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Stiffness matrices for simple analogous frames for shear wall analysis
Computers & Strucrures Vol. 43, No. 4. pp. 613633, Printed in Great Britain.
1992 0
0045-7949192 S5.00 + 0.00 1992 Pergamon Press Ltd
STIFFNESS MATRICES FOR SIMPLE ANALOGOUS FRAMES FOR SHEAR WALL ANALYSIS V. K. KOUMOUSIS~ and G. AG. PEPPAS Institute of Structural Analysis and Aseismic Research, National Technical University of Athens, Athens GR-157 73. Greece (Received
2 May 1991)
Abstract-The global stiffness matrix of two types of analogous frames developed by Stafford Smith and Girgis for shear wall analysis are presented. These matrices can be used in the analysis of multistorey buildings using 2D or 3D frame programs. The matrix entries are given in closed-form relations which are determined using symbolic manipulation. For the 3D element they express the transformation from the local to global system as well as the modification due to the remote displacements of the diaphragms. Numerical results are presented by using full-width elements in every storey. Discrepancy of the results, compared with that obtained using the standard wide-column analogy, are presented.
In this paper closed-form relations for the stiffness coefficients for modified frame modules are developed. They relate the nodal forces to the nodal displacements of regular 3D columns. This modified module behaves as a shear wall in the one principal plane and as fixed at both column ends in the other plane. In this form these elements can be directly incorporated into frame programs adequate for the analysis of multistorey buildings.
1. INTRODUCTION
The analysis of planar shear walls having a heightto-width ratio less than five is not considered satisfactory if based on classical bending theory. In order to avoid a finite element solution for the shear walls of multistorey buildings various attempts have been made to approximate their behaviour. Coull et al. [l-3] and Heidebrecht and Swift [4] analysed simple and coupled shear walls using discrete and continuous elements. The important problem of the interaction of shear walls and 2D frame systems was analysed by Khan and Sbarounis [5], McLeod [6], Gould [7] and others. This problem was the subject of extensive research until methods and programs for forming the stiffness of a complete 3D structure were established. The programs TABS and ETABS developed at Berkeley by Wilson and co-workers [8,9] and the work of Weaver and Brandow [lo, 1I], were both based on the work of Clough et al. [ 121which offered the general framework for the discrete element approach in building analysis. Rutenberg et al. [13] and others have analysed the behaviour of shear walls under dynamic loading. Stafford Smith and Girgis [14] developed two frame models for shear wall analysis, the braced wide-column analogy and the braced frame analogy. The efficiency and accuracy of these elements has been proved quite satisfactory in comparison with plane stress finite element analysis.
t To whom correspondence
should be addressed.
2. ANALYSIS OF BRACED WIDE-COLUMN
The braced wide-column (Fig. 1) is similar to the ordinary wide-column with additional diagonal braces. The member properties of the frame are determined1141 by simulating the bending, shear and axial stiffness of the braced wide module with that of the corresponding wall segment (l), (2), (3), respectively, as follows:
I, = I,, = $
i2EI, + 2EA,
COS* 0
=-
btG
I
h’
$S+
(1)
2EA, sin’ 0 I
(2)
h EA =L h
’
(3)
where E is Young’s modulus of the braces and the column, G = E/(2(1 + v)) is the shear modulus, v is the Poisson ratio, A, = bt the cross-sectional area, and 0 the angle between the horizontal beam and the 613
V. K. KOUMOUSIS and G.
614
AG. PEPPAS
Thus the validity of the model is verified for a ratio of height-to-width between the limits
The stiffness matrix in the major plane of the wall
L-----b------7
Fig. 1. A braced frame module.
diagonal bracing. Thus from relations (2) and (3) we obtain Ad= &
(4B - 0.5)
A, = bt (2 - 8B),
(4) (5)
where B is defined as B=
h2 16b2(1 + v)’
(6)
From relations (4) and (5) it becomes evident that the area of the cross-section A,, of the diagonal bracing becomes negative when h/b c (2( 1 + v))‘/~ while the area of the cross-section of the column A, is negative when h/b > (4(1 + v))“‘.
The nodal displacements and forces of the modified 3D module are numbered as shown in Fig. 2. This numbering groups first the degrees of freedom which correspond to the displacements of a grid and then the degrees of freedom of a diaphragm whose plane is normal to the z axis. This choice facilitates the process of stiffness formation following the concept introduced by Weaver and Brandow [ 10, 111. The stiffness matrix of the braced wide-column 2D module can be determined in terms of the nodal displacements and forces that correspond to a 2D column by using the definition of the stiffness coefficient. The coefficient S, expresses the component of force corresponding to the ith component of displacement which is required in order to displace the module in a deformation pattern involving a unit displacement Sj = 1 while all the other displacements are kept to zero. Thus each stiffness coefficient for the 2D module can be determined by subjecting to the previously described deformation patterns and computing the required components of force which must be applied at the nodes of the module in order to keep it in equilibrium. In computing the internal forces in the column and the braces of the module its underformed configuration is considered. The equivalent nodal forces are obtained by appropriate transformation.
Fig. 2. Nodal displacements and forces of wall element.
Shear wall analysis using stiffness matrices For 6,=1
and Sj=0,j#2
b/2
s,, = 0
$2 = 0
$+A,sinecos’B
S10,2= -E
6,
d2.
>+A,,sinBcos2B >
For 6, = 1 and 6, = 0, j # 3 s,, = 0
A
Ad
S,, = E L + 2 T sin3 0 ~5~ h
s,, = 0 h
s,, = 0
h.3 = 0.
For ~5,= 1 and di = 0, j # 7
61 . --f+AdsmBcos2~ h2
6,
s,, = 0 61
.
6,
~+Adslnecos2e h2
s,, = 0 121 2A, +cos3 e s,, = E h3 b
6,
6,.
-b------T
615
V. K. KOUMOUSIS and G. AG. PEPPAS
616
For 6, = 1 and 6, = 0, j # 5
:+%sin’Bcos0
a5 >
s,, = 0
I
S,, = E ?
+ y
sin2 9
cos
0
6,
h
>
s,, = 0 S,, = E $
1
+ A,, sin 0 cos’ 0 6, >
r--
s,o,s=-E
b
---
>
$+A,sinOc0s82
6,.
For 6, = 1 and 6, = 0, j # 6 s,, = 0
s,,=-E
$-Fsin30
6,
(
> h
s,, = 0 S, = E 2 + y
sin3 fI 6, >
.’
-I
i
s,, = 0
_ b
9)...
f
&e =l
I
S10.6 = 0.
For 6,, = 1 and S, = 0, j # 10
s,,,, = -E
S,,,, = -E
+ A, sin 0
$
COS'
0
~$0
+ A,sin 0 cos2 9 &, >
s,,,il = 0
s,,,,,=
S
,,,,o
=
-E
E
!$ f 7 2 +7
> e >
COS3 e &,
COS' al,,.
-b
,40=1
Shear wall analysis using stiffness matrices
Thus the local stiffness matrix of the module in the wall plane becomes s 2.2
0
s2.5
0
s2.1
S2.M
0
s3.3
0
S3.6
0
0
0
s5.5
0
s5,7
s5,,0
S6.3
0
S6.6
0
0
s 7.2
0
s7.5
0
s7.7
s7.10
S 10.2
0
SlO.5
0
s10.7
s,o.,o
s 5.2
PI =
0
(8) The behaviour of the other two modules introduced by Stafford Smith and Girgis [14] namely the braced frame module in symmetric and asymmetric form is identical to that of the braced wide-column. The main reason for the introduction of these modules was to render more efficient their use in programs for the analysis of planar frames by introducing fewer nodes and members. Thus, these modules were intended for more general cases where the wall is subdivided into fractional wall width modules and multi-wall assemblies.
s I.1
PI =
0
0
s,,4
0
617
The properties of the members of these modules are determined by equating the behaviour of the wall segment with that of the braced frame module [ 141. By choosing as nodal displacements those of the middle of the top and bottom beams of the module it turns out that the stiffness coefficients of the wide-braced column module and those of the braced frame module are identical and thus there is no reason to use two distinct modified modules. Consequently below only the relations describing the response of the brace wide column module are established. 3. THE
STIFFNESS MATRIX
IN THREE DIMENSIONS
In order to form the 3D stiffness matrix of a shear wall the stiffness matrix (8) is augmented to incorporate the response of the module in the plane perpendicular to the wall plane. The stiffness coefficients in this direction are those of a standard column. As nodes we choose the middle of the top and bottom rigid beams. Thus the stiffness matrix of the modified 3D module is
0
0
S1.8
0
0
Sk,,
0 0
o
s2.5
o
s 2.7
0
0
s,,,,
0
0
0
S3.6
0
0
0
0
00
0
s4,4
0
0
0
s43
0
0
s5,
o
o
s5.5
o
s 5.1
0
0
0
0
S6.3
0
0
0
s7,2
0
0
s7,5
0
s 7.7
s 8.1
0
0
s*,4
0
0
0
0
0
0
0
0
0
0
0
s,o.z
0
0
s,o.5
0
s 10.7
s 11.1
0
0
0
0
0
0
0
0
0
o
s2,
0
0
s 4.1
0
o
o s,,,
s,,, 0
0
0
0
s,,,,
000
&I
s4,,,
0 0
%*
0
00
s,,,,
0
St,,
(9) 0
0
0
0
0
SW
0
0
SW2
0
0
s,o,,o
0
0
0
S11.8
0
0
0
0
s,2.9
0
s,,.,,
0 0
s,2,,2
where the coefficients in excess of those given by relation (8) are
S,,,=E(,+),
8.4
=
&,4=
s 12.12
s,,,
-ss.,,
=
S9.9~
S4.,=E(+),
?
&
= E
=
,
12 $ (
SH.8
S4,4=s ,,,,
S,,9 = G 2 ,
)
-s&s*
&.,I
=
S8.8,
s~,,=E(+)
S,,,, = -SK,
s,,,,
=
-
$9
(10)
V.
618
K. KOIJMOUSIS and G. AG. PEPPAS
4. TRANSFORMAnON FROM LOCAL TO GLOBAL COORDINATE SYSTEM
In order to express the nodal displacements and forces of the modified 3D module into the global vertical coordinate system (Fig. 2), the horizontal components of translations and rotations and the horizontal components of forces and moments are transformed to the global horizontal directions. This is thieved by applying the following transformation [ 151
The above transformation is not the most general one [15] it is valid for vertical elements that can only be rotated about the z axis. In a 3D frame program the components of displacement of the diaphragms are the independent components of displacement. The displacements along the X, y axis and rotation about the z axis of the top and bottom nodes of the module are not considered as independent. Thus the nodal com-
1
IsI* = VI’ PI . [A I,
(11)
where sin4
0
0
0
0
0
0
0
0
0
0
-sin4
cosf#
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
cos.4
sin f#~ 0
0
0
0
0
0
0
0
0
0
-sinr$cos4
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-sin4
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
cosfj
PI=
C$is the angle of rotation which the global x axis must be rotated about the z axis in order to coincide with the local x axis.
cos I$ sin 4 cos4
cos qt sin4 -sin+ 0
.
(12)
0
COSC$ 0 0
1
ponents of displacement must be transformed to the corresponding components of displacement of the diaphragms. This is achieved by using the following transformation [lo]
Lf?I= [Tlz[~l*[Tl
(131
189= [~l’[~l’[~l[~l[Tl,
(14)
or where
ITI =
10000000
00
0
0
01000000
00
0
0
00100000
00
0
0
00010000
00
0
0
00001000
00
0
0
00000100
00
0
0
0
0
0
0
0
0
0
0
1 0
-y
0
00000001
x0
0
0
00000000
10
0
0
00000001
01
0
00000000
00
1
X
00000000
00
0
1
-Y
(1%
Shear wall analysis using stiffness matrices
619
relation (24) can be written as
s,,, $2 s,,, s-2 0
0
0 0
31.4 $5 s2.4 Q5
s,,
0
& % 31.9 SqJO%I s2.7 s2,* 32, s2.10 %I
0 0
s,,
0
0
0
0
0
%* ~2.12
0
0
34.1
s4,2 0
34,4 34,5 0
s4,7 34.8 s4,9 ~4,lO s4,,, s4,,2
3 5.1
s5,2
0
35.4
$5,5
s5,7
$5,8
0
s,,,
0
0
$6
0
0
0
s 7.1
%,2
0
%4
%,5
0
%7
$7.8
37.9
&JO
%,
%2
s 8.1
38.2
0
%4
38,5
cl
s,,,
S8,8
S8,9
S8.10
%,
%*
s 9.1
39.2
0
39.4
$9.5
0
s9,7
$9,8
s9,9
$8,,0
$9*,,
$9.12
s 10.1
$0.2
0
3,0,4
SO.5
0
$0.7
ZO,8
3,0,9
s,O,,ll
SO.,,
SO,,,
3 11.1
h.2
0
%,4
%5
0
L,,
%,,8
G.9
&JO
%,
%,,2
0
$,2.4
s,,,
0
$2.7
s2,*
3,2,9
Lo
&,,I
s,,,*
0
VI =
0
$5,9
s5,,0
35,,,
s5,,2
0
0
0 '
3 I?.., s2.2
(16)
where S,,, = C 1 cos2 cp + C2 sin’ q
S,,, = (C 1 - C2) sin cp cos cp
& = C 1 sin’ cp + C2 cos2 4
s,,, = c3
S,., = C4 cos’ cp + CS sin2 4
S,,, = (C4 - C5) sin cp cos 4
$4
%
s.,
=
34.2
$2
=
C4 sin2 f$ + C5 cos2 cp
32,,
G.5
=
32.2
s,,
=
-
=
35.4 =
c3
S,,$ = c3
g,,, = -(CO + C6) sin cp cos cp s,,2 = C6 cos2 cp - CO sin2 cp
37.4 =
%,
ST.5 =
s 8.1 =
-
&,5
s8,2
=
C6 sin2 $J + CO cos’ 4
g8, = -s,,,
s,,,
(C7 - C8) sin cp . cos 4
38.8 =
c7 sin2 cp + C8 COGfp
s9,
-Y&J
+
x38,,
-Ys8.7
+
x&,8
+ xs,,.l
s9,2 =
-Y&.2
sb.5 =
-Y%.2
-
39,7 =
-Y&.7
&
C7(y cos cp -x
=
$8,4 =
S,,, =
s,,, = -.&,
X&J
g,,, = C7 cos2 4 + C8 sin2 4
ST.2
-
+
x$,
xs8,,
sin cp)’ + CS(y sin 4 +x cos cp)*+ C9
=
s9,8 =
%*,
=
- %
3 10.2=
- %2
SO.4 =
- %,
so.5 =
&o.,
- %,
'%I8 =
-s,,
s,0.9 =
Y&.7
%
%,,
=
-s8,,
s
s,.,
s,,
&,,
=
so.,0 =
s II.4=
- s8,,
%5
=
s 11.8=
- 38,8
$1.9
= Ys8.7 -
s,,,,, =
s,,8
$2.,
=
xs8,8
1.2=
=
- 37.2
-
-s8,7
s,,,o =
$7
-s,,,
&2,2 =
- $9,2
3 12.4=
- 39.,
$2.5 =
- s9,2
s2.7 =
- 519.7
s 12.8=
- 39,
s,2,9 =
- $9.9
s,,o
39.7
s 12.11=
s9.8
5r12.12=
s.9.
=
xs8.,
(17)
V. K.
620
KOUMOUSIS and G. AC. PEPPAS
nodes=10
nodes=6
members=14
members=12
a)
braced
wide
b) braced
column
nodes=6 members=6 c) modified
frame
module
Fig. 3. Structural idealization using different modules. The constants C8
and
C9
CO,
Cl,
C2,
C3,
C4,
C5,
C6,
C7,
used in relations above are defined as
CO = E(61,,/h2) C 1 = E(41,,/h) C2 = E(41,/h
+ bAdsin 0 cos e/2)
C3 = E(A,/h
+ 2A,sin’f?/h)
C4 = E(2Z,,/h) C5 = ,5(21,/h + bAdsin
0 cos e/2)
C6 = - E(61,. /h2 + A, sin 0 cos2 0) C7 = E(121,,/h2+
in the routines WALL and WALL1 written in Fortran 77 which are given in the Appendix. These routines can be used to calculate the matrices in relations (16) and (19) which can be incorporated in a 3D frame program that introduces as independent components of displacement those of the diaphragms. These routines can be introduced to other programs for buildings with diaphragms such as ETABS and Super ETABS by appropriately renumbering of the degrees of freedom. In addition by eliminating the transformation described in relation (13) the resulting elements can be introduced to a general 3D frame program. Finally using only the coefficients that appear in relation (8) the resulting elements can be incorporated into a 2D frame program. 7. REFINEDMODELLING
2A,cos3 B/b)
C8 = E(12Z,,/h3)
C9 = GI;_/h.
(18)
The stiffness matrix of the module given in relation (16) depends on the selection of the reference point of the diaphragm. Thus this stiffness matrix characterizes the wall module as well as its position in the structure.
In order to establish a more refined solution [14], modules of fractional width and height (see Fig. 3) can be employed where the modules depend on the dimensions of the corresponding segment. For these cases the use of the modified module (Fig. 3c) presented in this paper results in a smaller number of nodes and members than those of the braced wide-columns (Fig. 3a) and the braced frame modules (Fig. 3b). 8. FURTHERGENERALIZATION
5. MEMBERFORCES Having established the components of displacement which characterize the response of a building for a given loading, the nodal actions at the top and bottom of each module can be determined on the basis of the following relation [ 151 {A } =
Pl[~lP-l{~},
In the case of adjacent modules the number of independent nodal displacements can be reduced further by considering the displacements at the two hinges as independent (Fig. 4). Assuming the beams
(19)
where (A} is the local matrix of nodal forces and {b} is the global matrix of nodal displacements. 6. COMPUTERIMPLEMENTATION In order to avoid the matrix multiplications in relations (14) these operations were performed in symbolic form. The results are presented in the closed form relations (17) and (18). Similar expressions are developed for relation (19). These relations were used
I-b-
Fig. 4. Nodal displacements and forces of wall element.
i1
621
Shear wail analysis using stiffness matrices
as rigid, the nodal displacements - u”
are expressed as
VU 8”
0
000
0
112
l/200
0
0
l/6
0
0
010
VO
0
0
0
0
l/2
. o”
0
0
0
0
l/6
U@
Moreover the components relation
1
0
=
-I/b
0
0
0
(20)
0 l/2 -l/b
of nodal actions are transformed to those at the two hinges using the following
0 0
N”
0
M”
0
Q0
-l/b00 010
or
Q”
000 I/b00
0
0
l/2
0
0
l/2
f/b -I/b
-
) .
(21)
NO
_MO
the nodal components of displacement and the nodal components of forces at the hinges {a) = [R]‘{A >.
@I =
Substituting relations (20) and (21) into relation (8) we obtain the stiffness matrix of the module relating
~~l~~l[Rl~~~.
PI
The nodal components of displacement 16) are the same as those used for the panel element in the
r1.8-44.2----7--SJi~3.9-4.s~
Fig. 5. A ten-storey building.
V. K. KOUMOUSIS and G. AG.
622
F'EPPAS
Table 1. Dimensions of the cross-section of vertical elements Floor
10,9,8, 7 40 x 160 x 30 x 30 x
40 (cm2) 25 70 50
6,534
332
I
50 x 50 160x25 35 x 70 30 x 60
55 x 55 180x30 40 x 70 35 x 60
60 x 60 200x30 45 x 70 40 x 60
Fig. 6. Bending moments of shear walls and their percentage difference.
Shear wall analysis using stiffness matrices TABS, ETABS and Super ETABS program. However, the stiffness coefficients are different. Following the above transformation the stiffness of the module in the plane of the wall is determined with respect to three displacements at the top and bottom of the module. In a plane stress finite element eight components of displacement are intro-
623
duced two for each corner node of a rectangular element. 9. NUMERICALRESULTS The stiffness matrix of this modufe was used in a program for the analysis of multistorey buildings. to 6_
___
Well
Column
Fig. 7. Shearing forces of shear walls and their percentage difference.
V. K. Kou~ousrs and G. AC. F%PPAS
624
This program considers the floors as diaphragms by introducing directly their three components of displacement. The general method was developed by Weaver and Brandow introduced in [lo] and [l 11. Moreover, various modifications have been made in order to extend the method to non-canonical topologies. The vertical elements of every storey are treated as 3D elements and their stiffness matrices are determined from relations (16) and (17).
Using relation (23) the horizontal seismic forces for the example are: H 1 = 44 kN, H2 = 87 kN, H3 = 131 kN, H4= 175kN, H5=218kN, H6=262kN, H7 = 305 kN, H8 = 350 kN, H9 = 393 kN, HlO = 436 kN. The dead load corresponds to a specific weight y = 24 kN/m3 while the floor thickness is d = 0.18 m. The external walls for this storey height weight q1 = 10 kN/m and the internal wall q2 = 5 kN/m. The live load is p = 2 kN/m2.
Example 1
The loading cases are:
Using this program, the ten-storey building of Fig. 5 was analysed twice. First by using the standard wide-column analogy and then by the modified 3D module established in this paper. The storey height is 3.2 m. The dimensions of the vertical elements change according to Table 1. The foundation consists of a grid and footings on elastic foundation with k = 300,000 kN/m3. The grid is of the same plan as Fig. 5 and the dimensions of its beams are 30/60. The building was analysed for five different load cases. The total seismic force 6N is distributed along the height of the building in a triangular way. According to the Greek Code the equivalent horizontal static loads are determined by the following relation
1. 1.35 x dead load + 2. 1.00 x dead load + + 1.40 x earthquake 3. 1.OOx dead load + + 1.40 x earthquake 4. 1.00 x dead load + - 1.40 x earthquake 5. 1.00 x dead load + - 1.40 x earthquake
c NJ, i=,
(23)
where 6 is the seismic coefficient, N is the total vertical load at the base, N, is the vertical load at level z and hl is the height of level z starting from the foundation level.
1.50 x live 0.50 x live along X. 0.50 x live along Y. 0.50 x live along X. 0.50 x live along Y.
load. load load load load
In Fig. 6 the bending moments M, of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide column analogy are presented. In Fig. 7 the shearing force Qx of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. Example 2
The five-storey building of Fig. 8 was analysed using the same program. The vertical elements of this
r1.8-4.2----r-----50---r---33,9-
I
5.50 K840x30
I 2.10 K1040x30
i
4.05
i
Fig. 8. A five-storey building.
625
Shear wall analysis using stiffness matrices
building are considered fixed at the base. The height of every storey is 3.2 m. The columns are 40 x 40 in the first storey and 35 x 35 in the other storeys. The remaining vertical elements are kept constant in all storeys. The horizontal forces are H 1 = 67.67 kN, H2 = 135.33 kN, H3 = 203 kN, H4 = 270.67 kN, HS =
338.33 kN. The vertical loads and the loading cases are the same as for example 1. In Fig. 9 the bending moments My of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide-column analogy are presented.
Difference
s
Diffgence
%
Oe
Fig. 9. Bending moments of shear walls and their percentage difference.
V. K.
626
KOUMOUSIS
In Fig. 10 the shearing force QX of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. As a third example a three-storey building of the same plan as example 2 was used in this analysis.
and G. AC.
F’EPPAS
In Fig. 11 the bending moments MY of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide-column analogy are presented. In Fig. 12 the shearing force QX of shear walls 2, 5 and 13 obtained on the basis of the method
,
1
4-I
To
$
4
ii f? :
E 02 i?i 1
0_ -
-0
ti
Difference
I
Fig. IO. Shearing forces of shear walls and their percentage difference.
1
1
627
Shear wall analysis using stiffness matrices
_a
II
0-U
I
0
-20
-10
Difference
Difference
Fig. Il. Bending moments of shear walls and their percentage difference.
presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. The stress resultants and the percent error of their difference are given in Figs 7-12 side-by-side in order to distinguish the logistic error and the significant one for the design of shear walls. CAS 43/4-B
IO.
CONCLUSIONS
The stiffness matrix of a 2D frame type module introduced by Stafford Smith and Girgis for shear wall analysis is presented as a modified 3D column module. This module can be used in 3D programs for analysing buildings. Moreover the stiffness matrix of
628
V. K. Kou~ous~s and G. AG. PEPPAS
6
iFi
_I
0
-IS
- 0
4
Difference
s
I
5
oJ Differ&e
1
Fig. 12. Shearing forces of shear walls and their percentage difference.
these modules was modified to take into account the influence of the remote nodal displacements and forces of the diaphragms. From the numerical results obtained in this investigation the following conclusions become evident: 1. The difference between the stress resultants determined using the modified element and the
wide-column analogy are important for medium rise buildings. 2. The difference between the stress resultants and determined using the modified element presented in this analysis and the wide-column analogy is of the order of 20% in the shearing forces of a five-storey building.
Shear wall analysis using stiffness matrices 3. The differences appear on the unsafe side as well as to the safe side depending on the floor level without a specific pattern.
8.
4. The difference is more pronounced in the shearing force in the direction of the wall while the
9.
difference in bending moments is not so critical. 5. The max significant difference which occurred was 21% on the unsafe side in the base shear of a three-storey building.
10.
In general the use of the more accurate modified module determines larger values for t’ L base shear of shear walls than the wide-column analogy.
12.
11.
13. REFERENCES I.
2. 3. 4.
5.
6. 7.
A. Cot111and A. W. Irwin, Model investigations of shear wall structures. J. Srrucf. Div., ASCE 98, 1223-1237 (1972). A. Coull and J. R. Choudhury, Stresses and deflections in coupled shear walls. ACZ Jr1165-72 (1967). A. Co1111and J. R. Choudhury, Analysis of coupled shear walls. ACZ Jnl September, 587-593 (1967). A. C. Heidebrecht and k. D. Swift, Analysis of asymmetrical coupled shear walls. J. Sfrucr. Div., ASCE 97, 1407-1422 (1971). F. R. Khan and H. A. Sbarounis, Interaction of shear walls and frames. J. Struct. Div., AXE 90, 285-335 (1984). I. A. McLeod, Connected shear walls of unequal width. ACZ Jnl May, 408412 (1970). Ph. L. Gould, Interaction of shear wall-frame systems
APPENDIX:
C
CODE IN FORTRAN
14.
15. 16.
17.
18.
in multistorey buildings. J. Am. Concrete Inst. January, 45-69 (1965). E. L. Wilson and H. H. Dovey, Three-dimensional analysis of building systems-TABS. Report No. EERC 72-8, December (1972). E. L. Wilson, J. P. Holhngs and H. H. Dovey, Threedimensional analysis of building systems (extended version). Report No. EERC 75-13, April (1975). W. Weaver, Jr, Three-dimensional analysis of tier buildings. J. Struct. Div., ASCE 92, 385404 (1966). W. Weaver, Jr and G. E. Brandow, Tier buildings with shear cores, bracing, and setbacks. Compuf. Sfruct. 1, 57-83 (1971). R. W. Clough, I. P. King and E. L. Wilson, Structural analysis of multistorey building. J. Sfrucf. Div., ASCE, August (1963). A. Rutenberg, W. K. Tso and A. C. Heidebrecht, Dynamic properties of asymmetric wall-frame structures. Earthquake Engng Struct. Dynamics 5, 41-51 (1977). B. Stafford Smith and A. Girgis, Simple analogous frames for shear wall analysis. ASCE J. Sfruct. Engng 110, 2655-2666 ( 1984). A. E. Armenakas, Modern Structural Analysis. The Matrix Method Approach. McGraw-Hill (1991). L. A. Bejar, Seismic analysis of multistory building systems using microcomputers. ASCE J. Computing Civil Engng 3, 1-17 (1989). Al. Carpinteri and An. Carpinteri, Lateral loading distribution between the elements of a three-dimensional civil structure. Comput. Strucr. 21, 563-580 (1985). C. F. Neuss and B. F. Maison, Dynamic analysis of a forty-four storey building, ASCE J. Strucr. Engng 111, 1559-1572(1985).
77. SUBROUTINES
WALL AND WALL1
SUBROUTINEWALL(E,G,AR,IX,IY,IZ,H,CF,SF,X,Y,S) REAL MI,IC,IX,IZ,S(I2,12) MI=(E/(2*6))-1 B=SQRT(12. *IY/AR) ______________________~~_~~~~~~~~~~~~~~~~~~~~~~~~~~~~_~~ H = HEIGHT B = WIDTH T = THICKNESS E = ELASTIC MODULUS G = SHEAR MODULUS MI = POISSON RATIO AC, AD = CROSS SECTIONALAREA OF COLUMN AND DIAGONAL BRACES IC = MOMENT OF INERTIA OF COLUMN SECTION (Iyy) IX = MOMENT OF INERTIA OF WALL (Ixx) = TORSIONAL MOMENT OF INERTIA ABOUT Z AXIS ZETA = ANGLE OF DIAG BRACES BETA = AUXILIARY CONSTANT [S] = STIFFNESSMATRIX ____________________~_~~_~~~~~_~~~~~~~~~~~~~~~~~~~~~~~~~ THETA=ATAN(H/B) BETA=H*H/(16*B*B*( 1tMI)) **************************************** **tk**********
WALL
629
*****************
*******¶t********************************
WALL PROPERTIES IC=AR*B*B/12. AC=AR*(2.-8.*BETA) ADsAR*(4.*BETA-.S)/(SIN(THETA))**3
V. K. KOUMOUSIS and G.
630
Ac. PEPPAS
CSF=COS(F)*SIN(F) SF=SIN(F), CF=COS(F), CFZ=COS(F)**2, SF2=SIN(F)**2 Y = Y COORD. X = X COORDINATE CSF=CF*SF CF2=CF*CF SF2=SF*SF FREQUENTLY-USED PARAMETERS CO=E*(6.*IX/H**2) Cl=E*(4.*IX/H) C~IE*(~.*IC/H+B*AD*(SIN(THETA))**~*COS(THETA)/~.) C3=E*(AC/Ht2.*AD*(SIN(THETA))**3/H) C4=E*(2.*IX/H) C5=E*(2.*IC/HtB*AD*(SIN(THETA))**2*COS(THETA)/2.) C6=-E*(6.*IC/H**2tAD*SIN(THETA)*(COS(THETA))**2) C7=E*(12.*IC/H**3+2.*AD*(COS(THETA))**3/6) ~~=~*~;~H*IX/H**J) z * MODIFIED GLOBAL STIFFNESS MATRIX : DO 10 K1=1,12 DO 10 K2=1,12 S(Kl,K2)=0 10 CONTINUE
C S(l,l)=Cl*CF2tC2*SF2 S(2,1)=(Cl-C2)*CSF S(2,2)=Cl*SF2tC2*CF2 S(3,3)=C3 S(4,1)=C4*CF2tC5*SF2 S(4,2)=(C4-C5)*CSF S(4,4)=S(l,l) S(5,1)=S(4,2) S(5,2)=C4*SF2tC5*CF2 S(5,4)=S(2,1) S(5,5)=S(2,2) S(6,3)=-C3 S(6,6)=C3 S(7,1)=-(COtC6)*CSF S(7,2)=C6*CF2-CO*SF2 S(7,4)=S(7,1) S(7,5)=S(7,2) S(7,7)=C7*CF2tC8*SF2 S(8,1)=-C6*SF2tCO*CF2 S(8,2)=-S(7,l) S(8,4)=S(8,1) S(8,5)=S(8,2) S(8,7)=(C7-C8)*CSF S(8,8)=C7*SF2tC8*CF2
S(9,1)=-Y*S(7,1)tX*S(8,4) s(9,2)=-Y*s(7,2)-x*s(7,1) S(9,4)=-Y*S(7,1)tX*S(8,1) s(9,5)=-Y*s(7,2)-x*s(7,1) S(9,7)=-Y*S(7,7)tX*S(8,7) S(9,8)=-Y*S(8,7)tX*S(8,8)
S(9,9)=C7*(Y*CF-X*SF)**2+ lC8*(Y*SFtX*CF)**2tC9 s(10,1)=-s(7,1) S(10,2)=-S(7,2) S(10,4)=-S(7,l) S(10,5)=-S(7,2) S(10,7)=-S(7,7) S(10,8)=-S(8,7) S(lO,S)=Y*S(7,7)-X*S(8,7)
[Sc]=[Tc]'[Lc]'[S][Lc][Tc]
Shear wall analysis using stiffness matrices S(10,1O)=S(7,7) S(ll,l)=-S(8,l) S(11,2)=S(7,1) S(11,4)=-S(8,l) S(11,5)=S(7,1) S(11,7)=-S(8,7) S( 11,8)=-S(8,8)
S(ll,S)=Y*S(8,7)-X*S(8,8) S(ll,lO)=S(8,7) S(ll,ll)=S(8,8) S(12,1)=-S(9,l) S(12,2)=-S(9,2) S( 12,4)=-S(9,l) S(12,5)=-S(9,2) S(12,7)=-S(9,7) S( 12,8)=-S(9,8) S(12,9)=-S(9,9) S(12,1O)=S(9,7) S(12,11)=S(9,8) S(12,12)=S(9,9) C
DO 15 K1=1,12 DO 15 K2=Klt1,12 S(Kl,KP)=S(KZ,Kl)
15 CONTINUE RETURN END
==I=E===========I===D=-========I===P========================~=========
SUBROUTINE WALLl(E,G,AR,IX,IY,IZ,H,CF,SF,X,Y,S)
C REAL MI,IC,IX,IZ,S(12,12) MI=(E/(2*G))-1 B=SQRT(l2.*IY/AR)
____________________~~__~~~_~~__~~~~~__~~~_~~_~~__~_____ H - HEIGHT 8 - WIDTH T = THICKNESS E = ELASTIC MODULUS G = SHEAR MODULUS MI = POISSON RATIO AC, AD = CROSS SECTIONAL AREA OF COLUMN AND DIAGONAL BRACES IC - MOMENT OF INERTIA OF COLUMN SECTION (Iyy) IX = MOMENT OF INERTIA OF WALL (Ixx) IZ - TORSIONAL MOMENT OF INERTIA ABOUT Z AXIS THETA = ANGLE OF DIAG. BRACES BETA = AUXILIARY CONSTANT [S] = STIFFNESS MATRIX THETA=ATAN(H/B) BETA=H*H/(16*B*B*(ltMI))
**************************************** id***********
WALL-1
*****************
****************************t*************
WALL
PROPERTIES
IC=AR*8*8/12. AC=AR*(2.-8.*BETA) AD=AR*(s.*BETA-.5)/(SIN(THETA))**3 CF=COS(F), SF=SIN(F) X = X COORDINATE Y - Y COORD. FREQUENTLY USED PARAMETERS CO=E*(6.*IX/H**2)
631
V. K. K~UMOUSIS and G. AG.
632
PEIPPAS
Cl=E*(4.*IX/H) C2=E*(4.*IC/HtB*AD*(SIN(THETA))**2*COS(THETA)/2.) C3=E*(AC/Ht2,*AD*(SIN(THETA))**3/H) C4=E*(2.*IX/H) C5=E*(2.*IC/Ht8*AD*(SlN(THETA))**2*COS(THETA)/2.) ~6=-E*(6.*IC/H**2tAD*SIN(THETA)*(~OS(THETA))**2) C7=E*(l2.*~C/H**3t2.*AD*(&OS~THETA))**3/6~ C8=E*(12.*IX/H**3) C9=G*IZ/H MATRIX : C
IScl=[Sl [Lcl[Tcl
DO 10 K1=1,12 DO 10 K2=1,12 S(Kl,K2)=0 10 CONTINUE
s(l,l)=cl*cF S(1,2)=Cl*SF S(1,4)=C4*CF S( 1,5)=C4*SF s(1,7}=-co*sF S( 1,8)=co*cF s(1,9)=co*Y*sF S(l,lO)=-S(1,7 S(l,ll)=-S(1,8 S(1,12)=-S(1,9 S(2,1)=-C2*SF S(2,2)=C2*CF S(2,4)=-C5*SF S(2,5)=C5*CF S(2,7)=C6*CF S(2,8)=C6*SF
tCO*X*CF
S(2,9)=-C6*Y*CFtC6*X*SF S(2,10)=-S(2,7) S(2,11)=-S(2,8) S(2,12)=-S(2,9) S(3,3)=C3 S(3,6)=-C3 S(4,l)=C4*CF S(4,2)=C4*SF S(4,4)=S(l,l) S(4,5)=S(1,2) S(4,7)=S(1,7j S(4,8~=S{l,8) S(4,9)=S(1,9) S(4,10)=-S(1,7) S(4,11)=-S(1,8) S(4,12)=S(1,12) S(5,1)=S(2,4) S(5,2)=S(2,5) S(5,4)=S(2,1) S(5,5~=S~2,2~ S(5,7)=S(2,7) S(5,8)=S(2,8) S(5,9)=S(2,9) S(5,10)=-S(2,7) S(5,11)=-S(2,8) 5(5,12)=5(2,12) S(6,3)=-C3 S(6,6f=C3 S(7,1)=-S(2,8) S(7,2)=S(2,7) S(7,4)=-S(2,8) S(7,5)=S(2,7) S(7,7)=C7*CF S(7,8)=C7*SF S(7,9)=-C7*Y*CFtC7*X*SF
Shear wall analysis using stiffness matrices S(7,10)=-S(7,7) S(7,11)=-S(7,8) S(7,12)=-S(7,9) S(8,1)=S(1,8) S(8,2)=-S(1,7) S(8,4)=S(8,1) S(8,S)=S(8,2) S(8,7)=-C8*SF s(8,8)=c8*cF
S(8,9)=C8*Y*SFtC8*X*CF S(8,10)=-S(8,7) S(8,11)=-S(8,8) S(8,12)=-S(8,9) S(9,9)=C9 S(9,12)=-c9 S( lO,l)=S(2,8) S(10,2)=-S(2,7) S(10,4)=S(2,8) S(10,5)=-S(2,7) S(10,7)=-S(7,7) S(10,8)=-S(7,8) S(10,9)=-S(7,9) s(10,10)=s(7,7) S(lO,ll)=S(7,8) S(10,12)=S(7,9) S(ll,l)=-S(8,l) S(11,2)=-S(8,2) S(11,4)=-S(8,l) S(11,5)=-S(8,2) S(11,7)=-S(8,7) S(11,8)=-S(8,8) S( 11,9)=-S(8,9) S(ll,lO)G(8,7) S(ll,ll)=S(8,8) S(11,12)6(8,9) S(12,9)=-c9 S(12,12)=C9 RETURN END
633
1992 0
0045-7949192 S5.00 + 0.00 1992 Pergamon Press Ltd
STIFFNESS MATRICES FOR SIMPLE ANALOGOUS FRAMES FOR SHEAR WALL ANALYSIS V. K. KOUMOUSIS~ and G. AG. PEPPAS Institute of Structural Analysis and Aseismic Research, National Technical University of Athens, Athens GR-157 73. Greece (Received
2 May 1991)
Abstract-The global stiffness matrix of two types of analogous frames developed by Stafford Smith and Girgis for shear wall analysis are presented. These matrices can be used in the analysis of multistorey buildings using 2D or 3D frame programs. The matrix entries are given in closed-form relations which are determined using symbolic manipulation. For the 3D element they express the transformation from the local to global system as well as the modification due to the remote displacements of the diaphragms. Numerical results are presented by using full-width elements in every storey. Discrepancy of the results, compared with that obtained using the standard wide-column analogy, are presented.
In this paper closed-form relations for the stiffness coefficients for modified frame modules are developed. They relate the nodal forces to the nodal displacements of regular 3D columns. This modified module behaves as a shear wall in the one principal plane and as fixed at both column ends in the other plane. In this form these elements can be directly incorporated into frame programs adequate for the analysis of multistorey buildings.
1. INTRODUCTION
The analysis of planar shear walls having a heightto-width ratio less than five is not considered satisfactory if based on classical bending theory. In order to avoid a finite element solution for the shear walls of multistorey buildings various attempts have been made to approximate their behaviour. Coull et al. [l-3] and Heidebrecht and Swift [4] analysed simple and coupled shear walls using discrete and continuous elements. The important problem of the interaction of shear walls and 2D frame systems was analysed by Khan and Sbarounis [5], McLeod [6], Gould [7] and others. This problem was the subject of extensive research until methods and programs for forming the stiffness of a complete 3D structure were established. The programs TABS and ETABS developed at Berkeley by Wilson and co-workers [8,9] and the work of Weaver and Brandow [lo, 1I], were both based on the work of Clough et al. [ 121which offered the general framework for the discrete element approach in building analysis. Rutenberg et al. [13] and others have analysed the behaviour of shear walls under dynamic loading. Stafford Smith and Girgis [14] developed two frame models for shear wall analysis, the braced wide-column analogy and the braced frame analogy. The efficiency and accuracy of these elements has been proved quite satisfactory in comparison with plane stress finite element analysis.
t To whom correspondence
should be addressed.
2. ANALYSIS OF BRACED WIDE-COLUMN
The braced wide-column (Fig. 1) is similar to the ordinary wide-column with additional diagonal braces. The member properties of the frame are determined1141 by simulating the bending, shear and axial stiffness of the braced wide module with that of the corresponding wall segment (l), (2), (3), respectively, as follows:
I, = I,, = $
i2EI, + 2EA,
COS* 0
=-
btG
I
h’
$S+
(1)
2EA, sin’ 0 I
(2)
h EA =L h
’
(3)
where E is Young’s modulus of the braces and the column, G = E/(2(1 + v)) is the shear modulus, v is the Poisson ratio, A, = bt the cross-sectional area, and 0 the angle between the horizontal beam and the 613
V. K. KOUMOUSIS and G.
614
AG. PEPPAS
Thus the validity of the model is verified for a ratio of height-to-width between the limits
The stiffness matrix in the major plane of the wall
L-----b------7
Fig. 1. A braced frame module.
diagonal bracing. Thus from relations (2) and (3) we obtain Ad= &
(4B - 0.5)
A, = bt (2 - 8B),
(4) (5)
where B is defined as B=
h2 16b2(1 + v)’
(6)
From relations (4) and (5) it becomes evident that the area of the cross-section A,, of the diagonal bracing becomes negative when h/b c (2( 1 + v))‘/~ while the area of the cross-section of the column A, is negative when h/b > (4(1 + v))“‘.
The nodal displacements and forces of the modified 3D module are numbered as shown in Fig. 2. This numbering groups first the degrees of freedom which correspond to the displacements of a grid and then the degrees of freedom of a diaphragm whose plane is normal to the z axis. This choice facilitates the process of stiffness formation following the concept introduced by Weaver and Brandow [ 10, 111. The stiffness matrix of the braced wide-column 2D module can be determined in terms of the nodal displacements and forces that correspond to a 2D column by using the definition of the stiffness coefficient. The coefficient S, expresses the component of force corresponding to the ith component of displacement which is required in order to displace the module in a deformation pattern involving a unit displacement Sj = 1 while all the other displacements are kept to zero. Thus each stiffness coefficient for the 2D module can be determined by subjecting to the previously described deformation patterns and computing the required components of force which must be applied at the nodes of the module in order to keep it in equilibrium. In computing the internal forces in the column and the braces of the module its underformed configuration is considered. The equivalent nodal forces are obtained by appropriate transformation.
Fig. 2. Nodal displacements and forces of wall element.
Shear wall analysis using stiffness matrices For 6,=1
and Sj=0,j#2
b/2
s,, = 0
$2 = 0
$+A,sinecos’B
S10,2= -E
6,
d2.
>+A,,sinBcos2B >
For 6, = 1 and 6, = 0, j # 3 s,, = 0
A
Ad
S,, = E L + 2 T sin3 0 ~5~ h
s,, = 0 h
s,, = 0
h.3 = 0.
For ~5,= 1 and di = 0, j # 7
61 . --f+AdsmBcos2~ h2
6,
s,, = 0 61
.
6,
~+Adslnecos2e h2
s,, = 0 121 2A, +cos3 e s,, = E h3 b
6,
6,.
-b------T
615
V. K. KOUMOUSIS and G. AG. PEPPAS
616
For 6, = 1 and 6, = 0, j # 5
:+%sin’Bcos0
a5 >
s,, = 0
I
S,, = E ?
+ y
sin2 9
cos
0
6,
h
>
s,, = 0 S,, = E $
1
+ A,, sin 0 cos’ 0 6, >
r--
s,o,s=-E
b
---
>
$+A,sinOc0s82
6,.
For 6, = 1 and 6, = 0, j # 6 s,, = 0
s,,=-E
$-Fsin30
6,
(
> h
s,, = 0 S, = E 2 + y
sin3 fI 6, >
.’
-I
i
s,, = 0
_ b
9)...
f
&e =l
I
S10.6 = 0.
For 6,, = 1 and S, = 0, j # 10
s,,,, = -E
S,,,, = -E
+ A, sin 0
$
COS'
0
~$0
+ A,sin 0 cos2 9 &, >
s,,,il = 0
s,,,,,=
S
,,,,o
=
-E
E
!$ f 7 2 +7
> e >
COS3 e &,
COS' al,,.
-b
,40=1
Shear wall analysis using stiffness matrices
Thus the local stiffness matrix of the module in the wall plane becomes s 2.2
0
s2.5
0
s2.1
S2.M
0
s3.3
0
S3.6
0
0
0
s5.5
0
s5,7
s5,,0
S6.3
0
S6.6
0
0
s 7.2
0
s7.5
0
s7.7
s7.10
S 10.2
0
SlO.5
0
s10.7
s,o.,o
s 5.2
PI =
0
(8) The behaviour of the other two modules introduced by Stafford Smith and Girgis [14] namely the braced frame module in symmetric and asymmetric form is identical to that of the braced wide-column. The main reason for the introduction of these modules was to render more efficient their use in programs for the analysis of planar frames by introducing fewer nodes and members. Thus, these modules were intended for more general cases where the wall is subdivided into fractional wall width modules and multi-wall assemblies.
s I.1
PI =
0
0
s,,4
0
617
The properties of the members of these modules are determined by equating the behaviour of the wall segment with that of the braced frame module [ 141. By choosing as nodal displacements those of the middle of the top and bottom beams of the module it turns out that the stiffness coefficients of the wide-braced column module and those of the braced frame module are identical and thus there is no reason to use two distinct modified modules. Consequently below only the relations describing the response of the brace wide column module are established. 3. THE
STIFFNESS MATRIX
IN THREE DIMENSIONS
In order to form the 3D stiffness matrix of a shear wall the stiffness matrix (8) is augmented to incorporate the response of the module in the plane perpendicular to the wall plane. The stiffness coefficients in this direction are those of a standard column. As nodes we choose the middle of the top and bottom rigid beams. Thus the stiffness matrix of the modified 3D module is
0
0
S1.8
0
0
Sk,,
0 0
o
s2.5
o
s 2.7
0
0
s,,,,
0
0
0
S3.6
0
0
0
0
00
0
s4,4
0
0
0
s43
0
0
s5,
o
o
s5.5
o
s 5.1
0
0
0
0
S6.3
0
0
0
s7,2
0
0
s7,5
0
s 7.7
s 8.1
0
0
s*,4
0
0
0
0
0
0
0
0
0
0
0
s,o.z
0
0
s,o.5
0
s 10.7
s 11.1
0
0
0
0
0
0
0
0
0
o
s2,
0
0
s 4.1
0
o
o s,,,
s,,, 0
0
0
0
s,,,,
000
&I
s4,,,
0 0
%*
0
00
s,,,,
0
St,,
(9) 0
0
0
0
0
SW
0
0
SW2
0
0
s,o,,o
0
0
0
S11.8
0
0
0
0
s,2.9
0
s,,.,,
0 0
s,2,,2
where the coefficients in excess of those given by relation (8) are
S,,,=E(,+),
8.4
=
&,4=
s 12.12
s,,,
-ss.,,
=
S9.9~
S4.,=E(+),
?
&
= E
=
,
12 $ (
SH.8
S4,4=s ,,,,
S,,9 = G 2 ,
)
-s&s*
&.,I
=
S8.8,
s~,,=E(+)
S,,,, = -SK,
s,,,,
=
-
$9
(10)
V.
618
K. KOIJMOUSIS and G. AG. PEPPAS
4. TRANSFORMAnON FROM LOCAL TO GLOBAL COORDINATE SYSTEM
In order to express the nodal displacements and forces of the modified 3D module into the global vertical coordinate system (Fig. 2), the horizontal components of translations and rotations and the horizontal components of forces and moments are transformed to the global horizontal directions. This is thieved by applying the following transformation [ 151
The above transformation is not the most general one [15] it is valid for vertical elements that can only be rotated about the z axis. In a 3D frame program the components of displacement of the diaphragms are the independent components of displacement. The displacements along the X, y axis and rotation about the z axis of the top and bottom nodes of the module are not considered as independent. Thus the nodal com-
1
IsI* = VI’ PI . [A I,
(11)
where sin4
0
0
0
0
0
0
0
0
0
0
-sin4
cosf#
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
cos.4
sin f#~ 0
0
0
0
0
0
0
0
0
0
-sinr$cos4
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-sin4
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
cosfj
PI=
C$is the angle of rotation which the global x axis must be rotated about the z axis in order to coincide with the local x axis.
cos I$ sin 4 cos4
cos qt sin4 -sin+ 0
.
(12)
0
COSC$ 0 0
1
ponents of displacement must be transformed to the corresponding components of displacement of the diaphragms. This is achieved by using the following transformation [lo]
Lf?I= [Tlz[~l*[Tl
(131
189= [~l’[~l’[~l[~l[Tl,
(14)
or where
ITI =
10000000
00
0
0
01000000
00
0
0
00100000
00
0
0
00010000
00
0
0
00001000
00
0
0
00000100
00
0
0
0
0
0
0
0
0
0
0
1 0
-y
0
00000001
x0
0
0
00000000
10
0
0
00000001
01
0
00000000
00
1
X
00000000
00
0
1
-Y
(1%
Shear wall analysis using stiffness matrices
619
relation (24) can be written as
s,,, $2 s,,, s-2 0
0
0 0
31.4 $5 s2.4 Q5
s,,
0
& % 31.9 SqJO%I s2.7 s2,* 32, s2.10 %I
0 0
s,,
0
0
0
0
0
%* ~2.12
0
0
34.1
s4,2 0
34,4 34,5 0
s4,7 34.8 s4,9 ~4,lO s4,,, s4,,2
3 5.1
s5,2
0
35.4
$5,5
s5,7
$5,8
0
s,,,
0
0
$6
0
0
0
s 7.1
%,2
0
%4
%,5
0
%7
$7.8
37.9
&JO
%,
%2
s 8.1
38.2
0
%4
38,5
cl
s,,,
S8,8
S8,9
S8.10
%,
%*
s 9.1
39.2
0
39.4
$9.5
0
s9,7
$9,8
s9,9
$8,,0
$9*,,
$9.12
s 10.1
$0.2
0
3,0,4
SO.5
0
$0.7
ZO,8
3,0,9
s,O,,ll
SO.,,
SO,,,
3 11.1
h.2
0
%,4
%5
0
L,,
%,,8
G.9
&JO
%,
%,,2
0
$,2.4
s,,,
0
$2.7
s2,*
3,2,9
Lo
&,,I
s,,,*
0
VI =
0
$5,9
s5,,0
35,,,
s5,,2
0
0
0 '
3 I?.., s2.2
(16)
where S,,, = C 1 cos2 cp + C2 sin’ q
S,,, = (C 1 - C2) sin cp cos cp
& = C 1 sin’ cp + C2 cos2 4
s,,, = c3
S,., = C4 cos’ cp + CS sin2 4
S,,, = (C4 - C5) sin cp cos 4
$4
%
s.,
=
34.2
$2
=
C4 sin2 f$ + C5 cos2 cp
32,,
G.5
=
32.2
s,,
=
-
=
35.4 =
c3
S,,$ = c3
g,,, = -(CO + C6) sin cp cos cp s,,2 = C6 cos2 cp - CO sin2 cp
37.4 =
%,
ST.5 =
s 8.1 =
-
&,5
s8,2
=
C6 sin2 $J + CO cos’ 4
g8, = -s,,,
s,,,
(C7 - C8) sin cp . cos 4
38.8 =
c7 sin2 cp + C8 COGfp
s9,
-Y&J
+
x38,,
-Ys8.7
+
x&,8
+ xs,,.l
s9,2 =
-Y&.2
sb.5 =
-Y%.2
-
39,7 =
-Y&.7
&
C7(y cos cp -x
=
$8,4 =
S,,, =
s,,, = -.&,
X&J
g,,, = C7 cos2 4 + C8 sin2 4
ST.2
-
+
x$,
xs8,,
sin cp)’ + CS(y sin 4 +x cos cp)*+ C9
=
s9,8 =
%*,
=
- %
3 10.2=
- %2
SO.4 =
- %,
so.5 =
&o.,
- %,
'%I8 =
-s,,
s,0.9 =
Y&.7
%
%,,
=
-s8,,
s
s,.,
s,,
&,,
=
so.,0 =
s II.4=
- s8,,
%5
=
s 11.8=
- 38,8
$1.9
= Ys8.7 -
s,,,,, =
s,,8
$2.,
=
xs8,8
1.2=
=
- 37.2
-
-s8,7
s,,,o =
$7
-s,,,
&2,2 =
- $9,2
3 12.4=
- 39.,
$2.5 =
- s9,2
s2.7 =
- 519.7
s 12.8=
- 39,
s,2,9 =
- $9.9
s,,o
39.7
s 12.11=
s9.8
5r12.12=
s.9.
=
xs8.,
(17)
V. K.
620
KOUMOUSIS and G. AC. PEPPAS
nodes=10
nodes=6
members=14
members=12
a)
braced
wide
b) braced
column
nodes=6 members=6 c) modified
frame
module
Fig. 3. Structural idealization using different modules. The constants C8
and
C9
CO,
Cl,
C2,
C3,
C4,
C5,
C6,
C7,
used in relations above are defined as
CO = E(61,,/h2) C 1 = E(41,,/h) C2 = E(41,/h
+ bAdsin 0 cos e/2)
C3 = E(A,/h
+ 2A,sin’f?/h)
C4 = E(2Z,,/h) C5 = ,5(21,/h + bAdsin
0 cos e/2)
C6 = - E(61,. /h2 + A, sin 0 cos2 0) C7 = E(121,,/h2+
in the routines WALL and WALL1 written in Fortran 77 which are given in the Appendix. These routines can be used to calculate the matrices in relations (16) and (19) which can be incorporated in a 3D frame program that introduces as independent components of displacement those of the diaphragms. These routines can be introduced to other programs for buildings with diaphragms such as ETABS and Super ETABS by appropriately renumbering of the degrees of freedom. In addition by eliminating the transformation described in relation (13) the resulting elements can be introduced to a general 3D frame program. Finally using only the coefficients that appear in relation (8) the resulting elements can be incorporated into a 2D frame program. 7. REFINEDMODELLING
2A,cos3 B/b)
C8 = E(12Z,,/h3)
C9 = GI;_/h.
(18)
The stiffness matrix of the module given in relation (16) depends on the selection of the reference point of the diaphragm. Thus this stiffness matrix characterizes the wall module as well as its position in the structure.
In order to establish a more refined solution [14], modules of fractional width and height (see Fig. 3) can be employed where the modules depend on the dimensions of the corresponding segment. For these cases the use of the modified module (Fig. 3c) presented in this paper results in a smaller number of nodes and members than those of the braced wide-columns (Fig. 3a) and the braced frame modules (Fig. 3b). 8. FURTHERGENERALIZATION
5. MEMBERFORCES Having established the components of displacement which characterize the response of a building for a given loading, the nodal actions at the top and bottom of each module can be determined on the basis of the following relation [ 151 {A } =
Pl[~lP-l{~},
In the case of adjacent modules the number of independent nodal displacements can be reduced further by considering the displacements at the two hinges as independent (Fig. 4). Assuming the beams
(19)
where (A} is the local matrix of nodal forces and {b} is the global matrix of nodal displacements. 6. COMPUTERIMPLEMENTATION In order to avoid the matrix multiplications in relations (14) these operations were performed in symbolic form. The results are presented in the closed form relations (17) and (18). Similar expressions are developed for relation (19). These relations were used
I-b-
Fig. 4. Nodal displacements and forces of wall element.
i1
621
Shear wail analysis using stiffness matrices
as rigid, the nodal displacements - u”
are expressed as
VU 8”
0
000
0
112
l/200
0
0
l/6
0
0
010
VO
0
0
0
0
l/2
. o”
0
0
0
0
l/6
U@
Moreover the components relation
1
0
=
-I/b
0
0
0
(20)
0 l/2 -l/b
of nodal actions are transformed to those at the two hinges using the following
0 0
N”
0
M”
0
Q0
-l/b00 010
or
Q”
000 I/b00
0
0
l/2
0
0
l/2
f/b -I/b
-
) .
(21)
NO
_MO
the nodal components of displacement and the nodal components of forces at the hinges {a) = [R]‘{A >.
@I =
Substituting relations (20) and (21) into relation (8) we obtain the stiffness matrix of the module relating
~~l~~l[Rl~~~.
PI
The nodal components of displacement 16) are the same as those used for the panel element in the
r1.8-44.2----7--SJi~3.9-4.s~
Fig. 5. A ten-storey building.
V. K. KOUMOUSIS and G. AG.
622
F'EPPAS
Table 1. Dimensions of the cross-section of vertical elements Floor
10,9,8, 7 40 x 160 x 30 x 30 x
40 (cm2) 25 70 50
6,534
332
I
50 x 50 160x25 35 x 70 30 x 60
55 x 55 180x30 40 x 70 35 x 60
60 x 60 200x30 45 x 70 40 x 60
Fig. 6. Bending moments of shear walls and their percentage difference.
Shear wall analysis using stiffness matrices TABS, ETABS and Super ETABS program. However, the stiffness coefficients are different. Following the above transformation the stiffness of the module in the plane of the wall is determined with respect to three displacements at the top and bottom of the module. In a plane stress finite element eight components of displacement are intro-
623
duced two for each corner node of a rectangular element. 9. NUMERICALRESULTS The stiffness matrix of this modufe was used in a program for the analysis of multistorey buildings. to 6_
___
Well
Column
Fig. 7. Shearing forces of shear walls and their percentage difference.
V. K. Kou~ousrs and G. AC. F%PPAS
624
This program considers the floors as diaphragms by introducing directly their three components of displacement. The general method was developed by Weaver and Brandow introduced in [lo] and [l 11. Moreover, various modifications have been made in order to extend the method to non-canonical topologies. The vertical elements of every storey are treated as 3D elements and their stiffness matrices are determined from relations (16) and (17).
Using relation (23) the horizontal seismic forces for the example are: H 1 = 44 kN, H2 = 87 kN, H3 = 131 kN, H4= 175kN, H5=218kN, H6=262kN, H7 = 305 kN, H8 = 350 kN, H9 = 393 kN, HlO = 436 kN. The dead load corresponds to a specific weight y = 24 kN/m3 while the floor thickness is d = 0.18 m. The external walls for this storey height weight q1 = 10 kN/m and the internal wall q2 = 5 kN/m. The live load is p = 2 kN/m2.
Example 1
The loading cases are:
Using this program, the ten-storey building of Fig. 5 was analysed twice. First by using the standard wide-column analogy and then by the modified 3D module established in this paper. The storey height is 3.2 m. The dimensions of the vertical elements change according to Table 1. The foundation consists of a grid and footings on elastic foundation with k = 300,000 kN/m3. The grid is of the same plan as Fig. 5 and the dimensions of its beams are 30/60. The building was analysed for five different load cases. The total seismic force 6N is distributed along the height of the building in a triangular way. According to the Greek Code the equivalent horizontal static loads are determined by the following relation
1. 1.35 x dead load + 2. 1.00 x dead load + + 1.40 x earthquake 3. 1.OOx dead load + + 1.40 x earthquake 4. 1.00 x dead load + - 1.40 x earthquake 5. 1.00 x dead load + - 1.40 x earthquake
c NJ, i=,
(23)
where 6 is the seismic coefficient, N is the total vertical load at the base, N, is the vertical load at level z and hl is the height of level z starting from the foundation level.
1.50 x live 0.50 x live along X. 0.50 x live along Y. 0.50 x live along X. 0.50 x live along Y.
load. load load load load
In Fig. 6 the bending moments M, of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide column analogy are presented. In Fig. 7 the shearing force Qx of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. Example 2
The five-storey building of Fig. 8 was analysed using the same program. The vertical elements of this
r1.8-4.2----r-----50---r---33,9-
I
5.50 K840x30
I 2.10 K1040x30
i
4.05
i
Fig. 8. A five-storey building.
625
Shear wall analysis using stiffness matrices
building are considered fixed at the base. The height of every storey is 3.2 m. The columns are 40 x 40 in the first storey and 35 x 35 in the other storeys. The remaining vertical elements are kept constant in all storeys. The horizontal forces are H 1 = 67.67 kN, H2 = 135.33 kN, H3 = 203 kN, H4 = 270.67 kN, HS =
338.33 kN. The vertical loads and the loading cases are the same as for example 1. In Fig. 9 the bending moments My of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide-column analogy are presented.
Difference
s
Diffgence
%
Oe
Fig. 9. Bending moments of shear walls and their percentage difference.
V. K.
626
KOUMOUSIS
In Fig. 10 the shearing force QX of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. As a third example a three-storey building of the same plan as example 2 was used in this analysis.
and G. AC.
F’EPPAS
In Fig. 11 the bending moments MY of shear walls 2, 5 and 13 obtained on the basis of the method presented in this investigation and the percent difference of these moments with those determined using the standard wide-column analogy are presented. In Fig. 12 the shearing force QX of shear walls 2, 5 and 13 obtained on the basis of the method
,
1
4-I
To
$
4
ii f? :
E 02 i?i 1
0_ -
-0
ti
Difference
I
Fig. IO. Shearing forces of shear walls and their percentage difference.
1
1
627
Shear wall analysis using stiffness matrices
_a
II
0-U
I
0
-20
-10
Difference
Difference
Fig. Il. Bending moments of shear walls and their percentage difference.
presented in this investigation and the percent difference of these forces with those determined using the standard wide-column analogy are presented. The stress resultants and the percent error of their difference are given in Figs 7-12 side-by-side in order to distinguish the logistic error and the significant one for the design of shear walls. CAS 43/4-B
IO.
CONCLUSIONS
The stiffness matrix of a 2D frame type module introduced by Stafford Smith and Girgis for shear wall analysis is presented as a modified 3D column module. This module can be used in 3D programs for analysing buildings. Moreover the stiffness matrix of
628
V. K. Kou~ous~s and G. AG. PEPPAS
6
iFi
_I
0
-IS
- 0
4
Difference
s
I
5
oJ Differ&e
1
Fig. 12. Shearing forces of shear walls and their percentage difference.
these modules was modified to take into account the influence of the remote nodal displacements and forces of the diaphragms. From the numerical results obtained in this investigation the following conclusions become evident: 1. The difference between the stress resultants determined using the modified element and the
wide-column analogy are important for medium rise buildings. 2. The difference between the stress resultants and determined using the modified element presented in this analysis and the wide-column analogy is of the order of 20% in the shearing forces of a five-storey building.
Shear wall analysis using stiffness matrices 3. The differences appear on the unsafe side as well as to the safe side depending on the floor level without a specific pattern.
8.
4. The difference is more pronounced in the shearing force in the direction of the wall while the
9.
difference in bending moments is not so critical. 5. The max significant difference which occurred was 21% on the unsafe side in the base shear of a three-storey building.
10.
In general the use of the more accurate modified module determines larger values for t’ L base shear of shear walls than the wide-column analogy.
12.
11.
13. REFERENCES I.
2. 3. 4.
5.
6. 7.
A. Cot111and A. W. Irwin, Model investigations of shear wall structures. J. Srrucf. Div., ASCE 98, 1223-1237 (1972). A. Coull and J. R. Choudhury, Stresses and deflections in coupled shear walls. ACZ Jr1165-72 (1967). A. Co1111and J. R. Choudhury, Analysis of coupled shear walls. ACZ Jnl September, 587-593 (1967). A. C. Heidebrecht and k. D. Swift, Analysis of asymmetrical coupled shear walls. J. Sfrucr. Div., ASCE 97, 1407-1422 (1971). F. R. Khan and H. A. Sbarounis, Interaction of shear walls and frames. J. Struct. Div., AXE 90, 285-335 (1984). I. A. McLeod, Connected shear walls of unequal width. ACZ Jnl May, 408412 (1970). Ph. L. Gould, Interaction of shear wall-frame systems
APPENDIX:
C
CODE IN FORTRAN
14.
15. 16.
17.
18.
in multistorey buildings. J. Am. Concrete Inst. January, 45-69 (1965). E. L. Wilson and H. H. Dovey, Three-dimensional analysis of building systems-TABS. Report No. EERC 72-8, December (1972). E. L. Wilson, J. P. Holhngs and H. H. Dovey, Threedimensional analysis of building systems (extended version). Report No. EERC 75-13, April (1975). W. Weaver, Jr, Three-dimensional analysis of tier buildings. J. Struct. Div., ASCE 92, 385404 (1966). W. Weaver, Jr and G. E. Brandow, Tier buildings with shear cores, bracing, and setbacks. Compuf. Sfruct. 1, 57-83 (1971). R. W. Clough, I. P. King and E. L. Wilson, Structural analysis of multistorey building. J. Sfrucf. Div., ASCE, August (1963). A. Rutenberg, W. K. Tso and A. C. Heidebrecht, Dynamic properties of asymmetric wall-frame structures. Earthquake Engng Struct. Dynamics 5, 41-51 (1977). B. Stafford Smith and A. Girgis, Simple analogous frames for shear wall analysis. ASCE J. Sfruct. Engng 110, 2655-2666 ( 1984). A. E. Armenakas, Modern Structural Analysis. The Matrix Method Approach. McGraw-Hill (1991). L. A. Bejar, Seismic analysis of multistory building systems using microcomputers. ASCE J. Computing Civil Engng 3, 1-17 (1989). Al. Carpinteri and An. Carpinteri, Lateral loading distribution between the elements of a three-dimensional civil structure. Comput. Strucr. 21, 563-580 (1985). C. F. Neuss and B. F. Maison, Dynamic analysis of a forty-four storey building, ASCE J. Strucr. Engng 111, 1559-1572(1985).
77. SUBROUTINES
WALL AND WALL1
SUBROUTINEWALL(E,G,AR,IX,IY,IZ,H,CF,SF,X,Y,S) REAL MI,IC,IX,IZ,S(I2,12) MI=(E/(2*6))-1 B=SQRT(12. *IY/AR) ______________________~~_~~~~~~~~~~~~~~~~~~~~~~~~~~~~_~~ H = HEIGHT B = WIDTH T = THICKNESS E = ELASTIC MODULUS G = SHEAR MODULUS MI = POISSON RATIO AC, AD = CROSS SECTIONALAREA OF COLUMN AND DIAGONAL BRACES IC = MOMENT OF INERTIA OF COLUMN SECTION (Iyy) IX = MOMENT OF INERTIA OF WALL (Ixx) = TORSIONAL MOMENT OF INERTIA ABOUT Z AXIS ZETA = ANGLE OF DIAG BRACES BETA = AUXILIARY CONSTANT [S] = STIFFNESSMATRIX ____________________~_~~_~~~~~_~~~~~~~~~~~~~~~~~~~~~~~~~ THETA=ATAN(H/B) BETA=H*H/(16*B*B*( 1tMI)) **************************************** **tk**********
WALL
629
*****************
*******¶t********************************
WALL PROPERTIES IC=AR*B*B/12. AC=AR*(2.-8.*BETA) ADsAR*(4.*BETA-.S)/(SIN(THETA))**3
V. K. KOUMOUSIS and G.
630
Ac. PEPPAS
CSF=COS(F)*SIN(F) SF=SIN(F), CF=COS(F), CFZ=COS(F)**2, SF2=SIN(F)**2 Y = Y COORD. X = X COORDINATE CSF=CF*SF CF2=CF*CF SF2=SF*SF FREQUENTLY-USED PARAMETERS CO=E*(6.*IX/H**2) Cl=E*(4.*IX/H) C~IE*(~.*IC/H+B*AD*(SIN(THETA))**~*COS(THETA)/~.) C3=E*(AC/Ht2.*AD*(SIN(THETA))**3/H) C4=E*(2.*IX/H) C5=E*(2.*IC/HtB*AD*(SIN(THETA))**2*COS(THETA)/2.) C6=-E*(6.*IC/H**2tAD*SIN(THETA)*(COS(THETA))**2) C7=E*(12.*IC/H**3+2.*AD*(COS(THETA))**3/6) ~~=~*~;~H*IX/H**J) z * MODIFIED GLOBAL STIFFNESS MATRIX : DO 10 K1=1,12 DO 10 K2=1,12 S(Kl,K2)=0 10 CONTINUE
C S(l,l)=Cl*CF2tC2*SF2 S(2,1)=(Cl-C2)*CSF S(2,2)=Cl*SF2tC2*CF2 S(3,3)=C3 S(4,1)=C4*CF2tC5*SF2 S(4,2)=(C4-C5)*CSF S(4,4)=S(l,l) S(5,1)=S(4,2) S(5,2)=C4*SF2tC5*CF2 S(5,4)=S(2,1) S(5,5)=S(2,2) S(6,3)=-C3 S(6,6)=C3 S(7,1)=-(COtC6)*CSF S(7,2)=C6*CF2-CO*SF2 S(7,4)=S(7,1) S(7,5)=S(7,2) S(7,7)=C7*CF2tC8*SF2 S(8,1)=-C6*SF2tCO*CF2 S(8,2)=-S(7,l) S(8,4)=S(8,1) S(8,5)=S(8,2) S(8,7)=(C7-C8)*CSF S(8,8)=C7*SF2tC8*CF2
S(9,1)=-Y*S(7,1)tX*S(8,4) s(9,2)=-Y*s(7,2)-x*s(7,1) S(9,4)=-Y*S(7,1)tX*S(8,1) s(9,5)=-Y*s(7,2)-x*s(7,1) S(9,7)=-Y*S(7,7)tX*S(8,7) S(9,8)=-Y*S(8,7)tX*S(8,8)
S(9,9)=C7*(Y*CF-X*SF)**2+ lC8*(Y*SFtX*CF)**2tC9 s(10,1)=-s(7,1) S(10,2)=-S(7,2) S(10,4)=-S(7,l) S(10,5)=-S(7,2) S(10,7)=-S(7,7) S(10,8)=-S(8,7) S(lO,S)=Y*S(7,7)-X*S(8,7)
[Sc]=[Tc]'[Lc]'[S][Lc][Tc]
Shear wall analysis using stiffness matrices S(10,1O)=S(7,7) S(ll,l)=-S(8,l) S(11,2)=S(7,1) S(11,4)=-S(8,l) S(11,5)=S(7,1) S(11,7)=-S(8,7) S( 11,8)=-S(8,8)
S(ll,S)=Y*S(8,7)-X*S(8,8) S(ll,lO)=S(8,7) S(ll,ll)=S(8,8) S(12,1)=-S(9,l) S(12,2)=-S(9,2) S( 12,4)=-S(9,l) S(12,5)=-S(9,2) S(12,7)=-S(9,7) S( 12,8)=-S(9,8) S(12,9)=-S(9,9) S(12,1O)=S(9,7) S(12,11)=S(9,8) S(12,12)=S(9,9) C
DO 15 K1=1,12 DO 15 K2=Klt1,12 S(Kl,KP)=S(KZ,Kl)
15 CONTINUE RETURN END
==I=E===========I===D=-========I===P========================~=========
SUBROUTINE WALLl(E,G,AR,IX,IY,IZ,H,CF,SF,X,Y,S)
C REAL MI,IC,IX,IZ,S(12,12) MI=(E/(2*G))-1 B=SQRT(l2.*IY/AR)
____________________~~__~~~_~~__~~~~~__~~~_~~_~~__~_____ H - HEIGHT 8 - WIDTH T = THICKNESS E = ELASTIC MODULUS G = SHEAR MODULUS MI = POISSON RATIO AC, AD = CROSS SECTIONAL AREA OF COLUMN AND DIAGONAL BRACES IC - MOMENT OF INERTIA OF COLUMN SECTION (Iyy) IX = MOMENT OF INERTIA OF WALL (Ixx) IZ - TORSIONAL MOMENT OF INERTIA ABOUT Z AXIS THETA = ANGLE OF DIAG. BRACES BETA = AUXILIARY CONSTANT [S] = STIFFNESS MATRIX THETA=ATAN(H/B) BETA=H*H/(16*B*B*(ltMI))
**************************************** id***********
WALL-1
*****************
****************************t*************
WALL
PROPERTIES
IC=AR*8*8/12. AC=AR*(2.-8.*BETA) AD=AR*(s.*BETA-.5)/(SIN(THETA))**3 CF=COS(F), SF=SIN(F) X = X COORDINATE Y - Y COORD. FREQUENTLY USED PARAMETERS CO=E*(6.*IX/H**2)
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V. K. K~UMOUSIS and G. AG.
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PEIPPAS
Cl=E*(4.*IX/H) C2=E*(4.*IC/HtB*AD*(SIN(THETA))**2*COS(THETA)/2.) C3=E*(AC/Ht2,*AD*(SIN(THETA))**3/H) C4=E*(2.*IX/H) C5=E*(2.*IC/Ht8*AD*(SlN(THETA))**2*COS(THETA)/2.) ~6=-E*(6.*IC/H**2tAD*SIN(THETA)*(~OS(THETA))**2) C7=E*(l2.*~C/H**3t2.*AD*(&OS~THETA))**3/6~ C8=E*(12.*IX/H**3) C9=G*IZ/H MATRIX : C
IScl=[Sl [Lcl[Tcl
DO 10 K1=1,12 DO 10 K2=1,12 S(Kl,K2)=0 10 CONTINUE
s(l,l)=cl*cF S(1,2)=Cl*SF S(1,4)=C4*CF S( 1,5)=C4*SF s(1,7}=-co*sF S( 1,8)=co*cF s(1,9)=co*Y*sF S(l,lO)=-S(1,7 S(l,ll)=-S(1,8 S(1,12)=-S(1,9 S(2,1)=-C2*SF S(2,2)=C2*CF S(2,4)=-C5*SF S(2,5)=C5*CF S(2,7)=C6*CF S(2,8)=C6*SF
tCO*X*CF
S(2,9)=-C6*Y*CFtC6*X*SF S(2,10)=-S(2,7) S(2,11)=-S(2,8) S(2,12)=-S(2,9) S(3,3)=C3 S(3,6)=-C3 S(4,l)=C4*CF S(4,2)=C4*SF S(4,4)=S(l,l) S(4,5)=S(1,2) S(4,7)=S(1,7j S(4,8~=S{l,8) S(4,9)=S(1,9) S(4,10)=-S(1,7) S(4,11)=-S(1,8) S(4,12)=S(1,12) S(5,1)=S(2,4) S(5,2)=S(2,5) S(5,4)=S(2,1) S(5,5~=S~2,2~ S(5,7)=S(2,7) S(5,8)=S(2,8) S(5,9)=S(2,9) S(5,10)=-S(2,7) S(5,11)=-S(2,8) 5(5,12)=5(2,12) S(6,3)=-C3 S(6,6f=C3 S(7,1)=-S(2,8) S(7,2)=S(2,7) S(7,4)=-S(2,8) S(7,5)=S(2,7) S(7,7)=C7*CF S(7,8)=C7*SF S(7,9)=-C7*Y*CFtC7*X*SF
Shear wall analysis using stiffness matrices S(7,10)=-S(7,7) S(7,11)=-S(7,8) S(7,12)=-S(7,9) S(8,1)=S(1,8) S(8,2)=-S(1,7) S(8,4)=S(8,1) S(8,S)=S(8,2) S(8,7)=-C8*SF s(8,8)=c8*cF
S(8,9)=C8*Y*SFtC8*X*CF S(8,10)=-S(8,7) S(8,11)=-S(8,8) S(8,12)=-S(8,9) S(9,9)=C9 S(9,12)=-c9 S( lO,l)=S(2,8) S(10,2)=-S(2,7) S(10,4)=S(2,8) S(10,5)=-S(2,7) S(10,7)=-S(7,7) S(10,8)=-S(7,8) S(10,9)=-S(7,9) s(10,10)=s(7,7) S(lO,ll)=S(7,8) S(10,12)=S(7,9) S(ll,l)=-S(8,l) S(11,2)=-S(8,2) S(11,4)=-S(8,l) S(11,5)=-S(8,2) S(11,7)=-S(8,7) S(11,8)=-S(8,8) S( 11,9)=-S(8,9) S(ll,lO)G(8,7) S(ll,ll)=S(8,8) S(11,12)6(8,9) S(12,9)=-c9 S(12,12)=C9 RETURN END
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