2. Chapter 2 - Structural Analysis

8/13/2019 2. Chapter 2 - Structural Analysis

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STRUCTURAL ANALYSIS

With Wisdom We Explore

www.uthm.edu.my

By:

Masni A. [email protected]


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The primary purpose of structural analysis is to establish thedistribution of internal forces and moments over the wholepart of a structure and to identify the critical designconditions at all sections.The type of analysis should be appropriate to the problembeing considered. The following may be used: linear elasticanalysis, linear elastic analysis with limited redistribution, andplastic analysis.

Linear elastic analysis may be carried out assuming crosssections are uncracked (i.e. concrete section properties), usinglinear stress-strain relationships, and assuming means valuesof elastic modulus.


Introductionwww.uthm.edu.my


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Actions that applied on a beam may consist of beams self-weight , permanent and variable actions from slabs, actionsfrom secondary beams and other structural or non-structuralmembers supported by the beam.The distribution of slab actions on beams depends on the slabdimension, supporting system and boundary condition .Beam supporting slabs designed as spanning one-way can beconsidered to be uniformly load as shown in figure below.


Analysis of Actionswww.uthm.edu.my


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l x

l y

A B

C D

l y

w = 0.5.n.l x kN/m

Beam AC and BD

l x

w = 0 kN/m

Beam AB and CD


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For beam supporting a two-way slab panel freely supported alongfour edge.



l x

l y

AB

C D

l y

mkN l l nl

w y

x x / 36

2

Beam AC and BD

l x

mk N nl

w x / 3

Beam AB and CD


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There are alternatives methods which consider various supportconditions and slab continuity. The methods are, (i). Slab shearcoefficient from Table 3.15 BS 8110 , (ii). Yield line analysis and (iii).Table 63 Reinforced Concrete Designers Handbook by Reynold.




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l x

l y

A B

C D

l y

w = vx.n.l x kN/m

Beam AC

l x

Beam CD

w = vy.n.l x kN/m


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Example 2.1www.uthm.edu.my

3000 4500

2 5 0 0

4 0 0 0

1 2 3

A

B

C

200 x 500

200 x 500 200 x 500

200 x 500 200 x 500

2 0 0

x 5

0 0

2 0 0

x

5 0 0

2 0 0

x 5

0 0

2 0 0

x

5 0 0

FS1 (150 thk.)

FS2 (150 thk.) FS3 (150 thk.)

Determine the characteristic permanent and

variable action act on beam B/1-3.

Weight of concrete = 25 kN/m 3

Finishes, ceiling = 2.0 kN/m 2 and services

Variable action = 3.0 kN/m 2 (All slab)


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Action on slabSelfweight = 0.15 x 25 = 3.75 kN/m 2

Finishes, ceiling and services = 2.0 kN/m 2

Chac. Permanent action, G k = 5.75 kN/m2

Chac. Variable action, Q k = 3.0 kN/m 2

Distribution of actions from slabs are as follows;FS1 : l y /l x = 7.5 / 2.5 = 3 > 2.0, One-way slabFS2 : l y /l x = 4.0 / 3.0 = 1.33 < 2.0, Two-way slabFS3 : l y /l x = 4.5 / 4.0 = 1.13 < 2.0, Two-way slab


Solution of Example 2.1www.uthm.edu.my


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3000 4500

2 5 0 0

4 0 0 0

1 2 3

A

B

C

w1 = 0.5.n.l x

w2 = vy.n.l x w3 = vx.n.l x


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Action from slabw1 Gk = 0.5 x 5.75 x 2.5 = 7.19 kN/mw1 Qk = 0.5 x 3.00 x 2.5 = 3.75 kN/m

From Table 3.15: BS 8110: Part 1: 1997

w2 Gk = 0.4 x 5.75 x 3.0 = 6.90 kN/mw2 Qk = 0.4 x 3.00 x 3.0 = 3.60 kN/m

w3 Gk = 0.44 x 5.75 x 4.0 = 10.12 kN/mw3 Qk = 0.44 x 3.00 x 4.0 = 5.28 kN/m




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Actions on beamBeam selfweight = 0.20 x (0.5 0.15) x 25 = 1.75 kN/m

Span 1-2Permanent action,G k = 7.19 + 6.90 + 1.75 = 15.84 kN/mVariable action, Q k = 3.75 + 3.60 = 7.35 kN/m

Span 2-3Permanent action, G k = 7.19 + 10.12 + 1.75 = 19.06 kN/mVariable action, Q k = 3.75 + 5.28 = 9.03 kN/m



3000 4500

Gk = 15.84 kN/mQk = 7.35 kN/m

G k = 19.06 kN/mQ k = 9.03 kN/m


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Combination of action is specifically used for the definition of themagnitude of actions to be used when a limit state is under theinfluence of different actions.For continuous beam , Load cases is concerned with the

arrangement of the variable actions to give the most unfavourable conditions.If there is only one variable actions (e.g. Imposed load) in acombination, the magnitude of the actions can be obtained bymultiplying them by the appropriate factors.

If there is more than one variable actions in combination, it isnecessary to identify the leading action( Qk,1) and otheraccompanying actions (Qk,i). The accompanying actions is alwaystaken as the combination value.


Combination of Actionswww.uthm.edu.my


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Design values of actions, ultimate limit state-persistent andtransient design situations


Combination of Actionswww.uthm.edu.my


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In considering the combinations of actions, the relevant casesshall be considered to enable the critical design conditions tobe established at all sections, within the structure or part ofthe structure considered.For simply supported beam , the analysis for bending andshear force can be carried out using statically determinateapproach. For the ultimate limit state we need only considerthe maximum load of 1.35G k + 1.5Q k on the span.While for continuous beam a simplification in the number ofload arrangements for use in a Country is required, referenceis made to its National Annex . The following simplified loadarrangements are recommended for buildings:


Load Cases and Combination www.uthm.edu.my Clause 5.1.3 : MS EN 1992-1-1


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Load set 1: Alternate or adjacent spans loaded (ContinuousBeam)Alternate span carrying the design permanent and variable

load (1.35 Gk + 1.5Qk), other spans carrying only the design permanent loads (1.35Gk)

Any two adjacent spans carrying the design permanent and

variable loads (1.35 Gk + 1.5Qk), all other spans carrying onlythe design permanent load (1.35Gk)


Load Cases and Combination www.uthm.edu.my Section 5.1.3 : MS EN 1992-1-1


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Alternate spans loaded


Load Cases and Combination www.uthm.edu.my

1.35Gk + 1.5Qk1.35Gk

1.35Gk + 1.5Qk1.35Gk

1.35Gk + 1.5Qk1.35Gk

1.35Gk + 1.5Qk1.35Gk


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Load set 2: All or alternate spans loaded (Continuous Beam)All span carrying the design permanent and variable loads(1.35 Gk+ 1.5Qk)

Alternate span carrying the design permanent and variableload (1.35 Gk+ 1.5Qk), other spans carrying only the design

permanent loads (1.35Gk)


Load Cases and Combination www.uthm.edu.my UK National Annex & Malaysia National Annex


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All span loaded

Alternate span loaded



1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk

1.35Gk + 1.5Qk 1.35Gk + 1.5Qk1.35Gk 1.35Gk

1.35Gk1.35Gk + 1.5Qk 1.35Gk + 1.5Qk

1.35Gk


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Load Case 3

SFD

BMD

SHEAR FORCE DIAGRAM ENVELOPE

BENDING MOMENT DIAGRAM ENVELOPE


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Three analysis methods may be used in order to obtain shearforce and bending moment for design purposes. There are;

Elastic analysis using moment distribution method

(Modified Stiffness Method)Simplified method using shear and moment coefficientfrom Table 3.6: BS 8110: Part 1.Using commercial analysis software such as Staad Pro,

Esteem, Ansys, Lusas, etc.


Method of Structural Analysiswww.uthm.edu.my


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Moment distribution method is only involving distributionmoments to joint repetitively.The accuracy of moment distribution method is dependent to

the number repeat which does and usually more than 5repeat real enough. Right value will be acquired when nomore moments that need distributed.In general the value is dependent to several factor as :

Fixed end moment Carry over factor Member Stiffness Factor (distribution factor)


Moment Distribution Methodwww.uthm.edu.my


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Fixed end moment (FEM)The moment at the fixed joints of a loaded member are calledfixed-end moment.

This moment can be determined from table below, dependingupon the type of loading on the member.

Carry over factor (CO)

The carry-over factor to a fixed end is always 0.5, otherwise itis 0.0.




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Moment Distribution Methodwww.uthm.edu.my Fixed End Moment


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A series of 250 x 400 mm reinforced concrete beams spaced at 3 mcentres and spanning 7.5 m support a 175 mm thick reinforced concreteslab as shown in Figure 2.1. If the variable floor action is 3 kN/m 2 and theload induced by the weight of concrete is 25 kN/m 3, calculate the

maximum shear force and bending moment of beam B/1-2.




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Figure below shows the first floor layout plan of commercialbuilding. If all beams size are 300 x 500 mm, determine thefollowing;

1. Characteristic permanent and variable action act on the beam 2/A-Eif all slab thickness are 150 mm and the brickwall heights is 3m.

2. Shear force and bending moment envelope of beam 2/A-E.

Given the following data;Variable load on slab = 4.0 kN/m 2

Finishes, ceiling & services = 1.5 kN/m 2

Unit weight of concrete = 25 kN/m 3




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2) Shear force and bending moment envelope of beam 2/A-E.Loading




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Moment of Inertia, II = bh 3/12 = 300 x 500 3/12 = 3.125 x 10 9 mm4

Stiffness, KA-B K AB = K BA= 3I/L = 3 x 3.125 x 10 9/8000 = 1.17 x 10 6 mm3

B-C K BC = K CB = 4I/L = 4 x 3.125 x 109/8000 = 1.56 x 10

6 mm

3

C-D K CD = K DC = 4I/L = 4 x 3.125 x 10 9/8000 = 1.56 x 10 6 mm3

D-E K DE = K ED= 3I/L = 3 x 3.125 x 10 9/8000 = 1.17 x 10 6 mm3

Distribution Factor, DFJoint A & E ~ DF AB & DF ED = K AB / (K AB + 0) = 1.17 / (1.17 + 0) = 1.0

Joint B ~ DF BA = K BA / (K BA + K BC) = 1.17 / (1.17 + 1.56) = 0.43DF BC = K BC / (K BA + K BC) = 1.56 / (1.17 + 1.56) = 0.57

Joint C ~ DF CB = K CB / (K CB + K CD) = 1.56 / (1.56 + 1.56) = 0.50DF CD = K CD / (K CB + K CD) = 1.56 / (1.56 + 1.56) = 0.50

Joint D ~ DF DC = K DC / (K DC + K DE) = 1.56 / (1.17 + 1.56) = 0.57DF DE = K DE / (K DC + K DE) = 1.17 / (1.17 + 1.56) = 0.43




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Fix End Moment, FEM

- M AB = M BA = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

- M BC = M CB = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

- M CD = M DC = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

- M DE = M ED = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

Load Case 1


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Fix End Moment, FEM

- M AB = M BA = wL 2 /12= 24.72 x 82 /12= 131.84 kNm

- M BC = M CB = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

- M CD = M DC = wL 2 /12= 24.72 x 82 /12= 131.84 kNm

- M DE = M ED = wL 2 /12= 33.72 x 82 /12= 179.84 kNm

Load Case 3


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SFD Envelope

BMD Envelope


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The analysis using moment distribution method is timeconsuming and is more conveniently carried out usingstandard computer technique.

Therefore, as a simplification BS 8110 cl. 3.4.3 can be use.Table 3.5 are given in BS 8110 which enable a conservativeestimate of shear force and bending moment values to bedetermined for the design of continuous beam.

There are conditions which must be satisfied in each casebefore these tables can be used. They are:


Simplified Methodwww.uthm.edu.my


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The beams should be approximately equal span. The characteristic variable action Qk may not exceed the characteristic

permanent action Gk. Load should be substantially uniformly distributed over three or more

spans. Variation in span length should not exceed 15% of the longest span.



(1.35Gk + 1.5Qk)


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-0.11FL

0.09FL

-0.08FL -0.08FL

0.07FL

0.45F

0.60F

0.55F

0.55F

End Span Interior Span

BendingMoments

ShearingForces


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By using simplified method, analyze the beams shown below.

F = 1.35G k + 1.5Q k = 1.35(18.31) + 1.5(6.00) = 33.72 kN/m x 8 m = 269.75 kN




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Shear force and bending moment diagrams



0.45F =121.39 kN

0.60F =161.85 kN

0.55F =148.36 kN 0.55F =148.36 kN

0.55F =148.36 kN 0.55F =148.36 kN

0.60F =161.85 kN

0.45F =121.39 kN

0.09FL =194.22 kNm

0.11FL =2 37.38 kNm 0.08FL =172.64kNm 0.11FL =237.38 kNm

0.07FL =151.06 kNm 0.07FL =151.06 kNm 0.09FL =194.22 kNm


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EC2: Section 5.5 permit the moment redistribution with thefollowing requirement;

The resulting distribution remains in equilibrium with the load.The continuous beam are predominantly subject to flexural.

The ratio of adjacent span should be in the range of 0.5 to 2

There are other restrictions on the amount of momentredistribution in order to ensure ductility of the beam such as gradeof reinforcing steel and area of tensile reinforcement and hence the

depth of neutral axis.Class A reinforcement; redistribution should 20%Class B and C reinforcement; redistribution should 30%


Moment Redistributionwww.uthm.edu.my


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Load Case 1



33.72 kN/m

8 m

184.97 kNm

V A

VB1

MB = 0VA(8) 33.72(8) 2/2 + 184.97 = 0VA = 894.07 / 8 = 111.76 kN

Fy = 0111.76 + V B1 33.72(8) = 0VB1 = 158.0 kN

MC = 0VB2(8) 33.72(8) 2/2 + 123.21 - 184.97 = 0VB2 = 1140.8 / 8 = 142.60 kN

Fy = 0142.60 + V C1 33.72(8) = 0VC1 = 127.16 kN

33.72 kN/m

8 m

123.21 kNm

VC1VB2

184.97 kNm

Span A - B

Span B - C


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33.72 kN/m

8 m

184.97 kNm

VC2

VD1

MD = 0VC2(8) 33.72(8) 2/2 - 123.21 +184.97 = 0VC2 = 1017.28 / 8 = 127.16 kN

Fy = 0127.16 + V D1 33.72(8) = 0VD1 = 142.60 kN

ME = 0

VD2(8)

33.72(8)2

/2 - 184.97 = 0VD2 = 1264.01 / 8 = 158.0 kN

Fy = 0158.0 + V E 33.72(8) = 0VE = 111.76 kN

33.72 kN/m

8 m

VEVD2

184.97 kNm

123.21 kNmSpan C - D

Span D - E


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Load Case 2



33.72 kN/m

8 m

184.97 kNm

V A

VB1

MB = 0VA(8) 33.72(8) 2/2 + 184.97 = 0VA = 894.07 / 8 = 111.76 kN

Fy = 0111.76 + V B1 33.72(8) = 0VB1 = 158.0 kN

MC = 0VB2(8) 24.72(8) 2/2 + 123.21 - 184.97 = 0VB2 = 852.8 / 8 = 106.60 kN

Fy = 0106.60 + V C1 24.72(8) = 0VC1 = 91.16 kN

24.72 kN/m

8 m

123.21 kNm

VC1VB2

184.97 kNm

Span A - B

Span B - C


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33.72 kN/m

8 m

184.97 kNm

VC2

VD1

MD = 0VC2(8) 33.72(8) 2/2 - 123.21 +184.97 = 0VC2 = 1017.28 / 8 = 127.16 kN

Fy = 0127.16 + V D1 33.72(8) = 0VD1 = 142.60 kN

ME = 0

VD2(8)

24.72(8)2

/2 - 184.97 = 0VD2 = 976.01 / 8 = 122.0 kN

Fy = 0122.0 + V E 24.72(8) = 0VE = 75.76 kN

24.72 kN/m

8 m

VEVD2

184.97 kNm


Span D - E


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Load Case 3



24.72 kN/m

8 m

184.97 kNm

V A

VB1

MB = 0VA(8) 24.72(8) 2/2 + 184.97 = 0VA = 606.07 / 8 = 75.76 kN

Fy = 075.76 + V B1 24.72(8) = 0VB1 = 122.0 kN

MC = 0VB2(8) 33.72(8) 2/2 + 123.21 - 184.97 = 0VB2 = 1140.8 / 8 = 142.60 kN

Fy = 0142.60 + V C1 33.72(8) = 0VC1 = 127.16 kN

33.72 kN/m

8 m

123.21 kNm

VC1VB2

184.97 kNm

Span A - B

Span B - C


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24.72 kN/m

8 m

184.97 kNm

VC2

VD1

MD = 0VC2(8) 24.72(8) 2/2 - 123.21 + 184.97 = 0VC2 = 729.28 / 8 = 91.16 kN

Fy = 091.16 + V D1 24.72(8) = 0VD1 = 106.60 kN

ME = 0

VD2(8)

33.72(8)2

/2 - 184.97 = 0VD2 = 1264.01 / 8 = 158.0 kN

Fy = 0158.0 + V E 33.72(8) = 0VE = 111.76 kN

33.72 kN/m

8 m

VEVD2

184.97 kNm


Span D - E


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SFD EnvelopeAfter 20%Redistribution

BMD EnvelopeAfter 20%Redistribution


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