This document discusses the equivalent frame method for analyzing two-way slabs. It introduces the equivalent frame method, which transforms a 3D structural system into a 2D system by representing the stiffness of slab and beam members as Ksb, and the modified stiffness of columns as Kec. This allows the 3D behavior to be analyzed using conventional 2D frame analysis methods. The document then covers determining the values of Ksb and Kec to represent the slab and column stiffness in the equivalent frame.

1. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Lecture 13
Lecture-13
Equivalent Frame Method
By: Prof Dr. Qaisar Ali
Civil Engineering Department
NWFP UET Peshawar
drqaisarali@nwfpuet.edu.pk
Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Topics Addressed
Introduction
Stiffness of Slab-Beam Member
Stiffness of Equivalent Column
Stiffness of Column
Stiffness of Torsional Member
Examples
Prof. Dr. Qaisar Ali 2
1

2. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Topics Addressed
Moment Distribution Method
Arrangement of Live Loads
Critical Sections for Factored Moments
Moment Redistribution
Factored Moments in Column and Middle Strips
Summary
Prof. Dr. Qaisar Ali 3
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (ACI 13.7)
Introduction
Consider a 3D structure shown in figure. It is intended to transform this 3D
system into 2D system for facilitating analysis. This can be done by using
the transformation technique of Equivalent Frame Analysis (ACI 13.7).
Prof. Dr. Qaisar Ali 4
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3. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (ACI 13.7)
Introduction
First, a frame is detached from the 3D structure. In the given figure, an
interior frame is detached.
The width of the frame is same as mentioned 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.
Prof. Dr. Qaisar Ali 5
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
Interior 3D frame detached from 3D structure.
Prof. Dr. Qaisar Ali 6
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4. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness of
laterally present members (slabs and beams).
Prof. Dr. Qaisar Ali 7
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness of
laterally present members (slabs and beams).
Prof. Dr. Qaisar Ali 8
4

5. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness of
laterally present members (slabs and beams).
Prof. Dr. Qaisar Ali 9
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
Ksb represents the combined stiffness of slab and longitudinal beam (if any).
Kec represents the modified column stiffness. The modification depends on lateral
members (slab, beams etc) and presence of column in the storey above.
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 10
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6. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
Therefore, the effect of 3D behavior of a frame is transformed into a 2D frame in terms of
these 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
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 11
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Introduction
Next the procedures for determination of Ksb and Kec are presented.
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 12
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7. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
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 in literature (Nilson and MacGregor) for determination
of k for various conditions of slab systems.
c1
l2 c2
l1
Prof. Dr. Qaisar Ali 13
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Slab Beam member (Ksb):
( )
Determination of k
Prof. Dr. Qaisar Ali 14
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8. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Slab Beam member (Ksb):
( )
Isb determination
Prof. Dr. Qaisar Ali 15
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Slab Beam member (Ksb): Values of k for usual
( )
cases of structural systems.
Column l1 l2 c1/l1 c2/l1 k
dimension
12 × 12 10 10 0.10 0.10 4.182 As evident from the
15 15 0.07 0.07 4.05 table, the value of k for
20 20 0.05 0.05 4.07 usual cases of structures
15 × 15 10 10 0.13 0.13 4.30 is 4.
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
Prof. Dr. Qaisar Ali 16
8

9. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Equivalent Column (Kec):
q ( )
Stiffness of equivalent column consists of stiffness of actual columns
{above (if any) and below slab-beam} plus stiffness of torsional members.
Mathematically,
nKc × mKt
1/Kec = 1/nKc + 1/mKt OR Kec =
nKc + mKt
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
Prof. Dr. Qaisar Ali 17
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
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.
Prof. Dr. Qaisar Ali 18
9

10. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Column (Kc):
( )
Prof. Dr. Qaisar Ali 19
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Column (Kc):
( )
Determination of k
Prof. Dr. Qaisar Ali 20
10

11. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Column (Kc):
( )
Determination of k: Values of k for usual cases of structural
systems.
ta tb ta/tb lc lu lc/lu k
As evident from the
table, the value of k for
3 3 1.00 10 9.5 1.05 4.52
usual cases of structures
4 3 1.33 10 9.4
94 1.06 4.56
is 5.5.
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
Prof. Dr. Qaisar Ali 21
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
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.
Two conditions of torsional members (given next).
Prof. Dr. Qaisar Ali 22
11

12. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Torsional Member (Kt):
( )
Condition (a) – No transverse beams framing into columns
Prof. Dr. Qaisar Ali 23
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):
( )
Condition (b) – Transverse beams framing into columns
Prof. Dr. Qaisar Ali 24
12

13. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):
( )
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
analysis,
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
Prof. Dr. Qaisar Ali 25
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):
( )
Cross sectional constant, C:
Prof. Dr. Qaisar Ali 26
13

14. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Equivalent Frame
q
Finally using the flexural stiffness values of the slab-beam
and equivalent columns, a 3D frame can be converted to 2D
frame.
Ksb Ksb Ksb
Kec Kec Kec Kec
Ksb Ksb Ksb
Kec Kec Kec Kec
Ksb Ksb Ksb
Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 27
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Example: Find the equivalent 2D frame for 1st storey of the E-W interior
frame of fl t plate structure shown b l
f f flat l t t t h below. Th slab i 10″ thi k and LL i
The l b is thick d is
144 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 floor
top to slab top)
Data:
l1 = 25′ (ln = 23.83′)
l2 = 20′
Column strip width = 20/4 = 5′
Prof. Dr. Qaisar Ali 28
14

15. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 01: 3D frame selection.
20′
Prof. Dr. Qaisar Ali 29
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 01: 3D frame extraction.
20′
10′
10′
25′
25′
10′
25′
Prof. Dr. Qaisar Ali 25′ 30
15

16. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 02: Extraction of single storey from 3D frame for separate analysis.
20′
25′
25′
10′
25′
25′
Prof. Dr. Qaisar Ali 31
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03a: Slab-beam Stiffness calculation.
Table: Slab beam stiffness (Ksb).
l1 and l2 and k
Spa
Span c1/l1 c2/l2 I =l h 3/12
l / Ksb=kEIs/l
c1 c2 ( bl A-20) s 2 f
(table A 20)
25' & 20' and
A2-B2 0.05 0.06 4.047 20000 270E
14" 14"
The remaining spans will have the same values as the geometry is same.
Prof. Dr. Qaisar Ali 32
16

17. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Prof. Dr. Qaisar Ali 33
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of torsional member stiffness (Kt)
Table: Kt calculation.
Column
l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4)
0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3}
location
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 column
dimensions.
Prof. Dr. Qaisar Ali 34
17

18. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM): A
lu
Solution:
B
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
kAB CAB
Ic (in4)
Column (from (from
lc lu = (lc – hf) lc / lu for 14″ × 14″ ta/tb ΣKc = 2 × kEIc/lc
location table table
column
A23) A23)
10′ 120/110 = 14 × 143/12 = 2×(5.09Ecc×3201/ 120)
A2 110″ 5/5 = 1 5.09 0.57
(120″) 1.10 3201 = 272Ecc
Note: For flat plates, ∑Kc term is multiplied with 2 for interior storey with similar columns
above and below. For top storey, the ∑Kc term will be a single value (multiplied by 1)
Prof. Dr. Qaisar Ali 35
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Prof. Dr. Qaisar Ali 36
18

19. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03b: Equivalent column stiffness calculation
(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 = 1/272Ecc + 1/231Ecs
Because the slab and the columns have the same strength
concrete, Ecc = Ecs = Ec.
Therefore, Kec = 124.91Ec
As all columns have similar dimensions and geometric
conditions, the Kec value for all columns will be 124.91Ec
Prof. Dr. Qaisar Ali 37
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Equivalent Frame; can be analyzed using any method of analysis
Prof. Dr. Qaisar Ali 38
19

20. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: To analyze the frame in SAP, the stiffness values are multiplied by
lengths.
Ksblsb = 270×25×12=81000E
Keclec = 124.91×10×12=14989E
10′
Prof. Dr. Qaisar Ali 39
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Load on frame:
Solution: As the horizontal frame element
Step 04: SAP results (moment at center). represents slab beam, load is
computed by multiplying slab load
with width of frame
wul2 = 0.3804 × 20 = 7.608 kip/ft
Prof. Dr. Qaisar Ali 40
20

21. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at center).
Prof. Dr. Qaisar Ali 41
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at faces).
Prof. Dr. Qaisar Ali 42
21

22. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison with SAP 3D model results.
Load on model = 144 psf (LL)
Slab thickness = 10″
Columns = 14″× 14″
Prof. Dr. Qaisar Ali 43
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison of SAP 3D model with EFM.
Prof. Dr. Qaisar Ali 44
22

23. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Example: Find the equivalent 2D frame for 1st storey of the E-W interior
frame of b
f f beam supported f
t d frame structure shown b l
t t h below. Th slab i 7″
The l b is
thick with LL of 144 psf so that ultimate load on slab is 0.336 ksf. All
columns 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′
Column strip width = 20/4 = 5′
Prof. Dr. Qaisar Ali 45
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 01: 3D frame selection.
20′
Prof. Dr. Qaisar Ali 46
23

24. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 01: 3D frame extraction.
20′
10′
10′
25′
25′
10′
25′
Prof. Dr. Qaisar Ali 25′ 47
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 02: Extraction of single storey from 3D frame for separate analysis.
20′
25′
25′
10′
25′
25′
Prof. Dr. Qaisar Ali 48
24

25. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03a: Slab-beam Stiffness calculation.
Table: Slab beam stiffness (Ksb).
l1 and l2 and k
Span c1/l1 c2/l2 Isb Ksb=kEIs/l1
c1 c2 (table A 20)
A-20)
25' & 20' and
A2-B2 0.0467 0.058 4.051 25844 349E
14" 14"
The remaining spans will have the same values as the geometry is same.
Prof. Dr. Qaisar Ali 49
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of torsional member stiffness (Kt)
Table: Kt calculation.
Column
l2 c2 C = ∑ (1 – 0 63x/y)x3y/3 (i 4)
0.63x/y)x (in Kt = ∑ 9EcsC/ {l2(1 – c2/l2)3}
location
A2 20′ 14" 11208 3792.63Ecs
B2 20′ 14" 12694 4295.98Ecs
Prof. Dr. Qaisar Ali 50
25

26. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM): A
lu
Solution:
B
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
Ic (in4)
kAB (from
Column location lc lu lc / lu for 14″ × 14″
14 14 ta/tb Kc
table A23)
column
10′ 120/100 = 14 × 143/12 = 16.5/3.5 =
A2 (bottom) 100″ 7.57 201.9Ecc
(120″) 1.20 3201 4.71
10′ 120/100 = 14 × 143/12 = 3.5/16.5=
A2 (top) 100″ 5.3 141.39Ecc
(120″) 1.20 3201 0.21
∑Kc = 202Ecc + 141Ecc = 343Ecc
Prof. Dr. Qaisar Ali 51
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM): A
lu
Solution:
B
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
Ic (in4)
kAB (from
Column location lc lu lc / lu for 14″ × 14″
14 14 ta/tb Kc
table A23)
column
10′ 120/100 = 14 × 143/12 = 16.5/3.5 =
B2 (bottom) 100″ 7.57 201.9Ecc
(120″) 1.20 3201 4.71
10′ 120/100 = 14 × 143/12 = 3.5/16.5=
B2 (top) 100″ 5.3 141.39Ecc
(120″) 1.20 3201 0.21
∑Kc = 202Ecc + 141Ecc = 343Ecc
Prof. Dr. Qaisar Ali 52
26

27. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03b: Equivalent column stiffness calculation (Column A2)
(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 = 1/343Ecc + 1/3792.63Ecs
Because the slab and the columns have the same strength
concrete, Ecc = Ecs = Ec.
Therefore, Kec = 315Ec
Prof. Dr. Qaisar Ali 53
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 03b: Equivalent column stiffness calculation (Column B2)
(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 = 1/343Ecc + 1/4295.98Ecs
Because the slab and the columns have the same strength
concrete, Ecc = Ecs = Ec.
Therefore, Kec = 318Ec
Prof. Dr. Qaisar Ali 54
27

28. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Equivalent Frame; can be analyzed using any method of analysis
Prof. Dr. Qaisar Ali 55
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: To analyze the frame in SAP, the stiffness values are multiplied by
lengths. Ksblsb = 349×25×12=104700E
Keclec = 315×10×12=37800E
Keclec = 318×10×12=38160E
Prof. Dr. Qaisar Ali 56
28

29. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Load on frame:
Solution: As the horizontal frame element
Step 04: SAP results (moment at center). represents slab beam, load is
computed by multiplying slab load
with width of frame
wul2 = 0.336 × 20 = 6.72 kip/ft
Prof. Dr. Qaisar Ali 57
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at center).
Prof. Dr. Qaisar Ali 58
29

30. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at faces).
Prof. Dr. Qaisar Ali 59
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison with SAP 3D model results.
Load on model = 144 psf (LL)
Slab thickness = 7″
Columns = 14″× 14″
Beams = 14″× 20″
Prof. Dr. Qaisar Ali 60
30

31. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison of beam moments of SAP 3D model with beam
moments of EFM by SAP 2D analysis.
Prof. Dr. Qaisar Ali 61
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Distribution Method:
The original derivation of EFM assumed that moment distribution would be
the procedure used to analyze the slabs, and some of the concepts in the
method are awkward to adapt to other methods of analysis.
In lieu of computer software, moment distribution is a convenient hand
calculation method for analyzing partial frames in the Equivalent Frame
Method.
Once stiffnesses are obtained from EFM, the distribution factors are
conveniently calculated.
Prof. Dr. Qaisar Ali 62
31

32. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Kct
Ksb1
1 Kt
2 Ksb2 lc
l1 Kt
Kec
l1 3
Kcb
K = kEI/l lc
Prof. Dr. Qaisar Ali 63
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Slab Beam Distribution Factors:
Ksb1
DF (span 2-1) =
Ksb1 + Ksb2 + Kec
Ksb2
DF (span 2-3) =
Ksb1 + Ksb2 + Kec
Prof. Dr. Qaisar Ali 64
32

33. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Equivalent Column Distribution factors:
Kec
DF =
Ksb1 + Ksb2 + Kec
Prof. Dr. Qaisar Ali 65
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
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
Prof. Dr. Qaisar Ali 66
33

34. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison of SAP 3D model with EFM done by Moment
distribution 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
119.955 279.148 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
Prof. Dr. Qaisar Ali 67
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison of SAP 3D model with EFM.
Prof. Dr. Qaisar Ali 68
34

35. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 69
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM):
Solution:
Step 04: Comparison of beam moments of SAP 3D model with EFM analysis
results obtained by moment distribution method.
Prof. Dr. Qaisar Ali 70
35

36. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
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 maximum
factored moment
Prof. Dr. Qaisar Ali 71
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Prof. Dr. Qaisar Ali 72
36

37. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Critical section for factored moments (ACI 13.7.7):
( )
Interior supports
Critical section at face of rectilinear support but ≤ 0.175l1 from center of
the 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.
Prof. Dr. Qaisar Ali 73
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Prof. Dr. Qaisar Ali 74
37

38. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Moment Redistribution (ACI 13.7.7.4):
( )
Mu2
Mu1
Mo
Mu3
ln
c1/2 c1/2
l1
Prof. Dr. Qaisar Ali 75
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Factored moments in column strips and middle strips:
p p
Same as in the Direct Design Method
Prof. Dr. Qaisar Ali 76
38

39. Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
Equivalent Frame Method (EFM)
Summary of Steps required for analysis using EFM
y p q y g
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 any 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.
Prof. Dr. Qaisar Ali 77
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
The End
Prof. Dr. Qaisar Ali 78
39