Lecture-08 Gravity Load Analysis of RC Structures

1

Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Lecture-08

Gravity Load Analysis of RC Structures

1

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

www.drqaisarali.com



Contents

Analysis Approaches

Point of Inflection Method

Equivalent Frame Method

Case Study

Limit Analysis

Plastic Analysis

References

2

2



Analysis Approaches

3

Analysis Approaches

Elastic AnalysisBased on elastic properties of structure.

Presents the behavior of the structure within certain limitations.

ExactFEM, Slope Deflection,

Moment Distribution,

Stiffness methods, Equivalent Frame etc.

ApproximateACI

Coefficients, Direct

Design, Point of Inflection, Portal Frame Methods etc

Plastic AnalysisBased on inelastic capacities of structure,

presents the behavior of the structure more accurately.

Non-Linear pushover

Analysis etc.



Analysis Approaches

The approximate analysis methods such as ACI Coefficients

and Direct Design Method have been discussed in detail in

earlier lectures.

In this lecture, another approximate method known as Point of

Inflection Method will be briefly discussed.

The exact analysis methods such as Slope Deflection,

Moment Distribution and Stiffness method etc. have already

been studied. The Equivalent Frame Analysis method will be

discussed in detail in this lecture.

4

3




In this method, points of inflection are located on the frame

and the members are assumed separate determinate

members at point of inflection.

The individual members can be analyzed by statics as

shown next.

5




6

1. Indeterminate frame 2. Point of inflectionlocated over frame byinspection

3. Indeterminate framebroken intodeterminate members

4.Analysis of determinatemembers by statics

5. Determinate member’sBMD combined to formfinal approximate BMD ofindeterminate structure

4



Equivalent Frame Method (ACI 8.11)

Introduction

Consider a 3D structure shown in figure. It is intended to transform this

3D system to 2D system for facilitating analysis. This is done using the

transformation technique of Equivalent Frame Method (ACI 8.11).

7




Introduction

The width of the frame is same as in DDM. The length of the frame

extends up to full length of 3D system and the frame extends the full

height of the building.

8

5




Introduction

This 3D frame is converted to a 2D frame by taking the effect of

stiffnesses of laterally present members (slabs and beams).

9

In EFA method, thehorizontal members of theconverted 2D frame arecalled slab-beam membersand the vertical membersare called equivalentcolumns.




Introduction

The modified stiffnesses Ksb and Kec are calculated and assigned to

slab-beam and equivalent columns as shown in the figure.

10

Ksb representsthe combinedstiffness of slaband longitudinalbeam and Kecrepresents themodified columnstiffness.

6




Stiffness of Slab Beam member (Ksb):

The stiffness of slab beam (Ksb = kEIsb/l) consists of combined

stiffness of slab and any longitudinal beam present within.

For a span, the k factor is a direct function of ratios c1/l1 and c2/l2

Tables are available for determination of k for various conditions of

slab systems.

l2

l1

c2

c1

Plan view of a floor of width l2 which is to beconverted to slab beam line element

11





Determination of k

12

7





Isb determination

13




Stiffness of Equivalent Column (Kec):

Stiffness of equivalent column consists of stiffness of actual columns

{above and below the slab} plus stiffness of torsional members.

Mathematically,

Kec =∑Kc ∑Kt

∑Kc + ∑Kt

Where,

∑Kc = sum of flexural stiffnesses of columns above and

below the slab.

∑Kt = Torsional stiffness of attached torsional members

1/Kec = 1/∑Kc + 1/∑Kt OR

14

8




Stiffness of Column (Kc):

General formula of flexural stiffness is given by K = kEI/l

Design aids are available from which value of k can be readily obtained

for different values of (ta/tb) and (lu/lc).

These design aids can be used if moment distribution method is used

as method of analysis.

15





16

9





Determination of k

17




Stiffness of Torsional Member (Kt):

Torsional members (transverse members) provide moment transfer

between the slab-beams and the columns.

Assumed to have constant cross-section throughout their length.

18

10





Stiffness Determination: The torsional stiffness Kt of the torsional

member is given as:

If beams frame into the support in the direction of analysis, the torsional

stiffness Kt needs to be increased.

Ecs = modulus of elasticity of slab concrete; Isb = I of slab with beam;

Is = I of slab without beam = l2h3/12

19





Cross sectional constant, C:

20

11




Moment Distribution Method:

The original derivation of EFM assumed that moment distribution would

be the procedure used to analyze the slabs.

In lieu of computer software, moment distribution is a convenient hand

calculation method for analyzing partial frames in the Equivalent Frame

Method.

Next slides discuss the application of Moment Distribution Method to

complete the analysis using EFM.

21





Distribution Factors:

Slab Beam Distribution Factors:

DF (span 2-1) =Ksb1

Ksb1 + Ksb2 + Kec

DF (span 2-3) =Ksb2

Ksb1 + Ksb2 + Kec

22

12





Distribution Factors:

Equivalent Column Distribution Factor:

DF =Kec

Ksb1 + Ksb2 + Kec

23





Distribution of unbalanced moment to columns:

Portion of unbalanced moment to upper column =

Kct

Kcb + Kct

Portion of unbalanced moment to lower column =

Kcb

Kcb + Kct

24

13




Arrangement of Live loads (ACI 8.11.1)

ACI 8.11.1 states that when the loading pattern is

known, the equivalent frame shall be analyzed for that

load.

When LL 0.75DL

Maximum factored moment when Full factored LL on all spans

Other cases

Pattern live loading using 0.75(Factored LL) to determine maximum

factored moment.

25




26

14




Summary of Steps required for analysis using EFM

Extract the 3D frame from the 3D structure.

Extract a storey from 3D frame for gravity load analysis.

Identify EF members i.e., slab beam, torsional member and columns.

Find stiffness (kEI/l) of each EF member using tables.

Assign stiffnesses of each EF member to its corresponding 2D frame member.

Analyze the obtained 2D frame using Moment Distribution method of analysis to

get longitudinal moments based on center to center span.

Distribute slab-beam longitudinal moment laterally using lateral distribution

procedures of DDM.

Slab analysis can be done using DDM.

27




28

Analysis by EFM

Now 3D model will be converted into equivalent 2D model using EFM.

20 ft

20 ft

20 ft

25 ft 25 ft 25 ft 25 ft

Beam

15




29

Analysis by EFM

Sizes: Sizes will be as in 3D line model.

Slab: Slab thickness, hf = 7″

Columns: All columns are 14″ 14″ as before.

Beams: All beams are 14″ wide and 20″ deep.




30

Analysis by EFM

Step a: 3D frame selection

20′20 ft

20 ft

20 ft

25 ft 25 ft 25 ft 25 ft

16




31

Analysis by EFM

Step b: 3D frame extraction

20′

25′

25′

25′

10.5′

10.5′

10.5′

25′




32

Analysis by EFM

Step c: Extraction of single storey from 3D frame for separate analysis

20′

10.5′

25′

25′

25′

25′

According to ACI 8.11.2.5, Itshall be permitted to assumethat the far ends of columnsbuilt integrally with thestructure are considered tobe fixed for gravity loadanalysis.

17




33

Analysis by EFM

Step d (i): Slab-beam Stiffness calculation

Table: Slab beam stiffness (Ksb).

Spanl1 and

c1

l2 and c2

c1/l1 c2/l2 k Isb Ksb=kEIs/l1

AB25' & 14"

20' and 14"

0.0467 0.058 4.051 25844 349E

The remaining spans will have the same values as the geometry is same.Table A-20 (Reinforced concrete: Mechanics and Design, 3rd Ed)

BA C D E




34

Analysis by EFM

Step d (ii): Equivalent column stiffness calculation

(1/Kec = 1/∑Kc +1/Kt)

Calculation of torsional member stiffness (Kt)

Table: Kt calculation.

Column location

l2 c2 C = ∑ (1 – 0.63x/y)x3y/3 (in4) Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3}

C2 20′ 14" 11208 3792.63Ecs

C1 20′ 14" 12694 4295.98Ecs

C2 C1 C1 C1 C2

18




35

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)

Calculation of column stiffness (Kc) for Column C2

Kc = 192Ecc + 135Ecc = 327Ecc

Table: ∑Kc calculation.

Column location lc lu lc/ lu

Ic (in4) for 14″ 14″

columnta/tb kAB Kc

C2 (bottom)10.5′

(126″)106″

126/106 ≈ 1.20

14 143/12 = 3201

16.5/3.5 = 4.71

7.57 192Ecc

C2 (top)10.5′

(126″)106″

126/106 ≈ 1.20

14 143/12 = 3201

3.5/16.5= 0.21

5.3 135Ecc

lu

A

B




36

Analysis by EFM

19




37

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)

Calculation of column stiffness (Kc) for Column C1

Kc = 192Ecc + 135Ecc = 327Ecc



Ic (in4) for 14″ 14″

columnta/tb kAB Kc

C1 (bottom)10.5′

(126″)106″

126/106 ≈ 1.20

14 143/12 = 3201

16.5/3.5 = 4.71 7.57 192Ecc

C1 (top)10.5′

(126″)106″

126/106 ≈ 1.20

14 143/12 = 3201

3.5/16.5= 0.21 5.3 135Ecc




38

Analysis by EFM

Step d (ii): Equivalent column stiffness calculation (Column C2)

(1/Kec = 1/∑Kc +1/Kt)

Calculation of column stiffness (Kc)

Equivalent column stiffness calculation (1/Kec = 1/∑Kc +1/Kt)

1/Kec = 1/∑Kc +1/Kt = 1/327Ecc + 1/3792.63Ecs

Because the slab and the columns have the same strength

concrete, Ecc = Ecs = Ec.

Therefore, Kec = 301Ec

20




39

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)



1/Kec = 1/∑Kc +1/Kt = 1/327Ecc + 1/4295.98Ecs







40

Analysis by EFM

Step e: Equivalent Frame; can be analyzed using any method of analysis.

21




41

Analysis by EFM

As the ground storey is same as 1st one, therefore the stiffness calculated

shall also be assigned to ground storey.

For the top storey, the slab beam stiffness will be same as lower stories.

However the equivalent stiffness of the top storey column is computed

next.




42

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)


Kc = 192Ecc

Similarly for interior column, Kc = 192Ecc



Ic (in4) for 14″ 14″

columnta/tb kAB Kc

C2 (bottom)10.5′

(126″)100″

126/106 ≈ 1.20

14 143/12 = 3201

16.5/3.5 = 4.71

7.57 192Ecc

22




43

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)



1/Kec = 1/∑Kc +1/Kt = 1/192Ecc + 1/3792.63Ecs







44

Analysis by EFM


(1/Kec = 1/∑Kc +1/Kt)



1/Kec = 1/∑Kc +1/Kt = 1/192Ecc + 1/4295.98Ecs




23




Analysis by EFM

Step e: Equivalent Frame

45




Analysis by EFM

Step f: Analysis using moment Distribution Method

Load on frame for Bending Analysis:As the horizontal frame elementrepresents slab beam, load iscomputed by multiplying slab load withwidth of framewDL = 0.0875 × 20 = 1.75 kip/ftwLL = 0.144 × 20 = 2.88 kip/ft

46

24




Analysis by EFM Step f: Analysis results for dead load (interior storey) using moment

distribution method.

47




Analysis by EFM Step f: Analysis results for dead load (top storey) using moment

distribution method.

48

25




Analysis by EFM Step f: Analysis results for dead load (Values at centerline).

49




Analysis by EFM

Step f: Analysis results for dead load

Distribution of moments to slab and beam for top storey.

Table: E-W Interior frame analysis (Top Storey).

LengthLongitudinal

moment section

Longitudinal

moments (LM)

Column Strip Moment %age factor (Graph

A4)

Column Strip slab Moment

(CSSM) =

0.15CSM

Column strip Beam Moment

(BM)=

0.85CSM

Middle Strip slab Moment

25‘-0“(Ext)

E - 34 0.93 5 27 2.38

+ 67 0.8 8 46 13.4

I - 111 0.8 13 75 22.2

25‘-0“(Int)

- 105 0.8 13 71 21

+ 42 0.8 5 29 8.4

- 85 0.8 10 58 17

50

26




Analysis by EFM


Distribution of moments to slab and beam for interior storey.

Table: E-W Interior frame analysis (Interior Storey).

LengthLongitudinal

moment section

Longitudinal

moments (LM)


A4)


(CSSM) =

0.15CSM


(BM)=

0.85CSM


25‘-0“(Ext)

E - 45 0.93 6 36 3.15

+ 64 0.8 8 44 12.8

I - 109 0.8 13 74 21.8

25‘-0“(Int)

- 101 0.8 12 69 20.2

+ 43 0.8 5 29 8.6

- 87 0.8 10 59 17.4

51




Analysis by EFM


Analysis of columns for DL (factors for moment distribution)

The computed unbalanced longitudinal moments shall be

transferred to columns and shall be distributed to top and

bottom columns as follows:

Portion of unbalanced moment to upper column =Kct

Kcb + Kct

Portion of unbalanced moment to lower column =Kcb

Kcb + Kct

52

27




Analysis by EFM


Analysis of columns for DL (factors for moment distribution)

34 6

C2 C1 C1 C1 C2

45 8

45 8

Kcb3=192E

Kct2=135E

Kcb2=192E

Kct1=135E

Kcb1=192E

Kct2/(Kct2+Kcb2)=0.41

Kcb2/(Kct2+Kcb2)=0.59



Kcb3/(Kcb3=1.00

53




Analysis by EFM


Analysis of columns for DL

C2 C1 C1 C1 C2

Kcb3=192E

Kct2=135E

Kcb2=192E

Kct1=135E

Kcb1=192E





Kcb3/(Kcb3=1.00

34×1.00=34

45×0.41=18.4545×0.59=27

6×1.00=6

8×0.41=3.3

8×0.59=4.7

45×0.41=18.4545×0.59=27 8×0.41=3.3

8×0.59=4.7

54

28




Analysis by EFM

Step f: Analysis results for live load

55




Analysis by EFM


Distribution of moments to slab and beam

Table: E-W Interior frame analysis (Top Storey).

LengthLongitudinal

moment section

Longitudinal

moments (LM)


A4)


(CSSM) =

0.15CSM


(BM)=

0.85CSM


25‘-0“(Ext)

E - 56 0.93 7.8 44 3.92

+ 111 0.8 13.3 75 22.2

I - 183 0.8 22.0 124 36.6

25‘-0“(Int)

- 172 0.8 20.6 117 34.4

+ 69 0.8 8.3 47 13.8

- 140 0.8 16.8 95 28

56

29




Analysis by EFM


Distribution of moments to slab and beam

Table: E-W Interior frame analysis (Interior Storey).

LengthLongitudinal

moment section

Longitudinal

moments (LM)


A4)


(CSSM) =

0.15CSM


(BM)=

0.85CSM


25‘-0“(Ext)

E - 73 0.93 10.2 58 5.11

+ 105 0.8 12.6 71 21

I - 179 0.8 21.5 122 35.8

25‘-0“(Int)

- 166 0.8 19.9 113 33.2

+ 70 0.8 8.4 48 14

- 144 0.8 17.3 98 28.8

57




Analysis by EFM


Analysis of columns for LL (factors for moment distribution)

Interior column moment is due to pattern load.

The computed unbalanced longitudinal moments shall be

transferred to columns and shall be distributed to top and

bottom columns as follows:

Portion of unbalanced moment to upper column =Kct

Kcb + Kct

Portion of unbalanced moment to lower column =Kcb

Kcb + Kct

58

30




Analysis by EFM


Analysis of columns for LL (factors for moment distribution)

56 151

C2 C1 C1 C1 C2

73 151

43 151

Kcb3=192E

Kct2=135E

Kcb2=192E

Kct1=135E

Kcb1=192E





Kcb3/(Kcb3=1.00

59




Analysis by EFM


Analysis of columns for LL

C2 C1 C1 C1 C2

Kcb3=192E

Kct2=135E

Kcb2=192E

Kct1=135E

Kcb1=192E





Kcb3/(Kcb3=1.00

56×1.00=56

73×0.41=30.00

73×0.59=43

151×1.00=151

151×0.41=62

151×0.59=89

73×0.41=30.00

73×0.59=43

151×0.41=62

151×0.59=89

60

31



Case Study: Comparison of the Results of

Equivalent Frame Method, ACI Coefficient Method,

Direct Design Method & SAP 2D Model with respect to

SAP2000 3D Line Model

61



Case Study

Dead Load Bending Moment in Beam

62

5417.9819.23023

6256.24464044

8769.07656260

4634.53293028

6664.14485250

6664.14625857

5417.98323930

6256.24543840

8769.07786060

4634.5330.53029

6664.14605250

6664.14725756

ACI Coefficient MethodDirect Design MethodEquivalent Frame Method by SAP 2D modelSAP 2D ModelSAP 3D Model

C2 C1 C1 C1 C2

32



Case Study

Dead Load Bending Moment Column

63

5420344942

2066.06.95.4

271018.453835

1033.34.23.7

2710273331

1034.72.63.0

C2 C1 C1 C1 C2

271018.453834

1033.34.24.0

1034.72.42.6

2710272319




Case Study

Live Load Bending Moment in Beam

64

7926334930

9079756762

1259711210182

6448474939

9490958668

94901269677

7926476340

9079716356

125971099982

6448484739

9490858668

9490999275C2 C1 C1 C1 C2


33



Case Study

Live Load Moment in Columns

65

78.733568159

861331516845

39.3516.5306248

39.3516.5306247

4366.5625539

4366.5625438

C2 C1 C1 C1 C2

39.3516.5435543

39.3516.5433727

4366.5893121

4366.5894935




Limit Analysis

Introduction

Most RC structures are designed using following approach:

Moments, shears, and axial forces in RC structures are found by

elastic theory.

The actual proportioning of members is done by strength methods, in

which inelastic section and member response is considered.

Although this design approach is safe and conservative but

is inconsistent to total analysis-design process.

66

34



Limit Analysis

Redistribution of Moments

A frame normally will not fail when the nominal moment

capacity of just one critical section is reached:

A plastic hinge will form at that section

Large rotation at constant resisting moment will occur.

Load transfer to other locations (having more capacity) along the span

will occur.

On further increase in load, additional plastic hinges may form at other

locations along the span.

As a result, structure will collapse, but only after a significant

redistribution of moments.

67



Limit Analysis

Redistribution of Moments

Full use of the plastic capacity of reinforced concrete beams

and frames requires an extensive analysis of all possible

mechanisms and an investigation of rotation requirements

and capacities at all proposed hinge locations.

On the other hand, a restricted amount of redistribution of

elastic moments can safely be made without complete

analysis, yet may be sufficient to obtain most of the

advantages of limit analysis.

68

35



Limit Analysis

Negative Moments Redistribution in Continuous

Flexural Members (ACI 6.6.5)

A limited amount of redistribution is permitted by ACI Code

6.6.5. depending upon a rough measure of available

ductility, without explicit calculation of rotation

requirements and capacities.

69



Limit Analysis



The net tensile strain in the extreme tension steel at nominal strength εt

given in eq. below, is used as an indicator of rotation capacity.

Accordingly, ACI Code 6.6.5 provides as follows:

Except where approximate values for moments are used, it shall be

permitted to increase or decrease negative moments calculated by

elastic theory at supports of continuous flexural members for any

assumed loading arrangement by not more than 1000εt percent, with

a maximum of 20 percent.

70

εt = εu(d – c)/c)

36



Limit Analysis



71

εt = εu(d – c)/c)

εu = 0.003

As example, for given As if:

d = 16.5″ ; c = 4″

εt = 0.009

1000εt = 9 % < 20 %

εu

εt

c

d

As



Limit Analysis



Graphical representation of ACI code provision

72

37



Limit Analysis



6.6.5.1 — The modified negative moments shall be used

for calculating moments at sections within the spans.

6.6.5.1 — Redistribution of negative moments shall be

made only when εt is equal to or greater than 0.0075 at the

section at which moment is reduced.

73



Limit Analysis

Negative Moments Redistribution (Example)

For the beam shown, find moment redistribution.

74

38



Limit Analysis

Negative Moments Redistribution (Example)

Solution: To obtain maximum moments at all critical design

sections. it is necessary to consider three alternative

loadings.

It will be assumed that 20 % adjustment of support

moment is permitted throughout.75

Mmax+, ext

Mmax+, int

Mmax-, int


Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 76

82+0.2 ×82 = 98

78+0.2 × 78 = 93

134−0.2 × 134 = 107

Mmax+, ext

Mmax+, int

Mmax-, int

Decrease in exterior positivemoment:Negative moment increased 20 %,which results in decrease in Mmax+,ext

from 109 to 101

Decrease in interior positivemoment:Negative moment increased 20 %,which results in decrease in Mmax+,int

from 72 to 57

Decrease in interior negativemoment:Negative moment decreased 20 %,without exterior positive and interiorpositive moments exceeding Mmax+,ext

and Mmax+,int respectively.

Limit Analysis

39



Limit Analysis

Conclusion on Redistribution of Moments

It can be seen, then, that the net result is a reduction in design moments

over the entire beam. This modification of moments does not mean a

reduction in safety factor below that implied in code safety provisions;

rather, it means a reduction of the excess strength that would otherwise

be present in the structure because of the actual redistribution of

moments that would occur before failure.

It reflects the fact that the maximum design moments are obtained from

alternative load patterns, which could not exist concurrently. The end

result is a more realistic appraisal of the actual collapse load of the

indeterminate structure.

77



Plastic Analysis

• Non-Linear Static (Pushover) Analysis

• Points on the structure whose performance (when it yields,

cracks or fails) is required to be monitored are selected.

• The structure is pushed at the top.

• Top Drift (D) and corresponding base-shear (V) is calculated

and plotted on V-Δ curve.

• Structure is further pushed in steps and V-Δ curve is plotted.

Also performance of the selected points is monitored and

marked on the V-Δ curve.

• Therefore a single chart that shows the performance of the

whole structure (or separate charts for all points of interests)

is obtained.

• These charts can be used to identify points where

strengthening of structure is required (i.e points that fail or

start to fail in the start of the curve.

78

BeamCrack Failure

Cracking in Column

Beam yield

D

V

PUSHD

V(base shear)

Points of interest

40



References

ACI 318-14

PCA Notes on ACI 318

RC Design Teaching Aids by PCA

Example by Dr. Qaisar Ali and Engr. Umer

79



The End

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Lecture-08 Gravity Load Analysis of RC Structures

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