- Email: [email protected]
Two-stage solution for framed tube buildings
Compurers& S~rurrurrv Vol. SO.No. 5. pp. 655663. 1994 Copyright 8’) 1994Elsevier kience Ltd Printed in Great Britain.All rights remrved 0045-794919416.00+ 0.00
Pergamon
TWO-STAGE SOLUTION FOR FRAMED TUBE BUILDINGS Y.
SINGH
and A. K.
NAGPAL
Department of Civil Engineering, IIT Delhi, Hauz Khas, New Delhi-110016, India (Received 11 September
1992)
Abstract-A two-stage solution procedure is presented for the lateral load analysis of framed tubes. In the first stage dual shape functions and Gauss elimination solution are used to obtain an approximate solution. In the second stage a Gauss-Seidel iterative solution is used to obtain an accurate solution. The proposed two-stage solution procedure only requires a small fraction of the CPU time needed in the standard Gauss elimination solution. Computer storage locations required are also smaller than those in the standard solution. With the use of the two-stage solution procedure analysis of framed tubes of practical sizes can he carried out on PCs.
1. INTRODUCTION
Tubular systems are economical for resisting loads for mediumto high-rise buildings.
lateral
In its basic form, a tubular building is designated as a framed tube and consists of closely spaced columns on the periphery interconnected by deep spandrel beams (Fig. 1). Under lateral loading the framed tube undergoes shearing and flexural deformations. Closely spaced columns on the periphery provide large flexural stiffness. With the increase in stiffness of spandrel beams, the behaviour of the framed tube approaches that of the closed tube; the framed tube may also thus be viewed as a perforated tube. The computational work and computer storage in the displacement method of analysis for a framed tube modelled as a space frame is very large due to a large number of joints. Use of an equivalent plane frame model [l] reduces the computational and storage requirements. In this model web frames and flange frames interact through only vertical degrees of freedom (DOF) at common joints; to realize this, horizontal fictitious members are introduced between flange frames and web frames (these only transfer vertical shear forces). The continuum approach, in which the perforated tube is converted into an equivalent continuum, has also been used. A method in which web frames and flange frames are modelled as equivalent orthotropic plates is available for framed tubes with uniform properties along the height [2]. Closed-form solutions for stress resultants are obtained. The method is subsequently extended for bundled tubes[3]. In another continuum method, flange displacements are assumed to vary in a parabolic manner to cater for shear lag effects, shearing deformation is incorporated and the frame tube is converted into an equivalent cantilever tube [4, 51. The method predicts
deflections quite well but not the stresses and has been recommended for ascertaining only overall behaviour at a preliminary design stage. The continuum approach is unsuitable for predicting deflections and stresses for frame tubes with varying member properties at the final design stage. Equivalent plane frame idealization is well suited for practical problems as it caters for varying member properties. Generally, equations resulting from plane frame idealization are solved by the Gauss elimination solution. The solution procedure will from now on be referred to as the ‘standard’ solution procedure. Computational time and storage requirements increase rapidly with the increase in the number of storeys and column lines. Here, a two-stage solution procedure is proposed for the equivalent plane frame idealization in which the computational work and storage requirements are only a fraction of those in the standard solution procedure. The procedure involves use of the Gauss elimination solution in the first stage and a Gauss-Seidel solution in the second stage. 2. TWO-STAGE SOLUTION PROCEDURE Figure 2 shows the plane frame idealization of a framed tube in which the symmetric quarter of the building is considered with appropriate boundary conditions indicated on the end column lines. Web frame and flange frame are shown interconnected at floor levels by fictitious horizontal members. In the first stage the vector of deformations at the joints is estimated by solving an auxiliary problem of reduced size by a Gauss elimination solution. In the second stage the displacement vector of the first stage is taken as the initial vector in the Gauss-Seidel iterative solution for the frame idealization at actual size. 655
Y. SINCH and A. K. NAGPAL
656
Fig. 3. Level I shape functions for members: (a) lateral translations and rotations and (b) vertical translations.
Fig. 1. Symmetric quarter plan of a framed tube building.
2. I. Stage 1 solution The auxiliary problem is formed by dividing the equivalent plane frame into a number of segments, each segment spanning a number of storeys (Fig. 2). Within each segment three reference floors are
defined, one each at the bottom floor level and the top floor level and the remaining at an intermediate floor. Two levels of DOF and corresponding shape functions are identified. Level 1 DOF, as usual, comprise a translational DOF at each floor level and a vertical and a rotational DOF at every joint of floor levels. Level 1 shape functions define member deformations locally over member lengths (Fig. 3). Level
r
Fictitious members
T 9
I 2
r 1
1
-
1
21
Fig. 2. Equivalent plane frame, segments and reference floors.
657
Two-stage solution for framed tube buildings Iavoll d.of.: (ot. vi, +--a) Leved2d.of.: (IB~,~&--83)
Fig. 4. Level 2 shape functions for segments. 2 DOF are identified at selected reference floors only, as shown in Fig. 2. Each reference floor also has a translational DOF and every joint on it has a vertical and rotational DOF. Corresponding level 2 shape functions define the overall behaviour of a segment and are presently assumed to be quadratic as shown in Fig. 4. These shape functions define the relationship between level 1 and level 2 DOF which may be expressed as 3 u’=
Npf
c
(la)
Fig. 5. Level 1 DOF of a column element and associated level 2 DOF of the segment.
j-1
v’=
1 N,vf
where L,, L2 and x are the distances of intermediate reference floor, top reference floor and an intermediate floor of the segment, respectively. Figure 5 shows a column mn of storey length within a segment height. The relation between the vector of level 1 DOF at ends 1 and 2
(lb)
j=l
8’= t: N,ef, j=l
where u’, v I, 0 ’ are level 1 DOF (lateral displacement, vertical displacement and rotation, respectively, at a joint of a floor), uf, vj, 0; are the corresponding level 2 DOF at thejth reference floor of the segment and Nj are the shape functions given as
N
I
p,
-xl@*-x)
and the vector of level 2 DOF of the segment on the same column line (8) +:
(24
LlL2
v:
e:
u: v:
e:
e:)
0:
U:
can be expressed as N2 =
-45 -x)
(2b)
L,&-Ll)
N
x(x-4)
_
3-L,(L,-L,)’
-
N,h)
0
0 0
[Tel=
N162)
0 _
0
N,(x,) 0 0 N,(xz)
0
where [7’,] is the transformation column given by
(24
0 0 N,(x,)
0 0 N,(x,)
N2h)
0
0 0 N2@2)
0 0
N2h)
0 0
0 0
N2h)
0
N2@2)
0
0
N262)
N3@,)
0 0
matrix
0
0
N,(x,)
0
0
N,(x,
for the
-
I
(4) N3(x2)
0 0
o N3(x2)
0
0
’
o N3tx2)
-
Y. SINGHand A. K. NAGPAL
658
where IV,(x,) and Nj(x2) (j = 1,2,3) are shape functions evaluated at the bottom (X = x,) and at the top (x = x2) of the column. Similarly, for the beam element (Fig. 6) the relation between the vector of level 1 DOF at ends 1 and 2,
and the vector of level 2 DOF on the two adjacent column lines {CP}= {Vi e:
0:
e:
v:
e:
v: 0:
0: e;
vi
e;}
can be represented as
matrix for the beam
Fig. 6. Level 1 DOF of a beam element and associated level 2 DOF of the segment.
where N, , N,, N3 are evaluated at the beam level (x =x+). The stiffness matrix, [K:] for a column element with respect to its level 1 DOF (see Fig. 5) is given
The stiffness matrix of a segment is obtained by adding the contributions of all the beams and columns in it. The segment lateral load vector due to lateral loads at floor levels is expressed as
where [T,,] is the transformation given by
[Tc+l=
0 EA/L 0 [K:l=
-EA/L
12EI/L’ 0 6EIIL2 0
6EI/L2 0 4EI/L 0
0 - EA/L 0
- 12EIIL’ 0 - 6EIIL2
EA/L
0
6EI/L2 0 2EIIL 0
0
-12EIjL’
-6EI/L*
0
12EIIL’
-6EIIL2
0
6EIIL2
2EIIL
0
-6EIIL2
4EIIL
where E is the modulus of elasticity, Z, A and L are the moment of inertia, area and length of the column, respectively. Deleting the first and fourth rows and first and fourth columns, and interpreting the properties I, A, and L as that of a beam, gives the stiffness matrix [K:] of the beam element with respect to its level 1 DOF (Fig. 6). Stiffness matrix [Kf] of a column and [Ki] of a beam with respect to their respective level 2 DOF are given by
_
1 ’
(7)
[NI(Xi)) (9)
where F, and xi are the lateral load at the ith floor and the distance of the ith floor from the bottom of the segment, respectively, and the summation extends over the number of floors in the segment. The stiffness matrix and the load vector of the Ktl= [T,.lTIK:~IV-,.l @a) auxiliary model are assembled from the stiffness @b) matrices and the load vectors of the segments, Kil= JT/JTKiIITJ respectively. The equilibrium equation may be Explicit expressions for terms of [KT] and [K,?,] written as can be obtained. Alternatively these matrices may be obtained numerically as triple matrix products. [K2]{D2j = {P2), (10)
659
Two-stage solution for framed tube buildings where [K2], {O’} and {P’} are the stiffness matrix displacement vector, and the load vector, respectively. The number of reference floors required to be chosen in the present method is generally smaller than the number of column lines. It is, therefore, economical to number the level 2 rotational and vertical DOF, S, at reference floors consecutively along a column line followed by numbering on the subsequent column lines in succession as shown in Fig. 2. Level 2 translational DOF, p, are numbered consecutively at reference floors. In eqn (10) translational DOF are grouped separately and the equation is then written in a partitioned form. A lateral stiffness matrix can then be obtained which could be used to evaluate free vibration characteristics. A lateral stiffness matrix is also required when the framed tube resists the lateral loading together with other stiffening elements, such as an interior core. Equation (10) is now written in the partitioned form as
Fig. 7. Level 1 DOF contributing to joint/floor equilibrium in stage 2.
Equations for the evaluation of vertical translations and rotations at nodes at floor levels and lateral translations at floor levels can be expressed as (Fig. 7) u;“+
where subscripts s and p refer to vertical and rotational DOF at joints and to translational DOF at reference floors, respectively. From eqn (11) the following relations can be written
ww:1=
{CI
UW
WI = EJ - vww’E,ol
U2b)
ICI = {PiI - ~f$IKYv3
(124
I#> = Wisl-‘({PfI - [$JP:D
(124
in which [KS] is the lateral stiffness matrix. For numerical evaluation of [Kf], [K:;] is not inverted. Recognizing the banded nature of [K:;], we can obtain [Kzz]-‘[KfP] by Gauss elimination in which columns of [Ki,] are treated as load vectors. The standard Jordan method can be used in eqn (11) as well to eliminate s DOF and yield [Kf], but in this process a greater bandwidth is required. Equations (12a) and (12d) are solved for CO’). The level 1 displacement vector is then obtained from eqn (1). 2.2. Stage 2 solution The level 1 displacement vector differs from the true displacement vector, the difference depending on the number of reference floors chosen. The former vector can be used as a starting vector in the GaussSeidel solution procedure to have an accurate displacement vector. The constraints imposed on level 1 DOF by level 2 shape functions are now relaxed at this stage.
u’j+’ c
I =
=
$i
-/vQ:+Rb+‘)/C,
ufj + j(j’,
-
Q: - Rt+‘)/C,,
R:+l=C,v~+l+C,v:'tl+c,e~'+l
(13)
(1%
(17)
Q; = C,fl;*i+ Clout’+ Cllufj + c,#+
Cl&
+ c,,ef.i
(18)
Rb+‘=C,5u~+~+~,6e~+~+~,7U)‘i+I +
Q; = C2&;.’ + R:+’ = C,,up+
w:.i+ I + c,9e;.i+l
(19)
c*,uy+ c,,ey
(20)
+ c24ey+r + c,,e;,f+r
(21)
where subscripts I and r refer to nodes at the left and right of a node C; subscripts b and t refer to nodes to the bottom and the top of the node C, or to the floors at the bottom and the top of a floor C; superscripts i and i + 1 refer to the values at the beginning and at the end of an iteration; coefficients . , C,, involve member properties; /I is the c,,c*,.. over-relaxation factor; and P, is the lateral load at the floor c. Iterations are carried out until an error norm ((E 11 is less than a specified value a. The error norm chosen [6] is defined as
Adi
I I
IIt 11= max -
di,,,,
(22)
Y. SINGH and
660
in which Ad, is the difference between two values of a displacement (i.e. at the beginning and at the end of an iteration) and di.,, is the maximum displacement of the similar type (either vertical, translational or rotational). 3. COMPUTER STORAGE AND COMPUTATIONAL WORK
In the first stage of this method the major portion of the storage required is for [K,] and [&,I (the superscript 2 has now been dropped). For q segments in the equivalent plane frame, the number of reference floors, 1 is 2q + 1 and with k column lines in the symmetric quarter plan, the number of p and s DOF are I and 2k1, respectively. Storage locations for [K,] and [J&J together are 6kl(l+ 2). In the second stage two cases arise. In case A, for each vertical DOF coefficients C, to C,, for each rotational DOF coefficients C, to C,9 and for each translational DOF coefficients C,, to C,, are stored. These require N(6 + 19k) (N = number of floors including the ground) storage locations for all the level 1 DOF. These storage locations are not needed simultaneously with those of stage 1. The storage requirement is therefore the greater of the two quantities 6kl(l+ 2) and N(6 + 19k). In case B, coefficients C, to C,, are not stored but are evaluated at every iteration from the member properties. This may have to be resorted to when working with PCs. In this case the storage requirement is that of stage 1 i.e. 6kl(l + 2). If N is greater than k, in the standard solution procedure, it is advantageous to number all the s DOF on a floor consecutively followed by s DOF at the subsequent floors. Similarly p DOF are numbered sequentially at floors. The number of s and p DOF are 2kN and N, respectively. A major portion of storage locations in the standard solution procedure is also for [K,] and [K,] and is equal to 2KN(2K + N + 2). If, on the other hand, N is smaller than k, the numbering is done in a manner similar to that for the present method. The number of storage locations for [K,] and [K,] are 2kN(3N + 2). In the present method, CPU time, T, is the sum of the times in both stages. Taking more reference levels results in increase in T, in the first stage with a decrease in the second stage. There is an optimum number of reference levels for which the total T,. of both stages is minimum. If more than one loading case is to be analysed, a higher number of reference levels (in the first stage) than that for a single loading case may be desirable. This increases time in the first stage, consequently less time would be needed in the second stage. As T, for the first stage, for most practical loading cases, is nearly equal to that for a single loading case, an increase in T,. for more than one loading case is confined to the second stage only. Again, an optimum number of reference floors may be arrived at, which results in a smaller total Tc.
A. K. NAGPAL Table 1. Cross-sectional properties of members of the 66storey example building
Cross-sectional properties Columns
Beams
Storeys/ floors
A (m*)
I (m4)
I (m4)
l-10 1l-20 21-30 31-40 41-50 51-60
0.700 0.580 0.520 0.440 0.370 0.370
0.200 0.190 0.170 0.150 0.130 0.130
0.200 0.180 0.150 0.120 0.100 0.100
4. NUMERICAL
EXAMPLES
First a 60-storey building with storey heights of 3 m, plan dimensions of 42 m (web) x 54 m (flange), and column spacing of 3 m is considered. Member properties I and A for columns and I for beams are given in Table 1 and E = 2.5 x 10” N/m2. The building is subjected to a lateral loading 234 kN at floors l-59 and 117 kN at floor 60. Results from the present method are first taken for 1 = 7 (reference floors: ground, 2, 7, 15, 27, 42, 60) and for tl = 5 x lo-‘. The spacing between the reference floors increases along the height. Lateral deflections from the present solution procedure at the first and second stages and from the standard solution procedure are shown in Fig. 8 where it is seen that deflections from the first stage are close to those from the standard solution procedure. Iterations of the second stage are not needed to estimate the deflections. Maximum member forces, the axial force, N, and bending moment M, in the columns of the first, 30th and 60th storeys, and bending moment M and shear force S, in the beams at the first, 30th and 60th floors are given in Tables 2 and 3, respectively. For columns the maximum N, occurs in the corner column at all floors and the maximum M occurs in 60’
’
123456 DoflofJtion (em)
Fig. 8. Lateral deflections for the 60-storey example building.
661
Two-stage solution for framed tube buildings
Table 2. Maximum axial force, No, and maximum bending moment, M in web columns of the 60-storey example building (a = 5 x 10-r in stage 2, I = 7) Solution procedure Present stage 1
Standard storey
I
M (kNm x lo*)
(kN?lO’) 2.160 0.324 0.009
30 60
(kN:lO’)
8.680 4.320 0.122
Stage 2
(kN&
1.986 0.256 0.050
IOr)
9.680 5.020 0.438
(kN?tO”)
(kN rn\
2.160 0.320 0.009
10’)
8.610 4.240 0.122
Table 3. Maximum bending moment, M, and maximum shear force, S, in web beams of the 60-storey example building (a = 5 x 10e5 in stage 2, I = 7) Solution procedure Present Standard
Stage 2
Stage I M
Storey 1 30 60
(kN2
10’)
(kN ,s l@)
(kN m x 10’)
(kN : 102)
(kN n?x 102)
(kN : 102)
4.660 2.830 0.079
3.010 4.860 0.502
2.010 3.240 0.359
6.950 4.160 0.122
4.630 2.780 0.079
6.990 4.250 0.122
the central web column at the first and 30th floors but in the corner column at the 60th floor. For the beams the maximum M and S occur in the central web beam at the first and 30th floors, but in the comer end beam at the 60th floor. For the present solution procedure, the forces are shown at both stages, those at stage 1 differ from those of the standard solution. When the Gauss-Seidel iterative process is applied in stage 2, these forces approach those of the standard solution. The convergence of N, and M for the first storey column with number of iterations, Ni, is shown in
Fig. 9. The value of the over-relation factor, j?, has been chosen as 1.6. Table 4 shows the variation with the number of reference levels I of CPU time, T, needed in stages 1 and 2 and of the total time which is expressed as a percentage of that in the standard solution. The CPU time is on an EL-3980 mainframe computer at IIT Delhi. Also shown in the table are the number of iterations needed in the second stage. A check on the error norm IIt 11is made after every five iteration cycles. In the table we imply by reference floors 61 that the auxiliary structure is the actual structure; the first stage solution becomes the standard solution and the second stage solution is absent. As the number of reference floors is increased, T, for the first stage increases and decreases for the second stage. For seven reference levels the total T, is minimum, equal to 27 set which is about 2.59% of that (1041 set) in the standard solution. The time
Table 4. Variation of the number of iterations, N,, in stage 2 and of CPU time, T,., with the number of reference floors in the present method for the 60-storey example building (a = 5 x 10ms)
T
0’
I
1
sa 2s No.ofltadaas
1
75
Fig. 9. Convergence of axial force, N,,, and bending moment, M, in the comer column of first storey for the 60-storey example building (a = 5 x 10es, I = 7).
I
Ni
Stage 1 (set)
5
180 90 55 40 35 3s -
10 13 19 28 41 58 1041
7 9 11 13 15 61
Stage 2 (set) 23 14 10 9 8 8 -
Total (W 33 27 29 37 49 66 1041
Percentage 3.17 2.59 2.78 3.55 4.71 6.34 -
662
Y. SINGH and A. K. NAGPAL Table 5. Cross-sectional properties of members of the 40-storey example building Cross-sectional properties Columns
Beams-
Storeys/ floors
A (mr)
I (m4)
I (m4)
I-IO I l-20 21-30 31-40
0.520 0.440 0.370 0.370
0.170 0.150 0.130 0.130
0.150 0.120 0.100 0.100
needed in the present solution procedure is thus only a small fraction of that needed in the standard solution procedure. In the table, T, for the first stage are for the case when explicit expressions for terms of [Kf] and [Ki] are used. Similarly in the second stage, the T,in Table 4 are for the case when sufficient storage locations are available, as in the mainframe computer, and thus coefficients C,-C,, are stored at the beginning of the iterative procedure in the second stage. If sufficient storage locations are not available and coefficients are evaluated in every iteration, T,in the second stage is increased; for I = 7 it increases from 14 to 22 sec. The variation of percentage errors in the maximum column force, N,, of the first storey and in total T,
other buildings, one is 40 storeys and the other is 80 storeys, both of these are of the same planform as the 60-storey example building. The member properties of the 40-storey building are shown in Table 5 and for the 80-storey building in Table 6. For these buildings the lateral load at an intermediate floor is also 234 kN and that at the respective top floors is 117 kN. Forces N, and M in columns for the 40- and 80-storey example buildings are shown in Tables 7 and 8, respectively. Optimum 1for which minimum T, is required are 7 and 9, and the number of iterations in the second stage for convergence are 55 and 65 for the 40- and 80-storey example buildings, respectively. The increase in efficiency of the present solution with a number of storeys is shown in Table 9. As the number of storeys increases, the T, required in the standard solution increases rapidly from 389 set for the 40-storey building to 2194sec for the 80-storey building. On the other hand, the T, required in the present solution increases from 17 set for the 40storey building to only 37 set for the 80-storey building. In the present solution T, required is about 4.37%, 2.59% and 1.69% of the corresponding T,
in the present solution with specified error norm c( is shown in Fig. 10. For c1= 5 x 10e5, the error in N, remains sufficiently small in all the storeys. For two loading cases, T, of stage 1 is nearly the
Table 6. Cross-sectional properties of members of the 80-storey example building
“0.5
50
5.0
500
a(x lWs) Fig. 10. Variation of precentage error in No and T,,with 11 t )I for the 60-storey example building.
same as for one loading case, i.e. 19 set for 1 = 9. For stage 2, T,is nearly doubled to 20 sec. The minimum total T, is thus 39 set and occurs for I = 9. For the present solution procedure, the minimum total T, is
Cross-sectional properties Columns Storeys/ floors
A (m2)
l-20 21-40 41-60 61-80
1.000 0.730 0.518 0.370
increased from 27 set for one loading case (1 = 7) to 39 set for two loading cases (I = 9). The efficiency of the present method with a varying number of storeys is studied by considering two
Beams I (m4)
1 (m4) 0.230 0.196 0.150 0.130
0.260 0.195 0.150 0.130
Table 7. Maximum axial force, N,,, and maximum bending moment, M, in web columns of the 40-storey example building (a = 5 x 10m5in stage 2, I = 7) Solution orocedure Present Standard Storey
(kN:lOr)
(kN
Stage I
A 10’)
(kN%O’)
Stage 2
(kN u?x IO’)
(kN?lO’)
(kNmT
2:
0.131 I.104
2.920 5.820
0.086 I .034
3.360 6.380
0.131 I .096
2.900 5.780
40
0.009
0.113
0.051
0.210
0.009
0.1 I4
IO’)
663
Two-stage solution for framed tube buildings
Table 8. Maximum axial force, N,, and maximum bending moment, M, in web columns of the 80-storey example building (a = 5 x 10m5in stage 2, I = 9) Solution procedure Present Standard Storey
(kN?lO’) 3.620 0.610 0.008
1 40 80
Stage 1
M (kNm x 102)
(kN ,“,
(kN:lO’)
11.420 5.760 0.115
Stage 2
3.420 0.672 0.054
102)
(kN:lO’)
12.520 6.340 0.072
3.600 0.602 0.008
M (kNm x 102) 11.320 5.740 0.115
Table 9. Comparison of CPU time, r,, for the 40-, 60- and IO-storey example buildings Tr (set)
Present No. of storeys
Optimum 1
Standard
7 7 9
40
60 80
389 1041 2194
Table 10. Storage location for [K,] and [I$,,] in the standard solution procedure and the present solutton procedure and for coefficients C, to C,, in the present solution procedure for 4O-, 6& and-80-storey example buildings
KS1 andK,,l
G
to
c25
No. of storeys
No. of levels
Standard ( x 104)
Present (xl@)
Present (x 104)
40 60 80
7 7 9
11.66 21.74 34.70
0.680 0.680 1.069
1.427 2.123 2.819
in the standard solution for the 40-, 60- and IO-storey example buildings, respectively. Comparisons of storage locations for [K,] and [I&,] for the two solutions and storage locations for coefficients C, to C,, are shown in Table 10. In the present solution for case A (coefficients Cr to Czs, stored) as well as for case B (coefficients C, to CZS, not stored) the storage locations required are only a fraction of those needed in the standard solution. Using the present solution results are also obtained on a PC/AT with a 286 microprocessor. Coefficients C, to C,, are generated in every cycle of iteration in the second stage to keep the storage requirements within limits. The time needed for 40-, 60- and 80-storey buildings are 11, 21 and 24 min, respectively. Software based on the standard solution could not be loaded due to insufficient storage space available. required
Stage 1
Stage 2
Total
T< (%)
11 13 21
6 14 16
17 27 37
4.37 2.59 1.69
5. CONCLUSIONS
1. The proposed two-stage solution only requires CPU time which is a small fraction of that needed in the standard single-stage Gauss elimination solution. Computer storage locations required are also a fraction of the space for those in the standard solution procedure. Work is in progress to apply the proposed two-stage solution procedure to other types of tubular buildings and other large systems. 2. The number of reference levels needed in stage 1 which yields smallest total CPU time is about 7-9. 3. The efficiency of the method increases with the increase in size of the buildings. 4. The proposed solution can be implemented on PCs to analyse buildings of practical sizes. REFERENCES
1. A. Coull and N. K. Subedi, Framed tube structures for high rise buildings. J. Sfrucr. Din, ASCE97,2097-2105 (1971). 2. A. Coull and B. Bose, Simplified analysis of frame-tube structures. J. Strucr. Div., ASCE 110,22232239 (1975).
3. A. Coull, B. Bose and A. K. Ahmed, Simplified analysis of bundled-tube structures. J. Strucr. Div., ASCE 108, 1140-1153 (1982). 4. P. Chang and D. Foutch, Numerical solution of tube buildings modelled as a continuum. Compur. Strucr. 21, 771-776 (1985). 5. P. C. Chang, Analytical modelling of tube in tube structures. J. Srruct. Din, ASCE 111,1326-1337 (1985). 6. P. G. Bergan and R. W. Clough, Convergence criteria for iterative processes. AIAA Jnl 10, 1107-l 108 (1972).
Pergamon
TWO-STAGE SOLUTION FOR FRAMED TUBE BUILDINGS Y.
SINGH
and A. K.
NAGPAL
Department of Civil Engineering, IIT Delhi, Hauz Khas, New Delhi-110016, India (Received 11 September
1992)
Abstract-A two-stage solution procedure is presented for the lateral load analysis of framed tubes. In the first stage dual shape functions and Gauss elimination solution are used to obtain an approximate solution. In the second stage a Gauss-Seidel iterative solution is used to obtain an accurate solution. The proposed two-stage solution procedure only requires a small fraction of the CPU time needed in the standard Gauss elimination solution. Computer storage locations required are also smaller than those in the standard solution. With the use of the two-stage solution procedure analysis of framed tubes of practical sizes can he carried out on PCs.
1. INTRODUCTION
Tubular systems are economical for resisting loads for mediumto high-rise buildings.
lateral
In its basic form, a tubular building is designated as a framed tube and consists of closely spaced columns on the periphery interconnected by deep spandrel beams (Fig. 1). Under lateral loading the framed tube undergoes shearing and flexural deformations. Closely spaced columns on the periphery provide large flexural stiffness. With the increase in stiffness of spandrel beams, the behaviour of the framed tube approaches that of the closed tube; the framed tube may also thus be viewed as a perforated tube. The computational work and computer storage in the displacement method of analysis for a framed tube modelled as a space frame is very large due to a large number of joints. Use of an equivalent plane frame model [l] reduces the computational and storage requirements. In this model web frames and flange frames interact through only vertical degrees of freedom (DOF) at common joints; to realize this, horizontal fictitious members are introduced between flange frames and web frames (these only transfer vertical shear forces). The continuum approach, in which the perforated tube is converted into an equivalent continuum, has also been used. A method in which web frames and flange frames are modelled as equivalent orthotropic plates is available for framed tubes with uniform properties along the height [2]. Closed-form solutions for stress resultants are obtained. The method is subsequently extended for bundled tubes[3]. In another continuum method, flange displacements are assumed to vary in a parabolic manner to cater for shear lag effects, shearing deformation is incorporated and the frame tube is converted into an equivalent cantilever tube [4, 51. The method predicts
deflections quite well but not the stresses and has been recommended for ascertaining only overall behaviour at a preliminary design stage. The continuum approach is unsuitable for predicting deflections and stresses for frame tubes with varying member properties at the final design stage. Equivalent plane frame idealization is well suited for practical problems as it caters for varying member properties. Generally, equations resulting from plane frame idealization are solved by the Gauss elimination solution. The solution procedure will from now on be referred to as the ‘standard’ solution procedure. Computational time and storage requirements increase rapidly with the increase in the number of storeys and column lines. Here, a two-stage solution procedure is proposed for the equivalent plane frame idealization in which the computational work and storage requirements are only a fraction of those in the standard solution procedure. The procedure involves use of the Gauss elimination solution in the first stage and a Gauss-Seidel solution in the second stage. 2. TWO-STAGE SOLUTION PROCEDURE Figure 2 shows the plane frame idealization of a framed tube in which the symmetric quarter of the building is considered with appropriate boundary conditions indicated on the end column lines. Web frame and flange frame are shown interconnected at floor levels by fictitious horizontal members. In the first stage the vector of deformations at the joints is estimated by solving an auxiliary problem of reduced size by a Gauss elimination solution. In the second stage the displacement vector of the first stage is taken as the initial vector in the Gauss-Seidel iterative solution for the frame idealization at actual size. 655
Y. SINCH and A. K. NAGPAL
656
Fig. 3. Level I shape functions for members: (a) lateral translations and rotations and (b) vertical translations.
Fig. 1. Symmetric quarter plan of a framed tube building.
2. I. Stage 1 solution The auxiliary problem is formed by dividing the equivalent plane frame into a number of segments, each segment spanning a number of storeys (Fig. 2). Within each segment three reference floors are
defined, one each at the bottom floor level and the top floor level and the remaining at an intermediate floor. Two levels of DOF and corresponding shape functions are identified. Level 1 DOF, as usual, comprise a translational DOF at each floor level and a vertical and a rotational DOF at every joint of floor levels. Level 1 shape functions define member deformations locally over member lengths (Fig. 3). Level
r
Fictitious members
T 9
I 2
r 1
1
-
1
21
Fig. 2. Equivalent plane frame, segments and reference floors.
657
Two-stage solution for framed tube buildings Iavoll d.of.: (ot. vi, +--a) Leved2d.of.: (IB~,~&--83)
Fig. 4. Level 2 shape functions for segments. 2 DOF are identified at selected reference floors only, as shown in Fig. 2. Each reference floor also has a translational DOF and every joint on it has a vertical and rotational DOF. Corresponding level 2 shape functions define the overall behaviour of a segment and are presently assumed to be quadratic as shown in Fig. 4. These shape functions define the relationship between level 1 and level 2 DOF which may be expressed as 3 u’=
Npf
c
(la)
Fig. 5. Level 1 DOF of a column element and associated level 2 DOF of the segment.
j-1
v’=
1 N,vf
where L,, L2 and x are the distances of intermediate reference floor, top reference floor and an intermediate floor of the segment, respectively. Figure 5 shows a column mn of storey length within a segment height. The relation between the vector of level 1 DOF at ends 1 and 2
(lb)
j=l
8’= t: N,ef, j=l
where u’, v I, 0 ’ are level 1 DOF (lateral displacement, vertical displacement and rotation, respectively, at a joint of a floor), uf, vj, 0; are the corresponding level 2 DOF at thejth reference floor of the segment and Nj are the shape functions given as
N
I
p,
-xl@*-x)
and the vector of level 2 DOF of the segment on the same column line (8) +:
(24
LlL2
v:
e:
u: v:
e:
e:)
0:
U:
can be expressed as N2 =
-45 -x)
(2b)
L,&-Ll)
N
x(x-4)
_
3-L,(L,-L,)’
-
N,h)
0
0 0
[Tel=
N162)
0 _
0
N,(x,) 0 0 N,(xz)
0
where [7’,] is the transformation column given by
(24
0 0 N,(x,)
0 0 N,(x,)
N2h)
0
0 0 N2@2)
0 0
N2h)
0 0
0 0
N2h)
0
N2@2)
0
0
N262)
N3@,)
0 0
matrix
0
0
N,(x,)
0
0
N,(x,
for the
-
I
(4) N3(x2)
0 0
o N3(x2)
0
0
’
o N3tx2)
-
Y. SINGHand A. K. NAGPAL
658
where IV,(x,) and Nj(x2) (j = 1,2,3) are shape functions evaluated at the bottom (X = x,) and at the top (x = x2) of the column. Similarly, for the beam element (Fig. 6) the relation between the vector of level 1 DOF at ends 1 and 2,
and the vector of level 2 DOF on the two adjacent column lines {CP}= {Vi e:
0:
e:
v:
e:
v: 0:
0: e;
vi
e;}
can be represented as
matrix for the beam
Fig. 6. Level 1 DOF of a beam element and associated level 2 DOF of the segment.
where N, , N,, N3 are evaluated at the beam level (x =x+). The stiffness matrix, [K:] for a column element with respect to its level 1 DOF (see Fig. 5) is given
The stiffness matrix of a segment is obtained by adding the contributions of all the beams and columns in it. The segment lateral load vector due to lateral loads at floor levels is expressed as
where [T,,] is the transformation given by
[Tc+l=
0 EA/L 0 [K:l=
-EA/L
12EI/L’ 0 6EIIL2 0
6EI/L2 0 4EI/L 0
0 - EA/L 0
- 12EIIL’ 0 - 6EIIL2
EA/L
0
6EI/L2 0 2EIIL 0
0
-12EIjL’
-6EI/L*
0
12EIIL’
-6EIIL2
0
6EIIL2
2EIIL
0
-6EIIL2
4EIIL
where E is the modulus of elasticity, Z, A and L are the moment of inertia, area and length of the column, respectively. Deleting the first and fourth rows and first and fourth columns, and interpreting the properties I, A, and L as that of a beam, gives the stiffness matrix [K:] of the beam element with respect to its level 1 DOF (Fig. 6). Stiffness matrix [Kf] of a column and [Ki] of a beam with respect to their respective level 2 DOF are given by
_
1 ’
(7)
[NI(Xi)) (9)
where F, and xi are the lateral load at the ith floor and the distance of the ith floor from the bottom of the segment, respectively, and the summation extends over the number of floors in the segment. The stiffness matrix and the load vector of the Ktl= [T,.lTIK:~IV-,.l @a) auxiliary model are assembled from the stiffness @b) matrices and the load vectors of the segments, Kil= JT/JTKiIITJ respectively. The equilibrium equation may be Explicit expressions for terms of [KT] and [K,?,] written as can be obtained. Alternatively these matrices may be obtained numerically as triple matrix products. [K2]{D2j = {P2), (10)
659
Two-stage solution for framed tube buildings where [K2], {O’} and {P’} are the stiffness matrix displacement vector, and the load vector, respectively. The number of reference floors required to be chosen in the present method is generally smaller than the number of column lines. It is, therefore, economical to number the level 2 rotational and vertical DOF, S, at reference floors consecutively along a column line followed by numbering on the subsequent column lines in succession as shown in Fig. 2. Level 2 translational DOF, p, are numbered consecutively at reference floors. In eqn (10) translational DOF are grouped separately and the equation is then written in a partitioned form. A lateral stiffness matrix can then be obtained which could be used to evaluate free vibration characteristics. A lateral stiffness matrix is also required when the framed tube resists the lateral loading together with other stiffening elements, such as an interior core. Equation (10) is now written in the partitioned form as
Fig. 7. Level 1 DOF contributing to joint/floor equilibrium in stage 2.
Equations for the evaluation of vertical translations and rotations at nodes at floor levels and lateral translations at floor levels can be expressed as (Fig. 7) u;“+
where subscripts s and p refer to vertical and rotational DOF at joints and to translational DOF at reference floors, respectively. From eqn (11) the following relations can be written
ww:1=
{CI
UW
WI = EJ - vww’E,ol
U2b)
ICI = {PiI - ~f$IKYv3
(124
I#> = Wisl-‘({PfI - [$JP:D
(124
in which [KS] is the lateral stiffness matrix. For numerical evaluation of [Kf], [K:;] is not inverted. Recognizing the banded nature of [K:;], we can obtain [Kzz]-‘[KfP] by Gauss elimination in which columns of [Ki,] are treated as load vectors. The standard Jordan method can be used in eqn (11) as well to eliminate s DOF and yield [Kf], but in this process a greater bandwidth is required. Equations (12a) and (12d) are solved for CO’). The level 1 displacement vector is then obtained from eqn (1). 2.2. Stage 2 solution The level 1 displacement vector differs from the true displacement vector, the difference depending on the number of reference floors chosen. The former vector can be used as a starting vector in the GaussSeidel solution procedure to have an accurate displacement vector. The constraints imposed on level 1 DOF by level 2 shape functions are now relaxed at this stage.
u’j+’ c
I =
=
$i
-/vQ:+Rb+‘)/C,
ufj + j(j’,
-
Q: - Rt+‘)/C,,
R:+l=C,v~+l+C,v:'tl+c,e~'+l
(13)
(1%
(17)
Q; = C,fl;*i+ Clout’+ Cllufj + c,#+
Cl&
+ c,,ef.i
(18)
Rb+‘=C,5u~+~+~,6e~+~+~,7U)‘i+I +
Q; = C2&;.’ + R:+’ = C,,up+
w:.i+ I + c,9e;.i+l
(19)
c*,uy+ c,,ey
(20)
+ c24ey+r + c,,e;,f+r
(21)
where subscripts I and r refer to nodes at the left and right of a node C; subscripts b and t refer to nodes to the bottom and the top of the node C, or to the floors at the bottom and the top of a floor C; superscripts i and i + 1 refer to the values at the beginning and at the end of an iteration; coefficients . , C,, involve member properties; /I is the c,,c*,.. over-relaxation factor; and P, is the lateral load at the floor c. Iterations are carried out until an error norm ((E 11 is less than a specified value a. The error norm chosen [6] is defined as
Adi
I I
IIt 11= max -
di,,,,
(22)
Y. SINGH and
660
in which Ad, is the difference between two values of a displacement (i.e. at the beginning and at the end of an iteration) and di.,, is the maximum displacement of the similar type (either vertical, translational or rotational). 3. COMPUTER STORAGE AND COMPUTATIONAL WORK
In the first stage of this method the major portion of the storage required is for [K,] and [&,I (the superscript 2 has now been dropped). For q segments in the equivalent plane frame, the number of reference floors, 1 is 2q + 1 and with k column lines in the symmetric quarter plan, the number of p and s DOF are I and 2k1, respectively. Storage locations for [K,] and [J&J together are 6kl(l+ 2). In the second stage two cases arise. In case A, for each vertical DOF coefficients C, to C,, for each rotational DOF coefficients C, to C,9 and for each translational DOF coefficients C,, to C,, are stored. These require N(6 + 19k) (N = number of floors including the ground) storage locations for all the level 1 DOF. These storage locations are not needed simultaneously with those of stage 1. The storage requirement is therefore the greater of the two quantities 6kl(l+ 2) and N(6 + 19k). In case B, coefficients C, to C,, are not stored but are evaluated at every iteration from the member properties. This may have to be resorted to when working with PCs. In this case the storage requirement is that of stage 1 i.e. 6kl(l + 2). If N is greater than k, in the standard solution procedure, it is advantageous to number all the s DOF on a floor consecutively followed by s DOF at the subsequent floors. Similarly p DOF are numbered sequentially at floors. The number of s and p DOF are 2kN and N, respectively. A major portion of storage locations in the standard solution procedure is also for [K,] and [K,] and is equal to 2KN(2K + N + 2). If, on the other hand, N is smaller than k, the numbering is done in a manner similar to that for the present method. The number of storage locations for [K,] and [K,] are 2kN(3N + 2). In the present method, CPU time, T, is the sum of the times in both stages. Taking more reference levels results in increase in T, in the first stage with a decrease in the second stage. There is an optimum number of reference levels for which the total T,. of both stages is minimum. If more than one loading case is to be analysed, a higher number of reference levels (in the first stage) than that for a single loading case may be desirable. This increases time in the first stage, consequently less time would be needed in the second stage. As T, for the first stage, for most practical loading cases, is nearly equal to that for a single loading case, an increase in T,. for more than one loading case is confined to the second stage only. Again, an optimum number of reference floors may be arrived at, which results in a smaller total Tc.
A. K. NAGPAL Table 1. Cross-sectional properties of members of the 66storey example building
Cross-sectional properties Columns
Beams
Storeys/ floors
A (m*)
I (m4)
I (m4)
l-10 1l-20 21-30 31-40 41-50 51-60
0.700 0.580 0.520 0.440 0.370 0.370
0.200 0.190 0.170 0.150 0.130 0.130
0.200 0.180 0.150 0.120 0.100 0.100
4. NUMERICAL
EXAMPLES
First a 60-storey building with storey heights of 3 m, plan dimensions of 42 m (web) x 54 m (flange), and column spacing of 3 m is considered. Member properties I and A for columns and I for beams are given in Table 1 and E = 2.5 x 10” N/m2. The building is subjected to a lateral loading 234 kN at floors l-59 and 117 kN at floor 60. Results from the present method are first taken for 1 = 7 (reference floors: ground, 2, 7, 15, 27, 42, 60) and for tl = 5 x lo-‘. The spacing between the reference floors increases along the height. Lateral deflections from the present solution procedure at the first and second stages and from the standard solution procedure are shown in Fig. 8 where it is seen that deflections from the first stage are close to those from the standard solution procedure. Iterations of the second stage are not needed to estimate the deflections. Maximum member forces, the axial force, N, and bending moment M, in the columns of the first, 30th and 60th storeys, and bending moment M and shear force S, in the beams at the first, 30th and 60th floors are given in Tables 2 and 3, respectively. For columns the maximum N, occurs in the corner column at all floors and the maximum M occurs in 60’
’
123456 DoflofJtion (em)
Fig. 8. Lateral deflections for the 60-storey example building.
661
Two-stage solution for framed tube buildings
Table 2. Maximum axial force, No, and maximum bending moment, M in web columns of the 60-storey example building (a = 5 x 10-r in stage 2, I = 7) Solution procedure Present stage 1
Standard storey
I
M (kNm x lo*)
(kN?lO’) 2.160 0.324 0.009
30 60
(kN:lO’)
8.680 4.320 0.122
Stage 2
(kN&
1.986 0.256 0.050
IOr)
9.680 5.020 0.438
(kN?tO”)
(kN rn\
2.160 0.320 0.009
10’)
8.610 4.240 0.122
Table 3. Maximum bending moment, M, and maximum shear force, S, in web beams of the 60-storey example building (a = 5 x 10e5 in stage 2, I = 7) Solution procedure Present Standard
Stage 2
Stage I M
Storey 1 30 60
(kN2
10’)
(kN ,s l@)
(kN m x 10’)
(kN : 102)
(kN n?x 102)
(kN : 102)
4.660 2.830 0.079
3.010 4.860 0.502
2.010 3.240 0.359
6.950 4.160 0.122
4.630 2.780 0.079
6.990 4.250 0.122
the central web column at the first and 30th floors but in the corner column at the 60th floor. For the beams the maximum M and S occur in the central web beam at the first and 30th floors, but in the comer end beam at the 60th floor. For the present solution procedure, the forces are shown at both stages, those at stage 1 differ from those of the standard solution. When the Gauss-Seidel iterative process is applied in stage 2, these forces approach those of the standard solution. The convergence of N, and M for the first storey column with number of iterations, Ni, is shown in
Fig. 9. The value of the over-relation factor, j?, has been chosen as 1.6. Table 4 shows the variation with the number of reference levels I of CPU time, T, needed in stages 1 and 2 and of the total time which is expressed as a percentage of that in the standard solution. The CPU time is on an EL-3980 mainframe computer at IIT Delhi. Also shown in the table are the number of iterations needed in the second stage. A check on the error norm IIt 11is made after every five iteration cycles. In the table we imply by reference floors 61 that the auxiliary structure is the actual structure; the first stage solution becomes the standard solution and the second stage solution is absent. As the number of reference floors is increased, T, for the first stage increases and decreases for the second stage. For seven reference levels the total T, is minimum, equal to 27 set which is about 2.59% of that (1041 set) in the standard solution. The time
Table 4. Variation of the number of iterations, N,, in stage 2 and of CPU time, T,., with the number of reference floors in the present method for the 60-storey example building (a = 5 x 10ms)
T
0’
I
1
sa 2s No.ofltadaas
1
75
Fig. 9. Convergence of axial force, N,,, and bending moment, M, in the comer column of first storey for the 60-storey example building (a = 5 x 10es, I = 7).
I
Ni
Stage 1 (set)
5
180 90 55 40 35 3s -
10 13 19 28 41 58 1041
7 9 11 13 15 61
Stage 2 (set) 23 14 10 9 8 8 -
Total (W 33 27 29 37 49 66 1041
Percentage 3.17 2.59 2.78 3.55 4.71 6.34 -
662
Y. SINGH and A. K. NAGPAL Table 5. Cross-sectional properties of members of the 40-storey example building Cross-sectional properties Columns
Beams-
Storeys/ floors
A (mr)
I (m4)
I (m4)
I-IO I l-20 21-30 31-40
0.520 0.440 0.370 0.370
0.170 0.150 0.130 0.130
0.150 0.120 0.100 0.100
needed in the present solution procedure is thus only a small fraction of that needed in the standard solution procedure. In the table, T, for the first stage are for the case when explicit expressions for terms of [Kf] and [Ki] are used. Similarly in the second stage, the T,in Table 4 are for the case when sufficient storage locations are available, as in the mainframe computer, and thus coefficients C,-C,, are stored at the beginning of the iterative procedure in the second stage. If sufficient storage locations are not available and coefficients are evaluated in every iteration, T,in the second stage is increased; for I = 7 it increases from 14 to 22 sec. The variation of percentage errors in the maximum column force, N,, of the first storey and in total T,
other buildings, one is 40 storeys and the other is 80 storeys, both of these are of the same planform as the 60-storey example building. The member properties of the 40-storey building are shown in Table 5 and for the 80-storey building in Table 6. For these buildings the lateral load at an intermediate floor is also 234 kN and that at the respective top floors is 117 kN. Forces N, and M in columns for the 40- and 80-storey example buildings are shown in Tables 7 and 8, respectively. Optimum 1for which minimum T, is required are 7 and 9, and the number of iterations in the second stage for convergence are 55 and 65 for the 40- and 80-storey example buildings, respectively. The increase in efficiency of the present solution with a number of storeys is shown in Table 9. As the number of storeys increases, the T, required in the standard solution increases rapidly from 389 set for the 40-storey building to 2194sec for the 80-storey building. On the other hand, the T, required in the present solution increases from 17 set for the 40storey building to only 37 set for the 80-storey building. In the present solution T, required is about 4.37%, 2.59% and 1.69% of the corresponding T,
in the present solution with specified error norm c( is shown in Fig. 10. For c1= 5 x 10e5, the error in N, remains sufficiently small in all the storeys. For two loading cases, T, of stage 1 is nearly the
Table 6. Cross-sectional properties of members of the 80-storey example building
“0.5
50
5.0
500
a(x lWs) Fig. 10. Variation of precentage error in No and T,,with 11 t )I for the 60-storey example building.
same as for one loading case, i.e. 19 set for 1 = 9. For stage 2, T,is nearly doubled to 20 sec. The minimum total T, is thus 39 set and occurs for I = 9. For the present solution procedure, the minimum total T, is
Cross-sectional properties Columns Storeys/ floors
A (m2)
l-20 21-40 41-60 61-80
1.000 0.730 0.518 0.370
increased from 27 set for one loading case (1 = 7) to 39 set for two loading cases (I = 9). The efficiency of the present method with a varying number of storeys is studied by considering two
Beams I (m4)
1 (m4) 0.230 0.196 0.150 0.130
0.260 0.195 0.150 0.130
Table 7. Maximum axial force, N,,, and maximum bending moment, M, in web columns of the 40-storey example building (a = 5 x 10m5in stage 2, I = 7) Solution orocedure Present Standard Storey
(kN:lOr)
(kN
Stage I
A 10’)
(kN%O’)
Stage 2
(kN u?x IO’)
(kN?lO’)
(kNmT
2:
0.131 I.104
2.920 5.820
0.086 I .034
3.360 6.380
0.131 I .096
2.900 5.780
40
0.009
0.113
0.051
0.210
0.009
0.1 I4
IO’)
663
Two-stage solution for framed tube buildings
Table 8. Maximum axial force, N,, and maximum bending moment, M, in web columns of the 80-storey example building (a = 5 x 10m5in stage 2, I = 9) Solution procedure Present Standard Storey
(kN?lO’) 3.620 0.610 0.008
1 40 80
Stage 1
M (kNm x 102)
(kN ,“,
(kN:lO’)
11.420 5.760 0.115
Stage 2
3.420 0.672 0.054
102)
(kN:lO’)
12.520 6.340 0.072
3.600 0.602 0.008
M (kNm x 102) 11.320 5.740 0.115
Table 9. Comparison of CPU time, r,, for the 40-, 60- and IO-storey example buildings Tr (set)
Present No. of storeys
Optimum 1
Standard
7 7 9
40
60 80
389 1041 2194
Table 10. Storage location for [K,] and [I$,,] in the standard solution procedure and the present solutton procedure and for coefficients C, to C,, in the present solution procedure for 4O-, 6& and-80-storey example buildings
KS1 andK,,l
G
to
c25
No. of storeys
No. of levels
Standard ( x 104)
Present (xl@)
Present (x 104)
40 60 80
7 7 9
11.66 21.74 34.70
0.680 0.680 1.069
1.427 2.123 2.819
in the standard solution for the 40-, 60- and IO-storey example buildings, respectively. Comparisons of storage locations for [K,] and [I&,] for the two solutions and storage locations for coefficients C, to C,, are shown in Table 10. In the present solution for case A (coefficients Cr to Czs, stored) as well as for case B (coefficients C, to CZS, not stored) the storage locations required are only a fraction of those needed in the standard solution. Using the present solution results are also obtained on a PC/AT with a 286 microprocessor. Coefficients C, to C,, are generated in every cycle of iteration in the second stage to keep the storage requirements within limits. The time needed for 40-, 60- and 80-storey buildings are 11, 21 and 24 min, respectively. Software based on the standard solution could not be loaded due to insufficient storage space available. required
Stage 1
Stage 2
Total
T< (%)
11 13 21
6 14 16
17 27 37
4.37 2.59 1.69
5. CONCLUSIONS
1. The proposed two-stage solution only requires CPU time which is a small fraction of that needed in the standard single-stage Gauss elimination solution. Computer storage locations required are also a fraction of the space for those in the standard solution procedure. Work is in progress to apply the proposed two-stage solution procedure to other types of tubular buildings and other large systems. 2. The number of reference levels needed in stage 1 which yields smallest total CPU time is about 7-9. 3. The efficiency of the method increases with the increase in size of the buildings. 4. The proposed solution can be implemented on PCs to analyse buildings of practical sizes. REFERENCES
1. A. Coull and N. K. Subedi, Framed tube structures for high rise buildings. J. Sfrucr. Din, ASCE97,2097-2105 (1971). 2. A. Coull and B. Bose, Simplified analysis of frame-tube structures. J. Strucr. Div., ASCE 110,22232239 (1975).
3. A. Coull, B. Bose and A. K. Ahmed, Simplified analysis of bundled-tube structures. J. Strucr. Div., ASCE 108, 1140-1153 (1982). 4. P. Chang and D. Foutch, Numerical solution of tube buildings modelled as a continuum. Compur. Strucr. 21, 771-776 (1985). 5. P. C. Chang, Analytical modelling of tube in tube structures. J. Srruct. Din, ASCE 111,1326-1337 (1985). 6. P. G. Bergan and R. W. Clough, Convergence criteria for iterative processes. AIAA Jnl 10, 1107-l 108 (1972).