ELEMENT DESIGN TO SHAPE A STRUCTURE SABAH SHAWKAT
I
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Reviewer: Cover Design: Software Support: Printing/ Binding:
Prof. Dipl. Ing. Ján Hudák, PhD. Richard Schlesinger asc. Applied Software Consultants, s.r.o., Bratislava, Slovakia Tribun EU, s.r.o., Brno, Czech Republic
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.
Element Design to Shape a Structure I. ©
Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD. 1. Edition, Tribun EU, s.r.o. Brno 2017 ISBN 978-80-263-1190-4
Sabah Shawkat Zahawi is the Head of Engineering Room at the Academy of Fine Arts and Design in Bratislava, Slovakia. He teach students of architecture several structural engineering subjects. Moreover, he regularly organize workshops for students and exhibitions of their projects and construction models. He also actively practise in projecting and building constructions as well as reconstructions and modernizations of buildings. Sabah Shawkat is also a passionate expert in reinforced concrete, prestressed concrete structures and structural design. He has published numerous articles in professional journals and wrote several books.
STRUCTURAL ENGINEERING ROOM
Department of Architecture About the Author
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Example 3.1-5 Determine the required tension
Introduction 1. Materials
1 1
Example 3.1-6 Determine the required tension
1.3 The behaviour of hardened concrete
6
Example 3.1-7 Determine the required tension
1.4 High-strength concrete
7
1.1 Concrete mix design
2. Limit states
Example 3.1-8 Determine the required tension
9
13
2.3 Stress-strain Diagram for Reinforcing Steel
14
2.4 Basic Values for the Design Resistance to Moments and Axial Forces
16
Example 3.1-9 Determine the required tension Example 3.1-10 Determine the required tension
17
Example 2.1
17
3.2.1 Flat Slabs
Example 2.2
20
3.2.1-1 Analysis and Design of Flat Plate
21
properties of concrete and steel 23
perform load test of reinforced concrete beam. 2.6 Shrinkage
69 71 72
Example 3.2-2 Assessment of the punching according to EC 2
80
Example 3.2-3 Assessment of the punching according to EC 2
84
Example 3.2-4 The calculation of the maximum bending
88
moment of reinforced concrete rectangular slab Table D 3 Coefficients for calculating bending moments
30
2.7-1 The investigation of the effects of shrinkage of
68
Example: 3.2-1 Design and calculation of Flat Plate
29
2.7 Creep
59
reinforcement to the section
2.5 Verification at the Serviceability Limit States
Example 2.4 The beam has a theoretical span 2.50 m,
57
reinforcement to the section.
3.2 Reinforced Concrete Slabs
Example 2.3 The determination of actual and statistical
55
reinforcement to the section
10
2.2 Stress - strain Diagram for Concrete
53
reinforcement to the section
10
2.1 Strain Diagrams in the Ultimate Limit State
52
reinforcement to the section
1.2 Steel fibre reinforced concrete (SFRC).
1.4-1 Components of cement-based matrices
50
reinforcement to the section
88
over between the supports
30
Table E3 Calculation of deflection and moments circular plates
a composite concrete structure
90
depending the method of support and load 3. Structural system
37
3.1 Reinforced Concrete Beams Example 3.1-1
37
3.3 Staircases
93 Example 3.3-1 Design a straight flight staircase in a residential building
39
3.1-1 Continuous beams Example 3.1-2 Determine the required tension
41
3.4 Reinforced concrete column 48
reinforcement to the section
98 100
Example 3.4-1 Design reinforced concrete columns
110
rectangular cross section
reinforcement to the section Example 3.1-4 Determine the required tension
Example 3.3-2 Circular-staircase
46
reinforcement to the section Example 3.1-3 Determine the required tension
96
49
Example 3.4-2 Determine the carrying capacity of a rectangular column
111
118
Example 3.5-2 The Assessment of the cross section for the
121
interaction of torsion moment and shear force Example 3.5-3 Spreading of box girders beams
of reinforced concrete rectangular column
113
Example 3.5-1 Failure due to torsion
122
Example 3.7-8 Determination of the tension reinforcement
128 Example 3.6-1: Reinforced concrete wall subjected
Example 3.7-9 Ultimate limit state of the cross-section
169
Example 3.7-10 Determination of the ultimate bending moment
169
and stresses at the upper and lower of cross-section Example 3.7-11 Determine the ultimate bending moment Example 3.7-12 Determine the ultimate bending moment
173
and concrete stresses
134
3.8 Calculation of stiffness of concrete members
with openings subjected to vertical load
171
and concrete stresses.
132
to horizontal load Example 3.6-2: Solution of reinforcing concrete walls
167
of reinforced concrete rectangular column
Example 3.5-4 Dual-chamber section or 2 box girder cross-section 125 3.6 Shear walls
166
175
139
Example 3.8-1 Calculation of bending and shear stiffness
175
3.7.1 Pre-tensioning
139
Example 3.8-2 Calculation of bending and shear stiffness
177
3.7.2 Pre-tensioning
139
Example 3.8-3 Cross section with cracks
181
3.7.3 Partial prestressing
141
Example 3.8-4 Calculation of initial bending and axial stiffness
189
3.7.4 In post-tensioned
141
Example 3.8-5: Calculation of curvature of rectangular cross-section
198
3.7 Pre-tensioned Prestress Concrete Beam
Example 3.7-1: Pre-tensioned Prestress Reinforce Concrete I Beam
3.7.5 Loss of Prestress Due to Creep and Shrinkage
142 148
3.9 Concrete Foundations
203
3.9.1 Shallow Foundations
205
3.7.6 Ultimate limit state of failure due to bending moment
152
3.9.2 Strap Footing
206
3.7.7 Limit state stress limitation
154
3.9.3 Combined Footing
207
3.7.8 Cracking limit state
155
3.9.4 Strip/continuous footings
207
3.7.9 Deflection
157
3.9.5 Mat or Raft footings
208
157
3.9.6 Pile foundations
208
Example 3.7-2 Determination of required tension reinforcement to the cross-section Example 3.7-3 Partial prestressed beams
160
Example 3.7-4 Determination of required tension reinforcement
163
to the section.
Example 3.9-1 Assessment of slab foundation to punching
211
Example 3.9-2 Determination of the design bearing capacity
213
of the soil at depth Example 3.9-3 Endless beam
213
Example 3.7-5 Determination of the tension reinforcement to the section
164
Example 3.9-4 Design of reinforcement in footing
217
Example 3.7-6 Determination of the tension reinforcement
165
Example 3.9-5 Static calculation of extreme square isolated footings
223
of reinforced concrete rectangular column Realization Projects
227
References
257
STRUCTURAL ENGINEERING ROOM
3.5 Mode of failure of reinforced concrete members subjected to torsion
Example 3.7-7 Determination of the tension reinforcement
Department of Architecture
Example 3.4.3 Determine the carrying capacity of circular columns 112
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Introduction
The visual expression of efficient structural function is thus a fundamental criterion of
This textbook is designed to teach students but it can serve as a reference for practising
elegance in structure design. It is one of the primary distinguishing factors between structural
engineering or researchers as well and accessible to a larger number of engineers. The text
engineering art and architecture. All design examples given in this book are produced in the
offers a simple, comprehensive, and methodical presentation of the basic concepts in the
form of worksheet. The examples are fully self-explanatory and well annotated and the authors
analysis of members subjected to axial loads, and bending. Many practical worked examples
are confident that the readers, whether practising design engineers, course instructors or
are included and design aids in the form of tables are presented.
students, will find them extremely useful to produce design solutions or prepare course
Primary considerations in structural design are safety and serviceability. Design rules of this textbook are based on the concept of Concrete Structures Euro-Design. For greater clearness, the survey follows the sequence of practical concrete members design process by dividing itself into three main parts, namely, Design, structural analysis, and dimensioning and structural detailing, each section with its individual numbering examples. The major quantities which are considered in the design calculations, namely the loads,
handouts. In particular, the worksheets will allow design engineers to undertake sensitivity analyses to arrive at the most suitable/economic solution(s) very quickly. I kindly 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. Finally, I am grateful to my colleague Richard Schlesinger for his help, constructive criticism, patience and encouragement that have made this project possible.
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. This textbook contains design aides such as tables and diagrams which enable very simple and rapid determination of reinforcement in a given rectangular concrete cross-section subjected to bending moment or bending moment with axial force. The elements of a structure are normally classified, by consideration of their nature and function, as beams, columns, slabs, walls, foundation etc. Rules are provided for the analysis of the commoner of these elements and of structures consisting of combinations of these elements. The purpose of analysis is to establish the distribution of either internal forces and moments, or stresses, strains and displacements, over the whole or part of a structure. Linear analysis of elements based on the theory of elasticity may be used for both the serviceability and ultimate limit states. Reinforced concrete structures 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).
Bratislava 2017
Sabah Shawkat
STRUCTURAL ENGINEERING ROOM
Department of Architecture
STRUCTURAL ENGINEERING ROOM
Department of Architecture
1
The choice of a structural detail can significantly influence the performance of a structure, bond, anchorage, bearing, fire, corrosion and other properties of concrete have to be taken into consideration individually or in combination in the design and detailing of concrete structures.
1. Materials 1.1 Concrete mix design The mechanical properties of the concrete are given by the properties of the aggregates and cement matrix and by the proportions between them, and by the interfacial physical and
Since concrete is weak in tension, it is important that where bond is most important,
chemical bond.
structural details must ensure the force transfer between steel and concrete. In the case of
The cement paste is the continuous matrix in the material. Its contribution to the
bearings, structural details must be adequate to transfer forces from one member to another with
properties of the concrete depends on the progress of the hydratation process, the nature of the
a limited area of high local stresses. Structural details must cater for thermal expansion and
hydratation products and their microstructural arrangement. These characteristics are functions
creep where they are expected. The fire endurance of concrete structures is an essential part of
of the composition of the cement and the water/cement ratio.
a good design. The endurance period is generally taken as one hour plus. For high seismic risk,
The - curve for concrete is highly non-linear. This non-linearity is due, in part, to the
a certain minimum number of bars must be provided as continuous longitudinal steel for beams.
composite action of the concrete, since there is a highly imperfect bond between the aggregate
For beams framing into both sides of a column, these bars must extend through the column for
and the hcp.
at least twice the beam depth without splices. The structural engineer must indicate quantities
The - curve for concrete in compression may be divided into the four regions. Below
of reinforcement, cut-off points and length and location of splices to satisfy code requirements.
about 30% of the ultimate stress (ult) there is very little extension of these cracks for normal
Where concrete structures are subject to impact, reinforcement details must take into account potential ones of spalling, scabbing and cracking. The bars must span these areas. Verification of stress on reinforced concrete cross - section subjected to bending and bending moment with axial load in serviceability limit state can easily be performed. Three different diagrams, bi-linear ( ´ 0), bi-linear ( ´
0)
, and parabola-rectangle stress -
strain for reinforcing steel and two diagrams, bi-linear ( ´ 0 ), parabola-rectangle
rates of loading, and the - curve essentially linear. Beyond this point, the - curve becomes increasingly non-linear as the interfacial cracks begin to grow under load, partly due to differences in the elastic constants for hcp and aggregate, and partly due to the high stress concentrations which occur in these regions.
for
concrete stress distribution are used for the determination of reinforcement to concrete crosssection for any concrete strength. Calculations are demonstrated by examples in order to check the sufficient bearing capacity of reinforced concrete rectangular column in centric compression for axial force taking into account the existing reinforcement.
Figure 1.1-1 : Stress vs strain curve for concrete in compression
Materials
At about 0.5ult, in addition to further growth of the interfacial cracks, cracks begin to
paste, but also affects the quality of the hydration products. Low w/c ratios can lead to stronger
extend through the cement matrix, bridging between the coarse aggregate particles,
bonds and thus improve the strength of the concrete.
approximately parallel to the axis of loading, but still in a stable fashion.
The strength is consequently not only a function of the porosity.
Finally beyond about 0.75ult, the matrix cracks begin to form a much more extensive network, though there is still enough reductancy in the system for it to remain relatively stable under short-term loads. Eventually, this network becomes so extensive that failure occurs. Static fatigue in concrete is generally associated with loading within this region. It should be noted that, even at ult, the crack pattern still leaves a structure that has some load-carrying capacity. The aggregates are traditionally considered as the more inert component of the concrete, with respect to the development of the properties of the hardened material. Yet, the choice of aggregate and their grading is of prime importance in controlling the properties of the fresh and hardened concrete. Hardened cement paste is a porous, inhomogeneous material. The degree of porosity and heterogeneity of the material is a decisive factor controlling its properties. According to Professor S.Mindess, the strength of cement paste depends on:
Based on the strength controlling parameters given by S. Mindess three corresponding paths might be taken to increase the strength of cement paste and cement-based materials: a) decrease of total porosity b) improved pore size distribution, i.e. reduction of maximum and average pore size, and also elimination of large Griffith flaws c) improved quality of CS Hydration products As a consequence, the pillars of practical mix design for HSC are: reduced w/c ratio extensive use of plasticizers application of cement with a high strength potential application of silica fume (SF) 1.2 Steel fibre reinforced concrete (SFRC).
a) total porosity
The use of fibres to strengthen materials which are much weaker in tension than in
b) pore size distribution
compression goes back to ancient times. when probably the oldest written account of such a
c) the nature of the solid phase
composite material, clay bricks reinforced with straw occurred.
The porosity is mainly a function of the w/c ratio and the degree of hydration. This forms
The concept of using fibers to improve the characteristics of construction materials is
the basis for the well known empirical relation between strength and w/c ratio as formulated
very old. Addition of fibers to concrete makes it more homogeneous and isotropic and
by R. Feret and D. Abrams. Several investigators such as J. Jambor and Zaitsev have also established relations between the pore size distribution and the strength of the concrete, showing that reduced average pore size and reduced maximum pore size lead to increased compressive strength. According to classical fracture mechanics, i.e. Griffith theory, the maximum pore size might be considered as controlling the strength of brittle materials. The solid phase of cement paste is a very complex system of amorphous hydration products (CSH), crystals of calcium hydroxide, Ca(OH)2, un hydrated cement and minor quantities of other components. The CSH-products amount to about 70 % of the paste. The
transforms it from a brittle to a more ductile material. Historically, reinforcement has been in the form of continuous reinforcing bars, which could be placed in the structure at the appropriate locations to withstand the imposed tensile and shear stresses. Fibres, on the other hand, are discontinuous, and are most commonly randomly distributed throughout the cementitious matrix. Due to these differences, there are certain applications in which fibre reinforcement is better than conventional reinforcing bars. These include: 1.
structure and chemical composition of the CSH vary locally, and the different forms have only partly been resolved.
Thin sheet materials, in which conventional reinforcing bars cannot be used, and in which the fibres therefore constitute the primary reinforcement. In thin sheet materials, fibre concentration are relatively high, typically exceeding 5 % by volume.
Materials
STRUCTURAL ENGINEERING ROOM
J. Jambor has pointed out that the w/c ratio controls not only the porosity of the cement
Department of Architecture
2
2.
STRUCTURAL ENGINEERING ROOM
Department of Architecture
3 Components which must withstand locally high loads or deformations, such as tunnel linings, blast resistant structures, or precast piles which must be
1
hammered into the ground. 3.
h
z( )
Components in which fibres are added primarily to control cracking induced by
1
humidity or temperature variations, as in slabs and pavements. In these applications, fibres are often referred to as secondary reinforcement.
2.04 fc
2 o 3 3
b ( )
The strength of the fiber reinforced concrete can be measured in terms of its maximum resistance when subjected to either compressive, tensile, flexural and shear loads.
4 h
h 2.04 fc
lf
3 3 2
o
1 2 o 3 3 h
2.04 fc
There are a number of factors that influence the behavior and strength of FRC in flexure.
Fibres do little to enhance the static compressive strength of concrete, with increases in
These include: type of fiber, fiber length (L), aspect ratio (L/df) where df is the diameter of the
strength ranging from essentially nil to perhaps 25%. Even in members, which contain
fiber, the volume fraction of the fiber (Vf), fiber orientation and fiber shape, fiber bond
conventional reinforcement in addition to the steel fibres, the fibres have little effect on
characteristics (fiber deformation).
compressive strength. However, the fibres do substantially increase the post-cracking ductility, or energy absorption of the material.
When concrete cracks, the randomly oriented fibers arrest a microcracking mechanism and limit crack propagation, thus improving strength and ductility.
In the simplest case concrete may be modeled as a composite material consisting of
The addition of fibers to such matrices, whether in continuous or discontinuous form,
two phases : hardened cement paste and aggregates. It is evident that the aggregate as well
leads to a substantial improvement in the tensile properties of the FRC in comparison with the
as the hardened cement paste exhibit a brittle behaviour. The - curve of concrete, on the
properties of the unreinforced matrix. The enhancement of the properties is particularly
other hand, deviates from linearity even at low loads and has a decresing slope after
noticeable.
maximum load.
The stress–strain or load–elongation response of fiber composites in tension depends mainly on the volume fraction of fibers. The addition of fibers generally improves the shear strength and ductility of concrete. It has been reported that the stirrups as shear reinforcement in concrete members can be partially or totally replaced by the use of steel fibers [Lim et al. 1987; Mansur et al. 1986]. Most of the work has been limited to concretes of normal strength. The influence of fibres on the composite behaviour of elements subjected to loadings is complex. In the limits of elastic deformation the fibres are not active and their role may be derived from the law of mixtures. When the micro-cracks are open, the fibres act as crack-
Figure 1.2-1: Fracture energy
arrestors and control their propagation. The load corresponding to the first crack appearance is 2 3
b
fc Gf MPa
0.6
the same or is increased when a specially high volume of fibres is added. The main fibre effect f lf
2
b MPa
12 df 7850
o
f lf b MPa
2 df 7850
( u)
u o 1 2 l f
2
is a considerable increase in deformability and in the amount of energy of the external load which may be accumulated before the rupture occurs. The load-displacement diagram is completely different and quasi-ductility of the material behaviour is observed.
Materials
Experimental studies [Fanella and Naaman 1985; Shah et al. 1978] have shown that the addition of fibers have only a slight effect on the ascending branch (modulus of elasticity) of the stress–strain curve of the composite The same factors that influence the shrinkage strain in plain concrete influence also the shrinkage strain in fiber reinforced concrete; namely, temperature and relative humidity, material properties, the duration of curing and the size of the structure. The addition
Figure 1.2-2: A loaded concrete beam with a crack and a fracture zone
of fibers, particularly steel, to concrete has been shown to have beneficial effects in
The fibre-reinforced concretes and mortars offer increased toughness and that advantage is decisive for their applications The modulus of elasticity of a material, whether in tension, compression, or shear, is a fundamental property that is needed for modeling mechanical behavior in various structural applications. The role of fibres depends also on their form, i.e. whether they are used as short chopped fibres, continuous single fibres, distributed at random in the matrix or arranged in a more regular way. The fibre-matrix bond is assured by different process: by adhesion, mechanical anchorage and by friction, depending on the chemical and mechanical properties of both phases. The quality of bond may vary with the intensity of load and its duration.
counterbalancing the movements arising from volume changes taking place in concrete, and tends to stabilise the movements earlier when compared to plain concrete. The primary advantage of fibers in relation to shrinkage is their effect in reducing the adverse width of shrinkage cracks [Swamy 1985; Lim et al. 1987; Shah and Grzybowski 1989]. Shrinkage cracks arise when the concrete is restrained from shrinkage movements. The presence of steel fibers delays the formation of first crack, enables the concrete to accommodate more than one crack and reduces the crack width substantially [Swamy 1985 The ability to withstand relatively large strains before failure, the superior resistance to crack propagation and the ability to withstand large deformations and ductility are characteristics that distinguish fiber-reinforced concrete from plain concrete. These characteristics are generally described by "toughness" which is the main reason for using fiberreinforced concrete in most of its applications.
Materials
STRUCTURAL ENGINEERING ROOM
Figure 1.2-3: Effect of hooked-steel fibre content on compressive stress-strain curves
Department of Architecture
4
Cementitious matrices such as mortar and concrete have low tensile strength relative to
STRUCTURAL ENGINEERING ROOM
Department of Architecture
5 applications such as pavement (highways, roads, parking areas, runways, and bridge desks),
their compressive strength, and fail in a brittle manner. One way to improve their fracture
industrial floors, shear failure zones in structures, shotcrete, repair of concrete structures and
properties is to reinforce them with randomly distributed fibers.
lining of tunnels.
Toughness is a measure of the capacity of a material to absorb energy and resist fracture when subjected to static or dynamic loads. Flexural toughness is defined as the total energy absorbed
However, Bentur and Mindess showed that, with a 1.5% fibre volume, partial fibre
prior to complete separation of the specimen.
reinforcement (to 1/2 of the beam depth) increased the ultimate load by 32%, while full depth
In many applications, particularly in pavements, bridge deck overlays, and offshore structures,
fibre reinforcement increased the ultimate load by about 55%. Thus, there are benefits to having
the flexural fatigue strength and endurance limit are important design parameters because these
fibres even in the top (compression) half of a beam.
structures must be designed for fatigue load cycles.
With increasing crack opening, more and more fibers are fully out and , for u = lf / 2, the
Ramakrishnan et al. [1989b] observed that the fatigue strength and endurance limit (to achieve two million cycles) increased with the addition of fibers and increasing volume fraction of fibers.
stress drops to zero. Because the number of bounded fibers decreases linearly with increasing u, for 0<u<lf/2.
One of the major advantages of SFRC is that SFRC will not support the classic galvanic corrosion cells which are often the cause of corrosion and deterioration in reinforced concrete. The fibres, being non-continuous and discrete and protected by an alkaline matrix, provide no mechanism for propagation of corrosion activity. This phenomenon is well established from examination of numerous SFRC structures subjected to aggressive exposure environments. Since the role of the steel fibres is primarily to provide tensile capacity in the bottom portion (tension side) of the elements, it has been suggested that it may not be necessary to provide steel fibres throughout the full depth of a reinforced concrete elements. For reasons of economy, it may be sufficient simply to use SFRC in the bottom half of the elements. Figure: 1.2-5
(u)
o 1
2 u lf
2
f lf b
o
2 d f
Figure 1.2-4: Stress vs strain curve, Distribution of stresses, when the ultimate load is reached When fibres are wisely used, they can help us to produce concrete with increased tensile
( u)
strength and strain capacities, failure and impact resistance, energy absorption, crack resistance and durability. However, fibre give us the opportunity to utilize the concrete for a variety of
Materials
b La ( u ) f ( u ) 2
b La f 2
df
df
1 2
lf u
2
o 1 2
lf u
2
Properties of hardened concrete 1. Short term properties
Strength in compression, strength in tension, strength in bond, modulus of elasticity The strength of concrete depends on a number of factors including the properties and proportions of the constituent materials, degree of hydration, rate of loading, method of testing and specimen geometry. The properties of the constituent materials, which affect the strength are the quality of fine and coarse aggregate, the cement paste and paste-aggregate bond characteristics (properties of the interfacial, or transition zone). These, in turn depend on the macro- and microscopic structural features including total porosity, pore size and shape, pore distribution and Figure: 1.2-6
morphology of the hydration products, plus the bond between individual solid components. 2. Long-term properties
Creep, shrinkage, behaviour under fatigue, durability (porosity, permeability, abrasion La ( u )
La 1 2
u
lf
f
f 1 2
resistance)
u
Two dominant constituent materials that are considered to control maximum concrete
lf
strength are coarse aggregate and paste characteristics. The important parameters of coarse aggregate are its shape, texture and the maximum size. Since the aggregate is generally stronger than the paste, its strength is not a major factor for normal strength concrete. However, the aggregate strength becomes important in the case of higher-strength concrete. Surface texture and mineralogy affect the bond between the aggregates and the paste as well as the stress level at which microcracking begins. The use of larger maximum size of aggregate affects the strength in several ways. First, since larger aggregates have less specific surface area and the aggregate–paste bond strength is less, the compressive strength of concrete is reduced. It is generally accepted that the most important parameter affecting concrete strength is the Figure 1.2-7: Behavior of the concrete in compression and tension before the formation of the crack, Stress-crack opening diagram used in service, and Characteristic stress-crack opening diagram used in ultimate limit state calculation.
w/c ratio, sometimes referred to as the W/B (binder) ratio. Even though the strength of concrete is dependent largely on the capillary porosity or gel/space ratio, these are not easy quantities to measure or predict. The capillary porosity of a properly compacted concrete is determined by the w/c ratio and degree of hydration. Most high performance concrete are produced with a w/c ratio of 0.40 or less. The practical use of very low w/c ratio concretes has been made possible by use of both conventional and high-range water reducers, which permit production of workable concrete with very low water contents [Fiorato 1989; Zia et al. 1991; Burg and Ost 1992].
Materials
STRUCTURAL ENGINEERING ROOM
1.3 The behaviour of hardened concrete
Department of Architecture
6
STRUCTURAL ENGINEERING ROOM
Department of Architecture
7 Fly ash, slag and silica fume have been used widely as supplementary cementitious
High-strength concrete is a brittle material, and as the concrete strength increases the
materials in high performance concrete. Although fly ash is probably the most common mineral
post-peak portion of the stress-strain diagram almost vanishes or descends steeply. The increase
admixture, on a volume basis, silica fume (ultraâ&#x20AC;&#x201C;fine amorphous silica, derived from the
in concrete strength reduces its ductility - the higher the strength of concrete, the lower is its
production of silicon or ferro silica alloys) in particular, used in combination with high-range
ductility. This inverse relation between strength and ductility is a serious drawback for the use
water reducers, has increased achievable strength levels dramatically [Ezeldin et al. 1989;
of high-strength concrete and a compromise between these two characteristics of concrete can
Baalbaki et al. 1993; Zia et al.1993a, 1993b; Farny and Panarese 1994].
be obtained by adding discontinuous fibers.
Collepardi et al. [1990] studied the effect of combined addition of silica fume and
Addition of fibre to concrete makes it more homogeneous and isotropic and transforms it from
superplasticizer on concrete compressive strength by taking into account such parameters as:
a brittle to a more ductile material. When concrete cracks, the randomly oriented fibres arrest a
(a) type and dosage rate of superplasticizer, (b) type and content of portland cement, and (c)
microcracking mechanism and limit crack propagation, thus improving strength and ductility.
way of silica fume utilization (as additional component or as cement replacement). They
The addition of fibres has a minor effect on the improvement of the compressive strength
concluded that in the presence of silica fume, for both type I and type III portland cement, the
values.
melamine sulphonated polymer superplasticizer performs better than the naphthalene
Strength development and strength potential in HSC depend on the choice of cement.
sulphonated polymer, particularly when a high dosage such as 4% is used. A change from 2 to
The clinker composition and the fineness are factors that influence both early and final strength.
4% superplasticizer dosage rate in general does not modify or reduce compressive strength in
The clinker minerals C3S, C2S and C3A have the greatest influence on the strength development
the absence of silica fume, whereas significantly increases compressive strength in the present
in cement paste. C3S contributes both to a rapid early age strength development and a high final
of silica fume.
strength. C2S hydrates somewhat slower, but can contribute significantly to the final strength. C3A has particular influence on the early strength.
1.4 High-strength concrete
The hydratation of the clinker minerals may be influenced by the fineness of the cement.
The influence of fibre content on the compressive strength, modulus of rupture,
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.
toughness and splitting tensile strength is presented. The increased sensitivity of HPC compared to normal concrete relates to the mixing,
Other components in the cement may also influence the development of strength and
transport, placing, compaction and curing processes. HPC requires an experienced and
heat. Especially, a high content of alkalis will result in an increased early strength and reduced
competent workforce and high quality workmanship to achieve the potential benefits, but this
final strength potential.
is not always available on site. HPC concrete in the better" end of the above scale can be very difficult to place and compact, and the risk of honeycombs, particularly in the cover concrete, increases. In addition, the quality and efficiency of compaction is extremely dependent on the individual person handling the vibrator. Hence, the better the concrete mix, from a durability point of view, the greater the risk of having inferior or bad execution leading to reduced quality in the final structure. This fact is seldomly respected on site. This inconsistency is due to the dominating influence of the execution process on the final quality and performance of the structure
Figure 1.3-1: Cement paste in NSC and in HPC
regardless of the good quality of the initial concrete mix.
The amount of aggregate or filler is usually between 60 and 85 % of volume fraction in all kinds of concrete. Therefore their influence on mechanical and other properties is important. In concrete natural stone aggregate is used in the form of gravel or crushed rock and sand.
Materials
tension-stiffening effect, for the control of crack width, and, thus, the necessary minimum
1. shape and texture
amount of reinforcement.
2. compressive strength
It is generally agreed that the elastic modulus of concrete increases with its compressive
3. other mechanical properties
strength. The modulus is greatly affected by the properties of the coarse aggregate, the larger
4. distribution of grain size
the amount of coarse aggregate with a high elastic modulus, the higher would be the modulus
According to their shape, rounded grains of natural gravel and sand are distinguished from
of elasticity of concrete. The modulus also increases with concrete age.
angular grains obtained from crushed rocks.
Generally speaking, the measurement of dynamic modulus corresponds to a very small instantaneous strain. Therefore the dynamic modulus is approximately equal to the initial tangent modulus which is appreciably higher than the static (secant) modulus. The difference between the two moduli is due in part to the fact that heterogeneity of concrete affects the two moduli in different ways. For low, medium, and high strength concretes, the dynamic modulus is generally 40%, 30% and 20% respectively higher than the static modulus of elasticity [Mehta 1986]. These changes in the load-response are a consequence of improved aggregate-paste bond for HSC. The more linear stress-strain relationship reflects the reduced amount of
Figure 1.3-2: The effect of aggregate and cement on the concrete
microcracking at lower levels of loading for these concretes.
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. The tensile strength is determined either by direct tensile tests or by indirect tensile tests such as flexural or split cylinder tests. The direct tensile strength is difficult to obtain. Due to the difficulty in test, only limited and often conflicting data is available. It is often assumed that direct tensile strength of concrete is about 10% of its compressive strength. The most commonly used tests for estimating the indirect tensile strength of concrete are the
Figure 1.3-3: CEB/FIB MC 90 - The examples of the stress-strain curve in compression
splitting tension test and the third-point flexural loading test. The bond behaviour of a reinforcing bar and the surrounding concrete has a decisive importance regarding the bearing capacity and the serviceability of reinforced concrete members. This knowledge is an indispensable requirement to give design rules for anchorage and lap lengths of reinforcing bars, for the calculation of deflections taking into account the
Materials
STRUCTURAL ENGINEERING ROOM
Important characteristics of aggregate grains are:
Department of Architecture
8
1.4-1 Components of cement-based matrices
STRUCTURAL ENGINEERING ROOM
Department of Architecture
9 High alumina cement is used in mortars for refractory brick walls and in refractory
The main groups of inorganic binders:
concretes when mixed with special aggregate, like corundum.
1./ hydraulic cements (pozzolans, slag cement, naturel cements, high alumina cement and
Portland Cements are by far the most important binder for concrete. They are obtained by
Portland cement)
grinding and mixing raw material which contain argils and lime: clays, shales and slates
2./ non-hydraulic cements (gypsum, lime, magnesium cement)
together with limestones and marls. The ground materials are fused in temperature of about
3./ hydrothermal cements
1400 oC into clinker which is ground again into powder of particles below 100 ď m and gypsum
4./ sulphur cement
is added.
The adjective hydraulic means that the product when hardened is water-resistant. Non-
Not only composition but also fineness of cement has a considerable influence on the
hydraulic materials are those that decompose when subjected to water and cannot harden under
hydration process and properties of the hardened product. The fine cements offer a higher
water.
hydration rate and higher final strength. On the other hand, the higher water requirement of fine
Pozzolans were obtained initially from volcanic rocks and ashes. After grinding, a kind of
cement increases its shrinkage.
cement was produced. In present times â&#x20AC;&#x2DC;pozzolanicâ&#x20AC;&#x2122; admixtures are obtained also from fly ash or burnt clays and shales. These admixtures are added to Portland cements as a les active binder
Expansive cements are special kinds of PC which contain particular constituents that increase
which partly plays the role of a fine aggregate of low cast.
their volume during hydration and hardening. By appropriate composition, the shrinkage of PC may be entirely compensated for, or even a final increase of total volume may be obtained.
Slag cements are obtained from rapidly cooled blast furnace slag which is then subjected to
processes of grinding and mixing with lime and Portland cement in different proportions.
Gypsum is a kind of non-hydraulic cement obtained from mineral of the same name. The
Besides being at a lower cost than Portland cement, the addition of slag cement improves
production is based on crushing the raw material and heating it to temperatures between +130
resistance against sulphate corrosion of mortars and concretes.
and +170 C for dehydration. If burnt is higher temperatures, a gypsum of higher strength is obtained. The last stage of production is the grinding into a fine powder and mixing with
High alumina cements are obtained from bauxite Al2O3 fused together with limestone at a
additives which delay and control the setting time to enable effective mixing with water and
temperature of 1600 oC to form clinker which is then ground into powder.
easy placing in forms.
High alumina cement is about three times more expensive than PC, mainly because of
the high energy consumption involved in grinding the hard clinkers.
Lime is obtained from natural limestone or dolomite burnt at temperatures between 950 and
1100 oC. The hardening of the lime is a slow process called carbonation, which is based on the
Hydratation of high alumina cement requires more water and mixes of better workability are obtained. The hydratation process starts later, after mixing with workability are obtained.
absorption of CO2 from the atmosphere. The lime is then transformed again into limestone CaCO3 and free water is evaporated.
The hydratation process starts later, after mixing with water, but occurs more quickly with high rates of liberated heat. Also, hardening is quick and usually within 24 hours about 80% of final
Magnesium oxychloride is produced by burning of magnesite MgCO3 at temperature between
resistance is obtained. The hardened cement has both lower porosity and permeability.
600 and 800 oC. The hardening process is quick and already after 28 days a compressive strength of about 100 MPa may be reached. The application is limited to the interiors of
Mortars and concretes made with high alumina cement are resistant against sulphate
buildings because its durability when exposed to moisture is insufficient.
attack and also resist better CO2 from ordinary water. In comparison, the resistance against alkalis is lower than of PC.
Materials
The purpose of structural analysis is the establishment of the distribution of either
internal forces and moments, or stresses, strains and displacements over the structure as a whole or a part of it. It is the aim of each design procedure - on condition that the structure or member is actually constructed according to the design - to minimize the probability of failure without unreasonable expenditure. Further more serviceability during the lifetime of the structure should be guaranteed with adequate probability.
- failure by excessive deformation, transformation of the structure or any part of it into a mechanism, rupture, loss of stability of the structure or any part of it, including supports and foundations, - failure caused by fatigue or other time-dependent effects. When considering an ultimate limit state of rupture or excessive deformation of a section, member or connection (fatigue excluded), it should be verified that Sd Rd
Where Sd is the design value of internal forces or moments (e.g. M, N, V, T for
Therefore two limit states are defined:
bending moment, axial force, shear force, torsional moment) produced by the actions F (loads, prestressing forces and so called indirect actions such as imposed deformations from the
a) Serviceability Limit State (SLS) b) Ultimate Limit State (ULS)
settlement of supports or from temperature and shrinkage effects) and Rd is the corresponding
a) Serviceability Limit State (SLS)
design resistance, associating all structural properties with the respective design values. The index d makes clear that we have to calculate the corresponding design values of S and R
Analyses carried out in connection with serviceability limit states will normally be based on the linear elastic theory. In this case it will be satisfactory to assume the stiffness for members based on the stiffness of the uncracked cross-section, when cracking of the concrete has a significant unfavourable effect on the performance of the structure or member considered,
. In cases where the ultimate state is characterised by the loss of statical equilibrium, the condition is given in the form: Ed dst Ed stb
it shall be taken into account in the analysis.
Where Ed dst stands for the destabilising influences and Ed stb for the stabilising influences.
The serviceability requirements concern:
Examples for such cases are piers during the construction phase of symmetrical free cantilever
- the functioning of the construction works or parts of them,
beam or the buckling of slender columns.
- the comfort of people, - the appearance.
2.1 Strain Diagrams in the Ultimate Limit State The allowable strain diagrams which are possible in accordance to the stress - strain
b) Ultimate Limit State (ULS)
diagrams are shown in figure 2.1-1.
Depending on the specific nature of the structure, the limit state are considered and on the specific conditions of design or execution, analysis for the ultimate limit states may be linear elastic with or without redistribution, non-linear or plastic. Ultimate limit states concern: - the safety of the structure and its contents - the safety of people. Ultimate limit states which may require consideration include:
The strain diagrams are based on the assumption of Bernoulli s hypothesis. The strain limits of the stress-strain diagrams for concrete and steel result in different domains for the strain diagrams in the design of cross sections. The stress state in a cross section is defined by the chosen strain diagram of the materials. With the assumption of an ideal bond the strain diagram in figure 2.1-2 governs not only the concrete compression stresses but also the steel stress for reinforcement in the cross-section. The compressive strain of reinforcing steel caused by creep and shrinkage of concrete are normally negligible in the ultimate limit state.
- loss of equilibrium of the structure or any part of it, considered as a rigid body,
Limit states
STRUCTURAL ENGINEERING ROOM
2. Limit states
Department of Architecture
10
In the following a short explanation of the five domains shown in figure 2.1-2 is given:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
11
Figure 2.1-2 Strain distributions of reinforced concrete cross - section with respect to the limitation of the steel strains Assumption: s2 ( A)
0.01
Zone 1: The entire section is in tension, the neutral axis lies outside the section, normally reinforcement yields. Zone 2: The neutral axis lies within the section, there exists thus a compression and a tension zone. The maximum strain in the concrete is less than the limiting value of 0.0035, thus the strength of the concrete is not exhausted. The boundary between zone 2 and 3 is defined by a strain diagram where both the maximum concrete strain ( c strain s
0.01
0.0035)
and reinforcement
are present.
Zone 3: The concrete compression strain at the upper fibre is 0.0035 (point B). The steel strain lies between 0.01 and yd, the strain corresponds to the design strength of the steel f yd.
This means that the stress of reinforcing steel is fully exhausted in tension. The boundary
between 3 and 4 is called the balanced condition. Zone 4: The strain in the steel at failure lies between yd and 0.00 that means the steel Figure 2.1-1: Bending – Axial – Load Interaction
stress at the ultimate limit state is thus less than f yd. Zone 4a: All the steel (except prestressing steel) is in compression. Some small part of the section remains in tension. Zone 5: The entire section (with exception of possibly existing prestressing steel) is in compression.
Limit states
is between 0.0035 - with c
0
Department of Architecture
All strain profiles pass through point C. The maximum compressive strain of concrete at the lower line and 0.002 for centric axial compressive
load. Point C is the point where the line BO (which defines the boundary between the sections partially in tension and sections in compression only) intersects the line characterized by c
0.002
const. The distance of this point from the upper compressive fibre is equal to 3 / 7
of the total depth of the section. Both assumption a and b for the steel strain limitation (point A) (in figure 2.1-2) result in slightly different strain diagrams. Zone 4 and parts of zone 3 characterize the transition from dominant bending moments to dominant axial compression force. These areas cover the cases in which strong pure bending and moments with compression force require a deep situation of the neutral axis. In cases of pure bending or dominant bending with compression force this uneconomic area can be avoided by placing compression reinforcement which leads to a higher neutral axis and thus to full exhaustion of the tension reinforcement. Zone 3 with relatively low neutral axis cannot be used for dominant bending if the rules for the limitation of the relative depth of the compression zone (x / d) depending on the chosen redistribution of moments are obeyed. Thus the necessity of additional compression reinforcement depends also on the limits for (x / d).
Figure 2.1-3: Bending – Axial – Load Interaction
Limit states
STRUCTURAL ENGINEERING ROOM
12
2.2 Stress - strain Diagram for Concrete
STRUCTURAL ENGINEERING ROOM
Department of Architecture
13
Studies on the development of strength show that after 24 hours 70% of the compressive strength can be reached. The cement used is an important factor. These material properties offer new possibilities in the production of structural members. The stress - strain diagram for concrete (the conventional stress distribution in the compression zone) is presented normally by the parabola - rectangle curve. According to the design concept with partial safety factors the maximum stress value is f cd figure 2.2-1. The factor 0.85 Figure 2.2-2: Bilinear stress-strain diagram for concrete
takes account of the concrete strength under sustained load which is smaller than the strength under short term loading. The non-proportional interdependence between stress and strain of the concrete in the compression zone according to figure 2.2-1 represents an approximation to
The design concrete strength is defined by f cd
the real stress distribution in the ultimate limit state. For simplification reasons this distribution can also be used in cases
f ck c
.
The design diagram is derived from the chosen idealized diagram by means of a
- with the strain diagrams acc. to figure 2.1-2 for the neutral axis in high positions.
reduction of the stress ordinate of the idealized diagram by a factor ( / c ), in which c is
In these cases only a part of the parabola - rectangular diagram gives the stress distribution in
the partial safety factors for concrete c 1.5
the compression zone which is limited by the relevant strain at the upper fibre.
is a coefficient taking account of long term effects on the compressive strength and of
unfavourable effects resulting from the way the load is applied. The additional reduction factor
for sustained compression may generally be assumed to be |0.85|.
A rectangular stress distribution as given in figure 2.2-2 may be assumed. The - factors as given for idealized diagram are valid, except that it should be reduced to |0.80|, when the compression zone decreases in width in the direction of the extreme compression fibre. The behaviour of hardened concrete can be characterized in terms of its short-term (essentially instantaneous) and long-term properties. Short-term properties include strength in compression, tension, bond and modulus of elasticity. The long-term properties include creep, shrinkage, behaviour under fatigue and durability characteristics such porosity, permeability, Figure 2.2-1: Parabola - rectangular diagram for concrete stress distribution
freeze-thaw resistance and abrasion resistance.
Other idealized stress-diagrams may be used, provided they are effectively equivalent
The strength of concrete depends on a number of factors including the properties and
to the parabola-rectangular diagram, e.g. the bi-linear diagram figure 2.2.-2 or the rectangular
proportions of the constituent materials, degree of hydration, rate of loading, method of testing
stress block figure 2.2.-2. The following design aids are based exclusively on the parabola-
and specimen geometry. The properties of the constituent materials, which affect the strength
rectangular diagram because of its universal applicability and its good approximation to test
are the quality of fine and coarse aggregate, the cement paste and paste-aggregate bond
results. For quick calculation and for predesign of members the rectangular stress block is a
characteristics (properties of the interfacial, or transition zone). These, in turn depend on the
useful tool.
macro- and microscopic structural features including total porosity, pore size and shape, pore
In this diagram cu max is taken as 0.0035
Limit states
-reduced ultimate strain may be observed for medium HSC
components. The most linear stress-strain relationship reflects the reduced amount of microcracking at lower levels of loading for these concretes. These aggregate-paste bond failures first started to propagate at cca.90% of the ultimate load resulting in a linear stress-strain relationship up to this level. A less developed microcrack pattern also results in a more sudden failure, because the ability of redistributing stress is reduced. 2.3 Stress-strain Diagram for Reinforcing Steel The characteristic value f yk for the yield stress is defined as 5% - fractal of the yield stress ( f y). The design values f yd and f td for ultimate limit state result from the characteristic values, divided by the safety factors s:
f yd
f yk s
f td
f tk s
According to EC 2 the safety factor s
1.15
in case of basic combinations and s
1.0
in
case of accidental combinations.
Figure 2.2-3: Bending – Axial – Load Interaction For HSC the long-term sustained strength of concrete is less than that determined by short-term loading. This is of fundamental importance in the design since it means a reduction in the safety factor with regard to strength, which is usually based on short-term test. In MC-90
Figure 2.3-1: Stress-strain diagrams for reinforcing steel
we find the relation 0.85. The stress-strain behaviour of HSC in uniaxial compression is reported by many research centres. The main differences between the stress-strain curves of normal and HSC 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
Three different stress-strain diagrams for reinforcing steel are used figure 2.3-2. a) Assumption of a horizontal ideal-plastic branch ( ´ b) Assumption of a branch ascending from
f yk s
to
f tk s
0) with a steel strain limitation to uk
but with a steel strain limitation to uk
c) Assumption of parabolic-rectangle with a steel strain limitation uk
Limit states
0.01
0.01 0.01
STRUCTURAL ENGINEERING ROOM
distribution and morphology of the hydration products, plus the bond between individual solid
Department of Architecture
14
STRUCTURAL ENGINEERING ROOM
Department of Architecture
15
Figure 2.3-2: Three different stress-strain diagrams for reinforcing steel
Figure 2.3-4: Bi-linear diagram for steel and bi-linear diagram for concrete stress distribution for pure bending or bending moment with axial forces.
Figure 2.3-3: Bi-linear diagram for steel and bi-linear diagram for concrete stress distribution for pure bending or bending moment with axial forces.
Figure 2.3-5: Parabolic-rectangular diagram for steel and bi-linear diagram for concrete stress distribution for pure bending or bending moment with axial forces.
Limit states
For clearly under-reinforced beams the flexural strength is determined by the tensile
2.4.1 Basic Assumptions figure 2.4.-1
steel ratio, and the influence from the compressive stress distribution is of less importance.
a) Plane sections remain plane (hypothesis of Bernoulli), that means strain distributions in the
The flexural strength of over-reinforced beams is more dependent on the stress-strain curves.
section are linear.
The position of the concrete compression resultant becomes more important as the compression
b) Perfect bond between concrete and (reinforcing and bonded prestressing) steel. In each fibre
zone in beams increases. Flexural capacities under different steel ratios and different steel ratios
is valid cs s with cs as "smeared" concrete strain (including crack opening) and s as steel
and different assumptions of compressive stress distributions are outlined by Nielsen
strain.
Comparison between test results from flexural strength of beams (fc=100 N/mm2) and
c) The tensile strength of concrete including the tension stiffening effect is neglected (so called
calculations with triangular stress block assumption referred by Bernhardt, show improved
"naked" state II).
agreement for over-reinforced beams compared to the parabola curve usually assumed in
d) Simplified design stress - strain diagrams for concrete and steel which take account of the
standards and codes.
elastic - plastic behaviour of the materials approximately. In EC 2 different diagrams for concrete and steel are given.
Flexural rotation capacity
The ultimate compressive strain in concrete (3.5 Promile) for bending and for normal force with
For reasons of safety a ductile behaviour of the structural elements is necessary.
great eccentricity must be reduced to (2 Promile) for centric normal force.
Cracking and visible deformations give obvious warnings and announce the imminent failure.
The strain diagram is to be turned around point C in figure 2.1-2, which distance from the most
Besides a sufficient ductility enables an economical design by allowing moment redistribution
compressed fibre is 3 / 7 of the total height of the concrete section.
and by taking into account the energy dissipation as a result of the plastic deformations in the case of cyclic excitations, e.g. earthquakes. The ductility of beams is characterised by its rotational capacity, which is usually defined as the plastic hinge rotation at maximum bending moment. For small reinforcement percentages, leading to tension failures, the rotational capacity is controlled by the stress-strain behaviour of the steel and increases with smaller amount of reinforcement. In the case of a concrete failure for over-reinforced beams the curvature and by that the rotational capacity and by that the rotational capacity decrease due to the reduced strain of the reinforcement. Concerning the behaviour of the compression zone, the shape of the curve is of a higher importance than the magnitude of the ultimate shortening. therefore, the steep descending part of the -curve for HSC leads to a decrease of the rotation capacity.
Figure 2.4-1 The compression stress distribution for HSC is different from normal strength concrete. The steep and linear ascending part of the curve, and the sudden drop after the ultimate stress peak is reached, influence both the position and the magnitude of the concrete stress resultant in beams subjected to flexure.
Limit states
STRUCTURAL ENGINEERING ROOM
2.4 Basic Values for the Design Resistance to Moments and Axial Forces
Department of Architecture
16
Example 2.1:
2.5 Verification at the Serviceability Limit States
STRUCTURAL ENGINEERING ROOM
Department of Architecture
17
Limitation of Stress under Serviceability Conditions The provisions at the ultimate limit states in Eurocode 2 may lead to excessive stresses in concrete, reinforcing steel. These stresses may, as a consequence, adversely affected the appearance and performance in service conditions and the durability of concrete structures. Key-words in this context are:
Maximum stress fc_ 103 MPa
Initial slope Ec of the stress-strain Ec 3320
Factors
- non-linear creep of concrete due to excessive compressive stresses, - increased permeability of the concrete surface due to micro - cracking of concrete around the reinforcing bars,
fc_ MPa
6900
n 6.858824
k 2.33129
Strain when fc reaches fc_
c_ 0.00297
- yielding of steel in service condition leading to cracks with unacceptable width, - cracks parallel to the reinforcing bars, - excessive deflection of concrete members caused by high stresses, cracking and creep of the
Expression, which describes the shape of a concrete cylinder stress-strain curve fc c
c
c_
fc_
Members. In reinforced concrete structures, longitudinal cracks parallel to the reinforcing bars may
1.210
occur if the stress level in the concrete under the rare combination of actions exceeds a critical
1.0810
value. Such cracking may lead to a reduction in durability.
9.610
In detail EC 2 requires the verification of stresses, associated with the following permissible
8.410
values, for the following reasons:
7.210
fc c
610 4.810
1.13 c_ fc_
7 7 7 7 7 7
under rare combinations of actions are limited to 0.6 f ck in absence of other methods (for
2.410
7 7
0 0
c 0.6 f ck
c 0 0.0001 0.007
n k
8
1) For preventing the development of longitudinal cracks, the concrete compressive stresses example, increasing the concrete cover of confinement of the compression zone).
c c_
8
3.610 1.210
n
n 1
710
4
1.410
3
2.110
3
3) for reinforcing steel subjected to loads and restraints, s 0.8 f yk 4) for reinforcing steel subjected to restraints only, s f yk
3
3.510
3
3
4.210
4.910
3
5.610
3
6.310
3
710
3
c
2) For preventing excessive creep, the concrete compressive stresses under quasi - permanent combinations of actions G k j Pk 2 i Qk i are limited to 0.45 f ck, c 0.45 f ck
2.810
Figure 2.1-1 The axial force Pmax that can by resisted by a rectangular section with the linearly varying strain the max. force is 0,6 .fc_.b.h
1.13 c_
0
fc c d c 0.451661fc_ c_ 1.513
fc 1.13 c_ fc_
Limit states
0.600001
fc 1.13 c_ 61.800061MPa
The characteristic cube strength of concrete (MPa): fc 20
fc_ 21 MPa
The stress of the concrete at the end of the diagram:
Initial slope Ec of the stress-strain fc_
Ec 3320
Department of Architecture
Maximun stress
MPa
bcu 0.73 fc MPa
- corresponding strain:
6900
bcu 0.0035
Maximum stress of the concrete (strength of prism)
Factors
n 2.035294
bc1 0.85 fc MPa
k 1.00871
corresponding strain:
Strain, when fc reaches fc_
bc1 0.002
c_ 0.001867
Limit quasi-linear stress concrete: bco 0.5 fc MPa
corresponding strain
Expression, which describes the shape of a concrete cylinder stress-strain curve c
fc c
c_
2.510 2.2510 210 1.7510
fc c
1.510 1.2510 110 7.510 510 2.510
fc_
4
n n k c n 1 c_
Instant (secant) modulus: Eij
7
1.99 c_
7
fc_
Eij 3.257 10 MPa
Tangent module for short-term stress:
7
1.65
7
Eijo28 4
1 3 12000 fc MPa
Eijo28 5.375 10 MPa
6 6
Region 1 (quasi-linear region) supposition
6
710
4
1.410
3
2.110
3
the max. force is 0,8. fc_.b.h fc c d c 0.800103fc_ c_ 1.99
fc 1.99 c_
4
3
7
2.810
3
3.510
3
4.210
3
3
4.910
5.610
c
0
4
Eij 3.257 10 MPa
bco
Eij 12000 fc MPa
7
0
bco 1
fc_
7
0
1.99 c_
bco 3.07 10
c 0 0.0001 0.007
0.787593
Figure 2.1-2 fc 1.99 c_ 16.539453MPa
3
6.310
3
710
3
1.65
Calculation of the coefficient: n 1.435
Tangent module for short-term stress:
1 3
Eijo28 12000f c MPa
60000 psi 413.685438MPa
Limit states
4
Eijo28 5.375 10 MPa
STRUCTURAL ENGINEERING ROOM
18
Tangent modulus = 0: 4
4
for comparison
E( 0) 5.375 10 MPa
Eijo28 5.375 10 MPa
The proof assuming= 1.65: _ 1.65
Terms for bc () can only be managed more complicated: n n n 2 n bco bco 1 bco bco bc( ) bco 2 ( n 1) bco
Region 2:
Figure 2.1-3 Equation an up curve (quasi-linear) for the region 1:
4
0 0.00001 bco
bc bco 3.1 10
( n 2) bc( ) bco n 2 ( n 1) bco bco n
1
bc_ ( )
bco bco
bc1
n 1 1.215
An up curve equation (linear) for the region1:
bc2 bc
bco b
bc1 bco b
210
5
1.510
bco
6
Stress of concrete
6
n1 bc bco bc bco n 1 n 1 1 bc1 bco bc1 bco
1
Region 1 and 2
7
7.5 bc( ) 10
bc_ ( ) 10
Region 1
10
Stress of concrete
STRUCTURAL ENGINEERING ROOM
Department of Architecture
19
2.5
0
0.1
0.2
0.3
0.4
bc1
7
bc2 bc bc( )
110
7
510
6
bco
3
10
Strain of concrete
0
510
110
3
Strain of concrete
Figure 2.1-4
Region 2 Region 1
The equation of tangent modulus-in:
Figure 2.1-5
1 d bco ( n 2) n 2 ( n 1) d bco bco n
Limit states
3
1.510
bc
for a = 1.65 for a = 1
E( )
4
210
3
c1 0.002
2 c c c1 c1
fcs c bc1 2
c 0 0.0001 c1
Stress
bc2 bc
210
7
1.510
7
110
7
510
6
bc( )
fcs c
The values of n for different strain of the concrete:
bco
no
n o
bc1 c
bc3 c
bc1 b
no c c n o bc1 n o 1 bc1 1
bc1
for
510
4
110
3
1.510
3
210
3
n 2.092
c 0.0005
bco
0
c 1 d bc1 d c b n o 1 bc1
Eb c
Diagram of concrete
for
c 0.001
n 2.092
for
c 0.0015
n 2.092
bc c Strain
Example 2.2: The determination of actual and statistical properties of concrete
Sh 2 Sh 1 HOGNESTAD
Testing of cube compressive strength of concrete Cubes dimensions:
Figure 2.1-6 0.5
Stress of concrete
fcs c
1.510
7
110
7
510
6
bc3 c
v 150.7 mm
b 149.6 mm
v 150.5 mm
a 156.4 mm
b 150.2 mm
v 151.8 mm
1
2
2
3
3
Weight: mbet 7.92 kg 1
bco
bc1
bet i
4
510
110
3
1.510
3
210
mbet 7.98 kg
mbet
i
a b v
i i i
3
Strain of concrete
bet i
2323.36 m 3 kg 2355.04
Average bulk density:
bet
Figure 2.1-7
3
2327.55
c Diagram of concrete HOGNESTAD
mbet 8.3 kg
2
Density:
bco
0
1
3
Diagram of Concrete
7
b 150.3 mm
2
no bc1 c c 1 n o bc3 c b n o 1 bc1 bc1 210
a 150.5 mm a 150.5 mm
1
n o 2.092
i
bet i
3
3
Limit states
bet 2335.32m
kg
STRUCTURAL ENGINEERING ROOM
Hognestad:
Department of Architecture
20
Compressive surface of the test cube:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
21
A a b i
Example 2.3: The determination of actual and statistical properties of concrete and steel:
A i
i i
0.02262 m2 0.02251
Test cube compressive strength of concrete Cubes dimensions:
0.02349
Maximum compressive force in press F 639 kN
F 753 kN
1
Cube strength for each cube:
fcub
F
i
fcub
A
i
F 714 kN
2
MPa
i
he true characteristics of concrete:
fcub.p
i1
v2 150.5 mm
a 3 156.4 mm
b3 150.2 mm
v3 151.8 mm
1
30.394
mbet 7.98 kg
bet i
mbet
i
a i bi vi
i
2323.36 m 3 kg 2355.04
fcub.p 30.696MPa
2327.55
Prismatic compressive strength:
fccyl fcub.p 0.855 0.005 fccyl fcub.p
Average Density:
fcub.p
MPa
fccyl 21.534MPa
beti bet
0.702
fct 0.274
2
3
kg
Ai
0.02349
fct 2.686 MPa
Max. compressive force in the jack: F1 639 kN
3 fcub.p 12710 10610 MPa MPa
F2 753 kN
4
Eco 2.919 10 MPa
Cube strength for each cube:
Classification of concrete: 1
bet 2335.32 m
0.02262 m2 0.02251
fcub.p MPa MPa
fcub 28.249MPa
3
Ai a i bi
Modulus of elasticity: Eco
i
Compression area of the test cube:
Tensile strength of concrete: 3
3
bet
i
3
mbet 8.3 kg
2
Density:
3
fcub
v1 150.7 mm
b2 149.6 mm
mbet 7.92 kg
28.249
33.445
Average cube strength
b1 150.3 mm
a 2 150.5 mm
Weight:
3
i
a 1 150.5 mm
Rbk i
fcub 33.445 MPa 2
fcub 30.394 MPa 3
Fi
Rbk
Ai
MPa
i
28.249 33.445 30.394
fcubm 25 MPa
Limit states
F3 714 kN
Safety factor of the concrete:
Average cube strength:
mb 1.3
3
Rbk.p
Rbk
i1
Nominal diameter of reinforcement:
i
Rbk.p 30.696 MPa
3
D 14 mm
Specific initial length: Lo 5 D
Prismatic- cylinder compressive strength:
Rbh Rbk.p 0.855 0.005 Rbh Rbk.p
mbt 1.5
Test of tendon by means of tensile
Rbk.p
MPa
Lo 70 mm
Density of reinforcement: Rbh 21.534 MPa
o 7850 kg m
3
Modulus of elasticity of steel:
0.702
Es 210000 MPa
length samples:
or
L1 354 mm
fccub_ f fccyl 0.855 0.005 MPa ccub_
fccyl 21.53 MPa
fccyl fcd 0.85 1.5
fcd 12.2025 MPa
L2 358 mm
Sample weight: M1 0.435 kg
Tensile strength of concrete:
Breaking force:
3
Fm 110 kN
2
Rbk.p Rbt 0.274 MPa MPa
Rbt 2.686 MPa
L3 360 mm
or
1
M2 0.435 kg
M3 0.440 kg
Fm 110.5 kN
Fm 106.5 kN
2
3
force obtained yield strength when deformed. 0.2%: fccub_ fctm 0.274 MPa
0.66
MPa
fctm 2.63 MPa
Fe.0.2 74.25 kN 1
Fe.0.2 76.83 kN
Fe.0.2 76.89 kN
Lu 91.5 mm
Lu 92.8 mm
2
3
Final measured length: Ebt 32000 MPa
Lu 93.4 mm 1
Modulus of elasticity: E bo
3 R bk.p 12710 10610 MPa MPa
The actual cross-sectional area: 4
Ebo 2.919 10 MPa
Ao i
Mi
Ao
o Li
mm
CLASSIFICATION OF CONCRETE Rbk 28.249 MPa
>
Rbk 33.445 MPa
>
Rbg 25 MPa
=> C25
1 2
2
i
2
156.537 154.788 155.697
Rbk 30.394 MPa 3
Limit states
3
STRUCTURAL ENGINEERING ROOM
The true characteristics of concrete:
Department of Architecture
22
Example 2.4: The beam has a theoretical span 2.50 m, perform load test of reinforced concrete beam.
Elongation:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
23
As
Lu Lo i
As
Lo
%
i
i
1. Determine the real and statistical properties of concrete and steel at the time of testing
33.429
2. Theoretical calculation with the actual properties of materials
30.714
3. Proposal test load and load process
32.571
4. Realization of load test
Tensile strength of reinforcement: Rm i
Fm
Rm
Ao
MPa
i
5. Evaluation of load test
i
i
1. Determine the real and statistical properties of concrete and steel at the time of testing
702.71 713.881
Real properties:
684.02
The agreed yield strength in permanent deformation of 0.2% Fe.0.2
Re.0.2
Ao
MPa
i
Re.0.2 i
i
i
fccub_ 30.696 MPa
average cube strength short prismatic strength:
fccub_ f fccyl 0.855 0.005 MPa ccub_
474.329 496.357 493.844
tensile strength of concrete:
The average value of the measured characteristics:
Re.0.2i i
Re.0.2.p
3
fccub_ fctm 0.274 MPa
Re.0.2.p 488.177 MPa
i
3
0.66
MPa
fctm 2.63 MPa
modulus of elasticity:
Rmi Rm.p
fccyl 21.53 MPa
Ect 32000 MPa
Rm.p 700.204 MPa
statistical properties
Asi As.p
i
3
fyk 488.177 MPa
As.p 32.238 %
fyk fyd 1.15
fyd 424.5017 MPa
Figure 2.4-1: View of tested beam
Limit states
fccyl fcd 0.85 1.5
The lower area of reinforcement:
fcd 12.2025 MPa
Ast 3
modulus of elasticity:
( 0.014 m)
2
Ast 4.62 cm
4
2
Force at the bottom reinforcement:
Ec 30000 MPa
Nst Ast fyd
Steel:
real properties
Nst 196.04 kN
The depth of compression zone of concrete:
yield stress of bars 14 fyk
fyk 488.177 MPa
fyd 1.15
yield stress of bars 6
x u fyd 424.5017 MPa
Nst
x u 0.134 m
b fcd
Ultimate bending moment: Re6
Re6 191.0 MPa
fyst 1.15
modulus of elasticity:
fyst 166.087MPa
xu Mut Nst d 2
Mut 21.77 kN m
Shear resistance
E s 210000 MPa
Coefficients:
2. Theoretical calculation with the actual properties of materials
q 1
n 1
Force which bearing by concrete: The actual section width of the beam:
Vcu
b 12.0 cm
The actual amount of the cross-section of the beam: h 20.0 cm
Real Densities: bet 23.31 kN m
b h q fctm
Vcu 21.00 kN
Area of stirrups: ( 0.006 m)
2
4
A ss 0.57 cm
Force in the stirrups:
a 1 m
Nss Ass fyst
Nss 9.39 kN
3
Distance between the stirrups:
The actual resistance of the beam
s s 0.160 m
Equivalent section height d h 0.015 m
3
Ass 2
The actual length of the beam: l 2.5 m
1
0.014 m 2
Force per meter in the stirrups d 0.178 m
Nss fss ss
Limit states
fss 58.70 kN m
1
2
STRUCTURAL ENGINEERING ROOM
Design of compressive strength
Department of Architecture
24
Length the inclined cracks:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
25
c d
Deflection at the time of cracking
1.2 n b fctm
c 0.452 m
fss
x r
Force which transferred by stirrups: Vssu c fss
Shear resistance load capacity Vut Vcu Vssu
n As1 As2
1
b
Vssu 26.52 kN
Ac b x r 2 n As2
Vut 47.52 kN
z r
Moment of cracking
2
x r as2
x r 0.069 m
Ac 0.0095 m
xr
2 As2 n 2
2 b As1 d As2 as2 n As1 As2
xr
b xr d
1
x r as2 xr
d as2
2
z r 0.144 m
Ac
Area of reinforcement: As1 3 As2 2
( 0.014 m)
2
As1 4.62 cm
4 ( 0.010 m)
2
As2 1.57 cm
4
2
2
as1 0.015 m as2 0.025 m
0.014 m 2 0.01 m 2
r
as1 0.022 m as2 0.030 m
Brb
modulus ratio of steel and concrete: n
Es
Br
n 6.56
Ect
ideal cross-sectional characteristics:
Ai b h n As1 As2
bh x i
h 2
Ai 0.0281 m
n As1 d As2 as2
Ai 2
3
5
M rt M rt
1
r 1
d zr
Brb 1477.7 kN m
2 1 E A E A c c s s1 1
Br 2788.8 kN m
1 r r Bra Brb
q0 b h bet
q0 0.559 kN m
Bending moment by self-weight:
x i 0.106 m 4
Ii 1.025 10
m
M0
4
1 8
q0 l
2
M0 0.44 kN m
Applied force which cause cracking
Moment of cracking: Ii Mrt fctm h xi
Bra 0.85 Ect Ii
Frt
Mrt 2.86 kN m
Limit states
Mrt M0 a
Frt 2.42 kN
2
2
Self-weight:
2
h 2 2 b h b h x i n As1 d x i n As2 x i as2 Ii 2 12 1
4 1
1
Bra 2788.8 kN m
2
5 q0 l frt 384 Br
4
Frt a 24 Br
2
2
3l 4a
Department of Architecture
Proposal test load and load process
Deflection in cracking:
frt 0.64 mm
Deflection and crack width at base load Base load: Ms 0.5 Mut
1
Br
Mrt 1 r 5 4 Ms
Ms 10.89 kN m
B 1534.2 kN m
r 1 r r Bra Brb Force which produce the base load:
1
r 0.0783
2
Figure 2.4-2 Limit value of arm forces:
Fs
Ms M0
Mut ahr Vut
Fs 10.45 kN
a
Deflection for base load: frt
st
5 384
q0 l Br
As1
4
Selected value of lever arm: Fs a
24 B r
3 l
2
4 a
2
a
frt 4.37 mm
Ms
sr fyd M
st 1.92 %
bh
ahr 0.458 m
ut
0.458m
1m ahr
failure occurs of the moment
Self-weight of devices: Dynamometer sr 212.25 MPa
Fz1 0.170 kN
Hydraulic press
Fz2 0.350 kN
tb 6
as1
tb max 1 tb
h
tb 1
1.2
k 2500
Piece for bearing load Fz3 0.600 kN
Crack width for base load:
Steel plate 1
sr 3 wa k tb 0.039 st ( 1.2) mm Es
1
wa 0.0637 mm
Fz4 2 ( 10 mm 150 mm 100 mm) 78.5 kN m
3
Fz4 0.024 kN
Rollers Fz5 2
( 2 cm )
Limit states
4
2
14 cm 78.5 kN m
3
Fz5 0.007 kN
STRUCTURAL ENGINEERING ROOM
26
Together
STRUCTURAL ENGINEERING ROOM
Department of Architecture
27
Fz Fz1 Fz2 Fz3 Fz4 Fz5
Fz 1.150 kN
Mi kN m
Bending moment of self-weight equipment and beams: M0
1
Fz
2
q0 l
8
2
M0 1.012 kN m
a
the initial stage of loading: k 0
M0
k 0 0.09
Ms
the degree of load at crack formation: k rt
Mrt
k rt 0.26
Ms
Load levels:
k k 0 0.15 0.2 0.3 0.4 0.5 0.6 k rt 0.7 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.5 3.0 n 18
T
i 1 n
Fl
i
pl i
kN
MPa
i
mm
ki
1.0123
0
0
0
1.6329
1.2413
0.1951
0.0248
2.1772
2.3299
0.3663
0.0466
3.2658
4.5071
0.7086
0.0901
4.3545
6.6843
1.0508
0.1337
5.4431
8.8616
1.3931
0.1772
6.5317
11.0388
1.7354
0.2208
2.8593
3.694
0.5807
0.0739
7.6203
13.216
2.0777
0.2643
8.7089
15.3933
2.4199
0.3079
10.8861
19.7477
3.1045
0.395
13.0634
24.1022
3.7891
0.482
15.2406
28.4566
4.4736
0.5691
17.4178
32.8111
5.1582
0.6562
19.5951
37.1656
5.8427
0.7433
21.7723
41.52
6.5273
0.8304
27.2154
52.4062
8.2387
1.0481
32.6584
63.2923
9.9501
1.2658
0.093 0.15 0.2 0.3 0.4 0.5 0.6 0.2627 0.7 0.8 1 1.2 1.4 1.6 1.8 2 2.5 3
load moment: Mi k i Ms
Applied force produce by press: Fl 2
Mi M0 a
i
area of the press: Al 63.61 cm
2
4. Realization of load test test took place according to the specified scheme, measured by the rotation of all, deflection, rotation at mid-span and development of cracks 5. Evaluation of load test Graf deflection in the center of the beam on the degree of load:
pressure in the press:
n 15
Fl i pl i Al
constant force measuring device: k s 50 kN mm
i 1 n
Reading the deformometer:
1
T
P ( 6.20 6.29 6.351 6.485 6.591 6.711 6.936 7.052 7.24 7.61 8.083 8.589 9.6 10.44 11.52) mm
Read on force measuring device: i
Fl
i
The resulting deflection:
ks
wi Pi P1
Limit states
deflection at the mid span of the beam
0
0.093 0.15
0.151
0.2
1.6
0.285
0.3
1.4
0.391
0.4
1.2
0.511
0.5
1
0.736
0.6
0.852
0.2627
1.6 1.4
0.8
1.04
0.6
1.41
0.4
1.883
0.2 0
2.389 0
1
2
3
4
deflection [mm]
5
6
3.4 4.24 5.32
1.8
Loading level [-]
1.8
Rotation on the end of the beam
2
ki
0.09
2
Loading leve [-]
0.7
1.2 1 0.8 0.6
0.8
0.4
1 1.2
0.2
1.4
0
1.6
0
1
2
1.8
3
4
5
6
Rotation [mm/m]
Figure 2.4-4 Figure 2.4-3
L1 i
Graf rotation supporting cross-sectional view of the degree of load:
mm m
Reading on the left side of the beam: T
1
L1. ( 11.24 10.81 10.76 10.62 10.48 10.33 10.12 9.99 9.78 9.41 8.91 8.34 7.24 6.31 5.11 ) mm m
reading to the right side of the beam: T
L2. ( 6.60 6.69 6.75 6.875 7.01 7.16 7.385 7.52 7.71 8.02 8.53 9.05 10.18 11.09 12.26 ) mm m
the resulting rotation of the left end: L1 L1. L1. i i 1
the resulting rotation to the right end: L2 L2. L2. i i 1
1
L2 i 1
mm m
1
0
0
0.43
0.09
0.48
0.15
0.62
0.275
0.76
0.41
0.91
0.56
1.12
0.785
1.25
0.92
1.46
1.11
1.83
1.42
2.33
1.93
2.9
2.45
4
3.58
4.93
4.49
6.13
5.66
ki 0.093 0.15 0.2 0.3 0.4 0.5 0.6 0.2627 0.7 0.8 1 1.2 1.4 1.6 1.8
Graf of strain in the center of the beam for moment 0.5 Mrt Mrt 0.5 Mut: n 4
i 1 n
The original length of the measuring elements: l 20 cm
Limit states
7
STRUCTURAL ENGINEERING ROOM
wi mm
Department of Architecture
28
2.6 Shrinkage
height each measuring elements from the top of the beam
Shrinkage is the decrease of concrete volume with time. This decrease is due to changes in the
T
h ( 1.4 8.2 14.9 22.5 ) cm
moisture content of the concrete and physicochemical changes, which occur without stress attributable
Reading the individual measuring elements:
to actions external to the concrete. Swelling is the increase of concrete volume with time. Shrinkage and
T
D1 ( 0.338 0.408 0.929 0.948 ) mm
swelling are usually expressed as a dimensionless strain (in./in. or mm/mm) under given conditions of
T
D2 ( 0.345 0.4081 0.9252 0.9398 ) mm
relative humidity and temperature. Concrete immersed in water does not shrink but may swell.
T
D3 ( 0.3765 0.4129 0.901 0.8785 ) mm
Shrinkage of high performance concrete may be expected to differ from conventional concrete
T
D4 ( 0.388 0.4129 0.8792 0.8425 ) mm
in three broad areas: plastic shrinkage, drying shrinkage, and autogenous shrinkage. Plastic shrinkage
resulting Strain: 1
D1 D1 i i l
i
occurs during the first few hours after fresh concrete is placed. During this period, moisture may 3 i
D1 D3 i i l
2 i
D1 D2 i i l
4 i
D1 D4 i i l
evaporate faster from the concrete surface than it is replaced by bleed water from lower layers of the concrete mass. Paste-rich mixes, such as high performance concretes, will be more susceptible to plastic shrinkage than conventional concretes. Drying shrinkage occurs after the concrete has already attained
Average strain in the middle of the beam 1.4
its final set and a good portion of the chemical hydration process in the cement gel has been accomplished. Drying shrinkage of high strength concretes, although perhaps potentially larger due to higher paste volumes, do not, in fact, appear to be appreciably larger than conventional concretes. This
3.51
Depth [cm]
STRUCTURAL ENGINEERING ROOM
Department of Architecture
29
is probably due to the increase in stiffness of the stronger mixes. Data for VES and HES mixes is limited.
5.62
Autogenous shrinkage due to self-desiccation is perhaps more likely in concretes with very low W/CM
7.73
ratio, although there is little data outside indirect evidence with certain high strength concrete research [Aitcin et al. 1990]. Shrinkage should not be confused with thermal contraction which occurs as concrete
9.84
loses the heat of hydration.
11.95
Shrinkage is a function of the paste, but is significantly influenced by the stiffness of the coarse
14.06
aggregate. The interdependence of many factors creates difficulty in isolating causes and effectively
16.17
predicting shrinkage without extensive testing. The key factors affecting the magnitude of shrinkage are:
18.28
20.39
Aggregate. The aggregate acts to restrain the shrinkage of cement paste; hence concrete with higher aggregate
22.5 Average strain [-]
Figure 2.4-5
content exhibits smaller shrinkage. In addition, concrete with aggregates of higher modulus of elasticity or of rougher surfaces is more resistant to the shrinkage process.
Water-cementitious material ratio. The higher the W/C ratio is, the higher the shrinkage. This occurs due to two interrelated effects. As
W/C increases, paste strength and stiffness decrease; and as water content increases, shrinkage potential increases.
Limit states
Member size.
shrinkage varies according to the sequence of occurrence of carbonation and drying process. If both
Both the rate and the total magnitude of shrinkage decrease with an increase in the volume of the
phenomena take place simultaneously, less shrinkage develops. The process of carbonation, however,
concrete member. However, the duration of shrinkage is longer for larger members since more time is needed for shrinkage effects to reach the interior regions.
2.7 Creep
Medium ambient conditions.
Creep is the time-dependent increase in strain of hardened concrete subjected to sustained stress. It is
The relative humidity greatly affects the magnitude of shrinkage; the rate of shrinkage is lower at
usually determined by subtracting, from the total measured strain in a loaded specimen, the sum of the initial instantaneous strain (usually considered elastic) due to sustained stress, the shrinkage, and any
higher values of relative humidity. Shrinkage becomes stabilized at low temperatures.
is dramatically reduced at relative humidities below 50 percent.
thermal strain in an identical load-free specimen, subjected to the same history of relative humidity and
Admixtures. Admixture effect varies from admixture to admixture. Any material which substantially changes the
pore structure of the paste will affect the shrinkage characteristics of the concrete. In general, as pore
temperature conditions. Tests indicate a reduction in the creep coefficient for HPC. At the same time, the creep coefficient is constant up to a higher stress ratio for high strength than for low strength concrete.
refinement is enhanced, shrinkage is increased. Pozzolans typically increase the drying shrinkage, due to several factors. With adequate curing,
Experimental results reported in by Ngab indicate an even higher reduction of the creep
pozzolans generally increase pore refinement. Use of a pozzolan results in an increase in the relative
coefficient. Testing of Specimens of concrete strength of 60-70 MPa under unsealed conditions showed
paste volume due to two mechanisms; pozzolans have a lower specific gravity than portland cement
that the creep coefficient was about 50 to 75% of that of normal strength concrete. Under sealed
and, in practice, more slowly reacting pozzolans (such as Class F fly ash) are frequently added at better
conditions the creep coefficient of HPC increased to 75 to 90% of that of normal strength concrete. The
than one-to-one volume replacement factor, in order to attain specified strength at 28 days. Additionally,
stress-creep relation for HPC was found to be approximately linear over a range from 0 to about 70% of
since pozzolans such as fly ash and slag do not contribute significantly to early strength, pastes
the ultimate strength, while normal strength concrete has a linear stress-creep relation only up to about
containing pozzolans generally have a lower stiffness at earlier ages as well, making them more
30 to 50%.
susceptible to increased shrinkage under standard testing conditions. Silica fume will contribute to strength at an earlier age than other pozzolans but may still increase shrinkage due to pore refinement. Chemical admixtures will tend to increase shrinkage unless they are used in such a way as to reduce
2.7-1 The investigation of the effects of shrinkage of a composite concrete structure
the evaporable water content of the mix, in which case the shrinkage will be reduced. Calcium chloride,
Time - dependent analysis of prestressed and partially prestressed concrete composite beams is
used to accelerate the hardening and setting of concrete, increases the shrinkage. Air-entraining agents,
difficult and complicated, because the cross-section can be cracked or without cracks,
however, seem to have little effect.
depending on the degree and level of load. Solving methods that are available are not
Cement.
sufficiently simple that they can be commonly used in practice.
Type the effects of cement type are generally negligible except as rate-of-strength-gain changes.
If there is no available a good building model, it is virtually impossible to perform any useful
Even here the interdependence of several factors make it difficult to isolate causes. Rapid hardening cement gains strength more rapidly than ordinary cement but shrinks somewhat more than other types, primarily due to an increase in the water demand with increasing fineness. Shrinkage compensating cements can be used to minimize shrinkage cracking if they are used with appropriate restraining reinforcement.
time - dependent analysis of structures. It is also necessary to understand the basic facts and physical mechanism of creep and shrinkage. Examine the cross section of two coupled concrete varying quality ( Ec) And also of different ages, which will be reflected on the effect of stress such cross-section.
Carbonation. Carbonation shrinkage is caused by the reaction between carbon dioxide (CO2) present in the
atmosphere and calcium hydroxide (CaOH2) present in the cement paste. The amount of combined
Limit states
STRUCTURAL ENGINEERING ROOM
Department of Architecture
30
STRUCTURAL ENGINEERING ROOM
Department of Architecture
31
Figure: 2.7.1-2 Figure: 2.7.1-1 A) Because the cross-section of external forces do not act, we can write the following
E
equilibrium conditions:
E
or when the external forces do not act on the cross-section, we can write the following
1
0
N
1
i
N
2
M
2
equilibrium conditions: V
E J m
M M
1
2
M
M
0
1
i
1 1
E J
E J
M
2 2
Because
2
E J
E J
Nr
N
1 1
1
Nr
2 2
1 1
E J
M
2 2
E J
1
E J
M 1
1
M
Nr M
2
1 1
will be
1
E1 J 1 E J 2 2
Nr
2 2
then
E J
1 1
B) When Composite member should have two of the same curvature (the difference and 1 Is
2
M
negligible at large values of the radius of curvature R)
1
2 2
Nr
1
M
dx
( z ) d d
z
dx
d
z
z E dF
z
E
E
M
E
M
2
M
z dF
E
1
E E
J
1 1
thus
1 2
1 1
We gate zm
E J E J
1 1
2 2
E J
2 2
E Nr
Nr
E J
E
z dF M
E J
E J
1
J
2
1
E J
1
E
m J M
2 2
J
2
1
Nr
J
1
m J J
1
M
2
2
Nr
2
m J J
1
2
C) Beam curvature and avoiding the displacement of joint offset in part 1 and 2 can be provided
1
M
1
M
E J
E J
1
1 1
M
to merge continuity cross- sections before and after deformation (for simplicity we introduced 2
E J
2 2
an element of length L = 1 m, then the deformation l will soon be relative deformation) The equation for curvature dx
Limit states
d
1
dx
d
also
1
M
1
E J
1 1
M
2
E J
2 2
M r
r
M r
1
2
E J
Deformation and flow from Hook's law 1 2
E J
1 1
2 2
N
1
M
N F
1
J
1
z
1
1
N M2 z 2 F J 2 2
2
E F
When investigating continuous structures will be advantageous to calculate the support bending moments and use them to construct a diagram of additional bending moments for continuous beam. Values of moments can be determined using the equation of the force method, where the outer deformation io is obtained by the relationship derived here:
Figure: 2.7.1-3
Then the total relative deformation as shown we can write (beware of the forces only affects
Figure: 2.7.1-4
deformation by shrinkage of the balance shrinkage Parts 1 and 2):
1
zm
1zm 2zm
1
2
M´ r N N1 2 1 E F E J E F 2 2 1 1 1 1
N
1
E F
N
1 1
2
E F
2 2
Nr
2
2 2
E J E J
1 1
1zm 2zm
N
l
1 E F
1 1
then
1 E F
2 2
r
1
Nr
2
1 F
E J E J
1 1
m J M
N
1
m F
2
and
2
M
2
Nr
m r 2
m J J
1
III
M
1
1 1
2
E J
2 2
2
II
3
2
III
l
M 2 1
2 E1 J 1 II l
M 3 1
2 E1 J 1 III
l
M 2 2
2 E2 J 2 II l
l I
1
2
I
M 3 2
2 E2 J2 III
2
Then
2
m J J
2
l
2zm
1
m J
1
m J J
1
2 2
1zm
E J
II
thus
M
On this basis, we can determine the rotation of the numbers of individual parts as follows:
Then we can express the unknown force N:
1
M l M 1 1 2 1 2 E1 J 1 2 E1 J 1 I II
l
2
10
From these results, we can calculate the stress level of each section, and therefore the entire
the investigation of the fiber after the neutral axis of the respective cross section - Part 1 and 2:
l M M 2 1 3 1 2 E1 J 1 2 E1 J 1 II III
l
cross-section of the eccentric patterns for strain pressure, where z and z are the distance from 1 2
20
Limit states
M l M 1 2 2 2 2 E2 J 2 2 E2 J 2 I II
l
l M M 2 2 3 2 2 E2 J2 2 E2 J2 II III
l
l
M 1 1
2 E1 J 1 I
l
M 1 2
2 E2 J2 I
STRUCTURAL ENGINEERING ROOM
then the strain will be
Department of Architecture
32
From the equations of the force method:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
33
X
1( t ) 11
X
2( t ) 12
10
0
X
1( t ) 21
X
2( t ) 22
20
Differential equations
0
= d . When the concrete of the two beams is the same age, then=1, ´ = , ´ 11 11 10( t ) 10( t )
at any time . The finale effect of shrinkage We can calculate the unknown moments X , X 1( t ) 2( t )
because, go = constant. then d = Const. and further will be: 10( t )
is determined by summing together the two effects. Then we can determined the stresses in different cross sections due to moments X , X , as in the classic cross-section composed 1( t ) 2( t ) of two materials with different modulus of elasticity. In our case, when we introduced
E E
1
10( t )
dx
1( t )
d
11
X
1( t ) 11
10
dx
0
1( t )
d
X
1( t )
10 11
M
then stresses at the interface of Parts 1 and 2, intended for the same deformation (from Hook's
Where M is the bending moment above the support in a monolithic beam
law).
the equation has solution as follows.
1
2
1
E
1
E
2
E
2
1
E
1 2
2
2
X
transformed cross-section (introduction
E E
1 2
) Ji W i z the
M ( t)
1( t )
M
stress will be 2
M
1 e
where
Wi
0
( t) t
0
Stress in cross-section we find, by means of the calculation of section Properties of
0
d
,
2
d
B). simplified procedure Is based on the following equation:
Creep of concrete in the calculation of structures X
A) Solution using differential equations
11( t ) 1( t )
10( t )
0
X
1( t )
10( t )
10
Where
Ec( t )
10( t ) 11( t )
then
1 e Ec
11( t )
M M
1
0 0
We use the L'Hospital's rule (derivative by numerators, denominators fraction)
Figure: 2.7.1-5
Limit states
lim 0
1 ( ) 1 ( ) e 2 1
ds
1 1 e
If we neglect the reinforcement, then
1
Ec( t ) J c
lim 0
1 e lim ( ) 2 0 1
if =0, then = then next
Ec( t )
Ec
1 e( )
Then (see Bending moment diagrams):
1
11( t ) Ec
11( t )
1 e( )
Jc
M M ds
1
1
1 2 2 l ( ) J c 2 3 Ec 1 e
2 l ( ) J c 3 Ec 1 e
Figure: 2.7.1-6
10( t )
10( t )
10
10
Ec J c
1. Span M M ds
1
1
t1
3 go l Ec Jc 12
2 1 2 1 2 l go l E J 2 c c 3 8
1
( )
0
1
t1 0.083
12
t4
1month
t2
2 12
t4 8.333
12
t2 0.167
t3
305
3 12
t3 0.25
10
And X
1( t )
M ( t)
Ec J c
go l
3
12
Ec J c 1 e
o
2 l 3
X
1( t )
M ( t)
1 8
2
go l 1 e
( )
M 1 e
( )
t 3
5
3 12
k
or
4
Ec
1 MPa
l
1m
J
1 m4
go
1 kN m
t 3 0.25
Rocsh 1mes
1e
1mes 0.501
Limit states
t1
2mes
1e
2mes 0.579
t2
3mes
1e
3mes 0.627
t3
1
STRUCTURAL ENGINEERING ROOM
c) The simplified procedure in the construction of two unknown moments
1 2 2 e 2 1
Department of Architecture
34
STRUCTURAL ENGINEERING ROOM
Department of Architecture
35
t4
1e
nekone
nekone
11t 1.706
1
i=1.3
1 e 2
2
t4
1 e
1 e 1
t2 o
2.105
1 e 3
1 e
Ec
o
1e
3
2.496
1
Ect Ec 1
i
1 e
1.864
3
1
1e
1 e 3
Ect Ec 3
22t
t3
i
Ecti
t4
t4
1
2
1 e
1 e 1
3
12t
t1 o
Ect Ec 2
1
1 e
1 e 2
0.368
2
2
0.417
d
10( t )
10
2( t ) 12( t )
0
10( t )
X
1 21( t )
X
2( t ) 22( t )
0
20( t )
We Calculate the coefficient ij as in elastic state, but we consider a change of modulus of elasticity of concrete (see figure) 1
11( t ) Ec
11t
1e
1
1
Jc
then
M M d
1
1
E c J c
s
1
M M ds
1
1
Ec
1e
2
10t
1 2 1 1 2 1 2 1 2 1 l go l l go l 1000 1 2 E J E J 3 8 2 3 8 2 c c
10t 0.192
20t
1 2 1 1 2 1 2 1 2 1 l go l l go l 1000 2 3 E J E J 3 8 2 3 8 2 c c
20t 0.165
2
We substitute calculated values (all applicable to Part t = ) mentioned in Condition
M M ds
1
1 2 1 1 2 1 2 1 2 1 l go l l go l 2 1 Ec Jc 3 8 2 2 Ec J c 3 8
101 1 102 2
System of conditional equations: 1 11( t )
e
10t
d
e
1.86364327
0.4533558
X
1.534
2.10523659
0.41714226
X
22t
0.4
2.49628063
MPa
0.36759053
1 1 1 2 l 1 103 kN m 2 3 2 3 1 e E 1 e J Ec J c 2 3
21t
we know that:
i
i
21t 12t
We get them by multiplying the respective values of elastic coefficients of creep, because
Ect
12t 0.4
Coefficients i0 ( t ) represent creep deformation caused by external loading (self-weight go).
0.453
3
1 1 1 3 l 1 10 kN m 2 3 2 E 1 e J c 2
1
Equation system of conditional equations:
Jc
X1t
1 1 2 1 1 2 3 l 1 1 l 10 kN m 2 3 2 3 1 2 J 1 e J E 1 e Ec c 1 2
Limit states
5 kN m
X2t
10 kN m
Given X1t 11t X2t 12t 10t kN m
0
X1t 21t X2t 22t 20t kN m
0
If
Ec
11t
X1t 0.093 m kN
32000 MPa
go
30 kN m 1
l
X2t 0.084 m kN
27 m
I
2.57 m
4
1 1 2 1 1 2 3 l l 10 kN m 2 3 2 3 1 2 J 1 e J E 1 e E c c 1 2
11t 0.00143935
12t
22t
1 1 1 3 l 1 10 kN m 2 2 3 E 1 e J c 2
12t 0.00034
21t 12t
21t 0
1 1 1 2 l 1 103 kN m 22t 0.001 2 3 2 3 1 e E 1 e J Ec J c 2 3
10t
1 2 1 1 2 1 2 1 2 1 l go l l go l 1000 1 2 E J E J 3 8 2 3 8 2 c c
10t 3.538
20t
1 2 1 1 2 1 2 1 2 1 l go l l go l 1000 2 3 E J E J 3 8 2 3 8 2 c c
20t 3.052
The system of conditional equations: X1t
5 kN m
X2t
10 kN m
Given X1t 11t X2t 12t 10t kN m
X1t Find X1t X2t X2t
0 X1t 2029.749 m kN
X1t 21t X2t 22t 20t kN m
0
X2t 1828.576 m kN
Limit states
STRUCTURAL ENGINEERING ROOM
X1t Find X1t X2t X2t
Department of Architecture
36
3. Structural system consists of the primary load-bearing structure, including its members and
STRUCTURAL ENGINEERING ROOM
Department of Architecture
37
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.
Figure 3.2: Structural system of reinforced concrete members 3.1 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 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
Figure 3.1: Perspective view of RC frames
then called a doubly reinforced concrete beam. As beams usually support slabs, it is possible to
Loads are forces that act or may act on a structure. For the purpose of predicting the
make use of the slab as part of a T-beam. In this case the slab is generally not doubly reinforced.
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. 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 analyzing the system will only be gained
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
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
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.
flexural buckling or excessive lateral deformation.
Structural system
Figure 3.1-1 Figure 3.1-2
Structural system STRUCTURAL ENGINEERING ROOM
Department of Architecture
38
Example 3.1-1: The cross - section dimensions of a beam shown on figure 3.1.1-1 are
STRUCTURAL ENGINEERING ROOM
Department of Architecture
39
subjected to bending moment Msd. Determine the required tension reinforcement to the crosssection.
Msd
Material data:
0.09368
2
b d fcd
Characteristic value of concrete cylinder compressive strength (MPa): fck 40 MPa
1) To apply the design, the bending moment Msd has to be brought into a dimensionless form:
From the design diagram B3-B3.3, for fyk = 300 MPa we obtain the reinforcement ratio: 0.03833
fckcyl 0.8 fck
2
The required tension reinforcement A2 ( cm ) is as follows:
Design value of concrete cylinder compressive strength (MPa): 2
fcd 0.85
fckcyl 1.5
3
fcd 18.13333 MPa
fckcyl MPa fctm 3.04015 MPa 10 MPa
fctm 1.4
Characteristic yield stress of reinforcement (MPa):
fyk 300 MPa
fyd
Ecm 35 GPa
e
fyk 1.15 Es Ecm
MN
Ast 6
e 5.71429
( 1.8 cm )
2
Ast 0.00141m
2
Ast 0.00153m
4
0.12
2
we provide
2
x d
x 0.0696m
x u 0.8 x
L 6 m
The required compression reinforcement Asc ( cm ) is as follows:
Asc
dst 0.02 m
Ast fyd x u b fcd
Asc 4
d 0.58 m
Msd 0.200 MN m
Asc 0.00017m
fyd ( 1.2 cm )
Design maximum bending moment ( MN m):
4
2
2
Asc 0.00045m
we provide
2
w
Ast Asc bh
w 0.00942
Compression forces acting in the reinforcement resp. in concrete and tension force in tension reinforcement are calculated as follows:
Fsc Asc fyd
Fst Ast fyd
Fc x u b fcd
Fsc 118.01461kN
Fst 398.29931kN
Fc 353.3824kN
Fsc Fc 471.39701kN 2
Area of the transformed uncracked cross-section ( m )
x u 0.05568m
2
h 0.60 m
Effective depth of a cross-section (m): d h dst
100 cm
Es 200 GPa
Cover of reinforcement (m): dsc 0.02 m
b d fcd
The depth of compression zone: fyd 260.87 MPa
Cross-section ( m): b 0.35 m
Ast
Figure: 3.1.1-1
Ai b h e Ast Asc
Structural system
Ai 0.22131m
2
further determine the stiffness of the beam with the complete exclusion of tension in concrete, then the depth of the compression zone x r we determined from the conditions of the balance
agi
forces in cross section, which after the treatment can be written in the form:
b h 0.5 h e Asc dsc Ast h dst
agi 0.30777m
Ai
2
Moment of inertia of the transformed uncracked cross-section ( m ) 2
3 1
Ii b h
x r
12
h a A a d 2 A h d a 2 gi e sc gi sc st st gi 2
bh
Ii 0.00717m
c1 8.58089 MPa
e Ast Asc
2
x r 0.14163m
fcd c1
Ac b x r 2 e Asc
x r dsc
Ac 0.05401m
xr
2
Arm of internal forces:
Msd
c2 8.14774 MPa
Ii h agi
2
b xr
4 e A sc d sc
z r d
x r d sc xr
z r 0.51336m
2 A c
b
2 b Ast d Asc dsc
1
where b xr 2 the static moment of area A c based to the upper compression edge of the crosssection, bending stiffness of the beam with total exclusion of tension in concrete will:
fctm c2
1
first, we determine:
Msd
Ii agi Stress in tension (MPa): c2
4
Stress in compression (MPa): c1
e Ast Asc
Fsc agi dsc Fc agi
xu 2
Fst d agi
Brb
Msd
The ultimate bending moment is found using the following equation:
d zr
2
Brb 68720.14187m kN
2 1 E A E A s st cm c
The resulting bending stiffness:
xu Mu Fc d Fsc d dsc 2
Mu 261.21181m kN
Mu Msd
ok
r
Mcr
Ii h agi
fctm
Mcr 74.62552m kN
Ms
Msd 1.12
Ms 178.57143m kN
Br
b) Determine the bending stiffness
1 4
Mcr
5
Ms
1
1 1 r r Brb Bra
- element has no cracks the bending stiffness we determine using the following equation: Br 0.85 Ecm Ii
2
Br 213406.5593m kN
- the element with the expected cracks first determine the stiffness of the beam without crack:
Structural system
r 0.27238
r 0
2
Br 84284.9175m kN
STRUCTURAL ENGINEERING ROOM
The distance of the extreme fibre from the neutral axis (m):
Department of Architecture
40
Excessive deflections are unacceptable in building construction, as they can cause
STRUCTURAL ENGINEERING ROOM
Department of Architecture
41
cracking of plaster in ceilings and can result in jamming of doors and windows. Most building codes limit the amount of allowable deflection as a proportion of the member’s length, i.e. 1/180, 1/240 or 1/500 of the length. It can be seen that deflection is greatly influenced by the span L, and that the best resistance is provided by beams which have the most depth (d), resulting in a large moment of inertia.
flim
L 500
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
Deflection: 5 Ms 2 f L 48 Br
3.1-1 Continuous beams
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.
f 7.9449904 mm
If the number of reactions exerted upon a beam exceeds the number of equations in static flim 0.012 m
equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of
f flim
the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam. 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. According to figure 3.1.1-1, we determine the reactions and sketch the shear diagrams. Then we compute the values of maximum vertical shear V and maximum positive bending moment M. 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. 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 Figure: 3.1.1-2
top of a beam.
Structural system
Figure 3.1.1-3: equal spans of continuous beam A uniform load is carried over the more than 3 equal spans with different shapes of crosssection as shown in figure 3.1.1-4, figure 3.1.1-5.
Figure 3.1.1-1 A continuous beam carries a uniform load over two equal spans as shown in figure 3.1.1-1. A beam carrying the loads shown in figure 3.1.1-2 is composed of four spans. It is supported
Figure 3.1.1-4: four spans continuous beam
by five vertical reactions. We determine the values of the bending moments over supports as follows.
Figure 3.1.1-5
Figure 3.1.1-2
Structural system
STRUCTURAL ENGINEERING ROOM
A uniform load is carried over three equal spans as shown in figure 3.1.1-3.
Department of Architecture
42
symmetrical T-beams shall not exceed 0.40 of the span length of a simply supported beam or
STRUCTURAL ENGINEERING ROOM
Department of Architecture
43
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.
Figure 3.1.1-8
Figure 3.1.1-6: Floor T-Beam Reinforcing Elevation According to figure 3.1.1-7 a simple beam shown of length L that carries a uniform load of gd (kN/m) throughout its length and is held in equilibrium by reactions Ra and Rb. Assume that the beam is cut at a distance of Sm from the left support and the portion of the beam to the right of Sm be removed. The portion removed must then be replaced by vertical shearing force V together with a couple M to hold the left portion of the bar in equilibrium under the action of Ra and gd Sm.
Figure 3.1.1-9: cross-section of T-beam Without shear reinforcement the beam would have a catastrophic failure due to shearweb 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). figure 3.1.1-8 shows Figure 3.1.1-7: Simple supported rectangular reinforced concrete beam
a sample factored shear diagram for the floor load of reinforced concrete beam.
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
Structural system
connections in perfect fixed. This frame becomes very strong, and must resist the various loads that act on a structure during service load figure 3.1.1-11 and figure 3.1.1-12.
Figure 3.1.1-11: Internal column beam multi-storey frame
Figure 3.1.1-10: The shear diagram of reinforced concrete beam 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.
Figure 3.1.1-12: corner column – beam connection
The column is the most important, as it is the primary load-carrying element of the building. 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.
Figure 3.1.1-13
The design principle of Strong Beam-Column Joints is essential for building structure to resist horizontal load such as wind or earthquakes figure 3.1.1-11. 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
Figure 3.1.1-14 Redistribution procedures for frames
Structural system
STRUCTURAL ENGINEERING ROOM
the two members are firmly connected to each other. In concrete frame structures have moment
Department of Architecture
44
STRUCTURAL ENGINEERING ROOM
Department of Architecture
45 Structural floor systems are, of course, influenced by the material used, but in all cases
consider throughout the design process, but especially during the initial planning stages. In
they are a combination of slabs and joists or secondary beams (floor beams in the case of larger
general, span lengths, floor loads, and geometry of a floor panel all play a key role in the
spacing). The characteristic element, for the whole floor structure, is the floor slab whose
selection process The section fails if the design moment exceeds the resistance moment
thickness and reinforcement is dependent upon the span, the loading and the support conditions.
Figure 3.1.1-16: Continuous reinforced concrete slabs Additional parameters must be considered when selecting an economical floor system. The corners of the slab lift, if they are not loaded by vertical forces of constructions above, resulting in torsional moments.
Figure 3.1.1-15: Diagram of reinforcement in RC frame They are structural elements with a small thickness comparable to their dimensions in the other two directions. Used for floor, roofs and bridge decks. Maybe supported by edge beams or walls, or they may be supported directly by columns, flat slab. When two-way slab
Figure 3.1.1-17
systems are supported directly on columns, shear around the columns is critically important,
The most efficient floor plan is rectangular, not square, in which main beams span the
especially at exterior slab-column connections where the total exterior slab moment must be
shorter distance between columns and closely spaced floor beams span the longer distance
transferred directly to the column.
between main beams. The spacing of the floor beams is controlled by the spanning capability
For a typical continuous RC slab as shown in figure 3.1.1-16, is a flexural member that
of the concrete floor construction.
requires flexural reinforcement in addition to the concrete strength. Concrete, reinforcement,
The structure rests on foundations, which transfer the forces â&#x20AC;&#x201C; from the building and on the
and formwork are the three primary expenses in cast-in-place concrete floor construction to
building â&#x20AC;&#x201C; to the ground.
Structural system
Example 3.1-2: The cross - section dimensions of a beam shown on figure 3.1.2-2 are subjected
connections, which are used in steel structures figure 3.1.1-18 and figure 3.1.1-19.
to bending moment Msd and Nsd Determine the required tension reinforcement to the section. Msd 0.2198 MN m
Nsd 0.120 MN
fck 20 MPa
Characteristic value of concrete cylinder compressive strength (MPa): fckcyl 0.8 fck
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
fckcyl
fcd 9.06667 MPa
1.5
Es 200 GPa
Ecm 29 GPa
e
Es Ecm
e 6.89655
Characteristic yield stress of reinforcement (MPa): fyk 412 MPa
fyd
. Figure 3.1.1-18
fyk 1.15
fyd 358.26 MPa
Cross sectional dimensions: b 0.30 m
h 0.55 m
Cover of reinforcement (m): dst 0.03 m
dsc 0.03 m
Effective depth of a cross-section (m): d h dst
d 0.52 m
Design maximum bending moment ( MN m): Msd 0.2198 MN m
Figure 3.1.1-19
1) To apply the design diagram B3-B3.3 the bending moment Msd has to be brought into a dimensionless form: h Mtot Msd Nsd dst Mtot 190.4 m kN 2
Structural system
STRUCTURAL ENGINEERING ROOM
There are other types of connections, including hinged connections, respectively fixed
Department of Architecture
46
STRUCTURAL ENGINEERING ROOM
Department of Architecture
47
Mtot
Stress in compression (MPa)
0.25888
2
b d fcd
c1
c2
2
The required tension reinforcement A2 ( cm ) is as follows: MN
2
100 cm
c1 10.49229 MPa
Ii
Stress in tension (MPa)
0.083
b d fcd
Mtot
Ai
agi
From the Design Table we obtain the reinforcement ratio:
Ast
Nsd
Nsd Ai
Mtot
c2 10.95 MPa
Ii h a gi
Nsd
Ast 0.00151m
fyd
2
The depth of compression zone: 0.3829
x d
x u 0.8 x
x 0.19911m
x u 0.15929m
2
The required compression reinforcement Asc ( cm ) is as follows: fyd Asc x u b fyd
Asc
Ast fyd
Ast fyd x u b fcd
Asc 0.0003m
fyd
2
Figure: 3.1.2-1
Compression forces acting in the reinforcement resp. in concrete and tension force in tension reinforcement are calculated as follows: Fst Ast fyd
Fc x u b fcd
Fsc 107.32206kN
Fst 540.58106kN
Fc 433.25901kN
Fsc Fc 540.58106kN
Mu Fc d
xu
2
Area of the transformed uncracked cross-section ( m )
Mu1 Fc
Ai 0.17747m
xu 2
Fst d agi
Msd
The ultimate bending moment is found using the following equation:
Fsc Asc fyd
Ai b h e Ast Asc
Fsc agi dsc Fc agi
2
2
xu 2
Fsc d dsc
dsc Ast fyd d dsc
Mu 243.37636m kN
Mu Msd
Mu1 243.37636 m kN
Mu1 Msd
ok
The distance of the extreme fibre from the neutral axis (m) agi
b h 0.5 h e Asc dsc Ast h dst Ai
agi 0.28651m
2
Moment of inertia of the transformed uncracked cross-section ( m ) 3 1
Ii b h
12
2
h a A a d 2 A h d a 2 gi e sc gi sc st st gi 2
bh
Ii 0.00488m
4
Structural system
Figure: 3.1.2-2
1) To apply the design diagram B 3 – B 3-3 the bending moment Msd has to be brought into a
to bending moment Msd. Determine the required tension reinforcement to the section.
dimensionless form:
Characteristic value of concrete cylinder compressive strength (MPa): fck 20 MPa
Msd
0.33094
2
b d fcd
fckcyl 0.8 fck
From the Design table, we obtain the reinforcement ratio:
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
fckcyl 1.5
Es 200 GPa
0.11637 fcd 9.067 MPa Ecm 29 GPa
e
Es Ecm
Characteristic yield stress of reinforcement (MPa): fyk 412 MPa
fyd
fyk
fyd 358.26 MPa
1.15
e 6.89655
The depth of compression zone: 0.52113
x d
x u 0.8 x
x 0.24441m
2
The required tension reinforcement A2 ( cm ) is as follows: Ast
b d fcd MN
100 cm
2
Ast 0.00148m
2
Cross sectional dimensions:
b 0.30 m
2
The required compression reinforcement Asc ( cm ) is as follows:
h 0.5 m
fyd Asc x u b fyd
Cover of reinforcement (m): dst 0.031 m
Ast fyd
Asc
Ast fyd x u b fcd fyd
Asc 0.0001486 cm
2
dsc 0.031 m
Effective depth of a cross-section (m): d h dst
x u 0.19553m
d 0.469 m
Design maximum bending moment ( MN m):
Compression forces acting in the reinforcement resp. in concrete and tension force in tension reinforcement are calculated as follows: Fsc Asc fyd
Fst Ast fyd
Fc x u b fcd
Fsc 0.00532kN
Fst 531.84142kN
Fc 531.83609kN
Fsc Fc 531.84142kN
Msd 0.198 MN m 2
Area of the transformed uncracked cross-section ( m ):
Ai b h e Ast Asc
Ai 0.16024m
2
The distance of the extreme fibre from the neutral axis (m):
Figure: 3.1.3-1
agi
b h 0.5 h e Asc dsc Ast h dst
Structural system
Ai
agi 0.26399m
STRUCTURAL ENGINEERING ROOM
Example 3.1-3: The cross - section dimensions of a beam shown on figure 3.1.3-1 are subjected
Department of Architecture
48
Example 3.1-4: The cross - section dimensions of a beam shown on figure 3.1.4.-1 are subjected to bending moment Msd and compression force Nsd . Determine the required tension
2
STRUCTURAL ENGINEERING ROOM
Department of Architecture
49
Moment of inertia of the transformed uncracked cross-section ( m ): 3 1
Ii b h
12
2
h a A a d 2 A h d a 2 gi e sc gi sc st st gi 2
bh
Ii 0.00358m
4
reinforcement to the section. Characteristic value of concrete cylinder compressive strength (MPa): fck 20 MPa
Stress in compression MPa: c1
Msd
c1 14.58172 MPa
Ii agi Stress in tension MPa: c2
Msd
Ii h agi
fckcyl 0.8 fck
Design value of concrete cylinder compressive strength (MPa):
fcd c1
fcd 0.85
fckcyl 1.5
fcd 9.06667 MPa
Characteristic yield stress of reinforcement (MPa): c2 13.03598 MPa
fctm c2
fyk 412 MPa
fyd
fyk
fyd 358.26 MPa
1.15
Figure: 3.1.3-2
Cross-Section dimensions (m): b 0.45 m xu Fsc agi dsc Fc agi Fst d agi 2
Msd
h 0.45 m
Cover of reinforcement (m): dsc 0.03 m
The ultimate bending moment is found using the following equation:
dst 0.03 m
Effective depth of a cross-section (m): xu Mu Fc d Fsc d dsc 2
ok
xu dsc Ast fyd d dsc 2
Mu1 Fc
Mu 197.43904m kN
Mu Msd
Mu1 197.43904 m kN
Mu1 Msd
d h dst
d 0.42 m
Design maximum bending moment ( MN m): Msd 0.05 MN m
Nsd 2.8 MN compression force
L 4 m
To apply the design table the bending moment Msd has to be brought into a dimensionless form:
h d st 2
Mext Msd Nsd
Mext 2
b d fcd
Figure: 3.1.3-3
Structural system
0.82811
Mext 596m kN
entire section is subjected in compression
w
Next 0.8 b d fcd
Ast bd
Ast 0.00399m
fyd
2
w 2.11061
100
Figure: 3.1.4-1 Eccentricity due to action effects: e
Msd
h
e 0.01786m
Nsd
6
h
0.075 m
6
e
Centric compression force or compression force with small eccentricity: The minimum amount or longitudinal reinforcement is given by: Nsd f yk 1.15 Stress of concrete (MPa):
Nsd 2.8 MN
Asmin 0.15
c
Asmin 0.00117m
2
Figure: 3.1.4-2 Example 3.1-5: The cross - section dimensions of a beam shown on figure 3.1.5-1 are subjected
Nsd
to bending moment Msd a Normal force Nsd (compression). Determine the required tension
c 18.40894 MPa
h 2dst b 2 dsc
reinforcement to the section. Characteristic value of concrete cylinder compressive strength (MPa):
Slenderness ratio:
fck 20 MPa
0.7 L
21.60494
0.288 b
Design value of concrete cylinder compressive strength (MPa):
From the design diagram B3-B3.3 we obtain the following dimensionless parameters: 2.5
2
2
Ast h 2dst b 2 dsc 10
Ast 0.0038m
Ns Ast fyd
Next 2800 kN
2
The required tension reinforcement A2 ( cm ) is as follows: Ns
Next Nc
fckcyl 0.8 fck
Ast fyd
Next 0.8 b d fcd
fcd 0.85
fckcyl
Nc 0.8 b d fcd
1.5
fcd 9.06667 MPa
Characteristic yield stress of reinforcement (MPa): fyk 410 MPa
fyd
fyk 1.15
fyd 356.52 MPa
Cross-section dimensions: b 0.40 m
Structural system
h 0.45 m
L 4 m
STRUCTURAL ENGINEERING ROOM
Ast
Department of Architecture
50
Cover of reinforcement (m):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
51
dsc 0.03 m
Eccentricity due to action effects:
dst 0.03 m
e
e 0.1 m
Nsd
h 6
h
0.075 m
6
e
Effective depth of a cross-section (m): d h dst
Msd
Centric compression force or compression force with small eccentricity:
d 0.42 m
Design maximum bending moment ( MN m):
The minimum amount or longitudinal reinforcement is given by:
Msd 0.150 MN m
Nsd 1.5 MN
compression
To apply the design table 3.1-1 the bending moment Msd has to be brought into a dimensionless form: h d st 2
Mext Msd Nsd
Mext 2
Nsd f yk 1.15
Asmin 0.15
Asmin 0.00063m
2
Stress of concrete (MPa): Mext 442.5 m kN
0.69168
Nsd
c
entire section is subjected to compression
b d fcd
h 2dst b 2 dsc
c 11.31222 MPa
Slenderness ratio:
0.7 L
24.30556
0.288 b
From the design diagram B3-B3.3 we obtain the following dimensionless parameters:
0.1 2
2
A ( h b) 10
A 0.00018m
Ns Nsd 0.8 b d fcd
Ns 281.44kN
Nc 0.8 b d fcd
2
The required tension reinforcement A2 ( cm ) is as follows: Ns
Nsd Nc
Ast
Figure: 3.1.5-1
w
Ast fyd
Nsd 0.8 b d fcd
Ast bd
fyd 100
Structural system
Nsd 0.8 b d fcd
Ast 0.00079m
w 0.46988
2
Nc 1218.56kN
to bending moment Msd and normal force Nsd (tension) Determine the required tension reinforcement to the section. Characteristic value of concrete cylinder compressive strength (MPa): fck 20 MPa
fckcyl 0.8 fck
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
fckcyl
fcd 9.06667 MPa
1.5
Characteristic yield stress of reinforcement (MPa): fyk 410 MPa
fyd
fyk
fyd 356.52 MPa
1.15
Cross sectional dimensions (m): b 0.20 m
h 0.35 m
dsc 0.03 m
dst 0.03 m
Figure: 3.1.6-1 Forces in kN in tension, compression reinforcement and concrete in compression zone (kN):
Cover of reinforcement (m):
Effective depth of a cross-section (m): d h dst
Nsc Asc fyd
Nst Ast fyd
Nc d 0.8 b fcd
Nst 490.36924kN
Nc 108.85803kN
Nsc 381.51122kN
w 3.4936
w wmax
Nsc Nc 490.36924kN
d 0.32 m
Design maximum bending moment ( MN m): w
Msd 0.087 MN m
Tension force
Nsd 0.380 MN
if
To apply the design diagram B3-B3.3 the bending moment Msd has to be brought into a dimensionless form: h dst 2
Mext Msd Nsd
Mext 31.9 kN m
0.05335
Mext 2
Ast Asc
0.1718
fyd
100
fyk 190 MPa
Mext
0.1718
2
0.17728
b d fcd
fyk
fyd 165.22 MPa
1.15
2
The required tension reinforcement Ast ( cm ) is as follows:
b d fcd Ast
0.2345
bh
b d fcd MN
2
100 cm
Nsd fyd
Ast 0.00333m
2
2
The required tension reinforcement Ast ( cm ) is as follows: Ast
b d fcd MN
2
100 cm
Nsd fyd
2
Ast 0.00138m
2
The required compression reinforcement Asc ( cm ) is as follows:
2
The required compression reinforcement Asc ( cm ) is as follows: Asc
Ast fyd 0.8 d b fcd fyd
Asc Asc 0.00107m
2
Ast fyd 0.8 d b fcd fyd
w wmax
Structural system
Asc 0.00267m
2
w
Ast Asc bh
100
w 8.56931
STRUCTURAL ENGINEERING ROOM
Example 3.1-6: The cross - section dimensions of a beam shown on figure 3.1.6-1 are subjected
Department of Architecture
52
Example 3.1-7: The cross - section dimensions of a beam shown on figure 3.1.7-1 are
STRUCTURAL ENGINEERING ROOM
Department of Architecture
53
subjected to bending moment Msd . Determine the required tension reinforcement to the section. Characteristic value of concrete cylinder compressive strength (MPa):
Design maximum bending moment ( MN m): Msd 0.192 MN m
Nsd 0.125 MN
compression
1) To apply the design diagram B3-B3.3 the bending moment Msd has to be brought into a dimensionless form:
h d st 2
Mext Msd Nsd Mext
0.10568
2
e
Mext 226.375m kN
Es Ec
0.03103
b d fcd
Figure: 3.1.7-1 fckcyl 35 MPa
Ec 35 GPa
Es 200 GPa
fck 42 MPa
2
The required tension reinforcement A2 ( cm ) is as follows:
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
fckcyl 1.5
Ast
b d fcd MN
fcd 19.833 MPa Ast 4
( 1.6 cm )
fyd
fyk 1.15
fyd 358.26 MPa
b 0.30 m
h 0.65 m
Cover of reinforcement (m): dsc 0.05 m
2
we provide
Ast 0.000804m
2
Msd
e 1.536 m
Nsd
h 6
h
0.10833m
6
e
Centric compression force or compression force with small eccentricity: 2
The required compression reinforcement Asc ( cm ) is as follows:
dst 0.05 m
Effective depth of a cross-section (m): d h dst
L 4 m
Ast 0.00076m
fyd
Eccentricity due to action effects (m): e
Cross-section dimensions (m):
Nsd
2
4
Characteristic yield stress of reinforcement (MPa): fyk 412 MPa
2
100 cm
d 0.6 m
Asc
Ast fyd 0.8e d b fcd
Asc 2
fyd ( 1.2 cm )
Structural system
4
Asc 0.04475 m
2
2
Asc 0.000226m
2
we provide
e 5.71429
h2
Ii b h
3
c
Nsd
c 1.04167 MPa
b 2 dsc h dst
c 0.45 fck
b 2 dsc h dst
2
10
2 Asc xi dsc2
Ii 0.00731m
4
2 3
MPa 10 MPa
fctm 1.4
0.45 fck 18.9 MPa
Ast
or
c fcd
The reinforcement ratio ρ is given by:
x i x i h e Ast d x i
Ms
0.67021
fckcyl
Msd
Mcr fctm
fctm 3.22731 MPa
Ii h xi
Mcr 73.58168m kN
Ms 171.42857m kN
1.12
b) Determine the bending stiffness 2
Br 0.85 Ec Ii
Br 217377.77269m kN
Slenderness ratio:
0.7 L
- element has no cracks and have bending stiffness:
32.40741
0.288 b
From the table VI.4
2
Bra 0.85 Ec Ii
Bra 217377.77269m kN
- the element of expected cracks, first determine stiffness of the beam without crack:
c 20.61 MPa
further we determine the stiffness of the beam with total eliminate of the tension in concrete, Design ultimate capacity of a cross-section may be determined:
where the depth of concrete zone x r of the cross-section determined from the conditions of the balance of forces in cross section, which after adjustment we can be written in the form:
Nud c b 2 dsc h dst
Nud 2473.2kN
Nsd 125kN
Nud Nsd
x r
The minimum amount of longitudinal reinforcement is given by:
Asmin 0.15
Nsd
Asmin 0.00005m
fyk
2
e
1.15
x i
0.5 b h e Ast d Asc dst b h e Ast Asc
Ec
e 5.71429
b
1
1
2 b Ast d Asc dsc
e Ast Asc
2
bending stiffness of the beam with total eliminate of the tension in concrete will: compression area of the cross section
x i 0.32952m
x r 0.11893m
where b xr 2 the static moment of area A c , based to the upper compression edge, the first, we determine:
We decide whether the expected cracking 2
Es
e Ast Asc
Ac b x r 2 e Asc
x r dsc
Structural system
xr
Ac 0.03718m
2
STRUCTURAL ENGINEERING ROOM
Stress of concrete (MPa):
Department of Architecture
54
Example 3.1-8: The cross - section dimensions of a beam shown figure 3.1.8-1 are subjected
Arm of internal forces
STRUCTURAL ENGINEERING ROOM
Department of Architecture
55
2 4 e Asc dsc
b xr z r d
Brb
to bending moment
x r dsc xr
2 Ac
The material properties are assumed as follows:
d zr
2
Brb 41855.73425m kN
1 2 E A s st Ec Ac
1 4
Br
Mcr
5
Ms
1
Characteristic value of concrete cylinder compressive strength (MPa): fck 31.25 MPa
r 0.28653
1
Brb Bra
c1
Mext Ii
fcd 0.85
r 0
Br 54454.38842m kN
c1 10.20903 MPa
f
5 Ms 2 L 48 Br
Mext Ii
f 5.24686 mm
fckcyl 1.5
fctm 3.22731 MPa
0.45 fck 18.9 MPa
fcd 14.16667 MPa
Characteristic yield stress of reinforcement (MPa): fyk 325 MPa
fyd
fyk 1.15
fyd 282.61 MPa
Cross-sectional dimensions:
xi
c2
fckcyl 0.8 fck
Design value of concrete cylinder compressive strength (MPa):
2
1 r
r
Determine the required tension reinforcement to the
section.
z r 0.54092m
The resulting bending stiffness: r
Msd.
c2 9.92885 MPa
fctm 3.22731 MPa
h xi
bd 0.4 m
hd 0.4 m
hs 0.8 m
Ac bd hd bs hs bh hh
bh 0.3 m
hh 0.3 m Ac 0.45 m
H 1.5 m
Cover of reinforcement (m):
dsc 0.03 m
dst 0.03 m
Effective depth of a cross-section (m): d H dst
d 1.47 m
Msd 410 kN m
Figure: 3.1.7-2
Structural system
Ms 365 kN m
Msd Ms
1.12329
bs 0.25 m 2
H hd hs hh
agi
Ac agc e Ast d Asc dsc
agi 0.80862m
Ai
Modulus of elasticity of cross-section (without any reinforcement): Ic
1
bs H bh bs hh bd bs hd 3
12
Figure: 3.1.8-1 Tension reinforcement:
st 25 mm nst
2
4
Ast1 0.00196m
sc 16 mm
2
Ast2 sc
2
3
2 bs H 0.5 H agc
hh
Ii Ic Ac agi agc
nsc 4 nsc 4
Ast2 0.0008m
Ii 0.11245m
2
Ast Ast1 Ast2
Ast 0.00277m
2
2
Asc 0.00188496m
4
Es 200 GPa
Ecm 30.5 GPa
2
Atot Asc Ast
e
Atot 0.00465m
4
2 e Ast agi dsc2 Ast H dst agi 2
Determination of μ from diagram B3‐B3.3
Compression reinforcement: Asc 6 ( 20mm)
Modulus of inertia of transformed cross-section (with reinforcement):
nst 4
Ast1 st
4
2 2 hd bd bs hd H agc 2 2
bh bs hh agc Ic 0.0935m
3
Msd b Ab fcd
Ast
Ab fcd
Atot Ac fcd
MN 100 2 m
0.07298
2
Design ultimate capacity of a cross-section may be determined:
Es
e 6.55738
Ecm
Mu Ac d fcd
Mu 1733.68125m kN
Mu Msd
Stress in concrete - compression zone;
Area of concrete net cross-section: Ac bs H
bh bs hh bd bs hd
Ac 0.45 m
2
c1
Ai 0.48051m
2
c2
Distance of the neutral axis (centroid) of the net section: 0.5 bs H bh bs hh 2
agc
hd bd bs hd H 2
2
Ac
Msd
c1 2.9483 MPa
Ii agi Stress in concrete - tension zone:
Area of concrete transformed cross-section: Ai Ac e Ast Asc
0.185
agc 0.80333m
M sd
c2 2.52083 MPa
Ii H a gi
0.8fck 10 MPa
c2 fctm
0.666
fctm 1.4
fcd c1
Structural system
MPa
fctm 2.57725 MPa
Msd 410m kN
STRUCTURAL ENGINEERING ROOM
Distance of the neutral axis (centroid) of the transformed section:
Department of Architecture
56
STRUCTURAL ENGINEERING ROOM
Department of Architecture
57 Example 3.1-9: The cross - section dimensions of a beam shown on figure 3.1-9 are subjected to bending moment Msd . Determine the required tension reinforcement to the section. Characteristic value of concrete cylinder compressive strength (MPa): fck 45 MPa
fckcyl 0.8 fck
fckcyl 36 MPa
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
fckcyl 1.5
fcd 20.4 MPa
fctm 3.2 MPa
Characteristic yield stress of reinforcement (MPa): fyk 325 MPa
fyd
fyk 1.15
fyd 282.61 MPa
Cross-section dimensions (m):
Figure: 3.1.8-2
b 0.4 m
h 1.0 m
Ac b h
Ac 0.4 m
ap 0.15 m
e 15
Cover of reinforcement (m):
dsc 0.05 m
dst 0.05 m
Effective depth of a cross-section (m):
d h dst
d 0.95 m
H ap 1.35 m
Figure: 3.1.8-3
Structural system
Figure: 3.1.9-1
2
Np 2.8 MN
Stress in concrete - tension zone:
compression
2
Ast 4 ( 1.6cm )
Ast 0.000804m
4
Np
c1
c1 fctm
Tension reinforcement:
Ai
or
2
Np
c1
c1 fctm
Ai
agi
Ii
1
Asc 0.001257m
4
c1 3.25598 MPa
Ii
Stress in concrete - compression zone:
2
c2
Prestress tendon: 2 2 Ap 1 ( 0.55cm ) 6 ( 0.5cm ) 4.5 4 4 4
Np Ai
h agi
Ii
1
Ai eext
c2 14.41501 MPa
Ai b h e Asc Ast Ap
c2 fctm
or Ap 0.002548m
2
c2
c2 fctm
Np Ai
Msd Np h ap agi
c2 14.41501 MPa
Ii h agi
Area of concrete transformed cross-section:
c1 3.25598 MPa
agi
2
LA 15.5 (1 fi 5.5+6 fi 5)
Msd Np h ap agi
Compression reinforcement: Asc 4 ( 2cm )
Ai eext
Ai 0.46914m
2
Distance of the neutral axis (centroid) of the transformed cross - section:
agi
b h 0.5 h e Asc dsc Ast h dst Ap h ap Ai
agi 0.52201m
Modulus of inertia of transformed cross-section (with reinforcement):
Method of verification see B3-B3.3: For
2
h 2 2 2 Ii b h b h agi e Asc agi dsc Ast h dst agi Ap agi h ap 12 2 3 1
Ii 0.04405m
4
Figure: 3.1.9-2
fyk 325 MPa
Ast Ap b d fcd
MN 100 2 m
0.04325
0.115
Np 2800 kN
dp
Mext Msd Np h ap agi
h ap agi
2
Mu b d fcd
dp 0.32799m
Mext Np
eext 0.27799m
Mext 778.37885 m kN
For
Mext 778.37885 m kN
eext
Mu 846.906m kN
fyk 520 MPa
0.18
Mext 778.37885 m kN
Mext Mu
Structural system
2
Mu b d fcd
Mu 1325.592m kN
STRUCTURAL ENGINEERING ROOM
Msd 140 kN m
Department of Architecture
58
STRUCTURAL ENGINEERING ROOM
Department of Architecture
59 Example 3.1-10: The cross - section dimensions of a beam are subjected to bending moment
To apply the design diagram B3-B3.3 the bending moment Msd has to be brought into a
Msd.
dimensionless form:
Determine the required tension reinforcement to the section.
Msd
Characteristic value of concrete cylinder compressive strength (MPa): fck 30 MPa
From the Design Table, for fyk= 325 MPa we obtain the reinforcement ratio:
fckcyl 0.8 fck
0.09382 2
The required tension reinforcement A2 ( cm ) is as follows:
Design value of concrete cylinder compressive strength (MPa): fcd 0.85
0.22812
2
b d fcd
Material data:
fckcyl
Ast
fcd 13.6 MPa
1.5
b d fcd MN
Ast 0.0019m
100 cm
2
Ast 0.00182m
2
we provide
Ast 5
( 2.2 cm ) 4
2
2
2 3 fckcyl MPa 10 MPa
fctm 1.4
The depth of compression zone (m):
fctm 2.51 MPa
0.3314
Characteristic yield stress of reinforcement (MPa): fyk 325 MPa e
Es Ecm
fyd
fyk
fyd 282.61 MPa
1.15
h 0.50 m
dst 0.025 m
Effective depth of a cross-section (m): d h dst
d 0.475 m
Design maximum bending moment ( MN m): Msd 0.210 MN m
Es 200 GPa
Ecm 35 GPa
we provide ( 1.2 cm )
Ast fyd
Asc
Ast fyd x u b fcd
2
4
Asc 0.00023m
Asc 0.00008m
fyd 2
w
Ast Asc bh
2
w 0.01418
Compression forces acting in the reinforcement resp. in concrete and tension force in tension reinforcement are calculated as follows:
Cover of reinforcement (m): dsc 0.025 m
x u 0.12593m
2
Asc 2
Cross-section ( m):
x u 0.8 x
x 0.15741m
The required compression reinforcement Asc ( cm ) is as follows: fyd Asc x u b fyd
e 5.71429
b 0.30 m
x d
Fsc Asc fyd
Fst Ast fyd
Fc x u b fcd
Fsc 63.92458kN
Fst 537.14405kN
Fc 513.80256kN
Fsc Fc 577.7271kN
2
Area of the transformed uncracked cross-section ( m )
Ai b h e Ast Asc
Structural system
Ai 0.16215m
2
b h 0.5 h e Asc dsc Ast h dst
agi
Department of Architecture
A 3 Design of reinforced concrete members (c = 0,0035, s = 0,01)
The distance of the extreme fibre from the neutral axis (m):
agi 0.26328m
Ai
2
Moment of inertia of the transformed uncracked cross-section ( m ):
2
h a A a d 2 A h d a 2 gi e sc gi sc st st gi 2
3 1
Ii b h
bh
12
Ii 0.00371m
4
Stress in compression (MPa):
c1
Msd
c1 14.89569 MPa
Ii agi
fcd c1
1. Bending moment and tension axial force h M sds M sd N sd d 2 2
Stress in tension (MPa): c2
Msd
c2 13.39334 MPa
Ii h agi
Fsc agi dsc Fc agi
fctm c2
xu 2
Fst d agi
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
Msd
s
Ast d
x d
1
x
s 115.29396 MPa
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
where coefficient obtained from the graph according to coefficient obtained from M.sd ...... in MNm h, d2,d....... in m M sds Nsd ........ in MN Ac ............. in m2 A c d f cd fyd .........in MPa fcd ..............in MPa
Stress in tension reinforcement (MPa): Msd
2. Bending moment and compression axial force h M sds M sd N sd d 2 2
s fyd
Determination of the concrete cross-section area Ac :
d
The ultimate bending moment is found using the following equation:
Mu Fc d
xu
Fsc d dsc 2
2
Mu 240.47019m kN
Mu Msd
ok
Ac = b . d
Ac
D
4
Ac
D 1 D 2 4
3. Pure bending In case of pure bending can be used previous relations, where Nsd is equal zero.
Structural system
STRUCTURAL ENGINEERING ROOM
60
STRUCTURAL ENGINEERING ROOM
Department of Architecture
61 A 3.1 Design of reinforced concrete members (c = 0,0035, s = 0,01)
1. Bending moment and tension axial force h M sds M sd N sd d 2 2
A 3.2 Design of reinforced concrete members (c = 0,0035, s = 0,01)
2. Bending moment and compression axial force h M sds M sd N sd d 2 2
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
1. Bending moment and tension axial force h M sds M sd N sd d 2 2
2. Bending moment and compression axial force h M sds M sd N sd d 2 2
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
The required reinforcement [cm2]: N sd A2 A c f cd 100 104 f yd req
where coefficient obtained from the graph according to coefficient obtained from M.sd ...... in MNm h, d2,d....... in m M sds Nsd ........ in MN Ac ............. in m2 A c d f cd fyd .........in MPa fcd ..............in MPa
where coefficient obtained from the graph according to coefficient obtained from M.sd ...... in MNm h, d2,d....... in m M sds Nsd ........ in MN Ac ............. in m2 A c d f cd fyd .........in MPa fcd ..............in MPa
Determination of the concrete cross-section area Ac :
Determination of the concrete cross-section area Ac :
2
Ac = b . d
Ac
D
4
Ac
2
D 1 D 2 4
3. Pure bending In case of pure bending can be used previous relations, where Nsd is equal zero.
Ac = b . d
Ac
D
4
Ac
D 1 D 2 4
3. Pure bending In case of pure bending can be used previous relations, where Nsd is equal zero.
Structural system
1. Bending moment and tension axial force h M sds M sd N sd d 2 2
2. Bending moment and compression axial force h M sds M sd N sd d 2 2
The required reinforcement [cm2]: A2 req
1
f yd
4
Department of Architecture
A 3.3 Design of reinforced concrete members (c = 0,0035, s = 0,01)
4
A c f cd 10 N sd 10
The required reinforcement [cm2]: A2 req
1
f yd
A c f cd 104 N sd 104
where coefficient obtained from the graph according to coefficient obtained from M.sd ...... in MNm h, d2,d....... in m M sds N Ac ............. in m2 sd ........ in MN A c d f cd fyd .........in MPa fcd ..............in MPa Determination of the concrete cross-section area Ac :
2
Ac = b. d
Ac
D
4
Ac
D 1 D 2 4
3. Pure bending In case of pure bending can be used previous relations, where Nsd is equal zero.
Structural system
STRUCTURAL ENGINEERING ROOM
62
STRUCTURAL ENGINEERING ROOM
Department of Architecture
63
Design of reinforced concrete members
0,50
fyk
0,48
520 MPa
0,45
490 MPa
0,43
450 MPa
0,40
412 MPa
0,38
410 MPa 392 MPa
0,35
375 MPa
0,33
325 MPa
0,30
300 MPa
0,28
245 MPa
0,25
235 MPa
0,23 Stress-strain curve of steel
0,20
206 MPa
Stress-strain curve of concrete
190 MPa
0,18 0,15 0,13 0,10 0,08 0,05 0,03 0,00 0
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
0,45
0,48
Figure B 3: Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel and bi-linear diagram for concrete
Structural system
Design of reinforced concrete members
fyk
0,50 0,48
190 MPa
0,45
206 MPa
0,43
235 MPa
0,40
245 MPa 300 MPa
0,38
325 MPa
0,35
375 MPa
0,33
392 MPa
0,30
410 MPa
0,28
412 MPa
0,25
450 MPa 490 MPa
0,23
Stress-strain curve of steel
0,20
520 MPa
Stress-strain curve of concrete
0,18 0,15 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
0,33
0,36
0,39
0,42
0,45
0,48
Figure B 3.1: Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel and bi-linear diagram for concrete
Structural system
STRUCTURAL ENGINEERING ROOM
Department of Architecture
64
STRUCTURAL ENGINEERING ROOM
Department of Architecture
65
m
Design of reinforced concrete members
fyk
0,50
190 MPa 206 MPa
0,45
235 MPa 245 MPa 300 MPa
0,40
325 MPa 375 MPa
0,35
392 MPa 410 MPa 412 MPa
0,30
441 MPa 450 MPa
0,25
490 MPa Stress-strain curve of steel
0,20
520 MPa
Stress-strain curve of concrete
0,15
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
0,33
0,35
0,38
0,40
0,43
0,45
0,48
r
Figure B 3.2: Sections without compression reinforcement for pure bending or bending moment with axial force, based on the parabolic-rectangular diagram for steel and a bi-linear diagram for concrete
Structural system
h
2
d 2
2
b d f cd
M sd N sd
h
2
d 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
h
0,30 0,28
req d2
0,25
d
0,33
0,23 0,20 0,18 0,15
N sd
b d f cd 100
Bending w ith compression
A req
Pure bending
N sd 4 b d f cd 100 10 f yk
A req
b d f cd 100
f yk
0,10 0,08 0,05 0,03
4
A req
Bending w ith tension
0,13
10
h, d, b, d2 fcd, fy d Nsd
in MN
Msd
in MNm
Areq
in cm2
0,00 0,00
0,04
0,08
0,12
0,16
0,20
0,24
0,28
0,32
0,36
0,40
0,44
0,48
0,52
0,56
0,60
0,64
in m in MPa
0,68
0,72
r
Figure B 3.3: Sections without compression reinforcement for pure bending or bending moment with axial force, based on the bi-linear diagram for steel and parabolicrectangular diagram for concre
Structural system
STRUCTURAL ENGINEERING ROOM
M sd N sd
Department of Architecture
66
STRUCTURAL ENGINEERING ROOM
Department of Architecture
67
C 3: Preliminary design cross-sectional thickness of reinforced concrete slabs structures uniform loading (dimensions rounded to 10 mm) One way reinforced slab
C 3.1: Preliminary design cross-sectional dimension of reinforced concrete beams and frames structures uniform loading (dimensions rounded to 50 mm)
- freely supported hs
T - beam
1 L 25
- freely supported ht
bt
1 L t 20
1 2 h 2 3 t
- perfectly cantilevered 1 1 h s L 30 35
- perfectly cantilevered ht
- cantilevered landings hs
Two way reinforced slab
1 L t 25
1 L 10
- freely supported hs
1 L x L y 75
- cantilevered landings ht
1 L t 5
hp
1 1 L 10 15 p
- perfectly cantilevered hs
1 L x L y 105
Beams bp
1 2 h 2 3 p
Structural system
Slabs are divided into suspended slabs. Suspended slabs may be divided into two groups:
D-3 Positioning the neutral axis of reinforced concrete The moment of inertia of ideal cross-section: hs
d1
bc
v1 h d
M sd
2 A 2 Â&#x2DC; d v 1 2º¼
bw
¬ 12
§ ©
¨v 2
2º ¸ » 2 ¹ ¼
hs·
2 º ¸» ¹¼
ht· 2
M sd
Nsd
(2)
slabs supported directly on columns without beams and known as flat slabs. Supported
directions). In one-way slabs the main reinforcement is provided along the shorter span. In order to distribute the load, a distribution steel is necessary and it is placed on the longer side. Oneway slabs generally consist of a series of shallow beams of unit width and depth equal to the slab thickness, placed side by side. Such simple slabs can be supported on brick walls and can
3
slabs supported on edges of beams and walls
direction only) and two-way slabs (slabs supported on four sides and reinforced in two
2 ª h s2 § hs· º I gg´ Â&#x2DC; v 1 v 2 b c b w Â&#x2DC; h s Â&#x2DC; « ¨v ¸ » 3 ¬ 12 © 1 2 ¹ ¼ 2 2 n Â&#x2DC; ª A 1 Â&#x2DC; v 1 d 1 A 2 Â&#x2DC; d v 1 º ¬ ¼
bw
(1)
slabs may be one-way slabs (slabs supported on two sides and with main reinforcement in one
The moment of inertia of ideal cross-section:
A1
A2
ª h t2
§ ©
¨v 1
2 º» ª« b w Â&#x2DC; h 2 hs ht· § v1 Â&#x2DC; b t b w Â&#x2DC; h t Â&#x2DC; ¨ h ¸ b c b w Â&#x2DC; » 2 2 A « 2 ¹ © « n Â&#x2DC; A 1 Â&#x2DC; d 1 A 2 Â&#x2DC; d
» ¬ ¼ v2 = h â&#x20AC;&#x201C; v1 A = bw. h + (bc - bw). hs + (bt - bw). ht + n. (A1+ A2) area:
d1 v1
2
1
bc
v2
ªh
b c b w Â&#x2DC; h s Â&#x2DC; «¬ 12s
3
bt bw Â&#x2DC;htÂ&#x2DC;«
Nsd ht
v2 d2
3
¬
bw
d2
Â&#x2DC; v1 v2
n Â&#x2DC; ªA 1 Â&#x2DC; v 1 d 1
bt
hs
3
A1
A2
h d
bw
I gg´
3
be supported on reinforced concrete beams in which case laced bars are used to connect slabs to beams.
where
ª b w Â&#x2DC; h2
1
v1
Â&#x2DC;«
b c b w Â&#x2DC; h s 2
nÂ&#x2DC; A Â&#x2DC;d
A 2Â&#x2DC; d
1 1 2 ¬ 2 v2 = h â&#x20AC;&#x201C; v1 A = bw. h + (bc - bw). hs + n. (A1+ A2) area: A
º
»¼
If M sd N sd
I gg´
M sd
AÂ&#x2DC;v2
N sd
I gg´ AÂ&#x2DC;v2
Mt = bc. h.s2. (d â&#x20AC;&#x201C; hs /3). 0,8. fyk / 30 .( d â&#x20AC;&#x201C; hs )
The entire section is in compression
If Mt < Msd
A1
Mt > Msd A1
A1
A2
M sd
Nsd
A1
A2
M sd
Nsd A2
A1
A2
M sd
A2
M sd
Nsd
A1
Nsd
M1 = (bc-bw).hs.fcd.(d-0,5.hs) Md = Msd â&#x20AC;&#x201C; M1 As = U . bw . d . fcd . 100 b c b w Â&#x2DC; h s Â&#x2DC; f cd A 2req As f yd
!
P
M sd
Nsd A2
Md 2
b w Â&#x2DC; d Â&#x2DC; f cd
P
M sd
Nsd
Figure 3.2-1: One â&#x20AC;&#x201C;way slab, two-way slab, ribbed slab, flat slab, solid flat slab with drop panel, waffle slab
M sd 2
b c Â&#x2DC; d Â&#x2DC; f cd
according to the graph of Figure B-3 until B-3-3 we obtaine U A2req = U . bc . d . fcd .100
In R.C. Building construction, every floor generally has a beam/slab arrangement and consists of fixed or continuous one-way slabs supported by main and secondary beams.
Structural system
STRUCTURAL ENGINEERING ROOM
3.2 Reinforced Concrete Slabs
Department of Architecture
68
3.2.1 Flat Slabs
STRUCTURAL ENGINEERING ROOM
Department of Architecture
69
Flat plate is defined as a two-way slab of uniform thickness supported by any combination of columns, without any beams, drop panels, and column capitals. Flat plates are most economical for spans from 4,5 to 7,5m, for relatively light loads, as experienced in apartments or similar building. -A flat slab is a reinforced concrete slab supported directly on and built monolithically with the columns, the flat slab is divided into middle strips and column strips. The size of each strip is Figure 3.2-2: Solid flat slab, solid flat slab with drop panels The usual arrangement of a slab and beam floor consists of slabs supported on crossbeams or secondary beams parallel to the longer side and with main reinforcement parallel to the shorter side. The secondary beams in turn are supported on main beams or girders extending from column to column. Part of the reinforcement in the continuous is bent up over the support, or straight bars with bond lengths are placed over the support to give negative bending moments.
defined using specific rules. The slab may be in uniform thickness supported on simple columns. These flat slabs may be designed as continuous frames. However, they are normally designed using an empirical method governed by specified coefficients for bending moments and other requirements which include the following: 1. There should be not less than three rectangular bays in both longitudinal and transverse directions. 2. The length of the adjacent bays should not vary by more than 10 %.
Figure 3.2.1-1: Post punching behaviour of slab- critical section The general layout of the reinforcement is based on the both bending moments (in spans) and bending moments in addition to direct loads (on columns).
Figure 3.2-3: Types of the reinforced concrete slab systems Figure 3.2.1-2: Combined punching shear and transfer of moments
Structural system
Figure 3.2.1-4
Figure 3.2.1-3
Structural system STRUCTURAL ENGINEERING ROOM
Department of Architecture
70
3.2.1-1 Analysis and Design of Flat Plate
STRUCTURAL ENGINEERING ROOM
Department of Architecture
71 Uniform. The procedure generally adopted is to divide the slab into column strips (along the
To obtain the load effects on the elements of the floor system and its supporting members using an elastic analysis, the structure may be considered as a series of equivalent
column lines) and middle strips and then apportion the moment between these strips and the distribution of the moment within the width of each strip being assumed uniform.
plane frames, each consisting of vertical members – columns, horizontal members - slab. Such plane frames must be taken both longitudinally (in x-direction) and transversely (in y direction) in the building, to assure load transfer in both directions. For gravity load effects, these equivalent plane frames can be further simplified into continuous beams or partial frames consisting of each floor may be analysed separately together with the columns immediately above and below, the columns being assumed fixed at their far ends. Such a procedure is described in the “Equivalent Frame Method”. When frame geometry and loadings meet certain limitations, the positive and negative factored moments at critical sections of the slab may be calculated using moment coefficients, termed “Direct Design Method”. These two methods differ essentially in the manner of determining the longitudinal distribution of bending moments in the horizontal member between the negative and positive moment sections. However, the procedure for the lateral distribution of the moments is the same for both design methods.
Figure 3.2.1.1-2: Moments and frames
Figure 3.2.1.1-1: Steel shear –heads, steel plats joined by welding
Since the outer portions of horizontal members (slab) are less stiff than the part along the support lines, the lateral distribution of the moment along the width of the member is not
Structural system
Figure: 3.2.1.1-3
Live load (apartments):
Geometric Shapes
vd 2.0
hd 300mm The geometry of the building floor plans:
1.5
qd qdo q1d vd
l2 3.6m
ly 7.7m
kv 2.850m Dimensions columns:
vd 3
kN m
2
qd 17.325
kN 2
m Force load Peripheral masonry thickness of 400 mm YTONG: F1 10
kN
kv ly 400mm1.35 3 m Total load acting on the console:
bs 400mm hs bs The peripheral dimensions of the beam: ho 0.5m
2
m Total load on 1 m 2 of slab:
Slab thickness
l1 7.7m lk 2.3m Construction height of object:
kN
F1 118.503kN
F1d 118.503kN F1d F1 Investigation replacement frame in the X-axis Frame 1:
bo 0.30m
Calculation model
Figure: 3.2.1-1 Load calculation Load per area Reinforced concrete slab thickness of 300 mm qdo hd 25
floor layer: q1d 3
kN m
2
kN m
3
1.4
qdo 10.125
1.35
q 1d
4.2
kN m
kN m
2
Figure: 3.2.1-2 load calculation Load width in a direction perpendicular to the x:
zsx ly 2
Structural system
STRUCTURAL ENGINEERING ROOM
Example: 3.2-1 Design and calculation of Flat Plate
Department of Architecture
72
Load in the x-direction:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
73
qdx qd zsx
qdx 133.403
Calculation of internal forces
kN m
Moment on a console:
lk Mk F1d lk qdx 2
Mk 625.407kN m
2
Moment of inertia:
Figure: 3.2.1-3
Transverse replacement frame:
Ip
ly hd
3
Ip 0.017m
12
9
Central girders replacement of frame:
1 12
3
Is 5.208 10 3 m
Primary moments in node 10:
4
M109o
Bending stiffness:
Ip l1
1000
kN
Kp 2.25kN
2
rad m Central girders replacement frame: Kst
Ist
l2
1000
Column
Ks
Is kv
1000
1 12
M1010ò
Transverse replacement frame:
Kp
M910o
M97o 0kN m bs hs
kN rad m kN
rad m
2
2
M108 1 M911 1
M109 1 M97 1
M1012 1
Primary moments in node 9:
Ist Ip column: Is
M910 1 M10'10 1
1 10 1 M1010' 1
4
m rad
Kst 4.813kN
Ks 1.827kN
m rad
qdxl1 1 12
2
qdxl2
1 12
qdxl1
2
M108o 0kN m 2
M911o 0kN m
M1012o 0kN m
M10'10o M1010ò
Given M97kN m M97o Ks 3 9rad M911kN m M911o Ks 2 9rad M910kN m M910o Kp 2 9rad 10rad M108kN m M108o Ks 3 10rad M109kN m M109o Kp 2 10rad 9rad M1012kN m M1012o Ks 2 10rad M1010'kN m M1010ò Kst 10rad M10'10kN m M10'10o Kst 10rad
m
Equilibrium conditions:
rad
Node 9
Mk M97kN m M910kN m M911kN m Node 10:
0 kN m
M109kN m M1012kN m M1010'kN m M108kN m
Structural system
0 kN m
The calculated moments of individual members of equilibrium conditions: M910 v ( 1.0) kN m M10'10 v ( 9 0) kN m M1010' v ( 5.0) kN m M911 v ( 2.0) kN m M109 v ( 3.0) kN m
M910 691.408kN m M10'10 282.659kN m M1010' 282.659kN m M911 26.401kN m M109 545.787kN m
The computation of shear forces in the individual members: V910o qdx
l1
0
2
V109o V910o V910 V910o
M910 M109
0
39.601
1
-691.408
2
26.401
3
l1
V910 532.511kN M910 M109 V109 V109o l1 V109 494.688kN l1 V1010ò qdx
545.787
v 4
-105.251
5
-282.659
6
-157.877
7
7.223
8
-28.797
9
282.659
4
V1010' V1010ò V1010' 256.8kN
Transformation moments for the part columned strip and between the columns Ma M910 Mb M109 Ma 691.408m kN Mb 545.787m kN Mc' Mmax1.25 Mc Mstr 1.25 Mc' 464.278kN m Mc 5.539kN m Moments over support: 0.75 M1a p Ma M2a 172.852kN m M2b 1 p Mb p
M1a 518.556kN m M1b p Mb M2b 136.447kN m
m 0.60 M3c Mc' m
l1
M3c 278.567kN m
M4c Mc' 1 m
a 3.992m
V910 V109 2
a Mmax V910a M910 qdx 2
Mmax 371.423kN m
Maximum moment between 10-10 Mstr
Mstr
l2 l1 2 V1010' M1010' qdx 4
M2a 1 p Ma M1b 409.34kN m
Positively support moments:
Maximum moment between 9-10 Mmax a V910
Figure: 3.2.1-4
2
2
Mstr 4.431kN m
Figure: 3.2.1-5
Structural system
M4c 185.711kN m
STRUCTURAL ENGINEERING ROOM
v Find M97 M910M911M109M1012M1010'M108 9 10 M10'10
Department of Architecture
74
Column strip M 1b:
Dimensioning of the reinforcement:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
75
M1b 409.34kN m
Material characteristic of concrete f ckcyl and steel fyk fyd 375MPa
effective height:
M1b
Ast
width, which act the moment
b d fcd MN
2
M1a 518.556kN m
M1a 0.518MN m fcd 12MPa
d 0.27m
M1a
2
b d fcd
Ast b d fcd
Ast
b 3.85m
0.0475
b d fcd MN
2
0.154
Ast
2
100cm
Ast 59.252cm
Ast 44.857cm
3.85
M2a 0.172MN m fcd 12MPa
M2a
2
b d fcd
Ast
b d fcd MN
2
100cm
0.121
Ast
2
11.651cm
3.85
M2b 0.136MN m
2
15.39cm
b 3.85m
M2b
2
b d fcd
b d fcd MN
2
100cm
d 0.27m
0.01145
2
Ast 14.283cm
0.04
Ast 3.85
2
3.71cm
The lower reinforcement for moments: Column strip M 3c:
M2a 172.852kN m d 0.27m
Ast
Among the columned strip M 2a:
2
100cm
fcd 12MPa
0.03596
M2b 136.447kN m
Column strip M 1a:
b 3.85m
Among the columned strip M 2b:
b 3.85m
2
2
b d fcd
d hd 3cm
ly
fcd 12MPa
d 0.27m
fcd 12MPa
The top reinforcement for moments:
b
M1b 0.409MN m
2
Ast 17.95cm
0.051
Ast 3.85
M3c 0.278MN m
b 3.85m
b 3.85m
0.01439
M3c 278.567kN m
2
4.662cm
M3c
2
b d fcd
Ast
b d fcd
Structural system
fcd 12MPa
d 0.27m
MN
2
100cm
0.0249
2
Ast 31.06cm
0.083
Ast 3.85
2
8.068cm
a magnification between support:
M4c 185.711kN m
M4c 0.185MN m fcd 12MPa
d 0.27m
Mc 453.412kN m
b 3.85m
Transformation moments for the part columned strip and among columned
M4c
2
b d fcd
Ast
Mc Mc' 1.25
0.055
0.01588
support t of Ma2 p
b d fcd
2
100cm
2
Ast 19.809cm
MN Investigation replacement frame in y Frame 2 Calculation Model
Ma1 p Ma
0.75
Ma1 362.73kN m
Ma2 1 p Ma
Ma2 120.91kN m
Between the support of M c m
Mc1 mMc
0.6
Mc2 1 m Mc
Mc1 272.047kN m
Mc2 181.365kN m
Dimensioning of reinforcement Upper reinforcement of moment: Effective depth: d hd 3cm Column strip M 1a:
The width on which acting the moment:
b Figure: 3.2.1-6
q2d qd l1 l2 1 2
Calculation internal forces
q2d 97.886
kN m
Support part: 1
2
Ma 483.64kN m
Among the supports: Mc'
16
q2d ly
l2
b 2.825m
4
M1a 0.518MN m
q l 12 2d y
1
4
Column strip M 1a:
Load calculation
Ma
l1
2
M1a
2
b d fcd
Ast Mc' 362.73kN m
d 0.27m
b d fcd MN
2
100cm
Structural system
0.06687
fcd 12MPa
b 2.825m
0.21
2
Ast 61.206cm
STRUCTURAL ENGINEERING ROOM
Among the columned strip M 4c:
Department of Architecture
76
STRUCTURAL ENGINEERING ROOM
Department of Architecture
77 Between the column strip M 2a:
Investigation extreme frame replacement
Calculation Model: M2a 0.172MN m
d 0.27m
M2a
2
b d fcd
Ast
b d fcd MN
0.02037
fcd 12MPa
b 2.825m
0.07
2
Ast 18.645cm
2
d 0.27m
fcd 12MPa
100cm
Column strip M 1a:
Mc1 0.272MN m
Mc1 2
b d fcd
Ast
b d fcd MN
0.03277
b 2.825m
0.11
Figure: 3.2.1-7 Calculation of load:
2
2
100cm
Ast 29.994cm
From the slab: l1 q3do qd lk 2
Between column strip M 2a:
Mc2 0.181MN m
Mc2 2
d 0.27m
b d fcd
Ast
b d fcd MN
2
100cm
0.0216
fcd 12MPa
b 2.825m
0.073
2
kN m
Peripheral masonry thickness of 400 mm YTONG:
F1 10
Ast 19.77cm
q3do 106.549
kN m
3
kv 400mm1.4
F1 15.96
kN m
Total load replacement frame: qkd q3do F1
Structural system
q kd 122.509
kN m
Design the reinforcement to the reinforced concrete slab
Moment of the end strip:
The top reinforcement for moments:
Support bending moment:
effective height: d d 3cm
Mka
1 12
qkd ly
2
Mka 605.295kN m
Column extreme strip Mexta:
width which act moment
Between the column bending moment:
b lk Mkc
1 16
qkd ly
2
Mkc 453.971kN m
b 2.3 m
Column extreme strip Mexta: see diagram B3-B3.3
Transformation moments for the part columned bands and among columned
Mexta 0.265MN m
d 0.24m
fcd 12MPa
b 2.3 m
columned strip width: bp3 lk
l1 2
bp3 6.15m
Mexta
Moments over support: Ast Mexta
Mka lk 1 2 4 b p3
MN
2
0.166
2
100cm
Ast 33.716cm
Column strip inside M k4a:
Mk4a p Minta
Mk3a 1 p Minta
Mk3a 85.197kN m
Mk4a 255.59kN m
width, which acts moment, see diagram B3-B3.3 b
Between the column moments:
Mkc lk 1 2 4 b p3
b d fcd
0.0509
Mexta 264.509kN m
Minta Mka Mexta
Mextc
2
b d fcd
l1
b 1.925m
4
Mk4a 0.256MN m Mextc 198.382kN m
Mintc Mkc Mextc
Mk4c mMintc
Mk3c 1 m Mintc
Mk3c 102.236kN m
Mk4c 153.354kN m
d 0.24m
Mk4a
2
b d fcd
Ast
b d fcd MN
2
100cm
Structural system
0.05965
fcd 12MPa
b 1.925m
0.192
2
Ast 33.07cm
Ast 1.925
2
17.179cm
STRUCTURAL ENGINEERING ROOM
Calculation of internal forces
Department of Architecture
78
STRUCTURAL ENGINEERING ROOM
Department of Architecture
79 Among the columned strip M k3a:
Mk4c 0.153MN m
width, which acts moment b
l1
b 1.925m
4
Mk3a 0.085MN m
2
b d fcd
b d fcd MN
fcd 12MPa
d 0.24m
Mk3a
Ast
0.01739
b 1.925m
2
2
100cm
b lk
Ast 9.641cm
Ast 1.925
b 2
5.008cm
Mextc 0.198MN m
d 0.24m
l1
2
b d fcd
4
Ast 19.049cm
0.03758
2
100cm
b d fcd
b 2.3 m
0.125
2
Ast 24.893cm
Ast 2.3
2
b d fcd
Ast fcd 12MPa
d 0.24m
Mk3c
2
10.823cm
width, which acts moment l1
Ast
2
100cm
Mk3c 0.102MN m
Column strip inside M k4c:
b
2
0.115
1.925
2
9.896cm
b 1.925m
4
Mextc
MN
MN
See diagram B3-B3.3
b 2.3 m
b d fcd
b d fcd
0.03436
width, which acts moment
width, which acts moment, see diagram B3-B3.3
Ast
2
b d fcd
b 1.925m
Among the columned strip M k3c:
0.064
Column extreme strip Mextc:
Mk4c
Ast
The lower reinforcement for moments:
fcd 12MPa
d 0.24m
b 1.925m
Structural system
MN
2
100cm
0.02189
fcd 12MPa
b 1.925m
0.077
2
Ast 12.136cm
Ast 1.925
2
6.304cm
Carrying capacity of the concrete section
thickness is hd = 0.3m, Column diameter (round column) d =0.50 m, the maximum force applied one column at Nd= 1800 kN. d
0.5 m
hd
0.3 m
b
1 m
Nd
17 MPa
0.42 hd g b fctm
1.2 MPa
fyd
1
kN
Maximum force per columns Vcd 1800 kN
Vcd Nd
fctm
qbu 262.86 m
Assess the resistance of the concrete section
1800 kN
Material characteristics: fcd
qbu
Department of Architecture
Example 3.2-2: In the example we are considering reinforced concrete slab flat, floor slab
375 MPa
Basic critical perimeter
ucr 2.51 m
Shear force on the critical perimeter qd
Vcd
qd 716.2 m
ucr
1
kN
Shear resistance of concrete qbu
1
qd 2 qbu
stw
Ast
As1
1
4
n
5
Ast n As1
Ast 0.00127 m
h
1.4
correct proposal
Proposal of hidden head
Maximum critical perimeter with hidden head
1 fctm 3 fyd
1
qd qbu
2 qbu 525.72 m 1 kN
qd 2 qbu
On 1m plate
b
kN
2
in both directions
stmin
1
Incorrect design, head to be designed so that they apply condition:
2
stw 0.004241 m
b hd
716.2 m
Coefficient of shear strength
18 mm
qd
1
kN
We suggest shear reinforcement
Figure: 2.3.2‐1
262.86 m
stmin s
2 hd 3 m
g s n h f
0.001067
1.159
h 1.2 g
1.74
Ucrmax
n
1.0
f
1.25
b
1
1.9 ucr
Ucrmax 4.78 m
qda
Vcd Ucrmax
qda qbu
qda 376.95 m
1
2 qbu 525.72 m
Structural system
qbu 262.86 m
kN 1
kN
1
kN
STRUCTURAL ENGINEERING ROOM
80
If we want to make a proposal without head, subject to the following parameters:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
81
d
1.0 m
fctm40
1.40 MPa
ucr40 d
Carrying capacity of the concrete section qbu40
0.42 hd g b fctm40 ucr40
qbu40
2 qbu40 2504.94 kN
2
hd
2
ucr40
4.08 m
hh
2qbu40 Vcd
qd2
dh
Vcd
2.0 m
qd2
Ucr2
8 mm
diameter of one profile
1
2
Asssku Ass
design value of shear resistance of plate without shear reinforcement (per unit length of critical perimeter)
sin ( ) 0.71
0.6 m
1
2 Asssku 0.00025133 m 4 Assessment of the punching according to EC 2
Geometry head
45 deg
number of bars in one bin / m 'ss = 0.25m
Asssku n1
Proposal visible head
5
Vcd 1800 kN
1252.47 kN
n1
286.48 m
v Rd1
cos ( ) 0.71
1
kN
qbu
262.86 m
1.2 40 1 d
shear resistance
Ucr2 6.28 m
Ucr2 dh
Rd k
1
kN
qd2 qbu
Rd
0.3
MN m
hd
2
0.3 m
k
1.6
hd m
k 1.3
average width tension section
bt
1 m
fyk
410
m
min2
min
Figure: 2.3.2‐2
Proposal shear reinforcement - reinforced by bins q
bu
fyd
qsu
190 MPa
qsu ss
n A ss ss s fyd
1
0.0015 bt
min1 min2
s
1
m
0.6 bt
hd kN 4 m
fyk
min2 0.00045
2
min 0.00045
Ac 0.3 m
2
The maximum degree of reinforcement
q su q d2 q bu A ss
hd
min max min
Ac hd bt
qd2 2 qbu
d2
min1
2
concrete area
We suggest shear reinforcement
q
MN
1 m
2
n
q su
23.62 m
1
kN
max
0.04
m
1
Ac
2
max 0.01
The average degree of reinforcement
n is the number of bins reinforced, Ass area of reinforcement to a bin
Given
qsu
n Ass ss s fyd m
1
Ass Find Ass
1 Ass 0.000124 m
2
min max
2
vRd1 Rd k
Structural system
1.2 40 1 hd
1 0.01 vRd1 169.53 m
1
kN
min1 0.00000044
The load effects Vsd
length of critical perimeter) vRd2
1.6 vRd1
vRd2 271.25 m
1
Computing shear force
1800 kN
1.15
internal columns
kN
Design value of shear resistance of plate with shear reinforcement (per unit length of critical perimeter)
A f
s yd sin ( )
vRd1
vRd3
i
u
Column diameter Ps
Diameter of critical perimeter Pu
2 1.5 hd Ps
vsd
Pu 1.4 m
Critical perimeter
Acw
2
4
2
Ps
lh
Acw 1.34 m
4
2
concrete shear area
0.9 m
incorrect design, design head
vsd vRd3
As 0 m
2
As fyd sin ( ) u
2
hh
0.6 m
d2crit
3.11 m
ucrit 9.77 m
fyd
vRd3 255.28 m
360
1
MN m
kN
d2crit
2
2
Ps
2 Acwh 7.4 m 4 4 The expected level of reinforcement by shear reinforcement Acwh
2
Ash 0.01 m
Ash ´w Acwh
carrying capacity
vRd3 vRd1
Ash fyd sin ( ) ucrit
2
vsd
A w
sw sin ( )
i
Ax
Structural system
Vsd
vsd 211.87 m
ucrit
vRd3 382.21 m
Slab with shear reinforcement to a void punching.
vsd vRd3
Concrete shear area
0.0013 0.6
vRd3 vRd1
kN
Critical perimeter with head
As ´w Acw
1
ucrit d2crit
Assumption degree of shear reinforcement
´w
vsd 470.64 m
u
Geometry head u 4.4 m
u Pu Pu
Vsd
as being applicable condition
Figure: 2.3.2‐3
0.5 m
1
kN
vsd vRd3
1
kN
STRUCTURAL ENGINEERING ROOM
The maximum design value of shear resistance of plate with shear reinforcement (per unit
Department of Architecture
82
Space inside the critical perimeter less the contact surface
STRUCTURAL ENGINEERING ROOM
Department of Architecture
83
Ax
d2crit
2
4
2
Ps
Ax 7.4 m
4
w
Asw sin ( ) Ax
wtab
w 0.00001528
0.0013
wmin 0.6 wtab
wmin 0.00078
2
Necessary degree of reinforcement EC2:
For dimensioning elements requiring shear reinforcement
0.5 fcd bw 0.9 d 1 cotg ()
VRd2
fck
0.7
200 MPa
fck
if
0.58
25 MPa
0.5 0.5
fcd
0.58
13.3 MPa
0.5
w if w wmin wmin w
w 0.00078
Minimum design values of moments on columns in contact with the plate at the eccentric load
x 0.125
Internal Column, top moment
Smallest section width in the range of effective height bw
y 0.125
1.0 m
Internal Column, top moment
Vsd 1800 kN
acting shear
Height of the floor slab hd
force msdx x Vsd
msdx 225 kN
msdy y Vsd
msdy 225 kN
0.3 m
cot ( )
0
VRd2
0.5 fcd bw 0.9 hd 1 cot ()
VRd2 1032.41 kN
Maximum distance of stirrups
s max
0.3 hd
s max 0.09 m
s max if s max
0.2 m smax 0.2 m
s max
0.09 m
2 VRd2 688.27 kN 3 Maximum diameter of reinforcement stirrups with a smooth surface Vsd 1800 kN
s
0.012 m Figure: 2.3.2‐4
Sectional area of shear reinforcement in the length range Asw
s
4
2
Asw 0.00011 m
2
2
Structural system
of thickness hs at a critical cross-section carries a full load slab, shear force Vcd = 400 kN,
Bending moment and shearing forces: Vcd1 400 kN
Vcd 325 kN
Mcd 20 kN m
shear force from accidental load Vcd = 325 kN and the bending moment Mcd = 20 kNm (moment transmitted from slab to reinforced column).
hs
Uc1 as
2
Ucr 2 Uc1 Uc2 Vcd1 qdmax Ucr
Ucr 2.4 m
Mkontr 0.2 Vcd hs
Mkontr 13 m kN
Material characteristics:
fckcyl 0.8 fckcub
fckcyl
fcd 0.85
1.5
1
qdmax 166.667 m
hs 2
2
Uc2 0.6 m
kN
fckcyl 16 MPa
Icr
U c1
3
6
U c2 U c1
2
Icr 0.144 m
2
3
fcd 9.067 MPa 2 3
fckcyl MPa 10 MPa
fctm 1.4
Uc2 bs
If Mkontr less than the Mcd, should be respected Mcd
Figure: 2.3.3-1
fckcub 20 MPa
Uc1 0.6 m
2
fyk 345 MPa
fctm 1.915 MPa
fyd
fyk 1.15
fyd 300 MPa
Figure: 2.3.3-2
where fctk is the characteristic tensile strength of concrete (5-percent fractile), fctm is the mean tensile strength and fck is the characteristic compressive strength of concrete measured on cylinders.
1
n 1
The depth of reinforced concrete slabs hs 0.2 m
dmax
2
V cd U cr
n 0.4
U c2 U c1
M cd n 0.5 U c1
dmax
Icr
1
152.08m
kN
Calculation of Qbu
Dimension columns: as 0.40 m
3
bs 0.40 m
2
d
18 mm
Structural system
As1
d
4
2
As1 0.00025447 m
n 6
STRUCTURAL ENGINEERING ROOM
Example 3.2-3: In the middle columns of dimensions as x bs from adjacent reinforced flat slab
Department of Architecture
84
STRUCTURAL ENGINEERING ROOM
Department of Architecture
85
c1 as \
stmin
sty
Perimeter displaced the critical cross section
2
Astd n As1
Astd 0.00152681 m
c2 bs
stx
1 fctm 3 fyd
stmin
Ucrp 2 c1 hs 2 ss c2 hs 2 ss
Astd
stx
c2 4 hs hs
0.00636173 qdmaxp
Vcd1
1
qdmaxp 77.04451232 m
Ucrp
It is less than qbu, that is, the cross section satisfies without shear reinforcement. Alternative we suggest shear reinforcement consisting of a flexible conduit at an angle
Astd
0.00636173
sty
c2 4 hs hs
s
stx sty
stm
1 50 b stm stmin
qbu 0.42 s h n fctm 1
qbu 246.91096132 m
=60.deg.
hs m
kN
s
qbu 246.911 m
0.00636173
1.212
h
1.4 m 2
qbur 0.42 fctm
hs
hs 3
h
1.267 m
Asb
1
qbur 160.87453706 m
oh
2
qbur 160.87453706 m
1
qbu 246.911 m kN
qdmax 0.5 qbu Ucr
kN
Asoh
oh
4
2.59304223
kN 2
qbueur 160.87453706 m
kN
I suggest shear reinforcement in the form of welded of mesh
ss
s
8 mm
ss Ass1 s fyd
qdmax 0.5 qbu
Ass1
s
2
4
2
The cross-section without shear reinforcement does not comply
1
kN
Asb 0.00039917 m
sin ( ) s fyd
14 mm
Asoh
kN
qdmax 166.667 m
2
2
Asb
qdmax 166.667 m
1
kN
The proposal
m
The reliability condition
ss
kN
0.00212797
1
stm
Ucrp 5.19180391 m
2
Ass1 0.00005027 m
ss 0.34897549 m
Figure: 2.3.3-3
Structural system
2
Asoh 0.00015394 m
60 deg
Column of 400x 500 extreme
hd 25 cm
fctm 1.2 MPa
fyd 375 MPa
P1 856 kN
bs 50 cm
hs 50 cm
hd 25 cm
fctm 1.2 MPa
bs 40 cm
hs 50 cm
fyd 375 MPa
Qbu1 0.42 hd fctm ucr1
Qbu1 226.8 kN
Step 1: step 1:
ucr1 bs hd hs hd 2
Qbu1 0.42 hd fctm ucr1
P P1 0.5 Qbu1
P 667 kN
Asb
As1
P 0.86 fyd
2
Asb 0.00206822 m
2
4
0.5 Qbu1 113.4 kN
When applied to the plate even bending moment, then we take 0.5 qbu
25 mm
2
V1 0.42 hd fctm ucr1
ucr1 bs 0.5 hd 2 hs hd
Qbu1 378 kN
P1 577 kN
Asb
As1 0.00049087 m
n
V1 378 kN
P 667 kN
As1
n 4.21333718
As1
ucr2 bs 3 hd hs 3 hd 2
ucr2 5 m
Qbu2 0.42 hd fctm ucr2
Qbu2 630 kN
P P1 0.5 Qbu2
n
Asb As1
n 3.42
P 541 kN
25 mm
As1
P 0.42 hd fctm
ucr 4.2937 m
Asb 0.00143752m
4
As1 0.00031416m
2
2
2
2
4
V1 226.8 kN
20 mm n
Asb
n 4.575766
As1
P 463.6 kN V1 P does not comply
Step 2
2
Asb 0.00167752 m
2
As1 0.00049087 m
ucr2 hs 3 hd bs 1.5 hd 2
ucr2 2.8 m
Qbu2 0.42 hd fctm ucr2
Qbu2 352.8 kN
We expand the circumference in order to prevent the creation of a new shear crack
V2 0.42 hd fctm ucr2 V2 630 kN
ucr
P 0.86 fyd
V1 0.42 hd fctm ucr1
P Asb 0.86 fyd
P 463.6 kN
For P we calculate the required shear reinforcement.
Asb
Step 2
P P1 0.5 Qbu1
ucr
4
1.0734 m
P P1 0.5 Qbu2
20 mm
Structural system
P 400.6 kN
As1
2
4
Asb
P 0.86 fyd
2
Asb 0.00124217 m 2
As1 0.00031416 m
STRUCTURAL ENGINEERING ROOM
Internal column of 500 x 500
Department of Architecture
86
STRUCTURAL ENGINEERING ROOM
Department of Architecture
87
n
Asb As1
V2 0.42 hd fctm ucr2
n 3.95395164 V2 P
P 400.6 kN
V2 352.8 kN
The bending moment calculation of rectangular slabs loaded due to distributed load
does not comply
When calculating the maximum bending moments are used coefficients C, in the tables D-3. The choice of table depends on the method of support of rectangular slab. Kind of support:
- simple supported - cantilevered or fixed
The following tables also shown ways of reinforcing various kinds of slabs
Figure 2.3.3-4: Shear reinforcement at slab-column connection Step 3:
:
ucr3 hs 5 hd bs 2.5 hd 2 Qbu3 0.42 hd fctm ucr3
Asb
Qbu3 478.8 kN
P 0.86 fyd
25 mm
2
4
Ma
C aq q a
2
2 - In direction b M b C bq q b
where q = g + v = dead load + variable loads
P P1 0.5 Qbu3
P 337.6 kN
2
- In direction a
ucr3 3.8 m
Asb 0.00104682 m
As1
Calculation of the maximum bending moments over the supports
Calculation of maximum bending moments between the supports
- the effects of the dead load g - In direction a
2
As1 0.00049087 m
n
Asb As1
n 2.13
Ma
C ag q a
2
-In direction b
Mb
C bg q b
2
where q = g = dead load - from the variable load v
V3 0.42 hd fctm ucr3
V3 478.8 kN
P 337.6 kN V3 P OK
V3 is greater than P, thus the determination of the reinforcement to avoid the punching in
2 - In direction a : M a C av q a
where q = v = variable loads
reinforced concrete slab flat over the column is o
Structural system
2 - In direction b M b C bv q b
Table: D 3
Loads on slab: - dead load
g = 9 kN/m2
-variable loads
v = 3 kN/m2
- total load
q = g + v = 12 kN/m2
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv between the a/b supports where Cav q=v Cbv
1,00 0,70 1,00 0,036 0,036 0,70 0,068 0,016 1,00 0,036 0,036 0,70 0,068 0,016
0,95 0,65 0,95 0,040 0,033 0,65 0,074 0,013 0,95 0,040 0,033 0,65 0,074 0,013
0,90 0,60 0,90 0,045 0,029 0,60 0,081 0,010 0,90 0,045 0,029 0,60 0,081 0,010
0,85 0,55 0,85 0,050 0,026 0,55 0,088 0,008 0,85 0,050 0,026 0,55 0,088 0,008
0,80 0,50 0,80 0,056 0,023 0,50 0,095 0,006 0,80 0,056 0,023 0,50 0,095 0,006
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv a/b between the supports where Cav q=v Cbv
1,00 0,045 0,045 0,70 0,074 0,017 1,00 0,018 0,018 0,70 0,030 0,007 1,00 0,027 0,027 0,70 0,049 0,012
0,95 0,050 0,041 0,65 0,077 0,014 0,95 0,020 0,016 0,65 0,032 0,006 0,95 0,030 0,025 0,65 0,053 0,010
0,90 0,055 0,037 0,60 0,081 0,010 0,90 0,022 0,014 0,60 0,034 0,004 0,90 0,034 0,022 0,60 0,058 0,007
0,85 0,060 0,031 0,55 0,084 0,007 0,85 0,024 0,012 0,55 0,035 0,003 0,85 0,037 0,019 0,55 0,062 0,006
0,80 0,065 0,027 0,50 0,086 0,006 0,80 0,026 0,011 0,50 0,037 0,002 0,80 0,041 0,017 0,50 0,066 0,004
0,75 0,069 0,022
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv between the a/b supports where Cav q=v Cbv
1,00 0,076 0,70 0,050 1,00 0,018 0,027 0,70 0,046 0,016 1,00 0,027 0,032 0,70 0,057 0,016
0,95 0,072 0,65 0,043 0,95 0,021 0,025 0,65 0,054 0,014 0,95 0,031 0,029 0,65 0,064 0,014
0,90 0,070 0,60 0,035 0,90 0,025 0,024 0,60 0,062 0,011 0,90 0,035 0,027 0,60 0,071 0,011
0,85 0,065 0,55 0,028 0,85 0,029 0,022 0,55 0,071 0,009 0,85 0,040 0,024 0,55 0,080 0,009
0,80 0,061 0,50 0,022 0,80 0,034 0,020 0,50 0,080 0,007 0,80 0,045 0,022 0,50 0,088 0,007
0,75 0,056
Coefficients for calculating _ bending moments over the supports where q=g+v
rectangular slab subjected to uniform load at two directions.
span of the slab: - In the direction a: a = 5,4 m - In the direction b: b = 7,2 m
Coefficients for calculating _ bending moments over the supports where q=g+v
Figure 3.2.4-1: Slab acting in two directions When calculating the maximum bending moments are used coefficients C, in Table D-3.
0,75 -
0,75 0,061 0,019
0,75 0,061 0,019
0,75 0,028 0,009
0,75 0,045 0,014
Maximum moment over the supports: Mac- a MbcMac- = Caq. q a2 = 0,076 .12. 5,4 2 = 26,59 kNm/m
Mbc- = Cbq q b2 = 0,024 .12. 7,2 2 = 14,92 kNm/m
The maximum bending moment between supports Mac+ a Mbc+ - the effects of the dead load g Mag+ = Cag g a2 = 0,043 .9
5,4 2 = 11,28 kNm/m
Mbg+ = Cbg g b2 = 0,013 .9 .7,2 2 = 6,06 kNm/m
- the effects of variable loads v Mav+ = Cav v .a2 = 0,052 .3
5,4 2 = 4,55 kNm/m
Mbv+ = Cbv. v b2 = 0,016 .3 .7,2 2 = 2,49 kNm/m
- total maximum bending moments between supports Mac+ a Mbc+ Mac+ = Mag+ + Mav+ = 15,83 kNm/m
Mbc+ = Mbg+ + Mbv+ = 8,55 kNm/m
Structural system
Coefficients for calculating _ bending moments over the supports where q=g+v
0,75 0,040 0,018
0,75 0,051 0,019
STRUCTURAL ENGINEERING ROOM
Example 3.2-4: The calculation of the maximum bending moment of reinforced concrete
Department of Architecture
88
Table: D 3
STRUCTURAL ENGINEERING ROOM
Department of Architecture
89
Coefficients for calculating _ bending moments over the supports where q=g+v Coefficients for calculating bending moments between the supports where q=g Coefficients for calculating bending moments between the supports where q=v
Coefficients for calculating _ bending moments over the supports where q=g+v Coefficients for calculating bending moments between the supports where q=g Coefficients for calculating bending moments between the supports where q=v
Coefficients for calculating _ bending moments over the supports where q=g+v Coefficients for calculating bending moments between the supports where q=g Coefficients for calculating bending moments between the supports where q=v
a/b Caq Cbq a/b Caq Cbq a/b Cag Cbg a/b Cag Cbg a/b Cav Cbv a/b Cav Cbv
1,00 0,050 0,050 0,70 0,081 0,019 1,00 0,027 0,027 0,70 0,046 0,011 1,00 0,032 0,032 0,70 0,057 0,014
0,95 0,055 0,045 0,65 0,085 0,015 0,95 0,030 0,024 0,65 0,050 0,009 0,95 0,035 0,029 0,65 0,062 0,011
0,90 0,060 0,040 0,60 0,089 0,011 0,90 0,033 0,022 0,60 0,053 0,007 0,90 0,039 0,026 0,60 0,067 0,009
0,85 0,066 0,034 0,55 0,092 0,008 0,85 0,036 0,019 0,55 0,056 0,005 0,85 0,043 0,023 0,55 0,072 0,007
0,80 0,071 0,029 0,50 0,094 0,006 0,80 0,039 0,016 0,50 0,059 0,004 0,80 0,048 0,020 0,50 0,077 0,005
0,75 0,076 0,024
a/b Caq Cbq a/b Caq Cbq a/b Cag Cbg a/b Cag Cbg a/b Cav Cbv a/b Cav Cbv
1,00 0,075 0,70 0,086 1,00 0,027 0,018 0,70 0,035 0,005 1,00 0,032 0,027 0,70 0,051
0,95 0,079 0,65 0,087 0,95 0,028 0,015 0,65 0,036 0,004 0,95 0,034 0,024 0,65 0,055
0,90 0,080 0,60 0,088 0,90 0,029 0,013 0,60 0,037 0,003 0,90 0,037 0,021 0,60 0,059
0,85 0,082 0,55 0,089 0,85 0,031 0,011 0,55 0,038 0,002 0,85 0,041 0,019 0,55 0,063
0,80 0,083 0,50 0,090 0,80 0,032 0,009 0,50 0,039 0,001 0,80 0,044 0,016 0,50 0,067
0,75 0,085 -
0,011
0,009
0,007
0,005
0,004
a/b Caq Cbq a/b Caq Cbq a/b Cag Cbg a/b Cag Cbg a/b Cav Cbv a/b Cav Cbv
1,00 0,071 0,70 0,091 1,00 0,033 0,027 0,70 0,051 0,009 1,00 0,035 0,032 0,70 0,060 0,013
0,95 0,075 0,65 0,093 0,95 0,036 0,024 0,65 0,054 0,007 0,95 0,038 0,029 0,65 0,064 0,010
0,90 0,079 0,60 0,095 0,90 0,039 0,021 0,60 0,056 0,006 0,90 0,042 0,025 0,60 0,068 0,008
0,85 0,083 0,55 0,096 0,85 0,042 0,017 0,55 0,058 0,004 0,85 0,046 0,022 0,55 0,073 0,006
0,80 0,086 0,50 0,097 0,80 0,045 0,015 0,50 0,061 0,003 0,80 0,051 0,019 0,50 0,078 0,005
Table: D 3
0,75 0,043 0,013
0,75 0,052 0,016
0,75 0,033 0,007
0,75 0,047 0,013
0,75 0,088 -
0,75 0,048 0,012
0,75 0,055 0,016
Structural system
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv between the a/b supports where Cav q=v Cbv
1,00 0,071 0,70 0,038 1,00 0,027 0,033 0,70 0,058 0,017 1,00 0,032 0,035 0,70 0,063 0,017
0,95 0,067 0,65 0,031 0,95 0,031 0,031 0,65 0,065 0,014 0,95 0,036 0,032 0,65 0,070 0,014
0,90 0,062 0,60 0,024 0,90 0,035 0,028 0,60 0,073 0,012 0,90 0,040 0,029 0,60 0,077 0,011
0,85 0,057 0,55 0,019 0,85 0,040 0,025 0,55 0,081 0,009 0,85 0,045 0,026 0,55 0,085 0,009
0,80 0,051 0,50 0,014 0,80 0,045 0,022 0,50 0,089 0,007 0,80 0,051 0,023 0,50 0,092 0,007
0,75 0,044
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv a/b between the supports where Cav q=v Cbv
1,00 0,033 0,061 0,70 0,068 0,029 1,00 0,020 0,023 0,70 0,040 0,011 1,00 0,028 0,030 0,70 0,054 0,014
0,95 0,038 0,056 0,65 0,074 0,024 0,95 0,022 0,021 0,65 0,044 0,009 0,95 0,031 0,027 0,65 0,059 0,011
0,90 0,043 0,052 0,60 0,080 0,018 0,90 0,025 0,019 0,60 0,048 0,007 0,90 0,035 0,024 0,60 0,065 0,009
0,85 0,049 0,046 0,55 0,085 0,014 0,85 0,029 0,017 0,55 0,052 0,005 0,85 0,040 0,022 0,55 0,070 0,007
0,80 0,055 0,041 0,50 0,089 0,010 0,80 0,032 0,015 0,50 0,056 0,004 0,80 0,044 0,019 0,50 0,076 0,005
0,75 0,061 0,036
a/b Caq Cbq a/b Caq Cbq a/b Coefficients for Cag calculating bending moments Cbg between the a/b supports where Cag q=g Cbg a/b Coefficients for Cav calculating bending moments Cbv a/b between the supports where Cav q=v Cbv
1,00 0,061 0,033 0,70 0,081 0,011 1,00 0,023 0,020 0,70 0,033 0,006 1,00 0,030 0,028 0,70 0,050 0,011
0,95 0,065 0,029 0,65 0,083 0,008 0,95 0,024 0,017 0,65 0,034 0,005 0,95 0,032 0,025 0,65 0,054 0,009
0,90 0,068 0,025 0,60 0,085 0,006 0,90 0,026 0,015 0,60 0,036 0,004 0,90 0,036 0,022 0,60 0,059 0,007
0,85 0,072 0,021 0,55 0,086 0,005 0,85 0,028 0,013 0,55 0,037 0,003 0,85 0,039 0,020 0,55 0,063 0,006
0,80 0,075 0,017 0,50 0,088 0,003 0,80 0,029 0,010 0,50 0,038 0,002 0,80 0,042 0,017 0,50 0,067 0,004
0,75 0,078 0,014
Coefficients for calculating _ bending moments over the supports where q=g+v
Coefficients for calculating _ bending moments over the supports where q=g+v
Coefficients for calculating _ bending moments over the supports where q=g+v
0,75 0,051 0,020
0,75 0,056 0,020
0,75 0,036 0,013
0,75 0,049 0,016
0,75 0,031 0,007
0,75 0,046 0,013
Mr
-Radial
moment
Mt -tangential moment
slab stiffness: D
h -slab thickness
E h
The deflection at any distance r from
3
12 1
2
where E - modulus
the center of plate
of elasticity
z -deflection plates
- Poisson ratio
z( r)
g 64 D
2
2
2
a r
Table E3: Plate simply supported, loaded by uniform load g [kN/m2] The deflection at any distance r from the center of plate
z( r)
2 2 g a r
64 D
The radial moment at any distance r from the center of plate 5 a2 r2 1
M r( r)
g 16
a 1 r 3 2
2
The radial moment at any distance r from the center of plate
M r( r)
g 16
2
3 a r
2
Tangential moment at any distance r from the center of plate
Tangential moment at any distance r from
M t( r)
the center of plate
M t( r)
g 16
a 3 r 1 3 2
2
Radial moment in the middle of plate
M r( r
0)
M t( r
0)
ga 16
2
3
Structural system
g 16
a 1 r 1 3 2
2
Radial moment in the middle of plate
M r( r
0)
M t( r
0)
ga 16
2
1
STRUCTURAL ENGINEERING ROOM
Table E3: Cantilevered circular plate subjected to uniform load g [kN/m2]
Calculation of deflection and moments circular plates depending the method of support and load
Department of Architecture
90
STRUCTURAL ENGINEERING ROOM
Department of Architecture
91 Table E3: Simply supported circular slab loaded due to two concentrated loads P [kN / m]
Table E3: Plate around the perimeter of cantilevered, loaded with continuous circular load P
acting at a distance ro
[kN / m] acting at a distance ro
Deflection at any distance r > ro from the center of plate
z r ro
z r ro
P
2
a r 1 2
1
1
2
a ro
2
2 2 1 a r 2 2 r o r ln a The deflection at any distance r < ro from the center of plate
8 D
ro 2 2 r o r ln 8 D a 3 a2 1 r2 2 2 a r o 2 2 1 a
P
Deflection at the middle of plate
Radial and tangential moment at any distance
z( r
P
0)
8 D
Mt r ro
Cantilevered radial moment
r < ro from the center of plate
Mr r ro
1 2 r o 2 2 a r o r o ln 2 a
1 a
2
8 a
ro 2
M r( r
P 1 P ln r o
2
4
a
a)
P 4
a
2
ro a
2
2
Cantilevered tangential moment
M t( r
Structural system
a)
P 4
a
2
ro a
2
2
acting on the plate of the center distance C
The radial moment at any distance r> c of the center of plate a 1 P c 1 1 2 2 16 r r a Tangential moment at any distance r> c of the center of plate M r( r c)
M t ( r c)
1 P 4
P 4
2
ln
2 1 P c 1 16
1 ln
a
r
1
1
2
r
a
2
2
Deflection plate at the edge of the load (in the distance c)
z( r
c)
3
P 16 D
1
2
a c
c2 3 ln c 1 1
2
a
2
1
2
a c a
2
Radial moment on the edge of the load (In the distance c)
M r( r
c)
1 P 4
ln
a
c
1 P a 2 c 2 16 a
2
Tangential moment on the edge of the load (In the distance c) M t( r
c)
P 4
1 ln
2 2 1 P a c 1 2 16 a
a
c
Deflection in the middle of plate 3 a2 c2 ln c 7 3 c2 a 4 1 1 Radial moment in the middle of plate 1 c2 P a M r( r 0) M t( r 0) 1 ln 1 2 4 c 4 a z( r
0)
P
16 D
Structural system
STRUCTURAL ENGINEERING ROOM
Table E3: Simple supported circular slab loaded due to two partial uniform load W [kN / m2]
Department of Architecture
92
3.3 Staircases
STRUCTURAL ENGINEERING ROOM
Department of Architecture
93 Minimum number of risers except for stairs within a dwelling unit, at least 3 risers shall be
Staircases provide means of movement from one floor to another in a structure.
provided in interior flights.
The effective span of a simply supported slab should normally be taken as the clear distance between the faces of supports plus one-third of their widths. However, where a bearing pad is provided between the slab and the support, the effective span should be taken as the distance between the centres of the bearing pads. The span/effective depth should not exceed the appropriate value from table C3. There is normally no need to calculate shear stresses in staircases supported on beams or walls. The effective span is the distance between centre-lines of supporting beams or walls. The initial design should be checked, to obtain the final sizes of the stair slab and to calculate the amount and dimensions of the reinforcement. Types of staircases 1. Straight stairs – simplest form of stair layout and consists of one straight two levels. The
Figure 3.3-1: Continuous supported stairs, Reinforcement details for flight
width and the length of the landings should be equal = width of flight +10cm. 2. L Shaped stair (or sometimes called quarter turn stairs) – L shaped stair may have either
A straight staircases can be defined as one having a single, straight flight of stairs that connects two levels or floors in a building a straight staircase can be simple but quite elegant. Because a straight staircase offers a clear view of the stairs, there is a lower chance of falls or
equal or unequal flights. 3. U shaped stairs (or sometimes called half turn stairs or switchback stairs) 4. Winder stairs – stairs refer to stairways that make a turn without including an
misplaced steps. In contrast, curved staircases often require a lot of attention when you are going up or down one. Straight staircases require more space as compared to curved or platform staircases.
intermediate landing or platform to provide a flat rectangular turning space. 5. Spiral stairs – have tread which turn and rise around a central column. 6. Curved stairs – as winder stairs. Some of the functional requirement of staircases are, stability, protection from fire, suitable dimensions, and appearance. Staircases consist of components, flight, landing, tread, riser. In a flight of stairs all steps should have the same riser and same tread. Relationship between riser and tread can be shown as 2h+b = 63cm. Figure 3.3-2: Simply supported stairs, design of reinforcement, Reinforcement details
The vertical height between any landings shall not exceed 3.7m.
Structural system
in the longitudinal direction, while shrinkage reinforcement runs in the transverse direction. Special attention has to be paid to reinforcement detail at opening joints, as shown in
Loading: Dead Load: The dead load, which can be calculated on horizontal plan, includes: 1. Own weight of the steps.
figure 3.3-2.
2. Own weight of the slab. For flight load calculations, this load is to be increased by dividing it by cos to get it on horizontal projection, where a is the angle of slope of the flight. 3. Surface finishes on the flight and on the landings. For flight load calculations, the part of load acting on slope is to be increased by dividing it by cos to get it on horizontal projection.
Live Load: Live load is always given on horizontal projection. Longitudinally supported stairs may be supported in any of the following manners: 1. Beams or walls at the outside edges of the landings. 2. Internal beams at the ends of the flight in addition to beams or walls at the outside edges of
Figure 3.3-3: Steps cantilevering from a wall or a beam
the landings.
Figure 3.3-3 shows a stairs cantilevered from a reinforced concrete beam. The effective length of a cantilever reinforced concrete stairs and beams where this forms the end of a
3. Landings which are supported by beams or walls running in the longitudinal direction figure 3.3-5, figure 3.3-6, figure 3.3-7, figure 3.3-8.
continuous slab is the length of the cantilever from the centre of the support. Where the slab is an isolated cantilever the effective length is the length of the cantilever from the face of the support.
Figure 3.3-5: Longitudinally supported stairs due to walls, beams
Figure 3.3-4: Longitudinally supported stairs, bending moment and shear forces diagram
Structural system
STRUCTURAL ENGINEERING ROOM
The stairs slab is designed for maximum shear and flexure. Main reinforcement runs
Department of Architecture
94
STRUCTURAL ENGINEERING ROOM
Department of Architecture
95
Figure 3.3-6
Figure 3.3-7
Figure 3.3-8: Simply supported steps supported by two walls, beams and a combination of both
Structural system
Figure 3.3-9
reinforced concrete walls (center-to-center) on both sides and carries a live load of 3kN/m2. The risers are 163 mcm and goings are 300 cm. Stair thickness required to satisfy deflection requirements is given by h = 220mm The depth of landing slab is hp = 220 mm = 29 deg Calculation model: Load calculation The section b-c: Calculation of replacement shoulder height: v 0.143m
v cos ( ) h v
h 1
v
h 1 0.071m
2
h h 1 h d h 0.321m
Thickness staircase shoulders h = 0.321 m q1s h 25
kN
1
q1d q1s 1.35
3 cos ( )
m
q1 s 9.183 kN m
2
2
q1d 12.398kN m
floor layer shoulders 2
q2s 2.106 kN m
2
q2d q2s 1.4
Section a-b, c-d:
q2d 2.948kN m
Thickness staircase landings h p 0.02m
q3s h p 25
Layer floor landing: 2
q4s 1.504 kN m
kN 3
m cos ( )
2
q3d q3s 1.35
q3s 0.572kN m
2
q3d 0.772kN m
2
q4d q4s 1.35
q4d 2.03kN m
Imposed loads Circulation areas v 1s 3
kN
v 1d v 1s 1.5
2
m
2
v 1d 4.5kN m
The total surface load on board: q bcs q1s q2s v 1s 2
Figure 3.3-10
q bcd 19.846kN m
2
q cds 5.076kN m
Structural system
2
q bcd q1d q2d v 1d
q bcs 14.289kN m
q cds q3s q4s v 1s
q cdd q3d q4d v 1d
2
q cdd 7.302kN m
STRUCTURAL ENGINEERING ROOM
Example 3.3-1: Design a straight flight staircase in a residential building that is supported on
Department of Architecture
96
Sectional forces:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
97
L 6200 mm
Dimensioning: L2 2550 mm
L1 2100 mm
L3 1550 mm
M a 68.573kN
The calculation of bending moments: 2
M a
q cdd L 12
2
q cdd L
L2 q bcd q cdd L3 2 2 L2
q bcd q cdd
L2
L1
M a 68.573kN
L2
12 2 2 L3 L2 2 q cdd L 2 L2 M ad q bcd q cdd L2 0.5 L1 2 L2 24 L1 2 M d
Bending moment over the support:
m m
M d 77.369kN
M ad 56.877kN
m
m
b 1m h 0.22m See diagram B3-B3.3
m m
m
fcd 11.5MPa
Ma
2
b d fcd
fyd 375MPa
0.0475
fyk 410MPa
0.154
2
Astd b d fcd 100
Astd 12.01cm
m
We provide 7 16
0.21167
0.91533
z d
xu 0.033m
Ast = 14.07 cm2
16mm
n 7 2
Ast 14.074cm
x d
xu 0.8x
x 0.042m
z 0.18m
As1
2
2
As1 2.011cm
4
Mu Ast fyd z
Ast n As1
Mu 95.171kN m
Mu Ma
b. Cross section between the supports Mad = 56.877 kN m
Figure 3.3.1-1
The calculation of reactions A, B A
B
q cdd L 2
q cdd L 2
L3 q bcd q cdd L2 L1
q bcd q cdd L2 1
16mm
ast 20mm
Mad
m
2
b d fcd
0.134
0.5 L2 2 2
A 36.024
kN m
B 41.237
kN m
2
2
4
Mu Ast fyd z
Structural system
d h ast
2
h 0.22
0.03679
0.182
0.9272
Astm b d fcd
z 0.178m
8.3cm
As1
L2
L3 2 0.5 L2 L1 2
Astm
b 1 m
z d
L2
d 0.197m
we provide
2
As1 2.011cm
Mu 67.113kN m
16mm
Ast n As1
n 5 2
Ast 10.053cm Mu Mad
Example 3.3-2: The supporting structure of this type is a spiral reinforced concrete slab anchored at both ends in thick reinforced concrete slab or wall. Diagram of bending moments and forces in such reinforced concrete slab is plotted in figure 3.3.2-2. Structural height
Angle o
H
o
Number widths
ns
Width of stairs
b
Step height
vs
Plaster thickness
ho
3.9 m
2 270 360
4.712
b1
25
3
22 kN m
nv
nv
25 kN m
26
thickness of staircase boards
1.2 m H
b2
3
hd
g4
13.486kNm
g1 g2 g3 g4 p
r
- inside radius stairs
r1
1.2 m
- external radius of stairs
r2
2.4 m
o r1
- density
0.015 m
o
3
O1
s1
5.655m
O2
1 -1
1.8 m
o r
0.03 m
0.015 m 0.15 m
1
b b1 0.967kNm
0.339 m
g1
-1
0.967kNm
0.5 v s b b1 h d b b2 0.914
1
1.98kNm
1
g2
1.98kNm
g3
6.565kNm
ho o b 0.914
g4
vs
tan
8.482m
O3
1
b
cos
o r2
o cos 2
2 2
o 2 2 r k tan 3 sin o 1
0.339m
11.31m
s3
0.915
sin cos
0.196 6
hd
ns
23.86 deg
0.405
2
O2
s2
0.442
s2
11
1
0.374kNm
0.226m
ns
19 kN m
11
- Plaster
q
q
- combination of loading - radius of the axis of stairway
k
g3
3.6kNm
3 kN m b
1
- calculation step width
- Steps
- Self - weight of reinforced concrete slab
p
p
sin
g2
2
variable load
tan
g1
3.6kNm
3 kN m b
0.15m
Permanent load - Surface treatment
p
p
O1
0.2 m
2
variable load
0.452m
1
cos o
sin
0.164
cos
2
3
0.354
2
o sin 2
0 0.836 2
0.5
0.707
3 o cos o
3 2 1 tan 2 sin 2 o 3 sin o o 2 2 k 3 cos o o 6 cos 2 sin o o
2 o 7 2 k sin 3 sin o 5 2 12 9 cos o o 2 o 2 o 3 4 2 6 sin o cos o sin o
3 o
2
1
2
11
445.165m
r
137.397
2
ns
sin o
o cos 2
0.37
O3
12
r tan 3 k sin o o cos o 1.75 k cos 3 sin o o cos o 2 o 3 1 sin 2 o cos o sin o 2 cos 2 o sin o k
12
68.769m
2
12 r
Structural system
38.205
STRUCTURAL ENGINEERING ROOM
Circular-staircase
Department of Architecture
98
STRUCTURAL ENGINEERING ROOM
Department of Architecture
99
3.5 k cos 2 o sin o 6 1k sin 2 o sin o
22
6 k o sin o
22
234.913
3 o o o o 3 10 q r tan 3 k 5 o cos 8 sin 2 o cos cos sin o 2 2 2 2 o o o 2 2 1.75 k cos 24 sin 2 2 o sin 2 15 o cos 2 2 o o o sin 3 cos sin o 2 o cos 2 2 2 o 1 2 2 3 sin 2 o sin 2 k 2 o o o 9 cos 2 o cos sin o 2 2 2 o o sin cos 16 sin o 2 2 o o o 2 cos 6 o cos 8 sin 2 cos sin o 2 2 2
10
Calculation of moments, shearing force and longitudinal forces
10 3
19060.183kNm 2
q r
Calculation A, A, A2 A
2
11 22 12
1
4 5 4
5
2
6
A2
10 12 20 11
7002102.731kN m
Mx
i
2 6 4
3
7
3 4 7 4
4
i
My
Tx
i
Ty Ni
i
i
0
Rc
q r
o 2
57.196kN
1 r tan sin cos i i cos sin i 2 sin sin i o 2 q r i cos sin i sin 2 1 cos i 1 sin sin i q r i cos 1 cos sin i q r i sin
Figure 3.3.2-1: Moment diagrams
A
Rb
37.603kN
436.215
A2
1
1 r sin cos i i sin i 2 cos sin i o 2 q r i cos sin i cos 2
2
2
Ra
o 2 1 r tan i sin i 2 cos i q r 1 cos cos i 2
Calculation 1,2
A
1
148.421kNm
2
3754528.482kN m
37.603kN
sin
2
20 12 10 22
A1
2
99845.761m
A1
1
o
Moments and forces in any cross-section (according to ) the axis of the spiral are calculated as follow:
274.156
o o o 2 20 q r 6 k o cos 4 sin sin o cos 2 2 2 o o 1 o 2 2 4 sin sin o cos 3.5 k cos 6 sin 3 o cos k 2 2 2 20
q r
1
Mt
2
21562.414kN m
q r
20
o 2 2.669kNm 2 o o o o 2 M b q r sin cos 1 r tan 2 2 2 2 o M c 1 r sin 47.861kNm 2 2
Ma
70.129kN m
Structural system
Design of reinforcement in reinforced concrete elements loaded concentric and eccentric compression with a small eccentricity 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. 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.
Determining the buckling length of the column lo
Figure 3.3.2-2: Dimensioned drawing of reinforcement spiral staircase
Table: 3.3.2-1 0,785 1,571 2,356 3,142 3,927 4,712 5,498
Mx 21,557 -3,332 -0,673 4,462 -0,673 -3,332 21,557
cos() 0,707 0,000 -0,707 -1,000 -0,707 0,000 0,707
My -54,018 -73,795 -53,156 0,000 53,156 73,795 54,018
sin() 0,707 1,000 0,707 0,000 -0,707 -1,000 -0,707
Mt 3,807 -0,489 1,859 0,000 -1,859 0,489 -3,807
-2,356 -1,571 -0,785 0,000 0,785 1,571 2,356
2,356 1,571 0,785 0,000 -0,785 -1,571 -2,356
Tx 26,590 0,000 -26,590 -37,603 -26,590 0,000 26,590
Ty -41,552 -19,661 -6,680 0,000 6,680 19,661 41,552
Figure: 3.4-1 1. One end fixed in direction and position, the other free k = 2 N -47,453 -49,814 -32,029 0,000 32,029 49,814 47,453
2. Both ends pinned k = 1 3. One end pinned, the other fixed in direction and position k = 0.707 4. Both ends fixed in direction and position k = 0.5 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). 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.
Structural system
STRUCTURAL ENGINEERING ROOM
3.4 Reinforced concrete column
Department of Architecture
100
STRUCTURAL ENGINEERING ROOM
Department of Architecture
101
However, in practice this type of end condition is almost never perfect and therefore
Design of reinforcement in reinforced concrete columns
somewhat higher values for k are used and can be found in building codes. In fact,
Stress in concrete [MPa] :
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
c = Nsd / Ac’
direction.
Total required area of reinforcement v [cm2] :
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
Arequired = c‘ .10 2
the end conditions are the same, it is sufficient to check for buckling in the direction
where can be obtained from the following graph:
that has the least radius of gyration). Although the buckling of a column can be compared with the 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
=5
=7,5
12.84472 13.14803
12.76732 13.06879
13.75465
13.67173
0.3 0.4
Determination of basic characteristics
slenderness:
i
radius of gyration: Ac‘= b‘ . h‘
lo i I A c´
Minimum amount of reinforcement:
=10
13.28918
=c
13.5888
• Determination of the maximum carrying capacity of cross-section The maximum capacity of the cross section can be determined using the following graph:
- rectangle:
Aprovided . 10 2 / (Ac‘)
- where I is the moment of inertia of
I = 1/12.b.h3
the cross section
0.1 0.2
determine the minimum cross-sectional dimension columns in terms of
=5
=7,5
0.1 0.2
12.84472 13.14803
12.76732 13.06879
13.75465
13.67173
0.3 0.4
buckling
=10
13.28918
=b < 0,6 . fck
13.5888
sdv Ac‘ v lo v i v Areq, Aprovided v
MN m2 m m cm2
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. Figure 3.4-2
The relationship between the length of the column, its lateral dimensions and the end fixity conditions will strongly affect the column’s resistance to buckling.
Structural system
and to the column above will vary with manufacturer. Foundation connection may be via a
documents which perform a similar function. It is necessary for a designer to become familiar
base plate connected to the column or by reinforcing bars projecting from the end of the
with local requirements or recommendations in regard to correct practice.
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. 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 contra-flexure, 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.
Figure 3.4-3 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 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.
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
Structural system
Figure 3.4-4
STRUCTURAL ENGINEERING ROOM
Many countries have their own structural design codes, codes of practice or technical
Department of Architecture
102
STRUCTURAL ENGINEERING ROOM
Department of Architecture 103
Figure 3.4-5 Figure 3.4-6
Structural system
Figure 3.4-7
Structural system
Department of Architecture STRUCTURAL ENGINEERING ROOM
Figure 3.4-7 104
Figure 3.4-8
STRUCTURAL ENGINEERING ROOM
Department of Architecture 105
Figure 3.4-8
Structural system
Figure 3.4-9
Structural system
Department of Architecture STRUCTURAL ENGINEERING ROOM
Figure 3.4-9
106
Figure 3.4-10
STRUCTURAL ENGINEERING ROOM
Department of Architecture 107
Figure 3.4-10
Structural system
Figure 3.4-11
Structural system
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Figure 3.4-11
108
Figure 3.4-12
STRUCTURAL ENGINEERING ROOM
Department of Architecture 109
Figure 3.4-12
Structural system
part of a statically indeterminate structures. Extreme loads induces normal force Nd and bending moment Mf. The actual length of the column is l and the effective length le. 2800 kN
Nd
Mf
50 kN m
l
le
4.0 m
0.7 l
le
2.8m
Asd
11.5 MPa
0.9 MPa
fctm
Ec
1
lim 1.25
0.96
Nd 1 b h 0.8 fcd u fyd
Asd
2
0.00281m
1
1
22 mm
As1
As
n As1
As
0.00304m
Ast
4 As1
Ast
15.2cm
Asc
4 As1
Asc
15.2cm
2
As1
4 2
2
0.00038m
n
8
As Asd
2
0.467
lim
fyd
u
50
2
210 MPa
Es
20 h
Suggestion:
27 GPa
Steel: 375 MPa
1
The necessary amount of reinforcement:
Concrete:
fyd
u
mm
Material characteristics:
fcd
Geometry factor:
Department of Architecture
Example 3.4-1: Design reinforced concrete columns rectangular cross section. The column is
420 MPa
Design cross-sectional dimension: M f
ef
Acd
N d
Nd 0.8 fcd vystu fyd
Abd
2
0.16766m
Abd
0.40947m
Design the cross-section: b
0.45 m
0.45 m
h
Design the bearing reinforcement to the cross-section: Determining the influence of slenderness:
12
le h
21.55
ef
ed
ee
Assessment: Determination of slenderness:
35
The calculation of eccentricities - statically indeterminate structures: ee
Figure: 3.4.1-1
ed
0.018m
h
le 12
h
h
21.55
h 35
The calculation of eccentricities ed: ee
ef
ef
Structural system
M f N d
ef
0.018m
ed
ee
ed
0.018m
STRUCTURAL ENGINEERING ROOM
110
Example 3.4-2: Determine the carrying capacity of a rectangular column shown in
Percentage reinforcement:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
111
Asc
sc
stmin
0.00563
sc
bh
0.03
scmax
scmin sc scmax
1 fctm 3 fyd le
scmin
h
4
10
stmin
0.0008
scmin
0.00062
Extreme normal force which influence collapse of the cross section Neu: Neu
u 0.8 b h fcd As fyd
2883.26kN
Neu
Nd
2800kN
Nd Neu
satisfies section
The design of reinforcement to cross-section will be
c
0.02 m
fck
b´
20 MPa
c
b 2 c
0.41m
b´
Nd
16.65 MPa
c
b´ h´
h´
b´
21.55
Design of reinforcement: Areq
b´ h´ 100
Areq
2
28.577cm
2
As1
1
As1
4
Aprovided
n As1
As
2
0.00038m
2
0.00304m
1
n
22 mm
8
Aprovided Areq
h´
0.41m 1.7
figure 3.4.2-1. Cross-section and material characteristics of the column: concrete cover: Section height:
h
0.60 m
h´
h 2 c
h´ = 0,56.m
Section width:
b
0.30 m
b´
b 2 c
b´ = 0,26.m
c
0.02 m
Characteristic cylinder compressive strength of concrete:
f ck.cyl
Characteristic strength of steel:
f yk
The extreme value of the compressive forces acting on the column:
N sd
Overall height of column:
l
Buckling length of the
lo
0.7 l
25 MPa
410 MPa 3250 kN
3.70 m
2.59 m
column: radius of gyration:
i
slenderness column:
Stress in concrete:
c
From diagram the value at a
0.288 b lo i
0.0864m
30
N sd b´ h´
22.32 MPa
we obtaine given
3.23846
stress in steel s:
Minimum area of reinforcement can be calculated as follows A smin :
A smin
0.15
N sd
4
f yk 1.15
10
2
13.67378cm
The necessary amount of longitudinal see reinforcement in the column A
Figure: 3.4.1-2
Structural system
b´ h´ 100
2
47.151 cm
Number profiles:
2.5 cm
The actual amount of reinforcement designed to column:
of
n
Cross-cutting and material characteristics of the column: 2
As
n
Example 3.4.3: Determine the carrying capacity of circular columns, shown in figure 3.4.3-1
10
2
49.08739cm
4
The real amount of reinforcement must be greater than the value of A:
As A
reinforcement ratio is calculated as follows:
As 100 b´ h´
concrete cover:
c
0.03 m
column diameter:
D
0.5 m
D´
Characteristic cylinder compressive strength of concrete:
f ck.cyl
Characteristic strength of steel:
f yk
The extreme value of the compressive forces acting on the column:
3.37139
again, from the graph of can be obtained c
Buckling length of the column:
22.67275MPa
radius of gyration: Multiplying the amount of stress in concrete c with net area
Nud = c . b´.h´=3,3 MN
lo
i
D 64
D 4
2
D
25 MPa
410 MPa 3200 kN
3.5 m
2.45 m
0.125 m
4
ultimate capacity of column:
Nsd:
0.7 l
0.44 m
4
of concrete cross-section (b´. h´) then we Obtained the
which must be greater than the force exerted on the column
N sd l
Overall height of column:
in reverses way the stress in concrete:
D 2 c
N ud N sd
3.25 MN
slenderness column:
Stress in concrete:
lo
of
c
i
N sd
2
D
19.6
21.045 MPa
4
From the graph we obtain the value of at stress
2.2
in steel s: Minimum area of reinforcement A smin can be
A smin
0.15
calculated as follows: Figure: 3.4.2-1
N sd
f yk 1.15
2
13.46341cm
The necessary amount of longitudinal reinforcement in the column can be determined by using the obtained coefficients :
Structural system
A
2
D´ 4
100
2
33.452 cm
STRUCTURAL ENGINEERING ROOM
- diameter:
Department of Architecture
112
STRUCTURAL ENGINEERING ROOM
Department of Architecture
113
- diameter reinforcing
25 mm
profiles:
- Number profiles:
of
n
The real amount of reinforcement designed to
n
Torsion: twisting of a structural member, when it is loaded by couples that produce rotation about its longitudinal axis. Under twisting deformation, it is assumed
2
As
3.5 Mode of failure of reinforced concrete members subjected to torsion
8
2
39.26991cm
4
1. plane section remains plane 2. radii remaining straight and the cross sections remaining plane
column:
The task of the design engineer is to avoid torsional stresses, either to eliminate them
The real amount of reinforcement must be greater than the value of A:
As A
reinforcing coefficient calculated as follows:
Again, from the graph we can be obtained
torsion of concrete is a problem of less importance than its bending or compression, there are many structural members in which twisting forces occur and such forces must not be ignored.
As
reverses the stress in concrete:
entirely, or to greatly reduce them by suitably arranging the layout of the structure. Although
c
2
D´ 4
2.5826
100
22.2 MPa
Multiplying the amount of stress in concrete c with a net area of the concrete section of the column is obtained by
N ud
D´2 4
c
3.376 MN
carrying capacity:
which must be greater than the force exerted on the column Nsd :
N ud N sd
3.20 MN
Figure 3.5-1: The notation of balanced and unbalanced stability Experiments with plain and reinforced concrete were recorded as early as 1903 at Stuttgart, Germany. Professor Morsh tested to failure eight cylindrical specimens, 26 cm in diameter of plain concrete, and some cylinders reinforced with spirals. He noted that the fracture was on helical surface, approximately at 45 degrees with surface elements parallel to the axis and that the torsional moment at initial cracking was higher than that for unreinforced specimens. The Stuttgart experiments by Prof. Bach (1911) of thirty-seven test –pieces (circular 40 cm in diameter, square 30cm by 30cm, and rectangular 42cm by 21cm) of plain concrete and
Figure: 3.4.3-1
Structural system
In practice, torsion usually occurs as a secondary effect of bending. Such torques occur when a
longitudinal bars and spirals, constituted the first thorough investigation of the problem.
beam is loaded in a plane which does not pass through the shear centre of the cross-section, or
The ratio of the torsional shear strength to compressive strength varied from 0.07 to
the beam is curved in a plane at right angles to the applied loads.
0.13. the ratio of the torsional shear strength to the tensile strength varied from 0.92 to 1.75. the
Longitudinal balcony a girders that support cantilever beams, frame beams are also subject to
highest values obtained for the rectangular sections and the lowest for the hollow circular
torsion because of eccentricity of the load.
sections.
For Reinforced concrete frame if we used it in vertical position, be subject to combined bending
Cowan (sydney) 1960, presented formulas for determining the torsional shear stresses,
and compression, used in horizontal position as balcony beam, would be subject to combined
diagonal tensile stresses and angle of rotation for several section. Most of these formulas form
bending, shear and torsion.
a part of the new Australian concrete beams subject to torsion with due consideration to both
In the rigid space frame, the end moments of a loaded beam give rise to bending in the column
safety and economy.
and torsion in the beams which frame in the same joint at right angles to the loaded beam.
Cowan proposed a dual criterion of failure for the behaviour of concrete under stresses,
The problem of combined bending and shear with torsion arises essentially out of the monolithic
particularly combined bending and torsion. Although Cowan has contributed a great deal to the
character of reinforced concrete structures. In a beam and slab floor, any asymmetry of the
subject by extending his theoretical and practical investigations to rectangular reinforced and
loading from the slab produces torsion in the supporting beams, the extreme case being a
prestressed concrete sections, the number of tests performed by him in each case was limited.
continuous beam with alternate spans loaded.
The distribution of lateral loads to several shear walls depends on the rigidity of the floor and the rigidity of the shear wall. A rigid floor with flexible shear walls is one extreme
In seismic region where earthquakes are frequent and severe, torsional stresses occur in all parts of the framed structure as secondary stresses during the earthquake tremors.
case and a flexible floor with rigid shear walls is another extreme case. In the first case, the lateral force is distributed to the shear walls depending on their relative levels of rigidity. In a case where the floor is supported by three shear walls of equal rigidity, each of these walls carries a third of the lateral load. However, if the inner wall is not located at the centre, a torsion component is also developed figure 3.5-2.
Figure 3.5-2: The notation of balanced and unbalanced stability Figure 3.5-3
Many structural elements in building and bridge construction are subjected to significant torsional moments that affect the design. Spandrel beams in buildings, beams in eccentrically
For the box-girder beam the webs of a box girder must not only handle shear but also
loaded frames of multi-deck bridges, and box girder bridges are examples of such elements.
torsion and transverse bending. If as in the case of bridges constructed with the cantilever
The effects of the torsional moment are accounted for by superimposing the amount of
method, the compression chord is inclined, smaller tensile forces are produced for the
transverse and longitudinal steel and the intensity of the shearing stresses required for torsion
determination of the stirrups, but larger forces occur in the tension chord. The design of the web
resistance to those required for shear resistance.
for transverse bending has up until now normally been carried out separately for shear and torsion, and the reinforcement for each has been fully superimposed.
Structural system
STRUCTURAL ENGINEERING ROOM
concrete reinforced with (a) longitudinal bars, (b) longitudinal bars and rings and (c)
Department of Architecture
114
This force is equal to the shear flow cross-section, it is shear force per unit length corresponding
Torsion the closure of one or more cell-type cross-sections
STRUCTURAL ENGINEERING ROOM
Department of Architecture
115
to perimeter of the cross section. Then the average shear stress is inversely perimeter ofsectional thickness it is. T
t On the center line lengths ds account for then torque.
dMk
Tds
t
ds
The entire torque integration we get all around Figure 3.5-4
Mk t ds Because
Basic relations theory of torsion: From Hook's law
G
1
the relative angle of twist per unit length attributable Because dF
dMk
1
dFs
2
2 TFs
ds
t ds
t G ds
T
ds
2 dFs
a ds
Tds
t ds
t
t ds
T
Mk
2 dF
t G ds
Mk
Mk
2 F s 2 TFs
2
2 t Fs
F s
2
ds
G ds
1
G
s
2 F s
t Fs
Then for a moment torsional stiffness we get
then will
Mk GJk
G
Mk Jk
Jk
Mk
Jk
G
2 1 2 Fs
Fs 2
t Fs
4
ds
Hollow-chamber section Twisting a thin straight bar that corresponds to the relative angle of twist , formed by
And further
longitudinal shear force. Let us define the size of the force attributable to unit length of tendon T.
2 dFs
will be
2
G dF
2d dMk G F G Jk Where we have identified
Jk
T ds Tds
distance is investigated location of the center of torsion
2 F s
G
Where G is the shear modulus
dMk
ds
T ds
Mk Jk G
Structural system
Mk
ds
t G 4 F 2 s
ds t
Jk
Fs 2
4
ds t
ds
2 G
ds
F s
2 F s
If the cross-section of more chambers, the individual shear flows uncertain variables. We use them to determine the hydrodynamic analogy. If the node know two outlet quantities we can determine the amount of the third (shear force). For bypass (shear rate) of each chamber has use Derivatives of Formulas, wherein the relative twist each well the same as the ratio of the pitch of the entire cross section. So we obtain cross-section, from which we can calculate the individual shear flows T, torsional stiffness momentJ or relative twist angle . For each chamber will apply these relationships:
i
ds
2 G
F i
2 G
Mk GJk
Fi
2
Mk Jk
F i
Diameter Jk moment torsional stiffness of the entire cross-section. The picture drawing more pre-chamber cross section, and there are marked the shear flow A. Let's pick as the third chamber (i = 3) to the surface F i = F 3, we can write: D
A
ds D
B
ds A
C
ds B
2 G
D
A
B
C
1 1 1 1 ds T3 T2 ds T3 ds T3 T4 ds T3 t 3h t 23 t 3d t 34 C D A B 2 G
F 3
F 3
2
Mk Jk
2 G
F3
Mk
F 3 Jk After writing and receive treatment (removal of brackets) 2
1 1 1 ds T2 ds T4 ds T3 t t t
In this formula mean Mk torque applied to the cross section, Fi area closed chamber.
ds C
If we substitute for relationship = (T/t) then will be
2 G
F3
2
Mk Jk
F 3
The relationship can be generalized and such. for i - th chamber will be:
1 ds Ti 1 Ti t
1 ds T( i 1) ds t t 1
2 G
Fi
2
Mk Jk
F i
What we express in short hand as follows: F 3
Ti Where
1
t
ds
k
Tk 1 ds t
2 G
Fi
2
Mk Jk
Fi
The box girder i = 1,2 ..... n k = The cells adjacent to the box girder and, i.e. (I-1), (i + 1) i, k = considered, the walls of the chamber, and the chambers adjacent to the (i k = -1, k = i + 1) For n of cells, we can write n equations of type
1 1 ds T2 ds T1 t t Figure 3.5-5 D
ds C
A
ds D
B
ds A
C
ds B
2 G
F3
2
Mk Jk
F 3
2 G
F1
2
1 1 1 ds T1 ds T3 ds T2 t t t
Structural system
Mk Jk
F 1
2 G
F2
2
Mk Jk
F 2
STRUCTURAL ENGINEERING ROOM
More cell-type hollow section
Department of Architecture
116
STRUCTURAL ENGINEERING ROOM
Department of Architecture
117 1 1 ds T( n 1) ds Tn t t
2 G
Fn
2
Mk Jk
The condition is composed of the cross-section shape F n
n
Mk 2
Ti Fi
i1
n
Jk
If the label clarity integrals, which are dimensionless coefficients such as: ii
1 ds t
i = 1,2.......n
ik
1 ds t
4
i1 k = i -1, i+1
i i k
Then,
2 Mk
Jk
A quantity 2 G
2
Mk
Jk We can write this (modified, the order due to the unknown shear flows Ti) 11 T1
12 T2
( ) ( n n 1) Tn 1 n n Tn
G
F n
section, or the relative angle of twist unknown. We see that when we get chambers n + 1 unknowns. Because all equations contain a common factor 2 G = 2 (Mk / Jk) = of these equations we can determine the ratio of shear flows
Ti 2 G
Ti
Ti
Ti 2 G
( 1 )
2 G
E
E 2 ( 1 )
The reactions at the supports of box-girder bridges cannot be directly introduced into the webs in most cases and can also not be carried by means of transverse bending of the bottom flange. In addition, often room between the bridge bearings must be provided for hydraulic jacks needed to replace the bearing at a later time. The transverse diaphragm, which is favourable in stabilizing the cross-section under torsion anyway, can be used in solving this problem as well.
Jk
The forces in the diaphragm can be simply determined with the truss analogy, whereby the
2 Mk
transition between the truss of the diaphragm and that of the web must be maintained without
After solving the equations of them multiplying factor determine the actual shear flows, thus Ti
Where we took shear modulus
F 2
On the right side of the set of equations the torque stiffness to torsion Jk the whole cross-
Ti Ti
A relative angle of twist
F1
( ) 21 T1 22 T2 23 T3
Ti Fi
a break. With sufficient frame stiffness the torsional moment can be handled without a diaphragm.
2Mk Ti Jk
If 2 or more box girders 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
Quantities T´i we obtain the equations where the place Ti on the left we write T´i and on the
flanges are connected, they will be highly stressed due to bending moments in the transverse
right side will act only known value Fi (we actually shared factor )
direction without being able to effectively distribute the stresses in this direction. It is then better to separate the two box girder.
Structural system
fctm 1.92 MPa
Example 3.5-1: A simple example for predicting the ultimate strength and mode of failure of reinforced concrete beams subjected to pure torsion. Where the torsional strength is related to the amounts of transverse and longitudinal reinforcement and to the concrete strength. Torsional moments in reinforced concrete are typically accompanied by bending moments and shearing forces. An example of the design and assessment of reinforcement for the torsion sided cantilevered beam with a cross-section b(m), h(m) see figure 3.5.1-1. Extreme load causes in face of support bending moment M(kN.m) shearing force V (kN) and torsion moment T(kN m).
fcd 9.07 MPa
Department of Architecture
Failure due to torsion
Cross-sectional dimension: b 0.35 m
h 0.45 m
Beam span: l 4.35 m
Concrete cover: ast 0.03 m
Effective section height: d h ast
d 0.42 m
Torsion moment: Td 9 kN m
Uniform load: qd 73 kN m
Figure 3.5.1-1: Reinforced concrete beam subjected to torsion
1
Bending moment: Md 115 kN m
q 1
Shearing force: Vd 159 kN
Figure: 3.5.1-2
Md
0.205439319
2
0.07111
b d fcd
material characteristics fctkom 1.40 MPa
fcko 10 MPa
fck 16 MPa
fyk 375MPa
2
fctm
fck fctkom fcko
3
fcd 0.85
fck 1.5
fyd
fyk 1.15
Required reinforcement: Astd b d fcd
fyd 326.09 MPa
1 1 100 MPa
Structural system
Astd 9.478 cm
2
STRUCTURAL ENGINEERING ROOM
118
Torsional moments acting on a cross section of a beam cause shearing stresses, which circulate
STRUCTURAL ENGINEERING ROOM
Department of Architecture
119
near the periphery. For this reason, torsion design has typically been linked to shear design. Design of reinforcement: As1
Ast n As1
1
As1 0.000314159m
4
Ast 0.001256637m
stmax 0.03 b h
2
Vd
n 4
Vc
2
stmax 0.004725m
2
stmin
1 fctm bh 3 fyd
1 3
b h q fctm
Vumax
1 3
stmin 0.000308343m
c max 0.18
fcd q fctm
h
assd 0.00115311m
choose stirrups with area Ass s s 1 m
2
ass1d
4
Tc 8.933177419m kN
3 Td W t fcd
is less than 1, that is, the concrete cross section satisfies
0.568127251
Tc 8.933177419m kN
Td Tc
Abt 0.1131m
bt b 2 ast 2
ut 2 bt ht
ht 0.39 m
bt 0.29 m
Abt bt ht
ut 1.36 m
assd n
n4
longitudinally inserts As1d
Td
ut
2 Abt fyd
as1d
Td 2 Abt fyd
1m
as1d 0.000122016m
2
Design of cross-sectional area - additional reinforcement at the top (bottom) surface
ass1d 0.000576555m
2
ss
3 Vd
as1d
To one side of the beam (stirrups n = 2)
Ass1
W t k n fctm
It is greater than 1
2.588836802
ht h 2 ast
c max 0.383464041m
fcd c
n 2
3
Vd1 125.925kN
Vd Vc s s
assd
1
Proposal reinforcement for torsion
Vumax 476kN
Vd1 ( 0.5 l h) qd
Tc
Td 9 m kN
greater than 2.5 and Vc less than Qumax -To propose shear reinforcement
Vd 159kN
Td
b h fcd
Vc 100.55kN
b h fcd
Tc
Verification of the cross-sectional dimension 2
shear reinforcement, shearing force Carrying by concrete: Vc
k n 1.1
The reliability condition is not satisfied it is necessary to determine torsion reinforcement
2
1 20 mm
2 h 3 m
k n 1.4
Ass1 0.000628319m
2
2
as1d bt 2 s s 0.25 m Ass1 assd
ss 10 mm
satisfies
0.000059178m3
4
on the side of the beam as1d
Collapse due to torsion
ht
ht 2
0.000023793m
3
remains at the top surface W t
1 3 2
2
h
h b
W t 0.012721154m
3
Ast Astd 0.000308883m2
b
Structural system
is greater than
0.000064324 m
2
Ast Astd bt
ht
m
as1 6.37 cm
2
0.2 stmin bt s s 0.00000244m
4
2
10 mm
Ass1
4
is less than Ass1 0.79 cm 2
2
Checking conditions of reliability Stirrups on the side of the beam is designed 10 2
10 mm
As
A s 0.79 cm
4
2
as1
At the bottom side of the cross-section
asst ass ass1d
As
as1 0.000402768m
ht 2
asst 1.63 cm
2
from the longitudinal reinforcement we can determine the balance:
Ast Astd 3.09 cm 2
2
2
n 2
As
20 mm
4
That the condition
As 0.002166616m
bt
asst fyd
b) Stirrups for torsion
as1 fyd
Because inequality must be satisfied 0.5
Asst fyd ut 2.0 As1 fyd s st
asstd
Td 2 Abt fyd 2
m
then will be asstd 0.86 cm
asstd
Asst
Td
s st
2 Abt fyd 2
0.5
2 bt ht
2
Tu
Td Ast Astd m m fyd ass ass1d fyd ht 2 Abt fyd bt 2
Is greater than
m
Td 9 m kN
reinforcement as1 2 asstd
as1 1.73 cm
2
is less than
0.000187561 0.000332637
Stirrups for shear and torsion ass1d asstd 0.000662833m
2
10 mm
Stirrups a 1 m
at distances 0,19 m 2
ass
s s 0.19 m
a 4 ss
ass 4.13 cm
2
Assessment of the reinforcement for torsion s s 0.19 m
bt 0.29 m
stmin
1 fctm bt ht 3 fyd
stmin 0.00022142m
2
Structural system
Tu 21.38i m kN
The proposed reinforcement satisfies.
STRUCTURAL ENGINEERING ROOM
as1
Department of Architecture
120
Example 3.5-2: The reinforced concrete section is rectangular in shape with dimensions bt, ht
STRUCTURAL ENGINEERING ROOM
Department of Architecture
121
We want to assess the cross section for the interaction of torsion moment Td and shear force
Vd Vcu
Td
3.36
Tcu
Vd. It is greater than 1, that is, the tensile reinforcement is needed.
Material characteristics, concrete and steel: fcd
11.5 MPa
fctm
0.9 MPa
fyd
Verification of the cross-sectional size:
300 MPa
Cross-sectional dimension:
Vd
Section width: b
1 3
0.30 m
ast 0.025 m bt b 2 ast
1
Tcu
Concrete cover: bt
0.25m
ht
h 2 ast
Torsion moment: 1.4 kN m
ht
0.35m
3
2.4 kN m
Td
Ttq
0.5 Vd bt
Ttq
3
b h fctm
Vcu
36kN
Section modulus in torsion (rectangular section): 2
Wt
b h 3 2
b
Wt
0.008m
T cu
3
Td Tcu
12.5kN m
Td Ttq
Tcu
Td
Vtd
Ttq
116.8kN
The greatest shearing force of long-acting service load does not exceed 80% of the shear force from overall service load Vcu
Vcu
d
h
W t fctm
1.4kN m
Vcu
0.8
45kN
Vss
Vtd Vcu
3
Calculation torsion moment transmitted by concrete: 1
Vd 1 1.5
Vtd
Shear force transmitted by concrete: Vcu
W t fctm
Both conditions are fulfilled, then effect of torsion of the increased value of shear forces will be:
100 kN
1
3
0.26
W t fcd
Tcu
Shear force: Vd
1
Calculation torsion moment transmitted to the violation of concrete:
0.40 m
Td
b h fcd
Td
It is less than 1, the size of the cross section satisfies.
Section height: h
2.4kN m
h ast
d
c
0.375m
1.2
b fctm Vd Vcu
Vss
2
d
c
71.8kN
0.828m
Maximum distance double shear stirrups str
8 mm
Structural system
n
2
Ass
n str 4
2
Ass
0.000100m
2
Vss
s smax
Ass fyd c
0.347m
ss
Example 3.5-3: Spreading of box girders beams
0.30 m
The procedure for constructing lines to the influence of spreading of the cross-section:
Verification of the distribution and size of the stirrups: Ass
0.000050m
2
Ass 2
1 fctm 0.5 b s 3 fyd t s
2
a) calculate the deflection of structure from concentric acting unit force P = 1 - ordinates 0.0000375m
2
sagging w b) establishes the deflection lines ordinates of the applied bending moment M = P.y (where y is the distance from the center of the cross-section of torsion to the end of beam see
1 fctm 0.5 b s 3 fyd t s
figure 3.5.3-1, for that we want to construct a line to the influence of spreading of the cross-section) so that we find first relative angle of twist (torsional rigidity of closed
The longitudinal reinforcement at 1m:
cross-section) and through that we obtain ordinates w c) determine the resulting deflection lines of force P = 1 acting over respective beams
as1min
1 fctm b m 0.5 3 fyd t
as1min
0.000125m
which will be the sum of ordinate w´ and w´´ therefore w = w´ + w´´
2
d) by means of deflection lines we get the influence of spreading line (condition qi=1, where qi=1, are ordinates of the influence lines of spreading under each beam)
Perimeter ut
2 bt ht
ut
1.2m
q i
1
k w i
k
1 w i
a
q i
k w i
The longitudinal reinforcement: as1min
As1min
m
ut
As1min
0.00015m
Beam of one box girder of length L and cross-section dimensions shown in figure 3.5.3-1 is
2
given. The girder is torsionally fixed at both ends.
Proposal longitudinal reinforcement in cross-section: l
8 mm
A sl
l
Calculation of cross-sectional characteristics:
2
4
Asl
0.0000502m
2
Span: L = 25m
For the symmetric load, Vw is obtained by dividing the total shear by V/2. The web bending
The number of required longitudinal profile in cross-section: n
A s1min A sl
n
Modulus of elasticity of concrete: Ec = 32000GPa
moment Mw is obtained from the total bending moment M.
2.98
Suggestion 4 J 8 (one in each corner profile).
Figure: 3.5.2-1
Structural system
Figure 3.5.3-1: Cross-section of one-box girder
STRUCTURAL ENGINEERING ROOM
1
s smax
Department of Architecture
122
STRUCTURAL ENGINEERING ROOM
Department of Architecture
123 Fbi
Width of individual parts of the cross-section:
Sbi
1.95
b 1 13.00m
b 2 2.8m
b 6 6.40m
b 3 1.75m
h 1 0.15m
h 5 0.15m
h 2 0.2m
h1
r 1
r 1 0.075m
r 2 0.21667m
r 5 h 1
2
h 3 0.2m
b 5 0.4m
h 4 1.35m
h 6 0.15m
Fbi b i h i
h5
b 4 1.0m
0.56
r 2 h 1
2
h2
r 3 h 1
3
r 4 h 1
r 3 0.21667m
h 4 h 5 h 6
r 6 h 1 h 4
r 5 1.275m
h6
h3 3
h4 2
m
2
0.07583
0.21667
1.35
1.11375
0.825
0.06
0.0765
1.275
0.96
1.368
1.425
SBi
Jboi
0.01097 0.02629
m
4
0.01721 1.12388
0.09754
0.00008
0.09761
1.9494
0.0018
1.9512
2
6
Fb
6
Jbo1 0.00366m
4
Jbo3 0.00078m
1
b 1 h 1
Jbo4 0.20503m
1 3 Jbo5 b 5 h 5 2
36
Sb
Fbi
i1
4
Jbxi Jboi Fbi r i
Jbo6
4
36
4
1 12
b 6 h 6
Jbo6 0.0018m
1 3 Jbo3 b 3 h 3 2
Jbo5 0.00008m 2
3
m
4
3
3
4
SBi
4
Fb 5.23m
2
6
Jbx
Jbxi
i1 4
Jbx 3.23205m
4
Position of the neutral axis from the equation:
1 3 Jbo2 b 2 h 2 2
Jbo2 0.00124m
Sbr
i1
Sbr 3.01947m
Jboi
Jbo 0.21258m
4
36
6
Sbi
i1
6
12
0.02753
0.20503
i1
Jbo1
0.01462
0.00078
Moment of inertia: 3
4
0.91884
SBi Sbi r i
b 4 h 4
m
0.01643
Jbo 1
m
Jbxi
0.00366 0.00124
Sb 2.90167m
12
0.075 0.21667
0.35
Static moment of individual parts of the cross-section:
Jbo4
m
0.12133
3
r 4 0.825m
r 6 1.425m
Sbi Fbi r i
ri
0.14625
Fb t
Sb
t
Sb Fb
t 0.55481m
The resulting inertia we get from the relationship: Jbx
Jbt Fb t
Jb Jbt
Structural system
2
Jbt Jbx Fb t Jb 1.62217m
4
2
Jbt 1.62217m
4
Jk
span axis of symmetry of the cross section:
4 Fs
2
Jk 3.56342 m
hs bs 2 b4 h1
4
Considering
Mk 1 Jk
Gc
P 1 MN
3
Ec
4
Gc 15652173.91304 m
2 ( 1 )
2
0.15
kN
Figure: 3.5.3-2
P L f 48 Ec Jb 1
Mk 3.56342 m
f 0.00557m
Figure 3.5.3-4: Cross-section and loads, girder and equivalent torque Mk, torsional moments T Both ends of the girder are torsionally fixed. The torsional moments will be resisted by a combination of St.Venant and Warping torsion. Torsional moment T in a single-cell box girder are normally considered to be equilibrated by a state of pure shear. This phenomenon is called St.Venant torsion. The torsional shear stresses in the deck slab cantilevers are small relative to those in the walls of the box and are therefore neglected. Because the webs and slabs are thin relative to the overall box dimensions, the shear stresses can be assumed constant across the wall thickness. The torsional stresses can thus be expressed as a constant closed shear flow
Figure: 3.5.3-3
around the box, (kN/m). The value of is obtained from the equation of moment equilibrium about one of the corners of the box
Calculation ordinates deflection lines: Moment stiffness in torsion: bs b 6 0.5 b 4
hs h 4
Fs bs hs
bs 6.9 m
hs 1.35 m
h1
y
h1 0.15 m
b4 0.5 m
y 3.45 m
b4
b4 2
h1 h6 2 2
Fs 9.315 m
bs
2
= (T/2Ao )
Ao = b h
Relative angle of twist will be:
2
Mk Gc Jk
Structural system
2
0 m kN
1
STRUCTURAL ENGINEERING ROOM
Calculate the deflection of a simple beam from force P = 1 MN acting at the middle of the
Department of Architecture
124
STRUCTURAL ENGINEERING ROOM
Department of Architecture
125 Torsion in box girder bridges is rarely equilibrated by shear stresses alone.
Dual-chamber section or 2 box girder cross-section
and the angle of twist at mid-span
Example 3.5-4: Calculation of cross-sectional characteristics:
L
3
10 kN m 2
3
Span L = 35m
Modulus of elasticity of concrete: Ec = 36000GPa
0.0008 rad
Ordinates deflection lines from this rotation under the beam will be:
y
0.00276 m
The resulting ordinates we get by counting: l
f
l
0.00282 m
p
f
p
0.00833 m
Line cross to the influence of spreading of the affine with crease lines coefficient of affinity: ql l k
ql 0.25286 m
qp p k
Figure: 3.5.4-1
qp 0.74714 m
Width of individual parts of the cross-section:
check: ql qp 1 m We were based on the assumption that the cross-section is perfectly rigid and therefore also to the influence of spreading of cross-line has a linear course.
b 1 18.0 m
b 2 2.9 m
b 3 3.6 m
b 4 0.50 3 m
b 5 0.6 m
b 6 10.7 m
h 1 0.15 m
h 2 0.2 m
h 3 0.2 m
h 4 1.35 m
h 5 0.15 m
h 6 0.15 m
ah
b 1 2 b 2 2 b 3 b 4 2
ah 1.75 m
r 3 h 1
Figure: 3.5.3-5
Fbi b i h i
r 5 h 1
h3 3 h5 2
r 1
h1 2
r 3 0.21667 m
h 4 h 5 h 6
r 1 0.075 m r 4 h 1
r 2 h 1
h4 h6
Static moment of individual parts of the cross-section: Sbi Fbi r i
Structural system
SBi Sbi r i
3
r 2 0.21667 m
r 4 0.825 m
2
r 6 h 1 h 4
h2
2
r 6 1.425 m
r 5 1.275 m
Jbo6
1 12
Position of the neutral axis from the equation:
b 6 h 6
Jbo1 0.00506 m
3
4
1 b h 3 2 2 2 36 4
Jbxi Jboi Fbi r i Sbi 0.2025 0.12567
m
12
b 4 h 4
Jbo4 0.30755 m
Jbo2
Jbo2 0.00129 m
1
Jbo4
3
4
1 b h 3 2 3 3 36
Jbo3
Jbo3 0.0016 m
4
1
Jbo1
12
b 1 h 1
Jbo6 0.00301 m
1 b h 3 2 5 5 36
Jbo5 0.00011 m
Fbi
3
2.7 0.58
m
ri
2
0.075
0.825
0.11475
0.09
1.275
2.28712
1.605
1.425
SBi 0.01519
m
0.02723
4
0.00506 0.00129
0.0354
0.0338
0.0016
1.37827
0.30755
0.14642
0.14631
0.00011
3.26216
3.25915
0.00301
6
6
Fbi
Sb
i1
Fb 7.72 m
Sb 4.55667 m
Jboi
Jbx
Sbr
3
6
Jbo 0.31862 m
Jbxi
2
Jbt Jbx Fb t
2
Fb t
Sb
Jc Jbt
Jc 2.48903 m
4
Determine the deflection of a simple beam from force P = 1 MN acting in the middle of the
f
3
1 P L 48 Ec Jc
f 0.00997 m
finally: Moment stiffness in torsion:
m
4
SBi
Figure 3.5.4-2
Jbx 5.17856 m
bs 5.075 m
hs 1.35 m
Fs bs hs 1 1 1
Fs 6.85125 m
1
Sbr 4.85994 m
4
b4
i1 4
Jbt Fb t
for the calculation of shear flow, as the resulting stiffness and torsional twist angle proportional
i1
2
i1
Jbx
6
Sbi
i1
6
Jbo
The resulting inertia we get from the relationship:
m
Jboi
1.68581
Fb
4
Because this case is a two-chamber section, we have to write two equations with two unknowns
2.025
4
Jbt 1.62217 m
t 0.59024 m
axis of symmetry of the cross section:
0.21667
1.67063
m
Sb Fb
P 1 MN
0.21667
0.02852
4
2
0.72
0.02025
4
t
Jbo5
0.156
Jbxi
3
4
b4 3
b4 0.5 m
Structural system
b 4 1.5 m
1
1
y
2 bs
1
2
y 5.075 m
2
Fs Fs 2 1
Fs 6.85125 m
2
2
h1 h6 2 2
h1
2
h1 0.15 m
STRUCTURAL ENGINEERING ROOM
Moment of inertia:
Department of Architecture
126
To quantify factors
STRUCTURAL ENGINEERING ROOM
Department of Architecture
127
hs 1
11
2
12
b4
bs 1
h1
hs 1 b4
11
73.06667
22
11
12
2.7
21
12
22
21
73.06667
2.7 Figure: 3.5.4-3
Write conditional equation: 11T1
12T2
Proportional twist angle is:
11 12 T
Fs1
F
22T2
21T1
Fs2
Mk
rad m
0.00006 m Gc Jk A twist angle in the middle span:
After partitioning the equations factor and after the introduction of marking:
L 2 0.00106 ( rad ) m 2 Ordinates deflection lines from the rotation under the beams:
T´1
T1
11T1´m
T2
T´2
2
12T2´m
2
Fs
2
Jk
4
T
i1
Mk yP
0.15
2
Fs
T1´ 0.09736
1´ Fs i i
l
T2´ 0.09736
Jk 4 T1´Fs T2´Fs 1
f f
Jk 5.33658 m
l p
0.00457 m 0.01536 m
s
f
0.0054 m
Line cross to the influence of spreading of the extreme beam: k
2
0.0054 m
Ordinates the resulting lateral deflection in the center span from the force P = 1 MN:
2
p
T1´ Find T1´ T2´ T2´ n
21T1´m 22T2´m
1
y
2
1 m l
ps
k 33.43857
Ordinates to the influence of spreading of transversal lines: Mk 5075 m kN
Gc
Ec 2 ( 1 )
2 Mk Jk
1901.96887 m
Gc 15652173.91304 m
2
kN
1
kN
ql l k qp p k check:
ql 0.1529 m qp 0.51377 m
qp p k ql qp qs 1 m For middle beam
qp 0.51377 m
ql qp qs 0.3333 The result is correct.
Structural system
qs s k
qs 0.33333 m
Department of Architecture
3.6 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. 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 contains 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 shearwall 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.
Figure 3.6-2: Shear walls subjected to bending moment and vertical load 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.
Figure 3.6-1: Building with shear walls 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. 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.
Structural system
Figure 3.6-3
STRUCTURAL ENGINEERING ROOM
128
STRUCTURAL ENGINEERING ROOM
Department of Architecture
129 The system (Hull - Core Structures) has been used for very tall buildings in both steel and
greatest degree of protection against non-structural damage in moderate earthquakes, while
concrete. Lateral loads are resisted by both the hull and the core, their mode of interaction
assuring survival in case of catastrophic seismic disturbances, on account of their ductility.
depending on the design of the floor system.
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 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.
Figure 3.6-4: Shear walls subjected to horizontal load and vertical load
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.
Figure 3.6-6: Diagonal reinforcement in coupling beams, beam cross-section and possible mechanisms involving openings A single cantilever shear wall, such as the one shown in figure 3.6-7, can be expected to behave in the same way as a reinforced concrete beam. The shear walls will be subjected to bending moments and shear forces originating from lateral loads, and to axial compression induced by Figure 3.6-5: side view of shearing wall shows the thickness of bearing wall in accordance with boundary conditions of the members
gravity.
Normally, for wind loading, the governing design criterion or limit state will be
At the base of the wall, where yielding of the flexural reinforcement in both faces of the
deflection. Shear walls, when carefully designed and detailed, hold the promise of giving 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
Structural system
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.
Figure 3.6-9: Precast reinforced concrete walls
Figure 3.6-10: Shear subjected to lateral load Figure 3.6-7: Geometry and reinforcement of typical shear wall
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.
Figure 3.6-8: Geometry and reinforcement of shear wall in tall building
Figure 3.6-11: Shear walls with flexible coupling beams
Structural system
STRUCTURAL ENGINEERING ROOM
strength of the concrete. Sectional area of the concrete and should be equally divided between
Department of Architecture
130
STRUCTURAL ENGINEERING ROOM
Department of Architecture 131
Figure 3.6-12 Figure 3.6-13
Structural system
Example 3.6-1: Reinforced concrete wall subjected to horizontal load Ho or Wo Construction height
H = 27.5 m
Storey height
l = 2.75 m
Sectional area of the first pillar 1
and A1 or 1 = 2 m2
Sectional area of the second pillar 2 and A2 or 2 = 1.6 m2 Moment of inertia of the first pillar I1 = 4 m4 Moment of inertia of the first pillar I2 = 2 m4 Moment of inertia of the cross-sectional area Structures weakened openings I = 39 M4 Moment of inertia of girders IPR = 0.006 m4 Modulus of elasticity of pillars E = 10 GPa Modulus of elasticity of girder E´ = 20 GPa
Figure 3.6.1-2: Geometry calculated reinforcing walls subjected to horizontal loading Ho Data: E 10000 MPa 3
S 5.42 m
Walls weakened with openings S = 5.42 m3
1 1 S 1 2 c
I1 4 m 4
I 39 m
i 0.006 m
2
2 1.6 m
4
I2 2 m
l 2.75 m
c 3.049m
2
The distance between the center of gravity of the pillars 2c = 6.10 m
4
1 2 m
4
Static moment of sectional area
Shear force applied in base Construction Ho = 354 kN
2
E´ 20000 MPa
Width of the window type openings 2a = 2m Z 10 l
3 E´ i
E I1 I2
Z
I S
Ho l S
a 1 m
Ho 354 kN c
2
2
135.292kN
1
0.048m
3
I 0.219m
a l
6.016
Determination the value of 6.016
d
2
d
2
2
T( ) T( )
Figure 3.6.1-1: Shear walls contains openings
Structural system
2
( 1 )
T( 0)
0
T'( 1)
0
Odesolve ( 1)
STRUCTURAL ENGINEERING ROOM
Calculation of sectional forces and moments of structures
Department of Architecture
132
Diagram () vs
STRUCTURAL ENGINEERING ROOM
Department of Architecture
133
J
0.6
S I
I1 I2
2 c
v 2 e2
1 2
I1 I2 1 v 1 e1 1 2 c
3 1
J 1.893 10 m
0.54 0.48
i 2 12
j 1 11
0
1
0.42
i
i 1
0.1
( )
Ho l S I
kg
( )
0.36
Determination the values of M1 and M2
( ) 0.3 0.24
M 1( )
0.18 0.12 0.06 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
M 2( )
1
I1 I1 I2
Ho Z
I2
Ho Z I1 I2
(1 )2 S 2 c ( ) I 2 (1 )2 S 2 c ( ) I 2
where the value of the function can read from the graf, based on the coefficient for paying the relationship H ( )
j2
1
2
j
1 kN
j
(1 )
1
j
0
21.828
0
1
0.161
5·10-3
25.068
0.017
0.9
0.185
0.02
32.611
0.038
0.8
0.241
0.045
42.226
0.066
0.7
0.312
0.08
52.451
0.101
0.6
0.388
0.4
0.125
62.055
0.143
0.5
0.459
0.36
0.18
69.575
0.192
0.4
0.514
0.32
0.245
72.767
0.245
0.3
0.538
0.28
0.32
67.778
0.297
0.2
0.501
0.405
47.701
0.341
0.1
0.353
0.5
7.526·10-14
0.361
0
0
1
0 0.01 1
( ) d
Diagram () vs
0.24 ( ) 0.2 0.16 0.12 0.08 0.04 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2 c
1
v 1 30
tonne l
v 2 20
Solving the values of K and J
tonne l
e1 0.5 m
I1 I2 I1 I2 1 1 K v 2 e2 v1 e1 2 c I 2 c 2 1 S
e2 1 m
3 1
K 1.893 10 m
kg
S I
1
j
j2 2 c S 2
I
1 j
j
q
0
0
21.828
0.014
-9.383·10-3
25.068
0.032
-0.012
32.611
0.056
-0.011
42.226
0.085
-5.313·10-3
52.451
0.121
3.761·10-3
62.055
0.163
0.017
69.575
0.208
0.037
72.767
0.252
0.068
67.778
0.289
0.116
47.701
0.306
0.194
7.526·10-14
Structural system
kN
j
2
2
2 c m
I
( )
Ho l S I
2 c m
q ( ) ( ( 1 ) )
( )
I
Example 3.6-2: Solution of reinforcing concrete walls with openings subjected to vertical load
The reinforcing walls, as mentioned in the introduction, other than the horizontal load and the 1.5 10 110 M 1 j
510
6
810 610
6
M 2 j
5
0 510
410 210
vertical load are transmitted as well.
5 5
This chapter is about solving the stiffening of reinforced walls in terms of a vertical load defined
5
the basic assumptions that in dealing with all three types of reinforcing walls.
5
0
5
0
0.2
0.4
0.6
0.8
210
1
j
5
0
0.2
0.4
0.6
0.8
j
1
L - floor height H - total height of the wall A1A2 - cross-sectional area of each pillar
External load w acting on the structure induces in the individual pillars bending moments.
2c - distance between pillars 2 - width of openings
M1 1
j 0
M2 1 kN m
j
0
kN m
N - normal force acting in the pillar
shear force applied in the girders E - modulus of elasticity of the walls
-60.893
-30.447
-79.765
-39.883
-69.196
-34.598
-34.479
-17.239
24.407
12.204
112.802
56.401
243.358
121.679
E1 - eccentricity at which the load acts v1
440.758
220.379
750.922
375.461
e2 - eccentricity at which the load applied v2
1.258·103
629.068
E'- modulus of girders V1 - vertical loads on pillar 1 at level each floor v2 - vertical loads on the Pillar 2 at the level of each floor
Figure: 3.6.2-1
Structural system
STRUCTURAL ENGINEERING ROOM
(1 )
Department of Architecture
134
Data:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
135 Diagram () vs 2
E 10000 MPa E´ 20000 MPa 3
S 5.42 m
4
4
I 39 m
1 1 S 1 2 c
i 0.006 m
c 3.049m
2
4
2
I1 4 m
1 2 m
l 2.75 m
v 1 300 kN
4
2 1.6 m
Z 10 l
I2 2 m
a 1 m
Ho 354 kN
v 2 200 kN
e1 0.5 m
e2 1 m
( )
1
( ) d
0 0.01 1
0.4 0.36 0.32
Ho l S
135.292kN
I
I1 I2 I1 I2 1 1 K v 2 e2 v 1 e1 I 2 c 2 1 2 c
3 E´ i
E I1 I2
I S
0.28
S
c
2
2
0.048m
3
0.24 ( ) 0.2 0.16
1
K 52.06kN
0.219m
0.12
a l
0.08 0.04
Z
6.016
i 2 12
j 1 11
0
1
i
i 1
0
0.1
0
0.1
0.2
0.4
0.5
0.6
0.7
Figure: 3.6.2-3
( ) K ( ) M 1( )
Diagram () vs
d
6.016 Odesolve ( 1)
2
d
2
2
T( ) T( )
2
( 1 )
T( 0)
0
T'( 1)
M 2( )
0
I1 I1 I2 I2 I1 I2
Z l
Z l
( 1 ) v 1 e1 v 2 e2 2 c K ( )
( 1 ) v 1 e1 v 2 e2 2 c K ( )
Z N1 ( ) v ( 1 ) K ( ) l 1
0.6 0.54
Z N2 ( ) v ( 1 ) K ( ) l 2
0.48 0.42 0.36
Diagram () vs
( ) 0.3
1
0.24 0.18
( ) ( ) d
0.12
0 0.01 1
0.06 0
0.3
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure: 3.6.2-2
Structural system
0.8
0.9
1
j
1
0
0.81
j
39.606
(1 )
j
kN
1 j
1
-0.995
-0.099
39.57
0.9
-0.994
-0.199
39.447
0.8
-0.991
-0.298
39.194
0.7
-0.985
-0.396
38.714
0.6
-0.973
-0.492
37.829
0.5
-0.95
0.36
-0.585
36.21
0.4
-0.91
0.27
-0.673
33.251
0.3
-0.835
0.72 0.63 0.54 ( ) 0.45
0.18 0.09 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
-0.75
27.85
0.2
-0.7
-0.809
17.992
0.1
-0.452
-0.834
2.654·10-14
0
0
Department of Architecture
1
0.9
1
Figure: 3.6.2-4
6
510
5
5
410
5
310
110
5
810 M 1 j
Diagram () vs
610
M 2 j
5
410
5
2
d
K 1
2
0
110 0
0.2
0.4
0.6
0.8
T( )
K
T( 0)
0
T'( 1)
0
j
0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
M2 1
kN m
72.384
0.6
j
0
kN m
N1 1 j
0
N2 1 j kN
0.8
0
36.192
260.406
239.594
145.081
72.54
520.889
479.111
218.519
109.259
781.554
718.446
293.404
146.702
1.043·103
957.424
370.982
185.491
1.304·103
1.196·103
453.501
226.75
1.567·103
1.433·103
545.055
272.527
1.832·103
1.668·103
653.105
326.553
2.101·103
1.899·103
791.268
395.634
2.378·103
2.122·103
984.388
492.194
2.668·103
2.332·103
Figure: 3.6.2-5 ( 0.4) 0.91
0.4
Figure 3.6.2-6
0.1
0.2
Odesolve ( 1)
M1 1
0
0
j
1 0.9 0.8 0.7 0.6 ( )0.5 0.4 0.3 0.2 0.1 0
0
1
j
T( ) 2
5 5
210
d
5
210
( 0.5) 0.95
Structural system
kN
1
STRUCTURAL ENGINEERING ROOM
136
STRUCTURAL ENGINEERING ROOM
Department of Architecture
137
Diagram vs Diagram vs
Figure: 3.6.2-8 Figure: 3.6.2-7
Structural system
Diagram vs 1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
0
0,1
0,2
0,3
Figure: 3.6.2-10 Figure: 3.6.2-9
Structural system
0,4
STRUCTURAL ENGINEERING ROOM
Diagram vs
Department of Architecture
138
Designs must ensure adequate behaviour at ultimate limit state and under service conditions,
3.7 Pre-tensioned Prestress Concrete Beam
STRUCTURAL ENGINEERING ROOM
Department of Architecture
139
Prestressing is a special state of stress and deformations which is induced to improve structural behaviour. Structures can be prestressed either by artificial displacements of the supports or by steel reinforcement that has been pre-strained before load is applied. The forces induced by the former method are sharply reduced by creep and shrinkage and are generally ineffective at ultimate limit state. Due to these inherent disadvantages, support displacements are rarely used for prestressing. The forces induced by pre-strained reinforcement can, however, survive the effects of shrinkage and creep, provided the initial steel strain is sufficiently larger than the anticipated shortening in the concrete. The required pre-strains are best achieved using
both of which must be verified directly. Structures can be prestressed either by pre-tensioning or post-tensioning. 3.7.1 Pre-tensioning is used primarily for the prefabrication of concrete components. The prestressing steel is stressed between fixed abutments, forms are installed around the steel, and the concrete is cast. After the concrete has hardened, the prestressing steel is detached from the abutments. Anchorage of the steel, and hence the transfer of prestressing force from steel to concrete, is achieved entirely through bond stresses at the ends of the member.
high-strength steel. High-strength steel that has been pre-strained can normally be stressed to its full yield strength at ultimate limit state.
Figure:3.7-2 3.7.2 Pre-tensioning is usually more economical for large-volume precasting operations, since the costs of anchors and grouting can be eliminated. it is quite common,
Figure: 3.7-1
however, to combine both methods of prestressing in a given structures.
Prestressing can be full, limited, or partial. Full prestressing is designed to eliminate concrete tensile stresses in the direction of the prestressing under the action of design service loads, prestressing, and restrained deformations. In structures with limited prestressing, the calculated tensile stresses in the concrete must not exceed a specified permissible value. Behaviour at ultimate limit state must nevertheless be checked in both cases. Partial prestressing places no restrictions on concrete tensile stresses under service conditions. Concrete stresses need not, therefore, be calculated. Partial prestressing encompasses the entire range of possibilities from conventionally reinforced to fully prestressed concrete.
Prestressing with pre-strained steel reinforcement induces a self-equilibrating state of stress in the cross-section. The tensile force in the steel and the compressive force in the concrete, obtained from integration of the concrete stresses, are equal and opposite. In statically determinate structures, the two forces act at the same location in the cross-section. The sectional forces in the concrete due to prestressing can thus be easily determined from equilibrium. The following expressions are based on the assumption that the direction of the prestressing force deviates only slightly from a vector normal to the cross-section.
Structural system
The maximum tensile force in the tendons at tensioning should generally not exceed the
Figure:3.7-4
lower of the following values after transfer or prestressing to the concrete.
Losses occurring before prestressing (pretensioning)
ď łpo,max=0.75 fptk
The following losses should be considered in design
ď łpo,max = 0.85 fpo,1k The minimum concrete strength required at the time when tensioning takes place is given in the approval documents for the prestressing system concerned.
Where particular rules are not given, the time when prestressing takes place should be fixed
2. safety with respect to the compressive strength of the concrete 3. safety with respect to local stresses 4. early application of a part of the prestress, to reduce shrinkage effects.
c- Loss due to relaxation of the pretension tendons during the period which elapses between the tensioning of the tendons and prestressing of the concrete.
the permanent actions presents at tensioning.
1. deformation conditions of the component
b- Losses due to drive-in of the anchoring devices (at the abutments) when anchoring on a prestressing bed.
The initial prestress (at time t = 0) is calculated taking into account the prestressing force and
with due regard to:
a- Loss due to friction at the bends (in the case of curved wires or strands)
d- The prestressing force at a given time t is obtained by subtracting from the initial prestressing force the value of the time dependent losses at this time t. e- These losses are due to creep and shrinkage of concrete and relaxation of 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. h- Possible formation of small cracks in the anchorage zone may not impair the permanent efficiency of the anchorage if sufficient transverse reinforcement is provided. i- This condition is considered to be satisfied if stabilization of strains and cracks widths is obtained during testing.
Structural system
STRUCTURAL ENGINEERING ROOM
Figure:3.7-3
Department of Architecture
140
STRUCTURAL ENGINEERING ROOM
Department of Architecture
141
Figure:3.7-5 3.7.3 Partial prestressing is generally more economical than full or limited prestressing. Although structures that are partially prestressed require a significant portion of mild reinforcement for crack control and distribution, this steel contributes to the ultimate resistance of the section. Whatever mild steel is added to improve bahaviour under service conditions thus reduces the amount of prestressing steel necessary for safety at ultimate limit state. The prestressing force must always be carefully monitored during construction, since deviations from the prescribed prestressing force can led to cracking, deformations, and fatigue. 3.7.4 In post-tensioned construction, the prestressing steel is only stressed after the concrete has been cast and hardened. The steel must therefore be enclosed in ducts and anchored using special devices. The ducts are most commonly embedded in the concrete and filled with grout after stressing to bond steel to concrete and to provide protection against corrosion. The
Figure:3.7-7
ducts can also be located outside of the concrete section and left unbonded for the entire life of the structure. In the such cases, grout is only a means of protecting the steel.
Figure:3.7-8
Figure:3.7-6
Structural system
Department of Architecture
Tension and compression reinforcement in cross-section:
Example 3.7-1: Pre-tensioned Prestress Reinforce Concrete I Beam
Compression reinforcement:
asc 0.05 m
ast 0.05 m
Asc 0.001571 m
2
c
c 2 4
nc 5 Asc nc
20 mm
Tension reinforcement: t
16 mm
nt 6
t 2 4
Ast nt
Ast 0.001206 m
2
Figure 3.7.1-1: Static scheme Total span of the prestress reinforced concrete I beam: L =30 m
The space between the prestress beams:
Lx = 5 m
Cross-section near the support: hd1 0.15 m
hh1 0.15 m
Hp 1.1 m
hh 0.20 m
hd 0.20 m
app Hp Cp
app 0.9 m
dp Hp ast
dp 1.05 m
Hpk 1.1 m
hsp Hp hh hh1 hd hd1
hsp 0.4 m
Figure: 3.7.1-2
Hpk hh hh1 hd hd1 hsp
hs Hp hh hh1 hd hd1
Hpp hh hh1 hs hd1 hd
Hpp 1.1 m
Cross-section at mid span of the beam:
Cp 0.2 m
H 1.8 m
bs 0.20 m
bh 0.50 m
bd 0.50 m
aps H Cp
aps 1.6 m
ds H ast
ds 1.75 m
Hk 1.8 m
hs 1.1 m
Hk hs hh hh1 hd hd1
hs H hh hh1 hd hd1
hs 0.4 m
hsm H hh hh1 hd hd1 Area of prestress tendons: np 20 Ap np Ap1
Structural system
Ap1
hsm 1.1 m
4
2
2
2
5.5 6 5 mm 2
Ap 28.314 cm
2
Ap1 1.416 cm
STRUCTURAL ENGINEERING ROOM
142
Design value of concrete strength (MPa):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
143
Concrete:
fck 50 MPa bmax
0.0035
Ecm 35 GPa
fckcyl
fckcyl 0.8 fck
fcm fck 8 MPa
fcd 0.85
fckcyl 40 MPa
fcm 58 MPa
fcd 22.667 MPa
0.45 fckcyl 18 MPa
c
25 kN m
1.5
3
2 3
fckcyl MPa 10 MPa
fctm 1.4
fctm 3.528 MPa
Ecm
Ecd
1.5
Ecd 23.3333 GPa
Characteristic stress of reinforcement (MPa): fyk 510 MPa n
Es
s
1.15
Es 200 GPa
fyd
fyk s
fyd 443.478 MPa
n 5.714
Ecm
Characteristic stress in prestress tendon: fpk 1720 MPa
fpd
fpk s
fpd 1495.652 MPa
Ep 200 GPa
0.9
The maximum stress in prestress tendon at time = 0 at tensioning: fpd
fpk
fpd 1495.652 MPa
s
The maximum stress in concrete, hence the transfer of prestressing force from steel to concrete: pmax
fpk
pmax
1.15
1346.087 MPa
Load due to roof covering: Figure: 3.7.1-3
gstrs 0.5
Structural system
kN m
2
Lx
gstrs 2.5 m
1
kN
gstrd gstrs 1.5 gstrd 3.75 m
1
kN
hpanel 0.30 m
The distance of the extreme fibre from the neutral axis (m):
bpanel 1.2 m
gpanels hpanel bpanel
5 c 4
2
2
n 5
gpanels 8.961 m
1
gpaneld 12.097 m
gpaneld gpanels 1.35
bd hs 2 ch1 hd1 2
20 mm
hd 2 2 0.5 bs Hp bh bs hh bd bs hd Hp 2 1 2 0.5 bh bs hh1 hh1 hh bd bs hd1 hd1 hs hh 0.5 3 3 agcp Acp
kN
1
kN
agcp 0.541 m
2
bh hs 2 cd1 hh1 2
ch1 0.158 m
cd1 0.158 m
Perimeter cross sections - section in the middle and section near the support beam: u1 bd 2 hd 2 cd1 2 hs 2 ch1 2 hh bh
u1 3.232 m
u2 bd 2 hd 2 cd1 2 hsm 2 ch1 2 hh bh
u2 4.632 m
Cross-section at the support:
Area of cross-section near the support: Acp bs Hp bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1 Acp 0.385 m
2
Area of the transformed uncracked cross-section:
Aip Acp n Ast Asc Ap
Aip 0.417 m
2
Figure: 3.7.1-4
Structural system
STRUCTURAL ENGINEERING ROOM
self -weight of the panels SIPOREX:
Department of Architecture
144
The distance of the extreme fibre transformed cross-section from the neutral axis (m):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
145
Acp agcp n Ast dp Asc asc Ap app
agip
Section modulus at the lower chord of cross-section:
Wdp
agip 0.553 m
Aip
Iip Hp agip
Wdp 0.113 m
3
Moment of inertia of uncracked cross-section: Cross-section at the mid span of I beam: Icp
1
bs Hp bh bs hh bh bs 3
12
3
hh1
3
3
bd bs hd bd bs
bs Hp 0.5 Hp agcp bh bs hh agcp 2
bd bs hd Hp
2
hh
2
Acs bs H bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1 2
2
2
Area of transformed cross-section (at mid span of beam): Ais Acs n Ast Asc Ap
Ais 0.557 m
2
Area of cross-section:
2
Ap app agip 2
4
epp Hp agip Cp
epp 0.347 m
Section modulus at the upper chord of cross-section: agip
4
2
Iip
Area of concrete cross-section (at mid span of I beam)
3
Acs 0.525 m
Ac
Iip Icp Acp agip agcp n Asc agip ast Ast Hp ast agip
Whp
3
2
Moment of inertia of the transformed uncracked cross-section:
Iip 0.062 m
hd1
1 1 agcp bh bs hh1 agcp hh hh1 2 2 3
hd
1 1 bd bs hd1 Hp hd hd1 agcp 2 3 Icp 0.056 m
3
Whp 0.112 m
3
2
Ais Aip 2
Ac 0.487 m
2
The center of gravity of the concrete section from the upper edge (at mid-span):
hd 2 2 0.5 bs H bh bs hh bd bs hd H 2 1 2 0.5 bh bs hh1 hh1 hh bd bs hd1 hd1 hs hh 0.5 3 3 agcs Acs agcs 0.864 m
The center of gravity of the transformed cross-section from the upper edge:
agis
Acs agcs n Ast ds Asc asc Ap aps
Structural system
Ais
agis 0.883 m
3
3
hh1 hd1 1 3 3 3 Ics bs H bh bs hh bh bs bd bs hd bd bs 12 3 3
2 2 hd hh agcs bd bs hd H 2 2 2 2 1 1 1 hh hh1 bd bs hd1 H hd hd1 agcs 3 2 3
bs H 0.5 H agcs bh bs hh agcs 2
1 bh bs hh1 agcs 2
Ics 0.199 m
Iis Ics Acs agis agcs n Asc agip asc Ast H ast agip
Ap aps agip 2
Iis 0.229 m
2
Whs 0.26 m
agis
Iis H agis
Acp Acs 2
g0d g0s 1.35
c
kN
sd sk 1.5 1
gd gstrd gpaneld g0d sd
gd 44.703 m
sd 13.5 m
1
g0d 15.356 m
1
kN
kN
1
kN
The bending moment from the total load: Mk
1 8 1
8
2
Mk 3581.52 m kN
gs L
2
gd L
Md 5029.114 m kN
Mk1 2771.52 m kN Md1 3814.114 m kN
Shear force at the supports from total load: 1 gs L Vs 477.536 kN 2 1 Vd gd L Vd 670.549 kN 2 We calculate the value of initial prestress force ppo (kN) near the support at the lower side of the beam: Vs
kN
1
kN
The bending moment from the action of self-weight at the mid span of the I beam
1 2 g0s L 8
1
gs 31.836 m
hIp
Mg0k
sk 9 m
1 2 gstrs gpanels g0s 0.2 sk L 8 1 2 Md1 gstrd gpaneld g0d 0.2 sd L 8
3
g0s 11.375 m
Lx
Mk1
Self-weight of the prestress concrete beam:
g0s
m
2
The bending moment due to Combinations load
3
Wds 0.25 m
kN
gs gstrs gpanels g0s sk
eps 0.717 m
Section modulus at the lower edge of the cross section: Wds
Section modulus at the top edge cross-section: Whs
sk 1.8
Md
Iis
Mg0d 1727.578 m kN
Load due to snow:
4
eps H agis Cp
2
g0d L
The moment of inertia of the transformed concrete cross-section: 2
8
4
2
1
Mg0d
Mg0k 1279.688 m kN
c
pm0k Aip
pm0k epi
18 MPa
Structural system
Wdp
0.45 fckcyl
Ppo 100 kN
c
0.45 fckcyl
eps 0.537 m
STRUCTURAL ENGINEERING ROOM
The moment of inertia of the concrete cross-section:
Department of Architecture
146
The loss of prestress due to relaxation, is a function of the properties of the steel and the
Given
STRUCTURAL ENGINEERING ROOM
Department of Architecture
147
1.333 Ppo
c
Acp
1.333 Ppo epp Wdp
P1 fpd Ap1 P1 211.74 kN
po
Ppo Ap
po
Ppo Find Ppo
ratio of initial stress to tensile strength po/fpi. Relaxation progresses more rapidly than creep
Ppo 2386.451 kN
28 days. This same percentage of the final creep and shrinkage strain is achieved only after 90 days. In spite of this fundamental difference, a detailed calculation of the interaction between
Ppo
11.27 we suppose number of cables will be 14
P1
relaxation, creep, and shrinkage is usually not necessary. The interaction between relaxation, creep, and shrinkage can then be considered by calculating the loss of prestress due to creep and shrinkage using the following reduced initial prestress:
842.864 MPa
p = p0 – 0.5 p,rel,∞
We calculate the value of initial prestress force at the mid span of the beam in state of decompression at lower chord of the beam where the stress in lower part of the beam is σos = 0 kN:
Where p,rel,∞ denotes the long – term loss to relaxation.
Relaxation loss of prestressing tendons: tendon with a limit 0.2..... =0.9
dIs
and shrinkage in concrete. Whereas 50 percent of the final relaxation strain is reached at roughly
pm0k pm0k epi Mk 1 Ais Wds Wds
Pso 100 kN
eps 0.537 m
pd 0.266
0
decompression
Wds 0.25 m
3
0 MPa
Acs 0.525 m
2
Pso Find Pso
relative relaxation losses arising between tensioning and inserts by introducing pre-stressing in concrete: 1r
p p fpd
Relaxation loss of prestressing tendons:
Given decompression
pe 0.5
1.333 Pso Acs
1.333 Pso eps
Pso 2052.225 kN
Wds
so
Pso Ap
tendon with a limit 0.2..... =0.9
Mk1
pd
Wds
so
724.819 MPa
0.266
pe
0.5
Relative relaxation losses arising between tensioning and inserts by introducing pre-stressing in concrete: 1r
p p fpd
Time period 0-3 days: t 3 24 60 p
t 4320
fpk
fpd
pd 1 pe
Structural system
t0 0 24 60 p
0.113
t0 0
t1,t2 in minutes
0.18
pr2 ( 28 3)
1 7
log( t )
3.7.5 Loss of Prestress Due to Creep and Shrinkage
0.699
p
Time-dependent plastic shortening in the concrete due to creep and shrinkage causes
p p pm0
1r
p p fpd
1r
118.249 MPa
shortening in the prestress steel and thus a loss of prestressing force. The reduction in
1346.087 MPa
and prestressing. Several different cases must therefore be distinguished:
prestressing force due to creep and shrinkage is a function of the state of stress due to dead load r1
fpd 1r
r1
1377.403 MPa
pmax
Uncracked Section, Bonded Prestressing Steel. This is the most common case for
0.9 fpd 1346.087 MPa
pm0
pm0max
prestressed concrete bridges. The concrete at the level of prestressing steel is precompressed
As the maximum stress introduced into the tendon choose which is more than 1456 MPa: fpdtrans 1456 MPa p
pr1
transferred directly to the prestressing steel. The loss of prestressing force due to creep and shrinkage can therefore be calculated at a given section based on the compatibility of strain in
pd 1 pe
under dead load plus prestressing. It is assumed that any change in concrete strain can be
fpk
p
fpdtrans
p p fpdtrans
pr1
steel and concrete, assuming an appropriate creep law.
0.109
These concepts can be used to drive an expression for the loss of prestress, P, in a member subjected to initial prestressing force, P0, and external axial compression, N. Both forces act
110.873 MPa
concentrically the time-varying strain in the concrete, c(t), is computed from the following creep law.
r1
fpdtrans 1r
r1
1337.751 MPa
r1
0.9 fpd
0.9 fpd 1346.087 MPa
c = (co / Ec) (c / Ec) (1+cs Where is the concrete creep coefficient and cs shrinkage strain, the parameter can be taken as 0.8 the initial stress in the concrete, co, and the change in stress, c, are given by the following expression:
b) Time period 3-28 days:
co = ((-P) +N) / Ac
t2 28 24 60 p
t2 40320
fpk
r1
pd 1 pe
p
0.18
t1,t2 in minutes
p
0.095
c = (-P) / Ac Where Ac is the area of the concrete section, compatibility requires that p = c . P = p Ep Ap = c Ep Ap
1 1 log t2 0.18 log t1 7 7
r328 p p r1
t1 3 24 60 t1 4320
r328
17.61 MPa
p
0.139
Substituting for c and solving for P, the following expression is obtained: P = n ((co Ac cs Ec Ac) / (1+n ) (1+)) n = Ep / Ec and Ap / Ac.
Structural system
STRUCTURAL ENGINEERING ROOM
p
Department of Architecture
148
STRUCTURAL ENGINEERING ROOM
Department of Architecture
149 where equation for P also gives reliable results for prestressing losses in members subjected
t-ts - the actual duration of untreated shrinkage, respectively. swelling in days shrinkage
to both flaxture and axial load. In such cases, co and c are computed, at the elevation of the
coefficient during the time, we can determine from the relationship:
prestressing tendon and must include the flexural and axial components of stress. s25
Cracked Section, bonded prestressing steel. When cracks penetrate up to or beyond the
t25
2
ho 0.035 t25 mm
level of the tendons, the stress in the prestressing steel will be whatever value is necessary to
s25
maintain internal equilibrium, regardless of creep and shrinkage. This stress is always greater
than the initial prestress. The effect of creep and shrinkage is thus limited to additional
Strain from the effect of shrinkage:
shortening and deflections in the girder, without reduction in steel stress. The concrete stress-
cs
free at the level of the tendons and thus does not deform due to creep. The strain in the
t ts cso s25
cs
prestressing steel is likewise not reduced by shrinkage of the concrete at the level. Although the
t ts cs t ts Ep
0.107 ho - replacment element dimension in mm
cs
t ts
cs
0.00003375
t ts
6.751 MPa
stress in the prestressing steel is reduced somewhat by shortening of the flexural compression
Loss due to creep of concrete:
zone due to creep and shrinkage, this effect is small and can be neglected.
Time period 28 to ∞ days: Loss due to shrinkage of concrete:
s RH
cs
RH 100
1
0
for rapid hardening cement
8
3 s RH
0.488
RH
1.55 s RH
RH
0.756
RH
fcm 6 sfcm 160 cs 90 10 MPa
sfcm
0.000416
cso
2 Ac
u1 u2
t25 25
2
0.00031466
t ts
0.1 3 ho
t 28
ts 3
t25 t ts
fcm
16.8 fcm
coefficient which express the effect of age of concrete at the time of load introduction to the theoretical factor: Creep:
s
100
ho 0.248 m
1
RH
Coefficient which express the influence of strength on theoretical creep coefficient:
Replacement element dimension in mm, where Ac is the cross-sectional area: ho
fcm t0
1 RH
Theoretical coefficient of shrinkage: sfcm RH
t t0
Coefficient which express the influence of relative humidity on the theoretical creep coefficient:
Coefficient which expressed the influence of the strength of concrete on shrinkage:
cso
0 c
Theoretical coefficient of creep:
Relative humidity of environment: RH 80
t t0
t0
s25
Structural system
1
0.1 t00.2
RH
RH
1
fcm
1 100
The resulting equation for the creep coefficient:
t0 3
t 28 RH
3 ho 0.1 mm
16.8
fcm
fcm
1.031
c 25
t25
0
p c s r
s ( t t0) Ep
t0
t0T
9
2 t0T
for quick hardening cement with high strength = 1
1.2
0.5
1
pr2 n t t0 28 cp0
From the above interval we appoint an interval of Investigation:
a) Time Period 3-28 days:
Age of concrete in days at the time of application load: 9 t0 t0T 1 1.2 2 t0T
0.407
Ap Ac 2 zcp 1 0.8 t t0 1 n 1 Ic Ac
Age concrete regulation: t0T 28
t25 25
Stress due to relaxation, shrinkage and creep:
2.206
1
c 25
H t25
MPa
t25 t t0
0.3
t0 32.458
n
Es Ecm
c 25
n 5.714
Coefficient which express the age of concrete at the time of applying the load on the theoretical creep coefficient:
0.407
t 28
t0 3
cs
t t0
0.0000338
Stresses in the adjacent concrete from its self-weight, other permanent load and from the
prestress on an ideal cross-section at the middle span of the beam:
t0
1
0.1 t00.2
t0
0.475
Theoretical coefficient of creep: 0
RH fcm t0
0
pm0
1.08
1345.75 MPa
Time Coefficient of creep RH v % a ho v (mm)
Pm0k
During the time factor creep in HR and% h (mm) c
18
RH ho H 1.5 1 0.012 250 100 mm
H
621.561
fpk 1.15
Pm0k Ais
Ap
Pm0keps
Iis eps
Structural system
fpk 1.15
1346.087 MPa
Pm0k 3811.257 kN
Mg0k
Iis eps
c
8.638 MPa
STRUCTURAL ENGINEERING ROOM
Coefficient which express the influence of relative humidity on the theoretical creep coefficient:
Department of Architecture
150
Stress by relaxation, shrinkage and creep:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
151 Strain from shrinkage:
cs cs
t ts
0.00003375 c 25
n 5.714 Acs 0.525 m pcsr
2
Es 200000 MPa
0.407
c
4
r328
eps 0.537 m n c 25 c
Ap Acs 2 eps 1 0.8 c 1 A I cs cs
1 n
17.611 MPa
Ap 0.00283 m
8.638 MPa
Ics 0.199184 m
cs t ts Es
r328
cs
pm0 pcsr
p28
t ts cs t ts Ep
t ts
cs
0.00031466
t ts
g is the self-weight of the panel to be placed on the prestressed beam:
41.479 MPa
25
Np28 3692.864 kN
Mg0k
Np28 p28 Ap
Np28 3692.864 kN
Mp28 Np28 eps
Mp28 1983.068 m kN c28
1304.272 MPa
Np28 Ais
Mp28
Iis eps
Mg0k
1 8
gstrd gpaneld L
2
adjacent fibers of the concrete at t = 28 days.
for fast drying cement with high strength = 1:
t ( 28)
2 28
fpk
p28
pd 1 pe
p
1 0.18
pr2nekon
1 7
p
log t2
p p p28
skon
2
ho 0.035 ( 28) mm
1 1.2
t 32.458
coefficient:
p
0.162
pr2nekon
19.153 MPa
t0
1
0.1 ( t) 0.2
t0
0.475
Theoretical creep coefficient: 0
RH fcm t0
0
1.08
t
( 28)
Coefficient During the time factor creep: skon
1
4.101 MPa
Factor express concrete effect of age at the time of applying the load to the theoretical creep
0.091
Loss due to shrinkage of concrete:
ts 28
9
c28
Iis eps
p
1 2 gstrd gpaneld L 8
Np28 Ap p28
Loss of prestress tendon due to relaxation, therefore we have to determine the stress of the
t2 40320
62.932 MPa
It should also be determine the stress of the adjacent fibers to the concrete at t = 28 days
Stress in the tendon at t = 28 days the p28
cs
2
0.8 c 25 0.326
pcsr
t ts cso skon
H
1.5 1 0.012
Structural system
18 RH ho 250 100 mm
H
621.561
ckon
28 H 28
0
p
cs
pcsr
0.3 ckon
t ts Ep pr2nekon
1.08
Ac Ec
n ckon c28
0.75 fpk
0.9 p 1086.06 MPa
97.541 MPa
pcsr
1206.73 MPa
p
0.75 fpk 1290 MPa
fpk s
p 139.356 MPa
The total loss can be calculated from the difference stresses 0.9
0.9
fpk 1.15
0.9 p
The total loss of stress in tendon
0.9
tendon.
1.15
a 0.9 pnekon as
0.000326
2
Iip
Ecm
agip
2
0.0001488 0.0002916 0.0005539 0.00021182
o
1 2
o
0.00454
o
Ep o
0.00045422 Ecm
2
o
0.0003253 907.928 MPa
3.7.6 Ultimate limit state of failure due to bending moment
compressive strain reaches cu, or when the strain in the outer layer of reinforcing steel reaches s,max. These ultimate states of strain define the effective resistance of the cross-section.
fpk
0.193
19.3 % the total loss of stress in the
1. Plane section remain plane after deformation 2. Strain of tension and compression reinforcement and prestressing of concrete reinforcement,
1.15
fpk
1.15 maximum stress in tendon (see working diagram). thus, deformation will be:
spmax
Iip
gstrd gpaneld L
Calculation assumptions is based on the following assumption:
The deformation of the tendon, which will correspond to the stress 0.9
0.9
0.0001463
8
The ultimate resistance of the cross-section is achieved when the extreme fibre
fpk
260.029 MPa
Mg0k agip
Ap Acs 2 eps 1 0.8 ckon 1 n 1 A I cs cs
p28 pcsr
pnekon
1
Ppo
is equal to strain the fiber adjacent concrete interaction with the concrete is assured by bond. It represents the
3. Rectangular stress distribution of concrete in compression 4. The tensile strength of concrete can be neglected 5. Idealized stress-strain diagrams can be used for both steel and concrete, upper strand of the
fpk
diagram of prestress tendon, respectively. concrete reinforcement is horizontal, ie stress in the
1.15
spmax
Ep
0.00673 prestressing respectively. Reinforcing steel is limited to
and the deformation of the tendon after the losses will be:
0.9 fpk fyk , resp. , and Strain of steel s s
is not limited. sp
1
0.9 p
spmax
Ep
Ppo Ap Ep
1
sp 0.001
0.004
sp
0.005
Ppo 2386.451 kN
Ec 33500 MPa
Ppo eps Iip agip
0.0003265
Md 5029.114 m kN
Np p Ap
Np 0.9 Np
Np 3075.02 kN
Ecm
Structural system
Np 3416.689 kN
STRUCTURAL ENGINEERING ROOM
The resulting equation for the creep coefficient:
Department of Architecture
152
The area of compression concrete surface on the carrying capacity of the:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
153
1
Abc
Nb
Abc 0.168 m
fcd
2
hh 0.2 m
hh1 0.15 m
The area of the upper flange: Ahh bh hh
Ahh 0.1 m
2
or that x > hh1
The entire surface of the flange, including start-up:
Ah bh hh bs hh1
bh bs 1 hh1 2 2 2
x will be approximately as follows:
x
hh hh1
From the figure we determine the height of the compression zone of concrete H v 1.55 m spmax 0.007
Hv pmax sp bmax
0.0035
xmax 0.907 m
xu 0.8 xmax
sp
0.00425
bmax
Mb Abc fcd agis
0.0035
( H v ) 0.0035 xmax spmax sp bmax
M
Ap eps
MRd Mb M
Abc
0.907
xu 0.309 m
xu
Asc fyd agis asc
Mb 3356.501 m kN
2
M
395.359 m kN
MRd 3751.86 m kN
Mext 3377.828 m kN Mext MRd
So if x xmax this means that in all layers of the prestress tendon it should be the limitation 0.9 fpk . stress s
Nb Ap Np
Ah
xu 0.726 m
hh hh1 0.35 m
Moment on the load-bearing capacity:
Bending moment of the overall load
xmax
2
Abc
Figure 3.7.1-5: Diagram of the prestress tendon
v 0.25 m
area ratio will be
xu 0.8 x
x 0.386 m
Ah
Ah 0.153 m
Nb 3811.257 kN
Structural system
Mext Md Np eps
satisfies
NRd
Acp fcd
Acp
Force to the load-bearing capacity: NRd Acp fcd
Nsd fcd
Acp 0.206 m
2
NRd 4658.203 kN
The entire surface of the flange, including start-up Ad bd hd bs hd1
Figure 3.7.1-6: Cross-section of the mid span of the beam
Acp
Cross-section of the support:
0.742
hd hd1
x
Nsd 4658.203 kN
N0max 4234.73 kN
xu 0.8 x
x 0.472 m
Ad
2
xu 0.377 m
Acp
t tr, (the time of the transfer stress):
N0max Ap fpd
Ad 0.153 m
Ad t
hd hd1 0.35 m
i.e. that x is less than hd + hd1 and is greater than hd
Acp Ad
bd bs 1 hd1 2 2 2
NRd xbd fcd
Nsd 1.1 N0max
epp 0.347 m
NRd 5345.479 kN NRd Nsd
vyhovuje
Nsd 4658.203 kN
Cross-section of the middle span of the beam:
We are looking for a part of the entire cross section of I-profile, which has a center of
Nsd 1.1 Ap fpd
gravity at a distance from the epp-sectional center of gravity. Therefore that the force on the
Nsd 4658.203 kN
bearing capacity was on the same line as the force the load, which in this case is the force of
prestress. This requirement is fulfilled part of the cross section of a line in the distance 0.377m
Moment difference will be: M
from the lower chord of the cross-section.
Mpd Mg0k
M
Mpd Nsd eps
Mpd 2501.455 m kN
1221.77 m kN
The greatest moment in cross‐section can bear it if the force is applied Nsd
M Whs
Nsd Ais
0
M
Nsd Ais
Whs
M 2171.635 m kN M M satisfies section
3.7.7 Limit state stress limitation
Excessive compressive stress in the concrete for the any service loads can induce the formation of longitudinal cracks and may lead to the development of microcracks in the Figure 3.7.1-7: Cross-section of the support
concrete, respectively, greater creep. Therefore, it is necessary to adequately limiting this stress. In accordance with EC2 will check stress in element in quasi-permanent load, the value of which is a combination of
Structural system
STRUCTURAL ENGINEERING ROOM
Nbp
Nsd
Department of Architecture
154
a) Service permanent load
STRUCTURAL ENGINEERING ROOM
Department of Architecture
155
Stress at the top surface
b) Service prestress tendon c) multiple variable load
State prestressing t Mgo
t tr
Npm0 Ap fpd pr1
0 kN m
Npm0
h
3.859 MPa
Ais
Npm0eps
h
Whs h
Mg0k
Whs 0.45 fckcyl
Npm0 3920.809 kN Stress in the lower surface
agip agcp 0.012 m d
Stress at the top surface Npm0
h
h
2.32 MPa
Aip
Npm0epp agip agcp
fctm
d
Npm0 Aip
d
Npm0epp agip agcp
20.996 MPa
Satisfies
d
Wds
d
10.342 MPa
0.45 fckcyl
0.75 fpk 1290 MPa
p
1206.731 MPa
p
0.75 fpk
Figure: 3.7.1-8
3.7.8 Cracking limit state
The cross-section in the middle of the beam
Cracks must be limited so as not to disrupt the proper function of the structure,
t tr
respectively. unacceptable degradation of their appearance. Durability of prestressed elements
eps 0.537 m
0.45 fck 22.5 MPa
Npm0 Ap fpd pr1
Wds
The tension in the tendon and the offsetting of losses does not exceed further value 0.75 fpk
Wdp
0.45 fck
State prestressing t
Mg0k
Stress in the lower surface d
Ais
Npm0eps
0.45 fck 22.5 MPa Satisfies
Whp h
Npm0
Npm0 3920.809 kN
agis agcs 0.019 m
can be greatly affected cracks. If the requirements are not entered more precisely, it is required for the pretensioned prestress elements placed in an environment of class 2 to 4 fulfill decompression limit, which means that all the liners prestress tendon frequently charged in the combination of at least 25 mm inside the concrete compression. In our case, we can simplistically be required to lower the fiber cross-section was zero stress. Calculation of stress once again showcased the ideal crosssection.
Structural system
Decompression request is granted
a) Service load values
The cross-section of the support
b) any service loads of prestress force
Stress at the top surface:
c) times the value of variable loads = 0.2 for snow load
h
Cross-section of the middle span:
Np p Ap
d
Bending Moment for frequent load combinations: Mk1
gstrs gpanels g0s 0.2 sk L
Mk1 2771.52 m kN
1 2 gstrd gpaneld g0d 0.2 sd L 8
Md1 3814.114 m kN
8
Md1
2
Np Ais
Np eps Whs
Np epp Whp
h
2.38 MPa
h
fctm
Np Aip
Np epp Wdp
d
18.651 MPa
d 0.45 fckcyl satisfies 0.45 fckcyl 18 MPa The limit state of decompression is defined as the state where all axial concrete stresses are
below or equal to zero. Compressive stresses in concrete
Stress at the top surface:
h
Np Aip
Stress in the lower surface:
Np 3416.689 kN
1
Excessive compressive stress in the concrete under service load may lead to longitudinal Mk1 Whs
cracks and high and hardly predicable creep, with serious consequences to prestress losses.
9.741 MPa
h
When such effects are likely to occur, measures should be taken to limit the stresses to an appropriate level.
h
0.45 fckcyl
If the stress does not exceed 0.6fck
stress in the lower surface:
d
Np Ais
Np eps Wds
Mk1 Wds
0.45 fckcyl 18 MPa satisfies
1. Under the rare combination, longitudinal cracking is unlikely to occur 2. Under the quasi-permanent combination, creep and the corresponding prestress losses can be correctly predicted.
d
2.386 MPa
If under the quasi-permanent combination, the stress exceeds 0.4fck, the non-linear model shall be used for the assessment of creep.
d
0.45 fckcyl
Steel stresses
Tensile stresses in the steel under serviceability conditions which could lead to inelastic deformation of the steel shall be avoided as this will lead to large, permanently open cracks. Stress verifications should be carried out for partially prestressed members because there may be fatigue problems.
Structural system
STRUCTURAL ENGINEERING ROOM
Frequent load value is a combination of:
Department of Architecture
156
STRUCTURAL ENGINEERING ROOM
Department of Architecture
157 When prestressed and non-prestressed types of steel are simultaneously used, since the bond
Example 3.7-2: The cross - section dimensions of a beam shown on figure 3.7.2-1
behaviour of prestressing tendons is different from the bond behaviour of deformed reinforcing
are subjected to bending moment Msd . Determine the required tension reinforcement to the
bars, different steel stress will be developed in each type of steel.
cross-section.
Both equilibrium and compatibility should be respected for the calculated stresses in prestressing and reinforcing steel.
Characteristic value of concrete cylinder compressive strength (MPa):
For single cracks, different transmission lengths ls and lp should be calculated for reinforcing
fck 60 MPa
fckcyl 0.8 fck
fckcyl 48 MPa
and prestressing steel respectively. For reinforced or prestressed slabs subjected to bending without significant axial
Design value of concrete cylinder compressive strength (MPa):
tension, no special measures to control cracking are needed, provided that the overall depth of
fcd 0.85
the slab does not exceed 160mm.
fckcyl
fcd 27.2 MPa
1.5
fckcyl 10 MPa
0.66667
fctm 1.4
MPa
fctm 3.98374 MPa
Characteristic yield stress of reinforcement (MPa): 3.7.9 Deflection
Stress in any cross-section does not exceed the tensile strength of concrete, some sections remain even in compression, we can therefore count on the entire cross-section,
fyk 325 MPa
Ecm 33.5 GPa
because we do not expect cracking. When constructing buildings generally considered sufficient deformation of quasi-permanent load combinations and consider all load being
e
persistent. Np eps L 1 5 gstrs gpanels g0s sk L ftot 8 Ecd 384 Ecd Ics Ics 1 ckon 1 ckon 2
4
ftot 0.024 m
L 500
fm 0.06 m
ftot fm
Es 200 GPa
Ep 200 GPa
e 5.97015
hh 0.20 m hs 0.450 m
hh1 0.10 m
bd 0.60 m
h hh hs hd hh1 hd1
hd 0.20 m
hd1 0.15 m
h 1.1 m
Cover of reinforcement (m):
Limit state of deformation
dsc 0.05 m
In-service deformations (deformations and rotations) may be harmful to
dst 0.05 m
ap 0.15 m
from lower chord of the beam
Effective depth of a cross-section (m):
2. The integrity of non-structural parts
d h dst
3. The proper function of the structure or its equipment.
d 1.05 m
To avoid harmful effects of deformations appropriate limiting values should be respected.
Es
bs 0.30 m
1. The appearance of the structure
fyd 282.61 MPa
fpk 1620 MPa
Ecm
bh 0.6 m
deflection satisfies.
fyk 1.15
Cross-sectional dimensions (m):
Serviceability limit deflection think of as fm
fyd
Msd 2350 kN m
Structural system
Nsd 1200 kN
Tension
Np 4.2 MN
compression
2
Ast 10 ( 2cm )
The center of gravity of ideal cross-section (m):
Ast 0.003142m
4
2
2
Asc 0.003142m
4
The moment of inertia of the concrete cross section (m4):
2
Prestress tendon (m2):
2 2 Ap 1 ( 0.55cm ) 6 ( 0.5cm ) 4.5 7 4 4
LA 15.5 (1 fi 5.5+6 fi 5)
agi 0.56885m
Ai
Compression reinforcement (m2): Asc 10 ( 2cm )
Ac agc e Ast d Asc dsc Ap h ap
agi
Ap 0.004459m
Ic
2
1
3
12
3
bs h bh bs hh bh bs 2
hh1 3
3
3
bd bs hd bd bs 2
hd1 3
3
bs h 0.5 h agc
2
2 hd 1 1 bd bs hd h agc bh bs hh1 agc hh hh1 2 3 2 2 2 1 1 bd bs hd1 h hd hd1 agc 2 3
bh bs hh agc
Ic 0.06634m
hh
4
The moment of inertia of ideal cross-section (m4):
Ii Ic Ac agi agc Ii 0.07979m
2 e Asc agi dsc2 Ast h dst agi 2 Ap h ap agi 2
4
Mtot Msd Nsd agi
Ntot Nsd Np etot
Figure: 3.7.2-1
Area of concrete net cross-section (m2):
bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1 Area of concrete transformed cross-section (m2): Ac bs h
Ai Ac e Asc Ast Ap
Ai 0.55163m
Ac 0.4875 m
2
Ac
agc 0.54949m
Mtot 726.55533m kN
Ntot 3000 kN
Mtot
etot 0.24219 m
Ntot
Ast Ap Asc Ac fcd
MN 100 2 m
Mtot Ntot etot 0.08101
Mtot 726.55533m kN 0.21
Mu 2923.83m kN
for
hd 1 bd bs hd h 2 0.5 bh bs hh1 3 hh1 hh 2 bd bs hd1 hd1 hs hh 0.5 3 agc 2
Np h ap agi
2
Mu Ac d fcd
2
The center of gravity of the cross section of the concrete from the upper chord (m):
0.5 bs h bh bs hh
2
h
fyk 520 MPa
Mu Ac d fcd
Mu 4316.13m kN
s2 20.71986 MPa
0.8 fyk 416 MPa
0.31
Stress in Ast reinforcement (MPa):
s2
Ntot Ai
Mtot Ii
e
d agi
Structural system
Mu Mtot
ok
STRUCTURAL ENGINEERING ROOM
Tension reinforcement (m2):
Department of Architecture
158
Stress in prestress tendon (MPa):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
159
Ntot p Ai
p
Np Ap
c2
Ii h ap a gi
p 15.28323 MPa
is in tension
c2 0.60156 MPa
Ii
Nsd
Mtotcomp Msd Nsd agi p
p 957.11554 MPa
0.75 fpk 1215 MPa
Ntotcomp Nsd Np
h
Np h ap agi
2
is compression
Np 4200kN
Mtotcomp 771.79959m kN Ntotcomp 5400 kN
Cracks cannot be excluded even in prestressed concrete structures. Crack width and etotcomp
alternating stress from fatigue loading have to be limited to obtain adequate durability. Therefore the knowledge of steel stresses under service load is necessary.
c1comp
means ignoring the tension stresses in concrete between the cracks and at the ends of the cracks. In prestressed concrete beams, especially in case of partial prestressing the bond produced c1comp
tension stresses in concrete between the cracks cannot be neglected because the bond behaviour of normal reinforcing steel and of prestressing steel is very different.Therefore the stress s at the cracks is normal reinforcing steel is much higher than the increase sp of the stress of
etotcomp 0.14293 m
Ntotcomp
Ntotcomp Ai
Ntotcomp Ai
agi
Ii
1
Ai etotcomp
c1comp 15.29183 MPa
Ntotcomp etotcomp
c1comp 15.29183 MPa
Ii a gi
prestressing steel, although the different steel members are in one layer.
c2comp
Ntotcomp Ai
h agi
Ii
1
Ai etotcomp
c2comp
Nsd Np agi 1 Ai etot c1 Ai I i
c1 10.61855 MPa
Nsd Np Mtot
c1 10.61855 MPa
Ii
or
Ntotcomp Ai
Ntotcomp etotcomp Ii h a gi
c1 fctm
agi
Nsd Np Ai
h agi
Ii
1
c2comp 4.65106 MPa
or
Stress in concrete (MPa):
Ai
Mtotcomp
The steel stress in reinforced concrete beams may be calculated for “pure state II”, that
c2
fctm c2
If
c1
Mtot
Ai
h a gi
e
Mtot
Nsd Np
Ai etot
c2 0.60156 MPa
Figure: 3.7.2-2
Structural system
c2comp 4.65106 MPa
concrete C35 / 45 f ck, f ctm , Ecm. Reinforcement As1, As2, Es. prestressing tendon Ap, Ep. The cross-section is subjected for exceptional load combinations Bending Moment:
Msd = 3.55 MNm
Normal force:
Nd (tension) = 1.60 MN.
Value of prestressing force
Np (compression) = 5.2 MN
The initial prestressing force in a tendon is the force existing in this tendon at the end of the prestressing operation. The initial prestressing force on a prestressed element is obtained by considering all the forces existing in the tendons, at the end of the last prestressing operation. Under service load conditions the limitation of stresses may be required for
Figure 3.7.3-2: Cross-sectional dimension
1. Tensile stresses in concrete 2. Compressive stresses in concrete 3. Tensile stresses in steel.
H 1.2 m bs 0.20 m
The limitation of tensile stress in concrete is an adequate measure to reduce the probability of cracking.
ap 0.20 m as1 0.05 m bh 0.50 m bd 0.3 m
hh1 0.10 m hd1 0.10 m hd 0.30 m
as2 0.05 m hh 0.25 m
dp H ap dp 1 m hs 0.45 m
d H as2
hh hd hh1 hd1 hs 1.2 m Material Characteristics: Concrete 40/50: fck 40 MPa
fctm 3.5 MPa
Ecm 35 GPa fcd 0.85
Ep 200 GPa
fyk 520 MPa
Steel: Es 200 GPa 2
22 mm
As2 n A2
1
Figure 3.7.3-1: Values of ideal cross-section and stress distribution
16 mm
As1 n A1
Structural system
A2
2
2
1
fcd 22.667 MPa
fyk 1.15
fyd 452.174 MPa
A2 0.0003801 m
4
As2 0.0019007 m
A1
fyd
fck 1.5
2
2
2
4
As1 0.0010053 m
n 5
A1 0.0002011 m 2
2
n 5
STRUCTURAL ENGINEERING ROOM
Example 3.7-3: Partial prestressed beams according to figure 3.7.3-1 is proposed as,
Department of Architecture
160
STRUCTURAL ENGINEERING ROOM
Department of Architecture
161
1 ( 5.5mm) 2 6 ( 5mm) 2 Ap1 4 4 Ap1 0.0001416 m
The moment of inertia of the concrete cross section (m4): 3
np 25
Ic
2
bs H 0.5 H agc 2
Ap np Ap1
Ap 0.0035392 m
As Ap As2
As 0.0054399 m
2 2 hd agc bd bs hd H 2 2 2 1 1 bh bs hh1 agc hh hh1 2 3 2 1 1 bd bs hd1 H hd hd1 agc 2 3
2
2
bh bs hh agc
e
Es Ecm
e
5.71
Cross-section area of concrete (m2): Ac bs H bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1
Ac 0.365 m
3
hh1 hd1 1 3 3 3 bs H bh bs hh bh bs bd bs hd bd bs 3 3 12
Ic 0.0594 m
2
hh
4
The moment of inertia of ideal Cross section (m4):
Ideal cross-section area (m2):
Ai Ac e As1 As2 Ap
2 2 2 2 Ii Ic Ac agi agc e As1 agi as1 As2 H as2 agi Ap dp agi
Ai 0.4018 m
2
Ii 0.0689 m
The center of gravity of the concrete cross section from the upper extremity (m):
4
Calculation the values of Mtot and Ntot .
hd 2 2 1 0.5 bs H bh bs hh bd bs hd H 0.5 bh bs hh1 hh1 2 3 2 bd bs hd1 hd1 hs hh 0.5 3 agc Ac agc 0.5287 m The center of gravity ideal cross-section (m):
agi
Ac agc e As2 d As1 as1 Ap dp Ai
agi 0.5623 m
Ntot Np Nsd
Ntot 3600 kN
H a N d a gi p p gi 2
Mtot Msd Nsd Mu
b d d fcd As
0.255
Ac fcd
100 MPa
Mu 2109.7 m kN
Structural system
As
Mtot 1334.3774598 m kN
b d fcd
0.0657518
Mu Ac dp fcd Mu Mtot
ok
and durability of the structure.
H a N d a gi p p gi 2
Msd Nsd etot
It should be ensured that, with an adequate probability, cracks will not impair the serviceability
etot 0.3707 m
Nsd Np
c
f ctm , Where f ctm, is the mean value of the tensile strength of concrete. The section without
H a N d a gi p p gi 2
Mtot Msd Nsd
Mtot 1334.377 m kN
Ntot Nsd Np
Ntot 3600 kN
The ultimate resistance of partially prestressed structures is a function of the combined yield force of prestressing and reinforcing steel. The appropriate amount of prestressing to be
crack assumes the full effect of the concrete section and flexible behaviour of concrete reinforcement in tension and compression. There-fore we begin the calculating of the limit state stress limitation by evaluate the values of ideal cross-section. Because EC 2 considers the value of unity modulus for post-tensioning prestress concrete and reinforcement Es
provided, therefore, cannot be directly determined on the basis of behaviour at ultimate limit state. Supplementary criteria for the design of prestressing in partially prestressed structures are therefore required. These criteria, which will be called prestressing concepts, are established to account for aspects of serviceability, economy, and construction in the design of the
The upper edge: The limitation of compressive stresses in concrete should avoid excessive compression,
Es Ecm
term loading components in the calculation according to limit state stress limitation considered only in cases where more than 50% of the stresses induced Quasi-permanent load 2 1 Q k 1 simplifying
value e
15.
c1 19.844 MPa fctm 3.5 MPa This competence means cracks do not arise:
Calculation of stress in concrete extreme fibers of Cross section:
c1
e
Where Ecm the value of the secant modulus of elasticity of concrete. The effects of long
prestressing as they apply to various types of structures.
producing irreversible strains and longitudinal cracks. Nsd Np agi c1 1 Ai etot I Ai i
Ep,
.c1
fctm
Checking formation of cracks on the lower edge of Cross section c2
3.38 MPa
fctm 3.5 MPa
Limitation of concrete compression stress regarding the formation of longitudinal crack:
19.84 MPa
Using conditions of reliability, we can check this condition.
or
c
0.6 f ck
Satisfies the compressive stress in absolute value, this condition or not
c1
Ntot Ai
Mtot
Ii agi
c1
19.8436 MPa
fck Is the characteristic compressive strength of concrete. c c
The stress at lower edge:
H agi Nsd Np c2 1 Ai etot Ai I i
c1
c2
3.38 MPa
Ntot Ai
Mtot Ii Hagi
19.8436 MPa
0.6 fck 24 MPa
The reliability condition is satisfied. In prestress beam is assessment should be done
or
c2
is the greatest compression stress of concrete under exceptional load combinations or G k j P Qk 1 resp. G k j P 0.9 Q k 1.
especially in the stage of introducing preload into the concrete. It assessed the fulfillment of c2
the conditions of reliability for compressive stress in the concrete at the quasi-permanent load
3.38 MPa
combinations.
c
Structural system
0.45 fck
STRUCTURAL ENGINEERING ROOM
Distance resultant of horizontal forces from the center of gravity of ideal Cross section (m):
Department of Architecture
162
Example 3.7-4: The cross - section dimensions of a beam shown on figure 3.7.4-1
STRUCTURAL ENGINEERING ROOM
Department of Architecture
163 The required tension reinforcement A2 (m2) is as follows:
are subjected to bending moment Msd. Determine the required tension reinforcement to the
Astrequired b d fcd
section.
1 2 10 MPa
Material data: Characteristic value of concrete cylinder compressive strength (MPa):
fccyl 0.8 fck fcd 0.85
fccyl
fcd 11.333 MPa
1.5 Characteristic yield stress of reinforcement (MPa): fyk 412 MPa e
Es
fyd e
Ec
fyk
2
A1 0.0003142 m
Astprovid 0.0015708 m
Astprovid n A1
Ec 29 GPa
2
2
n 5
2
Tension force in reinforcement (kN):
fyd 358.261 MPa
1.15
A1
4 Provided tension reinforcement (m2):
Design value of concrete cylinder compressive strength (MPa): fck 25 MPa
20 mm
Astrequired 0.001457 m
Es 210 GPa
7.2413793
Fst Astprovid fyd
Fst 562.755 kN
x d Fc xu b fcd
x 0.193 m Fc 524.852832 kN
xu 0.8 x
xu 0.1544 m
Cross-section (m), cover of reinforcement (cm), effective depth of a cross-section (m): b 0.30 m d h ast
h 0.45 m
asc 3 cm
ast 3 cm
d 0.42 m
Figure: 3.7.4-1 Design maximum bending moment Msd (kN.m), to apply the Design figure B3-B3.3 the bending moment Msd has to be brought into a dimensionless, we obtain the reinforcement ratio: Msd 180 kN m
Msd 2
b d fcd
Figure: 3.7.4-2
0.30012
0.10203
0.45943
Fst Fc
Ascrequired 0.0001058 m fyd Diameter, number of bars, provided: Ascrequired
12 mm
Structural system
A1
2
4
2
A1 0.0001131 m
2
n 3
Ascprovid n A1
Ascprovid 0.0003393 m
Fsc Ascprovid fyd
Fsc 121.555 kN
to bending moment Msd. Determine the required tension reinforcement to the section.
2
Material data: Characteristic value of concrete cylinder compressive strength (MPa):
Mu Fc d
Design value of concrete cylinder compressive strength (MPa):
xu
Fsc d asc 2
Mu 227.3342917 m kN
Mu Msd fck 20 MPa
Msd 180 m kN
Ai b h e Astprovid Ascprovid
Ai 0.1488317 m
2
agi
b h 0.5 h e Ascprovid asc Astprovid h ast
agi 0.2366841 m
Ai
fccyl 0.8 fck
fcd 0.85
fccyl 1.5
fcd 9.067 MPa
Characteristic yield stress of reinforcement (MPa): fyk 412 MPa
fyd
fyk 1.15
fyd 358.261 MPa
Cross-section ( m), cover of reinforcement (m), effective depth of a cross-section (m): 4
Moment of inertia of the transformed cross-section about neutral axis ( m ): 3 1
Ii b h
12
h a A 2 2 gi e scprovid agi asc Astprovid h ast agi 2
b h
Ii 0.0027838 m
b 2.63 m d h ast
2
4
asc 3 cm
h 0.50 m d 0.47 m
ast 3 cm
Design maximum bending moment (kN.m), to apply the Design, figure B3-B3.3 the bending moment Msd has to be brought into a dimensionless form, from the Design we obtain the reinforcement ratio:
Compression and tension concrete stress (MPa):
Msd 198 kN m Msd
Ii agi
Msd
15.3041944 MPa
Ii
13.7931827 MPa
h agi
Msd 2
b d fcd
0.03759
0.0103
2 The required tension reinforcement A2 ( cm ) is as follows:
Astrequired b d fcd
Figure: 3.7.4-3
Structural system
1 2 10 MPa
Astrequired 0.0011544 m
2
0.05104
STRUCTURAL ENGINEERING ROOM
Example 3.7-5: The cross - section dimensions of a beam shown on figure 3.7.5-1 are subjected
Provided compression reinforcement:
Department of Architecture
164
Reinforcing steel (MPa):
STRUCTURAL ENGINEERING ROOM
Department of Architecture
165
fyk 412 MPa
fyd
fyk
fyd 358.261 MPa
1.15
It is assumed that (m), cover of reinforcement (m), effective depth (m):
b 0.45 m d h ast
asc 3 cm
h 0.45 m d 0.42 m
ast 3 cm
Design value of bending moment (kN.m), design value of axial compression force (kN), combined actions have to be transformed in relation to the centre of gravity of the tension reinforcement (kN.m):
Nsd 2800 kN compression
Msd 50 kN m
Figure: 3.7.5-1
h Mext Msd Nsd ast 2
Mext 596 m kN
To apply the design diagram figure B3-B3.3 the transformed actions Msds has to be brought into a dimensionless form (-):
Mext
2
b d fcd
Example 3.7-6: Determine the tension reinforcement of a reinforced concrete rectangular
Use the bi - linear diagram for steel ( ´
0)
and bi - linear diagram for concrete.
entire cross-section is in compression
Figure: 3.7.5-2
column shown on figure 3.7.6-1, subjected to bending moment Msd and compression force Nsd.
0.82811
Fc 0.8 b d fcd
Fc 1370.88 kN
Fstrequired Nsd Fc
Fstrequired 1429.12 kN
2
The required tension reinforcement A2 ( cm ) is as follows:
Material data: Concrete (MPa): fck 20 MPa
fccyl 0.8 fck
fcd 0.85
fccyl 1.5
Astrequired fcd 9.067 MPa
Structural system
Fstrequired fyd
Astrequired 0.003989 m
2
location is reduce the possibility of the reinforcement buckling with the resulting inability to take the load for which the column is supposed to be designed. Such buckling is prevented by lateral reinforcement in the form of ties or closely-spaced spirals. The minimum number of longitudinal bars in a tied column should be four and the minimum diameter of bar is 12mm.
Example 3.7-7: Determination of the tension reinforcement of reinforced concrete rectangular column shown on figure 3.7.7-1, subjected to bending moment Msd and compression force, Nsd by
the bi - linear diagram for steel ( ´
0)
and bi - linear diagram for concrete.
Material data: Concrete (MPa): fck 20 MPa
Figure: 3.7.6-1
22 mm
Astprovid n A1
A1
fyk 412 MPa
2
4
A1 0.0003801 m
Astprovid 0.0045616 m
2
n 12
2
Astprovid b d
0.0241354
fcd 0.85
fccyl
Reinforcing steel(MPa):
Provided, diameter, number of bars:
fccyl 0.8 fck
fyk
fyd
1.15
It is assumed that (m):
1.5
fcd 9.067 MPa
fyd 358.261 MPa
asc 3 cm ast 3 cm b 0.40 m h 0.45 m d h ast d 0.42 m Design value of bending moment (kN.m), design value of axial compression force (kN), Combined actions have to be transformed in relation to the centre of gravity of the tension
max
reinforcement (kN.m):
Columns are usually under compression and they are classified as short columns or long and slender columns. Long columns are liable to buckle under axial load. The length of the
Msd 150 kN m
h
Mext Msd Nsd
2
ast
Nsd 1500 kN compression Mext 442.5 m kN
column is the distance between the supports at its end or between any two floors. The effective
To apply the Design figure B3-B3.3 the transformed actions M sds has to be brought into a
length of a column is governed by the condition of fixity at its ends in position and in direction.
dimensionless
Codes give the values of the effective length for a number of cases by multiplying the actual
form (-):
length by a factor. Columns may also loaded eccentrically or there may be a bending moment imposed on them in addition to concentric (axial) and eccentric loads.
Mext 2
Reinforcement in the form of longitudinal bars is provided both in short and long columns.
b d fcd Fc 0.8 b d fcd
These reinforcements are necessary to withstand both tension and compression loads. The most
Fstrequired Nsd Fc
Structural system
0.69168
entire cross-section in compression
Fc 1218.56 kN Fstrequired 281.44 kN
STRUCTURAL ENGINEERING ROOM
efficient location of the longitudinal reinforcement is near the faces of the columns. Such
Department of Architecture
166
Example 3.7-8: Determination of the tension reinforcement of reinforced concrete rectangular
STRUCTURAL ENGINEERING ROOM
Department of Architecture
167
column shown on figure 3.7.8-1, subjected to bending moment by the bi - linear diagram for steel ( ´
0)
M sd and
compression force
and bi - linear diagram for concrete.
Material data: Concrete (MPa): fck 20 MPa
fccyl 0.8 fck
fcd 0.85
fccyl
fcd 9.067 MPa
1.5
Reinforcing steel(MPa): fyk 412 MPa
fyd
fyk
fyd 358.261 MPa
1.15
It is assumed that (m), cover of reinforcement (m), effective depth (m): b 0.20 m d h ast
asc 3 cm
h 0.35 m
ast 3 cm
d 0.32 m
Design value of bending moment (kN.m), design value of axial compression force (kN): Combined actions have to be transformed in relation to the centre of gravity of the tension
Figure: 3.7.7-1 The required tension reinforcement Astrequired
A2
Fstrequired
reinforcement (kN.m):
2
( cm ) is:
Msd 8.7 kN m
Astrequired 0.0007856 m
fyd
2
16 mm
A1
Astprovid n A1
A1 0.0002011 m
4
2
n 4
Mext 2
b d fcd A2
Astprovid 0.0008042 m
0.34359
The required tension reinforcement
( cm 2) is: 2
Astprovid Astrequired
Astrequired b d fcd
M sds has
to be brought
into a dimensionless form (-), from the design diagram figure B3-B3.3 we obtain:
2
The provided tension reinforcement
Mext 63.8 m kN
To apply the design diagram figure B3-B3.3 the transformed actions
Provided, diameter, number of bars:
Nsd 380 kN
h Mext Msd Nsd ast 2
16 mm
Structural system
A2
Nsd 1 fyd MPa
A1
2
4
0.12579
0.554403
( cm 2) is: Astrequired 0.0719311 m A1 0.0002011 m
2
n 4
2
Nsd
Astprovid 0.0008042 m
Astprovid n A1
Ascprovid 0.0002262 m
Ascprovid n A1
( cm 2) is:
A2
2
Fsc Ascprovid fyd
2
Fsc 81.0366995 kN
Mu Fc d
xu
Fsc d asc
Mu 87.593 m kN
2
Ai b h e Astprovid Ascprovid
b h 0.5 h e Ascprovid asc Astprovid h ast Ai 3 1
Ii b h
12
etoti
Astprovid b d
x d
Fc xu b fcd
12 mm
Fst Astprovid fyd
xu 0.8 x
x 0.1774 m
Fc 257.3612646 kN
A1
2
4
4
Mext Nsd
etoti 0.1679 m
Fst 288.1304873 kN
A2
Ascrequired
Fst Fc fyd
Nsd agi 1 Ai etoti Ai Ii
c2
h agi Nsd 1 Ai etoti Ai Ii
2
c1
8.5532 MPa
c1
fctm
( cm 2) is:
A1 0.0001131 m
c1
xu 0.1419 m
2
The provided compression reinforcement
2
Compression and tension concrete stress (MPa):
0.0125664
Ascrequired 0.0000859 m
agi 0.1828 m
h a A 2 2 gi e scprovid agi asc Astprovid h ast agi 2
b h
Ii 0.0008667 m
The reinforcement ratio is given by (‐):
2
agi
Figure: 3.7.8-1
Ai 0.0774618 m
Mu Msd
n 2
Structural system
c2
17.2109 MPa
c2
fctm
STRUCTURAL ENGINEERING ROOM
The provided tension reinforcement
Department of Architecture
168
Example 3.7-9:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
169 Example 3.7-10:
Material data:
Concrete (MPa):
fccyl 25 MPa
fcd 0.85
Reinforcing steel(MPa): fyk 325 MPa 2
25 mm
A2
Ast n2 A2 1
2
fyd
2
A1
Asc n1 A1
1
fyk
are subjected to bending moment Msd, Np (compression), Nsd (compression). Determine the
fcd 14.167 MPa
1.5
ultimate bending moment and stresses at the upper and lower of cross-section.
fyd 282.609 MPa
1.15
A2 0.0004909 m
4
2
n2 4
A1 0.0000503 m
Asc 0.0002011 m
2
fyk 520 MPa
n1 4
20 mm
A2
Ast n A2
1
Figure: 3.7.9-1 The provided compression and tension reinforcement
h 110 cm
ast 5 cm
b=0.20m
A2
16 mm
asc 5 cm
d h ast
To apply the design diagram figure B3-B3.3 the transformed actions
Ast Ac fcd
100 MPa
Mu Ac d fcd
Msd Ac d fcd 0.047793
1
A1 0.0002011 m
2
2
Asc 0.0008042 m
4
2
M sds has
Ast
b d fcd
2
6 ( 5mm) 4 2
Ap1 0.0001416 m
Ap np Ap1
Ap 0.0033976 m
As Ap Ast Asc
As 0.0054585 m
2
np 24
2
2
Cross-sectional dimension:
Ac 0.29 m
Mu Msd
n 4
to be brought
2
0.135
Mu 582.35625 m kN
n 4
2
4
A2 0.0003142 m
d 1.05 m
into a dimensionless form (-), from the design diagram figure B3-B3.3 we obtain:
2
2
Msd 365 kN m
fyd 452.174 MPa
1.15 2
fcd 19.833 MPa
1.5
4
A1
1 ( 5.5mm) 2
Ap1
( cm 2) is:
fccyl
fyk
Ast 0.0012566 m
Asc n A1
As 0.0021646 m
fyd
2 2
As Asc Ast
fcd 0.85
Steel:
2
4
Material Characteristics: fccyl 35 MPa
2
Ast 0.0019635 m
8 mm
fccyl
The cross - section dimensions of a beam shown on figure 3.7.10-1
ok
ap 0.15 m
asc 0.05 m
ast 0.05 m
bs 0.40 m
bh 0.60 m
bd 0.4 m
hh 0.30 m
hs 0.90 m
hd 0.30 m
hh1 0.10 m
hd1 0.05 m
H hh hd hh1 hd1 hs
Structural system
H 1.65 m
dp H ap
dp 1.5 m
The moment of inertia of ideal cross-section:
d 1.6 m
Gross area of concrete cross-section:
2 2 2 2 Ii Ic Ac agi agc e As1 agi as1 As2 H as2 agi Ap dp agi
Ac bs H bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1
Ii 0.2027 m
4
2
Ac 0.73 m Area of the transformed uncracked cross-section: Ai Ac e Ast Asc Ap
Msd 4350 kN m
Ai 0.7612 m
Ntot Np Nsd
hd 2 2 1 0.5 bs H bh bs hh bd bs hd H 0.5 bh bs hh1 hh1 hh 2 3 2 bd bs hd1 hd1 hs hh 0.5 3 agc Ac
H a N d a gi p p gi 2
etot
Distance of the neutral axis (centroid) of the transformed cross-section:
agi 0.7852 m
Ai
3
Mtot
3
2
hd hh bh bs hh agc agc bd bs hd H 2 2 2 1 1 bh bs hh1 agc hh hh1 2 3 2 1 1 bd bs hd1 H hd hd1 agc 2 3 Ic 0.1821 m
As
b d d fcd
b d fcd
As Ac fcd
100 MPa
Mu 5537.962 m kN
0.0377013 0.255
Mu Mext
ok
Stress in concrete:
2
2
etot 0.0736969 m
Ntot
hh1 hd1 1 3 3 3 bs H bh bs hh bh bs bd bs hd bd bs 3 3 12
bs H 0.5 H agc
Mtot 545.357006 m kN
Mtot
Mu Ac dp fcd
The moment of inertia of the concrete cross-section: Ic
Ntot 7400 kN
Mtot Msd Nsd
agc 0.7628 m
Ac agc e Ast d Asc as1 Ap dp
Compression Nsd 2200kN Compression
2
Distance of the neutral axis (centroid) of the net cross-section:
agi
Np 5200 kN
c1
Nsd Np Ai
agi
Ii
1
Ai etot
c1
11.83 MPa
or c1
Ntot Ai
Mtot
Ii agi
4
Structural system
c1
11.8337 MPa
STRUCTURAL ENGINEERING ROOM
d H ast
Department of Architecture
170
Example 3.7-11: The cross - section dimensions of a beam shown on figure 3.7.11-1
The lower edge
STRUCTURAL ENGINEERING ROOM
Department of Architecture
171
are subjected to bending moment Msd, Np (compression), Nsd (tension). Determine the c2
Nsd Np Ai
H agi
Ii
1
Ai etot
ultimate bending moment and concrete stresses at the upper and lower of cross-section. c2
7.4 MPa
Material Characteristics Concrete:
or c2
N tot Ai
fccyl 35 MPa
Mtot
c2
Ii
7.4 MPa
fcd 0.85
Steel:
fyk 520 MPa
Hagi
2
20 mm
fyd
A2
Ast 0.0012566 m
1
16 mm
A1
Figure: 3.7.10-1
fyd 452.174 MPa
2
A2 0.0003142 m
4
2
n 4
Ast n A2
1 4
2
A1 0.0002011 m
2
n 4
Asc n A1
2
1 ( 5.5mm) 2
fyk 1.15
2
fcd 19.833 MPa
1.5
2
Asc 0.0008042 m Ap1
fccyl
4
6 ( 5mm) 4 2
Ap1 0.0001416 m
Ap np Ap1
Ap 0.0033976 m
As Ap Ast Asc
As 0.0054585 m
2
np 24
2
2
Cross-sectional dimension:
ap 0.15 m
asc 0.05 m
ast 0.05 m
bs 0.150 m
bh 0.40 m
bd 0.3 m
hh 0.20 m
hs 0.45 m
hh1 0.10 m
hd1 0.05 m
hd 0.20 m
Structural system
H hh hd hh1 hd1 hs
H 1m
The moment of inertia of ideal cross-section:
dp 0.85 m dp H ap Gross area of concrete cross-section:
2 2 2 2 Ii Ic Ac agi agc e As1 agi as1 As2 H as2 agi Ap dp agi
Ac bs H bh bs hh bd bs hd 0.5 bh bs hh1 0.5 bd bs hd1
Ac 0.24625 m
2
Ii 0.0355 m
Area of the transformed uncracked cross-section: Ai Ac e Ast Asc Ap
4
Ai 0.2774 m
2
Msd 2350 kN m
Distance of the neutral axis (centroid) of the net cross-section:
Np 4200 kN
Compression
Ntot Np Nsd
hd 2 2 1 0.5 bs H bh bs hh bd bs hd H 0.5 bh bs hh1 hh1 hh 2 3 2 bd bs hd1 hd1 hs hh 0.5 3 agc Ac
Nsd 1200kN Tension
Ntot 3000 kN
H a N d a gi p p gi 2
Mtot Msd Nsd
Mtot 866.4714747 m kN
agc 0.4568 m etot
Distance of the neutral axis (centroid) of the transformed cross-section:
Ac agc e Ast d Asc as1 Ap dp Ai The moment of inertia of the concrete cross-section: agi
3
3
hh1 hd1 1 3 3 3 Ic bs H bh bs hh bh bs bd bs hd bd bs 3 3 12 bs H 0.5 H agc 2
2 2 hd hh bh bs hh agc agc bd bs hd H 2 2 2 1 1 bh bs hh1 agc hh hh1 2 3 2 1 1 bd bs hd1 H hd hd1 agc 2 3
Ic 0.0294 m
4
etot 0.2888238 m see diagram B3-B3.3
Ntot Mtot
agi 0.4955 m
Mtot
b d d fcd
0.375
As
b d fcd
Mu Ac dp fcd
As Ac fcd
100 MPa
0.1117642
Mu 1556.762 m kN Mu Mext
Stress in concrete:
c1
agi Nsd Np 1 Ai etot Ai I i
Structural system
c1
22.89 MPa
ok
STRUCTURAL ENGINEERING ROOM
d H ast
Department of Architecture
172
Example 3.7-12: The cross - section dimensions of a beam shown on figure 3.7.12-1
or
STRUCTURAL ENGINEERING ROOM
Department of Architecture
173
c1
Ntot Ai
Mtot
c1
Ii agi
22.8916 MPa
are subjected to bending moment Msd, Np (compression). Determine the ultimate bending moment and concrete stresses at the upper and lower of cross-section. Material Characteristics: Concrete:
The lower edge
fccyl 35 MPa
Nsd Np H agi c2 1 Ai etot Ai I i
Steel: c2
1.49 MPa
c2
Ai
Mtot
fyk 520 MPa
1
c2
Ii
20 mm
fcd 19.833 MPa
1.5
fyd 452.174 MPa
1.15 1
2
A1 0.0003142 m
4
1.49 MPa
Asc n A1
Hagi
A1
fccyl
fyk
fyd
Compression reinforcement:
Or
Ntot
fcd 0.85
Asc 0.0012566 m
2
e
15
n 4
2
Tension reinforcement:
2
22 mm
A2
Ast n A2
2
2
A2 0.0003801 m
4
Ast 0.0015205 m
2
n 4
2
Tendon prestress cables:
1 ( 5.5mm) 2
Ap1
Figure: 3.7.11-1
4
Ap1 0.0001416 m
2
6 ( 5mm) 4 2
np 18
Ap np Ap1
Ap 0.0025482 m
Cross-section dimensions: h 120 cm
b 45 cm
d h ap
d 1.05 m
Structural system
ap 15 cm
asc 5 cm
2
As 0.0053254 m
w
As b h
h a p 2
Mext Msd Np
w 0.0098618
Msd 240 kN m
Np 2500 kN
Msd
As
b d d fcd
Nd = 0 kN
As b d fcd
100 MPa
Mu b d d fcd
b d fcd
Mext 885 m kN
0.0568269
Mu 3542.3325 m kN
Mtot Msd Np h agi ap
c1
c2
N p Ai
N p Ai
Mtot
Ii agi Mtot Ii
c1
1.992 MPa
c2
9.461 MPa
Transformed cross-section area:
Ai b h e Ast Asc Ap
Ai 0.6198809 m
2
The center of gravity of the cross section of the concrete from the top of edge
agi
b h 0.5 h e Asc asc Ast h ast Ap h ap Ai
agi 0.6312603 m
The moment of inertia of ideal cross-section 3 1
Ii b h
12
4
Figure: 3.7.12-2
h a A a a 2 A h a a 2 A h a a 2 gi sc st st gi p p gi e sc gi 2
b h
Ii 0.0845359 m
2
Structural system
0.36
Mu Mext
Mtot 806.8493483 m kN
h agi
Figure: 3.7.12-1
ok
STRUCTURAL ENGINEERING ROOM
As Ap Ast Asc
2
Department of Architecture
174
3.8 calculation of stiffness of concrete members
STRUCTURAL ENGINEERING ROOM
Department of Architecture
175
Example 3.8-1: Cross-section in tension Ms 0 N s 0 , we determine input data: b
Ii
rc
rc
A i agi
0.086m
From the diagrams, we determine the reinforcement to the cross-section as follows:
0.40 m
h
0.50 m
d2
ast
d
Bending moment: Ms
fckcube
150 kN m
fcdcyl.
20 MPa
he
fckcube
b
0.8 0.85
fcdcyl.
9.07 MPa
Axial force: Ns
fyd
Material properties: fckcube
1.15
s
275 kN
20 MPa
fctm
1.4 MPa
Ec
375 MPa
fycd
375 MPa
c
18 mm
nt
210 GPa
12 mm
nc
A st
2
A sc
n t
n c
c
t
2
4
0.00153m
2
A sc
4
A 2required
2
A st
2
M sds
M sds
89.5kNm
2
0.11172
b d fcd
0.03277
6
h M s Ns d 2 2
M sds
Tension and compression reinforcement: t
s
We calculate coefficient , and . then we determine the required reinforcement to the section see B3-B3.3
27 GPa
Es
fyk
fyd
410 MPa
356.52MPa
Area and position of reinforcement: fyd
fyk
1.5
b
A 1required
2
b d fcd 10
3
Ns 10 fyd
4
10
A 2required
Ns A 2provided fyd
A 1required
fyd
0.00023m
2
13.299cm
A 2provided
2
0.00076m
A 1provided
Fully ideal acting cross-section asc
2.8 cm
ast
agi
1 b h 2 n A st h e A sc asc 2 b h n A st A sc
2
3 cm
he
h ast
agi
n
Es
rt
b h n A st A sc
Ii
Ai
7.77778
r
t
i
ef
Ns
0.545
rt
- has sign of normal force
( I ) - sectional area (moment of inertia) of ideal cross-section i
c
4
0.00481m
a) First, we determine whether the expected cracking
2
0.21363m
0.0938m
f
( r ) -core segment due to tension (compression) edge is determined as follows:
N r1
f ctm
A i 1
A i h agi
e
- is characteristic value of the tensile strength of concrete.
ctm
A
Ii
Ms
ef
f
0.2604m
h 2 2 2 Ii b h agi agi h n A st h e agi A sc agi asc 3 Ai
n
Eb
Ns
275kN
Structural system
ef rt
Nr2
0.6 fck
Ai 1
ef rc
A st
A sc
Nr1
72.48kN
Nr2
361.74kN
cracking is expected
a1
4 n A st 0.5 h ef h e A sc 0.5 h ef asc b
ao
b
b) Determination of stiffness - unless cracking is not expected: the bending stiffness without cracks Bfla
Bfla
0.85 Ec Ii
4
a2
2 0.5 h ef
n A st h e 0.5 h ef h e A sc asc 0.5 h ef asc
a2
1.59091m
2
a1
0.05215m
ao
0.01854m
3
2
110275.24kNm
The value of x we determine according to Newton numerical methods, i.e. setting it from the r
- axial stiffness of the beam without crack igi
where
0
relationship:
1. iteration
Ii
igi
Ai
0.14997m
xri
xr1
xr
xri
0.5 h
2 xri3
ao a1 xri a2 xri
xr1
2
a1 2 a2 xri 3 xri
xr1
0.1314m
xr
0.1314m
using as initial value (i=1) we can be considered x 0.5 h (i is number of iteration) and r i repeat the operation until the condition will be satisfies.
Figure: 3.8.1-1 B axa
B fla
igi
2
Baxa
e f 0.5 h a gi
6556750.69kN
further we provide stiffness of the beam with total elimination of concrete in tension
eccentricity etbal
etbal
A sc asc agi asc A st h e h e agi A sc asc A st h e
0.2098m
etbal
0.2098m
ef
0.54545m
ef etbal
Figure 3.8.1-2 2. iteration
if it is ef et bal, there is a cross-sectional area of compression if it is ef et bal, first determine the stiffness of the beam without crack
xr2
xr1
2 xr13
ao a1 xr1 a2 xr1
2
a1 2 a2 xr1 3 xr1
Structural system
xr2
0.099m
STRUCTURAL ENGINEERING ROOM
Ns Nr1
xr 3 a2xr 2 a1xr ao
Department of Architecture
176
3. iteration
STRUCTURAL ENGINEERING ROOM
Department of Architecture
177
xr3
xr2
Bending:
2 xr23 2 a1 2 a2 xr2 3 xr2
ao a1 xr2 a2 xr2
xr3
Bfl
0.095m
4. iteration xr4
xr3
1 1 r r B B flb fla
2
Bfl
39125.28kNm
Bax
444573.14kN
- Axial:
2 a1 2 a2 xr3 3 xr3
2 xr43
ao a1 xr3 a2 xr3
2
xr3
3
xr4
Bax
0.095m
1 1 r r B B axb axa
5. iteration xr5
xr4
ao a1 xr4 a2 xr4
xr5
2
a1 2 a2 xr4 3 xr4
0.095m
xr
xr5
xr
0.095m
Example 3.8-2: Eccentrically loaded elements of bending moments and axial force M 0 s
where P is the number of digits that in two consecutive iterations agree further determine he asc P b xr 2 n A st 1 A sc 1 ef xr xr
P
3
0.029m
N 0 s
input data Cross-sectional dimension: b
0.40 m
stiffness of the beam with a fully excluded concrete in tension
Bending moment
Bending stiffness:
Axial force
Eb P xr
Bflb
Bflb
2
2
37060.57kNm
Eb = Ec
tension , N 0 compression - effects from working load s
When compression cross-sections:
Ms
130 kN m
Ns
460 kN
asc
2.8 cm
h
0.50 m
ast
3 cm
he
h ast
- Axial stiffness: Eb P
Baxb
Baxb
agi 2 ef 1 xr
Eb = Ec
411453.99kN
The resulting stiffness:
r
1 4
Nr1
5
Ns
1
r
0.0795
0 r
Figure: 3.8.2-1
Structural system
he
0.47m
fckcube
20 MPa
fyk
fcdcyl.
0.8 0.85
fckcube 1.5
206 MPa
d2
fcdcyl.
ast
d
9.06667MPa
Ec
d
he 27 GPa
Es
h 2
b h
Ii
0.47m
210 GPa
3
Ai
According to diagrams we calculate the value of , then :
agi agi h
b h n A st A sc
n Ast he agi2 Asc agi asc 2
Ai
Ii
4
0.00455m
2
0.2080m
c ) core segment respectively on the tension, (compression) edges, setting it from the
r r t
relationship:
h M s Ns d 2 2
from the graf we finde out 0.194
0.29
2
rt
b d fcd 3
2
A 2required
b d fcd 10
Ns 10
4
10
fyk
A 2required
2
f
7.39 cm
1.15
t
c
16 mm
nt
12 mm
nc
r A st
4
A sc
2
nt
2
t
nc
A st
4 c
2
0.0008m
2
4
A sc
t
i
ef
ef
Ns
0.28261m
e
f
ef
6 h
bg Rbtn
- has sign of normal force
Ai
Nr1
ef
277.22kN
where bg Rbtn = fctm
rt
axial force provided that decisions driven by compression in concrete
12 m
1
-1
ef
l e
f
6 h
cracks
Nr2
0.6 Rbn
Ai
6 h
ef
Nr2
436.37kN
Rbn = fck
rc
If it is Nr1 Nr2, then Nr
Nr1 ,
if Ns Nr cracking is not expected,
The calculation of the ideal characteristics of a fully-acting sectional Distance of center of gravity of ideal cross-section of the upper extremity: 2
0.08589m
axial force provided that decisions driven by tension in concrete
1
Cracks formed
agi
rc
c
1 -1
Ii A i agi
2
not formed 3.53846 m
rc
0.0892m
( r ) -core segment due to tension (compression) edge is determined as follows:
0.00023m
First, we determine whether the expected formation of cracks, (if it is N s 0 ,
1
rt
( Ii) - sectional area (moment of inertia) of ideal cross-section
Nr1 Ms
Ii
- is characteristic value of the tensile strength of concrete.
ctm
A
Area and position of reinforcement:
A i h agi
1 b h 2 n A st h e A sc asc 2 b h n A st A sc
if Ns Nr cracking is expected: b) determination of stiffness
agi
0.254m
xi
agi
- If the cracking is not expected: the bending stiffness without cracks
Structural system
STRUCTURAL ENGINEERING ROOM
The moment of inertia about an axis through the center of gravity of ideal cross-section:
Material properties:
Department of Architecture
178
STRUCTURAL ENGINEERING ROOM
Department of Architecture
179
Bfl
0.85 Ec Ii
Bfl
Further we provide stiffness of the beam to the total exclusion of the concrete strength) when
2
104455.92kNm
compression cross-section ( Ms 0Ns 0 ) in cross-section with cracks then occur compression area, the depth of this area x We r
radius of gyration igi
determined from the equation:
Ii
igi
Ai
0.147m
xr 3 a2xr 2 a1xr ao
where
0
a2
2 0.5 h ef
a2
0.06522m
- axial stiffness of the beam without crack
4 n A 0.5 h e h A 0.5 h e a b st f e sc f sc
a1 B axa
Baxa
B fl
igi 2
e f 0.5 h a gi
4498629.29kN
If cracking is expected: first determine the stiffness without cracks - Bending Bfla
0.85 Ec Ii
Bfla
2
b
n A st h e 0.5 h ef h e A sc asc 0.5 h ef asc
0.03251m
ao
3
0.01481m
The depth x determine the value of numerical methods, ie iterated according to a relationship: r
xri
0.5 h
xrii
xri
104455.92kNm
2 xri3
ao a1 xri a2 xri
xrii
2
a1 2 a2 xri 3 xri
0.198m
xr
xrii
xr
0.198m
using as initial value (I = 1) we can be considered xr i 0.5 h (i is number of iteration) and
Axial: Baxa
4
ao
2
a1
Bfla
igi2 ef 0.5 h agi
repeat the operation until the condition will be satisfies Baxa
4498629.29kN x
r i 1
x
1
P
10
r i
where P is the number of digits to the two subsequent iterations coincide. Next, we determine
P
he
asc 1 A sc 1 ef xr xr
b xr 2 n A st
P
3
0.01845m
stiffness of the beam with a fully excluded tension in concrete Figure 3.8.2-2
- bending Bflb
Structural system
Eb P xr 2
Bflb
2
49431.0kNm
Eb = Ec
The stiffness went out with negative sign, indicating that the sign of the axial deformation not Eb P
Baxb
Baxb
agi 2 ef 1 xr
Eb = Ec
3107654.19kN
conforming the sign axial force.
The resulting stiffness r
1 4
Nr1
5
Ns
1
r
0.50333
r 0
- Bending: Bfl
1
Bfl
1 r r B Bflb fla
2
67266.17kNm
- axial: B ax
Figure: 3.8.2-4
1
r B axa
1 r B axb
Bax
20861377.5
Figure: 3.8.2-5 Figure: 3.8.2-3
Structural system
STRUCTURAL ENGINEERING ROOM
- axial
Department of Architecture
180
Example 3.8-3: Cross section with cracks
STRUCTURAL ENGINEERING ROOM
Department of Architecture
181 Force in concrete reinforcement at the bottom of the cross section Ns2
If the condition of formation of cracks is fulfilled at one edge in the plan of bending of the cross-section and the opposite side is at the same time in compression, we determine for
N p
calculation the stresses values of cross section with cracks as shown in figure 3.8.3-1, where the tension zone of concrete does not work (neglected) and the stress in compression zone of the section and in reinforcement is proportional to the plane of deformation of the cross section. For a rectangular cross section of width b and d effective depth of the upper (lower)
horizontal forces and bending requirement of the upper edge of the cross section: Nd Npd
the cross-sectional area Ap which is subject by bending moment Md, normal force from an
c
x a s1 p c x
ap x x
Ncc Ns1 Ns2 N p
Nd Npd N p Ns2 Ncc Ns1 0 Nd Npd e Ncc x3 Ns1 a s1 Ns2 d N p a p
external load Nd and from the prestress Npd (Service value), shall apply according to figure:
d x s1 c x
Ap p Ep
The cross section which is in the picture, we can also write the equilibrium condition of the
with concrete reinforcement cross-sectional area As1 item (AS2) and the prestress tendon with
s2
As2 s2 Es
.
If we exclude the axial forces due to external load from equations, we get substituting the expressions: d x s1 x c
s2 c
x a s1 p x c
ap x x
And modified cubic equations for calculation the depth of the compression zone in the cross section: 6 e 3 2 As1 e a s1 As2 ( e d ) Ap e a p x x 3e x b
6 e
b
0
As1 a s1 e a s1 As2 d ( e d ) Ap a p e a p
Area of transformed cross-section: Ai
Figure: 3.8.3-1 Values of cross-section with cracks in bending or tension and compression with high eccentricity
Distance of center of gravity of ideal cross-section of the upper edge:
For compression area shall valid: Force in compression part of cross-section
Ncc 0.5b x cm Ecm Force in concrete reinforcement on the top edge of the cross section Ns1
As1 s1 Es
Ac e As1 As2 Ap
a gi
Ac a c e As2 d As1 a s1 Ap a p Ai
The moment of inertia of ideal cross‐section
Ii
2 2 2 2 Iby Ac a gi a c e As1 a gi a s1 As2 a gi d Ap a gi a p
Ac - sectional area of the concrete
Structural system
Nd Npd
As2- sectional area of reinforcement at the lower edge of concrete section
Nd Npd e
Ap - sectional area of prestress tendon Iby - moment of inertia of the concrete cross section of the y axis 15, e -is the distance of resultant forces N d N pd from the top of the cross section. To calculate the cross-sectional values with crack A a g I where instead of the entire surface of
the concrete section Ac is now using its compression part Acc (for rectangular cross-section xb ).
For searching the stress in concrete resp. in reinforcement in cross-section with cracks apply by analogy:
x Ncc Ns1 a s1 Ns2 d N p a p 3
Npd a N p = 0 , in equation
where e
Acc
Ncc Ns1 Ns2 N p
6 e 3 2 As1 e a s1 As2 ( e d ) Ap e a p x x 3e x b
6 e
b
As1 a s1 e a s1 As2 d ( e d ) Ap a p e a p
fall away members of the cross-sectional area Ap,
in members without compression
reinforcement also we omitted the cross-sectional area As1. c1
Nd Npd ag 1 A a g e A I
s1
Nd Npd a g a s1 1 A a g e e A I
s2
s2
p
p
Nd Npd ag d 1 A a g e e A I
For rectangular cross-section with a crack, subject only to bending moment from the service load Md (without prestress tendon) then the force equilibrium conditions will be Ncc Ns1 Ns2
where using a relationship and Ncc Ns1 Ns2 N p , we obtained after
adjustments quadratic equation for calculating the depth of compression zone of concrete crosssection x: or
x
Nd ag e Nd Npd Npd ag e e I I A A d ag d ag Nd Npd ag ap 1 A a g e e A I
0,
b
As1 As2 1
1
2 b e
As1 a s1 As2 d
As1 As2 2
The depth of compression part of concrete x, at the same time is the core axis of weakened cross section a g
x , So we can directly determine the moment of inertia of the weakened
cross-section:
or
Nd ag e Nd Npd Npd ag e e I I A A ag ap ag ap
e
I
1 3
3 2 2 b x e As1 x a s1 As2 ( d x)
Then the stress of concrete and the stress of reinforcement will be: c1
M x I
s1
Structural system
e M
I
a s1 x
2
e M
I
( d x)
STRUCTURAL ENGINEERING ROOM
For non- prestressed elements will be in relations:
As1- sectional area of reinforcement at the upper edge of concrete section
Department of Architecture
182
The cross-section of the general form, the compression zone should be divided by surface
STRUCTURAL ENGINEERING ROOM
Department of Architecture
183
portions, respectively. strips and the compression part of cross-section determined by the relationship, s2 c
d x s1 c x
x a s1 p c x
ap x x
,
Ncc Ns1 Ns2 .
and
If the condition c f ctm on both edges of the cross section c1 , c2 are fulfilled, then it goes
that the cross-section subjected by eccentric tension with small eccentricity and the ideal cross section for calculating the stress s1 it consists only reinforced. Figure 3.8.3-2: Values of ideal cross-section and stress distribution
on the upper edge
c1
The stresses will vary from maximum compression at the top to maximum tension at
Nd Npd a gi 1 Ai etot i Ii Ai
the bottom. Where the stress changes from compressive to tensile, there will be one layer that remains unstressed and this is called the neutral layer or the neutral axis (NA).
on the lower edge
This is why beams with an I-section are so effective. The main part of the material is
concentrated in the flanges, away from the neutral axis. Hence, the maximum stresses occur c2
where there is maximum material to resist them. If the material is assumed to be elastic, then
Nd Npd h a gi A e 1 i tot i Ii Ai
Nd Npd
the stress distribution can be represented by two triangular shapes with the line of action of the resultant force of each triangle of stress at its centroid. The couple produced by the compression and tension triangles of stress is the internal-reaction couple of the beam section.
Ntot
Characteristics of T- shaped cross section, and I inverted T cross-section loaded by
c1
Ntot Ai
Npd Ai
Nd
Ii agi
etot.i
Npd Ii agi
bending moments. Dimension an I - beam in the illustration, the following procedures can, of etot.i
course be used to cross-sections rectangular (appoint the b 2 - shaped cross section or inverted T-shape ( b 1
b 2, h 1
c2
Nd Ai
Npd Ai
Nd
Ii h agi
etot.i
Npd Ii h agi
etot.i
Figure 3.8.3-3
Structural system
0)
b 1, b 3
b 1, h 2
0,
h3
0 )T
3
Ac
area As tot static moment Ss0 . And inertia Is0 (both of the upper edge of the cross section),
b k h k
if necessary. Distance of the center of gravity of reinforcement a gs from the top edge.
k1
Characteristics of an ideal fully-acting sectional get then the following applies
Static moment of area of the concrete section to the upper edge:
A
h hk b k h k i 2 i1 k 1
3
Sc0
Sectional area:
k1
a gc
Ic0
k1
I
Where as k = 1 substitutes
Ecm 1
Sc0 n Ss0 A
Ic0 n Is0 A a g
2
The rectangular cross section with dimensions b,h is Ac
b h , Sc0
b h 2
2
, Ic0
b h
3
3
The depth of the compression zone xr, and moment of inertia to the neutral axis Ixr cross-
k 1
Ec eff
Moment of inertia of the cross-section to the centre of gravity of cross section:
2 k 1 2 b k h k h k h i h k 2 12 i1
Es Ec eff
Distance from the centre of gravity of the compression edge of the cross-section:
The moment of inertia to the upper edge
3
n
Ac n As tot
hi
section is completely excluding concrete in tension. If we mark
0
i1
Acx as a surface of
compression concrete, Scx is static moment of this area to the compression edge, then the These relationships also apply in cross-sections consisting of multiple rectangles (and
summation must be extended to their count). In the following we will limit on I cross-section composed of 3 rectangles.
conditions which shall be counted the depth of compression area xr, will be of the form: Axc xr Scx n As tot xr Ss0
0
These relationships will use in calculating the replacement thickness by the relationship h0
Ac up
.. half perimeter
up exposed to the environment can be determined from the
section, however, Acx
relationship
up
k h 1
In the general case, we need to solve this equation numerically. When a rectangular cross
b1 2
k h 2 h 1 k h 3
where coefficients k h k, k
b1 b2 2
k h 4 h 2 k h 5
1 ............7 are
not or is exposed to the environment.
b3 b2 2
k h 6 h 3 k h 7
b xr, Scx
b xr 2
2
, leading to quadratic equations with solution:
b3 2
equal 0 or 1 depending on whether given sides is
xr
k s 1
1 2
Structural system
a gs
ks
where k s
n
As tot b1
, a gs
Ss0 As tot
STRUCTURAL ENGINEERING ROOM
Reinforcement layers may be more in the cross section. In the calculation applies total
Area of the concrete section is then:
Department of Architecture
184
To calculate the moment of inertia about the neutral axis of the cross-section with excluding
STRUCTURAL ENGINEERING ROOM
Department of Architecture
185
the tension zone in concrete is valid a general relationship Ixr
Icx Scx xr n Is0 Ss0 xr
where Icx
b 1 h 1 b 2 h 2 2b 1 h 1 h 2 2n As tot a gs h 1 h 2 passes through the flange 2
b xr
3
3
, for rectangular cross-
section
2
2
1
Ec eff Ixr
There, the neutral axis passes through the wall. Conditions the first and the second in the order
b h
n
If it is satisfied b 1 h 1 2n As tot a gs h 1 , but the conditions 2
The characteristics of the ideal full acting cross section ( A a g I ) area, the distance of the centre of gravity from the top edge, the moment of inertia about the centre of gravity
A
2
b 1 h 1 b 2 h 2 2b 1 h 1 h 2 2n As tot a gs h 1 h 2
Flexibility sectional weakened due to cracks will then be: CII
If is valid:
b h n As tot
Es Ec eff
2
ag Ec eff
2
n Ss0 A
Ecm 1
I
b h 3
3
n IS0 A a g
2
as listed. If condition b 1 h 1 2n As tot a gs h 1 do not pay, we determine the coefficients k s1 k s2 from the equations: 2
k s1
creep coefficient
n
As tot
k s2
b1
2 k s1 a gs
a gs
Ss0 As tot
If condition b 1 h 1 2 n As tot a gs h 1 is valid, but do not valid condition 2
Full flexibility sectional acting CI will be: CI
1
b 1 h 1 b 2 h 2 2 b 1 h 1 h 2 2 n As tot a gs h 1 h 2 then: 2
Ec eff I
Moment of the cracking Mcr is then:
k s1
I h a g f ctm is the mean value of the concrete tensile strength Mcr
2
b 1 b 2 h 1 n As tot b2
f ctm
k s2
b 1 b 2 h 1 2 2n Ss0
Where there are two conditions simultaneously, then:
When calculating the characteristics of the cross section with the complete exclusion of
the tension in concrete we must first determine who frequently cross the neutral axis. This can be done according to the following two conditions. If is valid:
k s1
k s2
b 1 h 1 2 n As tot a gs h 1 2
b 1 b 2 h 1 b 3 b 2 h 1 h 2 n As tot b3
b 1 b 2 h 1 2 b 3 b 2 h 1 h 2 2 2n Ss0
neutral axis is below the top flange. If the condition does not apply the neutral axis passes through the upper flange or right on the edge between the top flange and the wall.
Structural system
b3
b2
- Bending moment Ms the entire service load (considered as a positive value). The calculation of the ideal characteristics of a fully cross-section of the applied bending
xr
k s1
k s1 2 k s2
Hcx
b 1
xr
3
moment at cracking.
6
The largest of ordinate zbk we mark h, i.e h
maxzbk. sectional area ideal Ai, static moment
t Si0 to the upper edge and the moment of inertia Ii0 the upper edge of the cross section If the neutral axis passes through the upper flange Hcx
b 2
xr 3 6
b 1 b 2 h 1 2
h1 3
calculated from:
xr 2
Ai
If the neutral axis passes through a wall Hcx
b 3
xr 3 6
x h 1 xr 2 h 1 h 2 r b 3 b 2 h 1 h 2 3 2 3 2
b 1 b 2 h 1 2
If the neutral axis passes through the lower flange Ixr CII
Icx Scx xr n Is0 Ss0 xr
Icx Scx xr
Si0
Hcx Ii0
1
Ec eff Ixr
plane load. In the calculation of the initial bending stiffness of the cross-section of a of the multiple layers (see figure) to be introduced the following input data - Dimensions of the cross-section: the cross sections limited by line segments can be introduced
2
n
1 6
Es
k1
Es
k1
b ( k 1) zbk b k z( bk 1) zbk 1 zbk Ec
Modulus of elasticity of steel Es (Populated with the Es The standard tensile strength of concrete f ctm..
210000MPa )
Asjzsj
1 n b ( k 1) zbk b k z( bk 1) zbk 1 2 z( bk 1) zbk zbk 2 12 k 1 n E s Asj zsj 2 Ec j1
moment of inertia Ii to the ideal center of gravity of the cross section according to formulas
xi Ii
- Material properties: modulus of elasticity of concrete Ec ,
Based on these cross-section values can be calculated height of compression area xi and
procedure for the award of the coordinates shown by the arrows in figure 3.8.3-4 - Position of a surface of the reinforcement layer ( Asj, Zsj ), j = 1,..m,
m
j1
by the cross-section dimensions of the coordinates of the polygon vertices of the contour ( zb, zbk), K = 1, n ., wherein for a = 0, is given b 0 = 0, zb0 = 0, and for k = n must b n = 0, the
Asj
j1
n
m
b ( k 1) zbk b k z( bk 1) Ec
The calculation of the initial bending stiffness of the cross-section of a symmetrical by symmetrical loading of the plane with reinforcement, which may be distributed along the height
1
Si0 Ai Ii0 Ai xi
2
Moment at cracking is then Mr
fctm
Ii h xi
Structural system
If
Ms Mr
STRUCTURAL ENGINEERING ROOM
The height of compression zone is then:
Department of Architecture
186
Calculation of stiffness, where the cracking is not expected
STRUCTURAL ENGINEERING ROOM
Department of Architecture
187
If the condition Ms Mr, The initial flexural stiffness Br is calculated from equation: Br
0.85Ec Ii
Calculation of stiffness, where the cracking expected If the condition Ms Mr, it is expected that the formation of cracks perpendicular to the axis of the element. In this case, the first identifying the characteristics of the cross section
Figure 3.8.3-4
(depth of compression area xr,. The ideal compression area of the cross section Ac, And the arm of internal forces zr, Provided that the tension concrete does not act drawn. We determine the depth of the compression zone according to numerical solution of equation Where F xr
F xr
0
Abc xr 2
Es Ec
m
0
Abc xr 2
Es Ec
m
Asj zsj xr
j1 The new value xr i is given by:
Asj zsj xr
xr i
j1
In the formula, Abc is a significant compression areas of concrete. When numerical solution of the equation F xr
F xr
can be use this procedure by (regula falsi method): In
calculating the surface of the compression area of the cross section Abc at a given xr is first
r x d F xr h xr h F xr d F xr h F r x d
The index i in the formula is a number of iterations, e.g. at the start of the calculation will be i = 1 The value xr i calculated is then calculated the value F xr i . if then F xr i
of all determined the coordinates of vertices of the polygon that creates the compression area (see figure). Assume that the coordinates of which are ( b l zbl), l
1 .........n ),where
the b l
xr.
0 is
zbl
and for placed l
n l become b nl
0, zbnl
l
l
0 , and
We substitute to another calculation xr d
Surface of the compression area for a given xr will be: Abc
1 2
nl
b l 1 zbl b l zbl 1
xr i , F xr d
F xr i (Values xr h, F xr h
F xr i
F xr h
0 does not comply, the
next calculation shall be substitutes
l1
0.4h , xr h
0
Are left out previous operating process). If the condition
In numerical solutions are chosen at the beginning of the two values xr that indicate xr d and xr h ( xr d
F xr h
xr h xr i: F xr h F xr i and values xr d, F xr d leaving unchanged. With the new values thus determined xr d, F xr d, xr h, F xr h given by
0.6h ,). For both of these values can be determined from the formula
Abc area Abc d, Abc h and then the function F xr d, F xr h from equation.
xr i
r x d F xr h xr h F xr d
Structural system
F xr h F r x d
new value xr i 1 .
Furthermore, we determine the values
subsequent iteration process in value xr i repeat. If then xr i 1
p
1 10 p
xr i
Abc zs max a bc 2
calculated and static moment of compression part of the cross section to the upper
Br
(compression) edge of the cross section of the formula.
6
j1
zsj Asj 1 zs max zsj x r
Bending stiffness Br in the case that the expected formation of cracks, is then given by:
1
Ec
n
Iteration is considered as completed and hence are also known values xr, Abc . This can be
Sbc
Es
Ec zs max p
zs max 2 1 s b xr
Calculating the coefficients br, sr , b, s to work around so that the calculated value from
b l 1 zbl b l zbl 1 zbl 1 zbl
the relationship p and substituting in b
l 1
and therefore the distance of the center of gravity of the compression area from the compression
s
1 in
relation to the calculation Br bending
stiffness Brc , i. j.
edge. Sbc
a bc
and therefore the distance zs max of the center of gravity of the compression area of concrete from compression edge, coefficients br, sr can then be determined from formulas.
xi
br
1.7Ii
Es
Ec
Abc zs max a bc 2
n
j1
zsj Asj 1 zs max zsj xr
2
From the formula for calculation r we calculate the parameter level load r and the resulting bending stiffness in the interval 0 r 1 will then ( Br
Calculation of deflection
n xi zs max Es zsj Abc zs max a bc 2 Asj 1 zs max zsj zs max Ec 1.7Ii xr 1 j1 xr 1
sr
Ec p xr
Brc
Abc
f
5 48
Ms Br
2
L
Furthermore, the calculated parameter r from the formula: r
1 4
Mr 1 Ms
5
and coefficients b, s from the formula b
1
r 1 br
s
1
r 1 sr
Structural system
1
r 1 r Brc Bra
)
STRUCTURAL ENGINEERING ROOM
We choose an accuracy further process by calculating the number of digits to be in two
Department of Architecture
188
STRUCTURAL ENGINEERING ROOM
Department of Architecture
189 Example 3.8-4: Calculation of initial bending and axial stiffness of the cross-section symmetrical about a plane load subjected to bending moments and axial force input data
compression cross-section
Ns 5100 kN
2
Material characteristics: fcyl 22 MPa
fctm 3.15 MPa
Ec 32.5 0.9 GPa
Es 210 GPa
b 3 0.25 m
b 6 0.9 m zb 0 m
b 2 0.8 m b 7 0.0 m zb 0.30 m
zb 1.60 m
zb 1.60 m
h zb
1 6
2
7
3 4
Cross-sectional dimension: k 1 7 b 1 0.8 m
j 1 8 1
Bending moment with axial force Ms 3900 kN m
Area and location of reinforcement:
zb 0.45 m 3
7
h 1.6 m
b 4 0.25 m zb 1.15 m 4
b 5 0.9 m
5
28 mm 25 mm 16 mm 16 mm 20 mm
n1 4 n2 2 n3 2 n4 2 n5 2
As n1
2 1
4
1
As n2 2 4 As n3 3 4
As n5
4 4
20 mm
n6 2
As6 n6
7
28 mm
n7 4
As n7
zb 1.30 m 5
8
28 mm
asc 5.0 cm
n8 4
7
As n8 8
ast 5.0 cm
0 0.24 0.285 2 b ( k 1) zbk b k zb( k 1) 0.175 m 0.71 0.27 1.44
Figure: 3.8.4-1
Structural system
6
2
zs 0.30 m
As 0.0004 m
2
As 0.0004 m
2
2
zs 0.45 m 3
zs 0.80 m 4
As 0.00063 m
2
zs 1.15 m
5
2
4
2 7
4
As6 0.00063 m
As 0.00246 m
5
2
zs 1.25 m 6
2
zs 1.35 m
7
2 8
4
As 0.00098 m
4
2 5
6
1
3
2 4
5
zs 0.05 m
2
2 3
4
2
1
2 2
As n4
As 0.00246 m
As 0.00246 m
7
2
zs 1.55 m
8
d h ast
zbk
1
zb
8
d 1.55 m
k
0 0.3 0.75 1.6 m 2.45 2.9 3.2
0.00246 0.00098 0.0004 0.0004 2 As m j 0.00063 0 0.00246 0.00246
0
0.00012 0.00029 0.00018 0.00032 3 m As j zs j 0.00072 0 0.00333 0.00382
0.00001 0.00009 0.00008 As zs 2 0.00026 m4 j j 0.00083 0 0.00449 0.00592
b ( k 1) zbk b k zb( k 1)
1 Ai 2
1.7 m
8
2
7
k1
Es
b ( k 1) zbk b k zb( k 1) Ec
As 0.0098 m
j1
8
As
j1
j
Ai 0.92038 m
j
Si 0
4
Si
0
xi 0.83221 m Ai Moment at cracking: xi
Mcr fctm
Ii
h xi Bending stiffness:
agi xi
1
3
ef
1 6
k1
Es
b ( k 1) zbk b k zbk 1 zbk 1 zbk Ec 3
8
A
s
j1
j
zs
j
Ii Ii Ai xi 0
2
Ii 0.28543 m
2
Br 7096429.91035 m kN
Ms
ef 0.76471 m
Ns
2.15157
1.30769 m
l 6 cracks h ef
bg
10.5 ef h
n
6 ef h
Es Ec
n 7.17949 1
ef 0.76471 m
6
h
3.75 m
1
1
ef
6
h
rt, rc core segment corresponding to tension (compression) edge, setting it from the relationship: rt rc
4
Mcr 1171.016 m kN
First, we determine whether the expected formation of cracks, (as Ns 0,
ef
4.21725 m
cracks are formed 7
Si 0.76595 m 0
b ( k 1) zbk b k zbk 1 zbk 1 zbk
0
bg
7
k1
2
are formed
2
Static moment:
Ii 0.92286 m
2 2 b ( k 1) zb b k zb zb zb zb zb k k 1 k ( k 1) ( k 1) k As zs 2 j j
Br 0.85 Ec Ii
7
k1
Area ideal cross-section:
1 7 12 k 1 8 Es Ec j1
Ii
Ii
Ai h agi Ii Ai agi
Structural system
rt 0.40391 m rc 0.37264 m
STRUCTURAL ENGINEERING ROOM
2 zb zb zb zb ( k 1) k k ( k 1)
The moment of inertia of the upper edge of the cross section:
0 0.09 0.4275 2 2 2.0425 m 4.5075 6.33 7.68
Department of Architecture
190
STRUCTURAL ENGINEERING ROOM
Department of Architecture
191 axial:
fctm (fcyl) - Is the tensile strength of concrete (compression) Ai, (Ii) - sectional area (moment of inertia) of ideal cross-section
Baxa
rt, (rc)- core segment due to tension (compression) edge is determined as follows.
Nr1 fctm
1
rt axial force, provided that the decisions concrete compression zone Ai
Nr2 0.6 fcyl
1
Nr2 3980.54417 kN
ef
xrd 0.1 h
xrh 0.6 h
0.1 h 0.16 m
0.6 h 0.96 m
xrd 0.16 m b 3 0
xrh 0.96 m
b 1 0.8 m
b 2 0.8 m
zb 0 m
zb xrd 2
zb xrd 3
Abcrd
rc
If Nr1 Nr2, Then Nr
Sbcrd
Nr1 ,
If Ns Nr cracking is not expected,
abcrd
If Ns Nr cracking is expected:
Abcrh
Ii
Bfl
igi
2
ef 0.5 h agi
Sbcrh
Baxa 21199298.17231 kN abcrh
If the cracking is expected: first we determine the stiffness without any cracks - Bending: Bfla 0.85 Ec Ii
bl z
1 b
l1
b l zb
l
bl z
1 b
l1 Sbcrd
l
Abcrd 0.128 m
l 1
3
b l zb
zbl
l 1
1
zb l
2
Sbcrd 0.01024 m
3
abcrd 0.08 m
Abcrd
b 2 0.8 m zb 0.3 m
1
igi 0.55688 m Ai axial stiffness of the beam without crack Baxa
1 6
3
b 1 0.8 m zb 0 m
2
Bfl 7096429.91035 m kN
Radius of gyration igi
1 2
1
Calculation Abcrh, Abcrh
Determination of stiffness Bfl 0.85 Ec Ii
Baxa 21199298.17231 kN
ef 0.5 h agi
Let's do it using the method regula falsi
Nr1 3245.63928 kN
ef
igi
further we provide stiffness of the beam with a full exclusion of concrete strength
axial force provided that decisions tension zone in concrete Ai
Bfla 2
2
1 2 1 6
2
Structural system
3
5
bl z
1 b
l1
b l zb
l
bl z
l1 Sbcrh
Abcrh
1 b
l
b l zb
zbl
l 1
b 4 0.25 m zb xrh 4
b 5 0.0 m zb xrh 5
Abcrh 0.44625 m
l 1
5
agi ef 0.0675 m
Bfla 7096429.91035 m kN
b 3 0.25 m zb 0.45 m
1
zb l
Sbcrh 0.15439 m abcrh 0.34597 m
2
3
8
a
gi
e f zs
j 1 8
j1
j
As zs agi ef zs j j j
A s j
0.00812 m
As j zs j agi ef zs j 0.01108 m
Fxrd Abcrd xrd agi ef abcrd 2 Fxrd 0.14014 m
Ec
4
Fxrh Abcrh xrh agi ef abcrh 2 Fxrh 0.07222 m xr1
Es
4
xrd Fxrh xrh Fxrd
Es Ec
j1
j1
As j zs j xrd agi ef zs j
xr2
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb xr1
zb xr1
4
Area of compression zone: 1 b
l1
l
b l zb
zbl
l 1
1
zb l
Sbc 0.09835 m
Sbc
b l zb
l 1
3
abc 0.26001 m
Abc
Es Ec
4
F xr1 F xrh
0.3977
0
8
j1
As j zs j xr1 agi ef zs j
to the next calculation should be substitute
5
xrd 0.68794 m
xrd Fxrh xrh Fxrd
Fxrd Fxr1
xr2 0.76535 m
Fxrh Fxrd
Calculation Abc, Abc b 1 0.8 m
b 2 0.8 m
b 3 0.25 m
b 4 0.25 m
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb xr2
zb xr2
1
2
3
4
Surface compression area:
5
bl z
l
xrd xr1
b 4 0.25 m
1 2
l1
b 3 0.25 m
Abc
1 b
As j zs j xrh agi ef zs j
b 2 0.8 m
3
bl z
Fxr1 0.02872 m
F xrh
b 1 0.8 m
2
5
compression concrete edge
F xr1
Calculation Abc, Abc
1
1 6
Distance of the center of gravity of the compression zone of concrete from
abc
xr1 0.68794 m
Fxrh Fxrd
Sbc
Fxr1 Abc xr1 agi ef abc 2
8
Static moment of compression zone of cross-section on the upper (compression) edge.
3
4
8
0 0.00007 0.00007 0.00024 4 m 0.00078 0 0.00426 0.00566
Abc 0.37823 m
2
Abc
1 2
5
bl z
1 b
l1
Structural system
l
b l zb
l 1
Abc 0.39759 m
2
5
STRUCTURAL ENGINEERING ROOM
0.00004 0.00023 0.00015 0.00029 3 agi ef zs As m j j 0.00068 0 0.00316 0.00365
Department of Architecture
192
STRUCTURAL ENGINEERING ROOM
Department of Architecture
193 Static moment of compression zone of cross-section on the upper (compression) edge crosssection Sbc
1 6
5
bl z
1 b
l1
l
b l zb
zb l l
zb 1 l
1
Sbc 0.11241 m
3
Distance of the center of gravity compression surface of concrete from compression concrete edge abc
Sbc
abc 0.28605 m
Abc
Distance of the center of gravity compression surface of concrete from compression concrete edge abc
Sbc
Fxr3 0.00187 m
abc 0.28272 m
Abc
Fxr2 Abc xr2 agi ef abc 2
Fxr2 0.00429 m xrd xr2 xr3
Es Ec
xr4
8
j1
Fxrd Fxr2
xrd Fxrh xrh Fxrd
Fxrd 0.00429 m
4
b 1 0.8 m
b 2 0.8 m
b 3 0.25 m
b 4 0.25 m
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb xr3
zb xr3
4
5
Surface compression area Abc
1 2
bl z
1 b
l1
b l zb
l
Abc 0.40032 m
l 1
1 6
1 b
l1
Fxrd Fxr3 Fxrd 0.00187 m
4
xr4 0.7809 m
b 3 0.25 m
b 4 0.25 m
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb xr4 4
zb xr4 5
1
2
3
1 2
5
bl z
1 b
l1
b l zb
l
Abc 0.40147 m
l 1
2
Static moment of compression zone of cross-section on the upper (compression) edge crosssection 1 6
5
bl z
1 b
l1
l
b l zb
zbl
l 1
1
zb l
Sbc 0.11541 m
3
Distance of the center of gravity compression surface of concrete from compression concrete edge
5
bl z
xrd 0.77627 m
Fxrh Fxrd
2
Static moment of compression zone of cross-section on the upper (compression) edge crosssection Sbc
b 2 0.80 m
Sbc
5
As j zs j xr2 agi ef zs j
Surface compression area
Calculation Abc, Abc
3
j1
b 1 0.80 m
Abc
2
xrd xr3
8
Calculation Abc, Abc
xr3 0.77627 m
Fxrh Fxrd
4
Ec
xrd Fxrh xrh Fxrd
As j zs j xr2 agi ef zs j
4
xrd 0.76535 m
1
Es
Fxr3 Abc xr3 agi ef abc 2
l
b l zb
zb l l 1
zb 1 l
Sbc 0.11451 m
3
abc
Sbc
abc 0.28747 m
Abc
Fxr4 Abc xr4 agi ef abc 2
Fxr4 0.00082 m
Structural system
4
xrd xr4
Es Ec
8
j1
As j zs j xr2 agi ef zs j
xrd 0.7809 m
Fxrd Fxr4
Fxrd 0.00082 m
4
The resulting stiffness
xr5 0.78291 m
Fxrh Fxrd
r
1 4
Calculation Abc, Abc b 1 0.8 m
b 2 0.8 m
b 3 0.25 m
b 4 0.25 m
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb xr5 4
zb xr5 5
1
Abc
Sbc
abc
1 2 1 6
2
3
5
bl z
1 b
l1
b l zb
l
bl z
1 b
l1
l
b l zb
zbl
l 1
1
zb l
Sbc 0.11581 m
Sbc
Fxr5 0.00036 m xrd xr5
sr
3
sr
Ec
4
8
j1
xrd Fxrh xrh Fxrd
Fxrd 0.00036 m
1 r 1 br
4
Bax
8
j1
xr 0.78379 m
zs j As 1 j xr
0.00141 m
zsmax h ast 2
Abc 0.40198 m
zs Es 8 As j 1 ef P Abc 2 Ec j xr j1
br
0.74474
0.19219
The calculation is repeated until you achieve the desired accuracy.
xr xr6
zs ef Es 8 As j 1 Abc 2 rc Ec j xr j1
0
s
1 r 1 sr
Ec zsmaxP
2
zsmax 2 1 s b x r
Bfl 4701985.94096 m kN
Axial stiffness:
xr6 0.78379 m
Fxrh Fxrd
r
8 zs j A 2 Es As 1 bc Ec j xr ef zsmax agi 1 j1 1 2 1.7 Ai zsmax i gi 1 xr
Bfl
Fxrd Fxr5
1.7 Ai
1
0.5455
Bending stiffness:
As j zs j xr2 agi ef zs j
xrd 0.78291 m
1
r
and coefficients b
Es
Ns
1
2
abc 0.28809 m
Abc
Fxr5 Abc xr5 agi ef abc 2
xr6
Abc 0.40198 m
l 1
5
br
Nr1
5
zsmax 1.55 m
Bax 53163385.822 kN
agi zsmax agi 2 ef 1 s 1 b zsmax xr zsmax
Stiffness Bfl and Bax also can be calculated without calculating the coefficients s, b Bending stiffness
2
P 0.29196 m
Ec P
3
Bflb
Ec P xr 2
Structural system
2
Bflb 3346675.19031 m kN
STRUCTURAL ENGINEERING ROOM
xrd Fxrh xrh Fxrd
xr5
Department of Architecture
194
Axial stiffness
STRUCTURAL ENGINEERING ROOM
Department of Architecture
195
Baxb
From the concrete section active part with a height hsh. Which is determined as follows: Ec P
in the section in which we do not expect cracking hsh = h. where h is a whole section height
Baxb 90381434.25457 kN
agi 2 ef 1 xr
- In the section in which we expect cracking introduced into the calculation of the smaller of hsh = 2xr, hsh = 0.6h.
Calculation of bending and axial stiffness when cracking expected hsh 2 xr
Bending stiffness 2
Bfl 0.85 Ec Ii
Bfl 7096429.91035 m kN
Axial stiffness Bax 0.85 Ec
Ii igi ef 0.5 h agi 2
Bax 21199298.17231 kN
The resulting stiffness will then
Bf´
Bax´
1
2
r 1 r Bfla Bflb 1
r 1 r Baxa Baxb
Bf´ 4701985.94096 m kN
Bax´ 48301511.12233 kN
hsh 1.56758 m
0.015 t1 0.15 0.8 e
t1 14
0.07 t1
t1
1 e
t2
1 e
0.07 t2
t1
0.23042
t2
0.60904
5.5
dry environment
bf2
3.8
normal environment
bf1 bf2
2
t2 t1
hsh 0.93 m
t2 180
bf1
hsh 0.6 d
t2 t1
0.37862
1.76056
Modulus of elasticity of concrete is introduced value Where Ebo it is a basic module of elasticity of concrete Eco Ec
Ect
Eco
Ect 14860478.73568 m
1 0.55
2
kN
So that we can identify Ns,sh. We need to determine: - Acting on concrete section of height hsh: Cross-sectional dimension:
k 1 5 b 1 0.8 m
b 2 0.8 m
b 3 0.25 m
b 4 0.25 m
b 5 0.0 m
zb 0 m
zb 0.3 m
zb 0.45 m
zb 0.6 d
zb 0.6 d
1
Figure: 3.8.4-2
Structural system
2
3
4
5
2 zb zb zb zb ( k 1) k k ( k 1)
zbk
1
zb
k
0 0.3 0.75 m 1.38 1.86
ab,sh sectional center of gravity distance from the top edge.
absh
absh 0.33573 m
Absh
Ib,sh sectional moment of inertia of the center of gravity axis
0.00246 0.00098 0.0004 0.0004 2 As m j 0.00063 0 0.00246 0.00246
0 0.09 2 2 0.4275 m 1.4859 2.5947
Sbsh
Department of Architecture
0 0.24 2 b ( k 1) zbk b k zb( k 1) 0.285 m 0.12 0.2325
1 Ibsho
2 2 b ( k 1) zb b k zb zb zb zb zb ( k 1) ( k 1) k k k 1 k
5
12 k1
Ibsho 0.07708 m
Ibsh Ibsho Absh absh
4
2
Ibsh 0.02763 m
4
the cross-section consisting only of reinforcement:
0.00012 0.00029 0.00018 0.00032 3 m As j zs j 0.00072 0 0.00333 0.00382
0.00001 0.00009 0.00008 As zs 2 0.00026 m4 j j 0.00083 0 0.00449 0.00592
Total area of reinforcement in cross-section 8
Assh
As
j1
Assh 0.0098 m
j
8
Sssh
A
s
j1
j
zs
j
2
Sssh 0.00879 m
3
The distance in the overall reinforcement from the top of the cross section Ab,sh cross-sectional area without reinforcement 5
1 Absh 2 k1
b ( k 1) zbk b k zb( k 1)
Absh 0.43875 m
assh
2
Or
Sssh
assh 0.89619 m
Assh
Issh moment of inertia of the reinforcement to the center of gravity axis Static moment:
8
1 b ( k 1) zb b k zb zb zb k k 1 k 1 k 6 k 1 5
Sbsh
Sbsh 0.1473 m
3
Issh
j 1
C1
A s zs 2 j j
8
j 1
asc assh Es asc absh Issh Ect Ibsh
Structural system
A s a ssh j
2
Issh 0.0038 m
C1 368.98722 m
3
4
STRUCTURAL ENGINEERING ROOM
196
STRUCTURAL ENGINEERING ROOM
Department of Architecture
197
C2
d1
d2
d assh Es d absh Issh Ect Ibsh
1
Assh 1
Assh
Es Ect Absh Es Ect Absh
C2 793.19651 m
assh asc assh Issh assh d assh Issh
Es absh asc absh Ect Ibsh
Es absh d absh
Ect Ibsh
sh1
3
0.00013
sh2
0.00019
force or load Ns,sh and the eccentricity from the top edge calculated from: d1 382.99836 m
2 sh1 C2
Nssh d2 228.59165 m
sh2 C1
C2d1 C1d2
Nssh 162.50055 kN
Es
2
esh
sh2 d1
sh1 d2
sh1 C2
sh2 C1
esh 0.59334 m
Suppose that the shrinkage of the two surfaces is not equal, there is a difference bs (differential shrinkage). The concrete part a section of height hsh subjected by force Nb,sh. Equally great force, but opposite in sign, loaded reinforcement is placed across, then Ns,sh = - Nb,sh.
Curvature of shrinkage: 1
sh
rsh sh
The cross-section is exposed on the upper surface of the permeability of the current environment, the lower the surface to a dry environment. If not used for the production of concrete with increased content of mixing water will be the values of free shrinkage at the top surface bsfh = -0.33 10-3. At the bottom surface bsfd = -0.5 10-3. t1 14
Shrinkage - proportional measuring length deformation caused by shrinkage of concrete, is given by: bsh
0.00012
bsd
bsfd t2 t1
bsd
0.00019
The reduced curvature of shrinkage:
The calculation will introduce bsh = -0.00012 bsh
bs
sh1
bs
bsd bsh
bs bs
asc h
0.00012
bs
sh2
0.00006
bs bs
sh
0.00006 m
Nssh 1 esh assh agc assh Es Assh Issh
shredukovane
bs
esh assh
1
agc 0.5 h gsh
bsfh t2 t1
Es Issh
And axial deformation (shortening)
t2 180
bsh
Nssh
d h
Structural system
d
.bs ( t1) .bs ( t2)
gsh
0.00008
moment: Reinforced concrete beam:
The consistency of concrete mix: S2
Department of Architecture
Example 3.8-5: Calculation of curvature of rectangular cross section subjected by bending
Relative humidity of: rh
50 100
Input data: b
0.3 m
h
0.5 m
asc
0.038 m
ast
Age of concrete at the beginning of the load:
asc
t0
60 days
Bending moment from service load: Ms
145 kN m
M sd
M s 1.25
M sd
181.25kNm
Environmental influence is exposed to only the bottom surface ms
Span:
1
d
7.0 m
d
h ast
kh = (0,0,1)
0.462m
Concrete C25/30: Ecm
30500 MPa
fctm
2.6 MPa
Figure: 3.8.5-2
The characteristic yield strength of reinforcement fyk
The duration of humid treatment of fresh concrete:
410 MPa
ttr
Modulus of elasticity of reinforcement: Es
10 days
We propose reinforcing the cross-section of the following:
200000MPa
b
1.5
fck
30 MPa
s fyk
Figure: 3.8.5-1
fck
b
fcdcyl.
0.8 0.85
fyd
356.52MPa
fcdcyl.
13.6 MPa
1.15 410 MPa
fyd
fyk s
b d fcdcyl.
0.066
A required
b d fcdcyl. 100
Structural system
A required
M sd 2
2
12.44074cm
0.20813
STRUCTURAL ENGINEERING ROOM
198
Reinforcement:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
199 Bending element is acceptable in terms of serviceability limit state if the condition is satisfied
14 mm
sc
n sc
14 mm
st1
n st1
20 mm
st2 A st
2
A sc
0
n st2
n sc
A st1
5
Assessment by defining bending slenderness:
A st
2
A sc
4 st1
n st1
A st2
A st1 A st2
sc
2
A st1
4
2
2
A st2
4
2
0.00157m
( 1825)
tab 1
18
tab1
0.00157m
A st b d
tab2
25
2
A provided
A st
kh1
0
h0
Provided area of reinforcement of cross-section (should assumed that A s prov A s req ) A provided
b h
ht
Area of reinforcement required to carry over ultimate moment: 12.44 cm
Ac
Ac
2
0.15m
kh2
0
kh3
kh1 kh3 b2 kh2 h
up
1
up
0.15m
Table values replacement cross-sectional thickness:
Ast is area of tension reinforcement, b is the width of cross-section
A required
Calculation of deflection:
Perimeter of the concrete section that is exposed to the environment:
1.13333
100
15.15
The entire area of the concrete section:
Percentage of reinforcement:
d
Calculation of creep coefficient
2
Type of construction:
beam is fits in terms of serviceability limit state. It is not necessary to calculate the
The ratio of span to effective depth of cross-section.
0m
l
st2
n st2
l d d deflection.
2
0.00031m
2
0.00157m
( 0.050.150.60.050.150.6) m
Ac
h0
up
1m
potom
h 0 0.6
h0
0.6 m
0.05 h 0 0.6
Tabulated values of the age of the concrete at the beginning of a long-term load operation: t0
60
t0
60
t
(days)
( 1 7 2890365)
t
ht
( 0.050.150.60.050.150.6) m
Determining coefficients: c1
1
c2
7 8
c3
410 MPa
c3
A required fyk A provided
d
Defining bending slenderness:
d
c1 c2 c3 tab1 ( 0.5) tab2 ( 1.5)
1.2627
d
t
22.72 dt
t
t i
t0 tt i1 tt i tt i1 t0 28 ( 90 28)
Structural system
90
t
d
h
dt
t i 1
h 0 h t j 1 h t j h t j 1 0.51613
28
0.05 h 0 0.6
dh
h0 0.15 m ( 0.6 0.15) m
h0
dh
0.6m
1
The characteristics of the ideal full acting cross section (A, ag, I) area, distance from the center of gravity the upper edge, the center of gravity of the moment of inertia
c
h0
0.6m
cs
1 d t d h d t d h t i1 j 1 d t d h t ij 1 1 d h d t t ij d t d h
t i 1 j 1
t0
60
t
90
ti
4
j
3
t i 1 j 1
2.5
2.0
t i 1 j
ccs
b h
t
3.2 2.5 2.5 2.1 2.5 2.0 1.9 1.7 2.1 1.6 1.6 1.4 1.6 1.2 1.2 1.0
1.6
t i j
2
0.18441m
e Ss0
agi
Ai
0.2766m
2.1
t i j 1
The moment of inertia of ideal cross-section:
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
Ai
2
2
agi
1.9 1.5 1.2 1.0
A c e A stot
Distance of center of gravity of ideal cross-section of the upper edge:
1
5.4 4.4 3.6 3.5 3.0 2.6 3.9 3.2 2.6 2.0
Ai
Tabulated values of the coefficient of creep: i
Sectional ideal area:
1.79 b h
I
3
Calculation of bending flexibility of full acting cross-section:
2
I
e Is0 A i agi
3
4
0.00454127m
Full flexibility sectional acting C1 will be: A stot
A st A sc
2
A stot
0.00188m
CI
Effective elastic modulus: Ecm
Eceff
Eceff
1
18.318
e
Eceff
M cr
Ecm is the initial mean (secant) modulus As
1
j
A si
A sc
( 1 3) 3
Eceff
j 1
2
zs 1
Es
3
Is0
As
j 1
As
j
A z 2 s j s j
A si
-1
0.00002017kN m
I fctm h agi
M cr
52.85165kNm
fctm is the mean value of the tensile strength of concrete
A st1
As
asc
zs 2
A st2
3
d
zs 3
The depth of compression area xr , and moment of inertia about the neutral axis Ixr , the cross-
d
section without the action of concrete in tension. If we denote as Acx area of compression
3
2
0.03441m
Ss0
j 1
Is0
CI
At the moment of cracking Mcr is then
Es
e
10918.01MPa
1 Eceff I
A s j zs j
Ss0
3
0.00073741m
concrete, Scx static moment of the surface edge, the conditions which shall be counted the depth of compression area of concrete xr will be in the form:
4
0.00033572m
A x S xc r
cx
A e
s tot
Structural system
xr Ss0
0
STRUCTURAL ENGINEERING ROOM
Then we get the creep coefficient from the relationship:
Department of Architecture
200
This equation solve numerically (eg. by method Polen or by regula falsi method). When a
STRUCTURAL ENGINEERING ROOM
Department of Architecture
201
rectangular cross section, however,
A
cx
ags
b x , S r
cx
r 2 resulting in a quadratic
b x 2
Scx
Ss0
ags
A stot
b xr
sr Is the stresses in reinforcement in the same cross section of the load between the crack equation with the solution: The resulting curvature of bending moment:
ags ks 1 1 2 ks
xr
xr
0.39252m
0.20655m
2
Scx
2
ks
e
A cx
A stot
ks
b
b xr
A cx
m
2
0.06197m
3
3
b xr
Icx
3
4
0.00088m
Ixr
Icx Scx xr e Is0 Ss0 xr
Ixr
4
0.00292m
-1
CII
Eceff Ixr
t
( 1 7 2890365)
t
1
1.0
Coated bars
2
0.5
long-acting load
t
sr
s
M
cr
M
s
d
h0 0.15 m
h
dh
( 0.6 0.15) m
cs
2
0.93357
-1
0.00003kN m
ht h0
( 0.050.150.60.050.150.6) m
t
28
t i
0.6m
h 0 h t j 1 h t j h t j 1
ttr 7
dt
( 28 7)
dt
0.14286
1
1 d t d h d td h t i1 j1 d td h t ij 11 d h d t t ijd td h
t i 1 j 1
Tabulated values of the coefficient of creep:
t i 1 j 1
When bending elements the stress ratio replaces the ratio of moments:
(days)
0.05 h 0 0.6
ttr tt i1 tt i tt i1
c
Mcr 1 1 2 Ms
C
Creep coefficient we get then from the relationship:
Effects of reducing flexibility of the action of of the tension stiffening between cracks:
7
t i 1
dh
Partition coefficient
-1
0.00444168m
Tabulated values for the age of at the beginning of a long-term load action
d
0.00003138kN m
m
m
Calculation of boundary curvature of shrinkage:
t
1
M s ( 1 ) CI CII
m
( 1 ) CI CII
Flexibility sectional weakened due to cracks will then be: CII
r
The resulting tension stiffening C (considered constant over the length of the beam) C
0.0064m
operations of the tensile concrete is a general relationship:
1
0.11471m
To calculate the moment of inertia about the neutral axis of the cross-section excluding
Icx
s Is the tensile stress in the reinforcement calculated under load effect on the weakened section due to cracks
t
5.4 3.9 3.2 2.6 2.0
Structural system
3.2
t i 1 j
2.5
ccs
1
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
The final values of the relative shrinkage cs(%o)
Calculation of boundary bending flexibility of full acting cross-section: sh1 0.0006 Ecm
Eceff
Eceff
1
e
8895.83MPa
Es Eceff
e
22.48
sh2 0.0005
ccs sh1 sh2 sh1
cs
h 0 0.15 m 0.45 m
cs
0.0005
The characteristics of the ideal full acting cross section (A, ag, I ) area, distance from the If h 0.15 m, sat in this relationship h 0 0
center of gravity the upper edge, the moment of inertia
Curvature due to shrinkage Area of ideal cross-section: Ai
2
0.19224m
ag
h 2
e A st d
2
0.19224m
ag
I
b h
2
e Is0 A i agi
3
I
1
shI
xr
Ss0
4
1 2
shI
rshII
ags
A stot
ks 1
1
shII
rshI
ags
0.27994m
The moment of inertia of ideal cross-section 3
1 rs hII
cross-sectional
Ai
acting in full cross-section and
weakened due to crack then are: Ai
Distance of center of gravity of ideal cross-section of the upper edge: b h
1 rshI
0.15 m
0.39252m
ags ks
cs e A stot
ags ag I
shI
A stot
ks
e
xr
0.22025m
ks
b
A cx
b xr
-1
0.00044513m
0.14079m
A cx
2
0.06607m
0.00534075m
Scx
2
b xr
Scx
2
3
0.00728m
Full flexibility sectional acting C1 will be: To calculate the moment of inertia about the neutral axis of the cross-section without the CI
1 Eceff I
CI
action of the concrete in tension there is a general relationship:
-1 -2
0.00002105kN m
Free shrinkage cs is determined depending on the thickness of the replacement ho and the relative humidity rh
Icx Ixr
rh
50 100
h0
3
b xr 3
4
Icx
0.00107m
Hcx
Icx Scx xr
Ixr
0.00336228m
Icx Scx xr e Is0 Ss0 xr
Hcx 4
0.6m shII
cs e A stot
ags xr
Structural system
Ixr
shII
-1
0.00108203m
4
0.00053418m
STRUCTURAL ENGINEERING ROOM
Department of Architecture
202
203
STRUCTURAL ENGINEERING ROOM
Department of Architecture
3.9 Concrete Foundations A foundation is a integral part of the structure which transfer the load of the superstructure
Flexibility sectional weakened due to crack will then be: 1
CIIsh
CIIsh
Eceff Ixr
to the soil without excessive settlement. A foundation is that member which provides support
-1 -2
0.00003343kN m
for the structure and it's loads.
The resulting curvature of shrinkage:
sh
1 r
sh
sh
It also provides a means by which forces or movements within the ground can be resisted by
( 1 ) shI shII
sh
-1
the building. In some cases, foundation elements can perform a number of functions: for
0.00103972m
example, a diaphragm wall forming part of a basement will usually be designed to carry loading from the superstructure.
Deflection in the middle of span due to bending effect (uniform load): fm
5 48
m l
2
fm
If new foundations are placed close to those of an existing building, the loading on the ground
0.02267m
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
Deflection in the middle of span due to shrinkage: fsh
1
l 8 sh
2
fsh
supported. This is particularly important with sands and gravels which derive their support from lateral restraint.
0.00637m
The choice of foundation type or the type of foundation selected for a particular structure is influenced by the following factors:
The resulting deflection: f
fm fsh
f
0.02904m
fmedz
l 250
Does not comply deflection or deflection does not satisfies.
fmedz
0.028m
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.
Figure: 3.9-1 Foundations of tall building
Structural system
204
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.
Square or rectangular footing supporting a single column. Strip footing Long footing supporting a continuous wall. Combined footing
For example, strip footings are usually chosen for buildings in which relatively small loads
Footing supporting two or more columns.
are carried mainly on walls. When the spread footings occupy more than half the area covered
Balanced footing
by the structure and where differential settlement on poor soil is likely to occur a raft foundation
Footing supporting two columns, one of which lies at or near one end.
is found to be more economical. Pad footings, piles or pile groups are more appropriate when
Raft
the structural loads are carried by columns. If differential settlements must be tightly controlled,
Foundation supporting a number of columns or loadbearing walls so as to transmit
shallow strip or pad footings (except on rock or dense sand) will probably be inadequate so
approximately uniform loading to the soil.
stiffer surface rafts or deeper foundations may have to be considered as alternatives.
Pile cap
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.
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
Â
Â
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
Figure: 3.9-2
c. When the foundation is eccentrically loaded, the reactions vary linearly across the
Types of foundations
footing or across the pile system. Footings should generally be so proportioned that zero
These are generally supporting columns and may be square or rectangular in plan and
pressure occurs only at one edge. It should be noted that eccentricity of load can arise
in section, they may be of the slab, stepped or sloping type. The stepped footing results in
in two ways: the columns being located eccentrically on the foundation; and/or the
a better distribution of load than a slab footing. A sloped footing is more economical although
column transmitting a moment to the foundation. Both should be taken into account and
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
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.
bottom of the pad parallel to its sides. 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.
Structural system
STRUCTURAL ENGINEERING ROOM
An essential requirement in foundations is the evaluation of the load which a structure can
Department of Architecture
Pad footing
205
STRUCTURAL ENGINEERING ROOM
Department of Architecture
3.9.1 Shallow Foundations A shallow foundation distributes loads from the building into the upper layers of the ground. 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 (â&#x20AC;&#x2DC;bearing capacityâ&#x20AC;&#x2122;) to carry the load with an acceptable margin of safety and tolerable settlement over the design life. The different types of shallow foundation are: a)
Strip footing
Shallow Foundations
b) Spread or isolated footing c)
Figure: 3.9.1-2
Combined footing Strap or cantilever footing
d) Mat or raft Foundation.
Spread Footing Design of Reinforcement
Punching in Spread Footing Figure: 3.9.1-3 Figure: 3.9.1-1
Structural system
Figure: 3.9.1-4
Figure: 3.9.1-6
3.9.2 Strap Footing It consists of two isolated footings connected with a structural strap or a lever, as shown in figure 3.9.2-1. 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.
Figure: 3.9.1-5
Figure: 3.9.2-1
Structural system
STRUCTURAL ENGINEERING ROOM
Department of Architecture
206
207
STRUCTURAL ENGINEERING ROOM
Department of Architecture
3.9.3 Combined Footing It supports two columns as shown in fig. 3.9.3-1. 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.
Figure: 3.9.3-1 Combined Footing 3.9.4 Strip/continuous footings A strip footing is another type of spread footing which is provided for a load bearing Figure: 3.9.2-2
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 footing is also known as “continuous footing”.
Figure: 3.9.2-3 Figure: 3.9.4-1
Structural system
208
Figure: 3.9.5-2
Figure: 3.9.4-2 3.9.5 Mat or Raft footings
Â
It is a large slab supporting a number of columns and walls under entire structure or a
3.9.6 Pile foundations
large part of the structure. A mat is required when the allowable soil pressure is low or where
Deep foundations are used when the soil at foundation level is inadequate to support the
the columns and walls are so close that individual footings would overlap or nearly touch each
imposed loads with the required settlement criterion. Where the bearing capacity of the soil is
other. Mat foundations are useful in reducing the differential settlements on non-homogeneous
poor or the imposed load are very heavy, piles, which may be square, circular or other shapes
soils or where there is large variation in the loads on individual columns.
are 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. They can have bases that are flat, pointed or bulbous, and shafts that are vertical or raked. In some circumstances, piles can be constructed of other materials, such as timber or
Figure: 3.9.5-1
plastics. Piled walls or sheet piles are used to resist lateral movements, such as in forming a basement.
Structural system
STRUCTURAL ENGINEERING ROOM
required to bridge over soft spots at movement joints or changes in founding strata.
Department of Architecture
A traditional strip foundation consists of a minimum thickness of 150 mm of concrete placed in a trench, typically 0.8â&#x20AC;&#x201C;1 m wide. Reinforcement can be added if a wider strip is
209
STRUCTURAL ENGINEERING ROOM
Department of Architecture
The piling technique used to install the piles will be determined by the ground conditions, loading requirements for the final pile as well as other factors such as access or proximity to other buildings and the need for noise reduction. Pile types There are two basic types of piles: â&#x2014;? cast-in-place (or replacement) piles and â&#x2014;? driven (or displacement) piles.
Figure: 3.9.6-3 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
Figure: 3.9.6-1
anchorage past the centreline of each pile.
Figure: 3.9.6-2 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.
Structural system
Figure: 3.9.6-4
Figure: 3.9.6-5
Figure: 3.9.6-6
Figure 3.9.6-4: Main reinforcement in slab foundation Figure 3.9.6-3: collapse of unbearable soil
Structural system
STRUCTURAL ENGINEERING ROOM
Department of Architecture
210
211
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Example 3.9-1: Assessment of slab foundation to punching
P P1 Qbu
P 1188 kN
Required surface area of reinforcement to punching:
Depth of the reinforced slab foundation: hd 80 cm
Asb
Tensile strength of concrete:
P 0.86 fyd
2
Asb 0.00368372 m
Reinforcement diameter:
fctm 0.9 MPa Width of column:
bs 50 cm
25 mm
As1
2
4
2
As1 0.00049087 m
Number of profiles:
Height of column: n
hs 40 cm
Asb
n 7.504
As1
Q 0.42 hd fctm ucr
Q 1512 kN
Design strength of reinforcement:
fyd 375 MPa
Figure: 3.9.1-2 Data of rolled I profiles:
Figure: 3.9.1-1
I28
Perimeter of critical cross-section: ucr bs hd hs hd 2
3
b1 119 mm
Shearing force carrying by concrete:
Qbu 0.42 hd fctm ucr
Qbu 1512 kN
2
A1 6.10 10 mm
ucr 5 m
P1 2700 kN
b3 119 mm
Structural system
6
4
J1y 75.8 10 mm b2 119 mm
h1 280 mm
212
2
6
4
J1y 157 10 mm
h1 340 mm b1 137 mm
b2 137 mm
b3 137 mm I38 3
2
6
A1 10.7 10 mm b1 149 mm L 1.45 m
b2 149 mm p1
P 8
4
h1 380 mm
J1y 240 10 mm
b3 149 mm
p1 148.5 kN
h2 20 mm
M 161.49375 m kN
M p1 0.75 L 3
A2 b2 h2
A3 b3 h3
J2
3
b2 h2
J3
12
h2 h1 h3 h2 A2 A3 h1 h2 2 2 2
A1 e2
12
Section modulus: J e1
3
Wd 0.00227904 m
Wh
J e2
3
Wh 0.00227904 m
Stress control:
e1 H e2
H 0.42 m
Figure: 3.9.1-3
Wd
b3 h3
e2 0.21 m
A1 A2 A 3
H h1 h2 h3
h3 20 mm
d
M Wd
h
M Wh
d
h
70.86 MPa 70.8604 MPa
s
210 MPa
s
210 MPa
e1 0.21 m a1
h1 2
h2 e2
a2 e2
h2
a3 H
2
h3 2
a1 0 m
e2
The total moment of inertia of composite section: 2
2
2
J J1y J2 J3 A1 a1 A2 a2 A3 a3
4
J 0.0004786 m
J 1.99416111 J1y
Structural system
STRUCTURAL ENGINEERING ROOM
3
A1 8.67 10 mm
Department of Architecture
I34
213
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Example 3.9-2: Determination of the design bearing capacity of the soil at depth
Example 3.9-3: Endless beam
dp =1,5 m
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
Soli classification F6:
cef
16 kPa
21 deg
ef
cef
cd
2
M at any distance x from the point of application P can be expressed as follows:
ef
d
m
P( x)
C Y( x)
( x)
Bearing coefficient of the soil: Nd m
tan 45 deg
d
2
tan d
e 2
1.5 Nd 1 tan d
Nb
ef
21
1
ef 4 deg
Nc
C 15000
2
2
3
m
1
Base area of the footing:
Width:
bp = 1 m
1 12
dd
1 0.1
dp bp
1
sin 2 d
bp lp id
sin d 1
ic
sb
1
1 0.3
ib
bp lp
dc
1 0.1
dp
4 E I C B
h 1 m
4
L 22.2 m
Le
4 E I C B
Le 4.949m
L Le
( )
1 4
0 0.05 6
(cos() sin())
e
( )
1
2
e
( cos ( ) )
( cos ( ) sin ( ) )
0.6 0.4 ( )
0.2
( )
( )
cd N c sc dc ic 1 dp N d sd dd id 2
I 0.125m
x
1
Design bearing capacity of the soil:
Rd
4
3
Le 1
bp
B h
() e 2
sd
B 1.5 m
Le
E = modulus of elasticity of the base strip
Coefficient of the shape of the footing:
lp
E 27000 MPa
3
( x) P Le
Where
bp
M ( x)
The coefficients , , are functions of the ratio
1 0.2
Le B
4
( x) P
I = the moment of inertia of the cross section of the base strip
Length: Lp = 6 m
sc
T( x)
m
I
kN
kN
P
bp 2
0
2
4
0.2
N b sb db ib
Rd
243.077 k
0.4 0.6
Figure: 3.9.3-1
Structural system
6
4.486
Q1 ........ Le / 2 from the cross-section A => = P / 2 => (P / 2) = -0.052 and (P / 2) = 0
i 1 n
P 2265 kN
x 2.1 m
P 1821 kN
P 2265 kN
x 20.1 m
L 22.2 m
1 4
1 4
2
x 9.3 m 2
P 1821 kN 3
x 12.9 m 3
=> 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 =>
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
A signed convention:
=>
Shear
force
applied
force
Q3
in
cross-section
B
is
equal
to
0
- Coefficients , apply equally to both positive and negative ratio, ie If the = 0.140 ->
Q4 ........ L / 4 the cross section B => = p / 2 => (P / 4) = 0, and (P / 4) = 00:16
= 0.185 as well as the value of the ratio = - 0.140 -> = 0.185 Coefficients are symmetrical
=> Bending moment deduced force Q4 in cross-section A is equal to 0
- Coefficient is asymmetrical.
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: 1. condition: MA
Fi iA
V Le 2 1
0
n
2. condition: VA
Fi iA
0
V 2 1
n
V Le FSd i Le i A 4 2 n
V 4 2
3. condition: MB
Fi iB
0
V Le 2 3
n
4. condition: Figure: 3.9.3-2
VB
Fi iB
0
V 2 3
n
FSd i iA n
V Le FSd i Le i B 4 4 n
V FSd 4 4 i iB n
B - Final beam The final beams are called real beams. In this case, the final beam computed as part of
( )
the endless, which in the extreme cross-sections A, B satisfies the following boundary
( )
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:
4
e
( cos ( ) sin ( ) ) if 5.1
0 otherwise
conditions: MA = 0, MB = 0, TA = 0, TB = 0,
1
1 e ( cos ( ) ) if 5.1 2 0 otherwise
Structural system
STRUCTURAL ENGINEERING ROOM
n 4
Department of Architecture
214
From 3 Q3 conditions can be calculated. Q4 product. (P / 4) = 0, while (p / 2) = -0052 =>
STRUCTURAL ENGINEERING ROOM
Department of Architecture
215
n
n
Q3
i1
Pi Bi
3
Q3 1.03 10 kN
0.052
V
i1
FSd i Bi
4
0.052
From 4 conditions can be calculated Q4, Q3 product. (p / 2) = 0, while (p / 4) = -0160 => Traces of bending moments M along the entire beam of length L xL 0 m0.1 m L
M xL Le
Figure: 3.9.3-3 From 1. Condition the first conditions can be calculated Q1, Q2 product. (P / 4) = 0, while
(p / 2) = -0052 => n
xi A Le i
B i
Lx
i
Le
i1
Q1
Pi A i
3
Q1 1.03 10 kN
0.052
( Q1) Q3
n x x 4 Le xL P L i Q2 i Le Le Le i 1 L L x Le L xL L 4 e 2 Q4 Le Le
2
Le xL
2 794.011 kN m
M x
n
B
A
i 4.061
i 0.424
2.606
1.879
1.879
2.606
0.424
4.061
V
i1
FSd i A i
1
d 1 xL
0.052
xL x if xL x 0 1 1
d 2 xL
0 otherwise
d 3 xL
0 otherwise
xL x if xL x 0 3 3
d 4 xL
0 otherwise 4
Q2
i1
i
2.265·103
n
Pi A i 0.160
q
P
n
3
Q2 3.521 10 kN
V
2
i1
FSd i A i 0.160
xL x if xL x 0 4 4 0 otherwise
From 2. Condition the second condition can be calculated Q2, Q1 product. (p / 2) = 0, while (p / 4) = -0160 =>
xL x if xL x 0 2 2
kN
P
i1
L
i
xL 0 m0.1 m L
1.821·103 1.821·103 2.265·103
2
xL P d 1 xL P d 2 xL P d 3 xL P d 4 xL MA xL q 1 2 3 4 2
Structural system
Beam width = 1.5m L = 22.2m
h=1.0m
6
110
MA xL M xL
0
10
20
6
110
30
1 P. xL Le B
6
210
( Q1) Q3
L x n 4 e L xL xi Q2 P i Le Le Le i 1 Le L xL 4 Le L xL 2 Q4 Le Le
2
Le xL
6
310
P.( 9.3 m) 229.001 kPa
xL
Figure: 3.9.3-4 3
5
MA ( 5.7 m) 2.174 10 kN m 2
3.210
xL MA xL q P d 1 xL P d 2 xL P d 3 xL P d 4 xL 1 2 3 4 2
xL
5.2 m
-1.066·103
kN m
L
0 m
M xL
-2.045·103
5.3
-1.022·103
-2.078·103
5.4
-974.464
-2.107·103
5.5
-923.759
-2.133·103
5.6
-869.735
-2.156·103
5.7
-812.388
-2.174·103
5.8
-751.714
-2.189·103
5.9
-687.711
-2.2·103
6
-620.372
-2.208·103
6.1
-549.693
-2.211·103
6.2
-475.668
-2.211·103
6.3
-398.293
-2.208·103
6.4
-317.559
-2.201·103
6.5
-233.462
-2.19·103
6.6
-145.995
-2.175·103
...
...
...
x4
310
5
P. xL dxL L
MA xL
x2
5
P. xL
2.810
5
2.610
5
2.410
kN m
5
2.210
0
13.333
26.667
40
xL
Figure: 3.9.3-5
The course of shear forces along the beam of length L
n
i 1
P
i
d 1 xL
8.172 10
3
kN
1 if xL x 0 1
L 0m
P . xL L
d 2 xL
1 if xL x 0 3 0 otherwise
Structural system
245.3
kPa
1 if xL x 0 2 0 otherwise
0 otherwise d 3 xL
dx L
d 4 xL
1 if xL x 0 4 0 otherwise
STRUCTURAL ENGINEERING ROOM
The course of the contact stress in the ground along the entire beam of length L 6
210
Department of Architecture
216
STRUCTURAL ENGINEERING ROOM
Department of Architecture
217
T xL
( Q1) Q3
L x n 4 e L xL xi xL xi Q2 P i Le xL x Le Le i 1 i L L x Le L xL L 4 e 2 Q4 Le Le
2
Le xL
TA xL q xL P d 1 xL P d 2 xL P d 3 xL P d 4 xL 1 2 3 4
210
110
Example 3.9-4: Design of reinforcement in footing
Material characteristics Concrete
fccub_ 30 MPa fccub_ f fccyl 0.855 0.005 MPa ccub_
3
T( 13.2 m) 1.095 10 kN
6
fccub_ fctm 0.274 MPa
x2
x4
0.66
MPa
fctm 2.59MPa
fccyl fcd 0.85 1.5
6
fccyl 21.15MPa
fcd 11.985MPa
T xL
TA xL
0 110 210
10
20
30
6
xL
Figure 3.9.3-6
P. xL B 456.628 454.779 452.928 451.073 449.215 447.352 445.482 443.605 441.718 439.819 437.906 435.975 434.023 432.047 430.044 ...
Steel
6
kN m
T xL
-4.22
kN
TA xL
0
41.394
36.811
86.829
73.622
132.086
110.432
177.164
147.243
222.064
184.054
266.785
220.865
311.326
257.676
355.688
294.486
403.86
331.297
447.732
368.108
491.411
404.919
534.898
441.73
578.188
478.541
621.28
515.351
...
...
kN
fyk 410 MPa
fyk fyd 1.15
ly 6 m l1 7.2 m
lk 2.16 m
fyd 356.522 MPa bs 400 mm
Distributed loads Distributed loads of the floor:
qdx 11.75
kN 2
m Surface roof load: qstr 8.22
Load area ZS1
kN m
2
l1 zs1 ly lk 2
zs1 34.56 m
Load area ZS2
l2 l1 zs2 ly 2 2
Structural system
zs2 32.4 m
2
2
l2 3.6 m
kv 2.85 m
Force load from the outer walls:
h
F1d 82 kN
kN
F2d bs kv 25 1.35 3 m
F2d 15.39 kN
Service force acting on the end column: N1s
Number of floors: 4 kN 2 N1d qdxzs1 F1d F2d n qstrzs1 bs ( 2.85 m 1.5 m ) 25 1.35 3 m
3
N1d 2.321 10 kN
d 1.2 m
N1d 2.321 103 kN
Total load on the foundation will be:
h 1.575 m
Extreme force applied at the end column:
Force load of self-weight of the column: 2
H 10
2
N2d qdxzs2 F2d n qstrzs2 bs ( 2.85 m 1.5 m ) 25
kN m
3
1.35
N2d 1.874 103 kN
N1d
N1s 1.857 103 kN
1.25
Extreme force applied in the central column: N2d 1.874 103 kN
Service force acting on central columns: N2s
N2d
N2s 1.499 103 kN
1.25
Preliminary design of footing dimensions: Max. Distance between columns: l1. 7.2 m 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
Figure 3.9.4-1: Calculation model
The depth of the footing:
Design dimensions of reinforced concrete STRIP – footing. Building height:
H n 2.85 m ( 2.85 m 1.5 m )
h 1.44
H 15.75 m
Structural system
N1s 0.6 fcd
5 cm
h 0.782 m
STRUCTURAL ENGINEERING ROOM
The degree of constraint of building:
Force load
Department of Architecture
218
STRUCTURAL ENGINEERING ROOM
Department of Architecture
219
h
Ns.max
1.44
0.6 Rbd
Dimension checking of the foundation:
Nd.max
5 cm
1.44
f
0.6 Rbd
5 cm
The actual weight footing:
0.8 m
18
kN m
3
zs1
650.849 kN
zs2
250.669 kN
kN m
3
Weight of soil below the foundation:
The tabulated carrying capacity of the soil at a depth of footing bottom = 1 m for S4 Rdt 260kPa
zs2
(BL) (d h )
Force in footing bottom:
Tabulated load bearing capacity of the soil at a depth of footing bottom = t + h: R
BLh 25
zs1
Earth weight:
dt Rdt 2.5 ( d 1 m)
R
dt 269 kPa
The sum of all forces from the columns to the base strip – footing: Vs 2 N1s 2 N2s
3
Vs 6.713 10 kN
Vskut Vs zs1 zs2 Maximum tension in footing bottom: skut
Vskut
skut
B L
228.664 kPa
<
R
Overall extreme force (weightlessness strip) n
Vd 8.391 103 kN
Vd
i1
Gzs 0.1 Vs
Preliminary ground plan of the foundation strip L x B = Aef Vs Gzs Aef R dt The width of the foundation: B
Aef
L I suggest: B 1.5 m
B 1.237 m
Aef 27.451 m
2
satisfies
dt 269 kPa
Calculation sectional forces
Vd 2 N1d 2 N2d
Preliminary weight of foundation:
Vskut 7.615 103 kN
Nd
i
The stress in footing bottom: d1
Vd
d1
B L
251.99 kPa
Vd
1
B L
n
Ndi xi
e´
i1 n
i1
Vd
2
Nd
BL
6 Vd e 2
BL
6 Vd e 2
BL
i
The reaction in footing bottom: fd d1 B
Structural system
kN fd 377.985 m
E I
d4
4
dx
s ( x) BC s ( x)
f ( x)
Q
e
E I
x Le x C1 cos C2 sin e Le Le
x
d3
3
dx
Le
s ( x)
4
C3 cos
4 EI
x C4 sin Le Le x
M
BC
EI
d2
2
dx
s ( x)
The bending moment calculation: lv Ma fd lv 2
Ma 833.456 kN m
l1 lv Mab fd lv 2 2
Mb fd lv l1
lv l1 2
l1 2
N1d
l1
Mab 2.217 103 kN m
2
N1d l1
l2 lv l1 Mbb fd lv l1 2 2
Mb 368.511 kN m
l2 2
N1d l1
l2 2
l
N2d 2 2
Mbb 980.846 kN m
Figure 3.9.4-2: Bending moments diagram vs Shear forces diagram due to design loads Dimensioning
The computation of shear forces:
Design of reinforcement for the maximum Bending moment between the columns : Design moment:
Vk fd lv
Vk 793.768 kN
Vka N1d Vk
Vka 1.528 103 kN
M Mab
Cover
Vab fd lv l1 N1d
Vab 1.194 103 kN
Vbb N2d Vab
Vbb 680.373 kN
cst 5 cm effective height:
d h c st
d 0.732 m
Compression depth of concrete: xu d
2
d
M 0.5 Bfcd
Structural system
xu 0.079 m
STRUCTURAL ENGINEERING ROOM
s ( x)
x Le
Department of Architecture
220
Or we can be calculated according to the design of reinforced concrete beam as follows:
The compression height limit
STRUCTURAL ENGINEERING ROOM
Department of Architecture
221
420 d MPa xulim 525 MPa fyd
xulim 0.349 m
Required area of reinforcement: xuBfcd As fyd
As As1
( 25 mm)
2
As1 4.909 cm
4
Mab
0.86742
d h cst
2
Bd fcd
x d
h 0.782 m
0.23
x 0.243 m
Designed of reinforcement:
2
Amsk 39.27 cm
The depth of compression zone of concrete: xureal 0.078 m
xus 0.194 m
As
97.86 cm
Mult As zfyd
Mult 2.263 MN m
2
As 100 cm
M
h 0.782 m
d 0.732 m
0.102
0.03118
0.94441
x d
x 0.102 m
xu 0.081 m
z d
z 0.691 m
Mult As zfyd
Mult 1.011 MN m
As Bd fcd 100
As
2
Bd fcd
Mu xureal B fcd d
0.13898
xu 0.8 x
2
41.03 cm
Design of reinforcement for bending moment between the columns:
Carrying capacity of cross-section:
xus 0.8 x
we provide
As Bd fcd 100
2
Bfcd
0.07437
z 0.635 m
M 0.980 MN m
Assessment of the Design:
fyd Amsk
d 0.732 m
Or we can be calculated according to the design of RCB B3-B3.3 as follows:
n 8.094
Amsk As1 8
0.33144
z d
2
I suggest 8 V25 to the entire width of the strip foundation b:
xureal
fcd 11.985 MPa 2
Number of profiles: n
B 1.5 m
As 39.733 cm
Area 1 of 25 profile: As1
Mab 2.217 MN m
xureal 2
Mu 970.046 kN m
M 980.846 kN m
M 0.833 MN m
h 0.782 m
fyd 356.522 MPa
Structural system
M 2
Bd fcd
d 0.732 m
0.087
fcd 11.985 MPa
0.0265
x d
x 0.086 m
xu 0.8 x
xu 0.069 m
z d
z 0.697 m
As Bd fcd 100
As
Mult As zfyd
Mult
0.11808
2
34.87 cm
Sectional area of the stirrups: 2
dss Ass 4
0.866 MN m
The force that is transferred by stirrup: Nss Ass 4 fyd
Design of shear reinforcement
Nss 71.683 kN
Area A
Distance between the stirrups:
Sectional shear resistance:
c ss Nss ss 3.79 m Vs Suggestion 4-cutting stirrups profile dss = 8mm at a distance of 400 mm
1
Vcu 1.011 103kN
Vcu Bh fctm 3
Shear force applied at a given cross-section: Vk 793.768 kN Force transferring by stirrups;
Vs Vk Vcu
Vs 217.152 kN Figure 3.9.4-3: Shear forces diagram with side view of foundation
The projection of the cracks: c
1.2 Bfctm d
2
Vs
c 11.48 m
maximum. the projection of the cracks: Cmax 0.18
fcd q fctm
Cmax 0.834
q
1
min. the projection of the cracks: Cmin h 0.04 m
Cmin 0.742 m
Suggestion 4-cutting stirrups diameter:
Figure 3.9.4-4: Load calculation areas on plate - load from one floor
dss 8 mm
Structural system
STRUCTURAL ENGINEERING ROOM
0.95277
Department of Architecture
222
Example 3.9-5: Static calculation of extreme square isolated footings
STRUCTURAL ENGINEERING ROOM
Department of Architecture
223 The total load on the foot:
Number of floors: n = 4
Nd qd zs F1d F2d n qstr zs bs ( 2.85m 1.5m ) 25 2
kN m
3
1.35
Nd 2.48 103 kN Foundation dimensions: Building height:
H n 2.85m ( 2.85m 1.5m )
H 15.75m
Figure: 3.9.5-1 Material properties – design value of concrete cylinder compressive strength: fcd 14.16MPa
fctm 2.01MPa
fyd 443MPa
The span among the columns: ly 6m
l1 7.2m
lk 2.16m Figure: 3.9.5-2
Distributed loads:
The degree of constraint of buildings:
Distributed loads of the ceiling: qd 13kN m
2
h
Flat roof load: qstr 7.8kN m
H
h 1.575m
10
The force acting in the contact columns and footing foundation (extreme values): 2
Nd 2.48 103 kN Service value applied force:
Load area: l1 zs ly lk 2 Force load:
zs 34.56m
2
Ns
Nd 1.2
Ns 2.066 103 kN
Force load from the peripheral walls:
Tabulated load bearing capacity of the soil in the bed joints of the depth = 1 m:
F1d 82kN Force load of self-weight of columns:
Rdt 400kPa Earth gravity:
2
F 2d b s k v 25
kN m
3
1.35
F2d 15.39kN
18
kN m
Structural system
3
Ns 5cm 0.6fcd
d 1.44 I suggest:
Weight footings:
d 0.76m
2
Nzs1 b d 25
N zs1 92.45 kN 3 m Weight of soil under the foundations of structures:
d 0.8m
Rdt 2.5( h d 1m )
R dt
461.875kPa
Vs Ns 0.1Ns Vs 2.273 103 kN
Nzs Nzs1 Nzs2
Preliminary calculation of the effective surface which is able to transfer force and stress not exceed Rtd: R dt
Nzs2 126.512kN
Vsk Ns Nzs
Vsk 2.285 103 kN
The stress in the bed joints and assessed for bearing capacity of foundation soil:
Aef 4.921m
2
Dimensions of square foundation: b Aef b 2.218m Design dimensions of the foundation: 2
Aef b b 2.15m Assessment of foot in terms of soil bearing capacity:
Vsk
z 494.41kPa < R dt 461.875kPa 2 b Dimensioning (of extreme values weightless footing): z
Vs
2
The resulting force applied at the footing bottom (with gravity base):
Preliminary calculation of the force applied in footing bottom (if self-weight Ns = 10%):
Aef
2
Nzs2 b bs ( h )
The tabulated bearing capacity of soil to the depth of the bed joints = d + h: R dt
kN
geometry: a
b bs
a 0.875m
2
Length of console Projection: ak a 0.15bs
ak 0.935m
The stress in footing bottom of extreme load: Nd
d 536.45kPa Aef Maximum moment acting on the footing: d
M
ak
2
2
b d
Vmax ak b d
M 504.151kN m Vmax 1.078 103 kN
Arm of internal forces determined approximately: ast 6.5cm deff d ast Force in reinforcements: Ns Figure 3.9.5-3: Determination of bending moment and shear force
M zb
Ns 806.964kN
Structural system
zb 0.85 deff
zb 0.625m
STRUCTURAL ENGINEERING ROOM
Height of the foundation:
Department of Architecture
224
Required area of reinforcement:
STRUCTURAL ENGINEERING ROOM
Department of Architecture
225
As
Ns
For square footing and column:
As 1.822 10 3 m
fyd
uc2 uc1 Peripheral length of the critical cross-section:
2
ucr 4.8 m ucr 4 uc1 Stress from acting load-of bearing structure acting on footing bottom (without sel-weight of footing):
Area 1 18: d1 18mm
2
A1 d1
4
2
A1 2.545cm
n
As
n 8.442
A1
we propose 10 V18 the entire width of the footing => distance between the reinforcement will be 200 mm: smin
1 3
Rbtd Rsd
Ask s d b
smin
s
8 10 4
1.479 10 3
Ask A1 10
2
Ask 25.447cm
> smin 8 10 4
Nd
d 0.536MPa 2 b Shear force of the extreme loads acting on the critical cross section: d
2
Nqd d b uc1
2
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
factor of reinforcement: s 1 50 s smin factor for height of section (for d > 0.6m): h 1 normal force coefficient: n
qcu 0.42h s h n fctm
1
qdmax 355.677
kN m
<
qcu 1.327 103
kN m
Figure 3.9.5-4: Critical area for punching The reliability condition to avoid punching: Sectional dimensions critical for punching: uc1 d bs
uc1 1.2 m Figure: 3.9.5-5
Structural system
=> satisfies
STRUCTURAL STRUCTURALENGINEERING ENGINEERINGROOM ROOM
Department of Architecture
226
STRUCTURAL ENGINEERING ROOM
Department of Architecture 227
Realisation Projects
Apartment Buildings- Panské in Stupava Author: Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD.
STRUCTURAL STRUCTURALENGINEERING ENGINEERINGROOM ROOM
Department of Architecture
228
STRUCTURAL ENGINEERING ROOM
Department of Architecture
229 Assumptions.
Variables
Concrete simple, non-structural C 16/20 – X0
snow load
Reinforced concrete C 25/30 – XC2, XF1
Height above sea level
175.00 + 15.00 ≈ 190.00, zone 1.
Steel B 500 B - 10 505 – R and welded networks KARI B 490 B – SZ
coefficients zone – a = 0.454; b = 970
Steel flat rolled bars and S 235
Sk = 0.454 + 190/970 = 0.65 KN/m2
Masonry YTONG
coefficients:
μ1 = 0.80; Ce = 1.00; Ct = 1.00
snow load
0.80 x 1.00 x 1.00 x 0.65
0.55 x 1.50 = 0.80 KN/m2
Loading and operating effects. vertical load.
Ceiling interior - typical floor – RC slab 180 mm
A roof over the last floor
Dead load
Dead load 2
The floor
≈
1.80 x 1.35 = 2.45 KN/m²
0.18 x 25.00
gravel embankment
0.05 x 18.00
0.90 x 1.35 = 1.25 KN/m
Self-weight of the RC slab
4.50 x 1.35
6.10
Water insulation
≈
0.25 x 1.35 = 0.35
Plaster, ceiling, wiring professions ≈
0.35 x 1.35
0.45
Thermal insulation with gradients
≈ 0.30 x 1.35
∑
6.65
9.00 KN/m²
Vapor barrier, geotextiles
0.10 x 1.35
0.15
Variables
Self-weight of the RC slab 0.18 x 25.00
4.50 x 1.35
6.10
Utility – Apartments
Plaster, ceiling, wiring professions ≈
0.35 x 1.35
0.45
∑
6.40
8.70 KN/m²
≈
0.40
social and assembly areas utility room - boiler room
Variables
partitions
snow load
I wonder weighing
Height above sea level
175.00 + 15.00 ≈ 190.00, zone 1.
coefficients zone
Sk = 0.454 + coefficients: snow load
970
5.00 x 1.50 = 7.50 1.20 x 1.35 = 1.65 KN/m2
Dead load 2
= 0.65 KN/m
μ1 = 0.80; Ce = 1.00; Ct = 1.00 0.80 x 1.00 x 1.00 x 0.65
3.00 x 1.50 = 4.50
Balconies, balcony
a = 0.454; b = 970 190.00
2.00 x 1.50 = 3.00 KN/m2
0.55 x 1.50 = 0.80 KN/m2
The floor
6.10
Plaster soffit
0.35 x 1.35
0.45
6.35
8.60 KN/m²
∑
Dead load
Variables 0.05 x 18.00
0.90 x 1.35 = 1.25 KN/m2
Water insulation
≈
0.25 x 1.35 = 0.35 0.40
1.50 x 1.35 = 2.05 KN/m² 4.50 x 1.35
Roof above the elevator shaft – RC slab150 mm gravel embankment
≈
Self-weight of the RC slab 0.18 x 25.00 ≈
4.00 x 1.50 = 6.00 KN/m2
Utility - balconies, loggias RC slab interior - the 1 st level- 200 mm slab
Thermal insulation with gradients
≈ 0.30 x 1.35
Vapor barrier, geotextiles
≈
0.10 x 1.35
0.15
Dead
Self-weight of the RC slab
0.15 x 25.00
3.75 x 1.35
5.10
The floor
≈
1.80 x 1.35 = 2.45 KN/m²
Self-weight of the RC slab
0.20 x 25.00
5.00 x 1.35
6.75
Plaster soffit
≈
0.35 x 1.35
0.45
Plaster, ceiling, wiring professions ≈
0.35 x 1.35
0.45
∑
5.65
7.70 KN/m²
Realisation Projects
230 9.65 KN/m²
Horizontal load Wind
2
Utility – Apartments
2.00 x 1.50 = 3.00 KN/m
Wall partitions
I wonder how static load in two directions (X and Y) Basic wind speed - 26 m / s
1.00 x 1.50 = 1.50 KN/m2
I wonder weighing
Terrain category – III I am not dividing the load along the height, then we considering the same, which will have
Arm of the staircase - 240 mm slab depth
mean wind speed and peak wind pressure: Z e = 15.50 m; vm , ze1 = 21.60 m/s; q p, ze, 1 = 0.85 KN/m2
The slope of the staircase ≈ 35º The average thickness of the slab with steps
h priem = 0.181 x cos 35 / 2 + 0.24 = 0.32 m
Dead load on plan view plane Surface treatment
0.03 x 23.00 0.18 x 0.03 x 23.00 / 0.28
RC slab + levels
0.32 x 25.00 / cos 35
Plaster treatment
≈
∑
Wind - Wind actions
0.85 x 1.50 = 1.30 KN/m2
The coefficients of the external pressure, - a suction pressure in vertical: 2
0.70 x 1.35 = 0.95 KN/m
pressure- Cpe,10 = + 0.80
0.45 x 1.35
suction -
0.60
9.80 x 1.35 13.20 0.45 x 1.35 11.40
0.60
Cpe,10 = - 0.70
Horizontal load scheme pressure suncion
15.35 KN/m²
all
Variables Utility - staircase accommodation
3.00 x 1.50 = 4.50 KN/m2
Stair-case – RC slab 240 mm Dead load The floor
≈
0.70 x 1.35 = 0.95 KN/m²
RC slab
0.24 x 25.00
6.00 x 1.35
Plaster ≈
0.35 x 1.35
0.45
∑
7.05
8.10 characteristic
9.50 KN/m²
Variables Utility - staircase accommodation
3.00 x 1.50 = 3.00 KN/m2
Seismic load The area is classified in 7º seismic MSK-64; Category ground B. Source areas of the seismic risk of a seismic acceleration agr = 0.86 m / s 2 coefficient significance class γ = 1:00; subsoil factor B → S = 1:35 Behaviour factor for horizontal seismic load: The compliance structure – DCM Behaviour factor q 0 = 3:00; The effect of the dominant mode of failure to K w = 1.00
Realisation Projects
STRUCTURAL ENGINEERING ROOM
7.15
Variables
Department of Architecture
∑
231
STRUCTURAL ENGINEERING ROOM
Department of Architecture
Coefficient of behaviour construction q = 3.00 x 1.00 = 3.00 The magnitude of the earthquake: I 0 = 70; the depth of focus h ≈ 8 Km
The epicenter intensity Magnitude Mp spectra
= 00:55 0.95 x 7 = 5:50 → 4.80 ≤ elastic response type 2
- 0.05 s → 2.91 m / s 2
For these loads I am considering the following combinations: Combinations ULS: 1. combination 1.35x1 + 1.35x2 + 1.50x3 + 1.35x4 + 1.50x5 2. combination 1.35x1 + 1.35x2 + 1.35x4 + 1.50x0.70 x (3+5) + 1.5x6 3. combination 1.35x1 + 1.35x2 + 1.35x4 + 1.50x0.70 x (3+5) + 1.5x (-6) 4. combination 1.35x1 + 1.35x2 + 1.35x4 + 1.50x0.70 x (3+5) + 1.5x7 5. combination 1.35x1 + 1.35x2 + 1.35x4 + 1.50x0.70 x (3+5) + 1.5x (-7)
Magnitude earthquake: The epicenter intensity Magnitude
0
I 0 = 7 ; depth of focus h ≈ 8 Km
Ms = 0.55 x 7 + 0.95 = 4.80 ≤ 5.50 → the elastic response of the type 2
spectra – 0.05 s → 2.91 m/s2
6. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 1.00x8 + 0.30x9 7. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 1.00x8 + 0.30x (-9) 8. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 1.00x (-8) + 0.30x9 9. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 1.00x (-8) + 0.30x (-9)
0.20
2.91
10. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 0.30x8 + 1.00x9
0.50
1.46
11. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 0.30x8 + 1.00x (-9)
1.00
0.72
12. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 0.30x (-8) + 1.00x9
1.50
0.39
13. combination 1.00x1 + 1.00x2 + 0.30x3 + 1.00x4 + 0.30x5 + 0.30x (-8) + 1.00x (-9)
2.00
0.22
3.50
0.11
combinations SLS:
5.00
0.07
1. combination 1.00x1 + 1.00x2 + 1.00x3 + 1.00x4 + 1.00x5
These values are considered for the calculation of seismic loading. Actions and combinations thereof. In the calculation, we considered these basic load: 1 - self weight parts of the structure 2 – dead load 3 - snow 4- the partition and outer walls 5 – variables load 6- wind load direction Y 7 - wind direction X 8 - seismicity the Y – direction 9- seismicity X – direction
Horizontal bearing structures For the design of horizontal bearing structures based on the results of the spatial 3D model, which are documented internal forces, respectively, directly required reinforcement and deformation on concrete with cracks. Reinforced concrete flat slab Level 6 - roof lift. The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below. Reinforced concrete flat slab Level 5. The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below. Reinforced concrete flat slab Level 4. The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below.
Realisation Projects
232 Reinforced concrete flat slab Level 3.
along the two edges horizontal 0.001 x 20 x 100 = 2.00 cm2 along the two edges
Reinforced concrete flat slab Level 2. The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below.
Graphic attachments are classified according to the numbering of the drawing plans Foundation.
Reinforced concrete flat slab Level 1.
I take the load on the foundation of spatial 3D model basis to consider blanket, base slab
The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below.
foundation with a thickness of 500 mm. In a computational model it considers the infra-structure in interaction with the upper structure. The base gap will be located in a layer of clay sandy F4 to F6-EN-CI. The effects of groundwater
staircase.
on the foundation not considered.
The angle slab stairs thickness of 240 mm. The reinforced concrete slab is supported in the level 1 to the base slab foundation and the upper floors to the ceiling tile. Deformation of the concrete with cracks and required minimal reinforcement - see below.
These soils have the following mechanical properties: E def= 5.00 until 7.00 MPa C ef= 0.05 until 0.01 MPa φ ef= 200
Vertical support structures.
γ = 19.00 KN/m3
The vertical support structures form is reinforced concrete monolithic, reinforced concrete wall thickness of 180 to 200 mm, the wall thickness precision Ytong 300 mm, steel columns Q 150/150/5.
ν = 0.40 β = 0.50
partial coefficients – γ R = 1.40; γ c = 1.00; γ φ = 1.00 C d = 1.00 x 0.01 = 0.01 MPa
Results taken from the spatial model. On the walls and columns they are documented in
φ d = 1.00 x 20.00 = 20.000
graphic form the resulting internal forces for the steel columns and on each wall marked the
depth of slab foundation D min = 0.80 m; width of slab foundation B = 1.00 m; Length of slab foundation
necessary reinforcement in cm2; in the areas of reinforcement is taken into account minimum
L = 10.00 m
percentage of reinforcement under the current guidelines of EC2 Minimum % reinforcement:
The design resistance Subsoil : Coefficients of bearing capacity of foundation soil for φ d = 20.000
- vertical reinforcement A s, vmin = 0.002 x b x h
N q = tg2 (45 + 20.00/2) . e3.14 x tg 20.00 = 6.40
- horizontal reinforcement A s, hmin = 0.001 x b x h, or 0.25 x A s,vmin -
N c = (6.40 – 1) x 1/tg 20.00 = 14.83
- The greater of maximum reinforcement%: A s, vmax = 0.04 x b x h
N γ = 1.50 (6.40 – 1) x tg 20.00 = 2.95 2
Minimum reinforcement for a wall thickness of 180 mm - vertical 0.002 x 18 x 100 = 3.60 cm along the two edges horizontal 0.001 x 18 x 100 = 1.80 cm2
Coefficients of the shape of the base: s c = 1 + 0.20 x 1.00/10.00 = 1.02 s q = 1 + 1.00/10.00 x (sin 20.00) = 1.03
Realisation Projects
STRUCTURAL ENGINEERING ROOM
The reinforced concrete flat slab - 180 mm thick, deformation of the concrete with cracks a necessary minimum reinforcement - see below.
Department of Architecture
along the two edges Minimum reinforcement for a wall thickness of 200 mm - vertical 0.002 x 20 x 100 = 4.00 cm2
233
STRUCTURAL ENGINEERING ROOM
Department of Architecture
s Îł = 1 â&#x20AC;&#x201C; 0.30 x 1.00/10.00 = 0.97
Coefficients deep foundation: d c = 1 + 0.10 x â&#x2C6;&#x161;
0.80
1.00
d q = 1 + 0.10 x â&#x2C6;&#x161;
= 1.09
0.80 đ?&#x2018;Ľđ?&#x2018;Ľ đ?&#x2018; đ?&#x2018; đ?&#x2018; đ?&#x2018; đ?&#x2018; đ?&#x2018; 2 đ?&#x2018;Ľđ?&#x2018;Ľ 20.00
d Îł = 1.00
1.00
= 1.07
â&#x2C6;&#x161; (0,8 x sin2 x 20.00/ 1.00)
Coefficients of skewness load: i c = i q = i Îł = 1.00
Coefficients of skewness terrain, slope of terrain β = 0.00 0 : j c = j q = j γ =1.00
Calculation of design resistances Subsoil: R d = (0.01 x 103 x 14.83 x 1.02 x 1.09 x 1.00 x 1.00 + 19.00 x 0.80 x 6.40 x 1.03 x 1.07 x 1.00 x x 1.00 + 19.00 x 1.00/2 x 2.95 x 0.97 x 1.00 x 1.00 x 1.00) / 1.40 = 214.00 KN/m2
For reasons difficult foundation conditions speculating
R d = 200.00 KN/m2
In calculating the spatial subsoil will restore springs type WINKLER For the following base ratios speculating modulus of subsoil reaction C = 9000 KN/m3 Approximate load on subsoil: Ceiling - level 5
8.70 + 0.8
9.50 KN/m2
Ceilings - Level 4; 3; 2
3 (9.00 + 3.00 + 1.65)
41.85 Ceiling
- Level 1
9.65 + 3.00 + 1.50
14.15
The walls
7.00
Base foundation
0.55 x 25.00 x 1.35
18.60 KN/m2
â&#x2C6;&#x2018;
â&#x2030;&#x2C6;
91.10 KN/m2
assessment. Load on subsoil - â&#x2030;&#x2C6;
Bratislava 04. 2016.
91.10 KN/m2 â&#x2030;¤ 200.00 KN/m2
Sabah Shawkat
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
234
STRUCTURAL ENGINEERING ROOM
Department of Architecture 235
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
236
STRUCTURAL ENGINEERING ROOM
Department of Architecture 237
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
238
STRUCTURAL ENGINEERING ROOM
Department of Architecture 239
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
240
STRUCTURAL ENGINEERING ROOM
Department of Architecture 241
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
242
STRUCTURAL ENGINEERING ROOM
Department of Architecture 243
Realisation Projects
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
244
STRUCTURAL ENGINEERING ROOM
Department of Architecture 245
STAIRWAY WALL IN AXIS B- ST19
Realisation Projects
ST19
Realisation Projects STRUCTURAL ENGINEERING ROOM
Department of Architecture
246
1.2.9
1.3.8
1.3.5
1.3.4
1.3.6
1.3.9
1.2.10
Zsch1
1.3.2
H
1.3.7
1.3.10
17
16
15
14
13
12
11
10
9
8
Realisation Projects
7
6
5
4
3
2
1
18 x 257 x 183
1.3.3
1.3.1
STRUCTURAL ENGINEERING ROOM
Department of Architecture 247
PLAN GROUND FLOORBLOCK B
Realisation Projects
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
1
2
3
4
5
6
7
8
9
10
11
1.3.1
1.3.3
1.2.1
1.2.3
Department of Architecture
Zsch1
1.3.2
1.3.7
1.3.10
1.2.7
Zsch1
1.2.2
12
13
14
STRUCTURAL ENGINEERING ROOM
1.3.8
1.3.5
1.3.4
1.3.6
1.3.9
1.2.8
1.2.5
1.2.4
1.2.6
15
16
18 x 257 x 183 18 x 257 x 183
17
H
H
1.2.9
248
3,66 m2
2,50 m2
3,66 m2
2,50
m2
2,50 m2
3,66 m2
2.H.4
2.G.6
2.F.6
2.E.4
2.E.3
2.D.3
2.C.3
2.B.3
2.A.4
2.D.2
2.F.4
2.G.5 2.G.4
2.G.1
2.A.5
H
2.1.1
2.C.4
2.B.1
2.A.1
2.G.7
2.F.7
2.C.2
2.B.4
2.G.3
2.F.3
2.F.1
2.E.2
2.E.1
2.D.1
2.C.1
2.B.2
2.A.2
2.G.2
2.F.2
2.F.5
2.A.3
16
15
14
13
12
Realisation Projects
11
10
9
8
7
6
5
4
3
2
1
2.A.6
16 x 181 x 268
5,89 m2
12,66 m2
PAVLAÄ&#x152;
STRUCTURAL ENGINEERING ROOM
Department of Architecture 249
PLAN 1,2,3 FLOOR BLOCK B
Realisation Projects 3,66
m2
3,66 m2
2,50 m2
3,66 m2
3,66 m2
2.K.5
2.N.5
2.M.4
2.M.3
2.L.3
2.L.4
2.J.6
2.I.5
2.I.3
2.H.3
2.N.6
2.N.3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
2.N.2
2.3.1
2.K.7
1
2
3
4
5
6
7
8
9
2.N.1
5,89 m2
2.J.5
H
H
8,03 m2
Department of Architecture
2.M.2 2.M.5
2.M.1
2.L.1
2.K.6
2.J.2
2.J.1
2.K.1
2.L.2
2.K.3
2.J.3
2.I.4
2.2.1
10
11
12
13
14
15
16
STRUCTURAL ENGINEERING ROOM
2.N.4
2.K.4
2.K.2
2.J.4
2.I.2
2.I.1
2.H.1
2.H.2
2.G.1
16 x 181 x 268 16 x 181 x 268
2.G.5 2.G.4
250
Realisation Projects
ZS
1
2
3
4
5
6
7
8
9
18 x 257 x 183
H
10
11
12
13
14
15
16
17
1.1.3
1.1.1
Zsch1
1.2.1
H
1.A.1
1.A.7
1.1.2
1.1.5
1.B.3
1.C.2
1.C.1
1.D.4
Zsch1
1.A.X
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
1.C.X
1.D.3
1.D.1
1.1.6
1.A.4
1.B.1
1.A.3
1.A.5
1.B.2
1.1.4
1.C.3
1.C.5
1.D.6
1.D.2
1.C.4
1.E.5
1.D.5
1.E.2
1.E.6
1.1.8
1.1.7
1.A.2
1.A.6
1.B.4
1.B.5
1.C.6
1.C.7
1.D.7
1.D.8
1.E.8
1.E.7
1.1.9
4,20 m2
4,20 m2
4,20 m2
4,20 m2
4,20 m2
STRUCTURAL ENGINEERING ROOM
1.E.4
1.E.1
1.E.3
18 x 257 x 183
Department of Architecture 251
PLAN GROUND FLOORBLOCK A
15
2.D.X
2.2.1
16 x 181 x 268
2.A.4
28,95 m2
8,98 m2
2.A.X
2.B.X
2.C.X
1
2
3
4
5
6
7
8
9
10
11
12
13
2.E.3
2.1.1
2.I.4
2.J.4
3,09 m2
3,09 m2
3,09 m2
3,09 m2
3,09 m2
3,09 m2
3,09 m2
Department of Architecture
2.A.8
2.B.5
2.B.4
2.C.3
2.D.4
2.E.5
2.F.5
2.G.5
2.H.5
2.I.5
3,86 m2
PLAN 1,2,3 FLOOR BLOCK A
STRUCTURAL ENGINEERING ROOM
2.A.7
2.A.5
2.A.6
2.B.1
2.B.2
2.D.5
2.C.1
2.C.2
2.D.3
2.E.2
2.E.4
2.F.4
2.F.2
2.G.4
2.G.3
2.H.3
2.H.4
2.I.2
2.B.3
2.A.2
2.A.1
H
2.D.2
2.D.1
2.G.2
2.H.2
2.F.3
2.E.1
2.F.1
2.G.1
2.H.1
2.J.3
2.I.3
2.J.2
2.J.5
2.I.1
2.J.1
16 x 181 x 268
14
16
15
14
13
12
11
10
9
8
7
6
5
4
3
Realisation Projects 2
H 1
16
252
STRUCTURAL ENGINEERING ROOM
Department of Architecture
253
PHOTODOCUMENTATION- FOUNDATIONS, GROUND FLOOR
Realisation Projects
PHOTODOCUMENTATION- ROOF REINFORCEMENT
Realisation Projects
STRUCTURAL ENGINEERING ROOM
Department of Architecture
254
STRUCTURAL ENGINEERING ROOM
Department of Architecture
255
PHOTODOCUMENTATION- ROOF REINFORCEMENT , 3rd FLOOR IN PROCESS
Realisation Projects
SITE INSPECTION AND STUDENT FIELD TRIP .
Realisation Projects
STRUCTURAL ENGINEERING ROOM
Department of Architecture
256
STRUCTURAL ENGINEERING ROOM
Department of Architecture
257 References
[18] CEB - Bull. 124/125 - F: Code modéle CEB - FIP pour les structures en béton. CEB,
[1] ACI: Cracking of concrete members in direct tension. ACI Journal, Vol. 83, January –
Paris, 1980
February, 1986
[19] CEB - Bull. 156 - F: Fissuration et déformations. École Polytechnique Fédérale de
[2] ACI Committee 318 (1995), Building Code Requirements for Reinforced Concrete, ACI
Lausanne,1983.
318-89, and Commentary, ACI 318R-89, American Concrete Institute, Detroit, MI, USA
[20] CEB - FIP Model Code 1990, Comité Euro - International du Béton, 1991
[3] Aide - mémoire: Composants en béton précontraint. Bordas, Paris, 1979
[21] CEB - Bull. 159: Simplified methods of calculating short term deflections of reinforced
[4] Allen, H. G.: “The strength of thin composites of finite width, with brittle matrices and
concrete slabs. Paris - Lausanne, 1983
random discontinuous reinforcing fibres, J. Phys. D. Appl. Phys. (1972)
[22] Cholewicki, A. ´Shear Transfer in Longitudinal Joints of Hollow-Core Slabs´ Beton
[5] Aveston, J.; Mercer, R. A. & Sillwood, J. M.: “Fibre reinforced cements – scientific
Fertigteil Technik n. 4, Wiesbaden 1991
foundations for specifications. In Composite – Standards, Testing and Design, Proc. National
[23] Comité Euro-international du Béton, Bull. n. 161, T.P. Tassios, Reporter, Ch. 4: Highly
Physical Laboratory Conference, UK, 1974
Sheared Normally Loaded Linear Elements, Paris 1983
[6] Azizinamini, A.: “Design of Tension Lap Splices in High Strength Concrete.” High
[24] Composite structures, FIP 1994
Strength Conference, First International Conference, Proc. ASCE, 1997, Kona, Hawaii
[25] Cholewicki, A: Shear Transfer in Longitudinal Joints of Hollow-Core Slabs´ Beton
[7] Balaguru, P. & Kendzulak, J.:“ Mechanical properties of Slurry Infiltrated Fiber Concrete
Fertigteil Technik n. 4, Wiesbaden 1991
(SIFCON). American concrete Institute, Detroit, 1987
[26] Composite Dycore Office Structures, Company literature-Finfrock Industries, Inc.,
[8] Balaguru and Shah 1992, Fiber-Reinforced Cement Composites. McGraw-Hill Inc. 1992.
Orlando, FL, 1992.
[9] Bentur, Mindness 1990, Fiber Reinforced Cementitious Composites. Elsevier Applied
[27] CCBA 68: Régles Techniques de conception et de calcul des ouvrages et constructions
Science, 1990, 449 p
en béton armé. D.T.U. Paris, 1975
[10] Bazant, Z. P. & Oh, B. H.: “Crack band theory for fracture of concrete.”, Materials and
[28] Consenza, E., Greco, C.: Comparison and Optimization of Different Methods of
Structures (RILEM), 1993
Evaluation of Displacements in Cracked Reinforced Concrete Beams. Materials and
[11] Beeby, A. W.: The Prediction of Crack Widths in Hardened Concrete, Cement and
Structures, No. 23, 1990
Concrete Association, London, 1979
[29] Coates, R. C., Coutie, M. G., Kong, F. K.: Structural analysis, Second Edition, Hong
[12] Belgian standard NBN B15-238 (1992)
Kong, 1980
[13] Brandt, A. M.: “Cement – Based Composites”, 1995 E & FN SPON
[30] Cox, H.L:“ The elasticity and strength of fibrous material”, Br. J, Appl. Phys. (1952)
[14] Bjarne, Ch. J.: Lines of Discontinuity for Displacements in the Theory of Plasticity of
[31] Davidovici, V.: Béton armé, aide - mémoire. Bordas, Paris, 1974
Plain and Reinforced Concrete, Magazine of Concrete Research, Vol. 27, No. 92, September,
[32] DBV-Merkblatt ”Bemessungsgrundlagen fur Stahlfaserbeton” (1992)
1975
[33] Design of Concrete Structure, Norwegian Standard NS3473 (1992), Norwegian Council
[15] Boulet, B.: Aide - mémoire du second oeuvre du batiment. Bordas, Paris, 1977
for Building Standardization, Oslo, 1992
[16] Brooks, J. J., Neville, A. M.: A comparison of creep, elasticity, and strength of concrete
[34] Deutscher Ausschuss fur Stahlbeton – DafStb (1994): Richlinie fur Hochfesten Beton,
in tension and in compression. Magazine of Concrete Research, Vol. 29, 1977
Supplement a DIN 1045, DIN 488 and DIN 1055, Berlin, Germany
[17] Chan, S.Y.N.; Anson, M.; Koo, S. L.:” The development of very High Strength
[35] Eibl, J.: Concrete Structures. Euro - Design Handbook. Karlsruhe, 1994 – 96
Concrete” Concrete in the Service of Manking, Radial Concrete Technology – Proceeding of
[36] Edward, G., Nawy, P.E.: Prestressed Concrete A fundamental Approach. Part 1, New
the International Conference, University of Dunde, Scotland, UK, 27-28 June 1996.
Jersey, 1989
References
[53] Hognestad, E., A Study of Combined Bending and Axial Load in Reinforced Concrete
members. ACI Journal, Vol. 67, 1970
Members, University of Illinois Engineering Experiment Station, Bulletin Series No. 399,
[38] Elliot, K. S.; Torey, A. K.: Precast concrete frame buildings, Design Guide. British
Bulletin No. 1, 1951.
Cement Association 1992
[54] Hsu, T. T. C.: Torsion of reinforced concrete. Van Nostrand Reinhold, New York, 1984
[39] FIP Recommendations ´Design of multi-storey precast concrete structures´. 1986
[55] Ismail, M. A., Jirsa, J. O.: Bond deterioration in reinforced concrete subject to low cycle
[40] Goto, Y.: Cracks Formed in Concrete Around Deformed Tension Bars, Journal of the
loads. ACI Journal, Vol. 69, June, 1972
ACI, No. 68, April, 1971
[56] Klink, S. A.: Actual Elastic Modulus of Concrete. ACI Journal, September - October,
[41] Gregor, J. G.: Reinforced Concrete, New Jersey, 1988
1985
[42] Gvozdev, A. A.: Novoje v projektirovanii betonnych i železobetonnych konstrukcij.
[57] Kelly, J. M., Skinner, R. I., and Heine, A. J., ‘Mechanisms of Energy Absoption in
Moskva, 1978
Special Devices for use in Earthquake Resistant Structures, ‘Bulletin of the New Zealand
[43] Goulet, J.: Résistance des matériaux. aide - mémoire, Bordas, Paris, 1976
Society for Earthquake Engineering. Vol. 5, No. 3, Sept. 1972.
[44] Grandet, J.: „ Durability of High Performance Concrete in Relation to ‚External‘
[58] Laws, V.: “The efficiency of fibrous reinforcement of brittle matrices.” J. Phys. D. Appl.
Chemical Attack“, High Performance Concrete: From material to structure, 1992 E & FN
Phys., (1971)
Spon
[59] Lawrence, P.:” Some theoretical considerations of fibre pull-out from an elastic matrix”,
[45] Grigorian, C. E., Yang, T.-S., and Popov, E. P., ‘Slotted Bolted Connection Energy
Journal of Material Science 7 (1972)
Dissipators,’ Earthquake Engineering Research Center, University of California,
[60] Leonhardt, F.: Reducting Shear Reinforcement in Reinforced Concrete Beams and Slabs,
Berkeley, Report No. UCB/EERC-92/10, July, 1992.
Magazine of Concrete Research, Vol. 17, No. 53, December, 1965
[46] Gupta, A. K.: Unified Approach to Modelling Postcracking Membrane Behavior of
[61] Leonhardt, F.: Recommendations for the Degree of Prestressing Prestressed Concrete
Reinferced Concrete, Journal of Structural Engineering, Vol. 115, No. 4, April, 1989
Structures. FIP Notes 69, July - August, 1977
[47] Gupta, A. K.: Postcracking Behavior of Membrane Reinforced Concrete Elements
[62] Leonhardt, F.: Crack Control in Concrete Structures. ACI Journal, July - August, 1988
Including Tension-Stiffening. Journal of Structural Engineering, Vol.115, No. 4, April, 1989
[63] Leonhardt, F.: Vorlesungen uber Massivbau. Vol. 4, 1978
[48] Grigorian, C. E., Yang, T.-S., and Popov, E. P., ‘Slotted Bolted Connection Energy
[64] Lenkei, P.: Deformation capacity in reinforced concrete slabs. In: IABSE Colloquium -
Dissipators,’ Earthquake Engineering Research Center, University of California, Berkeley,
Plasticity in reinforced Concrete, Copenhagen, 1979
Report No. UCB/EERC-92/10, July, 1992.
[65] Leong, T. W., Warner, R. F.: Creep and shrinkage in reinforced concrete beams. Journal
[49] Hanečka, Š., Križma, M., Ravinger, J., Shawkat, S.: Contribution to Limit State of the
of the Structural Division, Vol. 96, March, 1970
Second Group of Beams Subjected to Moving Load. First Slovak Conference on Concrete
[66] Lin, C.S. and Scordelis, A.C., Nonlinear Analysis of RC Shells of General Form,
Structures, Bratislava, September, 1994.
Journal of Structural Division, ASCE, 101(1975) 523-538.
[50] Hassanzadeh, M.; Haghpassand, A.:”Brittleness of Normal and High-Strength Concrete”,
[67] Maidl, B., “Stahlfaserbeton”, Ernst & Sohn, Berlin, 1991
Utilization of High Strength Concrete, Proceedings, Symposium in Lillehammer, Norway,
[68] Mindes, S. & Bentur, A.:”Fibre Reinforced cementitious Composites”, 1990 Elsevier
June 20-23, 1993
Science Publishers Ltd
[51] Hillerborg, P. E., Int. J. Cem. Comp., 1980
[69] Moranville-Rgourd, M.:“Durability of High Performance Concrete : Alkali-Aggregate
[52] Holzmann, P.: “High strength concrete C 105 with increased Fire-resistance due to
Reaction and Carbonation“, High Performance Concrete: From material to structure, 1992 E
polypropylen Fibres”, Utilization of High Strength /High Performance Concrete, Proceedings,
& FN Spon
Symposium in Paris, France, May 29-31 1996
References
STRUCTURAL ENGINEERING ROOM
[37] Elvery, R., Shafi, M.: Analysis of shrinkage effect on reinforced concrete structural
Department of Architecture
258
STRUCTURAL ENGINEERING ROOM
Department of Architecture
259 [70] Muller, H. S.: “Creep of High Performance Concrete – Characteristics and Code-Type
[85] Schlaich, J., Scheef, H.: Concrete box - girder bridges. Structural Engineering
prediction Model.” Fourth International Symposium on the Utilization of High Strength /
Documents 1e. Stuttgart, January, 1982
High Performance Concrete, 1996, Paris, France
[86] Shah, S.P.: “Material aspects of High Performance Concrete.”
[71] Naaman, A. E.: „Fasern mit verbesserter Haftung.” Beton- und Stahlbetonbau 95, 2000,
Conference, First International Conference, Proc. ASCE, 1997, Kona, Hawaii
Heft 4
[87] Shah, S. P.: “Advanced Cement-Based Composites”, Concrete in the Service of
[72] Ontario Precast Concrete Manufactures Association and PCA, Duotek, Portland Cement
Manking, Radial Concrete Technology – Proceeding of the International Conference,
Association, Skokie, IL.
University of Dunde, Scotland, UK, 27-28 June 1996.
[73] Placas, A., Regan, P.E.: Limit - state design for shear in rectangular and "T" beams.
[88] Shawkat, S.: Deformácie konštrukcie. Medzinárodná vedecká konferencia - 60. Výročie
Magazine of Concrete Research, December, 1970
Stavebnej fakulty STU v Bratislave. Zborník vedeckých prác. Bratislava, 18. - 20. november
[74] Placas, A., Regan, P.E.: Shear Failure of Reinforced Concrete Beams. Journal ACI,
1998.
October, 1971
[89] Shawkat, S.: Deformation of reinforced concrete beams. Proceedings of the RILEM
[75] Pitoňák, A., Nürnbergerová, T., Shawkat, S.: Pretvárne charakteristiky betónových
International Conference Concrete bridges, Štrbské pleso, 22. - 24. september 1997
nosníkov vystužených predpätými tyčami. Stavebné Materiály a Skúšobníctvo, Zborník
[90] Low, S-G., et al.: Minimization of Floor Thickness in Precast Prestressed Concrete
Podbanské, Vysoké Tatry, 27.- 29. máj 1998
Multistory Buildings, PCI Journal, Vol. 36, No. 4, July-August 1991
[76] Prekop, L., Shawkat, S., Králik, J.: Modelovanie experimentálnych priehybov prútových
[91] Shawkat, S., Križma, M., Cesnak, J., Bartók, A.: Moment and Shear Deflection for
betónových prvkov. Vybrané problémy stavebnej mechaniky. Zborník prednášok z vedeckej
Reinforced Concrete Beams. Slovak Journal of Civil Engineering, Volume II, No. 2-3, 1994
konferencie pri príležitosti 70. narodenín Prof. Ing. J. Sobotu, DrSc., Bratislava, September,
[92] Shawkat, S., Cesnak, J.: Crack Development and Strain Energy of Reinforced Concrete
1996
Beams. First Slovak Conference on Concrete Structures, Bratislava, September 1994
[77] PCI Technical Report no. 2. Connections for Precast Prestressed Concrete Buildings
[93] Shawkat, S., Cesnak, J.: Deflection of Reinforced Concrete Beams due to Actions of
including earthquake resistance. March 1992.
Shearing Forces. Proceedings of an International Conference RILEM. Failures of Concrete
[78] Prior, R. C., Pessiki, S., Sause, R., Slaughter, S., van Zyverden, W.: Identification and
Structures, Štrbské pleso, 1993
Preliminary Assessment of Existing Precast Concrete Floor Framing System, ATLSS Report
[94] Shawkat, S., Križma, M, Cesnak, J.: Determination of Strain Energy on Reinforced
93-07, August 1993, 141 pp.
Concrete Beams. Slovak Journal of Civil Engineering, Volume II, 1994
[79] Rehm, G., Eligehausen, R., Mallee, R.: Limitation of Shear Crack Width in Reinforced
[95] Shawkat, S., Cesnak, J.: Deformations of Reinforced Concrete Beams Subjected to
High Strength
Concrete Construction. University of Stuttgart, Heft 6, 1983
Stationary Loading. Inžinierske stavby 43, č. 9 -10, 1995
[80] Rehm, G.: Berechnung der Breite von Schubrissen in Stahlbetonbauteilen. Ausfsatz fur
[96] Shawkat, S., Bolha, Ľ.: Internal Energy of Concrete Elements by Moving Load,
CEB. Stuttgart, 1977
Inžinierske stavby, 43, č. 4, 1995
[81] Recommendations on precast prestressed Hollow-Core Floors Th. Telford Publ., London
[97] Shawkat, S., Nürnbergerová, T., Pitoňák, A.: Ohybová tuhosť prútových betónových
1988
prvkov. Betonárske dni 1996, Zborník prednášok, Bratislava, September, 1996
[82] Russell, H.& Thomas, M. D.:“Durability of Concrete containing Fine Pozzolan“
[98] Shawkat, S., Križma, M.: Šmyková tuhosť betónových prvkov po vzniku trhlín.
Utilization of High Strength / High Performance Concrete, Florida, USA,
Vybrané problémy stavebnej mechaniky. Zborník prednášok z vedeckej konferencie pri
[83] Sargin, M: Stress - strain relationships for concrete and the analysis of structural
príležitosti 70. narodenín Prof. Ing. J. Sobotu, DrSc., Bratislava, September, 1996
concretesections. University of Waterloo, Study No. 4, 1971
[99] Shawkat, S., Križma, M., Šuchtová A.: Deformation of Reinforced Concrete Beams.
[84] Saliger, R.: Der Stahlbetonbau. 8. Auflage. Franz Deuticke, Wien, 1956
Slovak Journal of Civil Engineering, č. 3-4, 1996
References
Bratislava, SvF STU, 1993 [101] Shawkat, S.: Structural design I. - 1. vyd. - Brno : Tribun EU, 2015. - 210 p. - ISBN 978-80-263-0977-2 [102] Shawkat, S.: Structural design II. - 1. vyd. - Brno : Tribun EU, 2015. - 125 p. - ISBN 978-80-263-0978-9 [103] Shawkat, S.: Structural design III. - Brno : Tribun EU, 2016. - 356 p. - ISBN 978-80-263-10113-6 [104] Neville 1995, Properties of Concrete, Fourth Ed., Longman Group, 1995, 844 p. [105] 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. [106] Vagner, V. W., Erlhof, G.: Praktische Baustatik. Teil 1, 2, 3, Stuttgart, 1977 [107] Vecchio, F. J.: Reinforced Concrete Membrane Element Formulations. Journal of Structural Engineering, Vol. 116, No. 3, March, 1990 [108] 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. [109] Wecharatana, M.; Shah, S. P.:”Fracture in Concrete”, Proc. ASCE Session, Hollywood, Florida, Oct. 27-31 1980 [110] Yee, A.A.: Design Considerations for Precast Prestressed Concrete Building Structures in Seismic Areas, PCI Journal, Vol. 36 No. 3, May-June 1991.including earthquake resistance. March 1992. [111] Zhou, F. P.; Barr, B. I. G.:” Effect of coarse aggregate on elastic modulus and compressive strength of High Performance Concrete”, Cement and Concrete Research, Vol. 25, No. 1, 1995 Elsevier Science Ltd.
References
STRUCTURAL ENGINEERING ROOM
[100] Shawkat, S.: Vplyv priečnych síl na priehyb prútových nosníkov [Dizertačná práca].
Department of Architecture
260
Department of Architecture
STRUCTURAL ENGINEERING ROOM Element Design to Shape a Structure I. ©
Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD. 1. Edition, Tribun EU, s.r.o. Brno 2017 ISBN 978-80-263-1190-4
9 788026 311904