lecture 12 equivalent frame method
TRANSCRIPT
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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]
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Topics AddressedIntroduction
Stiffness of Slab-Beam Member
Stiffness of Equivalent Column
Stiffness of Column
Stiffness of Torsional Member
Prof. Dr. Qaisar Ali 2
Examples
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Topics AddressedMoment Distribution Method
Arrangement of Live Loads
Critical Sections for Factored Moments
Moment Redistribution
Factored Moments in Column and Middle Strips
Prof. Dr. Qaisar Ali 3
Factored Moments in Column and Middle Strips
Summary
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
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).
Prof. Dr. Qaisar Ali 4
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
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.
Prof. Dr. Qaisar Ali 5
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Introduction
Interior 3D frame detached from 3D structure.
Prof. Dr. Qaisar Ali 6
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness oflaterally 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 SlabEquivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness oflaterally present members (slabs and beams).
Prof. Dr. Qaisar Ali 8
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Introduction
This 3D frame is converted to a 2D frame by taking effect of stiffness oflaterally 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 SlabEquivalent Frame Method (EFM)
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
Prof. Dr. Qaisar Ali 10
Ksb Ksb Ksb Ksb Ksb
Ksb Ksb Ksb Ksb Ksb
Kec
Kec
Kec
Kec
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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
Kec Kec Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 11
Ksb Ksb Ksb Ksb Ksb
Ksb Ksb Ksb Ksb Ksb
Kec
Kec
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Kec
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Kec
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
IntroductionNext the procedures for determination of Ksb and Kec are presented.
Ksb Ksb Ksb Ksb Ksb
Ksb Ksb Ksb Ksb Ksb
Kec Kec Kec Kec Kec Kec
Prof. Dr. Qaisar Ali 12
Ksb Ksb Ksb Ksb Ksb
Ksb Ksb Ksb Ksb Ksb
Kec
Kec
Kec
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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.
Prof. Dr. Qaisar Ali 13
l2
l1
c2
c1
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Slab Beam member (Ksb):( sb)
Determination of k
Prof. Dr. Qaisar Ali 14
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Slab Beam member (Ksb):( sb)
Isb determination
Prof. Dr. Qaisar Ali 15
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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
Prof. Dr. Qaisar Ali 16
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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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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
Prof. Dr. Qaisar Ali 17
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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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.
Prof. Dr. Qaisar Ali 18
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Column (Kc):( c)
Prof. Dr. Qaisar Ali 19
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Column (Kc):( c)
Determination of k
Prof. Dr. Qaisar Ali 20
11
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Column (Kc):( c)
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
Prof. Dr. Qaisar Ali 21
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.
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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).
Prof. Dr. Qaisar Ali 22
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Torsional Member (Kt):( t)
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 SlabEquivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):( t)
Condition (b) – Transverse beams framing into columns
Prof. Dr. Qaisar Ali 24
13
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):( t)
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
Prof. Dr. Qaisar Ali 25
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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Stiffness of Torsional member (Kt):( t)
Cross sectional constant, C:
Prof. Dr. Qaisar Ali 26
14
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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
Prof. Dr. Qaisar Ali 27
Kec
Kec
Kec
Kec
Kec
Kec
Kec
Kec
Ksb Ksb Ksb
Ksb Ksb Ksb
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
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′
Prof. Dr. Qaisar Ali 28
2
Column strip width = 20/4 = 5′
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 SlabEquivalent Frame Method (EFM):
Solution:
Step 01: 3D frame extraction.
20′
Prof. Dr. Qaisar Ali 30
25′25′
25′
10′
10′
10′
25′
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 02: Extraction of single storey from 3D frame for separate analysis.
20′
Prof. Dr. Qaisar Ali 31
10′
25′25′
25′25′
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 32
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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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Prof. Dr. Qaisar Ali 33
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 l C ∑ (1 0 63x/y)x3y/3 (i 4) K ∑ 9E C/ {l (1 c /l )3}
Prof. Dr. Qaisar Ali 34
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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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution: lu
A
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
Column Ic (in4) kAB(from
CAB(from
B
Prof. Dr. Qaisar Ali 35
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)
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Prof. Dr. Qaisar Ali 36
19
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
Prof. Dr. Qaisar Ali 37
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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 04: Equivalent Frame; can be analyzed using any method of analysis
Prof. Dr. Qaisar Ali 38
20
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 39
10′
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 40
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 SlabEquivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at faces).
Prof. Dr. Qaisar Ali 42
22
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 SlabEquivalent Frame Method (EFM):
Solution:
Step 04: Comparison of SAP 3D model with EFM.
Prof. Dr. Qaisar Ali 44
23
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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′
Prof. Dr. Qaisar Ali 45
2
Column strip width = 20/4 = 5′
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 01: 3D frame selection.
20′
Prof. Dr. Qaisar Ali 46
24
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 01: 3D frame extraction.
20′
Prof. Dr. Qaisar Ali 47
25′25′
25′
10′
10′
10′
25′
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 02: Extraction of single storey from 3D frame for separate analysis.
20′
Prof. Dr. Qaisar Ali 48
10′
25′25′
25′25′
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 49
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.
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 l C ∑ (1 0 63x/y)x3y/3 (i 4) K ∑ 9E C/ {l (1 c /l )3}
Prof. Dr. Qaisar Ali 50
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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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution: lu
A
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
Column location l l l / lIc (in4)
for 14″ × 14″ t /tbkAB (from K
B
Prof. Dr. Qaisar Ali 51
∑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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution: lu
A
Step 03b: Equivalent column stiffness calculation
(1/Kec = 1/∑Kc +1/Kt)
Calculation of column stiffness (Kc)
Table: ∑Kc calculation.
Column location l l l / lIc (in4)
for 14″ × 14″ t /tbkAB (from K
B
Prof. Dr. Qaisar Ali 52
∑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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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
Prof. Dr. Qaisar Ali 53
Because the slab and the columns have the same strengthconcrete, Ecc = Ecs = Ec.
Therefore, Kec = 315Ec
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
Prof. Dr. Qaisar Ali 54
Because the slab and the columns have the same strengthconcrete, Ecc = Ecs = Ec.
Therefore, Kec = 318Ec
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 56
29
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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 = 0.336 × 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 SlabEquivalent Frame Method (EFM):
Solution:
Step 04: SAP results (moment at center).
Prof. Dr. Qaisar Ali 58
30
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 SlabEquivalent 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
31
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 04: Comparison of beam moments of SAP 3D model with beammoments 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 SlabEquivalent Frame Method (EFM)
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.
Prof. Dr. Qaisar Ali 62
Once stiffnesses are obtained from EFM, the distribution factors areconveniently calculated.
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Kt
Kt
K
Ksb1
Ksb2l1lc
1
2
Kct
Prof. Dr. Qaisar Ali 63
Kec
K = kEI/l
l1
lc
3Kcb
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Slab Beam Distribution Factors:
DF (span 2-1) =Ksb1
Ksb1 + Ksb2 + Kec
Prof. Dr. Qaisar Ali 64
sb1 sb2 ec
DF (span 2-3) =Ksb2
Ksb1 + Ksb2 + Kec
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Moment Distribution Method:
Distribution Factors:
Equivalent Column Distribution factors:
DF =Kec
K + K + K
Prof. Dr. Qaisar Ali 65
Ksb1 + Ksb2 + Kec
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
Prof. Dr. Qaisar Ali 66
done.
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
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
Prof. Dr. Qaisar Ali 67
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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 04: Comparison of SAP 3D model with EFM.
Prof. Dr. Qaisar Ali 68
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
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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
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM):
Solution:
Step 04: Comparison of beam moments of SAP 3D model with EFM analysisresults obtained by moment distribution method.
Prof. Dr. Qaisar Ali 70
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 maximumfactored moment
Prof. Dr. Qaisar Ali 71
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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 ofthe support
Exterior supports
At exterior supports with brackets or capitals, the critical section < ½ the
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pp p ,projection of bracket or capital beyond face of supporting element.
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way Slab
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
Moment Redistribution (ACI 13.7.7.4):( )
Mu1Mu2
Mo
Prof. Dr. Qaisar Ali 75
Mu3
l1
ln c1/2c1/2
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent 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
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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
Two Way SlabEquivalent Frame Method (EFM)
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
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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.
Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar
The End
Prof. Dr. Qaisar Ali 78