gam distributionofloads 2 horizontal torsion
TRANSCRIPT
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Distribution of Forces
in Lateral Load Resisting Systems
Many slides from 2009 Myanmar Slides of Profs Jain and Rai 1
Part 2. Horizontal Distributionand Torsion
IITGN Short CourseGregory MacRae
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Reinforced Concrete Cast-in-Situ Slabs
• Moment of inertial for bending in its own plane
( Very large quantity!!)
• Practically, floor is infinitely stiff for bending
deformation in its own plane.
12
3tbI
• The slab is subject to horizontal load.
b
t
2Sudhir K Jain
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Floor Diaphragm Action
Plan of a one-storey building
with shear walls
t = floor thickness; width of the beam representing floor diaphragm
b = floor width; depth of the beam representing floor diaphragm
L = span of the beam representing floor diaphragm
Springs represent lateral
stiffness walls / frames
k/2k k
b
L L
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Vertical load analogy for floor diaphragm action
Lateral earthquake force, EL
Beam representing
floor diaphragm
Ibeam = tb3/12
K K/2 K
4
Floor Diaphragm Action
Sudhir K Jain
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In Plane Force Out of Plane Force
In Plane Deformation of
Floor
Out of Plane Deformation
of Floor5
In-plane versus out-of-plane deformation of floor
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Floor Deformations
In-Plane Floor Deformation Out of Plane Floor Deformation
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Rigid-body movements of a rigid floor diaphragm
Rotation about z-axis
Longitudinal
Translation
Translation in x-direction Translation in y-direction
Transverse
Translation
Angle of
rotation
Resultant Translation
Combination of translations and rotation
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Example 1: Effect of floor diaphragm action
Slab thickness = 150 mm
E = 25,500 N/mm2
k = 2300 103 N/mm
4123
104.612
8000150mmI diaphragm
• Actual Analysis 440 120 440
• Rigid Diaphragm 455 90 455
• Tributary Area 250 500 250
Force in Springs
k k0.2k
20 m 20 m
8m
1000 kNI
E I
1000 kN
k 0.2k k
8Sudhir K Jain
Rigid Diaphragm
Assumption is generally
used for RC floors
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Example 2: Centre of Mass
• Given floor plan and lumped masses per unit area
• Locate centre of mass of the floor
Force in Springs
700 kg/m2
1000 kg/m2
8m
5 m
30 m
10 m
Force in Springs
A
X
CM (A)
B
CM (B)
C
CM
CM (C)
z
y
y
9
F
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Example 2: Centre of Mass
• Locate centre of mass of segments A, B, C as:
– CM(A) = (4.0, 7.5); CM (B) = (4.0, 2.5);
– CM(C) = (19.0, 5.0)
• Calculate centre of mass of floor asyx,
mm
ymy
mm
xmx
i
ii
i
ii
1.5
4.14
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Centre of Stiffness (of a Single-storey Building)
• Point on the floor through which a lateral load should pass in
order to have only rigid body translation (i.e., no rigid body
rotation).
• Use the above definition to locate the centre of stiffness.
• Example:
k1.5k0.5k
30 m
10m
14 m
1.5kA
B
321
1.2k
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k ∆1.5k ∆0.5k ∆
F
∆
X
Force equilibrium:
Moment equilibrium:
kkkkF 35.15.0
mk
kx
kkkkxF
3.173
52
52305.1145.00.
Example 3. Centre of Stiffness
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Force equilibrium:
Moment equilibrium:
kkkF 7.22.15.1
mk
ky
kkyF
6.57.2
15
105.102.1.
∆
F 1.5k ∆
y 1.2k ∆
Example 3. Centre of Stiffness
13Sudhir K Jain
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ELkkk
kFEL
kkk
kFEL
kkk
kF
321
33
321
22
321
11 ;;
Building Plan
k 2=0.3kWall stiffness
k 1=kk 3=k
EL
Wall stiffness
Definition of lateral stiffness
Fk
F
ELFFF
kFkFkF
321
332211 ,,
Lateral load distribution due to rigid floor diaphragm
(Symmetrical case – no torsion)
F3F2F1
EL
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k 1.2k0.5k
30 m
10m
9m
4
k
y
23
5
1
x
kNkkk
kF
kNkkk
kFkN
kkk
kF
9.882.15.0
2002.1
0.372.15.0
2005.0;1.74
2.15.0
200
3
21
Example 4: 200 kN applied along y-direction
• Locate centre of stiffness : (15m, 5m)
• Locate centre of mass : (15m, 5m)
• Hence, no torsion
• Wall 1, 2, 3 share load proportional to stiffness
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k The centre of stiffness
(CS) is at the centre
of the building. If the
centre of mass is also
here then the building
undergoes translation
but no torsion
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Load at centre of mass = Load at centre of stiffness + Twisting
moment about the centre of stiffness
• CM
• CS
CM
ELe.EL
EL
ELCS
M = e.EL
ELCS
CM
Eccentric Systems
e
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r3
r2
r1
r5r1 r3
r5
r2
1
2
3
5
4
r4
ki = Lateral stiffness of the ith element
ri = Perpendicular distance of the ith element from centre of stiffness
= Rotation of the floor diaphragm in its own plane
Analysis of force induced by twisting moment (rigid floor diaphragm)
17Sudhir K Jain
CS
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• Displacement of ith element, in its own plane,
due to rotation about centre stiffness
• Resisting force in ith element
• Restoring moment by force in ith element
• By moment equilibrium
• Force in the ith element
ii r
iii rkF
2
iiiii rkrFM
2
iit rkM
t
ii
iii M
rk
rkF
2
Analysis of force induced by twisting moment (rigid floor diaphragm)
18Sudhir K Jain
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Example 5: Load distribution in eccentric system
k 1.5k0.5k
30 m
10m
14m
4
1.2k
y
2 3
5
1
x
1.5k
(a) (b)
17.33m
15m
5m5.56m
ex=2.33m
ey=0.56m
200kN 200kN
1 34
200kN
•CS•
CM5
F3
21 34
•
CM5
F1F2 F3
2
F4
F5
•CS466kNm
466kNm = 200kNx 2.33m
•CS
(c) Forces
(d)
F1F2
=• CM •CM
•
CM
•CS
Sudhir K JainTranslational Forces Torsional Forces
Walls CS and CM
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Analysis for 200 kN force acting at centre of stiffness
This force is resisted by walls 1, 2, and 3 in proportion to their lateral stiffness. This gives:
Analysis for 466 kN-m moment acting on the diaphragm at CS:
The twisting moment of 466 kN-m is resisted by all the walls (including walls 4 and 5).
kNkkk
kF 7.66
5.15.0
2001
kNkkk
kF 3.33
5.15.0
2005.02
kNkkk
kF 100
5.15.0
2005.13
Example 5: Load distribution in eccentric system
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1 k 17.33 17.33k 300.3k 13.2
2 0.5k 3.33 1.67k 5.5k 1.3
3 1.5k 12.67 19.00k 240.8k 14.4
4 1.5k 4.44 6.66k 29.6k 5.1
5 1.2k 5.56 6.67k 37.1k 5.1
t
ii
ii Mrk
rkF
21
k3.613
ikWallir iirk 2
iirk
Example 5: Load distribution in eccentric system
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The total force resisted by the walls (= translational force + torsional force):F1 = 66.7+13.2 = 79.9 kN
F2 = 33.3+1.3 = 34.6 kN
F3 = 100 – 14.4 = 85.6 kN
F4 = 5.1 kN = 5.1 kN
F5 = 5.1 kN = 5.1 KN
Example 5: Load distribution in eccentric system
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1 34
•
CM5
F1F2 F3
2
F4
F5
•CS
Total Forces
Translational Forces Torsional Forces
1 34
200kN
•CS•
CM5
F3
21 34
•
CM5
F1F2 F3
2
F4
F5
•CS466kNm
+F1F2
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(a) Without Torsion
All frames must follow through the same displacements at each level.
(b) With Torsion
All frame displacements at each level must be compatible with level translational and torsionaldisplacements.
Multistorey Frames
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F4
None of the floors undergoany rotation as forces passthrough the CS (i.e. CS =CM)
(CS)i
F3
(CS)i
F2
(CS)iF1
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Multistorey Frames
Important:
• First calculate lateral load at different floors for the entire building
• Then distribute to different frames/walls as per floor diaphragm behavior
Do Not
• Calculate seismic design force directly for individual frames of the building
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Multistorey Frames
Plan of a building with space frame: this may be thought of as four 2-bay frames in the y-direction, and three 3-bay frames in the x-direction
A B C D1
2
3
Planx
y
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A
B
C
1 2 3
Plan
• Frames 1&3 same
• Frame spacing same
The requirement is:(a) Displacement in frames 1, 2 & 3 are equal at floor 1.
(b) Displacement in frames 1, 2 & 3 are equal at floor 2.
(c) Displacement in frames 1, 2 & 3 are equal at floor 3.
Design force in y-direction
on the entire building
1000
400
100
• Missing column
• Symmetric system
26Sudhir K Jain
Example 6: 3-storey symmetric building
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36
52
91
9
8
710
11
12 15
14
13
18 21
2017
1916
29
23
28
Impose the conditions
This will ensure proper load distribution.
1613107
1714118
1815129
1000
400
100
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Frame A Frame B Frame C
Example 6: 3-storey symmetric building
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Simple calculate the member forces
100
400
1000
Imaginary rigid
links
to ensure floor diaphragm action
Think of the translational problem as:
28Sudhir K Jain
Example 6: 3-storey symmetric building
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2-D Frame with Rigid Lines
Direction of
Earthquake force
A B C D E
1
2
3
4
5(a)
Frame 1 Frame 2 Frame 3 Frame 4
Link bars
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Approximate Lateral Load Distribution
• Exact distribution requires computer analysis
• How do we carry out approximate hand calculations for
buildings up to 4 stories without torsion?
– Assume that all 2-D frames have same displacement profile
(shape only) for lateral loads
– Now match roof displacement only
– If assumption is exactly valid, analysis will still be exact
30Sudhir K Jain
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Example 7. Approximate Distribution, No Torsion
A B C
PlanDesign force in y-direction
on the entire building
1000
400
100
301000
400
100
A
151000
400
100
B
301000
400
100
C
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unitsk
unitsk
unitsk
C
B
A
5030
1500
10015
1500
5030
1500
unitski 2005010050
fffk
kf
fffk
kf
fffk
kf
C
C
BB
AA
25.0200
50
50.0200
100
25.0200
50 500
200
50
Frame B
250
100
25
Building Frames A and C
1000
400
100
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Example 7. Approximate Distribution, No Torsion
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Evaluate , , such that roof displacement is same
+ + = 1.0
A B C
Plan Entire Building
1000
400
100
. 1000
. 400
. 100
. 1000
. 400
. 100
. 1000
. 400
. 100
A B C
EQ
(No Torsion Case)
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Example 7. Approximate Distribution, No Torsion
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• Illustration
– Parts of building in double height
– Symmetric1000
700
400
50
Portion in double
height Plan Elevation
1 2 3 4
1000
700
400
50
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Example 8. Approximate Distribution, No Torsion
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Further Simplification
• For load distribution, relative lateral stiffness is
needed
1 2 31 1 1
1
1
1k
2
2
1k
3
3
1k
Relative terms only are required
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Approximate Lateral Load Stiffness of Frames
• Number of approximate methods, e.g.,
– McLeod’s Method
– Computer methods for analysing frames
• Caution
– Do not believe in storey stiffness as
– This assumes beams are infinitely rigid!
– Never happens
!!L
EI3
12
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Torsion in Multistory Buildings
• Centre of stiffness at different floors
• Number of definitions
• Depends on usage
• Implementation
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None of the floors undergoany rotation
(CS)i
F3
(CS)i
F2
(CS)iF1
F3
F2
F1
Design lateral load profile
Centre of Stiffness for multistory buildings
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Torsion in Multistory Buildings . . .
The requirement on design eccentricity can be fulfilled by applying
earthquake force away from centre of mass by a distance 0.5 times the
calculated eccentricity, such that eccentricity between centre of stiffness and
the load becomes 1.5 times the calculated.
CM = centre of mass
CS = centre of stiffness
1.5ex
ELy
.CM
.CS1.5ey
ELx.CM
.CS
ey.CM
.CS
ex
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Torsion in Multistory Buildings
jsj
jsjdj
be
bee
Q1
Q2
Q3
Q4
40
• Typical building code specifies design eccentricity in terms of
• Static eccentricity esj
• Accidental eccentricity bj
is typically 1.5
is typically 0.05 to 1.0
(5% to 10% of plan dimension bj)
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jb
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Torsion in Multistory Buildings . . .
41
Goel and Chopra
(ASCE, Vol.119; No:10)
Sudhir K Jain
CM CS
Can Conduct Analyses Directly Using Computer
Program with Rigid Diaphragm (e.g. ETABS)
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• For buildings generally uniform with height
• Centre of stiffness for different floors on the same vertical line
• Treatment similar to that for single storey building
• Example: Earthquake force in X-direction
6m
3m
4m
4.5m 4.5m 4.5m 4.5m
333
287
141
48
All columns 400x400
Exterior Beams 250x600
Interior Beams 300x450
Approximation in Torsion Calculations
42Sudhir K Jain
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Earthquake force in X-direction
6m
3m
4m
4.5m 4.5m 4.5m 4.5m
333
287
141
48
All columns 400x400
Exterior Beams 250x600
Interior Beams 300x450
43Sudhir K Jain
Example 9. Approx. Analysis Torsion
A
B
C
D
1 2 3 4 5 1 2 3 4 5
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Eccentricity e = 6.5 – 6.1 = 0.4m
Design eccentricity = 1.5e = 0.6m (Dynamic eccentricity)
Design force profile V acting at CM
= Force profile V at CS
+ Twisting moment profile (Mt = 0.6m x V)
m.
,
,,,y
m/kN,kk
m/kN,kk
Pk
CB
DA
16
22053
1322016739010439010
39010
22016
P
44Sudhir K Jain
Example 9. Approx. Analysis Torsion
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• Force profile V at CS
– Frames A, D =
– Frames B, C =
• Twisting moment profile Mt
V.V,
,
V.V,
,
195022053
39010
305022053
22016
t
jj
ii Mrk
rk2
45Sudhir K Jain
Example 9. Approx. Analysis Torsion
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Frame
(x 103)
(kN/m)
ri
(m)(x 103)
(kN)(x 103)
Fd
(torsion)
Fd
(Direct)
FTO
(Total)
A 16.22 6.9 111.92 772.23 0.0293 0.0170V 0.304V 0.321V
B 10.39 0.9 9.35 8.42 0.0024 0.0014V 0.195V 0.196V
C 10.39 -2.1 -21.82 45.82 -0.0057 -0.0034V 0.195V 0.195V
D 16.22 -6.1 -98.94 603.55 -0.0259 -0.0156V 0.304V 0.288V
1 12.70 -9.0 -114.30 1028.70 -0.0299 -0.0180V . . . . . . . .
2 8.18 -4.5 -36.80 165.65 -0.0096 -0.0057V . . . . . . . .
4 8.18 4.5 36.80 165.65 0.0096 0.0057V . . . . . . . .
5 12.70 9.0 114.30 1028.70 0.0299 0.0018V . . . . . . . .
2
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46Sudhir K Jain
Example 9. Approx. Analysis Torsion
From frame
analysis with
point load at
roof
Dist from
CS
![Page 47: GAM DistributionOfLoads 2 Horizontal Torsion](https://reader034.vdocuments.net/reader034/viewer/2022052302/577cc9e61a28aba711a4e7c8/html5/thumbnails/47.jpg)
Level
Total Level
Design Force
(kN)
Design Force For Frame
A
(kN)
B
(kN)
C
(kN)
D
(kN)
4 333 106.90 65.27 64.94 101.20
3 287 92.13 56.25 55.97 87.25
2 141 45.26 27.64 27.50 42.86
1 48 15.40 9.41 9.36 14.60
47Sudhir K Jain
Example 9. Approx. Analysis Torsion
From beginning
of example
![Page 48: GAM DistributionOfLoads 2 Horizontal Torsion](https://reader034.vdocuments.net/reader034/viewer/2022052302/577cc9e61a28aba711a4e7c8/html5/thumbnails/48.jpg)
Thank you!!
48