ABSTRACT

EFFECT OF WALLS ON
STRUCTURAL RESPONSE TO EARTHQUAKES

By

Palamadai S. Natarajan

The development of an optimal numerical model for
filler walls in building frames is described. A method
of elasto—plastic dynamic analysis of plane building
frames with filler walls is presented. A computer program
written in FORTRAN IV for use on the Michigan State
University CD3 6500 computer system was prepared to
accomplish the dynamic solution of such structures sub-
jected to earthquake ground motion.

The filler walls are treated as finite elements in
plane stress, interacting with the moment-resisting frame
such that the translational displacements of the joints
are compatible. Both triangular and rectangular elements
are considered. The effect of mesh size and modelling of
the cracking phenomenon are studied. Computed results of
load-displacement relations, crack propagation pattern and
ultimate load capacity have been compared with known

experimental results.

Palamadai S. Natarajan

In the dynamic analysis the mass of the system is
handled by a lumping procedure, which accounts for rotary
as well as translational inertia. Mass-proportional
viscous damping has been taken into account. The equations
of motion have been formulated in terms of joint displace-
ments relative to supports. Two methods of analysis have
been derived. In the first, all the three degrees of
freedom of a joint are considered. In the second (modified)
method the degrees of freedom associated with the axial
deformation of frame members and rotation of joints are
eliminated. The modified method makes possible the use
of a much larger time step for the numerical integration
procedure, resulting in a considerable saving of compu-
tation time.

Numerical results of the study of a three story steel
frame with concrete filler walls subjected to selected
portions of the El Centro Earthquake of 19u0, are presented.
The results of the study highlight the importance of the
effect of walls on the lateral stiffness and dynamic
response of infilled frames. It is also shown that there
is no significant loss of accuracy in using the modified

methOdo

EFFECT OF WALLS ON

STRUCTURAL RESPONSE TO EARTHQUAKES

By

Palamadai S. Natarajan

A THESIS

Submitted to
Michigan State University
in partial fulfilment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Civil Engineering

1970

G'é#fl33
/0-.2'§"70

To my parents

ii

ACKNOWLEDGEMENTS

The author wishes to express his gratitude to his
major professor, Dr. Robert K. Wen for his invaluable
advice and guidance during all phases of this work.

The author wishes to thank the members of his guidance
committee, Dr. C. E. Cutts, Dr. W. A. Bradley and

Dr. J. S. Frame for their guidance and encouragement.
Thanks are also due to Dr. J. L. Lubkin for his helpful
suggestions.

Acknowledgement is made to the National Science
Foundation and the Division of Engineering Research for
their financial support which made this work possible.

The author also wishes to thank his wife Sundari

for her assistance and cooperation.

iii

TABLE OF CONTENTS

LIST OF TABLES . . . . . . . . . . . . . vi
LIST OF FIGURE 0 . I O O O O O I O O O O Vii

CHAPTER
I. INTRODUCTION . . . . . . . . . . . 1
1.1 General . . . . . . . . . . 1
1.2 Scope and Outline of the Investigation 2
1.3 Literature Review . . . . . . 3
1.4 Organization of this Report . . . . 6
1.5 General Definitions . . . . . . 6
1.6 Notation . . . . . . . . . . 9

II. STATIC LOAD-DISPLACEMENT RELATIONS . . . . 12

2.1 General . . . . . . . . . . 12
2.2 Member Stiffness Matrix . . . . . 13
2.2.1 Frame elements . . . 13
2.2.2 Triangular Wall Elements . . 14
2. 2. 3 Rectangular Wall Elements . 18
2.3 Joint Stiffness Matrix . . . . . 20
2.4 Static Analysis . . . 22
2.4.1 Effect of Reduction Factor . 23
2.4.2 Ultimate Load Capacity . . 24
2.4.3 Effect of Mesh Size . . . 24
2.5 An Optimal Numerical Model . . . . 25
2.6 Effect of Openings in Wall Panel . . 25

III. METHOD OF DYNAMIC ANALYSIS . . . . . . 27

301 General 0 o o o o o o o I o o 27
3.2 Lumping of Masses . . . . . . . 27
3.3 Numerical Integration Procedure . . 28
3.4 Modified Procedure in the Reduced

Degrees of Freedom . . . . . . 30
3.5 Computer Program . . . . . 34

3.5.1 Limitations of the Program . 37

Iv. NUMERICAL RESULTS 0 o o o o o o o o 38
“'01 The StrUCture o o o o o o o o 38

4.1.1 Loading . . . . . . . 38
4.1.2 Time Increment . . . . . 38

iv.

Numerical Examples . . .
Effect of Walls . . . .

Effect of Axial and Rotational

Interaction Procedure for
Acceleration of Joints .

V. SUMMARY AND CONCLUSIONS . . .

TABLES .
FIGURES .

LIST OF REFERENCES . . . . . . .

APPENDIX I.
APPENDIX II.

MEMBER STIFFNESS MATRICES
COMPUTER PROGRAM . . .

odes

Page

39

39
42

43
45
48
51
64
66

7O

Table

LIST OF TABLES

Page
Ultimate load capacity of model
infilled frame . . . . . . . . 48
Effect of mesh size on load-displacement
relation using triangular elements . . 48
Effect of mesh size on load-displacement
relation using rectangular elements . 49
Computed values of fundamental periods . 50

vi

LIST OF FIGURES

Figure Page
1 Structure Global Coordinates . . . . 51
2 Joint Coordinates . . . . . . . . 51
3 Member Coordinates . . . . . . . 52
4 Wall Elements . . . . . . . . . 53
5 Static Load-Displacement Relations . . 54
6 Crack Propagation in Wall . . . . . 55
7 Effect of Wall Openings . . . . . . 56
8 Model Three Story Structure . . . . 57
9 Linear Response . . . . . . . . 58

10 Nonlinear Response:- Ground Motion 1 . 59
11 Nonlinear Response:- Ground Motion 2 . 6O
12 Nonlinear Response:— Ground Motion 3 . 61

13 Effect of Axial and Rotational Modes:-
Without Wall-Stiffness . . . . . 62

14 Effect of Axial and Rotational Modes:-
With Wall-Stiffness o o o o o o 63

vii

CHAPTER I

INTRODUCTION

1.1 General

The effect of frame-wall interaction in a structure
subjected to lateral loads, such as those resulting from
earthquake shocks, has received much attention in recent
years. In order to design structures that are safe and
economical, the importance of more accurate analysis has
long been recognized. Although the loading under con-
sideration is dynamic in nature, earlier investigations
were based on statics only and often in conjunction with
gross simplifying assumptions leading to a crude modelling
of the structure. Nevertheless, those investigations were
significant contributions considering the importance of
the problem and the limited computational facilities avail-
able then. With the availability of modern high speed
digital computers a more exact analysis of the problem
seems warranted. The work reported in this thesis repre—
sents an effort in this direction.

It has often been claimed that the aseismic design of
framed structures without considering the effect of walls
would produce a conservative design. If this is true,
considering the effect of walls would lead to a more

economical design. However such an observation is made

purely from a statics point of view. It is well known that
the fundamental frequency of a structure is a major factor
in its response to earthquakes. Corresponding to an in-
crease in the fundamental frequency of a structure, the
Code (16)* calls for an increase in the percentage of its
gravity load to be applied as seismic lateral load. The
fundamental frequency is substantially increased by the
presence of walls even if cracked. This according to the
Code would increase the lateral loads on the structure.
Thus, a design neglecting the effect of walls might not be
conservative after all. It must, however, be recognized
that the inclusion of the effect of walls adds substantially

to the lateral stiffness of the structure.

1.2 Scope and Outline of the Investigation

The purpose of this investigation is to study, using
a more accurate formulation of the stiffness characteristics
of structures, the effect of walls on the dynamic response
of infilled frames subjected to earthquakes. The analysis
extends beyond the elastic range. The nonlinearity of the
structural response, due to the formation of plastic hinges
in the frame members and crack propagation in the wall
elements, is taken into account.

Only plane frames are considered. The bending moment-
curvature relationship is assumed to be elastic-perfectly

plastic. The filler walls, treated as finite elements

 

* Numbers refer to items listed in the List of References.

in plane stress, interact with the frame such that the
translational displacements of the joints are compatible.
It is also assumed that the wall elements can not transmit
moments at the joints. A wall element is assumed to have
cracked when the principal tensile stress exceeds a pre-
scribed "cracking stress". The overall stiffness of a
cracked wall element is approximated by that of the original
uncracked element except that the elastic modulus is reduced.
In the dynamic analysis the structure is treated as a
discrete system of masses lumped at the joints. The equations
of motion are formulated using the displacements of the joints
:relative to the supports as variables. The transient response
.is obtained by a numerical integration of the equations of
nuytion. A modified procedure in which the axial and the
Instational modes are eliminated, is also derived. This
nuikes possible the use of a much larger time step for the

rnxmerical integration, saving considerable computation time

\Nithout significant loss of accuracy.

1-:3 Literature Review

Smith (13) presented a procedure for predicting the
approximate lateral stiffness of infilled frames, by
aSEBurning an equivalent diagonal strut to replace the infill.
Thfé effective width of the strut was derived theoretically
arui checked by model experiments. Frischman, Prabhu and
TOIXPler (5) used the influence coefficient method and the

egllivalent column method for the analysis of multistory

frames with interconnected shear walls. The frame was
replaced by a column whose stiffness equals the sum of all
column stiffnesses and restraints were applied at each
floor level equivalent to the beam stiffnesses. Beck (1)
presented an approximate method of analysis, replacing the
discontinuous frame system by a continuous system. He took
(into account shear wall deformations due to normal forces.

Cardan (2) proposed that the problem of flexural
deformations of the wall, could be reduced to a second degree
differential equation, with a few simplifying assumptions.
Roseman (11) presented an approximate method of analysis of
walls of multistory buildings with openings. The basic idea
of his approach consisted in the replacement of the connect-
ing beams with a continuous connection. To obtain the
distribution of lateral forces on the elements of a frame-
Wall system, Khan and Sbarounis (8) considered the structure
as:separate parts with certain conditions for compatibility.

Gould (7) studied shear wall-frame interaction by
remucing the problem, with some simplifying assumptions, to
that of a cantilever beam supported by concentrated elastic
rmartions. He represented each story of the frame as an
elastic:spring connected to the shear wall at each floor by
a l"igid bar and a (rotational spring, and connected to each
0f the adjacent floors through a rigid joint.

Kokinopoulos (9) investigated the seismic response

0f a-Inultistory system, treating it as a cantilever with

masses concentrated at the floor levels. Coull and

Choudhury (3) analyzed coupled shear walls by replacing

the discrete system of connecting beams by an equivalent

continuous medium. Fedorkiw and Sozen (4) proposed a lumped

parameter model to simulate the response of reinforced

concrete frames with concrete filler walls. Their study

included load-displacement characteristics of infilled

frames and crack propagation in the filler walls.
Experimental work on the ultimate lateral load capacity

of concrete frames with masonry filler walls has been

reported by Sachanski (12). Yorulmaz and Sozen (15) made

experimental studies on the lateral load.capacityand.load-
displacement characteristics of reinforced concrete frames
with concrete filler walls.

The preceding works were related to the static
anaLysis of-frame-wall systems. Goldberg and Herness (6)
Smuiied the vibration of multistory buildings. The floor
and wall deformations were studied by use of a generalized
slope-deflection equation. Saghera (17) investigated the
effect:of shear walls on the frequencies of vibration of
a.stzuacture. His study included experimental verification
OfC(Dmputed results. The results of his study indicate
that the filler walls, as expected, increase the stiffness

Ofii frame, resulting in considerably higher frequencies.

1.4 Organization of this Report

The formulation of the joint stiffness matrix of the
structure is presented in Chapter II. A comparative study
is also made, treating the walls by the finite element
method. Computed load-displacement characteristics are
compared with known experimental results. The effect of
wall openings on the lateral stiffness of infilled frames
is also discussed. Chapter III deals with the numerical
solution of the governing differential equations of motion.
Two methods of analysis are derived. In the first, all the
three degrees of freedom of a joint are considered. In the
second (modified) method, the degrees of freedom associated
with the axial deformation of frame members and rotation
of joints are eliminated. In this chapter a section on the
cmmputer program has also been included. In Chapter IV
Cmnparative results of dynamic analysis of a three story
stee1.frame with concrete filler walls, with and without
'Uue effect of walls are presented. Comparative responses
of the structure and the computation time involved, with
andWithout the axial and rotational modes are discussed.

A Summary and conclusions are presented in Chapter V.

1-5 Generalggefinitigns

The joints of the structure consist of supports and
free joints. The free joints are classified as frame
jotnts and interior wall joints (see Figure 1). At a

franks joint two or more frame members are incident. At an

interior wall joint only wall elements are incident. At a
frame joint three components of forces or displacements can
be specified and at an interior wall joint only the two
translational displacements or the corresponding forces can
be specified since it is presumed that the wall elements can
not transmit moments at the joints. The joints of the
structure are numbered consecutively starting with 1. The
ordering is arbitrary, but once assigned it remains fixed
during the analysis,

The incidence of a frame or wall element is defined
by the Joint Numbers of its ends or corners. A frame member
whose incidence is IJ has its initial or positive end at I
and its final or negative end at J. The incidence of a
triangular wall element is given by IJK where I, J and K
are the Joint Numbers of its corners. The choice of the
initial end I is arbitrary but IJK is ordered counter-
clocflcwise. The incidence of a rectangular wall element
wifii its corners at Joints I, J, K and L is given by IJKL.
I ks the left hand bottom corner of the rectangle and IJKL
is Ordered clockwise.

Three coordinate systems are used in the analysis.
(See Figures 1.2 and 3)}

1) Structure Global Coordinate System:

This system consists of a single set of cartesian
axes with origin at any chosen point. Since the
analysis is confined to plane frames, the set of

cartesian axes implies a two dimensional system

2)

3)

with the X axis horizontal and Y axis vertical.
Rotation is defined to be positive in the counter-
clockwise sense.

Joint Coordinate System:

This system consists of one set of cartesian axes
for each joint with its axes parallel to the
structure global axes. This system is used to
describe the displacements of the joint and forces
acting on it. At a frame joint the displacements
are represented by a vector with three elements,
the first two elements being the translational
displacements in the X and Y directions and the
third element being the rotation in the XY plane.
The force vector is made up of forces correspond-
ing to these displacements. In the case of an
interior wall joint, this vector is made up of

two elements, the rotation or the moment component
being absent.

Member Coordinate System:

This consists of a set of cartesian axes for each
end of a frame member and for each corner of a
wall member. The origins are located at the ends
or corners as the case may be, and the axes are
parallel to the global axes. These axes are used
to describe the end or corner displacements of

the elements, and also the forces acting at the

ends or corners of the elements. For the frame

members there will be three components of dis-

placements or forces at each end and for the wall

elements there will be two components of displace-

ments or forces at each corner.

1.6 Notation

The notation shown below has been used in this

report:

a, b

dt

the dimensions of the rectangular wall
element in the x and y directions;

area of the triangular wall element;

the matrix relating strain and corner
displacements of a rectangular wall element;
the matrix relating strain and corner
displacements of a triangular wall element;
damping matrix;

damping constant;

the matrix relating stress and strain in a
plane stress formulation;

time interval for the numerical integration
procedure;

the strain vector at any point in the wall
element;

Young's Modulus;

vector of forces acting at the ends of a
frame element or the corners of a wall
element;

element of the joint flexibility matrix F;

10

joint flexibility matrix of the unreduced
system;

joint flexibility matrix of the reduced
system;

a submatrix of K;

joint stiffness matrix of the unreduced
system;

joint stiffness matrix of the reduced system;
stiffness matrix of the frame element;
stiffness matrix of the rectangular wall
element;

stiffness matrix of the triangular wall
element;

mass matrix of the unreduced system;
mass matrix of the reduced system;
Poisson's ratio of the wall material;
number of free joints;

vector of joint loads of the unreduced
system;

the ratio of the height to breadth of a
rectangular wall element;

internal joint resistance vector of the
unreduced system;

internal joint resistance vector of the
reduced system;

vector of end or corner displacements of

a frame or wall element;

(:0

ll

11

joint displacement vector of
system;
joint velocity vector of the
system;
joint acceleration vector of
system;
joint displacement vector of
system;
joint velocity vector of the
joint acceleration vector of

system;

- the coordinates of Joint i;

the unreduced

unreduced

the unreduced

the reduced

reduced system;

the reduced

a vector formed from prescribed ground

accelerations;

dimensionless member coordinates of the

rectangular wall element; and

stress vector at any point in the wall

element.

CHAPTER II

STATIC LOAD-DISPLACEMENT RELATIONS

2.1 General

The main objective of this first phase of the investi-

gation is to determine an "optimal" numerical model to re-

present the stiffness characteristics of infilled frames.

The following approach is used.

1.

3.

Only plane frames are considered. Failure of frame
elements is assumed to occur due to flexure only. The
bending moment-curvature relationship is assumed to be
elastic-perfectly plastic. It is assumed that when the
yield moment is reached an abrupt transition takes place
from a purely elastic state to a state where all the
fibres are stressed to the yield limit and unrestricted
plastic deformation can occur. This is tantamount to
assuming a shape factor equal to unity.

The wall elements are treated as finite elements in
plane stress, interacting with the frame such that the
translational displacements of the joints are compatible.
Both triangular and rectangular elements are considered.
Cracking is assumed to occur in the wall material when
the principal tensile stress exceeds a prescribed
"cracking stress". The stiffness characteristics of a

wall element after cracking are computed as usual except

12

7.
L
Q

 

13

that the value of the elastic modulus for that element

is reduced.

2.2 Member Stiffness Matrix

The member stiffness matrix of a frame element is a
6 x 6 symmetric matrix which relates the end displacements
to the forces acting at the ends of the member. The member
stiffness matrix of a wall element is also a symmetric
matrix which relates the corner displacements to the forces
acting at the corners of the wall element. Since it is
presumed that the wall elements can not transmit moments
at the joints, only the two translational components of the
displacements or the corresponding forces can be specified
at the corners. Thus the member stiffness matrix has a
dimension of 6 x 6 for a triangular element and 8 x 8 for
a rectangular element.

The member stiffness matrices are generated with
their axes of reference parallel to the global axes. This
has a distinct advantage in that the joint stiffness matrix
of the structure can be assembled in a very efficient and
quick manner by the direct summation of the appropriate

stiffness coefficients of the member stiffness matrices.

2.2.1 Frame Elements.--The relation between the end dis-
placements and the forces acting at the ends of a frame

member is given by

 

 

 

 

rulI rfl‘r
u2 f2
1113 f3

Kf 0 ‘ L: j ? 0000(18.)
uh f4
U5 1‘5
L116» L136;

or in matrix notation
KfU’= f 0000(1b)

where Kf is the member stiffness matrix of the frame
element, f the vector of end forces and u the vector of

end displacements. For the analysis a consistent procedure
has been adopted to specify the incidence of the frame
elements. The lower end of the vertical members and the
left end of horizontal members are taken as positive ends.
The member stiffness matrices of frame elements for

different cases of moment releases are given in Appendix I.

2.2.2 Triangular Wall Elements.--The corner displacements
u and forces f acting at the corners of a triangular wall

element are related by
Kwtu = f 0000(2)

where Kwt is the stiffness matrix of the triangular wall
element, u the vector of corner displacements and f the
vector of forces acting at the corners of the element.

The vector u contains only the two translational components

15

of the displacements at each corner and f the corresponding
force components. The matrix Kwt is computed by a method
suggested by Zienkiewicz and Cheung (15).

Consider a typical triangular wall element shown in
Figure 4. A displacement function is chosen such that the
displacement components are compatible at the interfaces of
the various finite elements. This condition is satisfied
by assuming linearly varying boundary diSplacements of the

form

ux = kl + kZX + k3y 0000(38)

Uy k4 + kSX + kéy 0000(3b)

where ux and uy are the displacements in the x and y
directions at any point (x,y) and k1, k2 etc. are constants
which can be determined from the known values of the corner

displacements. Solving, the following relations are

obtained.
1=1
- 1
uy ‘ 51- (ai + biX + 01301121 ....(4b)
i=1
where A = the area of the triangular element,
a1 = X2y3 — x3y2 0000(53)
bl = y2 - YB 0000(5b)

Cl = X3 - X2 0000(5c)

16

x1,y1, x2,y2 and X3,y3 being the coordinates of the corners.
Other constants a2, b2, 02 etc. can be obtained by cyclic
permutations of the relations given above.

The relation between strains and displacements is

 

 

given by

r 1

exx 22x
ex

9 = <eyyg 2" ally 0000(6)

by

exy 32X + 223/

L J oy 6X 1

 

where e is the strain vector, exxo eyy, and exy are the
components of direct strain in the x direction, direct
strain in the y direction and the shear strain in the xy

plane respectively. Using Eq.(4) for u and uy the following

X

relation is obtained.
e = Bt u ----(7)

where

Bt = O 01 O 02 0 C3 ....(8)

L01 b1 02 b2 03 b3 ....

 

 

It is obvious that the strain field is constant
within a given triangular element. Since a state of plane

stress has been assumed, the stress components can now be

17

computed from the following relationship.

 

 

 

 

 

 

fimx‘ '1 n olegl
_ E
fidyy>’ - I:;2 n 1 O fieyy? 0000(9)
o o o llfle
\ XyJ .L 2.11ka

where Oxx and ny are the normal stress components in the
x and y directions and ny the shear stress component in
the xy plane. E is the Young's modulus and n is the
Poisson's ratio of the wall material. In matrix notation

we have
C = De 0000(10)

where

D = n 1 O E, 0000(11)
1-n2

 

 

and o is the stress vector.

By the principle of virtual work, equating the work
done by the external forces to the work accomplished by

the internal stresses the following relationship is obtained.
uT f = eT dAt ....(12)

where t is the thickness of the wall element and the

superscript T denotes the transpose of the matrix.

18

Substituting for e and o and grouping terms to one side

of the equation, we obtain,

T T _

u (f - Bt D Bt AtU) — O0 0000(13)
Since the corner displacements u are arbitrary,

f = B¥ D Bt Atu = KWL u coo-(1”)

where

B¥DBt At. ....(15)

Kwt

2.2.3 Rectangular Wall Elements.--Consider a typical

rectangular element shown in Figure 4. The origin of the
local coordinate system is taken at the lower left corner
of the rectangle. The following dimensionless coordinates

are introduced.

9
fl
ANN

=‘X
and B b 0000(16)

where a and b are the dimensions of the element and, a and

B are the dimensionless coordinates corresponding to the x
and y directions. Let u be the vector of corner displace-
ments again; it contains eight components, i.e. two for each
of the four corners. Simple displacement functions,
assuming linearly varying boundary displacements are given

by the relations,

ux 013 + 0298 + 039 +- cu ....(17a)

Uy C5a + 06GB + C78 + 08 0000(17b)

19

where the constants c1, c2 etc. can be determined from the
known values of displacements at the four corners. Solving,

the displacements at any point (x,y) are given by

ux (1-a)(1-B)u1 + (1-a)Bu3 + aBu5 + a(1-B)u7 ...(18a)

uy (1-a)(1-B)u2 + (1-a)Bu4 + a8u6 + a(1-B)u8 ...(18b)

From the known relations between the strain and displacement

components. we obtain

 

e = Bru 0000(19)
where
l”-(1;§) o “.3. 0 -§ 0 1:12 0‘
‘ a a a a
Br" 0 41:11.) 0 1:2 0 a 0 -g ”(20)
b b b b

41:91.) -(1_-&) (1:92) -§ 2:. B ~91 1_-E

b a b a b a b a 1

L. _.

 

The corner displacements and forces acting at the

Corners of the wall element are given by the relationship

Kwru = f 0000(21)

Where Kwr is the member stiffness matrix of the rectangular
wall element. To obtain Kwr we proceed in a very similar
manner and equate the work done by the external forces to
the work accomplished by the internal stresses. It should

be noted that the matrix Br, unlike Bt is a function of the

20

position variables, and as such, the work accomplished by
the internal stresses should be obtained by integration.
The internal stress components are determined by using
Eq.(9). The member stiffness matrix Kwr is given in
Appendix I. The method described above for computing the
stiffness matrix of a rectangular element is according to
Przemieniecki (10).

The assumption of linear edge displacements ensures
compatibility of displacements at the interfaces of the
finite elements. It can be observed that the displacement.
components represented by Eqs.(17) are second degree
functions similar to a hyperbolic paraboloid and, when
compared with the previously treated case of a triangular
element, are suggestive of a better representation of the
state of stress. This, in fact, is verified during the

course of this investigation.

2.3 Joint Stiffness Matrix
The equilibrium equation for a statically loaded

framed structure can be expressed in matrix form as
KU = P 0000(22)

Where K is defined as the joint stiffness matrix of the
Sturucture, U the vector of joint displacements and P the

JO int load vector .

21

In partitioned form

K = 0000(23)

 

 

LJN.1 JN,N (

where the J's are the submatrices corresponding to the
various free joints and N the number of free joints. Since
the member stiffness matrices of the frame and wall elements
have been generated with reference to global axes, the
submatr1ces Ji,j

of the contributions by the frame and/or wall members

are easily computed by the direct summation

incident at Joint i and Joint j. Since wall elements can
not transmit moments at the joints, displacement or force
components corresponding to joint rotations can not be

specified at the corners of a wall element. As such, the
number of rows in a typical submatrix Ji,j will be 3 or 2
according as i is a frame joint or an interior wall joint
and the number of columns in Ji,j

as j is a frame joint or an interior wall joint. Thus Ji,j

will be 3 or 2 according

will be a rectangular matrix when i and j are different

type of joints and the extra row or column will be made up
of zero entries, to make the matrix multiplication defined
in Eq. (22) possible. It may be noted that Ji'j = Jin.

Referring to Figure 1, the submatrix J1,9 would have

three rows and two columns, since Joint 1 is a frame joint

22

and Joint 9 is an interior wall joint. By definition, J1,9
multiplied by the column vector of displacements at Joint 9,
gives the force components at Joint 1 contributed by members
incident at both Joints 1 and 9. Since wall elements can
not transmit moments at the joints, irrespective of the
magnitude of the translational displacements of Joint 9,

the contribution from J1'9 to the moment at Joint 1 is zero.
Therefore the third row of J1’9 consists of zero elements.
Also, any rotation of Joint 1 can not affect the force
components at Joint 9. Therefore the submatrix J9’1 which
has two rows and three columns, contains all zero elements

in its third column.

2.4 Static Analysis
For a given set of static joint loads, the joint

displacements are given by

U = K-1P. ....(24)

A computer program was written to obtain the load-
displacement relation and crack propagation pattern in the
filler wall for steadily and linearly increasing joint
loads. Initially, loads are specified at one or more
joints. The magnitude of the loads are such that no cracks
develop in the wall elements and no plastic hinges form in
the frame members. The maximum tensile stress and the

wall element in which this is developed are determined.

The loads are increased proportionately such that this wall

element just cracks. The joint stiffness matrix of the

23

structure is now modified as explained in section 2.1, and
the procedure is continued until the entire wall panel is
cracked. The sequence in which the wall elements crack
gives the crack propagation. At every step it is also
checked whether plastic hinges form in any frame member.
If any plastic hinge forms, the joint stiffness matrix is

modified accordingly.

2.4.1 Effect of Reduction Factor.--The model structure
considered for this part of the study is a single story
reinforced concrete frame with concrete filler walls,
experimental work on which has been reported by Yorulmaz
and Sozen (14). Steadily increasing horizontal loads were
applied at the quarter points of the beam. The load-
displacement relation and crack propagation pattern in the
wall were obtained for various values of a "reduction
factor," which is defined to be the ratio of the elastic
moduli of the wall material after and before cracking.
Figure 5 shows the computed and experimental load-displace-
ment plots. In this analysis the wall panel was divided
into an 8 x 8 mesh with 64 triangular elements as shown in
Figure 6. Results are given for two extreme values of the
reduction factor assumed for the analysis. The crack
propagation pattern is also shown in Figure 6. The location
of the initialcrack and the general pattern of crack
propagation agreed well with the results reported by

Fedorkiw and Sozen (4).

24

2.4.2 Ultimate Load Capacity.--A second type of model
structure considered is a concrete frame with masonry filler
walls, experimental work on which has been reported by
Sachanski (12). The wall panel was divided into 8 tri-
angular elements and a steadily increasing load was applied
at the story level. The load capacity is given by the
magnitude of the applied load which causes all the wall
elements to crack. It may be mentioned here that this load
will be far greater than the load capacity of the same frame
without filler walls, the ultimate load in this case cor-
responding to the collapse mechanism. Table 1 shows the
experimental values of load capacity and computed values for
a reduction factor of 0.01. It is seen that the computed
values agree well with the experimental ones. Results

of Figure 5 and Table 1 would indicate that a reduction
factor of 0.01 is a reasonable value for concrete and masonry

filler walls.

2.4.3 Effect of mesh sige.--The horizontal displacements

of the single story structure referred to in Figure 5,

under horizontally applied loads of 10 kips each at the

ends of the beam, for several mesh sizes are shown in

Tables 2 and 3, respectively, for triangular and rectangular
elements. It is observed that there is convergence of
results with increasing mesh size, which is an essential
requirement of any finite difference or finite element

formulation. It is also observed that the size of the mesh

does not affect the displacement a great deal. Comparing

25

the results corresponding to the finest and coarsest mesh

in Table 3, the variation in the value of displacements is
about 15%. On the other hand the computation time required
is 45 seconds for the finest mesh and 0.16 second for the
coarsest mesh. It may also be noted that for the same

amount of computation effort, which depends on the size of
the joint stiffness matrix, rectangular elements are superior

to triangular elements.

2.5 An Optimal Numerical Model

It is obvious that there are bound to be some uncertain
elements due to the complex nature of frame-wall interaction.
Based on the data presented here and on the assumptions made
in the course of the analytical formulation, it is assumed
that an optimal numerical model to represent the lateral
stiffness characteristics of a frame wall system could be
obtained by treating the wall panel as a single rectangular
element. This minimizes computation effort and seems to

yield reasonably accurate results.

2.6 Effect of openings in the wall panel

The presence of openings in the wall panel reduces the
lateral stiffness of infilled frames. Figure 7 shows the
effect of openings on the lateral stiffness of a single
story frame. A constant horizontal load is applied at
story level. For various dimensions of the Openings, the
horizontal displacement at the story level has been plotted.

It may be noted that the lateral stiffness is not

26

appreciably affected for normal sized openings, i.e., for
values of h/H between 0.45 to 0.55 and b/B between 0.2 and
0.3. (See Figure 7 for definition of symbols used here.)
In the dynamic analysis the effect of openings has not
been considered. If desired this can be easily included

by a suitable modification of the computer program.

CHAPTER III
METHOD OF DYNAMIC ANALYSIS

3.1 General

The structure considered here is a multistoried
plane frame with filler walls in plane stress. The dynamic
loading consists of the horizontal and vertical components
of ground motion in the plane of the frame. In the
modified method of dynamic analysis described later in this
chapter, dynamic loading consists of horizontal components

of ground motion only.

3.2 Lumping of Masses

In the formulation of the equations of motion the
structure is treated as a discrete system of masses lumped
at the joints. Dead and live loads from each floor are
assumed to be uniformly distributed over the beams. The
mass of a beam or column is lumped equally at the ends of
the member. The mass of a wall panel is lumped equally at
the corners. The rotary inertia of a joint is taken to
be equal to the sum of the moments of inertia about the
joint as contributed by the columns and beams incident at
that joint. The contribution of each member is taken to
be that of a rigid bar with half of the mass of the member
distributed uniformly over a quarter of its length. Walls

are ignored in the calculation.cfi'the rotary moment of

27

28

inertia since they do not add to the rotary stiffness of
the joints. In the modified procedure, the mass elements
are first computed as indicated above and then all those
mass elements which correspond to the joints of the same
floor are summed up and lumped at one point. Rotary moment

of inertia is ignored in the modified procedure.

3.3 Numerical Integration Procedure

The Analysis is formulated in terms of the joint
displacements relative to the supports. The transient
response of the structure is obtained by a numerical
integration of the equations of motion of the joint masses.
Assuming a velocity damping, proportional to mass inertia,

the equations of motion are given in matrix form by
MU + of] + KU = P ....(25a)

In modified form this can be written as

00 1

U + Cf] + M- KU = 'fi 0000(25b)

S
where M is the diagonal mass matrix, U, U and U are the
vectors of joint accelerations, velocities and displace-
ments respectively, K is the joint stiffness matrix, c is

a damping constant and Ug is a vector formed from the
prescribed ground accelerations. In Eq. (25a) C is the
damping matrix. Elements of Ué corresponding to the
horizontal motion of joints will be equal to the horizontal

component of ground acceleration, those corresponding to

the vertical motion of joints will be equal to the vertical

29

component of ground acceleration and those corresponding
to rotation of joints will be equal to zero.

The initial values of the displacements and velocities
of the joints are taken as zero. Theinitialvalues of the
forces in the various members are those corresponding to
the static loads before the commencement of dynamic loading.
The initial accelerations of the joints are readily computed
from Eq. (25b). The displacements, velocities and
accelerations of the joints at any time after the ground
motion commences, are computed by a step by step numerical

integration procedure, using the following relations.

. 2“
U1 3 U0 + dtUO + E5)- UO 0000(263)
.0 - -1 o 00
1 " “M KUl - CUO - Ugl 0 0 0 0 (26b)
131 = 1'10 + 0.5dt(Uo + '61) ....(26c)

where dt is the time interval used, the subscript 0 denotes
known conditions at the beginning of any given time step
and subscript 1 denotes quantities to be computed at the
end of the time step.

To minimize computation time, the largest possible
time interval should be used. To obtain a stable solution
this is limited to a certain fraction of the smallest period
of the system. The theoretical value of this fraction is
of the order 1/h, and for the present investigation a value
of 1/6 is used. The smallest period is easily estimated

from the largest eigenvalue of the matrix M'IK. Due to

30

the nonlinear nature of the response, K may vary with time
and so would the optimum time interval. Whenever a crack
forms in a wall element there is a substantial increase

in the value of the optimum time interval. The computer
program automatically calculates the best time interval
whenever there is a change in K due to formation of cracks
in wall elements and uses it for further analysis.

For any step in the numerical integration procedure,
the changes in the joint displacements cause changes in
the stresses in the frame and wall elements. For given
values of joint displacements these are easily computed
from known stiffness properties of the elements. If the
stress in any member at the end of a given time step
exceeds the elastic limit values, causing a wall element to
crack or a plastic hinge to form in a frame member, a
smaller time interval estimated by interpolation, is used
for the present step so that the crack in the wall element
or the plastic hinge in the frame member forms just at the
end of the time step. The joint stiffness matrix K is now
modified accordingly and the procedure continued. At the
end of each time interval, the energy absorption due to the
incremental rotation of each plastic hinge is also checked.
If it is negative, the elastic state is reinstated for the

concerned hinge.

3.4 ModifiedPgocedgre With Reduced Degrees of Freedom
In the procedure discussed before there are three

degrees of freedom for each of the N free joints and the

31

structure has 3N degrees of freedom. In this section a
modified procedure is presented. In this procedure the
axial and rotational modes are eliminated and consequently
the smallest period of the modified system would be much
larger than that of the previous system with 3N degrees
of freedom. This would make possible the use of a much
larger time interval for the numerical integration and a
considerable saving of computation time. For the sake of
simplicity, the previous system with 3N degrees of freedom
and the modified system will henceforth be called "the
unreduced system" and "the reduced system", respectively.
The number of degrees of freedom of the reduced system
is equal to the number of floors. The equations of motion

are given by

*-1

fi* + CU* + M K*U* = '1}; 0 0 0 0 (27)

where the superscript * refers to the reduced system. The
generalized coordinates of this system correspond to the
horizontal displacements of the joints along a line of
exterior columns. Consequently, there is one coordinate
for each floor. The elements of M* are the sum of the
elements of M corresponding to all joints at the same
floor level. The joint stiffness matrix of the reduced
system, K*, can be computed from K as follows.

Let F and F* be the joint flexibility matrix of the
unreduced and reduced systems respectively; i.e., F = K"1

and F‘eK*'1. From physical considerations it is apparent

32

that F* is a submatrix of F. Thus to obtain K*, F is first
computed by inverting K. The appropriate elements of F are
picked to form F*, the inversion of which yields K*. To
illustrate this,l the structure shown in Figure 8 is now
considered. The unreduced system corresponding to this
structure has eighteen degrees of freedom and the joint

flexibility matrix is given by

" f1,1 - - - - - f1,18

F = _ l’j ' 0000(28)

 

 

_f18,1- - - - - f18,18

and the joint flexibility matrix of the reduced system is

given by

l— — 1F _

'36 * if
f1,1 f1,2 f1.3 f1,1 f1,4 f1,7

* * * *
F = f2,1 f2,2 f2,3 = fu,1 fu,u fu,7 ----(29)

 

 

 

 

*- * *-
_f3.1 f3.2 £3.34 Lf7.1 f7.4 f7.7

The damping constant c is assumed to be the same in
both systems. The displacements, velocities and accelera-
tions of the joints of the reduced system are computed by a
iitep by step numerical integration procedure using the

following relations which are similar to Eqs. (26).

33

U; = U3 + dtfig + 0.5(dt)zfig ....(30a)
“-x» *-1-* at '4;- "-5

U1 = "M K U1 -' CUO - Ugl 0 0 0 0 (30b)
1.]; = fig + 005dt(fi3 + fi:)0 0000(30C)

Because of the nonlinear nature of the response, it
is necessary to compute the internal forces in each element
at every step of the numerical integration procedure. This
requires the determination of the vector U, the displace-
ments of the unreduced system. These displacements are
computed as follows. The vector R*, defined as the internal
resistance at the joints due to displacements U*, is given

by the relation
R* = K*U*. 0000(31)

Let R be the unreduced resistance vector corresponding to
U. The vector R may be constructed from R* by setting

each element equal to zero except those corresponding to
the horizontal displacements of the joints of the unreduced
system; and those corresponding to the horizontal displace-
ments will be taken to be the same as those of R*. This

is tantamount to assuming that the inertia and damping
forces in the axial and the rotational coordinates are
:negligible. Subsequent numerical results show that this is
a reasonable assumptions. The vector U can now be computed

from

U = K'lR ....(32)

34

The rest of the procedure is the same as before. Whenever
there is a change in the stiffness properties of an element,
due to cracking of a wall element, or the formation or
disappearance of a plastic hinge in a frame member, K*
must be recomputed.

It should be noted that the computation of K*
involves the inversion of K, and as such the procedure
will not operate once the joint stiffness matrix K becomes
singular. However this does not appear to be a serious
disadvantage, since in most cases, if a structure becomes
statically unstable, collapse would often follow shortly,
even in dynamic response. Besides, if necessary, one can
always proceed with the analysis by reverting to the

original procedure when K becomes singular.

3.5 Computer nggram

An outline of the program developed for the study is
presented in this section, the program itself is given in
Appendix II. The important steps in the program are
described in the order in which they are executed.

1) Input information includes location of joints,
the incidence, geometrical and physical
properties of all structural elements, number
of floors, number of degrees of freedom, time
limit up towhich analysis is to be continued,
etc. Both methods of analysis discussed in

the preceding sections are incorporated in the

2)

3)

4)

5)

35

program. For a given structure, the value of
the number of degrees of freedom specified
directs the program to select the apprOpriate
method of solution.

From the information input in 1, the stiffness
matrices of the frame and wall elements are ,
computed. The joint stiffness matrix K is next
assembled. If necessary, K* is computed.

The elements of the member stiffness matrices
and the joint stiffness matrix K are stored as
one dimensional arrays. Only the upper or lower
triangular elements are stored. This makes
possible a considerable saving of computer
memory.

The mass matrix is next assembled using the
procedure outlined in section 3.2.

The optimum value of the time interval namely
1/6th of the smallest period of the system is
computed from the largest eigenvalue of the
matrix M-lK obtained by an iteration procedure.
Information regarding earthquake loading is
input. At any desired time the ground accelera-
tion components are computed by a straight line
interpolation between the appropriate discrete
points. The input motion of supports is adopted
from the two components (North-South and Up-Down)

of ground acceleration records of the El Centro

6)

7)

36

Earthquake of May 18, 1940. This is accomplished
by a subroutine which can also be easily modified
to include dynamic loading on the joints.

The dynamic analysis is accomplished using the
procedure outlined in sections 3.3 and 3.4.
Information is output at designated intervals to
permit a running record of joint diSplacements,
forces in the frame members, maximum tensile
stress in the wall elements and location of
plastic hinges. In addition, the above informa-
tion is given as output, whenever there is a
change in the structural properties of an element,
i.e., whenever there is a crack formation in a
wall element or the formation or disappearance

of a plastic hinge in a frame member.

The program checks joint displacements at every
time increment against prescribed maximum values
and will exit once the prescribed maximum values
are exceeded. At the end, the final status of
joint displacements, velocities and accelerations,
forces in the frame members, maximum tensile
stress in the wall elements, location of plastic
hinges and the energy absorbed due to the
rotation of plastic hinges are furnished as
output. This data can be used to continue the
analysis. The displacement of the top story and

the corresponding time are furnished as punched

37

output. This information is used to plot the
horizontal diSplacement of the top story versus
time in the computer plotter CALCOMP 643. This
procedure was necessary because the program has
been written for the CDC 6500 computer system

and the plotter forms part of the CDC 3600 system.

3.5.1 Limitations of the Program.--The program has been
dimensioned to analyse structures with a maximum of 40
joints, 40 frame members and 40 wall elements. The frame
should also be rectangular and there is no restriction to
the number of bays or number of stories. Dynamic loading
consists of components of ground accelerations due to
earthquake motion. However, the program can be modified
easily to handle different types of plane frames, more
number of joints, frame members and wall elements and also

other types of dynamic loading.

I»!

CHAPTER IV
NUMERICAL RESULTS

4.1 The_§tructure

The structure considered for the present study is a
three story steel frame with concrete filler walls, six
inches thick. The dimensions and sectional properties are
shown in Figure 8. Values of other parameters are; elastic
modulus of steel 30 x 106 psi, elastic modulus of wall
material 2.5 x 106 psi before cracking and 2.5 x 104 psi
after cracking, Poisson's ratio of concrete zero, yield
stress for steel 33000 psi and cracking stress for concrete
150 psi. A.10% critical damping (based on the fundamental

mode) is assumed in obtaining the following results.

4.1.1 Loading.—-For initial static loading, a uniformly
distributed load of 2400 lb./ft. is assumed on each beam.
Dynamic loading consists of selected portions of the

El Centro Earthquake of May, 1940. Failure of the structure
is assumed to occur if the horizontal displacement exceeds
2" per story height or the rotation of any joint exceeds

0.2 radians.

4.1.2 Time Increment.--As mentioned previously, for the
numerical integration procedure, the choice of time incre-
ment depends on the smallest period of the system. For the

unreduced system the time interval used is 0.002 second.

38

I")

(ll

39

For the reduced system the time step used is 0.004 second
before the first floor wall cracks and 0.01 second after-
wards. These values were estimated from the largest eigen-

1K. In the final version of the

value of the matrix M-
computer program given in the Appendix, the Optimum time
interval to be used for the analysis is computed by the
program itself and modified whenever a wall element develops

cracks.

4.2 Numerical Examples

The results presented in this chapter may be divided
into two parts. In the first part, four examples of earth-
quake loading are considered for the study of the effect of
walls on the dynamic response of the structure. The
responses with and without wall-stiffness are presented in
the form of graphs, with the horizontal displacement of
the top story plotted against time. In the second part,
the dynamic response is studied by the modified procedure
in which the axial and rotational modes are eliminated.
The results are compared with those obtained by the previous
procedure in which all the modes are retained. Items
compared include the horizontal displacement of the top

story and the computation time.

4.3 Effect of Wal;§
The linear responses of the structure with and without
the effect of wall-stiffness are shown in Figure 9. The

structure is subjected to the ground motion of El Centro

40

Earthquake of May, 1940, from 1.5 to 5.0 seconds with a
scaling factor of 0.005 to ensure that the moments in the
frame members and the maximum tensile stress in the wall
elements do not exceed the elastic limit and that the
response would be entirely in the linear range. It is
seen that, if the wall-stiffness is taken into account,
the structure vibrates with a high frequency and low
amplitude. Without wall-stiffness the response has a
large amplitude and low frequency.

Figure 10 represents the first example of nonlinear
responses with and without wall-stiffness. The loading
is the ground motion of the same El Centro Earthquake
from 1.5 to 4.0 seconds, with a scaling factor of 0.5.
If the wall-stiffness is considered, the first floor wall
cracks soon after loading commences. The response has a
relatively low amplitude and high frequency. The vibration
is damped out after the ground motion ceases and the
structure remains stable. When wall stiffness is neglected
the response has a large amplitude and low frequency, and
the structure fails due to excessive rotation of Joint 5.

In the next example shown in Figure 11 the nonlinear
response to a more severe earthquake is considered. The
ground motion of the same El Centro Earthquake from 1.5 to
4.0 seconds with a scaling factor of 1.0 is used. With
wall-stiffness taken into account, soon after loading
commences, the walls crack one by one starting from the

ground floor. The vibrations are damped out after the

41

ground motion ceases and the structure remains stable.

The amplitude is substantially larger and the frequency
lower when compared with the previous example in which two
of the three walls did not develop cracks. When the wall-
stiffness is ignored the structure fails within 1.5 seconds
of loading, due to excessive rotation of Joint 3.

Figure 12 shows the last example of nonlinear re-
sponse, the loading being the ground motion of the same
earthquake from 24.0 to 30.0 seconds with a scaling factor
of 1.0. With wall-stiffness, the walls crack one by one
starting from the ground floor soon after loading commences.
The response is quite similar to the previous example,
except that the ground motion is present during the entire
period for which the response is plotted. Without wall-
stiffness the structure vibrates with a relatively large
amplitude and low frequency. For a given interval of time
the energy absorption due to the rotation of plastic hinges
is also greater with wall-stiffness ignored than that with
wall-stiffness considered.

The above results indicate that even when all the;
wall elements are cracked, the contribution of walls to
the lateral stiffness of the structure and its dynamic
response is too considerable to be neglected. The computed
fundamental periods of vibration of the structure with and
without wall-stiffness are shown in Table 4. The funda—
mental periods,in seconds, of the structure with the effect

of walls ignored, with all walls cracked and with all walls

42

intact are 2.170, 0.855 and 0.200, respectively. Expressed
as a percentage of the fundamental period with walls
ignored, the fundamental period with all walls cracked

and with all walls intact are, 40% and 9%, respectively.

4.4 Effect of Axial and Rotational Modes

Figure 13 shows the dynamic responses of the struc-
ture without considering wall-stiffness by the modified
procedure with three degrees of freedom and by the previous
procedure with eighteen degrees of freedom. It may be seen
that there is hardly any difference in the response. The
loading conditions are the same as for the previous example,
i.e., Figure 12. The computation time required by the
modified method is only about 25% of that required by the
previous procedure.

The graphs shown in Figure 14 represent the responses
of the structure wigg wall-stiffness by the two methods,
the conditions of loading being the same as in the previous
example. This is representative of cases in which somewhat
larger discrepencies are found between the responses
computed by the two methods. In the unreduced system
cracks develop in all the walls, whereas in the reduced
system cracks develop in the first and second floor walls
only. (Actually the maximum tensile stress developed in
the third floor wall falls just short of the cracking
stress so that it remains intact.) Thus the modified

procedure would appear to be slightly less conservative.

43

It would seem reasonable to expect this considering the
fact that the axial and rotational modes have been elimi-
nated and also the structure is not subjected to the
vertical components of ground motion. There is no appre-
ciable difference in the amplitudes and the general trends
of the response. With one wall still uncracked the
response by the modified procedure has a slightly higher
frequency. The computation time required by the modified
procedure is again only about 25% of that required by the
other procedure with eighteen degrees of freedom. The
periods of the responses in the various graphs presented

above agree well with the computed values given in Table 4.

4.5 Egeration Procedure for Acceleration of Joints
From the relation specified in Eq.(25b) the accelera-

tion vector at a given time, say, t1 should be

U1 2 —M-1KU1 - fig]. "' CU]. 0 0 O 0 0 (33)

In the numerical integration procedure, Eq.(26b) was
actually used, the difference being,UO was used in place

of U1 in order to avoid iterations. This procedure amounted
to assuming that the damping force is proportional to the
velocity at the previous time step in stead of the current
velocity. It was anticipated that this would be a justi-
fiable assumption because the time interval was usually

so small that it would hardly make any difference in the

solution. Two numerical problems were actually solved

44

with and without iteration. The time-displacements were
almost identical, thus confirming the validity of the use

of Eq.(26b) in stead of Eq.(33).

CHAPTER V
SUMMARY AND CONCLUSIONS

The development of an "optimal" numerical model and
its application to the elasto-plastic dynamic analysis of
building frames with filler walls have been presented.
The nonlinearity of the structural response, due to the
formation of plastic hinges in the frame members and/or
crack propagation in the wall elements has been taken into
account. The dynamic analysis has been embodied in a
computer program written in FORTRAN IV. This has been
used to study the dynamic response of a three story steel
frame with concrete filler walls subjected to an earth-
quake loading. The study also includes the effect of the
axial and rotational modes on the dynamic response.

For the study of the modelling of the stiffness
characteristics of the structure, the filler walls are
treated as finite elements in plane stress, interacting
with the frame such that the translational displacements
of the joints are compatible. After formation of crack
in a wall element, the stiffness properties of the struc-
ture are computed by assuming a reduced elastic modulus
for that wall element. Both triangular and rectangular
elements are considered. The effect of mesh size and the

cracking phenomenon in the wall panel are studied.

45

46

Computed results of load-displacement relations, crack
propagation pattern and ultimate load capacity have been
compared with known experimental results.

In the dynamic analysis the mass of the system is
handled by a lumping procedure. Two methods of analysis
have been derived. In the first, all the three degrees
of freedom of a joint are considered. In the second
(modified) method, the degrees of freedom associated with
the axial deformation of frame members and the rotation
of joints are eliminated. Mass-proportional viscous
damping has been included in the analysis which is formu-
lated using the joint displacements relative to supports
as variables. The transient response of the structure is
obtained by a numerical integration of the equations of
motion of the joint masses. The time interval used for
the numerical integration procedure is based on the
smallest period of the system at any instant. For the
dynamic analysis, the initial values of the forces in the
various members are those corresponding to the static
loads before the commencement of dynamic loading.

Based on the assumptions made during the course of
analysis and on the limited numerical results presented in
this report, the following observations may be made.

1. The modelling of the wall panel of a floor as a single
rectangular finite element interacting with the moment-
resisting frame could reasonably satisfactorily

account for the contribution of walls to the overall

2.

3.

4?

lateral stiffness of the structure.

The contribution of walls to the lateral stiffness and
consequently to the fundamental frequency and the
dynamic response of a structure appears to be too
considerable to be ignored, even when the walls are
cracked. In general it would be appear that the
inclusion of the effect of walls would lead to a
substantially conservative design.

Elimination of the axial and rotational modes makes
possible the use of a much larger time interval for
the numerical integration procedure, and thus resulting
in aconsiderable saving of computation time without
any appreciable loss accuracy.

Future extensions of this investigation could include;

(1) experimental study of the dynamic response of frame-

wall systems in the laboratory to form a basis of compari-

8011!

(2) contribution of floors to the rotational stiffness

of joints and the effect of floors on the dynamic response

of the structure.

48

Table 1. Ultimate load capacity of model infilled frame.

 

 

 

Frame Length Height of Cross Thickness Collapse load in tons
.Of be am columns section If wall. Sachanski's Computed
1n centi- 1n centi- of frame in can 1- tests values
meters meters 1n centi- meters

meters
1 350 280 15 x 60 3O 23. 2 22. 70'
2 350 280 15 x 60 15 8. 6 8. 12

 

 

 

 

 

 

 

Table 2. Effect of mesh size on load-displacement relation
using triangular elements

 

 

 

Pattern Dimension Number of Number of Horizontal
of joint frame wall displacement
stiffness members members of deck in
matrix inches x 10

8 3 4 0. 11965

 

 

 

XX 13 4 8 0.13025

L‘KABQL‘VAK‘VA
H§Q§VA§KAE 87 16 64
fi’AfiVAL‘VABVA

VA§VAEVA§VAB

      

0. 13884

    

 

 

 

 

 

PAL F’L E DUNE

49

 

 

 

 

 

 

 

 

 

 

 

 

Table 3. Effect of mesh size on load-displacement relation
using rectangular elements
Pattern Dimension Number of Number of Horizontal
of joint frame wall displacement
stiffness members members of deck in
matrix inches x 10
6 3 1 0. 12074
I l 9 4 2 0. 13091
ll ll I {J1 27 8 8 0.13960
87 16 32 0. 14017

 

 

 

 

 

 

Table 4.

50

Computed values of fundamental periods.

 

Condition of W alls

Fundamental Period

 

1.
Z.

3.

4.

5.

All walls intact
First floor wall cracked

First and second floor
walls cracked

All walls c racked

Effect of walls ignored

 

0. 200 sec.

0. 636 sec.

0. 806 sec.
0. 855 sec.

2.170 sec.

 

51

 

 

 

 

 

 

 

3 6
‘/”—‘Structure
1,2,3 - frame joints
2 5 9 - interior wall
.joint
1 4
9
7 8
////7///7/

 

FIGJ STRUCTURE GLOBAL COORDINATES

 

 

no.2 JOINT coono mugs

52

 

 

Frame Elements: (Incidence IJ)
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fiyi

  
 

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y
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0 IX.
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Triangular Wall Elements: (Incidence IJK)

 

 

 

 

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I x x
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Rectangular Wall Elements: (Incidence IJKL)

no.3 MEMBER COORDINATES

53

3 (XBOYB)

2
(X2:YZ)

leyl)

 

 

Triangular Finite Element: (Incidence 123)

 

 

 

 

 

 

bk 2
t—* a T71 3
(-—_——-'X _____fiiT_ 'b x/a = a
y/b=B
y
b/a = r
1 l 4

 

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LIST OF REFERENCES

Beck, H., "Contribution to the Analysis of Shear Walls,"
A.C.I. Journal, Vol. 59, Aug. 1962.

Cardan, B., "Concrete Shear walls Combined with Rigid
Frames in Multistoried Buildings," A.C.I. Journal, Sep.

1961.

Coull, A., and Choudhury, J.R., "Analysis of Coupled
Shear Walls," A.C.I. Journal, Sep. 1967.

Fedorkiw, J.P., and Sozen, M.A., "A Lumped Parameter
Model to Simulate the Response of Reinforced Concrete
Frames with Filler Walls," A Report to the Department
ofégefense, Office of the Secretary of the Army, June,
19 .

Frischman, Prabhu and Toppler, "Multistoried Frames
and Interconnected Shear Walls subjected to Lateral
Loads," Concrete and Construction Engineering, June,

1963.

Goldberg, J.E., and Herness, E.D., "Vibration of Multi-
storied Buildings, Considering Floor and Wall Deforma-
tions," Bulletin of the Seismological Society of America,
Vol. 55, pages 181- 200.

Gould, P.L., "Interaction of Shear Wall-Frame System,"
A.C.I. Journal, Vol. 62, 1965.

Khan, F.R., and Sbarounis, J.A., "Interaction of Shear
Walls and Frames," Journal A.S.C.E. Structural Division,
June, 1964.

Kokinopoulos, F. E., "Aseismic Dynamic Design of Multi-
story systems," Journal A. S. C. E., Structural Division,
June, 1966.

Przemieniecki, J.S., "Theory of Matrix Structural
Analysis," Mo. Graw Hill Book Co., New York, 1968.

Roseman, E., "An Approximate Analysis of Walls of Multi-
storied Buildings," Civil Engineering and Public Works
Review, Vol. 59, 196M.

Sachanski, 3., "Analysis of Earthquake Resistance of
Framed Buildings taking into Account the Carrying
Capacity of Masonry," Second World Conference oxlEarth-
quake Engineering, 1960.

64

13.

14.

15.

16.

17.

65

Smith, B.S., "Lateral Stiffness of Infilled Frames,"
Journal A.S.C.E., Structural Division, Dec. 1962.

Yorulmaz, M., and Sozen, M.A., "Behavior of Single
Story ReinforcedConcrete Frames with Filler Walls,"
Interim Report to the Department of Defence, Office
of the Secretary to the Army, July, 1967.

Zienkiewicz, O.C., and Cheung, Y.K., "The Finite Element
Method in Structural and Continuum Mechanics," Mo Graw
Hiél Publishing Company, Limited, Maidenhead, England,
19 7.

"Joint Committee Code for Lateral Forces," Proceedings
of the A.S.C.E., Vol- 77. separate no. 66, April, 1951.

Saghera, 8.8., "Effect of Shear Walls on Dynamic
Response of Frames," Conference on Finite Element
Methods and Applications in Structural Engineering,
Nashville, Tennessee, 1969.

A1.1

APPENDIX I

MEMBER STIFFNESS MATRICES

Stiffness Matrix of a Frame Element

 

H ['11 t“ >
ll

Case I:

 

O

Young's Modulus

Length of member

Moment of Inertia

66

Area of cross section

I
O\
1'11
P1

out-4
14 N

N
.l

 

Y1 Y°
ii Xj
Symmetric
A;
L
O lZEI
L3
O ~6EI 4E1
L2 L

No moment release at either end.

 

67

 

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L
0 El...
L3
O O O Symmetric
Kf ’
11:2 0 o 11::
L L
O - E1 0 O 3E1
L3 L3
O 2E1 O O - BI
1.2 II2

Case II: Moment release at positive end i.

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L3
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K - L2
f _
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L L
O - EI -§EI O fiEI
L3 L2 L3
0 O O O O

 

Case III: Moment release at negative end j.

 

 

68

 

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L

O O

O O O

Kf =

24.1 o 0
L

O O O
O O 0

Case IV: Moment releases

LE:
L

O O

O O O

 

at both ends.

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APPENDIX II

COMPUTER PROGRAM

A2.1 General

The analysis is accomplished by a main program called
FINELEM, twelve subroutines and two function subprograms.
All the basic information concerning the structure is read
by the main program. The subroutine FRSTIF generates the
stiffness matrices of the frame members and the subroutine
WSTIFF generates the stiffness matrices of the wall elements.
The subroutine JSTIFF assembles the joint stiffness matrix
of the structure. As already explained in Chapter III
the choice of the method of analysis is made by the number
of degrees of freedom declared in the input. In case the
modified method of analysis is chosen, the joint stiffness
matrix assembled by JSTIFF is inverted by the subroutine
INVERSE to yield the joint flexibility matrix of the un-
reduced system. The subroutine FLEX then forms the joint
flexibility matrix of the reduced system and the subroutine
MATIN computes the joint stiffness matrix of the reduced
system by inversion.

The subroutine MASS reads data regarding dead and
live loads on the structure and generates the mass matrix
of the unreduced or reduced system as the case may be.

Data pertaining to ground motion is read by the subroutine

7O

71

GROUND which generates the vector Ug or U; as the case may

be, at any desired time. by linear interpolation between
discrete points. The subroutine EIGEN and function TSTEP
determine the best time interval to be used for the
numerical integration procedure. All matrix multiplica-
tions are accomplished by the subroutine MATMULT. Function
IPOS maps elements of a symmetric matrix to an one dimen-
sional array and vice versa. The joint stiffness matrix

of the unreduced system, and the member stiffness matrices
of the frame and wall elements are stored as one dimen-
sional arrays to save computer memory.

The subroutine FRAMST computes the incremental forces
in the frame members due to incremental displacements
during any time interval. The maximum tensile stress in
the wall elements at any time is computed by the subroutine
WALST.

The dynamic solution is accomplished by a step by
step numerical integration procedure as already described

in the section on computer program in Chapter III.

A2.2 Variables used in the Computer Program
The variable names used in the program are listed

below in the alphabetical order:

Program FINELEM

AFR(J) = Area of section of frame member J:
COEF = ratio of elastic modulus of wall after and

before cracking;

72

CRSTW = cracking stress for wall;
CTIME = computer time elapsed in seconds;
DAMP = damping constant;
DPN = incremental rotation of the J end of a
frame member;
DPS = incremental rotation of the I end of a
frame member;
DWL = computed value of maximum tensile stress in a
wall element during any time step;
EF = elastic modulus of frame material;
EN = initial value of elastic modulus of wall
material;
FACT = scaling factor for earthquake loading;
FACTOR =_ interpolating factor used when crack
develops in a wall element;
FFR(I,J) = Jth element of internal force vector of
frame member I;
FLONG = length of a frame member considered;
FRAMEK(I,J) = Jth element of the member stiffness
matrix of frame member I stored as an
one dimensional array;
FRMI(J) = moment of inertia of frame member J;
FWL(I) = maximum tensile stress in wall element I;
ICOUNT = counter to keep track of endless looping;
IDEXN = variable identifying transition of plastic

to elastic state of J end of a frame member;

73

IDEXP = variable identifying transition from plastic

to elastic state of I end of frame member;

IDFN(I) = variable identifying transition from
plastic to elastic state of J end of
frame member I;
IDFP(I) = variable identifying transition from
plastic to elastic state of I end of
frame member I;
IFLAG = variable identifying wall element when
crack develops;
IFR(J) = joint number of I end of frame member J;
INDEX = variable controlling printing of results;
INQ = a common parameter for the main program and
subroutine ground which causes earthquake
data to be read on the first call only;
INTX = variable identifying a frame or wall element
when structral properties change during
any time interval;
IWALL(J) = joint number of I end of wall element J;
JFLAGN = variable identifying frame element when
plastic hinge develops at the J end;
JFLAGP = variable identifying frame element when
1 plastic hinge develops at the I end;
JFR(J) = jbint.number of J end of frame member J:
JWALL(J) = joint number of J end of wall element J:
KWALL(J) = joint number of K end of wall element J;
LWALL(J) = joint number of L end of wall element J;

7fl

MODEF(J) = variable identifying moment releases in
frame member J; MODEF = 1 for no releases,
2 for release at I end, 3 for release at
J end and 4 for releases at both ends;

NDEG = number of degrees of freedom;

NFLOOR = number of floors;

NFRAME = number of frame members;

NI = number of time steps after which results are

printed;

NJF = number of frame joints;

NJFREE = number-of.free joints;

NJOINT = number of joints including supports;

NJW = number of interior wall joints;

NN = number of degrees of freedom of the unreduced

system;

NSUP = number of supports;

NWALL = number of wall elements;

PLINCN(I) = incremental energy absorbed due to
rotation of plastic hinge at the J
end of frame member I;

PLINCP(I) = incremental energy absorbed due to
rotation of plastic hinge at the I
end of frame member I:

PLMOM(J) = yield moment in frame member J:

POISS = Poisson's ratio of wall material;

PROOF(J) = bending energy stored in frame member

J at yield;

75

PWORKN(I) = energy absorbed due to rotation of
plastic hinge at J end of frame
member I;

PWORKP(I) = energy absorbed due to rotation of
plastic hinge at the I end of frame
member I;

RU = incremental joint displacement vector of the

reduced system;

RUDF = joint displacement vector of the reduced

system at the end of the time step;

RUDI = joint displacement vector of the reduced

system at the beginning of the time step;

RUVF = joint velocity vector of the reduced system

at the end of the time step;

RUVI = joint velocity vector of the reduced system

at the beginning of the time step;

RUXF = joint acceleration vector of the reduced

system at the end of the time step;

RUXI = joint acceleration vector of the reduced

system at the beginning of the time step:

S = mass matrix:

SMJ = stiffness matrix of the unreduced system in
one dimensional array; (in case the modified
method is chosen, this is inverted and stored
as joint flexibility matrix of the unreduced
system)

STARK = stiffness matrix of the reduced system;

 

76

TINT = time interval used for the current step of
numerical integration;

TIME = time at any stage of the dynamic analysis;

TLIMIT = time upto which analysis is to be

carried out;

TMH = time interval for the numerical intergration;

TSTART = starting time of earthquake loading;

U = incremental joint displacement vector of the

unreduced system;

UDF = joint displacement vector of the unreduced
system at the end of the time interval;

UDI = joint displacement vector of the unreduced
system at the beginning of the time step;

UVF = joint velocity vector of unreduced system at
the end of the time interval;

UVI = joint velocity vector of the unreduced system
at the beginning of the time step;

UXF = joint acceleration vector of the unreduced
system at the end of the time interval;

UXI = joint acceleration vector of the unreduced
system at the beginning of the time step;

UXLIM = maximum permissible horizontal displacement

of joints;
UXLG = vector of accelerations Ug;
UZLIM = maximum permissible rotation of joints;
WALLK(I,J) = Jth element of the member stiffness

matrix of wall element I, stored as

77

WT = thickness of wall in inches;
XJ(J) = X coordinate of joint J;
YJ(J) = Y coordinate of joint J;
ZW(J) = ratio of current value of elastic modulus

to the initial value of elastic modulus of
wall element J; (for wall openings this

value is declared as zero).

Subroutine IPOS
IPOS(I,J) = Integer value defining the position
of the I,J element of a symmetric

matrix mapped on to an one dimensional

 

array.
Subroutine FRSTIF
A = Matrix used for temporary storage of stiffness

matrix during computation;
I = variable identifying the frame member whose

stiffness matrix is being currently computed;

MODE = variable identifying moment releases;
R = rotation matrix;
RT = transpose of the rotation matrix.

.3 ubr out ine VISTIFF
BATA = Ratio of the height to the breadth of a
rectangular wall element;
I = variable identifying the wall element whose

stiffness matrix is being currently computed;

78

W(I.J)

Jth element of the member stiffness matrix
of the wall element I, stored as one

dimensional array.

Subroutine MASS
ALFWiA = Fractional length of frame member over
which one half of the mass of the frame

member is lumped;

DENSE = density of frame material;

DNS = density of wall material;

NBAY = number of bays in the frame;

WLOAD = total dead and live load on beams in lb./in.

Subroutine EIGEN

B = Derived normalized eigen vector;

C = derived eigen vector;

EPSI = tolerence specified for the iteration;
procedure;

INDEX = counter for the number of iterations;

TMH = time interval computed from the largest

eigen value.

Subroutine GROUND
AH = Interpolated value of horizontal component of
ground acceleration;
AV = interpolated value of vertical component of

ground acceleration;

AXLH horizontal component of ground acceleration;

79

AXLV = vertical component of ground acceleration;
INQ = a parameter which causes the subroutine to

read data cards on first call only;

NOH = number of data cards for horizontal ground
acceleration;

NOV = number of data cards for vertical ground
acceleration;

QTH = time corresponding to any given value of AXLH;

QTV = time corresponding to any given value of AXLV;

TIME = time at which U or U* is to be generated by

g g
linear interpolation between discrete points;

UXLG

*-
vector U or U .
g g

Subroutine FRAMST

X = Incremental end displacement vector.

Subroutine WALST

ALFA = Dimensional x coordinate of a point under
consideration in the wall element;

BETA = dimensional y coordinate of a point under
consideration in the wall element;

PMAXT = variable used for temporary storage of
maximum stensile stress;

UU = vector of corner displacements in a wall

element.

Subroutine MATIN
N = Dimension of S or STARF;

80

S = stiffness matrix of the reduced system;

STARF = flexibility matrix of the reduced system.

Subroutine INVERSE

A = Joint stiffness matrix of the unreduced system
stored as one dimensional array;
N = dimension of the matrix to be inverted.

81

A2.2 Computer Program

000

000

90

100
110

III

PROGRAM FINELEM (INPUTcOUTPUToPUNCH)

COMMON IFR(40).JFR(40)~IWALL(40).JWALL(40).KWALL(40).
1FRAMEK(40.21).wALLK(4n.3e>.NWALL.NFQAMF.NJF.NJw.Nsup.

PLWALL(4O)oZW(4O)
COMMON/1/5MJc7120).Fc150).uc180).xcew.V(e).

166(308) 9HH(308)
COMMON/P/STAQFI20.20).STAQK<20.2of.s(120>.Noec
COMMON/B/ XJ(50)oYJ(50)oAFR(40)oFRMI(40)oR(6o6)oRTI6o6)o

18(606)QBB(396)OBT(603)OBC(693)oDDI3O3IOEFoEWvWTQPOISSo

?A(606)0NMAX9PMAXT
COMMON/a/orpcao.4).owL<40).FFQ(40.4:.FWLtao:
COMMON/GRND/UXLGIlzo)cAHoAVoINQ.FACT
DIMENSION uo:(120).uv1(120).ux1(120).UDF<120;.UVF(120).
1UXF<120:.pLM0Mc40).MODEF(40>.PwORKD<40).PwonKN<4o:.

2PLINCP<40).PLINCN(40>.IDFP(40).IDF~(40:.PR00F(40)

3.RU(?O>. RUDI(20)oRUVI(2O)oRUXI(?0)09UOF(203oRUVF(20)c
4RUXF(20)

READ NO. OF FRAME AND WALL JOINTS AND SUPPORTS
FORMATIléFSoO)

READ IIO,NJFoNJWoNSURoNDEGoNIoNFLOOR

FORMATIIéIS)

PRINT IIIoNJFvNJWqNSUP

FORMATIIH19* NO. OF FRAME JOINTS =*0139* NO. OF INTER*
1*10R WALL JOINTS c*oI30* NO. OF SUPPORTS =*oI3/)
PRINT IIZQNDEGoNFLOOR

II? FORMAT(* NO. OE DEGREES OF FREEDOM =*.[3.* NO. OF*

120

IRO

1* FLOORS 2*13/3,

NPUNCH = NFLOOQ 4 3 - 2
NJOINTzNJF+NJW+NSUP
NJFREE = NJF + NJW

NN = 3*NJF + 2*NJW

READ JOINT COORDINATES

DO 120 ItlcNJOINT

READ IBOQJQXJCJI¢YJIJI
FORMATIISCZFIOoO)

PRINT JOINT NO. AND COORDINATES

PRINT 140

140 FORMATI // * JOINT NO. X COORD. Y COORD. *

I//)
DO ISO I=IoNJOINT

ISO PRINT IGO.I¢XJ(I)0YJ(I)

000

000 000 GOO

000

82

16a FORMAT:7X.13.7x.2F12.2/>
READ NO. OF FRAME AND WALL ELEMENTS
READ 11O.NFRAME.NwALL
READ FRAME ELEMENT NO..1NCIDENCE. ETC.
DO 156 1:1.NFRAME
I6R READ I7OQJOIFRIJ)QJFQ(J)OMODEEIJIQAFR(JIqFRMI(J)oPLM0M(J)
17o FORMAT¢415.3F10.O:
READ wALL ELEMENT NO.. AND INCIDENCE
DO 180 I=IoNWALL _
180 READ llOoJoIWALLIJ)oJWALL(J)sKWALL(J)oLWALL(J)
READ 90. (zw(::.1=1.NwALL)
READ E OF FRAME AND wALL. THICKNESS OF wALL ETC.
READ 19n.EF.Ew.wT.RO15$.CRSTw.TL1M1T.TMH.COEF
READ 190.FACT.DAMR.TRTART
READ IROQUXLIMoUZLIM
1am FORMAT<8FI0.0)

ERINT FRAME AND WALL MEMBER INCIDENCE

PRINT I91

191 FORMAT(/* FRAME ELEMFNT I NODE J NODE *
1* AREA Mo! MAXoMOMo */)
DO 192 IzloNFRAME

10? PRINT 1930IoIFRII)¢JFR(I)9AFR(I)9FRMIII)0PLMOM(I)

191 FORMAT(3(IOXOIBIOBFIEO?/I
PRINT 104

194 FORMAT(/* WALL ELEMENT I NODE J NODE *
1*K NODE L NODE*//)
DO I95 I=IONWALL

195 PRINT I969IOIWALL‘IIOJWALLIIIOKWALLII)oLWALLIII

196 FORMATIEI9XQI3I/I
pRINT 197

197 FORMAT(//* E OF FRAME E OF WALL WALL THICKNESS *)
pRINT 1980EF9EWOWT

198 FORMATI/BIE120502XII
PRINT IQQQTMHQTLIMITQPOISS

199 FORMAT(/* TIME INTERVAL 3*0F8059* TIME LIMIT =*9E8.50
1* POISSONS RATIO =*oE805/I
PRINT ZOOOEACToDAMP

zoo FORMAT(/* AMPLIFICATION FACTOR =*.Es.5.* DAMPING *
1*COEEEICIENT =*9E805/)
PRINT 201

0

0(30

0(30

0(1()

201

202

206

24“

25°

320

500

.83

FORMAT(/* LIMITING VALUES OF JOINT DISPLACEMENTS FOR *
I*COLLAPSE*/)

PRINT 2020UXLIMOUZLIM

FORMAT(* MAX-HORLO JOINT DISPLACEMENT IN INCHES =*9F8.4v

1 * MAX. JOINT ROTATION IN RADIANS =*9F8-4/)

PRINT 2060CRSTW

FORMAT(/* CRACKING STRESS FOR WALL IN PSI*0F8o2/)
GENERATE MASS MATRIX

CALL MASS(O-25)

GENERATE STIFFNESS MATRIX OF FRAME AND WALL ELEMENTS

DO 240 I=IONFRAME

CALL FRSTIFIIoMODEFIIII

PROOFII) = PLMOM(I)**2/FRAMEKII¢2I) / 2.0
CONTINUE

DO 250 I=IONWALL

CALL WSTIFFII)

DO 300 IBIO3

DO 300 J=103

DDIIOJ) = 0.0

DDIIoI) = 1.0

DD(?02) 8 1.0

DDIIv?) 8 POISS

DDI291) = POISS

DDI3o3) 8 O.5*(1¢O - POISS)

DO 320 I=103
DO 320 J=193
DDIIoJ) = DDIIcJ)* EW/(loO - POISS)

GENERATE JOINT STIFFNESS MATRIX
CALL JSTIFF
COMPUTE BEST TIME INTERVAL TO BE USED

CALL EIGENINNQTMH)
INTIALIZE VARIABLES
ICOUNT = O

IDEXP 3 0 $ IDEXN = 0
TIME 3 TSTART

ING=O

INDEX=I

DO 500 I=IOIQO

FII) 3 000

UIIIBOOO

UVI(I)=0.0

UDI‘II3000

DO 501 I81020

RU(I) 3 OO 5 RUDI‘I) = O. S RUVIII) 2 Do

non

(30(1

501

511

S?1
522
5?3
«can
560
57n

575

5580

SF”
‘58?

4584

‘S85

55863

814

RUXIII) = 0.0

RUDFIII=OoO $ RUVE(I)=0.0 S RUXFII)=0.0
CONT I NUE

DO 510 I=1040

IDFP(I) = 0 $ IDENII) = 0

FWL‘I)=0.0

PWORKPII):0.0 $ PWORKNIII=000
DO 510 J=Io4

FERIIoJ)=0oO

DO 511 I=10NERAME

READ 1900(FFRIIQJ)0J=194)

SET INTIAL ACCELERATION

CALL GROUNDITIME)
IFINNoGToNDEG) GO TO S?1
DO 520 I=I¢NN
UXI‘I)=-UXLG(I)

GO TO 523

DO 52? I=IONFLOOR
RUXIII) = -UXLG(I)
CONTINUE

TINT = TMH
TIMF=TIME+TINT

CONTINUE
FACTOR =
IFLAG=O
JELAGP=O
JFLAGN=O
CONTINUE

Ion

COMPUTE INCREMFNTAL DISPLACEMENTS

IFINNoGToNDEG) GO TO 581
DO 580 I = IQNN
UII)=TINT*UVIII)+O.5*TINT*TINT*UXI(I)
F(I) = UDIII) + UII)
GO TO 590

DC 582 I=10NFLOOR
RUII) = TINT*RUVI(I)
DO 585 I=IQNFLOOR
SUM = 00

DO 584 ngONFLOOQ
SUM = SUM + STARKIIOJ) * RUIJ)
K = 3*I - 2

UVIIK) = SUM

DO 587 I=IoNN

SUM = 00

DO 586 J=IoNN

SUM = SUM + SMJIIPOSIIQJI)

+ 005*TINT*TINT*RUXI(I)

* UVI(J)

n<w(wnrwn

SR7
FOO

600

61”
650

700

85

UII) = SUM
F(I) = UDIII) + UII)
CONTINUE

COMPUTE FORCES IN FRAME AND WALL ELEMENTS

CHECK FOR CRACKING OF WALL ELEMENTS. FORMATION OF
pLASTIC HINGES IN FRAME ELEMENTS

ADJUST TIME INTERVAL IF NECESSARY

CALL WALST

CALL FRAMST

SUM=OOO

DO 600 I=IQNWALL
IEIDWLIIIoLFoOoO) GO TO 600
ONT 8 DWLII)

IFISUMoGTgoNT) GO TO 60O
SUM=ONT

INTX=I

CONTINUE

IFISUMoLToCRSTW) GO TO 650
IFLAG=INTX

INDEX = NI

TIME=TIME-TINT

FACTOR = CRSTW/SUM

TINT : TINT*(F(I)*FACTOR - UDIIIII/U(1)
IFITINToGToTMH) GO TO 610
IFITINTQLTOOOO’ GO TO 6?O
GO TO 630

TINT=TMH

GO TO 630

TINT=OOO

FACTOR=IOO

TIMF=TIME+TINT

CONTINUE

TTIN = TMH

DO 700 I=10NFRAME
IFIMODEFII).EO.2oOP.MODEF(I).FO.4) GO TO 700
ONT=ABS(FFR(I93)+DFR(I93),
IFIONToLEopLMOMIII) GO TO 700
STIME=TMH*(PLMOM(I) - ARSIFERIIO3)))/(QNT-ARS(EFR(I93)I)
IEISTIMEoGToTTIN) GO TO 700
TTIN = STIME

INTXBI

CONTINUE

IFITTINoGEoTINT) GO TO 750
IFLAGBO

JFLAGPSINTX

INDEX = NI
TIME=TIME-TINT+TTIN
TINTaTTIN

0(3()O

730

800

Ran

860

870

880

an:

eon

86

CONTINUE

TTIN=TMH

DO 800 I=loNFRAME

IEIMODEFII).EO.1.0D.MODEF(I).EO.A) GO TO 800
QNT=ABSIFFRIIQAI+DERI104))

IFIONTQLEQPLMOMIIII GO To 800

STIMF=TMH*(PLMOM(I) - ARSIEER(194))I/(ONT-ARSIFFR(194)))
IF(STIME.GT.TTIN) GO TO 800

TTIN=$TIME

INTX=I

CONTINUE

IF(TTIN.GE.TINT) GO TO BRO '

IELAG:O

JFLAGP=0

JFLAGN=INTX

INDEX = NI

TIME=TIME-TINT+TTIN

TINT=TTIN

CONTINUE

IFIIDEXP.NE.O-AND.IDFXP.EO.JFLAGP) GO TO 860
IFIIDEXNONEQOQANDOIDEXNOEQOJFLAGN) GO TO 870

GO TO 890

PRINT IOBSOIDEXP

IDFP(IOEXP) = 1
IFIMODEEIIDFXP).EOoII MOOEFIIDEXP)
IF(MODEF(IDFXP)oFOo3) MODEF(IDEXPI = 4
CALL FRQTIFIIOFXRoMODFFIIDFXP))

GO TO 880

PRINT InéqQIDEXN

IDFN(IDEXN) = 1
IFIMODEFIIDEXN).EO¢1) MODEFIIDEXN)
IF!MODEF(IDEXN).EO¢2) MODEFIIDEXNI
CALL FRSTIF(IDEXN¢MODEF(IDEXN))
CALL JSTIFF

TINT = TMH

IOEXP = 0 $ IDFXN = 0 $ JFLAGP = n S JFLAGN = 0

TIME = TIME + TINT

PRINT 885

FORMAT(/* ASSUME PLAQTIC HINGE TO AvOID ENDLESS LOOPING*/)
GO TO 575

CONTINUE

II
V

II II
b "

CHECK WHETHER ENERGY ABSORRED DUE TO ROTATION OF PLASTIC
HINGE IS NEGATIVE. RESTORE ELASTIC STATE IE WARRENTED.

DO 1000 I=19NFRAME

pLINCPII):OoO $ PLINCN(I)=0.0
IFIIDEP(I)¢NE.O¢OR.IDEN(I).NE.O) GO TO 1000
IFIMODEFIIIoEOoI) GO TO 1000

IP = IERII)

87

IN = JFQIII
FLONG = SOQTI(XJ(IN)-XJ(Ip))**?+IYJ(IN)-YJ(ID))**?)
CPSXJIIFRII))-XJ(JFR(I)I
ICp=3*IFQIII'?
ICN=3*JER(I)-?
IE‘CPQEOoOoO) GO TO 900
DDS=IUIICP+II ' UIICN+I)I/FLONG + UIICD+PI
DPN=IUIICP+II - UIICN+I))/FLONG + UIICN+2I
GO TO 910
GOO DPS=-(U(ICPI‘UIICNII/FLONG + UIICD+PI
DPN=~IUIICPI‘UIICNII/FLONG + U(ICN+2)
010 IEIMODEFII).NF¢2) GO TO 930
ON=DPS*FFR(I93)
IFIONOGEoOcOI pLINCDIII=ON
IF‘ONOGFO-OOIO) GO TO Infifi
IDEXP = I
pRINT QISOI
915 FORMAT(/* PLASTIC HINGE DISAPPEAPS AT + END OF FRAME *
1*MEMREQ *0I4)
INDFX = NI
970 MOOEFIII=I
925 CALL FRSTIFIIQMODEF(I))
TIME=TIME~TINT
CALL JSTIFF
GO TO EGO
OSO IFIMODEFIIIQNE03’ GO TO 950
QN=DPN*EER(IOAI
IFIONoGFoOoO) pLINCNIII=ON
IFIONQGFo-qoln) GO TO IOOO
IOEXN = I
PRINT QBSOI
935 FORMAT(/* pLASTIC HINGE DISAPPEARS AT - END OF FRAME *
1*MEMRER *0I4)
INDEX = NI
GO TO 920
QQO ON=DPS*FFRII93)
IFIONOGFQOOO) PLINCP(I)=ON
IF‘ONOGF0-00IO) GO TO IOOO
IDEXP = I
PRINT QISOI
INDEX : NI
MODEEIII=3
GO TO 9?5
Q60 QN:DPN*FFPIIO4)
IEIONoGEoOoO’ pLINCNIII=QN
IF‘ONOGFo-OQIO) GO TO IOOO
IDEXN = I
pRINT 9359I
INDEX = NI
MODFEIII=2

10nd
IOIn

1013

10pm
1030

IOBS

104m
qun

106A

106:

10An
1090
1160

IIIS

IISO

1160

(3530

1170

1171
117?

88

Go To 9P5

CONTINUE

IFIIFLAG.EO.6) GO TO 163G

PRINT IOISoIFLAG

FORMAT(/* CRACK FORMS IN WALL ELFMFNT NO. *oIS/I
ZWIIFLAG) = COEF*7W(IELAGI

DWLIIFLAG) = COEF * DWLIIFLAG)

Do 1020 I=Io36

WALLKIIFLAGoI) = COFF * WALLKIIFLAGoI)
IFIJFLAGP.EO.n) GO To 1060

PRINT IOBSQJFLAGP

FORMAT(/* PLASTIC HINGE FORMS AT + END OF FRAME MFMBER*

1I4)

IFIMOOEFIJFLACP).NF.I) GO TO 1040

MOOFFIJFLAGP)=9

GO TO 1050

MODEFIJFLAGD>=4

CALL FPSTIEIJELAGPoMOPFFIJFLAGP))

GO TO 1100

IFIJFLAGNoE0.0) GO To 11mg

PRINT 106%.JFLAGN

FORMAT(/* PLASTIC HINGE FORMS AT — FND OF FRAME MFMRFP*

IIA)

IEIMODEEIJELAGNIoNEoI) GO TO IORO
MODEEIJELAGNI = 3

GO TO 1090

MODEFIJFLAGNI=4

CALL ERSTIEIJELAGNQMODEE(JELAGN)I
CONTINUE

IOEXP = O S IDEXN : n

DO 111; I=IoNFRAME

IDFP(I) O

IDFN(I) O
ONTY=(E(1)-U(1)+TINT/TMH*U(1))/E(1)
IEIIFLAG.EQ.O) EACTOR=ONTY

DO 1150 I=19NWALL

DWLII) = DWL(I)*EACTOR

EWL(II = DWLII)

DO 1160 I=10NERAMF

DO 1160 J=Io4

DERIIOJI = DERIIOJI * TINT/TMH
FERII9J1=EFRIIQJI+DERIIQJ)

BEGIN NUMERICAL INTEGRATION

DO 1170 ISIQNN

UDFII)=UDI(I)+U(I)* TINT/TMH
IF(NN¢EO¢NDEG) GO TO 1173

DO 1172 IaloNELOOQ

RUDFII) = RUDIII) + RUII)*TINT/TMH

1173

IIRO
1190
1131
11R?
1191
110:

C

C

C

1300

140m

1401
1405

1410

144R
14:0

P960

89

CONTINUE

CALL GROUNDITIME)

IFINNoGToNDEG) GO TO 1181

DO 1190 I=19NN

SUM=OQO

DO 1180 JIIoNN
SUM=SUM+SMJ(IDOQIIQJ))*URF(J)

UXF(I)= ~SUM/SII) - UYLGIII - UVIII)*OAMP
UVFIII=UVIII)+O.S*TINT*(UXI(II+UXFIII)

GO TO 1195

DO 1191 I=IoNFLOOR

SUM 8 O.

D0 1182 J=loNFLOOR

SUM = SUM + STARKIIQJ) * RUOFIJ)

RUXFII) = -SUM/S(I) - UXLGII) -RUVI(I)*DAMp
RUVFII) = RUVIII) + 0.S*TINT*(RUXI(I)+RUXE(I))
CONTINUE

NUMERICAL INTEGRATION COMPLETED FOR ONE STEP

FORMATI1H19* TIME = *0E10.50* COMPUTER TIME ELAPSED =*

IEIOoS/I

DO 1400 IanNN

UOIIII=UDFII1

IFINNQGTQNDEGI GO TO 1400

UVIII)=UVF(I)

UXIIII=UXF(I)

CONTINUE

IF(NN.EO.NDEG) GO TO 1403

DO 1401 I=1oNFLOOR

RUOIII) = RURE(I)

RUVI(II = RUVEII)

RUXIII) _ RUXFII)

CONTINUE

DO 1410 IanNFRAME

PWORKPII) = PWORKPII) + PLINCP(I) * TINT/TMH
PWORKNII) = PWORKNII) + PLINCN(I) * TINT/TMH
IFIINDEX.NE.NI> Go TO 2100

CTIME = SECOND(O)

INDEX = 0

PUNCH IAAOoTINFoUDFINPUNCH)

FORMATIPFlfio5)

PRINT I300.TIMF.CTIMF

LOOK FOR ENDLESS LOOPING AND EXIT IF NECESSARY
IFITINT.GT.0.000?) ICOUNT = o
IFITINT.LE-O-OOOOII ICOUNT = ICOUNT + I
IFIICOUNT.EO.6) Go To 2200

PRINT 1950

FORMATI/l/ * JOINT NO. DIPPL. X DISPL. Y *

1*ROTATION Z */I

90

DO 1960 I=19NJF

K1 = 3*1 - 2
K? = K1 + 1
K3 = K1 + 2

1°60 PRINT 197OOIOUDFIK1)OUDF(K?)0UDF(K3)
1070 FORMATI/2X0IBQPX03FIS.S)
PRINT 1°7S
197: FORMAT(/* WALL ELFM. MAX.TENSILF STRESS *
1*COEEEICIENT */I
DO 1000 I=IQNWALL
PRINT IQBEQIOFWLIIIO7WII)
198? FORMATIAX01306X02EIE.S/)
1900 CONTINUE
PRINT 1995
1993 FORMATI//* END FORCES IN FRAME ELEMENTS */)
PRINT 2000
P000 FORMAT(* NO. AXIAL FORCE CHEAR MOM. AT *
1*+END MOM. AT -END MODE */)
DO 2010 I=IONFQAMF
FFRI = -FFR(101) S EFR2 = -FFR(I¢?)
CP = XJ(IERIII) - XJ(JFRIIII
IEICPoEQ.0.0) GO TO 2004
PRINT ZOOSOIOFFRIOFFPIICZIOFFRIIQ3)OFFQIIQ4IQMODEFII)
GO TO 2010
2004 PRINT 2OOGOIQFFR20FFRIOFERII9319FERII0410MODEFII)
2003 FORMATI2X913v4E1505016/)
P010 CONTINUE
DO 2011 I:1.NJF
J = 3*1 - 2
K = 3 * I
IF(ABS(UDF(J)).GT.UXLIM) GO TO 2013
.IF(ARS(UDE(K))oGT.U7LIM) GO TO 201:
9011 CONTINUE
GO TO 2019
2013 PRINT 20140I
2014 FORMAT(/* COLLAPSE DUE TO EXCESSIVE HORL. DISPLACEMENT OF
1 JOINT N0. *914/)
GO TO 2200
p01: PRINT 2016.!
P016 FORMAT(/* COLLAPSE DUE TO EXCESSIVE ROTATION OF JOINT
1N00*014/1
GO TO 2200
P019 CONTINUE
IFICTIMEOGT02QOOO) GO TO 2200
IFITIMEoGToTLIMIT) GO T0 2200
P100 CONTINUE
CTIME = SECONO(O)
IEICTIMEoGTo2QOOOI GO TO 1450
IF(TIME.GT.TLIMIT) GO TO 1450
IEIIFLAGoNEoO) GO TO 2110

91

IF(JFLAGP0NEQO1 GO TO 2110
IFIJFLAGNQNEOO1 GO TO 2110
GO TO 2120

2110 CALL JSTIFF
IF‘NJFOEQOO, GO TO 2200
IFIIFLAGONEOO1 GO TO 2112
GO TO 2120

PECOMPUTE TIME INTFQVAL IF A WALL FLEMFNT CPACKQ

0(30

2112 CALL FIGFNINNoTMH)
P190 CONTINUE
INDEX=INDFX+1
GO TO 0R0
2200 CONTINUE
IFINNOGTONDEG) 60 TO 2236
PPINT 2210
P210 FORMATIIH19* FINAL STATUS OF JOINT VELOCITY oACCFLFPA*
1*TION ETC. */)
PPINT 2220
2290 FORMAT(/// * JOINT N0. VFL. X VFL. Y *
1*ANG. VFLo Z*/)
00 2225 I=IoNJF
KI = 3*! - 2
K2 = KI + 1
K3 = K1 + 2
2226 PRINT 1°7OOI‘UVFIK1)OUVPCK210UVF(K3)
POINT 2230
2230 FORMAT(/// * JOINT NO. ACLN. X ACLN. Y *
1*ACLN. 7*/)
DO 2230 1=IoNJF
K1 3*! - 2
K2 K1 + 1
K1 2 K1 + 2
PPQG PRINT IQ7OOIOUXPIK1)QUXF(K2)QUXF(K1)
2206 CONTINUE
POINT 2240
2240 FORMATI/* ENERGY ABSORBTION DUE TO ROTATION 0F PLASTIC*
1* HINGES IN INCH KIDS* )
PRINT 2241
2241 FORMAT(/* FRAME MEMBER POSITIVE END PERCENTAGE OF *
1* NEGATIVE END PERCENTAGE 0F*)
PPINT 2242
224? FORMAT(29X0*PRO0F PFSILIENCF*14XQ*PPOOF PFSILIFNCE*)
DO 224% IsloNFPAMF
PDCP c PWORKD‘!) /PDOOF(I) * 1mm.
PQCN : PWOPKN(I) /PQO0F(I) * 100.
PWORKPII) = PWORKPII) / 1000.
PWORKNII) = PWORKNIII / 1000.
224") PRINT 2250019PWORKPI I ) oPRCPoPWORKN( I ) oPPCN

2230

2320

?22q

7310

?11=
2:100

1n

11

92

FORMATI/4x9I303XQF100306XOF10.39?XOF10.39GX9F1003)
IEINNOEOONDEG) GO TO 2500

PRINT 2320

FORMATI// * JOINT NO. VEL. X */1
DO 232: IBIOWFLOOR

PRINT 1070010RUVE(11

PRINT 2330

FORMAT(/// * JOINT N0. ACLN. X */1
DO 2113 I=IQNELO0R

pRINT IOTOQIvRUXEII)

CONTINUE

END

FUNCTION IPOS(JOK1

THIS MAPS ELEMENTS OF A SYMMETRIC MATRIX ON TO AN ONE
DIMENSIONAL ARRAY

IFIJ-K) 10910011

IPOS=(K*(K-I))/2 +J

RETURN

IP09 3 (J*(J-11)/2+K

RETURN

END

FUNCTION TSTEPIX)

CALCULATE TIME INTERVAL FROM EIGENVALUEP

X: 1.0/SORTIX) * 6028318/600

ROUND OFF TO TWO RIGNIEICANT DIGITQ

2:100000

DO 5 131920 5 Y=1000/Z $ IFIXoGEoZ) GO TO 10 $ 7=Z*0.10
CONTINUE

X=Y*X $ I=X $ X31 $ X=X/Y $ TSTEP=X

RETURN

END

SUBROUTINE FRSTIFIIoMODEI
THIS COMPUTES THE STIFFNESS MATRIX OF THE FRAME ELEMENTS
COMMON IFRI4019JFRIAOIoIWALLI40)9JWALL(40)vKWALL¢4O)o

IFRAMEK(400211oWALLKI40¢361vNWALLoNFRAMFQNJFoNJWqNSUPo
2LWALLI4010ZWIA01

COMMON/3/ XJ(0010YJ(=010AFR(A0)vFRMII40)0R(60619RT(6Q619

18(60610981306198TI69?)09C(603)9DD(193)QEEQEWoWToPOISSQ
2A160619NMAX

DO 200 11:196

200

mn

211

?19

21%

93

DO 200 JJ=196

Q‘ IIQJJ1=00

AIIIQJJ1=00 $ RIIIQJJ1=00 $ FQAMFKIIQIPOSIIIQJJ11=OO
RTIIIOJJ1=00

IP = IERII)

IN 3 JERII)

FLONG = SORT((XJ(IN)-XJ(1P))**2+(YJ(IN)-YJ(IP))**2)
RIIQI) = (XJ(IN) - XJ(IP))/ FLONG
RI2021=RIIOII

RI4041=R(1911

RIROR)=R(191)
RI1921=IYJIIN1-YJ(IP)1/ FLONG
RI2011= -R(192)

RI4051=RIIO21

R(5¢41= fiRI495)

R(3931=100

RI6961=100

DO 210 11:196

DO 210 JJ=196

RTIJJQIII=RIIIQJJ1

GO TO (211.212.213.214) MODE
A(1911=AER(I1*EE/FL0NG

A‘IOA): ~AI1011

AI4941=A(1911
AI2921=IP200*EE*ERMI(I))/(EL0NG**3)
AI30013AI2921

AI2901= -A(2921
AI2031=(600*FF*FRMI(I))/(FLCNG**2)
AI296) = A(2031

A(1051 = -A(2931

AI5961= ~AI203)
AI3931=(400*EF*FRMIII))/EL0NG
A(6.6$=A<3.3)

AI1061=AI3031*003

GO TO fijé

AI 191 ) AFR(1)*EP/FL0NG

A(1~4) = -A(1911

AI494) = AIlvl)

A(2~2) = 1.0*EF*FPMI(I)/(FLONG**3)
AIZQS) = -A(202)

AISOS) = AI292)

A(2961 = 3o0*FF*FRMI(I)/(FLONG**21
AIQQ6) = -A(2961

AI6961 =.1.0*FF*ERMI(I)/FLONG

GO TO 215

AI1011 = AFR(I)*FF/FLONG

AI104) = -A(1911

A(a.4) = AIIOI)

AI2'2) = 300*EE*FDMI(11/(ELONG**11
AIZOS) = -A(202)

94

A(:.:) = A(?.?)
AI203) = 3o0*FF*FRMI(I)/(ELONG**21
AI30g1 = -A(2931
AIBCB) = 3.0*EF*FRMI(I)/FLONG
GO TO 215
?1a A(1011 = AFRII)*EF/FLONG
A(194) = -A(101)
AI494) = AIIOI)

21? DO 220 11:106
DO 220 JJ=106
200 A(JJQII)=A(IIQJJ)
CALL MATMULTIRTOA98060696)
CALL MATMULTIRoRvofiofioé)
DO 230 II=106
DO 230 JJ=1oII
710 FRAMEKIICIPOSIIIQJJ))= A(IIoJJ)
RETURN
END

SUBPOUTINE WSTIEF(I1

THIS COMPUTES THP STIEENEEQ MATRIX OE RECTANGULAR
WALL ELEMENTS.

COMMON IER‘4019JFR(40)9I‘NALLI40)oJ't’ALLI‘IO)9KWALLI40)9
IFRAMEK(40921)OW (40036)ONWALLQNFRAMEQNJEQNJWQNSUP.
2LWALLI40107WI401

COMMON/3/ XJ(5010YJIEO)oAFRI40)vFRMI(40)9R(696)0RT(606)O
18(6061988I396)QBTI693198C(6Q3)000(39319EFQEW0WTQPOISSQ
2AI69619NMAX
BETA=IYJIJWALLI111-YJIIWALL(1111/(XJ(LWALL(I))'
IXJIIWALLIII1)

BETIN : 1.0/BETA

GAMA = (1.0 _ DOIQC) * QETA

GAMIN = (1.n - DOIQQ) * PFTIN

=0 W(Ic1) 400*BETA + 200*GAMIN
WIIQ?) 1.: * (1.0+DOIQQ)

WIIO31 400*RCTIN + 200*GAMA
WIIO4) = 200*RETA- 2.0*GAMIN
W(I¢E) = -Io5*(100-1.0*POIP9)
WIIoé) = WII'I)

WIIQ71 = -W(qu1

WIIQB) = -400*PETIN + GAMA

WIIQQ) =-W(1921

WIIQIO) a WIIQB)

WIIOII) -200*BFTA - GAMIN
W(1012) -W(Ic21

W(19131 -400*RETA + GAMIN
ch.14) W(Io=)

WIIQI5) WIIQI)

95

WIIOIé) = -W(Io21

W(1917) = -2-0*BETIN - GAMA
WIIQ18) = ~W‘IoE)

WIIOI9) = 200*BETIN - 200*GAMA
WIIOZO) = WIIQ?)

W(I!21) = WIIQB)

WIIQ221 = ~400*RETA + GAMIN
WII923) = -W(Ic9)

W(I924) = -200*RETA - GAMIN
WIIQ2E1 = W(Io21

WII¢261 = 2.0*DETA - 200*GAMIN
WIIQ27) = WIIQE)

WIIv28) = W(Iol)

WIIv29) = WIIQ‘)

WIIoBO) = 200*RETIN - 2o0*GAMA
W(Ic31) = W(Ic21
W(I~32)=-2.0*RFTIN - GAMA
WIIOBB) = -W(I95)

WI1934) = ~400*RETIN + GAMA
WIIQB‘E) = -‘AI(I.P)

W(Io16) = WIIofi)
DO 60 J=1o36
60 WIIoJ) = WIIoJ) * FW*WT/(19.O*(1.n-DOIQQ**a)) * ZW(I)
RETURN
END

SURQOUTINE JSTIFF
THIS COMPUTES THE JOINT STIFFNESS MATRIX
COMMON IFP(40).JFP(40)oIWALL<4OIoJWALLIAO).KWALL¢40)o
1FPAMEK(ano21)owALLK(4fi.36)oNWALL.NPQAMF.NJF.NJw.N<uP.
2LWALLIAD)o7WI40>
COMMON/I/CMJI7??O).FII‘n)oU(1=0)oX(6)oYIE)
COMMON/P/QTARF(PO.?0).QTARK(?0.?0).Q(12019NHEG
COMMON/?/ XJ(50)0YJ(€O)oAFPI40)oFPMI(40)¢P(6o6).QTI6o6Io
18(606)098(3o61QBT(693)09C(693)ODD(3931oEEoEWqWToPOISSQ
2A(696IoNMAx
NN=3*NJF + 2*NJW
NJFPEE=NJF+NJW
00 ago I=IoNN
DO 350 J=IOI
3:0 SMJIIPOSII.J))=0.0
DO 810 I=1oNJF
DO 810 J=19NJF

11=3*I-2
13=H+2
JI=3*J-2
J3=Jl+2

IFIIoGToJ) GO TO 810

660 Do 710 II:I.NFDAME
«=0
00 730 K=11013
KK = KK+1
LL30
DO 730 L=J10J3
LL = LL+1
IEILOLTOK) GO TO 730
IFIIaEOoIFRIII1) GO TO
IFIIaNEoJERIII11 G0 T0
KK1 = KK + 3
GO TO 690

670 KK! = KK

680 IFIJoEOoIFRIIIII GO TO
IFIJ.NE.JFR(II)) GO TO
LL1 3 LL + 3
GO TO 700

600 LL1 = LL

700 9MJ(1POS(K.L)) = <MJ¢IPO<tKoL))

1LL1)1
730 CONTINUE
810 CONTINUE
00 1100 I=10NJERPP
DO 1100 J=19NJEREP
IEIIQGTQJI GO TO 1100
I:(IOGTONJF) GO TO 030
I1 = 3*1 - 2
I? = 11 + 1
GO TO 960
0:0 11 3 NJF + 2*1 ‘ 1
I? = 11 + 1
96h IEIJQGTONJF1 GO TO 070
J1 = 3*J ‘ 2
J2 = J1 + 1
GO TO GPO
070 J1 = NJE + 2*J * 1
J2 = J1 + 1
980 D0 1090 11:1 ONWALL
KK = 0
DO 1090 K:11012
KK = KK + 1
LL = 0
DO 1090 L=J1‘J?
LL = LL + 1
IFILOLTOK) GO TO 1000
IEIIOEOOIWALLIIII1 GO
IFIIQEQOJWALLIIII1 GO
IFIIQEOQKWALLIII11 GO
IEIIONEQLWALLIIIII GO
KKI = KK+6 5 GO TO 10

96

67m
71m

690
730

T0
T0
T0
T0
an

1000
1010
10?0
1000

+ CRAMPK(IIOIPOR(KK1.

mm “In... if. tat”

 

TL

0(30

97

1000 KK1 = KK $ a0 TO 1030
1010 KKI = KK+2 $ GO TO 1030
IO?n KKI = KK+A

10‘3n IFIJ.EQ.IWALL(II)) GO TO 1040
IF(J.EO.JWALL(II)) Go To 10:0
IE(J.EO.KWALL(II)) GO TO 1060
IFIJ.NE.LWALL(II)1 GO TO 1000

LL1 = LL+6 $ 60 TO 1070

1040 LL1 = LL $ GO TO 1n7n

10=fi LL1 = LL+P $ GO TO 1070

1060 LL1 = LL+4

10‘7n SMJIIPOSIKoL)) = FMJIIPOQ(K.L)) + WALLK(IIoIDOS(KKI.
1LL1)>

1090 CONTINUE

I100 CONTINUE
IFINNoGToNDEG) GO T0 1200
RETURN

1200 CALL INVERSEINNI
IFINNoEOoO) NJF=0
IFINNoEOoO) RETURN
CALL FLEX
CALL MATININDEG)
RETURN
END

SUBDOUTINF MASSIALDHA)
THIS COMPUTFS THF MAQS MATRIX

COMMON IERI4010JPR(4010IWALL(401OJWALLI40)OKWALLIAO)Q
IERAMEK(400211OWALLKI40936)cNWALLqNERAMEoNJFoNJWcNSUPo
2LWALLI40)Q7W(AO)
COMMON/2/9TARFI20020)0RTARKI2OQ2O)02(12019NDEG
COMMON/1/ XJ(R010YJIRO)OAER(4O)9ERM1(40)QR(69610PT(696)9
18(616)oBBI3o6)QBT(603)QRC(603)0DD(3¢3)vEEoEW9WTcPOISSo
2AI69619NMAX
NN=3*NJE+2*NJW
10 READ 200WL0AD¢DEN§EQDNRQNBAY
20 FORMATI3E10000151

DO 30 1:10120
10 QC11=000

DO 80 I=10NJF

J=3*I-2
DO 70 II=10NERAME

IEIIFRIIIIQNEQIQANDOJER(II10NEQI1 GO TO 70
FLON=SORTIIXJIIERIII11-XJ(JFR(IIII1**2+(YJ(IER(1111‘
1YJ(JFR(II)1)**21

IW=IERIII1

98

JW=JFR(II)
IF(XJ(IW)-XJ(JW)) 40.60.40
40 DLOAD=0.S*FLON*(AFR(II)*DENSE + WLOAD)/386o4
GO To 60
“A DLOAD=0-=*FLON*AFD(1I)*0FNSF/386.4
60 9¢J>=DL0AD+9(J1
<1J+11=0L0A0+S¢J+1>
<1J+?)=(DLOAD/1.0)*(=LON**?)*(ALDHA**?) + 9(J+?)
70 CONTINUF
an CONTINUF
00 RS 1:1.NwALL
II:3*IWALL(I)-?
JJ=3*JWALL(I)-?
KK=3*KWALL(I)-?
LL=3*LWALL(I)-?
WLD=(YJ(JWALL(I))-YJ(IWALL(I)))*(XJ(KWALL(1))-XJ([WALL
1(111)*w7*0Ns*0.2a/286.4
1=(11.GT.NN) 00 To 81
$1111=S(11)+WL“
9(11+1)=9(11+1)+WLD
n1 IF(LLoGT.NN) 60 TO 99
S(LL)=<(LL)+WLD
:1LL+1)=<(LL+1>+WLO
qa S(JJ):S(JJ)+WLD
S(JJ+1)=S(JJ+1)+WLD
S(KK)=S(KK)+WLD
S1KK+1)=S(KK+1)+WLD
as CONTINUE
IF(NN.GT.NDEG) GO TO 00
60 TO 120
on N = NDFG*(NRAY+1)
00 100 1:1.N
J : 1*!-?
100 S(I)=S(J)
00 110 I=loNRAV
00 110 J=loNDEG
K=J+I*NDFG
110 S(J)=S(J)+S(K)
170 PRINT IPI
1¢1 FORMAT(/* MASS MATRIX*)
PRINT 12?.(St1).1=1.N0FG)
1?? F09MAT¢/10€12.=)
RETURN
END

SUBPOUTINE EIGEN(NN9TMH)
THIS COMPUTES THE BEST TIME INTERVAL TO BE USED
COMMON/1/SMJ‘7320)

II

9O

91

201
??

?1

P4

?6
P6

07
”1
R0

99

COMMON/?/GTARE(POoPO)oCTARK(?Oo?O)o°(I?O)oNDEG
DIMENSION B(1?0)oC(1?0)oX(170)

N=NDEG

INDEX :0

THE ITERATION PROCEDURE FOR EIGENVALUES
CONTINUE

DO ?0 I=10N

XIII=IoD

CALCULATE CONDONENTQ OE EIGENVECTODQ
DO ?? I:19N

C(II = 0.0

DO 9? J=10N

IE(NN-NE.NDEG) GO TO POI

C(I) = C(I) + QMJIIPOQIIQJ)) * X(J) / QII)
GO TO 2?

C(I) = C(I) + 9TARK(I¢J) * X(J) / C(I)
CONTINUE

DO 23 I=IoN

8(1) = C(I) / C(l)

CHECK EOR ACCURACY

F091 = 0.000001

DO ?4 I=ION

DIEF:X(I) - RI!)

IE(ARS(DIEE)-ERGI) 74,7q,?:

CONTINUE

GO TO P7

DO 26 I=10N

X(I) = RII)

INDEX = INDEX + 1

IE(INDEX.GT.50) GO TO ?7

GO TO ?I

TMH=TSTEP(C(1))

DRINT ?QOTMH

EORMAT(//* TIME INTERVAL = *FR.E/)
RETURN

END

SUBROUTINE GROUND(TIME)

THIS COMPUTES THE ACCELEQATION DUE TO GROUND MOTION AT(TIME)
COMMON IFRI40) OJFQ(40) O IWALL‘40) OJWALLJQO) OKWALLI40) 0
IFPAMEK(4092II9WALLK‘4OO36)ONWALLQNFQAMEQNJFQNJWQNSUPQ
9LWALLIAOIQZWI4O)

COMMON/2/STAQEIPOOQO)QQTAPK(20.2fi),C(1pm,,NncG
COMMON/GQND/UXLG(1?O)cAHoAVoINOqEACT

DIMENSION AXLH(4nfi).AYLV(afiO).QTH(4nn),QTV(4nn)
NN=3*NJE + ?*NJW
IE(INGoEOol) GO TO lafi
READ IOQNOHQNOV

100

10 FORMAT(?IS)
DO 80 I=1oNOH
J1=4*I-3 $ J2=J1+1 s J3=JI+P S J4=J1+3
Rfi, DEAD IOOOQTH‘J1)QAXLHIJ1)OOTHIJa)QAXLHIJEIQQTHIJ3)Q
1AXLH(J3)oOTH(J4)oAXLH(J4)
00 GO I=19NOV
J1=4*1—1 $ J2=Jl+l $ J3=J1+? $ J4=J1+1
Q“ RFAD IOOoQTV(J1)oAXLV(J1)~QTV(J2)oAXLV(J2)oQTV<J31o
IAXLV(J3)«GTVIJ41oAXLV(J4)
1mm FORMAT(3X94(ERoOoE°oO))
140 00 ?no 1:1.400
IF(OTH(I).GT.TIME) GO TO ?10
P0“ CONTINUF
71“ TG=QTH(1)
AG=AXLH(I)
TL=QTH(I-1)
AL=AXLH(I-l)
CF:(TIMF-TL)/(TG-TL)
AH=AL+CF*(AG-AL)
DO 300 1=1c400
IP(OTV(11.GT.TIMF) GO TO 11“
finn CONTINUE
11" TG:QTV(I)
AG=AXLV(I)
TL=OTV(I-])
AL=AXLV(I-1>
CF=(TIMF-TL)/(TG-TL)
AV=AL+CF*(AG-AL)
IF(NDEG.LT.NN) GO TO 800
00 400 I=1oNJE
J=3*I-?
UXLG<J1=AH*FACT * 396.a
UXLG(J+1)=AV*FACT * 095.4
400 UYLC(J+P)=0.O
GO TO 600
:00 00 RR“ I=IoNDEG
Ran UXLG(I) = AH*EACT*QRAo4
600 CONTINUF
INO=1
RETURN
END

§URROUT1NE FDAMQT

THIS COMPUTES THE INPRFMFNTAL FORCFQ IN FRAMF FLFMFNTQ
COMMON 1FR(40).JFR(40).IWALL(40).JwALL(40).KWALLcan).
IFRAMFKIQOOEI I QWALLKI40036) QNWALL QNFRAMEQNJFQNJW .NSUPQ
2LWALL(afi)o7W(AO)
COMMON/I/SMJ(712O)9F(160).U(1=0).X(6).Y(6)

pan

3OO

100
11"

mm
110

101

COMMON/3/ XJ(50)OYJIRO)OAFQI4A)QPDMI(40)QD(606)OQTI616).
18(696IOBBI306)oBT‘693)oBC(693)oDD(393)QEFQEWQWTQPOISSv
?A(696I9NMAX

COMMON/4/DER(4O04)OOWL(4O)QEEQ(4O04)oEWL(4O)

DO 300 I=IONERAME

KI=3*IFR(I) ‘ 2 $ K?=KI+I 3 K3 = KI + ?

K4 = 3*JFQ‘I) - P $ Kg = K4 + I $ K6 = K4 + ?

X(1)=U(K1) $ X(2)=U(K?) 5 X(3)=U(K3) ‘5 X(4)=U(K4)

X(5) = UIKE) % X(6) = U(K6)

OO ?OO II=196

00 ?OO JJ=196

AIIIQJJI=ERAMEKIIQIDOQIIIQJJ))

CALL MATMULT (A9X0Y069691)

DER(IOI)=Y(I)

DER(IQ?)=Y(?)

DER(IQ3)=Y(3)

DER(I¢4I=YI6)

CONTINUE

QETUQN

END

QURROUTINE WALQT

THIS COMPUTEé THE MAY TENEILE QTREQQ IN THE WALL ELEMENT:

DIMENSION UU(8)9EE(1IQK(P)

COMMON IEQI40) QJEDCAO) 0 I‘IIALL(4O) oJWALL(40) 9K‘MALLI40) c
IFRAMEKI40921)CWALLKI40936)ONWALLONFRAMFQNJFQNJWQNSUPI
ZLWALLIAO)QZW(40)

COMMON/I/SMJ(73RO)9U(I?O)QEII?O)qX(6)9Y(6)o
166(398)9HH(3!8)

COMMON/3/ XJ(SOIOYJ(EO)QAEQ(4O)9ERMI(4O)0Q(696)oPT(606).
18(6v6)098(396)OBTI693)ORC(693)ODD(303)oEFoEWvWTQPOISSo
2A(606)9NMAX¢DVAXT

COMMON/a/DFQI4094) QH‘AILIQO) QFFD'(4OQ4) OFWLIAO)

DO 6OO N=IOMWALL

PMAXT = 0.0

ALEA = XJ(LWALL(N)) - XJ(IWALL(N))

BETA = YJCJWALLIN)) - YJ(IWALL(N))

IF (IWALLIN)0GT0NJE) GO TO 100

K(1) = 3*IWALL(N) - 7

GO TO 110

K‘l) = NJE + 9*IWALLIN) - I

K(?) = KII) + I

IF (JWALLINIoGToNJE) GO TO IPO

K(3) = 1*JWALL(N) - R

60 TO 130
K13) = NJE + ?*JWALL(N) - 1
K(a) = K11) + 1

IF (KWALLIN) o GToNJF) GO TO 140

102

KIEI = 1*KWALL(N) — 9
GO TO 1:0
140 K(q) = NJF + ?*KWALL(N) ‘ 1
IE0 K(6) = KI“) + I
IE (LWALL(N)oGToNJE) GO TO 160
K(7) = 3*LWALL(N) - ?

GO TO 170
I6n K(7) = NJE + ?*LWALL(N) - I
I7O K(8I = K(7) + I

DO ZOO 1:198
?OO UUII) = U(K(I))
DO EDD II=I¢?
DO “OD JJ=I¢P
XEE = (II-I)
ETA = (JJ-l)
DO 330 I=I¢3
DO 330 J=108

31” GG(I0J) = 0.0
GGIIOI) = -(loO - FTAI/ALFA
66(103) = -ETA/ALEA
GGIIQF) = -GG(193)
GG(I¢7) = -GG(IOI)
66(292I = “(IOO - XFEI/pETA
GG(?¢4) = -GG(?9?)
GG(?96) = XEE/RETA
GG(?98) = -GG(?06)
66(301) = GG(?0?)
GG(3.2) = GGIIQI)
GG(393) =-GG(29?)
66(304) = 66(103)
66(305) = GG(206)
66(396) = - GGIIOBI
GG(397) = -GG(206)
GG(3.8) = -GG(191)

CALL MATMULTIDD.GG.HHo3o3oP>
1:0 CALL MATMULT(HH.UU.FF.1.R.11
$1 = n.=*(pc(1) + =F(?)+ CORTIIFF(1)-EF(2))**? + 4.0*
FF(3)**P))
IFISIOLTOPMAXT) GO TO 40D
PMAXT = 51
400 CONTINUE
500 CONTINUE

DWLIN) = PMAXT * 7W(N)
600 CONTINUE

RETURN

END

fl

SURROUTINE MATMULT (AqRoCCLQMoN)

GOO

1n

3O

30“
400
40?
401
404
401

1‘;

I6

103

THIS MULTIPLIES MATRICES A AND R AND STORFS IN C
DIMENSION AILQM)9R(M9N)9C(L9N)

DO 10 I=IQL $ OO 1O K=IQN S C(IQK)=O0 T DO 1O J=IOM
CIIOK)=C(IQKI+AII0J)*RIJ0KI

RE, TURN

END

EUBROUTINE FLEX

THIS PICKS OUT THE RFRTINENT ELEMENTS OF THE JOINT
FLEXIBILITY MATRIX OF THE UNREDUCED EYQTFM AND FORMS
THE JOINT FLEXIBILITY MATRIX OF THE REDUCED QYSTEM
COMMON/I/SMJI71?O)oF(1:0)oUIIRO)oX(6)qYI6)
COMMON/?/A(90.p0).CTARK(¢O.¢0).CI190).NDFG

DO 30 I=IONDEG

DO 30 J=IoNDEG

NI = 3*1 - ?

NJ = 3*J - ?

AII.J)=SMJ(IPOS(NI.NJ))

RETURN

END

SURROUTINE MATININ)

THIS COMPUTES THE STIEENESS MATRIX OE THE REDUCED SYSTEM
COMMON/P/STAREIQOQ?O)OS(?O0?O)

DO 300 I=ION

DO 3OO J=ION

SIIOJ)=§TARE(IOJ)

DO 401 I=I0N $YX=SII9II $ S(I9II=IoD $ DO 40? J=10N
SIIOJI=SII0J)/XX$ DO 401 K=IQN $ IE (K-I) 4O194OIQ403
XX=SIK9II$ SIKQII=O0O $ O0 404 J=10N

S(KvJ)=S(K9J) -XX*9(I9J)

CONTINUE

RETURN

END

SURROUTINF INVFRR¢(N)

THIS INVERTS THE JOINT STIEENESS MATRIX OE THE
UNREDUCED SYSTEM

DIMENSION RIIPDIOOII?OIOIR(I?D)
COMMON/I/AI73?OI

IE (N‘II 1491:016

AIII=Io/A(1)

RETURN

N4=(N*(N+III/9

1O

2OI

EOD

7".

?II

?1

1:1

301
BO

471
47
4%
=50

100

14

101+

DO IO I=IcN

IR(II=O

GRAND LOOP STARTS

DO 100 IleN

BIGAJ =Oo

DO 2O J=ION

IE (IRIJII EOOPOIOPO
M=(J*(J+I)I/2

Z=ABS (AIMII ’
IE (7-RIGAJ) 2O02OOPO2
BIGAJ=Z

K=J

CONTINUE

IE (RIGAJI PIO?IIO?I
PRINT 6

EORMATI///* STRUCTURE COLLAPSES*//I
N = D

RETURN

PREPARATION OE ELIMINATION STEP I
IRIKI=I

M=(K*(K+I ) I/2
OIK)=Io/A(MI

pIKI=I o

AIMI=OO

L=K-I

I: (L) 7SQ3SvSSI
M=IK*II("I I I/?

DO 3O J=IQL

M=M+I

P(J)=A(MI
OIJ):A(M)*OIK)

IE (IRCJ)) 309301930
O(J)=-O(JI

AIMI=OQ

IE (K+I-N) 3:915cgn
L=K+I

DO 45 J=L9N
M=IJ*(J-I)I/2+K
RIJ)=A(M)

IE (IR(JI) 4710470471
P(J)=-P(J)
OIJ)=-A(M)*O(KI
AIMI=OO

DO IOO J=19N

DO 100 K=J9N
M=(K*(K-I))/2+J
A(M)=A(M)+P(J)*D(KI
END OF GRAND LOOP
RETURN

END

   

WCIHHIIIIIIHIIIIII IUIWIIIISIIII WIT IT
3 1293 03174 4315