088 0.094 0.100 0.107 0.112 0.117 0.121 0.124 0.126 0.128 0.132 0.133 0.133

E  h3

b 2

Mx2

0.8100

 

E  h3

M y2

 0.227 0.319 0.428 0.500

g  b4

 1  g  a2  2  g  a2

 0.039 0.055 0.059 0.064 0.069 0.074 0.080 0.085 0.090 0.094 0.098 0.101 0.113 0.122 0.125

 0.022 0.030 0.032 0.034 0.036 0.037 0.039 0.040 0.041 0.042 0.042 0.042 0.041 0.039 0.037

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My2

M y1

>2h

 z x 

>2h

   

0.5100 0.4100

0.1750 0.1600 0.1450 0.1300

0.3100 0.2100

0.1000

0.1100

0.0850

0.0100 0.30

   

0.1150

0.0700 0.40

0.50

0.60

0.70 b/a

0.0550 0.0400 0.0250 0.0100 0.0

0.5

1.0

1.5

2.0

Determination of Bending Moments in RC Slabs

2.5

3.0

b/a

131

z( x

0y

 

y My1 Mx1

My2

Mx2

a/2

b 2

ga

  g  a

M y1

  1  g  a2

x b 2

a/2 a

b/a 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

 0.0138 0.0164 0.0188 0.0209 0.0226 0.0240 0.0251 0.0260 0.0267 0.0272 0.0277

 0.0513 0.0581 0.0639 0.0687 0.0726 0.0757 0.0780 0.0799 0.0812 0.0822 0.0829

z( x

Eh

M x2

 2  g  a2

M y2

 3  g  a2

 0.0513 0.0538 0.0554 0.0563 0.0568 0.0570 0.0571 0.0571 0.0571 0.0571 0.0571





0.0264 0.0299 0.0327 0.0349 0.0365 0.0381 0.0392 0.0401 0.0407

0.0231 0.0228 0.0222 0.0212 0.0203 0.0193 0.0182 0.0174 0.0165

0y

 

y My1

b b

M x1

Mx1

b 2

a/2

M y1

  1  g  a2

0.0850

0.0550

0.0400

My1 Mx1

b 2

0 My2

a/2

0.9

1.1

1.3

1.5

1.7

1.9

2.1

z( x

0y

 

Mx2

b/a

 3  g  a2

a/2

0.0250

0.0100

M y2

    

 2  g  a2

0.0700

M x2

b 2

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E  h3

x My2

  g  a

Mx2

M x1

ga

b x

b 2

a/2 a a

Determination of Bending Moments in RC Slabs

M x1

  g  a

M y1

  1  g  a2

ga

E  h3 M x2

 2  g  a2

M y2

 3  g  a2

132

Moment representation of concrete constitutive data for moment capacity calculations

fcmax  30 MPa

The constitutive parameters o, 1, and 2 are based on in situ data and are given as follows:



 0.7678

Three non dimensional constitutive parameters based on stress-strain data are presented. In conjunction with the max. stress and ultimate strain, these constitutive parameters can be used

A s  4  

b  0.3  m

fy  400 MPa 

d  530  mm

d s  25  mm

 0.4589

3 2

A s  1.963495 10

 cu  0.0038

T  fy  A s

T  785.4 kN

to compute the capacities of rectangular, triangular, of linearly tapering cross sections. In addition to the usual constitutive models (linear, Whitney stress block, Hognestad, etc.), it

c 

was desired to construct a general constitutive model class based on either experimental uniaxial stress-strain data or other constitutive models.

c  113.658 mm

b  fcmax  o





1



o

M n  T d   1 



One of the advantages of objected-oriented programming is that the internal representation of an object is hidden from the rest of the program. Thus, the stress-strain data can be stored internally in any manner. This text is a presentation of an efficient way to store stress-strain

 

   o

data using five numerical values. These five value allow the ultimate moment capacity Mn and curvature at failure of a rectangular or linearly tapering section to be computed without any loss

   c 

1 o

 380.35 kN  m

  0.7678   0.597682 d  0.465 m

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of accuracy. Not only is there a saving in storage space by replacing the entire series of stressstrain pairs with five numbers, but the modelling obviated the need for breaking a cross section into strips to compute Mn.

1

A  fcmax   b  ( 1   )

A  2.7801  10 m

B  fcmax   b  d

B  3.213243  10 kN

M  258 kN m

C  M

Z 

B 

B  4 C A 2 A

Z   86.813348 mm

Parameter definitions

- cu

ultimate in situ concrete compressive strain

- fcmax in situ compressive strength axial strain

Z 

B  B  4 C A 2 A

T    b  Z  fcmax

- f  axial stress 1 

o 1  2  1   o 

Z   1.068988 m

A s 

Z  fcmax   b fy

T  5.75023 10 kN

1  0.95422

Determination of Ultimate Moment Capacity of RC members

A s  0.018467 m

133

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Shear Failure of RC beams

134

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RC Stairs

135

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RC Stairs

136

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Connections in RC

137

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One-way and Two-way RC Slabs

138

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Realisation of RC Flat Slab

139

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Realisation of RC Flat Slab - Building Houses

140

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Realisation of RC Flat Slab - Family House

141

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Calculation of Bending Moments in RC Slabs

142

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Realisation of RC Roofs

143

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Calculation of Bending Moments in RC Slabs

144

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Realisation of RC Roofs

145

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RC Beams – Transform of Cross-Section Forces

146

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RC Beams – Transform of Cross-Section Forces

147

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RC Beams – Transform of Cross-Section Forces

148

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RC Beams – Transform of Cross-Section Forces

149

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Calculation of Bending Moments and Deflections in RC Slabs

150

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Calculation of Bending Moments in RC Slabs

151

Shear Walls Walls carrying vertical loads should be designed as columns. Basically walls are designed in the same manner as columns, but there are a few differences. A wall is distinguished from a column by having a length that is more than five times the thickness. Plain concrete walls should have a minimum thickness of 120 mm. Where the load on the wall is eccentric, the wall must have centrally placed reinforcement of at least 0.2 percent of the cross-section area if the eccentricity ratio exceeds 0.20. This reinforcement may not be included in the load-carrying capacity of the wall.

The equilibrium and compatibility equations at each level produces a set of simultaneous equations which are solved to give the lateral deflection and rotation at each level. If a tall building has an asymmetrical structural plan and is subjected to horizontal loading, torsional as well as bending displacements will occur, and hence a full three-dimensional analysis is required. In many tall building shear wall provide most, if not all, of the required strength for lateral loading resulting from gravity, wind, and earthquake effects. The system (Hull - Core Structures) has been used for very tall buildings in both steel and concrete. Lateral loads are resisted by both the hull and the core, their mode of interaction depending on the design of the floor system.

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Shear walls should be designed as vertical cantilevers, and the reinforcement arrangement should be checked as for a beam. Where the shear walls have returns at the compression end, they should be treated as flanged beams.

If the walls contain openings, the assumption for beams that plane sections remain plane is no longer valid. Shear walls connected by beams or floor slabs. The stability of shear-wall structure is often provided by several walls connected together by beams or floors. Where the walls are of uniform section throughout the height and are connected by regularly spaced uniform beams. Many shear walls contain one or more rows of openings. When walls are used to brace a framed structure, it may be acceptable to disregard the lateral stiffness of the frame and assume the horizontal load carried entirely by the walls.

A floor slabs of multi-story buildings, when effectively connected to the wall, acting as stiffeners, provide adequate lateral strength. As essential prerequisites, adequate foundations and sufficient connection to all floors, to transmit horizontal loads, must be assured. Normally, for wind loading, the governing design criterion or limit state will be deflection. Shear walls, when carefully designed and detailed, hold the promise of giving the greatest degree of protection against non-structural damage in moderate earthquakes, while assuring survival in case of catastrophic seismic disturbances, on account of their ductility. Yielding of the flexural bars will also affect the width of diagonal cracks. The shear strength of tall shear walls may also be controlled by combined moment and shear failure at the base of the wall. Door and service openings in shear walls introduce weaknesses that are

Shear Walls

152

not confined merely to the consequential reduction in cross-section. Stress concentrations are developed at the corners, and adequate reinforcement needs to be provided to cater for these concentrations. This reinforcement should take the form of diagonal bars positioned at the corners of the openings. The reinforcement will generally be adequate if it is designed to resist a tensile force equal to twice the shear force in the vertical components of the wall as shown, but should not be less than two 16mm diameter bars across each corner of the opening. At the base of the wall, where yielding of the flexural reinforcement in both faces of the section can occur, the contribution of the concrete towards shear strength should be disregarded where the axial compression on the gross section is less than 12% of the cylinder crushing strength of the concrete. Sectional area of the concrete and should be equally divided between the two faces of the wall. The maximum area of vertical reinforcement should not exceed 4% of the gross cross-sectional area of the concrete. Horizontal reinforcement equal to not less than half the area of vertical reinforcement should be provided between the vertical reinforcement and the wall surface on both faces. The spacing of the vertical bars should not exceed the lesser of 300mm or twice the wall thickness. The spacing of horizontal bars should not exceed 300mm and the diameter should not be less than one-quarter of the vertical bars.

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The prime function of the vertical reinforcement, passing across a construction joint, is to supply the necessary clamping force and to enable friction forces to be transferred.

Shear Walls

153

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RC Walls

154

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Diagram x vs F

155

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Diagram x vs G

156

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Diagram x vs D

157

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Diagram x vs Y

158

Solution of reinforcing walls with openings loaded by vertical loading

(A 1,E)

(i,E')  =1

e 1 v1

e 2 v2

( A2,E)

 =z/H

v1 z

l – floor height H – total wall height A1,A2 - cross-sectional areas of individual pillars 2c – distance between pillars 2a – hole width N – normal force acting in the pillar  shear force acting in the crossbar E – modulus of elasticity of walls E´ – transverse elastic modulus v1 – vertical load applied to pillar 1 at the level of each floor v2 – vertical load acting on pillar 2 at the level of each floor e1 – eccentricity on which the load v1 is applied e2 – eccentricity on which v2 is applied

v2 v2

 =0

e1 2a 2c

b/ Walls with very stiff transvers above the openings

Defining geometry

The transvers above the holes are so stiff that the wall can be considered solid, without holes. 2

v1 e1



e2 v2 e2

v1 e1



v2 e2

v1 e1



v2 e2

Bending moment at the base of wall due to load application v1 and v2:



M = H / l. (v1. e1 + v2. e2 ) Axial force acting at the base of the wall N = H / l. (v1 + v2) Along the entire height of the structure the shearing forces which acting in the transvers are constant



(v1 + v2) / l

Static effect

Sabah Shawkat © c/ Walls with medium stiff walls above the openings

a/ Walls with very soft transvers above the openings

ex = in

In this case, the total moment M is divided into individual pillars in proportion to their stiffness. v1

v1 e1

e 2 v2

N2v1

Static effect

I1  I2







v1 e1



 v 1 e1  v 2 e 2 l

e2 v2

ex

m v =(v1 .e1 +v2 .e2 ).2c l l

ex2



ex1 

 ex3

Bending moment in the pillar 2:

M2 N2



ex = ex1 + ex2 - ex3

Bending moment in the pillar 1: I1

Deformations ex from the external loads V1 and V2 acting on eccentricities e1 and e2

I2 I1  I2



H l





 v 1 e1  v 2 e 2

Determination of deformation ex from the external loads V1 and v2 acting on eccentricities e1 and e2

Axial force: - pillar 1: N1 = H . v1 / l 1

- pillar 2: N2 = H . v2 / l Shear forces acting in transvers

 ex1

2 c



Shear Walls

  m v   d 





E I 1  I 2

v2 v1   e1  e 2 1 l  l   2 c   d





E I1  I2



159

 v 2    d  l 

 ex3

E A 2

v2 l

    d E  A 2 

- displacement from the centric load v1 1

 v 1    d  l 

 ex2

E A 1

 l    d E  A 1 

- displacement from the centric load v2

Total displacement from external load v1 and v2

- construction height H = 27,5 m - floor height l = 2,75 m - cross-sectional area of the first pillar A1 = 2 m2 - cross-sectional area of the 2nd pillar A2 = 1,6 m2 - moment of inertia of the first pillar I1 = 4 m4 - moment of inertia of the second pillar2 = 2 m4 - moment of inertia of the cross-sectional area of the structure weakened by holes I = 39 m4 - moment of inertia of transvers ipr = 0,006 m4 - modulus of elasticity of pillars E = 10 GPa - modulus of elasticity of transvers E´ = 20 GPa - static moment of cross-sectional area of the structure weakened by holes S = 5,42 m3 - transverse force acting at the foot of the structure Wo = 354 kN - distance between the centroids of the pillars 2c = 6,10 m - width of window openings 2a = 2 m

 =1

10.NP

1.NP W o  =0

- displacement from eccentric loads V1 and v2

0.NP 2c

A1 G1

Geometry of the computed reinforcing wall subjected to horizontal loading Wo

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v2   v 1  v1 v2  1  e1  e 2  2c l l l l      ex  ex1   ex2   ex3     d  E I  I  E  A E  A  1  2      1 2

The course of axial forces in individual pillars along the height of the structure is calculated as

compressive effect of external vertical load in individual pillars diminished resp. Increased by The effects of total shear forces N() = Nex () ET().

N 1 

N 2 

H l

 v 1   1    K    





 v 2   1    K    





Results of the example solution: NP







0.161

0.9

8 7

1  2

 1   2

2 c  S

M1 M2 N [kN.m] [kN.m] [kN]



21.736

0.185

25.018

5.00E-03

0.017

0.014

-9.48E-03

-61.551

-30.775

46.754

0.8

0.241

32.58

0.02

0.038

0.032

-0.012

-80.007

-40.003

79.334

0.7

0.312

42.2

0.045

0.066

0.056

-0.011

-69.393

-34.696

121.534

0.6

0.387

52.422

0.08

0.101

0.085

-5.32E-03

-34.547

-17.274

173.956

0.5

0.458

62.017

0.125

0.143

0.121

3.77E-03

24.459

12.23

235.973

0.4

0.514

69.521

0.18

0.192

0.163

0.017

113.068

56.534

305.494

0.3

0.537

72.694

0.245

0.207

0.038

243.756

121.878

378.188

0.2

0.5

67.69

0.32

0.297

0.252

0.068

441.891

220.946

445.878

0.1

0.352

47.62

0.405

0.341

0.289

0.116

750.808

375.404

493.498

6.00E-14

0.5

0.36

0.305

0.195

1.26E+03

632.008

493.498

Examples of Shear Walls

2

cS I







160

2c2

2a 1 I1

2

2a 3

1 N 2  2

1

2c 3

2a 2

3

N4 M4

- construction height H = 27,5 m  =1 10.NP - floor height l = 2,75 m v1 e 2 v2 - cross-sectional area of the first pillar A1 = 2 m2 e1 v1 1 2 v2 - cross-sectional area of the 2nd pillar A2 = 1,6 m2 - moment of inertia of the first pillar I1 = 4 m4 - moment of inertia of the second pillar2 = 2 m4 v2 v1 - moment of inertia of the cross-sectional area of 1.NP 4 the structure weakened by holes I = 39 m v2 v1  =0 0. ND - moment of inertia of transvers i = 0,006 m4 - modulus of elasticity of pillars E = 10 GPa 2c A1 - transvers elastic modulus E´ = 20 GPa A2 G2 G1 2a - static moment of cross-sectional area of the structure weakened by holes S = 5,42 m3 v1 – vertical load applied to pillar 1 at the level of Geometry of reinforcing wall loaded by each floor = 300 kN on floor level vertical load v2 – vertical load applied to pillar 2 at level each floor = 200 kN on floor level e1 – v1 load acts on eccentricity = 0,50 m e2 – the load v2 acts on eccentricity =1,00 m H

2c1 G1

Reinforcing wall with multiple rows of holes



  i pr  c 12 i pr  c 22  1 2     ....  3 I 1  I 2  I 3  ......  l  a 13 a2  6

l

I i



 i

 i pr  c i2   i   a3  i  

1

W o  l

i pr  c 1 1



 we get from graph

Sabah Shawkat ©

 i 1  c 12 i 2  c 22    .... 2 a1   3  a 3  a2  1  3

Similarly, the values of shear forces in transvers can be calculated 2, 3.

Normal forces N acting in the individual pillars can be obtained as follows: N1 =  1

N2 =  2 -  1 Ni =  i -  i-1 , The bending moments in the individual pillars caused by the external load Wo are calculated:

  1   2   W o  H    2 I 1  I 2  I 3  ....  

N1 [kN]



-0.995

39.605

0.000

0.9

-0.994

39.568

-0.099

72.397

36.198

260.409

239.591

0.8

-0.991

39.445

-0.199

145.100

72.550

520.894

479.106

0.7

-0.985

39.189

-0.298

218.552

109.276

781.563

718.437

0.6

-0.973

38.707

-0.396

293.461

146.730

1043.000

957.410

0.5

-0.950

37.819

-0.492

371.070

185.535

1304.000

1196.000

0.4

-0.909

36.194

-0.585

453.650

226.825

1567.000

1433.000

0.3

-0.835

33.230

-0.673

545.243

272.621

1832.000

1668.000

0.2

-0.699

27.824

-0.750

653.509

326.754

2102.000

1898.000

0.1

-0.451

17.968

-0.809

791.304

395.652

2378.000

2122.000

0.000

-0.833

986.187

493.093

2669.000

2331.000

I2 I 1  I 2  I 3  ....

M3 = ......

  1  

 W o  H 



N2 [kN]



M 1  M2



M1 M2 [kN.m] [kN.m]





I1 I1  I2











  1    v 1  e 1  v 2  e 2  2  c  K      

 

 M 2 

I2 I1  I2

  1    v 1  e 1  v 2  e 2  2  c  K    



Examples of Shear Walls



161

Reinforcement design • Simple supported wall - main tensile reinforcement if a/2 < b < a A

0.90 

if b > a A

1.5 

Mo bs

1/3.A h

2 b 

  1 

 3 a 



0,15b

as

- upper reinforcement in both cases A´= 0 where Mo = q . a2 / 8

0.25 

g hd

Placement of reinforcement in RC wall

Toa

- vertical reinforcement At

bs To

s

4 7 4

 

where

bs To

s

R btn

1.5 MPa

R sd

375 MPa

R scd

qa

Self weight of the wall

Total load

• Continuous wall

g rd

R sd

kN  -1 g o h s  b   25  1.1 123.75kNm 3 m   g o  3 g rd  g dd  g hd

g cd

g rd

g dd

Sabah Shawkat © 7

g hd

g rd

- steel

Toa

60

m kN g dd 100 m g rd g dd R bd 11.5 MPa

- from the first ceiling - typical floor Material characteristics concrete

- horizontal reinforcement

b>a

2/5b

2/3.A h

a/2<b<

Industrial building - reinforced concrete bearing wall, based on stressx,z, v design the reinforcement and assess the quality of the reinforced concrete wall. h s 0.25 m Wall thickness b 18 m Wall height a 13.5 m Axial span of the wall c 0.5 m Column width: kN Wall load - from the roof

1

583.75kNm

Calculation of internal forces

0,10b

1/3.A h

2/3.A h

0,4b

0,15b

Distribution of reinforcement in continuous wall Extreme fields

Internal fields

- lower reinforcement Mo b  a/2 < b < a A 0.70  b     1  a  s

b>a

1.40 

- upper reinforcement a/2 < b < a b>a

A´

0.60  2.40 

as Mo

1.40  1.20 

Mo as Mo

as





0.75

0.03704

- resultant of tensile at the same time Nx = . gcd . a  > 0.05 =>  > 0.05 stresses = 1568,244 kN - distance of the resultant z´= ´. a = 0,876 from the lower edge = z´



0.232      0.1667   0.3057   0.2968  0.237    0.184  

´

0.012      ( 0.0667)    1.6667 10    0.1138  0.12    0.074   a Q d g cd  3940.31kN Transverse force in 2

( a  3  b)   s

This is the same as for extreme fields

0.0649 g hd

z

supports 

Maximum shear stresses in the support area where k = min(a,b) Maximum vertical compression stress

 max 3

z

g hd hs

Qd 2 k h s

1751.25kPa 

X

765.88kPa

This is the same as for extreme fields

Examples of Shear Walls

z



max

240kPa g cd

Cross-sectional assessment of cross-sectional stresses

bs

g cd

0.199

3

Nx z'

 b  R bd

9775kPa

=> ok

162

z

240 MPa

 b  R bd

=> ok

9775kPa

Vertical

reinforcement As zv

design 2x6/m‘ = 2,827 cm

 max 1751.25kPa 

 b  R btn

1.8 MPa

3

2.66667

1275 kPa

=> It does not ok becausemax is not less than Rt, it means, that shearing cracks arise in the cross-section, therefore we Rt

propose to change the quality of the concrete and to express the

R btn

g dd  s  R sd

calculation Anchorage length lbd

wall thickness as follows:

bt

lbf

2 k R btn

0.24323m

sf  

 0.5

cm m

ss

25 mm

ef

0.25

Rbtd  1.2 MPa

ss

ss 

Asxsku

sf

ss max 1

11.775

s s  ds

20 cm

1.5

  Rsd   ef     d s 1.20 m bt  Rbtd  tss max 1 MPa   ds

We suggest hs=30 cm go

h s  b  25

gc d 

 max 3

kN 3

 1.1

1

g cd

148.5kNm

g o  3 g rd  g dd  g hd

According to CCBA 68

1

608.5kNm

a 2 Qd

2 k h s

 g cd  a

13862.39kNm

200 MPa

Rbd

 b  R bd

4107.38kN

 auo

To  2 h o h t 3

17 MPa

Rbtd

1.2 MPa

30 MPa

fckcube

c

1140.93kPa

1.5

=> ok

9775kPa

 b  R bd

9775kPa

=> ok

 b  R btn

1530 kPa

=> ok

max

g cd 

The minimum wall thickness is calculated as follows:

1521.25kPa

665.29333 kPa

z

Sabah Shawkat © To

4107.38kN

Assessment of stresses X

f ckcube   f cdcyl 0.8  0.85  13.6 MPa c  

h o1

lt 2



g cd



100 f cdcyl  h t



  

1 3

0.19699m

Or  max 1521.25kPa 

h o2 construction reinforcement

Calculation of tensile force N x   g cd  a 1634.735kN transmitted by reinforcement Nx

0.00436m

A szv

2.5

Design of tensile A sx.sku 10  4 reinforcement 1025 *

z

g hd hs

g cd f cdcyl  h t

0.00249 1

 0.01923

n is less than 1/52, that is, formula ho1 and not

 s  R sd

A sx

of A sx axis

0.05034m

konštrukèná výstuž

0.5K

Calculation reinforcement direction x

g cd  l t  2 f cdcyl  h t 3

49.087cm 

0.2K

ho2 Design of main reinforcement ht > lt

0.3K

200 kPa

Examples of Shear Walls

s

f yk

1.15

 Mo A 1 1.5   f yk  l t  s 

    

0.00432m

410 MPa

163

Solution of reinforcing walls in terms of horizontal deformations

2c=7 m 2a=2 m

When the reinforcing walls are connected to rigid ceiling panels, the horizontal load is transmitted to the individual vertical elements in proportion to their relative stiffness,

l=2,5 m

1,5 1m

depending on the deformation of the reinforcing walls from the bending and shear effects of

0,3 m

the horizontal load. Relative deformation H/L

c = 3,5 m

from bending

from shear

Relative deformation of walls, many times

effects

exceeds the shear effect. In such cases, it is

0,50

sufficient to consider bending effects. The table

0,80

0,20

shows the relative deformations depending on

0,90

0,10

0,94

0,06

0,96

0,04

A1 = A2 = 0,3.5 = 1,5 m2

0.3  5

2 c 1

7 1



1  2     1.5 

3.13  m

I = 3,13+ 3,13+ 2. 5,25. 3,5= 3

0.3  1

i pr

the ratio of wall height to width (H / L). 

3  i pr

I1  I2



I S

3  0.025



6.26

a l



3.5

5.25 1  2.5 3

0.025  m

= 43 m4 =>  = 0,37 according to graph (0)=0,44 => =0,37 . 40 =14,8

0.139

where the total height of the wall = 40 m

Sabah Shawkat ©

The horizontal load stresses the wall as a cantilever bracket

WoH

according to the relationship:

8 E Ie

16  5.25  3.5 0.44  1 6.26 220

39.2  m

And the deflection can be approximately determined as the

 wall with multiple holes

deflection of the cantilever bracket according to the relationship: where

- equivalent moment of inertia:

Ie - equivalent moment of inertia of the wall

Wo = w . H

Characteristics of the different wall types needed to calculate deflections  solid wall without holes f

- equivalent moment of inertia::

1 12

8 I

I 1  I 2  ......



(4.4.1.6.5)

o 

1

hsL

h =0,3 m

12 m

1 12

 0.3  12

2,5

1,5

43.20  m

30 x 100

c1 =1,75 m, c2 = 2 m, c3 = 1.75 m

ipr = 0,025 m4

A1 = A4 = 0,3 m2

I = 23,4 m4

A2 = A3 = 0,6 m2 I1 = I4 = 0,3. 13 /12 = 0,025 m4

 wall with one row of holes

I2 = I3 = 0,3. 23 /12=0,2 m4

- equivalent moment of inertia: H

Ie I1

I 16  S  c

I 1  I 2



o 

1

kde I=I1 + I2+ 2.S.c and o = (0) podľa obr. 4.4.1.4.9



l

 i

 i pr   c j 2   j      a j 2  j 1  Ii 3



1

Examples of Shear Walls

6 2  ( 0.025  0.2)  2.5



 2 



0.025  1.75 3



0.025  2 3

  1.35 

5.25  m

164

=>  = 1,16 from the graph. (0)=0,48 => =1,16 . 40 =46,4 where the total height of the wall = 40 m 23.4

8  23.4 2  ( 0.025  0.2)



0.48 2153

2.28  m 1

 frame construction with rigid transvers  - equivalent moment of inertia: l l'



 Ip

where n - number of floors p- number of vertical elements in one height level

Ip = hs . b3 / 12  = l´. l

3 p  n

Cross-sectional area-

30 x 70

Sabah Shawkat ©

30 x 30

of columns:

1,8

1,5

3  9  16

 0.00068 12.6  m

0.72

0,3 . 0,3 - transvers: 0,3 . 0,7

n =16 p=9 Ip=0,3 . 0,33 / 12  = 1,8 / 2,5 = 0,72

 wall supported on columns

Ie  H''

A col  I  L  H

8  I  H  H´  ( H  H´)  A col  L  ( H  H´)

(4.4.1.6.7)

where Acol – Cross-sectional area- of columns:  l

Acol = 0,5 . 0,6 = 0,3 m2 I = 0,3.123 / 12=43,2 m4

0.5 x 0.6

40 5

H = 40 m, H´= 5 m H - H´ = 35 m L= 8 m

0,3

0.3  43.2  8  40

16.4  m

8  43.2  40  5  35  0.3  8  35

Examples of Shear Walls

165

0.8

Data:

0.72

I  39  m

0.64

2  1.6  m

i  0.006  m

0.56

l  2.75  m

0.48 ( ) 0.4

 1  1 S    1 2  c 

0.32

c  3.049m

I1  4  m 4

S  5.42  m

I2  2  m

Ho  l  S

1  2  m

E´  20000  MPa

E  10000  MPa

Z  20  l

a  1  m

Ho  1200  kN

0.24 0.16

v 1  30 

 458.615  kN

tonne l

v 2  20 

tonne l

e 1  0.5  m

0.08

e 2  1  m

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9



K 

I1  I2 I1  I2    1  1   v 2   e2    v 1   e1     2 c 2 c 2 1 I     

3 1

K  1.893 10 m

 kN

  0 0.01 1 0.45 0.405

J 

I1  I2 I1  I2    1  1   v 2   e2    v 1   e1     I 2 c 2 c 2 1

3 1

J  1.893 10 m

0.36

 kN

Sabah Shawkat © 









0.315 0.27

 ( )0.225

 

3  E´  i







E  I1  I2

    Z



2

  0.048m

0.18

1

  0.219m

0.135

a l

0.09

0.045

  12.033

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9





 0

   i

i 1

 0.1

 (1   )2  S M 1(  )   Ho  Z   2 c   (  ) I1  I2 I  2  I1

(  ) 

Ho  l  S

M 2(  ) 

 (  )

 Ho  Z  I1  I2

 (1   )2  S  2 c   (  ) I  2 

 j2  2

2 c

S I



 1



 j2  2  c  S   2

1   j



 1

 

38.138

5·10-3

8.335·10-3

-3.335·10-3

57.305

0.02

0.022

-2.273·10-3

95.131

0.045

0.044

1.183·10-3

138.516

0.08

0.074

6.449·10-3

183.42

0.125

0.112

0.013

228.281 271.468

0.18

0.158

0.022

0.245

0.212

0.033

308.62

0.32

0.271

0.049

325.539

0.405

0.328

0.077

275.036

0.5

0.359

0.141

5.595·10 -13

Examples of Shear Walls

 kN

166



 1

 



(1   ) 

 1

 

0.083

9.836·10-3

0.9

0.125

0.026

0.8

0.207

0.052

0.7

0.302

0.087

0.6

0.4

0.132

0.5

0.498

0.186

0.4

0.592

0.25

0.3

0.673

0.319

0.2

0.71

0.387

0.1

0.6

0.424

1.22·10 -15

  0   (  )   

Horizontal load   Odesolve (  1)

0.8

0.6

0.6 0.4

0.4

0.2

 ( )

2 c m





(  ) 

 0.4

Ho  l  S I

2 c m

q (  )  ( ( 1   ) )

 (  )

 0.6



 0.2

(1  )

 (  ) d 

0.8

( )

 0.8 1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9





M1 1  

 



M2 1  

 

 j 

q 

Sabah Shawkat © 0

 kN  m

38.138

-146.738

-73.369

-100.03

-50.015

95.131

52.052

26.026

138.516

283.77

141.885

183.42

589.756

294.878

228.281

Vertical load K=1

976.252

488.126

271.468

1.471·10 3

735.646

308.62

2.17·10 3

1.085·10 3

325.539

3.392·10 3

1.696·10 3

275.036

6.199·10 3

3.099·10 3

5.595·10 -13

810 610 M 1  j 

 kN

57.305

410

210

310 M 2  j 

0  210

210 110

 (  )    (  ) d  

  Odesolve (  1)

1 0.8 0.6 0.4 0.2  ( ) 0  0.2  0.4  0.6  0.8 1

6 6 6 6

1 0.8 0.6 0.4 0.2  ( )

 0.4  0.6  0.8 1

0.1

0.2

0.3

0.4

0.5

0.6



0.2

0.4

0.6 j

0.8

 110

0.2

0.4

0.6

0.8

j

d

 0.2

T(  ) 2

 T(  )

K



Examples of Shear Walls

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 

0.6

0.7

0.8

0.9

167

  2

Horizontal load

d

  1 d

d

T(  )    T(  )

  ( 1   )

  (  )   

0.12

  (  )   

  ( 1   )

0.25

0.225

0.225 0.2

0.175

0.15

( ) 0.125

 ( )0.125



0.07

0.05

0.072

0.06  ( )0.05

0.025

( ) 0.06 0.048

0.04

0.036

0.03

0.024

0.02

 (  ) d 

0.075

0.08

0.084

0.1

0.09

0.096

T(  )    T(  )

0.2

 (  ) d

0.1

0.108

0.05 0.025 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9



0.01

0.012 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9



Horizontal load

Sabah Shawkat © d

K  1 d

d

 (  )    (  ) d 

 T(  )

T(  )

 T(  )

 (  )    (  ) d 

K







T(  )

K

  Odesolve (  1)



0.4 0.36 0.32 0.28 0.24  ( ) 0.2 0.16 0.12 0.08 0.04 0

1.6 1.44 1.28 1.12 0.96  ( ) 0.8 0.64 0.48 0.32 0.16 0

0.25

0.225 0.2 0.175

0.15  ( ) 0.125 0.1 0.075 0.05

0.1

0.2

0.3

0.4

0.5 

0.6

0.7

0.8

0.9

0.025 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.2

1.08 0.96 0.84

0.72  ( ) 0.6 0.48 0.36 0.24 0.12

0.1

0.2

0.4

0.5 



Vertical load

0.3

K  2

Vertical load

Examples of Shear Walls

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 

0.6

0.7

0.8

0.9

168

  10

Horizontal load   (  )   

  5 d

d

T(  )    T(  )

 (  ) d

d



  ( 1   )

T(  )    T(  )

  (  )   

  ( 1   )

0.7

0.5

0.35 0.315

0.4

0.28

0.35

0.63

0.45

0.56

0.405 0.36

0.49

0.245

0.3

0.315

0.42

0.21

0.27  ( )0.225

( ) 0.25

 ( )0.175

( ) 0.35

0.2

0.14

0.28

0.18

0.15

0.105

0.21

0.135

0.07

0.1

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

d



5 4.5 4 3.5 3  ( )2.5 2 1.5 1 0.5 0

T()

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9



0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

10 9 8 7 6  ( ) 5 4 3 2 1 0

0.6

0.7

0.8

0.9

 (  )    (  ) d 

K

10 9 8 7 6

 ( ) 5 4 3 2 1

0.1

0.2

0.3



0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3



3.6

K  10

3.15 2.7  ( ) 2.25

1.8 1.35

d

0.9 0.45

0.2

0.3

0.4

0.5 

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

T(  ) 2

 T(  )

K





Vertical load Vertical load

Examples of Shear Walls

0.4

0.5 

4.5

0.1

0.5 

4.05

0.4



Sabah Shawkat ©  T()

0.045

0.07



K  5

0.09

0.14

0.035

0.05

 (  ) d 

Horizontal load

0.45

 (  )    (  ) d  

0.6

0.7

0.8

0.9

169

Reinforced concrete column Although the column is essentially a compression member, the manner in which it tends to fail and the amount of load that causes failure depend on: 1. The material of which the column is made. 2. The shape of cross-section of the column. 3. The end conditions of the column.

bending of a beam, there is an important difference in that the designer can choose the axis about which a beam bends, but normally the column will take the line of least resistance and buckle in the direction where the column has the least lateral unsupported dimension

The consideration of the two end conditions together results in the following theoretical values for the effective length factor (the factor usually used in practice).

When the load on a column is not axial but eccentric, a bending stress is induced in the column as well as a direct compressive stress. This bending stress will need to be considered when designing the column with respect to buckling.The relationship between the length of the column, its lateral dimensions and the end fixityconditions will strongly affect the column’s resistance to buckling. Many countries have their own structural design codes, codes of practice or technical documents which perform a similar function. It is necessary for a designer to become familiar with local requirements or recommendations in regard to correct practice.

Columns and struts with both ends fixed in position and effectively restrained in direction would theoretically have an effective length of half the actual length. However, in practice this type of end condition is almost never perfect and therefore somewhat higher values for k are used and can be found in building codes. In fact, in order to avoid unpleasant surprises, the ends are often considered to be pinned (k = 1.0) even when, in reality, the ends are restrained or partially restrained in direction.

In low and normal strength concrete, significant non-linearities in the stress-strain behaviour start to develop at about 0.001 strain and the slope of the curve is close to zero at about 0.002 strain. The steel is therefore still in the elastic range, and is able to carry an increasing part of the load, when the non-linearities in the concrete start to develop. The usual range of the yield strength of ordinary reinforcement is 400 to 500 N/mm2. The reinforcement thus starts to yield at about the same strain level as the concrete reaches its maximum strength.

In this case, the end conditions for buckling about the x-x axis are not the same as about the y-y axis. There-fore both directions must be designed for buckling (Where the end conditions are the same, it is sufficient to check for buckling in the direction that has the least radius of gyration). Although the buckling of a column can be compared with the

In high strength concrete the stress-strain curve is more linear, and the strain at maximum stress is higher compared to lower strength concrete. The reinforcement in HSC columns will therefore yield before the concrete reaches its maximum strength and will continue to yield at about the same stress level until the concrete reaches its ultimate strain level.

As the loads on columns are never perfectly axial and the columns are not perfectly straight, there will always be small bending moments induced in the column when it is compressed.

Sabah Shawkat ©

RC Columns

170

Precast Concrete Columns can be circular, square or rectangular. For structures of five storeys or less, each column will normally be continuous to the full height of the building. For structures greater than five storeys two or more columns are spliced together. Precast concrete columns may be single or double storey height. The method of connection to the foundation and to the column above will vary with manufacturer. Foundation connection may be via a base plate connected to the column or by reinforcing bars projecting from the end of the column passing into sleeves that are subsequently filled with grout. Alternatively, a column may be set into a preformed hole in a foundation block and grouted into position.

Sabah Shawkat ©

Column-column connections may be by threaded rods joined with an appropriate connector; with concrete subsequently cast round to the dimensions of the cross-section of the column. Alternatively, bars in grouted sleeves, as described above, may be used. This results in a thin stitch between columns while the previous approach requires a deeper stitch. Connections may be located between floors, at a point of contraflexure, or at floor level. Columns are provided with necessary supports for the ends of the precast beams (corbels or cast-in steel sections). There will also be some form of connection to provide beam-column moment connection and continuity. For interior columns this may be by holes through which reinforcing bars pass from one beam to another. For edge columns, some form of bracket or socket is required. During erection columns must be braced until stability is achieved by making the necessary connections to the beams and slabs.

RC Columns

171

Design of reinforcement for reinforced concrete elements

slenderness:



Radius of inertia:

lo i I

h´ = 0,56.m

Characteristic strength of steel:

f yk

Extreme value of the compressive force on the column: Column height: l o 0.7 l 2.59 m Column length: i 0.288 b 0.0864m  Inertia radius: Column slenderness: lo

N sd



A c´

Stress in concrete: - where I is the - rectangle: moment of section I = 1/12.b.h3 Ac‘ = b‘ . h‘ inertia Determination of the minimum dimension of the cross-section of columns in terms of buckling

c

From the graph we get the value  at a given stress in concretec:

b´ = 0,26.m 25 MPa

f ck.cyl

410 MPa 3250 kN

3.70 m

Determination of basic characteristics

Cross-sectional and material characteristics of the column: c 0.02 m Reinforcement cover: h 0.60 m h´ h  2 c Section height: b 0.30 m b´ b  2 c Cross-sectional width: Characteristic cylindrical compressive strength of concrete:

h h'



Determining the buckling length lo

N sd b´  h´

b' b

Sabah Shawkat © • Design of reinforcement for reinforced concrete column Stress in concrete [MPa]: c = Nsd / Ac’ Total required reinforcement area in [cm2] : Arequired = c‘ .10 2 where  can be obtained from the graph as follows: Minimum amount of =5 =7,5 =10  0.1 12.84472 12.76732 reinforcement: 0.2

13.14803

13.06879

13.75465

13.67173

0.3 0.4

13.28918

=c

13.5888

Determining the maximum load-bearing capacity of the cross-section The maximum load-bearing capacity of the cross-section can be determined using the graph as follows: Aprovided . 10 2 / (Ac‘) =5 =7,5 =10 sdv MN  0.1 12.84472 12.76732 Ac‘ v m2 0.2 13.14803 13.06879 lo  v m =b < 0,6 . fck 0.3 13.28918 i  v m 0.4 13.75465 13.67173 13.5888 Areq, Aprovided v cm2 



3.23846

The minimum reinforcement area can be calculated as follows: Required amount of longitudinal reinforcement in the column can be determined using the coefficient obtained: - average:



N sd

cgs

 25

22.32  MPa



Determine the load-bearing capacity of the rectangular column shown in the figure bellow

2.5  cm

The actual amount of reinforcement designed for the column: Actual amount of reinforcement must be greater than A value: The reinforcement coefficient  is calculated as follows: Again, from the graph we can be obtained In reverse way the stress in concrete:

0.15

A smin

N sd

 f yk     1.15 

  b´  h´  100

-number profiles:

 10

13.67378cm 

47.151 cm

n  



49.08739cm 

As A

 c

As 100  b´  h´

3.37139

22.67275MPa 

By multiplying the value of concrete stress c with the net concrete cross-sectional area (b´. H´) we obtain the loadbearing capacity of the column:

Nud = c . b´.h´=3,3 MN

which must be greater than the value of Nsd applied to the column:

N ud  N sd

Examples of RC Columns

3.25 MN

172

Determine the load capacity of the circular column shown in the figure Cross-sectional and material characteristics of the column: c 0.03  m Reinforcement cover: D 0.5  m D´ D  2 c 0.44 m Column diameter: f ck.cyl 25 MPa Characteristic cylindrical compressive strength of concrete: f yk 410 MPa Characteristic strength of steel: N sd 3200 kN Extreme value of the compressive force on the column: l 3.5 m Column height: l o 0.7 l 2.45 m Strut length of column: radius of inertia:

 D 64

 D

0.125 m

Column slenderness:



Stress in concrete:

c

N sd

cgs

8  25 19.6

N sd 2



21.045 MPa

D' D

Sabah Shawkat © 4

From the graph we get the value  at a given stress in concretec: The minimum reinforcement area can be calculated as follows: Required amount of longitudinal reinforcement in the column can be determined using the coefficient obtained:



2.2

A smin

0.15

N sd

D´

  

- diameter of reinforcing -number  25 mm profiles: profiles: 2 The actual amount of reinforcement designed for  A s n   the column: 4 Actual amount of reinforcement As A must be greater than A value: The reinforcement coefficient  is calculated as follows:

c

 100

2.5826

D´

33.452 cm

39.26991cm 

 

Again, from the graph we can be obtained in reverse way the stress in concrete:

13.46341cm 

 f yk     1.15 

 100

22.2 MPa

By multiplying the concrete stress c by the net area of the concrete cross-section we obtain the load-bearing capacity of the column: Which must be greater than the value of Nsd applied to the column:

N ud

 D´2    4 

 c  

N ud  N sd

3.376 MN

3.20 MN

Examples of RC Columns

173

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 25

l=7,5 l=17,5 l=27 5 l=37 5 l=47,5

Slenderness ratio

l=57,5 l=67 5 l=77 5 l=87,5 l=97,5



l=5 l=15 l=25 l=35 l=45

where lo effective length in m i radius of gyration in m

l=55

in MPa

l=65 l=75 l=85 l=95

Stress of concrete

c

N sd

Sabah Shawkat © l=100 l=90 l=80 l=70 l=60

4,0

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

4,0

l=92,5 l=82,5 l=72 5 l=62 5

l=52 55 l=50 l=40 l=30 l=20 l=10

l=42 5

l=32 5 l=22,5 l=12,5

f ck.cyl

16  MPa

A c´

Design of Reinforcement for Reinforced Concrete Columns

Ac' in m2 Nsd in MN

Area of reinforcement in cm2

A req

  A c´  10

174

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 25

l=7 5 l=17 5 l=27,5 l=37,5 l=47 5

Slenderness ratio l=5 l=15 l=25 l=35 l=45

l=55

l=57,5

l=77,5 l=87 5 l=97,5

l=70

l=75 l=85 l=95

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

where lo effective length in m i radius of gyration in m

Stress of concrete in MPa c

N sd

2,0

2,5

3,0

3,5

4,0

A c´

l=82 5 l=72,5

l=62,5 15

l=30 l=20 l=10

l=92 5

l=60

l=50 l=40

Sabah Shawkat © 5

4,0

l=100 l=90 l=80

l=65

l=67 5



l=52,55

20 f ck.cyl

20  MPa

l=42,5 l=32,5 l=22 55 l=12 5

Ac' in m2 Nsd in MN Area of reinforcement

in cm2

A req

Design of Reinforcement for Reinforced Concrete Columns

  A c´  10

175

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 30

l=7 5 l=17,5 l=27 5 l=37,5 l=47 5

l=57,5

l=67 5

Slenderness ratio 

l=5 l=15 l=25 l=35 l=45

where lo effective length in m i radius of gyration in m

l=55

Stress of concrete in MPa

l=65

Sabah Shawkat © l=77 5 l=87,5 l=97,5

4,0

l=100 l=90 l=80 l=70 l=60 l=50 l=40 l=30 l=20 l=10

l=75 l=85 l=95

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

2,5

3,0

3,5

4,0

N sd A c´

l=92 5 l=82 5

l=72 5 l=62,5

Ac' in m2 Nsd in MN

l=52,55

l=42 5

f ck.cyl 30

2,0

c

25  MPa

l=32,5 l=22 55 l=12 5

Design of Reinforcement for Reinforced Concrete Columns

Area of reinforcement

in cm2

A req

  A c´  10

176

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 35

l=10 l=20 l=30 l=40 l=50

Slenderness ratio 

l=5 l=15 l=25 l=35

l=55

l=80 l=90 l=100 4,0

l=97 5 l=87 5 l=77 5 l=67 5

where lo effective length in m i radius of gyration in m

Stress of concrete in MPa

l=65

Sabah Shawkat © 10

l=75 l=85 l=95

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

4,0

c

A c´

l=92 5 l=82,5

l=72,5

l=62 5

l=52,5

l=42 5 l=32,55

30 35

N sd

Ac' in m2 Nsd in MN

l=57,5 l=47,5 l=37 5 l=27,5 l=17 5 l=7 5

l=45

l=60 l=70

f ck.cyl

30  MPa

l=22 55 l=12 5

Area of reinforcement in cm2 A req

Design of Reinforcement for Reinforced Concrete Columns

  A c´  10

177

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 35

l=7,5 l=17,5 l=27,5 l=37,5 l=47,5

l=5 l=15 l=25 l=35 l=45

25 20

l=57,5

l=55

l=67,5

Slenderness ratio

l=65



lo i

where lo effective length in m i radius of gyration in m

Stress of concrete in MPa

Sabah Shawkat © 5

4,0

l=100 l=90 l=80 l=70

l=75 l=85 l=95

l=77,5 l=87,5 l=97,5

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

4,0

l=62,5 l=52,5

A c´

N sd

l=92,5 l=82,5 l=72,5

l=60

l=50 l=40 l=30 l=20 l=10

c

f ck.cyl

35  MPa

l=42,5 l=32,5 l=22,5 l=12,5

Design of Reinforcement for Reinforced Concrete Columns

Ac' in m2 Nsd in MN Area of reinforcement in cm2 A req

  A c´  10

178

Design of reinforcement for reinforced concrete column in centric compression force and compression force with small eccentricity 40

l=7,5 l=17 5 l=27,5 l=37,5 l=47,5

Slenderness ratio

35 30 25

Stress of concrete in MPa

l=55

l=67 5 l=77,5 l=87,5 l=97,5

l=65

l=100 l=90 l=80 l=70 l=60

l=50 l=40 l=30 l=20 l=10

i where lo effective length in m i radius of gyration in m

l=57,5

4,0



l=5 l=15 l=25 l=35 l=45

Sabah Shawkat © l=75 l=85 l=95

10 5

3,5

3,0

2,5

2,0

1,5

1,0

0,5

0,0

0,5

1,0

1,5

2,0

2,5

3,0

3,5

4,0

c

N sd A c´

l=92,5 l=82 5 l=72,5

10 15

l=62,5

20 25

l=52,5

l=42,5 l=32 5 l=22,5 l=12 5

35 40

f ck.cyl

40  MPa

Ac' in m2 Nsd in MN Area of reinforcement in cm2 A req

Design of Reinforcement for Reinforced Concrete Columns

  A c´  10

179

Determine a reinforced concrete column of rectangular cross-section. The column is part of a

Design of reinforcement:

statically indeterminate structure. Extreme load induces normal force Nd and bending moment

Geometry factor:

Mf.

The actual length of the column is l and the effective length is le. u

Assumptions:

2800 kN

50 kN  m

0.7 l

u

 50

Design values of bending moment and compression force: Nd

1

0.96

Required amount of reinforcement:

2.8m

Asd

 Nd  1  b  h  0.8 Rbr    u  Rscr

Asd

0.00281m

Material characteristics, concrete, steel 10 425: Rbd

11.5 MPa

Rscd

375 MPa

b

Rbr

s

Rscr

11.5 MPa

Rbtd

375 MPa

0.9 MPa 210 MPa

Eb Rsd

27 GPa 375 MPa

Proposal: 2

1

22 mm

0.00304m

 1 

As1 2

As  Asd

As1

0.00038m

Asc

3  As1

Ast

n  As1

3  As1

Sabah Shawkat © 1

 lim

1.25 

 lim

Rsd

0.467

Asc

420 MPa

0.00114m

Ast

0.00114m

Assessment:

Design of cross-sectional dimensions: Mf

Abd

 vystu.

0.01786m

Abd

0.02

0.8 Rbr   vystu  Rscr

Determination of slenderness:

Abd

0.16766m

h

le  12

h

0.40947m

 h  35

21.55

Calculation of eccentricity ed: ee

Then the dimensions of the column we suggest: h

0.45 m

M f

N d

0.018m

  ee

0.018m

Percentage of reinforcement:

Slenderness ratio of the column will be: 

le 

12 h



  35

21.55



 sc 1

Calculation of Eccentricity - statically indeterminate construction ee

  ee

0.018m

Asc

 sc

bh

scmin  sc  scmax  scmin

le h

4

 10

Examples of RC Columns

0.00563

 scmax  scmin

0.03

0.00062

 stmin

1 Rbtd  3 Rsd

 stmin

0.0008

180

Design ultimate capacity of a cross-section may be determined as follows: Neu



 u  0.8 b  h  Rbr  As  Rscr



Neu

2883.26kN

2800kN

Nd  Neu

see diagrams, we design the reinforcement to the given cross-section as follows: c

0.02 m

c

b´

c

b´  h´

b  2 c

16.65 MPa

b´

0.41m

h´

b´



21.55



1.7

h´

0.41m

fck

20 MPa

Determination of reinforcement to the cross-section: Areq

  b´  h´  100

Areq

28.577cm

As1

0.00038m

1

22 mm

As1

 1 

Aprovided

Sabah Shawkat © n  As1

0.00304m

Aprovided  Areq

Examples, Column - Beam Joints

181

Reinforced concrete rectangular column on Figure bellow subjected to bending moment Msd and axial

To apply the design graphs the transformed actions Msds and Nsd have to be brought into a

compression force Nsd is given. Determine the required tension reinforcement for the cross - section. Use

dimensionless form:

the bi - linear diagram for steel ( ´

fcd 0.85

fck 25

fyd

fyk 300

Modulus of elasticity of reinforcement:

1.5

fyk 1.15

Design value of axial compression force:

Nsd 0.1

h 0.4

Nsd 0

b 0.3

lim

fyk MPa 1.15 Es

The required tension reinforcement A2 ( cm ) is as follows:

Nsd 104 A2 b d fcd 100

fyd

A2 8.81824

Determination of the tension reinforcement of reinforced concrete rectangular column shown on Figure

d 0.36

Nsd

bellow, subjected to bending moment Msd and tension force Nsd, use the bi - linear diagram for steel

( ´

0.06667

0) and bi - linear diagram for concrete

Assumptions:

Material data:

Concrete:

fck 25

Reinforcing steel:

fyk 300

fcd 0.85 fyd

Modulus of elasticity of reinforcement:

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement:

h Msds Msd Nsd d2 2

Msds 0.106

Nsd 0

1.5

1.15

Design value of axial tension force:

Nsd 0.1

b 0.20

Effective depth:

h 0.50

d2 0.04

d h d2

fcd 14.16667

fyd 260.86957

Es 200 GPa Msd 0.06

Cover or reinforcement:

fyk

fck

Design value of bending moment:

It is assumed that:

lim 0.72851

Sabah Shawkat © Msd

e 0.9

0.27

Eccentricity due to action effects:

0.08269

The compression reinforcement is not necessary.

d2 0.04

Effective depth: d h d2

0.0035

Es 200 GPa

0.0035

lim

fyd 260.86957

0.19245

From the Design graphs we get:

fcd 14.16667

Msd 0.09

Cover of reinforcement:

fck

Design value of bending moment:

It is assumed that:

Msds b d fcd

Material data

Reinforcing steel:

Assumptions:

Concrete:

0) and bi - linear diagram for concrete.

Nsd 0

d 0.46

Eccentricity due to action effects:

Msd

Nsd

Examples of RC Columns

0.08333

h 6

182 Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement:

Msds Msd Nsd

Msds 0.039

Eccentricity due to action effects: Msd

Nsd

0.09167

To apply the Design graphs the transformed actions Msds, Nsd have to be brought into a dimensionless form:

Combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement:

Msds

0.06505

From the graph we obtain:

0.0035

0.0258

0.08408

lim 0.72851

lim

fyk MPa 1.15 Es

h d2 2

Msds 0.19

To apply the Design graphs the transformed actions Msds and Nsd have to be brought into dimensionless form:

The compression reinforcement is not necessary.

0.0035

lim

Msds Msd Nsd

b d fcd

Msds 2

b d fcd

0.26824

Sabah Shawkat ©

The required tension reinforcement A2 ( cm ) will be: A2 b d fcd 100

Nsd 104 fyd

From the Design graphs we get:

A2 7.19593

0.12235

The required tension reinforcement A2 ( cm ) is as follows:

Determine the tension reinforcement of a reinforced concrete rectangular column shown on Figure bellow, subjected to bending moment Msd and compression force Nsd. Use the bi - linear diagram for steel ( ´

Nsd 4 10 fyd

A2 b d fcd 100

0) and bi - linear diagram for concrete.

Assumptions:

Material data Concrete:

fck 25

fcd 0.85

Reinforcing steel:

fyk 300

fyd

fck

fyk

Msd 0.1

Design value of axial compression force:

Nsd 0.4

b 0.2

Cover of reinforcement: Effective depth:

Nsd 0

h 0.55

d2 0.05

d h d2

fyd 260.86957

1.15

Design value of bending moment:

It is assumed that:

fcd 14.16667

1.5

d 0.5

Examples of RC Columns

A2 1.99958

183

Sabah Shawkat ©

Interaction Diagrams of RC Members

184

Calculation of Integration Diagram for column of rectangular cross-section according to ACI 318-89

1fc'

Fs2 C

h/2 Neutral axis

r s1

cgc

s2 c'

r s2

ds2

y c'

. 1 c

A g A s2

 

fc'

ds1

Characteristic value of concrete cylinder compressive strength: fc’ = 30 MPa Design value of concrete cylinder compressive strength: Rc = 0,85. fc’ = 25.5 MPa Characteristic yield stress of the steel: fy = 420 MPa Cross-section: b = 0.5 m h = 0.55 m Lower reinforcement: - distance of the lower reinforcement centroid from the lowest fibre of the cross-section: ds1 = 0.028 m - diameter:  = 16 mm - number of bars: n=4 2 - the entire area of the lower reinforcing bars :    2 A s1  n    8.042 cm 4   - lever arm of the reinforcement to the cgc rs1 = 0.247 m Upper reinforcement: - distance of the upper reinforcement centroid from the uppest fibre of the cross-section: ds1 = 0.028 m - diameter:  = 16 mm - number of bars: n=4 2 - the entire area of the upper reinforcing bars :    2 A s2  n    8.042 cm 4   - lever arm of the reinforcement to the cgc rs2 = 0.247 m The ACI Code stress block parameters: 1 = 0.85 1 = 0.85 – (fc‘ – 30) 0.008 = 0.83244   1 f Compression  F



s2 fy

As1

Fs1

s1 y s1 Es

fy s1

 Pure compression



Sabah Shawkat © c

f c'

 c'

Compression

c'

3%o

c



1fc

3%o

4200

2800

 rs2 r c

Fs2 C

rs1

Fs1

3% o c

3 BALANCED CONDITION

1 fc

y

b COMPRESSION FAILURE

Fs2 C

rs2 rc

Fs1

1 fc

3%o

N [kN]

rs1

1 f c'

5600

rs2

 rs2 rc

Fs2 C

rs1

Fs1

3.  y

1 fc

5 1400

TENSION FAILURE

100

200

300

400

M [kN.m]

rs1

Fs2 rs2

1400 Tension

Fs1

y

rs1

y

Fs1

-compression force acting in the reinforcement Fs1 = - As1 . fy = -337.78404 kN Fs2 = - As2 . fy = -337.78404 kN -area of the compression zone Ag = b. h = 0.275 m2 -net area of concrete compression zone A = Ag – As1 – As2 = 0.27339 m2 -concrete compressive force C = - Rc . Ag = -6971.48 kN -the axial load capacity Fs1 + Fs2 + C = -7647.05 kN -the moment capacity M = Fs1 .rs1 + Fs2 . rs2 + C.rc = 0 kNm N =0,7. N = -5352.94 kN M1 = M =0,7. M = 0 kNm

  -depth of compression zone c = h – ds1 = 0,522 m -equivalent depth of compression zone =1 .c = 0.43454 m -area of the compression zone Ag = b . = 0.21727 m2 -concrete compressive force C = - Rc . Ag = -5519.82 kN -the strain of the upper renforcement

Interaction Diagrams of RC Members



rs2

Fs2 C

rs1

3%o

Fs1 1 fc



-lever arm of concrete compressive force A s1 r s1  A s2 r s2

0 m

where is the strength reduction factor

3%o Ag

Fs2 C

r  s2rc

1fc

-lever arm of concrete compressive force

185

 s2

c  d s2 c

0.003



0.00284

-compression force in upper reinforcement Fs2 Fs2



if s2  y  s2 Es  A s2  fy  A s2



 -net area of concrete compression zone A = Ag – As2 = 0.21646 m2

0.003



0.00273

1fc



-lever arm of concrete compressive force

337.78404 kN

 

0.5 A g   h  A s2 r s2

-depth of compression zone 0.003   d 0.16839m  c   0.003  3  y   -the strain of the upper reinforcement c  d s2

0.003

0.0025



-tension force in lower reinforcement Fs1 = fy . As1 = 337,78 kN

0.14656m

 -net area of concrete compression zone A = Ag – As2 = 0.127 m2    N = 0,7 . N = -2266.95 kN M3 = M = 0,7 . M = 449.05 kNm

lever arm of concrete compressive force 0.5 A g   h  A s2 r s2

-the axial load capacity  Fs1+ Fs2 + C = -1766.69 kN -the moment capacity M = Fs1 . rs1 + Fs2 . rs2 + C.rc = 528.02 kNm

  -tension force in lower reinforcement Fs1 = fy . As1 = 337,78 kN -the axial load capacity  Fs1 = 337,78 kN -the moment capacity M = Fs1 . rs1 = 83.43 kNm N = 0,9 . N = 304.00 kN M5 = M = 0,9 . M = 75.08 kNm

0.20442m

   N = 0,7 . N = -1236.68 kN M4 = M = 0,7 . M = 369.61 kNm 

Sabah Shawkat ©

if s2  y  s2  Es  A s2  fy  A s2



-tension force in lower reinforcement Fs1 = fy . As1 = 337,78 kN -equivalent depth of compression zone =1 .c = 0.25561 m -area of the compression zone Ag = b . = 0.1278m2 -concrete compressive force C = - Rc . Ag = -3238.51 kN -the axial load capacity  Fs1+ Fs2 + C = -3238.51kN -the moment capacity M = Fs1. rs1 + Fs2 . rs2 + C.rc = 641.51 kNm

 s2

Fs2 C

r  s2rc



-compression force in upper reinforcement



3%o

-net area of concrete compression zone A = Ag – As2 = 0.06928 m2

N = 0,7 . N = -4100.32 kN M2 = M = 0,7 . M = 278.75 kNm

-depth of compression zone  0.003   h  d c  s1 0.30706m   0.003   y  -the strain of the upper renforcement c  d s2

-equivalent depth of compression zone =1 .c = 0.14017 m -area of the compression zone Ag = b . = 0.07009 m2

-concrete compressive force C = - Rc . Ag = -1766.69 kN

 Balance condition

Fs2 Fs2

0.05703 m

337.78404 kN

-the axial load capacity  Fs2 + C = -5857.6 kN -the moment capacity M = Fs2 . rs2 + C.rc = 398.22 kNm

 s2

0.5 A g   h  A s2 r s2

1 fc

3% o

 rs2 rc

h b

Fs2 C

 Pure tension

-tension force in upper reinforcement Fs2 = fy . As2 = 337,78 kN -tension force in lower reinforcement Fs1 = fy . As1 = 337,78 kN -the axial load capacity  Fs1 + Fs2= 675.56 kN -the moment capacity M = Fs1 . rs1 + Fs2 . rs2 = 0 kNm N = 0,9 . N = 608.01 kN M6 = M = 0,9 . M = 0 kNm

rs1

Fs1

3.  y

 -compression force in upper reinforcement F if       E  A  f  A  s2 Fs2

s2 y s2 s 337.78404 kN

Interaction Diagrams of RC Members

rs1

y

Fs1

 

h b



Fs2

rs2 rs1

y

Fs1

186

Concrete Foundations A foundation is an integral part of the structure which transfer the load of the superstructure to the soil without excessive settlement. A foundation is that member which provides support for the structure and it‘s loads. It also provides a means by which forces or movements within the ground can be resisted by the building. In some cases, foundation elements can perform a number of functions: forexample, a diaphragm wall forming part of a basement will usually be designed to carry loading from the superstructure. If new foundations are placed close to those of an existing building, the loading on the ground will increase and movements to the existing building may occur. When an excavation is made, the stability of adjacent buildings may be threatened unless the excavation is adequately supported. This is particularly important with sands and gravels which derive their support from lateral restraint.

An essential requirement in foundations is the evaluation of the load which a structure can safely bear. The types of foundation generally adopted for building and structures are spread (pad), strip, balanced and cantilever or combined footings, raft and pile foundations. For example, strip footings are usually chosen for buildings in which relatively small loads are carried mainly on walls. When the spread footings occupy more than half the area covered by the structure and where differential settlement on poor soil is likely to occur a raft foundation is found to be more economical. Pad footings, piles or pile groups are more appropriate when the structural loads are carried by columns. If differential settlements must be tightly controlled, shallow strip or pad footings (except on rock or dense sand) will probably be inadequate so stiffer surface rafts or deeper foundations may have to be considered as alternatives.

Sabah Shawkat ©

The choice of foundation type or the type of foundation selected for a particular structure is influenced by the following factors: 1. The imposed loads or deformations, the magnitude of the external loads 2. Ground conditions, the strength and compressibility of the various soil data 3. The position of the water table 4. Economics 5. Buildability, and the depth of foundations of adjacent structures 6. Durability.

This type of foundation viewed as the inverse of a one-storey beam, slab and column system. The slab rests on soil carrying the load from the beam/column system which itself transmits the loads from the superstructure.

Types of foundations

These are generally supporting columns and may be square or rectangular in plan and in section, they may be of the slab, stepped or sloping type.The stepped footing results in a better distribution of load than a slab footing. A sloped footing is more economical although constructional problems are associated with the sloping surface. The isolated spread footing in plan concrete has the advantage that the column load is transferred to the soil through dispersion in the footing. In reinforced concrete footings, i.e. pads, the slab is treated as an inverted cantilever bearing the soil pressure and supported by the column. Where a two-way footing is provided it must be reinforced in two directions of the bending with bars of steel placed in the bottom of the pad parallel to its sides.

RC Foundations

187

Foundations under walls or under closely spaced rows of columns sometimes require a specific type of foundation, such as cantilever and balanced footings and strip footings. Pad footing Square or rectangular footing supporting a single column. Strip footing Long footing supporting a continuous wall. Combined footing Footing supporting two or more columns. Balanced footing Footing supporting two columns, one of which lies at or near one end.

c. When the foundation is eccentrically loaded, the reactions vary linearly across the footing or across the pile system. Footings should generally be so proportioned that zero pressure occurs only at one edge. It should be noted that eccentricity of load can arise in two ways: the columns being located eccentrically on the foundation; and/or the column transmitting a moment to the foundation. Both should be taken into account and combined to give the maximum eccentricity. d. All parts of a footing in contact with the soil should be included in the assessment of contact pressure e. It is preferable to maintain a reasonably similar pressure under all foundations to avoid significant differential settlement. Shallow Foundations A shallow foundation distributes loads from the building into the upper layers of theground. Shallow foundations are susceptible to any seismic effect that changes the ground contour, such as settlement or lateral movement. Such foundations are suitable when these upper soil layers have sufficient strength (‘bearing capacity’) to carry the load with an acceptable margin of safety and tolerable settlement over the design life.

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Raft Foundation supporting a number of columns or loadbearing walls so as to transmit approximately uniform loading to the soil. Pile cap Foundation in the form of a pad, strip, combined or balanced footing in which the forces are transmitted to the soil through a system of piles.

The plan area of the foundation should be proportioned on the following assumptions: a. All forces are transmitted to the soil without exceeding the allowable bearing pressure b. When the foundation is axially loaded, the reactions to design loads are uniformly distributed per unit area or per pile. A foundation may be treated as axially loaded if the eccentricity does not exceed 0.02 times the length in that direction

The different types of shallow foundation are: a) Strip footing b) Spread or isolated footing c) Combined footing Strap or cantilever footing d) Mat or raft Foundation.

RC Foundations

188 Strap Footing It consists of two isolated footings connected with a structural strap or a lever. The strap connects the footing such that they behave as one unit. The strap simply acts as a connecting beam. A strap footing is more economical than a combined footing when the allowable soil pressure is relatively high and distance between the columns is large. Combined Footing It supports two columns; it is used when the two columns are so close to each other that their individual footings would overlap. A combine footing may be rectangular or trapezoidal in plan. Trapezoidal footing is provided when the load on one of the columns is larger than the other column. Strip/continuous footings A strip footing is another type of spread footing which is provided for a load bearing wall. A strip footing can also be provided for a row of columns which are so closely spaced that their spread footings overlap or nearly touch each other. In such a cases, it is more economical to provide a strip footing than to provide a number of spread footings in one line. A strip footingis also known as “continuous footing”.A traditional strip foundation consists of a minimum thickness of 150 mm of concrete placed in a trench, typically 0.8–1 m wide. Reinforcement can be added if a wider strip is required to bridge over soft spots at movement joints or changes in founding strata.

Pile foundations Piles are individual columns, generally constructed of concrete or steel, that support loading through a combination of friction on the pile shaft and end-bearing on the pile toe. The distribution of load carried by each mechanism is a function of soil type, pile type and settlement. They can also be used to resist imposed loading caused by the movement of the surrounding soil, such as vertical movements of shrinking and swelling soils. Piles can be installed vertically or may be raked to support different loading configurations. All pile caps should generally be reinforced in two orthogonal directions on the top and bottom faces and the amount of reinforcement should not be less than 0.0015bh in each direction. The bending moments and the reinforcement should be calculated on critical sections at the column faces, assuming that the pile loads are concentrated at the pile centres. This reinforcement should be continued past the piles and bent up vertically to provide full anchorage past the centreline of each pile.Deep foundations are used when the soil at foundation level is inadequate to support the imposed loads with the required settlement criterion. Where the bearing capacity of the soil is poor or the imposed load are very heavy, piles, which may be square, circular or other shapesare used for foundations. If no soil layer is available, the pile is driven to a depth such that the load is supported through the surface friction of the pile. The piles can be precast or cast in situ. Deep foundations act by transferring loads down to competent soil at depth and/or by carrying loading by frictional forces acting on the vertical face of the pile. Diaphragm walls, contiguous bored piles and secant piling methods are covered later in this chapter. Short-bored piles have been used on difficult ground for low-rise construction for many years.They can be designed to carry loads with limited settlements, or to reduce total or differential settlements.

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Mat or Raft footings It is a large slab supporting a number of columns and walls under entire structure or a large part of the structure. A mat is required when the allowable soil pressure is low or where the columns and walls are so close that individual footings would overlap or nearly touch each other. Mat foundations are useful in reducing the differential settlements on non-homogeneous soils or where there is large variation in the loads on individual columns.

RC Foundations

189

Sabah Shawkat ©

RC Foundations

190

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RC Foundations

191

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RC Foundations

192

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RC Foundations

193

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Realisation of Foundations

194

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Calculation of Bending Moments in RC Foundations

195 A signed convention:

Endless beam In the case of infinite beams subjected to concentrated loads P, the diagram of the contact stresses in the soil P, shear forces on the beam T and the bending moment on the beam M at any distance x from the point of application P can be expressed as follows: P

C Y

P( x)

C  5000  kPa

C  Y( x)

 ( x) 

I  0.0535  m 4

4 Le 

E C kPa



0.185 as well as the value of the ratio  = - 0.140 -> = 0.185 Coefficients are symmetrical Coefficient is asymmetrical.

Y  10  mm

T( x)

Le  B B  2  m

MPa

- Coefficients ,  apply equally to both positive and negative ratio, if the  = 0.140 ->  =

 ( x) P

M ( x)

E  210  MPa

( x)  P  Le

4 E I C B

I  0.0535  m



m B

 10

Le  2.589156

Where

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E = modulus of elasticity of the base strip

I = the moment of inertia of the cross section of the base strip The coefficients , ,  are functions of the ratio x



Le 1 

 ()   e 2  (  ) 

1 4

  0 0.05 6

 (cos()  sin())

 e

 

 (  )  

2

 e



 ( cos (  ) )

B - Final beam



The final beams are called real beams. In this case, the final beam computed as part of

 ( cos (  )  sin (  ) )

the endless, which in the extreme cross-sections A, B satisfies the following boundary

0.6

conditions:

0.4

MA = 0, MB = 0, TA = 0, TB = 0,

 (  ) 0.2

Assume that the beam is loaded not only real forces P1 ... Pn but fictitious Q1, Q2 Q3, Q4 active

 ( )  ( )

outside the actual beam distance and the locus of the edge of the beam is as follows:

 0.2

Q1 ........ Le / 2 from the cross-section A =>  = P / 2 => (P / 2) = -0.052 and  (P / 2) = 0

 0.4

=> shear force applied force Q1 in cross section and is equal to 0 Q2 ........ .Le / 4 from the cross section A => = p / 2 => (P / 4) = 0, and (P / 4) = 00:16

 0.6 

i  1

x  1  m 1

Bending moment deduced force Q 2 in cross-section A is equal to 0

Q3 ........ Le / 2 from the cross-section B =>  = P / 2 =>  (P / 2) = -0.052 and  (P / 2) = 0 =>

Shear

force

Examples of RC Foundations

applied

force

cross-section

equal

196 Q4 ........ Le / 4 the cross section B => = p / 2 =>  (P / 4) = 0, and  (P / 4) = 00:16 => Bending moment deduced force Q4 in cross-section A is equal to 0

On the basis of the simplified, above-mentioned boundary conditions MA = 0, MB = 0, TA = 0, TB = 0 to write the following: and then can be calculated from the sizes of these conditions fictitious forces Q 1, Q 2, Q 3 and Q 4: Q1  1000

Q2  1000

Q3  1000

L  2  m

Le  2.94  m

P  1000

Q4  1000

L    L   e 4    1.46567

P1 = Nd1

Given Q1  



 2

 Q2  



 4

 

 L      Q3        Le   

 P    1



L     Q4       

 4

 Le  



 

Traces of bending moments M along the entire beam of length L 0

xL  0  m0.01  m L

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L   L   2        Q1      Q2      P      Q3      2  4  1   Le    L    Q1   

 Le  



 Le  

L     Q2       



 Le  

L     Q4        



 Le  



 

 Le  

 L    P    2    Q3      Q4      1  L      2 4   e 

      M  xL  Le   Q     1        Q     3   

 x L         4  Le  xL    L 2  Q     P       2 1 Le Le      Le      L  L  x     Le   L  xL  L     4 e 2    Q4    Le Le    

 Le  xL

   

 



    

M ( 1  m)  417.104651m

500

L   L  L   L     e  e    2 4   Q2       P     2    Q3       Q4          L      1 Le Le 2 4      e  



Q1   

400

300

 

M xL 200 100

Q  Find ( Q1Q2Q3Q4)



Q  2.073712 10

5.320358 10

1.794347 10

5.940857 10



0.5

 100

Examples of RC Foundations

1.5

197 The settlement of foundations

The course of the contact stress in the ground along the entire beam of length L

Characteristic of embankment:

B  1.2  m

      P. xL   ( Q1)     Le  B         Q3          1



     1  x x    4  Le  xL    P    L i    Q2       i   Le Le     i  1   Le           Le   L  xL    4  Le   L  xL   2   Q4     Le Le    2

 Le  xL

4 1

P.( 25.52  m)  4.010588 10

1

 kg



        

Eoedn  2.2  MPa

 n  17 



kN 



 sun  1  nn    sn  10 

 s kPa



kN m

3



Characteristic of other soils: EdefS3  19  MPa

290

 sn  22 

c efn  17  kPa

 sun  8.16

EdefF4  5  MPa

 

P . xL

 efn  28.5  deg

kN m

For S3

 S3  0.74

EoedS3 

 F4  0.62

EoedF4 

EdefS3

 S3  17.5 

 S3

For F4 289

nn  0.32

EdefF4

 F4  18.5 

 F4

kN m

 suS3   S3 0.55

kN m

288

Sabah Shawkat © Original stress at individual depths:

287

0.5

1.5

1.  or1   n  2  m

2.  or2   n  2  m   sun  2  m

3 up.  or3h   n  2  m   sun  2  m   suS3  3  m

The course of shear forces along the beam of length L

       1     2  Le  xL     4  Le  xL    x x P    L i T xL  ( Q1)      Q2         i  Le Le Le         i  1      L   L  x       L  L  x     e L  L    4 e   Q3    2  Q4         Le Le     

 or1  34 kPa





 x x     L i   xL  xi  

      

 or3d   n  2  m   sun  2  m   suS3  3  m  10 

Contact stress:

 or2  50.32 kPa

 or3h  79.195 kPa

kN m

 or3d  129.195 kPa

5 m

 ds  0.95  80  kPa

additional load:  ol  42 kPa

 ol   ds   or1

600

Base dimensions:

400

b  2  m

200

 

T xL

0.5

 200

1.5

I ( l b z) 

 400

I ( l b z)

 600

4

l  4  m 1 2  1 2 

l b   2 2 2  z l  b  z 

 atan 

 

l b   2 2 2 z l   b  z  

 atan 

 

l b z 2

 

l  b  z l  z l b z 2

l b z

1 2

l  z





2 

b z





2 

b z



 ol  ( I ( 0.87  l 0.87  b z)  I ( 0.87  l 0.13  b z)  I ( 0.13  l 0.13  b z)  I ( 0.13  l 0.87  b z) )

Examples of RC Foundations

198

I1 ( z)  ( I ( 0.87  l 0.87  b z)  I ( 0.87  l 0.13  b z)  I ( 0.13  l 0.13  b z)  I ( 0.13  l 0.87  b z) )

V1 

I1 ( 2  m)  0.3079

Vertical stress in the subsoil foundation:

 s 4    

 z ( z)  I1 ( z)   ol 4

6 10

3 10

0.5

1.5

2.5

3.5

4.5

 or1  0.2  6.8 kPa

 z ( 2  m)  12.931 kPa

 or2  0.2  10.064 kPa

K  1

 or1

C  1

 or2A

Calculation of active depth:

 or3h  0.2

K 2  C

kPa





 or.m ( z)   V1  z 

kPa

V  Find( K C)

K 5  C

 V2  kPa



 z ( z )   or.m1

E oedn

0m

C 

 ol

6 10

 or2

kPa

3 10





0.5

 1.925     6.214 

kPa

 dz    

1.5

2.5

3.5

4.5

K1 0  C1

K1  1  or2  0.2

kPa

V1 



2.77  m

 z ( z )   or.m ( z )

E oedS3

2m

3 kN 3

s 4  14.005 mm

K1 2  C1





 V12  kPa 



 oR2 ( 2  m)  50.32 kPa

K1 2  C1

z = 10mm..5m

   oR1 ( 2  m)   oR2 ( 2  m)   17.68 kPa  or2

kPa

K1 2  C1

V1  Find( K1 C1)

Examples of RC Foundations

C1  1

Given V1  Find( K1 C1)  8.16  V1     34  K1  1 C1  1 Given V1  Find( K1 C1) V1 

C1  1

 17     34 

K1  1

K1 0  C1

 oR2 ( z)   V11  z  0

 V12  kPa 

 or1  0.2

( z)

K1  1 C1  1 Given V1  Find( K1 C1)

K1 2  C1

m  oR1 ( 2  m)  68 kPa

kPa

V



C  2.999  10

K1 0  C1

 oR1 ( z)   V11  z 

 or1

 or.m( z)

 V12  kPa 

 or2A   or1   n  2  m

kPa

 z ( z)



Sabah Shawkat ©

 or3d  0.2  25.839 kPa

kPa

Calculation of settlement when (upper groundwater level) reduction in the embankment (S3):

 or2  0.2

2m

s4  0.014m 0



 or.m1 ( z)   V11  z 

Calculation of settle S4:

z  10  mm 20  mm  5  m

 z ( z)

 1.632     6.8 

 24.66     1 

K1  1 C1  1 Given V1  Find( K1 C1)

199

 or3h

kPa

V1 

K1 5  C1





 oR4 ( z)   V11  z 

 V12  kPa 

Settle of subsoil embankment S5:

 9.625     48.75 

s 2  s S3  s F4 2



kPa

  0.06

Specific weight:

K1 2  C1

 or3h  

 s   sn

Weight after removal of organic matter:

K1 5  C1

kPa

 d   n

The calculation of settlement S3:

s 5 





 oR3 ( z)   V11  z 

S2 = 30.838 mm

  0.4

Organic matter content:

 oR4 ( 2  m)  50.32 kPa  or2A

 V12  kPa 



   d s

4 m

s 5  74.182 mm

Total settlement:

Sabah Shawkat © 5 m

h = 4m

-4,000

s 1 

 or2  h

S3, S - F

The original stress at the interface of embankment and gravel:  or2  50.32 kPa

s 1  45.745 mm

2  Eoedn

-7,000

F4, CS

Settle of subsoil embankment S3: Height S3, height F4: hS3  3  m

hF4  4  m

s S3  2

 or2  hS3

EoedS3

s F4  2

 or2  hF4

EoedF4

Examples of RC Foundations

ds = 0,95 . Rtd

d=2000

-2,000 = HPV

Pokles

Settle of subsoil embankment S2: The depth of embankment:

S  172.238 mm

0,000

s 3  7.467 mm

NÁSYP

 oR3 ( 2  m)  68 kPa

S  s 1  s 2  s 3  s 4  s 5

Aktivna hlbka

2 m

     oR1 ( z)   oR2 ( z)  oR3 ( z)   oR4 ( z)  dz   dz s 3      Eoedn EoedS3  0 m  2 m  

200 Replacement of unsupported Needle Subsoil with gravel

Conditions: 1 (z) < Rdt

kN m

6 10

3 10 4

1 10

Eudiometric elastic modulus of gravel: Eoed  145  MPa

 2.5 10 0 y ( z)

foundation width:

0.35

0.7

1.05

1.4

1.75

2.1

2.45

2.8

3.15

Rdt

0 5

3.5

 3 10

 1.5 10

 6 10  10  8  6  4  2

 2.75 10

Load on strip:

Gravel width: bv  bmax  2  hv

bv  7.69 m

Condition 2:

 448.276 kPa >

0 z

=> minimum depth of gravel: hv  3.12  m

 4 10

Vde  650  kN Vde

 ( z)

dy ( z)

bmax  1.45  m

b max  1  m

 Vde   d   bmax1 m 

 ( z)    d   nv  z  I2 ( z)  

z =2.5m…4m

Bulk weight of gravel:  nv  21.5 

(z) < Rdt

Horizontal bearing of subsoil:

y < Rh

Rdt  206  kPa  de 

Weight of soil:   21 

kN m

Vde bmax 1  m

 hv   z

1 ( z)  atan 

z  1  mm 0.01  m  hv

Sabah Shawkat © d  1  m

l  10  m

b  bmax

 hv  bmax  z  

 ( z)  atan 

l b  I ( l b z)   atan   2   2 2 2   z l  b  z 

l b z

1 1    2 2  2 2 2  l2  z2 b  z  l b z 

y ( z) 

 de 

 

  1 ( z) 

k  0.5

Rh  k  Rdt

(for clay)

1 1   sin ( 2  1 ( z) )   ( z)   sin ( 2   ( z) )  2 2 

dy ( r)

dy ( z) 

d ( y ( z)  m) dz

I2 ( z)  ( I ( 0.87  l 0.87  b z)  I ( 0.13  b 0.87  l z)  I ( 0.87  l 0.87  b z)  I ( 0.13  b 0.87  l z) )

 Vde   d   bmax1 m 

y ( 2.15  m)

z= 1m…7m

 ( z)    d   nv  z  I2 ( z)  

 35.054 kPa 4

1 10

6 10

 2.5 10 0

3 10

 ( z)

=> 2. condition is ok

Rh  103 kPa

Rdt

y ( z) dy ( z)

hv 0.35

0.7

1.05

1.4

1.75

 1.5 10

 2.75 10

 4 10

 3 10

z 5

 6 10  10

5

Examples of RC Foundations

2.1

2.45

2.8

3.15

3.5

201 Design of reinforcement in footing

Force load

Material characteristics

Force load from the outer walls: F1d  82 kN

Concrete fccub_  30 MPa

 fccyl   0.855  

0.005 

 fccub_   fctm  0.274   MPa 

Force load of self-weight of the column:

fccub_  MPa

 fccub_ 

fccyl  21.15MPa

F2d  bs kv 25 

kN m

1.35

F2d  15.39 kN

0.66

MPa

fctm  2.59MPa

Total load on the foundation will be: Number of floors: 4

fccyl

fcd  0.85  1.5







N1d  qdxzs1  F1d  F2d n  qstrzs1  bs ( 2.85 m  1.5 m ) 25 

fcd  11.985MPa



kN m

N1d  2.321  103 kN

Steel

Sabah Shawkat © fyk  410 MPa

fyk fyd  1.15

ly  6 m l1  7.2 m

lk  2.16 m

kN   2 N2d  qdxzs2  F2d n  qstrzs2  bs ( 2.85 m  1.5 m ) 25  1.35 3 m  



fyd  356.522 MPa

bs  400 mm

l2  3.6 m

kv  2.85 m



N2d  1.874  103 kN

Distributed loads

Distributed loads of the floor: qdx  11.75 

kN 2

m Surface roof load: qstr  8.22 

kN m

Load area ZS1

l1   zs1  ly  lk   2  

zs1  34.56 m

Calculation model Design dimensions of reinforced concrete STRIP – footing.

Load area ZS2

 l2 l1  zs2  ly    2 2

Building height: zs2  32.4 m

H  n 2.85 m  ( 2.85 m  1.5 m )

Examples of RC Foundations

H  15.75 m



1.35



202

The degree of constraint of building: h 

h  1.575 m

Nd.max

d  1.2m

Extreme force applied at the end column: N1d  2.321  10 kN

Ns.max

1.44 

0.6 Rbd

 5 cm

1.44 

f

0.6 Rbd

 5 cm

0.8 m

Earth weight:



 18 

kN m

Service force acting on the end column: N1s 

N1d

N1s  1.857  103 kN

1.25

The tabulated carrying capacity of the soil at a depth of footing bottom = 1 m for S4 Rdt  260kPa

Extreme force applied in the central column: N2d  1.874  103 kN

Tabulated load bearing capacity of the soil at a depth of footing bottom = t + h: R

dt  Rdt  2.5 ( d  1 m ) 

R

dt  269 kPa

Sabah Shawkat ©

Service force acting on central columns: N2s 

N2d

N2s  1.499  103 kN

1.25

The sum of all forces from the columns to the base strip – footing: Vs  2 N1s  2 N2s

Vs  6.713  103 kN

Preliminary design of footing dimensions:

Preliminary weight of foundation:

Max. Distance between columns: l1.  7.2 m

Gzs  0.1 Vs

Preliminary ground plan of the foundation strip L x B = Aef

Unloading strip: 1

lv  l1 4

lv  2.1 m

The entire length of strip- foundation: L  2 lv  2 l1  l2

L  22.2 m

N1s 0.6 fcd

Vs  Gzs R

The width of the foundation: Aef L I suggest: B 

The depth of the footing: h  1.44 

Aef 

B  1.237 m

B  1.5 m  5 cm

h  0.782 m

Examples of RC Foundations

Aef  27.451 m

203

Dimension checking of the foundation: s ( x) BLh 25 

 zs1 

kN m

 zs1

 zs2

 250.669 kN

Vskut

 skut

B L

4 EI

BC

E I 

Vskut  7.615  103 kN

 228.664 kPa

d2 2

s ( x)

lv Ma  fd lv  2

Vskut  Vs   zs1   zs2 Maximum tension in footing bottom: 

s ( x)

The bending moment calculation:

Force in footing bottom:

 skut

EI 

Weight of soil below the foundation: (BL) (d  h ) 

Le  x x  x x   e  C1 cos   C2 sin   e  C3 cos   C4 sin  Le Le Le Le     Le

 650.849 kN

Q  zs2 

x

The actual weight footing:

l1  lv   Mab  fd  lv    2  2 

dt  269 kPa

R

Ma  833.456 kN m

satisfies





Mb  fd  lv  l1 

lv  l1

l1 2

 N1d 

Mab  2.217  103 kN m

Mb  368.511 kN m

 N1d l1

Sabah Shawkat © 2

l2  lv  l1   Mbb  fd  lv  l1    2  2 

Calculation sectional forces

Overall extreme force (weightlessness strip)

l2 2

 

 N1d  l1 

l2  2

  N2d  2 2 

Vd  2 N1d  2 N2d

Vd  8.391  103 kN



i1

Mbb  980.846 kN m

The computation of shear forces:

The stress in footing bottom:  d1

Vd BL



 d1

 251.99 kPa

1

6 Vd e Vd  2 BL BL

2

6 Vd e Vd  2 BL BL

Vk  793.768 kN

Vka  N1d  Vk

Vab  fd  lv  l1  N1d



Vab  1.194  103 kN

Vbb  N2d  Vab

Vbb  680.373 kN

Vk  fd lv

  Ndi xi

e´

i1



i1



The reaction in footing bottom: fd   d1 B

kN fd  377.985 m

E I 

d4 4

s ( x)  BC s ( x)

f ( x)

Examples of RC Foundations

Vka  1.528  103 kN

204

The compression height limit

 420 d MPa  xulim     525 MPa  fyd 

xulim  0.349 m

Required area of reinforcement: As 

xuBfcd

As  39.733 cm

fyd

Area 1 of 25 profile: As1   

( 25 mm)

As1  4.909 cm

Number of profiles: As As1

Sabah Shawkat © n 

n  8.094

I suggest 8 V25 to the entire width of the strip foundation b: Assessment of the Design:

Figure 3.9.4-2: Bending moments diagram vs Shear forces diagram due to design loads

Designed of reinforcement:

Dimensioning

Design of reinforcement for the maximum Bending moment between the columns :

Amsk  As1 8

Amsk  39.27 cm

Design moment: The depth of compression zone of concrete:

M  Mab Cover

xureal 

cst  5 cm effective height:

d  h  c st

d  0.732 m

d 

xureal  0.078 m

Carrying capacity of cross-section:



Mu  xureal  B  fcd   d  

Compression depth of concrete: xu  d 

fyd Amsk Bfcd

M 0.5 Bfcd

xureal 

xu  0.079 m

Examples of RC Foundations

 

Mu  970.046 kN m

M  980.846 kN m

205 Or we can be calculated according to the design of reinforced concrete beam as follows:

Mab  2.217 MN m

B  1.5 m

fcd  11.985 MPa



 0.33144

z   d





Mab



 0.86742

d  h  cst 

h  0.782 m

 0.23



d  0.732 m

 0.95277

x   d

x  0.086 m

xu  0.8 x

xu  0.069 m

z   d

z  0.697 m

As   Bd fcd 100

Mult  As zfyd

Mult



 0.11808



34.87 cm

0.866 MN m

 0.07437

Bd fcd

Design of shear reinforcement

x   d

x  0.243 m

xus  0.8 x

xus  0.194 m

Area A

z  0.635 m

Sectional shear resistance: 2

As   Bd fcd 100

97.86 cm

Mult  As zfyd

Mult  2.263 MN m

we provide

As  100 cm

Vcu  Bh fctm 3

Vcu  1.011  103kN

Shear force applied at a given cross-section: Or we can be calculated according to the design of RCB B3-B3.3 as follows:

Sabah Shawkat © M  0.980 MN m 



M 2

h  0.782 m 

Vk  793.768 kN Force transferring by stirrups;

d  0.732 m

 0.102



 0.03118

Bd fcd



 0.13898

Vs  Vk  Vcu

Vs  217.152 kN

The projection of the cracks:



 0.94441

x   d

x  0.102 m

xu  0.081 m

z   d

Mult  As zfyd

Mult  1.011 MN m

As   Bd fcd 100

xu  0.8 x

c 

z  0.691 m

1.2 Bfctm d

c  11.48 m

maximum. the projection of the cracks:

41.03 cm

Cmax  0.18 

fcd  q fctm

Cmax  0.834

Design of reinforcement for bending moment between the columns: min. the projection of the cracks: M  0.833 MN m

h  0.782 m

fyd  356.522 MPa





M 2

Bd fcd

d  0.732 m 

 0.087

fcd  11.985 MPa 

 0.0265

Cmin  h  0.04 m

Cmin  0.742 m

Suggestion 4-cutting stirrups diameter: dss  8 mm

Examples of RC Foundations

q

 1

206

Sectional area of the stirrups:

Static calculation of extreme square isolated footings

dss

Ass    4

The force that is transferred by stirrup: Nss  Ass 4 fyd

Nss  71.683 kN

Distance between the stirrups: c ss  Nss  ss  3.79 m Vs Suggestion 4-cutting stirrups profile dss = 8mm at a distance of 400 mm

Material properties – design value of concrete cylinder compressive strength: fcd  14.16MPa

fctm  2.01MPa

fyd  443MPa

The span among the columns:

Sabah Shawkat © ly  6 m

l1  7.2m

lk  2.16m

Distributed loads:

Distributed loads of the ceiling:

Shear forces diagram with side view of foundation

qd  13kN m

2

Flat roof load: qstr  7.8kN m

2

Load area:

 

zs  ly  lk 

l1  2

 

zs  34.56m

Force load: Force load from the peripheral walls: F1d  82kN Force load of self-weight of columns:

Load calculation areas on plate - load from one floor

F 2d  bs kv 25 

kN m

1.35

Examples of RC Foundations

F2d  15.39kN

207

The total load on the foot:

Number of floors: n = 4

kN  2  Nd   qd zs  F1d  F2d  n  qstr zs  bs ( 2.85m  1.5m ) 25 1.35 3 m   Nd  2.48  103 kN

Height of the foundation:

 Ns    5cm  0.6fcd 

d  1.44 

d  0.76m

I suggest: d  0.8m

Foundation dimensions:

The tabulated bearing capacity of soil to the depth of the bed joints = d + h:

Building height:

R dt

H  n 2.85m  ( 2.85m  1.5m )

 Rdt  2.5( h  d  1 m ) 

R dt

 461.875kPa

H  15.75m Preliminary calculation of the force applied in footing bottom (if self-weight Ns = 10%): Vs  Ns  0.1N s Vs  2.273  103 kN

Preliminary calculation of the effective surface which is able to transfer force and stress not exceed Rtd:

Sabah Shawkat © Aef 

Aef  4.921m

R dt

Dimensions of square foundation:

The degree of constraint of buildings:

h 

b  Aef b  2.218m Design dimensions of the foundation:

h  1.575m

The force acting in the contact columns and footing foundation (extreme values):

Aef  b b  2.15m Assessment of foot in terms of soil bearing capacity:

Nd  2.48  103 kN Service value applied force:

Ns 

Nd 1.2

Ns  2.066  103 kN

Tabulated load bearing capacity of the soil in the bed joints of the depth = 1 m:

Rdt  400kPa Earth gravity: 

 18

kN m

Determination of bending moment and shear force

Examples of RC Foundations

208

Required area of reinforcement: Weight footings: 2

Nzs1  b d 25

As 

N zs1  92.45  kN 3 m Weight of soil under the foundations of structures:





Nzs2  b  bs  ( h )

As  1.822  10 3 m

fyd

Area 1  18:

Nzs2  126.512kN

d1  18mm

2 

A1  d1  4

A1  2.545cm

n 

n  8.442

The resulting force applied at the footing bottom (with gravity base): Nzs  Nzs1  Nzs2

Vsk  Ns  Nzs

Vsk  2.285  103 kN

The stress in the bed joints and assessed for bearing capacity of foundation soil: Vsk

 z  494.41kPa < R dt  461.875kPa 2 b Dimensioning (of extreme values weightless footing): z



geometry: a 

b  bs 2

we propose 10 V18 the entire width of the footing => distance between the reinforcement will be 200 mm:  smin s



1 3



Rbtd Rsd

Ask d b

 smin s

 8  10 4

 1.479  10 3

Ask  A1 10 >  smin  8  10 4

Sabah Shawkat © a  0.875m

Length of console Projection: ak  a  0.15bs

ak  0.935m

The stress in footing bottom of extreme load: Nd  d  536.45kPa Aef Maximum moment acting on the footing: d



M 

b  d

Vmax  ak b  d

M  504.151kN m Vmax  1.078  103 kN

Arm of internal forces determined approximately: deff  d  ast ast  6.5cm Force in reinforcements: Ns 

M zb

Ask  25.447cm

Ns  806.964kN

zb  0.85 deff 

zb  0.625m

Critical area for punching

The reliability condition to avoid punching: Sectional dimensions critical for punching: uc1  d  bs

uc1  1.2 m

Examples of RC Foundations

209

For square footing and column:

Determination of the design bearing capacity of the soil at depth:

uc2  uc1 Peripheral length of the critical cross-section:

dp =1,5 m

ucr  4 uc1 ucr  4.8 m Stress from acting load-of bearing structure acting on footing bottom (without sel-weight of footing):

Soli classification F6:

 d  0.536MPa 2 b Shear force of the extreme loads acting on the critical cross section: d





Nqd   d  b  uc1



Nqd  1.704  103 kN

Maximum shear force caused by extreme stress relating to the unit of the critical cross section: Nqd

kN qdmax  355.053 m ucr Computing shear force that carries by means of concrete qcu: qdmax 

cef

16 kPa

21 deg

 ef

cef

 ef

d

 m

Bearing coefficient of the soil: Nd  m



tan 45 deg 



d 

tan  d

 e 2 

 ef  ef  4 deg

Base area of the footing:

 

1.5  Nd  1 tan  d

1

21 

2

Width: bp = 1 m

s

Coefficient of the shape of the footing:

 1  50  s   smin

factor for height of section (for d > 0.6m):  1 normal force coefficient:

1  0.2 

1  0.1 

h

n

 1

qdmax  355.677

qcu  0.42h  s  h  n fctm kN m

1

Length: Lp = 6 m

Sabah Shawkat ©

factor of reinforcement:

2 

qcu  1.327  103 

kN m

=> satisfies

1





sin 2  d

bp lp id

 

sin  d

1  0.3 

bp lp

1  0.1 

dp bp

Design bearing capacity of the soil:

cd Nc sc dc ic   1 dp Nd sd dd id   2 

Examples of RC Foundations

bp 2

Nb sb db ib

243.077 kPa

210

[1] ACI: Cracking of concrete members in direct tension. ACI Journal, Vol. 83, January – February, 1986

[10] Bazant, Z. P. & Oh, B. H.: “Crack band theory for fracture of concrete.”, Materials and Structures (RILEM), 1993

[2] ACI Committee 318 (1995), Building Code Requirements for Reinforced Concrete, ACI 318-89, and Commentary, ACI 318R-89, American Concrete Institute, Detroit, MI, USA

[11] Beeby, A. W.: The Prediction of Crack Widths in Hardened Concrete, Cement and Concrete Association, London, 1979 [12] Belgian standard NBN B15-238 (1992)

[3] Aide - mémoire: Composants en béton précontraint. Bordas, Paris, 1979

[13] Brandt, A. M.: “Cement – Based Composites”, 1995 E & FN SPON

[4] Allen, H. G.: “The strength of thin composites of finite width, with brittle matrices and random discontinuous reinforcing fibres, J. Phys. D. Appl. Phys. (1972)

[14] Bjarne, Ch. J.: Lines of Discontinuity for Displacements in the Theory of Plasticity of Plain and Reinforced Concrete, Magazine of Concrete Research, Vol. 27, No. 92, September, 1975

Sabah Shawkat ©

[5] Aveston, J.; Mercer, R. A. & Sillwood, J. M.: “Fibre reinforced cements – scientific foundations for specifications. In Composite – Standards, Testing and Design, Proc. National Physical Laboratory Conference, UK, 1974 [6] Azizinamini, A.: “Design of Tension Lap Splices in High Strength Concrete.” High Strength Conference, First International Conference, Proc. ASCE, 1997, Kona, Hawaii

[7] Balaguru, P. & Kendzulak, J.:“ Mechanical properties of Slurry Infiltrated Fiber Concrete (SIFCON). American concrete Institute, Detroit, 1987 [8] Balaguru and Shah 1992, Fiber-Reinforced Cement Composites. McGraw-Hill Inc. 1992. [9] Bentur, Mindness 1990, Fiber Reinforced Cementitious Composites. Elsevier Applied Science, 1990, 449 p

[15] Boulet, B.: Aide - mémoire du second oeuvre du batiment. Bordas, Paris, 1977 [16] Brooks, J. J., Neville, A. M.: A comparison of creep, elasticity,

and strength of concrete in tension and in compression. Magazine of Concrete Research, Vol. 29, 1977 [17] Chan, S.Y.N.; Anson, M.; Koo, S. L.:” The development of very High Strength Concrete” Concrete in the Service of Manking, Radial Concrete Technology – Proceeding of the International Conference, University of Dunde, Scotland, UK, 27-28 June 1996. [18] CEB - Bull. 124/125 - F: Code modéle CEB - FIP pour les structures en béton. CEB, Paris, 1980 [19] CEB - Bull. 156 - F: Fissuration et déformations. École Polytechnique Fédérale de Lausanne,1983.

References

211

[20] CEB - FIP Model Code 1990, Comité Euro - International du Béton, 1991

[30] Cox, H.L:“ The elasticity and strength of fibrous material”, Br. J, Appl. Phys. (1952)

[21] CEB - Bull. 159: Simplified methods of calculating short term deflections of reinforced concrete slabs. Paris - Lausanne, 1983

[31] Davidovici, V.: Béton armé, aide - mémoire. Bordas, Paris, 1974

[22] Cholewicki, A. ´Shear Transfer in Longitudinal Joints of HollowCore Slabs´ Beton Fertigteil Technik n. 4, Wiesbaden 1991 [23] Comité Euro-international du Béton, Bull. n. 161, T.P. Tassios, Reporter, Ch. 4: Highly Sheared Normally Loaded Linear Elements, Paris 1983

[32] DBV-Merkblatt ”Bemessungsgrundlagen fur Stahlfaserbeton” (1992) [33] Design of Concrete Structure, Norwegian Standard NS3473 (1992), Norwegian Council for Building Standardization, Oslo, 1992

[24] Composite structures, FIP 1994

[34] Deutscher Ausschuss fur Stahlbeton – DafStb (1994): Richlinie fur Hochfesten Beton, Supplement a DIN 1045, DIN 488 and DIN 1055, Berlin, Germany

[25] Cholewicki, A: Shear Transfer in Longitudinal Joints of Hollow-Core Slabs´ Beton Fertigteil Technik n. 4, Wiesbaden 1991

[35] Eibl, J.: Concrete Structures. Euro - Design Handbook. Karlsruhe, 1994 – 96

[26] Composite Dycore Office Structures, Company literature-Finfrock Industries, Inc., Orlando, FL, 1992.

[36] Edward, G., Nawy, P.E.: Prestressed Concrete A fundamental Approach. Part 1, New Jersey, 1989

[27] CCBA 68: Régles Techniques de conception et de calcul des ouvrages et constructions en béton armé. D.T.U. Paris, 1975

[37] Elvery, R., Shafi, M.: Analysis of shrinkage effect on reinforced concrete structural members. ACI Journal, Vol. 67, 1970

[28] Consenza, E., Greco, C.: Comparison and Optimization of Different Methods of Evaluation of Displacements in Cracked Reinforced Concrete Beams. Materials and Structures, No. 23, 1990

[38] Elliot, K. S.; Torey, A. K.: Precast concrete frame buildings, Design Guide. British Cement Association 1992

Sabah Shawkat ©

[29] Coates, R. C., Coutie, M. G., Kong, F. K.: Structural analysis, Second Edition, Hong Kong, 1980

[39] FIP Recommendations ´Design of multi-storey precast concrete structures´. 1986

References

212

[40] Goto, Y.: Cracks Formed in Concrete Around Deformed Tension Bars, Journal of the ACI, No. 68, April, 1971 [41] Gregor, J. G.: Reinforced Concrete, New Jersey, 1988

[49] Hassanzadeh, M.; Haghpassand, A.:”Brittleness of Normal and High-Strength Concrete”, Utilization of High Strength Concrete, Proceedings, Symposium in Lillehammer, Norway, June 20-23, 1993

[42] Gvozdev, A. A.: Novoje v projektirovanii betonnych i železobetonnych konstrukcij. Moskva, 1978

[50] Hillerborg, P. E., Int. J. Cem. Comp., 1980

[43] Goulet, J.: Résistance des matériaux. aide - mémoire, Bordas, Paris, 1976

[51] Holzmann, P.: “High strength concrete C 105 with increased Fireresistance due to polypropylen Fibres”, Utilization of High Strength / High Performance Concrete, Proceedings, Symposium in Paris, France, May 29-31 1996

[44] Grandet, J.: „ Durability of High Performance Concrete in Relation to ‚External‘ Chemical Attack“, High Performance Concrete: From material to structure, 1992 E & FN Spon

Sabah Shawkat ©

[45] Grigorian, C. E., Yang, T.-S., and Popov, E. P., ‘Slotted Bolted Connection Energy Dissipators,’ Earthquake Engineering Research Center, University of California, Berkeley, Report No. UCB/EERC92/10, July, 1992. [46] Gupta, A. K.: Unified Approach to Modelling Postcracking Membrane Behavior of Reinferced Concrete, Journal of Structural Engineering, Vol. 115, No. 4, April, 1989

[47] Gupta, A. K.: Postcracking Behavior of Membrane Reinforced Concrete Elements Including Tension-Stiffening. Journal of Structural Engineering, Vol.115, No. 4, April, 1989 [48] Grigorian, C. E., Yang, T.-S., and Popov, E. P., ‘Slotted Bolted Connection Energy Dissipators,’ Earthquake Engineering Research Center, University of California, Berkeley, Report No. UCB/EERC92/10, July, 1992.

References

213

03. Design of Prestress Concrete Prestressing is a special state of stress and deformations which is induced to

Pre-tensioning is used primarily for the prefabrication of concrete

improve structural behaviour. Structures can be prestressed either by artificial

components. The prestressing steel is stressed between fixed abutments, forms

displacements of the supports or by steel reinforcement that has been pre-

are installed around the steel, and the concrete is cast. After the concrete has

strained before load is applied. The forces induced by the former method are

hardened, the prestressing steel is detached from the abutments. Anchorage of

sharply reduced by creep and shrinkage and are generally ineffective at ultimate

the steel, and hence the transfer of prestressing force from steel to concrete, is

limit state. Due to these inherent disadvantages, support displacements are

achieved entirely through bond stresses at the ends of the member.

rarely used for prestressing. The forces induced by pre-strained reinforcement can, however, survive the effects of shrinkage and creep, provided the initial

Pre-tensioning is usually more economical for large-volume precasting

steel strain is sufficiently larger than the anticipated shortening in the concrete.

operations, since the costs of anchors and grouting can be eliminated. it is

The required pre-strains are best achieved using high-strength steel. High-

quite common, however, to combine both methods of prestressing in a given

strength steel that has been pre-strained can normally be stressed to its full yield

structures.

Sabah Shawkat ©

strength at ultimate limit state.

Prestressing with pre-strained steel reinforcement induces a self-

Prestressing can be full, limited, or partial. Full prestressing is designed

equilibrating state of stress in the cross-section. The tensile force in the steel and

to eliminate concrete tensile stresses in the direction of the prestressing under

the compressive force in the concrete, obtained from integration of the concrete

the action of design service loads, prestressing, and restrained deformations.

stresses, are equal and opposite. In statically determinate structures, the two

In structures with limited prestressing, the calculated tensile stresses in the

forces act at the same location in the cross-section. The sectional forces in the

concrete must not exceed a specified permissible value. Behaviour at ultimate

concrete due to prestressing can thus be easily determined from equilibrium.

limit state must nevertheless be checked in both cases. Partial prestressing

The following expressions are based on the assumption that the direction of

places no restrictions on concrete tensile stresses under service conditions.

the prestressing force deviates only slightly from a vector normal to the cross-

Concrete stresses need not, therefore, be calculated. Partial prestressing

section.

encompasses the entire range of possibilities from conventionally reinforced to

The maximum tensile force in the tendons at tensioning should generally not

fully prestressed concrete.

exceed the lower of the following values after transfer or prestressing to the concrete.

Designs must ensure adequate behaviour at ultimate limit state and under service conditions, both of which must be verified directly.Structures can

rpo,max=0.75 fptk

be prestressed either by pre-tensioning or post-tensioning.

rpo,max = 0.85 fpo,1k

214

The minimum concrete strength required at the time when tensioning

h- Possible formation of small cracks in the anchorage zone may not impair the

takes place is given in the approval documents for the prestressing system

permanent efficiency of the anchorage if sufficient transverse reinforcement is

concerned.The initial prestress (at time t = 0) is calculated taking into account

provided.

the prestressing force and the permanent actions presents at tensioning.

This condition is considered to be satisfied if stabilization of strains and

cracks widths is obtained during testing. Where particular rules are not given, the time when prestressing takes place Partial prestressing is generally more economical than full or limited

should be fixed with due regard to: 1. deformation conditions of the component

prestressing. Although structures that are partially prestressed require a

2. safety with respect to the compressive strength of the concrete

significant portion of mild reinforcement for crack control and distribution, this

3. safety with respect to local stresses

steel contributes to the ultimate resistance of the section. Whatever mild steel is

4. early application of a part of the prestress, to reduce shrinkage effects.

added to improve bahaviour under service conditions thus reduces the amount

Losses occurring before prestressing (pretensioning)

of prestressing steel necessary for safety at ultimate limit state. The prestressing

Sabah Shawkat ©

force must always be carefully monitored during construction, since deviations

The following losses should be considered in design

from the prescribed prestressing force can led to cracking, deformations, and

a- Loss due to friction at the bends (in the case of curved wires or strands)

fatigue.

b- Losses due to drive-in of the anchoring devices (at the abutments) when anchoring on a prestressing bed.

In post-tensioned construction, the prestressing steel is only stressed

c- Loss due to relaxation of the pretension tendons during the period which

after the concrete has been cast and hardened. The steel must therefore be

elapses between the tensioning of the tendons and prestressing of the concrete.

enclosed in ducts and anchored using special devices. The ducts are most

d- The prestressing force at a given time t is obtained by subtracting from the

commonly embedded in the concrete and filled with grout after stressing to bond

initial prestressing force the value of the time dependent losses at this time t.

steel to concrete and to provide protection against corrosion. The ducts can also

e- These losses are due to creep and shrinkage of concrete and relaxation of

be located outside of the concrete section and left unbonded for the entire life of

steel.

the structure. In the such cases, grout is only a means of protecting the steel.

f- The finale value of the prestressing force is obtained by subtracting from the initial prestressing force the maximum expected value of the time-dependent losses. g- The strength of the anchorage zones should exceed the characteristic strength of the tendon, both under static load and under slow-cycle load.

215

The calculation of the ultimate flexural strength Mu of the rectangular section. Characteristics of the steel tendon:

n  10

- diameter

  12.7  mm

- effective prestress

P  1200 kN

-elastic modulus

E p  195 10  MPa

  0.801

b  350 mm

The concrete strain at the prestressed steel level corresponds to the ultimate load condition, when the  dn  of the extreme compressive fibre strain  extrernely moment is applied, can be expressed dinp terms cu   pt  cu  (2) and the depth to the neutral axis at failure dn is  dn   

h  750 mm

-area of the section:

A  b  h

- second moment of area:

I 

The concrete strain at the prestressed steel level corresponds to the ultimate load condition, when the extrernely moment is applied, can be expressed in terms of the extreme compressive fibre strain  cu and the depth to the neutral axis at failure dn is

E c  29800  MPa

fc  35  MPa

Ap  1000 mm

Characteristics of the concrete:

- number

Sabah Shawkat © 1

3

 pe  6.15385  10

Ep  Ap

The eccentricity of the tendon to axis of the centroid:

e  275 mm

1 Ec

P



A



4

2

 ce  4.00895 10

P e

 I 

 ce   pt



n 

1 dnP  P  e   Ec  A I

(2)  dp  dn     cu     pe  pu  dn   d  n  

2

(3)

2

dp  dn  1 C carried P Pby ethe concreteon The magnitude of resultant compressive force is    section  by cu  the rectangular  pu rectangular  ce   pt stress block   and  is given the volume of the idealized pe  pu dn Ec

A



0.85  f  b    d

as a function of thecneutraln axis dn

 

 

 

(3)

(4)

The tensile force T of the prestressing steel is

Where  pu is the stress in the prestressing steel at ultimate.

(4)

0.85 A fc  b    dn

According to this fact, we can obtain 

(6)(5)

A p  pu

as a function of the neutral axis dn as follow

pu Where  pu is the stress in the prestressing steel at ultimate. pu

0.85  fc  b    dn

The horizontal equilibrium must be satisfied, which means that the compressive force in the concrete C must be equal to the tensile force in the steel The tensile force T of the prestressing steel is T and hence

 cu  

The magnitude of resultant compressive force C carried by the concrete section is (5) on the rectangular T A p  pu the volume of the idealized rectangular stress block and is given by

In the stage (a), when the externally applied moment is zero, the concrete strain caused by the effective prestress is elastic. The strain in the concrete at the steel level is compressive and is equal to

 ce 

 pu

 pt

The final strain in the prestressing is given as asteel function of the by neutral axis d

dp  650  mm

 pe 

 cu  0.003

The initial strain in the tendons:

The final strain in the prestressing steel is given d  d by

bh

0.85  fc  b    dn

 

pu dn

(7)

Ap The horizontal equilibrium must be satisfied, which means that the compressive force in the concrete C must be equal to the tensile force in the steel T and hence

The iterative procedure can0.85 start after of the (7) important for the fc  b definition    dn A   equations (3) and (6) determination of the ultimate flexural strength Mu.p pu

According to this fact, we can obtain  pu as a function of the neutral axis dn as follow

pu

0.85  fc  b    dn

Examples of Prestress Concrete Beams

 

pu dn

(7)

The iterative procedure can start after definition of the equations (3) and (7) important for the determination of the ultimate flexural strength Mu.

pu 

0.85  fc  b    dn Ap

Co-ordinate of the point 3 after third step are

216

 pu  0.01242 3

pu  1.83489  10 MPa

  dn 

  pu Ap   dp  2  

T z

Point 3 plotted on Figure bellow lies sufficiently close to the stress-strain curve for the tendon and  fc  b    dto therefore the correct value for dn0.85 is close n 220 mm.

(8)

pu 

According to the Figure above, the curvature at ultimate is

(1) Select an appropriate trial value of dn and determine  pu and  pu by means of equations (3) and (7) and plotted on the stress-strain curve for the prestressing steel. If the point falls on the curve, then the value of dn selected in step 1 is the correct one. (2) If the point is not on the curve, then is required to repeat step 1 with a new estimate of dn. A larger value for dn is required if the point plotted in the step 1 is below the stress-strain curve and smaller value is required if the point is above the curve.

Co-ordinate of the point 3 after third step are

 u 

 cu

pu  1.83489  10 MPa

 u  0.01364m

About three iterations are usually required to determine a good estimate of dn and hence Mu.

1. step :

 P P e d  d     cu   p n    pe  pu     dn  I  Ec  A  



  dn 



Mu  pu Ap   dp 

 cu

  dn    Mu  pu Ap   dp  The ultimate moment is found 2  (8):  using Equation  u 

2

 

 u  0.01364m

1

Mu  1.03101  10 kN  m 3

Mu  1.03101  10 kN  m

Sabah Shawkat ©

dn  230  mm

1

Point 3 plotted on Figure n bellow lies sufficiently close to the stress-strain curve for the tendon and therefore the correct value for dn is close to 220 mm.

The ultimate According momentto is found using Equation (8): the Figure above, the curvature at ultimate is

With the correct values of  pu and dn obtained from a few iteration steps, the ultimate flexural strength Mu can be calculated by means of equation (8).

 pu  0.01242

pu 

0.85  fc  b    dn

Co-ordinate of the point 1 after first step are

 pu  0.01203

pu  1.91829  10 MPa

Point 1 plotted on Fig. 2 is above the curve, therefore the value dn must be decreased. 2. step:

dn  210  mm

 pu 

pu 

P



A



2

P e I

0.85  fc  b    dn

 dp  dn     cu     pe  dn    

Co-ordinate of the point 2 after second step are

 pu  0.01284

pu  1.75149  10 MPa

Point 2 plotted on Fig. 2 is below the curve, therefore the value dn must be increased. 3. step:

dn  220  mm

 P P  e2   dp  dn     cu    pu      pe  dn  I  Ec  A   1

Examples of Prestress Concrete Beams

217

Continuous members enjoy certain structural and aesthetic advantages. Maximum bending moments are significantly smaller and deflections are substantially reduced, by comparison

Design geometry in general:

Transverse arrangement of the bridge. The box girder cross-section shown below has proven

with simply-supported members. The reduced demand on strength and the increase in overall

Transverse arrangement of the bridge. The box girder cross-section shown below has proven

stiffness permit a shallower member cross-section for any given serviceability requirement, and

itself with regard its form and structural characteristics. Many variations of this standard cross-section areto possible.

this leads to greater flexibility in sizing members for aesthetic considerations.

cross-section are possible.

itself with regard to its form and structural characteristics. Many variations of this standard

As a statically indeterminate structure is prestressed, reactions are usually introduced at the supports. The supports provide restraint to the deformations caused by prestress (both axial shortening and curvature). The supports also provide restraint to volume changes of the concrete caused by temperature variations and shrinkage. The reaction induced at the supports during the prestressing operation cause so-called secondary moments and shears in a continuous member, and these actions may or may not be significant in design. Methods for determining the magnitudes of the secondary effects and their implications in the design for both strength and serviceability are discussed in this example.

Sabah Shawkat © The highway bridging is designed as a monolithic prestressed concrete structure concreted on a supporting structure. standardized rails must be fitted from the outside of the pavements.

Considering the fact that the speed of driving on the bridge is limited.

The lower structure consists of supports and two intermediate supports. The substructure must

be designed from a monolithic concrete load transfer from the top construction to the substructure is using bearings.

The type of bearing depends on the load-bearing response to the supports and supports. Longitudinal arrangement of bridge

l  32 m

Examples of Concrete Box Girder Bridge

218

H 

1 18

l

H  1.778 m

we suppose

H  2  m

Transverse arrangement of the bridge: If available depth of the girder, dslab=(14,200m – 2x2m) / 6= 1,7m , is greater than from 1/6 to 1/5 of the bridge width, b slab, a single-cell box girder is in order, if d/bslab < 1/6, a 2-cell or multiple-cell box girder is more sensible. For wider bridges the vehicle loads acting on the cantilever slab can be distributed longitudinally by means of a pronounced edge beam, enabling the cantilever length to be increased. The number of cells should be kept as small as possible even for wider bridges with a small depth in order to minimize problems in construction. For more economic, today more than 2 cells are rare. If 2 or more box girder are placed next to each other it is advantageous to connect their top and bottom flanges in order to achieve a better transverse load distribution. If only the top flanges

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

0 14,2 2,0 12,2 2,3 11,9 2,5 6,85 3,95 5,4 2,74 6,85 2,94 6,65 7,35 11,7 8,8 10,25 7,35 11,46 7,55 11,26

0,15 0,15 0,3 0,3 2,0 2,0 0,3 0,3 0,15 0,15 1,65 1,65 1,85 1,85 0,3 0,3 0,15 0,15 1,65 1,65 1,85 1,85

0 14,2 2,0 12,2 2,3 11,9 2,7 6,75 4,15 5,3 2,89 6,75 3,10 6,55 7,45 11,5 8,9 10,05 7,45 11,3 7,65 11,1

0,15 0,15 0,3 0,3 2,0 2,0 0,3 0,3 0,15 0,15 1,14 1,4 1,6 1,6 0,3 0,3 0,15 0,15 1,4 1,4 1,6 1,6

Sabah Shawkat ©

are connected, they will be highly stressed due to bending moments in the transverse direction

without being able to effectively distribute the stresses in this direction. It is then better to

separate the two box girder.

Cross-section properties, in the middle of the span and over the support:

Calculation of the cross-sectional variables of the bearing structure: Material properties: Data: B = 14,2m

1 2

H=2,0m Cross‐section at the middle of span yi (m) zi (m) 0 0 14,2 0

Cross‐section over the support yi (m) zi (m) 0 0 14,2 0

Characteristic properties of cross-section area Ay Az lt Iy Iz Iw wely wply welz wplz

Cross-section in the middle of the span 6.941m2 6.941m2 6.941m2 0,535m4 4.0335m4 108,378m4 0 3,6259m3 4,779m3 15,2645m3 23,3822m3

Examples of Concrete Box Girder Bridge

Cross-Section over the support 9.892m2 9.892m2 9.892m2 1,186m4 5,409m4 131,699m4 0 4,5463m3 6,623m3 18,549m3 30,531m3

219

ypm Zhm resp. Zhp Zdm resp. Zdp

7,10m -0,832m 1,168m

Crosssection widths B(m)

The depth of crosssection H(m) 14,2 2,0 2 A (m ) Jy (m4) Jz (m4)

Load of pavement:

7,10m -1,02m 0,98m

Cross-section at the middle of the span -yl=yp Zh Zd (m) (m) (m)

Cross-section over the support -yl=yp (m)

Zh (m)

Zd (m)

7,10

-1,02 9.892m2 5,409m4 131,699m4

0,98

bpav  1.5m

qpav  2hpav bpav  pav

qpav  17.94kN m

We estimate 7,10

-0,812 6.941m2 4.0335m4 108,378m4

1,168

A mm  6.941 m

q om  A mm 1.0  pB

q om  180.466

pav  23kN m

3

1

qz  0.6kN m

1

Calculation of total load:

Self-weight of the structure, in the middle span:  pB  26kN m



Load of handrail:

Calculation of dead load:

3

hpav  0.26m

1

qtotal  qom  qz  qpav  qd

qtotal  235.243kN m

qtotalp  qop  qd  qpav  qz

qtotalp  311.969kN m

1

A mp  9.892 m

Live load:

Sabah Shawkat © F1 = 80kN, F2 = 240kN

kN m

Intensity of uniform load at the point of vehicle loading:

Self-weight of the structure, over the support:

bo  5.5m

b  11.5m

q op  A mp 1.0  pB

q op  257.192



V1  vo  b  bo



asfalt  12kN m

hconc  0.05m 

mastix  0.001kN m

V1  15 kN m

1

Intensity of uniform load off-site of vehicle loading:

Dead load: b  11.5m



vo  2.5kN m

3

hasfalt  0.02m

hasfconc  0.08m 2



conc  23kN m



3

asfconc  22kN m

3

V2  28.75kN m

hmastix  0.01m 1

qasfalt  hasfaltb  asfalt

qasfalt  2.76 kN m

qasfconc  hasfconc asfconcb

qasfconc  20.24kN m

qmastix  mastixb

qmastix  0.012kN m

qlevconc  b  conchconc

qlevconc  13.225kN m

qd  qasfalt  qasfconc  qmastix qlevconc

V2  vo b

B  14.2m

qd  36.236kN m

1

Examples of Concrete Box Girder Bridge

1

2

220

5  L1 i 1 sk5   i 1000 m

sk10  i

0 i sk0  1 i 1000

10  L1 i 1



1000

sk15  i

10p i sk10p  1 i 1000

15  L1 i 1

1000

sk10z  i



m L1

10z  i L2

1000

1

Table of lines influence: 5  i

Load of sidewalk: vsidewalk  4 kN m

2

bsidewalk  1.25m

V sidewalk  v sidewalk  b sidewalk  2

Vsidewalk  10 kN m

1

10  i

15  i

0  i

10p  i

10z  i

1·103

-23.9

-8.6

876.1

123.9

30.5 59.1

76.8

-46.3

-16.8

753.7

246.3

117.1

-65.8

-23.8

634.2

365.8

159.5

-81

-29.2

519

481

103.5

204.8

-90.4

-32.6

409.6

590.4

115.5

153.8

-92.5

-33.4

307.5

692.5

118.2

107

-86

-31

214

786

109.9

65.3

-69.4

-25

130.6

869.4

88.7

29.4

-41.2

-14.9

58.8

941.2

52.6

1·103

Sabah Shawkat ©

In the design and analysis of continuous prestress concrete members, it is usual to make the

sk5  i

following simplifying assumptions.

sk15  i

sk0  i

sk10p  i

sk10z  i

The concrete behaves in a linear elastic manner within the range of stresses considered.

0.912

-0.574

-0.206

0.876

0.124

0.024

Plane sections remain plane throughout the range of loading considered.

1.843

-1.111

-0.403

0.754

0.246

0.047

The effects of external loading and prestress on the member can be calculated separately

2.81

-1.579

-0.571

0.634

0.366

0.067

3.828

-1.944

-0.701

0.519

0.481

0.083

and added to obtain the final conditions, i.e. the principle of superposition is valid.

4.915

-2.17

-0.782

0.41

0.59

0.092

The magnitude of the eccentricity of prestress is small in comparison with the member

3.691

-2.22

-0.802

0.308

0.692

0.095

2.568

-2.064

-0.744

0.214

0.786

0.088

1.567

-1.666

-0.6

0.131

0.869

0.071

0.706

-0.989

-0.358

0.059

0.941

0.042

length, and hence the horizontal component of the prestressing force is assumed to be equal to the prestressing force at every cross-section. Determination of dynamic coefficient:

L1  24m 1



L2  30m





Ld   2L1  L2 3 r

sk10  i

1 0.95  ( 1.426)

 0.6

Ld  26 m r

sk5  j

L1  24  m

1000



sk10  j

10  L2 j 1

1000



sk15  j

15  L2 j 1

1000

 1.199

L1=24 (m), L2=30 (m), L3=24(m) i  0  10

5  L2 j 1

L2  30  m

L3  24  m

0 j L2 sk0   j 1000 L1

10p j L2 sk10p   j 1000 L1

Examples of Concrete Box Girder Bridge

10z j sk10z  j 1000



221

Table of influence lines: 0

10p



10z

5  j



10  j

15  j

Table of influence lines:

-43.2

43.2

927.7

-21.6

-43.2

20.6

-70.6

70.6

836.9

-35.3

-70.6

47.8

-84.6

84.6

732.3

-42.3

-84.6

76.6

-87.4

87.4

618.4

-43.7

-87.4

121.8

-81.5

81.5

500

-40.8

-81.5

168.5

-69

381.6

-34.5

-69

121.8

-52.3

52.3

267.7

-26.2

-52.3

76.6

-33.7

33.7

163.1

-16.8

-33.7

47.8

-15.5

15.5

72.3

-7.8

-15.5

20.6

sk5  j

sk10  j

sk15  j

sk0



sk10p



sk10z

0

-0.648

-1.296

0.618

-0.054

0.054

0.928

-1.059

-2.118

1.434

-0.088

0.088

0.837

-1.269

-2.538

2.298

-0.106

0.106

0.732

-1.311

-2.622

3.654

-0.109

0.109

0.618

-1.224

-2.445

5.055

-0.102

0.102

0.5

-1.035

-2.07

3.654

-0.086

0.086

0.382

-0.786

-1.569

2.298

-0.065

0.065

0.268

-0.504

-1.011

1.434

-0.042

0.042

0.163

-0.234

-0.465

0.618

-0.019

0.019

0.072

10p



10z



5

10



15



11.4

-11.4

-52.6

5.7

11.4

-14.9

19.3

-19.3

-88.7

9.6

19.3

-25

23.9

-23.9

-109.9

23.9

-31

25.7

-25.7

-118.2

12.8

25.7

-33.4

25.1

-25.1

-115.5

12.6

25.1

-32.6

22.5

-22.5

-103.6

11.2

22.5

-29.2

18.2

-18.2

-84

9.1

18.2

-23.8

12.8

-12.8

-59.1

6.4

12.8

-16.8

6.6

-6.6

-30.5

3.3

6.6

-8.6

sk5  i



sk10  i

sk15  i

sk0  i

sk10p  i

sk10z  i

0.912

-0.574

-0.206

0.876

0.124

0.024

1.843

-1.111

-0.403

0.754

0.246

0.047

2.81

-1.579

-0.571

0.634

0.366

0.067

3.828

-1.944

-0.701

0.519

0.481

0.083

4.915

-2.17

-0.782

0.41

0.59

0.092

3.691

-2.22

-0.802

0.308

0.692

0.095

2.568

-2.064

-0.744

0.214

0.786

0.088

1.567

-1.666

-0.6

0.131

0.869

0.071

0.706

-0.989

-0.358

0.059

0.941

0.042

Sabah Shawkat © Table of influence:

sk5



sk10p

5  L3 k 1

1000



10p



1000

1

sk10

sk0



10  L3 k 1

1000

0

1000

1



sk15

sk10z



15  L3 k 1





1000

10z

k L1

1000



sk5  j



sk10p

-0.648

sk10  j -1.296

sk15  j 0.618

-0.054

0.054

0.928

-1.059

-2.118

1.434

-0.088

0.088

0.837

-1.269

-2.538

2.298

-0.106

0.106

0.732

-1.311

-2.622

3.654

-0.109

0.109

0.618

-1.224

-2.445

5.055

-0.102

0.102

0.5

-1.035

-2.07

3.654

-0.086

0.086

0.382

-0.786

-1.569

2.298

-0.065

0.065

0.268

-0.504

-1.011

1.434

-0.042

0.042

0.163

-0.234

-0.465

0.618

-0.019

0.019

0.072

Examples of Concrete Box Girder Bridge

sk0



sk10z



222

sk5

sk10z



sk10



sk15



sk0

sk10p



0.137

-0.042

0.274

-0.358

0.011

-0.011

0.23

-0.071

0.463

-0.6

0.019

-0.019

0.288

-0.088

0.574

-0.744

0.024

-0.024

0.307

-0.095

0.617

-0.802

0.026

-0.026

0.302

-0.092

0.602

-0.782

0.025

-0.025

0.269

-0.083

0.54

-0.701

0.023

-0.023

0.218

-0.067

0.437

-0.571

0.018

-0.018

0.154

-0.047

0.307

-0.403

0.013

-0.013 -6.6·10-3 0

0.079

-0.024

0.158

-0.206

6.6·10-3

A 15 



A 25 

1

3 2

 sk5 0  sk5 1  4  sk5 2  2  sk5 3  4  sk5 4  2    sk5 4  sk5 2  sk5 4  sk5 2  sk5 4  sk5  5 6 7 8 9 10   sk5 11  sk5 12 4  sk5 13 2  sk5 14 4    sk5 2  sk5 4  sk5 2  sk5 4  sk5  15 16 17 18 19 

  

A 25  23.076 A 35 

3

 sk5 20  sk5 21 4  sk5 22 2  sk5 23 4    sk5 2  sk5 4  sk5 2  sk5 4  sk5 2  sk5  24 25 26 27 28 29 

  

A 35  4.625

1  2.4

2  3.0

The analysis presuppose that the statical system, the loading, and the design cross-sectional

3  2.4

dimensions are known. Influence line for max Mx, longitudinal loading. Positioning of vehicle loading in the longitudinal direction

Sabah Shawkat ©

The box girder represents a special form of a folded plate structure. The loads are always carried three dimensionally. The analysis of the sectional forces to which the structure is subjected can be handled with the aid of the following: -analogy of a beam on an elastic foundation

-folded plate theory with the use of series expansions -finite strip method

-finite element method

M10

A 110 

1

  sk10 0  sk10 1  4  sk10 2  2  sk10 3  4  sk10 4  2  sk10 5  4  sk10 6  2     sk10 4  sk10 2  sk10 4  sk10  7 8 9 10  

A 110  34.706

A 210 

A 310  A 15  55.603

2

3

 sk10 11  sk10 12 4  sk10 13 2  sk10 14 4  sk10 15 2  A 210  46.149   sk10 164  sk10 172  sk10 184    sk10  19    sk10 20  sk10 21 4  sk10 22 2  sk10 23 4  sk10 24 2  A 310  9.247   sk10 254  sk10 262    sk10 4  sk10 2  sk10  27 28 29  

Examples of Concrete Box Girder Bridge

223

The loading of the top flange so as to produce the largest moment, M or the largest shear, V

A 315 

respectively.

3

With this approach, the box girder loaded symmetrically in the transverse direction can be analysed longitudinally as a beam and transversely as a frame independent of each other. This is valid if the box girders slenderness ratio, l/d > a, if the span lengths, l > 1.5 b, and if the

 sk15 20  sk15 21 4  sk15 22 2  sk15 23 4     sk15 242  sk15 254  sk15 262  sk15 274   sk15 2  sk15  28 29  

A 315  12.031

disturbance zones at the introduction of the applied forces are separately handled. The particular problematic nature of the simple analysis approach to the behaviour of a box girder under load arises wit unsymmetrical lading in the transverse direction. In this case the longitudinal and transverse stresses are connected with each other, a connection which shall be designated in what follows here as the folded plate action. The connection is taken account of through the formulation of the condition of compatibility between the separately analyzed longitudinal and transverse direction.

Sabah Shawkat ©

M15

A 115 

1

 sk15 0  sk15 1  4  sk15 2  2  sk15 3  4     sk15 42  sk15 54  sk15 62  sk15 74   sk15 2  sk15 4  sk15  8 9 10  

A 115  12.526

A 215 

2

 sk15 11  sk15 12 4  sk15 13 2  sk15 14 4  sk15 15 2    sk15 4  sk15 2  sk15 4  sk15   16 17 18 19  

A 215  61.242 M15

A 115 

A 215 

1

2

 sk15 0  sk15 1  4  sk15 2  2  sk15 3  4  sk15 4  2    sk15 54  sk15 62  sk15 74    sk15 2  sk15 4  sk15  8 9 10  

 sk15 11  sk15 12 4  sk15 13 2  sk15 14 4    sk15 2  sk15 4  sk15 2  sk15 4  sk15  15 16 17 18 19 

A 215  61.242

A 315 

A 115  12.526

  

3

 sk15 20  sk15 21 4  sk15 22 2  sk15 23 4  sk15 24 2    sk15 254  sk15 262  sk15 274    sk15 2  sk15  28 29  

A 315  12.031

Examples of Concrete Box Girder Bridge

224

Influence line for the largest shear force Vx

T0

B15 

Tp10

Sabah Shawkat © Bp10 

1

 sk0 0  sk0 1  4  sk0 2  2  sk0 3  4  sk0 4  2    sk0 4  sk0 2  sk0 4  sk0 2  sk0 4  sk0  5 6 7 8 9 10 

2

3

 sk10p 0  sk10p 1  4  sk10p 2  2  sk10p 3  4    sk10p 4 2  sk10p 54  sk10p 6 2   sk10p 4  sk10p 2  sk10p 4  sk10p 7 8 9 10 

   

Bp10  13.446

 sk0 11  sk0 12 4  sk0 13 2  sk0 14 4  sk0 15 2  3     sk0 164  sk0 172  sk0 184  sk0 19 

Bp25 

B25  1.923

B35 

  

B15  10.554

B25 

 1

 sk0 20  sk0 21 4  sk0 22 2  sk0 23 4  sk0 24 2    sk0 4  sk0 2  sk0 4  sk0 2  sk0  25 26 27 28 29 

  

 2 3

 sk10p 11 4  sk10p 12 2  sk10p 13 4     sk10p 142  sk10p 154  sk10p 162   sk10p 4  sk10p 2  sk10p  17 18 19  

Bp25  1.979

B35  0.385

Bp35 

 3  sk10p 20  sk10p 21 4  sk10p 22 2  sk10p 23 4  3     sk10p 24 2  sk10p 25 4  sk10p 26 2   sk10p 4  sk10p 2  sk10p  27 28 29  

Bp35  0.385

Examples of Concrete Box Girder Bridge

225

Idealization of permanent load on three – span beam, in box girder bridge design:

Bz35 

3

 sk10z 20  sk10z 21 4  sk10z 22 2  sk10z 23 4    sk10z 242  sk10z 254  sk10z 262    sk10z 4  sk10z 2  sk10z  27 28 29  

Bz35  1.419

Calculation of Internal forces using influence lines Calculation of Mmax, Mmin, Tmax, Tmin in cross-section 0 F2  160  kN

F1  480  kN

1

v1  15  kN m

sk0 0  1

sk0 5  0.41

sk0 8  0.131

sk0 6  0.308

sk0 9  0.059

B35  0.385

1

v2  28.75  kN m

B25  1.923

Tmax0I  F1   sk0 0  sk0 5  F2   sk0 1  sk0 6  v1  B15 m    sk0 8  sk0 9  ( 4.5  m  1  m)   v2  B35 m  ( v2  v1)  2   sk0  9     1  m 2   

B15  10.554 sk0 1  0.876

     

Sabah Shawkat © 3

Tmax0I  1.039  10 kN

Load from footpath:

1

Vch  10  kN m

Tmax0ch  Vch   B15  B35  m sk0 11  0.054

sk0 14  0.109

Tmax0ch  109.392 kN sk0 18  0.042

sk0 15  0.102

sk0 19  0.019

Tz10

Bz10 

Bz25 

Tmin0I  129.543 kN

1

 sk10z 0  sk10z 1  4  sk10z 2  2  sk10z 3  4  sk10z 4  2  Bz10  1.478 3     sk10z 5  4  sk10z 6  2  sk10z 7  4   sk10z 2  sk10z 4  sk10z  8 9 10  

2

Tmin0I  F1   sk0 14  sk0 18  F2   sk0 11  sk0 15  sk0 19  v1  B25 m  1

 sk10z 11 4  sk10z 12 2  sk10z 13 4  sk10z 14 2  Bz25  13.783   sk10z 154  sk10z 162  sk10z 174    sk10z 2  sk10z  18 19  

Tmin0ch   Vch  B25  m

Tmin0ch  19.229 kN

1

q1  235.3  kN m

1

q2  311.2  kN m

T0q  q1   B15  B25  B35  m  sk0 10  sk0 11 sk0 18  sk0 19 sk0 20  sk0 21   sk0 8  sk0 9  4  4  4  4  ( q2  q1)  m 2 2 2 2  

T0q  2.17  10 kN

Examples of Concrete Box Girder Bridge

226

sk0 19  0.019

sk10p 10  1

sk10p 14  0.109

Bp10  13.446

sk10p 15  0.102

sk10p 7  0.786

sk10p 11  0.054

Calculation of Mmax, Mmin, Mq in cross-section 10:

Bp25  1.979

sk10p 5  0.59

Tmax10p  F1   sk10p 10  sk10p 5  sk10p 14  F2   sk10p 7  sk10p 11  sk10p 15   sk10p 3  sk10p 0   v1   Bp10  Bp25  1  m  ( v2  v1)    7.5  m   2  sk10p  sk10p  16 17   10.5  m  2  

1

sk10 4  1.944

sk10 15  2.445

A 110  34.706

v2  v1  13.75 m

sk10 6  2.22

sk10 11  1.296

sk10 16  2.07

A 210  46.149

sk10 7  2.064

sk10 12  2.118

sk10 17  1.569

 kN

Mmax10I  F1   sk10 6  sk10 11  sk10 15  m 

 F2   sk10 7  sk10 12  sk10 6  m  v1   A 110  A 210  m   sk10 17   sk10 4 2   9.9   8.1  ( v2  v1)  m 2 2   2

Tmax10p  1.228  10 kN

Mmax10I  5.318  10 m kN

Calculation of Mmax, Mmin, Mq in cross-section 5: 1

v1  15 m

1

 kN

v2  28.75 m

 kN

A 15  55.603

A 35  4.625

Area of the triangle: Mmax10ch  Vch   A 110  A 210  m

Mmax10ch  808.549 m kN

Sabah Shawkat ©

Mmax5I  F1   sk5 5  sk5 0  m  F2   sk5 1  sk5 6  m  sk5 9  sk5 8 2 2 2  v1  A 15 m  ( v2  v1)   4.5  m  v2  A 35 m 2

sk10 22  0.463

sk10 27  0.437

A 310  9.247

sk10 23  0.574

sk10 28  0.307

Mmin10I  F1   sk10 22  sk10 27  m  F2   sk10 23  sk10 28  m  v1  m  A 310 2

Mmax5ch  Vch   A 15  A 35  m

Mmax5I  4.133  10 m kN

sk5 10  0

sk5 14  1.311

sk5 15  1.224

sk5 19  0.234

Mmax5ch  602.285 m kN

sk5 18  0.504

Mmin10I  711.629 m kN

sk5 11  0.648

1

q2  311.2 m

Mmin5I  F1   sk5 10  sk5 14  sk5 18  m  F2   sk5 11  sk5 15  sk5 19  m  v1  m  A 25  1 3

1

q1  235.3 m

q2  311.2 m

 kN

1

Vch  10 m

 kN

A 25  23.076

  sk5 9  sk5 8  sk5 20  sk5 21   sk5 18  sk5 11 2 2  

M5q  8.75  10 m kN

A 210  46.149

Mmin10ch  92.467 m kN

sk10 8  1.666

sk10 18  1.011

sk10 11  1.296

sk10 21  0.274

1

q2  q1  75.9 m

 kN

M5q  q1   A 15  A 25  A 35  m  ( q2  q1 )  2  m   2

 kN

Mmin5ch  230.76 m kN

Vch  10 m

M10q  q1   A 110  A 210  A 310  m  ( q2  q1)  2   sk10 8  sk10 21  sk10 18  sk10 11  m

Mmin5ch   Vch  A 25  m

A 15  55.603

1

 kN

A 310  9.247 2

Mmin5I  1.554  10 m kN 1

 kN

Mmin10ch   Vch  A 310  m

A 15  55.603

1

q1  235.3 m

A 110  34.706

M10q  1.741  10 m kN sk0 19  0.019

sk10p 10  1

Bp10  13.446

sk10p 15  0.102

sk10p 14  0.109 sk10p 11  0.054

Examples of Concrete Box Girder Bridge

sk10p 7  0.786

Bp25  1.979

227

sk10p 5  0.59

Tmax10L  F1   sk10p 10  sk10p 5  sk10p 14   F2   sk10p 7  sk10p 11  sk10p 15  v1   Bp10  Bp25  1  m    sk10p 3  sk10p 0  ( v2  v1)    7.5  m   2  sk10p  sk10p  16 17   10.5  m  2  

Bp10  13.446

 kN

sk10z 14  0.618

sk10z 19  0.072

sk10z 18  0.163

Tmax10pI  F1   sk10p 10  sk10z 14  sk10z 18   F2   sk10z 11  sk10z 15  sk10z 19  v1   Bz10  Bz25  m   ( v2  v1)   Bz10 m 3

Tmax10Lch  154.252 kN sk10p 28  0.013

1

Vch  10  kN m

sk0 18  0.042

Bz10  1.478

Bz25  13.783

Tmax10pch  Vch   Bz10  Bz25  m

Tmax10pch  152.609 kN

3 sk10p 29  6.6  10

sk10p 24  0.026

Sabah Shawkat © sk10z 23  0.088

Tmin10LI  F1   sk10p 23  sk10p 28  F2   sk10p 24  sk10p 29     ( sk10p ) 21 2.1  m  v1 Bp35m  ( v2  v1)    2    

1

q1  235.3  kN m

q2  311.2  kN m

Tmin10Lch   Vch  B35  m

sk10p 11  0.054

sk10p 20  0

Bp10  13.446

sk10p 8  0.869

sk10p 10  1

sk10p 19  0.019

sk10p 21  0.011

Bp25  1.979

Bp35  0.385

sk10p 18  0.042

T10Lq  q1   Bp10  Bp25  Bp35  m   sk10p 8  sk10p 9   sk10p 11  sk10p 10   ( q2  q1)     1    1  m 2 2     sk10p 1  sk10p   18 21    

sk10z 29  0.024

1

sk10z 9  0.042

sk10z 11  0.928

sk10z 18  0.163

Bz10  1.478

sk10z 8  0.071

sk10z 21  0.042

Bz25  13.783

Bz35  1.419

T10pq  3.348  10 kN

Bz10  1.478

sk10z 24  0.095

1

( v2  v1)  13.75 m

1

q1  235.3  kN m q2  311.2  kN m Tmin10pI  105.827 kN Tmin10pch   Vch  Bz10  m Tmin10pch  14.778 kN



sk10p 10  1

 kN

sk0 18  0.042

sk10z 20  0 sk10z 19  0.072

T10pq  q1   Bz10  Bz25  Bz35  m   ( q2  q1)  sk10z 8  sk10z 11  sk10z 18  ( sk10z ) 21  m

T10Lq  3.652  10 kN

sk10z 11  0.928

1

v1  15 m

sk10z 21  0.042

Tmin10pI  F1   sk10z 23  sk10z 28  F2   sk10z 24  sk10z 29   1   ( sk10z ) 21 2.1  m  v1 Bz35m  ( v2  v1)   2  

sk10p 9  0.941

Tmin10Lch  3.853 kN

sk10z 28  0.047

Bz35  1.419

Tmin10LI  28.728 kN

1

 kN

Tmax10pI  1.344  10 kN

sk10p 23  0.024

Bp35  0.385

v2  28.75 m

Bp10  13.446

Tmax10Lch  Vch   Bp10  Bp25  m sk10p 21  0.011

Bz10  1.478

1

v1  15 m

sk10z 8  0.071

1

Bz25  13.783

Tmax10L  1.228  10 kN Vch  10  kN m

sk10z 15  0.5

 kN

Examples of Concrete Box Girder Bridge



228

Calculation of Mmax, Mmin, Mq in cross-section 15: sk15 11  0.618

sk15 12  1.434 1

v2  v1  13.75 m

 kN

sk15 15  5.055

A 210  46.149

sk15 19  0.618

sk10 15  2.445 sk15 16  3.654

A 110  34.706 sk10 16  2.07

sk15 20  0

sk10 17  1.569

Mmax15I  F1   sk15 11  sk15 15  sk15 19  m 

 F2   sk15 12  sk15 16  sk15 20  m  v1  A 215 m  2

Mmax15I  4.752  10 m kN 2

Mmax15ch  Vch  A 215 m sk15 3  0.571

Mmax15ch  612.42 m kN

sk15 4  0.701

A 115  12.526

sk15 8  0.6

sk15 9  0.358

Cross- section 0:

Cross-section 5:

Crosssection T0q = 2.17 x 1030: kN

Cross-section 5:3 m.kN M5q = 8.75 x 10

3 T0q = 2.17 x 10 kN T0max = 81 . Tmax0lch

M5q = =8.75 x 103 m.kN M5max 1.Mmax5l +Mmax5chT

= 81 . xTmax0lch T0max = 1.256 103 kN

1.Mmax5l M5max = 5.562 x 103 +Mmax5chT m.kN

T0max==(1.Tmin0l 1.256 x 10+3 kN T0min Tmin0ch).-1

M5max== (1.Mnin5l 5.562 x 103+Mnin5ch).-1 m.kN M5min

(1.Tmin0l + Tmin0ch).-1 T0min = -174.68kN

(1.Mnin5l M5min = -2.096 x 103+Mnin5ch).-1 m.kN

T0min ==-174.68kN T0MAX T0q +T0max

M5min ==-2.096 x 103 m.kN M5MAX M5q +M5max

T0q +T0max T0MAX = 3.526 x 103 kN

= M5q +M5max M5MAX =1.431 x 104 m.kN

T0MAX == 3.526 x 103 kN T0MIN T0q +T0min

M5MAX =1.431 x 104 m.kN M5MIN= M5q +M5min

T0q +T0min T0MIN = 1.995 x 103 kN

M5MIN== M5q M5MIN 6.654+M5min x 103 m.kN

T0MIN = 1.995 x 103 kN

M5MIN = 6.654 x 103 m.kN

Cross- section 10:

Left side

Cross-=section M10q -1.741 x10: 104 m.kN

Left side T10Lq = -3.652 x 103 m.kN

M10q = -1.741 x 10 m.kN M10max = (1.Mmax10l +Mmax10ch).-1

T10Lq = =-3.652 x 103 m.kN T10max (71.Tmax10l + Tmax10Lch).-1

M10max = (1.Mmax10l +Mmax10ch).-1 -7.19 x 103 m.kN

T10max = -1.628 (71.Tmax10l + Tmax10Lch).-1 T10max x 103 kN

M10max==(1.Mmax10l -7.19 x 103 m.kN M10min +Mmax10ch)

T10max==(81. -1.628 x 103 kN T10min Tmin10Li + Tmin10Lch)

(1.Mmax10l M10min = 946.422 m.kN+Mmax10ch)

(81. Tmin10Li + Tmin10Lch) T10min = 38.326 kN

M10min ==946.422 m.kN M10MAX M10q +M10max

T10min ==38.326 T10MAX T10Lq kN +T10max

Sabah Shawkat ©

Mmin15I  F1   sk15 3  sk15 8  m 

   2 2  F2    sk15   sk15   m  v1  A   m  v2  A   m   4 9 115 115     3

1

q2  311.2 m

1

q1  235.3 m

 kN

1

Vch  10 m

 kN

A 215  61.242

sk15 19  0.618

Mmin15ch  245.568 m kN 1

q2  q1  75.9 m

 kN

sk15 11  0.618

sk15 21  0.358

M15q  q1   A 115  A 215  A 315  m 

3 kN T10Lq x+ 10 T10min T10MIN = -3.613

4 m.kN

T10MIN = -3.613 x 103 kN

Right side:

Reaction:

Right side: T10pq = 3.348 x 103 kN

Reaction: R10q = T10pq + (T10Lq).-1

T10pq = 3.348 x 103 kN +Tmax10pch) T10maxp = (1.Tmax10p

R10q = 7T10pq x 103 + kN(T10Lq).-1

T10maxp = 1.626 x 10

R10q = 7=xT10maxp 103 kN +(T10max).-1 R10max

T10maxp==(1. 1.626 x 103 kN+Tmin10pch) T10minp Tmin10pl

3 kN R10max == 3.253 T10maxp R10max x 10+(T10max).-1

(1. Tmin10pl +Tmin10pch) T10minp = 141.77 kN

3 kN R10max==T10minp 3.253 x 10 R10min +T10min

T10minp ==141.77 T10MAXp T10pq kN + T10maxp

T10minpkN +T10min R10min = 180.096

3 kN T10pqx+10 T10maxp T10MAXp = 4.974

R10min ==180.096 kN +(T10MAX).-1 R10MAX T10MAXp

T10MAXp==T10pq 4.974 +T10minp x 103 kN T10MINp

R10MAX == 1.025 T10MAXp m.kN R10MAX x 104+(T10MAX).-1

T10pq T10MINp = 3.49 x +T10minp 103 kN

4 m.kN R10MAX==T10MINp 1.025 x 10 R10MIN +(T10MIN).-1

T10MINp = 3.4915: x 103 kN Crosssection

3 kN R10MIN == 7.103 T10MINp R10MIN x 10+(T10MIN).-1

Cross-= section M15q 8.674 x 10

 ( q2  q1 )  2   sk15 8  sk15 11  sk15 19  sk15 21  m  3

M10q +x M10min M10MIN = -1.646 104 m.kN

15:3 m.kN

M15q  8.674  10 m kN

T10MAX==T10Lq -5.279+xT10min 103 kN T10MIN

3 kN+Tmax10pch) (1.Tmax10p

A 315  12.031

2 Mmin15ch  Vch   A 115  A 315  m 

sk15 8  0.6

T10MAX = -5.279 T10Lq x+T10max 103 kN

M10MAX==M10q -2.46 +x M10min 104 m.kN M10MIN M10MIN = -1.646 x 10

Mmin15I  1.28  10 m kN A 115  12.526

M10MAX = -2.46 M10qx+M10max 104 m.kN

R10MIN ==7.103 Mmax10l 5.318xx10 1033kN m.kN

M15q = 8.674 x 103 m.kN+ Mnmax15ch) M15max = (1.Mmax15l

3 m.kN Mmax10l = =5.318 x 10m.kN Mmax10ch 808.549

(1.Mmax15l + Mnmax15ch) M15max = 6.315 x 103 m.kN

Mmax10ch = 808.549 m.kN Mmin10l = 711.629 m.kN

M15max==(1.Mmin15l 6.315 x 103 m.kN M15min +Mmin15ch).-1

Mmin10l = =711.629 Mmin10ch 92.467 m.kN m.kN

(1.Mmin15l +Mmin15ch).-1 M15min = -1.781 x 103 m.kN

3 m.kN Mmin10ch= =4.752 92.467 m.kN Mmax15l x 10

103 m.kN

M15min ==-1.781 M15MAX M15q x+ M15max

3 m.kN Mmax15l = =4.752 x 10 Mmax15ch 612.42 m.kN

4 m.kN M15q x+ 10 M15max M15MAX = 1.499

Mmax15ch = 612.42 m.kN Mnin15l = 1.28 x 103 m.kN

M15MAX==M15q 1.499+M15min x 104 m.kN M15MIN

Mnin15l = 1.28 x 103 m.kN Mmin15ch = 245.568 m.kN

3 m.kN +M15min

M15q x 10 M15MIN = 6.893

M15MIN = 6.893 x 10

3 m.kN

Examples of Concrete Box Girder Bridge

Mmin15ch = 245.568 m.kN

229

The horizontal segments (box girder cross-sections) as a beam and vertical columns are

Determination of sectional forces by means of influence lines:

Determination of sectional forces by means of influence lines: 10 Cross-Section 0 5 Cross sectional T0 =R0 M5 Cross-Section 0(kN) 5 forces (kN.m) 0 0 5 M8750 Cross1-Dead sectional T =R load 2170 forces 2-Mmax (kN) Moving 1356 (kN.m) 5562 1-Dead load load3-Mmin 2170 -174,6 8750 -2096 Moving Max 2-Mmax 1356 5562 1+2 3526 14310 load Min 3-Mmin -174,6 1+3 1995 -2096 6654

Max 1+2 Min 1+3

3526 1995

14310 6654

M10 (kN.m) 10 M -17410 (kN.m) -7190 -17410 +946 -7190 -24600 +946 -16460

T10L 10 (kN) 10L T-3652 (kN) -1628 -3652 38,32 -1628 -5279 38,32 -3613

-24600 -16460

-5279 -3613

T10R (kN) 10R T3348 (kN) 1626 3348 141,77 1626 4974 141,77 3490

R10 (kN) R10 7000 (kN) 3253 7000 180,09 3253 10250 180,09 71030

15 M15 15 (kN.m) M15 8674 (kN.m) 6315 8674 -1781 6315 14990 -1781 6893

4974 3490

10250 71030

14990 6893

therefore usually stressed (box separately. The horizontal segments girder cross-sections) as a beam and vertical columns are

therefore usually stressed separately.

Sabah Shawkat ©

Examples of Concrete Box Girder Bridge

230

Forces along prestress tendon:

The moment induced by prestress on a particular cross-section in a statically indeterminate

  12 deg

Npoint1   Np  N 4  cos (  )

Npoint1  1388.37292 kN

structure may be considered to be made up of two components:

  12  deg

Npoint23   Np  N 1  N 2  cos (  )

Npoint23  1465.84221 kN

The first component is the product of the prestressing force Pm, and its eccentricity from the The moment induced by prestress on a particular cross-section in a statically indeterminate centroidalmay axis. is the moment that on components: the concrete part of the cross-section when the structure be This considered to be made upacts of two

Npoint3   Np  N 3 Npoint4  1511.69 kN

Npoint5   Np  N 2  cos (  )

  12  deg

Npoint6   Np  N 2  cos (  )

  12  deg

Npoint5  Npoint4 2

Npoint3  1511.69 kN Npoint4   Np  N 3

Npoint5  1574.33834 kN Npoint6  1574.33834 kN

 1543.01417 kN

geometric constraints by of thethe redundant supports The moment p.ethe is known The first component is imposed the product prestressing forceare Pm,removed. and its eccentricity from centroidal axis. moment. This is the moment that acts on the concrete part of the cross-section when as the primary the constraints is imposed by the caused redundant are removed. The moment p.e Thegeometric second component the moment by supports the hyper-static reactions, i.e. the additional is known as the primary moment. moment required to achieve deformations that are compatible with the support conditions of The second component is the moment caused by the hyper-static reactions, i.e. the additional the indeterminate structure. The moments caused by the hyper-static reactions are the secondary moment required to achieve deformations that are compatible with the support conditions of moments. the indeterminate structure. The moments caused by the hyper-static reactions are the In a similar way, the shear force caused by prestress on a cross-section in a statically indeterminate secondary moments. member can way, be divided in to force primary and secondary components. The primary force in the In a similar the shear caused by prestress on a cross-section in ashear statically

concrete is equal to thecan prestressing P, time the Θ ofcomponents. the tendon at theprimary cross-section indeterminate member be dividedforce, in to primary andslopes secondary The

Sabah Shawkat ©

  0  deg   0  deg

Npoint7   Np  N 3  cos (  )

Npoint7  1511.69 kN

Npoint8   Np  N 3  cos (  )

Npoint8  1511.69 kN

underforce consideration. For aismember only force, horizontal tendons (Θ =of 0), the shear in the concrete equal to containing the prestressing P, time the slopes the primary tendon shear

at the on cross-section under consideration. a member containing horizontalis tendons force each cross-section is zero. The For secondary shear force at only cross-section caused by the (), the primary shear force on each cross-section is zero. The secondary shear force at hyper-static reactions.

Npoint7  Npoint6 2

 1543.01417 kN

  12  deg

Npoint9   Np  N 4  cos (  )

Npoint9  1388.37292 kN

  12  deg

Npoint10   Np  N 4  cos (  )

Npoint10  1388.37292 kN

Npoint8  Npoint9 2

 1450.03146 kN

Calculation of reactions:

  0  deg

Npoint11   Np  N 5  cos (  )

Npoint11  1333.13 kN

  0  deg

Npoint12   Np  N 5  cos (  )

Npoint12  1333.13 kN

Npoint10  Npoint11 2

cross-section caused by the hyper-static reactions. The resultantisinternal actions at any section caused by prestress are the algebraic sums of the The resultant internal actions at any section caused by prestress are the algebraic sums of the primary and secondary effects. primary and secondary effects. Since the secondary effects are caused by hyper-static reactions at each support, it follows that Since the secondary effects are caused by hyper-static reactions at each support, it follows the secondary moments always vary linearly between the supports in a continuous prestressed that the secondary moments always vary linearly between the supports in a continuous concrete member and the secondary shear forces are constant in each span. prestressed concrete member and the secondary shear forces are constant in each span.

95.15  0.17  169.04  0.065  70.02  0.029  97.97  0.027  139.20  0.054  57.80  0.027  50.20929 88.90  0.1058  149.5  0.0529  56.56 0.1056  56.56 0.5095  149.5 0.02815  88.90 0.0654  62.12671

57.80  0.02105  139.20  0.0194  97.97  0.0097  70.02  0.0057  169.0  0.0193  8.52829  1360.75146 kN

53.41  0.0248  263.63  0.0128  17.17  0.01255  97.47  0.0182  150  0.0128  47.52  0.0097  9.06941

43.74228  50.20929  62.12671  8.52829  9.06941  56.20082 56.20082  14.05021 4

Examples of Concrete Box Girder Bridge

231

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Examples of Concrete Box Girder Bridge Secondary effect of prestress tendon:

232

Secondary effect of prestress tendon:

Sabah Shawkat ©

Examples of Concrete Box Girder Bridge

233

Sabah Shawkat © The calculation of stresses-for cross-section 1, in the middle of the span: Wbhm 

J ym

Wbhm  4.85072 m

Zhm

Wbdm 

J ym

Zdm

Wbdm  3.45531 m

Over the supports: Wbhp 

J yp

Wbhp  5.30294 m

Zhp

Wbdp 

J yp

Zdp

Wbdp  5.51939 m

The calculation of stresses due to ideal tendon: Cross-section 5: Width of cross-section: 14,2m

Depth of cross-section: 2,0m

Nk5  1584.22  kN

Wbhm  4.85072 m  kh5 

 kd5 

0.9  Nk5 Am 0.9  Nk5 Am



Mk5  1396.21  kN  m

Wbdm  3.45531 m

0.9  Mk5 Wbhm 0.9  Mk5 Wbdm

Examples of Concrete Box Girder Bridge

Am  6.941 m

 kh5  0.05364 MPa

 kd5  0.56909  MPa

234

Cross-section 10:

 dimd10 

Nk10  1511.69  kN

Wbhp  5.30294 m

Ap  9.892 m

Mk10  1344.7  kN  m

Wbdp  5.51939 m

 kd10 

0.9  Nk10 Ap 0.9  Nk10 Ap



0.9  Mk10 Wbhp

Mmax15  14990  kN  m

Wbdp

Wbhm  4.85072 m

Wbdm  3.45531 m

 kh10  0.36576  MPa

 dimh15 

0.9  Mk10

 dimd10  4.45702  MPa

Wbdp

Cross-Section 15:

.  kh10 

Mmax10

 kd10  0.08173  MPa

 dimd15 

Mmax15 Wbhm Mmax15 Wbdm

 dimh15  3.09026  MPa

 dimd15  4.33825  MPa

Cross-Section 15: Nk15  1333.13  kN

Wbhm  4.85072 m  kh15 

 kd15 

Am  6.941 m

Mk15  1032.0  kN  m

Wbdm  3.45531 m

Sabah Shawkat ©

0.9  Nk15 Am



0.9  Mk15 Wbhm

0.9  Mk15 Wbdm

 kh15  0.01862  MPa

 kd15  0.44166  MPa

The calculation of stresses due to permanent load: Cross-Section 5:

Mmax5  14310  kN  m  dimh5 

 dimd5 

Mmax5 Wbhm Mmax5 Wbdm

Wbhm  4.85072 m

Wbdm  3.45531 m

 dimh5  2.95008  MPa

 dimd5  4.14145  MPa

Cross-Section 10: Mmax10  24600  kN  m  dimh10 

Mmax10 Wbhp

Wbhp  5.30294 m

 dimh10  4.63894  MPa

Wbdp  5.51939 m

Number of prestress cable tendons in bridge design

Examples of Concrete Box Girder Bridge

235

Calculation of required number of cables- tendons:  dimd5  4.14145  MPa

 dimh10  4.63894  MPa

 dimd15  4.33825  MPa

 kd5  0.56909  MPa

 kh10  0.36576  MPa

 kd15  0.44166  MPa

n5 

 dimd5

n10 

 kd5

n5  7.27738

 dimh10

n15 

 kh10

n10  12.68314

 dimd15  kd15

n15  9.82254

It would be sufficient 13 cables but considering that we did not take into account the lateral distribution n= 16 cables. 9 Lp 15.5 mm End wall of 6 cables; middle wall 4 cables. Anchor:

Sabah Shawkat © a  260  mm

Nb  1825.09 kN

  2.5

 adm    13  MPa

Cable Lp15 5

1  5 5  6 5

Ap1 



 d1  6   d2  2

Ap1  0.00014 m

Prestress force for 1. Cable:

Np  0.935   02  n  Ap1

b  260  mm

fcd  0.85 

1.5

Ap1  0.00014 m

0.25 

fcd 0.85 Np

d2  5  mm

 adm  32.5  MPa

3  02  1.532  10  MPa

Eb  35.5  GPa

Ep  190  GPa

9  Lp15 5

Np  1825.06376 kN

 adm  32.5 MPa

30  MPa

d1  5.5  mm

 b 

Np a b

 b   dov

 b  26.99798  MPa

0.25 

fcd 0.85

 Am  34705 kN

Ap  n  Ap1  0.00127 m

Assessment of stresses according to permissible stress in individual stages: 1.Stage pre-tensioning 16 cables on the finished load-bearing structure-only instantly lose.

Np  0.935   02  Ap

Np  1825.06376 kN

 Am  19.01577

Examples of Concrete Box Girder Bridge

236

 kd5  0.56909  MPa

 kh5  0.05364 MPa

kd10  kd10  0.08173  MPa

 kh10  0.36576  kh10  MPa 0.36576 MPa

 kd15  0.44166  MPa  kd15 0.44166  MPa

 kh15  0.01862  kh15  0.01862 MPa MPa

 kd5  0.56909  MPa

 5h 

 kh5  16  kh5   16  5h

0.9

 5h  0.95351 MPa

 5h  0.95351 MPa  5d 

 5d 

 kd5  16

0.9

0.9  5d  10.11708 MPa

 5d  10.11708 MPa

 kh5  0.05364 MPa



10h  10h 

 kh10  16 kh10  16

0.9

 10h  6.50233 MPa

 10h  6.50233 MPa  10d 

 10d 

 10d

 kd10  16

 kd100.9  16

0.9 MPa  1.453

 10d  1.453 MPa

 0.08173  MPa



 kh15  16  kh15  16  15h  0.9

 15h 

0.9

 15h  0.33097 MPa

 15d 

 kd15  16

 15d 

The calculation of stresses due to self-weight of structure: Am  6.941 m

Ap  9.892 m

qo1  180.466 m

qo1  Am  1   pb

qo2  257.192 m

1

 pb  26  kN  m 1

 kN

Wbhp  5.30294 m

Wbdm  3.45531 m

Wbhm  4.85072 m

3

qo2  Ap  1   pb

Wbdp  5.51939 m

0.9  kd15  16

 15d  7.851790.9 MPa

 15d  7.85179 MPa

Sabah Shawkat © Cross-Section 5: A15  55.603

A25  23.076

A35  4.625

M5qo  qo1   A15  A25  A35  m   qo2  qo1  [ 2  ( 1.2800  0.7850  0.3240  0.1992 ) ]  m 2

M5qo  6700.34549 m  kN  qoh5 

 qod5 

M5qo Wbhm M5qo Wbdm

 qoh5  1.38131 MPa

 qod5  1.93915 MPa

Examples of Concrete Box Girder Bridge

237

 qoh15 

 qod15 

M15qo Wbhm M15qo Wbdm

 qoh15  1.38829 MPa

 qod15  1.94894 MPa

Cross-Section 10: A110  34.706

A210  46.149

A310  9.247

M10qo  qo1   A110  A210  A310  m   qo2  qo1  [ 2  ( 1.440  1.5700  0.6470  0.4000) ]  m 2

M10qo  13545.36409 m  kN

Sabah Shawkat ©  qoh10 

 qod10 

M10qo Wbhp

M10qo Wbdp

 qoh10  2.55431 MPa

 qod10  2.45414 MPa

 qoh5  1.38131 MPa

 qod5  1.93915 MPa

 qoh10  2.55431 MPa

 qod10  2.45414 MPa

 qoh15  1.38829 MPa

 qod15  1.94894 MPa

 10h  6.50233 MPa

 15h  0.33097 MPa

 5h  0.95351 MPa

 10d  1.453 MPa

 15d  7.85179 MPa

 5h   qoh5  0.42779 MPa

 5d   qod5  8.17794 MPa

 10d   qod10  1.00114 MPa

 10h   qoh10  3.94802 MPa

 15h   qoh15  1.05731 MPa

 qod15   15d  5.90285 MPa

Cross-Section 15: A115  12.526

A215  61.242

A315  12.031

M15qo  qo1   A115  A215  A315  m   qo2  qo1  [ 2  ( 0.5192  0.8900  0.8900  0.5192) ]  m 2

 5d   10.11708 MPa

M15qo  6734.19521 m  kN

Examples of Concrete Box Girder Bridge

238

Statically indeterminate cable: Ec  38.5  GPa

Ap1 



m  2 Tendon

Ep  190  GPa

 ( 5.5  mm)  6  ( 5  mm) 2

Ap  12.74112  cm

2



Ap1  1.41568  cm

Ap  0.00127 m

Ap  9  Ap1

02  1.532  GPa

Np  1825.06376 kN

 

Calculation of losses due to prestressing

Instant loss i  1  5

 



180

x1  12

x2  24

x3  36

x4  48

Eb  35.5 GPa

n  9

Ap  Ap1  n

Ep  190 GPa

ek  1028  mm

N 2  215.51926 kN

N2  1117.51221 kN

 2

Ep Ap  ek  Am   1  J ym  Ec Am 

N3  1391.33  kN

 f  x   i N i  Np  1  e 

friction

  0.00254

N 3      N2

N 3  1.42044 kN

0.75  Np  1368.79782 kN

k1  0.065

k1 final stress drop

Sabah Shawkat © for t3 will be  k2 = 1

N i  Np  N i

k 2  k2t3  k2

N i 

Ni 

111.14389 kN

1713.91987 kN

0.41888

215.51926

1609.54449

0.62832

313.53832

1511.52543

0.83776

405.58815

1419.4756

1.0472

492.03228

1333.03148

lk  2.88  m  3.96  m

ap lk

 Ep

N22   p22  Ap  9

lk  6.84 m

N22  2866.75238 kN

k 2  0.69

N 4  k1  k 2  N3

k2t3  1

k2  0.31

N 4  62.40115 kN

Loss due to shrinkage:

Shrinkage of concrete

t1  0.25

t1 je 3 month, thus we take 0.25 of year  t1

1e

ap  9  mm

 p22  250  MPa

for t1 be 10 minutes k 2= 0.31

0.20944

 p22 

Ap1  1.41  cm

Relaxation II state:

x5  60

Friction is not considered in straight sections

x i 

Am  6.941 m

N2  Np  N 5  N 2

Np  0.935  02  9  Ap1

x i  xi 

  0.5

N 5  492.03228 kN

Prestress force for 1. Cable:

f  0.3

m1

 

Lp 15,5 mm

  12  deg

   z   zo   1  e 

Loss due to slip at the anchor



1e

 0.62727

  N 5   zo    1  e 

1

 t1



1e

 z  0.00011

 t1 

1e

 Ep Ap

Gradually stretching the cables:

Creep of concrete:

Assumption:

 o  4

we have 2 pre-tensioning engine at one and the other end, then we pre-tensioning the cables.

 N3  N 4    Am 

 bk  

t2  

 bk   0.19146 MPa

Examples of Concrete Box Girder Bridge

N 5  26.96058 kN

 zo  3  10

4

239

N 6 

 bk Ec





 Ep  Ap   o   1  e



 t1

1e



N 6  1.78767 kN

Long term losses N  N 4  N 5  N 6

N  91.1494 kN

Assessment of stress in state II. Nt1  N3

k 

Nt1  1391.33 kN

Nt  Nt1  N

Nt  1300.1806 kN

k  0.93449

Nt1

In the following we will assume simplification that the coefficient k will influence

Determination of stresses due to prestress tendon in II. state

proportionally Nk, Mk even though in reality it is much more difficult.

 5hII 

Calculation of stresses due to permanent load in II. stadium

  kh5  16     0.93449  0.9 

 10hII 

  kh10  16     0.93449  0.9 

 15hII 

  kh15  16     0.93449  0.9 

Sabah Shawkat ©

 5hII  0.89105  MPa  10hII   6.07636  MPa  15hII  0.30929  MPa

Cross-Section 5:

 Mmax5  dimh5II    0.93449  Wbhm   Mmax5  dimd5II    0.93449  Wbdm 

 5dII 

 dimh5II   2.75682  MPa

  kd15  16     0.93449  0.9 

 10dII 

  kd10  16     0.93449  0.9 

 5dII  9.45431  MPa  15dII  7.33742  MPa  10dII  1.35782  MPa

 dimd5II  3.87015  MPa

0.93449 is the value obtained from the ratio of

 Mmax10   0.93449  Wbhp 

 15dII 

Nt1  N3

Cross-Section 10:  dimh10II 

  kd5  16     0.93449  0.9 

k 

 dimh10II  4.33504  MPa

Nt Nt1

Nt1  1391.33 kN

Nt  Nt1  N

Nt  1300.1806 kN

k  0.93449

 Mmax10  dimd10II    0.93449  Wbdp 

 dimd10II   4.16504  MPa

 dimh15II 

 Mmax15   0.93449  Wbhm 

 dimh15II   2.88782  MPa

 dimd15II 

 Mmax15   0.93449  Wbdm 

 dimd15II  4.05405  MPa

Cross‐Section 15:

Examples of Concrete Box Girder Bridge

240 A preliminary estimate is required of the thickness of a post-tensioned flat slab floor for an The results of stresses we obtained from the sum of stresses due to cables and stresses due to

office building. The supporting columns arethickness 400x400mm in sectionflat and are regularly A preliminary estimate is required of the of a post-tensioned slab floor for an spaced at

permanent load.

building. Thedirection. supporting columns are 400x400mm in section are regularly at over each 7600mmoffice centres in one Drop panel extending in eachanddirection arespaced located

interior column. The slab supports a dead load, (in addition to self-weight) and service live load.

A preliminary estimate is requiredDrop of the thickness of ainpost-tensioned flat located slab floor an 7600mm centres in one direction. panel extending each direction are overfor each office building. The supporting columns are 400x400mm in section and are regularly spaced at interior column. The slab supports a dead load, (in addition to self-weight) and service live load.

 5hII   dimh5II  1.86577  MPa

 5dII   dimd5II  5.58417  MPa

 10dII   dimd10II  2.80722  MPa

 10hII   dimh10II  1.74133  MPa

interior column. The slab supports a dead load, (in addition to self-weight) and service live load.

 15hII   dimh15II  2.57853  MPa

 dimd15II   15dII  3.28337  MPa

The three span slab shown below is to be designed.

The three7600mm span slab shown below isDrop totobe designed. in one direction. panel extending in each direction are located over each The threecentres span slab shown below is be designed.

Elevation of three-span Slab

Sabah Shawkat © Prestress tendon:

LSA

15.5mm

Prestress tendon:

w pl  0.5 LSA 15.5mm

Compression reserve:

LSA

Load: 15.5mm w pl  0.5

ht  280  mm Load: w pl  0.5

Load:

ht  280  mm

Elevation of SlabSlab Elevation ofthree-span three-span

Rpn  1800  MPa

Rpd  1440  MPa

w pd  0.133 Rpn  1800  MPa

Rpd  1440  MPa

Self- weight-reinforced concrete kNslab:  c  25  Determination the loads on the slab 32 g 0   c  d slab

dslab   200 mm 20 mm

 f  1.1

dslab  200  mm

 f  1.1

dslab  200  mm

   d

Moving load:

g  5m

gdd  g0   f

gdt  gst   f 2 gdt  8.25 m gdd  5.5 m  kN

g1  9.46 

  1.2

Self-weight-reinforced concrete 2 beam: m slab: Self- weight-reinforced concrete

Moving load:

g  g1   gst  d c  dbeam

m

 f  1.1

2

gdd  5.5 m

 kN

m

4

gdd  g0   f

0 c slab Determination the loads on 0thekNslab

Ap1  1.4157  10

g0  5 m m kN

Self- weight-reinforced concrete slab: Self-weight-reinforced concrete beam: gst   c  dbeam 2

m

Ep  200  4  2GPa

Ap1  1.4157  10

 p  20  mm

3 Determination the loads on the slab m

ht  280  mm

Ap1  1.4157  10 Ep  200  GPa

Rpd  1440  MPa

w pd  0.133 kN  c  25  w pd  0.133 3 m  c  25 

4

 p  20  mm

Rpn  1800  MPa

Ep  200  GPa

 kN

2

 kN

2

g  11.352 m  kN 2 gdt  8.25 m  kN gdt d gst   f

kN 2 2 kN 9.46 gv1  1.2 g  m mkN 2     2kN 1  v   gd  11.352 2.55 m v  3.25 v g sg  v  1.3g d  22  kN 0 dd d g0  s f v gddd  5.5 m  kN mm

load: g 0   Live c  d slab

2 2 Live load: vsconcrete  2.5   v  1.3 vs   v g vd Self-weight-reinforced beam: gst  vdc dbeam gst m  f kN gdt  8.25 m  kN dt  3.25 2

Cover: 15mm

The beam reinforced g8d  11.352 m 2  kN Moving load: g1  9.46by 12 at upper   and 1.2lowergchord,  d  g1stirrups

Cover: 15mm

The beam reinforced by 12 at upper and lower chord, stirrups  8 kN vs  2.5   v  1.3 vd  vs   v 2

Live load:

vd  3.25 m

Cover: 15mm

Example of Concrete Post-Tensioned Flat Slab The beam reinforced by 12 at upper and lower chord, stirrups  8

2

kN

241

Effective depth of slab: tb  15  mm

 xd  12  mm

Debx  ht  tb 

X direction:

 xd

 xt  12  mm

Effective depth of prestress cable:

Debx  ht  tb 

Direction y:

Deby  ht  tb   xt 

0.65   Lextl   7.2  0.4  0.2   m 2  

 

Lextr   7.2  0.4  0.2 

 yd

Desy  0.167 m

2  yt  12  mm

 xt

Direction x:

Effective depth of prestress tendon:

Debx  0.259 m

Desy  hd  tb   xd 

Y direction:

 yd  12  mm

The transverse load g causes moments and shear which usually tend to be opposite in sign to those produced by the external loads. If the slab is a two-way slab, with prestressing tendons which is our case in those examples placed in two orthogonal directions, the total transverse load caused by the prestress is the sum of g for the tendons in each direction.

 ystr  8  mm

Debx  0.259 m

2  yt

Deby  0.247 m

 k  20  mm

L2  7.6  m

0.65   m 2 

Lextr  7.475 m

L2  7.6 m

Parabola No. 1: The uniformally distributed transverse force caused by the prestress is

Sabah Shawkat © Direction y: In the middle of the span of beam:

 3  Vcabley  ht   2  tb  2   xd  2   yd    k 2 2   k

For extreme span: Vexty 

ht 2

 

  tb   ystr   xd 

l2  0.65 m

 

Lextr   7.2  m  0.4  m  0.2  m 

 2 

Vexty  0.095 m

l2 



2

Vcabley l2   1     Lextl  Lext 2

g1 

8  Vexty  Pm

g1  1.1875 m

2Lext2

1

 kN

The uniformally distributed transverse force caused by the prestress is

bcolumn  0.4 m

l2y  6.953 m

g2 

8  V3  Pm

l2y 2

V3  0.14821 m

g2  2.453 m

1

l2y  6.953 m

 kN

Lextr  7.4765 m

Parabola No.3: Lextl  7.4765 m

Lext  4  m

Checking of lengths:

V3 

 Pm

Parabola No.2:

k

l2y  L2  l2

l2   Lextl   7.2  m  0.4  m  0.2  m   2 

( 0.4  4  7.2  3)  m  23.2 m

Pm  100  kN

Deby  0.247 m

l2  bcolumn  Deby

Vcabley  0.162 m

Calculation of equivalent load: bcolumn  400  mm

8 p

V2  Vcabley  V3

V1  Vcabley  V3 g3 

Lextr  Lextr  l2y  2  l2  23.2 m V3  0.14821 m

The uniformally distributed transverse force caused by the prestress is

V2  0.01379 m

8  V1  Pm

l22

V1  0.01379 m

g3  26.356 m

1

l2  0.65 m

 kN

Where h V3, Vexty, and V1 are the sag of the parabolic tendon and l is the span. If the cable spacing is uniform across the width of a slab and Pm is the prestressing force per unit width of slab, then q is uniform upward load per unit area.

Example of Concrete Post-Tensioned Flat Slab

242 It has been shown that wherever a change in direction of the prestressing tendon occurs, a It has isbeen shownonthat change in direction of profile the prestressing tendon occurs, a transverse force imposed thewherever member.aFor a parabolic tendon such as that shown transverse force imposedalong on thethe member. For parabolic tendon profile suchimposed as that shown in figures,Itthe is isconstant tendon anda hence theprestressing transverse force hascurvature been shown that wherever a change in direction of the tendon occurs, a in figures, the curvature constant along the tendon and hence the transverse force imposed on the member is uniform along itsis transverse force is imposed onlength. the member. For a parabolic tendon profile such as that shown on the member is uniform along its length. in figures, the curvature is constant along the tendon and hence the transverse force imposed End forces:

z´( y)

4

Parabola No.1: pEnd2forces: x p 2z´( y) Parabola No.1: 2 l z ( y) 4  2  x

4 2 

x

z´(ly2)

Vexty  0.095 m x  Lext pLext  4 m 4 2  x Lext  4 m Vexty  0.095 m x  Lext 2 p l

pl 2 4   x z´( yV) exty4 2  x Lext  4 m x  Lext Vexty exty  0.095 m  z ( y) Vexty 2 2 Vextyz´( y) a  4  2  ( a ) x 4 2  l   x Vexty4 2  lV0.012 exty   24 2   0.012 a  4  22 z´( y) ( a ) x y)  x 2L 4 22   z´( 2L 2Lext ext ext 2 2 2   V V V exty  exty 2L   2Lext 0.012 a  4 2  exty2Lext z´( y) ( a ) x 4m 2  z´( y) 4 2   ext x m 2   2Lext 2 2L  2Lext 2 m ext m  

z  ( a )  x







z  0.0475 m

z  ( a )  x z  ( a )  x

z  0.0475 m

Immediately losses:







ap  4  mm

μ is a very low, loss is realized over the entire length cable type 1:

lcy  Lextl  l2  2  l2y  Lextr

lcy  23.2 m

 ap   Ep  lcy 

 p12  

 p12  34.48276  MPa

 ptr  Rpd   p12

 ptr  1474.48276  MPa

Loss due to friction:  p11

z  0.0475 m

V0y  Pm  sin ( 0.0475 V0y ) 4.74821 V0y  Pm )sin ( 0.0475 V0y kN 4.74821 kN V0y  Pm  sin ( 0.0475 )

Loss of slip in the anchor

 p12

End forces:

on the member is uniform along its length. Parabola No.1: z ( y)

Calculation of losses:

V0y  4.74821 kN

 p11



  0.06

 pin  1440  MPa

k  0.0005



 pin  1  e

     x k 

Sabah Shawkat ©

) ( 0.0475 H0x ) 99.88721 H0x  Pm  cos  Pm  cos H0x  kN 99.88721 kN H0x( 0.0475 H0x  Pm  cos ( 0.0475 )

H0x  99.88721 kN

Cable type 3:

Checking Checking of forces: of forces:

Checking of forces: l l2y l  2y  4 kN 43.60526 kN  24  g43.60526 g1  Lext  2 g g12L Lext 22y 2 4  43.60526 kN 1 ext2 2  g2  2

l2y  l2y

l2  0.65 m

r1y 

Lext  2  8 m

rexty 

g30y  l  2  V0y  2  43.60169 kN g3  l2 g2  kN l V 2 2V  2 43.60169  43.60169 kN 3 2

Details forillustrations: illustrations: for Details forDetails illustrations:

x

 ly    2 ry 

2  arcsin  

 l2    2 r2y 

 2y  2  sin 

 2 Lext    2 rexty 

 exty  2  sin 

 2  Lext 2  

 l2y    2 r1y 

 1y  2  sin 

rad

r2y 

l2  l2

8  V2

 2y  0.17032

rad

 exty  0.09496

rad

 1y  0.17032

 max  0.60592

No straight sections are designed





    max x k    p11   pin  1  e 

r2y  3.79414 m

rexty  84.21053 m

8  Vexty

 max  2 1y   2y   exty

x 0

r1y  40.77377 m

8  V3

 p11  51.41159  MPa

Idealized and actual tendon profiles in a continuous slab

Example of Concrete Post-Tensioned Flat Slab

243

Admissible stress of prestress tendon:  pin  1440  MPa

 tr  0.9

 ptr   tr  Rpd

 

 p to  0.18 

 tr  Rpd  1296  MPa

1 7

 log to

 

 p to  0.32286  p (  )

 

Short term losses:  ptr   pin   p12   p11

 ptr  1354.10565  MPa

 ptr   p2longterm  0.4  Rpd

0.4  Rpd  576  MPa

 p  1   p to

 p  0.67714

 p21   p   p   p

 p21  43.81876  MPa

Loss due to shrinkage: Loss of gradual cable tension:

 p13

Design of number tendons, x direction Loss in x direction

a  85 mm

The loss of stress in a tendon due to relaxation depends on the sustained stress in the steel. n  14

cables

faster rate than would occur due to relaxation alone. This decrease in stress level in the tendon

Cable No.1: Ab  0.28  m  0.96  m

 Ap1      pin    Ebo 



Owing to creep and shrinkage in the concrete, the stress in the tendon decreases with time at a

ep  0

p11  

 1 ep 2   p12      Ab Ib 

p11  

1 p12      Ab 

m1

Ep 

 p13

2 m

affects (reduces) the magnitude of the relaxation losses.  b  p22

 bs  Ep

 bs



 bsf   2   1



1

0.07 t

1e

 1  0.27442

Tendons are prestressed 21 days after concreting:

Sabah Shawkat © b

 Ap1   b      pin    Ebo 

 p11  51.41159  MPa

 p13  Ep 

n1 2 n

 b  2.19437657  10



 bs   bsf   2   1

 2  1



 p22   bs  Ep

 p12  34.48276  MPa

 b

days

t1  21

5

t  

 bsf  0.00033 common environment

 bs  0.000239

 p22  47.88845  MPa

 p13  2.03764  MPa

Loss due to creep of concrete:

Loss due to relaxation of steel:  p21

 p   p   p

 pin 

p

  p1i

Creep strain in the concrete at the level of the tendon depends on the stress in the concrete at

w pe  0.5

that level. Because the concrete stress varies with time, a reliable estimate of creep losses

i p

Rpn   w pd  1  w pe   p  

 p21

0.18 

p

1 7

requires a detailed time analysis. An approximate and conservative estimate can be made by

 log( t)  1

assuming that the concrete stress at the tendon level remains constant with time and equal to its initial (usually high) value, c (caused by Pi and the permanent part of the load). With this

   p ( t)   p ( t )    p   p

assumption, the creep strain at any time t after transfer (at ageo) may be calculated from an

Cable No.1:  p   pin   p11   p13

Rpn    p  w pd   1  w pe   p  

expression bellow:  p  1386.55077  MPa

 c (t)

 p  0.04667

After 10 minutes we prestressed the tendon again to

p

1440MPa

to  10

Example of Concrete Post-Tensioned Flat Slab

c   t  o Ec





Service stage:

 p23

Ep Eb

  bi  i



 

 

   bt   2   1

 p23

Ebo



 t1 t2



Plong  182.72338 kN

Pdlong  0.9  Plong

Pdlong  164.45104 kN



  2.75721



Design of prestress tendons:

b not to mention, so we estimate 4%

200  2.757   b 32.5

Plong    pin   long  Ap1

Plong  182.72338 kN

Service 0.9  Plong Pdlong  stage:

kN  PFdlong  164.45104 kN

n  Pshort ( kN)

Fptot Plong   pin   L ytot ( m)

  2.75721

    b   i 

 p23  

4 100

ptot 



long  Ap1

dlong

 p23  57.6  MPa

b

Fptot 

L ytot ( m) for y direction:

 n   Ldirectionx  debeamx   i  1 



b

y direction:

n 

 n   Ldirectiony  debeamy   i  1 



b

 m

d ( m)

 p23

    b   i 

 kN  Eb Fptot to  subtract  Then σbnis Psubstituted into the formula, the creep loss short ( kN) m Fptot b L ytot( m52.24 ) )Ep  m defbeam  0.85  dbeam 7  14 Lxtot m dbeam d( m0.28  p23     b   i  for y direction: Eb Then σb is substituted into the formula, to subtract the creep loss

n  7 14 Lxtot 52.24  m n  Pshort Fptot for   y direction:   Lxtot   n Pshort  F n ptot  7  14 L  52.24  m Fptot  Lxtot xtot   b  defbeam Fptotn  P short   b    F   ptot defbeam Lxtot   Ep  p23  E   b     Fbo E ptot p         

ncablesdirectionx Pshortdirectionx

d ( m)

Then σb is substituted into the formula, kN to subtract the creep loss n  Pshort ( kN)

ncablesdirectiony Pshortdirectiony

 m

Plong  182.72338 kN Pdlong  164.45104 kN

long

Design Fptot of prestress tendons:

x direction:

b

Design of 0.9 prestress tendons:  P P

  pin

244

Service stage:





   bt   2   1

 bt  3.8

Plong    pin   long  Ap1

dbeam  0.28E m p

defbeam  0.85  dbeam 1

Fptot  359.08121  p23     b   m i   kN

defbeam  0.238 m

Sabah Shawkat ©

Summarization of losses:  p12  34.48276  MPa

 p11  51.41159  MPa

 p13  2.03764  MPa

 p12   p11   p13  87.93199  MPa  short   p12   p11   p13

 short 

 short  pin

 short   87.93199  MPa

6.1%

 p22  47.88845  MPa

 long   p21   p22   p23

Ebo

defbeam

 p23 

 short  0.06106

 p21  43.81876  MPa

p23  b 

Ebo

    b  



1

Fdptot  kN 0.85  dbeam 0.28  m dmefbeam beam359.08121

 b   1.50874  MPa

Fb  1.50874  MPa m 1  kN 359.08121 ptot

defbeam  0.238 m

Pshort  191.41227 kN

 p23  25.59958  MPa

 p23 1.50874  25.59958   MPa  MPa b

Pshort  191.41227 kN

 p23  25.59958  MPa

 p23  57.6  MPa

 long  149.30722  MPa

 short   long  237.2392  MPa

 total 

 short   long  pin

 total  0.16475

The both losses present 16.5%

Forces between tendon and concrete girder due to prestressing the tendons with force Fp

Pretension stage:

considered as external forces on the concrete girder, and also the determination of 

 pin  Ap1  203.8608 kN

Pshort    pin   short   Ap1

Pid  Pshort  1.06

Pid  202.897 kN

Forces between tendon and concrete girder due to prestressing the tendons with force Fp Pshort  191.41227 kN

Forces between tendon and concrete girder due to prestressing the tendons with force F

p considered as external forces on the concrete girder, and also the determination of 

considered as external forces on the concrete girder, and also the determination of 

Example of Concrete Post-Tensioned Flat Slab

245

Live load:

vs  2.5 

kN m

The slab is supported on both direction x, y by rectangular columns or beams and contains

 v  1.3

vd  3.25 m

vd  vs   v

2

kN

Cover: 15mm

parabolic cables in both the x and y directions.

The beam reinforced by 12 at upper and lower chord, stirrups  8 Effective depth of slab: tb  15  mm x direction:

y direction:

Desx  hd  tb 

 xd  12  mm

 xd

Desx  0.179 m

Desy  hd  tb   xd 

 yd

The first step in the design of a post-tensioned slab is the selection of an initial slab thickness. Serviceability considerations usually dictate the required slab thickness.

Effective depth of prestress cable:

The second step in slab design is to determine the amount and distribution of prestress. Load balancing is generally used to this end. A portion of the load on a slab is balanced by the

Direction x:

Desy  0.167 m

 xt  12  mm

Debx  ht  tb 

 yd  12  mm

 xt

 yt  12  mm

 ystr  8  mm

Debx  0.259 m

Sabah Shawkat © transverse forces imposed by the draped tendons in each direction. To minimize serviceability

Direction y:

problems, a substantial portion of the sustained load should usually be balanced. Under the

Deby  ht  tb   xt 

 yt

Deby  0.247 m

balanced load, the slab remains plane (without curvature) and is subjected only to the resultant, longitudinal, compressive, Pm/A stresses. In practice, perfect load balancing is not possible,

since external loads are rarely perfectly uniformly distributed. However, for practical purposes, adequate load balancing can be achieved. Prestress tendon: LSA

15.5mm

w pl  0.5

Rpn  1800  MPa

Rpd  1440  MPa

w pd  0.133

Ep  200  GPa 4

Ap1  1.4157  10

 p  20  mm

m

Load ht  280  mm

Effect of changes in slab thickness  c  25 

kN m

dslab  200  mm

 f  1.1

At changes of slab thickness, such as occur in a flat slab with drop panels, the anchorage force Pm becomes eccentric with respect to the centroidal axis of the section, as shown in figure. The

Self- weight-reinforced concrete slab: g 0   c  d slab

g0  5 m

2

 kN

gdd  g0   f

Self-weight-reinforced concrete beam: gst   c  dbeam Moving load:

g1  9.46 

kN m

  1.2

gd  g1  

gdd  5.5 m

2

moments caused by this eccentricity, and also be considered in analysis. However, the moments

 kN

gdt  gst   f

gd  11.352 m

gdt  8.25 m 2

 kN

2

 kN

produced by relatively small changes in slab thickness tend to be small compared with those caused by cable curvature and, if the thickening is below the slab, it is conservative to ignore them. Effective depth of prestress tendon: Lextl  6.2  m

Example of Concrete Post-Tensioned Flat Slab

L2  7.4  m

Lextr  8.2  m  k  20  mm

246

 

Vcable  ht   2  tb  2   xd  2 yd  2 

V ext 

ht 2



  tb  

 xt   ystr



k

 k

 

Vcable  0.182 m

  

Vext  0.095 m

Parabola No. 1: g1 g1 

Direction y: In the middle of the span of beam

 

hely  ht   2  tb  2   xd  2   yd 

For extreme span

hely´ 

ht 2

 

  tb   ystr   xd 

k



 3  k 2 

k

 2 

hely  0.162 m

Pm  100  kN

Vext  0.095 m

g1  2.25922 m

2Lext2

Pm  100  kN g2 

1

Lext  2.9  m

 kN

V3  0.16469 m

8  V3  Pm

g2  2.899 m

2 l32

1

l3  Lextt  Lext

l3  3.3705 m

Lextt  6.2705 m

l2  0.66 m

Vcable  0.182 m

V3  0.165 m

 kN

Parabola No.3: e3  Vcable  V3

bcolumn3  424  mm

8  e3  Pm

e3  0.01731 m

Sabah Shawkat ©

l2  bcolumn  Debx

l2  0.66 m

Lextt  Lextl  bcolumn 

l2x  L2  l2

l2x  6.741 m

g3 

Lextt  6.2705 m

Lexttr  Lextr  bcolumn3 

l2 2

g3  31.884 m

l22

1

 kN

Cable No.4:

Lexttr  8.2945 m

Lextr  Lexttr  l3

The maximum cable sag V depends on the concrete cover requirements and the tendon dimensions. When V is determined, the prestressing force required to balance an external load g.

 Pm 2 l 8  Vext Pm

Parabola No.2:

hely´  0.095 m

Calculation of equivalent load: bcolumn  400  mm

8 p

g4 

8  Vext  Pm

2   Lextr 

Vext  0.095 m

g4  0.784 m

Pm  100 kN

1

l3  3.3705 m

Lextr  4.924 m

 kN

End forces:

Checking of lengths:

Parabola No.1: bcolumn  Lextt  L2  5  Lextr  bcolumn3  52.2945 m

V3 

Vcable l2   1   L extt  

Lextt  l2  6  l2x 5  Lexttr  52.224 m

z ( x)

4

p l

V3  0.16469 m

V2  Vcable  V3

V2  0.01731 m z´( x)

x

z´( x)

Vext    x 4 2    2Lext 2  

4 2 

p l

4 2 

x

Lext  2.9 m

Vext

2Lext

 0.023

Example of Concrete Post-Tensioned Flat Slab

Vext  0.095 m

a  4  2 

Vext

2Lext 2 m

x  Lext

247

( a ) x

z´( x)

z  ( a )  x  

z  0.06552 m V0x  Pm  sin ( 0.06552 )

g3 0.66 m 6  V0x  V0x  g1 2.9 m  g2 3.37 m 12  g4 4.924 m g1  2.9  m  g2  3.37  m  12  g4  4.924  m

 100

  4.94556

Care must be taken in the design of the areas of slab, where the prestress in one or both directions is not effective. It is good practice to include a small quantity of bonded non-

V0x  6.54731 kN

prestressed reinforcement in the bottom of the slab perpendicular to the free edge in all exterior spans. An area of non-prestressed steel of about 0.0015.b.dex is usually sufficient, where dex is

H0x  Pm  cos ( 0.06552 )

the effective depth to the non-prestressed steel. In addition, when checking the punching shear strength at the corner column, the beneficial effect of prestress is not available.

H0x  99.78543 kN

Details for illustration:

Parabola No.4:

Sabah Shawkat © z ( x)

4

z´( x)

x

z´( x)

4 2 

x

Vext    x 4 2    2Lextr  2  

Lextr  4.924 m

4 2 

Vext

2Lextr 2

Vext  0.095 m

x  Lextr

 0.008

a  4  2 

Vext

2Lextr 2

z´( x)

( a ) x

z  ( a )  x

z  0.03859 m

V0x  Pm  sin ( 0.03859 )

V0x  3.85804 kN

H0x  Pm  cos ( 0.03859 )

H0x  99.92555 kN

The total force imposed by the slab tendons that must be carried by the edge slab-beam is Checking of forces: g1  2.9  m  g2  3.37  m  12  g4  4.924  m  127.66342 kN g3  0.66  m  6  V0x  V0x  133.97709 kN

In the transverse direction, conventional reinforcement may be used to control shrinkage and Which is equal to the total upward force exerted by the slab cables. Therefore, for this twoway slab system, in order to carry the balanced load to the supporting columns, resistance must be provided for twice the total load to be balanced. This requirement is true for all twoway slab systems irrespective of construction type or material.

temperature cracking and to distribute local load concentrations. Minimum quantities of conventional steel for the control of shrinkage and temperature induced cracking, but as we know that the slab is prestressed in the transverse direction to eliminate the possibility of shrinkage cracking parallel to the span and to ensure a watertight and crack-free slab.

Example of Concrete Post-Tensioned Flat Slab

248

Calculation of losses: Admissible stress of prestress tendon: Immediately losses:

 pin  1440  MPa

Loss of slip in the anchor

 p12

ap  4  mm

 tr  0.9

 ptr   tr  Rpd

 ptr   pin   p12   p11

μ is a very low, loss is realized over the entire length

 tr  Rpd  1296  MPa

 ptr  1228.40866  MPa

Short term losses:  ptr   p2longterm  0.4  Rpd

0.4  Rpd  576  MPa

cable type 1: lcx  Lextt  l2  6  l2x 5  Lexttr

lcx  52.224 m

 ap    Ep  lcx

 p12  

 ptr  Rpd   p12

 ptr  1455.31863  MPa

Loss due to friction:   p11

k  0.0005



Ab  0.28  m  0.96  m

b

V2  0.01731 m

r3x 

r2x  3.13636 m

8  V2

 2  Lext 2   rextx  8  Vext

 lx    2  rx   l2   2x  2  sin    2 r2x x

  0.06

     x k 

l2  0.66 m

l2 l2

2  arcsin  

 2 Lext   extx  2  sin    2 rextx

ep  0

Ep 

 p13

r3x  34.48951 m

8  V3

r extrx  127.60935 m

 Ap1     pin    Ebo 

p11   p12  

 p13  Ep 

cables

m1  b 2 m



n1  b 2 n

 Ab



  A  

2

  

5  b  1.9906801  10

b

 p12  15.31863  MPa  p13  1.84849  MPa

rad

 l2x    2 r3x

 3x  2  sin 

 2x  0.20973

 extx  0.13094

rad

 2 Lextr   extrx  2  sin    2rextrx 

 3x  0.19514 rad

Loss due to relaxation of steel:  p21

 extrx  0.07715 rad

p





Rpn 



p

w pd  1  w pe 

 max  2.44217

 pin 

  p1i

 

p

0.18 

1  log( t)  1 7

   p ( t)   p ( t )    p   p

Cable No.1:

No straight sections are designed

    max x k    p11   pin  1  e 

p

 p   p   p

 p21



p11   p12  

 p11  196.27271  MPa

8  Vext

 1

 Ap1      pin    Ebo 



 b  

l2x l2x

 2  Lextr 2   rextrx 

rextx  44.26316 m

 max  6 2x  5 3x   extx   extrx

x 0

n  14

Sabah Shawkat © 

 pin  1  e

Cable type 1: r2x 

a  85 mm

Cable No.1:

 pin  1440  MPa

p11

 p13

Loss of gradual cable tension: Design of number tendons, direction x: Loss in x direction:

 p12  15.31863  MPa

 p   pin   p11   p13

 p  1241.8788  MPa

 p11  196.27271  MPa

Example of Concrete Post-Tensioned Flat Slab

w pe  0.5

249



Rpn 



p

 p  w pd   1  w pe 

 

 p  0.03661

b

y direction:

to  10

1440MPa

 

1  log to 7

 p to  0.18 

 

 p  1   p to

Summarization of losses: 

 

 p to  0.32286

 p  0.67714

 p ( )

 p21   p   p   p

 p21  30.7896  MPa

Loss due to shrinkage:  bs  Ep

 bs



 bsf   2   1



1

0.07 t

1e

 1  0.27442

p12   15.31863  MPa

 p11  196.27271  MPa

 p13  1.84849  MPa

 p12   p11   p13  213.43983  MPa

 short   p12   p11   p13

 short  213.43983  MPa

 short 

 p22

 n   Ldirectiony  debeamy   i  1 



After 10 minutes we prestressed the tendon again to p

ncablesdirectionx Pshortdirectionx

 short

 short  0.14822

 pin

 p21  30.7896  MPa

 p22  47.88845  MPa

 long   p21   p22   p23

 long  136.27806  MPa

 p23  57.6  MPa

 short   long  349.71789  MPa

Tendons are prestressed 21 days after concreting:

 short   long

Sabah Shawkat © days

t1  21



 bs   bsf   2   1



 p22   bs  Ep

 bsf  0.00033 common environment

t  

 2  1

 bs  0.00024

 total 

 total  0.24286

 pin

Pshort    pin   short   Ap1

 pin  Ap1  203.8608 kN

Pshort  173.64412 kN

 p22  47.88845MPa

Pretension stage:

 bs



 bsf   2   1



 bs   0.00024

b

0.07 t

1e

1

 p22   bs  Ep

Pid  Pshort  1.06

Pid  184.06277 kN

 p22  47.88845  MPa

Service stage:

not to mention, so we estimate 4%

Plong    pin   long  Ap1 Plong  184.56792 kN

Loss due to creep of concrete:  p23

Ep Eb



  bi  i



 

 bt  3.8



   bt   2   1

x direction:

Pdlong  0.9  Plong



b





 t1 t2

Pdlong  166.11112 kN

 Design of prestress tendons:

  2.75721

ncablesdirectiony Pshortdirectiony

   Ldirectionx  debeamx   i  1  n



Fptot

n  Pshort ( kN) L ytot ( m)

kN   m

Fptot  b

d ( m)

then σb is substituted into the formula, to subtract the creep loss for direction x  p23

Ep Eb

    b   i 

n  28

28  Ap1  39.6396  cm

Example of Concrete Post-Tensioned Flat Slab

Lytot  23.2  m

250

Band-beam floors have become an increasingly popular form of prestressed concrete

dbeam  0.28  m

construction over the past decade or so. The slab-bands, which usually span in the long

defbeam  0.9  dbeam

defbeam  0.252 m

 n Pshort    Lytot 

Fptot  

 b 

direction, have a depth commonly about two to three times the slab thickness and a width that

Fptot  209.57049 m

1

may be as a wide as the drop panels in a flat slab. If the slab is prestressed, the tendons are  kN

usually designed using a load balancing approach and have a constant eccentricity over the slab bands with a parabolic drape through the effective span. The depth and width of the band beams

Fptot

 b   0.83163  MPa

defbeam

should be carefully checked to ensure that the reaction from the slab, deposited near the edge of the band, can be safely carried back to the column line.

 p23 

Ep Ebo

    b  



 p23  14.11064  MPa

The most commonly used technique for the analysis of flat plates is the equivalent frame method. The structure is idealized into a set of parallel two-dimensional frames running in two

At the balanced load condition, when the transverse forces imposed by the cables exactly

orthogonal directions through the building. Each frame consists of series of vertical columns

balance the applied external loads, the slab is subjected only to the compressive stresses

spanned by horizontal beams. These beams are an idealization of the strip of slab of width on

imposed by the longitudinal prestress in each direction.

each side of the column line equal to half the distance to the adjacent parallel row of columns and includes any floor beams forming part of the floor system. In this example we are using a

Sabah Shawkat ©

At the end, when we want to analyse the problems for one way-slab, this type of structure is

linear-elastic frame analysis, to determine the number of prestress tendons to flat slab structure.

generally designed as a beam with cables running in the direction of the span at uniform centres. A slab strip of unit width is analysed using simple beam theory.

Flat slabs with drop panels behave and are analysed similarly to flat plates. The addition of drop panels improves the structural behaviour both at service loads and at overloads. Drop panels stiffen the slab, thereby reducing deflection. Drop panels also increase the flexural and shear strength of the slab by providing additional depth at the slab-column intersection. Building codes usually place minimum limits on the dimensions of drop panels. For example, on each side of the column centreline, drop panels should extend a distance equal to at least on sixth of the span in that direction.

Example of Concrete Post-Tensioned Flat Slab

251

Flat plates behave in a similar manner to edge-supported slabs expect that the edge beams are strips of slab located on the column lines. The edge beams have the same depth as the remainder of the slab panel and therefore the system tends to be less stiff and more prone to serviceability problems. The load paths for both the flat plate and the edge-supported slab, are however, essentially the same. In the flat plate slab, the total load to be balanced, the upward forces per

tb  15  mm

 xd  12  mm

 yd  12  mm

Debx  dbeam  tb 

x direction:

 xt

Debx  0.279 m

y direction: Deby  dbeam  tb   xt 

 yt

Deby  0.267 m

unit area exerted by the slab tendons in each direction.

Effective depth of prestress cable: prestress tendon LSA

15.5mm

w pl  0.5

Lextl  6.26  m Rpn  1800  MPa w pd  0.133

Rpd  1440  MPa

Ep  200  GPa 4

Ap1  1.4157  10

 p  20  mm

m

 c  25 

dbeam  300  mm

kN m

dslab  200  mm

Lextr  8.284  m

 

Vcable  dslab   2  tb  2   xd  2 yd  2  Vext 

Load

L2  6.72  m

dslab 2

 

  tb   xt   yt 

k

 

Vcable  0.102 m

k

 k  20  mm

 

Vext  0.051 m

Effective depth of prestress tendon:

 f  1.1

x direction:

Sabah Shawkat © In the middle of the span of beam:

 

hely  dslab   2  tb  2   xd  2   yd 

Self- weight-reinforced concrete slab: g 0   c  d slab

g0  5 m

2

 kN

gdd  5.5 m

gdd  g0   f

2

 kN

gdt  gst   f

hely´ 

gdt  8.25 m

2



 3  k 2 

hely  0.082 m

For extreme span

Self-weight-reinforced concrete beam: gst   c  dbeam

k

dslab 2

 

  tb   yt   xd 

k

 

hely´  0.051 m

 kN

Calculation of equivalent load:

Moving load g1  9.46 

kN m

  1.2

gd  g1  

gd  11.352 m

2

 kN

Live load vs  2.5 

l1k  6.72  m

 k  20  mm

l3  1.5  m

lx  7.4  m

l2x  bcolumn  Debx

l2x  0.68 m

l1x  lx  l2x

l1x  6.721 m

Lextl  Lext1  bcolumn 

kN m

bcolumn  400  mm bcolumn3  424  mm Lext1  6.2  m

 v  1.3

vd  vs   v

vd  3.25 m

2

 kN

l2x 2

Lextr  Lextr2  bcolumn3 

l2x 2

Lextr2  8.200  m

Lextl  6.2605 m Lextr  8.2845 m

bcolumn3  0.424 m

Cover: 15mm

For perfect load balancing. The column line tendons would have to be placed within the

The beam reinforced by 12 at upper and lower chord, stirrups  8

width of slab in which the slab tendons exert down-ward load due to reverse curvature.

Effective depth of slab:

Approximately 75% of the tendons in each direction are located in the column strips., the remainder being uniformly spread across the middle strip regions.

Example of Concrete Post-Tensioned Flat Slab

252

If the tendon layout is such that the upward force on the slab is approximately uniform, then Checking of lengths:

lx  7.4 m

Lextr2  8.2 m

Lextl  6.2605 m

at the balanced load the slab has zero deflection and is subjected only to uniform compression caused by the longitudinal prestress in each direction applied at the anchorages.

bcolumn  Lext1  lx 5  Lextr2  bcolumn3  52.224 m Vcable  0.102 m

V3 

Vcable l2x   1   2  l3  

Lextl  l2x 6  l1x 5  Lextr  52.224 m

End forces:

l1x  6.721 m V3  0.08317 m

V2  Vcable  V3

V2  0.01883 m

Parabola No.1: 4

z ( x)

p l

z´( x)

x

z´( x)

4 2 

p l

Vext     x1 4 2    2Lext 2  

x

4 2 

Lext  1 m

Vext

2Lext

Vext  0.051 m

 0.102

z´( x)

( a ) x

z  ( a )  x1

V0x  Pm  sin ( 0.102 )

M  Pm 

Parabola No. 1: 8 p l g1 

 Pm

Pm  100  kN

Vext

2Lext 2 m

z  0.102 m V0x  10.18232 kN

dbeam  dslab 2

H0x  99.48025 kN

M  5mkN 

Lext  1  m

Checking of forces:

8  Vext Pm

2Lext

Vext  0.051 m

g1  10.2 m

1

The prestressing forces at the slab tendon anchorages will also induce moments at the

 kN

change of depth from slab to slab-band.

g1 Lext 2  g2 l3 12  153.47964 kN

Parabola No.2: Pm  100  kN g2 

a  4  2 

Sabah Shawkat © H0x  Pm  cos ( 0.102 )

x1  Lext

8  V3  Pm

2 l32

V3  0.08317 m

l3  1.5 m  

g2  7.393 m

1

 kN

 g 1  Lext  2   g 2  l3   12  100 g 3  l2x  6  V0x  V0x

g3  l2x 6  V0x  V0x  153.44429 kN   100.02304

The slab-band is normally, designed to carry the full load in the transverse direction (usually Parabola No.3: V2  Vcable  V3

g3 

8  V2  Pm

l2x2

the long-span direction). The prestressing tendons in this direction are concentrated in the V2  0.01883 m

g3  32.666 m

1

l2x  0.679 m

Vcable  0.102 m

V3  0.083 m

slab-bands and are also designed by load balancing.

 kN

Example of Concrete Post-Tensioned Flat Slab

253

When the prestress is transferred to the concrete, immediate losses of prestress occur. The difference between the prestressing force imposed at the jack, Pj, and the force in the steel immediately after transfer at a particular section, Pi, is the immediate loss:

Cable type 1:

2 l32

r2x 

The gradual loss of prestress that takes place with time is called the time-dependent loss. If Pe x

is the force in the prestressing tendon after all losses, then Time-dependent loss=Pj-Pe

 Lext    2  rx 

2  arcsin  

 2l3    2 r2x

 2x  2  sin 

Both of these losses are made up of several components. The immediate losses are caused by elastic deformation of the concrete as the prestress is transferred, friction along the draped

r 3x 

r2x  13.52574 m

8  V3

Immediate loss = Pj-Pi

r1x 

Lext  1 m

2 Lext2

r 1x  9.80392 m

8  Vext

 l2x 2  

r3x  3.06132 m

8  V2

 Lext    2 r1x

 1x  2  sin 

rad

 2x  0.22135

 max  2 1x  6 2x  5 3x

 1x  0.10196

 l2x    2r3x

 3x  2  sin 

rad

 3x  0.22135

rad

 max  2.63871

tendon in a post-tensioned member, and slip at the anchorage. Other sources of immediate loss of prestress which may need to be accounted for in some situations include deformation of the forms of precast members, temperature changes between the time of stressing the tendons and

x 0

straight sections are designed

  0.06

 max  2.63871

k  0.0005

Sabah Shawkat © casting the concrete, deformation in the joints of the precast members assembled in sections, and relaxation of the tendons prior to transfer. The time-dependent losses are caused by the

gradual shortening of the concrete at the steel level due to creep and shrinkage, and by relaxation of the steel itself. Additional losses may occur due to time-dependent deformation of the joints in segmental construction.

Straight sections:

lstraight  ( 3.760  3.72  5  5.784 )





    max x k    p11   pin  1  e 

lstraight  28.144

x  lstraight

 p11  193.43385  MPa

Admissible stress of prestress tendon:  pin  1440  MPa

Calculation of losses:

Immediately losses:  p12 Loss of slip in the anchor realized over the entire length

ap  4  mm μ is a very low, loss is

 tr  0.9

 ptr   pin   p12   p11

 ptr   tr  Rpd

 tr  Rpd  1296  MPa

 ptr  1231.24752  MPa

Short term losses:

cable type 1:

 ptr   p2longterm  0.4  Rpd

0.4  Rpd  576  MPa

bcolumn  Lext1  lx 5  Lextr2  bcolumn3  52.224 m lcx  bcolumn  Lext1  lx 5  Lextr2  bcolumn3 

 ap  p12     Ep  lcx



 p11

 pin  1  e



     x k 

 pin  1440  MPa

simultaneously, the elastic deformation of the concrete occurs during the stressing operation

containing more than one tendon and where the tendons are stressed sequentially, the elastic deformation losses vary from tendon to tendon and are a maximum in the tendon stressed

 ptr  1455.31863  MPa

Loss due to friction:

For post-tensioned members with one cable or with two cables or more cables stressed

before the tendons are anchored. In this case, elastic shortening losses are zero. In a member

 p12  15.31863  MPa

 ptr  Rpd   p12

 p11

lcx  52.224 m

  0.06

k  0.0005

first and a minimum (zero) in the tendon stressed last. It is relatively simple to calculate the elastic deformation losses in any tendon provided the stressing sequence is known. However, these losses are usually small and, for practical purposes, the average elastic shortening loss is often taken as half the value obtained from equation:

Example of Concrete Post-Tensioned Flat Slab

254

p

 p

1440MPa

 

 Ep  0.5    cp  Ec 

 p to  0.18 

to  10 1  log to  p  to  0.32286 7

 

 p  1   p to

 p  0.67714

 p ( )

 p21   p   p   p

 p21  31.0608  MPa

 p13

Loss of gradual cable tension: Design of number tendons, direction x:

a  85 mm

Design of number tendons, direction y:

n  16

n  14

cables cables

Loss due to shrinkage:  p22

 bs  Ep

 bs



 bsf   2   1



1

0.07 t

1e

Identically in the y direction: Tendons are prestressed 21 days after concreting:

Loss in x direction:

Ab  0.30  m  1.0  m

b

ep  0

 Ap1      pin    Ebo 



Ep 

 p13



 1

p11   p12  



m1 2 m

 b

2

  

 Ab

 0.07 t 1  1  1  e

 1  0.27442

 p22   bs  Ep

 p22   47.88845  MPa

 p23

1   A 

p11   p12  



b

 p12  15.31863  MPa

n1  p13  Ep   b 2 n

 p13  1.67604  MPa

    b   i 

100



 bs   bsf   2   1



 bs   0.00024

 p21

 p   p   p

p

1 0.18   log( t)  1 7

 p   pin   p11   p13



Rpn 



p

 

 p  0.03685

b not to mention, so I estimate 4%

  pin

b

x direction

 p23   57.6  MPa

ncablesdirectiony Pshortdirectiony

 n   Ldirectionx  debeamx   i  1 



p

 pin 

  p1i

b

 p21

ncablesdirectionx Pshortdirectionx

   p ( t)   p ( t )    p   p

 n   Ldirectiony  debeamy   i  1 



w pe  0.5

Rpn   w pd  1  w pe   p  

 p  w pd   1  w pe 

y direction

Loss due to relaxation of steel:

Cable No.1:

 p23  

5  b  1.78777141  10

 p11  193.43385  MPa

p

 bsf  0.00033 common environment

t  

 2  1

Sabah Shawkat ©

 Ap1     pin    Ebo 

 b  

days

t1  21

Cable No.1:

Summarization of losses:  p12   15.31863  MPa

 p11   193.43385  MPa

 p13   1.67604  MPa

 p12   p11   p13   210.42851  MPa

 p  1244.89012  MPa

 short   p12   p11   p13

 short 

 short  pin

 p22   47.88845  MPa

 short   210.42851  MPa

 short  0.14613

 p21   31.0608  MPa

 p23   57.6  MPa

After 10 minutes we prestressed the tendon again to

Example of Concrete Post-Tensioned Flat Slab

255

 long   p21   p22   p23

 long   136.54926  MPa

span–to-depth ratios that had proved acceptable, in terms of both performance and economy,

 short   long   346.97777  MPa

 short   long

 total 

 pin

For the design of post-tensioned slabs, the post-tensioning institute (1977) suggested typical

for a variety of slab types. These recommendations are summarized in table bellow. Note that for flat plates and flat slabs with drop panels, the longer of the two orthogonal spans is used

 total  0.24096

in the determination of the span-to-depth ratio, while for edge-supported slabs, the shorter Pshort    pin   short   Ap1

 pin  Ap1  203.8608 kN

span is used. Pshort  174.07044 kN

Span-to-depth ratios (post-tension) Floor system

Pretension stage: Pid  Pshort  1.06

Pid  184.51466 kN

Service stage:

Plong    pin   long  Ap1

Pdlong  0.9  Plong

Pdlong  166.07657 kN

Span-to Depth ratio

Flat plate

Flat slab with drop panels

One-way slab

Edge-supported slab

Waffle slab

Sabah Shawkat © Band-beams (b=3D)

Design of prestress tendons:

Fptot

b

L ytot ( m)

subtract the creep loss

 p23

Lytot  23.2  m

insulation. The fire resistance period required for a particular structure is generally specified by the local building authority and depends on the type of structure and its occupancy. The

defbeam  0.9  dbeam

defbeam  0.27 m

Australian code AS 3600-1988 specifies the minimum effective thickness of a slab required to provide a particular fire resistance period for insulation and the minimum concrete cover to the

1

 kN

Fire resistance period (minutes)

 b   0.88925  MPa

defbeam

Ebo

resistance period. It must also be sufficiently thick to limit the temperature on one side when exposed to fire on the other side, i.e. it must provide a suitable fire resistance period for

Fptot  240.09715 m

Fptot

A slab exposed to fire must retain its structural adequacy and its integrity for a particular fire

bottom reinforcement in a slab in order to maintain structural adequacy.

 n Pshort    Lytot 

 p23 

then σb is substituted into the formula, to

for x direction

Fptot  

 b 

d ( m)

    b   i 

n  32

kN   m

Fptot 

n  Pshort ( kN)

    b  



 p23   15.0883  MPa

Minimum concrete cover to bottom reinforcement (mm) For simply-supported slabs

For continuous slabs

reinforcement

tendons

reinforcement

tendons

120

180

240

Example of Concrete Post-Tensioned Flat Slab

256

Prestress concrete beam

Un-prestressed reinforcement:

The details of the prestressed concrete beam are shown in Figure below. It is proposed to calculate the long term deflection under the permanent load g kN/m for the two following cases

fyk  460 MPa The ultimate moment for the beam is

using the method of global coefficients:

Mu 

The final prestress balances 80% of the permanent load

qu l

Mu  245m kN

The final prestress balances 40% of the permanent load

Prestress design to balance 80% of the permanent load.

qu is ultimate load, g is permanent load

The calculation is carried out as follows:

l  7 m

qu  40.0 kN m

e  0.2 m

b  0.3 m

d  0.54 m

d h

1

g 

h  0.6 m

g  20 m

2 c  50  mm

1

Calculate the load equivalent to the prestress: kN

d  h  c

 0.9

1

kN

Calculate the prestressing force at t= 0 and t = t and also the normal stress in the concrete prestress force at ∞ time

Sabah Shawkat ©

The following properties are assumed for the materials: Concrete

g´  16 m

g´  0.8 g

Ecm  30.5 GPa

P 

fct  2.5 MPa



 2.5

 c

8 e



Prestress tendons:

g´

P

P  490kN  c

Ac  b h

 2.72MPa

P0 

P

0.85

Ac  0.18 m

P0  576.47kN

calculate the area of prestressing steel:

f0lk  1600MPa

 P0

 0.85f0lk

 P0

 1360MPa

Asp 

Asp  0.00042 m

 P0

calculate the prestressed reinforcement ratio:



p

Asp

p

b d

 0.262%

The ultimate moment at mid-span is: 

 0.92

Geometry of reinforced prestress concrete beam s



 0

b d   s fyk   p f0lk  2

s



As  0 cm

The cables are grouted after stressing. it is assumed that the loss of prestress equals 15%. To simplify, we assume a constant prestressing force during the time P = P∞

Example of Post - Tensioned RC Beam

Mu   p f0lk  b d 

b d fyk

2 s

 0.00248

257

hence, the required area of un-tensioned steel is:

This leads to mechanical degree of prestress of 



Asp f0lk



Asp f0lk  As fyk

1 s

calculate the cracking moment from: b h Wc  6

Wc  0.018 m

P   Mr  Wc  fct   Mr  94m kN Ac  

 Mu    p f0lk   2   b d  

s

fyk

 0.20673% 2

As   s b d

As  3.34904 cm

Calculate the maximum moment in execs of that balanced by the prestress: M D 

( g  g´) 2 l 8

MD  Mr

M D  24.5 m  kN

The beam is thus un-cracked and will be in state I. Calculate the basic deflection due to a load of (g-g´)

( g  g´) l

Sabah Shawkat © Ic 

1 3 b h 12

g  g´  4 m

1

ac 

kN

384



ac  0.759 mm

EcmIc

Calculate the global correction coefficient for a beam without un-tensioned steel. This is equal to 1+φ, hence:

kt  3.5



 1

Finally, the probable long-term deflection is given by:

at   kt ac

a t  2.65745  mm

The mechanical degree of prestress then became:

prestress designed to balance 40% of the permanent load:

This leads to mechanical degree of prestress of The calculation follows the same steps as above to give: g´´  0.4 g

 c



Asp 

P Ac P0  P0

g´´  8 m  c

1



kN

P

 1.36111MPa

l  g´´ 8 e

P0  2

Asp  211.93772 mm

p

P 0.85



Asp ( b d )

P  245kN



Asp f0lk Asp f0lk  As fyk



 0.68761

Calculate the cracking moment from:

P0  288.23529kN

p

Wc 

b h 6

Wc  0.018 m

 0.13083%

Example of Post - Tensioned RC Beam



P 



Ac 

Mr  Wc  fct 



Mr  69.5m kN

258

Calculate the maximum moment in excess of that balanced by the prestress:

MD 

( g  g´´) 2 l 8 

MD  73.5m kN

MD  Mr

b d   rc fyk   prestress f0lk 2

1 8

qu l

during installation

0.85P0.

P

during service  0.94558

The global correction coefficients can be obtained from the graphs. The prestress tendons are

only taken into account if they are directly bonded. In this case it is assumed that the prestressing tendons consist of cables within ducts which were grouted later.

p0  prestress

 rc



P

Asp

 p0

 0.85  p0

b d

As b d

l 8 e

0.85

g

1  p0

l 0.8 g 8 e 1   p0 0.85

b d



Wc  fct 



P 

 Ac 

In this case, the prestressing tendons should not be taken into account. Thus: 

b d

1 8

( g  g´) l

  s

ac 



 0.20673 %

5 1 4  l ( g  g´´) 384 EcmIc

at   kt ac

for



 2.5

d h

 0.9

kt  7.5

ac  2.27782mm a t  17.08362  mm

Sabah Shawkat ©

5 ( g  g´) 4  l 384 EcmIc

0.95Pin

In principle, prestress, even moderate, will improve the service state of a structure. It should

always be noted that in the cracked state under the action of very high loads or imposed

deformations, it can have happened that a prestressed structure can behave much worse than a reinforced concrete beam of equivalent ultimate strength.

Example of Post - Tensioned RC Beam

259

Design reinforcement to pre-stress ceiling panel TT. consider the correctness of the proposal: Design reinforcement to pre-stress ceiling panel TT. consider the correctness of the proposal:

According to the limit state of crack formation

According to the ultimate limit limit statestate According to the ultimate

L  18 m i  1  5 b 1  125 mm

h 1  900 mm

b 2  465 mm

h 2  210 mm

of concrete thefollowing following dimensions: The panelThe is panel madeisofmade concrete classclass C40C40 andand hashasthe dimensions:

h 3  30 mm

b 4  30 mm

h 4  120 mm

b 5  20 mm

According to the limit state of crack formation

According to the limit deflections According to thestate limit of state of deflections

B  1180 mm

A11  b 1 h 1

A15 

A1 2  b 2 h 2

1 b 5 h 5 2

A1 3  b 3 h 3

A11  0.1125 m

A12  0.09765 m

a 

H  0.9 m

 A1i

A13  0.0006 m

a  0.21915 m

b 3  20 mm h 5  660 mm

A14 

1 2

b 4 h 4

A14  0.0018 m

A  2 a

A15  0.0066 m

A  0.4383 m

Sabah Shawkat © r 1  0.5 h 1

r 2  0.5 h 2

r 5  h 2  h 3 

r 3  h 2  0.5 h 3

r 4  h 2 

h 4

1 h 5 3

  A1i ri

ybh 

J11 

ybh  0.29432 m

b 1  h 1 

J11  0.00759 m J14 

J12 

8

Ac  b i h i i

 J1i

m

b 2  h 2 

J12  0.00036 m

1 3 h 4  b 4 36

J14  9  10

J 

J15 

ybd  H  ybh

1 3 b 5  h 5 36

J13 

J  0.00811 m

Example of Prestress Ceiling TT Panel

b 3  h 3  8

m

J15  0.00016 m

a i  ybh  r i 4

J13  4.5  10

Si  A1i r i

ybd  0.60568 m

260

Si 

Ac  i

m

0.1125

ri 

m

0.05063

0.01025

0.09765

0.00014

0.0006

0.00045

0.0036

0.00304

0.0132

Asi  A1i  a i

Live-load: m

-0.15568

0.105

0.18932

0.225

0.06932

0.25

0.04432

0.46

-0.16568

vs  2.0 kN m

vd  3.6 

Ac  

Asi  0.00273

ai  m

0.45

m

 

A

 2

S 

i



 Si

2

 1.5

f

v  vs Zs

v  2.4 

vd  v  f

kN m

g  go  g1  v

gd  god  g1d  vd

kN g1d  vd  9.27  m

kN gd  24.65433  m

Anchorage area length:

g1  v  6.6 

lbd  60 ds

ds  15.5 mm

kN m

lbd  0.93 m

0.0035 2.88279·10-6

The equalization length ld is considered to be the greater of the values:

3.53496·10-6

a11 max distance of edge of prestress cable edge

0.00018

As 

a22 max distance of two adjacent prestress cables

Sabah Shawkat ©

 Asi i

Ac  0.4551 m

As  0.00641 m

Wbh 

ybh

S  0.0645 m

Jc  2  J  As 

Wbh  0.09871 m

Jc  0.02905 m

Wbd 

ybd

a11, a22 are measured along the centre line

Wbd  0.04797 m

go  A  b

go  11.3958 

kN m

god  go  fo

god  15.38433 

kN m

gs  3.5 kN m

 1.35

g1d  5.67 

kN m

 

a2  2  H  a  ld1  1.5 a1

a  0.09 m akp  6.25 cm

a22 

a2  2.625 m   B  2 akl 2  ld1  1.215 m ld2  a2 ld2  2.625 m

2

g1  gs Zs Zs  1.2 m

g1  4.2 

kN m

 lbd 2   ld 2

ld  ld2 ld  2.625 m

g1d  g1 fs

lb  2.78487 m

Calculation of internal forces from vertical load: L lb  2 2 a  3

Dead load, Zs is width load:  fs

a  a11  a22

akl  6.25 cm

lb 

Self-weight of TT prestress beam  1.35

a22  5 cm

a1  0.81 m

Anchorage area length:

Determination of loads:

 fo

a11  4 cm

a1  H  a

i  1  4

a  2.53585 m

x1 

lb 2

xi  1.39244

3.92829 6.46415 9

Example of Prestress Ceiling TT Panel

x2 

lb 2

a

x3 

lb 2

 2 a

x4 

L 2

261

we should calculate the moments for self-weight at the distance xo lb/2 and x4 Design procedure:

self-weight + permanent loads + live-loads

 

permanent loads (without self-weight) + live-load 1. Choose epd

epd  ybd   a11 

2.we estimate ef

ef 

3. we calculate:

 bg

Calculation of moments from extreme load:

 god L

Mgod  

 2

xi  god 

  gd  L

Mgd   i

 2

 xi 2 2 



xi   gd  



Mgod 

  g1d  vd  L

Mgv   i



kN m

 xi 2  2 

 xi 2 

H 3





2 

epd  0.54068 m

ef  0.3 m

10.5  ef   H 6  ef   H

 bg

 2.5

4. Calculate minimal value of prestress force (Inequality 4)

2  Mgd 

177.88122



Mgv 



xi   g1d  vd  

a22 

 bg  b1 fctm

kN m

107.18432

285.06554

425.20575

256.21247

681.41822

573.60046

345.62937

919.22983

623.06536

375.435

998.50037



 Npin

 pp  2 

kN m





 Npin  epd 

 pp  2 

Wbd



Mgd

Wbd

Mgd  4

Sabah Shawkat © Mgod  623.06536 kN m

Mgod  177.88122 kN m

Mgd  998.50037 kN m

Mgv  375.435 kN m

Npin    bg fctm 



1  Wbd   0.9  2 0.9  2 epd     Wbd   Ac

Npin  1783.46505 kN

5. Determine the number of ropes, the strength in one rope, the number of cable Npin n  Np1  Ap1   pd  Np1  182.01857 kN n  9.79826  Np1

Design of prestressing force:

Proposal

n  10

Reliability Coefficient of prestressing:

 pp

In the installation stage (unfavourable): In service stage (favourable):

 1.06

 ppu

 0.94  2  0.72

True prestress force:

1

 0.9

NpinI  n  Np1

6. True static eccentricity:

 Mgd 4  0.9  2 NpinI epd   0.9  2 NpinI  

ef  

Material Properties: Concrete: C30  40 Ec  36000 MPa Steel prestress: LA15 5  pd

 pn

fctm  2.10MPa

 1620 MPa

 1285.71429 MPa

fckcyl  30MPa

fcd  22MPa

NpinI  1820.18571 kN

7. If ef > H/6 we repeat the calculation from point 3:

efkont 

H 6

ef  0.30588 m

efkont  0.15 m

Then the procedure must be repeated 2

Ap1  141.57 mm  pc

 0.5

 pd



 pn

1.26  pd  0.266

Ep  190000 MPa

3.  bg



10.5  ef   H 6  ef   H

 bg

Example of Prestress Ceiling TT Panel

 2.47173 bg <2.5

bg

 2.47173

262

a11  0.04 m

a22  0.05 m

Wbd  0.04797 m

 

epd  ybd   a11 

epd  0.54068 m

a22 



NpinrII  1820.18571 kN

Check v x = l/2



Npin4   b1 0.6  fckcyl  

 

 bg  b1 fctm

 fct

Mgd  4

1  Npin4  1790.26652 kN Wbd   0.9  2 0.9  2 epd     Wbd   Ac

 pp  2 



 Npin4



 Npin4

 pp  2 

 Npin4  epd  Wbd



Mgd



Npin2   b1 0.6  fckcyl  



 Npin4  epd 

Npin2  2309.06724 kN



Mgd

Npin4    b1 0.6 fckcyl 



Wbd

 Npin4  epd 

 pp  2 

Wbd



Mgd

Wbd



 pin4

Wbd

 2.41277 MPa



 2.41277 MPa

 fct

  pin4



5. Determine the number of ropes, the strength in one cable Np1  Ap1   pd 

n 

NpinrII  n Ap1   pd 

n  10

Npin3  3709.88389 kN

Np1  182.01857 kN

Number of ropes - cables

Npin

n  9.79826

 Np1

proposal

Npin4  Npin2

Mgd  4

1  Wbh    ppu  2  ppu  2epd     Wbh  Ac 

NpinrII  1820.18571 kN

 b1 0.6 fckcyl



 pp  1 N pinrII

  b1 0.6  fckcyl 



 pp  1 N pinrII epd

Wbd

 pp  1 N pinrII epd

Wbd c



 18 MPa

Mgod

Wbd c

Mgod

Data needed:

epd 

Wbd

 pin2

  pin2

 11.43887 MPa

Mgd  4

1  Wbh    ppu  2  ppu  2epd     Wbh  Ac 

Npin4  1790.26652 kN

Check the position of the prestressing force along the beam

check – inequalities 2



1  Wbd    pp  1  pp  1 epd     Wbd  Ac 

Npin4  1790.26652 kN

Npin3   b1 0.6  fckcyl  



 pin2

Mgod  4

Wbd

 5.19064 MPa

 pp  2 

1  Wbd    pp  1  pp  1 epd     Wbd  Ac 

 Npin4

 pp  2 

 fct

 pp  2 



  bg  b1 fctm

 pin4

c

Mgod  4



4. Npin4    bg fctm 

Npin4  1790.26652 kN

2 

epd 

NpinrII  1820.18571 kN

 Wbh    NpinrII Mgod 4    bg fctm   Ac  NpinrII   Wbh   Wbd

 NpinrII



 NpinrII



0.6  fckcyl  

Example of Prestress Ceiling TT Panel



Mgod  4



Wbd 

Npin3  Npin4

satisfy

263

epd 

Mgd  0.9  2 NpinrII  Wbh   4   0.6  fckcyl    0.9  2  NpinrII  Ac  Wbh  

epd2  i

Wbd  NpinrII Mgod i  0.6  fckcyl    NpinrII  Ac Wbd 

epd2  i

0.46668

m

0.60256

epd 



Wbd

 bg fctm 

0.9  2  NpinrII 

i  1  4 lb x1  2

0.9  2  NpinrII Ac



Mgd 

0.68409

 Wbd  4

0.71126

Mgd  0.9  2  NpinrII  Wbh   4   0.6  fckcyl    Ac 0.9  2 NpinrII   Wbh  

epd3 

x1  1.39244 m

x3  6.46415 m

x4 

Mgod 

x2 

lb 2

x2  3.92829 m

a

kN m

425.20575

 2 a epd3 

epd3 

Mgd 

177.88122

x4  9 m

Mgv 

x3 

kN m

285.06554

256.21247

0.6  fckcyl  

0.9  2NpinrII

0.9  2NpinrII 



Mgd  i



Wbh 

107.18432



Wbh

kN m

681.41822

-1.04788

m

-0.71184 -0.51022

345.62937

919.22983

623.06536

375.435

998.50037

NpinrII  1820.18571 kN

epd1 

epd1  i

epd1 

Wbh  0.09871 m

 bg fctm

 5.19064 MPa

 Wbh    NpinrII Mgod 4    bg fctm   Ac  NpinrII   Wbh  



Wbd

epd4 

 bg fctm 

0.9  2 NpinrII 

epd4  i

NpinrII Mgod i  Wbh     bg fctm   NpinrII  Ac Wbh 



Wbd

 bg  fctm  

NpinrII  1820.18571 kN

0.9  2 NpinrII

0.9  2NpinrII 



Mgd  4



Wbd 

Mgd  i



Wbd 

-0.07481

m

0.26123 0.46286 0.53006

0.73202 0.81355

NpinrII  1820.18571 kN

Static eccentricity Pretension stage

0.84072

 NpinrII



m

Wbd

0.9  2 NpinrII

epd4 

0.59614

epd2 

-0.44301



 NpinrII



0.6  fckcyl  



Mgod  4



Wbd 

efgod1   i

Mgod   pp  1NpinrIIepd1 i

 pp  1 N pinrII

efgod1  i

-0.49806 -0.49757 -0.49727 -0.49718

Example of Prestress Ceiling TT Panel

m

264

Final stage: efgd4  

Coefficient for LA ropes (cables): p Basic value of creep coefficient of prestressing reinforcement.

Mgd   ppu  2 NpinrIIepd4 i i

p stress of pre-stressing reinforcement generated by the pre-stressing device including

efgd4 

 ppu  2 N pinrII

losses p11 to p16 (we do not incur these losses)

m

-0.3165 -0.3165

pn characteristic tensile strength of prestressing reinforcement

-0.3165 -0.3165

Calculation of cross-sectional quantities in ideal cross-section



  1i

p

 pin

0

 pin

  pd

 1285.71429 MPa

 pin

i1

Ac  0.4551 m

Ap  n Ap1 Ep  p  Ec

ybd  0.60568 m

ybh  0.29432 m p

 5.27778 Ai  0.44904 m



 1i

  pin

p

t  60 24

p

 p (t)



 pn 



p

  pd  1   pc 

 0.18 

1 log( t ) 7

 

 p (t)

p

 0.09842

 0.63119

Loss due to relaxation of steel:

Distance of the centre of gravity of the ideal cross-section from the axis of the concrete cross-section

ti 



i1

Ai  Ac  Ap   p  1

Position of the ideal cross-section

Ap  0.00142 m

Ac 0 m  Ap   p  1 epd

 p17

t

Moment of inertia of the ideal cross-section to the centre of gravity of the ideal crosssection 2

 Ji  Wih      ybh  ti 

 p17

Ji  0.03085 m

 1 10

 p18

 ptr

Wih  0.10228 m

Wid 

 11

 0 MPa

 12

 0 MPa

 ybd  ti

Wid  0.05155 m

 13

 0 MPa

 15

 0 MPa

T2  70

  t  t Ep 

 0 MPa

lf l

 pd

 p18

  pin

Age of concrete to=0

t1=1 den

1.Loss by creep of reinforcement

 pc

 0.5

 pd

 1285.71429 MPa

 ptr

p 16

 0 MPa

 1 10

 p18

5

 0.266

 ptr

 1112.93083 MPa

T2  70

  t  t Ep 

 p18

 0.9  pd

Satisfy

 pd

 0.9  pd

 ptr

Losses from long term

T1  20

t

 T2  T1

 92.91209 MPa

  pin

  pin   p17   p18

common environment  pd

 T2  T1

2.Loss of temperature difference in concrete heat treatment:

 ptr

t2=

t

 92.91209 MPa

  pin   p17   p18

Losses from short term: lf  89 m

T1  20

t

 14

5

Short-term losses:

Quantification of pre-stress losses short-term losses:

Data: l  91 m

 79.87137 MPa

2. Loss of temperature difference in concrete heat treatment:

ti  0.00729 m

Ji  Jc  Ac ti  Ap   p  1 epd Cross-sectional module

  p  p ( t )  p

 p21

Example of Prestress Ceiling TT Panel

 pd

 1285.71429 MPa

 ptr

 1112.93083 MPa

satisfy

265

Loss due to shrinkage of concrete

Loss from creep  0.09842

p

 p (t)

 0.63119

 pt2

 1

Loss due to relaxation of prestress cable after pre-loading the concrete  p21

  p   pt2   p ( t )   ptr

 p21

  p122   p222

 p22

Resultant

 p22

 59.21987 MPa

Losses from creep concretep23 from long term load bf basic value of creep coefficient

 40.39697 MPa

Ep, Ec modulus of elasticity of prestressing reinforcement (concrete) Loss from concrete shrinkage:

b concrete stress from service values of initial pre-stressing force and long-term loads in

bs relative longitudinal deformation of concrete from shrinkage

the centre of gravity of all reinforcement t age of concrete in days , ttr age of concrete in the days at the moment of transferring of prestressing in pre-stressed construction, it is the age when the concrete reached 0.25 times the guaranteed compressive strength - chemically age

the hygrometric conditions moist -0.12

common -0.33

dry -0.5

1.6

2.2

3.8

5.5

 3.8

 bf

common environment

 5.5

dry environment

1. Time interval from ttr to t

bsf is the basic value of the relative longitudinal deformation from shrinkage depending on

wet 0.07

 1bf

 pin

 1285.71429 MPa

 p122

 12.37172 MPa

 p17

 79.87137 MPa

 p21

 40.39697 MPa

 92.91209 MPa

 p18

bf

turych  1

3

t  ( 75  1)  28

bsf

10

 bsf

 0.33 10

t

 0.50686

 p122

 t1

  bs Ep

 0.30955

 p122

 0.07  

 1  e

 p222

  bs Ep

 0.07  t

 1  e

 bs

t1  28

t  102

 t1

  bsf   t   t1

t´  28

t2  

 bs

 bsf  t2

1

 p222

 0.5 10  bs

  pin   p17   p18   p21   p122

Np  Ap  pt1

M1 

 0.07  t1

 1  e

 1b



5

 6.51143  10

  12.37172 MPa

2. Time interval from t to t2  t2

t´   tsku  turych  28

tsku  1

t

1. We read interval from ttr to t

 pt1

 p123

Np  1500.87155 kN

1 god 2  ( L  0.4 m )   Np epd  8 1.35 M1 Jc



epd 

Ep Ec

 pt1

 1bf   t   t1  1b

1.35

 10.23282 MPa

 p123

 40.49417 MPa

2. Time interval from t1 to t2 3

dry environment

  bsf  1   t

 46.84815 MPa

 bs

t

 0.50686

 0.00025

 pt2

  pt1   p222   p123

Np  Ap  pt2 Self-weight: Floor load:

 pt2

Np  1377.22102 kN god 1.35

 11.3958 

g1  3.5 kN m

Example of Prestress Ceiling TT Panel

kN m

1

 11.3958 

M1  370.25255 kN m

 1b

god

 1060.16214 MPa

 972.81982 MPa

kN m

266

v1 

Live load:

g 

god 1.35 1

M2   2b

 g1  v

g  17.2958 



epd 

Ep Ec

kN m

Wih  0.10228 m

kN m

 p223

v1  1.2 

g( L  0.4 m )   Np epd 

8 M2



v 2

h

 2b

 bf   t2   t  2b

 4.46181 MPa

 p223

 63.86955 MPa

Loss due to creep of concrete-the result  p23

  p123   p223

 pt2

 104.36372 MPa

Wih



 4.77673 MPa

 pp N 1potr

d

  b1 0.6  fckcyl 



Wid



h

Wih

 3.37045 MPa

  bg  pp fctm

Mgod

d

Wid

complies

 17.78531 MPa

 fckcyl 

 b1 0.6 

 18 MPa

N2potr  Ap  pt2

N2potr  1377.22102 kN



 ptr

 pin

1

 0.86561

2



h

 pt2

2

 pin

 0.75664

 1285.71429 MPa



N1potr  Ap  ptr



ef 

N1potr  1575.57617 kN

Mgod   pp N1potr epd 1  pp N 1potr

Mgod  177.88122 kN m

 ppu N 2potr



 ppu N 2potr epd

Wih



Mgd

 ppu N 2potr



 ppu N 2potr epd

Wid

 ppu N 2potr epd

 Mgd



Mgd

6 ef  H

H efk  6

efk  0.15 m d

 bg

 2.14588

 bg

Wid

h

 5.9704 MPa

h

 0.6  b1  fckcyl 

d

 3.60856 MPa

ef  0.26488 m

 ppu N 2potr



10.5 ef  H 6 ef  H

  bg  b1 Rbtn

 bg

 2.72925



 bg  b1  fctm



 2.5

 bg

 2.14588

Example of Prestress Ceiling TT Panel

 bg

we take

 5.25 MPa

ef is > efk ok 10.5 ef  H

Wih

bg

ef  0.43418 m

 670.1778 kN m

0.6  b1  fckcyl   18 MPa

d

Pretension stage, strength effects



N1potr  1575.57617 kN

complies

 972.81982 MPa

Assessment of limit state of crack formation, check of stress in place lf/2

 bg

Mgod

h

 pp N 1potr epd



 ppu N 2potr epd

 1112.93083 MPa

ef  

 pp N 1potr epd

d

 pt2

   pin   p17   p18   p21   p122   p222   p23  908.95027 MPa

Really values 1, 2  ptr



Ai  0.44904 m

Stage service check the stress in place l/2

 p23

Total loss pt2  pt2

 pp N 1potr

 bg  pp fctm

M2  74.94863 kN m



d

 2.5

  bg  b1 Rbtn

complies

267

Checking of limit deformation of an element

Deflection from the prestressing force Np, Instantaneous deflection at time t1 = tr when we

-For prestress beam, we calculate the deflection in the middle of the span

transferring the prestress force to the member

-At the transferring the prestress force to the member (t1 = tr), N potr, g0

po

Np  Ap po   ptr 

 1

-At the service stage (t2 =), Npotr, all loads are applied

Np  1575.57617 kN

 1 Np epd ( L  0.4 m) 2  fltinp1     8 Br

we expect the service values of the long – term load:



Self-weight

epd  0.54068 m

fltinp1  0.03494 m



Floor load (1/2) live load

Deflections at time t1 increase due to creep

Short-term load: (1/2) live-load (if snow then whole value)

 f.ltinp1

Prestress force (Npini, Npot, Npot2) long term load pp – 1, when no cracks occur, the bending stiffness of the beam can be considered similarly as for elastic materials.

 fltinp1  rl1

 f.ltinp1

 0.01217 m

Total deflection at time t2=

Br = 0.85 Ec we calculate for initial and short-term deflection

Ec  36000 MPa 4

Npin  1783.46505 kN

Ji  0.03085 m

N1potr  1575.57617 kN

Ap  0.00142 m

Br  9.43944  10 kN m

t  102

 bf

 5.5

 1bf

dry

 3.8 common

Br  0.85 Ec Ji

t1  28

t2  1  10

307

Npin  1783.46505 kN

 fltp2

fltp2  fltinp1   f.ltinp1   fltp2



1 Npin epd ( L  0.4 m )   rl2 8 Br

fltp2  0.08351 m god

Deflection due to self-weight g0s

1.35

 11.3958 

kN m

Immediate deflection in time 1 = t2

Investigate the deflection for the time interval t1 to ti, the coefficient for the creep calculation

god

for the time interval t1 to ti

 t1

 0.015 t1

 0.15  0.08 e



 rl1   1bf  t1 1  t2

fltingo1 

 0.07 

e

t t1 

(  0.015  t)

 0.15  0.08 e



 rl2   bf  t2 1

 0.07 

e

t2t 

 t1

 0.20256

 rl1

 0.34821

common

5 384



1.35

( L  0.4 m ) Br

fltingo1  0.01508 m

Deflection in time ti

 t2

 0.16732 dray

 rl2

 0.92028

Long term deflection increment  fltgo

  rl1fltingo1

 fltgo

 0.00525 m

Total deflection

fltg  fltingo1  1   rl1

Example of Prestress Ceiling TT Panel

fltg  0.02034 m

 fltp2

 0.0364 m

268

1 ( L  0.4 m ) flim  0.11733 m 150 Visible visual deflection t2 =  flim 

Deflection in time t2   rl2  0.92028 fltingo1  0.01508 m Long term deflection

 fltgo2

  rl2fltingo1

 fltgo2

 0.01388 m

fvis   fltp2   fltgo2  fltg2  fst

fvis  0.00899 m

Ultimate limit stat assessment, stress in reinforcement Total deflection fltgo2  fltingo1  1   rl1   rl2

 1285.71429 MPa

 pin

fltgo2  0.03422 m

 po

 po  pin

 po

 1285.71429 MPa

In time t2= Deflection from other long-term load g1s, vs/2 kN kN v v  2.4  g1  3.5  g1s  g1  m m 2

g1s  4.7 

kN m

 pt2

   pin   p17   p18   p21   p122   p222   p23

 pot2

 po  pt2

 pot2

 908.95027 MPa

Immediate deflection at the time of ti

Strain in reinforcement

5 g1s ( L  0.4 m ) flting1   384 Br

flting1  0.00622 m

 pot2

 po2

 0.00478

First step

Long term deflection

  rl2flting1



Average strain in concrete

Deflection in time t =

 ltg2

 po2

 ltg2

 0.00572 m

Total deflection

fltg2  flting1  1   rl2

 bd

x  0.2 m a11  0.04 m

we suggest

he  H  a11

Increment in strain fltg2  0.01195 m  p

Deflection due to live load v/2, immediate deflection v 4 ( L  0.4 m ) 5 2 fst  0.00159 m  fst  384 Br Requirements STN 73-1201 for assessment of deflection Total static deflection t2 = 

ftot  fltp2  fltgo2  fltg2  fst

 0.0025

he  0.86 m

ftot  0.03576 m



 bd

 he  x

 p

 0.00825

p

  p   po2

p

3. stress in tensile reinforcement -subtract p p

 pd

 1285.71429 MPa

 1157.14286 MPa

4. Force in reinforcement Np   p Ap

Np  1638.16714 kN

Example of Prestress Ceiling TT Panel

p

 0.9  pd

 0.01303 is < ��pd

269

5. force in concrete: b1  1180 mm

h1  50 mm

b2  30 mm

h2  120 mm

h3   xu  h1 

b4  125 mm

h4   xu  h1 

h4  0.11 m

A1  h1 b1 A1  0.059 m

b3  20 mm

A1  0.059 m

1 A2  2  b2 h2 2

1 A3  2  b3 h3 2

A3  0.0022 m

A2  0.0036 m

A1  h1 b1 2

1 A2  2  b2 h2 2

1 A3  2  b3 h3 2

A3  0.00092 m

A2  0.0036 m

Abc  A1  A2  A3  A4 Nc   b1 fcd Abc

Abc  0.07502 m

A4  2 b4 h4 2

A4  0.0115 m

Nc  1650.44 kN

6. Now we compare the force in concrete with the force in reinforcement A4  2 b4 h4 A4  0.0275 m

Abc  A1  A2  A3  A4

Abc  0.0923 m

N p  1638.16714 kN

N c  1650.44 kN

Conventional reinforcement

Then the force in concrete will be:

Load calculation

Nc  2030.6 kN

Np  1638.16714 kN

 b

 26 

See that the force in the concrete > like the force in the reinforcement it means that I have choose the value of x smaller then

kN m

 2

kN m

g 1  3.5 

kN m

Self-weight of RC slab

2end step

x  0.12 m  po2

 p2

 0.00478

 p



 bd

 he  x

god  b1 h1  b 1.35

 p2

 0.01542

 plim 

g od

 2.0709 

0.01

kN m

Dead load

  plim   po2

 p

 0.01478 <0.015 = pd

g1d  5.67 

3 The stress in the tensile reinforcement, subtraction  p  1157.14286 MPa

kN m

Live load 4. Force in the reinforcement Np   p Ap

vd  3.6 

Np  1638.16714 kN

gd  god  g1d  vd

5. Force in the concrete: xu  0.8 x

kN m

xu  0.096 m

x  0.12 m

b1  1180 mm

b2  30 mm

b3  20 mm

h2  120 mm

h4   xu  h1  h3   xu  h1 

b4  125 mm h4  0.046 m

b1  1.18 m

The calculation of bending moment h1  50 mm

Over support b4  0.125 m

b3  0.02 m

Example of Prestress Ceiling TT Panel

b4  0.125 m

gd  11.3409 

kN m

270

1 2 gd  b1  2 b4  2 b3  12

Mp 

Checking of limit deformation of an element

Mp  0.74859 kN m

-For prestress beam, we calculate the deflection in the middle of the span -At the transferring the prestress force to the member (t1 = tr), N potr, g0

At the middle span of slab

M str 

gd  b 1  2 b 4  2 b 3 

-At the service stage (t2 =), Npotr, all loads are applied

Mstr  0.3743 kN m

we expect the service values of the long – term load: Self-weight

The calculation of reinforcement u

 1

 b1

1

 1

fcd  22 MPa

fyd  420 MPa

b  1 m he  0.029 m

ast  h1  he

Ast 

s



 8 mm

ast  0.021 m

he 

xu  0 m

Ast  0.61461 cm

1

Floor load

fctm  2.1 MPa h1 2





(1/2) live load

Short-term load: (1/2) live-load (if snow then whole value) Prestress force (Npini, Npot, Npot2) long term load pp – 1, when no cracks occur, the bending stiffness of the beam can be considered similarly as for elastic materials.

zs  he

Br = 0.85 Ec we calculate for initial and short-term deflection

 5 mm

 u zs  s fyd

N pinrII  1820.18571 kN

Ec  36000 MPa

Npin  1783.46505 kN

Ji  0.03085 m

N1potr  1575.57617 kN

Ap  0.00142 m

 bf

 5.5

dry

As1 

  1

As1  0.19635 cm

Astsku  5 As1

Astsku  0.98175 cm

Then we should check that

Asre  5 As1  stmin



Asre  0.98175 cm

1 fctm  3 fyd

Nst  Asre fyd  s xu 

 stmin

 st



Asre

 st

b h1

 0.00196

 0.00167

xu  0.00187 m

b  b1 fcd

Mu   u Nst zs

zs  he 

Mu  1.15713 kN m

xu 2

zs  0.02806 m

Mp  0.74859 kN m

 t1

 0.015 t1

 0.15  0.08 e



 rl1   1bf  t1 1  t2

 0.07 

e

Fn  Ac  b L

Fn  212.9868 kN

t t1 

(  0.015  t)

 0.15  0.08 e

Calculation of handling force

 1.1

 3.8 common

Br  0.85 Ec Ji

t1  28

t2  1  10

307

for the time interval t1 to ti

Hanging eyes

 fg

t  102

 1bf

Investigate the deflection for the time interval t1 to ti, the coefficient for the creep calculation

Nst  41.2334 kN

Nst

Br  9.43944  10 kN m



 rl2   bf  t2 1

 0.07 

e

t2t 

Example of Prestress Ceiling TT Panel

 t1

 0.20256

 rl1

 0.34821

common

 t2

 0.16732 dray

 rl2

 0.92028

271

Deflection from the prestressing force Np, Instantaneous deflection at time t1 = tr when we transferring the prestress force to the member po

Np  Ap po   ptr 

 1

N p  1575.57617 kN

epd  0.54068 m

Long term deflection

 1 Np epd ( L  0.4 m) 2  fltinp1     Br 8 

fltinp1  0.03494 m

 fltinp1  rl1

 f.ltinp1

 fltgo2

  rl2fltingo1

 fltgo2

 0.01388 m

Total deflection fltgo2  fltingo1  1   rl1   rl2

Deflections at time t1 increase due to creep  f.ltinp1

Deflection in time t2   rl2  0.92028 fltingo1  0.01508 m

fltgo2  0.03422 m

Deflection from other long-term load g1s, vs/2 kN kN v v  2.4  g1  3.5  g1s  g1  m m 2

 0.01217 m

g1s  4.7 

kN m

Immediate deflection at the time of ti

Total deflection at time t2=

Npin  1783.46505 kN

 fltp2

fltp2  fltinp1   f.ltinp1   fltp2



1 Npin epd ( L  0.4 m )   rl2 8 Br

fltp2  0.08351 m god

Deflection due to self-weight g0s

1.35

 11.3958 

fltingo1 

5 384



1.35

( L  0.4 m ) Br

kN m

 ltg2

  rl2flting1

 ltg2

 0.00572 m

fltg2  0.01195 m

fltingo1  0.01508 m Deflection due to live load v/2, immediate deflection v 4 ( L  0.4 m ) 5 2 fst  0.00159 m  fst  384 Br  fltgo

 0.00525 m

Total deflection

fltg  fltingo1  1   rl1

Long term deflection

fltg2  flting1  1   rl2

Long term deflection increment

  rl1fltingo1

flting1  0.00622 m

Total deflection

Deflection in time ti

 fltgo

 0.0364 m

Deflection in time t =

Immediate deflection in time 1 = t2

god

 fltp2

5 g1s ( L  0.4 m )  384 Br

Requirements STN 73-1201 for assessment of deflection Total static deflection t2 = 

ftot  fltp2  fltgo2  fltg2  fst fltg  0.02034 m

Example of Prestress Ceiling TT Panel

ftot  0.03576 m

272

5. force in concrete:

1 ( L  0.4 m ) flim  0.11733 m 150 Visible visual deflection t2 =  flim 

fvis   fltp2   fltgo2  fltg2  fst

fvis  0.00899 m

b1  1180 mm

h1  50 mm

b2  30 mm

h2  120 mm

h3   xu  h1 

b4  125 mm

h4   xu  h1 

h4  0.11 m

Ultimate limit stat assessment, stress in reinforcement  1285.71429 MPa

 pin

 po

 po  pin

 po

 1285.71429 MPa

A1  h1 b1 A1  0.059 m

b3  20 mm

1 A2  2  b2 h2 2

1 A3  2  b3 h3 2

A3  0.0022 m

A2  0.0036 m

In time t2=  pt2

   pin   p17   p18   p21   p122   p222   p23

 pot2

 po  pt2

 pot2

A4  2 b4 h4 A4  0.0275 m

 908.95027 MPa

Abc  A1  A2  A3  A4

Abc  0.0923 m

Then the force in concrete will be:

Strain in reinforcement  po2



 pot2

Nc   b1 fcd Abc

 po2

 0.00478

 bd

 0.0025

2end step

x  0.2 m a11  0.04 m

we suggest

he  H  a11

he  0.86 m

 p



 he  x

-subtract p

 p

 0.00825

 pd

p

 1285.71429 MPa

 1157.14286 MPa

4. Force in reinforcement Np   p Ap

 p2

 0.00478

 p



 bd

 he  x

  plim   po2

 p2

 p

 0.01542

 plim 

 0.01478 <0.015 = pd

3 The stress in the tensile reinforcement, subtraction  p  1157.14286 MPa   p   po2

p

Np  1638.16714 kN

 0.01303 is < ��pd

4. Force in the reinforcement Np   p Ap

3. stress in tensile reinforcement

p

x  0.12 m  po2

Increment in strain  bd

Np  1638.16714 kN

See that the force in the concrete > like the force in the reinforcement it means that I have choose the value of x smaller then

Average strain in concrete First step

Nc  2030.6 kN

p

 0.9  pd

Np  1638.16714 kN

5. Force in the concrete: xu  0.8 x

xu  0.096 m

b1  1180 mm

b2  30 mm

b3  20 mm

h2  120 mm

h4   xu  h1  h3   xu  h1 

Example of Prestress Ceiling TT Panel

x  0.12 m b4  125 mm h4  0.046 m

h1  50 mm

0.01

273

A1  h1 b1 A1  0.059 m

1 A2  2  b2 h2 2

1 A3  2  b3 h3 2

A3  0.00092 m

A2  0.0036 m

Abc  A1  A2  A3  A4 Nc   b1 fcd Abc

Abc  0.07502 m

A4  2 b4 h4 2

A4  0.0115 m

Mp  0.74859 kN m

At the middle span of slab

M str 

Nc  1650.44 kN

gd  b 1  2 b 4  2 b 3 

Mstr  0.3743 kN m

The calculation of reinforcement u

6. Now we compare the force in concrete with the force in reinforcement N p  1638.16714 kN

1 2 gd  b1  2 b4  2 b3  12

Mp 

N c  1650.44 kN

 1

 b1

he  0.029 m

Conventional reinforcement Load calculation

ast  h1  he

Ast 

s

 1

fcd  22 MPa

fyd  420 MPa

b  1 m

1



 8 mm

ast  0.021 m

fctm  2.1 MPa he 

xu  0 m

Ast  0.61461 cm

1

h1 2





zs  he

 5 mm

 26 

kN m

 2

kN g 1  3.5  m

kN m

 u zs  s fyd

As1 

  1

As1  0.19635 cm

Astsku  5 As1

Astsku  0.98175 cm

Self-weight of RC slab god  b1 h1  b 1.35

g od

 2.0709 

kN m

Then we should check that

g1d  5.67 

Asre  5 As1

Dead load

kN m

 stmin



Asre  0.98175 cm

1 fctm  3 fyd

Nst  Asre fyd  s

Live load

xu 

kN vd  3.6  m gd  god  g1d  vd

b1  1.18 m

b4  0.125 m

gd  11.3409 

kN m

 stmin

b  b1 fcd

Mu   u Nst zs

Over support

Calculation of handling force  fg

 1.1

Asre

 st

b h1

 0.00196

 0.00167

xu  0.00187 m

Hanging eyes

b3  0.02 m



Nst  41.2334 kN

Nst

The calculation of bending moment

b4  0.125 m

 st

zs  he 

Mu  1.15713 kN m

Fn  Ac  b L

Example of Prestress Ceiling TT Panel

Fn  212.9868 kN

xu 2

zs  0.02806 m

Mp  0.74859 kN m

274

Bond post-tension prestress indeterminate reinforced concrete girder

Extreme load on beam

Dimensions of column

Installation stage:

c 1  60 cm

c 2  c 1

Self-weight

Light distance between the columns: Axis distance between the columns: c1  c2

lt  L 

kN 





qd  lw  q1  1.35  q2  1.35  vd  1.5  b  d  26

Loading width

qod  28.872

3



kN m

In service time



lt  15.6m



qod   f   q1  lw  b  d  26

 f  0.9

L  15 m

kN m

 f

q d  79.668 

kN m

Concrete cover- upper and lower reinforcement

lw  5.6 m

c cu  30 mm, ccl  30 mm

Design of beam dimensions:

Diameter of upper reinforcement Depth of reinforced concrete beam d 

d  0.78m

d  0.8 m

Width of reinforced concrete girder

psu  22 mm

Diameter of lower reinforcement psl  22 mm

Light distance among of cable prestress, concrete, reinforcement b 

3 4

d

b  0.6m

b  0.60 m

dd  30 mm

Diameter of prestress cable pk  20 mm

The effective depth

 

dp  d   c cu  psu  dd 

pk  2

pk      c cl  psl  dd   2  

dp  0.616m

Camber of parabola no.3: ep  3

Beam elevation and idealized tendon profile Determination of load

kN m

 

  c cl  psl  dd 

pk  2

 

ep  0.308m 3

Camber of parabola no.1 (estimate): ep  0.9 dp

Floor panel-self-weight of the girder q1  3.5

ep  0.554m

Length of parabola no. 2: l2  c 2  2 

l2  1.4m

p  6  deg

Length of parabola 1: Dead load-other load q2  3 

kN m

2  ep

 

tan p

l1  10.55m

3  4  deg

Length of parabola np.3: l3 

Live load vd  2 

l1 

2  ep

 

tan 3

l3  8.809m

kN m

Example of Bond Post- Tension Prestress Indeterminate RC Girder

275

Length of straight part: lp  L  c 1 

c2 2



At service stage:



 l1  l2  l3



 

Vp3ht2  Fpdt2  sin 3

lp  5.521m

Sag of parabola 1 (really= L1): ep  dp  1

Load – Balancing method

Originally introduced by T.Y. Lin, the name load balancing is due to attempt to balance a

ep  0.544m

l 1  l2

portion of the external load of the structure by the upward forces of the tendons the deviation

Sag of parabola 2 (really= L2): ep  dp  2

V p3ht2  20.33 kN

and anchor and friction forces of the tendons, which are activated when stressing the tendons

ep  0.072m

l 1  l2

ep  ep  0.616m 1

dp  0.616m

against the concrete, act vice versa on the plain concrete girder (including ordinary

The determination of equivalent load due to 1 cable- using cable LSA 15,5:

reinforcement) and on the tendons which are considered separately. These forces are called

The area of cable and the design strength of cable

balanced forces or balanced loads. The forces between concrete girder and tendon are:

Ap1  141.57 mm

, pd  1440 MPa, pin  pd

The design force of 1 cable, service stage, losses in stress at this time losses in stress at this time  2  0.8 , coefficient of reliability 

Anchor forces at the girder ends, where the tendons are stressed against the concrete and then anchored. The resulting forces act in the tangential direction of the tendons,

pp1

 1.06

their effects on bending moments and shear forces must not be overlooked, if the tendons are anchored eccentric to the centre –line of the beam axis is curved.

Fpdt2  146.78 kN

The design force of 1 cable, installation stage, losses in stress at this time losses in stress at this time  1  0.94 , coefficient of reliability,  Fpdt1  A p1   1   pp1  pin

pp1

 4 epi     li   

i

 

out when being stressed. In case of a parabolic tendon profile the vertical component of these deviation forces can be calculated as follows:

i 

0.203

0.139

ep  i

0.544 m 0.072 0.308

At installation stage: Fp3ht1  Fpdt1  cos 3

length. These forces are due to the tendons” curvature and tendons” attempt to straighten

 1.06

Fpdt1  203.127 kN

The real value of cable inclination: i  atan

(uniformly) distributed deviation forces acting at right angle to the tendons along their

1 8

Wp L

8 ep Fp 2

Whereby:

Wp distributed forces acting between concrete and tension Fp tendon force while prestressing = Fp.w+p Ep sag of the tendon with L

Fp3ht1  201.169 kN

L length of the considered section of the tendon At service stage:

 

Fp3ht2  Fpdt2  cos 3

For example, the respective values Wp1,Wp2 for the span and for the support region are given in Fp3ht2  145.365 kN

Figure above. As the tendons sag is small, these forces are considered to act vertical to the centre line of the girder.

The vertical component of prestressing force at the face of member:

Distribution forces acting parallel to the tendons along their length. These forces are due to

At installation stage

 

V p3ht1  Fpdt1  sin 3

Vp3ht1  28.134 kN

friction between tendon and duct while stressing the tendons. As usual they are not considered in this example. However, the reduction of tendon forces due to friction must be teken into

Example of Bond Post- Tension Prestress Indeterminate RC Girder

276

account. If the tendons are not curved much, an average constant tendon force Fp may be used in the calculation of the sectional forces.

Parabola 3: l3

Equivalent load due to horizontal projection of prestress force: At installation stage: wp1  i

8  ep

x  0  m 1  m 

 4.405m

x

y3 ( x )  4  ep  

wp1 

 Fp3ht1

li



l32

y3 ( x )  0 m

7.864 m 59.259

0.124

 y3( x) 0.2

1

0.216

 0.3

 kN

 0.4

0.277 0

0.8

1.6

2.4

3.2

0.305

Parabola 2:

At service stage: wp2 

2

6.387

( x)

 0.1

 li  2

8 ep

 l3 



ep  0.072m

wp2 

 Fp3ht2

x  0  m 0.2 m  l2

x

y2 ( x )  4  ep  

l2  1.4m

5.683 m 1  kN



 l2 

(x)

2



l22

y2 ( x )  0 m

42.821

0.08

4.616

0.055

-0.035

0.03

-0.059

510

Checking of vertical forces:

3

 0.02

1. Stage:

2  wp1  3

l3 2

 2  wp1 

-0.071

0.25

0.5

0.75

1.25

-0.071

1.5

 139.231 kN

-0.059

wp1  l2  2  Vp3ht1  139.231 kN

-0.035

2. Stage:

2  w p2  3

l3 2

 2  w p2  1

l1 2

wp2  l2  2  Vp3ht2  100.608 kN 2

 100.608  kN

Or wp1  i

8  ep

2 ( l) i

 

 Fp3ht1

wp2  i

8  ep

Parabola1: ep

2 ( l) i

 

 Fp3ht2

29.629 3.194

x  0  m 0.67 m 

x

y1 ( x )  4  ep  

 5.275m

wp2 

wp1  3.932

 0.544 m

kN m

2.841 21.41 2.308



wp2  l3  1  wp2  l1  50.304 kN 3

2



l12

y1 ( x ) 

kN m

0 m

0  0.15

-0.129

y1( x)  0.3  0.6

wp1  l2  1  Vp3ht1  69.615 kN 2

(x)



-0.241

 0.45

wp1  l3  1  wp1  l1  69.615 kN 3 1

 l1 



-0.335 0

wp2  l2  1  Vp3ht2  50.304 kN 2

The results should multiplying by 2

Example of Bond Post- Tension Prestress Indeterminate RC Girder

-0.412 -0.471 -0.513 -0.537

277

How to calculate the amount of prestress tendons to TT panel beam?

The distance of individual centre of gravity from the axis A: y1 

h1 2

y5  h 1  h 4 

y2  h 1 

y6  h 1 

y3  h 1 

y4  h 1 

The position of CG of cross-section:

Bi  A i yi

ybh 

B

ybh  265.137mm

A

The modulus of intertie for individual parts of cross-section:

I1  I4 

1 12 1 12

b 1  h 1

b 4  h 4

I2  I5 

1 36 1 36

b 2  h 2

b 5  h 5

I3  I6 

12 1

b 3  h 3

b 6  h 6

a 1  ybh  y1

a 2  ybh  y2

a 3   y3  ybh 

a 5  y5  ybh

a 6  ybh  y6

Di  A i  a i

bi 

Dimensions of individual parts of cross-section: i  1  6

hc  780mm

bc  1850mm

b 2  200mm b 4  20mm b 6  120mm

h 2  50mm h 4  30mm h 6  30mm

b 1 

bc 2

b 3  150mm b 5  40mm

0.07

A 2  b 2 

h2 2

A 3  b 3 h 3

A 4  b 4 h 4

A 5  b 5 

h5 2

A 6  b 6 

h6 2

0.23

0.178

0.69

0.104

0.02

0.03

6·10-4

0.18

0.04

0.66

0.013

0.055

0.12

0.03

1.8·10-3

0.185

Di  m

3.429·10-3 1.593·10-4

4.106·10-3

2.325·10-3

4.5·10-8

1.947·10-5

3.194·10-4

3.973·10-5

9·10-8

6.17·10-5

i  1  6 A1  b 1 h 1

0.15

6.944·10-7

Area of individual parts of cross-section:

5·10-3

0.05

2.644·10-5

h 3  690mm h 5  660mm

ai 

0.065

0.2

Ii 

h 1  70mm

Ai 

hi 

0.925

a 4  ybh  y4

Total area of cross-section:

Ab  2

A

Ab  0.378m

Example of Presstres Tendons to TT Panel Beam

0.15

h4 2

278

The total modulus of intertie: Jb  2 



  I

x1 

D

Jb  0.021m



Wbd  0.041m

 hc  ybh 

ybd  hc  ybh

Wbh 

Jb ybh

ybd  0.515m

Wbh  0.079m

gos  9.82

x4  8.983m

g o  g os  1.35

m Live load:

m m The calculation of anchor cable length, prestress cable 15,5:

d s  15.5  mm

x2  3.617m

x4  x1  3 c

gs  2 

x2  x1  c

x3  6.3 m

Dead load:

Self-weight of beam: kN

x1  0.933m

Load: self-weight, service coefficients

gos  Ab 26

 x0

x3  x1  2 c

Cross-sectional modulus: Wbd 

Ap1  141.6mm

 pd

 1285MPa

vs  2.0

b c

kN m

gd  gs  1.35

v d  v s  1.5

b c

vd  5.55

kN m

The calculation of anchor length of prestress post tension cable:

lbd  60ds

The determination of bending moments from the extreme load: i  1  4

 go l  xi 2  Mgo   xi  go 

lbd  0.93m





Equalizing length:

a1  100 mm  b 2 

 

a2  2  h 3 

b3 2



h1 2

a1  1.035m

 h 3  65 mm

h1  2

  gd  vd  l   xi 2  Mgv   xi   gd  vd  

  bc  2 100mm  b 2  b 3 



 

( q ) l Mq   xi  ( q ) 

 1.5a1  M     a2 

i 

0.933

Length of beam, length of unloading part:

3.617

l1 c 

 3

l  l1  2  wc  3  2



l  17.967m

 xi 2 

Mgo 

xi 

The calculation of internal forces:

l1  18.5m



a2  2.4m

6.3

8.983

Mgv 

105.381

83.821

c  2.683m

x0 

l1 2



l 2

kN m

189.202

344.021

273.638

617.659

487.205

387.529

874.733

534.933

425.492

960.425

lb 2

Mq 

kN m

x0  0.267m

Example of Presstres Tendons to TT Panel Beam

kN m

279

1 10

6 10

8 10

5 10

4 10

3 10

2 10

1 10

6 5

6 10

Mgo

4 10

2 10

 bg fctm

 Npin4

 ppu  2 



Wbd 2

Ap15  141.57mm

Np1   Ap15  pd15

 pd15

 Npin4 epd

 ppu  2 

 Wbd 

 pd15

 3.15MPa

 1285MPa

The proposal of prestress force LA 15,5: 2



Ab LA 15,5

Ap15  141.57mm

 3.15MPa

Np1  181.917kN

No. of cables

n 

N pin4 N p1

n  12.677 our proposal will be n  14

 1285MPa

Actual pre-tensioning force (proposal):

NpinrI   n Ap15  pd15

Losses of prestress force:

a 11  30  mm

a 22  40 mm

c  a11 

a22

c  0.05 m

NpinrI  2.547  103 kN

Bending moment at the mid-span of the beam:

epd  ybd  c

From the total load: Mq  960.425kN m 4

epd  0.465m

From self-weight:

Mgo  534.933kN m 4

Characteristic of concrete C30/37:

Actual static eccentricity:

fctm  2.1MPa fckcyl  30MPa The static eccentricity od’s external load to centre of cross-section:

ef  

 hc  bg  1.5  hc From 4. Minimal value of prestress force we obtain as follow:

ef 

 ppu

ef  0.26m

 0.9

1

 0.9



Npin4    bg fctm 



 bg fctm



 bg

2



10.5ef

 0.70

Mq  4  Wbd   0.9 2

   Ab

 Npin4

 ppu  2 

6 ef



 pp

 bg

 2.5

 bg



Mq  0.9 2NpinrI epd 0.9 2 NpinrI

10.5ef 6 ef



 1.06

Npin4    bg fctm 



1 0.9 2 epd

Wbd

 bg

LA 15,5



 27.999

Mq  4

 Wbd   0.9 2

   Ab

Npin4  2.306 103 kN

  

 Npin4 epd

 ppu  2 

 hc

Ap15  141.57mm

Np1   Ap15  pd15

ef  0.134m

 bg

 2.5

1 0.9 2 epd

Wbd  pd15

 0.13m

 bg

Npin4  2.069  103 kN

  

 1285MPa

Np1  181.917kN

No. of cables

our proposal is

n  12

 Wbd 

n 

N pin4 N p1

n  11.374

Example of Presstres Tendons to TT Panel Beam

 2.5

280

0.6fckcyl

Actual pre-tensioning force:

NpinrII   n Ap15  pd15  bg fctm



 NpinrII

 ppu  2 

Ab  bg fctm

 NpinrII



Ab 

 NpinrII epd

 ppu  2 



Wbd

 NpinrII epd

 ppu  2 

Wbd

 5.25MPa

 Npin3

 ppu  2 

NpinrII  2.183  103 kN

 ppu  2 

 Npin3 epd Mq   Wbh   Wbh 

 ppu  2 

 0.724MPa

NpinrII  Npin3

The proposal of prestress force according to 4:



 Wbd 

Npin4  2.069  103 kN

 4.241MPa

The checking of the prestress force position in section along the member:

Mgo 

Mq 

105.381

Maximum prestress force according to 2:

kN m

189.202

344.021

617.659

487.205

874.733

534.933

960.425

kN m



Npin2  0.6 fckcyl  



 NpinrII

 ppu  2 

Ab 0.6fckcyl



NpinrII  2.183  103 kN

 Wbd 

Checking the position of prestressing force:

 ppu  2 

 18MPa



Mgo  4

 Wbd   1 epd      Ab Wbd  1

 NpinrII epd

 ppu  2 

Wbd

 NpinrII epd

 ppu  2 

Wbd



Npin2  2.214  103 kN

NpinrII  2.183  103 kN

Mgo

 Wbd  Mgo

 Wbd 

 6.204MPa

NpinrII  2.183  103 kN

 18MPa

epd1 

Wbh  NpinrII Mgo i     bg fctm   NpinrII  Ab Wbh 

epd2 

Wbd  NpinrII Mgo i    0.6fckcyl   NpinrII  Ab Wbd 

 0.6fckcyl

NpinII  Npin2

Wbh  Wbh

epd3  i

Wbh



 0.6fckcyl 

NpinrII0.9 2 

NpinrII0.9 2 Ab



Mq  i



Wbh 

Prestressing force according to 3:



Npin3  0.6 fckcyl  



 Npin3

 ppu  2 

Mq  4    Wbh    0.9 2

1 0.9 2 epd 

Npin3  5.613  103 kN

   Wbh    Ab



 Npin3 epd Mq   Wbh   Wbh 

 ppu  2 

epd4  i

Wbd



  bg fctm 

NpinrII0.9 2 

NpinrII0.9 2 Ab



Mq  i



Wbd 

 0.6fckcyl

Example of Presstres Tendons to TT Panel Beam

281

epd2 

epd3 

0.276

epd1 

xi 

-0.688

0.386

-0.377

0.451

-0.19

0.473

-0.128

0.933

0.448

3.617

0.557

6.3

1 2 3

0.623

8.983

epd4 

i 

0.645

Installation stadium

-0.126

0.186

NpinrII  2.183  103 kN  pp  1.06

Mgo 

0.373

105.381

0.435

kN m

344.021 487.205 534.933

NpinrII  2.183  103 kN

i  1  4

efgod1  

Mgo   pp  1 NpinrIIepd1 i

 pp  1 N pinrII

efgod1  i

-0.397

-0.392

-0.389 -0.388

 epd4  epd3

0.263  m

Mq   ppu  2 NpinrIIepd4 i

 ppu  2 N pinrII

-0.263 -0.263

 epd2

 epd1

-0.263

1.333

2.667

 epd

1

i  pp

efgod4 

 1.06

1

 0.9

NpinrII  2.183  103 kN

Example of Presstres Tendons to TT Panel Beam

-0.263

282

Using external prestressing reinforcement has been proposed by Dishinger in 1934 already and recently be revived, especially in France and

through holes in the steel deviators, during construction, and the stressing of the steel itself caused problems due to friction in the deviation zones.

the USA, mainly for prefabricated segmentally constructed bridges. The advantages are mainly that the cables are prefabricated and therefore have a reliable corrosion protection and that they can be controlled and, if necessary, be replaced at any time.

The introduction of un-bonded single strand tendons made it necessary to develop a basic approach to the design of concrete structures without bond between the prestressing steel and the concrete, especially in the ultimate limit state.

Already in the early fifties the prestressing tendons in same concrete structures were not positioned within but outside the concrete cross-section. The well-known prof. G. Mangel from Belgium. One of the pioneers of prestressed concrete, designed several projects with such external prestress.

Some reasons for the development of this type of external cables are: -

The demand for methods to repair prestressed concrete bridges with corroded prestressing tendons in the concrete structures.

The aim of this construction procedure was to substitute uncontrollable

The development in practice of methods to strengthen concrete bridges or other structures already in use, due to increase of traffic loads.

grouting of ducts for concreting of the prestressed wires under full visual control. The role of the concrete was also to bond the prestressing steel to the concrete structure in order to ensure interaction of steel and concrete at overloading and. As a result, to ensure an acceptable factor of safety ahainst failure of the

structure.Prof. F. Leonhard introduced external cables in prestressed concrete

box-girder bridges, constructed with the so-called launching method. In France, external prestressing was also used in several structures.

New developments in bridges design and bridge construction are resulting in the use of external cables.

External prestressing implies the use of un-bonded prestressing

tendons outside the concrete section of a structural concrete member. Because of the substantial economic savings and dramatic increase in rapidity of construction possible with this technology, it is being increasingly considered in the construction of new concrete structures, particularly bridges. It is also

In the Duch recommendations for prestressed concrete, published

strengthening of existing structures.In seven years the experts build in France

in 1962, rules were given for the design of concrete structures with external

more than thirty bridges in prestress concrete with external tendons. This number

prestressing. In these recommendations much attention was paid to the need of

amounts to forty approximately if we add composite bridges- steel –prestressed

adequate bond between the external tendons and the concrete structure.

concrete-built with external tendons at la Ferte Saint-Aubin, Arbois, Cognac,

In the 1960s and 70s, the use of external prestressing did not break through as generally accepted construction method. The causes for it are not easy to ascertain, however, some draw-backs can be mentioned from practice. Several structures with external prestressing showed, some time after completion, corrosion problems due to insufficient protection of the prestressing steel by the compacted mortar. In other cases the pulling of prestressing steel t

Charolles, and recently near Compiegne. As the most important experts working in this field are M. Virlogeux and A.S.G. Bruggeling. Recently, some bridges have been built with external tendons in Belgium, with designs from René Greisch and Bruno Cremer (Ben Ahin and Wander bridges), and in Venezuela, with a Figg and Muller design. The interest in External prestressing is developing in Switzerland, in Germany and in Czechoslovakia.

External Un-Bonded Prestress Concrete

283

There are many different ways to strengthen a structure. Improving the load carrying capacity may be to change the static behaviour, or the change of physical appearance of the structure and in that way give it somewhat different properties in strength and stiffness. External prestressing refers to a posttensioning method in which tendons are placed on the outside of a structural member. However, there can be a problem with corrosion in the steel that forces the use of steel protection on the external tendons, for example by plastic sheeting. It is an attractive method in rehabilitation and strengthening operations because: -

It adds little weight to the original structure

Its application poses little disturbance to users

It allows the monitoring, re-stressing and replacement of tendons.

External prestressing, both for new and existing structures, has proven to be an effective technique, Picard et al. (1995) listed the following advantages: 1. Concreting of new structures is improved as there are no or few tendons and bars in the section. 2. Dimensions of the concrete section can be reduced due to less space needed for internal reinforcement. 3. Profiles of external tendons are simpler and easier to check during and after installation. 4. Grouting is improved because of a better visual control of the operation. 5. External tendons can be removed and replaced if the corrosion protection of the external tendons allows for the release of the prestressing force. 6. Friction losses are significantly reduced because external tendons are linked

bridges, but is today used for both strengthening and for new build structures. Prestressed concrete bridges with external prestressing are becoming popular because of their advantages, such as simplicity and cost-effectiveness. External prestressing is defined by un-bonded tendons that are placed, and prestressed,

outside a structure and anchored at the ends, sometimes with one or several deviators during the length of the structure. This reinforcement method is advantageous for strengthening of

structural members to obtain improved load-carrying capacity. The most commonly-used material for external tendons is steel, but also fibre reinforced polymer (FRP) materials can be used. With consideration to the need for upgrading the infrastructure the use of external prestressing provides one of the most efficient solutions to increase the load carrying capacity. The method can be used on concrete as well as steel and timber structures.

to the structure only at the deviation and anchorage zones.

7. The main construction operations, concreting and prestressing, are more independent of the another, therefore the influence of workmanship on the overall quality of the structure is reduced.

But it is also important to understand the weaknesses of the technique. The following disadvantages should be kept in mind, (Picard et al. 1995): 1. External tendons are more easily accessible than internal ones and, consequently, are more vulnerable to sabotage and fire. 2. External tendons are subjected to vibrations and, therefore, their free length should be limited. 3. Deviation and anchorage zones are cumbrous additions to the cross-section. These elements must be designed to support large longitudinal and transverse forces. 4. In the deviation zones, high transverse pressure acts on the prestressing steel. The saddles inside the deviation zones should be precisely installed to reduce friction as much as possible and to avoid damage to the prestressing steel.

External Un-Bonded Prestress Concrete

284

5. In the case of internal grouted tendons, the long-term failure of anchor heads has limited consequences because prestressing may be transferred to the

The beam is loaded by concentrated force in the middle of the span where the joint is located at that point

structure by bond. In the case of external tendons, the behaviour of anchor heads is much more critical. Therefore, anchor heads should be carefully protected against corrosion. 6. At ultimate limit states, the contribution of external tendons to flexural strength is reduced compared to internal grouted tendons. The stress variation between the cracking load and ultimate load cannot be evaluated at the critical section only, as is done for internal bonded tendons. 7. At ultimate states, failure with little warning due to insufficient ductility is a major concern for externally prestressed structures. 8. The actual eccentricities of external tendons are generally smaller compared

Data: the axial force is applied in point 1:

to internal tendons.

1.2m

10m

20kN

Angle calculation and its function:

 h     L  2 

0.24



13.495deg

sin    0.233

cos     0.972

tan    0.24

Calculation of reaction, in point A resp. B. under such a load, the bending stiffness of the beam part a-1 and 1-b is not used, since these beam parts assume only horizontal forces. Due to the symmetry of the beam we obtain reactions in supports a, i b.

10kN

Calculation of horizontal force S1H: M1L

0 kN m

M1L

Av 

L 2

 S1Hh

Calculation of vertical force S12:

S12

P

S12

20kN

Example of External Prestress Beam

Av  S1H

L 2

 M1L h

S1H

41.667kN

285

Calculation of axial forces SL, Sp: Av

tan  

Av

41.669kN

tan  

41.669kN

Forces in diagonals S1, S2: B

sin  

B

42.852kN

sin  

Calculation of force Na1, a N1b

Force in line S1: A

sin  

Av

sin  

N a1

42.852kN

N 1b

 S 1h

 S 2h

 20 kN

Force in line S12: S12

S1v  S2v

21.753kN

S12

Uniform loads due to beam symmetry and uniform load actions, then the calculation of the reaction in supports A, and B will be as follows.

We have shown that the rods a-1, and 1-b behave as simple beams with a span of

Data: Cross-sectional height, span, load

1/2. and their transverse forces in the middle of the span transmit to the system of

1.5m

 h     L  2 

10m



0.3

5 kN m

16.699deg

the design proposed in points 1 and 2.

1

sin    0.287

cos     0.958

If we analyse the bending stiffness and axial stiffness of the beams each separately,

tan    0.3

we obtain a system as sketched in figure below.

Calculation of reaction, Av, B: Av

q

25kN

Calculation of force S1h: M1

Av 

L 2

L L  q    S1h h

S1h

2 4

25kN

 A  L  q  L  L  2 4  v 2 h

S2h

S1h S2h

Calculation of force S1: tan    cos   

S 1v S 1h S 1h S1

S 1v S1

41.667kN

Calculation of bending moment on line a-1 S 1h  tan    S 1h

cos   

S 1v

12.5kN

S 2v

S 1v

La1 S1

43.501 kN

q

La1

Calculation of bending moments 2

M max

q

M max

62.5kN m

Example of External Prestress Beam

12.5kN

Ma1

q

La1 8

Ma1

15.625kNm

286

Let us draw a diagram of the bending moments, transverse forces and normal forces,

Calculation of axial force in the rod 2-4. where the first two parts of the equation

each part of the beam is from the right as a simple beam and the maximum bending

represent the calculated moments with respect to the joint, and therefore the position

moment is in the middle of the span of each part, see the figure below.

in the centre of the span is the moment Mog = q. l2 / 8. Thus, the tensile force in the rod 2-4 will be:



Av 

L L  q    S24 h 2 2 4

 A  L  q  L  L  2 4  v 2

S24

40kN

As the following the balancing equation, for the calculation of the horizontal force at point g:

H

Ng  S24

S24

40kN

V

Av  q 

L 2

 Vg

Av  q 

L 2

0kN

Accordingly, no transverse force is transmitted through the joint, the beam and the

Beam mounted on two supports and joint in the centre of the beam

not a load, and therefore the value of the transverse force in cross-sections on the

Data: Individual dimensions, height, length, load

1.8m

12m

load are symmetrical, and the joint is located on the axis of symmetry and it itself is axis of symmetry is zero.

1

4 kN m

Forces in diagonals

 h     L  3 

0.45



24.227deg

sin    0.41

cos     0.912

tan    0.45 Sa2

Calculation of reaction in supports A and B

q

L 2

24kN

S24

cos   

Sa2

43.863kN

Forces in vertical deviator (columns)

24kN

S12

S24 tan  

S12

17.999kN

As the load q acts vertically and the support at b is sliding then the horizontal force at the support b is zero Ah = 0. first we determine the force in the line segment 2-4.

Example of External Prestress Beam

S4b

Sa2

S4b

43.863kN

287

Va  

Mmax

Sa2v

2 3

L 1 L  q  

Sa2h

S24

Mo1

bending moments and transverse forces along the beam. The a-g rod is subjected to the vertical load, the reaction, the compressive force in the joint, the tensile force in

Mmax

Va  

L 1 L  q  

L 1 L3  q2    2 3

Mmax

2 3

q

Av 

L 3

1 L L  q  

u13

Mu13

12 3

19.999kNm Mo  Mo1

Mo1

2 3 3

M 19.999kNm L S 

Mmax

72kNm

M 2 3 1 L L 3L 1 2L 3 Mmax MVoa  q   q    Mo M72kNm max 19.999kNm 3 2 3 2 3 8

V1d

Av  S12

Va 

L 3

q

V1d

L L  q 

M1L

Va 

Mo1

9.999kN

M1P

L L 1 q   6 6 2

3 6

M1L M1P

q

V1d  S12

V1h

8kN

S12 

L L  q   3

2 3 3 Mu13

Mo  Mo1 Mu13

8kNm

71.997kNm

L MoMoMo1q  8 L o1

7.997kNm

64kNm

L L L 1 L M 71.997kNm Avu13   qS12  3  Mo1Mu13 64kNm 3 2 31 3L L 1 L 1 M Va    q    M 19.999kNm M max 3M2 3 8kNm LMo  M2o13 o1 max64kNm

Mu13 3 S12 

Av 

L 1 L L Av  Moq  72kNm   Mo1 3M 2 3 72kNm 3

L qM  o1

Mo1

6.001kN

Mu13 3

19.999kNm

S12

Now all normal forces are determined, and we can proceed to further calculate the

Mmax

3 2 3

Av 

Mo 8kNm

L 3

S12 

72kNm

71.997kNm

Mu13 1 L L

 q  

Mo1

2 3 3

Mo  Mo1

Example of External Prestress Beam

8kNm M u13

8kNm

64kNm

71.997kNm

8kNm

64kNm

71.997kNm

288

Double –T-Girder with Un-bonded Tendons

Sabah Shawkat © The simply supported double T girder is prestressed by un-bonded external tendons which are diverted locally by diaphragms (deviators) forming a trapeziform tendon profile. The stress-strain relationship for the prestressing steel is shown in adjacent Figure, and the initial elastic modulus, the properties of the section and other relevant material data are as Example of External Prestress in TT Girder follows:

The simply supported double T girder is prestressed by un-bonded external tendons which are diverted locally by diaphragms (deviators) forming a trapeziform tendon profile.

289

The stress-strain relationship for the prestressing steel is shown in adjacent Figure, and the initial elastic modulus, the properties of the section and other relevant material data are as

The simply supported double T girder is prestressed by un-bonded external tendons which are

follows:

diverted locally by diaphragms (deviators) forming a trapeziform tendon profile. fck  32 MPa

The stress-strain relationship for the prestressing steel is shown in adjacent Figure, and the initial elastic modulus, the properties of the section and other relevant material data are as

follows:

A c  4.79 m 2

P  16.18 MN

Ic  2.616 m

1

P  16.18 MN q  0.045 MN m g  0.208 A c  4.79 m Ic  2.616 m MN external y supported double T girder is prestressed by un-bonded tendons which are 2 m 2

pp  1270

A p  0.015 m

A s  0.015 m

2 pp  1270a  trapeziform A p  0.015 A s  0.015 m Es  200 GPa ocally by diaphragms (deviators) forming profile. m m tendon 2 MN

Ep  200 GPa

fcd  18.133MPa

1

q  0.045 MN 3 m fpd  1.27  10 MPa

fcd

Ac 0.25 Es  200n GPa  GPa 0.85 E p  200  n  1.761 9A p1 fpd

  160 mm fys  500  MPa 32 GPa  1.496 m zp132 fyp  1600 -strain relationship for the prestressing steel shown in adjacent and c   fEys 500Figure,  MPa Ecthe   GPa fyp MPais1600 MPa L  30 m

H  2.25 m

at  100 mm ac  50 mm tic modulus, the properties of the sectionL  and30other relevant material data are asap  200 mm m H  2.25 m Zp  H  ap  ac

Zp  2 m

Zs  H  ac  at

Zp  H  ap  ac 2

c  4.79 m

pp  1270

MN g  0.208  fpy fpd  m1.26

fck fcd  0.85 1.5

fpy  1600 MPa

P  16.18 MN

Ic  2.616 m

MN 2

p  1600 MPa

 30 m

p  H  ap  ac

Zp  2 m

g  0.208

Zs  H  ac  at

neglected under this action δ

at  100 mm Zs  2.1m

Fpp 

1

q  0.045 MN m

ap  200 mm 1

  0.556  1   Ec  Ic   Ec  A c     E A z 2    p p p1   Ep  A p

A s  0.015 m

Es  200 GPa

Fpg  P  Fpp

fys  500 MPa

Zp  2 m

 Ep  A p L

 Fpp  15.035MN

Ep  200 GPa

Fpg  1.145 MN

Increase of tendon force due to live load q assuming linear elastic behaviour:

Ec  32 GPa ac  50 mm

Zs  H  ac  at

  160 mm

at  100 mm Zs  2.1m

1.02  1.84

zp1  1.496 m

ap  200 mm

Fpq 

H  2.25 m



A p  0.015 m

 2.1m a c  50 mm

  160Tendon  mm force duezp1 to imposed deformation: the girder longitudinal deformation should not be  1.496 m



Ec  Ic

q L

Fpq  0.141 MN

8 zp1

Ep  A p  zp1

Bending moment due to live load: M q 

1 8

 q L

M q  5.063 MN m

The influence of tendon force due to live load q on bending moment: M pq  Fpq  zp1

M pq  0.211 MN m

The deference between them be: M cq  M q  M pq

M cq  4.851 MN m

The calculation of bending moment at (1/3L) due to dead load and tendon force Fpq A  g 

A  3.12 MN

M1  A 

L 3

L L  g   0.5  Fpg  zp1 3 3

Example of External Prestress in TT Girder

M 1  19.087  MN  m

290

q uP  M pg  Fpg  zp1

M pg  1.713  MN  m

M cp  Fpp  zp1

M cp  22.493 MN m

1 8

M support  3.405 MN m

 g L

M cg  21.687 MN m

girder: 1 8

 ( g  q) L

 0.427

 2.24

gq

 1.686

The ultimate flexural strength of the standardized double tee section shown in Figure is to be The ultimate flexural strength of the standardized double tee section shown in Figure is to be calculate.calculate.

The ultimate bending moment due to dead load and live load then will be, in mid span of the

Mu 

then the ratio and the degree q uP of reliability will be  1.686 gq g  qq uP qu qu

gq

M g  23.4  MN  m

M cg  M g  Fpg  zp1



8 M up

uP uP and2the degree of reliability then the ratio will be m L

The bending moment due to dead load q M g 

q uP  0.427

The difference between Mcp and M1 will be the value of bending moment subject at support M support  M cp  M 1

8 M up

M u  28.462 MN m

Design the amount of prestress tendons asasbonded pretension prestress concrete Design the amount of prestress tendons bonded tendons tendons ininpretension prestress concrete beam, the stress-strain relationship prestressing steel in adjacent Figure, and and beam, the stress-strain relationship forfor thetheprestressing steelisisshown shown in adjacent Figure, initial elastic modulus, the properties thesection section and relevant material data are as are as the initialtheelastic modulus, the properties ofofthe andother other relevant material data follows: follows:

The value of bending moment at any load stage due to equilibrium of the overall system should be

the bending moments of the ordinary reinforced cross sections due to ultimate load and assumed tendon force 2

( g  q) L 8

 Fp  zp1

where g+q is ultimate load

Assuming the prestress force reaching yield strength: Fpu  A p  fyp

Fpu  24 MN

M up  Fpu  Zp

M up  48  MN  m

is resisting moment due to eccentric tendons

assuming the prestress reinforcement reaching yield strength: Fsu  A s  fys

Fsu  7.5 MN

M us  Fsu  Zs

M us  15.75  MN  m

this moment must be assigned for traditional

reinforcement Zp, Zs are the lever arm of prestress un-bonded tendons and reinforcement M up  M us  63.75 MN  m

q u 

Mup  Mus  8 2

q u  0.567

gq

 2.24

neglecting all increase of force in the prestress reinforcement A p  pp  19.05 MN

M uP  A p  pp  Zp

M up  48  MN  m

Example of External Prestress in TT Girder

291

Dimensions of individual parts of cross-section:

i  1  6 bc

q  0.045 MN m MN pp  1270 2 m

h 1  200mm

bc  10000mm

b 1 

b 2  1000mm b 4  50mm b 6  1000mm

h 2  50mm h 4  50mm h 6  50mm

b 3  600mm b 5  50mm

h 3  2150mm h 5  2100mm

Area of individual parts of cross-section:

g  0.208

MN m

A2  b 2 

A1  b 1 h 1

A3  b 3 h 3

A4  b 4 h 4

A5  b 5 

The distance of individual centre of gravity from the axis A:

y2  h 1 

y6  h 1 

A s  0.015 m

Es  200 GPa

Ep  200 GPa

fyp  1600 MPa

fys  500 MPa

Ec  32 GPa

  160 mm

zp1  1.496 m

L  30 m

H  2.25 m

ac  50 mm

at  100 mm

ap  200 mm

bi 

y3  h 1 

y4  h 1 

hi  5

A p  0.015 m

Zp  H  ap  ac

i  1  6

y1 

P  16.18 MN

Ic  2.616 m 1

hc  2250mm

A6  b 6 

A c  4.79 m

0.2

Zp  2 m

Zs  H  ac  at

ai 

Ai  1

0.025

0.654

Zs  2.1m

Ii  m

3.333·10-3 3.472·10-6

0.05

0.537

0.6

2.15

1.29

0.521

0.497

0.05

2.5·10-3

0.529

5.208·10-7

0.05

2.1

0.053

0.196

0.013

0.05

0.025

0.537

3.472·10-6

y5  h 1  h 4 

h5 3

Di 

0.428

7.22·10-3

The position of CG of cross-section:

Bi  Ai yi

ybh 



0.35

6.998·10-4

2.015·10-3

ybh  754.08mm

A

7.22·10-3

The modulus of intertie for individual parts of cross-section: Total area of cross-section:

I1  I4 

1 12 1 12

b 1  h 1

b 4  h 4

I2  I5 

1 36 1 36

b 2  h 2

b 5  h 5

I3  I6 

1 12 1 36

b 3  h 3

b 6  h 6

Di  Ai  a i

a 2  ybh  y2

Ac  4.79m

I   D   

Jc  2 a 3   y3  ybh 

The total modulus of intertie:

The distance of individual parts from tb: a 1  ybh  y1 a 5  y5  ybh

A

Ac  2 

a 4  ybh  y4

Jc  2.616m

Cross-sectional modulus:

a 6  ybh  y6

Wbd 

 hc  ybh 

zp1  hc  ybh

Wbd  1.749m zp1  1.496m

Example of External Prestress in TT Girder

Wbh 

Jc ybh

Wbh  3.47m

292

Self-weight of beam:

gos  Ac 26

Dead load:

kN m

gos  124.54

gs  3 

m Live load:

The calculation of anchor cable length, prestress cable 15,5:

Ap1  9 141.6mm

 pd



1600 1.26

MPa

 pd

 1.27  103 MPa

bc

g d  g s  1.35

kN bc vd  45 v d  v s  1.5 2 m m The determination of bending moments from the extreme load: vs  3.0

ds  15.5mm

qs  gos  gs  vs

q s  184.54 

q  go  gd  vd

kN m

q  253.629

kN m

The calculation of anchor length of prestress post tension cable: lbd  60ds

i  1  4

lbd  0.93m

Equalizing length:

a1  100mm  b 2 



  gd  vd  l   xi 2 Mgv   xi   gd  vd  

 go l  xi 2 Mgo   xi  go 

 h 3  65mm

a1  3.585m







 ( q ) l   xi   Mq   xi  ( q ) 





h1     bc  2 100mm  b 2  b 3 a2  2 h 3  2  

 1.5a1  M     a2 

lb  max( M)



a2  11.1m



lb  11.1m

Mgo 

1.074·104

2

l  l1  2 

l1  30m

3

wc 

l  29.467m



xi 

5.283

8.433

11.583

14.733

Mgv 

The calculation of internal forces: wc  400mm

i 



Mq 

kN m

5.462·103

1.62·104

1.491·104

7.583·103

2.249·104

1.741·104

8.856·103

2.627·104

1.825·104

9.28·103

2.753·104

kN m

Length of beam, length of unloading part: l1 c  x 1 



lb 2

c  3.15m

lb 2

 x0

x0 

l1 2



l 2

x0  0.267m

x1  5.283 m

x2  x1  c

x2  8.433 m

x3  11.583 m

x4  x1  3c

x4  14.733m

The proposal of prestress force LA 15,5: A p15  9  141.57 mm

 pd15



1600 1.26

 MPa

Losses of prestress force:

x3  x1  2c

a11  30mm

a22  40mm

Load: self-weight, service coefficients

gos  124.54 

kN m

g o  g os  1.35

a22 c  a11  2

Eccentricity of prestress force: epd  zp1  c

epd  1.446m

Example of External Prestress in TT Girder

c  0.05m

293

Characteristic of concrete C30/37: fctm  2.1  MPa

 bg

fckcyl   30  MPa

ef  0.75m

 bg



10.5ef 6 ef

 hc

 1.5

 bg

 hc

10.5ef 6 ef

 hc

 bg

 hc

 bg

 2.5



Mq  4

Npin4    bg fctm 

 Wbd   0.9 2



 0.9

1

 0.9



2

 0.70

Mq  4

Npin4    bg fctm 

 Wbd   0.9 2



 pp

 1.06

Ap15  1.274  10 3 m

   Ac

0.9 2 epd

Wbd

Npin4  1.93  104 kN

  

 bg

 2.5

   Ac

From 4. Minimal value of prestress force we obtain as follow:  ppu

 2.405

Actual pre-tensioning force:

The static eccentricity od’s external load to centre of cross-section:

ef 



 pd15

Np1   Ap15  pd15 Npin4 n  n  10.128 Np1

0.9 2 epd

Wbd

Npin4  1.639 104 kN

  

 1.27  103 MPa

Np1  1.618 103 kN

n 

I suggest



 Npin4

 ppu  2 

 Npin4

 ppu  2 



n 

Wbd

 Npin4 epd

Wbd

n  11.928

Np1

 Npin4 epd

 ppu  2 

Ap15  1.274  10 3 m

Npin4



 Wbd 



 3.15MPa

Np1   Ap15  pd15

n  12

I suggest

 Wbd 

 bg fctm

Checking the position of prestressing force: NpinrII   n Ap15  pd15

 pd15

 1.27  103 MPa

 bg fctm

Np1  1.618 103 kN Ap15 n  0.015m

Actual pre-tensioning force (proposal):

NpinrI   n Ap15  pd15

 3.15MPa



Mq  0.9 2NpinrI epd 4

0.9 2 N pinrI



 NpinrII epd

 ppu  2 

Wbd

 NpinrII epd

 ppu  2 

Wbd



 Wbd 



 Wbd 

 5.185MPa

 bg fctm

 5.05MPa

Maximum prestress force according to 2:

NpinrI  1.942 104 kN



N pin2   0.6   fckcyl



Mgo  1.825  104 kN m

 ppu   2 

 N pinrII 

Actual static eccentricity:

ef  



 NpinrII

Bending moment at the mid-span of the beam:

Mq  2.753  104 kN m

 NpinrII

 ppu  2 

NpinrII  1.618 104 kN

ef  0.805m

hc 6

 0.375m

 ppu

  2   N pinrII Ac







M go 

 W bd   1 e pd     A W c bd   1

 ppu   2 

 N pinrII  e pd

W bd  ppu

  2   N pinrII

Example of External Prestress in TT Girder

W bd

  e pd



N pin2   2.746  104 kN

M go

 0.6  fckcyl

  0.121  MPa

 W bd  M go

 W bd 

294

0.6 fckcyl

N pinrII   1.618  104 kN

  18 MPa

N pinII  N pin2



Wbd

epd4 

NpinrII0.9 2  epd3  epd2  i i i

Prestressing force according to 3:

2.244



Npin3  0.6 fckcyl  



 Npin3

 ppu  2 

 Npin3

 ppu  2 



 

 Wbh  

 0.9 2

   Wbh    Ac

 Npin3 epd Mq   Wbh   Wbh 

 ppu  2 

 Npin3 epd Mq   Wbh   Wbh 

 ppu  2 

NpinrII  1.618 104 kN

Npin3  6.581 104 kN

0.9 2 epd 

NpinrII0.9 2

  bg fctm 

-3.813

2.502

-3.196

2.657

-2.825

2.708

-2.702



Wbd 

epd1 

xi  5.283

Mq 



2.471

8.433

2.729

11.583

2.883

14.733

2.935

NpinrII  1.618  104 kN

 pp

 1.06

1

 0.9

1.469

Installation stadium  0.689MPa NpinrII  1.618  104 kN

Mgo  i

 pp

 1.06

Npin4  1.639 10 kN 4

NpinrII  1.618 104 kN

Mq  i

1.074·104

kN m

1.62·104

1.491·104

2.249·104

1.741·104

2.627·104

1.825·104

2.753·104

kN m

1.491·104 1.741·104

efgod1  

Mgo   pp  1NpinrIIepd1 i i  pp  1 N pinrII

kN m

efgod1  i

-1.775 -1.763 -1.755 -1.753

Wbh  Wbh

epd1 

NpinrII Mgo i  Wbh     bg fctm   NpinrII  Ac Wbh 

epd2 

NpinrII Mgo i  Wbd    0.6fckcyl   NpinrII  Ac Wbd 

1.346

 0.6fckcyl 0.6fckcyl  18MPa

1.825·104

0.975

N pinrII  1.618  104 kN

The checking of the prestress force position in section along the member: i

0.358

i  1  4

NpinrII  Npin3

The proposal of prestress force according to 4:

Mgo 

epd4 

i 

Service stadium:

efgod4   i

Mq   ppu  2NpinrIIepd4 i

 ppu  2 N pinrII

Mq   NpinrII0.9 2 Wbh i  epd3   0.6fckcyl   i Ac NpinrII0.9 2  Wbh 

Example of External Prestress in TT Girder

efgod4  i

-1.232 -1.232 -1.232 -1.232

295

Beam is simple supported on two supports and joint in the centre of the beam 5

Data: Individual dimensions, height, length, load

 epd4

 epd3

 epd2

 epd1

0.263  m 0

1.333

2.667

 epd

h  0.9m

L  12m

L  2 a  b

L  12 m

h    a



0.225

g  8 kN m

 12.68deg

1

sin(  )  0.22

a  4m

b  4 m

cos (  )  0.976

Calculation of reaction in supports A and B

Av 

gL

Av  48 kN

B  Av

B  48 kN

As the load g acts vertically and the support at b is sliding then the horizontal force at the support b is zero Ah = 0. first we determine the force in the line segment 2-4.

Calculation of axial force in the rod 2-4. where the first two parts of the equation represent the calculated moments with respect to the joint, and therefore the position

5

in the centre of the span is the moment Mog = q. l2 / 8. Thus, the tensile force in the

rod 2-4 will be:

Installation stadium  pp  1.06

 pp   1 Npin(-) A b(+)

 1  0.9



 pp   1 Npin(-)  epd W h(-)

M g0  S h



 pp   1 Npin(-) epd(+) W d(+)

Mg o(+) W h(-)



  bg   b  Rbtn

Mg o(+) W d(+)

 0.6  b  Rbn

Mg0  144m kN

 pp   2 Npin(-)

 2  0.65  0.8

A b(+)  pp   2 Npin(-) A b(+)



 pp   2 Npin(-) epd(+) W h(-)  pp   2 Npin(-) epd(+) W d(+)

S 

Mq0

h Mg0

gL

Mg0 

1 8

gL

S  160kN

As the following the balancing equation, for the calculation of the horizontal force at point g:

Service stadium  pp  0.9

Vertical force at point g 



M g(+) W h(-) M g(+) W d(+)

 0.6  b  Rbn

  bg   b  Rbtn

Sv  S sin(  )

Sv  35.121kN

Sh  S cos (  )

Sh  156.098kN

Va  Av  Sv

Va  12.879kN

Vb  Va

Vb  12.879kN

V1left  Va  ga

V1left  19.121kN

Na1

N3b

Na1  S cos (  )

Example of External Prestress in TT Girder

V3r  Vb  ga Na1  156.098kN

V3r  19.121kN

296

N34  S sin(  )

N34  35.121kN

N12  N34 N12  35.121kN

V12

V34

V12  S 

  K sin  2

S 

   cos    2 2   cos   2

2 sin

   cos    2 2    cos   2

2 sin

V12  3.902kN

V34  3.902kN

Now all normal forces are determined, and we can proceed to further calculate the bending moments and transverse forces along the beam.

Accordingly,no notransverse transverse force through the joint, the beam theand the Accordingly, forceisistransmitted transmitted through the joint, the and beam load are symmetrical, and the joint is located on the axis of symmetry and it itself is load are symmetrical, and the joint is located on the axis of symmetry and it itself is not a load, and therefore the value of the transverse force in cross-sections on the

M1left  Va a  g

axis of symmetry is zero.

sin(  )

K S S

 sin 90  

N12

sin  ( )

sin 90 

N34



 2 2

Ma1  g

sin(  )



 

K  S 

sin( )

cos 

  2  cos  

   2

K cos 

 sin (  )

K cos S  2      cos  

2

   2

K cos 

sin(  )

2

   S sin(  ) 2    cossin  (  )   S   2  cos   S sin(  ) 2    cos  

S 

sin(  )

cos 

2

K  35.337kN

M1r  M1left  V12h

not a load, and therefore the value of the transverse force in cross-sections on the axis of symmetry is zero.

M1left  12.484m kN

M3r  M1left

M3r  12.484m kN

M1r  15.996m kN

M3left  M1r

M3left  15.996m kN

Ma1  16 m kN

M1  g

M3  g

M1r

M3  16 m kN

M1  16 m kN

M3left

Verificatio 2

 L   2   N  b  V h Mg  Va   g 12 12 L 2

Example of External Prestress in TT Girder

Mg  0.004m kN

297

The terminology for arch bridge, L is span, f is rise, support area is

The ratio of arch span to rise, l/f, should be chosen between 2:1 and

springing line, pick point of arch bridge is crown. We can have divided the arch

10:1. The sensitivity of arches to creep, shrinkage, temperature change, and

bridge as conventional arch bridge with road way supported above arch or tied

support displacements increases with increasing values of l/f if we are talking

arch with suspended roadway. The girder can be design to resist the horizontal

about reinforced concrete arch bridge. Stresses and deformations due to these

component of the arch reaction; this arrangement is called a tied arch. Suspending

actions are normally small when the ratio of span to rise is less than 4:1, regardless

the roadway from the arch may be an appropriate solution for low-level crossings

of the degree of statically indeterminacy of the arch. As l/f approaches 10:1, it

or when suitable foundation material for conventional arch abutments is not

may be necessary to reduce or eliminate redundant moments due to restrained

available. Economy and elegance are nevertheless difficult to achieve when this

deformations by providing hinges at the springing lines and at the crown. The

type of the bridge is built of reinforced and prestressed concrete. Arch bridges

long-term deformations of flat, hinged arches, in particular the angle break at the

with suspended roadway can be more successfully designed in structural steel.

crown hinge, will be large and must therefore be carefully checked. Excessive

For this reason, we show the calculation of an arch steel bridge as example

deformations are unavoidable when the span to rise ration is greater than 10:1,

to understand the process and to find the forces in relation with angles of the

even when full rotational restraint is provided at the arch abutments. Reducing

members. The arch, columns, and deck girder constitute a frame system. The

l/f below 2:1 results in an awkward appearance and substantial increases in

moments in the frame system can be divided into two components: fixed system

construction cost.

moments and flexible system moments. Fixed system moments are produced when vertical deformations of the arch are restrained and are thus equal to the

Arch stiffness affects not only the behaviour of the completed structure

continuous beam moments in the girder. Flexible system moments correspond

but also its behaviour during construction. The arch can be made sufficiently

to vertical displacements of the arch and are, in general, shared by arch and

stiff to carry the dead load of arch, columns, and girder without relying on girder

girder. Two idealized limiting cases are possible: stiff arches, which resist the

stiffness for global stability. This enables false work to be removed immediately

entire flexible system moment with no participation of the girder, and deck-

after completion of the arch, or construction of the arch using the cantilever

stiffened arches for which the entire flexible system moment is resisted by the

method. Girder depth should be constant for the entire length of the bridge. In

girder. Moments due to arch displacements can be further divided into dead load

addition, the approach spans should not differ markedly in the length from the

and live load components.

girder spans above the arch. The girder moments in the arch region, the sum of a fixed system moment and a flexible system moment, will thus normally be

The cost of false work and formwork for arch bridges is high in comparison to conventional cast-in-place girder bridge. As a result, arch bridges

greater than the moments in the approach spans. Stability of the frame is greatly enhanced by continuity in the deck girder.

are economical only under a limited range of topographical and geotechnical conditions. Arches may be appropriate for crossings of rivers, canyons, or steep valleys, where a single long span is required for the main obstacle and several short spans can be used for the approaches. The economical range of reinforced concrete arch spans extends from 50m to 200m. the higher costs of arch bridges may sometimes be justified by their superior aesthetic qualities.

Arch Bridges

298

Arch Bridge – Example

Langer beam is a mixed system, consists of the stapes of the lattice and bending-resistance sticks. As we can see, all parts of the Langer beam are therefore signs of normal forces than the corresponding parts of the reinforced beam. Here is an example to calculate the arc bridge as shown in the figure, the details of the individual elements and their slopes are in the figure. moments and forces in individual elements have to be calculated. The material and geometry characteristics as well as the dimensions are as follows q  15

kN m

f 

L  60m

n 

x



x

 10 m

f  12 m

The calculation of reactions:

Ah  0

Av  0.5q L

Av  450 kN

Bv  Av

Bv  450 kN

From the figure for point 3 we apply:

H V

S4cos   4

Qgre

N g

H

S4sin  4



Then we solve the equation, for point 3

Av 0.5L  0.5q L0.25L  Hf

Mgo

H  Ng

H V cos (  )

0.25q L

 0.125q L  Hf

0.125q L

 H f

Mgo  Hf

Ng 

q L

Ng  562.5kN

8 f

H  562.5kN 0

Si 1 cos   i 1  Si cos   i

sin(  )

sin(  )

Si cos   i

Si sin  i  Si 1 sin  i  1  Zi

cos (  )

From the figure for point 3 we apply:

Si 1 cos   i 1

Si 1 sin  i  1  Si sin  i

tg(  )

tg(  )

H V

S4cos   4

Qgre

N g

H

S4sin  4



Then we solve the equation, for point 3

Av 0.5L  0.5q L0.25L  Hf



Ng 

H  Ng

 0.125q L  Hf

0.125q L

 H f

Mgo  Hf

Mgo

0.25q L

q L 8 f

Ng  562.5kN

H  562.5kN 0

Si 1 cos   i 1  Si cos   i

Examples of Arch Bridges

Si cos   i

Si 1 cos   i 1

299

V

cos (  )

Si sin  i  Si 1 sin  i  1  Zi

sin(  )

cos (  )

sin(  )

Si 1 sin  i  1  Si sin  i

tg(  )

y1  y1 1 x

y5  y5 1

tg(  )

x

 0.67

 0.4

y2  y2 1 x

y6  y6 1 x

y3  y3 1

 0.4

x

tan  1  tan  1 1  0.27

tan  2  tan  2 1  0.27

the angle 4 to 6 will be negative

tan  3  tan  3 1  0.27

tan  4  tan  4 1  0.27

 H    cos   i 



and

H     cos   i  1 

H H   sin  i  1   sin(  ) cos cos   1     i i  

tan  i

H tg  i  tg  i 1 

 0.13

tan  5  tan  5 1  0.27



Si  1

x

 0.67

so on the left side of point 3 the angle 1 to 3 will be positive and on the right side

y4  y4 1

 0.13

yi  yi 1

Z1  H tan  1  tan  1 1 

Z1  149.59kN

Z2  H tan  2  tan  2 1 

Z2  149.63kN

Z3  H tan  3  tan  3 1 

Z3  150.71kN

Z4  H tan  4  tan  4 1 

Z4  149.63kN

Z5  H tan  5  tan  5 1 

Z5  149.59kN

i  1  5 x0  0 m

x1  10m

x2  20m

x4  40m

x5  50m

x6  60m

 H  Si     cos   i 

Zi  H tan  i  tan  i 1  1

 33.66deg

2

 21.80deg

4

 7.63deg

5

 21.80deg

xi 

i 

10.67

6.67

y5  y5 1  4 m

3

 7.63deg

6

 33.66deg

6.67

10.67

y2  y2 1  4 m y6  y6 1  6.67m

 H    cos   1 

S1  675.8kN

S4  

 H    cos   4 

S3  

S4  567.52kN

S3  567.52kN

S1  

 H    cos   5 

yi 

y1  y1 1  6.67m

x3  30m

y3  y3 1  1.33m y4  y4 1  1.33m

 H    cos   6 

S6  

S5  605.83kN

S6  675.8kN

Qgre  S4 sin  4

Qgre  75.35kN

Sh1  S1 cos   1

Sh1  562.5kN

Examples of Arch Bridges

S2  605.83kN

 H    cos   3 

S5  

or SV1  Htan  1

 H    cos   2 

S2  

SV1  374.57kN

Sv1  S1 sin  1

Sv1  374.57kN

300

Determination of shear forces along the structure:

Qa  Av  Sv1

Qa  75.43kN

Example: Unilateral right uniform load Q1L  q x  Qa

Q1L  74.57kN

Contrary to the procedure with a reinforced beam with a hinge in the centre, the load decomposition procedure shall be applied to symmetrical and ant metric load.

Q1r  q x  q x  Qa

Q1r  75.43kN

Q2L  q x  q x  Qa

Q2L  75.43kN

Q2r  q x  Q a

Q2r  74.57kN

Q3L  q x  Q a

Q3L  74.57kN

Q3r  q x  q x  Qa

Q3r  75.43kN

The determination of the bending moment: x M1   Av  Sv  x  q x  1 2 

M1  4.26m kN

M2   Av  Sv  2x  Z1 x  q



1

( x  x ) ( x  x )

M2  4.42m kN

We calculate the reactions in A, and B.

Mg   Av  Sv  3x   Z1 2x    Z1 x 1





q x

M0x 

( 3x )

Mg  0.48m kN

M0  6750m kN

Av xm  q 

Mm0  Hym

M0 f

 562.5kN

is the horizontal force

 Hym  Mm

ym  12m

Mm0  Hym  Mm

L  60m

f 

n  6

x



x

 10 m

Mm  M0x  Hym

Av  q xm  Htan  i  Qm

Av  ( 0.5q L) 

Bv 

3 4

 q L   



f  12 m

Av  75 kN

Bv  225kN

The horizontal force H we calculate from the Mgo. Mg0 L Mg0  Av  H  Mg0  2250m kN 2 f

H  187.5kN

Mm  6562.5m kN

M0x  Hym  Mm  0 m kN

V

Ah  0

q  10

M0x  187.5m kN



M0  q 

 q

( 3x )

Qm0  Htan  i

i  1  5 x0  0 m

x1  10m

x3  30m

x4  40m

x5  50m

x6  60m

y0  0m

y1  6.67m

y2  10.67m

Examples of Arch Bridges

x2  20m

y3  12m

301

y4  10.67m

y5  6.67m

y6  0m

1

 33.66deg

2

 21.80deg

5

 21.80deg

6

 33.66deg

3

 7.63deg

4

 7.63deg

Z4  H tan  4  tan  4 1 

Z4  49.88kN

Z5  H tan  5  tan  5 1 

Z5  49.86kN

 H    cos   1   H  S4     cos   4  S1  

From the equations we calculate the forces S and Z as follow:  H  Si   Zi  H tan  i  tan  i 1    cos   i 

 H    cos   3 

xi 

yi 

6.67

10.67

6.67

S3  189.17kN

 H    cos   5 

 H    cos   6 

S5  

S6  

S5  201.94kN

S6  225.27kN

10.67

S2  201.94kN

S3  

S4  189.17kN

i 

 H    cos   2 

S2  

S1  225.27kN

Av  75 kN

Bv  225kN

y2  y2 1  4 m

y3  y3 1  1.33m

V

y5  y5 1  4 m

y6  y6 1  6.67m

y4  y4 1  1.33m

Sv1  S1 sin  1

Sv1  124.86kN

Qa  Av  Sv1

Qa  49.86kN

y1  y1 1 x

y4  y4 1 x

 0.67

 0.13

y2  y2 1 x

y5  y5 1 x

 0.4

 0.4

y3  y3 1 x

y6  y6 1 x

tan  1  tan  1 1  0.27

tan  2  tan  2 1  0.27

tan  3  tan  3 1  0.27

tan  4  tan  4 1  0.27

Av  Sv1  Qa

 0.13

Av  Sv1  Qa  0 kN

 0.67

verification

V

Sh1  S1 cos   1

Sh1  187.5kN

Qb  S6 sin  6  B

Qb  Bv  S6 sin( 33.66deg)

Bv  S6 sin( 33.66deg)

Qb  100.14kN

Qb  S6 sin( 33.66deg)  Bv  0 kN

tan  5  tan  5 1  0.27

Z1  H tan  1  tan  1 1 

Z1  49.86kN

Z2  H tan  2  tan  2 1 

Z2  49.88kN

Z3  H tan  3  tan  3 1 

Z3  50.24kN

Calculation the shearing forces QL1  Qa

QL1  49.86kN

Qr1  ( QL) 1  Z1

QL2  Qr1

QL2  0.0054kN

Qr2  QL2  Z2

Qr2  49.88kN

QLg  Qr2

QLg  49.88kN

Qrg  QLg  Z3

Qrg  100.12kN

Examples of Arch Bridges

Qr1  0.0054kN

302

QL4  Qrg  q x

QL4  0.12kN

Qr4  QL4  Z4

Qr4  49.99kN

QL5  Qr4  q x

QL5  50.01kN

Qr5  QL5  Z5

Qr5  0.14kN

Qb  Qr5  q x

Qb  100.14kN

Design of Prestress concrete:

Calculation of bending moments along the construction M´1  Qa x  498.58m kN

M´1  498.58m kN

M´2  M´1  Qr1 x  498.53m kN

M´2  498.53m kN

M´g  M2  Qr2 x

M´g  503.25m kN

M´4  Qrg x  q 

( x )

M´4  501.18m kN

M´5  M´4  Qr4 x  q 

( x )

  0

M´5  501.12m kN

 

c c

 dA c   0

 

s s

 dA s   0

  dA p

p p

A A A  c  s  p   c   c    Zc  Z0 dAc    s   s    Zs  Z0 dA s    p   p   Zp  Z0 dA p 0 0 0

Mb  M´5  Qr5 x  q 

( x ) 2

Mb  0.3 m kN

c

A  c  E c   c dA 0

c

 0   0   Zc  Z0

  s   0  Zs  Z0

0

  0   Zs  Z0

s

0

E I

  0

2 Ec   0   0   Zc  Z0   Zc  Z0  dA

 EA ES    0       ES EI    0 

N   M N

A  c  Ec   0   0   Zc  Z0 dA 0

  0

Ec   0   0   Zc  Z0   Zc  Z0 dA

Recapitulation - Design Of Prestress Concrete Structures

303

o

1

Ap  Ep

 p

  o

 p 

p

M  yt Ic  Ec



Ac  Ec

zp  zo

 p

 o

 Zp  Z0

 p

 0   0p 

 s

 0

  0s   Zs  Z0

 c

 0

  0c   Zc  Z0

zs  zo

 s

 0p 

   o

 s

s

  p   zp  zo

  0

 Zp  Z0

A p   0

Np

2

The method of Neuman:

 0p

Np

1

 2

 o

ES EA

  s   zs  zo

  0

 Ec  Zc dA   0 Ac

  0

 E c dA   0

 Es  Zs dA   0

Ep  Zp dA

 E s dA   0

E p dA







L 2







Zc 2

 L

4  Zc



w  4  Zc

( EA )

 su

( ES )

Ncu

  cu

 cu

E c   cu

Ep  Ap 



  0

 t Ec (  z)  Zc dAc   0

 t Es (  z)  Zs dAs   0

Ep (  z)  Zp dAp

( ES ) ( EA )

Ec  A c   cu

The calculation of bending stiffness:

N pp

A A A  c  s  p t t t Ec (  z) dAc   Es (  z) dAs   Ep (  z) dAp  0 0 0

Zo  constant

The extension of the cable 

The calculation of stiffness in prestress concrete>

2  w Zc

at the middle of the span 

Npu

Zc   Np0  N pp

fpu  A p

( EJ)

A A A  c  s  p t 2 t 2 t 2  Ec (  z)   Zci  Z0 dAc   Es (  z)   Zsi  Z0 dAs   Ep (  z)   Zpi  Z0 dAp 0 0 0

Recapitulation - Design Of Prestress Concrete Structures

304

A A   Ac   s  p  Ec (  ) dAci   Es (  ) dAsi   Ep (  ) dApi   0 0  0 

 0



N M

0 







( EA )   0  ( ES )

Es (  )   0   0   Zsi  Z0   Zsi  Z0  Asi 

i 1

  0

 c ( ) 0

i

 c ( ) 

 0

 0   0

  0

  s (  ) dA s   0 

 Zci  Z0 dAc  

 Zs  Z0 Ac

  c (  ) dA c   0

E ci   ci s



Ep (  )   0   0   Zpi  Z0   Zpi  Z0  Api

i 1

 s ( ) 

  p ( )   0 dAp 

 Zsi  Z0 dAs  

 s ( )

E si   si

 p( )

Epi    pi   p0

s

0

m k    n  0   Ec (  )   Zci  Z0  Aci  Es (  )   Zsi  Z0  Asi  Ep (  )   Zpi  Z0  Api    i  1 i 1 i 1    n m k  2 2 2           E ( )  Z  Z  A E ( )  Z  Z  A E ( )  Z  Z            ci 0 ci  si 0 si  pi 0  Api  c  s  p  0  i 1 i 1  i  1 

      

  0   Zs  Z0

E pi   pi

 Zi  Z0

A A  s  p  Es (  )   0   0   Zsi  Z0 dAs   Ep (  )   0   0   Zpi  Z0 dAp 0 0

      



  p ( )   0   Zpi  Z0 dAp

Ec (  )   0   0   Zci  Z0 dAc 

m k    n  0   Ec (  )   Zci  Z0  Aci  Es (  )   Zsi  Z0  Asi  Ep (  )   Zpi  Z0  Api     i1 i 1 i 1    n m k  2 2     Ec (  )   Zci  Z0  Aci  Es (  )   Zsi  Z0  Asi  Ep (  )   Zpi  Z0 2  Api 0    i 1 i 1  i  1 



The calculation of internal forces>



Ec (  )   0   0   Zci  Z0   Zci  Z0  Aci 

i 1 m

  0

      

m k   n   0    Ec ( )  Aci    Es ( )  Asi    Ep ( )  Api     i  1 i 1 i 1    n m k      Ec (  )   Zci  Z0  Aci  Es (  )   Zsi  Z0  Asi    Ep (  )   Zpi  Z0  Api 0      i 1 i 1  i  1 

 0  ( ES )



  0  ( EJ)

N   M

 ( EA ) s ( ES ) s    0       s s  ( ES ) ( EJ)    0 

N   M

Eref  

 ( A ) s ( S) s    0      s s  ( S ) ( J)    0 

 o    0  o    0

 ( EA ) s ( ES ) s   N       s s M  ( ES ) ( EJ)   ( A ) s ( S) s   N      s s  ( S ) ( J)   M 

Eref  

Recapitulation - Design Of Prestress Concrete Structures

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[1] Picard A., Massicotte B. and Bastien J. (1995) „Relative Efficiency of External

[9] Nanni A., Bakis C.E., CNeil E.F. and Dixon T., (1996) „Performance of

Prestressing“ Journal of Structural Engineering, Vol. 121, No. 12, December

FRP tendon anchorsystems for prestressed concrete structures“ PCI Journal,

1995

January-February 1996

[2] Pincherira and Woyak (2001) „Anchorage of Carbon Fiber Reinforced

[10] Ng C. (2003) „Tendon Stress and Flexural Strength of Externally Prestressed

Polymer (CFRP) Tendons Using Cold-Swaged Sleeves“, PCI Journal, Vol. 46

Beams“ ACI Structural Journal, September-October 2003, pp 644-653.

no. 6, November-December 2001 [11] Tan K. and Ng C. K. (1997) „Effects of Deviators and Tendon Configuration [3] Pisani M.A. (1998), „A numerical survey on the behaviour of beams pre-

on Behaviour of Externally Prestressed Beams“, ACI Structural Journal, v 94, n

stressed with FRP cables“ Construction and Building Materials 12, 1998, pp 221-

1, January-February 1997.

232 [12] Tan K., Farooq A. and Ng C. (2001) „Behaviour of Simple-Span Reinforced [4] McKay K.S. and Erki M.A. (1993) „Flexural behaviour of Concrete Beams

Concrete Beams Locally Strengthened with External Tendons“ ACI Structural

Pretensioned with Aramid Fibre Reinforced Plastic Tendons“, Canadian Journal

Journal, March- April 2001, pp 174-183

of Civil Engineering vol 20

[13] Tan K. And Tjandra R.A. (2003) „Shear Deficiency in Reinforced Concrete

[5] Meier U (1998), US-Patent 5,713,169, Feb 3, 1998

Continuous Beams Strengthened with External Tendons“ ACI Structural Journal, September-October 2003, pp 565-572

[6] Meier U., Deuring H. and Schwegler G. (1992) „Strengthening of structures with CFRP laminates: Research and application in Switzerland“ Advanced composites materials in bridges and structures, Edt. Neale K.W. and Labossiere P., 1992 [7] Mutsuyoshi H., Machida A. and Sano M., (1991), „Behaviour of pretressed

concrete beams using FRP as external cable“ Japan Concrete Institute, Vol. 13, 1991 [8] Naaman A.E., Burns N., French C, Gamble W.L. and Mattock A.H. (2002) „Stresses in Unbounded Prestressing Tendons at Ultimate: Recommendation“ ACI Structural Journal, July-August, 2002

References

306

307

Influence Lines in Bridge Design

308

Post-Tensioned Prestress in Bridge Design

309

External Prestress in Bridge Design

310

Sectional Forces in Prestress Members

311

Post-Tension Prestress Box Girder in Bridge Design

312

Post-Tension Prestress Box Girder in Bridge Design

313

Analysis of Prestress Tendons

314

Flat Slab, Prestress Tendons

315

The Transfer of Post-Tensioned Prestress Forces in Flat Slab

316

Prestress Concrete

317

Structural System of External Prestress Tendons

318

319

04. Precast Concrete Precast concrete can be used in different kinds of structures, e.g. single storey, multi-storey and high-rise buildings both in non-seismic and seismic areas. It is one of the possible answers to the frequently heard and steadily increasing criticisms such as to build becomes uneconomical, or there are no volunteers any more for the difficult, dangerous and dirty building work, or the building activity has to be automated. Speed of construction is a major consideration in most building projects and it is here that the design of precast structures should be carefully considered. Building design is increasingly becoming a multi-functional process where the optimum use of all the components forming the building must be maximized. This advantage is maximized if the lay-out and details are not too complex.

principles which have to be respected to achieve the full profit which the prefabrication offers. A good design in precast concrete should therefore use details that are as simple as possible, since it is in the simplicity of the details that the advantages of precast concrete are inherit. Maximum economy of precast concrete construction is achieved when connection details are kept as simple as possible, consistent with adequate performance and ease of erection. Furthermore, complex connections are more difficult to design, to make and control and will often result in poor fitting in the field.

Designers are becoming more aware of the high quality finishes which are possible in prefabricated units, but changes are having to be made in the way that the traditional precast concrete structures are conceived and designed. The designers only have to be aware of these products and the basic design principles, i.e. how structural integrity using precast concrete can be achieved. The prefabrication has its own design approach and design

One of the most important principles in the design of connections is to keep them simple. The main difference between cast in-situ and precast frames and skeletal structures lies on one hand in the general design philosophy and connections between components, and on the other in the possibilities for larger spans and smaller cross sections of columns and beams. Skeletal construction is commonly used both in precast and in cast in-situ construction for low-rise and multi-storey buildings. Utility buildings normally require a high degree of flexibility. Interior load-bearing walls are therefore avoided. A column-beam solution is normally preferred when an interior vertical loadbearing structure is needed. There are several advantages with precast concrete construction. Precasting operations generally follow an industrial production procedure that takes place at a central precast plant. Thus, high concrete quality can be reliably obtained under the more controlled production environment. Since standard shapes are commonly produced in precasting concrete, the repetitive

320

use of formwork permits speedy production of precast concrete components at a lower unit cost. These forms and plant finishing procedures provide better surface quality than is usually obtained in field conditions. Precast concrete components may be erected much more rapidly than conventionally cast-in-place components, thereby reducing onsite construction time.

above will vary with manufacturer. Foundation connection may be via a base plate connected to the column or by reinforcing bars projecting from the end of the column passing into sleeves that are subsequently filled with grout. Alternatively, a column may be set into a preformed hole in a foundation block and grouted into position.

Precast concrete components can be designed as in situ forms for underwater construction so that the use of cofferdams may be eliminated or substantially limited. The precasting process is also sufficiently adaptable so that special shapes can be produced economically.

Precast and prestressed concrete often has 28-day compressive strengths in the range of 28 to 55 MPa. Such concrete can be produced with reasonable economy, provided proper care is taken in mixture proportioning and concreting operation. With proper use of water-reducing admixtures and pozzolanic materials, it is realistic and desirable to control the water-to-cementitious material ratio within the range of 0.35 to 0.43.

Precast Concrete Columns Precast Concrete Columns can be circular, square or rectangular. For structures of five storeys or less, each column will normally be continuous to the full height of the building. For structures greater than five storeys two or more columns are spliced together. Precast concrete columns may be single or double storey height. The method of connection to the foundation and to the column

Columns are provided with necessary supports for the ends of the precast beams (corbels or cast-in steel sections). There will also be some form of connection to provide beamcolumn moment connection and continuity. For interior columns this may be by holes through which reinforcing bars pass from one beam to another. For edge columns, some form of bracket or socket is required. During erection columns must be braced until stability is achieved by making the necessary connections to the beams and slabs.

Architectural structural precast concrete components are being used on an increasing number of prestigious commercial buildings. Designers are becoming more aware of the high quality finishes which are possible in prefabricated units, but changes are having to be made in the way that the traditional precast concrete structures are conceived and designed. Numerous proprietary and non-proprietary precast structural systems have been developed in the world in the last 30 years. This paper summarizes the essential elements of several precast structural systems that are suitable for office building, schools, healthcare facilities, parking garages, multi-story residential structures and multi-story commercial structures.

Survey of Precast Structural Systems Architectural structural precast concrete components are being used on an increasing number of prestigious commercial buildings. Designers are becoming more aware of the high quality finishes which are possible in prefabricated units, but changes are having to be made in the way that the traditional precast concrete structures are conceived and designed. Numerous proprietary and non-proprietary precast structural systems have been developed in the world in the last 30 years. This paper summarizes the essential elements of several precast structural systems that are suitable for office building, schools, healthcare facilities, parking garages, multi-story residential structures and multi-story commercial structures.

 U.S. Conventional System U.S. Conventional precast concrete system for office structures in the United States consists of precast inverted T- beams, L-shaped spandrel beams, multi-story columns, and hollow-core doublecast-in-place tees as floor members. The system generally uses cast-in-place and double slabs tees,or with connections between primary concrete only for a floor topping. In general, simple span members are employed, with beams andresisting columns. system connections shearThe and not moment.can be used for structures up to

five stories in height. Roof spans can be up to 24.4 m. Three types  Duotek System of beams are used with system. A is an inverted T-beam The Duotek system wasthis developed by Type the Ontario Precast Concrete Manufactures Association and the Ontario Division of the Portland The system was with horizontal openings of (711x356 mm)Cement on a Association. 1.52 m module. designed specifically for office or institutional structures. The system consists of three precast Type B, support slabs prestressed on one side andandalso include concrete elements: tee columns, beams, double tees, openings with cast-in-place TheType systemC canprimary be used forbeams structures up to of connections (711x356between mm)primary on a beams 1.52 and m columns. module. five stories in height. Roof spans can be up to 24.4 m. Three types of beams are used with this allow services to run over the beam and under the double tee floor system. Type A is an inverted T-beam with horizontal openings of (711x356 mm) on a 1.52 m module. Type B, support tee slabs one side is andintended also include to openings of (711x356 mm) on slab. The 1.22 m total flooron depth accommodate a 1.52 m module. Type C primary beams allow services to run over the beam and under the the precast structural members, HVAC, plumbing, and electrical double tee floor slab. The 1.22 m total floor depth is intended to accommodate the precast structural and members, HVAC, electrical services, and ceiling and lighting services, ceiling andplumbing, lighting and systems. systems (fig. 1a, 1b).

• U.S. Conventional System U.S. Conventional precast concrete system for office structures in the United States consists of precast inverted T- beams, L-shaped spandrel beams, multi-story columns, and hollow-core slabs or double tees as floor members. The system generally uses castin-place concrete only for a floor topping. In general, simple span members are employed, with connections resisting shear and not moment. • Duotek System The Duotek system was developed by the Ontario Precast Concrete Manufactures Association and the Ontario Division of the Portland Cement Association. The system was designed specifically for office or institutional structures. The system consists of three precast concrete elements: columns, prestressed beams,

Double tees slab as floor member

321

HVAC ducts

Figure 1a Inverted T-beam (Type Inverted T-beam (Type A) A).

Figure 1b Partial Inverted T-beam (Type C). Partial Inverted T-beam (Type C)  Dycore System The Dycore system has been used for office buildings, schools, healthcare facilities, and parking garages. Connections are composed of cast-in-place concrete. Columns may be castin-place or precast with multi-story precast columns containing block out cavities at the beam level to facilitate beam-to-column connections. Prior to placement of the cast-in-place concrete, negative moment beam reinforcement is tied to the precast soffit beam and to the column reinforcement. HVAC ducts, plumbing components, and electrical system conduits can be placed on the precast soffit and embedded in the composite cast-in-place concrete. Similarly, voids in Dycore slabs can be used to house electrical conduits and plumbing components (fig. 2).

Precast Concrete

322

• Dycore System • Dyna-Frame System The Dycore system has been used for office buildings, schools, The Dyna-Frame system is typically used in multi-story residential healthcare facilities, and parking garages. Connections are structures, office buildings, parking garages, and schools. The key composed of cast-in-place concrete. Columns may be castto this system is the column-to-column splice and the column-toin-place or precast with multi-story precast columns containing beam splice. The single-story precast columns are pretensioned Figure 1b Partial Inverted T-beam (Type C). block out cavities at the beam level to facilitate beam-to-column and reinforced with a structural steel tube running longitudinally in  Dyna-Frame System connections. Prior to placement of the cast-in-place concrete, the The centre of the column is used in theinsplice maderesidential at each structures, office Dyna-Frame systemthat is typically used multi-story  Dycore System negative moment beam reinforcement is tied to the precast soffit floor. Theparking tube orgarages, columnand coreschools. does not from eitheris end buildings, Theprotrude key to this system the column-to-column The Dycore system has been used for office buildings, schools, healthcare facilities, and splice and the column-to-beam splice. The single-story precast columns are pretensioned and beam parking and togarages. the column reinforcement. HVAC ducts, plumbing of the column. The inside diameter of the column core is held Connections are composed of cast-in-place concrete. Columns may be castreinforced with a structural steel tube running longitudinally in the centre of the column that is in-place or and precastelectrical with multi-story precast conduits columns containing block out cavities components, system can be placed onat the beam constant at 100 mm. used in the splice made at each floor. The tube or column core does not protrude from either level to facilitate beam-to-column connections. Prior to placement of the cast-in-place the precast and embedded in the composite cast-in-place concrete,soffit negative moment beam reinforcement is tied to the precast soffit beam and to end the of the column. The inside diameter of the column core is held constant at 100 mm (fig. 3). column reinforcement. HVAC ducts, plumbing and electrical system conduits concrete. Similarly, voids in Dycore slabscomponents, can be used to house can be placed on the precast soffit and embedded in the composite cast-in-place concrete. electrical conduits plumbing components. Similarly, voids and in Dycore slabs can be used to house electrical conduits and plumbing

components (fig. 2).

Upper precast column

Precast hollow-core slab (203 mm)

Cast-in-place

Beam sleeve

Mild steel reinforcement to carry the negative moment Floor members (203 mm)

Interior beam

Shear reinforcement

Composite cast-in-place

Spindle

Neoprene bearing gasket

Multispan cantilever beam

m 1,2

Precast beam

Pretensioned cocrete column

Lower precast column

Precast soffit beam 305 mm

Shear reinforcement

Figure 2 Composite Dycore Structural System. Composite Dycore Structural System

Figure 3 Dyna-Frame (Structural components The detail showing column-toDyna-Frame SystemSystem (Structural components andand The detail beam connection). showing column-to-beam connection).  Filigree Wideslab System The Filigree wideslab system was originally developed in Great Britain and is presently used there under the name of OMNIDEC. This method of construction is used throughout the United States and is also used widely in Japan. Though often used for parking garages, the system has been used in residential construction, office buildings, and other multi-story commercial structures. Electrical services can be placed within the cast-in-place portion of the floor system. HVAC system components are suspended beneath the precast floor components and passed vertically through performed block-outs in the floor members (fig. 4).

Precast Concrete

Cast-in-place

Mild steel reinforcement to carry the negative moment

323

Interior beam

Spindle

Neoprene bearing gasket

Multispan cantilever beam Shear reinforcement

Precast beam

Beam sleeve

Pretensioned cocrete column

Lower precast column

 PG Connection System This system is example of precast concrete construction in Jap Connection system, developed by the Obayashi Corporation Technical employs precast cruciform beam components that are placed at column lo cruciform component is placed over the column, and beam-to-beam conne welding or mechanical splices (fig. 5).

• Filigree Wideslab System The Filigree wideslab system was originally developed in Great Figure and 3 Dyna-Frame System components andofThe detail showing column-toBritain is presently used(Structural there under the name OMNIDEC. beam connection). This method of construction is used throughout the United States and is also used widely in Japan. Though often used for parking garages, the system has been used in residential construction,  Filigree Wideslab System office and other commercial Thebuildings, Filigree wideslab systemmulti-story was originally developed instructures. Great Britain and is presently used there under the name of OMNIDEC. This method of construction is used throughout the Electrical services can be placed within the cast-in-place portion United States and is also used widely in Japan. Though often used for thesplice bars Column of the floor system. HVAC system components are suspendedparking garages, system has been used in residential construction, office buildings, and other multi-story beneath precastElectrical floor components passed vertically commercialthe structures. services can beand placed within the cast-in-place portion of the through performed block-outs in the floor members. floor system. HVAC system components are suspended beneath the precast floor components and passed vertically through performed block-outs in the floor members (fig. 4).

Sleeve Shear reinforcement

Precast connection component component

Cast-in-place column

Typical thickness 57 mm

Reinforced precast floor panel

The steel truss ensures composite behaviour between precast and cast-in-place concrete

PG Connection System –cross System shaped beam-to-column Figure 5 PG Connection –cross shaped beam-to-column compone component.

Filigree with light steel Figure 4precast Filigreeslab precast slab with lighttruss. steel truss. • PG Connection System This system is example of precast concrete construction in Japan today. The PG Connection system, developed by the Obayashi Corporation Technical Research Institute, employs precast cruciform beam components that are placed at column locations. The precast cruciform component is placed over the column, and beam-to-beam connections are made by welding or mechanical splices.

 RC Layered Construction System This structural system, developed by the Taisei Corporation, cons • RC Layered Construction System with cast-in-place concrete. Components of the syst This members structural connected system, developed by the Taisei Corporation, single-story columns, precastconnected beams, and Filigree type slabs. Single-stor consists of single-span members with cast-in-place with reinforcing bars protruding from their upper face for the column-to-c concrete. Components of the system include precast single-story Single-span rest on Filigree the precast columns, precastbeams beams, and typecolumns. slabs. Single-story Once the precast slabs are positioned, negative moment reinforcin columns are cast with reinforcing bars protruding from their upper longitudinally along the top of the beam and through the beam-to-colum face for the column-to-column connections. Single-span beams Two-way reinforcement rest on the precast columns. is placed on the precast floor slab. The concrete s placed over the entire floor system, resulting in monolithic connection subsequent stories begins with precast columns being slipped over the pro bars from the lower column. The connection is then grouted.

 RPC-K System The RPC-K system, developed by Kabuki Construction Company precast beams that serve as stay-in-place forms for cast-in-place concrete of the system include cast-in-place columns and Filigree type floor mem concrete is used for all connections between components. Longi

Precast Concrete

anchorage. Additional longitudinal reinforcement is placed in the trough of the beam once the beam is in place. To facilitate the column-to column connection, the main reinforcement from cast-in-place 324 type slabs lower columns protrudes upward to tie in the cast-in-place upper column. Filigree rest on the precast beams and reinforcement is placed across the trough of the beam to tie the components together. Negative moment reinforcing steel is placed both longitudinally and transversely on the precast slab, and then embedded with cast-in-place concrete (fig. 6).

Once the precast slabs are positioned, negative moment reinforcing steel is placed longitudinally along the top of the beam and through the beam-to-column connection zone. Two-way reinforcement is placed on the precast floor slab.

Shear reinforcement

The concrete structural topping is placed over the entire floor system, resulting in monolithic connections. Construction of subsequent stories begins with precast columns being slipped over the protruding reinforcing bars from the lower column. The connection is then grouted.

• RPC-K System The RPC-K system, developed by Kabuki Construction Company, utilizes U-shaped precast beams that serve as stay-in-place forms for cast-in-place concrete. Other components of the system include cast-in-place columns and Filigree type floor members. Cast-in-place concrete is used for all connections between components. Longitudinal and shear reinforcement are embedded in the precast portion of the beam. The longitudinal reinforcement protruding from the precast portion is bent upward into the column to provide anchorage. Additional longitudinal reinforcement is placed in the trough of the beam once the beam is in place.

U shaped precast beam

Field placed reinforcing bars

 IMS System The IMS system, also known as the IMS-ZEZELJ system, originated in Serbia, Yugoslavia, at the Institute for Testing of Materials. Because the system can be designed for • IMS System seismic forces, it has gained acceptance in regions with frequent seismic activity. The State The IMS system, known asHungary, the IMS-ZEZELJ system, Building Co. ofalso Baranya County, further developed the system for longer spans and for greater flexibility in accommodating utilities. system originated in Serbia, Yugoslavia, at the Institute for The Testing of has been used in Cuba, Hungary, and Yugoslavia for schools, hospitals, administrative building, offices, and hotels. Materials. Because the system can be designed for seismic forces, The IMS system relies on friction and clamping action produced by post-tensioning to transfer it has gained acceptance in regions with frequent seismic activity. vertical loads and bending moments from floor units and edge beams to the column. The postThe State Building of Baranya County, tensioning runs inCo. the floor system and through Hungary, the columnsfurther in both principal directions (fig. 7, 8).system for longer spans and for greater flexibility developed the in accommodating utilities. The used in Cuba, 3.6 m 4.8 m system has been 6.0 m Hungary, and Yugoslavia for schools, hospitals, administrative building, offices, and hotels. The IMS system relies on friction and clamping action produced by post-tensioning to transfer vertical loads and bending moments from floor units and edge beams to the column. The post-tensioning runs in the floor system and Pretensioned floor members through the columns in both principal directions. 3.6 m

To facilitate the column-to column connection, the main reinforcement from cast-in-place lower columns protrudes upward to tie in the cast-in-place upper column. Filigree type slabs rest on the precast beams and reinforcement is placed across the trough of the beam to tie the components together. Negative moment reinforcing steel is placed both longitudinally and transversely on the precast slab, and then embedded with cast-in-place concrete.

Figure 7 RPC-K System – U shaped precast shell beam. RPC-K System – U shaped precast shell beam.

a) Single-unit systems

Precast Concrete

6.0

3.0 m 3.0 m

3.0 m

3.6 m

3.0 m

3.6 m

3.0 m

3.6 m

6.0 m 3.6 m

4.8 m

3.6 m

Pretensioned floor members

Single-unit systems

Pretensioned floor members c) Four-unit systems

a) Single-unit systems

c) Four-unit systems Four-unit systems

Figure 7 Modular composition possibilities for the IMS system.

Modular composition possibilities for the IMS system Figure 7 Modular composition possibilities for the IMS system.

2.4 m 2.4 m

3.0 m

3.6 m

3.0 m

3.6 m

Prestressing strand

Grout

Precast floor unit

6.0 m

Prestressing strand

Void

Pretensioned floor members

b) Two-unit systems Two-unit systems 3.6 m

3.0 m

3.0 m 3.6 m

3.6 m

Temporary Temporary supporting frame supporting frame

3.6 m

325

c) Four-unit systems

Precast columncolumn Precast

Figure 8 8IMS System-Joint between floorfloor units units and column duringduring construction. Figure IMS System-Joint and column constructi IMS System-Joint between floorbetween units and column during construction.

Precast Concrete Pretensioned floor members

Steel angles Steel angles

Reinforcement cage Cast-in-place footing

326

• Structurapid System • Thomas System Figure 9 Structurapid System- footing-to-column and beam-to-colum The structurapid system was first used in Italy for residential and The Thomas system is comprised of multi-story precast columns, commercial buildings. The system is comprised of precast column composite shell beams, and precast double tee floor members. System tubes and T-beams. The columns and beams are connected by This systemhasThomas been used in the mid-western United States The Thomas system is comprised multi-story precast columns means of a tongue and groove system of joining. Reinforcing for multi-story office building construction. The keyofstructural and of precast double teeshell floorbeam, members. This system has been used in steel is placed on the precast T-beam and bent down into the component the system is the the beam flanges  column Structurapid System forof multi-story office building construction. The key struc hollow core. With the placement of cast-in-place concrete supportStates the stems the double tees and the beam is supported The structurapid system was first used in Italy for residential and commercial buildings. in the column cavity, a monolithic beam-to-column and columnby precast column corbels. pretension single span support shell system is the shell The beam, the beam flanges the stems of the d The systemjoined is comprised of precast column tubes andmembers T-beams. rest The columns and beams are to-column is developed. Hollow-core floor beams also serve as forms for cast-in-place supported by precast columnconcrete. corbels. The pretension single span s connected by means of a tongue and groove system of joining. Reinforcing steel isisplaced on on the beam flange. Shear reinforcement protrudes from the top forms for cast-in-place concrete (fig. 10). the precast T-beam and bent down into the hollow column core. With the placement of castof the beam to achieve continuity with the floor members. Castin-place concrete in the column cavity, a monolithic beam-to-column and column-to-column Cast-in-place in-place is Hollow-core used as a floor topping to provide a rigid floor joined is concrete developed. floor members rest on the beam flange. Shear reinforcement protrudes from the top of the beam to achieve continuity with the floor members. Cast-indiaphragm. place concrete is used as a floor topping to provide a rigid floor diaphragm (fig. 9).

Prefabricated reinforcement cage

TT- beam

Precast T-beams

Column corbel

4 story precast column

Bearing pad

Precast beam

Figure 10 Thomas system-cross section of precast beam. Thomas system-cross section of precast beam.

Precast column tube

Reinforcement cage Cast-in-place footing

Structurapid System footing-to-column and beam-to-column connection.

Figure 9 Structurapid System- footing-to-column and beam-to-column connection.  Thomas System The Thomas system is comprised of multi-story precast columns, composite shell beams, and precast double tee floor members. This system has beenPrecast used in the Concrete mid-western United States for multi-story office building construction. The key structural component of the system is the shell beam, the beam flanges support the stems of the double tees and the beam is supported by precast column corbels. The pretension single span shell beams also serve as

 minimum amount of work at the site  uncomplicated to use, both in production and during erect  no protruding parts  must function within normal tolerances at a building site Based on these requirements Partek set about the task of finding a Central in the process was Mr. BjØrn Thoresen of Partek Østspenn A to the basic concept behind the system, the sliding knife. This allows movement due to temperature and shrinkage, and eliminates protrudi

327

• BSF-System The development of the hidden corbel system called the BSFSystem. Corbels are necessity in all construction work. In steel structures corbels are usually made of small brackets and do not represent much of a nuisance. For concrete construction however, corbels tend to be rather large and bulky details, and are not very popular among architects. In office buildings, hospitals, schools etc. they may obstruct windows or block the passage of ventilation ducts. In industrial buildings they can obstruct the passage of the roof drains or be in the way of overhead cranes. Another disadvantage with concrete corbels is that they are often rather difficult to manufacture, with much reinforcement of a fairly complicated shape.

Knife Column unit

Figure 11 BSF-System – corbel system.

The ideal connection detail will have to fulfil the following minimum requirements: • not visible • easily protected against faire • minimum amount of work at the site • uncomplicated to use, both in production and during erection • no protruding parts • must function within normal tolerances at a building site

BSF-System – corbel system.

This contribution has presented a view of the several precas

This contribution has presented a view of the several precast that are suitable for building construction. Precast concrete floors structural floor systems that are suitable for building construction. kinds of structures, e.g. single storey, multi-storey and high-rise buil Precast concrete floors can be used in different kinds of structures, and seismic areas. Whether the resulting building is stylish or dull d e.g. single storey, multi-storey and high-rise buildings both in and design and the capability of designers. non-seismic and seismic areas. Whether the resulting building is stylish or dull depends only on planning and design and the capability of designers.

Based on these requirements Partek set about the task of finding a solution to this challenge. Central in the process was Mr. BjØrn Thoresen of Partek Østspenn A.S., and his creativity led to the basic concept behind the system, the sliding knife. This allows for deviations at the site, movement due to temperature and shrinkage, and eliminates protruding parts

Precast Concrete

328

Precast Concrete

329

Precast Concrete

330

Precast Concrete

331

Precast Concrete

332

Precast Concrete Realisation

333

A1m Placing of Peripheral PC Beam on Lateral Column

334

A1m Placing of Peripheral PC Beam on Lateral Column

335

B´1m Placing of Peripheral PC Beam on Lateral Column in Place Dilatation

336

B´1m Placing of peripheral PC beam on Lateral Column in Place Dilatation

337

B1m Placing of Peripheral PC Beam on Lateral Column

338

B1m Placing of Peripheral PC beam on Lateral Column

339

C´1m Placing of Internal PC beam on Column in Place Dilatation

340

C1m Placing of Internal PC Beam on Column

341

D1m Placing of Peripheral PC beam on Lateral Column

342

E2m Placing of Internal PC Beam on Column

343

F1m Placing of Peripheral PC Beam on Lateral Column

344

F1m Placing of Peripheral PC Beam on Lateral Column

345

G1m Placing of Internal PC Beam on Column

346

G1m Placing of Internal PC Beam on Column

347

internal

periphetal

H I 1m Placing of TT-Slab on Internal and Peripheral PC Beam

348

K1m Placing Peripheral Bracing Beam on Lateral Intermediate Column

349

L1m Placing Peripheral Bracing Beam and Girder on Lateral Column

350

L1m Placing Peripheral Bracing Beam and Girder on Lateral Column

351

M1m Placing Girders and Beam on Intermediate Column

352

N1m Placing Girders and Beam on Intermediate Column

353

O1m Placing Peripheral PC Beam and Girder on Lateral Column

354

P1m Placing Peripheral PC Beam on Lateral Column

355

R1m Placing PC Girder on PC Girder

356

R1m Placing PC Girder on PC Girder

357

S1m Placing PC Girder on PC Girder

358

S1m Placing PC Girder on PC Girder

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Precast Realisation Projects

360

Precast Realisation Projects

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Precast Realisation Projects

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Precast Realisation Projects

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Precast Realisation Projects

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[1] Davidovici, V.: Béton armé, aide - mémoire. Bordas, Paris, 1974 Boulet, B.: Aide - mémoire du second oeuvre du batiment. Bordas, Paris, 1977

[8] Prior, R. C., Pessiki, S., Sause, R., Slaughter, S., van Zyverden, W.: Identification and Preliminary Assessment of Existing Precast Concrete Floor Framing System, ATLSS Report 93-07, August 1993, 141 pp.

[2] FIP Recommendations ´Design of multi-storey precast concrete structures´. 1986 Recommendations on precast prestressed Hollow-Core Floors Th. Telford Publ., London 1988

[9] Tadros, M. K., Einea, A., Low, S.-G., Magana, and Schultz, A. E., ‘Seismic Resistance of a Six-Story Totally Precast Office Building, ‘Proceedings, FIP Symposium, 93, Kyoto, Japan, Oct. 17-20, 1993.

[3] Vecchio, F. J.: Reinforced Concrete Membrane Element Formulations. Journal of Structural Engineering, Vol. 116, No. 3, March, 1990 PCI Technical Report no. 2. Connections for Precast Prestressed Concrete Buildings including earthquake resistance. March 1992.

[5] Elliot, K. S.; Torey, A. K.: Precast concrete frame buildings, Design Guide. British Cement Association 1992

[6] Yee, A.A.: Design Considerations for Precast Prestressed Concrete Building Structures in Seismic Areas, PCI Journal, Vol. 36 No. 3, MayJune 1991.including earthquake resistance. March 1992. Eibl, J.: Concrete Structures. Euro - Design Handbook. Karlsruhe, 1994 – 96 [7] Van Zyverden, W., Pessiki, S., Sause, R., and Slaughter, S., “Proposed Concepts for New Floor Framing Systems for Precast Concrete Office Buildings“ATLSS Report No. 94-05, Center for Advanced Technology for Large Structural System, Lehigh University Bethlehem, PA, March 1994, 49 pp.Ontario Precast Concrete Manufactures Association and PCA, Duotek, Portland Cement Association, Skokie, IL.

References

365

366

367

The experiments, the results of which are presented in this part were designed and realised with the aim of determining the deflections and strain energy, as well as the work of external load. A method of measuring the deformations was applied, making possible to separate the bending induced deformations from those induced by the shear force. At every loading level were the characteristics of cracking and the deformations of the basic fictitious truss system measured by means of mechanical strain deform meters. The principal member property that influences short-time deflection is the value of the flexural rigidity, EI. The modulus of elasticity of concrete, E=Ec depends on factors such as concrete quality, age, stress level, and rate or duration of applied stress. A knowledge of the modulus of elasticity of concrete is necessary for all computations of deformations as well as for design of sections by the working stress design procedure. The term Young’s modulus of elasticity has relevance only in the linear elastic part of the stress-strain curve.

Both concrete and steel stresses fall within the initial linear ranges of the stress-strain relations and hence the assumption of linear elastic behaviour is justified. The moment of inertia, I, of a section of a beam is influenced by the steel percentages as well as the extent of flexural cracking, and hence is not a constant along the span even for a beam of uniform cross-section. Flexural cracking will depend on the applied bending moment at a section and the modulus of rupture of the concrete. During the first time loading of a reinforced concrete beam, the portions of a beam where the applied moment is less than the cracking moment M < Mcr, remain uncracked and hence the moment exceeds Mcr, the concrete in tension fails at the outer fibres and develops flexural cracks at random spacing. However, the concrete close to the neutral axis and in between the cracks still carries some tension and contributes to the effective stiffness of the beam. The tensile stress in the steel in between cracks may be as low as 60 percent of the stress at the cracks due to the tension contribution of concrete - the so-called „tension stiffening“ effect.

The modulus of elasticity of the concrete, Ec, is affected by the modulus of elasticity of the cement paste and that of the aggregate. An increase in the water-cement ratio increases the porosity of the paste, reducing its modulus of elasticity and strength. This is accounted for in design by expressing Ec as a function of fc. Since the tensile stress in concrete is below the modulus of rupture, the assumption that the section is uncracked is correct.

In order to take into account this behaviour, it is convenient to introduce the average moment of inertia of the cross-section Im, which lies between the value I2 referring to the completely cracked cross-section and the value I1 referring to the uncracked crosssection, when computing the values I1, and I2 the contribution of reinforcing bars has to be taken into account by means of the coefficient n representing the ratio between steel and concrete Young’s module.

Investigation of Reinforced Concrete

368

The closer is the value of the bending moment M to the first cracking moment Mcr the closer will be Im to I1, on the other hand, the higher is the bending action with respect to Mcr the closer will be Im to I2. Stationary load causes an increase of the deflections of beams because of the interaction between the shear force and the bending moment. This phenomenon also effects the values of the strain energy.

The behaviour of reinforced concrete structures under service conditions can be considerably different from linearly elastic behaviour due to the cracking of concrete which, even though it arises only in a limited number of cross sections, modifies the stiffness of the structural elements and therefore leads to non-negligible redistribution of the stress resultants. Reinforced concrete elements subjected to bending are characterized when the bending moment is higher than the first cracking moment, by the formation of cracks which lead to transferring the tensile loads to the reinforcing bars while the concrete comprised between two consecutive cracks is still reacting this effect is usually referred to as tension stiffening.

In the case of a reinforced beam, a certain amount of the strain energy is dissipated in crack formation and propagation, and in other irreversible deformations. However, there are still not enough experimental results available for the calculation of strain energy. It is suitable if the system of the measured values is proposed in advance enabling the determination of the strain energy as exactly as possible.

Sabah Shawkat © Distribution scheme of interconnected measurement basis for deformation and distribution of deflection - meters: a - measurement basis, b - deflection meters The evaluation of the deflection of reinforced concrete beams is strongly affected by the cracking of concrete. In order to develop a correct calculation of deflection, it is necessary to

Deformation Behaviour of Reinforced Concrete Beams

369

define an adequate tension-stiffening model and the exact prediction of the values of shear and bending stiffening.





V GAc

 el

is shear deformation, the ratio of the shear force to the value of shear

The magnitude of the shear strain is highly dependent on the formation of inclined cracks.

stiffness GAc of the transformed cross-section

If no shear cracks exist (state I in shear), the shear deformations are usually small and can

The shape coefficient  has a value of 1.2 for the rectangular section.

in most cases be neglected. After the full development of inclined cracking (state II in shear), shear deformations can be quite large, even larger than deformations due to flexion. Theoretical and experimental analysis of the influence of the short term stationary load on deformation properties of reinforced concrete rectangular beams at different load levels are given.

The first member of relationship represents deflection due to the effects of bending moments, and the second member represents deflection due to the effects of shearing forces on the total deflection the effect of second member on the deflection is very small and can be ignored in most cases. The deflection am corresponds to the effects of the rotation, deflection av to the effects of

Deflections due to bending moment and shear force calculated from deformations of the fictitious truss system, using a method based on Willot-Mohr translocation polygons. Strains of the upper and lower chords and the diagonals of the truss analogy system were measured by means of deform meters, while the lengths of the bases were 140 mm for the upper and lower chords, 203 mm for the diagonals and 140 mm for the verticals.

shearing deformation, and atot are the total deflections. Moreover, we determined the relative value of the influence of shear forces (av/atot) and bending moments (am/atot) on the beam deflections due to the load. This influence was 17% for The series I, 21% for the series II, 18% for the series III and 23% for the series IV. This

From the ratio of the geometrical dimensions L/h = 5.6 (where L is the span, h is the depth of

the test beams) it is evident, that the influence of the shearing deformation on the total deflection will be significant.

The primary aim is to find such a coefficient that would describe the state in which is the effect

indicates that the beams with higher amount of shear reinforcement show smaller growth of deflection due to shear strain in comparison to the beams with lower amount of shear reinforcement. The coefficient  depends on the shape of the cross-section. Its value in the

linear state is 1,2 for the rectangular section.

of load and the consequent development of cracks causes reduction of beam stiffness and an increase of deflection.

These coefficients are supposed to facilitate an easy and simple determination of the influence of bending moments and shearing forces on deflection.

Deflection according to virtual work is usually calculated by means of the following expression.

    

   M M ds    Ec Ic  

 VV GAc

 where M , V are the values of the inner force from the unit force in the cross-section in which the deflection is calculated. M 1 I Ec Ic calculation

 el

is the curvatures (the flexural stiffness EcIc being taken into is that of the transformed section)

Deformation Behaviour of Reinforced Concrete Beams

370

avexp





 i Vi s



i1

where i are shear strains caused by loading force, s - length of measurement base, n - number of measurement bases along the beam. Coefficients for the average bending and shear stiffness were derived from the given deflections amexp, avexp and amtheor, avtheor, at each loading level  cr

amexp

avexp

 cr

amtheor

avtheor

      M M VV   s    d   ds cr  cr  Ec Ic GAc    

For the calculation of the reduction of flexural rigidities, we also used another approach based on the calculation of the flexural rigidities in linear state I and the flexural rigidities in state II after the development of cracks. The bending stiffness in state II was evaluated from the diagrams bending moment versus the curvature relationship

M EIc

 II

 pl

EIc



( - inclination)

Deflections of beams due to a) shearing deformation av, b) due to rotation am, c) total deflection atot The curvaturesi were determined by means of the deformation of the upper and lower chord of the truss analogy system. Deflection due to the bending moment in the middle of the beams was calculated as well and indicated by the index "exp" as we departed from the measured values n

amexp



  1 M    i s   i 

i1 From the horizontal and vertical displacements of the truss analogy along the beams, shear

strains were determined as well as deflections av, exp due to shear forces along the beams at all loading levels. Diagram bending moment versus curvature

Deformation Behaviour of Reinforced Concrete Beams

371

For the calculation of the reduction of shear strain, we used a similar approach departing from the calculation of the shear rigidities in state I and the shear rigidities in the state II after the

i1 1

 crII´

 GAc

 pl

 GAc



  cri  s    L1 



 crc

i1

 cr.i

 GAc G Ac

then

development of cracks from the diagram shear force versus the shear stain relationship.

  cri   s    L 



 cr

 crII´c

 cr

 crII´   crII´c

 crII

 crc

( - inclination)

After the arrangement of the calculated deflections according to virtual work, the deflection will be present in a new form. We can write:     

  d s    Ec Ic  crII  

 M M

 V V G Ac  crII

By means of the coefficients  m ,  m, which express the increase of the relative average of curvature and shear strain of the reinforced concrete tests beams in state II it’s possible to calculate the deflection after the full development of cracks.



k

i1 n



 ck

i1

k

m

Diagram shear force versus shear strain



k

   si   ci  Ec Ic   s   h Mi L1  

  ck

i1 n

i1

m



 ck

k

   i GAc  s     Vi L    i GAc  s     Vi L1 

  ck 2

The evolution of the deformations of the bases of the fictitious truss system enables to

Summarise of the evolution of the deflection at the mid span of the beam in state II

determinate the flexural stiffness  EJc of the test beams in the whole range of the loading

1. By means of curvature

force, using the diagrams "moment-curvature" as basis. A similar method was applied for

the calculation of the reduction of the shear stiffness- the shear stiffness of the transformed cross-section and the shear stiffness in the whole range of the loading force  GAc were applied.

 cr.i

 crII´

 EJc Ec Ic 1  cr

then

 cr



i1  crII´c

1  crc

  cri   s    L    crII

 crc



i1  crII´   crII´c 2

   cri s   L    1 





   i Mi s  

 i and

shear strain



 i along

the beam in state II



   i Vi s 

2. Through the coefficients  cr and  cr, which express the increase of the deflection due to bending moment and shear force of a reinforced concrete beam in state II    Mi  Vi     a  Mi   cr s    Vi    cr s   Ec Ic    GAc  



3. By means of the coefficients  m ,  m, which express the increase of the relative average of curvature and shear strain of reinforced concrete beam in state II

Deformation Behaviour of Reinforced Concrete Beams

372

 i



  Mi     m  Mi s     Ec Ic   

 i



  Vi     m  Vi   s  G  A c    

4. Through the coefficients  crII ,  crII, which express the reduction of the relative average of bending stiffness and shear stiffness of reinforced concrete beam in state II    Mi      Vi a  Mi  s    Vi   s   Ec Ic  crII    GAc  crII  



a) at failure (level load  s 1.55 ):  cr 2.95 , when the effect of bending moment on a total deflection was 79,30 % and on the total strain energy was 78,1 % respectively b) at service load (level load  s 1.0 ):  cr 2.57, when the effect of bending moment on a total deflection was 83,5 % and on the total strain energy was 82,4 % respectively The values of coefficient  cr versus of the level of loading shown on Figures bellow.

5. By means of the coefficients , which express the reduction of the moment of inertia of the section in state II, and the coefficient  1

0.85,

which express the effect of the

increase of the initial deformation at loading   Mi   a  Mi  s    1Ec Ic   

 i

6. By means of the coefficient , which express the reduction of the moment of inertia of the section in state II   Mi   a  Mi  s   Ec Ic   

 i

Average strains of the upper chord

7. By means of the coefficients ,  ck, which express the reduction of the moment of inertia and the reduction of the area of the cross-section of the beam in state II    Mi    Vi   a  Mi  s    Vi   s  E  I   G    A ck c  c c     



These coefficients express the state after crack development, while its values increase with an increased loading level, i.e., cracks development. The advantage of these coefficients are the fact that it permits the direct separation of the effect of shear forces on the deflections of

Average strains of the lower chord

reinforced concrete beams from the effect of bending moments. With these coefficients, the deflection due to bending moments and shear forces in state II will be: From the portion of the deflection amexp and amtheor respectively due to bending moment we determined the coefficient  cr which express the deflections of beams after the full development of cracks (state II).

Deformation Behaviour of Reinforced Concrete Beams

373

The value of coefficient  k we obtained as follows, first we determined the portion of curvatures  pl in state II to the curvatures  el in state I as  i along the beam then from the values of the portions  i we calculate the average value on span L, where L is the distance between the supports of beam i

 pl

 i



  L  s 

k

then

 el



a) at failure (level of loading  s 1.55 ):  k 3.29 b) at service load (level of loading  s 1.0 ):  k 2.75

Sabah Shawkat © Diagram of loading level versus deformations coefficients Average strains of the diagonals From the portion of the deflection avexp and avtheor respectively due to shear forces we determined the coefficient  cr which express the deflections of beams after the full development of cracks (state II) a) at failure (level of loading  s 1.55):  cr 8.82 , when the effect of shear forces on the total deflection was 20.71 % and on the total strain energy was 21.91 % b) at service load (level of loading  s 1.0):  cr 5.69 , when the effect of shear forces on a total deflection was 16.48 % and on the total strain energy was 17.60 %

The value of coefficient  k we obtained as follows, first we determined the portion of shear strain  pl in state II to the shear strain in state I  el along the whole of RC beams, then from the determined ratio of  i we calculated average value of the coefficient  k on span L, where L is the distance between the supports of beam

i

 pl  el

then

k



i

  L



 

s 

The values of coefficient  cr versus of the level of loading shown on Figures bellow.

Deformation Behaviour of Reinforced Concrete Beams

374

a) at failure (level of loading  s

1.55

b) at service load (level of loading  s

):  k 1.0

):  k

5.35

, 3.80

The value of coefficient  k versus load stage shown on Figures bellow.

The value of coefficient  crII´ we obtained as follows, first we determined the portion of bending stiffness in state II to bending stiffness in state I along the reinforced concrete

Diagram of loading level versus deformations coefficients

beam then from the ratio values we determine average value along the length of RC beam where L, is the span of the RC beam between the supports.

cr i

( EJ) c Eb Jb

then

 crII´

a) at failure (level of loading  s

 cr i

 

1.55

b) at service load (level of loading  s

 

s 

):  crII´ 1.0

):  crII´

0.27

0.314

The value of coefficient  crII´ versus load level is shown on Figures bellow.

Deformation Behaviour of Reinforced Concrete Beams

375

Diagram of loading level versus deformations coefficients

a) at failure (level of loading  s

1.55

b) at service load (level of loading  s

): 

1.0

0.308,

): 

0.359 .

The value of coefficient  i versus level load shown on Figure bellow.

cr i

GA

 GAi c

 cr

 rc i

  L



a) at failure (level of loading  s

1.55

b) at service load (level of loading  s

 

s  then

):  crII´ 1.0

 crII´

 cr

0.184 ,

):  crII´

0.27

Diagram of loading level versus deformations coefficients

Deformation Behaviour of Reinforced Concrete Beams

376

Diagram of loading level versus deformations coefficients

 Ji

s

Jck

s

then

a) at failure (level of loading  s

1.55

b) at service load (level of loading  s Acr

 ck

1.0

Jck



JbII

0.250,

): 

0.29 .

 ck Ab

 Ai Ak

): 

JbII



 Ai  s

Ack

Ab  s

then

AbII

Diagram of loading level versus deformations coefficients

Deformation Behaviour of Reinforced Concrete Beams

 2

Ab Ack

and

377

Diagram of loading level versus deformations coefficients

Deformation Behaviour of Reinforced Concrete Beams

378

Formation, development and width of cracks

We can also note that the maximum widths of shear cracks were in the dependence on loading

The bending cracks first appear in the region of the maximum moment, then follows the

growing more intensively than the widths of bending cracks.

development of cracks in the shearing forces region. These cracks slightly incline in the

On the basis of all diagrams we can state that immediately after the crack formation was the

direction of principle stress compression trajectories and finally a shear crack (from a new or

process of beam failure concentrated in to the regions influenced by the shear forces, especially

bending crack) is formed which prolongs to the compression region and to the tension beam

their middle parts.

end. In this cracks appears a failure which is called critical crack.

Evaluation of cracks

Character of shear crack formation in the zone of shear force acting can have two forms: a) Cracks begin from the tension side of elements as a normal crack in bending region and

Considering that the most calculation methods for widths of bending and shear cracks used the

they further develop in the direction of principle stress compression trajectories.

average crack width to determine the limit crack width, the average crack width at is the most

b) Shear cracks appear solitary in the middle zone along the element depth.

important evaluated parameter. The measured values of average bending crack widths at the

We observed the formation, development and widths of cracks at each loading level. by the means of optical device, we plotted course of cracks and each crack was labelled with a number as it was formed. A short perpendicular line indicated the development at the corresponding

level of tension reinforcement in the dependence on the moment were compared to the average values of bending crack widths calculated according to CSN 73 12 01- 86. The values calculated according to the CSN were higher than the measured once while the values measured on beams in the series II and III were higher than in the series I and IV.

loading level. The crack widths were measured at three levels: 1. At the level of tension edge concrete fibre.

We also evaluated the dependence of the sum of bending crack widths at the level of tension reinforcement on the moment, it is obvious that the maximum values were obtained on the

2. At the level of the centre of gravity of the tension reinforcement.

beams in series II and III where the distance between the loads was smaller than in the series I and IV. Further we compared the measured values of average widths of shear cracks at the level

3. At the middle of the beam depths.

of tension reinforcement on the left and right side of the beams in the dependence on shear force

Experimental beams of series I, II, III and IV were subjected to short term gradual loading. We

to these value calculated according to CSN 73 12 01 - 86.

plotted the process of the development of bending and shear cracks on the diagrams both on the

At the beginning of the loading and at the service load (= 1) were the values of average crack

level of tension reinforcement and in the middle of the beam depth.

widths smaller and less dependent on the shear reinforcement than at higher loading levels.

Inclinations of shear cracks located closest to the supports were measured on the left and right side of the beam. The length of inclination angle was given by the intersection of the crack with the longitudinal reinforcement and the end of the crack by the crack failure. After the formation of shear cracks an increase of loading caused the growth of the crack widths. In the dependence on the Percentage degree of shear reinforcement and the ratio a/h (where a is the distance of load from the support and h is the section depth) the crack width can gain the large values, even higher than the limit values. Maximum values of width of the bending and shear cracks wm and wq determined for all beams series at each loading level According to our investigations for all rectangular beams it is clear that the short-term gradually growing loading caused the formation and development of cracks along the whole beam span.

We also evaluated the dependence of calculated values of average shear crack widths at the level of tension reinforcement and in the middle of the beam depth on the left and right side on the loading. We found out that the maximum average values of shear crack widths were observed in the beam series II and IV on the left side of the beams, namely 0,22 mm and 0,24 mm at the maximum loading level before failure. We also came to the conclusion that the width of shear cracks is the bigger the smaller and less suitable the shear reinforcement and that the shear reinforcement influences the width of the shear cracks approximately in the middle of the shear span of beams and this only at the higher loading levels, at the level of tension reinforcement this influence is small. Further we investigated the dependence of the sum of shear crack widths at the level of tension reinforcement and in the middle of the beam depth on loading.

Deformation Behaviour of Reinforced Concrete Beams

379

Side view and finite crack pattern of tested beams

Deformation Behaviour of Reinforced Concrete Beams

380

25 20 15

Stresses bc

- The concrete quality had an influence on the inclination of shear cracks the higher the concrete

SERIE I

0,1 0,3 0,5 0,7

0,6.fck

5 0

0,9

1,2

4 5 6 7 8 Measurment basis

1,4

quality, the bigger the inclination of shear cracks. - The growth of the loading largely influenced the development of shear cracks and bending cracks along the whole beam span, as well as caused formation of new shear cracks. - The widths of shear cracks became bigger under the influence of loading force more intensively than the bending cracks and consequently grew the strain in the tensile diagonals in the shear span. - From the diagrams showing the dependence of concrete strain from the value of loading for each edge, also along the beam, is clear that the maximum values of strain had the tensile

Stresses bc

35 30 25 20 15 10 5 0

SERIE II

0,1 0,3

diagonals placed approximately in the middle of the shear span. This caused the increase of deflection with dependence on the effect of shear force.

0,5

0,6.fck

0,7

- The diagrams of dependence of average strain of lower chords along the beam from the load

0,9

and diagrams of dependence of curvature along the beam from the load have approximately

same form.

1,2

- The effect of shear force had also an influence on the strain of the upper chords, while

1,4

higher loading grades a decline of strain value was observed.

4 5 6 7 8 Measurment basis

Stresses of concrete versus measurement basis due to loading level

- For the reinforced concrete with given physical and mechanical specifications, as well as the cross-section type and load system, the average portion of influence of shearing forces on the

The effect of the short - term, gradually growing and on the deformation as well as the failure

beam deflection was 22% of the total deflection.

of the reinforced concrete beams can be summarized

- The higher the ratio a/h of shearing slenderness, the higher the values of average strain of

The testing elements were concrete reinforced rectangular beams subjected to the

diagonals. This fact has a big influence on the value of shearing forces, that means also on the

combination of bending and shear force The results obtained from indirect measurements of the

growth of the total beam deflection. The description of deformation of particular beams under

concrete strain by short-term gradually growing load according to the truss analogy method.

short-term gradual load shows that the process of beam deformation is concentrated into the

This method enables a separate determination of the effect of bending moments and the effect

regions of shear span were shear cracks appear.

of shear forces on deflection of reinforced concrete beams. Our results can be summarized as

After the formation of shear cracks appeared a growth of shear strain caused by shearing forces

follows:

and the moment that in this place caused a growth of curvature as a result of stiffness decline

- The perpendicular cracks did not get longer by higher loading levels, their number and widths

of the element with shear cracks. This is an interaction of the bending moment and shearing

remained relatively stable.

forces.

- At the level of tensile reinforcement were the widths of shear cracks smaller and less

- Maximum curvature was observed in the region of concentrated loads and maximum shear

dependent on the shear reinforcement (stirrups) than in the middle of the beam depth.

in the middle of the shear span.

- The stirrups had an influence on the width of shear cracks the weaker the stirrups, the bigger

- The bending slenderness l/h has a great influence on the proportion of shear forces and the

the widths of shear cracks. This effect was apparent especially in the middle of the beam depth

bending moments on the deflection, e.g. if the bending slenderness is larger than 10, the

and by higher loading levels.

influence of shear forces would be very small.

Deformation Behaviour of Reinforced Concrete Beams

381

- The average percentage ratio of the influence of shear forces on deflection was 22% for all beams. This ratio is a function of the bending slenderness. - The maximum obtained value of cr was 10 for reinforced concrete beams.

acrII 1,000

- The maximum value of cr for all the reinforced concrete beams was 2 - 3. -

Rigidity of beams can be defined by means of the diagram moment versus curvature

relationship and shear force versus shear strain relationship at the inclination angle and .

0,900

- Beam rigidities are dependent on the level of initial forces M, V, N. - Bending rigidity (EJ)c and shear rigidity (GA)c are neither a multiplication of the

0,800

modulus of elasticity E and the moment of inertia J of the gross concrete section, nor the shear modulus of concrete G and the cross-sectional area A. - The maximum average value of cr was determined for the reinforced concrete rectangular beams subjected to stationary load - 2.2 at the service load and 2.6 at the ultimate load. The average value of cr for the reinforced concrete rectangular beams was 9.4 at the ultimate

0,700

0,600

The portion of effect of the strain energy due to shear strain on total internal strain energy, was

0,500

within interval from 16 % to 24 %. This means that the average portion of energy due to

shearing forces is 20 % of the internal energy. We also determined the perceptual ratio of the

0,400

energy due to external forces. As the energy due to shearing forces we took 100 % and this ratio was than in the interval from 94 % to 106 %.

The evaluation of deflections of reinforced concrete beams is strongly affected by the cracking

0,300

of concrete. Between state I and state II in shear, there exists a gradual decline of stiffness (as in bending). In most practical cases where deformations are to be calculated for serviceability

0,200

checks, shear deformations are largely overestimated using the shear strain of state II. The development of the shear strain is a function of applied shear force. According to our results the amount of the effect of the shear forces on deflections of reinforced

0,100

y = -0,2018Ln(x) + 0,3174

concrete beams was 20% (Shawkat et all., 1993) on the average. The same value of 20% was determined by calculating the portion of strain energy due to shear forces on total internal energy.

0,000 0,00

0,50

1,00

1,50

2,00

2,50

The result reported in this paper are related to simply supported beams with rectangular sections. More tests are needed to investigate the applicability of the proposed models for continuous beams.

Diagram of loading level versus deformation coefficient acrII.

Deformation Behaviour of Reinforced Concrete Beams

382

bcrII

n av

1,200

i=1

rcr

Vi G cm. . A c .  crII i

.V . i

3,500

3,000

1,000

2,500 0,800

2,000

0,600

1,500

0,400

1,000

0,200

0,500

y = -0,2988Ln(x) + 0,2523

am i=1

0,000 0,00

0,50

1,00

1,50

2,00

2,50

0,000 0,00

i

Mi .  s

1,00

y = 0,7598Ln(x) + 2,5915

1,50

2,00

2,50

Diagram of loading level versus deformation coefficient

Diagram of loading level versus deformation coefficient bcrII.

0,50

 cr .

rcr.

Deformation Behaviour of Reinforced Concrete Beams

383

3,500

0,800

3,000

0,700

0,600

2,500

0,500

0,400

1,500

0,300

1,000

0,200

n 0,100

a tot i=1

0,000 0,00

0,50

Mi E cm. I c . i

. M . s i

0,500

y = 0,3012x-0,3976

1,00

i=1 1,50

2,00

0,000 0,00

2,50

Diagram of loading level versus deformation coefficient

0,50

 m. i

E cm. I c

1,00

i 1,50

Deformation Behaviour of Reinforced Concrete Beams

.M .

s y = 0,9125ln(x) + 2,5941

2,00

2,50

Diagram of loading level versus deformation coefficient Wm.

384

Strain energy in reinforced concrete beams

By the means of strain energy, we could more generally express also the condition of reliability, e.g. in the form:

Stationary load on a beam causes an increase of the deflection because of the interaction between the shear force and the bending moment. This phenomenon also effects

P  USS  USL  PREL

the values of the strain energy. For calculating the strain energy caused by external forces we used the results of measurements in the bases of the multiple truss analogy system (Williot-Mohr translocation polygons). It is clear that the crack development directly influences the calculation of strain energy and the deflection of reinforced concrete beams.

Where P

is the probability, that the limit state will occur

USS USL

is the strain energy needed for reaching the given state of strain is the strain energy needed for reaching the limit state

PREL is the given probability characterizing the reliability

Strain energy according to virtual work (state I) is usually calculated by means

U elastic

i 1

Cross-sections and side view of test beams

The method adopted for measuring the deformations enabled to calculate not only the



Ec Ic

 Mi  M i  s i    E  I  c c 

is curvature





GAc







i  1

is shear strain

 Vi  V i  s i   G  A c   Where

0.435Ec

work of the external forces but also the strain-energy of the internal forces (bending and shear).

For the determination of strain energy, we used tests on beams using stationary loading. Using

Due to the difficulties occurring by measuring and calculating the strain energy by the test on

a large number of measuring points, we could create suitable conditions for indirect evaluation

the reinforced concrete element this topic is discussed in literature only rarely. The

of strain energy.

development of the measuring and evaluation technique enables also investigations aimed on

At each loading level along the whole beam span we measured deflections as well as

the elaboration of a theory to judge the reliability of concrete structures by the means of strain

length changes at the nine bases on the compression and tensile chord and on diagonals using

energy.

a system of interconnected measurement bases. At the same time, we were investigating the

It would be possible to characterize by the means of strain energy the state of

process of crack development. By the means of curvature, we could evaluate the value of

construction, especially component strains, more suitably than by the values of strain or

strain energy of the bending moment Um at each load level. Then the bending-related energy

statically effects of loading. The suitability comes out of the fact that strain energy is a scalar

of the whole beam at each load level was obtained by the numerical integration along the

value what enables to express in a simple way also the reserve of construction in the given

length of the beam according to the formula:

strain state in the relation to the limit state.

Deformation Behaviour of Reinforced Concrete Beams

385

  1 M1   2 M2   3 M3   4 M4   5 M5   s    M   M   M   M  ..........  7 7 8 8 9 9  6 6 



respectively. Um

  i Mi s i

Where

i

i1

Vlk 1

shear forces) of the junctions at the lower 1 i

 si

  ci

chord of fictitious truss analogy, where k =

12,13,........k+1 Vi

is the curvatures along the reinforced concrete beam, i = 1,2...n

s i

is the base length (the same along the whole span of the beam)

is the number of the cross-sections considered including beam ends

are shear forces along the beam at the level of loading

is the bending moments along the beam at the loading level, i = 1,2, ........n

are the vertical displacements (caused by

Then we determine the amount of strain energy (internal energy) at the level of loading of reinforced concrete beams from the sum of strain energy due to curvatures and shear strain, respectively as follows: n

In the same way it’s possible to determine the strain energy due shear forces Uv by the

Uinternal



Um 

i1



i1

Sabah Shawkat © means of average of shear strain  i along the beam from the vertical displacements (caused by shear forces) of the fictitious truss analogy.

  1 V1   2 V2   3 V3   4 V4   5 V5   6 V6   7 V7   s    V   V  ..............   8 8 9 9 

The value of the work of the external forces at the load level was determined by the formula:

Wexternal

Fs2i ai

Fs2i (kN) is the loading force load at a given load level



ai (mm) is the value of deflection at a given load level and at a

  i Vi s i

given location under the loading force

i1

Average shear strain along the fictitious truss analogy at the middle distance of the depth of reinforced concrete beams was calculated according to the formula Vui 1  Vui i

s i



Vlk 1  Vlk

The amount of strain energy done due to bending moment Um and due to shear force Uv, as a part of the total strain energy (internal energy) versus load level is plotted in Figure bellow.

s k

The portion of strain energy due to shear forces and bending moment on total strain

Vui 1

are the vertical displacements (caused by shear forces) of the junctions at the upper chord of fictitious truss analogy, where i = 1,2,........n+1

energy can be given as follows:

Uv Um  Uinternal Uinternal The relative value of strain energy (internal) to external work at the level of loading is

Deformation Behaviour of Reinforced Concrete Beams

386

expressed according to the formula:

The portion of the total dissipated energy due to bending and shear after the unloading of

Uinternal

the external force to the total strain energy before unloading at the load level is expressed

Wexternal

as follows:

The evolution of the dissipated energy would also enable to determine the degradation of the stiffness under short-term, step-by-step increasing load. The bending induced strain

Udissipated Uinternal

energy U´m and the shear induced strain energy U´v after unloading of test beams will be

The relative value of dissipated energy due to bending and shear after the unloading of the

calculate according to the formula:

external force to the strain energy due to bending and shear before unloading at the load level

U´m



 ´ i M´i s i

i1

U´v



 ´ i V´i s i

i1

may be written as follows: U´m U´v  Um Uv

In case of the bending induced strain energy U´m the input quantities (data) are the

The value of the dissipated work after the unloading of the external loading force at the

curvatures ´ i determined from the formations of the upper and lower chord of the

load level can be obtained as:

fictitious truss system, in case of the shear induced strain energy U´v the input quantities

W´external

F´s2i (kN) is the unloading external force at a given load level a´i

are the shearing deformations ´ i determined from the horizontal and vertical

(mm) is the value of reversible deflection after unloading at a

displacements of the nodes of the chords of the truss system.

given load level and at a given location under the external

Then the total strain energy (internal energy) U´internal after unloading of the external

loading force

force at each of level of loading along the beam may be then calculated according to formula:

In order to compute the total dissipated external work at the load level we have to count

the amount of external work before unloading and after unloading of external force

U´internal

Wdissipated

U´v  U´m

Wexternal  W´external

The perceptual ratio of external work after unloading as well as dissipated work to The value of the dissipated energy after the unloading of the external loading force at the

external work before unloading may be then calculated according to the formula:

load level can be obtained according to the formula:

Udissipated

W´external Wdissipated  Wexternal Wexternal

Uinternal  U´internal

The portion of the dissipated energy due to bending and shear after the unloading of external loading force to the total internal energy before unloading of the external loading force at a load level can be obtained according to the formula: Um  U´m Uv  U´v  Uinternal Uinternal

Through the coefficients  cr and  cr, which express the increase of the strain energy due to bending moment and shearing force of a reinforced concrete beam in state II

Deformation Behaviour of Reinforced Concrete Beams

387



 cr



  i Mi s i

i1 n



i1

i1 n

 cr

 Mi  Mi s i    Ec Ic 







i1

  i Vi s i

Uplast



  E I  i

 Vi  Vi s i    G A c 

By means of the coefficients  k ,  k, which express the increase of the relative average

c c  crII









Mi  s  



  GA 

Vi c  crII









Vi   s 

5. By means of the coefficient , which express the reduction of the moment of inertia of the section in state II Uplast

 Mi

  E I 

c c

of curvature and shear strain of the reinforced concrete beam in state I









Mi  s 

6. By means of the coefficients ,  ck, which express the reduction of the moment of  si  ci

k

h Mi

inertia and the reduction of the area of the cross-section of the beam in state II

  si   ci  Ec Ic

i

k

h Mi



Ec  Ic



 i G A c 

Vi

Uplast

 Mi

  E I 

c c

G A c









Mi  s  



  G i

Vi  ck Ac









Vi   s 

Summarise of the evolution of the strain energy at the mid span of the beam in state II

1. By means of curvatures  i and shear strain  i along the beam in state II Uplast.

  i



 i Mi s

 

 



 i Vi s



2. Through the coefficients  cr and  cr, which express the increase of the strain energy due to bending moment and shear force of reinforced concrete beam in state II Uplast.

 i

  Mi  Mi   cr s     Ec Ic  

 i

  Vi  Vi    cr s    GAc  

Evaluation of strain energy by means of deformation coefficients Strain energy [kN.mm]

BEAM IA

Load [kN] 42 58

17,038 42,703

16,462 41,334

16,898 42,370

13,789 37,706

14,061 41,090

16,652 47,922

14,154 40,734

15,255 42,833

15,055 41,750

102

186,810

183,921

188,110

181,808

205,308

229,659

195,210

201,703

198,275

120

266,562

264,115

269,522

253,319

288,991

317,406

269,795

278,782

275,632

140

387,391

384,254

393,785

388,500

425,440

468,063

397,854

410,087

450,549

160

548,004

556,064

555,618

548,169

607,104

658,894

560,060

576,038

569,146

Load [kN]

0,553

0,591

0,490

0,773

0,992

0,655

0,895

0,784

0,645

Strain energy [kN.mm]

BEAM IB

1,317

0,900

1,170

3,064

2,372

1,451

1,233

1,509

1,336

44,833

44,457

44,519

42,991

52,648

62,125

52,806

54,906

53,689

3. By means of the coefficients  m ,  m, which express the increase of the relative

62,940

66,410

62,733

61,519

74,330

87,470

74,350

77,151

75,618

134,340

134,542

134,585

129,520

152,577

176,335

149,885

155,167

151,723

average of curvature and shear strain of reinforced concrete beam in state II

109

199,419

200,608

200,481

194,349

223,608

254,992

216,743

224,158

219,628

133

322,835

326,031

326,603

321,911

363,298

402,777

342,361

353,401

349,576

139

377,642

379,838

381,766

380,534

423,435

467,613

397,471

409,530

406,566

147

432,785

442,631

439,792

436,374

485,760

538,625

457,832

471,318

469,243

Uplast

 i

  Mi     m  Mi s     Ec Ic   

 i

    Vi   m  Vi   s  G  A c    

4. Through the coefficients  crII ,  crII, which express the reduction of the relative average of bending stiffness and shear stiffness of reinforced concrete beam in state II

Deformation Behaviour of Reinforced Concrete Beams

388 Strain energy [kN.mm]

BEAM IC Load [kN]

1,28766

1,18145

1,27366

1,30868

1,1226

1,32411

1,12549

1,21536

101,3781

3,16234

3,00538

3,04318

2,40984

3,56957

4,05826

3,44952

3,75159

3,73873

20,26162 19,9693 20,02829 17,5535

20,9363

24,8389

21,1131

22,1117

22,48131

39,3019

34,0288

41,2604

48,7293

41,4199

43,5194

48,46606

56,61149 55,5085 56,29882 48,1205

58,4852

69,1941

58,815

61,6167

64,62075

127,8839 124,941 127,5696 115,547

137,631

158,05

134,342

139,625

148,1752

100

153,3812 151,158 153,2395 140,312

166,065

189,052

160,694

166,935

178,1775

124

327,2136 328,432 329,0322 318,973

354,92

387,03

328,976

338,572

334,0537

38,4352

38,9394

Tested RC beam

Relationship between loading level and strain energy due to curvature and shear strain.

Tested RC beam

Relationship between loading level and strain energy due to curvature and shear strain.

RC beam Deformation Behaviour of ReinforcedTested Concrete Beams

389

Example

5 s   3  1   12 A s fyd 0.0035  cr  s  

x  0.010.05 2 y ( x)  0.1518 ln( x)  0.317

 b  0.0035 s  0.0018750.00188 0.01

Eb  b  d

1

 s  0.0035

1.2

1 0.8

0.8 y ( x)

 

 cr  s 0.6

0.6

0.4

0.5

1.5

0.2

3

210

 h  0.0035 d  0.01Pom  0.9

3

810

0.01

 steel 

 3 M s  b  2 A s fyd  d   b

3 Ms  3 A s fyd  d 

3

 steel  10  10 Ko 

A s fyd  Pom d b  d    h   d  fcd 2

Ko  16.75

0.19

0.18

cr  0.149

Pom  0.783

 d  0.001875

 

z  s 0.17

 h  0.0035

Ko 

A s fyd  Pom d b  d    h   d  fcd 2

cr  12 Ko

fcd Eb

0.012

 s  0.01 b  0.0035

1  1 M s  27.544kN m     s    3  1    b     

fcd

3

610

M s  A s fyd  d  

A s  4.02 cm fyd  375 MPa b  0.15 md  0.2 ml  1.12 m

cr  12 Ko

4 2

fyd  375 MPa A s  4.02  10

fcd  20 MPa Eb  27 GPa

A s fyd  0.9 d  27.135 kN m´

3

s

y ( 0.8)  0.351 y ( 0.6)  0.395 y ( 0.5)  0.422 y ( 0.4)  0.456 y ( 0.3)  0.5 y ( 0.2)  0.561

410

0.16

cr  0.325 0.15 0

3

410

810 s

Deformation Behaviour of Reinforced Concrete Beams

0.012

390

Maximum stress, initial slopes Ec of the stress-strain, factors

0.11458l   12 A s fyd  d  

1

s    3  1   0.0035   

fpl   s 

1   s    3  1   0.0035  12 A s fyd  3 b  d  Eb   Eb  b  d    0.0035 s 

 1 h   s 

1    

h s 3

h s 4

2 3

1  cr  s

810

n  0.8 

 6900

fc_

MPa

n  6.858824

0.012

fc_

e c_ 

k  0.67 

k  2.33129

Ec  MPa



ec_  0.00297

n  1

Expression, which describes the shape of a concrete cylinder stress-strain curve fc  ec 

3

410

3

410

810

ec ec_

 fc_ 

0.012

e c  0 0.0001  0.007

n k   ec   n  1   ec_      

s

0.01

1.210

1.13 ec_

1.0810

3

810

 

h   s 

 

fpl  s

cr  s

MPa

when fc reaches fc_

 

f c_

E c  3320 

fc_  103 MPa

 

fc_

9.610

8.410

3

610

 

7.210

fc ec

3

410

610

4.810

3.610

3

210

3

210

3

410

610

3

810

0.01

2.410

0.012

1.210

s

 b  0.0035

cr M s 

 st M s 

M s  27.8 kN m 12 M s

fpl  M s  0.11458M  s l 

 3 Ms  3 As fyd  d 

 

Ms kN m

b  d  Eb  cr M s 3

 

3

 0.539875  10

710

4

3

1.410

2.110

3

2.810

3.510

3

4.210

4.910

3

5.610

6.310

3

710

3

The axial force Pmax that can by resisted by a rectangular section with the linearly varying strain.

 st M s  0.011

cr M s  0.138

Eb  b  d    b   st M s  2

pr  M s   0.1268

 3 M s  b  2 A s fyd  d   b

  

fpl  M s  0.011m

1.13 ec_

3

 mpr  M s  2.985  10

fc  ec dec  0.451661fc_  ec_  1.513

fc  1.13 ec_   61.800061MPa

fc  1.13 ec_  fc_

Characteristic Behaviour of Concrete

=>the max. force is 0,6.fc.b.h

 0.600001

391

Maximum stress

Example:

fc_  21 MPa

Initial slopes Ec of the stress-strain Factors

E c  3320 

f c_ MPa

 6900

n  0.8 

fc_ MPa

k  0.67 

fcm



MPa

fc   c 

k  1.00871



c  c1

fc_ n  Ec  MPa n  1



 6900  MPa

 c1 



n n 1

ec_  0.001867

310

210

110

fcm

k  1.00871

62 MPa

fcm

n  Eci n  1

3

 c1  1.866863 10

 c  0 0.0001  clim



Strain when fc reaches fc_ ec_ 

k  0.67 

 clim  1.99  c1

n  2.035294

17 MPa

Eci   3320

fc_ MPa

fcm

n  0.8 

 c      c1 

n k

 fcm

Eci   c 

d  c   d c   c1

 



n n 1

 c      c1 

n k

e c  0 0.0001  0.007

  

 clim

Expression, which describes the shape of a concrete cylinder stress-strain curve fc  ec 

 fcm

fcm

 fc_ 

 n  1  

 ec  e   c_ 

n k

  

2.510

 

fc  c

1.99 ec_

2.2510

fc_

210

110

3

210

 

1.510

1.2510

110

7.510

510

2.510

0 0

4

710

1.410

3

2.110

3

2.810

3

3.510

4.210

3

4.910

3

5.610

  

1.99  ec_

fc  e c de c  0.800103 fc_  e c_  1.99

fc_  21MPa

fc  1.99 ec_   16.539453MPa

60000  psi  413.685438 MPa

=>the max. force is 0, 8.fc_.b.h fc  1.99 ec_  fc_

 0.787593

3

6.310

710

310

3

410

3

c

1.7510

fc ec

3

lf  30 mm

d f  0.5 mm

1   

 clim

 

fc  c d c

P max  clim   

 clim

 

f c  c d c

Pmax

fc  30 MPa

            

h  500 mm

if    0 0.01 1 2

 fc    MPa 

 b  0.6 

 b  5.793

Characteristic Behaviour of Concrete

Gf 

f  lf   b MPa 12 d f  7850

3 1

Gf  4.428  10 m

392

u  0 mm1 mm

o 

z(  )  1



3

2.04 fc



o  0.886MPa

2 d f  7850

z( 1)  7.132 10

o 3  3       

f  lf   b MPa

 b(  ) 



Calculation of the crack mouth opening displacement (CMOD)

z(  )  

Four –point bending beam test arrangement and beam dimension is illustrated in figure below.

h  z(  )

notch was cut perpendicular to the top surface at casting. The measurements which were taken during testing were the load applied, mid span displacements on both sides of the beam, the test

were carried out under CMOD control.

lf 2





Load-CMOD, Load-deflection relationship of the tested beam is illustrated below.

( u )  u du  1.6603957569650957619e7  Pa   mm  8.0   3.0   6.0

0 mm

( u )  o  1  2



u

The bending failure of concrete beam may be modelled by the development of a fictitious crack

in an elastic layer with a thickness proportional to the beam depth.

 lf 

110

0.025

810

0.02

 ( u)

610

z( )

410

0.015 0.01

210

510

3

510

0.01

0.015

0.2

0.4

0.015

0.6

0.8



2.510

210

0.01  b(  )

L  1120 h=200 h1 is a ligament height

m(  ) 1.5105

3

510

0.2

0.4

0.6 

0.8

510

t  150

ao  70

h1=130

=L/h E  210000 ft  3 ao = . h s  0.5h is shape function to be determined for every value of the span to depth ratio and is dependent

110

F is applied load

0.2

0.4

0.6

0.8

on the relative crack depth  = a / h



E = elastic modulus, ft = uniaxial tensile strength b1  1

b2  0.30

a2  0.20

a1  1.8

Where the dimensionless parameters i is defined by, if in this case I <1 the crack opening is restricted and a fictitious crack will develop in a controllable manner, otherwise the crack opening will be undefined and a brittle fracture will occur. 1

 ft a1 

s E

1

3

 1.671  10

Characteristic Behaviour of Concrete

2

 ft a2 

s E

2

4

 1.857  10

393

Here the constant c has been introduced as

c  ( 1  b2 ) 

( 1  1 ) ( 2  1 )

From the balance of the sectional stresses with external bending moment Mext can be evaluated relationship for m as a function of :

c  470.366

V2 is a function of the ratio between the length of initial notch a0 and distance from the top to bottom of specimen H:

  ao     H   v2     1  ao  H 

From the balance of the sectional stresses with external bending moment Mext can be evaluated relationship for  From the balance of the sectional stresses with external bending moment Mext can be evaluated relationship for 

v2  0.561

II. 2 3 4   ao   ao   ao   ao   5.58  19.57   36.82    34.94    12.77      H  H  H  H 

0_I

II

phase

 I_II  II_III

2   c  1  2 I_II   1  c  ( 1  c)  1  1  2 

I_II

( 1  2 ) 

 1  2 

II ( II )

 4  1  3 II ( II )  3 II ( II ) 

 1

I. phase

 ( 1  b2 ) 2 b2   2   4 II 2 ( 1  2 ) II   

1  b2  2 II

II ( II )



II ( II )

3

 II  1  2   2   c   3 II ( II ) 2        2 II     ( 6 II ( II )  3)  ( 1  b2 )  1   2  

 247.634

In terms of the point of transition from phase I to Phase II, may be found from the

condition that y1 = h, and the point of transition from phase II to III, may similarly be

found from y2 =h. these transition points, together with the point of transition between phase 0

From the balance of the sectional stresses with external bending moment Mext can be evaluated

and phase, are given by

relationship for 

II. phase

 1  b2 II_III    2  2 I. 1

From the balance of the sectional stresses with external bending moment Mext can be evaluated

2

( 1  b2 ) b2   1  2 2 

II_III

relationship for 

 821.961

III.

phase

III

phase

 II_III  120

 0_I  I_II

I ( 1 )

 1  1 

 1  1    1 

III ( III )

( 1  1 ) 

The displacement may be written as follow: I ( 1 )



 4  1  3 I ( 1 )  3 I ( 1 ) 



III ( III )

I ( 1 )

3

 1  ( 6 I ( 1 )  3)

1  1 

 1 



II_III



1   1 2 III 

 821.961 2

2

b2  ( 1  b2 )  1  2 2  2



 4  1  3 III ( III )  3 III ( III )  III ( III ) III  ( 6 III ( III )  3)  3 III ( III )  2  1  b2   b2   c   c     1   c  1   1  1        4 III 2  1  1   2 III   2   2    

Where u is the displacement of the load-point I in the direction of Fi be ui.

Examples of RC Beams

394

0  I ( I )

I ( I )  5

 10  15

2 4 6

8

I

1.5

I

0.99 0.98

 II ( II )

0.5

0 200

0.97 0.96 0.95

400

600

800

0.94 200

110

II

400

600

800

110

II

15 III ( III ) 10

0.95  III ( III )

0.9 0 0

200

400

600

800

110

0.85

III

200

400

600 III

Examples of RC Beams

800

110

395

18 16.2 14.4 12.6 I ( 1 ) 10.8 II ( II ) 9 III ( III ) 7.2 5.4 3.6 1.8 0

I. phase I_II

96 108 120

0.8  II ( II )



c  1  1 

 1  1    I_II 

( 1  1 ) 

 1  1 

I

 4  1  3 I  3 I 





I_II

I

 247.634

 0.95

  I_II  ( 6 I  3) 1  1  I

  h 3       L   H   2  h v2  1 I  u1 2 h  E I_II  2 3 s  s ft L  H s     h    

 I ( 1 )

( 1  c) 

I

Given

1 II III



1   1c 2 



0.6

 III ( III ) 0.4

I

 1.404

u1  1.5

u1  Find( u1)

u1  0.999

0.2

II. phase



1 II III

PI ( 1 )  I ( 1 ) 2 ft h 

26 23.4 20.8 18.2 PI ( 1 ) 15.6 PII ( II ) 13 PIII ( III ) 10.4 7.8 5.2 2.6 0

II

3 L1000



1  b2   2  2

 1  2 

2

( 1  b2 ) b2   1  2 2 

1  b2  2 II_III

II_III

 821.961

2  ( 1  b2 ) b2   2   4 II_III 2 ( 1  2 ) II_III   

( 1  2 ) 

PII( II )  II ( II ) 2 ft h 

3 L1000

PIII( III )  III ( III ) 2 ft h 

II_III

100

3 L1000

2  c  3 II 2    3   2  II   II_III 2    II_III  ( 6 II  3)  ( 1  b2 )   II  4  1  3 II  3 II  1  2  1  2 

u2  4.5 Given

  h 3      L h  v2 E H      2 II_III   1 II  u2 2 h  2 3 s  s  ft L  H s      h  0

1 II III

96 108 120

u2  3.29

Examples of RC Beams

u2  Find( u2)

396

I. phase 1  0_I  I_II

i  1  25

 I_II  I 0

I i

 I i 1 

I i

 1  1 

I i

 4 1  3 I i  3  I i 

I_II

 0_I

 247.634

0.8

 1  1    I i 

uIi

( 1  1 ) 





uI0  ft 

I 0

 I i 3 

1  1 

0.6 0.4 0.2

I i   6 I i  3

100

200

L  h  L  h         h v2 h E  H   6  H  

uIi  uIi 1 

300

I i

uI0  0.01

300

 u1  uI0

200

I j

0.9

I i

0.2

0.4

0.6

0.8

0.6

0.8

uIj

0.8

0.7

2.5

0.6

100

200

I j

300

I i

2 1.5

3 1 2.5 I i

0.4

Given

1.5

0.2

uIj

100

200 I i

300

  h 3       L   H   2  h v2  1 I  uI 2 h  E I  2 3 s  s ft L  H   s   h    

Examples of RC Beams

uI  Find( uI)

397 2

PIj  I j 2 ft h 

300

3 L1000

200

I i

10 PI j Y

100

8 6 4

0 0



 PIj uIj 

uIj

110

 112.199

j0

800

II. phase 1

II i

 0_I  I_II

II_III

II i

i  1  60

II 0

I_II

 I_II

400

 821.961

 II i 1 

( II_III  I_II )

uIIi  uIIi 1 

200

1  b2 II i  1  2   2 II i uII0  u1

600

 247.634

 u2  uII0 60

2  ( 1  b2 ) b2  ( 1  2 )   2   4  II  2 ( 1  2 ) II i  i  

110

800 uII0  0.999

2  c   3  II i 2      II   II 2 i i  2    II i   6 II i  3  ( 1  b2 )   II i  4 1  3 II i  3  II i  1  2  1  2  3

600 400 200

Given

  h 3      H L h  v2 E     II   2  1 II  uII2 h  2 3 s  s  ft L  H   s    h 

III i

0 0

uII  Find( uII)

Examples of RC Beams

20 i

398

I i

0.9

2.5 I i

0.8 0.7 0.6

2 1.5

0.99

1.4 1.2

0.98

 II i 0.97

II i 0.8

0.6

0.96 0.95

0.4

0.2

0.98 0.96

 III i 0.94

III i

0.92

0.9 0.88

20 i

Examples of RC Beams

399

110

0.8 0.6

uIi

800

I j II k

0.4

III l

0.2 0

600 400 200

uIj uII k uIII l

4 3

I j

II k

5.5

5 uIII i

4.5

I j

II k

3.5

III l

2 uIj uII k uIII l

PIj  I j 2 ft h 

uIj uII k

t 3 L1000

PIIk  II k 2 ft h 

t 3 L1000

Examples of RC Beams

400

PI j

PII k

2 0

uIj uII k

 4  III   i 

1

 2

2





2

PIIk 

PII6.354 k 

7.076 PIII l 

4.527 11.228

6.097 5.864

8.463

6.354

6.097 5.652



 2 III i  

7.723

7.076

7.723 9.315

5.864 5.457

8.463 10.303

5.652 5.278

9.315 11.456

11.059 11.313

5.113 5.457

12.815 10.303

10.882 11.207

4.959 5.278

14.43 11.456

10.686 11.059

4.817 5.113

16.371 12.815

10.475 10.882

4.684 4.959

10.686

4.817

10.253

Y PII k 15

4.559

21.642

16.371

4.441

25.287

9.784

4.331

29.938

9.541

4.226

36.001

10.253

4.684 4.559

10.022 4.441 INVESTIGATION 9.292 4.127

18.731 14.43

10.022

10.475

9.784... 9.541

4.331 ...

INVESTIGATION 4.226

18.731 21.642

25.287 44.11 29.938 ... 36.001

9.292 Deflection due 4.127 todue shearing force Deflection to shearing44.11 force

...

uIj uII k uIII l

III. phase



i  1  40

III 0

 III i 1 

uIIIi  uIIIi 1 

u2  3.29

 II_III

II_III

 821.961

uIII0  u2

( 142  II_III)

 5  uIII0

 1

V   G c A c

 1

Examples of RC Beams

b d



0.435

V E c A c





 fcd  4      

    0.5  L  

V E b d    G c A c 

0.435

M L

 bd

 fcd



0.435

L 4

 

   

2

V 0.435  E c  b  d

 fcd  d Ec



 fcd  4      

M L

2

  0.5  L    

 fcd  d

V  E    0.435  E c  A c 0.435



V E c b d

0.435

0.85

V     0.435  E  b  d   0.5  L c  

 1  0.5  L

4M

L

0.5



W 

 f cd

   v V  0.435  E  b  d   0.5  L c  

 b d

 

 a 

M 

 b d



W 

 1

 L  W  W 2 bd f

 1 1  0.5  L

uIII0  3.29

 

 a 

 f cd

L

W 

av 0.5  L

 W 



 L

W 

b d

III i

1  1 

PIIIl 

11.345 11.207

11.228 11.313

2

PIj 

10.685 11.345

PIII l

PIj 4.527 10.685

PI j

 ( 1  b2 ) 2 b2  2  ( 1  b2 )  b2   1  2 2 

2 3 2 III i  4  1  3 III i  3  III i 2  III i  3III i   6 III i  3  3  III i  2 III i  4  III i   III i  III i   6 III i  3  3  2 III i  1  31III i 3 b2 b2   c   c      1  b2  c  1  b2  1  1       c  2 1 c    2 2     4  III i  1  2   c  1    1  11 1  2III  i     2



1 1  III  1  III i i 1  2 III i 1  2 III i 

V   0.5  L    0.435  E c  b  d



4M



V 0.435  E c 

     

2

V 0.435  E c

 fcd  d

0.435

Ec

0.435

401

[1] ACI: Cracking of concrete members in direct tension. ACI Journal, Vol. 83, January - February, 1986

[10] Goto, Y.: Cracks Formed in Concrete Around Deformed Tension Bars, Journal of the ACI, No. 68, April, 1971

[2] Beeby, A. W.: The Prediction of Crack Widths in Hardened Concrete, Cement and Concrete Association, London, 1979

[11] Gupta, A. K.: Post cracking Behaviour of Membrane Reinforced Concrete Elements Including Tension-Stiffening. Journal of Structural Engineering, Vol.115, No. 4, April, 1989

[3] Bjarne, Ch. J.: Lines of Discontinuity for Displacements in the Theory of Plasticity of Plain and Reinforced Concrete, Magazine of Concrete Research, Vol. 27, No. 92, September, 1975 [4] Brooks, J. J., Neville, A. M.: A comparison of creep, elasticity, and strength of concrete in tension and in compression. Magazine of Concrete Research, Vol. 29, 1977

[12] Hsu, T. T. C.: Torsion of reinforced concrete. Van No strand Reinhold, New York, 1984 [13] Ismail, M. A., Jirsa, J. O.: Bond deterioration in reinforced concrete subject to low cycle loads. ACI Journal, Vol. 69, June, 1972

[5] CEB - Bull. 124/125 - F: Code modéle CEB - FIP pour les structures en béton. CEB, Paris, 1980

[14] Klink, S. A.: Actual Elastic Modulus of Concrete. ACI Journal, September - October, 1985

[6] CEB - Bull. 156 - F: Fissuration et déformations. École Polytechnique Fédérale de Lausanne,1983.

[15] Leonhardt, F.: Reducting Shear Reinforcement in Reinforced Concrete Beams and Slabs, Magazine of Concrete Research, Vol. 17, No. 53, December, 1965

[7] CEB - FIP Model Code 1990, Comité Euro - International du Béton, 1991

[16] Leonhardt, F.: Recommendations for the Degree of Prestressing Prestressed Concrete Structures. FIP Notes 69, July - August, 1977

[8] CEB - Bull. 159: Simplified methods of calculating short term deflections of reinforced concrete slabs. Paris - Lausanne, 1983

[17] Leonhardt, F.: Crack Control in Concrete Structures. ACI Journal, July - August, 1988

[9] Consenza, E., Greco, C.: Comparison and Optimization of Different Methods of Evaluation of Displacements in Cracked Reinforced Concrete Beams. Materials and Structures, No. 23, 1990

[18] Leonhardt, F.: Vorlesungen uber Massivbau. Vol. 4, 1978 [19] Lenkei, P.: Deformation capacity in reinforced concrete slabs. In: IABSE Colloquium - Plasticity in reinforced Concrete, Copenhagen, 1979

References

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[20] Placas, A., Regan, P.E.: Limit - state design for shear in rectangular and „T“ beams.Magazine of Concrete Research, December, 1970

[29] Shawkat, S., Cesnak, J.: Crack Development and Strain Energy of Reinforced Concrete Beams. First Slovak Conference on Concrete Structures, Bratislava, September 1994

[21] Placas, A., Regan, P.E.: Shear Failure of Reinforced Concrete Beams. Journal ACI, October, 1971

[30] Shawkat, S., Cesnak, J.: Deflection of Reinforced Concrete Beams due to Actions of Shearing Forces. Proceedings of an International Conference RILEM. Failures of Concrete Structures, Štrbské pleso, 1993

[22] Rehm, G., Eligehausen, R., Mallee, R.: Limitation of Shear Crack Width in Reinforced Concrete Construction. University of Stuttgart, Heft 6, 1983 [23] Rehm, G.: Berechnung der Breite von Schubrissen in Stahlbetonbauteilen. Ausfsatz fur CEB. Stuttgart, 1977

[31] Shawkat, S., Križma, M, Cesnak, J.: Determination of Strain Energy on Reinforced Concrete Beams. Slovak Journal of Civil Engineering, Volume II, 1994

[24] Sargin, M: Stress - strain relationships for concrete and the analysis of structural concrete sections. University of Waterloo, Study No. 4, 1971

[25] Saliger, R.: Der Stahlbetonbau. 8. Auflage. Franz Deuticke, Wien, 1956 [26] Schlaich, J., Scheef, H.: Concrete box - girder bridges. Structural Engineering Documents 1e. Stuttgart, January, 1982

[32] Shawkat, S., Cesnak, J.: Deformations of Reinforced Concrete Beams Subjected to Stationary Loading. Inžinierske stavby 43, č. 9 -10, 1995 [33] Shawkat, S., Bolha, Ľ.: Internal Energy of Concrete Elements by Moving Load, Inžinierske stavby, 43, č. 4, 1995 [34] Shawkat, S., Križma, M., Šuchtová A.: Deformation of Reinforced Concrete Beams. Slovak Journal of Civil Engineering, č. 3-4, 1996

[27] Shawkat, S.: Deformation of reinforced concrete beams. Proceedings of the RILEM International Conference Concrete bridges, Štrbské pleso, 22. - 24. september 1997 [28] Shawkat, S., Križma, M., Cesnak, J., Bartók, A.: Moment and Shear Deflection for Reinforced Concrete Beams. Slovak Journal of Civil Engineering, Volume II, No. 2-3, 1994

References

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404

Revitalization a Modernisation of Buildings – ”Pentagon” (Bratislava) The area around the residential complex „Petagon“ is currently in a very bad condition, it’s dirty and neglected. In this area we can find a lot of dark corners full of rubbish causing fear and danger. The green space between the buildings is neglected as well and it’s taken by the homeless. There are not enough parking spots and the ones present are in the middle of the yard with a very bad connection to the main road. The proposed solution tries to solve all these problems with a new site plan of the park and a new terrace connecting two buildings.

The terrace connects two floors of a sanatorium. It fills the void between two objects, it substitutes previous corridor and avoids the formation of dark and dangerous corners. Both terraces are outdoor and the upper one creates a roof for the lower one. Together they create a relaxing place for meeting, talking, reading, chill out or playing games for the patients.

Outdoor solution consists of opening and aeration of the whole place with a well-organized park with basketball fields, benches, skate-park and lightning. The slope near skatepark is used to create an auditorium. The parking lot was expanded and moved to the left side creating a better access to the main road. The whole proposal is raised by a water barrier around each building. Flowing water provides peaceful and calm atmosphere and makes the environment more pleasant.

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Situation - Present State

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Situation - Proposal

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Sober living Housing Sober living housing works in two phases. Phase I is for people freshly out of rehab. They live together in the two first floors of the building. They share bedrooms, bathrooms and the kitchen. Their schedule is full of activity, it is important to not throw them back into the real world right away and allow them to adapt slowly and under supervision. Phase II is when people move into the flats above and start living more independently, yet they still share the flat with people in similar situation and are still close to the Main House if they needed help. They still visit the Sunday meetings.

The House of Beginnings prompts the Phase I resident to start dealing realistically and effectively with their personal life issues. This process typically takes about 30 days to develop the skills and orientation to move to Phase II housing, though there are no pre-set time limits or schedules. As a house, the Phase I complex provides a range of settings to support and contain the residents’ daily journey into recovery. Support occurs through the flow of activities mediated by resident-administered rules for use of the setting. Containment occurs by restricting residents’ movements outside the Phase I complex.

1. An open circulation system invites easy entry into the building, provides high levels of visibility, encourages spontaneous socializing, and links all functions to each other for easy movement back and forth between dining/kitchen, meeting, and social areas. 2. The kitchen is the heart of the facility at the centre of the building and open to the circulation system. The smells of cooking fill the house.

At the organizational level, House of Beginnings is the central hub for the entire system of the Phase I complex and Phase II houses. The large meeting room, sized to accommodate approximately 100 occupants, is the site for a weekly Sunday Night meeting attended by managers and residents from all the houses, and for visits by friends. The meeting room also holds daily AA meetings and several kinds of classes for small groups.

3. A large meeting room located close to the street welcomes residents, visitors and guests. 4. Dining/social areas/outdoor areas thread through the house to connect and blend all uses. 5. Sleeping rooms (two persons per room) are included in the Main Building.

Sober Living

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SLEEPING ROOMS The bedroom is a place to rest but not to retreat. Phase I Sleeping Rooms are bedrooms designed and furnished solely for sleeping and storing limited personal belongings, and for limited contact between roommates. Residents are busy working on their personal recovery programs and participating in various house activities; they have limited time to rest and relax. 1. Two beds per sleeping room are provided for all residents. 2. Each room is the same size and with the same furnishings. 3. Sleeping rooms are not status indicators. Decorations and personal items are limited. 4. Sleeping rooms provide limited storage space for each resident. There is room for a few clothes and personal items such as a kit-bag and alarm clock. 5. Sleeping rooms share bathrooms.

Sabah Shawkat © Phase II residents have each their own sleeping room, and are allowed personal stuff and decorations. STAFF 1. Location: Next to the main entrance immediately accessible from the street and from the principal activity areas, inviting dropins and putting the staff in the heart of the action 2. Access policy: Residents have direct access when they need to talk with staff. 3. Privacy policy: Office functions operate to minimize barriers, formality, and closed-off space. Many contacts occur outside the office or with an open door, but when privacy is needed, the office door is closed.

Sober Living

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Sober Living - 1. Floor Plan

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Sober Living - 2. Floor Plan

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Stairways Proposal

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Housing Proposal Present State

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Housing Proposal Maisonette 4th Floor

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Housing Proposal Maisonette 5th Floor

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Housing Proposal Plan 10th Floor

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Sections

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Sections

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2st Floor

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Site Plan

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422

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Client Task

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Site Analysis

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The main topic was to create interconnected house with the new addition structure.The idea of making old and new house work separately or together depending on the day. Creating spots for both families to meet at, either inside or outside. Old house is extended by one room which allows for reorganization of the spaces and larger bathroom. Basement is opened up with small windows to allow for a new bedroom. New house addition has lowered floor height and angled roof in order not to block the sun coming into neighbouring plots. Digging down also creates enough space for a roofed frontal parking space and improves sunlight conditions for the basement room. Further light improvements are made by creating openings within the new roof structure.

3d Model of Existing Family House

426

Present State Plans

427

Variant 1 - Site Plan

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Variant 1 - Plan Underground, Plan 1st. Floor, Plan 2nd Floor

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Variant 1 - Section AA

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Variant 1 - Section BB, CC

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New House

Old House

Steel Structures

Timber Structures

Roof

Variant 1 - Diagrams

432

Variant 1 - Views

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Variant 2 - 1st Floor

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Variant 2 - 2st Floor

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Variant 2 - Site Plan

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Variant 2 - Views

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Variant 2 - Section AA

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Variant 2 - Section BB, CC

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Variant 3 - Site Plan

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Variant 3 - Plan 1st. Floor, Plan 2nd Floor

441

The house itself seems to be one object but in fact it is composed of 2 separate parts connected only by a mutual saddle roof. They have separate entrances and basement. Terrace and garden are mutual as well, allowing the two families to spend at least some time together. Even if the volumes of the objects are equal on the ground floor, the new part occupies 2/3 of the space on the 1st floor. Since the new family will need more space for them and their children, just one bedroom should be enough for the grandparents.

Sabah Shawkat © The reconstruction of the old part was mainly in the disposition of interior space. On the ground floor we can find a small hall and an entrance to a bathroom. Then we enter a corridor which leads us either upstairs, downstairs or straight into living room connected to the dining room up to the kitchen with small pantry and terrace. The staircase leads us upstairs to the 1st floor where there is a small library and one bedroom with bathroom and balcony. On the other hand, the new part was designed from the beginning to the end and is more complex. We enter through a separate entrance right into a small hall where we can find staircase leading downstairs into the basement where there are two rooms- gym and storage room.

Variant 3 - Section AA

442 Proceeding straight we get into small corridor with a door to the bathroom and then we walk into a big, bright and open space- it is the kitchen connected to living room with a big glass wall and terrace. On the left there is a staircase leading to a gallery which allows the connection of both floors making the house more airy and open. On the gallery there is a library and doors into 2 bedrooms (the one on the left is for parents with separate bathroom and balcony, the other one is for one child).

Then we continue to a small corridor leading into bathroom (mutual for the two children bedrooms) or to another bedroom. Both children’s bedrooms share one balcony. The most interesting thing about the project is the coexistence of the two families, living together and alone at the same time.

Variant 3 - Section BB, Views

443

Variant 4 - Section AA, BB, Views

444

Variant 4 - Plan 1st. Floor, Plan 2nd Floor

445

Variant 5 - Section AA, BB, Views

446

Variant 5 - Plan 1st. Floor, Plan 2nd Floor

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Variant 6 - Section AA, BB, Views

448

Variant 6 - Plan 1st. Floor, Plan 2nd Floor

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Variant 7 - Section AA, BB, Views

450

Variant 7 - Plan 1st. Floor, Plan 2nd Floor

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Variant 8 - Section AA, BB, Views

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Variant 8 - Plan 1st. Floor, Plan 2nd Floor

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Variant 9 - Section AA, BB, Views

463

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08.01

Pyrolysis Author: Sabah Shawkat

Pyrolysis is the thermal decomposition of materials at elevated temperatures in an inert atmosphere. It involves a change of chemical composition and is irreversible. The word is coined from the Greekderived elements pyro „fire“ and lysis „separating“. Pyrolysis is most commonly used in the treatment of organic materials. It is one of the processes involved in charring wood. In general, pyrolysis of organic substances produces volatile products and leaves a solid residue enriched in carbon, char. Extreme pyrolysis, which leaves mostly carbon as the residue, is called carbonization.

The process is used heavily in the chemical industry, for example, to produce ethylene, many forms of carbon, and other chemicals from petroleum, coal, and even wood, to produce coke from coal. Aspirational applications of pyrolysis would convert biomass into syngas and biochar, waste plastics back into usable oil, or waste into safely disposable substances.

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Pyrolysis

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Realisation - Renovation and Extension of Family House Author: Sabah Shawkat, Richard Schlesinger

479

Realisation - Renovation and Extension of Family House

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Realisation - Renovation and Extension of Family House

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Realisation - Renovation and Extension of Family House

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Realisation - Renovation and Extension of Family House

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Realisation - Renovation and Extension of Family House

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Realisation - Renovation and Extension of Family House

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486

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Realisation Projects - Devín

488

Realisation Projects - Dobrá Niva

489

Realisation Projects - Draškovec

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Realisation Projects - Fronce

491

Realisation Projects - Galvaniho

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Realisation Projects - Senec

493

Realisation Projects - Líščie Údolie

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Realisation Projects - Sunville

495

Realisation Projects - Marianka

496

Realisation Projects - Žilina

497

Other Books by Sabah Shawkat

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1.Deformation Behaviour of Reinforced Concrete Beams 2. Structural design I 3. Structural design II 4. Architektonika 5. Structural design III 6. Structural projects 7. Inžinierske drevené konštrukcie 8. The art of structural design 9. Element design to shape a structure I.

10. Element design to shape a structure II. 11. Lightweight steel structures

12. The art and engineering of Lightweight Structures 13. Art in/of Nature

14. Design of Reinforcement for concrete members 15. Železobetónové konštrukčné sústavy

Other Books by Sabah Shawkat

499

Reviewer: Cover Design: Editor: Software Support: Publisher: Printed and Bound:

Sabah Shawkat I IRichard RichardSchlesinger Schlesinger Sabah Shawkat 1. Edition, Tribun Tribun EU, EU,s.r.o. s.r.o. 1. Edition, Brno, Czech republic 2020 Brno, Czech republic 2020 ISBN 978-80-263-1561-2