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RJAV vol VII issue 1/2010 3 ISSN 1584-7284
Evaluating the Maximum Allowable Drift in a Shear Wall
with Variable Stiffness
Michel Farid Khouri, Ph.D.*
Lebanese University, Faculty of Engineering, Lebanon, mkhuri@ul.edu.lb
(Received 28 September 2010; accepted in revised form 21 January 2010)
Abstract: - In a reinforced concrete building, it is a common knowledge that the estimation of the drift
used by seismic codes as well as investigators in the field of wind analysis and earthquake engineering is
based on experience and logic. None of the drift formulas used by seismic codes take into consideration
the actual structural stiffness and concrete/steel properties. Maximum allowable drift ranges from h/50 in
some codes to h/2000 in others, where h is the height of a building. One of the main attempts to quantify
maximum allowable drift was done by the author who suggested a formula that established grounds to
start from in the estimation of the drift; it uses a constant lateral stiffness from the bottom to the top of a
building. Shear walls however are usually designed with variable stiffness where the stiffer sections are
at the bottom and the less stiff ones are as we go up the building. This paper analyses the building as a
shear building, and uses a variable lateral stiffness between the stories for the lateral load resisting
elements, and makes use of the finite element method along with structural dynamics and reinforced
concrete design to generate a formula that can be used by a designer to estimate the allowable drift for a
variable stiffness shear wall within elastic limits taking into consideration the effect of a cracked section
suggested by UBC and ACI. In comparing results with other seismic codes, the suggested formula tends
to be relatively conservative and close to the French code (PS92) and the Lebanese code.
Keywords: seismic codes, Finite Element Method, shear building, variable stiffness
1. INTRODUCTION
The maximum allowable drift suggested by
various investigators give a value that ranges from
h/50 to h/2000 [1-6]. This study makes use of the
finite element method [7,8,9] along with structural
dynamics [10] to evaluate the displacement at the
top of a shear wall with variable stiffness and relates
it to the maximum strain in the shear wall assuming
that the maximum occurs in a balanced section at the
point when tension reinforcing steel reaches the
maximum allowable strain and when compression
concrete crushes. This study then suggests an
equation that can be followed in determining the
maximum allowable drift; it is divided into five
parts:
1- In the first part, the stiffness matrix of a shear
wall is determined by making the assumption
that a shear wall is a vertical beam in flexure
with variable stiffness throughout its height.
2- In the second part, expressions for
displacement and relative displacement at any
level are determined.
3- In the third part, a general formula for the
strain at the jth story along the shear wall is
computed and maximum strain values are
found.
4- In the forth part, a combination of the results
of the previous parts is used to obtain a
relation between maximum displacement at
the top of the shear wall and maximum strains
at the bottom of every level of the shear wall.
Consequently, an expression for the allowable
drift at the top of the building is determined.
5- In the last part, an example is done using the
suggested formula and a comparison between
this formula and various selective seismic
codes is done.
2. ASSEMBLY OF MATRICES (Shear
Building)
As presented by Khouri [1], and after applying
boundary conditions, the stiffness matrix of an
element on the shear wall is:
[K] = ⎥
⎦
⎤
⎢
⎣
⎡
−
−
1212
1212
L
EI
3
1
From Figure 1, and taking into consideration the
boundary conditions presented in Khouri [1], matrix
∗ Prof. Michel F. Khouri received his Ph.D. from The University
of Michigan, Ann Arbor, in 1989. Since 1993, he is a professor
of Civil Engineering at the Lebanese University, Faculty of
Engineering, Branch II. He served as the Chairman of the
department between 1997 and 2002 and between 2004 and 2008.
He has written many articles in the area of Finite Element
Analysis, Structural Dynamics and Shell Analysis.
RJAV vol VII issue 1/2010 4 ISSN 1584-7284
of eqn(1) below can be assembled relative to degrees
of freedom 4, 7,…3j+1…3N+1. Also variable
stiffness is assumed in all stories where
(N)(2)(1) k ...kk ≠≠
Figure 1. Representation of an N-story building.
The stiffness matrix in this general case can be
written as:
The Stiffness Matrix = [A]; (1)
it can be represented in terms of the lateral stiffness
of each story in the following manner:
N
k
N
k000000000
N
k
N
k
1N
k
1N
k00000000
0
1N
k0000000
00
2i
k000000
000
2i
k
2i
k
1i
k
1i
k00000
0000
1i
k
1i
k
i
k
i
k0000
00000
i
k00
0
000000
4
k00
0000000
4
k
4
k
3
k
3
k0
00000000
3
k
3
k
2
k
2
k
000000000
2
k
2
k
1
k
A
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
+
−−
−
−
−
+
−
+
−
+
+
++
−
+
−
+
+−
−
−
−+−
−+−
−+
=
OO
OO
OO
OO
Where, k(i) is the stiffness of one story,
()
∑
=
=
n
1j
3
j
jj
i
L
IE
12k .
3. DISPLACEMENTS and RELATIVE
DISPLACEMENTS
The stiffness matrix for a constant stiffness shear
wall in the case where k(1) = k(2) = … = k(i) = ... = k(n)
= k , is as demonstrated in a reference [1]:
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−−− −
−−− −− −
−−− −− −
=
1100000
121000000
012000
010000
121
000121000
12
00010
00001210
0
00000121
0000012
kB
LLLL
LL
OOMOM
MOOLM
MMOOMOMM
LL
MMOMOOOMM
MLLOOM
OMO
LL
LLLL
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
++
+++ +++
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
−
N2i1ii321
2i2i1ii321
1i1i1ii321
iiii321
3333321
2222221
1111111
k
1
1
B
LLLL
MOMM
MOMMM
MOMMMOMMM
LLLL
LLLL
LLLL
MOMMMOMMM
MOMMMOMMM
LLLL
LLLL
LLLL
At this time matrix C need to be found such as: A
= C x B => C = A x B-1, =>
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
+
−
++
+
−
+
−
−−−
−−−−
−−−−−
=
N
k000000
N
k
1N
k
1N
k00000
0000
0000
2i
k
1i
k
1i
k0000
1i
k
i
k
1i
k
i
k
i
k
i
k
1-i
k
i
k
1-i
k
i
k
1-i
k000
000
3
k00
3
k
2
k
3
k
2
k
3
k
2
k
3
k
2
k
2
k0
2
k
1
k
2
k
1
k
2
k
1
k
2
k
1
k
2
k
1
k
1
k
k
1
C
LLLL
OLLL
MLOOMLL
MLOOMLL
LLLLL
LLLOOMM
M
LLLOO
MLLLMMOO
MLLLMMOO
LLLLL
LLLLL
The shear force Fi at any level i is computed as: F
= A x q’, and A = C x B => F = C x (B x q’).
RJAV vol VII issue 1/2010 5 ISSN 1584-7284
Therefore, once the values of the displacements q’
are known, the shear forces can be calculated.
Now, find matrix D as follows: A = B x D => D
= B-1 x A, and D = C T.
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−+
−
++
−−
−
−−
−
++
−−
−
−−
−
−
−−
−−
−
=
NNNiiii
N
iii
i
kk
1
k
2
k
1
k
1
k
i
k
i
k
1
k
3
k
2
k
2
k
1
k
0
1
k
0
000
1
k
1
k
i
k
i
k
1
k
3
k
2
k
2
k
1
k
00000
i
k
i
k
1
k
3
k
2
k
2
k
1
k
0000000
0000
0000
3
k
3
k
2
k
2
k
1
k
000000
2
k
2
k
1
k
000000
1
k
k
1
D
LLLL
OLMMMMMMM
MOOMMMMMMM
MMOO
MMMMMMM
OLL
OL
OOMM
MMMOOMM
MMMO
LLL
LLLL
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
∏
⎟
⎠
⎞
⎜
⎝
⎛−≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
∏
⎟
⎠
⎞
⎜
⎝
⎛−≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛−≠
=
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
∏
=
=
=
−
Ni
Ni
1i
i
k
1jj;i
1i
i
k
1-j
k
j
k
Ni
2;3i
1i
i
k
2
k
3
k
Ni
1;2i
1i
i
k
1
k
2
k
0
0
00
Ni
ji
1i
i
k
1jj;i
1i
i
k
1
-j
k
j
k
Ni
2;3i
1i
i
k
2
k
3
k
Ni
1;2i
1i
i
k
1
k
2
k
0000
Ni
1ji
1i
i
k
000000
Ni
3i
1i
i
k
Ni
2;3i
1i
i
k
2
k
3
k
Ni
1;2i
1i
i
k
1
k
2
k
00000
Ni
2i
1i
i
k
Ni
1;2i
1i
i
k
1
k
2
k
000000
Ni
1i
1i
i
k
Ni
1i
i
k
k
1
D
LLLLLLL
OOOOMOOOMM
MOOOOMOOOMM
MOOOMOOOMM
OOLLL
OOOOMM
OOOMM
MMOOMMOOOMM
MMOOMMOO
L
LLL
LLLL
Where
()
∏
=
=
Ni
1i
i
k = K(1). K(2)…… K(N).
F = A x q’ => q’ = A-1 x F, but A-1 = D-1 x B-1 =>
q’ = D-1 x [B-1 x F].
The matrix [B-1 F] = q represents the
displacement in the particular case where the
stiffness of all stories are the same as computed and
presented in reference [1].
(
)
()
() ()
()
()
()
()
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
Δ=
≡
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−−
−−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−
=−
+
+
+−
N
i
1i
2
1
1N3
1i3
11i3
7
4
2
2
2
2
2
q
q
q
q
q
q
q
q
q
q
X6
1N
1N
X6
1i
1i
X6
11i
11i
X6
12
12
X6
11
11
k
V
q
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
with X = 2
1)N(N
i
N
1i
+
=
∑
=
,
The displacement is calculated as:
(
)
qDq' 1×= −
and can be presented as follows:
()
()
() ()
[]
()
()
()
()
()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⋅−⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−= ∏
∑∏
∏
=
≠
=
−=
=
=
+≠
=
+
=
Ni
ji
1i
i
2
1ji
1i
Nl
1ii;l
1l
li1i
2
N
1i
i
'
jk
6X
1j
1jkkk
6X
1i
i
k
V
q (2)
Now, compute the value of the relative displacement
q’j – q’j-1:
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
∏
=
⎟
⎠
⎞
⎜
⎝
⎛−≠
=
⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
⎟
⎠
⎞
⎜
⎝
⎛−
−
⎟
⎠
⎞
⎜
⎝
⎛−−
∏
=
⎟
⎠
⎞
⎜
⎝
⎛≠
=
⎟
⎠
⎞
⎜
⎝
⎛
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
∏
=
⎟
⎠
⎞
⎜
⎝
⎛−≠
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡⎟
⎠
⎞
⎜
⎝
⎛−
−
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
⎟
⎠
⎞
⎜
⎝
⎛−
−
⎟
⎠
⎞
⎜
⎝
⎛−
∏
=
=
⎟
⎠
⎞
⎜
⎝
⎛
=
−
−Ni
1ji
1i
i
k
X6
1
2
1j
11j
Ni
ji
1i
i
k
X6
1
2
j
1j
Nl
j;1jl
1l
l
k
1j
k
j
k
X6
1
2
1j
11j
Ni
1i
i
k
V
'1j
q
'j
q
()
()()
()
()
()
()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−−
−−=− ∏∏
∏
=
≠
=
=
≠
=
=
=
−
Ni
ji
1i
i
2
Nl
jl
1l
l
2
Ni
1i
i
'
1j
'
jk
X6
1j
1jk
X6
11j
1j1
k
V
qq
()
()
()
() ()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−−
−−+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
−=− ∏
∏
=
≠
=
=
=
−X6
11j
1j1
X6
1j
1jk
k
V
qq
2
2
Nl
jl
1l
l
Ni
1i
i
'
1j
'
j
()
X6
X6j3j3
k
V
qq
2
j
'
1j
'
j
+−
=− −
RJAV vol VII issue 1/2010 6 ISSN 1584-7284
()
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡+−
=− −X2
jjX2
k
V
qq
2
j
'
1j
'
j (3)
4. COMPUTING MAXIMUM STRAIN
From reference [1], it was demonstrated that the
strain in x direction along a shear wall between the
two levels (j-1) and j is:
()
()
'
1-j
'
j
3
iqq6L12x
L
y
ε−−−= (4)
Where,
y is the algebraic distance measured from the
neutral axis to the extreme fiber of shear wall
section. y is considered to be positive in the
opposite direction of the deflection U. (Note that
maximum value of εi, is obtained when maximum
value of y is replaced), x is the abscissa of the
section along the shear wall between the two levels
(j-1) and j.
ε x is the strain in x direction.
By substituting the values of (q’j – q’j-1) from
eq.(3) into eq.(4), then
()
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−
−−= (j)
2
3
ik
V
2X
jj2X
6L12X
L
y
ε (5)
(5)
Now, find the position of the maximum value of
the strain εi between levels (i-1) and i.
[]
0
k
V
2X
jj2X
L
y
12ε
X(j)
2
3
i<
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−
−=
∂
∂
The function εi is decreasing which means that
the maximum strain in a shear wall, between levels
(j-1) and j, occurs at the bottom of the shear wall (x
= 0), and this maximum strain is equal to:
()
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−
=
=(j)
2
2
0x k
V
2X
jj2X
L
6y
ε (6)
5. MAXIMUM DRIFT
The maximum drift at the top of the shear wall is
reached when the strain in the reinforcement in the
tensile zone at the critical section of the shear wall is
equal to εst (maximum allowable strain in steel), and
the strain in the extreme fiber of the compression
zone in the same section is equal to εc = maximum
strain limit of concrete in compression = 0.003. So
the critical section in the shear wall is considered to
have the behavior described in figure D.1.
Figure 2. Balanced Reinforced Concrete Section.
From eq(2),
()
()
() ()
[]
()
()
()
()
()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⋅−⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−= ∏
∑∏
∏
=
≠
=
−=
=
=
+≠
=
+
=
Ni
ji
1i
i
2
1ji
1i
Nl
1ii;l
1l
li1i
2
N
1i
i
'
jk
6X
1j
1jkkk
6X
1i
i
k
V
q
=>
()
()
() ()
[]
()
()
()
()
()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⋅−⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−==Δ ∏
∑∏
∏
=
≠
=
−=
=
=
+≠
=
+
=
Ni
Ni
1i
i
2
1Ni
1i
Nl
1ii;l
1l
li1i
2
N
1i
i
'
Nk
6X
1N
1Nkkk
6X
1i
i
k
V
q
Now, let
()
()
() ()
[]
()
()
()
()
()
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−+
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⋅−⋅
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−
−= ∏
∑∏
∏
=
≠
=
−=
=
=
+≠
=
+
=
Ni
Ni
1i
i
2
1Ni
1i
Nl
1ii;l
1l
li1i
2
N
1i
i
k
6X
1N
1Nkkk
6X
1i
i
k
1
R
But
Δ
= R V (7)
Replace V in equation 6 =>
()
⎟
⎠
⎞
⎜
⎝
⎛Δ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−
=
=(j)
2
2
0x k R
2X
jj2X
L
6y
ε (8)
where D is the maximum displacement at the top of
the shear wall. From similar triangles of the
balanced section the following can be written:
d
εε
ε
y
ε
x
ε
y
cst
st
cst +
=⇒= (9)
The maximum strain in the shear wall presented
in eq(8) at the level of steel should be smaller than εy
(the yield strain of steel):
()
st
(j)
2
2
0x ε
k R
2X
jj2X
L
6y
ε≤
⎟
⎠
⎞
⎜
⎝
⎛Δ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−
=
= (10)
RJAV vol VII issue 1/2010 7 ISSN 1584-7284
Replace y from eq(9) ⇒
j)dj-X2(6
L)εε(k R X 2
Δ2
2
cst
(j)
+
+
≤ (11)
j ranging from 1 to N.
If the strain in the steel is considered to stay
within εy (the yielding strain of steel), the maximum
allowable displacement at the top of the shear wall
obeys the following equation,
d j)j-X2(3
L)εε(k R X
Δ2
2
cy
(j)
+
+
≤ (12)
In this case, the critical section of the shear wall
behaves as a balanced section; the limits are reached
in the reinforcement in tension and in the concrete in
compression at the same time, and at that point, the
maximum allowable displacement at the top of the
shear wall is reached.
Notice that as the height of a story increase, the
maximum allowable displacement increases for a
certain number of stories; and as d increases, the
maximum allowable displacement decreases since
any small movement tends to cause larger strain at
the critical section of the shear wall. On the other
hand, and as far as maximum displacement is
concerned and disregarding economical and
architectural issues, it is better to use larger number
of shear walls with small d than to use fewer shear
walls with large d; keeping in mind that the inertia
of a shear wall is increased cubically as a function of
d, and a bigger d will increase our stiffness
significantly in a certain direction.
6. MAXIMUM DRIFT CONSIDERING A
CRACKED SECTION
It is important to note that according to UBC
Modeling Requirement Section 1630.1.2 “Stiffness
Properties of Reinforced Concrete and masonry
elements shall consider the effects of cracked
sections”.[1]
This means that when a designer uses cracked
section analysis then according to ACI 318, section
10.10.4.1, the shear wall moment of Inertia would
have to be reduced by a factor 0.35 such that the
inertia would become 0.35Ig.[12] This will
consequently reduce the stiffness of the lateral load
resisting elements and will equally increase the drift
by the same factor, which means that the drift should
be multiplied by a factor of (1/0.35) to get a value
taking into consideration the effects of a cracked
section suggested by UBC-97 and ACI-318-08.
d j)j-X2(3
L)εε(k R X
)
35.0
1
(Δm2
2
cy
(j)
+
+
≤ (13)
7. EXAMPLE
In order to see how this formula works, let’s
consider an example of a 20 story building with
shear walls having variable lateral stiffness as we go
up the building.
Given that for a 20 story building, the lateral load
resisting stiffness varies as follows:
- from level 1 to 5 the stiffness is k,
- from level 6 to 10 the stiffness is 0.9 k,
- from level 11 to 15 the stiffness is 0.8k,
- and from level 16 to 20 the stiffness is 0.7k.
Yield Strength, fy = 414 Mpa, Modulus of
Elasticity, Es = 2.105 Mpa, Reinforcement Steel
Yield Strain, εy = 0.00207, Story Height = L = 3m;
Concrete Crushing Strain εc = 0.003.
* The shear walls have a constant width b.
* The biggest shear wall have h = 4m at the base,
so the depth d = 0.9h =3.6m at the base and
diminishes when moving upwards.
The results of the suggested equation are
presented in Table 1.
Table 1. Allowable drift for Example 1 using Eqn (13).
Story Stiff-
ness Displacement Strain e(x=0) D Dm
1 1 k 1.000 V/k 0.667 y*V/k 0.0327 0.0935
2 1 k 1.995 V/k 0.663 y*V/k 0.0329 0.094
3 1 k 2.981 V/k 0.657 y*V/k 0.0332 0.0949
4 1 k 3.952 V/k 0.648 y*V/k 0.0337 0.0963
5 1 k 4.905 V/k 0.635 y*V/k 0.0344 0.0982
6 1 k 5.937 V/k 0.688 y*V/k 0.0329 0.0939
7 1 k 6.937 V/k 0.667 y*V/k 0.0339 0.0969
8 1 k 7.900 V/k 0.642 y*V/k 0.0352 0.1006
9 1 k 8.820 V/k 0.614 y*V/k 0.0368 0.1052
10 1 k 9.693 V/k 0.582 y*V/k 0.0388 0.111
11 1 k 10.616 V/k 0.615 y*V/k 0.0382 0.1092
12 1 k 11.473 V/k 0.571 y*V/k 0.0411 0.1175
13 1 k 12.259 V/k 0.524 y*V/k 0.0449 0.1282
14 1 k 12.967 V/k 0.472 y*V/k 0.0498 0.1423
15 1 k 13.592 V/k 0.417 y*V/k 0.0564 0.1612
16 1 k 14.204 V/k 0.408 y*V/k 0.0602 0.1721
17 1 k 14.708 V/k 0.336 y*V/k 0.0732 0.2093
18 1 k 15.095 V/k 0.259 y*V/k 0.0951 0.2717
19 1 k 15.361 V/k 0.177 y*V/k 0.139 0.3971
20 1 k 15.497 V/k 0.091 y*V/k 0.271 0.7743
RJAV vol VII issue 1/2010 8 ISSN 1584-7284
As shown in Table 1, the maximum allowable
displacement is 0.09354 m = (1 / 641) ht, and the
critical section occurs at the base of the shear wall.
8. COMPARISON WITH VARIOUS
SELECTIVE SEISMIC CODES
In comparing the data obtained by equation (13)
with results obtained by various seismic codes, it can
be observed in Table 2 the following:
1- All codes do not take into consideration any
variation in stiffness of a shear wall, nor they care as
to weather the structure is relatively stiff or not.
2- All codes do not consider the geometry of the
shear wall.
3- All codes do not consider the cracking limit of
concrete and the yield limit of steel in a reinforced
concrete section.
Table 2. Comparison of the formula suggested by the
author and various selective seismic codes.
One
Story 20 Story
Building 20 Story
Building
Code Drift
Range*
Height
=300
cm Constant
Stiffness Variable
Stiffness
Height =
6000 cm Example
1
Drift
Values
(cm)
Drift
Values
(cm)
Drift
Values
(cm)
UBC91 0.004h -
0.005h 1.2 - 1.5 24 - 30 24. - 30.
UBC97 0.02 h -
0.25 h 6.0 - 7.5 120. - 150.
120. -
150.
IBC2006 0.01 h -
0.02 h 3.0 - 6.0 60. - 120. 60. - 120.
Lebanese
Code 0.0005h 0.15 3 3
Algerian
Code 0.0075h 2.25 45 45
French
Code -
PS92
0.0007h
-
0.0012h
0.21 -
0.36 4.2 - 7.2 4.2 - 7.2
Japanese
Code 0.005h -
0.0083h 1.5 - 2.5 30. - 50. 30. - 50.
New
Zealand
Code 0.0067h
- 0.010h 2.0- 3.0 40. - 60. 40. - 60.
Khouri (Eqn.
13) 1.1 14.8 9.354
4. Some codes consider the zone to be an issue in the
maximum allowable drift, while what we think is
that, regardless of the zone, the structure has a
displacement limit which if it is surpassed the
structure will be in danger.
5. The formula of equation 13 suggested by the
author in this article takes into consideration the
structural stiffness, the shear wall geometry, the
cracking limits of concrete and the yield limit of
steel and the height of the building.
9. CONCLUSIONS
In this study, the finite element method along
with structural dynamics and reinforced concrete
design was used to analyze a shear building with
variable lateral stiffness. The lateral stiffness matrix
was assembled for variable stiffness in the stories,
and the shear was obtained as a function of the
displacement. A value for the displacement at any
story was determined, and from which a function for
the relative displacement between two stories was
then obtained. Consequently, an equation for the
maximum strain was resolved and a limiting value
for the maximum allowable displacement within the
elastic limits was obtained as a function of the height
of a story, number of stories, depth of tension steel d
in a shear wall, the maximum allowable strain of
steel εst and the maximum allowable concrete strain
εc. Note that shear building was analyzed like a
beam with ignoring vertical loads and assuming
constant lateral stiffness in all stories.
From the above examples one can conclude that
the position of the critical section in the shear wall
cannot be predicted before calculations because it
depends on how the stiffness of the shear wall
decreases when going upwards. In addition, the
value of the maximum allowable drift at the top of
the shear wall also depends on how the stiffness
decreases with height.
This is so, keeping in mind that the values of the
maximum allowable drift obtained in the above
example can be increased if an allowable strain εst
larger than εy is used which means that the yielding
of steel is permitted.
What can be concluded from the comparison of
the results obtained by the author and the other
seismic codes is that when estimating drift one
should check actual structural behavior rather than
estimate the drift based on comfort of a person in
upper floors or other factors that have nothing to do
with the structural properties.
It is now left for the designing engineer to
evaluate his structure and decide/choose a maximum
allowable strain limit for concrete and for steel, and
determine the corresponding maximum allowable
drift values. Beyond these limits the designing
engineer would know that the shear wall in question
has passed the elastic limit in a shear building.
RJAV vol VII issue 1/2010 9 ISSN 1584-7284
ACKNOWLEDGEMENTS
The author thanks Optimal Engineering
Consulting and Contracting, (OECC) who sponsored
the major part of this work, and special thanks to the
Lebanese University, Faculty of Engineering,
Branch II for sponsoring part of this on-going
research. Also thanks to the engineer Bassam
Mazloum who helped in the set up of this paper.
REFERENCES
[1] Khouri, Michel, “Estimation of the Maximum
Allowable Displacement at the Top of a Shear Wall
(within Elastic Limits)” Earthquake Resistant
Engineering Structure, ERES 2009, Limasol, Cyprus.
[2] “Uniform Building Code”, by International
Conference of Building Officials, Whittier,
California, 1997.
[3] “International Building Code”, IBC-2006 Edition,
Published by the International Code Council, INC.,
2006.
[4] Fintel, M. “Handbook of Concrete Engineering”,
Published by Van-Nostrand Reinhold Company, 2nd
edition, New York, 1985.
[5] Searer, G., “Poorly Worded, Ill-Conceived, and
Unnecessary Code Provisions”, 2006 Annual
Meeting of the Los Angeles Tall Building Structural
Design Council, pp. 72-85, Los Angeles, 2006.
[6] Searer, G. and Freeman, S., “Design Drift
Requirements for Long-Period Structures”, 13th
World Conference on
[7] Bathe K.J., “Finite Element Procedures in
Engineering Analysis”, Pentice Hall, Englewood
Cliffs, New Jersey, 1982.
[8] Cook, R., Malkus, D., Plesha, M., “Concepts and
Applications of Finite Element Analysis”, 3rd Ed.,
John Wiley and Sons, New York, 1989.
[9] Zeinkiwicz, O, Taylor, R., “The Finite Element
Method”, 4th Ed., Vol.1, Mc-Graw Hill, London,
1989.
[10] Craig, R., “Structural Dynamics- An Introduction to
Computer Methods”, John Wiley and Sons, 1981.
[11] Naeim, F., “The Seismic Design Handbook”,
Structural Engineering Series, Van Nostrand
Reinhold, New York, 1989.
