CE 636 - Design of Multi-Story Structures
T. B. Quimby
UAA School of Engineering
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Behave as tall cantilever beams.
Deflection is predominately in flexure.
The wall elements have high shear stiffness,
mainly because the “beam” element has a
depth that is large compared to it's span
(level to level).
Economical for buildings up to 35 stories.
The walls can also be used to resist gravity
forces.
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Proportionate:
 Walls, at each level, maintain their relative
rigidities through out the height
 There is no redistribution of overturning moments
or shears at any level.
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Nonproportionate:
 The relative rigidities of the walls changes from
level to level.
 There is a redistribution of overturning moments
and shears at one or more levels.
All walls deflect the same when there is no twist.
Twisting takes place about the center of rigidity
Each wall on a given floor takes the Story Shear
in proportion to their relative rigidities.
 The author uses the bending stiffness of each
pier to determine the relative stiffness (see
section 9.2 of the text).
 Our text considers walls to be without openings,
which is generally true for tall buildings, but not
necessarily so for short buildings. Openings in
walls must be considered when computing
relative stiffness.
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
“Reinforced Masonry Engineering Handbook”
(by James Amrhein, published by the Masonry
Institute of America) has an excellent treatment
of shear wall analysis for low rise buildings.
 Piers are viewed as being fixed against rotation
at both ends (fixed-fixed) or at one end only
(cantilevered, fixed-pinned).
 Stiffness = 1/(deflection resulting from a unit
load)
 Amrhein's equations for deflection consider both
flexural and shear deflections.
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Author's equation only considers flexural
contribution to stiffness and considers only
cantilever beam action, OK for tall buildings
that have shear walls w/o openings.
See Amrhein's book, Table T-1i, for stiffness
calculation including shear deformation
effects and for different end fixity conditions.
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
Need to
determine the
deflection of the
wall under a unit
load, then invert
it.
Divide wall into
individual piers
and add up
deflections.
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Subdivide into piers
Find deflection of each vertical subdivision
and add together
All piers in a given vertical subdivision must
deflect the same.
 The stiffness of a vertical subdivision is the sum of
the stiffnesses of the individual piers.
 Deflection = 1/stiffness
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Create five piers
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Use Amrhein's formulas for Fixed-Fixed piers.
The stiffness is the force that will cause a unit
deflection in the pier.
Pier
A
B
C
D
E
h
10
4
4
6
10
d
60
16
10
46
4
deflect.
0.0505
0.0766
0.1264
0.0394
2.3125
stiffness
19.82
13.06
7.91
25.41
0.43
deflection A =
deflection BCDE =
0.0505
?
Determine the deflection of Piers B & C under a unit load.
Stiffness of B
13.06
Stiffness of C
7.91
Stiffness of BC
20.97
deflection BC for F(BC) = 1
0.0477
Add deflection of BC to deflection of D to get deflection BCD, then
invert to find the stiffness of BCD
deflection D for F(BC) = 1
deflection BCD for F(BC) = 1
Stiffness of BCD
0.0394
0.0870
11.49
Add deflections of A and BCDE to get total deflection of the wall under a
unit wall. Then invert the deflection to determine the stiffness of the wall.
Deflection ABCDE
Stiffness ABCDE
0.1343
7.44
It is interesting to compare this to the stiffness of the wall without the openings
h
solid wall
20
with openings
d
60
defl.
0.104
0.134
stiffness
9.643
7.444
22.80% reduction
For design, you will need to know how much shear force is on each of the piers
so that you can draw a free body diagram to determine the moments on the
piers and then design for the shear and moments.
Pier A will see 100% of the force
The full force will be divided between piers BCD and E in porportion to their stiffnesses.
BCD
E
Stiffness % Load
11.490 96.37%
0.432
3.63%
11.922
Pier D will see all of the load on BCD, or 96.37% of the total wall shear
This same load is divided between piers B and C in porportion to their stiffnesses.
B
C
Stiffness % BCD % Load
13.06 62.28% 60.02%
7.91 37.72% 36.35%
20.97 100.00% 96.37%
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Flexible diaphragms cannot transfer
moment.
Flexible diaphragms are considered to act as
simply supported beams between shear wall
support.
Normally wood or steel decking w/o
concrete.
Loads are distributed to walls according to
tributary loads.
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Story shear and Story Moment is distributed
to each wall according to relative stiffnesses.
Determine location of center of rigidity and
center of force to determine any eccentricity.
Story moment equals the story shear times
the distance between the centers of force
and rigidity.
Determine the Center of Stiffness
Wall
1
2
3
A
B
Totals
h
10
10
10
10
10
Center of Rigidity
Center of Force
Eccentricities
d
30
20
30
40
60
Kx
9.64
6.15
9.64
25.44
x
27.58
30.00
2.42
y
11.92
15.00
3.08
Ky
13.06
19.82
32.88
x
0
20
60
y
Kx*x
0.00
123.08
578.57
30
0
701.65
Ky*y
391.84
0.00
391.84
Let Qx =
M = Qx*y =
1.00
-3.08
Distribution of just the story shear, Q
Wall(j)
1
2
3
A
B
Totals
Ky
13.06
19.82
32.88
Q(j)
0.000
0.000
0.000
-0.397
-0.603
-1.000
Distribution of just the moment, Qx*y
Wall(j)
1
2
3
A
B
Totals
Wall(j)
1
2
3
A
B
Totals
Kx
Ky
9.64
6.15
9.64
x
0
20
60
13.06122
19.81651
y
30
0
c
27.58
7.58
-32.42
18.08
-11.92
25.44
Kc
265.96
46.65
-312.61
236.17
-236.17
Kcc
Kc/s(Kcc)
7335.43 0.0107
353.67
0.0019
10134.57 -0.0126
4270.48 0.0095
2814.71 -0.0095
24908.86
Qe(j)
-0.0329
-0.0058
0.0387
-0.0000
Qe(j)
-0.0292
0.0292
0.0000
Check
-0.9076
-0.0438
-1.2540
-0.5284
-0.3483
-3.0820
Wall(j)
1
2
3
A
B
Totals
Q(j)
0.000
0.000
0.000
-0.397
-0.603
-1.000
Qe(j)
-0.033
-0.006
0.039
-0.029
0.029
-0.000
Total
-0.033
-0.006
0.039
-0.426
-0.574
-1.000
Let Qy =
M = Qy*x =
1.00
2.42
Distribution of just the story shear, Q
Wall(j)
1
2
3
A
B
Totals
Kx
9.64
6.15
9.64
25.44
Q(j)
-0.379
-0.242
-0.379
0.000
0.000
-1.000
Distribution of just the moment, Qy*x
Use same distribution as for Qx*y
Wall(j)
1
2
3
A
B
Totals
Kc/s(Kcc)
0.0107
0.0019
-0.0126
0.0095
-0.0095
Qe(j)
0.0258
0.0045
-0.0304
0.0000
Qe(j)
0.0229
-0.0229
-0.0000
Check
0.7124
0.0343
0.9842
0.4147
0.2733
2.4190
Wall(j)
1
2
3
A
B
Totals
Q(j)
-0.379
-0.242
-0.379
0.000
0.000
-1.000
Qe(j)
0.026
0.005
-0.030
0.023
-0.023
0.000
Total
-0.353
-0.237
-0.409
0.023
-0.023
-1.000
Wall(j)
1
2
3
A
B
Totals
Qx
-0.033
-0.006
0.039
-0.426
-0.574
-1.000
Qy
Control
-0.353
-0.353
-0.237
-0.237
-0.409
-0.409
0.023
-0.426
-0.023
-0.574
-1.000
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Best solved by appropriate computer
modeling.
The author presents a method similar to
moment distribution to deal with
nonproportionate nontwisting structures. We
will not cover it.
If twisting is a factor, then the problem is too
complex for hand solutions.
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There is redistribution of moments and
shears at levels where changes occur.
Levels adjacent to change levels also
participate in the redistribution.
Take some time to model a disproportionate
system or two and explore the effects on
shear and moment distributions.