Degree Lecture Notes - Building Structure

Building Structures (ARC2522)

BEAMS (Part 1) By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design

Introduction to beam 

Usually a horizontal structural member that is subjected to a load that tends to bend it



Designed for moment, shearing forces and deflections brought about by the loads and spans of a building structure

Bending Moment - the bending force induced into the beam which is caused by:

 External loads i.e.: floor, wall, machineries, occupants, chandelier etc.  Own weight i.e.: weight of beam’s material, reinforcement, etc.

 External reactions

55kg

Different forms of beams

Types of Beam Continuous beam  Has more than two points of support along its length, usually in the same horizontal plane  It extends

without break in itself over two or more spans

Continuous beam  Provides resistance to bending when a load or force is applied  Commonly used in bridges and buildings

Fixed beam  Fixed at both ends

 Can resist vertical and horizontal forces as well as moment

Fixed beam  i.e. steel beams which are welded together & a cast-in-place concrete structure  Most common in building structures

Simply Supported Beam  Made up of beams that span between only

two supports  Both ends are free to rotate

Simply Supported Beam  Often found in zones of frequent seismic activity  Not often found in normal building structures  Commonly used in bridges

Cantilever Beam  A fixed beam projecting from a vertical support  The upper part is tension & the lower part in compression

Cantilever Beam  Examples of one-way and two-way span cantilever beams

Cantilever Beam

Shapes of Beam Rectangular  Commonly in concrete and timber beam  If the direction of loading can be predicted then the beam can be designed so that most of the material are placed in that plane

Rectangular

Minimum material in the direction of the force

Maximum material in the direction of the force

Steel beams  Different shape would have different stiffness and resistance to torsion Most stiff

Torsion ‘H’ channel square angle

Least stiff

‘tee’

Box girders  High resistance towards torsion

 Stresses tend to concentrate at the corners

Concrete box girder

stress

Steel box girder

Cylindrical tube  Best if the direction of the load could not be predicted  Resistance towards bending and torsion

Characteristics of Beam A beam should satisfy the following requirements:  Adequate strength to resist bending

moments  Least excessive amount of deflections under pressure

 Able to resist lateral buckling  Able to resist shear forces

Bending  The beam deflects under a heavy load  The amount of deflections are greater when the beam is flat  The beam deflects less under uniformly distributed load

 When a beam deflects under a load it, results in the formation of stresses such as tension and compression Sponge Test  Unbent sponge - the grid lines are spaced equally along the top and bottom edges of the sponge.

Unbent sponge  Bent sponge - the grid lines are closer along the top edge and farther along the bottom edge.

Bent sponge

Compression

Tension

 Cantilever beam the stresses are reversed with the top of the beam being in tension and the bottom in compression

Tension

Compression

Deflections  Deflection is the degree to which a structural element is displaced under a load  The amount of deflection depends on: • how the load is distributed - point load or UDL

• how the ends of the beam are supported – fixed or freely supported

Simply Supported Beam  Deflection becomes more of a problem as the span increases

Fixed Beam  Deflects less than a simply supported beam

 Assuming that the beams are the same size and material, then the deflection of the simply supported beam will be 8 times that of the fixed beam

Lateral or Sideways Buckling  A beam may buckle if its depth is much greater than its width when subjected to compressive stresses on top of the beam Force

Failure Due to Shear There are three different types of shear

Vertical  Imagine a beam made up of blocks

 When a force acts down on the beam, it deflects and causes the blocks to slide down one another

Horizontal  usually occurs when a beam is made up of layers  when a force acts down on the beam, it deflects and causes the layers to slide over each other

At an angle

Forces/Load on Beams Point Load  In point load, the load is focused in one point; depicted by a single arrow in diagrams  Examples: column sitting on a beam, connection points between secondary beam & primary beam (girder), a chandelier hanging down from a beam 50N

50N

Forces/Load on Beams Point Load In point load, the load is focused in one point – depicted by a single arrow in diagrams

Column

Point load

50N

Primary beam

Secondary beam

Uniformly Distributed Load (UDL)  In UDL, the load is spread evenly along the structure – depicted by a series of arrows or curves in diagrams  Examples of UDL on beams in building structures: beam self weight, wall on beam, slab on beam, etc.

10N/m

10N/m

Example of UDL on beams

Flat roof Brick wall Balcony wall

Floor slab

Self weight

Important Calculation 1) Beam self weight Assume the initial beam size, i.e. 200mm x 300mm = beam size x density of reinforced concrete = 0.2m x 0.3m x 24kN/m3 = 1.44kN/m (depends on initial beam size) 2) Brick wall weight = wall height x wall thickness x density of brick = 3m x 0.15m x 19kN/m3 = 8.55kN/m * The density of different materials can be obtained from schedule 4 of UBBL

Combination  In reality most times the total load acting on a structure is a combination between point load and UDL  i.e.: UDL from wall & beam’s self weight plus point load from beams carrying slab and columns carrying load from roof

20N 10N/m

Point load UDL

Load Transfer in Beams

beam UDL

 Beams generally carries

vertical forces  Can also be used to carry horizontal loads i.e. due to wind or earthquake  The load are transferred to columns, walls or girders which then transfer the force to adjacent structural

compression members

wall floor

PL

tension beam

compression

ground

Load transferred to columns

Load transferred to girders

Load transferred to wall

E

F

A

B

G

void C

D

H

A C

E

F

 Point Load  UDL  Combination E

B

D

Identify the type of load for each beam without considering the beam’s self weight

H G

F

G

A

B

C

D H

Free body diagram (FBD) A beam is converted into a free body diagram before the load is quantified A

B

C

D

A

B

C

D

5m

3m

1.5m

3m

Section

Plan 10kN/m

A

5kN/m

10kN/m

B B

D 1.5m C

3m

3m

25kN/m

A

D B

7.5m

C

√ X

Case study Calculate the load on Beam 2/B-D

Architectural plan

Steps: • Draw the structural plan & indicate all beams and columns • Determine the type of each slab whether one-way or two-way & draw the load distribution (refer to lecture notes on “Slabs”) • Identify which slab(s) contributes its load to the beam • Identify which part(s) of the beam carries a brick wall

Case study

Structural plan

Load distribution

Case study Beam 2/B-D carries: • Dead load from beam self weight: gridline B-D • Dead load from brick wall weight: gridline B-C only • Dead load from slab: gridline B-D/1-2, B-C/2-3 & C-D/2-3 • Live load from slab: gridline B-D/1-2, B-C/2-3 & C-D/2-3 • Point load at point C/2 from beam C/2-3 Steps: • Find the total amount of dead load according to grid B-C, C-D • Find the total amount of live load according to grid B-C, C-D • Include free body diagrams for reference • Find the ultimate load and draw the ultimate load diagram • Analise Beam C/2-3 using the same steps to know the reaction force at C/2 which would become the point load

Case study Dead Load • Beam self weight = 1.44kN/m (beam size is 200mm x 300mm)

• Brick wall weight = 8.55kN/m (wall height is 3000mm) • Dead load on slab B-D/1-2 = 7.2kN/m • Dead load on slab B-C/2-3 = 3.6kN/m • Dead load on slab C-D/2-3 = 3.6kN/m • Total dead load For B-C = 1.4 + 8.55 + 7.2 + 3.6 = 20.75kN/m For C-D = 1.4 + 7.2 + 3.6 = 12.2kN/m

Case study Live Load • Live load on slab B-D/1-2 = 4kN/m

• Live load on slab B-C/2-3 = 2kN/m • Live load on slab C-D/2-3 = 2kN/m • Total live load For B-C =4+2 = 6kN/m For C-D =4+2 = 6kN/m

Case study Ultimate Load

• • • •

Ultimate dead load at B-C = 20.75kN/m x 1.4 Ultimate dead load at C-D = 12.2kN/m x 1.4 Ultimate live load at B-C = 6kN/m x 1.6 Ultimate live load at C-D = 6kN/m x 1.6

= 29.05kN/m = 17.08 kN/m = 9.6kN/m = 9.6kN/m

• Total ultimate load at B-C = 29.05 + 9.6 = 38.65kN/m • Total ultimate load at C-D = 17.08 + 9.6 = 26.68kN/m • Include reaction force for beam C/2-3 at point C/2 in the diagram; in this case say 32.5kN

Quantifying the Internal Forces  The loads (and reactions) bend the beam and try to shear through it Bending

Shear

 When a beam is loaded by forces, stresses and strains are created throughout the interior of the beam

Quantifying the Internal Forces  To determine these stresses and strains, the internal forces that act on the beam must be identified: 1. Find all the forces (PL & UDL) 2. Make the beam into a free body diagram 3. Find the reactions using the conditions of equilibrium

Moment Moment M = Force F x Perpendicular distance d  It tends to rotate a structure (i.e.: beam or truss) about a point Distance d

Force F

+ve A

M=Fxd

B -ve

 However, it does not always cause rotation

Moment  i.e.:

A

MA = F x dA

dA

F

M B = F x dB

dB

 The force tends to rotate the structure clockwise about the support A but also tends to rotate the beam counter clockwise about the support B. Thus, there is no rotation, only bending  Rotation would occur if either support is removed

B

 Calculate moment about each point 1) 50kN A

2)

4m

2m

B

4m 2.5kN/m

A

B 10kN

 The total load is 2.5kN x 4m = 10kN

 This load is equivalent to a single point load of 10kN in the middle of the beam

Building Structures (BLD61003/ARC2523)

BEAMS (Part 2) By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design

Quantifying the Internal Force Reaction Force

 Applied to a structure when it rests against something

 Reaction forces can be solved by applying two simple rules in conditions of equilibrium: • A body in equilibrium has no resultant force in any direction • A body in equilibrium has no resultant turning moment in any direction

 In a state of equilibrium the sum of forces acting on a structure should be 0 Total load on structure + reaction forces = 0 ∑F = 0 2m

y

10kN/m

+ A 10kN

B 10kN

_

x

Reaction force on beam

 A uniform load on a beam is shown below with a value of 5kN per meter (kN/m) 6m 5kN/m

A Ra 15kN

B 30kN

Rb 15kN

 The total load is 5kN x 6m = 30kN  This load is equivalent to a single point load of 30kN in the middle of the beam  In this case the reactions must be equal to half the total load; 30kN ÷ 2 = 15kN both acting up

30kN _

M B1 3m

A Ra + 15kN

B

6m

Rb 15kN

M B2

1) Choose ONE reference point, in this example, point B 2) Calculate ALL moments from point B to get Ra 3) Balance ALL forces to get Rb ∑M = 0 = MB2 – MB1 = (Ra x 6) - (30 x 3) = 6Ra - 90 6Ra = 90 Ra = 15kN

∑Fy

=0 = Ra + Rb - 30 = 15 + Rb - 30 Rb = 30 - 15 Rb = 15kN

 Now consider a uniform load of 2.5kN/m over part of the beam 4m 1.5m

2.5kN/m

A

B

3.75kN

Ra 0.75m

Rb

 The total load is 2.5kN x 1.5m = 3.75kN & acts in the middle of the UDL length (0.75m) from A

_

MB1

4m

0.75m

3.75kN A Ra

B +

MB2

∑M = MB2 – MB1 = 0 = (Ra x 4) - (3.75 x 3.25) = 0 = 4Ra - 12.19 = 0 4Ra = 12.19 Ra = 3.05kN

Rb ∑Fy = 0 Ra + Rb - 3.75 = 0 3.05 + Rb - 3.75 = 0 Rb = 3.75 – 3.05 Rb = 0.7kN

Exercise 1 Calculate the reaction forces Ra & Rb 6m 2m

2.5kN/m

2m

1m

4kN/m

10kN

A

B

Ra ∑M = 0 = Ra(6) – 5(5) – 8(3) – 10(1) = 6Ra – 25 – 24 – 10 6Ra = 59 Ra = 9.8kN

Rb ∑Fy Rb

=0 = 9.8 + Rb – 5 – 8 – 10 = 13.2kN

Exercise 2 Calculate the reaction forces Ra & Rb 8m

3m

2kN/m

2m

5kN

3m

8kN

3kN/m

A Ra ∑M

=0 = Ra(8) – 6(6.5) – 5(5) – 8(3) – 9(1.5) = 8Ra – 39 – 25 – 24 – 13.5 8Ra = 101.5 Ra = 12.7kN

B Rb ∑Fy = 0 = 12.7 + Rb – 6 – 5 – 8 – 9 Rb = 15.3kN

Shear & Bending Moment

 Shear and moment occur internally in a beam  These forces can be identified in any section of a beam in the condition of equilibrium  For shears and moments calculations, uniformly distributed loads cannot be simply replaced with point load

 Sign conventions for shear force (V): +ve shear R.H. down L.H. up

-ve shear L.H. down R.H. up

 Positive shear: denoted by an internal shear force that causes clockwise rotation of the member on which it acts  Negative shear: opposite to the positive shear

 Sign conventions for bending moment (M): +ve (sagging/happy)

-ve (hogging/sad)

 Positive moment: denoted by an internal moment that causes compression, or pushing on the upper part of the member  Negative moment: opposite to the positive moment

Shear & Moment Diagrams  The magnitude of these internal shears & moments will vary depending on the location at which the beam is cut

 A shear & moment diagram shows the values of these forces at different locations along the beam

 Example of shear & moment diagrams in simply supported beam (point load) L x

L x

A 0

F

A

F B

C

Load Diagram

B C

• Identify and quantify all forces acting on the beam using the conditions of equilibrium • Use the information to produce shear force and bending moment diagrams

A

+

C

B

_

Shear Force Diagram C + A

B

Bending Moment Diagram

 Example of shear & moment diagrams in simply supported beam (uniformly distributed load) F/m A

F/m

B

Load Diagram A

B (+)

B

A

(-)

Shear Force Diagram A

B (+)

Bending Moment Diagram

 Example of shear & moment diagrams in cantilever beam F/m A

F/m

L

Load Diagram

B A

L

B

B

A (-)

Shear Force Diagram A

B (-)

Bending Moment Diagram

 Example of shear & moment diagrams in continuous beam (uniformly distributed load) F/m A

C

B

Load Diagram (+)

B

A

(+)

C (-)

(-)

Shear Force Diagram A

(+)

B (-)

Bending Moment Diagram

(+)

C

Easy steps in drawing load, shear force and bending moment diagrams: 3m

18kN

6m

A

B C

1. Identify and quantify all forces acting on the beam then draw the load diagram ∑MA = RB (9) – 18(3) = 9 RB – 54 RB = 6kN ∑Fx = 0 3m

∑Fy = 0 = RA – 18 + 6 = RA – 12 RA = 12kN

18kN

A

6m B

C 12kN

6kN

2. Plot the shear force diagram based on forces in the load diagram 3m

18kN

6m

A

B C

12kN

6kN

12kN

V

12kN (+)

0kN

A

C -6kN

B (-)

0kN -6kN

•

At point A there is a 12kN force acting upwards (+ve)

•

No other force placed between point A and C, thus the line remain constant at 12kN mark

•

At point C there is an 18kN force acting downwards (-ve), thus 12kN – 18kN = -6kN

•

No other force placed between point C and B thus the line remain constant at -6kN mark

•

At point B there is a 6kN force acting upwards (+ve), thus -6kN + 6kN = 0kN (the diagram is correct only if it ends with 0)

3. Plot the bending moment diagram based on the areas below the shear force diagram 3m

6m

12kN

V

12kN (+)

0kN

A

C

B (-)

-6kN

0kN -6kN

36kNm

V

(+)

M 0kNA

Relationship between lines in shear & moment diagrams

C

B

0kN

of the bending moment diagram isorequivalent to the M area underneiagram

• At point A there is only a line so no area = 0kN • At point C, M = area of rectangular between A and C = 12 x 3 = 36 kNm • At point B, M = area of rectangular AC (+ve) + area of rectangular CB (-ve) = 36 + (-6 x 6) =0

or

4. Redraw load, shear force and bending moment diagrams 3m

18kN

6m

A

B C

12kN

6kN

12kN

V

12kN (+)

0kN

A

C

B (-)

-6kN

0kN -6kN

36kNm

(+)

M

0

A

C

B

0

 Example 25N 2N/m B A

C 10m

10m 25N

2N/m A

B 10m

RAy = 27.5N Reactions: ∑Fy = 0; ∑MA = 0 = RCy (20m) – 25N(10m) – [2N(10m)(5m)] = 20RCy – 350 RCy = 17.5N

10m

C RCy = 17.5N

∑Fy = 0; = RAy + RCy – 25 – 2(10) = RAy + 17.5 – 45 = RAy – 27.5 RAy = 27.5

25N 2N/m B

A

10m

C

10m

17.5N

27.5N

27.5N

(+) 0N

7.5N (27.5 – 20) B

A

C

√

0N (-17.5 + 17.5)

(-) -17.5N (7.5 – 25)

-17.5N

Shear Force Diagram (V) Note: • UDL was converted to PL only for calculation of reaction forces • When drawing V diagram, it should be treated as UDL

27.5N 20kN (+) 0N

7.5N B

A 10m

-17.5N

C

0N

(-) 10m

-17.5N

175Nm

0 A

B

C

Bending Moment Diagram (M)

25N 2N/m B

A

C 17.5N

27.5N

Load Diagram 27.5kN

(+) 0kN

A

7.5kN B

C

0kN

(-)

-17.5kN

-17.5kN

Shear Force Diagram 175kNm

0 A

B

Bending Moment Diagram

C

 Example 8kN

16kN B

C D

A 2m

3m

3.5m 8kN

16kN B

A

C 3.5m

2m

D 3m

15.06kN

8.94kN

15.06 + (15.06 – 16 = –0.94)

0 A

B –0.94

– C

–

(–0.94 – 8 = –8.94)

–8.94

Shear Force Diagram (V)

0 D

Reactions: ∑MD = 0 = RA (8.5) – 16(6.5) – 8(3) = 8.5RA – 104 – 24 = 8.5RA – 128 RA = 15.06kN ∑Fy = 0 = RD + RA – 16 – 8 = RD + 15.06 – 24 = RD – 8.94 RD = 8.94kN

√

(–8.94 + 8.94 = 0)

–8.94

15.06 + 0 A

(15.06 x 2 = 30.12)

2m B –0.94

– 3.5m

D C

–

(–0.94 x 3.5 = –3.29) (–8.94 x 3 = –26.82)

–8.94

3m

Shear Force Diagram 30.12

0

A

B

[30.12 +(– 3.29) = 26.83]

26.83

C

D

√

0

[26.83 +(– 26.82) = 0.01 ≈ 0]

Bending Moment Diagram

Case study (continued from previous lecture on beams)

Steps • Find the reaction forces at B & D • Draw the shear force diagram • Draw the bending moment diagram

Case study

Reactions forces ∑MA = 0 = RB (6) – 115.95(4.5) – 32.5(3) – 80.04(1.5) = 6RB – 521.78 – 97.5 – 120.06 RB = 739.34/6 = 123.22kN ∑Fy = 0 = RD + 123.22 – 115.95 – 32.5 – 80.04 RD = 105.27kN

Case study

32.5kN 38.65kN/m 26.68kN A 105.27kN

123.22kN Load diagram 123.22kN

7.27kN 0kN -25.23kN

Shear force diagram

-105.27kN

Case study 123.22kN

7.27kN 0kN -25.23kN

Shear force diagram

-105.27kN

195.74kNm

0 Bending moment diagram

How to calculate the highlighted area?

2

3

Using triangular proportion

4

17kN/m 10.34

36.3kN/m

34kN/m

x 2.9m

2m

94.93 95.34kN/m

Load Diagram

94.93kN/m

95.34 2.9

+ V

x/2.9 = 10.34/(10.34+94.93) x = 0.28

95.34 – 68 = 27.34 27.34 – 17 = 10.34

-94.93 + 94.93 =0

_ 124

√

10.34 – 105.27 = -94.93

M

√

124 - 124= 0

Bending Moment Diagram Positive area – negative area in V +ve = [0.5(95.34 + 27.34)(2) + [0.5(10.34)(0.28)] = 122.68 + 1.45 = 124.13 ≈ 124 -ve

= 0.5(-94.93)(2.9-0.28) = -124.35 ≈ -124

Building Structures (ARC2522)

COLUMNS

By Norita Johar Taylor’s School of Architecture, Building, Design

Columns  A structural member usually inclined at 90º to the horizontal  Carrying an axial* compressive load  Found in buildings supporting beams floor & roofs

Columns  Axial load is a load applied through the center of gravity of a column’s cross section along an axis  Eccentric load is a load applied at any other point on a column’s cross section

Axial load

Eccentric load

Columns for Structure

Columns supporting roofs Columns supporting slab

Columns supporting beams

Columns supporting roofs

Columns for Structure  Strut - a structural member in any position & carrying an axial compressive load

Types of Columns ( Shape ) The main types of columns are: 1) Simple member with uniform crosssection 2) Simple member with non-uniform cross-section 3) Built-up columns

Types of Columns ( Shape ) 1)Simple member with uniform cross section

Types of Columns ( Shape ) 1)Simple member with uniform cross section  The connections to the circular section is expensive and difficult to design

Types of Columns ( Shape ) 1)Simple member with uniform cross section  Possible to use square or rectangular hollow sections whose geometrical properties are good  The connections are easier to design but rather expensive  Hot-rolled sections are the most common cross-sections used for compression members

Types of Columns ( Shape ) 2) Simple members with non- uniform cross-sections  In tapered members the cross- section geometry changes continuously along the length  Usually concrete columns  In steel it can be either open or box shapes formed by welding together i.e. tapered webs or flanges or both

Types of Columns ( Shape ) 2) Simple members with nonuniform cross-sections  Steel tapered members can be formed by welding together two reversed halves of a hot-rolled H or I-section whose web is cut

Types of Columns ( Shape ) 2) Simple members with non- uniform cross-sections  In stepped columns, the cross-section varies in steps  Usually found in industrial buildings with overhead travelling cranes  Can also be used in multistorey buildings to resist the loads in the columns at the lower levels

Types of Columns ( Shape )

Compression Members ( Shape ) 3)Built-up Columns  Fabricated from various different elements  Consist of two or more main components, connected together at intervals to form a single compound member  i.e.: I-section laced with small U-section or Usection battened with flat bars

Types of Columns ( Shape ) 3)Built-up Columns

Types of Columns ( Shape ) 3)Built-up Columns

Compression Members  Examples of built-up columns in buildings Stepped built-up columns

Elasticity vs Plasticity

Force (F)

shortening of the column

 An ideal column is one that is perfectly straight, homogeneous, and free from initial stress.  When an ideal column is subjected to a compressive axial force, the only deformation that takes place is a shortening of the column

Force (F)

Elasticity vs Plasticity Force (F)  For low values of F, if the column were to be deflected laterally by a perpendicular force and the force is then removed, the column would return to its straight position.  This indicates a condition of stability or elasticity

Force (F)

Elasticity vs Plasticity

Force (F)

 If the load F were increased, there is a value of F for which the column would remain in the deformed shape (plasticity)  Yield strength, or the yield point, is defined in engineering as the amount of stress that a material can undergo before moving from elastic deformation into plastic deformation.

Reaction Force (F)

Column Test  To measure the compressive strength of any material  Measured by plotting the applied force against deformation in a testing machine  Usually the tests are made on small specimens in the shape of cubes or circular cylinder

Column Failure 1) Crushing  If the force or load is gradually increased, the column will reach a stage when it will be subjected to the ultimate crushing stress  Beyond this stage the column will fail by crushing  The load corresponding to the crushing stress is called crushing or critical load/force

Column Failure 1) Crushing Main bars

Horizontal circular ties

 Short columns normally fail due to crushing  Failure of a column may be attributed to the insufficient number and spacing of horizontal circular ties.  i.e. during earthquake shaking the ties failed to confine the concrete and the vertical reinforcing main bars

Column Failure 1) Crushing  Stable if loaded within design limit  Failure occurs through crushing if heavily overloaded Unstable load Stable load

Column Failure 2) Buckling  Sometimes a compression member fails by bending, i.e. buckling  Buckling can also be described as bending or bowing of a column due to compressive load  Happens in the case of long columns. For a long (slender) column, buckling occurs way before the normal stress reaches the strength of the column’s material

Column Failure

Force (F)

2) Buckling Buckling

 Buckling is a failure mode; failure to react to the bending moment generated by compressive load  The load at which the column just buckles is called buckling load, critical load or crippling load.

Force (F)

Long compression member

Column Failure 2) Buckling  Stable when provided with lateral restraints  Without lateral restraints it would fails through buckling Stable load

Unstable load

Lateral restraints i.e. from beams & floors

Deformation

Critical Load The critical load depends on:  Material properties (i.e. yield strength, modulus of elasticity; can be found in reference books)  Length of member (can be determined by slenderness ratio)  Section properties

Critical Load Slenderness Ratio (SR or λ)  The Slenderness Ratio is the (effective) length Le of the column divided by radius of gyration  Is a means of classifying columns whether it is short, intermediate or long column  Sometimes expressed with lambda (λ)  A generally accepted relationship between the slenderness ratio (SR) and the type of column is as follows: 0 ≤ SR ≤ 60 = short column 60 ≤ SR ≤ 120 = intermediate column 120 ≤ SR ≤ 300 = long column

Critical Load Slenderness Ratio (SR or λ)  Formula for slenderness ratio:

SR = KL r Where

or

SR = Le r

SR = slenderness ratio K = column effective length factor KL Le L ==unsupported length of column Le = effective length of column r = radius of gyration

3000mm

SR = 3000 = 10 300 0 ≤ SR ≤ 60= short column

300mm

Critical Load Slenderness Ratio (SR or λ)

SR = 3000 = 100 30 60 ≤ SR ≤ 120= intermediate column

SR = 6000 = 200 30

3000mm

120 ≤ SR ≤ 300= long column

30mm

Critical Load Slenderness Ratio (SR or λ)  The end conditions of a column significantly influence the buckled shape  Identifying the end conditions helps in computing the buckling load Force (F)

Examples of pinjointed columns

Slenderness Ratio (SR or Ν)  Buckled shape of columns with different end conditions

Critical Load: Moment of Inertia  Is a property which measures the efficiency of a shape in respect of its resistance to bending  In general a shape is more efficient when the greater part of its mass is as far as possible from its center of gravity y

Center of gravity

y

y x

+

x

x

+

x

+

y

y

y

Column A (400cm2)

Column B (400cm2)

Column C (400cm2)

 i.e.: Column B & C have larger moment of inertia than column A thus they will be stronger/stiffer (because their area is spread out away from the center of gravity

Critical Load: Moment of Inertia  The strength of a column may be increased by distributing the material as far from the center of gravity as possible to increase the moment of inertia  At the same time the material needs to be thick enough to prevent local buckling  Therefore an I-section is much better in resisting bending compared to a rectangular section with similar crosssectional area

Column Load Calculation Basically there are 2 types of column load calculation methods: 1) Beams reaction method 2) Tributary area method

Column Load Calculation: Beams reaction method B3

RB3b

B1 RB1a

C3

RB3a

RB1b

C4

C2

C1

B4

B2 RB2a

RB2b

RB4a

RB4b

Which beams contribute to the column load for column C1?

RB1b RB3a RB2b RB4a

50kN

50kN

5m

100kN 2m

B3

B1 RB3a

RB1b

150kN

B4

B2 RB4a RB2b

C1 Calculate the column load for column C1 if beam self-weight is 10kN/m

∑FB1 = 10kN(2m) ∑FB1 = 20kN RB1b = 10kN# ∑FB3 = 50kN + 10kN(5m) ∑FB3 = 50kN + 50kN ∑FB3 = 100kN RB3a = 50kN# ∑FB2 = 10kN(2m) ∑FB2 = 20kN RB2b = 10kN# ∑FB4 = 75kN + 10kN(5m) ∑FB4 = 75kN + 50kN ∑FB4 = 125kN RB4a = 62.5kN# Column Load C1 = RB1b+RB3a+RB2b+RB4a = 10+50+10+62.5

= 132.5kN

100kN

100kN

8m

200kN 3m

B3

B1 RB3a

RB1b

400kN

B4

B2 RB4a RB2b

C1

Example 1

Calculate the column load for column C1 if beam self-weight is 10kN/m

100kN

100kN

8m

200kN 3m

B3

B1 RB3a

RB1b

400kN

B4

B2 RB4a RB2b

C1 Calculate the column load for column C1 if beam self-weight is 10kN/m

∑FB1 = 10kN(3m) ∑FB1 = 30kN RB1b = 15kN# ∑FB3 = 100kN + 10kN(8m) ∑FB3 = 100kN + 80kN ∑FB3 = 180kN RB3a = 90kN# ∑FB2 = 10kN(3m) ∑FB2 = 30kN RB2b = 15kN# ∑FB4 = 200kN + 10kN(8m) ∑FB4 = 200kN + 80kN ∑FB4 = 280kN RB4a = 140kN# Column Load C1 = RB1b+RB3a+RB2b+RB4a = 15+90+15+140

= 260kN

Column Load Calculation: Tributary Area Method A column does not support the entire weight of the whole floor. It only supports the load in its tributary area at all available levels. a) Live load= Tributary area x live load factor b) Dead Load= TA x weight of the material

Column Load Calculation: Tributary Area Method Calculate the tributary area labelled2A to I. 1 6000

3

8000

A 4000

B

A

B

C

D

E

F

G

H

I

3500

C

Column Load Calculation

In analyzing column load, few components need to be considered: Dead Load on roof Roof frame weight or Slab ( Flat Roof) Roof Beam Dead Load on 1st Floor Walls Slabs SW Beams SW Column SW Live Load on 1st Floor Dead Load on Grd Floor Walls Slabs SW Beams SW Column SW Live Load on Grd Floor

Column Load Calculation Tributary Area Method How to determine RC Column Self Weight?

Width x length x height of column (m) x weight of RC concrete ( 24 kN/m3) Example: Calculate the dead load of the RC columns provided that: 1) Column size = 400mm x 450mm Building height=4.75m Column = 0.45m x 0.4m x 4.75m x 24 kN/m3 = 20.52 kN 2) Column size = 300mm x 400mm Building height=3m Column = 0.3m x 0.4m x 3m x 24 kN/m3 = 8.64 kN

Tributary area method ( Live Load Only) • Half distance of slabs to adjacent columns • Also known as column area method • Column load = area of slab (xy) X load factor 4m

4.5m

3m

4m

Calculate the shaded area of slab x = 2m + 2m = 4m y = 2.25m + 1.5m = 3.75m Area of slab = xy = 4m x 3.75m = 15m2 Load factor: Office = 1.5kN/m2 Residence = 2.0kN/m2 Hospital/Factory = 4.0kN/m2

Exercise1: Find column load for column B2 of an office A

5m

B

5m

C

1

5.5m

x = 5m y = 6.25m

2 3.5m

Area of slab = 5 x 6.25 = 31.25m2

Column Load= 31.25m2 x 1.5kN/m2 =46.87kN

1500 1500

Load factor: Office = 1.5kN/m2 Residence = 2.0kN/m2 Hospital/Factory = 4.0kN/m2

Calculate column load for column A if the building is a residence Column load A =(2m x 6m) x 2.0kN/m2 = 24kN Calculate column load for column B if the building is a factory Column load B =[(2m x 3m) + (3m x 3.5m)] x 4kN/m2 = 66kN

• Column load in multi-storey building • Start calculation from the highest to the lowest level 5m

5m

5m

5m

5.5m

B

A

First Floor (gymnasium) Load factor: Office = 1.5kN/m2 Gymnasium = 4.0kN/m2

4m

4m

5.5m

B

A

Ground Floor (office)

5m

5m

5m

5m B

A First Floor – gymnasium (4.0kN/m2)

Column load for column 1A = (5m x 4.75m) x 4.0kN/m2 = 95kN Column load for column GA = (5m x 4.75m) x 1.5kN/m2 = 35.63kN Total Column load for column A = 95kN + 35.63kN = 130.63kN#

4m

4m

5.5m

5.5m

B

A Ground Floor – office (1.5kN/m2)

Column load for column 1B = (2.5m x 2.75m) x 4.0kN/m2 = 27.5kN Column load for column GB = (2.5m x 2.75m) x 1.5kN/m2 = 10.31kN Total Column load for column B = 27.5kN + 10.31kN = 37.81kN#

Exercise • Find the column load for each column • The load factor is 10kN/m2

Level 2

Level 1

• Identify the tributary area for each column on level 2 • Name all columns according to grid lines

A1

B1

C1

D2 A2

B2

B3 A4

Level 2

C3

• Identify the tributary area for each column on level 1 • Name all columns according to grid lines

A1

B1

A2

B2

C1

D2 B3 C3 A4

Level 1

• Calculate the column load based on the tributary area

D

D

D D

Column Area Load 2C A/1- 6 -60 2C A/2- 16.5 - 165 2C A/3- 10.5 - 105 2C B/1- 13.14 - 131.4 2C B/2- 25.5 - 255 2C B/3- 18.75 - 187.5 2C C/1- 7.14 -71.4 2C C/3- 8.75 - 87.5 2C D/2- 9 -90

Load Column Area 1c A/1 - 9 - 90 + 60 - 150 1C A/2- 24.75 - 247.5 + 165 -412.5 1C A/3- 15.75 - 157.5 + 105 - 262.5 1C B/1 - 10 - 100 + 131.4 - 231.4 1C B/2- 25.5 - 255 + 255 - 510 1C B/3- 23.75 -237.5 +187.5- 425 1C C/1- 7.14 - 71.4 + 71.4 - 142.8 1C C/3- 8.25 -82.5 + 87.5 - 170 1C D/2 - 9.5 - 95 + 90 - 185

Tributary area method ( Both Live & Dead Load ) The example given is the layout plan for a house.

Analyze the columns based on the specifications given.

Height of the floors=3m Brick wall 150mm thick Dead Load= 3.0m x 0.15m x 19 kN/m3= 8.55kN/m Beam size is 150 x 300mm Dead Load= 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m Slab thickness is 150mm, Dead Load= 0.15m x 24 kN/m3= 3.6 kN/m2 Column size=300 x 400mm Dead Load= 0.3m x 0.4m x 3m x 24 kN/m3 = 8.64 kN

The example begins analyzing column B1.

1

6000

2

8000

A 2000 4000

Living

Dining

7. 5m2

B

3750 3500

C GROUND FLOOR PLAN

Kitchen

3

1

6000

2

3

8000

A 2000 4000

Bedroom 1

Bedroom 2

7. 5m2

B

3750 3500

C FIRST FLOOR PLAN

Gymnasium

Toilet

1 A 4000

B 3500

C ROOF PLAN

6000

2

8000

3

1

6000

2

3

8000

A 2000 4000

Bedroom 1

Bedroom 2

7. 5m2

B

3750 3500

C STRUCTURAL PLAN

Gymnasium

Toilet

Determine the load acting on column B1.

Roof Level Dead Load Slab ( Flat Roof) 7.5m2 (total slab area) x 3.6 kN/m2 =27 kN Roof Beam 6.75 m (total beam length) x 1.08 kN/m = 7.29kN 1st Floor Dead Load Live Load Walls 6.75 m x 8.55kN/m= 57.7 kN Bedroom Slabs 7.5m2 x 3.6 kN/m2= 27 kN (3 X 2m) x 1.5 kN/m2 = 9 kN Beams 6.75m x 1.08 kN/m= 7.29 kN Gymnasium Column = 8.64 kN (3 X 1.75m) x 5 kN/m2= 26.25kN Total 100.63 kN Grd Floor Dead Load Live Load Walls 3.75 m x 8.55kN/m= 32 kN Livingroom Slabs 7.5m2 x 3.6 kN/m2= 27 kN (3 X 2m) x 1.5 kN/m2=9 kN Beams 6.75m x 1.08 kN/m= 7.29 kN Column = 8.64 kN Total 74.93 kN TOTAL Dead Load 209.85 kN TOTAL Live Load =44.25kN Apply 1.4 factor 293.79 kN Apply 1.6 factor 70.8kN So, ultimate load acting on column B1= 364.59 kN

Column for Residential Units

In order to know whether the designed column is able to hold the load, we may use the formula below in accordance to BS 8110;

N = 0.4fcuAc + 0.8 fyAsc N = capacity of concrete Fcu = concrete strength (N/mm2) Ac = cross section of concrete column fy = yield strength of steel (N/mm2) Asc = steel content in a column

Example :

Given that fcu = 30N/mm2 and fy = N = 0.4fcuAc + 0.8 fyAsc N = capacity of concrete 2 460N/mm . Assuming percentage of Fcu = concrete strength (N/mm2) Ac = cross section of concrete column steel reinforcement in a rectangular fcu = yield strength of steel (N/mm2) concrete column is 2%, determine Asc = steel content in a column the capacity of the column.

Ac = 300 x 400 = 120,000 Asc = 2 % x 120,000 = 2,400 N = 0.4fcuAc + 0.8 fyAsc = 0.4(30)(120000) + 0.8(460)(2400) = 1440,000 + 883,200 = 2323200 N = 2323.2kN

300mm

400mm

Thus, this column can sustain any ULTIMATE load below 26323kN

Forces

Introduction • The main function of a structure is to transfer load to the earth • What are load that act upon structure???

Accidental Load

100kg Force incurred to portal frame = mass x gravitational acceleration = 100 kg x 9.81 (ď ž 10) = 1000 N = 1 kN

Force System • In any structure, forces can be applied in a variety of directions • Coplanar forces – line of action of forces that acting on the same plane (2D arrangement) • Non coplanar forces – forces that do not act on the same plane (3D arrangement)

Coplanar Force (2D arrangement) Line of action lie in the same plane

F1

F3

F2

Non - Coplanar Force (3D arrangement) Line of action do not lie in the same plane

F1 F3

F2

When it comes to structural analysis, coplanar force or non coplanar force analysis is commonly adopted?

Although structures are three dimensional, but most assemblies are two dimensional.

F5

F4 F1

F2 F3

F6

F7

F8 Coplanar forces analysis is easier than non coplanar forces analysis

Non concurrent force – Forces that do not intersect

Collinear Force – Forces that have the same line of action

F4

F5

F2

Joint A F3

F6

F7

F8 Concurrent Force – forces intersect at a common point

A force is a vector quantity. It must have magnitude, direction and point of action. 

To indicate that it is a vector, it is usually indicated as F or F or F

F

 x

There can be many forces of different direction and magnitude acting on a particle.

These forces can be replaced by ONE single force that produce the same effect.

100kN

50kN

150kN

Both are equivalent

There can be many forces of different direction and magnitude acting on a particle.

These forces can be replaced by ONE single force (named as resultant force) that produce the same effect.

100kN

50kN

150kN

Both are equivalent

20kN + y -

50kN

30kN x +

Cartesian system - To simplify the direction of forces

250kN + y -

x +

20kN

Resultant force, Fr = -270kN in y direction

270kN

Fb = 100kN

45⁰ Triangle Rule Polygon Rule Resolution of Force

Fa = 200kN

35⁰ x

A single force can be replaced by two or more forces that produce the same effect. These forces are called as components of a forces There could be infinite pairs of components. In order to simplify structural analysis, force is usually resolved into components X and Y (similar to Cartesian system)

Similarly, two or more forces can be replaced by a single force that produces the same effect.

This process is named as resolution of force

Components can be in forces are called as components of a forces y

x

FA

FAX

FAY

y

x

Basic understanding of trigonometry is important!!!

In a right angled triangle, đ?‘œđ?‘?đ?‘?đ?‘ đ?‘–đ?‘Ąđ?‘’ đ??šđ??´đ?‘Œ sin đ?œƒ = = â„Žđ?‘Śđ?‘?đ?‘œđ?‘Ąđ?‘’đ?‘›đ?‘˘đ?‘ đ?‘’ đ??šđ??´ đ?‘Žđ?‘‘đ?‘—đ?‘Žđ?‘?đ?‘’đ?‘›đ?‘Ą đ??šđ??´đ?‘‹ cos đ?œƒ = = â„Žđ?‘Śđ?‘?đ?‘œđ?‘Ąđ?‘’đ?‘›đ?‘˘đ?‘ đ?‘’ đ??šđ??´ đ?‘œđ?‘?đ?‘?đ?‘œđ?‘ đ?‘–đ?‘Ąđ?‘’ đ??šđ??´đ?‘Œ tan đ?œƒ = = đ?‘Žđ?‘‘đ?‘—đ?‘Žđ?‘?đ?‘’đ?‘›đ?‘Ą đ??šđ??´đ?‘‹

FA

Also Pythagorean theorem, FR = đ??šđ?‘Ś 2 + đ??šđ?‘Ľ 2

FAX

FAY

Resolve the following forces into x and y components.

Fb = 100kN

Fa = 200kN

y

45â °

x

35â °

Fa = 200kN

y

35⁰ x

Fax

Fay

sin 𝜃 =

𝐹𝑎𝑦 𝐹𝑎 𝐹𝑎𝑦 200𝑘𝑁

sin 35 = 𝐹𝑎𝑦 = 200𝑘𝑁 × 𝑠𝑖𝑛35

=114.7kN

𝐹𝑎𝑥 cos 𝜃 = 𝐹𝑎 𝐹𝑎𝑥 cos 35= 200𝑘𝑁

𝐹𝑎𝑥 = 200𝑘𝑁 × 𝑐𝑜𝑠35

=163.8kN

sin 𝜃 =

Fb = 100kN

Fby

sin 45 = 𝐹𝑏𝑦 = 100𝑘𝑁 × 𝑠𝑖𝑛45

=70.7kN

y

45⁰ Fbx

𝐹𝑏𝑦 𝐹𝑏 𝐹𝑏𝑦 100𝑘𝑁

x

𝐹𝑏𝑥 cos 𝜃 = 𝐹𝑏 𝐹𝑏𝑥 cos 45= 100𝑘𝑁

𝐹𝑏𝑥 = 100𝑘𝑁 × 𝑐𝑜𝑠45

=-70.7kN

Fb = 100kN

Fby

Fa = 200kN y

y

45⁰ Fbx

x

35⁰

Fax

Fay

Fy = 185.4 kN x

Fx = 93.1 kN

𝐹𝑥 = Fax + Fbx = 163.8 + - 70.7kN = 93.1 kN 𝐹𝑦 = Fay + Fby = 114.7 + 70.7kN = 185.4 kN FR = 𝐹𝑦 2 + 𝐹𝑥 2 = 207.6 kN

tan đ?œƒ =

đ??šđ?‘Ś đ??šđ?‘Ľ 185.4 93.1

tan đ?œƒ = đ?œƒ = 63.33° y

63.33ď‚° x

Resolve the following forces into x and y components.

Fb = 250kN y

Fc = 150kN

40â °

x

25â °

Fa = 150kN

Fbx y

𝐹𝑏𝑦 = 250𝑘𝑁 × 𝑠𝑖𝑛40

Fby

= 160.7kN

𝐹𝑏𝑥 = −250𝑘𝑁 × 𝑐𝑜𝑠40

= -191.5kN

Fay

Fc = 150kN

40⁰

x

25⁰

Fa = 150kN Fax 𝐹𝑎𝑦 = −150𝑘𝑁 × 𝑐𝑜𝑠25

= -135.9kN

𝐹𝑎𝑥 = −150𝑘𝑁 × 𝑠𝑖𝑛25

= -63.3kN

FR = 107.7kN y

13.3⁰

x

24.8𝑘𝑁 −104.8

tan 𝜃 = 𝜃 = 13.3°

𝐹𝑦 = 160.7 − 135.9

= 24.8kN

𝐹𝑥 = −191.5 − 63.3 + 150 =−104.8kN 𝐹𝑅 = 24.8 2 + −104.8 2 = 107.7𝑘𝑁

An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).

F = ma = mass x acceleration For every action there is an equal and opposite re-action.

Building is a static structure, remain at rest all the time. Thus resultant force acting on a structure must always be ZERO! đ??šđ?‘Ľ = 0 đ??šđ?‘Ś = 0

đ?‘€ =0

Determine if the object shown is in equilibrium. đ??šđ?‘Ľ = 0 đ??šđ?‘Ś = - 100kN + 100kN = 0

đ?‘€đ?‘œ = -100kN(2m) + 100kN(3.5m) = +150kNm ď‚š 0 Moment is also a vector!

Clockwise moment = +ve

Anticlockwise moment = -ve

Determine resultant force for each of the following case.

To illustrate the idea of axial force, consider a simple column subject to a vertical load as shown.

Compression

Axial Force Internal External

Internal

Tension External

Restraint either direction

Restraint both directions

Unable to take moment!!

Restraint both directions & able to take moment

The End Thank you!

INTRODUCTION TO STRUCTURAL DESIGN By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design

Lecture outcomes At the end of this lecture students should be able to :  Know the basic issues that affect the design of building structures  Recall previous lesson on forces in Building Construction II  Understands how load is transferred within a building

Why learn building structures?

1. So that architects can solve structural problems 2. So that they can talk more intelligently to their engineering consultants

3. So that architectural design, becomes more rational and therefore, perhaps, more beautiful

Basic Concerns • All physical objects have structures • The design of structures is part of the general design problems

Basic Concerns Safety • Life safety is the major concern in structural design. Two critical considerations are: a) Fire resistance

b) Low likelihood of collapse under load

Basic Concerns a) Major elements of fire resistance: i) Combustibility of the structure ii) Loss of strength at high temperature iii) Containment of the fire

Basic Concerns b) Sustaining loads i) a building should have some margin of structural capacity beyond what is required for the actual task • The margin of safety is defined by the safety factor and calculated as follows actual capacity of the structure Safety Factor, SF =

required capacity of the structure

Basic Concerns Example: • If a structure is required to carry 20,000kg and is actually able to carry 35,000kg before collapsing, the safety factor is expressed as follows: SF = 35,000/20,000 = 1.75 • Building structures are generally designed with an average safety factor of about 2

Basic Concerns Other concerns of structural design apart from fire resistance & sustaining loads are: c) Feasibility d) Economy e) Optimization f) Integration Feasibility Structures must use materials & products which are available & can be handled by the production organization

Basic Concerns Economy • Structures are usually less appreciated unless it constitutes a major design feature • Expensive structures do not often add value in the way that expensive finishes or hardware may.

Basic Concerns Optimization • Building designers are subjected to conflicting needs • Often changes that tend to improve one factor would degrade others

Basic Concerns Integration • Good structural design requires integration of the structure into the whole physical system of the building

Building Construction II Revision Structural Fundamentals

Building Construction II Revision Forces

Building Construction II Revision Internal Forces

Building Construction II Revision External Forces

Structural Requirements To perform its function of supporting a building in response to whatever loads that being applied to it, a structure must : 1. be capable of achieving a state of equilibrium.

2. be stable.

3. have adequate strength.

4. have adequate rigidity.

Structural Requirements Equilibrium

• It is the state of rest under the action of applied load. • All applied loads are balanced exactly by the reaction forces generated at its foundations.

Structural Requirements Stable

• It is the property which preserves the geometry of a structure and allows its elements to act together to resist loads. • It is still possible for a structure to fall due to its being geometrically unstable even if it has sufficient strength & rigidity

Structural Requirements Strength

• It depends upon material properties. • The strength of a material depends on its capacity to withstand axial stress, shear stress, bending, and torsion.

Structural Requirements Rigidity

• It is a property that makes a structure does not bend or flex under an applied force. • Some structures are formed by collections of objects that are rigid bodies, with pairs of objects connected by flexible hinges.

Structural Failures Structure fails due to the following: 1. Stability failures • Relates to the overall structural system 2. Strength failures

• Relates to the strength of the individual structural members

Structural Failures

Structural Members Generally two types:

1. Horizontal members • Floors • Roofs • Beams • Girders 2. Vertical members • Columns • Loadbearing walls.

Load Path • The path that a load travels through the structural system

• “Tracing” or “chasing” the loads

• Each structural element must be designed for all loads that pass through it

HVAC

Load Path • Every load applied to the building will travel through the structural system until it is transferred to the supporting soil.

APPLIED LOAD

Load Path

Load Path

Load Path

beam

5kN

B 10kN

6m

A

1m

girder

1m

Tributary Area  Is the area of a slab that contributes to the load on a specific beam  The yellow area constitute the total load on Beam A  The green area constitute the total load on Beam B

Load Path 1

2 8m

C

1.5 1.5 1

2m

B

1

3m

A

Force on slab AB 1-2 = 100kN BC 1-2 = 80kN Beam A1-2 = 50kN or 6.25kN/m Beam B1-2 = 90kN or 11.25kN/m Beam C1-2 = 40kN or 5kN/m

FLOOR SLAB Mdm Norita Johar

Definition A flat piece of concrete, supported by the walls or beams and columns of a structure.

It serves as a walking surface and technically it serves as a load bearing member.

Functions of Flooring • To provide a uniform, flat surface to place building facilities • To act as sound, heat and fire insulator • Act as a vertical divider for the occupants • Upper slab became the ceiling for the storey below • To support dead & live load

Design Types • Flat slab • Ribbed / waffle • Hollow core Hollow core slab

Flat Slab

Waffle slab

Design Types

Flat slab with capital

Two-way slab with beams

Load Distribution & Units 20kN

(point) 40kN 5m

(point) 20kN/m

20kN/m2 (area) UDL 2m

(line) UDL

COLUMN

2m

BEAM & WALL

(point) 200kN

SLAB

Load Distribution & Units Slabs can be supported by load bearing walls, beams or straight to column. Loads are uniformly distributed to these supporting elements.

Single Path Slab on Columns

Load Distribution & Units 20kN (point)

COLUMN

Load Distribution & Units

Single Path Slab On Walls / Beams

Load Distribution & Units (point) 40kN

20kN/m (line) UDL

2m

BEAM & WALL

Load Distribution & Units

Dual Path Slab On Beams, Beams on Columns

Slab Analysis Factors • Types of Load – dead load + floor finishes – live load

• Types of system - one way -two way- trapezoid - triangle

Slab Analysis Factors -Types of Load- Dead Load The calculation for dead load from the slab transferred to the beam needs to consider the: • Thickness of the slab • Self weight of the slab • Weight of walls sitting on the slab • Self weight of the beam

Slab Analysis Factors -Types of Load- Dead Load The standard weight of the materials can be referred in UBBL.

Slab Analysis Factors -Types of Load- Live Load The calculation and analysis on a slab for live load is determined according to the functions of the rooms as stipulated in UBBL .

Generally , a room with heavy equipments or larger user like a hospital or gymnasium has a larger amount of standard live load.

Slab Analysis Factors -Types of Load- Live Load

Slab Analysis Factors -Types of Slab System Slab systems can be divided in 2 types: • One way • Two way Ly Ly=Longer side of slab Lx= Shorter side of slab When Ly / Lx > 2, it is a one way slab When Ly /Lx < 2 or =2,it is a two way slab

Lx

Slab Analysis Factors -Types of Slab System One Way System

Lx Ly

Ly / Lx > 2

One-way slabs carry load in one direction. Bending primarily about the long axis. Thus,main reinforcement is required in the one direction.

Slab Analysis Factors -Types of Slab System Two Way System Lx

Ly

Ly / Lx <= 2

Two-way slabs carry load in two directions. Bending will take place in the two directions in a dish-like form.

Accordingly, main reinforcement is required in the two directions.

Slab Analysis Factors -Types of Slab System Examples: one way slab Ly / Lx > 2 two way slab Ly / Lx <= 2 Ly=6m

Lx=2m

Ly=3m

Lx=2m

Slab Analysis Factors -Types of Slab System Exercise: Determine the system of the slabs below:

Ly=6m

Lx=5m

Ly=8m

Lx=3m

Ly=10m

Lx=5m

Slab Analysis Factors -Types of Slab System Exercise-Determine the system of the slabs below:

Ly=6m

Lx=5m

Ly / Lx < 2

two way slab

Ly=8m

Lx=3m

Ly / Lx > 2

one way slab

Ly=10m

Lx=5m

Ly / Lx = 2

two way slab

Load Distribution-Tributary Area Loads exerted on slabs are calculated per unit area kN/m2.

Loads are uniformly distributed on the slabs.

Many floor systems consist of a reinforced concrete slab supported on a rectangular grid of beams.

The distribution of floor loads on floor beams is based on the geometric configuration of the beams forming the grid.

When the loads are transferred to a beam, they are calculated per unit meter kN/m.

Load Distribution-One Way Slab When the length of slab is two times its width, almost all of the floor load goes to the longer beams AB,CD & EF B

Slab ABDC

D

Slab CDFE

F

one way slab

Ly / Lx > 2

50kN

20kN

30m

A

10m

C

8m

E

Load Distribution-One Way Slab The load transferred to these beams will depend on the area of the slabs. Beam CD will receive more slab load than beam AB and EF. B D Slab CDFE F Slab ABDC

1/2

1/2

1/2

1/2

30m

A

5m

5m

4m

4m

25kN

25kN

10kN

10kN

10m

C

8m

E

One Way Dead/Live Load (UBBL) x (Lx/2) Std weight of materials Sch Std Live Load Sch

Area of slab affected

Load Distribution-Two Way Slab Triangular form

When the ratio of the long span Ly to the short span Lx as Trapezoidal shown is less or more than 2, form 45° the floor load is carried in both directions to the four supporting beams around the two way slab panel.

Ly / Lx <= 2

Load Distribution-Two Way Slab Loads from slabs transferred to beams has 2 different patterns: • trapezoidal Trapezoidal • triangular pattern form

Triangular form

45°

two way slab

Ly / Lx <= 2

Two Way Dead/Live Load (UBBL) x (Lx/2) Std weight of materials Sch

Area of slab affected

Std Live Load Sch

[Dead/Live Load (UBBL) x (Lx/2)] x 2/3

Slab Analysis-Dead Load Area of the slab affected

Slab self weight (thickness x concrete density ) Eg, Assume slab thickness is 150mm, 0.15m x 24 kN/m3= 3.6 kN/m2

One Way Dead Load (UBBL) x (Lx/2)

Two Way ( Trapezoidal ) Dead Load (UBBL) x (Lx /2) Two Way ( Triangular ) [Dead Load (UBBL) x (Lx /2) ] 2/3 (need to times with 2/3)

Slab Analysis-Dead Load Exercise 1:

C 6m

A

5m 9m slab 2

4m slab 1

D

B

Calculate total dead loads transferred from slabs 1 & 2 onto beam CD. Beam CD receives load from: slab 1 (one way) Dead Load (UBBL) x (Lx/2) from slab 2 and the triangular pattern (2 way).

[Dead Load (UBBL) x (Lx /2) ] 2/3

Slab Analysis-Dead Load Dead load from slab 1 self weight Dead Load (UBBL) x (Lx/2) =3.6 kN/m2 x (4m/2) =7.2 kN/m

C 6m

A

5m 9m slab 2

4m slab 1

B

Dead load from slab 2 self weight

[Dead Load (UBBL) x (Lx /2) ] 2/3

=[3.6kN/m2 x (5m/2) ] 2/3 =[9 kN/m ] 2/3 =6 kN/m

D

Total dead load from both slabs: = 7.2 kN/m + 6 kN/m =11.2 kN/m

Slab Analysis-Dead Load Slab self weight is not the only dead load on the beam.

Slab self weight (thickness x concrete density ) Eg, Assume slab thickness is 150mm, 0.15m x 24 kN/m3= 3.6 kN/m2 There are other factors to be included that sit on the beam; Wall Self weight (height x thickness x brick density ) Eg, 3.0m x 0.15m x 19 kN/m3= 8.55kN/m

Beam Self weight (beam size x concrete density ) Eg, Assume beam size is 150mm x 300mm 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m

Slab Analysis-Dead Load Total dead load from both slabs: =11.2 kN/m

Beam Self weight =1.08kN/m Wall Self weight = 8.55kN/m Total dead load =

20.83 kN /m

Eg, Assume beam size is 150 x 300mm 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m Eg, 3.0m x 0.15m x 19 kN/m3= 8.55kN/m

Slab Analysis-Live Load One Way Live Load (UBBL) x (Lx /2) Two Way ( Trapezoidal ) Live Load (UBBL) x (Lx /2) Two Way ( Triangular ) [Live Load (UBBL) x (Lx /2) ] 2/3 (need to times with 2/3)

Slab Analysis-Live Load-One way Slab Example 1: The diagram is showing a one way slab in a house. Find the live load transferred from the slab to beam AB. A

C

12m

D

5m

B

From the UBBL, a house has 1.5 UDL standard live load amount.

So, live load from slab transferred to beam AB is: Live Load Factor (UBBL) x (Lx /2) = 1.5 kN/m2 x ( 5m / 2) =2.75 kN/m

Slab Analysis-Live Load-Two way Slab Example 2: The diagram is showing a two way slab in a house. Find the live load transferred from the slab to beam AB and AC. A

C

8m

D

5m

B

To beam AB and CD - trapezoidal form to beam AC and DB - triangular form

Slab Analysis-Live Load-Two way Slab Live Load Factor (UBBL) x (Lx /2)

To beam AB and CD - trapezoidal form So, live load = 1.5kN/m2 x ( 5m / 2) =2.75 kN/m [Live Load Factor (UBBL) x (Lx/2) ] 2/3

To beam AC and DB - triangular form (need to times with 2/3) So, live load = [1.5 kN/m2 x ( 5m / 2) ] x 2/3 = [ 2.75 kN/m ] x 2/3 =1.8 kN/m

Slab Analysis-The Ultimate Load The ultimate load is generally the amount of all loads after the consideration of safety load factor.

The purpose of load factor is to allow extra safety than the intended use. The load factor needs to be applied of 1.4 and 1.6 for dead load and live load respectively.

Example; Total Dead Load= 20.3 x 1.4= 29.1 kN/m Total Live load= 3.0 x 1.6 = 4.8 kN/m Ultimate Load= 33.9 kN/m

Truss

Truss â&#x20AC;˘ A truss is a structure built up of three or more members which are normally considered as being pinned or hinged at the various joints.

â&#x20AC;˘ Any loads which are applied to the truss are usually transmitted to the joints, so that individual members are in pure tension or compression.

Why Truss?

• Quick Installation – Can be installed quickly – Usually fabricated at factory and delivered to site for installation only.

• Longer Span – truss is able to span longer with minimal deflection • Accessibility – Triangular spaces allow installation of utilities

A Simple Truss â&#x20AC;˘ Members in compression - strut â&#x20AC;˘ Members in tension tie

Types of Truss • Perfect Frame, Imperfect Frame and Redundant Frame • Perfect frame – consists of member just sufficient for being stable • Imperfect frame – consists too few members to prevent collapse • Redundant frame – consist more than sufficient members for being stable

Number of members in Perfect Frame â&#x20AC;˘ The simplest perfect frame â&#x20AC;&#x201C; three (3) members in form of triangle â&#x20AC;˘ As long as triangles are added, the frame remains as perfect frame.

2đ??˝ = đ?&#x2018;&#x161; + 3 J = number of joints m = number of members

Which of the following is perfect frame?

Analysis of Perfect Frame • Resolution of forces at joints • Recall equilibrium – forces in any direction should be = 0 (fx=0 and fy=0 and M=0) • Thus, equilibrium of external and internal forces should be achieved at every joint of frame

Example 1 â&#x20AC;˘ Determine the force in each member. State that it is compression or tension.

Analysis Steps • Determine the reaction force • Analyze joint by joint – start with joint with least members – Assume direction of force, F in member – Resolve F into x and y components – Equilibrium @ every joint, Fx=0, Fx=0, M=0

Determine reaction force @ A & C

MA = +100kN (1.25/2) – RC(1.25) 0 = +62.5 – 1.25RC RC = 50kN Fy Ray

= Ray + Rc -100kN = 50 kN

At Joint â&#x20AC;&#x2DC;Aâ&#x20AC;&#x2122; FABx

0.75

tan đ?&#x153;&#x192; = 0.625 đ?&#x153;&#x192; = 50.2°

Resolve FAB into component x and y, FABy FABy = FAB sinď ą = -FAB sin50.2 FABx = FAB cosď ą = -FAB cos50.2

50kN

The structure must be in equilibrium, hence ď &#x201C;Fx = 0 and ď &#x201C;Fy = 0

ď &#x201C;Fx = 0 = -FABcos50.2 + RAX - FAC ď &#x201C;Fy = 0 = -FAB sin50.2 + 50kN

At Joint ‘C’ FCBx

FCBy

50kN

Forces in member

Example 2:

Space Frame â&#x20AC;˘ 3 Dimensional Trusses

BMW Welt in Munich, Germany