matrix computer analysis of curvilinear grid systems
TRANSCRIPT
Scholars' Mine Scholars' Mine
Doctoral Dissertations Student Theses and Dissertations
1967
Matrix computer analysis of curvilinear grid systems Matrix computer analysis of curvilinear grid systems
David Leon Fenton
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Department: Civil, Architectural and Environmental Engineering Department: Civil, Architectural and Environmental Engineering
Recommended Citation Recommended Citation Fenton, David Leon, "Matrix computer analysis of curvilinear grid systems" (1967). Doctoral Dissertations. 1872. https://scholarsmine.mst.edu/doctoral_dissertations/1872
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MATRIX COMPUTER ANALYSIS
OF CURVILINEAR GRID SYSTEMS
BY
DAVID LEON FENTON . ·
A
DISSERTATION
submitted to the faculty of
THE UNIVERSITY OF MISSOURI AT ROLLA
in partial fulfillment of the requirements for the
Degree of
DOCTOR OF PHILOSOPHY IN CIVIL ENGINEERING
Rolla, Missouri
1967
Approved by
PLEASE NOTE: Not original copy. Indistinct type on several pages. Filmed as received.
Universitv Hicrofilms
ABSTRACT
A general equilibrium-stiffness method of matrix
structural analysis was adapted and applied to the
ii
solution of the member end forces and moments of each of
the members in a curvilinear structural grid system. A
structural system of this nature is commonly used as the
supporting framework for a "steel-framed dome", in addition
to being a basic structural component in many aerospace
applications.
An integral part of the development of the analysis
was the development of a computer program to perform the
many complex operations required to obtain the solution.
The engineer, by supplying the appropriate structural
data and load data. to the computer program, is able to
obtain the forces and moments at the ends of each of the
members in the structural system corresponding to the
six possible degrees of freedom of each one of the joints,
or nodal points as they are referred to.
David Leon Fenton
ACKNOWLEDGEMENT
The writer wishes to express his appreciation to
Dr. J. H. Senne, his advisor, and to Dr. J. L. Best,
iii
for their careful and thorough review of this dissertation.
The writer also wishes to express his gratitude to
Barry Gregory for his valuable assistance in the develop
ment of the computer program.
Finally, the writer is deeply grateful to his wife,
Janice, who, in addition to typing this dissertation
while being full-time mother to his two wonderful children,
Scott and Susan, has shown so much consideration and
understanding during these years in school.
TABLE OF CONTENTS Page
ABSTRACT. • • • • • • • • • • • • • • • • • • • • • ii ACKNOWLEDGEMENT • • • • • • • • • • • • • • • • • • iii LIST OF FIGURES • • • • • • • • • • • • • • • • • • vi LIST OF TABLES. • • • • • • • • • • • • • • • • • • vii LIST OF SYMBOLS • • • • • • • • • • • • • • • • • • viii
Chapter 1.
Chapter 2.
2.1. 2.2. 2.3. 2.4. 2.5.
Chapter 3.
3.1.
3.2. 3.3. 3.4. 3. 5· 3.6.
Chapter 4.
4.1. 4.2. 4.3. 4.4. 4.5.
Chapter 5.
INTRODUCTION. • • • • • • • • • • • • •
GENERAL FORMULATION • • • • • • • • • •
Method of Formulation • • • • • • • • • Flexibility and Stiffness Matrices ••• Assembly of System Stiffness Matrix • • Pinned-End Rigid Foundation Attachment. Intermediate Span Loads • • • • • • • •
SEGMENTAL CIRCULAR MEMBER • • • • • • •
Geometric Properties of a Segmental Circular Member • • • • • • • • • • • • Load-Displacement Equation Derivation • General Energy Expression • • • • • • • Deflection Calculations • • • • • • • • Flexibility Matrix ••••••••••• Three Dimensional Axis Rotation Transformation Matrix • • • • • • • • •
APPLICATION OF ANALYSIS • • • • • • • •
Procedure • • • • • • • • • • • • • • • Mathematical Models • • • • • • • • • • Example Computer Analysis • • • • • • • Experimental Model ••••••••••• Experimental Results ••••••••••
COMPUTER PROGRAM FOR CURVILINEAR GRID SYSTEM • • • • • • • • • • • • • •
Introduction to the Program • • • • • • Input Data. • • • • • • • • • • • • • • Output Data • • • • • • • • • • • • • • MACS Subroutine Package • • • • • • • • MACS User Instructions ••••••••• Flow Diagrams and Computer Program. • •
1
6
6 7
16 20 22
24
24
25 27 28 33
36
44
44
~~ 57 60
67
67 68 69 71 73 76
iv
v
Chapter 6. CONCLUSIONS AND FUTURE INVESTIGATIONS • • 115
6.1. Conclusions • • • • • • • • • • • • • • • 115 6.2. Topics for Further Investigation. • • • • 116
BIBLIOGRAPHY. • • • • • • • • • • • • • • • • • • • • 120 VITA. . • • • • • • • • • . • • • • • • • • • • • • • 121 APPENDIX A. COMPUTER SOLUTIONS FOR THE MATHEMATICAL
MODEL. • • • • • • • • • • • • • • • • • 122 APPENDIX B. COMPUTER SOLUTIONS FOR THE EXPERIMENTAL
MODEL. • • • • • • • • • • • • • • • • • 157
Figure
2.1 2.2 2.3 2.4a
2.4b 3.1 3.2 3.3 3.4 4.1
4.2
4.3 4.4 4.5 4.6
4.7 4.8
LIST OF FIGURES
Fundamental member load displacement. Member end-load coordinates. Simple curvilinear grid. Fixed-end forces and moments on the loaded member. Equivalent nodal loads. Geometry of the segmental circular member. Moments and forces on major axes. Typical member orientation. The three required rotations. Mathematical Model-l h l , all
r=6 foundation attachments fixed. Mathematical Model-l h _ l , with both
y;-b pinned and fixed foundation attachments. Basic coordinate dimensions. Geometry of l, 3, 7, 9 node points. Required geometry for ~ • Experimental model with pinned foundation attachments. Free body diagram of Member-12. Photograph of Experimental Model
Page
8 9
16 23
23 24 26 37 39 46
47 48 49 51
59 61
63a
vi
vii
LIST OF TABLES Table Page
3.1 Partial derivatives for energy expressions. 28 4.1 Member coordinates for experimental model. 63 4.2 Summary of Test No. 1. 64 4.3 Summary of Test No. 2. 65 4.4 Summary of Test No. 3. 66 5.1 Table of Fortran symbols. 102
P'
K'
F
px' py' pz
mx' m ' m y z
H
*
K
T
K"
s
R
e
9
e
N
A
LIST OF SYMBOLS
Member end-loads in system coordinates.
Member end-displacements in system coordinates
Appropriate member stiffness matrix in system coordinates
Flexibility matrix
Member end-forces and moments in member coordinates.
Equilibrium matrix.
Rigid body end-displacements.
Appropriate member stiffness matrix in member coordinates.
Transformation matrix.
System stiffness matrix
One half of the cord length of the member.
Radius of curvature of the member.
Perpendicular distance from radius focus to the end of the member.
viii
One half of the total angle subtended by the member.
Member end rotation
General reference angle
Normal force on any cross-section of a member.
Cross-sectional area of a member.
M , M , M X y Z
MT' MYN
IN
IyN
I z
E
G
A
0(.
General moments at any point along a member with respect to member coordinate axes.
Moments with respect to the major axis of the cross-section.
Equivalent polar moment of inertia.
Moment of inertia about major "y" axis.
Moment of inertia about major "z" axis.
Modulus of elasticity
Shear modulus.
Direction cosine.
Rotation of major axis of a member about member x-axis.
Angle between system x-axis and the projection of the member x-axis on the system xz-plane
Vertical angle between the member x-axis and its projection on the system xa-plane.
ix
Chapter 1
INTRODUCTION
The appearance of the dome structure in today's
modern space age society is as accepted and as appro
priate as it was hundreds of years ago. Today's modern
materials and advancements made in construction methods
have made the dome structure one of the most economical
space-structure systems in use today.
There is an ever growing trend to enclose larger and
larger areas and in so doing man is endeavoring to control
his environment in order to eliminate the influence of
weather on his daily activities. By enclosing large areas
such as shopping centers the loss of business due to
unfavorable weather can be reduced. The same is true of
athletic stadiums, civic centers and auditoriums, green
houses, opera houses and a multitude of other applications.
The present demands, in addition to demands of the future
where domes spanning a mile or more may be desired, present
today's engineer with a tremendous challenge. The analysis
of these structural systems will require talented and
creative engineers with more than average ability, and,
in addition, there will be a continuing need for the
development of more efficient and sophisticated methods of
analysis.
1
Modern matrix analysis of large structural systems
together with the use of the high speed computer is
becoming as common to the structural engineer's world
today as the slope-deflection and moment-distribution
methods have been in the past. The analysis of space
frames, as well as flat grids, consisting of straight
members has been well documented. 1 ' 2 ' 3 On the other
hand it is felt that there is a definite need for a clean
straight forward approach to the analysis of space
structure systems composed of curved members.
Much of the work that has been done with curved
members deals with two dimensional structure such as arch
and multi-arch frames.5 In the area of members curved in
space the following approaches have been taken, a matrix
procedure using a trial and error approach was suggested
by, Baron,? in the paper by Eisemann, Woo, and Namyet1 ,
it is suggested that curved members or members with
variable cross-sections could be approximated by the
introduction of additional rigid joints along the length
of the member resolving the member into a series of
straight segments. This approach, however, would greatly
increase the number of equations required to solve and
would be highly inefficient for large systems. Another
approach to the solution of "curvilinear grid frames"
has been suggested4 , which involves the use of trigono
metric series to approximate moments and deflections.
2
In a paper on "Mettalic Dome-Structure Systems,"
by Shu-t'ien Li9 , it is suggested that any surface of
revolution supported on a horizontal thrust ring may,
if symmetrically loaded by distributed loads, be regarded
as supporting itself directly by tension or compression.
This paper then suggests that by adapting the established
theory of thin-shell spherical domes, the analysis o! the
lattice dome can be made quite simply. The compressive
stresses toward the pole and around lines o! latitude
when multiplied by the rib spacing will give the rib
stress. The shell approach, however, does not include
the bending and torsion taking place in the lattice
structure, which is necessary in order to describe more
accurately the behavior or the structure.
In a recent paper by J. Michalos10 , a general
approach to the analysis of "space networks' is described.
The method or analysis described, employs the use of both
the displacement and force methods of matrix analysis,
however, the method of formulation still uses an
iterative approach. The resisting forces and moments in
each branch of the network are first computed by the
force method, while the rotations and displacements of
the ends of a branch are prevented. Correction moments
and forces resulting !rom rotations and displacements or
the nodes are then computed !rom the displacement method
and finally the correction moments and forces are added to
the resisting moments and forces previously determined.
3
I ~
The number of operations and steps required by this
analysis would still encourage the development of a
more straightforward and simplified approach.
In view of the preceding discussion, a need !or a
more direct method of analysis for the curvilinear space
structure still exists. The fact that a given matrix
method of formulation has been developed does not
eliminate the possibility of the existence of a more
efficient and more direct method of matrix formulation.
The majority of the structures built are not
designed with unusual assortments of member shapes and
erose-sectional patterns, for the obvious reasons of
economy. In most cases the members are either all
straight or all curved and in some cases there may be a
combination of straight and curved members. In structures
where curved members are used, they are usually of the
same type, (i.e. all segments of circles or parabolas,
etc.).
It is for these reasons that this author favors
deriving the equations for the member flexibility
coefficients directly from energy considerations, in
terms of the geometric and elastic constants of a typical
member and then substitute the appropriate geometric
and elastic constants of each individual member into
these equations to obtain the member flexibility matrices.
The member stiffness matrices are then obtained simply by
inverting the member flexibility matrices.
4
It was with this philosophy that the following
method of analysis was developed. The method used to
formulate the analysis was a general equilibrium
stiffness formulation in which the load-displacement
equations for the individual members are used to
generate directly the system stiffness matrix for the
entire structure, from which the nodal displacements
are obtained. These displacements are then substituted
into the member load-displacement equations to obtain
the final member-end loads.
Finally, a computer program was developed to
effectively carry out the necessary computations, and
an experimental model was build to correlate the results.
5
Chapter 2
GENERAL FORMULATION
2.1 Method of Formulation
The equilibrium (or displacement) method has been
chosen as the most suitable approach to the formulation
of the curvilinear space grid. The load-displacement
equations are first derived for a single segmental cir
cular element, (or in general any other element for that
matter) in terms of the most convenient element coordinate
system, which in most cases is different from the coordi
nate system chosen to represent the entire structure.
Similar equations !or all the elements of the structure
are then combined to form a set of load-displacement
equations for the structure, the details, of which, will
be described later.
The elements of a structure may be oriented in any
6
manner relative to the structural system. It is, therefore,
necessary to express the member equations in terms of the
system coordinates before conditions of compatibility and
equilibrium can be applied. The following equations
(2-1), for example, are illustrative of the equations
which contain a complete description of the load-displace
ment characteristics of a single element in system
coordinates.
, , P' 1 = Kil 6• 1 + Ki2 6•
2
6• 6• (2-1)
P' = K21 + K22 2 1 2
Where P' and 6' represent end loads and displacements
respectively, K' represents the appropriate stiffness
matrix and the primes indicate that these quantities are
expressed in system coordinates.
One important advantage or this method is that
equations (2-1) can be derived !or a single element with
out regard to how this element will be oriented in the
structure. The formulation or the problem can be separated
into two separate parts, (1) developing the equations
(2-1) for a single element using the elastic properties
and geometry o! the element, (2) applying the conditions
of compatibility and equilibrium which are concerned with
the topology of the structure. Each of these parts are
entirely independent of one another which is a tremendous
advantage in the formulation of complex structural systems.
It is assumed in the formulation which follows that the
theory of small deflections holds true.
2.2 Flexibility and Stiffness Matrices ror a Single Element
It has been found in the case of a curved member,
that it is much easier to derive the flexibility matrix
and invert it to obtain the stiffness matrix than to
derive the stiffness matrix directly as would be the case
?
for a straight member. The inversion can be done quite
efficiently and rapidly on the computer since the maximum
size of an element flexibility matrix would be 6 X 6.
It will be necessary to consider a member as having
direction. As shown in Figure 2.1 the direction of a
member is from its l-end to its 2-end.
/ ' ,/,------ ............... L~
/ - ~ ' .-." 'J. a
End-1 End-2
Figure 2.1 Fundamental member load-displacement.
The flexibility matrix will be derived by fixing
the l-end of the member and computing the general
deformation vector "a" resulting !rom the general applied
load vector "p2". Each of these vectors will contain
six elements, the deformation vector having three trans
lations aDd three rotations and the load vector having
three forces and three moments. The relationship between
the deformation and the load vector can be expressed in
terms of the flexibility matrix, F, as follows,
a • Fp2 (2-2)
8
Since the l-end of the member is completely fixed against
any deformation, the deformations at the 2-end can be
expressed uniquely in terms of p2 thereby establishing the
validity for the existence of the inverse of F. p2 can
then be written in terms of "a" and F-l which equals K,
the stiffness matrix, as
-1 p 2 = F a = Ka, (2-3)
where both K and F are symmetric matrices.
If p2 can be expressed uniquely as in (2-3) it is a
simple matter to obtain p1 from equilibrium considerations.
Figure 2.2 shows a segmental circular member with forces
and moments at each end directed in a positive sense,
followed by the equilibrium equations for the member.
End Loads
Py1
Figure 2.2 Member end load coordinates
9
10
Pxl + Px2 = 0 mxl + mx2 = 0
Pyl + Py2 = 0 myl + my2 -2Spz2 = 0 (2-4)
Pzl + Pz2 = 0 mzl + mz2 +2Spy2 = 0
The equilibrium equations can be written in matrix form
as follows,
pxl 1 0 0 0 0 0 Px2
Pyl 0 1 0 0 0 0 Py2
Pzl 0 0 1 0 0 0 Pz2 + X = 0 (2-5)
mxl 0 0 0 1 0 0 mx2
myl 0 0 -28 0 1 0 my2
mzl 0 28 0 0 0 1 mz2
or,
(2-5a)
In equation (2-5a) the matrix H is introduced and
will be referred to as the "equilibrium matrix". H is a
function which depends entirely on the geometry of the
member, therefore, if the force vector at one end of the
member is known, the force vector at the other end can be
computed from simple matrix multiplication of H or H-l
by the appropriate force vector.
i.e.
or, H-1 • p2 = - pl
(2-5b)
The inverse of H can be obtained directly by matrix
11
inversion, or it can be obtained more easily from physical
considerations. This is accomplished by interchanging
the l-end and the 2-end, which only reverses the signs of
the coordinates. This affects the H matrix simply by
reversing the signs of the off-diagonal elements.
End Displacements
The equilibrium matrix, H, can also be used to relate
the displacements at one end of the member to the displace
ments at the other end.
Let, 0 *1 , c;2 equal rigid body end-displacements
of a particular member. Since rigid body displacements
by definition do not affect the equilibrium of the member,
the total work resulting from the rigid body displacement
is equal to zero. This can be expressed as follows,
~·1 t ~ + p 2 ~ -0 ~2- ' (2-6)
where pt1 and pt2 are the transpose of the force matrix.
Eliminating pt 1 from equation (2-6) by using (2-5b)
results in the following,
if,
then, t p 1 =
Substituting into (2-6) then yields,
t Ht 6* t * 0 -p 2 1 + p 2 6 2 = •
Finally,
(2-7)
If the rigid body displacements are now superimposed
on the displacements in Figure 2.1 of 6 1 = 0 and 6 2 = a
the results are,
(2-8)
6 = 6* + a 2 2
If the rigid body displacements are now eliminated from
equations (2-7) and (2-8) the following results are
obtained,
12
and, (2-9)
or, (2-9a)
where equation (2-9a) expresses a measure of the amount
of elastic strain associated with any arbitrary end
displacements C>1 and 6 2 • Finally, if equation (2-9a)
is substituted into equation (2-3) the equation for p2
13
becomes,
Using equation (2-5b), (p1 = -Hp2 ), p1 can now be defined
as,
This results in a set of equations in member coordinates,
(2-10)
'
which are similar to equations (2-1) in system coordinates.
These equations can be written in matrix form as follows,
= (2-11)
or, p = K6
where, K11 = HKHt, K12 = HK, K21 = KHt and
K22 = K, remembering that K = the inverse of F.
t It can also be shown that if K is symmetric that K12 = K 21 •
The symmetry of K can be verified by considering Maxwell's
Law.
14
Because individual members of a structural framework
can be connected in any orientation relative to the system
coordinates, member forces and displacements must be
expressed in terms of system coordinates before the condi
tions of compatibility and equilibrium can be applied in
assembling the complete stiffness matrix of the structure.
It therefore becomes necessary to derive a trans
formation matrix to transform both p and 6 in member
coordinates to p' and 6' in system coordinates. This
can be accomplished with a general three dimensional,
rotation of axis transformation matrix of the type,
Y, y \ \ \
-" \ -\ --\ ---- X
, z.
i\11 A12 i\13 0 0 0
i\21 ?\22 i\23 0 0 0
i\31 i\32 (\33 0 0 0
T = (2-12) 0 0 0 i\11 Al2 A15
0 0 0 A21 ?-.22 i\23
0 0 0 A21 i\32 /\33
where, A· . represents the cosines of the angles between lJ
the appropriate member and system axis. Since T is an
th 1 . . Tt -- T-1. or ogona matrlx lt follows that After T has
been defined,p' and 6' can be written as,
p' = Tp
(2-13) and, C) = T6.
Using (2-13), the equations (2-11) can be re-written as,
Pi t 6' t 6' = TK11 T - TK12T
1 2 (2-14)
-TK Tt 6' t 6' and, P2 = + TK22T
21 1 2
defining TK .. Tt = K! . results in an expression of the lJ lJ
form of Eq. (2-1)
The general form of the four system stiffness
matrices can now be written from equations (2-11) and
(2-14) as follows,
K]_1 = THKHtTt
K]_2 = -THKTt
(2-15)
K21 = -TKHtTt
K22 = TKTt.
15
It can be shown that
making K' symmetric.
and,
K't 12 and K21 are equal
If K' is symmetric,
K' 21 = K't 12
K't = -TKHtT, 12
thereby
(K is symmetric, K = Kt) and the symmetry is therefore
verified.
2.3 Assembly of System Stiffness Matrix and Solution of Member Forces
The assembly of the system stiffness matrix from the
member stiffness matrices makes no further reference to
the elastic behavior of the material, but depends only
16
on the way the members are connected together along with
considerations of equilibrium and compatibility. Figure 2.3
illustrates a simple curvilinear grid which will be used to
illustrate how the system stiffness matrix is assembled.
Figure 2.3 Simple curvilinear grid.
The principles involved are exactly the same for any
space frame. Consider any member "I" connecting nodes
Jl and J2, the member equations being,
17
(2-16)
Considering compatibility,
Equations (-16) can be re-written in terms of the displace
ments of nodes Jl and J2 as,
which expresses the forces and moments at each end of
member "I" in terms of nodal displacements. From
equilibrium considerations applied nodal loads can be
equated in terms of member end loads at a given node as,
pJl = P11 + the contributions of any other
member at node "Jl",
and, pJ2 = p21 + the contributions of any other
member at node "J2".
The load-displacement equations can be written for nodes
Jl and J2 as,
+ contributions of other members,
+ contributions of other members.
From the preceding equations the assembly scheme for
system stiffness matrix can be deduced.
Consider for example the structure in Figure 2.3.
Each node is capable of six degrees of freedom which
produces a displacement vector 6' having three trans-
lations and three rotations, a load vector p, having
three forces and three moments, and a 6 X 6 nodal stiff-
ness matrix, K' • ·rhe size of the system stiffness matrix
necessary to describe the four node structure in Figure
2.3 equals six times the number of nodes square or
24 X 24.
As an example, the load-displacement equation will
be written for node 1;
P1 = [CK22)l + (K22)3 + (K:i_1)4 + (Kll)6J 6].
+ ~K]_2)J 6 2 + 8Kl2)6] 63.
Similarly the equations for the other nodes are
p2 = EK21)4J 6i + ~K22)2 + (K22)4 + (K22)5
+ (K]_l )~ 6 2 + [CKl2)7] 64'
18
P3 =[CK21)6J 6i + [CK22)6 + (K22)8 + (K:i_l)9
+ (K22)11] 6 3 + ~Ki2)9] 64,
P4 =[CK21)7]62 +~K21)9]63 + [CK22)7 + (K22)9 + (K22)10 + (K22)12J 64·
The preceding equations written in matrix form would
appear as,
pl K" 11 K" 12 K" 13 0 6'1
p2 K" 21 Kn
22 0 K" 24 62 =
p3 K" 31 0 K" 33 K" 34 83
lp4 0 K" 42 Ku
34 K" 44 CJ4
or as,
P = K" 6 ' '
where K" is the system stiffness matrix. The contri-
but ion that a single member makes to system stiffness
matrix can now be seen more easily from the preceding
equations. Consider member [§] • The direction of
member [§] '
as well as all other members, is considered
to be from the smaller node number to the larger node
number. Member [§] , therefore, has its l-end at
node (!) and its 2-end at node Cl) . The contribution
of member [§] then to the system stiffness matrix will
be as .follows;
19
(1) (K{1 )6 will be added to Kll' the direct stiffness
of node @ ,
(2) (K~2 )6 will be added to K;3 , the direct stiffness
of node Cl) ,
(3) (Ki2 )6 = (K21 )~ will be added to the appropriate
off diagonal sti!!nesses K1 3 and K31 ,
which demonstrates the symmetry of the K" matrix. The
systematic nature o! the method of assembly of system
stiffness matrix lends itself well to computer
programming.
2.4 Pinned-End Rigid Foundation Attachment
The discussion thus far has been concerned only
with the fixed-end rigid foundation attachments with
no reference to the pinned-end or other types of
foundation attachments. The following discussion
explains the modifications in formulation, necessary
!or the treatment o! the non-fixed rigid foundation
attachment. The more common case of the pinned-end
rigid foundation attachment will be explained in detail.
The pinned-end case could be handled in a manner
similar to what is done in slope deflection solutions,
where modified end equations are developed by presolving
the general slope equations of a member so as to
eliminate the unknown displacements of the pinned-end.
This reduces the number of unknowns which one has to
20
solve for and simplifies the solution. In the case of
matrix methods, however, such as the equilibrium method
where the high speed computer is being used to solve
the problem it is desirable to preserve the general
nature of the solution as much as possible so that the
programming of the method of solution can be accomplished
in a straightforward manner. The load-displacement
equations (2-1) expressing the end-reactions of a member
in terms of the end-displacements of that member are
genera.l in nature and are still equally valid even though
some of the end displacements and end reactions may be
zero, as is also the case in the general slope deflection
equations. With this in mind the following treatment of
the pinned-end foundation attachment can be seen more
readily.
21
In the equilibrium method the "system stiffness
matrix• is generated as though the pinned-end foundation
attachment was like any other node in the structural
system. Then the rows corresponding to zero joint
displacements are altered. This is done simply by placing
l's on the leading diagonal of the system stiffness matrix
and setting all other coefficients in the rows equal to
zero. The l's on the diagonal are necessary in order to
prevent the equation solving routine trom being upset by
zeros on the diagonal. Symmetry of the system stiffness
matrix can be maintained by placing zeros in the appro
priate columns.
Similar adjustments in the system stiffness matrix
can be made to handle other types of foundation attach
ments.
2.5 Intermediate Span Loads
22
The treatment of loads applied at points other than
node points can be handled in a straightforward manner by
the use of the principle of superposition. It is, there
fore, necessary to assume linear behavior of the structure
since the principle of superposition is only valid for
structures which behave linearly.
The analysis would precede in the following manner,
the loaded span in question would first be considered
fixed at each end and the forces and moments necessary to
prevent any translation or rotation of the ends of the
member would then be computed as in Figure 2.4a. Next,
the sense of these forces and moments are reversed and
applied to the ends of the loaded member, thus replacing
the intermediate span loads with a system of "equivalent
fixed-end forces and moments• as in Figure 2.4b. These
fixed-end forces and moments are then combined with any
other forces or moments applied directly to the nodes at
the ends of the member, resulting in the final nodal load
vector. It is necessary not to forget that before the nodal
load vector can be incorporated in the solution of the
structural system it must be transformed to system
coordinates.
w k/.f:t.
_....,..,.._ __ _;,Hz
r~Mz Vz
Figure 2.4a Fixed-end forces and moments on the loaded member.
H, -~-----
2.4b Equivalent nodal loads
23
An additional assumption which can be made to simplify
the solution of the fixed end forces and moments is that
the intermediate span loads lie in the plane of the member,
however, the nodal loads may be applied in any orientation.
Chapter 3
SEGMENTAL CIRCULAR MEMBER
3.1 Geometric Properties of a Segmental Circular Member
24
Although the stiffness method will be used to analyze
the curvilinear system, the member flexibilities will be
derived first and then the member stiffness matrix will
be obtained ·by inverting the member flexibility matrix.
The geometry is as follows,
Pyz
myz
G .., l>xz
mxz
e
5 5
Figure 3.1 Geomerty of the segmental circular member
S = one-half member span
R = radius of member
e = vertical distance from focus to end of member
~ = one-half of the total angle subtended by the member
The variables x and y expressed in terms of geometric
constants are,
x = S - R sin9
y = R cose - e
ds = Rd9
S = R sin ~
e = R cos ~
3.2 Load-Displacement Equations Derivation
The load displacement equations will be derived
using Castigliano's Second Theorem,
au = u. J..
op. J..
which says that the first derivative of the energy
expression, 1J , with respect to p. equals the correJ..
(3-l)
spending displacement u .• Energy due to bending, torsion J..
and axial deformation will be considered in the following
derivation, neglecting the effects of shear since members
will generally be light and flexible.
The general expressions for moments at any point
along the member about an x, y and z axis are,
about the "x" axis, M = -p y + m X z X
M = -p X + m y z y about the "y" axis,
M = PxY + p X + m X y z
about the "z" axis,
25
In order to be able to write the general energy
expression for the member, the forces and moments must
be expressed with respect to the major axes of the cross
section at any point along the member. (Figure 3.2)
y
yN
Mx G X
N
Figure 3o2 Forces and moments on major axes.
The moments and forces with respect to the axis in
Figure 3.2, are,
a) Axial Force,
N = p cos9 - p sin9 X y
b) Torsion,
= M cose X
- :M y sin9
26
N
c) Moment about nzn axis,
Mz = p y + p X + m X y z
d) Moment about nyN II axis,
MyN = M sin9 + M cose • X y
The previous equations can now be written in terms of
the end forces and moments, and the geometric constants
as;
a) N = p cos a - p sine X y
b) MT = Pz (S sin9 + e cose -R) + m cose - m sinS X y
c) M = p (R cos9 - e) + p (S - R sine) + mz z X y
d) MyN = p (e sin9 - s cos~ ) + m sinS + m cose. z X y
3.3 General Energy Expression
The general energy expression can now be written as,
/M~ds + 2G!
N
The deflections corresponding to the six degrees
of freedom are,
au aP ' X
au ap ,
y
au oP ' z
27
au am '
X
au om ' y
au
The following Table 3.1 which tabulates the partial
derivatives of N, MT' Mz and MyN with respect to the
various coordinate forces and moments can be very helpful
in setting up the various integrals.
TABLE 3.1
c3N aMT oM z oMyN
apx cos a 0 (R case 0 - e)
aPy -sinS 0 (S 0 - R sin9 )
oPz 0 (S sine + e cos9 - R) 0 (e sine
- s case
am X
0 case 0 sine
om 0 -sin9 0 case y
am 0 0 l 0 z
3.4 Deflection Calculations
The deflections will now be calculated assuming
that members have constant cross sectional areas and
that IN is the equivalent polar moment of inertia for
the section.
28
)
1) The deflection in the x-direction is,
-JN oN ds 6x - c3Px AE +
6x = !E J (px cos2e - py sine cose) de
+ ~rzjr[Px<R cose -e) + py (S- R sine)
+ mJ ~ cose - e] d9
after integrating and simplifying, the results are
R p R3 6 = Px (~ + ~ sin 2~) + x (0 - -23 sin2¢
X IT' c:. EIZ
(2 sin¢ - 20 cos~).
2) The deflection in the y-direction is,
29
= !E f (px cose - py sine ) (-sine ) d9
+ ~Iz /!!x (R cose - e) + py (S - R sine ) + mJ
~ - R sine] de ,
after integrating and simplifying, the results are,
3) The deflection in the z-direction is,
JMToMTds J oMNds 6= +My __ , z aPz UIN yN aPz EiyN
~ = ~IN f~z (S sin9 + e cos9 - R) + mx cose
- my sineJ [s sin9 + e cose - R_] d9
+ my cos!) [e sine - s cose J d9
30
after integrating and simplifying, the results are,
3 m R2 = pzR (3¢ + ~ sin4¢ - 2 sin2¢) + GI (¢ cos¢ G~ N
m R2 - 2 sin¢ + ~ cos¢ sin2¢) - if:-C¢ sin¢
N
R3 - -21 sin¢ sin2¢) + Pz (¢ - l sin4¢)
EiyN 4
m R2 m R2
+ EI (¢ cos¢ - ~ cos¢ sin2¢) - ~(¢ sin¢ yN yN
+ ~ sin¢ sin2¢).
4) The rotation about the x-axis is,
+ e cose - R) + m cose X
- s cose ) + mx sine + my cose] [sine] de '
31
after integrating and simplifying, the results are,
p R2 = GI (~ cos~ - 2 sin~ + ~ cos~ sin2¢)
N
m R p R2 + rrf:C¢ + ~ sin2~) + EI (¢ cos~ - ~ cos¢ sin2~)
N yN
m R 1 + EI (¢ - 2 sin2¢).
yN
5) The rotation about the y-axis is,
ey
9y
J aM J oM~N ds = M __! ds + M Tamy m; yN amy ElyN '
= ~rNJ l_pzcs sinS + e cose - R + m cose X
- ~ sin!] [-sin~ dEl + ~I f Q> z ( e sine yN)I
- s cose ) + mx sin9 + my cos~ GoseJ de ,
after integrating and simplifying, the results are,
-p R2 m R GIN (0 sin0 - ~ sin0 sin20) + trf;C0 - ~ sin2~)
p R2 m R - EI (~ sin¢ + ~sin~ sin2~ + ~(¢ + ~ sin2¢). yN yN
32
6) The rotation about the z-axis is,
e = JM aMz ds z z 0 m rrz z
after integrating and simplifying, the results are,
R2 p R2 Px ~ 9z = EI (2 sin¢ - 2¢ cos¢) + ~(2¢ sin¢)
z z
m R - rl-C2¢).
z
3.5 Flexibility Matrix
Now that the deflections and rotations of end 2 of
the member have been found in terms of the six degrees
of freedom, the flexibilities f .. can be found, according l.J
to definition, by setting the j~ force equal to unity and
computing the i~ displacement (i = 1, 2, ••• 6)when the
other forces and moments are set equal to zero. Flexi-
bility f .. can also be found by, l.J
f .. l.J
(3-2)
33
which is equivalent to the definition previously stated.
Now for simplicity let;
Px = pl m :: p4 6x =~ e = e X X 4
Py = p2 m = p5 6 =6 ey =e 5 y y 2
Pz = p3 m = p6 6z = 63 9z =e z 6.
The elements of the resulting flexibility matrix will
then be;
R (~ - ~ sin~) + 3
- ~ sin2~ + 2~ sin2~) f22 R (~ = -AE Er z
R3 - 2 sin2~ + k sin4~) R3
(~ - l sin4~) f33 = GI:"(3¢ + ~I N yN 4
R + ~ sin2~) +
R (~ - l sin2~) f44 = GI (~ EiyN 2 N
R - ~ sin2~) +
R (~ + ~ sin2~) f55 = GI (~ EiyN N
R = Er(2~)'
z
which are the diagonal elements of the flexibility matrix.
34
The off diagonal elements of the flexibility matrix
are;
3 = ~(2 sin2~ - ~ sin2~
z
fl3 = f31 = fl4 = f f f 0 41 = 51 = 15 =
2 f 61 = f 16 = ~(2 sin~ - 2~ cos~)
z
2 f 34 = f 43 = ~IN(~ cos~ - 2 sin0 + ~ cos0 sin20)
R2 1 + EI (~ cos0 - 2 cos0 sin20)
yN
R2 1 = - ~(~ sin0 - 2 sin0 sin2~) N
R2 1 - EI (0 sin0 + 2 sin0 sin20). yN
As can be seen a considerable number of the off diagonal
elements of the flexibility matrix are equal to zero,
which makes finding the inverse of the six by six much
easier than if the matrix was completely full.
35
Finally, the assembled member flexibility matrix
written in matrix form would be,
fll fl2 0 0 0 fl6
f21 f22 0 0 0 f26
0 0 f33 f34 ~5 0
0 0 f43 f44 0 0
0 0 f53 0 f55 0
f61 f62 0 0 0 f66 '
and it is evident also that the flexibility matrix is
symmetrical. ~e K matrix in equation (2-3) can now
be computed by inverting F. This K corresponds to
K22 in Equation (2-11). If the equilibrium matrix is
known, K11 , K12 , K21 can be computed. These K's, how
ever, are still in member coordinates and before they
are transformed into system coordinates, the trans
formation matrix T must be derived.
3.6 Three Dimensional Axis Rotation Transformation Matrix
In order to keep the formulation of system stiffness
matrix general in nature, the transformation matrix, T,
will be derived for a member whose axis is skewed to all
three system axes, as indicated in Figure 3.2.
36
From
and,
y'
;-------~----------------~----x I I l I
....... I ....... ............... I
.............. I ........ ....... ........ I
-...J
Figure 3.3 Typical member orientation
equations (2.13) for example,
p' = Tp
6' = To
or in matrix .form,
p' X A11 A12 .i\13 Px
p' = "21 i\22 /\23 Py y
p' z A31 /t32 ?03 Pz
37
I
where i\11 , A12 , ••• etc. are direction cosines. It can
be shown that T is an orthogonal matrix thereby making
T-l = Tt so that
and,
Ttp, p = (3-1)
The derivation of the transformation matrix T can now
precede in a systematic manner by considering Tt as the
product of three individual rotation transformation
matrices Tcx , T,B and T7 corresponding to the three
rotations which must be performed required to transform
from member to system coordinates as shown in Figure 3.3.
In each row of Tt is located the direction cosines of the
corresponding member axis.
38
39
y I
Y, y"
x,x,..
Sid xi.
Figure 3.4 The three required rotations
Let a x' a and a equal the direction cosines of x.
y z
Then,
XJ2 - XJl a YJ2 - YJl . a =
ZJ2 - ZJl a = = ' X s y s z s
where,
t Row 1 of T equals a , a and a which are the
X y Z
direction cosines of the x axis and can be found directly
from the end coordinates of the member. The last two
rows can be found by considering j3 , 7f and Ol rotations.
Consider first the)3 rotation and T~ where,
cos.B 0 sin,.d
= 0 1 0 (3-2)
-sin,B 0 cosj!J
The elements cosfi and sin,8 can be expressed in terms
of a , a and a as follows, X y Z
sinfj a
z
The system forces have now been transformed to the axis
where,
P,.s = T,s p'
Next the /3 -axis will be rotated to the o-axis or
i.e. ' the /3 -forces will be transformed to r oriented
forces.
cosT sinr 0
-sinT cosT 0 (3-3)
0 0 1
40
where,
cosT = / ax2 2 '\/ + az ' sinY = a ,
y
so that the system forces have now been transformed to
the T-axis as,
p ')- = T "l" P,s
or,
Finally, the ex -transformation, where ex. is the rotation
of they-axis of the member with respect to Yy-axis
about the member x-axis and is part of the input data.
1 0 0
0 cos« sin ex. (3-4)
0 -sin ex coso<.
The member forces can now be expresses as,
or, using Equations (3-2), (3-3) and (3-4),
p = Toe TT T,s P'
where, Tt = TLX T# Tft
and, p = Ttp'
or, p' = Tp •
41
In a similar manner the member stiffness matrix can be
transformed to system stiffness matrix by,
K' = TKTt.
Now if the appropriate terms for cos~, sin,.d , cos "r,
sin"( are substituted into the appropriate matrices and
the indicated multiplications of T«. T't' T13 are carried
out the final form of Tt will be as follows,
where,
A11 = a X
a a COSO( - a sin()(
tu2 - X if. z
=
ja2+a2 X Z
a a sin ex - a COSO(
'A13 X l, z
=
j 2 2 + a ax z
A21 = a y
2' =~ ax2 + A22 a cos~ z
2' A23 ;j ax2 sinO(. = + a z
42
43
-a a cos~ + a sinO£ /\32 = -:I. z X
~ ax2 + 2
a z
a a sinO<: + a co sac::
/\33 = J. z X
,J ax 2 2
+ a • z
Chapter 4
APPLICATION OF ANALYSIS
In this chapter the information obtained from
Chapters 2 and 3 will be applied to the analysis of
"curvilinear space grids". The details of the computer
program, which performs the necessary calculations, are
discussed in the following chapter. The objective of
44
the material in this chapter is to demonstrate the validity
of the solution and to develop confidence in the analysis.
4.1 Procedure
Three different mathematical models were chosen to
demonstrate the analysis. These models are circular
framed dones having height to span ratios, (h/L), of one
sixth, one-fourth and one-half respectively, with each
one having a span of 72.0 inches.
Each of the structures were analyzed under the
influence of symmetric and unsymmetric loads. In one
case, all foundation attachments were fixed, while in
the other case half of the foundation attachments were
fixed and half were pinned. The details of the mathe
matical models can be seen clearly in the diagrams
which follow in (4.2).
Finally, an experimental model was build and tested
in order to provide the necessary correlation between the
analytical solution of the mathematical model and
experimental results from the physical model.
4.2 Mathematical Models
Each of the mathematical models are composed of
twenty-four segmental circular members, whose topological
arrangement describes a "lattice dome". All of the
members lie on the arc of a great circle. The model
dome has nine interior nodal points and twelve foundation
attachments.
a.) Mathematical Model 1- The computations involved
in obtaining the geometric data will be shown in
detail for this model as an example of the required
hand calculations, and subsequently deleted from
the remaining models as they are handled in the
same manner.
1.) Basic Geometry:
h = 1 L 72.0" h = 12.0" I:' 6 ' = '
R = ~ [1 + 4(~)2 J :r;
R = 60.0"
45
+l.
2.) Plan View- Figures 4.1 and 4.2 illustrate the
node and member numbering scheme used in the
analysis of the model with fixed foundation
attachments and combined fixed and pinned
foundation attachments.
z ~ 5 6
z --:-
8 I 9 ~ II 4 IZ 5 13 ~ >"' ---
/5 ~ 16 ~
.... 8 _.:a.
18 19 zo
t 1 Z4 2'3
+X
Figure 4.1 Mathematical Model-l (~ = ~) , all
foundation attachments fixed.
46
+~
z ~ 4 5 6 .,.. ~ 5 ~
8 t 9 I j 10
II 7 /Z. 8 /3 9 14 - >" >"'" .....
15 ~ /6 ~ J 11
II ..-::.
19 zo Zl
Z3 ~
+>'
Figure 4.2 Mathematical Model-l (~ = i) , with
both pinned and fixed foundation
attachments.
47
3.) Model profile:
h .. :: 12.0
e= 48.0 II
Figure 4.3 Basic coordinate dimensions
9 = s1"n-l L/2 = 36.87° R
hl = R sin (90° - 2¢) =
dl = d2 + d3
56.92
dl = R cos (90° - 2¢) = 18.97
d3 (h -dl
in. = e) n:- = 2.97 1 1
d2 = d - d = 16.0 in. (See 1 3
in.
in.
Fig. 4.1)
48
h,
4.) General coordinates of nodes 1 3 7 9 ' ' '
Figure 4.4 Geometry of 1, 3, 7, 9 node points.
OAB = 60.0 in.
AB = 60.000 - 53.066 = 6.934
Ax = Az = 16
6 • 934 53.066 =
48 Y = 6.934 53 •066 = 6.27 in.
in.
in.
2.09 in.
The resulting coordinates of B without regard to
sign are,
x = z = 16.0 + Ax = 18.09 in.
Y = 6.27 in.
49
50
5.) General coordinates of nodes 2, 4, 6, and 8 are,
x = z = d1 = 18.97 in. (See Fig. 4.3)
y = h1 - e = 8.92 in.
6.) Member-¢- Calculations
- = 4
¢1 = 0.160877 rad.
¢2 - Members 5, 6, 8, 10, 15, 17, 19, 20
where, s2 = one half the chord length of the
member.
Then,
2s2 = ,/c.8s)2 + (2.65)2 + (18.091)2 • 18.305 in.
¢2 = sin-1 18.gg5/2 = 8.774o
¢2 = 0.153135 rad.
¢ - Members 1, 3, 4, 7, 18, 21, 22, 24
where, 83 equals one half the chord length of
the member.
Then,
51
283 = ,j(l4.16)2 + (6.27) 2 + (2.09)2 = 15.625 in.
-1 = sin 15.625/2 • 7.480 60.0
~3 = 0.13055 rad.
7. ) Member - tX- computations
'
' \
Figure 4.5 Required geometry !or ~.
52
Computations continued,
(See Fig. 4.3)
a1 = 60 sin (81.23°) = 59.30 in.
c1 = 59.3 sin (18.43°) = 18.75
ri = Jc9.15)2 + (18.75)2 = 20.864
0( • -1 20.864 - 20 350 1 = sln 60.0 - •
()(. = • 3 5517 rad. 1
a2 = 60 sin (54.97) = 49.13 in.
b2 = 60 cos (54.97) = 34.44 in.
c2 = 49.13 sin (18.43) = 15.53 in.
r2 =~(34.44)2 + (15.53) 2 = 37.727 in.
~ = sin-1 37.727 = 38.96o 2 60.0
0(2 = .67998 rad.
4.3 Example Computer Analysis
The computer solution of Mathematical Model-l, with
fixed foundation attachments and symmetrically loaded is
presented here to illustrate the computer output.
The input data, such as coordinates, member
properties, nodal data and nodal loads is printed first.
The input data is followed by the computed data such as
the "Link" array, a sample flexibility, stiffness and
transformation mat~ix, the nodal displacements and the
member-end forces and moments in both system and member
coordinates.
The computer solutions for the other mathematical
models are presented in Appendix A.
53
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57
4.4 Experimental Model
For simplicity the experimental model was designed
with four interior nodes and eight foundation attachments,
while consisting of twelve members •. The dimensions given
for the experimental model are the true measured
dimensions, as, there was some discrepancy between the
original design dimensions and the finished model. This
can be expected, however, from a model of this nature.
a.) Material and Member Properties- The model was made
from one fourth inch square steel rods. The
geometric properties o! the members are as follows:
1.) Polar moment of inertia, Ix*' is
where 1 b b4
13 ~- - o. 21 -h (1 - ) ' 3 12h~
and since b = h for a square section it follows that
/3 ~.1418 '
and if h = 0.25 in.,
• 00055 in • 4 IX = •
*(See appendix c, of reference 2) Here Ix is the
equivalent polar moment of inertia and is the same as
the previously defined, ~·
2.) Moment of inertia about they and z axis is
b4 4 I = I = ~ = 0.000326 in. • y z ~~
2 Cross-sectional area equals b , or
Area= o.0625 in. 2•
4.) Young's modulus, E, was determined from a
tension test of the material. The results of
this test are as follows:
strain= 0.00061 in./in.,
load p = 1200.0 lbs.
therefore,
E = strainpx area = 31,475,000 psi
5.) The shear modulus, G, was assumed as,
12,000,000 psi.
58
b.) Geometry and Numbering Scheme for the Model
.3
6 7
10
~ IZ
Figure 4.6 Experimental model with pinned foundation attachments.
The coordinates for each of the members are given
in Table 4.1.
59
4.5 Experimental Results
Three tests were run on the experimental model in
which it was subjected to both symmetric and unsymmetric
loads, for both pinned and fixed end foundation attach
ments. In each of the tests the strain in member-12
at 6.5 inches from node 4, and the vertical displace
ment at each of the nodes were measured. Because o! the
difficulty encountered in measuring the true vertical
disp~acement of the node, these values are not too
accurate. The following is a summary of the three tests.
a.) Test No. 1
Foundation attachments: all pinned
Load: 30.0 1bs. at each node
Strain in mem.-12: -6 175.0 x 10 in./in., ten.
Nodal displacement: 0.027 in. ave.
2.) Theoretical calculations:
Member end forces on member-12 at End-1,
in member coordinates are
Px = 34.05 lbs. m • o.o X
Py • -1.32 lbs. m - 1.08 in.-lbs. y
Pz • -0.071 lbs. m • -20.06 in.-lbs., z
60
61
and at End-2,
Px = -34.05 lbs. m = o.o X
Py = 1.32 lbs. m = o.o y
p = 0.071 lbs. m = 0.0 z z
Nodal displacements: 0.022 in. ave.
3.) Correlation of experimental stress in member-12,
to the theoretical stress, neglecting Pz and
m· y'
1 /.3Z
34.05
e
Figure 4.7 Free body diagram of Member-12.
e' - ~ - ~ = .1857 = 10.6° - R - 35
f6 - a• = 12. 51 - 10. 6 = 1. 91 o
h' = 35 cos (1.91°) = 34.98 in.
e = 35 sin (77.49°) = 34.17 = 0.81 in.
x = R sin (12.51°) -sin (1.91°) = 6.41 in.
The moment at the strain gage is,
M = 20.06 - 34.05 (.81) - 1.32 (6.41)
M = 15.98 in.-1b.
Axial = 34.0 #, compression.
The resulting stress are,
34 Axial stress = •0625 = 544 psi, C
Bending stress = 15.98 (.125) = 6127 psi, T .000326
Total stress = 6127 - 544 = 5583 psi, T
4.) %-Deviation between experimental and
theoretical stress:
Exp. stress = 175.0 x 10-6 (31.475 x 106 )
II II = 5508 psi,
Therefore,
~ D 5583 - 5508 x 100 = 1.36% 10 ev. = 5508
62
Table 4.1 Member Coordinates
Mem. X(I,1) Y(I,l) Z(I,l) X(I,2) Y(I,2) Z(I,2)
1 -20.97 o.o 5.8? -7.29 6.25 7.22
2 -20.97 o.o -6.62 -?.23 6.25 -?.33
3 -6.06 o.o 20.9? -?.29 6.25 ?.22
4 -7.29 6.25 ?.22 -7.23 6.25 -7.33
5 -7.23 6.25 -7.33 -6.25 o.o -20.9?
6 -7.29 6.25 ?.22 7.22 6.25 7.29
7 -7.23 6.25 -?.33 ?.19 6.25 -7.19
8 6.06 o.o 20.9? 7.22 6.25 7.29
9 7.22 6.25 7.29 ?.19 6.25 -7.19
10 7.19 6.25 -?.19 6.12 o.o -20.97
11 7.22 6.25 ?.29 20.97 o.o 6.31
12 7-19 6.25 -7.19 20.97 o.o -6.31
b.) Summary of Tests- A summary of the experimental
results for Test No. 1, as well as Test No. 2 and
Test No. 3 are given in Tables 4.2, 4.3 and 4.4
respectively.
63
Table 4.2 Summary of Test No. 1 - Pinned-End Foundation Symmetric Loading
- Experimental -
Nodal Displacements
Load/node 1 2 3
30.0 lbs. -0.027 -0.029 -0.027
- Theoretical -
30.0 lbs. -0.021 -0.023 -0.022
Mem.-12- Strain= 175.0 x 10-6 in./in., T
Experimental Stress= 5508.0 psi., T
Computed Member-12 End Forces
Px py p m m End z X y
1 34.05 -1.32 -0.07 o.o 1.08
2 -34.05 1.32 0.07 o.o o.o
Theoretical Stress= 5585.0 psi., T
% Deviation = 5585 - 5508 X 100 • 1.36% 5508
4
-0.027
-0.022
m z
-20.06
o.o
64
Table 4.3 Summary of Test No. 2 - Fixed-End Foundation Symmetric Loading
- Experimental -
Nodal Displacements
Load/node 1 2 3 4
31.08 lbs. -0.015 -0.017 -0.017 -0.014
- Theoretical -
31.08 lbs. -0.012 -0.013 -0.012 -0.012
-6 Mem.-12- Strain= 135.0 x 10 in./in., T
Experimental Stress = 4280.0 psi.
Computed Member-12 End Forces
px py pz m m m End X y z
1 38.32 -0.15 -0.11 0.69 0.85 -21.06
2 -38.32 0.15 0.11 -0.69 -0.80 18.76
Theoretical Stress = 3729.0 psi.
% Deviation 4280 - 3722 X 100 = = 4280-- 12.9%
65
~able 4•4 Summary of Test No. 3 - Fixed-End Foundation Uns~etric Loading, Load Node 4 Only
----- Experimental -........._.,__ __
-.....__ __ Nodal Displacements
JJOad 1 2 3 ,_____ :;1.08 ll:>S. -0.0240 +0.030 +0.028 ~---
-Theoretical -.............,._
:;l.oB los. -o.o2o +0.024 +0.024 "-----
-6 Me~.-12 - strain ~ 55 X 10 in./in.
E~e~i~enta1 stress = 17~1 psi, T.
Computed Member-12 End Forces
:En<i Px p Pz m my y X
1 29·45 1·55 -0.14 o.o 1.06
2 -29·45 -1-55 0.14 o.o 1.08
~~o~etica1 Stress ~ 1800 psi
% be~iation 1800 - 1.2~~ X 100 = ~.98% ~ -' 1731
mz
-4.4
27.87
4
-0.051
-0.041
~e computer solUtions for the experimental model
-tS ~ p~~sented in Appendix B.
66
Chapter 5
COMPUTER PROGRAM FOR CURVILINEAR GRID SYSTEM
The computer program which performs the complex
computations required by this or any matrix solution to
a large structural system is an integral and extremely
important part of the development. This chapter contains
a detailed description of the development of the computer
program including a complete documentation of the program
including user instructions, flow diagrams, summary of
program identifiers, and a complete listing of the pro
gram and subroutines.
5.1 Introduction to the Program
Matrix Analysis of Curvilinear Systems (MACS), is
written in Fortran IV for IBM System 360/40. The
formulation of MACS is fundamentally an equilibrium
stiffness formulation. The program has been developed
in two main parts; first the main line which consists
of a general formulation of the equilibrium-stiffness
method adaptable to the solution of any space frame,
and second the subroutine package which consists of the
necessary routines for performing the required repeti
tive calculations as well as those routines which enable
MAcs to be quite flexible as a space frame solver. The
details of these routines will be discussed in (5.4).
67
In order for the engineer to use MACS it is only
necessary that he provide MACS with the appropriate
input data such as, Node data, geometry data, member
properties and loading data as described in (5.2).
5.2 Input Data
The following is a detailed description of the
necessary input data.
a) System Indices- NN, NB, LL (Fixed pt. no.)
NN - Number of nodes excluding fixed and foundation attachments.
NB - Number of members in the structure.
LL - Maximum number of members per node.
b) Member Indices
Nodeil - Node number at l-end of member I.
Node12 - Node number at 2-end of member I.
c) Member Coordinates
Xil' Y;l' Zil -Coordinates of the l-end of • member I in system coordinates.
Xi2 ' Y; 2 , z; 2 - Coordinates of the 2-end of • • member I in system coordinates.
d) Member Properties
IX., IY., IZ. - Moments of inertia about the 1 1 1 member X, Y and Z axis.
- Cross sectional area of member
e) Geometric Properties
- Radius of curvature of member
68
ALPH. 1
- One half of central angle described by the member
- The rotation of the member about the member x-axis with respect to a vertical plane through its end points.
f) Elastic Properties
E
G
g) Nodal Loads
D. J
5.3 Output Data
- Youngs Modulus
- Shear Modulus
- Six element vector which describes Px' Py' Pz' mx' my' mz' the six possible components of the applied nodal load vector.
The output data from MACS was designed to give the
engineer all the information necessary for establishing
the validity of the solution.
a) Input Data - Before any structure can be analyzed
correctly the correct input data must be used.
This is a common source of error in an analysis
of this type where large quantities of data are
assembled, coded and finally key punched onto
data cards. As a result it is quite easy for
human error to be a significant factor. For this
reason all of the input data is printed out in a
convenient form to enable the engineer to verify
that the data read by the computer is the correct
data.
69
b) 6 Link Array - The first computation performed by
MACS computes,
Link (J, L) = + I, where I is the L~ member to be attached to node J. The plus sign indicates that the l-end of member I is at node J, the minus sign indicates that the 2-end of member I is at node J.
Link (J, L) provides MACS with the necessary member
to node reference required for computing which
members contribute to the stiffness of a particular
node.
c) Member Flexibility, Stiffness and Transformation
Matrices - The flexibility, stiffness and trans
formation matrices for each member are printed out
to provide the engineer with information necessary
for developing a better understanding of the
behavior of the system. The member stiffness
matrix, for example, will show the contribution
of each of the quantities such as axial, torsional
and bending deformations to the overall member
stiffness.
d) System Stiffness Matrix - The system stiffness
matrix furnishes additional information on the
structural system as a whole, being the coef
ficients of the load-displacement equations for
the structural system.
?0
e) Nodal Displacements - After the system stiffness
matrix has been generated, the nodal loads and the
system stiffness matrix are then solved simul
taneously by S~LVE to obtain the nodal displace
ments. The six components of the displacement of
each node are written out as follows,
6y 6z By 9z
f) Member End Forces - The resultant force vector at
each end of the member are first printed out in
System Coordinates to facilitate checking the
statics of the structure and then printed out in
member coordinates· which is more convenient for
stress analysis.
5.4 MACS Subroutine Package
The following discussion describes the subroutines
required by MACS to perform the necessary repetitive
calculations. These subroutines have been appropriately
named STIFMA, MATRAN, MTXMUL, TRIPR¢ AND S~LVE.
a) STIFMA - The member stiffnesses are computed by
calculating the member flexibilities from the
equations derived in Chapter 2 and inverting the
resulting flexibility matrix. STIFMA also computes
the equilibrium matrix H, which is described in
Chapter 2.
71
b) MATRAN - The 3-axis rotation transformation
matrix T is computed by MATRAN from the member
end coordinates and ALPH which was described
earlier in this chapter.
c) MTXMUL - Matrix multiplication is performed by
MTXMUL; given two matrices A and B. MTXMUL
premultiplies B by A and places the resultant
AB in the B array making efficient use of storage.
d) TRIPR~ - Matrix Triple Product, premultiplies B
by A and then post multiplies AB by the transpose
72
of A to form ABAt. The resulting array is then
returned to the B array storage location whi.ch again
e)
makes efficient use of storage.
S~LVE - This subroutine is a "simultaneous
equation solver". S¢LVE uses a Jordan Elimination
and pivots on the largest element in the coefficient
matrix8 , thereby, reducing the roundoff error to a
minimum and preserving the accuracy of the solution.
As was mentioned previously in (5.1) this subroutine
package enables MAGS to be quite flexible as a space
frame solver. For example, if a space frame is composed
of straight members instead of segmental circular members,
the STIFMA which computes the stiffnesses of the circular
members can be replaced by a new STIFMA which computes the
l •!:,
"
73
stiffness of a straight member. In fact, as long as one
can derive the flexibility or stiffness matrix for a member,
regardless of its shape, and substitute the appropriate
STIFMA routine, MACS is capable of analyzing a system
composed of these members. In addition, there is also
the possibility that a structural system may be composed
of both straight and curved members. In this case a
stiffness subroutine could be included for each type of
member and properly labeled for ready reference.
5.5 MACS User Instructions
It is impossible for any computer program to obtain
the correct results unless the correct input data is
supplied to the program. For this reason the engineer
must be meticulous and systematic in preparing the input
data for the computer program.
The data can be divided into two main parts,
structure data and load data.
a) Structure Data - It is advisable to begin the
preparation of the structure data with a
convenient sketch of the structural system. Next,
the nodes and members can be numbered in any
arbitrary fashion, except that fixed-end foundation
attachments are labeled zero. Pinned-end foundation
attachments are labeled the same as any other node.
The location of the system coordinate axes can
now be placed at any convenient location. The data
cards can now be listed in the order in which they
appear in the data deck as follows:
1.) 1 Card- contains the system indices NB, NN
and LL respectively in an 110 Format.
2.) 2NB/12 Cards- Node (I,l) and Node (I,2) are
placed in a 12I6 Format. All values of
(I,l) are read first followed immediately by
the values of Node (I,2).
3.) NB Cards- The next NB cards contain the
joint coordinates at each end of a member
and ALPH (I) for each of the members. The
order of the data is X(I,l), Y(I,l), Z(I,l)
X(L,2), Y(I,2), Z(I,2) and ALPH (I) written
in a 7E10.2 Format.
4.) NB Cards- The next NB cards contain the
member properties IX(I), IY(I), IZ(I),
AREA(I), R(I) and PHI(I) respectively for
each member in a 6El0.2 Format.
5.) 1 Card- This card contains E and Gin a
2El0.2 Format.
74
b) Load Data - The preceding data cards were concerned
entirely with the structure data, next the load
data will be read.
1.) NN Cards- The last NN cards of the data deck
contain the nodal loads. There must be one
card for every non-zero node with the six
components of the load vector entered on the
card in a 6El2.4 Format. If a node is not
loaded a blank card can be used to represent
a zero load vector.
If a member is loaded between nodes the fixed end
forces and moments must be computed and then their
75
signs reversed and superimposed on the applied nodal loads
as explained in Chapter 2.
76
5·6 Flow Diagrams and Computer Program
In the development of large, complex computer programs
the flow diagram is a necessary and valuable tool. The
flow diagram presents to the programmer, in the appropriate
order, the various operations and decisions that the com
puter must make during the execution of the program.
The first flow diagram is a block diagram and
represents schematically the major operations performed
by the computer.
a.) Block Diagram
Solve For Generate Nodal ~-4------4 System Displmnts Stiffness
Solve For Member End Loa ds
Matrix
Write: Member End Loads
End
Write: Input Data
Generate ~--~---~Link
it~~tJ L)
b.) Main Line Flow Diagram.
REAL : T ( 6 , 6 ) , X ( 100 , 2 ) , Y ( 100 , 2) , Z ( 100 , 2 ) , ALPH ( 100 ) R(lOO),IX(lOO),IY(lOO) IZ(lOO),AREA(lOO), H(6,6),ST(6,6),PHI(lOO~~F(6,6)AK(6,6),D(lOO), A(lOO,lOO),AI(6,6),PP(6J,P(6)
INTEGER:NODE(l00,2),NBPN(25),LINK(25,6) COMMON: X,Y,A,ALPH,H,T,R,PHI,IX,IY,IZ,AREA,E,G,ST,F
1 Generate -----'Link
1 Array 1 Link(J ,L)
Read: Load Data
Set All NBPN(J) to Zero
+
Read: Structure Data
~ 20 I+l,NB
Jl• Node(I,l)
77
78
J2=NODE(I,2) LINK(Jl,Ll)•+I
0
..,.__..~ NBPN ( J2) ·NBPN ( J2) + 1
LINK(J2,L2)•-I L2•NBPN(J2)
continue
Write: LINK(J,L)
Generate System Stiffness Matri
Df" 21 Mlml,6 M2-=1,6
Clear H D~ 69 A( I ,J) . ' J==l NN I Storage 1 \_ ' _ __;
---------··--..----t I•IABS(LINK(J ,L))
0
+
Compute Member( I) Flexibilit Stiffness
Call
79
Write: >---4----t F(I ,J) & STIFMA (I )1>--~
ST(I,J)
...__.,:Ill____. AK ( M1 , M2 ) = S T ( M1 , M2 )
: Compute 1 Transfor-1 mation I Matrix
D~ 41 K•l,6
N=K+6*(J-l)
A(N,M)•AK(K,Kl)+A(N,M)
NJ•1
D~ 41 Kl•l,6
80
JJ=J
G~ T~ 52
DO 51 K=l,6
~-----~ A(N ,M)=AK(K ,Kl )+A(N ,M)
Call TRIPR9J (H,AK,6)
Call TRIPR9J (T,AK,6)
D~ 51 Kl=l,6
81
JJ=NODE(I,NJ)
DO 54 Kl=l,6
M=Kl+6*(JJ-l)
Call MTXl"'UL (H,ST,6)
Call TRIPR~ (T,ST,6)
DO 54 K=l,6
N=K+6*(J-l)
82
D~ 65 K=l,6
1 Modify I IF NODE IS ~---~---f 1 Pinned-End
D¢ 72 J=l,NN
N=K+6*(J-l)
D~ 65 Kl=l,6
83
D¢ 66 11=1,3 N=11+6*(J-1)
A(12,N)=O.
_ __,A (N, 12) =0. 1--~---t.
A(M,M)=1.0
c~ntinue
Write: System ~~~ffiess
D¢ 71 11=1,3
M=I1+6*(J-1)
84
1 Solve For ------"~~---'1 Nodal l Displmnts
I Solve For r-----~----~ Member
End Forces
Call Solve
(A,D,KKK)
Write: Dist>lmnts D(J)
85
D¢ 81 I=l,NB
l=NODE(I,l ~--~--~T2=NODE(I,2~---4------
Call MTXMUL
(H,AK,6)
Call TRIPR9J
(T,AK,6)
AI(J,K = ST(J,K
D¢ 82 J=l,6 K=l,6
Call TRIPR¢
(T,ST,6)
Call TRIPR¢
(T,AI,6)
Call TRIPR¢
(H,ST,6)
Df{J 84 K=1,6
J=L+6*(J2-1)
Df{J 83 K•l,6 L=1,6
D¢ 84 L=1,6
86
P(K)=O
D~ 85 L=l,6
,---......j PP(K)=O
D¢ 87 L=l,6
J=L+6*(Jl-l)
D~ 85 K=l,6
D¢ 87 K=l,6
PP(K)=PP(K)+ST(K,L)*D(J)
Write: 1 PP(J) ,in
1---............c::......__~ System Coord.
P(K)=P(K)+T(L,K)*PP(L)
Write: 1 P(K) ,in
>--........,----"~ Member Coord.
J=L+6*(J2-l
-----tpp(K)=PP(K)+AI(K ,L) *D(J)
87
PP(K)=PP(K)-AK(L,K)*D(J)
Write: 2 PP(K) ,in
~--...~ System Coord.
D~ 88 L=1,6
P(K)=P(K)+T(L,K)*PP(L)
END
D~ 86 K•l,6 L=1,6
J=L+6* (Jl-1)
D~ 88 K=1,6
P(K)=O
-Write: 2 P(K) ,in
t--~..., Member Coord.
G~ T~ 999
88
c.) Subroutine MTXMUL Flow Diagram
DIMENSION A(6,6),B(6,6),¢(6)
D¢ 1 J=l,N
¢(J)=¢(J)+A(J,K)*B(K,I)
89
END
d.) Subroutine TRIPR~ Flow Diagram
DIMENSION A(6,6),B(6,6),¢(6)
D¢ 2 K=l,N
~~ ~(J)=~(J)+A(J,K)*B(K,I)
D~ 4 J=l,N
¢(J)=O.
D¢ 3 I=l,N
D¢ 3 K=l,N
D¢ 1 I=l,N
D¢ 2 J=l,N
D¢ 1 J=l,N
B(J,I)•¢(J)
D¢ 3 J=l,N
)....--......--t ¢(J)=¢(J)+B(I ,K) * A(J ,K)
B(I,J)=¢(J) Return End
90
e.) Subroutine MATRAN (I)
REAL T(6,6),X(100,2),Y(100,2),Z(100,2),ALPH(100), R(lOO),PHI(100),IX(100),IY(100),IZ(100), AREA(100),H(6,6),ST(6,6),F(6,6)
COMMON X,Y,Z,ALPH,H,T,R,PHI,IX,IY,IZ,AREA,E,G,ST,F
T(M,N)=O.
SX=X(I,2)-X(I,1) ..___~ SY=Y(I ,2)-Y(I ,1)
SZ=Z(I,2)-Z(I,1)
I
T(1,1)= ••• T(1,2)= •••
• • •
T(3,3)= ••• See Chpt. 3
10~-(
D~ 10 M=1,6 N=1,6
CX=SX/S s---r--t CY =SY/S
CZ=SZ/S
D¢ 12 M=4,6 N=4,6
~ T(M,N)=T(M-3,N-3)
91
-
I
Dft' 20 JJ=l,6 KK=l,6
H(JJ,JJ)=l.O
H(5,3)=-S H(6,2)=S Return
End
H(JJ,KK)=O.
D¢ 25 JJ=l,6
92
f.) Subroutine STIFMA (I)
REAL T(6,6),X(l00,2),Y(l00,2),Z(l00,2), ALPH(lOO),R(lOO)~PHI(100),IX(100),IY(100), IZ(lOO),AREA(100 ,H(6,6),ST(6,6)
DIMENSION F(6,6),NN(6 ,MM(6) COMMON X,Y,Z,ALPHA,H,T,R,PHI,IX,IY,IZ,AREA,E,
G,ST,F
F(M,N)=O.
F(l,l)=··· F(l,2)=···
• •
F(3,4)=··· • •
F(6,6)=••• Defined in Chpt.2
D~ 10 M•l,6 N=1,6
D~ 31 M=1,6 N=1,6
------ts'l{ M,N~,N)
·~achine loaded subroutine
93
Write: Singular
Matrix
Call Exit
Return
End
94
95
g.) Subroutine S¢LVE Flow Diagram
DIMENSION A(lOO,lOO) ,B(lOO) ,IR(lOO)t----....._. __ _
D¢ 10 l=l,N
D¢ 20 I=l,N
XLARG=O.
D¢ 11 I=1,N J=l,N
B(I)=B(I)/100.
IR(J)=O
D¢ 110 L=l,N
~ 10 J•1,N
PN=N sos-o.
D~ 20 J-=1,N
SO=SQRT(SOS)/PN
96
0 +
XLARG•A(I,J)
97
__ ___..,.._-:_l_~i_:_£ _ ___,1 - (0 • [;;J~---
I=Il J•Jl
0
IR(Jl)•Jl
0
D¢ 68 K=l,N
D¢ 60 K=l,N
A(J,K)=A(J,K)+A(I,K) A(I,K)=A(I,K)-A(J,K) A(J,K)=A(J,K)+A(I,K)
BB=B(J) t-----=-----1 B ( J ) = B ( I )
B(I)=-BB
B(K)=B(K)-B(J)*A(K,J)/XLARG
98
99
0
0
A(K,I)=A(K,I)-A(J,I)*A(K,J)/XLARG
continue
D~ 90 J=1,N
D~ 110 K=l,N
DTM=l.
0
100
A(K,J)•O.
0
DTM•DTM*A(J,J)
Write: DTM
G(lJ T(lJ 92
DTM•O.
D(lJ 24 J•1,N
B(J)•B(J)/A(J,J)
End
101
0
Table 5.1 Table of Symbols
Program Symbol
Main Program
NN
NB
LL
NLC
NODE
X,Y,Z
ALPH
IX, IY, IZ
AREA
R
PHI
E
Problem Symbol
A
R
E
Definition
Number of nodes
Number of members
Maximum number of members per node
Number of the loading condition
Name for node numbers
Geometric Coordinates
Rotation of members about member x-axis
Principle moments of inertia
Cross-Sectional Area of member
Radius of curvature of member
one-half central angle of member
Modulus of elasticity
102
Table of Symbols (continued)
Program Sym.bol
Main Program
G
KKK
D
p
pp
NPPN
Jl
J2
Ll
LINX
Problem Symbol
G
,P
p
P'
Definition
Shear modulus
Dimension of system matrix
103
Used both as a label for nodal displacements and applied nodal loads
Member and forces and moments in member coordinates
Member end forces and moments in system coordinates
Counter used to count number of members attached to a node
Node number at end-1 of a member
Node number at end-2 of a member
The 1111 member attached to node
Name of member to node reference
Table of Symbols (continued)
Program Symbol
Main Program
A
ST,AK
F
T
H
NJ
PP
p
!:l_TXMUL
A
B
Problem Symbol
K"
F
T
H
104
Definition
System stiffness matrix
Member stiffness matrix in member coordinates
Member flexibility matrix
Transformation matrix
Equilibrium matrix
Member and reference
Member end forces in system coordinates
Member end forces in member coordinates
A general six by six matrix
A general six by six matrix
An operating vector used to store rows of AB back into the used columns of B
Table of Symbols (continued)
Program Symbol
TRIPR¢
A
B
MATRAN
sx
SY
sz
s
ex CY
cz
Problem Symbol
T
a X
ay a z
105
Definition
A general six by six transformation matrix
A general six by six matrix
An operating vector used to store rows of ABAt back into the used columns of B
The difference in X-coordinates of the 1-end and 2-end of the member
The difference in Y-coordinates of the 1-end and 2-end of the member
The difference in z-coordinates of the 1-end and 2-end of the member
Square root of the sum of the squares of SX, SY and SZ
CX,CY,CZ are the of direction cosines
the member X-axis
Table of Symbols (continued)
Program Symbol
c
T
H
S@LVE
A
B
IR
N
sos
so
TOL
XLARG
DTM
Problem Symbol
T
H
K"
P'
Definition
Square root of the sum of the squares of ex and cz
106
Transformation matrix
Equilibrium matrix
System stiffness matrix
Applied nodal loads
Operating vector
Size of system stiffness matrix which equals six times the number of nodes
Sum of the squares of the elements in the system stiffness matrix
Square root of the sum of the squares.
Tolerance
Largest eleaent in system matrix at beginning of each pivot
Deteraillant
10?
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',.. r T" •T "J'l110f,ll r"rt\ll'liiT~'t K'l',?!-= T*lll?, - -"Jf\-? nr ... r .. nr"!f1 'n ' nu
1\ ooPnPp!~TC ~YA 1'1f~T[f'1~ ~ f'1T 1Hr"T c;ryrr>Jr<:;<:: II''" ~1'1'1<"; TT T'1 Tl-4> T ~~ r c; v c:, T r '-' "' 1\ T " T 'It •)! 1\ r:r' ''~I •
40 C~Ll TRTPROIT,AK,~I N.l= 1 nn 4 1 K = 1 , f. "J=K+M(J-ll 1')'1 4 1 K 1 = l, 1'"'=Kl+fi*(J-11
41 ~(N,•·n=AK(K,I<l I+A!f\l,M) r;l") rn 52
r JF F"JO-l flF l't'!F~RFR(ll 1<; 1\T NnnFtJI rt1"lPIJTI= K'f1,11=Hti<H'T 1 1\~0
r /\!')~<; TT Tn THF APPOf'IPRT~TF 6X6 LflCI\Tlf'JN flN THF ~VSTF"' "'1\T~JX
r nTI\f,nN•L. 'i() CUI TPJPQf'lfY,AK,f>l
C~Ll TRtPRO(T,AK,~I
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109
,. TYF rnttOWJNr; 1~ STATrMFf\lTS tnraTE THE [~n nF ~F"'PFPf!l N!')T T~lr:Tf)FNT nN Nfln( !.1) ANn rnMPIJTt= THF 1\PPPOPT ATF flFF nrar;rll'lfll fl'fft
<;Tfft=NFSS MI\TPTX.
r
o:;., T r: Otnn E( J, N J I ) "i 1 , " 1 , c;, "i1 JJ:NnnFtT,I\IJ)
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., , n, c:; '• ¥ = 1 ' "' "'-=K+<,* I J-11 1''1'1 '>4 Kl=l 1 A "'=Kl +f>*( JJ-11
"i4 ~~~rMl=-ST(K,Kll r.f') Tn At i.'1MP1fTF -TKH' T •
!, 4 f)f1 "- "i K = 1 , 1'
"l=KH-•(J-ll
nn "'" Kl=l 1 A M: tq +A* ( J J-1 )
<,<; ~IN,M)=-ST{Kl 1 K) "1 rrwr J"-IIJF <,() rll!\!Tlr-Jilf
!)'1 .,, J.,l,~l~l
T'=(l T"JK(J,?)l7'>,1'.7,"7? <, 7 f"l'l " " T l = t , ~
1\bfl+A*f.l-11 IY1 Af, T?=l,KK!C'
1rt ar '"•"'~="· "., ~ I ~J, r ;J t = n •
f)f1 71 y ' :: l ' ~ ~= T 1+1'.*1 J-11
71 ~ra.,,~l=l.1 7 ') r '1"--T l "ltJF
W" TTf:( 1 9 71? l ?D J:'lPMq(') 1 "H'I. 1 SVST~~ STlFHJFSS "'UPTlfl/11
!)'1 ?11 f=I,KI<Il ?11 W'~JTff1,?14lJ,(t.(T,Jl,,J=l,KKK) '14 r-nRM~Tf/' o~w·r~/fnf:17.7JI
r ** c::nLVF FnP ~''lfll\1 nTC::DL~f.F"!FNTS ** r: "u c:; nt v F 1 a, n •"' K 1< 1 I~" TTF( 3 1 ?lf:>l
110
'1fl CflPM~T(40X•~tnnAL nT<:;r>t_Af:f:MfMTS'/16'1.'~H10F'7Y•nrt.H-X•qx•I"'FI r~-v•av *'"rt TJ\-]Iq)(tTH!=T~-'itPIJITHFrtl-Y'I\)(IT!-~FTh-7'/1
'1" ?17 T=l,IIJ~I
J::A*T
'P \.J? TTf{ 1, ?I)~} T, nt J-<;J ,nr J-41 ,01 J-11 ,!"lf.J-'1 1 1"11 J-t l ,1"11 II r TY!= OC!Ioll\ ["4f)FR fll: Tl~l"" nonr;P'IM Cl"lMPIJTF<; THF "''f"''l1FP P.,H) P!=~f"Tf'1N<;
r 10 TOST TN SY<:;TF.,. f:n'"IPOf"'l~TFC:: ~"111 THPI TPh"JSFilPM<:; HIF\1 Fl !Oir~otaf"P r rl'l!)f~~rrc;.
W')JTrf1,?1f1) '11\ CflPM~Tf•11~7XI~r:~qr:p F~n Fnorrc; ~"'n ~~Mf"JTSI//14V'PP,Il'Y'""V'1'v
* '0, 7 I 1 .., X ' ~p 'I( ' 1 ? )( I ... 0 vI 1 ? l( 'M 07 ' I ' 14PII P. r "~"' I, rw' 0 )( ' 11 y 'PV I 1 "H' 'n 1 '
* 1 1 'I'M'(' 11)(' MV I 1 ,_ l( I M7 'I" "'"~ P 1 T = 1 , 1\1 P
J1-=~,ln<W( J 1 11 .J'=~'"'lF ( J'? J '""II C::TfF"'IIIfl r"t'l D? J:) 1 f. 1"1'1 a? K = l 1 f-,
hl<f.J,'<l=STfJ,K I "' 'IT(.l, 11 1=STIJ 1 Kl
r\1.1 ... HP<\N!Tl '"~ll TRJnonr~,c::T,~I
~" 1\11 TR[Piln!T,C::T,t-1 r'lll MT)(MlJI (11 9 1\K,f.l 1.~11 ToJDQn(T,~K,!,I
r~tt TorponrT,-\T,t-1 "l'l A4 1(:: 1 t I.
""fl<l=O. 11'1 P4 L=J,I, 1=1 .,. ( J 7-1 l
Pit 011(1( I=PP('<)-111<' (It·'- )~11( II T"' I J 1 I 7f,, 7n, 77
77 "1 "" 1<=1,1'-1''\l'l p"ll, l=l,~'-
J =I +"*I J l -1 l 0 '\ 00 11< 1-::Ppfi(J+<;Tfl<' ,l 1*11( Jl 7,_, 1-1'> lTFf 1, 11".1('1 (PDf .II 9 .1=1.!,)
1qt'l c:lPM~TflOX?Hl' ,lt•l(!,l=l'i.!,/l ~.,orvc; IT ~~n-1 nv ... ru~rorfl ')"l A<; I(:),~
"f"l=n. r"\'1 fl <; l = , ' ,.,
"" 0 (K):Pflt' I+T(I ,t<Jllri'O(II \./'> T T I= I 1 , 1 ~ 7 I I P { J I , J = J , f. 1 ' T
I ~ 7 c 'l P ~ 1\ T r 1 n x • 1 • ,., r 1 c;. 61 T" 1 '')r"l 07 1(:::1,,.,
PO(KI=O. '1'1 P 7 l = 1 , A
J=l+">* LP-ll q7 PO(II')=PP{l()+!l.l (K,f l*l"lfJ)
ft=(Jll7R, 7R, 7!')
7fl r"l'l P"> K=l,A f111 P A L = 1 , A l=l+A*( J 1-11
ill, !>O(I<):PP(t()-11.!<'(1 ,KI*f'(J) 7 A W" T T I= ( 3, 1 <l 0 l I P 0 f .J I , .J = 1 , A I
lQfl ~'-"'1P~I\T(l0X?H?• ,14Y61=1"i.6!) r: F'l!H'TS AT Ff'M)-? f1F ~F~~H'P(Tl t•~ "'fM~FQ r:nnROTN<\TFS.
1)"1 P<l 1(:] 'I,
O(V):-1) 0
1'\flPilf=J,f... oq D(K)=Il(Kl+TfL,K)*IlO(f '11 Wt>.JT!=('\,lqPl(pfJI,J=l,f...)
lqll F'lR~II.T!lOX'''~~'=l"i.6//l r.r1 Til oooo C::'\10
t:;IJPPI'ltfTTNE MTX~lllf o\,A,N) ") ' ~f N S l n ~~ A ! f... , f... I , A I A , f, 1 , n! A I '1n ? T = 1 ,11.1 '1 11 I J=l,"l '11.11="· '1'1 1 K-= 1 , N nt J }.,'1 f J I +1\ I J, K l *o, (V, f l n'l ? J = 1 , N
., "IJ,fl=nt.Jl Rf:TltP~l
r: \!f)
c:; !J n IUl 1 JT 1 "'r:~ T n T n p n f h , P , ~I}
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., "I .I I =fl LJ 1 +~ f J, I< I*" I K, Tl 1'):1 I J =I , 'IJ '11 J, T l=fl( Jl 1')'1 ., T = l , "l n'l t,. J = 1 , N '1(.11=0. '1'1 4 K=l,N
4 'lfJI~n!J)+P(J,'<l*l\fJ,K) ')'1 ":\ J-=l,"J
., q f T, .I I= n ( I I RI=TJJP!IJ n1n
111
S' I'" o l'ltJ T 1 1\l F ~-' .1\ T P A~~~ T l R f l n n 1 P~ 1 I )!HII • , c /1 I T I h ' 1-, I ' X I l n" ' ., l ' y I 1 n n ' ;> I ' 7 I l n f1 • ? I ' " L r>H f 1 n 0: '" f ~-, F, I •
*TY!101)),I7f10f') 11,QI"~f111fll,l-lfl,,l,),c:;TI.;,f)),T'((ll)f'l 'r ' ,.. ' f7 ~PI"a t: r': <;T,.-
-,.,MI'l'I.J 'IC,Y,7,~1_P~,~,T,Q 9 PJ-lT,TY,TV, ·' Qt ~T~ ponr,q~"' rr"'PIITI'<: IJSTW: TI-ll= rnnPf"lPII\T!=C:, 1\T r:l\,(t-1 '11" II Ml=~l'IF T
r THF THRFF OfMFN~Tn~AL POTaTION TQAN~FnR~~TfnN ~aTRI~ WHJfH [ TqAN~FOPM~ FLFMFNT rnnPOJNaTES Tn ~VSTE~ cnnRniMATFS.
nn 1 n M= 1, 6 !)fl l n N= 1, A
10 TfM,t.Jl=O.O <;)(':)(fl,2l-Yf J,ll <;V::V(I,;>)-Y( I,l)
<\7=11 T,?l-7( T, ll C:..=~ORT(S~**'+<;Y**'+c;7**'l rX=S)(IS rY=SYIS r,7:S7/C:.. r.:c;Q9T(fY**'+f7**?) T{),l)=('(
Tf 1 , ., l = (-cur v *r n s f "L PH ( r l 1-r 7t~ P.H -'1 1 PH 1 n l ) tr T f 1 , ~) = ( f X* ( V* S l N I I\ I PH f f ) ) -( 7 *(()<;( 1\ lPH ( f ) II If Tf?,ll=fY T(?,'>l=C*f:f"l"f!\LPH( T) l T(?,~)=-f*SP.Jfl\l PHfTll T(~ 9 ll=f7 T ( 1, ? ) = ( - r: Y* r 7 "'r '1 c:: { 1\ L P 4 ( T ) ) +( Y *q 1'.1 ( ht PH f I ) I ) /( Tf~,~l=ICY*f?*SINfi\LPH(l)l+fX*rOS(ALPHCTllllr 1"'1'1 1? M= 4, o f)'l , ? "-1=4,,.,
1' TIM,t.JI=T(~-~,N-~) f"l'"' ?n JJ=l,A f"l"' ?() KK=1,f.
"!l H{JJ,KII'l='l.~'"'
!i'l '" JJ=1,6 ?'5 HIJJ,JJI=l.n
H(",-,.1=-~
IH "·,) =S t>CTIIQ~J
F"J~
112
SlloPniJT pJc: STl p.PI( Tl Q r: 1\ l T ( 6, 6) , )( ( 1 n n, ? I , y ( 1 1)0,?) , 7 ( 1 ()I),') l , A I PH ( 1 ()()I, PI l !'If) I , ll4l ( 1 '1 ()I •
* T Y ( 1 n !i l , T 7 ( 1 10 ) , 1\ P r 1\ ( 1 '1 n l , H ( f. , 6 l , c:: T ( ~ , A I , T )( ( 1 "'1 I n T M ~=" "' c:: T n N r 1 r-, , r, l , •p.J( f 1 , M ~ 1 ,.., ) r. 'lM M Or>.J X , Y, l • fll ~PH, ~~, T , P , PH T , l 'If , T V, T 7 , ~ P f h t F '~ • S T • F
'1'1 tn "1=1,,..,
'"'l'l '" "·'=l·" 10 rpo~,N}:().I')
Cll,ll~CP(JllfiDF~{ Tllfl*IPHJfTI+.~*ST~(?.•PHTITlll ~ ! 1 • l l = F ( l , 1 1 + ( P ( l ) * * ':\ IF I T 7 ( I ) l * I PH l ( T 1-1 • r:; * <; T "J ( ? • *PH f I T I 1 1 F I 1 ' 1 l = F ( 1 , 1 l+ I P f T 1 * * ~ I 1= 1 T 7 f T l l * ( ? • * P4 I I I I * f U'l S I PH J f J 1 I 1 **? 1 r I ? ' '1 = ( Q ( T ) II\ P F .-. ( T l IF ) * ( PH l f f ) -. 'i * <; l "' ( ? • * Pl-4 T I J ) l l F I :"' • :"" 1 = F I ? , ? l + ( P ( T I * * '1 IF I T 7 { l 1 I * ( PY T ( T l- • r:; * q "' (? • *PI-IT I T I !! , 1 F ' ., ' ., 1 -= F ( ., ' :> ' + r o r ' ' • * ~ I c ' T 7 I J l I * I ? • * o ~~ I I I l • I c; I "' ( o H ' ' ' l 1 I F I ~, 1 l = ( P ( r l * * ~ I ~ I r x r ' ' 1 * I ~. • o 4 ' r T ,_., • *" ' N I ' • • Pl-l r ' T , , 1 F I ~ ' ~ 1-= F ( ~' ~ ) + I P I l I * * 1 I r; I J 'If I l l l * ( • 'c; * <; J "' I 4 • * o H T ( T l I I I c: I 1 • 1 l = r ( 1, 1 I+ ( P I T I** -:t I r It v ( J 1 l * I PH T I I I- • ?"i *q "'I 4 • *PH I ( J l l F I 4 t 4 1 = ( R ( T l I r; I T 'I( f T l l * ! fl4 J ( T 1 + • r:; * <; T "l ( ? • *PH f I l 1 I l r. 1 4 • '• 1 = F 1 4, 4 1 + r r> 1 t 1 1 r 1 r v 1 r , 1 * 1 °H T 1 r 1-. "* c: r N r '· •PH I 1 ' 1 1 1
F ( c; • c; 1 = I 0 ( T 1 I r. IT X ( T l I * I PH T ( l 1- • '5 * S f t.J I '• *PH l I T I 1 I
F I 5, 5 l = F ( 'i, 5 l + I P f l I l f IT Y f I )l * f PH I f I l +. r; t S HH ? • *PH J (I J J J F ( 1:>, A l = ( R ( I l IF IT 7 f I I l *?. *PH t f I l r- ( 1, ? l = ( R ( I ltr:* V F I ll ( I l l * ( ? • * ( SIN I PH T f J) 1 J ** 21 C( 1 , ? 1 = F I 1, ? l+ f P I l l **~IF If 7 I T 1 1 * (-PH T I Tl *S J N (?. tPH I f Tl J J l=f?,l l=Ffl,?l
113
r- I 1 , 6 l = I R ( I 1 * * ? I F I T7 ( T J 1 *I ? • * <;t N I PH Y ( I ) l-7. *PH T I J 1 tC nc; f PH I fT ) I I Fl6tl t=F( l,f-1 F I ? , 6 l = ( P ( l ) * * ? IF I f l ( l l ) *I ? • tP H t ( I 1 * c; I N I PH I f T J l l F(6,J)=F(2,6l F ( ~, 4 l = ( R ( T I ** 21 r:. I TX ( I I l *I Pl-q I l) *C n c:;( PH I fT J l -? • * c; TN ( PH I f T I J l FI, 9 4J=F(3,4)+(PITl**?/~/IXITilt(.5tC~SIPHTfJJJ*SINf2*~HiftllJ F ( ~, 4 J = F ( 1, 4 J + ( P ( T J * *? /F 1J V f l J I* ( PHT I T 1 *COS (PH J I I ) Jl I= I 1, 4)-= F ( 1, 4 J + ( P ( T 1 **?/I' /J V( I 1 1 * (-. t; *COS (PH I ( I J l tq !If P. *PH J f T I J I C(4,~l=F(1,4l
~=" ( ":!., "i l =I -Q. ( I l ** '?fr, I T X f J I l * f PH T ( T J * q N f PH II T 1 1-. 5 •<: PH PH T fT l J **<:TNP.*PHT ( T1ll
!= ( 1, c; l = F I 1, 'i l +r- R ( l I**.., IF IT VI I ) l * ( PH T f rt *S T ~I PH I I J l l I F ( "', c; l = F ( 1, 5 I+ ( -P ( T l ** U FIT V ( l l) * ( • c; tS J N ( PI·H I T ) )tc; P-4 P. *PH J IT l l I Ffr;t31=FI1,51 11"1 11 M: 1 t A
l")fl "'1 1\1= 1 '6 11 ~Tf~,Nt=F{M,N)
~~ll ~TNV(~T,A,r~T,NN,MM}
11= rnFTJI, 2,1 .., l.lq ITF (1, J;l)())
<;<;() F'lRMAT(tSJI\If,!Jl~P MI\TPJY•I qu ~x rr r.>~TtlPN
F"Jn SlJ!tPflUTINE ~nt VF(tt,q,Nl
f ~~TRTX TNVfR~Jnl\l qy J~DD~N FlfMIN,Tl"~ r: PFPFflRMfN!"; f., ("!l~f'ltFTf" PJVnT
l''fMFN~TnN AflOfl,liV'l),qfl'1nl,fPfll10} 1")'1 1 0 T: 1 , "' "'I l=~f t )!lnn. [)f1 1 (") J= 1 '"'
10 ~fTdl-=A(l,J)!]('ln. DN:~l
snc;:n. 1)11 ? n J = 1 ' N l'l(J):f)
0'1 .? 0 T = 1 , N 'O S'lS=SnS+AII,Jl**'
Sn=so•ncsnc;, /PN Tl"ll = 1 • F-] I)
l")f1 1 l 0 l = 1 ' "' '<1.. AP (;:I) •
l)n 1 I T: 1 , ~~ l')f1 1 1 J = 1 , N T~" I Y-TP ( l) )f.h, 11 ,f.,
fo,A fl=fJ-TRIJ)) 1?, 11,1"' l 2 l r- ( ArtS f X l 4P r, ) -A fl S ( 1\ f T , J ) ) l 11, 1 l, 11 l"'\ Yli\Pf::A(J,J)
T l "' J
J1 ooJ lJ (ClNTJ"'IIF
l"Llll=Jl l'="f,I,'I~IXI ~R(.}~~n*T'lt 1\1\ 0?? 0 7?
? 7 I= T 1 J-=Jl Tr(l-Jif>!,<'.?,f.l
"'' nn f<r"1 K=l.~l ~fJ,KJ=AfJ,K)+~I 1,'<1 ~~ [ 0 K);~f{ 0 K}~~(J,'<J
1,1) ~I.J.K)=A{J,'<H~IT,Kl
O.O.=!>.(J)
'~IJl=O,fJI
<> r T I e--r>P.
f.? 'l'1 f..<l K'-'"lo'.! 1'("-))?1)1,,?1)''•?1).;
?n,_ C\(I()=R(I()-P(J)*~fi(,J)/)Il~<>r";
">'1<i ry-. '-<l l=lo'>i f"l!f- )lf,7 0 f,P,f-?
1,7 {IO(J-J)A<i 0 f,R 0 f'i A'i ~l!1 0 f\=h(lf,T)~H I,T\c&hfK.,JJIXL~f>r. hq ''l'!TT"JII>
!Yl l 1 '1 I'= 1, '~ T"'fK-Jl7n,110,7r
7n ~1¥, """~"• 1 10 rrvn r ~11!1' l\1 '"lTM=J,'1
"'l 0'1 J= 1 ,~1 l'f"'"'<::)I\{J,jl)-<::'l*'""!)nJ,n'1 0 <lf)
Ql) '"lT~=nT~*\IJ,Jl
,...., '" <l?
rq I)TM='),
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114
115
Chapter 6
CONCLUSIONS AND FUTURE INVESTIGATIONS
6.1 Conclusions
A method of analysis for space frameworks composed
of segmental circular members has been developed in this
paper along with the computer program to perform all the
computations required by the analysis. This analysis
is fundamental to a number of noteworthy investigations
that could be made in the future, which are discussed
in (6. 2).
An analysis of this nature would be highly
impractical without the aid of the computer, however,
with the computer, the analysis of structures such as
those described in this paper can be analyzed in a
relatively short time. Since analysis described in this
paper reflects more closely the true behavior of the
structure, a more efficient and economical design of
the "curvilinear space grid" should be expected.
An experimental model was built and tested in order
that the theoretical solution could be correlated to the
observed and measured physical results.
B encountered in measuring ecause of the difficulty
the true vertical displacement of the node points, the
correlation between the computed vertical nodal displace-
ment and the measured vertical nodal displacement is
116
rather poor, however, the measured stress at a given
point in a member correlates very closely with the stress
in that member calculated from the member-end loads
computed by the theoretical analysis. It can be concluded
from these results that the theoretical analysis does
compute the member end loads properly and if the nodal
displacements had been measured more accurately the
correlation between them would have been much closer.
The analysis of curved structures other than circular,
could be made by approximating the curves with circular
segments, which would increase the number of node points.
This procedure, however, is suggested only in the case
where the derivation of the actual flexibility is not
feasible, since the addition of nodal points increases the
number of equilibrium equations which must be solved and
reduces the efficiency of analysis.
6.2 Topics for Further Investigation
a.) Temperature Affect- The affect of temperature on
the structure can be included in the analysis with
little difficulty. If the change in temperature
is equal to 11 t", the change in length of 8 member
from its unstressed length and consequently the
change in internal stress can be computed if the . for the material
coefficient of thermal expans1on
is known. The internal forces resulting from this
117
change in length can then be included in the member
load-displacement equations in general as,
(6-1)
where ptl and pt2 are the member end-loads resulting
from the thermal expansion when the end-displacements
are prevented.
The load-displacement equations for the structure
can be assembled as was explained in Chapter 2 if
equations (6-l) are first transformed into system
coordinates as follows,
' - K' 6 I K' 6' p' pl - 11 l + 12 2 + tl'
and
It is often desired to study the affects of
temperature independent of any externally applied
loads. This can be done in a straightforward manner
from the following equation;
0 = }. I 6 I + Pt. (6-2)
The results of equation (6-2) can then be super
imposed on the normal solution.
118
b.) Foundation Movements -The effects of foundation
movements on the curvilinear space grid can easily
be taken into account in the analysis by modifying
the appropriate displacements in the load-displace
ment equations.
c.) Optimum Design- It would also be interesting to see
an optimum design study made for the lattice dome.
Since the method of analysis described in this paper
neglects only the shear energy in the general energy
derivation, more efficient use is being made of the
structural framework and the engineer is able to
preduct its behavior more accurately. It is there
fore possible to study the effect of the grid
spacing on the size of the material and arrive at a
more efficient, economical design for the lattice
d.)
dome.
the curvilinear space grid can now be studied.
generation of the system stiffness matrix is as
fundamental a step to the formulation of the equations
of motion for the dynamic system as it was to the
formulation of the equilibrium equations for the
static system. The loading vector is a function of
time and the mass of the structure can be treated
and rotational inertias as a series of lumped masses
concentrated at the node points. The problem can
finally be reduced to the general eligenvalue
problem.
119
e.) Computational Problems -A discussion of the com
putational problems involved in analysis of this
type are discussed in reference (3). There is,
however, additional work needed in developing methods
by which very large structural systems can be
handled effectively such as by the substructure
method, where the larger systems are divided into
smaller units which can be handled more efficiently. 3
Bibliography
1.) Eisem~nn, K., Woo, L., and Namyet, s., (1962) Space Frame Analysis By Matrices and
Computers", Journal of the Structural Division, ASCE, Vol. 88, No. ST6. December, 1962, pp. 245-268
2.) Gere, J. M., and Weaver, W. Jr. (1965) Analysis of Framed Structures. D. Van Nostrand, Inc., Princeton, N.J.
3.) Livesly, R. K. (1964) Matrix Methods of Structural Analysis. Pergamon Press Ltd. London. pp. 1-111.
4.) Hutton, C. R. , (1961) "Curvilinear Grid Frames", Engineering Journal, July, 1964, Vol. 1, No. 3, AISC.
5.) Celis, A. J., Unpublished material being prepared for text book.
6.) Fenves, s. J. (1967) Computer Methods in Civil Engineering. Prentice-Hall, N. J. PP• 76-83.
120
7 •) Baron, F. , (1961) "Matrix Analysis of Structures Curved in Space". Journal of the Structural Division, ASCE, Vol. 8?, No. ST3, March, 1961,
8.)
9.)
10.)
Fox,
Li,
pp. 17-38.
L. , (1965) An Introduction to Numerical Linear Algebra, Oxford University Press, N. Y. pp. 60-93.
Shu-t' ien, ( 1962) "Metallic Dome-Structure Systems", Journal of the Structural Division, ASCE, Vol. 88, No. ST6. December, 1962, pp. 201-226.
Michalos J (1966) "The Structural Analysis of Spa~e N~tworks"' Int. Conf • on Space Structures' Univ. of Surrey. September, 1966.
121
VITA
The author, David L. Fenton, was born on March 20,
1941 in Murphysboro, Illinois. He received his primary
and secondary education in the Dupo, Illinois Public School
System graduating from high school in May, 1959.
He entered the University of Missouri at Rolla in
September, 1959 and graduated in May, 1963 with a B.S.
Degree in Civil Engineering. While employed as a stress
analyst for McDonnell Aircraft Corporation he began
graduate study at St. Louis University night school in
September, 1963 and continued until February, 1964.
In September 1964 he entered the graduate school of
the University of Missouri at Rolla. He received an M.S.
Degree in Civil Engineering in January, 1966 and continued
work towards the Doctor of Philosophy Degree in Civil
Engineering. He has held a National Science Foundation
Engineering Traineeship during the period from September,
1964 to August, 1967.
He is married to the former Janice A. Biel1er and
they have two children, David Scott and Susan Lynn.
APPENDIX A
COMPUTER SOLUTIONS FOR THE MATHEMATICAL MODEL
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ME~RFR COOROINANTS ~11.11 Y(!,21
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MEMBER COORDINANTS Yll.ll Y(l,21
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MEMBER PROPERTIES I Z AREA RADIUS PHI 4L PH4
0.000326 0.062500 60.000000 gJ~ 8:S799RO 0.000327i o.o6l500 60.00001!0 0.000326 0.067500 60.000000 0.110550 -O.~Jqq~l'}
0.000126 0.062500 60.000000 0.130550 -o. b7q9f!ll
0.000326 0.062500 60.000000 g:mm -0.155170
0.000326 0.062500 60.000000 -0. 1'51 70
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0.000326 0.062~00 60.000000 r.J!',OA77 o. 0
0.000326 0.062500 60.000000 o.t'JHV> -g:~~~~~g 0.000176 0.062500 60.000000 o.no~'o
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MEMRFP END FORCE~ ANO MOMENTS
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APPENDIX B
COMPUTER SOLUTIONS FOR THE EXPERIMENTAL MODEL
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