lecture 12 equivalent frame method

1

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Lecture-13Lecture 13

Equivalent Frame MethodBy: Prof Dr. Qaisar Ali

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1

Civil Engineering Department

NWFP UET [email protected]


Topics AddressedIntroduction

Stiffness of Slab-Beam Member

Stiffness of Equivalent Column

Stiffness of Column

Stiffness of Torsional Member

Prof. Dr. Qaisar Ali 2

Examples

2


Topics AddressedMoment Distribution Method

Arrangement of Live Loads

Critical Sections for Factored Moments

Moment Redistribution

Factored Moments in Column and Middle Strips


Factored Moments in Column and Middle Strips

Summary


Two Way SlabEquivalent Frame Method (ACI 13.7)

Introduction

Consider a 3D structure shown in figure. It is intended to transform this 3Dsystem into 2D system for facilitating analysis. This can be done by usingthe transformation technique of Equivalent Frame Analysis (ACI 13.7).


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Two Way SlabEquivalent Frame Method (ACI 13.7)

Introduction

First, a frame is detached from the 3D structure. In the given figure, aninterior frame is detached.

The width of the frame is same as mentioned in DDM. The length of theframe extends up to full length of 3D system and the frame extends the fullheight of the building.



Two Way SlabEquivalent Frame Method (EFM)

Introduction

Interior 3D frame detached from 3D structure.


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Introduction

This 3D frame is converted to a 2D frame by taking effect of stiffness oflaterally present members (slabs and beams).




Introduction



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Introduction





IntroductionKsb represents the combined stiffness of slab and longitudinal beam (if any).

Kec represents the modified column stiffness. The modification depends on lateralmembers (slab, beams etc) and presence of column in the storey above.

Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb

Kec Kec Kec Kec Kec Kec


Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

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IntroductionTherefore, the effect of 3D behavior of a frame is transformed into a 2D frame in terms ofthese stiffness i.e., Ksb and Kec.

Once a 2D frame is obtained, the analysis can be done by any method of 2D frame analysis.

Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb



Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec



IntroductionNext the procedures for determination of Ksb and Kec are presented.

Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb



Ksb Ksb Ksb Ksb Ksb

Ksb Ksb Ksb Ksb Ksb

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

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Stiffness of Slab Beam member (Ksb):( sb)

The stiffness of slab beam (Ksb = kEIsb/l) consists of combined stiffness ofslab 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 in literature (Nilson and MacGregor) for determinationof k for various conditions of slab systems.


l2

l1

c2

c1




Determination of k


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Isb determination




Stiffness of Slab Beam member (Ksb): Values of k for usual( sb)cases of structural systems.

Column dimension

l1 l2 c1/l1 c2/l1 k

12 × 12 10 10 0.10 0.10 4.182

15 15 0.07 0.07 4.05

20 20 0.05 0.05 4.07

As evident from thetable, the value of k forusual cases of structures


15 × 15 10 10 0.13 0.13 4.30

15 15 0.08 0.08 4.06

20 20 0.06 0.06 4.04

18 × 18 10 10 0.15 0.15 4.403

15 15 0.10 0.10 4.182

20 20 0.08 0.08 4.06

is 4.

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Stiffness of Equivalent Column (Kec):q ( ec)

Stiffness of equivalent column consists of stiffness of actual columns{above (if any) and below slab-beam} plus stiffness of torsional members.

Mathematically,

Kec =nKc × mKt

nKc + mKt1/Kec = 1/nKc + 1/mKt OR


Where,n = 2 for interior storey (for flat plates only)

= 1 for top storey (for flat plates only)m = 1 for exterior frames (half frame)

= 2 for interior frames (full frame)Note: n will be replaced by ∑ for columns having different stiffness



Stiffness of Column (Kc):( c)

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

Design aids are available from which value of k can be readily obtained fordifferent values of (ta/tb) and (lu/lc).

These design aids can be used if moment distribution method is used asmethod of analysis.


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Determination of k


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Determination of k: Values of k for usual cases of structuralsystems.

ta tb ta/tb lc lu lc/lu k

3 3 1.00 10 9.5 1.05 4.52

4 3 1.33 10 9 4 1.06 4.56

As evident from thetable, the value of k forusual cases of structures


4 3 1.33 10 9.4 1.06 4.56

5 3 1.67 10 9.3 1.07 4.60

6 3 2.00 10 9.3 1.08 5.20

7 3 2.33 10 9.2 1.09 5.39

8 3 2.67 10 9.1 1.10 5.42

9 3 3.00 10 9.0 1.11 5.46

10 3 3.33 10 8.9 1.12 5.5

is 5.5.



Stiffness of Torsional Member (Kt):( t)

Torsional members (transverse members) provide moment transferbetween the slab-beams and the columns.

Assumed to have constant cross-section throughout their length.

Two conditions of torsional members (given next).


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Stiffness of Torsional Member (Kt):( t)

Condition (a) – No transverse beams framing into columns




Stiffness of Torsional member (Kt):( t)

Condition (b) – Transverse beams framing into columns


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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


If beams frame into the support in the direction of analysis, the torsionalstiffness Kt needs to be increased.

Ecs = modulus of elasticity of slab concrete; Isb = I of slab with beam; Is = I of slab without beam




Cross sectional constant, C:


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Equivalent Frameq

Finally using the flexural stiffness values of the slab-beamand equivalent columns, a 3D frame can be converted to 2Dframe.

Kec Kec Kec Kec

Ksb Ksb Ksb

K K K


Kec

Kec

Kec

Kec

Kec

Kec

Kec

Kec

Ksb Ksb Ksb

Ksb Ksb Ksb


Two Way SlabEquivalent Frame Method (EFM):

Example: Find the equivalent 2D frame for 1st storey of the E-W interiorf f fl t l t t t h b l Th l b i 10″ thi k d LL iframe of flat plate structure shown below. The slab is 10″ thick and LL is144 psf so that ultimate load on slab is 0.3804 ksf. All columns are 14″square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height = 10′ (from floortop to slab top)Data:l1 = 25′ (ln = 23.83′)l2 = 20′


2

Column strip width = 20/4 = 5′

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Solution:

Step 01: 3D frame selection.

20′




Solution:

Step 01: 3D frame extraction.

20′


25′25′

25′

10′

10′

10′

25′

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Solution:

Step 02: Extraction of single storey from 3D frame for separate analysis.

20′


10′

25′25′

25′25′



Solution:

Step 03a: Slab-beam Stiffness calculation.

Table: Slab beam stiffness (Ksb).

Span l1 and l2 and c1/l1 c2/l2k

( bl A 20) Is=l2hf3/12 Ksb=kEIs/l


Spa c1 c2c1/l1 c2/l2 (table A-20) s l2 f / sb s/l

A2-B2 25' & 14"

20' and 14" 0.05 0.06 4.047 20000 270E

The remaining spans will have the same values as the geometry is same.

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Solution:

Step 03b: Equivalent column stiffness calculation

(1/Kec = 1/∑Kc +1/Kt)

Calculation of torsional member stiffness (Kt)

Table: Kt calculation.Column l C ∑ (1 0 63x/y)x3y/3 (i 4) K ∑ 9E C/ {l (1 c /l )3}


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

A2 20′ 14" {1 – 0.63 × 10/ 14} × 103 × 14/3 = 2567 2 × [9Ecs×2567/ {20×12 (1–14/ (20×12))3}]=231Ecs

Note 01: Kt term is multiplied with 2 because two similar torsional members meet at column A2.

Note 02: Kt values for all other columns will be same as A2 because of similar columndimensions.

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Solution: lu

A


(1/Kec = 1/∑Kc +1/Kt)

Calculation of column stiffness (Kc)

Table: ∑Kc calculation.

Column Ic (in4) kAB(from

CAB(from

B


Column location lc lu = (lc – hf) lc/ lu for 14″ × 14″

columnta/tb

(from table A23)

(from table A23)

ΣKc = 2 × kEIc/lc

A2 10′(120″) 110″ 120/110 =

1.1014 × 143/12 =

3201 5/5 = 1 5.09 0.57 2×(5.09Ecc×3201/ 120)= 272Ecc

Note: For flat plates, ∑Kc term is multiplied with 2 for interior storey with similar columnsabove and below. For top storey, the ∑Kc term will be a single value (multiplied by 1)




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Solution:


(1/Kec = 1/∑Kc +1/Kt)


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

1/Kec = 1/∑Kc +1/Kt = 1/272Ecc + 1/231Ecs


Because the slab and the columns have the same strengthconcrete, Ecc = Ecs = Ec.

Therefore, Kec = 124.91Ec

As all columns have similar dimensions and geometricconditions, the Kec value for all columns will be 124.91Ec



Solution:

Step 04: Equivalent Frame; can be analyzed using any method of analysis


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Solution:

Step 04: To analyze the frame in SAP, the stiffness values are multiplied bylengths.

Ksblsb = 270×25×12=81000E

Keclec = 124.91×10×12=14989E


10′



Solution: Load on frame:As the horizontal frame element

Step 04: SAP results (moment at center).

As the horizontal frame element represents slab beam, load is computed by multiplying slab load with width of frame

wul2 = 0.3804 × 20 = 7.608 kip/ft


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Solution:





Solution:

Step 04: SAP results (moment at faces).


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Solution:

Step 04: Comparison with SAP 3D model results.

Load on model = 144 psf (LL)Slab thickness = 10″Columns = 14″× 14″




Solution:

Step 04: Comparison of SAP 3D model with EFM.


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Example: Find the equivalent 2D frame for 1st storey of the E-W interiorf f b t d f t t h b l Th l b i 7″frame of beam supported frame structure shown below. The slab is 7″thick with LL of 144 psf so that ultimate load on slab is 0.336 ksf. Allcolumns are 14″ square. Take fc′ = 4 ksi and fy = 60 ksi. Storey height =10′ (from floor top to slab top)Data:l1 = 25′ (ln = 23.83′)l2 = 20′


2

Column strip width = 20/4 = 5′



Solution:

Step 01: 3D frame selection.

20′


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Solution:

Step 01: 3D frame extraction.

20′


25′25′

25′

10′

10′

10′

25′



Solution:

Step 02: Extraction of single storey from 3D frame for separate analysis.

20′


10′

25′25′

25′25′

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Solution:

Step 03a: Slab-beam Stiffness calculation.

Table: Slab beam stiffness (Ksb).

Span l1 and c1

l2 and c2

c1/l1 c2/l2k

(table A-20) Isb Ksb=kEIs/l1


c1 c2 (table A 20)

A2-B2 25' & 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.



Solution:


(1/Kec = 1/∑Kc +1/Kt)

Calculation of torsional member stiffness (Kt)

Table: Kt calculation.Column l C ∑ (1 0 63x/y)x3y/3 (i 4) K ∑ 9E C/ {l (1 c /l )3}


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

A2 20′ 14" 11208 3792.63Ecs

B2 20′ 14" 12694 4295.98Ecs

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Solution: lu

A


(1/Kec = 1/∑Kc +1/Kt)



Column location l l l / lIc (in4)

for 14″ × 14″ t /tbkAB (from K

B


∑Kc = 202Ecc + 141Ecc = 343Ecc

Column location lc lu lc/ lu for 14 × 14 column

ta/tb table A23) Kc

A2 (bottom) 10′(120″) 100″ 120/100 =

1.2014 × 143/12 =

320116.5/3.5 =

4.71 7.57 201.9Ecc

A2 (top) 10′(120″) 100″ 120/100 =

1.2014 × 143/12 =

32013.5/16.5=

0.21 5.3 141.39Ecc



Solution: lu

A


(1/Kec = 1/∑Kc +1/Kt)



Column location l l l / lIc (in4)

for 14″ × 14″ t /tbkAB (from K

B


∑Kc = 202Ecc + 141Ecc = 343Ecc

Column location lc lu lc/ lu for 14 × 14 column

ta/tb table A23) Kc

B2 (bottom) 10′(120″) 100″ 120/100 =

1.2014 × 143/12 =

320116.5/3.5 =

4.71 7.57 201.9Ecc

B2 (top) 10′(120″) 100″ 120/100 =

1.2014 × 143/12 =

32013.5/16.5=

0.21 5.3 141.39Ecc

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Solution:

Step 03b: Equivalent column stiffness calculation (Column A2)

(1/Kec = 1/∑Kc +1/Kt)



1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/3792.63Ecs



Therefore, Kec = 315Ec



Solution:

Step 03b: Equivalent column stiffness calculation (Column B2)

(1/Kec = 1/∑Kc +1/Kt)



1/Kec = 1/∑Kc +1/Kt = 1/343Ecc + 1/4295.98Ecs



Therefore, Kec = 318Ec

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Solution:

Step 04: Equivalent Frame; can be analyzed using any method of analysis




Solution:

Step 04: To analyze the frame in SAP, the stiffness values are multiplied bylengths. Ksblsb = 349×25×12=104700E

Keclec = 315×10×12=37800E

Keclec = 318×10×12=38160E


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Solution: Load on frame:As the horizontal frame element


As the horizontal frame element represents slab beam, load is computed by multiplying slab load with width of frame

wul2 = 0.336 × 20 = 6.72 kip/ft




Solution:



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Solution:

Step 04: SAP results (moment at faces).




Solution:

Step 04: Comparison with SAP 3D model results.

Load on model = 144 psf (LL)Slab thickness = 7″Columns = 14″× 14″Beams = 14″× 20″


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Solution:

Step 04: Comparison of beam moments of SAP 3D model with beammoments of EFM by SAP 2D analysis.




Moment Distribution Method:

The original derivation of EFM assumed that moment distribution would bethe procedure used to analyze the slabs, and some of the concepts in themethod are awkward to adapt to other methods of analysis.

In lieu of computer software, moment distribution is a convenient handcalculation method for analyzing partial frames in the Equivalent FrameMethod.


Once stiffnesses are obtained from EFM, the distribution factors areconveniently calculated.

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Distribution Factors:

Kt

Kt

K

Ksb1

Ksb2l1lc

1

2

Kct


Kec

K = kEI/l

l1

lc

3Kcb





Slab Beam Distribution Factors:

DF (span 2-1) =Ksb1

Ksb1 + Ksb2 + Kec


sb1 sb2 ec

DF (span 2-3) =Ksb2

Ksb1 + Ksb2 + Kec

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Equivalent Column Distribution factors:

DF =Kec

K + K + K


Ksb1 + Ksb2 + Kec





These distribution factors are used in analysis.

The equivalent frame of example 02 shall now be analyzed using moment distribution method.

The comparison with SAP 3D model result for beam moments is also done


done.

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Solution:

Step 04: Comparison of SAP 3D model with EFM done by Momentdistribution method.

Joint A B C D E

CarryOver 0.5034 0.5034 0.5034 0.5034

DF 0.000 0.301 0.699 0.412 0.177 0.412 0.412 0.177 0.412 0.412 0.177 0.412 0.699 0.301 0.000

Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab

FEM 0.000 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 399.103 ‐399.103 0.000 0.000

Bal 0.000 ‐119.955 ‐279.148 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 279.148 119.955 0.000


Bal 0.000 119.955 279.148 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 279.148 119.955 0.000

Carry over 0.000 ‐140.529 0.000 0.000 0.000 0.000 140.529 0.000

Bal 0.000 0.000 0.000 57.838 24.854 57.838 0.000 0.000 0.000 ‐57.838 ‐24.854 ‐57.838 0.000 0.000 0.000

Carry over 29.117 0.000 0.000 29.117 ‐29.117 0.000 0.000 ‐29.117

Bal 0.000 ‐8.751 ‐20.365 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 20.365 8.751 0.000

Carry over 0.000 ‐10.252 0.000 0.000 0.000 0.000 10.252 0.000

Bal 0.000 0.000 0.000 4.220 1.813 4.220 0.000 0.000 0.000 ‐4.220 ‐1.813 ‐4.220 0.000 0.000 0.000

Total 0.000‐129.395 129.395 ‐488.302 26.810 461.492‐367.695 0.000 367.695‐461.492‐26.810488.302‐129.395129.395 0.000



Solution:

Step 04: Comparison of SAP 3D model with EFM.


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Solution of example 02 by Moment Distribution Method:p y

Step 04: Analysis using Moment distribution method.

Joint A B C D E

CarryOver 0.5034 0.5034 0.5034 0.5034

DF 0.000 0.474 0.526 0.344 0.313 0.344 0.344 0.313 0.344 0.344 0.313 0.344 0.526 0.474 0.000

Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab Slab Column Slab

FEM 0.000 0.000 351.891 ‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 351.891‐351.891 0.000 0.000


Bal 0.00 ‐166.90 ‐185.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 185.00 166.90 0.00

Carry over 0.00 ‐93.13 0.00 0.00 0.00 0.00 93.13 0.00

Bal 0.00 0.00 0.00 31.99 29.15 31.99 0.00 0.00 0.00 ‐31.99 ‐29.15 ‐31.99 0.00 0.00 0.00

Carry over 16.11 0.00 0.00 16.11 ‐16.11 0.00 0.00 ‐16.11

Bal 0.00 ‐7.64 ‐8.47 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 8.47 7.64 0.00

Carry over 0.00 ‐4.26 0.00 0.00 0.00 0.00 4.26 0.00

Bal 0.00 0.00 0.00 1.46 1.33 1.46 0.00 0.00 0.00 ‐1.46 ‐1.33 ‐1.46 0.00 0.00 0.00

Total 0. ‐174.900 174.900 ‐415.961 30.544 385.417‐335.012 0.000 335.012‐385.417‐30.544415.961‐174.900174.900 0.000



Solution:

Step 04: Comparison of beam moments of SAP 3D model with EFM analysisresults obtained by moment distribution method.


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Arrangement of Live loads (ACI 13.7.6):g ( )

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 maximumfactored moment



Two Way Slab


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Critical section for factored moments (ACI 13.7.7):( )

Interior supports

Critical section at face of rectilinear support but ≤ 0.175l1 from center ofthe support

Exterior supports

At exterior supports with brackets or capitals, the critical section < ½ the


pp p ,projection of bracket or capital beyond face of supporting element.


Two Way Slab


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Moment Redistribution (ACI 13.7.7.4):( )

Mu1Mu2

Mo


Mu3

l1

ln c1/2c1/2



Factored moments in column strips and middle strips:p p

Same as in the Direct Design Method


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Summary of Steps required for analysis using EFMy p q y gExtract 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 any method of analysis to get longitudinal moments


Analyze the obtained 2D frame using any method of analysis to get longitudinal momentsbased on center to center span.

Distribute slab-beam longitudinal moment laterally using lateral distribution procedures ofDDM.


The End


lecture 12 equivalent frame method

Design

university of engineering

d frame analysis

interior frame

equivalent frame methodby

stiffness of slab beam

d system

d structure

technology peshawar