NONLINEAR ELASTIC STABILITY ANALYSIS OF GABLED FRAME
by
Wang, YangCheng B.S., Chinese Military Academy, 1980
A thesis submitted to the Faculty of the Graduate School of the University of Colorado in partial fulfillment of the requirements for the degree of Master of Science Department of Civil Engineering
1987
This thesis for the Master of Science degree byWang, YangCheng has been approved for the Department of Civil Engineering
by
Andreas S. Vlahinos
Date
n  vt2.1
Ill
Wang, YangCheng(M.S., Civil Engineering)
Nonlinear Elastic Stability Analysis of Gabled Frame Thesis directed by Assistant Professor Andreas S. Vlahinos
Consider a gabled frame of singlespan, pinned base composed of four straight, slender, and piecewise prismatic bar which are rigidly connected. The four bars have the same length. The external loads applied to any element of this gabled frame include segments of a distributed transverse load, and concenstrated transverse forces.
The objective of this study is to compare the loadrotation behavior between the linear system and the nonlinear system. The linear solutions are found by using the PC program FEMUCD. The nonlinear solutions are found by using differential equations. With this approach, the equalibrium conditions, boundary conditions, and some assumptions are considered. The computational approach is used to find the nonlinear solutions in this study.
The results are much different between the linear and nonlinear systems. Under concentrated load conditions, the behaviors of the nonlinear and linear systems are similar when evaluating a column subjected to an
eccentric load.
IV
ACKNOWLEDGEMENTS
I wish to express my gratitude and appreciation to my adviser, Professor Andreas S. Vlahinos, for his guidance, direction, and assistance throughout the course of this work. I also wish to thank Professors John R. Mays, and NienYin Chang for their willingness to serve on my thesis committee and in particular, their constructive criticisms.
Particular gratitude is expressed to Major General PinMin Chen, the Dean of the School of Natural Science at Chinese Military Academy, Taiwan, Republic of China, who gave me this opportunity and encouraged me to study abroad. The gratitude also is to extend to the ChungShan Institute of Science and Technology, Taiwan, Republic of China, who provided my financial support.
CONTENTS
CHAPTER
I. INTRODUCTION...............................................1
1.1 The Gabled Frame........................................1
1.2 Objective of Study......................................4
1.3 Basic Assumptions.......................................6
1.4 The Behavior of an Eccentrically Loaded Column..........8
II. MATHEMATICAL FORMULATION..................................16
2.1 Description of the Notation of the Gabled Frame........16
2.2 Description of the Mathematical Formulations of Equations_17
2.3 The Formulation of Equation U(X) and W(X)..............21
2.4 Boundary Conditions....................................23
2.5 Equilibrium Conditions.................................26
2.6 Formulation of Equations for Joint Conditions..........28
2.6.1 Joint 2............................................28
2.6.2 Joint 3............................................29
2.6.3 Joint 4............................................36
III. COMPUTER IMPLEMENT.......................................40
3.1 Basic Matrix Form of a Bar.............................40
3.2 Oganization of Equations...............................45
3.2.1 Joint 1 and Joint 5................................45
3.2.2 Joints 2, 3, and 4.................................45
3.2.3 Forces.............................................46
3.2.4 Bending Moments and Rotations......................46
3.2.5 Displacements......................................47
3.3 The Linear Solutions of Aâ€™s............................49
3.3.1 The First Part of Linear Matrix At.................49
VI
3.3.2 The Second Part of Linear Matrix A2..............50
3.3.3 The Third Part of Linear Matrix A3...............53
3.3.4 The Nonlinear system.............................54
3.4 The Solution of the Nonlinear System.................55
3.4.1 Initial Guess of K...............................55
3.4.2 Finite Element Method for PC Program FEMUCD .....56
3.4.3 Computer Program.................................57
IV. RESULTS................................................59
4.1 Concentrated Load At Joint 3.........................59
4.2 Concentrated Loads At Joint 2 , 3 and 4..........64
4.3 Concentrated Load At Joint 3 and Horizontal Loads At
Joint 2 and 4........................................68
4.4 Concentrated Loads at Joint 2, 3, and 4. and Horizontal Loads
at Joints 2 and 4....................................72
4.5 The Different Directions Between FEMUCD and the Nonlinear
System...............................................76
V. CONCLUSION.............................................78
5.1 Summary..............................................78
5.2 Conclusion...........................................79
BIBLIOGRAPHY.................................................81
APPENDIX
A. FEMUCD input data file.................................83
B. FEMUCD output data file................................85
C. Computer Program for the Nonlinear System..........93
FIGURES
Figure
1.1 The Rigidly Connected Gabled Frame................
1.2. (a) Uniform Loads on the Gabled Frame................
1.2. (b) Concentrated Loads on the Gabled Frame...........
1.3. (a) The Load Conditions of the Eccentrically
Loaded Column.....................................
1.3. (b) The Directions of Displacements of the Eccentrically
Loaded Column.....................................
1.4 The LoadRotation Behavior of Linear and Nonlinear Systems...................................................
2.1 Directions of Each Element of the Gabled Frame....
2.2. (a) Member Loads and Sign Convention.................
2.2. (b) Free Body of Deformed Bar Element................
2.3 The Uniform Loads at Joint Condition..............
2.4 Joint 1 and Joint 5 are Pinned and No Translation ..
2.5. (a) The Force Equilibrium Condition at Joint 2.......
2.5. (b) The Moment Equilibriun Condition at Joint 2......
2.5. (c) The Rotation Equilibrium Condition at Joint 2....
2.5. (d) The Displacement Equilibrium Condition at Joint 2 ..
2.6. (a) The Force Equilibrium Condition at Joint 3.......
2.6. (b) The Moment Equilibrium Condition at Joint 3......
2.6. (c) The Rotation Equilibrium Condition at Joint 3....
2.6. (d) The Displacement Equilibrium Condition at Joint 3 ..
2.7. (a) The Force Equilibrium Condition at Joint 4.......
2.7. (b) The Moment Equilibrium Condition at Joint 4......
2.7. (c) The Rotation Equilibrium Condition at Joint 4....
. 2
.3
.3
.9
.9
13
15
16
16
27
30
31
31
32
32
34
34
35
35
38
38
39
Vlll
2.7.(d) The Displacement Equilibrium Condition at Joint 4.....39
4.1 Concentrated Load At Joint 3..........................60
4.2 LoadRotation Behaviors with Different 9..............60
4.3. (a) Comparison of Linear and Nonlinear Solutions with
9  1Â°...............................................61
4.3. (b) Comparison of Linear and Nonlinear Solutions with
9 = 10Â°..............................................61
4.3. (c) Comparison of Linear and Nonlinear Solutions with
9 = 20Â°..............................................62
4.3. (d) Comparison of Linear and Nonlinear Solutions with
9 = 30Â°..............................................62
4.3. (e) Comparison of Linear and Nonlinear Solutions with
9 = 40Â°..............................................63
4.4 Concentrated Loads At Joints 2, 3, and 4........64
4.5 LoadRotation Behaviors with different 9........65
4.6. (a) Comparison of Linear and Nonlinear Solutions with
9 = 1Â°...............................................65
4.6. (b) Comparison of Linear and Nonlinear Solutions with
9 = 10Â°..............................................66
4.6. (c) Comparison of Linear and Nonlinear Solutions with
9 = 20Â°..............................................66
4.6. (d) Comparison of Linear and Nonlinear Solutions with
9 = 30"..............................................67
4.6. (e) Comparison of Linear and Nonlinear Solutions with
9 = 40"..............................................67
4.7 Concentrated Load At Joint 2 and Horizontal Loads
At Joints 2 and 4....................................68
4.8 LoadRotation Behaviors with different 9........69
4.9.(a) Comparison of Linear and Nonlinear Solutions with
9  1Â°
69
IX
4.9. (b) Comparison of Linear and Nonlinear Solutions with
6 = 10Â°...............................................70
4.9. (c) Comparison of Linear and Nonlinear Solutions with
9 = 20Â°...............................................70
4.9. (d) Comparison of Linear and Nonlinear Solutions with
9 = 30Â°...............................................71
4.9. (e) Comparison of Linear and Nonlinear Solutions with
9  40Â°...............................................71
4.10 Concentrated Loads At Joints 2. 3. 4 and Horizontal
Loads At Joints 2 and 4...............................72
4.11 LoadRotation Behaviors with different 9.............73
4.12. (a) Comparison of Linear and Nonlinear Solutions with
9  1"...............................................73
4.12. (b) Comparison of Linear and Nonlinear Solutions with
9 = 10Â°..............................................74
4.12. (c) Comparison of Linear and Nonlinear Solutions with
9 = 20Â°..............................................74
4.12. (d) Comparison of Linear and Noidinear Solutions with
9 = 30"..............................................75
4.12. (e) Comparison of Linear and Nonlinear Solutions with
9 = 40Â°..............................................75
4.13 The Directions of the Element in FEMUCD..............77
CHAPTER I
INTRODUCTION
1.1 The Gabled Frame
The gabled frame of singlespan, pinned base is widely used in structures covering large areas with no obstructions, such as industrial buildings, auditoriums, warehouses, etc.. Its analysis is of considerable interest to practicing engineers. Numerous investigations of rigid frames of various types have been reported in open literature.
Consider a gabled frame composed of four straight, slender, and piecewise prismatic bars which are rigidly connected, [see Fig. l.l] The four bars have the same length, L. The angle, 0, between element 1 and element 4 [see Fig. 1.2.(a) and 1.2.(b)] can be arbitrary values between 0 and , excluding 0 and Joint 1 and joint 5 are pinned supports, which are concentrated to prevent translations. The properties of the materials used are E<, A*, h, which are the Young's Modulus, CrossSection Area, and moment of inertia of i(h bar, respectively.
The external loads applied to any ith member of this gabled frame include segments of a distributed transverse load q;, concentrated transverse force QV, and QH, in vertical and horizontal directions, respectively. In this gabled frame, the application of a uniform load will be limited to the case with a single uniform load over each element and just two concentrated loads per bar with QV,(0) and QV;(1) or QH,(0) and QH.(l) applied at distance eÂ°, and eL, respectively from the ends, [see Fig.2.2] The ratios,
2
ei/L,, of it/l element are sufficiently small so that the loading can be replaced with the static equilibrium loading of force QV* and QH* at the frame joint.
Fig. 1.1 The Rigidly Connected Gabled Frame
3
Fig. 1.2.(a) Uniform Loads on the Gabled Frame
QV3
Fig. 1.2.(b) The Concentrated Loads on the Gabled Frame
4
1.2 Objective of Study
A number of approaches have been successfully used in determining the behaviors of elastic structures. Some of the approaches are to find linear solutions and some of them use the function of X 1 to find the nonlinear solutions. The objective of this study is to determine the behavior of the nonlinear loadrotation. The equilibrium equations of axial and lateral displacements are derived in reference 2 2 The method of analysis consists of solving the boundary value problem associated with these typical equations.
These equations of this system are derived from differential equations. Structure analysis based on the differential equations has three easily recognized advantages.3
1. Distributed loads and elasticity can be dealt with directly so that the quality of a solution is not affected by the choice of elements and equivalentloads used in a Finite Element Method.
1 George D. Manolis, Dimitrios E. Beskos, and Bruce J. Brand.,
â€œElastoplastic Analysis and Design of Gabled Frameâ€,
â– Journal of Computers & Structures Vol. 22, No.4, pp. 693697,1986
2 Andreas S. Vlahinos, C.V. Smith, JR. and G. .J. Simitses, â€œA Nonlinear Solution Scheme Multistoryâ€, Multibay Plane Frames,
Journal of Computers & Structures Vol. 22, No.6. pp.10351045, 1986
3 Riyad K. Qashu. A.M. ASCE and Donald A. Dadeppo. â€œLarge Defiection and Stability of Rigid Framesâ€, Journal of Engineering Mechanics, Vol.109, No.3, June, 1983
5
2. Nodes accuracy naturally at points of discontinuity in the differential equations and the member would be employed in a finite element analysis that can be expected to yield solutions of comparable accuracy.
3. Highly accurate results obtained by integration of differential equations provide a standard of comparison for the evaluation of approximate methods of analysis. The difficulties are finding the relative complexities of analysis and also the many ways to organize the computational procedures.
One of the approaches of Finite Element Method applied to find the loadrotation behavior of this gabled frame is the PC program FEMUCD4. The FEMUCD will be used in determining the linear behavior of loadrotation. The differences between linear and nonlinear solutions will be compared in chapter four. Usually, the results of linear and nonlinear are much different, especially when the external load is close to the critical load.
4 This is a PC program for Finite Element Method derived by Professor John R. Mays in the Department of Civil Engineering of University of Colorado at Denver.
c
1.3 Basic Assumptions
Since only a relatively small number of problems of mechanics can be solved by means of the exact field equations of the theory of elasticity, the exact solution is presented within the limitation of the number of simplifying assumptions in dealing with these problems. According to these assumptions, all of the structural elements fall in one of the following six categories.
1. Frame members are initially straight, piecewise prismatic with a cross section principal axis, and are rigidly connected.
2. The plane frame is loaded by static concentrated or uniformly distributed loads only, with the loading plane in the plane of the frame.
3. All displacements are in the plane of the frame.
4. Material behavior is linearly elastic, and the moduli of elasticity in tension and compression are equal. Material yielding and effects of residual stress are neglected.
5. Cross sections perpendicular to the longitudinal fibers before deformation remain perpendicular to the deformed logitudinal fibers, and lines in the cross section are inextensional (the usual EulerBernoulli assumption which is listed below).
A). The material is homogenous and isotropic.
B). Plane sections remain plane after bending.
7
C) . The streestrain curve is identical in tension and compression.
D) . No local type of instability will occur.
E) . The effect of transverse shear is negligible.
F) . No appreciable initial curvature exists.
G) . The loads and the bending moments act in a plane
passing through a principal axis of inertia of the cross section.
H) . Hookeâ€™s law holds.
I) . The deflections are small as compared to the crosssectioal
dimensions.
6. The axial strain exz is less than one, and the term u,* and (w*)2 are of the order of the strain, where u, and w are axial and transverse displacements respectively for the points on the bar centrodial axis, and (,x) denotes differentiation with respect to the axial coordinate x. [see Fig.2.2.(a)]. The geometric nonlinearity is limited to moderate rotations with (92 Â« 1. [see Fig. 2.2. (b)]
8
1.4 The Behavior of an Eccentrically Loaded Column
In section 1.2, the different results between linear and nonlinear systems were mentioned. Now, the sample case of an eccentrically loaded column will be discussed. Usually, only ideal columns are discussed. The ideal column is defined as the following. The ideal columns are initially straight and geometrically perfect and are subjected to the action of a compressive force without eccentricity. In practice, no column is geometrically perfect and the applied load does not necessarily pass through the centroid of the column cross section. It is therefore necessary to study the behavior of geometrically imperfect columns and of columns for which the load is eccentrically applied.
Consider the case of a simply supported and eccentrically loaded column with the same eccentricity, e, at both ends shown in Fig. 1.3.(a). The directions of axial deformation, U, and the lateral deformation, W, are shown in Fig. 1.3.(b). The properties of this column are E, A, and I which are Youngâ€™s Modulus, CrossSection Area, and Moment of Inertia, respectively. The equilibrium equation of this case is
E1WXX  PW  0 (141)
Since this column has no uniform load on it, the value in the right hand side of equation (141) is equal to zero. On the other hand, the value in the right side shouldnâ€™t be equal to zero if a uniform load is on this column. The boundary conditions are shown as the following.
W(0) = 0.0 (142)
U'(l) = 0.0
(143)
9
Tâ€”^
p
Fig. 1.3.(a) The Load Conditions of the Eccentrically Loaded Column
u
A
Fig. 1.3.(b) The Directions of Displacements of the Eccentrically Loaded Column
10
M(0) = Pe M{ 1) = Pe
(144)
(145)
The solution for this equation is
W = A\SIN(K X) + A2COS{KX) + A3 X + A4
(146)
The terms W ,Y and W Xx can be derived from equation (146).
Wx = A\KCOS{KX)  A2KSIN(KX) + A3 (147)
Wtxx =  Ai K2SIN(KX)  A2K2COS{KX) (148)
The relationship between M and W is as follow.
M = EIW xx
(149)
The relationship between the axial force and the constant K is the following.
PL2
K
2 _
El
(1  4  10)
Combine equations (146), (148), (149), and (1410). The equations (142), (143). (144), and (145) can be modified respectively as the following.
M'(0) = .42 + 44  0.0 (1  4  11)
VV'(l) = AiSIN(K) + A2COS(K) + .43 + ,44 (1  4  12)
A/(0) = A2PL = Pe (1413)
M(l) = A\ PLSIN(K) + A2PLCOS(K) =  Pe
(1  4  14)
11
In the above equations, Ai, A2, A3, A4, and K are the constants for each bar. In equations (1413) and (1414), load, P, can be canceled out by both sides of equations (1413) and (1414) shown as the following.
M(0) = A2L = e (1415)
M(l) = AiLSIN(K) + A2LCOS(K) = e (1  4  16)
Equations (1411), (1412), (1415), and (1416) can be modified as the matrix form
.4 A' = B (1417)
shown as the following,
/ 0 SIN(K) 1 COS(K) 0 1 1\ 1 A> 1 (2)
0 L 0 0 a3 â€œ  e
VLSIN(K) LCOS(K) 0 0/ \.44 / KeJ
in which A is the coefficient matrix, X is the unknown vector, and B is the constant vector.
The solution of equation (1417) is
A' ~ A 'B. (1  4  19)
After vector X is found, substituting these constants into equation (146) the lateral displacements can be found at every point in this bar. The rotations of every point of this bar can be found by substituting the vector X into the equation (147). The results are plotted in Figure 1.4 which shows as e from 50. â€”> 100. The plot represents the behavior of the linear and
12
nonlinear systems. Note that the load P begins very small, and gradually increases. The corresponding critical load is shown in equation (1420). 5
x2EI
L2
(1  4  20)
The linear solution found by FEMUCD program is shown in Fig. 1.4 for comparison with the nonlinear results. In this case, the linear solution is much different from those of nonlinear. The more eccentricity there is, the more difference there is between linear and nonlinear solutions. On the other hand, if the eccentricity, e â€”> 0, this column is more like an ideal column.
5 George J. Simitses, â€An Introduction to the Elastic Stability of
Structuresâ€, PrenticeHall, Inc., Englewood G'hiff. New Jersey, 1976
LOADS
13
e increase from 50 to 100
W,.(0)
Fig. 1.4 The LoadRotation Behaviors of Linear and Nonlinear Systems
CHAPTER II
MATHEMATICAL FORMULATION
This chapter discusses how to derive the differential equations from boundary conditions and equilibrium conditions. The axial forces, transverse shear, bending moments, and rotations will be considered at each joint. Twenty four differential equations will be found by this approach. According to boundary conditions, six constants can easily be found by six of the twenty four differential equations. The others will be solved by using computer implement.
2.1 Description of the Notation of the Gabled Frame
The gabled frame is shown in Fig. 21 in which {n} is the nth joint, and [m] is the mth element, QV(z) is the zth. joint of the vertical load, and qfc is the uniform load on Kth element which is vertical to the axial line of this element . The length, L, of each element is the same as those of the other elements. The angle, 0, is between element 1 and element 2, and also is between element 3, and element 4. They are the same as each other. At this gabled frame joint 1 and joint 5 are pinned. Joints 2, 3, and 4 are rigidly connected. The directions of axial and lateral displacements are u and w. respectively. The undeformed freebody of each element is shown in Fig. 2.2.(a) and Fig. 2.2.(b). Fig. 2.3 is the freebody of the undeformed joint. P denotes the forces resultant in the undeformed axial force direction, V denotes the resultant in the undeformed transverse shear direction. M denotes the bending moment, i and j denote the if/l, or j(/l element, 0 denotes the end of a bar. and 1 denotes the other end of the bar.
15
\Jl U;
w,
u4
/K
Fig. 2.1 Directions of Each Element of the Gabled Frame
16
Xj 4
Qi
Q?
m
e?
IE
element i
Kâ€” l. â€”>
'if
W,
Fig. 2.2.(a) Member Loads and Sign Convention
Fig. 2.2.(b) Free Body of Deformed Bar Element
17
2.2 Description of the Mathematical Formulations of Equations
Consider each element of this gabled frame, the geometric relationship is given by equation (221)6
fxx = + zk (221)
where z units from the reference surface, e%x is the extensional strain on the reference plane (average strain) and k is the change in curve.7 The straindisplacement relationship is the following.
(222)
and k = â€”U)xz (2  2  3).
These equations are written on the deformation body with the restriction to the moderate rotations, [see Fig.2.2.(a)]
(NSwiX)x = 0 (224)
(S 4 Nw z) x = q (225)
M}X 4 5 = 0 (226)
in which N, S, and M are the internal axial force, transverse shear and bending moment, respectively, at any point in a bar. In equation (224), the term Sw , naturally appears from consideration of force equilibrium in the Xdirection for the free body in Fig.(2.2.(b)). However, it can be derived from virtual work consideration that this term is inconsistent with
6 The same as footnote 2
7 The same as footnote 5
18
the strain displacement relationship of equation (222). Therefore, in what follows, the term Sw * will be ignored in any equilibrium considerations. It may be helpful to introduce the following definition.
P  N  SW'X (227)
V = 5 + Nw]X (228)
In equation (227), after the term Sw * is ingorned, equation (227) can be modified as the following.
P = N (229)
where P is the force resultant in the undeformed axial direction and V denotes the resultant in the undeformed transverse direction. These definitions will be used to write equilibrium equations which have to be satisfied at each joint. The bar constitutive equations are well known
N = EAtxx (2  2  10)
M =  Elk
(2  2  11)
These formulations are completed with the requirements at the reaction point and the joints.
ForceDisplacement Relationship
The equations (2212) and (2213) can be derived from the combination of equations (222), (229). and (2210).
P  N = EAeÂ°xx
(2  2  12)
19
P = N = EA(ux + w%)
(2  2  13)
The equation (2214) can be derived from the combination of equations (2211), and (223).
M = EIw
Since
dM
dx
= S
(2  2  15)
Combine equation (2214), (2215) and derive S as the following.
S = ElW'XXX
(2  2  16)
From equation (228), the V can be derived as the following.
V = â€” Elwxxx I Pw x
(2217)
Equilibrium Condition
Since N is a constant, the equation (2218) can be derived by using equation (2213) shown in equation (2218).
EA{uiX + 'uijJ  (l (2218)
 â€™ . X
Equation (2219) can be derived by using equation (225).
EIw xxxx  Pui,â€ž = q (2219)
To use the nondimensional equations, the following symbols will be defined.
20
x= I A  L V(I)
c! II We II
w = f K2 = + EI (2220)
In forcedisplacement relationship, using equations (2220), (2213), (2214), and (2217), the above equations can be modified to the nondimensional equations as the following.
Â±K2  X2(Ux + \w*x)
(2  2  21)
M = EIW
(2  2  22)
S =
El w
jjWjxx
(2  2  23)
V
E7 L2
(tK2Wx
Wâ€™,.Y X X )
(2  2  24)
In equilibrium conditions, equations (2218) and (2219) can be modified to equations (2225) and (2226), respectively.
U
,x
7 2
,X
(2  2  25)
W'XXXX Â± A'2VT.y.Y = q
(2  2  26)
21
2.3 The Formulation of Equation U(X) and W(X)
These two following equations are used for compressive axial loads only. Equation U(X) represents the axial deformation, and equation W(X) represents lateral deformation for every point at a bar.
U(X)
As
K2 X A2
(231)
VF(A') = AxSINKX + A2COSK X + .43A' + .44 +
2 A2
(232)
These two following equations are used for tensile axial forces only.
U( X) = A& +
K2 X 1 _A2_
\L
di
(233)
W(A) = AiSlNh(KX) + A2COSh{KX) + .43A + A4  ~ (234)
2KZ
In the above equations, K. A!, A2, A3, A4, A5 are the constants for each bar. If any frame has nlh bar , this frame should have 6n unknowns in which on are Aâ€™s and n are k!s. The compressive axial force will only be considered in this section.
After integration, the equation (231) can be developed as the following:
22
U(X)=
A5 
K\x
A3
\[A\K\X + s/^xxjco^xx))]
\[A\K\K sin^x^kx)^
_1 42 v _ iAl ok*
+ \A1AjKSIN2(KX) AiA3SIN(KX)
_ ^ ^ CQS(KX) + KX^S1N(KJT)1 ^
[.42.43(COS(AX)  1)]
4 A jq( SIN(KX)KXCÂ°SC
1 A,qX>
2 if3
(235)
Also, W *(X), and WiJ;i(X) can be derived from W(X) as the following.
IV.x(A') = .4, KCOS(KX)  A2KSIN(KX) + .43
qX
K2
(236)
Wxx(X) = .4!K2SIN(KX)  A2K2COS(KX) + jL
(237)
23
2.4 Boundary Conditions
At joint 1, and joint 5, the boundary conditions will be considered. Because joint 1 and joint 5 are pinned, they canâ€™t be moved in any direction and also can not subjected to any moments so that the bending moments, M^O), and M4(0), and the axial displacements, U^O) of element 1, U4(0) of element 4 respectively, and the lateral displacements, W)(0), W4(0) of element 1 and 4 respectively, are equal to zero. The equations C/i(0), M7i(0), Mx(0), and t/4(0), H4(0), M4(0), are shown as the following.
At joint 1:
C/i(0) = 0.0
(241)
W',(0) = 0.0
(242)
Al\ (0) = 0.0
(243)
At joint 5:
t/4(0) = 0.0
(24â€”4)
H'4(0) = 0.0
(245)
a/4(o) = o.o
(246)
24
According to boundary conditions, use equations U(X), W(X), M = EIW A .\ and substitute 0 into theses equations. The following equations can be developed.
Ui(0) = Ais = 0.0 (247)
Wt(0) = Ai2 + At4 = 0.0 (2  4  8)
EiIiWiiXX(0) = EtliiAaKf + = 0.0 (2  4  9)
After these equations are found, the uniform loads of element 1 and element 4 will be discussed. Elements 1 and 4 have no uniform load on them. Apparently, the q, and q2 are equal to zero and their nondimensional expressions q\ and q2 are equal to zero. too. In equation (2222), the properties of material are E, Young's Modulus, and I, the moment of inertia shouldn't be equal to zero for any materials so that the equation (249) can be modified as equation (249.A)
VE.v.v(O) = ~(An2K?) + ^ (2  4  9..4)
In equation (229.A), K, is not equal to zero if any axial force appears in ith element. For element 1, the equations (2410), (2412), and (2214) are derived from equations (247), (248),and (249.A). For element 4, the equations (2411), (2413), and (2415) are derived from equations (247), (248). and (249.A).
Ui( 0) = 4,5 = 0.0
(2  4  10)
1^4(0) â€” .445 â€” 0.0
(2411)
25
VFi(O) = A12 + 4i4 = 0.0 (2412)
M'4(0) = A42 + .444 = 0.0 (2  4  13)
W',,..(0) = AX2K\ = 0.0 (2414)
Wâ€™4iÂ«(0) = A42K\ =0.0 (2  4  15)
From equations (2410) and (2411), the values of A15 and A45 which are equal to zero are easily found. In equations (2414) and (2415), when any axial force appears, i.e. K, and K4 are not equal to zero, the values of A12 and A42 are equal to zero as equations (2416) and (2417) shown. Substituting equations (2416) and (2417) into equations (2412) and (2413), the values of A14 and A44 should be equal to zero as equations (2418) and (2419) show.
.412=0 (2416)
.4,4 = 0 (2417)
.414 =0 (2418)
â€¢444 =
0
(2  4  19)
26
2.5 Equilibrium Conditions
Now, the relationship between axial forces, transverse shear, bending moment, Xdirection displacement, and Ydirection displacement will be considered in this section. According to equilibrium conditions at each joint, all of Fw, Â£ Fy, Â£ Dw, Dy, M, and Â£ W * for each element are described as the following.
In Fig. 23, the dLi and dL2 are relatively very short so that the values of qidL2 and q2dL2 are also very small. Any force or displacement due to q, dL; are small enough to be ignored. Similarly, for the other joints, the effect due to uniform load can be ignored when the equilibrium conditions are considered. When Y FH or Y. Fy is considered, the forces, axial forces and transverse shear, and the angle, 0, are very important factors. At joint condition, the summation of forces is equal to zero in the Xdirection and Ydirection. Concerning bending moments, the angle, 9, is not considered because the moments whose directions are positive in every element are only in the two ends of each joint excluding joints 1 and 5 of this system. Similar
to bending moments, when the rotations are discussed, the angle, 9, is not considered, either. The directions of rotations depend on the transverse forces so that the directions of rotations are different in both ends of each joint of this frame. When the displacements are considered, the joint conditions should be different from those of forces, bending moments, and rotations because the displacements of each joint are not equal to zero; but the forces, bending moments, and rotations at each joint are equal to zero. Since the joint is rigid, at each joint the displacements at both of the ends in each direction should be the same as those of the other joint.
Fig. 2.3 The Uniform loads at Joint Condition
28
2.6 Formulation of Equations for Joint Conditions
The solutions of joint conditions have already been found for joints 1 and 5. Now joints 2, 3, and 4, will be discussed in this section.
2.6.1 Joint 2.
At this joint, the angle, 0, between the horizontal direction and the axial direction of the element 2 will be considered and it is a very important factor in this frame. Pi(l) is the same direction as the axial direction of element 1. Vj (1) is perpendicular to the axial direction of element 1. The bending moment, Mj(l), is the positive direction of element 1. Pi(l), Vi(l), M] (1), and P2(0), V2(0), M2(0), are shown in Fig. 2.5.(a). The P2(0), V2(0), are related to the angle, 0. P2(0), and V2(0) include two components, the
horizontal and the vertical directions, so that the forces in the horizontal direction are QH(2), V^O), P2(O)COS(0) , and V2(O)SIN(0). In the vertical direction the forces are Pi(l), QV(2), P2(O)SIN(0), and V2(O)COS(0). The summation of these forces in each direction shoud be equal to zero, then the following equations can be found.
YIjoint? Fh = 0
F,(l)  QH{2)  V2{O)SIN{0) + P2(O)COS{0) = 0.0 Iljointi Fv â€” U
 /',(!) QV(2) r r2(0 )COS(0) + P2(O)SIN(0) ~ 0.0
[See Fig. 2.5.(a)]
The summation of the bending moment in the two ends of joint 2 is equal to zero. The bending moments and rotations do not include the angle, 0.
(261)
(262)
29
Â£
JOINT 2
M = 0
A/2(0) + M] (1) = 0.0
(263)
[See Fig. 2.5.(b)]
Y2.JOINT2 =
^1,^(1) Wr2ix(0) = 0.0 (264)
[See Fig. 2.5.(c)[
At joint 2, the displacements of element 1 and element 2 are the same each other in both of the Xdirection and Ydirection.
YIeLEM ENT\ Dh = YLeLEMENT2^H
W,(l) =  W2(0)SIN(0) + U2(0)COS(0) (265)
YYeLEM ENT\ Dv = YYeLEM ENT2
17, (1) = ir2(O)C:OS(0) + U2{O)SIN(0) (266)
[See Fig. 2.5.(d)]
2.6.2 Joint 3.
At this joint when the force equilibrium is considered, the angle, 9, should also be considered. In the Xdirection, the forces include P2(l)COS(0), V2(1)SIN(0), QH(3), V3(1)SIN(0), P3(l)SIN(fl). In the Ydirection, the forces include P2(l)SIN(fl), V2(1)COS(0), QV(3). V3(l)COS(fl), P3(1)SIN(0) so that equations (267) and (268) will be written as the following.
77777
Joint 1
77777
Joint 4
Fig. 2.4 Joints 1 and Joint 5 are Pinned and No Translation
o 1
o 1
Fig. 2.5.(a) The Force Equilibrium Condition at Joint 2
Fig. 2.5.(b) The Moment Equilibriun Condition at Joint 2
32
Fig. 2.5.(c) The Rotation Equilibrium Condition at Joint 2
Fig. 2.5.(d) The Displacement Equilibrium Condition at Joint 2
33
1LJOINT3 Fh â€” 0
P2(\)COS{6) + V2(l)SIN{0)  QH{3) + V3(1)SIN(6) + 7>3(1)COS(0) = 0 (267)
Â£JOINTS ~ Â®
P2(1)SIN(6)  V2(l)COS(8)  QV(3) + Va(l )COS(0)  P3(l)SIN{6) =0.0 (2  6  8)
[See Fig. 2.6.(a)]
The bending moments and rotations do not include the angle, 9, but the directions of the moments and the rotations will be considered. The equations of bending moments and rotations can be written as the following.
Â£
JOINT2
M
 0
M2(l)+ m3(i) = 0.0
(269)
[See Fig. 2.6.(b)]
Y.JOJNT3 WU = 0
H'2,.y(1) HV.y(1) = 0
(2  6  10)
[See Fig. 2.6.(c)[
The displacements of elements 2 and 3 are the same as each other in Xdirection and Y direction at each joint. The equation (2611) is in the Xdirection. The equation (2612) is in the Ydirection.
HeLEMENT2 Dh = ^2eLEMENT3 Dh
W2(1)SIN{6) +U2(l)CC)S{9) = l/3(l)COS(<9)  \V3(\)S1N(9) (26 11)
Fig. 2.6.(a) The Force Equilibrium Condition at Joint 3
Fig. 2.6.(b) The Moment Equilibrium Condition at Joint 3
35
Fig. 2.6.(c) The Rotation Equilibrium Condition at Joint 3
Fig. 2.6.(d) The Displacement Equilibrium Condition at Joint 3
36
Y.ELEM ENT2  YIeLEMEIVT3
W2(1)COS(0) + U2(\)SIN(0) = U3{1)SIN{9)  W3(1)COS(0) (2  6  12)
[See Fig. 2.6.(d)]
2.6.3 Joint 4
At this joint, the forces, bending moments, rotations, and displacements are similar to those at joint 2. The forces in the Xdirection are V4(l), QH(4), V3(O)SIN(0), and P3(O)COS(0). The forces in the Ydirection are P4(l), QV(4), V3(O)COS(0), and P3SIN(fl). The equations of the forces are shown in equations (2613) and (2614).
JLjoint* Fh ~ 0
V4(l) + QH(A)  r3(0)SIN(0)  P3(O)COS(0) = 0.0 (2  6  13)
ILjOINT4 = Â®
P4(l)  Ql7(4)  V3(O)COS(0) 4 P3(0)SIN(6) =0.0 (2  6  14)
[See Fig. 2.7.(a)]
The equations of bending moments and rotations are shown in equations (2615) and (2616), respectively.
'E.JOINTA M ~ 0
A/4(l) 4 A/3(0) = 0.0 (2615)
[See Fig. 2.7.(b)]
'iLjOINTA Â®
^3,x(0) â€” Wa,x (1) â€” 0.0
(2  6  16)
[See Fig. 2.7.(c)]
The equations of displacements in the Xdirection and the Ydirection are shown in equations (2617) and (2618), respectively.
^ELEMENTS  Y.ELEMENTA
Wâ€™4(l) = W3(0)SIN(8)  U3(0)COS(8) (2  6  17)
[See Fig. 2.7.(c)]
\ELEMENT3 ~ H.ELEM ENTA
UA(l) = U3(0)SIN(8)  W3(0)COS(8)
(2  6  18)
[See Fig. 2.7.(d)[
After these equations are solved, the solutions of Kâ€™s and Aâ€™s can be solved by the algebra of these equations.
38
Fig. 2.7.(a) The Force Equilibrium Condition at Joint 4
M,(0)
Fig. 2.7.(b) The Moment Equilibrium Condition at Joint 4
39
W,,.(0)
Fig. 2.7.(c) The Rotation Equilibrium Condition at Joint 4
Fig. 2.7.(b) The Displacement Equilibrium Condition at Joint 4
CHAPTER III
COMPUTER IMPLEMENT
Use the computer implement to find the eight constants of this gabled frame by solving the eighteen equations. All of them include linear and nonlinear systems. To solve them, all of the differential equations will be formed into a matrix for the linear system. The others will be found by using the IMSL subroutines. The computer implement will be discussed in this chapter.
3.1 Basic Matrix Form of a Bar
The solution of this gabled frame is a nonlinear solution. The ieh bar of this frame has six unknowns, K,, An, At2, Ai3, Al4, and Ai5, which can be defined if K< is known. K, depends on axial force, P,, because
K = T
PjL\
Eili
(311)
if P is compressive axial forces,
KJ = 
PM
EiU
(312)
if P is tensile axial force,
K? = +
PjL}
EiU
(313)
in equation (311), (312), and (313), K* is always positive. To obtain the relationship between Kâ€™s and Aâ€™s, the values of K are assumed to be known. According to Equations (232),(235),(236),(237), (311), and
41
subtituting 0, and 1 into these equations, respectively, the following equations can be derived from this procedure.
Ui( 0) â€” Aj5 (314)
Wi(0) = Mi + Aj4 (315)
Wi(l) = AnSIN(Ki) + Ai2COS(Ki) + Ai3 + AÂ« + Â® (316)
*Vi,.(0) = AÂ«Ki + A3 (317)
Wi,.(l) = AnKiCOS(Ki)  AiiKiSINiKi) + Ai3 + Â£ K (318)
$ ^ r 1 II (319)
, K?EiIi qiEili â€˜<(1) = A* L? L\ (3  1  10)
*<â€¢>=03? (3111)
, K2SIN(Ki)EiIi , K?COS(Ki)EiIi AÂ»Â«(1J A i A 2 (3  1  13)
t Lt
in which the axial force of i(h bar, P,, is defined as axial compresssive force. After equation (314) to equation (3113) are found, these equations can be formed into a matrix,
B  .4A' (3  1  14)
, to be used in computer operation. In equation (3114), B is the constant vector. A is the coefficient matrix, X is the unknown vector. Combine equations (314), (315), (316), (317), (318), (319), (3110), (3111), (3112), (3113). These equations can be expressed in the matrix form as the following.
42
/ ^<(0) \ Wi(0) Wi( 1) Wi,(0)
^i(O)
K(l) Afi(0) Mi( 1)
V Pi
( 0
0
SIN(K)
K
KCOS(K)
0
0
0
K3SIN(K)EI
L
V 0
0
1
COS(K)
0
K SIN(K) 0 0
k3ei
L
K3COS(K)El
L
0
0 0 1
0 1 0
1 1 0
1 0 0
1 0 0
k3ei V3 0 0
K3EI L3 0 0
0 0 0
0 0 0
0 0 0
0
0
2K~*
0
k3
â€” q EI L3 El q LK3 qEl K3L K3EI
lAn\
Ai2
^i3
â– 4'4
4Â»5
V 1 /
/
(3  1  15)
The coefficient matrix, A, is the following.
/ 0 0 0 0 1 0 \
0 1 0 1 0 0
SIN(K) COS(K) 1 1 0 2Tr3
K 0 1 0 0 0
KCOS(K) KSIN(K) 1 0 0 X3
0 0 K3EI L2 0 0 0
0 0  K*EI L3 0 0 qEl L3
0 k3ei 0 0 0 El q
L LK3
 K3SIN{K)El  K3COS(K)EI 0 0 0 qEl
L L K3L
0 0 0 0 0 k3ei L3
The unknown vector, X, is the following.
/ Ail \
Ai2
Ai 3 â€¢A 4 ^5
V i /
A' =
(3  1  17)
43
The constant vector, B, is the following.
/ Ho) x Wi(0) Wi( 1) WÂ«t.(0)
Vi(l) Mi( 0) M,(l)
V Pi /
(3  1  18)
44
Equation (3115) can be modified as the following:
/ IMO) \ ^(0) ^(1)
K(0)
V<(1)
M,(0)
m i)
V Pi
0 0 0 0 1 0 \
0 1 0 1 0 0
C05(A',) 1 1 0 2h * 0 /Ail \
Ki 0 1 0 0
KiCOS(Ki) KiSIN{Ki) 1 0 0 Si x3 A{2 Ai3
0 0 0 0 Pi Pi 0 0 0 0 0 Q.Eil, L3 An Ais
0 Pi Li 0 0 0 L.X3 \ 1 /
PiLiSIN{Ki) P^iCOSiK,) 0 0 0 X3L,
0 0 0 0 0 Pi )
OR
/ ^i(O) \ / o 0 0 ] 0 0 0 1 1 0 0 \ 0
Mâ€™(0) wt (1 SIN(Ki) COSiKt) 1 1 0 2^7
M^(0) K\ 0 1 0 0 0
K,COS(Ki) KiSIN(Ki) 1 0 0 &
V'i(0) â€” 0 0 Pi 0 0 0
Vi(l) 0 0 Pi 0 0 sÂ»s< L,
Mi( 0) 0 Pi Li 0 0 0 iiSj X3
A/' (1) \ Pi / PiLiSIN(Ki) PiLiCOS(Ki) 0 0 0 QjSj X3
\ 0 0 0 0 0 p
l 4il \
Ai 2 '4i3 Ai 4
Ais 1 /
In which S{ = ^
45
3.2 Oganization of Equations
This gabled frame has four elements. Six unknowns appear in each element so that this system has twenty four unknowns. Twenty four equations are required to solve the system if the solution can be solved. According to section 2.5, the twenty four equations are already found. How to organize these equations is a very important step because every selection will find the solution. The following discussion will follow joint by joint and element by element, the forces, bending moments, and rotations to find the relationships between the twenty four equations.
3.2.1 Joint 1 and Joint 5
According to boundary conditions, the constants, A12, A14, Al5 of element 1, and A42, A44, A45 of element 4, are found. They are equal to zero and are not changed even if the K, of i(/l element is changed. These six constants of the system solutions are a part of the twenty four unknowns. If the other solutions can be solved, only eighteen equations are required, because eighteen constants are still unknown.
3.2.2 Joints 2, 3, and 4
Six equations can be derived by using joint equilibrium conditions at each joint. The total are eighteen equations available in this frame. At joint 2, equation (261) is found by Â£ F h = 0, equation (262) is found by Â£ Fv = 0, equation (263) is found by Â£ M = 0.0, equation (264) is found by Â£W,* = 0.0, equation (265) is found by Â£Dw, and equation (266) is found by EDv. Similarily, equation (267) is found by SFW = 0.0, equation (268) is found by EFv = 0.0, equation (269) is found by Â£M = 0.0, equation
46
(2610) is found by sWif = 0.0, equation (2611) is found by EDand equation (2612) is found by EDv at joint 3. Equation (2613) is found by EFw = 0.0, equation (2614) is found by EFv = 0.0, equation (2614) is found by EM = 0.0, equation (2615) is found by EW * = 0.0. equation (2616) is found by EDW, and equation (2617) is found by EDv at joint 4.
3.2.3 Forces
Equations (261), (262), (267), (268), (2613), and (2614) are derived from the force equilibrium. According to (3120) these equations include four unknowns, A13, A23, A33, and A43. If these six equations are used to solve the four unknowns, the matrix should be linear dependent so that four of these equations used to solve A13, A23, A33, and A43. are good enough. The other two of these six equations should be put in the nonlinear system to solve the nonlinear solutions of Kâ€™s. Apparently, this selection is not the only one that will work. If any four of these equations are selected, one canâ€™t be sure that they are linear independent because their limitations havenâ€™t been considered, yet. The limitation is that at least one of the equilibrium joint conditions should be selected in this algebra and the other one can be arbitrarily selected at joints 2, 3, or 4, because at joint 2, Ai3, and A23 are only included, at joint 3. A23, and A33 are only included, at joint 4. A33, and A43 are only included. If any one joint is not included, one of A,3 will not be found and the other two or more equations will be linear dependent.
3.2.4 Bending Moments and Rotations
Equations (263), (269) and (2615) are derived by the bending
moment equilibrium conditions at joints 2, 3, and 4 respectively. These
47
three equations only include the constants Ai and A2 of each element. Equations (264), (2610) and (2616) are derived by the rotation equilibrium conditions at joints 2, 3, and 4, respectively. Although these three equations include the constants Ai, A2, and A3 of each element, A3 was found by force equations. In this algebra, the Ai3 will be considered as known. In these six equations , An, A2j, A22, A31, A32, and A4i are still unknown. Six equations are available in the bending moments and rotations. Fortunately, six unknowns can be found by solving these six equations. This selection is the only one that works beacuse six unknowns and six equations are available.
3.2.5 Displacements
Equations (265), (2611) and (2617) are found by EDw at joint 2, 3, and 4. respectively. Equations (266), (2612) and (2618) are found by EDv at joint 2, 3. and 4, respectively. At joint 2, the displacement of element 1 in the horizontal direction should be equal to that of element 2 in the same direction [see equation 321] and the displacement of element 1 in the vertical direction should be the same as that of element 2, [see equation 322]. Similary. at joint 3, the displacements of element 2 should be equal to those of element 3 in the horizontal and vertical directions, [see equation (323) and (324) of joint 3 in the horizontal and vertical directions.] At joint 4, the displacements of element 3 should be the same as those of element 4 in both directions. [see equations (325), (326) of joint 4 in the horizontal and vertical directions.]
Although those equations include all of the unknowns of Aâ€™s, these excluding A24, A25, A34, and A35 are found in forces and bending moment equilibrium conditions. They should be considered as known. Accord
48
ing to these procedures, these six equations include only four unknowns. Similar to those of force conditions, the other equations will be put in the nonlinear system. The limitations should be considered to avoid being linear dependent. The limitation is the same as that of force conditions. At least one of the equations of each joint should be put into the algebra so that the selection of this procedure is not the only one that works, either.
E D" E
ELEMENTl ELEMENTS
E *  E
ELEM ENTl ELEMENTS
E DÂ» = E
ELEMENTl ELEM ENTS
E Dv = E
ELEMENT2 ELEMENTS
E Dh  E
ELEMENTS ELEMENT4
E Dv = E
ELEMENTS ELEMENT4
Dh
Dv
Dh
Dv
Dh
Dv
(321) (322) (323) (324) (3  2  5) (326)
In this section, two equations are available in each joint,with a total of six equations. Only these four constants, A24, A25, A34, A35, are still unknown. The procedure to solve these unknowns is similar to that of forces. Four of these six equations will be selected to the linear system and the other two equations will be put in the nonlinear system. Unfortunately, this choice also has limitations. Although this choice is not the only one that works, the limitations which sometimes cause linear dependent have to be considered to avoid to this ill condition.
49
3.3 The Linear Solutions of Aâ€™s
If the constant K of every element of this frame which are defined in equation (311) is known, all of the solutions of Aâ€™s will be found because the Aâ€™s depend on K. The solution of Kâ€™s will be discussed in section 3.3.4. In sections 3.3.1, 3.3.2, and 3.3.3, K of every element will be assumed as known to find the solution of Aâ€™s related to K.
3.3.1 The First Part of Linear Matrix A,
This matrix based on equations (261), (267), (268), and (2613) which are all of the force equilibrium conditions. The other equations, (262) and (2614), are put in the nonlinear system. Although this choice is not the only one that works, the solutions of Ai3 of ith element will be found by this choice and the solutions should be the same as those of the other choices. These equations are formed as AiXj = Bj in which the matrix A! is the coefficient matrix, Xi is the unknown vector, and Bj is the constant vector.
The first coefficient matrix Ai is as the following.
(p' P2SIN{6) 0 0
0 P2COS{6) p3cos{0) 0
0 P2SIN{6) P3SIN(0) 0
\ 0 0 p3sin{0) p*
The first constant vector Bx is as the following:
50
/ + QH2  P2COS(6) \
P2SIN(6) + QV(3) + P3SIN(0)  ^ ^
P2COS(0) + 2^S/iV(0) + QP3 + *^fSIN(9)  P3COS(e)
V 2 iÂ±^QHi + p3cos(e) )
in which qx = qA = 0 , and P,(0) = P;(l) = P. because the axial force is constant in the member. Substituting qi = q4 = 0 and P;(0) = P;(l) = P; into equation (332), the first constant vector Bi can be modified as the following:
/ QH2  P2COS(0) \
P2SIN(6) +
P2COS(0) + ^SIN(0) + QH3 + *^fSIN(0)  P3COS{6)
\ 3 ~QH4 + P3COS(6)
The first unknown vector, Ah, is as the following:
/.4(l,3)x
4(2,3)
4(3,3)
\ .4(4, 3) /
(334)
3.3.2 The Second Part of Linear Matrix A?
This matrix uses the results derived by equation (331). The solutions of equation (331) are A13, A23, A33, and A43 which are considered
51
as a known in this section. This matrix is based on equations (263), (264), (269), (2610), (2615),and (2616) which are derived from bending moments or rotations. This choice is the only one that works from the first part of linear matrix Ai because this matrix includes six equations and has six unknowns to solve.
A2X2 = B2 (335)
in which the A2 is the coefficient matrix. X2 is the unknown vector. The B2 is the constant vector in which all of the variables are known and the results of equation (331) are included. The matrix form of equation (335) is the following.
The second coefficient matrix is the following.
â€¢42 =
( PL\SK\ 0 0
KXCKX
0
\ 0
0
pl2sk2
0
 k2 k2ck2 0
pl2
pl2ck2
0
0
k2sk2
0
0
pl3sk3
0
0
â€” k3c A 3
k3
0
pl3ck3
pl3
0
k3sk3
0
0 \
0
PL4SK4 0 0
k^ckJ (3
6)
in which
PL, = PiL,
CK, = COS(Ki)
SK, = SlN{Ki)
and P,(0) = P,(l) =P, because the axial force is constant in the member. The second constant vector B2 is as the following:
B2 =
/
ii s 1 _ ^2^2
t~2^2
Kl
Kl
.4(1,3)â€” + .4(2,3)
\
.4(2,3)â€” ^ + .4(3,3)+ ^
\ â€”.4(3,3)+ .4(4,3)+ ^ )
(337)
52
in which A13, A23, A33, A43 are known. Since qx = q4 = 0, the constantvector Bi can be modified as the following:
B2
/
K 3
>1(1,3)+ .4(2, 3) .4(2,3) f, + .4(3,3) + .4(3,3) + .4(4,3)
h.
Â«l
(338)
The second unknown vector is .V2 as the following:
A 2
/M i,i)\ .4(2,1) .4(2,2) 4(3,1) â– 4(3,2)
V 4(4,1)/
(339)
53
3.3.3 The Third Part of Linear Matrix Aa
This matrix uses the results of the first part of the linear matrix and the second part of the linear matrix. All of the Aâ€™s excluding A24, A25, A34 and A35 are considered as known because they were derived in previous sections. This matrix is based on equations (265), (266), (2612), and (2617). The other two equations, equations (2611) and (2618), developed from displacements will be put in the nonlinear system. Use these equations to form the matrix as equation (3310),
A3X3 = B3 (3310).
Although this selection is not the only one that works, this solution of equation (3310) should be the same as those of the other choices. The matrix form of equation (3310) is as the following.
The third coefficient matrix, .43, is as the following.
/SIN {9) COS(6) COS(6) SIN(9) COS(9) SIN (6)
\ 0 0
0 0 \
0 0
COS(6) SIN {9) SIN(9) COS(9) /
(3  3  11)
The third constant vector B3 is as the following:
/ A(2,2)SIN(8)  Wâ€™,(l) \
17,(1)  .4(2, 2)COS(9)
W2(l)COS{9)  U2{1)SIN{9) + U3{\)SIN{9)  IV'3(1 )COS(6)
\ W4(l)  A(3,2)SIN(9) )
(3  3  12)
54
In the third constant vector B3 the Ui and W{ havenâ€™t Ai4, A<5 because those constants are zero while calculating.
The third unknown vector is as the following:
A'3
/4(2,4)\ 4(2, 5) 4(3,4)
\ 4(3, 5) /
(3  3  13)
3.3.4 The Nonlinear system
The nonlinear system is based on equations (262), (2611), (2614), and (2618) which are described in sections 3.3.1 and 3.3.3. These equations use the IMSL to find the solutions. Because the constant Ks of each element have to be assumed as known. All of Aâ€™s can be found in previous procedures. This nonlinear equation set is not the only ones that work, either. Since the first part of the linear matrix A, and the third part of the linear matrix A3 are not the only one that works, this problem set should be changed if matrix A, or matrix A3 is changed. Although this selection is not the only one that works and the solution of Kâ€™s and Aâ€™s are different, the relationship is the same as the other selections, and the final results should be the same as each other selection.
55
3.4 The Solution of the Nonlinear System
In this section, the nonlinear solutions will be discussed. Since all of the solutons of Aâ€™s depend on the constant Kâ€™s of each element. The solutions of Aâ€™s can be solved with the linear system. In the previous sections, the K of each element is 3assumedas known. Now, how to find K is the object of this section.
3.4.1 Initial Guess of K
Since the relationship between the axial force of the element and the constant K of the same element was discussed in equations (311), (312), and (313), K can be found if the axial force P is known. Use those Ks as the initial guess in the IMSL program to find the nonlinear solutions. If the external loads are very small, the results of the linear system should be very close to the nonlinear solutions. Use this concept with the finite element method to find the linear solutions. After doing that, the axial forces of the linear system can be found. Using equation (312), find out the initial guess of K.
K
/u?
Eih
(341)
in equation (341), P, is compressive axial force and K< is positive. When the first intial guess are found, the system can be solved in the first step and the first nonlinear solutions is found. Use the ratio of load increment to find the next initial guess. This relation is shown as the following.
Am f l ^ _ ( Pm f l
K ' ~~ ' P
** m rm
(342)
Equation (342) can also be modified as the following.
Am+1
= K:
iPm+ 1 ,
iV 5 )
(343)
56
in which the Pm+i is the initial guess of the (m+l)fh step. Km is the nonlinear solution of the step. According to this procedure, the initial guess should be related to load conditions. This way is better than if the initial guess is derived from linear system for every step. Especially if the external load approaches the critical load. In each step, the initial guess should be known before calculating the solution of Aâ€™s.
3.4.2 Finite Element Method for PC Program FEMUCD
FEMUCD is a Finite Element PC program. The FEMUCD program includes many subprograms. It can be used to find the displacements and forces of frame or plate. The FEMUCD uses free format for inputing data but the data must be input in a specific order required by the program. The input data file includes the TITLE. SYSTEM, JOINT, RESTRAINTS, FRAME, ASOLID, and LOADS data blocks. They are presented as the following.
1) . The TITLE Data Block
The first line of every FEMUCD data file consist of up to 80 characters of information used to label each page of output.
2) . The SYSTEM Data Block
The SYSTEM data block gives the basic information required to describe the structural model to FEMUCD.
3) . The JOINTS Data Block
The JOINTS data block begins to describe the geometry of the system by locating in global coordinates.
4). The RESTRAINTS Data Block
57
The RESTRAINTS data block constrains the basic information on the fixity of the nodes.
5) . The FRAME Data Block
This data section defines the material properties, connectivity, and orientation of the threedimensional beam elements of the model. Twodimensional truss elements and threedimensional truss elements are special cases of this general form.
6) . The ASOLID Data Block
This data section defines the material properties connectivity, and orientation of the asolid element of the model axisymetric. Plane stress, plane strain and 3D plane stress are differnent cases for this element.
7) . The LOADS Data Block
This data block defines the loading systems for the structures.
The FEMUCD program includes the subprograms, FEMUCD, FRAME. OPT, SOLVEl, SOLVE2. FRMFRC, ASOLID whose functions are described in the class notes of Applied Finite Element Techniques for Personal Computer by Professor John R. Mays, January, 1987. All results of FEMUCD are linear. In each case of this system, the first initial guess is from FEMUCD but the loads are very small. The input and output data files of the section 4.4 at 6 equal to 10Â° are in Appendix A and Appendix B respectively.
3.4.3 Computer Program
The computer program uses the IMSL library to solve matrix algebra and the nonlinear system. Because all of the Aâ€™s solutions depend on the constant K of each element, the solution of Aâ€™s have to be found after
58
the constant K is found by the nonlinear system. Based on this concept, when the constant of K is defined in each step of the nonlinear procedure, the solutions of Aâ€™s will be defined, too. This procedure will continue until the nonlinear solution, Kâ€™s, are satisfied by the requirements of this subroutine of IMSL. After K is found from the nonlinear procedures, the solutions of Aâ€™s should be found. All of the solutions, displacements, rotations and forces are found at this load by using the constants Aâ€™s and K and substituting them into the formula of rotations, displacements, axial forces, transverse shear forces, and the bending moments at each point of any element. Use these Kâ€™s to find the next initial guess for the next step. The solutions at every load will be found. The computer program is in Appendix C.
CHAPTER IV
RESULTS
Gabled frames subjected to various load conditions behave differently. This chapter will discuss many different load conditions. Even if the frame is subjected to one load condition, the behaviors of linear and nonlinear systems are different. The different results between linear and nonlinear systems will also be discussed.
4.1 The Concentrated Loads At Joint 3
The first, load condition, the gabled frame is subjected to a concentrated load only at joint 3. This concentrated load begins very small, and gradually increases. [See Fig. 4.1]. The Fig. 4.2 represents the loadrotation behavior of element 1 at the end of joint 2. The behaviors are different when the angle 9 is changed. The different behaviors with different angles are also shown in Fig. 4.2. The difference between the solutions of linear and nonlinear systems is shown in Fig. 4.3(a* e).
LOAD
60
QV3
Fig. 4.1 Concentrated Load At Joint 3
Fig. 4.2 LoadRotation Behaviors with Different 6
LOAD LOAD
61
4.3.(a) Comparison of Linear and Nonlinear Solutions with 9 = 1
 i0
Fig. 4.3.(b) Comparison of Linear and Nonlinear Solutions with 6 = 10^
LOAD
62
Fig. 4.3.(d) Comparison of Linear and Nonlinear Solutions with 9 = 30^
LOAD
63
Fig. 4.3.(e)
Comparison of Linear and Nonlinear Solutions with 6 = 40
0
64
4.2 Concentrated Loads At Joint 2 , 3 and 4
In the second concentrated load condition, joints 2 and 4 are subjected to a vertically concentrated load which is not changed. The load at joint 3 is changeable, [see Fig.4.4] This load begins very small and gradually increases. The difference between section 4.1 and this section is that the load conditions of section 4.1 doesnâ€™t have any concentrated load at joints 2 and 4. The loadrotation behaviors with different angles, 6, are different as shown in Fig. 4.5. The different behaviors between linear and nonlinear systems which are subjected to the same load conditions as each other are shown in Fig. 4.6.(aâ€”>e).
qv3
Fig. 4.4 Concentrated Loads At Joints 2, 3, and 4
LOAD LOAD
65
Wu(l)
Fig. 4.5 LoadRotation Behaviors with different 0
W!,Â«(!)
Fig. 4.6.(a) Comparison of Linear and Nonlinear Solutions with 9 = 1^
LOAD
66
Fig. 4.6.(b) Comparison of Linear and Nonlinear Solutions with Q â€” 10^
Wi,.(l)
Fig. 4.6.(c) Comparison of Linear and Nonlinear Solutions with 9 = 20^
LOAD ^ LOAD
67
,6.(d)
Comparison of Linear and Nonlinear Solutions with Â»
 30Â°
Fig. 4.6.(e) Comparison of Linear and Nonlinear Solutions with 0 = 40Â®
68
4.3 Concentrated Load At Joint 3 and Horizontal Loads At Joint 2 and 4
The third load condition is the first load condition with two horizontal loads at joints 2 and 4. [see Fig. 4.7] The two horizontal loads are not changed. Similar to the first load condition, the load at joint 3 begins very small and gradually increases. The behavior of this load condition changes when the angle, 6, changes. The linear and nonlinear solutions are shown in Fig. 4.9(aâ€”>e).
Fig. 4.7 Concentrated Load At Joint 2 and Horizontal Loads At Joints 2 and 4
LOAD LOAD
69
Fig. 4.8 LoadRotation Behaviors with different 6
Fig. 4.9.(a) Comparison of Linear and Nonlinear Solutions with d â€” 1Â°
70
Fig. 4.9.(b) Comparison of Linear and Nonlinear Solutions with 0 = 10^
71
Fig. 4.9.(d) Comparison of Linear and Nonlinear Solutions with 6 = 30^
72
4.4 Concentrated Loads At Joints 2, 3 and 4, and Horizontal Loads At Joints 2 and 4
In order to compare the behaviors of the gabled frame with horizontal loads and without the horizontal loads. These loads will combine the second load condition with two horizontal loads at joints 2 and 4. This load condition is shown in Figure 4.10. Under this load condition, the vertical loads at joints 2 and 4 are not changed and the horizontal loads at joint 2 and 4 are not changed, either. Of course the vertical load at joint 3 begins very small and gradually increases. The gabled frame will have different angles, 0, which are 1Â®, 10Â®, 20Â®, 30Â®, and 40Â®. The different loadrotation behaviors with different angles, 6. are shown in Figure 4.11. The difference between the linear behaviors and the nonlinear behaviors with different angles, 6. will be shown in Figures 12.(aâ€”e).
QVa
qh<
ig. 4.10 Concentrated Loads At Joints 2,3,4 and Horizontal Loads At Joints 2 and 4
LOAD
73
21012
W,,(l)
Fig. 4.11 LoadRotation Behaviors with different 9
Wi,.(l)
Fig. 4.12.(a) Comparison of Linear and Nonlinear Solutions with 9 = 1Â°
LOAD
74
Wl,*(1)
Fig. 4.12.(b) Comparison of Linear and Nonlinear Solutions with 9 = l()lJ
WM(1)
Fig. 4.12.(c) Comparison of Linear and Nonlinear Solutions with 9 = 20(*
75
Wu(l)
Fig. 4.12.(d) Comparison of Linear and Nonlinear Solutions with 6 â€” 30(â€™
0.8
0./
0.6 
0.5 
Â£ 0.* ^ W >
o 0.3 
C.2 
0.!
0.0
2
nonlinear solution
linear solution
Wi..(l)
Fig. 4.12.(e) Comparison of Linear and Nonlinear Solutions with 6 = 40^
76
4.5 The Different Directions Between FEMUCD and the Nonlinear System
The PC program FEMUCD defines the directions of the element as shown in Figure 4.13. In the beginning end, all of the forces are compressive and the displacements will be defined as positive when the displacements are moved to the other end. In the other end, all of the forces will be defined as positive when these forces are tensile. The displacements will be defined as positive direction when the displacements are moved in the same direction as the beginning end. Most importantly, the directions of the rotations of each element are defined as positive when the rotations are the clockwise direction.
All of the directions of the forces and displacements of the gabled frame are defined in chapter 2. Figure 2.1 shows the directions of each element in forces or displacements of the gabled frame. The direction of the rotation of each element depends on the direction of the lateral displacements of each element. These two definitions just have different phases. Because of these two definitions the output data files of the PC program FEMUCD and the nonlinear system have different signs but they are the same direction as each other. Because they have different signs in the same direction, none of the figures can use different signs to express the same direction so the definitions will be consistent. All of the figures of this chapter use the definition of the nonlinear system. After the outpiit, data files of FEMUCD are found, the sign should be times negative sign then the sign and the directions are the same as those of the nonlinear system.
GLOBAL
Fig. 4.13 The Directions of the Element in FEMUCD
CHAPTER V
CONCLUSION
5.1 Summary
Some other procedures use X function to analyze the behaviors of this system. The procedure of this thesis uses trigonometric functions. If the behavior of any system is nonlinear, the X function may sometimes find it difficult to express these behaviors so the expression of trigonometeric function is more convenient to express the nonlinear behaviors.
This thesis found the differences between the linear and nonlinear solutions. The gabled frame with different angles, 6, will have different behaviors depending on the load conditions. On the other hand, although the gabled frame with the same angle, 9, as those of others, their behaviors will be different when they are subjected to different load conditions in the linear system. Of course the nonlinear behaviors wonâ€™t be the same, either.
Using finite element analysis, all of the distribution loads will be the same as the concentrated loads by using interpolation functions and applying the loads at both ends of a bar. The concentrated loads are only used for the examples in chapter four. Similar to the concept of finite element analysis, those concentrated loads can be considered as the uniformly distributed loads applied on elements 2 and 3. The results of the structure subjected to distribution loads should be same as those of the structures subjected to concentrated loads.
The limitation of the angle, 0, is not equal to 0Â°, 45", and 90".
79
When the angle, 9, is equal to 0Â° the gabled frame is the same as the singlespan frame and the linear matrices Ai, A2, and A3 will be linear dependent. When the angle, 9, is equal to 45Â°, although the gabled frame is symetric, the linear matrices are also linear dependent. Of course when the angle 9 is equal to 90Â° elements 1 and 4 would be bound together. In this case, this structure is a column, not a frame.
5.2 Conclusion
1. The loadrotation behaviors of elements 1 and 4 are very similar to those of reference 2 8 when the angle, 0, is close to 0 i.e. 9 * 0, but not equal to 0.
2. When the angle, 9, is changed from very small to 40Â°, the loadrotation behaviors are much different from those of the gabled frame with small angle. 9.
3. When the gabled frame is only subjected to a vertical load at joint 3 with the large angle, 9. the loadrotation behavior of elements 1 and 4 are similar to those of an eccentrically loaded column.
4. If the gabled frame is subjected to vertically concentrated loads at joint 2, 3. and 4, the behaviors of the loadrotation are similar to those of ideal columns.
5. If the gabled frame with a small angle, 9, is subjected to vertical load at joint 3 and the horizontally concentrated loads at joints 2 and 4, the loadrotation behaviors are more similar to those of an eccentrical column
8 The same as footnote 1
80
with large eccentricity, e. If the angle, 6, is larger, the behaviors are more different than those of the gabled frame with the small angle, 9.
6. The uniform loads q2 and q3 can be considered the same as the concentrated loads at joints 2, 3, and 4. Similarly, the concentrated loads at joints 2, 3. and 4 can be considered the same as the uniform loads on elements 2 and 3.
BIBLIOGRAPHY
W.F. Chen and T. Atsuta, â€œTheory of BeamColumns Volumn 1. inplane beheavior and design â€, McGrawHill Inc., 1976.
George D. Manocis, Dimitrios E. Beskos, and Bruce J. Brand, â€œElastoplastic Analysis and Design of Gabled Frame â€, Journal of Computers & Structures Vol.22, No.4. PP.693697, 1986.
George D. Manolis and Dimitrios, E. Beskos â€œPlasic Design Aids for PinnedBase Gabled Frame â€, First quater/1987.
John R. Mays, â€œClass Notes of Applied Finite Element Techniques for Personnal Computerâ€, University of Colorado at Denver, January, 1987.
Riyad K. Qashu, A.M. ASCE and Donald A. Dadeppo, â€œLarge Deflection and Stability of Rigid Frame â€, Journal of Engineering Mechanics, Vol.109, No.3, June, 1987.
George J. Simitese, â€œAn Introduction to the Elastic Stability of Structures â€, PrenticeHall, Inc.. Englewood Chiff. New Jersey, 1976.
J.G. Simitese, A.S. Vlahinos, G. J. Simitese, â€œSway Bucking of Unbraced Multistory Frames â€, Journal of Computers & Structures, Vol.22, No.6 PP.10471054, 1987.
A.S. Vlahinos and A.N. Kounadis â€œNonlinear Elastic Limit State Analysis of Rigid Jointed Frames â€, Int. J: Nonlinear. Vol.6. No.3/4 PP.379385,1987.
82
Andreas S. Vlahinos, C.V.Smith, JR. and G. J. Simitese, â€œA Nonlinear Solution Scheme for Multistory, Multibay Plane Frames â€, Journal of Computers fe Structures, Vol.22 No.6 PP.10351045, 986.
Andreas S. Vlahinos and George J. Simithese, â€œBucking and Postbucking of Multibay Frame â€. Proceeding at International Conference on Steel and Aluminum Structure, Vol. 3, pp.689708, Cardiff, UK 1987
K. C. Wang, â€œSecondOrder Analysis of Multistory Building Frames by Maching Techniqueâ€, Microcomputer in Civil Engineering. Vol.l, No.3, 1986.
APPENDIX A
FEMUCD INPUT DATA FILE
84
FIND_AXIAL_FGRCE_AND_LINEAR SOLUTION FOR THESIS
SYSTEM P=2 S=400 N=5 L=7
JOINTS
1 X=0.0 Y=0.0 1=0.0
2 X=0.0 Y=200.0
3 X=196.96 Y=234.73
4 X=393.92 Y=200.0
5 X=393.92 Y=0.0
RESTRAINTS
1 0 0 R=l,1,0,1,1,0
2 0 0 R=0,0,0,1,1,0
3 0 0 R=0,0,0,1,1,0
4 0 0 R=0,0,0,1,1,0
5 0 0 R=l,1,0,1,1,0
FRAME
NM=1
1 A=1 1=0.083333 E=29000
1 1 2 M=1 LP=1,0
2 2 3 M=1 LP=1,0
3 3 4 M=1 LP=1,0
4 4 5 11=1 LP=1,0
LOADS
2 L=1 F=0.02,0.02,0.0,0.0,0.0,0.0 4 L=1 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=1 F=0.0,0.02,0.0,0.0,0.0,0,0
2 L=2 F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=2 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=2 F=0.0,0.10,0.0,0.0,0.0,0.0
2 L=3 F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=3 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=3 F=0.0,0.20,0.0,0.0,0.0,0.0
2 L=4 F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=4 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=4 F=0.0,0.30,0.0,0.0,0.0,0.0
2 L=5 F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=5 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=5 F=0.0,0.40,0.0,0.0,0.0,0.0
2 Lb F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=fc F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=6 F=0.0,0.50,0.0,0.0,0.0,0.0
2 L=7 F=0.02,0.02,0.0,0.0,0.0,0.0
4 L=7 F=0.02,0.02,0.0,0.0,0.0,0.0
3 L=7 F=0.0,1.40,0.0,0.0,0.0,0.0
APPENDIX B
FEMUCD OUTPUT DATA FILE
86
SYSTEM P=2 S=400 N=5 L=7
IND AX IAL FORCE AND L INEAR SOLliT ION FOR THESI5 F E M U C D VERSION 86.0
t*iit*itÂ»**t**Â»i*Â»t****tt*i*iitt
hh ECHO OF SAP INPUT DATA t * * *
TOTAL NUMBER OF JOINTS = 5
TOTAL NUMBER OF LOAD CONDITIONS = 7
1 X=0.0 Y=0.0 Z=0.0
2 X=0.0 Y=200.0
3 X=196.96 Y=234.73
4 1=393.92 Y=200.0
5 X=393.92 Y=0.0
GENERATED JOINT COORDINATES
JOINT I X Y 7
1 .000 .000 .000
2 .000 200.000 .000
3 196.960 234.730 .000
4 393.920 200.000 .000
5 393.920 .000 .000
1 0 0 R=1,1,0, 1,1,0
2 0 0 R=0,0,0, ,1,1,0
3 0 0 R=0,0,0, 1,1,0
4 0 0 R=0,0,0. ,1,1,0
5 0 0 R=1,1,0, 1,1,0
G E N E P. A T E D J 0 I [ N T RES T R A 1 N T S
JOINT X Y 7 RX RY R2
1 1 1 0 1 1 0
2 0 0 0 1 1 0
3 0 0 0 1 1 0
4 0 0 0 1 1 0
5 1 1 0 1 1 0
NM=1
NM=1
1 A=1 1=0.083333 E=29000
87
H I H H H * H * H H H i H H H n H H H **** ECHO OF FRAME INPUT DATA***** hhhhuhhhhhhhhhhhh
MEMBER PROPERTIES
TOTAL NUMBER OF MEMBER PROPERTIES
MEMBER PROPERTY NUMBER  =
AXIAL AREA, A  =
TORSIONAL MOMENT OF INERTIA, 3 =
MOMENT OF INERTIA, 122=
MOMENT OF INERTIA, 133  =
SHEAR AREA, A2=
SHEAR AREA, A3=
MODULUS OF ELASTICITY, E  =
SHEAR MODULUS, S=
1 1 2 M=1 LP=1,0
2 2 3 Â«=1 LP=i,0
3 3 4 11=1 LP=t,0
4 4 5 M=1 LP=1,0
1
.lOOE+Oi .OOOE+OO â– OOOE+OO
.B33E01
.OOOE+OO
.112E+05(USED FOR TOR k SHEAR)
FRAME ELEMENT DATA
MEMB J1 J2 MEMB. SECT. PROP. LP1 LP2
1 i 2 1 1 0
2 2 3 1 t 0
3 3 4 1 1 0
4 4 5 1 1 n
E 0 U A T I 0 N NUMB E R S
NODE IX IY II IXX IYY III
1 0 0 11 0 0 12
0 Jm 13 14 15 0 0 16
3 7 8 9 0 0 10
4 3 4 5 0 0 6
5 0 0 1 0 0 2
ABSOLUTE MAXIMUM COLUMN HEIGHT(ANY EQUATION) = 10
NUMBER OF BLOCKS = 1
MAXC = 16
MAXT = 400
FORMATION OF BLOCK IN STIFFNESS MATRIX
BLOCK NUMBER = 1 OF 1
LOWEST EQUATION NUMBER = l
HIGHEST EQUATION NUMBER = 16
NUMBER OF TERMS IN THIS BLOCK = 64
LOWEST COUPLED BLOCK NUMBER = 1
HIGHEST EQUATION NUMBER = 16 NUMBER OF TERMS IN THIS BLOCK = 64 LOWEST COUPLED BLOCK NUMBER = 1
MC VECTOR: 1 3 6 10 15 21 26 32 39 4^ 46 50 5? 65 7'
J 0 I N T LOADS
LOAD NUMBER: 1
NODE XD1R YDIR 2DIR XX YY 77
2 .200E01 200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
4 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
3 .OOOE+OO .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
LOAD NUMBER: 2
NODE XDIR YDIR ZDIR XX YY 77
0 L. .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
4 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
3 .OOOE+OO 100E+00 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
LOAD NUMBER: 3
NODE XDIR YDIR ZDIR XX YY 77
2 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
4 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
J .OOOE+OO .200E+00 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
LOAD NUMBER: 4
NODE XDIR YDIR ZDIR XX YY 77
2 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
4 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
J .OOOE+OO .300E+00 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
LOAD NUMBER: 5
NODE XDIR YDIR ZDIR XX YY 77
2 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
4 .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
7 J .OOOE+OO .400E+00 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
89
LOAD NUMBER: h
NODE XDIR YDIR
2 .200E01 .200E01
4 .200E01 .200E01
3 .OOOE+OO .500E+00
2DIR XX YY .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
77
.OOOE+OO
.OOOE+OO
.OOOE+OO
LOAD NUMBER: 7
NODE XDIR YDIR
2 .200E01 .200E01
4 .200E01 .200E01
3 .OOOE+OO .140E+01
ZDIR
.OOOE+OO
.OOOE+OO
.OOOE+OO
xx
.OOOE+OO
.OOOE+OO
.OOOE+OO
YY
.OOOE+OO
.OOOE+OO
.OOOE+OO
77
.OOOE+OO
.OOOE+OO
.OOOE+OO
CALCULATED DISPLACEMENTS FOR LOAD NO. 1
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .88143E02
2 .46952E+00 .20690E03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 1058&E01
T â– j .18784E10 .26639E+01 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .23479E13
4 . 46952E+00 .20690E03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 10586E01
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .88143E02
CALCULATED DISPLACEMENTS FOR LOAD NO. 2
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .63284E01
2 .33716E+01 .4B27BE03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .75994E01
3 .134B7E09 .19123E+02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 1&85BE12
4 .33716E+01 .4B276E03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .75994E01
5 â€¢OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .63284E01
CALCULATED DISPLACEMENTS FOR LOAD NO. 3
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 13137E+00
2 .69991E+01 .B2759E03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 15775E+00
3 .27999E09 .39697E+02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .34996E12
4 .69991E+01 .B2759E03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .15775E+00
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .13137E+00
CALCULATED DISPLACEMENTS FOR LOAD NO. 4
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO ,OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .19946E+00
2 .10627E+02 .11724E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .23951E+00
3 .42511E09 .60271E+02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO â€¢53135E12
4 .10627E+02 . 11724E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .23951E+00
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .19946E+00
90
CALCULATED DISPLACEMENTS FOP LOAD HO. 5
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .26754E+00
n L .14254E+02 .15172E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .32128E+00
3 .57020E09 .80B45E+02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO . 71270E12
4 .14254E+02 .15172E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .3212BE+00
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .26754E+00
CALCULATED DISPLACEMENTS FOR LOAD NO. 6
JOINT XDEFL YDEFL ZDEFL XROT YROT ZROT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .33563E+00
2 17882E+02 .18621E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .40304E+00
3 .71534E09 .10142E+03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .89413E12
4 .17B82E+02 .18621E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .40304E+00
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .33563E+00
CALCULATED DISPLACEMENTS FOP LOAD NO. 7
JOINT XDEFL YDEFL ZDEFL XPOT YROT ZRDT
1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .94841E+00
2 .50530E+02 .49655E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .11389E+01
3 .20214E08 .28658E+03 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .25265E1l
4 .50530E+02 .49655E02 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .11389E+01
5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .94841E+00
ELEMENT FORCES AND MOMENTS FOR LOAD CASE t 1
ELEM P1J1 V2J1 V3J1 T1J1 M2J1 M3J1 P1J2 V2J2
1 .300E01 .234E02 .OOOE+OO .OOOE+OO .OOOE+OO .282E07 .300E01 .234E02
2 .237E01 .597E02 .OOOE+OO .OOOE+OO .OOOE+OO .469E+00 .237E01 .597E02
3 .237E01 .597E02 .OOOE+OO .OOOE+OO .OOOE+OO .725E+00 .237E01 .597E02
4 .300E01 .234E02 .OOOE+OO .OOOE+OO .OOOE+OO .469E+00 .300E01 .234E02
V3J2 T1J2 M2J2 M3J2
.OOOE+OO .OOOE+OO .OOOE+OO .469E+00
.OOOE+OO .OOOE+OO .OOOE+OO .725E+00
.OOOE+OO .OOOE+OO .OOOE+OO .469E+00
.OOOE+OO .OOOE+OO .OOOE+OO .733E09
ELEMENT FORCES AND MOMENTS FOR LOAD CASE 1 2
ELEM P1J1 92J1 V3J1 T1J1 M2J1 M3J1 P1J2 V2J2
1 â€¢ 700E01 . 168E01 .OOOE+OO .OOOE+OO .OOOE+OO .357E07 .700E01 .168E01
2 .449E01 .428E01 .OOOE+OO .OOOE+OO .OOOE+OO .337E+01 .449E01 .428E01
3 .450E01 .428E01 .OOOE+OO .OOOE+OO .OOOE+OO .520E+01 .450E01 .428E01
4 .700E01 .168E01 .OOOE+OO .OOOE+OO .OOOE+OO .337E+01 .700E01 .lbBE01
Y332 T1J2 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO
N232 H3J2 â– OOOE+OO .337E+01 .OOOE+OO .520E+01 .OOOE+OO .337E+01 .OOOE+OO .730E07
ELEMENT FORCES AND MOMENTS FOP LOAD CASE # 3
ELEH FIJI M2J1 0331 T131 M231 M331 P132 V232
1 . 120E+00 .349E01 .OOOE+OO .OOOE+OO .OOOE+OO .390E06 120E+00 .349E01
2 .714E01 .BB9E01 .OOOE+OO .OOOE+OO .OOOE+OO .699E+01 .714E01 .889E01
7 j .714E01 . 8B9E01 .OOOE+OO .OOOE+OO .OOOE+OO .108E+02 .714E01 .3B9E01
i .120E+00 .349E01 .OOOE+OO .OOOE+OO .OOOE+OO .699E+01 .120E+00 .349E01
0332 T132 M232 M332
.OOOE+OO â€¢OOOE+OO .OOOE+OO .699E+01
.OOOE+OO .OOOE+OO .OOOE+OO .10BE+02
.OOOE+OO .OOOE+OO .OOOE+OO .6998+01
.OOOE+OO .OOOE+OO .OOOE+OO .446806
ELEMENT FORCES AMD MOMENTS FOP LOAD CASE 1 4
ELEM P131 0231 V331 T131 M231 M331 P132 V232
1 .170E+00 .530E01 .OOOE+OO .OOOE+OO .OOOE+OO .695E06 . 170E+00 .530E01
2 .979E01 .135E+00 .OOOE+OO .OOOE+OO .OOOE+OO .106E+02 .979E01 .135E+00
7 J .9BCE01 .135E+00 .OOOE+OO .OOOE+OO .OOOE+OO 1648+02 .980E01 .135E+00
4 .170E+00 .530E01 .OOOE+OO .OOOE+OO .OOOE+OO .106E+02 .170E+00 530E01
V332 T132 M232 M332
.OOOE+OO .OOOE+OO .OOOE+OO .106E+02
.OOOE+OO .OOOE+OO .OOOE+OO .164E+02
.OOOE+OO .OOOE+OO .OOOE+OO .1068+02
.OOOE+OO .OOOE+OO .OOOE+OO .374EOb
ELEMENT FORCES AND MOMENTS FOR LOAD CASE i 5
ELEH P131 0231 V331 T131 H231 M331 P132 ''.'232
1 .220E+00 â– .711E01 .OOOE+OO .OOOE+OO .OOOE+OO .405E06 .220E+00 .71 IE01
2 . 125E+00 .1B1E+00 .OOOE+OO .OOOE+OO .OOOE+OO .142E+02 .125E+00 .181E+00
7 J .125E+00 â– .181E+00 .OOOE+OO .OOOE+OO .OOOE+OO .220E+02 .1258+00 .181E+00
4 .220E+00 .71 IE01 .OOOE+OO .OOOE+OO .OOOE+OO .142E+02 .220E+00 .71 IE01
IJ332 T132 M232 M332
.OOOE+OO .OOOE+OO .OOOE+OO .142E+02
.OOOE+OO .OOOE+OO .OOOE+OO .220E+02
.OOOE+OO .OOOE+OO .OOOE+OO 142E+02
.OOOE+OO .OOOE+OO .OOOE+OO .237E06

PAGE 1
NONLINEAR ELASTIC STABILITY ANALYSIS OF GABLED FRAME by Wang , YangCheng B.S. , Chinese Military Academy , 1980 A thesi submitted to the Faculty of the G raduate chool of the U niv e r sity of Colorado in part ial fulfillmen t of t h e r e quir eme nts for t h e degree of Master of Science Department of Civil Engineering 1987
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This thesis for the Master of Science degree by Wang, YangCheng has been approved for the Department of Civil Engineering by Andreas S. Vlahinos Date __ \ ' __\ _6__g._l __ _
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Wang, YangCheng(M.S. , Civil Engineering) Nonlinear Elastic Stability Analysis of Gabled Frame Thesis directed by Assistant Professor Andreas S. Vlahinos lll Consider a gabled frame of singlespan, pinned base composed of four straight, slender, and piecewise prismatic bar which are rigidly connected. The four bars have the same length. The external loads applied to any element of this gabled frame include segments of a distributed transverse load, and concenstrated transverse forces. The objective of this study is to compare the loadrotation behavior between the linear system and the nonlinear system. The linear solutions are found by using the PC program FEMUCD. The nonlinear solutions are found by using differential equations. With this approach, the equalibrium conditions, boundary conditions, and some assumptions are considered. The computational approach is used to find the nonlinear solutions in this study. The results are much different between the linear and nonlinear systems. Under concentrated load conditions, the behaviors of the nonlinear and linear systems are similar when evaluating a column subjected to an eccentric load .
PAGE 4
lV ACKNOWLEDGEMENTS I wish to express my gratitude and appreciation to my adviser, Professor Andreas S. Vlahinos , for his guidance, direction, and assistance throughout the course of this work. I also wish to thank Professors John R. Mays, and NienYin Chang for their willingness to serve on my thesis committee and in particular, their constructive criticisms. Particular gratitude is expressed to Major General PinMin Chen , the Dean of t he School of Natural S cience at Chinese Military Academy, Taiwan, Republic of China, who gave me this opportunity and encouraged m e t o study abroad. The gratitude also i s to extend to the ChungShan Institute of Sci ence and T ec hnolog y Taiwan, R epublic o f China, who provide d m y financial support.
PAGE 5
CONTENTS CHAPTER I. INTRODUCTION .................... .............. .. . . ..... .......... . .. . . 1 1.1 The Gabled Frame ..... . .. . ........ .. .......... .. .............. ..... . . .. 1 1.2 Objective of Study ................... . ......... . ............ ............ 4 1.3 Basic Assumptions .. .............. . . .. .. ..... .. . ................... ..... 6 1.4 The Behavior of an Eccentrically Loaded Column ................. 8 II . MATHEMATIC AL FORMULATION .. . .................. .. . ......... 16 2.1 Description of the Notation of the Gabled Frame . .. . . . . .. . . . . .. . . 16 2.2 Description of the Mathematical Formulations of Equations . . . . 17 2.3 The Formulation of Equation t (X) and W(X) ..... .. . . . .. . . .. . . . . 21 2.4 Boundary Conditions .............. ................. .. ................ 23 2.5 Equilibrium Conditions . .. . . . .. . . ............. . . .. . .. .. . . .. . . .. .. ..... 26 2 . 6 Formulation of Equations for Joint Conditions . . .. .... .. .......... 28 2 . 6.1 Joint 2 . ..... .. . ...... ....... . . . ...... .. .. . ...... . . .. . ..... . ........ . 28 2.6.2 Joint 3 . .. . . . . . ..... . . . . . .. . . . ....................................... 29 2.6.3 Joint 4 ...... . ...... . .. ...... . . . . ........................ .. ....... .. . 36 III. COMPUTER IMPLEMENT .... ...................... .. . . . .. ......... . 40 3 . 1 Basic Matrix Form of a Bar ...... .. . . . .................. . ...... . ..... 40 3 . 2 Oganization of Equations . ...... . . . ..... .......... . . . . . . . . . . . . ........ 45 3.2.1 Joint 1 and Joint 5 .. .. . . , . . ....... . . . .. .. . . ...... . . ........ . ..... 45 3.2.2 Joints 2 , 3, and 4 ....... . ................................ .. ....... 45 3 . 2.3 Forces . . . .. . . . . ............... .. . . . .. . . . . . . . . . ...... .. .. . .......... . 46 3 . 2.4 B ending Moments and Rotation ............ .. .. . . .. . .. ..... . . . 46 3.2.5 Displacements ......... . . .. . . . . .. . . ................... . ....... . . . . . 47 3.3 The Linear Solutions of A 's . . ....... . .. .. . ......... . . . . ...... ...... . . 49 3 .3.1 The First Part of Linear Matrix A1 . . â€¢..â€¢.â€¢â€¢.â€¢...â€¢.â€¢â€¢..... . â€¢ . . . 49
PAGE 6
Vl 3.3.2 The S econd Part of Linear Matrix A2 ............. . . . ......... 50 3.3.3 The Third Part of Linear Matrix A3 . ........... . . . ..... ....... 53 3.3.4 The Nonlinear system . . . . . ........... . . .. ............. .. ........ . 54 3.4 The Solution of the Nonlinear Sy stem . . . ....... ....... . . ...... . . . .. 55 3.4.1 Initial Guess of K . . .. .. . . . .. . . .. .. . .. ......... . ....... .. . . ........ 55 3.4.2 Finite Element Method for PC Program FEMU CD .. . ..... 56 3.4.3 Computer Program ....... . . .. ..... .. . . . .. . ....................... 57 IV . RESU LT S ................... ..... . ................ .. . . .. . ........ . ...... .. 59 4.1 Conce n trated Load At Joint 3 .................... ......... . ......... 59 4.2 Co n ce n trated Loads A t Joint 2 , 3 and 4 ................. . ....... .. 64 4.3 Co n ce n trated Load A t Joint 3 and Hori zontal Lo a d A t Joint 2 and 4 .. ...... .. ..... ............... .. .. ......................... 68 4.4 Con ce n trated Lo a d s at J o in t 2 , 3 , and 4 , and Hori zontal Load s a t Joints 2 and 4 .......... . .. ......................................... 72 4.5 The Diff e r e n t Dire c t ion s Betw ee n FEM C D and t h e N o nl inear S ystem ..... .. ............ . . . . . ........ . ...... . ....... .. . .. . .. .. .. .. ..... 76 V . C ONCL S IO N ............................................................ 7 8 5.1 Summary .................... . ........... ........ . . . . . .......... . . ...... 78 5 . 2 Co n cl u s ion ....................................... .. ..................... 79 BIBLIOGRAPHY .... . .. ................... . . . . . ............... ....... ......... . .. 8 1 APPENDIX A. FEMUCD input data file .. ................. .. . ......................... . 8 3 B . FEMl C D o utpu t data fil e ............................... .. ........ ...... 8 5 C . C omputer Program for t h e Nonli n e a r S ys tem ................. ....... 93
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FIGURES Figure 1.1 The Rigidly Connected Gabled Frame ......... . . . .. ........... . 2 1.2 . (a) Uniform Loads on the Gabled Frame . ............... ............ 3 1.2. (b) Concentrated Loads on the Gabled Frame ...................... 3 1.3 . (a) The Load Conditions of the Eccentrically Loaded Column .................................. . .................. 9 1.3. (b) The Directions of Displac ements of the Eccentrically Loaded Column ........ . .. . ......... . ................. . . . . .. . . . ..... 9 1.4 The LoadRotation Behavior of Linear and Nonlinear 2 . 1 2.2 . (a) 2.2.(b) 2.3 2.4 2.5.(a) 2.5.(b ) 2.5.(c) 2.5.(d) 2.6. (a) 2.6. ( b ) 2.6.(c) 2.6.( d) 2.7.(a) 2.7.(b) 2.7.(c) System s ...... .. ............. ......................... ............ .. 13 Directions of Each Eleme nt of the Gabled Frame .... ...... . . 15 Member Loads and ign Convention ..... . . ....... . .. .......... 16 Free Body of D eformed Bar Element .. . .. . ..................... 16 The Uniform Loads at Joint Condition . ........... ..... ....... 27 Joint 1 and .Join t 5 are Pinned and No Trans lation . ...... . . 30 The Forc e Equilibrium Condition at Joint 2 .. ........... . . . . . . 31 The Moment Equilibriun Condition at Joint 2 ................ 31 The Rotation Equilibrium Condition at Joint 2 ............... 32 The Displacement Equilibrium Condition at Joint 2 ......... 32 The Forc e Equilibrium Condition at Joint 3 ................... 34 The Moment Equilibrium Condition at Joint 3 . .. ............ 34 The Rotation Equilibrium Condition at Joint 3 ............... 35 The Di splacement Equilibrium Condition at .Join t 3 ....... . . 35 The Force Equilibrium Condition at Joint 4 ................... 3 8 The Moment Equilibrium Condition at Joint 4 ........ . ...... 38 The Rotation Equilibrium Condition at Joint 4 ....... . . .. .... 39
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Z.7.(d) 4.1 4.Z 4.3.(a) 4 . 3 . (b) 4 . 3.(c ) 4.3.( d) 4 . 3.(e ) 4.4 4.5 4 . 6 .(a) 4 .6.(b) 4. 6 .(c) 4.6.( d ) 4 . 6 .(e) 4. 7 4 .8 4. 9. (a) Vlll The Displacement Equilibrium Condit ion a t Joint 4 ......... 39 C onc entrate d Load A t Joint 3 ....... .. . . .............. . .. .. . .. .. 60 LoadRot a t ion Behaviors with Different 8 ..................... 60 Comparison of Line a r and Nonl inear Solutions with 8=1 ............................... . ............................... 61 Comparison of Linear and Nonlinear Solutions with 8 = 10 ........................... .... . . . . . . .......... . ............. 61 Comparison of Linear and Nonlinear Solutions with 8 = zoo ...... ......... . ..... .............. .... ... ... ................ 62 Comparison of Linear and Nonlinear Solutions with 8 = 30 ............... . . ....................... .... . .......... . . .... 6Z Comparison of Linear and Nonlinear Solutions wi t h 8 = 40 . . . ...................................... .................... 63 C onc e n t rated Load s A t Joints Z , 3 , and 4 .................... . 64 Lo adRotation B e h av ior s with diff e r e n t 8 ........... . .......... 65 Comparison of Linear and Nonlinear Solutions with 8 = 1 ........ ..... ............... ................ . ................ .. 65 Comparison of Linear and No nlinear :::>olu t ion s with 8 = lQO ..................... . ......................... . ............. 66 Comparison of Linear and N onlinear Solutions w i t h () = zoo . . ......... . .... . ...... . . . ........ . . ...... . . . ................ 66 Comparison of Linear and No nl i near Solu t ion s wi t h 8 = 30 ..... . . . . . .......... . . . ........... ...... . . ............. ... ... ()7 Compariso n of Linear and N onlinear Solu t i o n s wi t h 8 = 4 0 ............................................................. ()7 Conc e n t rated Lo a d A t Joint Z and Horizontal Load s A t J o ints 2 and 4 ................. . . ....................... .. ..... ()8 Lo a d Rotation B e h av ior s w i t h d iff e r ent 8 ......... ............. 69 Comparison o f Linear and Nonlinear Solutions wi t h 8 = 1 ............................. . .............. .... ............. .. 69
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IX 4.9 . (b) Comparison of Linear and Nonlinear Solutions with () = 10 ............. .. . .. .............. .... . .. . .. . . . ................ 70 4.9 . (c) Comparison of Linear and Nonlinear olutions with () = 20 ............. . ........ . . .. ........ ....... . ................... 70 4.9.( d) Comparison of Linear and Nonlinear Solutions with () = 30 ..... .. . . . .. . .. ......................... . ...... . . ........ .. . . 71 4.9.(e) Comparison of Linear and Nonlinear Solutions with () = 40 ........... . . ........... .. ........................ .. . . ....... 71 4.10 Concentrated Loads A t Joints 2 , 3 , 4 and Horizontal Loads At Joints 2 and 4 ................................. ........ 72 4.11 LoadRotation B e h avio r s with different 8 ..................... 73 4 . 12. (a) Compari on of Linear and N onlinear Solutions with () = 1 .. ........................... .. . .. ............................. 73 4.12. (b) Comparison of Linear and Nonlinear olutions with () = lQO ............................................................. 74 4.12.(c) Comparison of Linear and Nonlinear Solutions with () = 20 ............................................................. 74 4.12.(d) Comparison of Linear and Nonlinear olution with () = 30 ........... .. ........... .. ..................... . . . . . . ........ 75 4 . 12.( e) Comparison of Linear and Nonlinear Solutions with () = 40 ................ . . .. . .............. . ..... .................... 75 4 .13 The Directions of the Element in FEMUC D .... . .............. 77
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CHAPTER I INTRO D U CTI 0 N 1.1 The Gabled Frame The gabled frame of singlespan, pinned base is widely used in structures covering large areas with no obstructions, such as industrial buildings, auditoriums, warehouses, etc.. Its analysis is of considerable interest to practicing engineers. Numerous investigations of rigid frames of various types have been reported in open literature. Consider a gabled frame composed of four straight, slender, and piecewise prismatic bars which are rigidly connected. [see Fig. 1.1 ] The four bars have the same length, L. The angle , 8, between element 1 and element 4 [see Fig. 1.2.(a) and 1.2.(b)] can be arbitrary values between 0 and iâ€¢ excluding 0 and i Joint 1 and joint 5 are pinned supports, which are concentrated to prevent translations. The properties of the materials used are E ;, A ; I;, which are the Young 's Modulus, CrossSection Area, and moment of inertia of ith bar, re s pec t ively. The external loads applie d to any ith member of this gabled frame include segments of a distributed t r ansve r se load q ;, concentrated trans verse force QV; and QH; in vertical and horizontal directions, respectiv e l y . In this gabled frame, the application of a uniform load will be limited to the case with a s ingle uniform load over eac h element and just two concentrated loads per bar with QV;(O) and QVi(1) or QH;(O) and QH;(1) applied at distance e0 , and eL, re spectively from the ends. [see Fig.2.2] The ratios,
PAGE 11
2 edL;, of ith are s ufficiently s mall so that the loading can be replaced with the static equilibrium loading of force QV; and QH; at the frame joint. Fig. 1.1 The Rigidl y Connected Gabled Frame
PAGE 12
3 Fig. 1.2.(a) niform Loads on the Gabled Frame Fig . 1.2.(b) The Concentrated Loads on the Gabled Frame
PAGE 13
4 1.2 Objective of Study A number of approaches have been successfully u sed in determining the behaviors of elastic structures. Some of the approaches are to find linear solutions and some of them use the function of X 1 to find the nonlinear solut ions. The objective of this study is to determine the behavior of the nonlinear loadrotation. The equilibrium equations of axial and lateral displacements are derived in reference 2 2 The method of analysis consists of s olving the boundary value problem associated with these typical equations. These equations of this system are d e riv e d from differential eq u a tions. Structure analysis based on the differen t ial equations has three easily recognized advantages.3 1. Distributed loads and elasticity can be d ealt wi t h direc t l y s o that the quality of a solution is not affected by the c hoice of elements and equivalent load u se d in a Finite Element Method. 1 Geo r ge D . Manolis, Dimitrios E . B es ko s, and Bruce J. Brand. , "Elastoplastic Analys i s and Design of Gabled Frame", .Journal of Com uters & Structures Vol. 22 , No.4 , pp. 693697 , 1986 2 Andreas S. Vlahinos, C .V. mith, JR. and G. J. Simitses, " A Oillinear So lu tio11 Sche m e Multistory", Multibay Pla n e Frames , of Computer s & t ruc tures Vol. 22 , No.6 , pp.10351045 , 1986 3 Riyad K. Qashu, A.M. ASCE and Dona ld A . Dadeppo, "Large D e flection and Stability of Rigid Frames", Journal of Engineering Mechanics, Vo l.109 , No.3 , June , 1983
PAGE 14
5 2. Nodes accuracy naturally at points of discontinuity in the differential equations and the member would be employed in a finite element analysis that can be expected to yield solutions of comparable accuracy. 3. Highly accurate results obtained by integration of differential equations provide a standard of comparison for the evaluation of approximate methods of analysis. The difficulties are finding the relative complexities of analysis and also the many ways to organize the computational procedures. One of the approaches of Finite Element Method applied to find the loadrotation behavior of this gabled frame is the PC program FE MUCD4 . The FEMUCD will b e used in determining the linear behavior of loadrotation. The differences between linear and nonlinear solutions will be compared in chapter four . l sually, the re ults of linear and nonlinear are much different , especially when the external load is close to the critical load. 4 This is a PC program for Finite Element Method derived by Prof essor John R. Mays in the Department of Civil Engineering of University of Colorado at Denve r.
PAGE 15
6 1.3 Basic Assumptions Since only a relatively small number of problems of mechanics can be solved by means of the exact field equations of the theory of elasticity, the exact solution is presented within the limitation of the number of simplifying assumptions in dealing with these problems. According to these assumptions, all of the structural elements fall in one of the following s ix categories. 1. Frame members are initially straight, piecewise prismatic with a cross section principal axis, and are rigidly connected. 2 . The plane frame is loaded by static concentrated or uniformly distributed loads only, with the loading plane in the plane of the fram. e. 3. All displacements are in the plane of the frame. 4. Material behavior i s linearly elastic, and the moduli of elasticity in ten s ion and compression are equal. Material yielding and e ffect s of residual stress are neglected. 5. Cross section s p erpendicular to t h e longi tudina l fib e r s before deformation r emain p erpendicula r to t ht' deformed lo g i tudina l fib e r s , and lin es i n th< ' cross section are inexten s ional (the u s u a l EulerBernoulli assumption which i s listed b e low). A). The material is homogenous and isotropic. B). Plane sections remain plane afte r b ending.
PAGE 16
7 C). The streestrain curve is identical in tension and compression. D) . No local type of instability will occur. E). The effect of transverse shear is negligible. F). No appreciable initial curvature exists. G). The loads and the bending moments act in a plane passiRg through a principal axis of inertia of the cross section. H). Hooke's law holds. I). The deflections are small as compared to the crosssectioal dimensions. 6. The axial strain E"zz is less than one, and the term u , z and (w,z)2 are of the order of the strain, where u , and w are axial and transverse displacements respectively for the points on the bar centrodial axis , and ( , x) denotes differentiation with respect to the axial coordinate x. [see Fig.2 . 2.(a)J. The geometric nonlinearity is limited to moderate rotations with 02 :: 1. [see Fig. 2.2.(b)]
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8 1.4 The Behavior of an Eccentrically Loaded Column In section 1.2 , the different results between lin ear and nonlinear systems were mentioned. Now , the sample case of an eccentrically loade d column will be discussed. Usually, only ideal columns are discussed. The ideal column is defined as the following . The ideal columns are initially straight and geometrically perfect and are subjected to the action of a com pressive force without eccentricity. In practice, no column i s geometrically perfect and the applied load does not necessarily pass through the centroid of the column cros s section. It i s t herefore neces sary to study the behavior of geometrically imperfect columns and of columns for which the load is eccentrically applied. Con s id e r the c ase of a s imply supported and eccentrically load e d column with the same eccentricity, e, a t both ends shown in Fig. 1.3.(a). The directions of axial deformat ion , U, and the lateral deformation, W , a r e shown in Fig. 1.3.(b). The properties of t his column are E , A, and I which are Young 's Modulus, Cros Section Area, and Moment of Inertia, respectively. The equilibrium equation of this cas e i s EIW,xxPW = 0 (14 1 ) mcE> t hi s column has no uniform load on it, t h e valu e in t h e right hand s ide of equation ( 141 ) is equal to zero. On the other hand, the valu e in the right s id e shouldn' t be e qual to zero if a uniform load is on this column. The boundary conditions are shown as t h e following . W(O) = 0.0 = 0. 0 { 1 4 2) (14 3)
PAGE 18
9 p Fig. 1.3.(a) The Load Conditions of the Eccentrically Loaded Column u Fig. 1.3.(b) The Directions of Displacements of the Eccentrically Loaded Column
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10 M(O) = P e (14 4} M(1) = Pe (14 5) The solution for this equation is (14 6) The terms W , x and W ,xx can be derived from equation (146) . ( I 4 7) (14 8) The relationship between M and W i s as follow. M = EIW,xx (14 9) The relations hip between the axial forc e and the constant K is the following . (14 1 0) Combine equations (146) , (148), ( 149) , and ( 1410). The equat ion s (142), ( 143 ). (144). and ( 145 ) can be modified r es p ective l y as t h e following. (l4 ll) ( 1 4 12) ( 1 4 13) M(l) = .41PLSIN(K ) + A2P LCOS(K) = P e (l4 14)
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11 In the above equations, A1 , A2 , A3 , A4 , and K are the constants for each bar. In equations (1413) and (1414) , load, P, can be canceled out by both sides of equations (1413) and (1414) shown as the following. (14 15) M(1) = A1LSIN(K) + A2LCOS(K) = e (14 16) Equations (1411), (1412), (1415), and (1416) can be modified as the matrix form shown as the following, ( 0 SIN(K) 1 COS(K) L LCOS(K) AX= B 0 1 0 0 (14 17) (1 4 18) in which A is the coefficient matrix, X is the unknown vector, and B is the constant vector. The solution of equation (1417) is X = .4 I B. ( J 4 19 ) After vector X is found, substituting these constants into equation ( 146) the lateral displacements can be found at every point in this bar. The rotations of every point of this bar can be found by substituting the vector X into the equation (147). The results are plotted in Figure 1.4 which shows as e from 50 . + 100. The plot represents the behavior of the linear and
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12 nonlinear systems. Note that the load P begins very small, and gradually mcreases. The corresponding critical load is shown in equation (1420). 5 (14 20) The linear solution found by FEMUCD program is shown in Fig. 1.4 for comparison with the nonlinear results. In this case, the linear solution is much different from those of nonlinear. The more eccentricity there i s , the more difference there is between linear and nonlinear solutions. On the other hand, if the eccentricity, e + 0, this column is more like an ideal column. 5 George J . S imitses , "An Introduction to the Elastic Stability of Structures", PrenticeHall, Inc. , Englewood C hiff. New Jersey, 1976
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13 e increase from 50 to 100 0 . 4 t"'4 C.j 0 > t:l 0.2 en 0.1 0.0 +............L,......l O.B 0. 5 0. 4 0. 2 0 . 0 0.2 0.4 0 . 6 O . B 1 . 0 1.2 W ,z(O) Fig. 1.4 The LoadRotation Behaviors of Linear and Nonlinear ystems
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CHAPTER II MATHEMATICAL FORMULATION This chapter discusses how to derive the differential equations from boundary conditions and equilibrium conditions. The axial forces, transverse shear, bending moments, and rotations will be considered at each joint. Twenty four differential equations will be found by this approach. Accord ing t o boundary conditions, six constants can easily be found by six of t he twenty four differential equations. The others will be solved by using computer implement. 2.1 Description of the Notation of t he Gabled Frame The gabled frame is shown in Fig. 21 in which {n} is the nth joint, and [m] is the mth element, QV(z) is the Zth joint of the vertical load, and is the uniform load on Kth e lement which is vertical to the axial line of t hi s element . The length, L , of each e lement is the same as those of the other elements . The angle , 8, i s between e lement 1 and eleme n t 2 , and also is between element 3 , and element 4 . They are the same as each other. A t this gabled frame joint 1 and joint 5 are pinned. Joints 2, 3 , and 4 are rigidly connected. The directions of axial and lateral displacements are u and w, respectively . The unde form e d fre e body of each element is shown in Fig. 2.2.(a) and Fig. 2 . 2.(b). Fig. 2.3 i s t he fr eebody of t h e undeformed joint . P denotes t he forces resultant in the undeformed axial force direction, V denotes the re sultant in the undeformed tran sve r se s hear direction. M denotes t he bending moment, i and j denote the ith, or jth element, 0 denotes the en d of a bar, and 1 denotes the other end of the bar.
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15 {2} {4} [ 1 ] [ 4) {1} Fig . 2.1 Directions of Each Element of t h e Gabled Frame
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16 Fig. 2.2.(a) Member Loads and Sign Convention x,u z,w Fig. 2.2.(b) Free Body of Deformed Bar Element
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17 2.2 Description of the Mathematical Formulations of Equations Consider each element of this gabled frame, the geometric relationship is given by equation (221 )6 fzz = + zk (2 2 1) where z units from the reference surface, is the extensional strain on the reference plane (average strain) and k is the change in curve.7 The straindisplacement relationship is the following. o I 2 t:.,., = u z + w"' ' . 2 ' (2 2 2) and k = w ,.,., (2 2 3). These equations are written on the deformation body with the restriction to the moderate rotations. [see Fig.2.2.(a)] ( N Sw.,) = 0 ' , z (22 4) (S +Nwz) = q ' , z {22 5) M ,., + S = 0 (2 2 6) m which N, S , and M are the internal axial force, transverse shear and bending moment, respectively , at any point in a bar. In equation (224), the term Sw ,., naturally appears from consideration of force equilibrium in the Xdirection for the free body in Fig.(2.2.(b)). However , it can be derived from virtual work consideration that this term is inconsistentwith 6 The same as footnote 2 7 The same as footnote 5
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18 the strain displacement relationship of equation (222). Therefore, in what follows , the term Sw,z will be ignored in any equilibrium considerations. It may be helpful to introduce the following definition. P = N Sw,z (2 2 7) V = S + Nw,z (2 2 8) In equation (227) , after the term Sw,z is ingorned, equation (227) can be modified as the following. (2 2 9) where Pis the force resultant in the undeformed axial direction and V denotes the resultant in the undeformed transverse direction. These definitions will be used to write equilibrium equations which have to be satisfied at each joint. The bar constitutive equations are well known N = (2 2 10) M =Elk ( 2 2 11) These formulations are completed with the requirements at the reaction point and the joints. ForceDisplacement Relationship The equations (2212) and (2213) can be derived from the combination of equations (222), (229), and (22 10) . P = N = (22 12)
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19 P = N = EA(u, z + (2 2 13) The equation (2214) can be derived from the combination of equations (2211), and (223). M = EIW,zz (22 14) Since (22 15) Combine equation (2214) , (2215) and deriveS as the following. S = Elw,zzz (22 16) From equation (228) , the V can be derived as the following. \' = EfW,zzz + Pw, z (2 2 17) Equilibrium Condition Since N is a constant, the equation (2218) can be derived by using equation (2213) shown in equation ( 22 18) . I ' w ) (I 2 , Z , z EA(u, z ( 2 2 18) Equation (2219) can be derived by u s ing equation (225). EfW,zz:u Pw,zz = q (2 2 19) To use the nondimens ional equations, t h e following symbol will be defined.
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L L W= '.E. L A _ L _ Ml q = EJ K2 _ , (PL') T EI 20 ( 2 220) In forcedisplacement relationship, using equations (2220) , (2213), (2214), and (2217), the above equations can be modified to the nondimensional equations as the following. (22 21) M = E!W,xx (2 2 22) (2 2 23) (22 24) In equilibrium conditions, equation ( 2218) and (2219 ) can be modified t o equation s (2225) and (222 6 ), re s p ective ly . (22 25) W K2W , X X . \ X ,.\ X q (2 2 26)
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21 2.3 The Formulation of Equation U {X) and W (X) These two following equations are used for compressive axial loads only . Equation U (X) represents the axial deformation, and equation W (X) represents lateral deformation for every point at a bar. â€¢ K2X 1 ['y 2 U(.\) = As)2 2 l o [ W ,e()] d (2 3 1) ( 2 3 2) These two following equations are used for tensile axial forces only. K2 X 1 {x 2 U(X) = As + )22 J o [ W ,e{)] d (23 3 ) (23 4) In the above equations, K , A1 , A2 A3, A4 , A s are the constants for eac h bar. If any frame has nth bar , this frame should have 6n unknowns in which Sn are A 's and n are k ' The compressive axial force will only b e considered in this section. After integration, the equation (231) can be developed as the following :
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U(X)= AsK>.'jf + _ ) ) lA2X g_: 2 3 6Kâ€¢ A1 A3SI N(K X) .4,q( COS(K X)+K:, S/N( K X ) I) [ .42A3(COS(KX)1)] + A2q( SIN( K X ) :;coS ( K X ) ) _.!_ A,qx' 2 K1 (235) 22 Also , W , .,(X), and W ,.,.,(X) can be derived from W(X) as the following. (23 6 ) (2 3 7 )
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23 2.4 Boundary Conditions â€¢ At joint 1 , and joint 5 , the boundary conditions will be considered. Becaus e joint 1 and joint 5 are pinned, they can't be moved in any direction and also can not subjected to any moments so that the bending moments, M1 (0), and M4(0), and the axial displacements, U1(0) of element 1 , U4(0) of element 4 respectively , and the lateral displacements, W 1 ( 0) , W 4 ( 0) of element 1 and 4 respectively , are equal to zero. The equations U1(0), W1(0), Mt(O), and U4(0), 11'4(0), M4(0), are shown as the following . At joint 1: U1 (o) = o . o ( 2 4 I ) \1'1 (0) = 0 . 0 ( 2 4 2) M1 (0) = 0 . 0 ( 2 4 3) At joint 5: ( 2 4 t) (24 5) ( 2 4 6)
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24 According to boundary conditions, use equations U(X), W(X), M = EIW,xx and substitute 0 into theses equations. The following equations can be de veloped. (2 4 7) (24 8) E;li Wi, zz(O) = E;li( Ai2Kl + :\) = 0 . 0 t (2 4 9) After these equations are found, the uniform loads of element 1 and element 4 will be discussed. Elements 1 and 4 have no uniform load on them. Apparently, the q1 and q2 are equal to zero and their nondimensional expressions q1 and q2 are equal to zero , too. In equation (2222), the properties of material are E , Young' s Modulus, and I , the moment of inertia shouldn't be equal to zero for any materials so that the equation (249) can be modified as equation (249.A) W ,xx(O) = (An2Kl) + :i2 t (2 4 9 . .4) In equation (229 . A) , K i is not equal to zero if any axial forc e appears in ith element. For element 1 , the equations (2410) , (2412), and (2214) are derived from equations (247) , ( 248),and ( 249 .A). For e lement 4 , the equations (2411 ) (2413) , and (24 15 ) a r e derived from equations (24 7 ) , (248), and (249.A). Ut (0) = At5 = 0.0 (24 10) (24 1 L)
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25 (24 12) ( 2 4 13) ( 2414) (24 15) From equations (2410) and (2411) , the values of A15 and A45 which are equal to zero are easily found. In equations (2414) and (2415), when any axial force appears, i .e . K 1 and K4 are not equal to zero , the valu es of A12 and A42 are equal to zero as equations (2416) and (2417) shown. Substituting equations (2416) and (2417) into equations (2412) and (2413), the values of A14 and A4 4 should be equal to zero a s equations (2418) and (2419 ) show. .t,2 = 0 (24 1 6 ) .4,4 = 0 (2 4 17) .4,4 = 0 (24 !H) .444 = 0 (24 19)
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26 2.5 Equilibrium Conditions Now, the relationship between axial forces transverse shear , bending moment, Xdirection displacement, andYdirection displacement will be considered in this section. According to equilibrium conditions at each joint, all of 2: FH, 2: F v, 2: D H, 2: Dv, 2: M , and 2: W , z for each element are described as the following. In Fig. 23 , the d11 and d12 are relatively very short s o that the values of q1 d12 and q2dL2 are also very small. Any force or displacement due to q; dL; are small enough to b e ignored. Similarly, for the other joints, the e ffect due to uniform load can be ignored when the equilibrium conditions are considered. When L FH or L Fv i s considered, the forces , axial forces and transverse shear, and the angle, B, are very important factors. At joint condition, the summation of forc es is equal to zero in the Xdirection and Ydirection. Concerning bending moments, the angle, B, i no t cons idered because the moments whose directions are positive in every element are only in the two ends of eac h joint excluding joints 1 and 5 of this system. imilar to bending moments, when the rotation s a r e discussed, the a n g l e, B , i s not considered, either. The directions of rotations depend on the transverse forces so that the directions of rotations a r e different in both ends of each joint of this frame. When the displacements are considered, the joint conditions should be different from those of force s, bending moments, and rotations because the displacement of each joint are not equal to zero; but the forces, bending moments, and rotations at eac h joint are equ a l to zero. ince the joint is rigid , at each joint the displacements at both of the ends in each direction should be the same as those of t h e other joint.
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27 Fig. 2.3 The Uniform loads at Joint Condition
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28 2.6 Formulation of Equations for Joint Conditions The solutions of joint conditions have already been found for joints 1 and 5. Now joints 2 , 3, and 4, will be discussed in this section. 2.6.1 Joint 2. At this joint, the angle , B, between the horizontal direction and the axial direction of the element 2 will be considered and it is a very important factor in this frame. P 1 ( 1) is the same direction as the axial direction of element 1. V 1 ( 1) is perpendicular to the axial direction of element 1. The bending moment, M1 ( 1), is the positive direction of element 1. P 1 ( 1 ), V 1 ( 1 ), M1 (1), and P 2 (0) , V 2 (0), M2 (0), are s hown in Fig. 2.5.(a). The P2(0), V2(0) , are related to the angle , B. P2(0}, and V 2(0} include two components, the horizontal and the vertical directions, so that the forces in the horizontal direction are QH(2) , V1(0), P 2 (0}COS(B) , and V 2 (0}SIN(B). In the vertical direction the forces are P1(1), QV(2), P 2 (0}SIN(B) , and V 2 (0)COS(B). The summation of these forces in each direction shoud be equal to zero, then the following equations can be found. LJoi NT2 Fn = 0 V1 (1) QH(2)V2(0)S I N(B) + P2(0)COS(8) = 0.0 (2 6 1 ) (2 6 2 ) [See Fig. 2.5. (a)] The summation of the bending moment in the two ends of joint 2 is equal to zero. The bending moments and rotat ion s do not include the angle , 8.
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29 LJOJNT2 M = 0 (2 6 3) [See Fig. 2.5.(b )] LJOINT2 W , x = 0 . 0 W1,x(l)W2,x(O) = o.o (26 4) [See Fig. 2.5.(c)] At joint 2, the displacements of element 1 and element 2 are the same each other in both of the Xdirection and Y direction. (26 5) LELEM ENTI D v = LELEM ENT2 Dv ( 2 6 6 ) [See Fig. 2.5 . (d)J 2.6.2 Joint 3. At this joint when the force equilibrium is considered, the angle , 8, should also be considered. In the Xdirection, the forces include P 2(l)COS(e), V2{l)SIN(e) , QH(3), V3(l)SIN(8), P3(l)SIN(e). In the Ydirection , the forces include P2(l)SIN(8) , V2(l)COS(e), QV(3) , V3(l)COS(e), P3(l)SIN(e) s o that equations (267) and (268) will be written as the following .
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30 Joint 1 Joint 4 Fig. 2.4 Joints 1 and Joint 5 are Pinned and No Translation
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31 Fig. 2.5 . (a) The Force Equilibrium Condition at Joint 2 Fig. 2.5.(b) The Moment Equilibriun Condition at Joint 2
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, , I I Fig. 2.5.(c) The Rotation Equilibrium Condition at Joint 2 Fig. 2.5.( d) The Displacement Equilibrium Condition at Joint 2 32
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33 "L.JoJNTJ FH = 0 P2(l)COS(8) + V2(l)SIN(8)QH(3) + V3(1)SIN(8) + P3(t)COS(8) = 0 (26 7) "L.JoJNTJ Fv = 0 P2(l)SIN(8)V 2(l)COS(8)QV(3) + V3(l)COS(8)P3(l)SIN(8) = 0 . 0 (26 8) [ ee Fig. 2 .6.(a)] The bending moments and rotations do not include the angle, 8, but the directions of the moments and the rotations will be considered. The equations of bending moments and rotations can be written as the following . L J OINT2 M = 0 (2 6 9) [ eeFig. 2 .6.(b)] L J OINT3 W,x = 0 (26 1 0 ) [ ee Fig. 2 . 6.(c)] The di splacement of elements 2 and 3 a r e the same as each other in Xdirection and Ydirection at each joint. The equation (2611) is in the Xdirection. Th equation ( 2612) i s in theYdirection. (26 11)
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34 Fig. 2.6 . (a) The Force Equilibrium Condition at Joint 3 Fig. 2.6.(b) The Moment Equilibrium Condition at Joint 3
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35 Fig . 2 . 6.(c) The Rotation Equilibrium Condition at Joint 3 W2(l) U3(1) ?U2(l) t W3(1) Fig. 2.6.(d) The Displacement Equilibrium Condition at Joint 3
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36 LELEM ENT2 Dv = LELEM ENT3 Dv W2(1)COS(O) + U2(1)SIN(O) = U3(l)SIN(O) W3(1)COS(O) (2 6 12) [ eeFig. 2.6.(d)] 2.6 . 3 Joint 4 At this joint, the forc es, bending moments, rotations, and displacements are similar to those at joint 2. The force s in the Xdirection are V4(1} QH(4), V3( 0)SIN(o) , and P3( 0)COS(o). The forces in t h eYdirection are P4(1), QV(4), V3(0)CO (o), and P3SIN(o) . The equations of the forc es are shown in equations (2613) and (2614). LJOINT4 Fn = 0 V4(1) + QH(4) \1 3(0)5/N(O) P3(0)COS(O) = 0.0 (26 13) P4(1)Q\'(4)\ '3(0)COS(O) + P :1(0)SIN(O) = 0.0 (26 14) [.'ee Fig. 2.i.(a)] The equations of bending moments and rotation s a re shown in equation s (2615 } and (2616 ), respectively. L J O INT4 M = 0 M4(l) + . M3(0) = 0.0 (2 6 l !l) [ ee Fig. 2.7 .(b)]
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37 L J OTNT4 w,., = 0 w3,x (O) w4, x ( 1) = o.o (2 6 16) [See Fig. 2.7.(c)] The equations of displacements in the Xdirection and the Y direction are shown in equations (2617) and (2618), respectively. LELEMENT3 DH = LELEMENT4 DH (26 17) [See Fig . 2.7 .(c)] LELEM ENT3 Dv = LELE/11 ENT4 Dv (26 18) [See Fig. 2 . 7.(d)] After these equations are solved, the solution s of K 's and As can be solved by the algebra of these equations.
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38 Fig. 2.7 . (a) The Force Equilibrium Condition at Joint 4 Fig. 2.7.(b) The Moment Equilibrium Condition at Joint 4
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I I Fig. 2 . 7.(c) The Rotation Equilibrium Condition at Joint 4 Fig. 2. 7 .( b) The Displacement Equilibrium Condition at Joint 4 39
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CHAPTER III COMPUTER IMPLEMENT Use the computer implement to find the eight constants of this gabled frame by solving the eighteen equations. All of them include linear and nonlinear systems. To solve them, all of the differential equations will be formed into a matrix for the linear system. The others will be found by using the IMSL subroutines. The computer implement will be discussed in this chapter. 3 . 1 Basic Matrix Form of a Bar The solution of this gabled frame is a nonlinear solution. The ieh bar of this frame has six unknowns, K;, A;1 , A;2 , A;3 , A;4 , and A;s, which can be defined if K; is known. K; depends on axial force , P;, because if P is compressive axial forces , if P is tens ile axial force , K2 . P;L[ = =Ft E;l; in equation (311), (312) , and (313), K; is always positive. ( 3 1 1) ( 3 1 2) ( 3 1 3) To obtain the relationship between K ' s and A ' s , the values of K are assumed to be known. According to Equations (232) , (235) , {236) , {237) , {311), and
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41 subtituting 0, and 1 into these equations, respectively, the following equations can be derived from this procedure. (3 1 4) ( 3 1 5) (3 1 6) W;,,(O) = .4;t K ; + A3 ( 3 1 7 ) ( 3 1 8) K2EJ. V(O) = 4 ' ' ' l . 3 ( 3 1 9) \'( 1 ) = A K'f E;l; _ ij;E;l; l 3 l l (3 1 10) (3 1 11) ( ) K2SIN( K;)E;l ; KlCOS(K;)E;l; M i 1 = A I .4 2 '::"''; L ; (3 1 13) in which the axial force of ith bar, P ;, is defined as axial compresssive force. After equation (314) to equation (3113) are found, these equations can b e formed into a matrix, B::. AX (31 14) , t o be used in computer operation. In equation (3114) , B is the constant vector, A i s the coefficient matrix, X is the unknown vector. Combine equations (314 ), (315), (316), (317), (318), (319) , (3110) , (3111) ' ( 3112) , ( 3113). These equations can be expressed in the matrix form as the following.
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4 2 0 0 0 0 1 0 U;(O) 0 1 0 1 0 0 W;(O) S IN(K) COS(K ) 1 1 0 _j_ W;(1) 2K1 A;1 K 0 1 0 0 0 W;,z(O) KCOS(K ) KSIN(K) 1 0 0 _j_ A;2 W (1) K1 A ; 3 t,z = 0 0 K 1El 0 0 0 Vi(O) __1_ A;4 Vi(1) 0 0 K 1El 0 0 qEI A;s vM;(O) 0 K 1 E l 0 0 0 Elij 1 LLK1 M;(1) K 1SIN(K)EI K 1COS(K)El 0 0 0 qEI P ; L L K 1 L 0 0 0 0 0 K 1 E l v( 3 1 15) The c o effic ien t mat rix, A , i s t h e follo w in g . 0 0 0 0 1 0 0 1 0 1 0 0 S f N(K) C OS(K) 0 _j_ 2K1 K 0 0 0 0 KCOS(K) KSIN(K) 0 0 ...i... K' A= 0 0 K1 EI 0 0 0 (3116) L , 0 0 K1 El 0 0 ijEl L , 0 K1El 0 0 0 E lij LL K1 K1SIN(K)EI K1COS(K)EI 0 0 0 qEJ L L K1L 0 0 0 0 0 K1El L,The unknown vector, X , is t h e following. X = (3 1 17 )
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The constant vector, B, is the following. B = U;(O) W;(O) W;(l) W;,z(O) W;,z(l) V;(O) V;(l) M;(O) M;(l) P; 43 (3 1 18)
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44 Equat ion { 3115) can be modified as t h e following : 0 0 0 0 1 0 U;(O) 0 1 0 0 0 W;(O) SI N(K;) COS(K;) 1 0 W;(1) A;! W;,z(O) K; 0 1 0 0 0 A;2 K;COS(K;) K;SIN(K;) 1 0 0 _jJ_ W;,z(1) K? A;J = 0 0 P; 0 0 0 A;4 V;(l) 0 0 P; 0 0 q;E;l; A;s M;(O) 0 P;L; 0 0 0 Elq; 1 M;(1) P;L;Sl N(K;) P;L;COS(K;) 0 0 0 q;E,l; P; K?L, 0 0 0 0 0 P; OR U;(O) 0 0 0 0 1 0 0 1 0 0 0 W;(O) SI N(K;) COS(K;) 1 1 0 __.9i__ W;(l) 2Kl A .;J W (0 ) K; 0 1 0 0 0 A;2 t,z K;COS(K;) K;SIN(K;) 0 0 !A w (1) A;J t,Z = K ; F;(O) 0 0 P; 0 0 0 A;4 V; ( 1) 0 0 P ; 0 0 q;S; .4;s L , M;(O) 0 P ;L; 0 0 0 1 K, l\.1;(1) P,L;Sl N(K;) P ;L;COS(K;) 0 0 0 9.2i P ; K ? 0 0 0 0 0 P; In which S; =
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45 3 . 2 Oganization of Equations This gabled frame has four elements. Six unknowns appear in each element so that this system has twenty four unknowns. Twenty four equations are required to solve the system if the solution can be solved . According to section 2 . 5, the twenty four equations are already found. How to organize these equations is a very important step because every selection will find the solution. The following discussion will follow joint by joint and element by element, the forces , bending moments, and rotations to find the relationships between the twenty four equations. 3 . 2 . 1 Joint 1 and Joint 5 According to boundary conditions, the constants, A12 , A14 , A15 of element 1 , and A42, A44, A45 of element 4 , are found. They are equal to zero and are not changed even if the K; of i0 , element is changed. These six constants of the system solutions are a part of the twenty. four unknowns. If the other solutions can be solve d , only eighteen equations are required, because eighteen constants are still unknown. 3.2.2 .Joints 2, 3, and 4 S ix equations can be derived by using joint equilibrium conditions at each joint. The total are eighteen equations available in this frame. At joint 2, equation ( 261) is found F H = 0 , equation ( 262) is found F1 . = 0 equation (263) i found M = 0.0, equation (264 ) is found by = 0.0 , equation (265) i s found by and equation (266) is found by S imilarily, equation (267) is found by H = 0 . 0 , equation (268) i s found by I:F 1 = 0.0 , equation (26 9 ) is found by = 0.0, equation
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46 (2610) is found by Y:W, z = 0 . 0 , equation (2611) is found by Y:DH, and equation (2612) is found by Y:Dv at joint 3 . Equation (2613) is found by Y:F H = 0.0 , equation (2614) is found by Y:F v = 0.0 , equation (2614) is found by Y:M = 0.0 , equation (2615) is found by Y:W, z = 0.0 , equation (2616) is found by Y:DH, and equation (2617) is found by Y:Dv at joint 4. 3.2.3 Forces Equations (261) , (262) , (267) , (268) , (2613) , and (2614) are derived from the force equilibrium. According to (3120) these equations include four unknowns, A1 3 , A23, A33, and A4 3 â€¢ If these s ix equation are used to solve the four unknowns, the matrix should be liriear dependent so that four of these equations used to solve A13, A23, A33, and A43, are good enough. The other two of these six equations should be put in the nonlinear system to solve the nonlinear solutions of K ' s. Apparently, this s election i s not the only one that will work. If any four of these equations are selected , one can' t be sure that they are linear independent because their limitations haven' t been considered, yet. The limitation i s that at least one of the equilibrium joint conditions should b e se lected in this algebra and the other one can be arbitrarily selected at joints 2 , 3 , or 4, because at joint 2, A13, and A23 are only included, at joint 3 , A23, and A33 are only include d , at joint 4 , and A43 are only include d . If any one joint is not include d, one of A,:1 will not b e fou n d and the o t h e r t wo o r more equations will be linear d e p ende n t . 3 . 2.4 Bending Moments and Rotations Equations ( 263) , (269) and (2615) are derived by the bending moment equilibrium conditions at joints 2 , 3 , and 4 r espectively . These
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47 three equations only include the constants A1 and A2 of each element. Equations (264) , (2610) and (2616) are derived by the rotation equilibrium conditions at joints 2 , 3 , and 4, respectively. Although these three equations include the constants A 1 , A2 , and A 3 of each element, A 3 was found by force equations. In this algebra, the A; 3 will be considered as known. In these six equations , A11, A21 , A22, A31, A32, and A41 are still unknown. Six equations are available in the bending moments and rotations. Fortunately, six unknowns can be found by solving these six equations. This selection is the only one that works beacuse six unknowns and s ix equation are available. 3 . 2.5 Displacements Equations ( 265) , (2611) and ( 2 617) are found by H at joint 2 , 3 , and 4 , respectively. Equations (266), (2612) and (2618 ) are found by at joint 2 , 3 , and 4 , respectively . At joint 2 , the displacement of element 1 in the horizontal direction should b e equal t o that of element 2 in the sam e direction [see equation 321 ] and t h e displacement of element 1 in the vertical direction should be the sam e as that of element 2 , [see equation 322] . Similary, at joint 3 , the di splaceme nts of element 2 should be equal to those of element 3 in t h e horizontal and vertical directions . [see equation ( 3 23 ) and (324) of joint 3 in the horizontal and ve r t ical direction .] A t joint 4 , the displac e ments of e l ement 3 should be t h e same as those of e lement 4 in both directions. [see equations (32 5) , ( 326) of joint 4 in the horizontal and vertical directions.] Although those equations include all of the unknowns of A's, these excluding A24, A25, A34, and A 3 5 are found in forces and b ending moment equilibrium conditions. They should be cons id e red as known. Accord
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48 mg to these procedures, these six equations include only four unknowns. Similar to those of force conditions, the other equations will be put in the nonlinear system. The limitations should be considered to avoid being linear dependent. The limitation is the same as that of force conditions. A t leas t one of the equations of each joint should be put into the algebra so that the selection of this procedure is not the only one that works, either. I: DH = I: DH (32 1) ELEMENT! ELE/11 ENT2 I: Dv = I: D v ( 3 2 2) ELEMENT! E LEMENT2 I: DH = I: DH (3 2 3) ELEM ENT2 ELE111ENT3 I: Dv = I: D v ( 3 2 4) ELEM ENT2 ELEMENT3 I: DH = I: DH (32 f1) ELEMENT3 ELEJ\1 ENT4 I: Dv = I: D v ( 3 2 6) E L EMENT3 ELE/11 ENT4 In t hi s section , two eq uation s a r e availabl e in eac h joint,with a total of six equat ion s. Only these four constants, A24, A25, A34, A35, a r e still unknown. The procedure to s olve these unknowns i s similar to that of force s . Four of t h ese six equations will be se lected to t h e lin ear system and the other t wo equation s will be put in the nonlinear system. U nfor tunately, t his choice a l so has limitations. Although this choice i s not the onl y one that wo rk s . the limitations which sometimes cau se linear dependent have to be cons id e r ed to avoid to this ill condition.
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49 3.3 The Linear Solutions of A's If the constant K of every element of this frame which are defined in equation (311) is known, all of the solutions of A's will be found because the A's depend on K. The solution of K 's will be discussed in section 3.3.4. In sections 3.3.1, 3.3.2, and 3.3.3 , K of every element will be assumed as known to find the solution of A ' s related to K. 3.3.1 The First Part of Linear Matrix A1 This matrix based on equations ( 261), ( 267), ( 268), and ( 2613) which are all of the force equilibrium conditions. The other equations, (262) and (2614), are put in the nonlinear system. Although this choice is not the only one that works, the solutions of Aia of ith element will be found by this choice and the solutions should be the same as those of the other choices . These equations are formed as A1X1 = B1 in which the matrix A1 is the coefficient matrix, X1 is the unknown vector, and B1 is the constant vector. The first coefficient matrix A1 is as the following. P2SIN(8) P2COS(8) P2SI N(8) 0 0 P3COS(8) P3Sl N ( 8) P3SIN(8) The first constant vector 81 is as the following: (33 1 )
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50 ( + QH2 P2COS(8) ) L, P2SIN(8) + QV(3) + P3SIN( 8 ) + B , â€¢ 1 P 2COS(8) + + QH3 + SIN(O)P3COS(8) QH4 + PaCOS(O) ( 3 3 2 ) in which til = q4 = 0 , and P ;(O) = P ;( l ) = P; because the axial force is constant in the member. Substituting t/1 = ij4 = 0 and P ;(O) = P ;(l) = P ; into equation (332) , the first constant v ector B 1 can be modified a s t h e following : B = P 2S l N(O ) Q V( 3 ) + P3S J N(O ) + ( QH2 P2COS(8 ) ) 1 P2COS( 8 ) + Sf N ( O ) + QH3 + 51 N(O)PaCOS(O ) QH4 + P3COS(8) ( 3 3 3 ) The firs t unknown vector, X1 , is as the following : (A(l ,3)) \" _ A(2, 3) â€¢ 1 A(3 , 3) A(4 , 3 ) ( 3 3 t) 3.3 . 2 The Second Part of Lin ear Matrix A 2 This m a trix u ses t h e r esults d e ri ve d b y equation ( 331 ) . The s o lu tions of equation ( 331 ) a r e A13, A23, A 3 3 , and A43 which are cons id e r e d
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51 as a known in this section. This matrix i s based on equations (263), (264) , (269), (2610), (26lS),and (2616) which are derived from bending moments or rotations. This choice is the only one that works from the first part of linear matrix A1 because this matrix includes six equations and has six unknowns to solve. (33 5) in which the A2 is the coefficient matrix. X2 is the unknown vector. The B2 is the constant vector in which all of the variables are known and the results of equation (331) are included. (335) is the following. The matrix form of equation The second coefficient matrix is t he following. .42 = in which PLtSKt 0 0 PL2SK2 0 0 K1CK1 K2 0 K 2CK2 0 0 PLi =PiLi CKi = COS(K i) SKi = SIN(Ki) PL2 0 PL2CK2 PL3SK3 0 0 0 0 K2SK2 K3CK3 0 K3 0 0 PL3CK3 0 PL3 PL4SK4 0 0 K3SK3 0 0 K4CK4 (3 3 6) and Pi(O) = Pi(l) =Pi because the axial for ce is constant in the member. The second constant vector B2 i s as t he following: _ q,s, _ q,s, K ; K i _ q,s , _ q.s, Ki K; .4(1, 3)+ A(2, 3) I (3 3 7) .4(2 , 3)fl, + .4(3 , 3) + !!> 2 3 .4(3 , 3) + A(4 , 3) + fl, â€¢
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in which A 1 3 , A23, A33, A 4 3 are known. Since l _q,s, _ q,s, K : q,s, K > l A(1, 3) + .4(2,3) .4( 2 , 3 ) .;;. + .4(3 , 3 ) +.;;. 2 l .4(3 , 3 ) + .4( 4 , 3 ) The s eco nd unknown vector is X 2 a s t h e following: X 2 = A(1 , 1 ) .4(2 , 1 ) .4 (2, 2 ) .4(3 , 1 ) .4(3 , 2) .4(4 , 1 ) 52 0 , the constant ( 3 3 8 ) ( 3 3 9)
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53 3.3.3 The Third Part of Linear Matrix Aa This matrix uses the results of the first part of the linear matrix and the second part of the linear matrix. All of the A ' s excluding A24 , A25 , A34 and A35 are considered as known because they were derived in previous sections. This matrix is based on equations (265), (266) , (2612), and (2617). The other two equations, equations (2611) and (2618) , developed from displacements will be put in the nonlinear system. Use these equations to form the matrix as equation (3310) , (3 3 10). Although this selection is not the only one that works, this solution of equation (3310) should be the same as those of the other choices. The matrix form of equation (3310) is as the following . The third coefficient matrix, A3 , is as the following. _ COS(O) ( SIN(O) .43 co:(o) COS(O) 51 N(O) 51 N(O) 0 0 0 COS(O) SIN(O) The third constant vector B3 is as the following: (3 3 11) _ ul (1) .4(2 , 2)COS(O) ( A.(2,2)5JN(0) 11'1( 1 ) ) B3 l12( l)COS(O) U2(l)SIN(O) + U3( l)SIN(O) l V3(l)C05(0) ( 3 3 12) W4(l) A( 3,2)SIN(O)
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54 In the third constant vector B3 the U; and W; haven't A ;4 , A ;5 because those constants are zero while calculating. The third unknown vector is as the following: ( A{2,4)) X _ A{2,5) JA{3,4) A{3,5) {3 3 13) 3 . 3.4 The Nonlinear system The nonlinear system is based on equations (262), (2611) , (2614) , and (2618) which are described in sections 3.3.1 and 3.3.3. These equations use the IMSL to find the solutions. Because the constant Ks of each element have to be as sumed as known. All of A 's can be found in previous procedures. This nonlinear equation set is not the only ones that work, either. S inc e the first part of the linear matrix A1 and the third part of the linear matrix A3 are not the only one that works, this problem set should be changed if matrix A1 or matrix A3 is changed. Although this selection is not the only one that works and the solution of K 's and A 's are different , the r elationship is thf' same as the other se lection , and the final r esults should be the same as each othe r se l ection.
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55 3.4 The Solution of the Nonlinear System In this section, the nonlinear solutions will be discussed. Since all of the solutons of A ' s depend on the constant K's of each element. The solutions of A 's can be solved with the linear system. In the previous sections, the K of each element is 3assumedas known. Now , how to find K i s the object of this section. 3.4.1 Initial Guess of K S ince the relationship between the axial force of the element and t h e constant K of the same element was discussed in equations (311) , (312) , and (313) , K can be found if the axial forc e P is known. Use those K s a s the initial guess in the IMSL program to find the nonlinear solutions. If the external loads are very small, the results of the linear system should be very close to the nonlinear solutions. Use this concept with the finite element method t o find the linear solutions . Afte r doing that, the axial force s of the linear system can be found. Us ing equation (312) , find out the initial guess of K. ( 3 4 1 ) in equation (341) , P i i s compressive axial force and Ki i s positive. When the first intial guess are found, the system can b e s olv e d in t h e first step and t h e first nonlinear so lu t ion s is found . Usc t h e ratio of load increment to find the next initial guess . This r elation is s hown as the following. ( 3 4 2) Equation (342) can al s o b e modifi e d as the following. ( 3 4 3)
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56 in which the p m + 1 is the initial guess of the ( m + 1) th step. Km i s the nonlinear solution of the m1 h step. According to this procedure, the initial gue ss should be related to load conditions. This way is better than if the initial guess is derived from linear system for every step. Especially if the external load approaches the critical load. In each step, the initial guess should be known before calculating the solution of A ' s. 3.4 . 2 Finite Element Method for PC Program FEMUCD FEMUCD is a Finite Element PC program. The FEMUCD program includes many subprograms. It can be used to find the displacements and forces of frame or plate. The FEMUCD uses free f9rmat for inputing data but the data must be input in a specific order required by the program. The input data file includes the TITLE, SYSTEM, JOINT, RESTRAINTS , FRAME, ASOLID , and LOADS data blocks. They are presented a s the following. 1). The TITLE Data Block The fir s t line of e very FEMUCD data file consist of up to 80 characte r s of information used to label each page of output. 2). The S Y STEM Dat a Blo c k The S Y STEM d ata blo c k giv es th<' b as i c information r e quir e d to desc rib e t h e struc tura l mode l to FEMUC D . 3). The .JOINT S Data Block The .JOINT data block b e gins to de s crib e t he geometry of the sy stem b y locating in global coordinates . 4). The RESTRAINTS Data Block
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57 The RESTRAINTS data block constrains the basic information on the fixity of the nodes. 5). The FRAME Data Block This data section defines the material properties, connectivity, and orientation of the threedimensional beam elements of the model. Twodimensional truss elements and threedimensional truss elements are special cases of this general form. 6) . The ASOLID Data Block This data section defines the material properties connectivity, and orientat ion of the asolid element of the mode l axisymetric. Plane stress, plane strain and 3D plane stress are differnent cases for this element. 7). The LOAD S Data Block This data block defines the loading systems for the structures. The FEMU CD program include s t he subprograms, FEMUCD , FRAME, OPT, OLVEl, S OLVE2 , FRMFRC, A S OLID whos e functions are described in the class notes of Applied Finite Elemen t Techniques for P e r sonal Co mputer by Profe ssor John R. Mays, January, 1987. All r esults of FEMUCD are linear. In eac h case of t hi s system, the first initial gues s i s from FEMUCD but the load s are very s mall. The input and output data files of t h e section 4.4 at() equal to 10 a r c in Appendix A and Appendix B r espective ly. 3.4.3 Computer Program The computer program u ses the IM S L library to s olv e matrix algebra and the nonlinear system. Because a ll of t he A 's solutions depend on t h e constant K of each element, the so lu t ion of A s have to b e found afte r
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58 the constant K i s found by the nonlinear system. Based on this concept , when the constant of K is defined in each step of the nonlinear procedure, the solutions of A 's will be defined , too. This procedure will continue until the nonlinear solution, K 's, are satisfied by the requirements of this subroutine of IMSL. After K is found from the nonlinear procedures, the solutions of A 's should be found. All of the solutions , displacements , rotations and forces are found at this load by using the constants A 's and K and substituting them into the formula of rotations, displacements , axial forces , transverse shear forces , and the bending moments at each point of any element . se these K 's to find the next initial guess for the next step. The solutions at every load will be found. The computer program i s in Appendix C.
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CHAPTER IV RESULTS Gabled frames subjected to various load conditions behave differently. This chapter will discuss many different load conditions. Even if the frame is subjected to one load condition, the behaviors of linear and nonlinear systems are different. The different results between linear and nonlinear systems will also be discussed. 4.1 The Concentrated Loads At Joint 3 The first load condition , the gabled frame is subjected to a concentrated load only at joint 3. This concentrated load begins very small , and gradually increases. [See Fig. 4.1]. The Fig. 4 . 2 represents the loadrotation behavior of element 1 at the end of joint 2. The behaviors are different when the angle B is changed. The different . behaviors with different angles are also shown in Fig . 4.2. The difference between the solutions of linear and nonlinear systems is shown in Fig. 4.3(a+ e).
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t1 0 > I:' .). U 2 . 5 2 . 0 ! I I 1.5 1 1 . 0 j 0.5 Fig. 4.1 Concentrated Load At Joint 3 (}are 1, 10 , 20 , 30 and 40. ! I I I : 0.0 42 0 2 4 5 8 1 0 12 14 '5 Fig. 4.2 LoadRotation Behaviors with Different 8 60
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61 I 1 . 2 1 linear solution I 1.0 nonlinear solution 0 . 8 0 > d 0 . 6 0 . 4 0.2 0 . 0 . 2 0 2 6 8 10 Fig. 4.3.(a) Comparison of Linear and Nonlinear Solutions with J = 10 .3 2 / nonlinear solution / 0 4 2 0 2 4 5 8 1 0 1 ? Fig. 4.3 . (b) Comparison of Linear and Non linear Solutions with J = 100
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62 3.0 linear solution 2 . 5 i 2 . 0 t' 1.5 0 nonlinear solution > t:1 1 . 0 . OS j 0 .0 4 2 0 2 4 6 8 1 0 12 Fig. 4.3.(c) Comparison of Linear Nonlinear Solutions with () = 200 3 . 0 2. 5 linear solution l l 2 . 0 t' 1.5 0 > nonlinear solution t:1 1 . 0 0 . 5 0 . 0 4 2 0 2 6 8 1 0 12 Fig. 4.3.(d) Comparison of Linear and Nonlinear Solutions with()= 300
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63 .3. 0 linear solution 25 2 . 0 t"" 0 > 1 . 5 t::::' nonlinear solution 1.0 0.5 0 . 0 4 2 0 2 4 6 8 1 0 1 2 Fig. 4.3.(e) Comparison of Linear and Nonlinear Solutions with() = 400
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64 4.2 Concentrated Loads At Joint 2 , 3 and 4 In the second concentrated load condition , joints 2 and 4 are subjected to a vertically concentrated load which is not changed. The load at joint 3 is changeable. [see Fig.4.4] This load begins very small and gradually increases. The difference between section 4 . 1 and this section is that the load conditions of section 4.1 doesn' t have any concentrated load at joints 2 and 4. The loadrotation behaviors with different angles , B, are different as shown in Fig. 4.5 . The different behaviors between linear and nonlinear systems which are subjected to the same load conditions as each other are s hown in Fig. 4.6.(a>e) . Fig. 4.4 Concentrated Loads At Joints 2 , 3 , and 4
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0.5 8 are 1 , 10 , 20 , 30 and 40. () = 40 t"'4 0 > 0 . 4 0 . 3 d 0.2 _______ () = 30 0 = 20 () = Fig. 4.5 LoadRotation Behaviors with different 0 0 .!:> 0 . 4 t"'4 0 . 3 0 > d J.2 0.1 I I 1 linear solution nonlinear solution 1/ I 0 4 5 5 65 7 Fig. 4.6.(a) Comparison of Linear and Nonlinear Solutions with()= 10
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0.5 ,r.;, 0 .40 . 3 t""! 0 > d 0 . 2 0.1 1 0 linear solution nonlinear solut ion 2 4 5 5 66 Fig . 4.6 . (b ) Comparison of Lin ear and Nonlinear Solutions wi t h e = 100 0 . 5 ,....,,..., 0 . 4 t"" 0.3 0 > d 0 2 0 . 1 1 0 lin ear s olu t ion nonlinear solution I 4 5 5 Fig. 4 . 6 .(c) Comparison of Lin ear and Nonlinear Solutions wi t h 0 = 200
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0 . 5 1 linear solution 0 .40 . 3 0 > t:l 0.2 0.1 nonlinear solution 1 0 2 4 5 5 67 Fig. 4 . 6. (d) Compari on of Linear and Nonlinear 'olution with" = 300 0.5 .,.rr: 0 . 4 0.3 0 > t:l 0.2 0 . 1 1 0 linear solution nonlinear solution 2 4 5 5 W l ,z(l) Fig. 4.6.(e) Comparison of Linear and Nonlinear Solutions with 8 = .toO
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4.3 Concentrated Load At Joint 3 and Horizontal Loads At Joint 2 and 4 68 The third load condition is the first load condition with two hor izontal loads at joints 2 and 4. [see Fig. 4 .7] The two horizontal loads are not changed. Similar to the first load condition , the load at joint 3 be gins very small and gradually increase s . The behavior of this load condition changes when angle , 8 , changes . The linear and nonlinear s olutions are s hown in Fig. 4.9(a___.e). Fig. 4.7 Concentrated Load At Joint 2 and Horizontal Loads At Joints 2 and 4
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1 . 4 8 are 1 , 10 . 20 , 30 and 40u. 1.2 1.0 t"" 0 0.8 > I::' 0 . 5 0 . 4 0 . 2 :. o 6 4 2 6 8 1(. Fig. 4 .8 LoadRotat ion Behaviors with diff erent 8 \ \ . 2 1.0 """ : . 8 '' > v ). 5 2. 5 lin ear so lu t ion nonlinear solution 0 . 0 2 . 5 5 . 0 75 a. o 1 2 . s 15.o 1 7 . 5 zo:J w 1 â€¢ .<( 1 ) 69 Fig. 4.9. ( a) Comparison of Linear and Nonlinear Solutions with 8 = 10
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1. 4 . 1. 2 1. 0 0 . 8 0 > t:' 0. 5 0 . 4 0 . 2 linear solution nonlinear solution 0 . 0 0 . 0 2 . 5 5 . 0 7 . 5 1 0 . 0 1 2 . 5 1 5 . 0 1 7 . 5 2 0.C W t.z ( 1 ) 70 F i g. 4. 9 .(b) Comparison of Linear and Nonlin ear S olutions with B 100 1 . 4 ,:,,1.2 .0 0 . 8 0 > t:' Q . 5 0 . 4 â€¢ l i near s olu t ion nonlinear solut ion F ig . 4 .9 . ( c ) Comparison of Linear and Nonli near ''elutions with B = 2 0 0
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1.2 t. O t"" 0 . 8 0 > t:j 0 . 6 0 . 4 0 . 2 6 71 nonlinear solution 5 4 J 1 0 2 Fig. 4 . 9 .( d) Comparison of Linear and Nonlinear elutions with(} = 300 . . 4 1 . 2 nonlinear solution linear solutiou 1.0 t"" J . B 0 > t:j J . 6 0.4 0 . 2 0 . 0 6 5 j ? 1 0 Fig. 4 . 9. (e) Comparison of Linear and Nonlinear Solutions with(} = 400
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4.4 Concentrated Loads At Joints 2, 3 and 4, and Horizontal Loads At Joints 2 and 4 72 In order to compare the behaviors of the gabled frame with horizontal loads and without the horizontal loads. These loads will combine the second load condition with two horizontal loads at joints 2 and 4 . This load condition is shown in Figure 4.10. Under this load condition, the vertical loads at joints 2 and 4 are not changed and the horizontal loads at joint 2 and 4 are not changed, either. Of cours e the vertical load at joint 3 begins very small and gradually increases. The gabled frame will have different angle s, B, which are 10, 100 , zoO, 300 , and 400. The diff erent loadrotation b e havior s with differ e n t angl es, B, are shown in Figure 4.11. The differenc e b e twe e n the linear behaviors and the nonlinear behaviors with different angl es, 8 . will b e s hown in Figures 12. ( a>e). QV3 QV2 l QV 4 J l QH2 < QH4 1g. 4 . 10 Concentrat ed Loads At Joints 2 , 3 , 4 and Horizontal Lo a d s A t .Join ts 2 and 4
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t"' 0 .810 . 7 0 . 6 0 . 5 ! I 0 0 . 4 > d 0 . 3 0 .2.L! 0.1 0 . 0 .2 J Fig. 4.11 LoadRotation Behaviors with different (} , ! I "')8 0 7 linear solution 0 . 6 t"' \ I 0 0.4 2 > 0 nonlinear so lu t ion 1 0 3 0 2 0 . 1 U. Q 2 1 Q W l,z(l } 73 Fig. 4.12 .( a ) Comparison of Linear and Non linear Solutions with (} = 10
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74 0 . 8 0 . 7 linear solution 0 . 6 0 . . ') t"'1 nonlinear solution 0 0.4 > t:j 0 . 3 0 . 2 0 . 1 0 . 0 (/ ! ....... _____ J !r1,..2 1 0 2 Fig. 4.12.{b) Comparison of Linear and Nonlinear Solutions with 8 = 100 0 . 8 0.7 linear solution 0 . 5 0 . 5 t"'1 0 0.4 > t:j 0 . 3 nonlinear solution 0 . 2 0.1 0 . 0 2 1 0 2 W l , z(l) Fig. 4 . 12. (c) Comparison of Linear and Nonlinear S olu t ion s with 8 = zoO
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75 0.8 0.7 linear solution 0.6 0.5 0 0.4 . > tj 0. 3 nonlinear I 0 . 2 l 0 I 1 0 . 0 ( 2 2 Fig. 4.12 .( d) Comparison of Linear and Nonlinear Solutions with () = 300 0 . 8 ,0.! linear solution 0.6 0.5 I 0.2 0 . 1 nonlinear solution ! ! I I i I I I . I' \ i / 0 . 0 2 1 0 2 Fig. 4.12.(e) Comparison of Linear and Nonlinear Solutions with() = 400
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76 4.5 The Different Directions Between FEMUCD and the Nonlinear System The PC program FEMUCD define s the directions of the element as shown in Figure 4.13. In the beginning end, all of the forces are compressive and the displacements will be defined as positive when the displacements are moved to the other end. In the other end, all of the forces will be defined as positive when these forces are tensile. The displacements will be defined as positive direction when the displacements are moved in the same direction as the beginning end. Most importantly, the directions of the rotat ion s of each element are defined as pos itive when the rotations are the clockwise direction. All of the directions of the forces and displacements of t h e gabled frame are defined in chapter 2. Figure 2.1 s hows the direction s of eac h e l e ment in forces or displacements of the gabled frame. The direction of the rotation of each element depends on the direction of the lateral displac ements of each element. These t wo definitions just have different phases . Becau se of these two definition s the output data file s of the PC program FEMUCD and the nonlinear system have different s ign s but they are t h e sam e direction as each other. Because they have different s igns in the same dire ction , none of the figures can use different signs to express the same direction s o t h e d efinitions will be consistent. All of the figur es of this chapter u se the definition of t h e nonlinear system. Afte r the output data file s of FEMUCD are found. the s i g n should be times negative s i gn t h en t h e s i g n and the direc t ion s ar<' the same as those of the nonlinear system.
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GLOBAL y local M,J,f local 1 local 3 GLOBAL X Fig. 4.13 The Directions of the Element in FEMUCD i7
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CHAPTER V CONCLUSION 5.1 Summary Some other procedures use X function to analyze the behaviors of this system. The procedure of this thesis uses trigonometric functions. If the behavior of any system i s nonlinear , the X function may sometimes find it difficult to express these behaviors so the expression of trigonometeric function is more convenient to expr ess the nonlinear behaviors. This thesis found the differences between the linear and nonlinear solutions. The gabled frame with different angles, 8, will have different behaviors depending on the load conditions. On the other hand, although the gabled frame with the same angle, 8, as those of others, their behavior will be different when they are subjected to different load conditions in the linear system. Of course the nonlinear behaviors won ' t be the same, either. Usi ng finite element analysis, all of the distribution loads will be the same as the concentrated loads by u sing interpolation functions and applying the loads at both ends of a bar. The concentrated loads are only used for the examples in chapter four. imilar to the concept of finite element analys i s , those concentrated loads can be considered as the uniformly distributed loads applied on e lements 2 and 3. The results of the structure subjected to distribution load s should be same as tho. e of the tructures subjected to concentrated loads. The limitation of the angle, 8 , 1 s not equal to 0 , 45 , and 90 .
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79 When the angle , B, is equal to 0 the gabled frame is the same as the singlespan frame and the linear matrices A1 , A2 , and A3 will be linear dependent. When the angle, B , is equal to 45 , although the gabled frame is symetric, the linear matrices are also linear dependent. Of course when the angle B is equal to 90 elements 1 and 4 would be bound together. In this case, this structure is a column, not a frame. 5.2 Conclusion 1. The loadrotation behaviors of elements 1 and 4 are very similar to thos e of reference 2 8 when the angle, B , i s close to 0 i.e . e _, 0 , but not equal to 0. 2. When the angle, B , is changed from very small to 40 , the loadrotation behaviors are much different from those of the gabled frame with small angle , B. 3 . When the gabled frame is only subjected to a vertical load at joint 3 with the large angle , B. the loadrotation behavior of elements 1 and 4 are similar to those of an eccentrically loaded column. 4. If the gabled frame is subjecte d to vertically concentrated load s at joint 2. 3 , and 4, the behaviors of the loadrotation are similar to thos e of id e al columns. 5. If the gabled frame with a small angle, B , i s subjected to vertical lo a d at joint 3 and the horizontally concentrated load s at joints 2 and 4 , the loadrotation behaviors are more similar to those of an e ccentric:al column 8 The same as footnote 1
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80 with large eccentricity, e. If the angle, 8, i s larger, the behaviors are more different than those of the gabled frame with the small angle, 8. 6. The uniform loads q2 and q3 can be considered the same as the concen trated loads at joints 2 , 3, and 4. Similarly, the concentrated loads at joints 2, 3 , and 4 can be considered the same as the uniform loads on elements 2 and 3.
PAGE 90
BIBLIOGRAPHY W .F. Chen and T. Atsuta, "Theory of BeamColumns Volumn 1. inplane beheavior and design", McGrawHill Inc., 1976. George D. Manocis, Dimitrios E. Beskos, and Bruce J. Brand, "Elastoplastic Analysis and Design of Gabled Frame ", Journal of Computers & Structures Vol.22, No.4. PP.693697 , 1986. George D. Manolis and Dimitrios , E . Beskos " Plasic Design Aids for Pinned Base Gabled Frame", First quater/1987. John R. Mays , " Class Notes of Applied Finite Element Techniques for Per sonna! Computer", University of Colorado at Denver , January, 1987 . Riyad K. Qashu, A.M. ASCE and Donald A . Dadeppo, " Large Deflection and Stability of Rigid Frame", Journal of Engineering Mechanics, Vol.109, No.3, June, 1987. George J. Simitese, "An Introduction to the Elastic Stability of Structures ", PrenticeHall, Inc. , Eng lewood Chiff. New Jersey, 1976 . J .G. Simitese , A.S. Vlahinos, G. J. Simitese , "Sw ay Bucking of Unbraced Multistory Frames", Journal of Computers & Structures, Vol.22 , No.6 PP.l0471054 , 1987. A.S. Vlahinos and A.N. Kounadis "No nlinear Elastic Limit State Analysis of Rigid Jointed Frames", Int. J: Nonlinear , Vol.6. No.3/4 PP.379385 , 1987.
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82 Andreas S. Vlahinos , C.V.Smith, JR. and G. J. S imitese , " A Nonlinear Solu tion Scheme for Multistory, Multibay Plane Frames ", Journal of Computers & Structures, Vol.22 No.6 PP.10351045 , 986. Andreas S. Vlahinos and George J. Simithese, "Bucking and Postbucking of Multibay Frame" . Proceeding at International Conference on Steel and Aluminum Structure, Vol. 3, pp.689708 , Cardiff, UK 1987 K. C. Wang , "SecondOrder Analysis of Multistory Building Frames by Mach ing Technique", Microcomputer in Civil Engineering , Vol.1 , No.3 , 1986 . â€¢
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APPENDIX A FEMUCD INPUT DATA FILE
PAGE 93
FIHD_AXIAL_FOP.CE_AND_LlHEAP._SOLUTION_FOP._THESIS SYSTEH P=2 S=400 N=5 L=? JOINTS 1 X=O.O Y=O.O I=O.O 2 X=O.O Y=200.0 3 X=19b.9b Y=234.?3 4 1=393.92 Y=200.0 5 X=393.92 Y=O.O RESTRAINTS 1 0 0 R=1,1,0,1,1,0 2 0 0 P.=0,0,0,1,1,0 3 0 0 R=0,0,0,1,1,0 4 0 0 R=0,0,0,1,1,0 50 0 P.=l,1, 0,1,1,0 FP.AHE HH=1 1 A=1 1=0.083333 =29000 1 1 2 H=1 LP=1,0 2 2 3 H=1 LP=l,O 3 3 4 H=1 LP=1,0 4 4 5 H=l LP=1,0 LOADS 2 l=l F=0.02,0.02,0 .0,0.0,0. 0 ,0. 0 4 L=1 F=0.02,0.02,0. 0 ,0.0, 0 .0,0. 0 3 L=1 F=0.0,0.02,0.0,0.0,0. 0 , 0 . 0 2 L=2 F=0 .02,o.o2,o. o,o.o,o.o,o.o 4 L=2 F=0.02,0.02,0.010.0,0.0,0.0 3 L=2 F=0.0,0.10,0.0,0.0,0.0,0.0 2 L=3 F=0.02,0.02,0.0,0.0,0.0,0.0 4 L=3 F=0.02,0.02,0.0,0.0,0.0,0.0 3 L=3 F=0.0,0.20,0.0,0.0,0.0,0.0 2 L=4 F=0.0 2 ,0.02,0.0, 0 .0,0. 0 , 0 . 0 4 L=4 F=0.02,0.02,0.0,0.0, 0 .0,0. 0 3 L=4 F=0.0,0.30,0.0, 0 .0,0.0, 0.0 2 L=S F=0 .02,0.02,0.0,0.0,0.0,0. 0 4 L=S F=0.02,0.02,0.0,0.0,0.0,0.0 3 L=S F=0. 0,0.40,0.0,0.0,0.0,0.0 2 L=b F=0.02,0.02,o.o,o.o,o.o,o.o 4 L=b F=0.02,0.02,0.0,0.0,0.0,0.0 3 L=o F=o.o,o.so,o.o,o.o,o.o,o.o 2 L=? F=0.02,0.02,0.0, 0 .0,0. 0 , 0 . 0 4 l=? F=0.02,0. 02,0.0,0.0,0.0,0.0 3 L=7 F=0.0,1.40,0 . 0 , 0 .0,0.0,0.0 84
PAGE 94
APPENDIX B FEMUCD OUTPUT DATA FILE
PAGE 95
SYSTEM P=2 5=400 N=S L=7 IND_AXIAL_FOP.CE_AND_LIHEAR_SOLUTIOH_FOP._THESIS F E H U C D 86.0 â€¢ â€¢ * * * â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ * â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ â€¢ * â€¢ * â€¢ â€¢ I t t I E C H 0 0 F S A P I N P U T D A T A t t I I t f f t I t f t f f f f t t f f I f f f f t f f f f f f f f f f TOTAL NUMBER OF JOINTS = 5 TOTAL NUIIBER OF LOAD CONDITIDtlS = ? 1 X=O.O Y = O . O Z =O.O 2 X=O.O Y=200. 0 3 X=196.96 Y=234.73 4 X=393.9 2 Y =200. 0 5 X=393..92 Y=O. O 6 E M E P. A T E D J 0 I M T C 0 0 P. D I M A T E S JOINT t X 1 .000 2 .000 1 196.960 " 4 393.920 5 393.920 1 0 0 P.=1,1,0,1,1,0 2 0 0 R =O,O,O,l,l,O 3 0 0 P.=O,O,O,l,l,O 4 0 0 P.=O,O,O,l,l,O 50 0 R=l,l,O,l,l,O 6 E M E P. A T E D J 0 I M T JOINT HH=l Nil=! 1 2 3 4 5 0 0 0 y 7 1 0 0 0 0 0 0 0 0 1 A=l 1 = 0 .083333 E=29000 RX y 7 .000 .000 200.000 .000 234.?30 .000 200. (100 . 000 .000 . 000 P. E S T P. A I M T S RY R! 0 0 0 0 (\ 86
PAGE 96
t t t t I f I f f f f f f f f f f f i f f i t f f I I 1 t 1 f f t f I I I E c H 0 0 F F R A E I N p u T D A T A f I I I I I f I f f f f f f f f f f f f f f f f f f f f I f f t t f I t 1 t t " E N B E P. P P. 0 P E P. T I E S MEMBER PROPERTY NUMBER = AXIAL AREA, A = TORSIONAL MOMENT OF INERTIA, J = MOMENT OF INERTIA, 122 = l'IOMENT OF iNERTIA, 133 = SHEAR AREA, A2 = SHEAP. AREA, A3 = MODULUS OF ELASTICITY, E = SHEAR MODULUS, 6 = 1 1 2 11=1 LP=1,0 2 2 3 !1=1 LP=1,0 3 3 4 11=1 LP=1 ,0 4 4 5 11=1 LP=1,0 F P. A " E E L E " E M T A T A .100E+01 ,I)(JI)E+I)I) â€¢ (100E+(1(1 â€¢ B33Ei)1 .OOOE+QO(USEF FOP. SHEAP. DEFL.) .OOOE+OO !USED FOP. SHEAR DE FL. l .290E+05 .112E+05(USED FOP. SHEAP.l 11EMB J1 J2 !1E!1B. SECT. PROP. LP1 LP2 1 2 1 1 0 2 2 3 1 1 (l 3 3 4 1 0 4 4 5 0 E Q U A T I 0 N N U 11 B E P. S NODE IX IY II IXX IVY IE 0 0 11 0 0 12 '1 13 14 15 0 0 16 ... 3 ? B 9 0 0 10 4 3 4 5 0 0 b 5 0 0 1 0 0 ., <)<L 87
PAGE 97
ABSOLUTE MAXIMUM COLUMN HEIGHT(ANY EQUATION) = 10 NUMBER OF BLOCKS = MAXC = 16 t!AXT = 400 FORMATION OF BLOCK IH STIFFNESS MATRIX BLOCK NUMBER = 1 OF LOWEST EQUATION NUMBER = HIGHEST EQUATION tlUMBEP. = 16 NUMBER O F TEP.I1S IN THIS BLOCK = 84 LOWEST COUPLED BLOCK NUMBER = l'IC \.lECTOR: 1 3 b 1 0 15 2 1 26 32 39 47 48 50 5? 65 74 84 J 0 I N T L 0 A D S LOAD NUMBER: 1 NODE XDIP. YDIR ZDIR n YY E 2 .20QE01 .200E0 1 .OOOE+OO â€¢ OOOE+('0 .OOOE+OO .OOOE+OO 4 .200E01 .200E0 1 .OOOE+OO .000+00 . 001)+00 .OOOE+OO 3 .OOOE+OO .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .000+0(\ LOAD NUMBER: 2 XDIP. YDIP. H.'IR n VY E ') .2000 1 .20001 .000+00 , (1(1QE+(t(\ .000+00 .0(1(1+(1(\ L 4 .20001 .200EOI ,I)I)OE+OO ,l)(li)E+OO .OOOE+OO 3 .OOOE+OO .100E+OO â€¢ 0(10E+(1(1 .OOOE+('0 .000+(10 .OOOE+O(I LOAD HUI1BEP.: 3 HODE XDIP. YDIP. ZDIP. u yy E 2 .200E01 .20001 .000+00 .000+00 .OOOE+OO ,(I(IOE+OO 4 .200E01 .200E01 .000+00 .000+00 .OOOE+OO â€¢ .000+(1(1 .200+0 0 .OOOE+OO .OOOE+OO .OOOE+OO â€¢ 000+!)(\ J LOAD tiU!1BEP.: 4 NODE XDIP. YDIP. z râ€¢IP. n yy Z1 2 .200E01 .200E01 .OOOE+OO ,(I(I(IE+OO .0(1(1+00 â€¢ (11)0+00 4 .20001 .200E01 ,I)I)Q+00 .001)+00 .000+1)0 .OOOE+OO 3 .OOOE+OO . 300+(10 .OOOE+OO .000+(10 â€¢ 000+(10 â€¢ 0(10E+(10 LOAD NUI1BEP.: 5 HODE XDIP. YDIP. ZDIP. n YY E 2 .200E01 .2(1001 .OOOE+(IO â€¢ 000+0(1 .000+00 .OOOE+OO 4 .2001)1 .200E01 . i)I)0+1)0 .OOOE+I)I) .000+1)1) , 1)01)E+I)I) .000+00 .400E+OO . 0(1(1+00 .OOOE+OO .i!OOE+OO .O(IOE+(\0 J 88
PAGE 98
LOAD NUI'\BEP.: 6 NODE 2 XDIP. YDIR IDIP. XX YY I1 4 "! J .200E0 1 .200EO l .OOOE+OO .OOOE+OO .OOOE+OO .0(10E+OO .200E01 .20001 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .SOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO LOAD NUIIBEP.: ? XDIP. YDIP. IDIP. YY NODE 2 4 .200EOI .20001 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .200E01 .200E01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO .140+01 .OOOE+OO .OOOE+OO .OOOE+OO .OOOE+OO CALCULATED FOP. LOAD MO. JOINT 1 2 4 5 XDEFL YDEFL â€¢ OOOOOE +00 . OOOOOE + 0 0 .46952+00 .20690E03 .18784E10 .26639+0 1 .46952+00 i20690E03 .00000+00 .00000+00 ZDEFL .00000+00 .00000+00 .OOOOOE+OO .OOOOOE+OO .00000+00 CALCULATED FOP. LOAD MO. 2 JOINT XDEFL YDEFL 1 .OOOOOE+OO .OOOOOE+OO 2 .33?16E+01 .4827603 3 .13487E09 .19123+02 4 .33716+01 .482760 3 5 .00000+00 .00000+00 ZDEFL .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO .00000+00 .OOOOO E+OO CALCULATED FOP. LOAD NO. 3 JOINT 1 2 3 4 5 XDEFL YDEFL .00000+00 .00000+00 .69991+01 .8275903 .2799909 .39697+02 .69991E+01 .8275903 .00000+(10 . 00000+00 1DEFL .00000[+00 .OOOOOE+OO .00000+00 .00000+00 .OOOOOE+OO CALCULATE! niSPLACEftEMTS FOP. LOAD MO. 4 JOINT 1 2 3 4 5 YDEFL HlEFL . OQOOOE +(1(1 . OC+00 â€¢ OOOOOE+OO .10627+02 .1172402 .OOOOOE+Oi) .42511(19 .602?1+02 .0(10(10+(11) .10627+02 .11?240 2 .00000+00 .00000+(1(\ ,(1(1(100[+(\Q .000(!0[+00 X P.OT .00000+00 .00000+00 .00000+00 .00 000+00 .00000+0 0 XROT .OOOOOE+OO .00000+00 .OOOOOE+OO .OOOOOE+O O .00000+00 XROT .00000+00 .OOOOOE+OO .OOO O OE+OO .OOOOOE+OO .00000+()0 XRGT . (1(11}\oOE +(10 . (JI}i)I)(J+(11) . (1(1(1(\(1[ +00 . (J(J(Ji)i) +()I) ,(1(1(1(1(1[+(!(1 Y ROT ZROT . O OOOOE+OO .881430 2 .OOOOOE+OO .1 05860 1 .OOOOOE+OO .234 79Ei3 .OOOOOE+OO .1059601 . OOOOOE+O O .8814302 YP.OT 1P.OT .00000+00 .632840 1 .00(100+00 .759940 1 .00000+00 .1685812 .OOOOOE+OO .759940 1 .OOOOOE+OO .6328401 YROT 1P.OT â€¢ 0000(1+(10 â€¢ 13137E +00 .00000+00 .15775+00 .00000+00 .3499612 .00000+00 .15775+(10 .00000+0 0 . 13137+00 YROT 1 P.OT ,Q(IQQQ+0 0 .19946+00 .00000+00 .23951+00 ,(o(l(l(l{\+(!(1 .5313512 .00000+00 .23951+00 .0(1(1(1(\+0{\ .19946+(1(1 89
PAGE 99
CALCILATED DISPLACEitEJITS HF. LOAD Ill. 5 JOINT XDEFL YDEFL ZDEFL 1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO 2 .14254+02 .151?202 .OOOOOE+OO 3 .5?02009 .80845+02 .OOOOOE+O O 4 .14254+02 .1517202 .OOOOOE+OO 5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO CALCII.ATED DISPlACDtEMTS FOP. LOAD MO. 6 JOINT XDEFL YDEFL ZDEFL 1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO 2 .1?882+02 .1862102 .OOOOOE+OO "t .?15340 9 .10142+03 .OOOOOE+OO v 4 . 1 ?882E+02 .18&2102 .00000+00 5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO CALCULATED DISPLACEitENTS FOP. LOAD MO. ? JOINT X DEFL YDEFL IOEFL 1 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO 2 .50530+02 .4965502 .OOOOOE+OO 3 .2021408 .2865BE+03 .00000+00 4 .50530+02 .4965502 .OOOOOE+OO 5 .OOOOOE+OO .OOOOOE+OO .OOOOOE+OO ELEt!EJIT FOP.CES AMD "O"EMTS FOP. LOAD CASE I ELE!'I PlJl \12J 1 V3J1 TlJ 1 1 .300E0 1 .23402 .OOOE+OO .OOOE+OO 2 .23?01 .59?02 .OOOE+OO .OOOE+OO 3 .23?E01 .59702 .OOOE+OO .OOOE+O O 4 .30001 .23402 .OOOE+OO .OOOE+OO V3J2 T1J2 1'12J2 113J2 .000+00 .OOOE+(lO .000+00 .4b9E+OO .OOOE+OO .OOOE+OO .OOOE+OO .725E+OO .OOOE+OO .OOOE+OO .OOOE+OO .469+00 .000+00 . OOOE+OO . 000+00 .733E0 9 ELEftEMT FOP.CES AMI FOP. LOAD CASE I 2 ELEI'I PlJl U2J1 U3Jl T1J1 .70001 .16BE01 .OOOE+OO .000+00 2 .4490 1 .42BE01 .OOOE+O O .OOOE+OO "t .4500 1 .42BE01 .OOOE+OO .OOOE+OO v 4 .700E0 1 . 168E01 .OOOE+OO .000+00 YROT !ROT .OOOOOE+OO .OOOOOE+OO .26?54+0(1 .OOOOOE+OO .OOOOOE+O O .3212BE+OO .OOOOOE+OO .OOOOOE+OO .712?012 .OOOOOE+OO .OOOOOE+OO .32128+00 .OOOOOE+OO .OOOOOE+(lO .26?54E+OO XROT YROT IROT . OOOOO E+OO . 000(10+00 . 33563E+OO .OOOOOE+OO .OOOOOE+OO . 4031)4+00 .OOOOOE+OO .00000+00 .89413E12 .00000+00 . OOOOOE+â€¢)O . +00 .OOOOOE+OO .OOOOOE+OO .33563E+Co0 XROT YROT 1ROT . OOOOOE+OO .OOOOOE+OO .94841E+OO . OOOOOE+OO .OOOOOE+OO .11389+0 1 .OOOOOE+OO .OOOOOE+OO .2526511 .OOOOOE+OO . OOOOOE+OO .11389+01 .OOOOOE+OO .OOOOOE+OO .94841E+OO M2Jl H3Jl P1J2 V2J2 .OOOE+OO .282EO? .30001 .234E0 2 .000+00 .4&9E+OO .23 701 .59?E02 .OOOE+OO .725E+OO .23701 .59702 .OOOE+OO .4&9E+OO .30001 .23402 112J1 l13Jl P1J2 Y2J2 .OOOE+OO .357EO ? .?OOE01 .168E0 1 .OOOE+OO .337+01 .449E01 .42BE01 .000+00 .520+0 1 .45 00 1 .428E0 1 .OOOE+OO .33?+01 .?OOE01 .1680 1 90
PAGE 100
1!3J2 T1J2 M2J2 M3J2 .000+00 .000+00 .OOOE+OO .33?+01 . OOOE+OO .000+00 .000+00 .520+01 .000+(10 .000+00 .OOOE+OO .337+01 .OOOE+OO .000+00 .OO'JE+OO . 73007 EWIEMT FOP.CES AND "llftENTS FOP. LOAD CASE I 3 ELEI'I P1J1 nJ1 1!3J 1 T1J1 1 .120+(\(\ .34901 .OOOE+t'O â€¢ (\(1(\+(\(l 2 .?1401 .88901 .000+1)1) .001)+1)0 "t .71401 .88901 .(1{10+00 .000 +(10 .) 4 .120+1)1) â€¢ 34901 ,1)1)1)+01) ,1)1)1)+1)1) V 3J2 T1J2 1'12J2 1'13J2 â€¢ (1(1(1 t(l(l â€¢ (1(10+(\i) ,QCI0+(1(1 .699+(11 ,1)0(1+00 .(101)+1)1) ,1)1)1)+1)1) . 1 %E +02 .ll00+(10 .000+(\(\ .00(1+(10 .699+(11 . i)l)(l +l)( l ,(\1)1)+01) .001)+00 .4460 6 ELEJtEMT FOP.CES AMI) llmt81TS FOP. LOA[I CASE I 4 ELEM P1J1 V2J1 1!3J 1 TlJ 1 1 .1?0+(\(\ .53001 .0(10+(1(\ .0(1(1+(10 2 .9?91)1 â€¢ 135+00 .1)01)+1)1) .000+1)1) "T .980(11 .135+(1(1 .0(10+00 .000+(10 .) 4 . 170+00 .53001 .1)00+0 1 ) .(11)0+01) V3J2 T1J2 1'12J2 lf!,J2 .OOQ+00 .000+00 .000+(1(\ .106+0 2 .000+00 .01)1)+1)1) . 1)01)+1)(! .164+02 ,(\(11)+00 .0(10+(10 â€¢ (1(10 +(1(1 . 106 +(12 . 001)+1)1) .1)1)0+00 .01)1)+00 .3?40b ELEIOT FOP.CES AIUI troltEJITS FOP. LOAD CASE I S ELEII P1J1 1!2J1 V3J1 TlJ 1 1 .220+(10 .?11(11 .000+(10 â€¢ 000E+(I0 2 .125+1)1) .181+1)0 .OOOE+OO ,l)l)(i+1)(1 1 â€¢ 125E+OO .181E+OO , 0(1(tE +(IQ .000+00 J 4 â€¢ 220E+0fJ .?1101 .OOOE+OO . (ll)l)+1)(! IJ3J2 T1J2 112J2 113J2 .000+(1(\ .01)0+0(1 .000+00 .142+1)2 . OOOE+OO â€¢ fJOOE+OO . 01)0+00 .220+0 2 .000+00 .OOOE+OO .000+00 .142+(12 .01)0+01) . 0 0 0+00 .OOQE+OO .23?E06 M2J1 M3J1 P1J2 1!2J2 â€¢ 000+(10 â€¢ 390(16 .120+00 â€¢ 349 (il .000+01) .699+01 .71401 .8890 1 .0(1(1+1)(\ .1(18+02 .714(11 .889(11 ,1)1)0+(!1) .699+1)1 .121)+1)1) . 3491)\ 1'12J1 H3J1 P1J2 '!2J2 . OOOE+OO â€¢ 6 9506 . 53001 ,1)0(1+01) .106+02 .9 ?91)1 .135+00 â€¢ (1{10 +(10 . 164 +1)2 . 980 (11 â€¢ 135 +1)0 ,(11)0+1)1) .106+02 ,llfJ+01) .531)1)1 1'12J1 113J1 P1J2 U2J2 .000+00 . 4(15 06 . 220+(10 .711(11 , i)l)i)E+OO .142E+02 .181E+OO .QOQE+OO .220+02 .125E+OO .181+(10 . (11)1)+1)1) . 142E+02 .22 0+00 .711 1 ) 1 91
PAGE 101
92 ELEftEJtT FOP.CES AMD tmftBITS FOP. LllAl' CASE I 6 EL PlJl 1!2J1 1!3J1 T1J1 112J1 113J1 P1J2 V2J2 1 .2?0+00 .89301 .000+(1(\ .000+(10 .000+00 .4?60 6 .2?0+0 0 .8930 1 2 .151+1)1) .227+01) ,l)l)l)+i)l) , I)Q(JE+I)O ,1)1)(1+1)1) .1?9+1)2 .151+1)1) .22?+1)1) 3 .151+(1(\ .227+00 â€¢ (1(10+(1(1 .(100+(1(1 ,(1(10+0(1 .2?6+02 .151+0(1 .227+(10 4 .2?1)+1)1) .8931)1 .1)1)1)+1)1) ,1)1)1)+1)1) .OOOE+OQ .179+1)2 .270+00 .89301 1!3J2 T1J2 112J2 l'\'!.J2 .000+(10 ,(.\(\(\+()(.\ .000+1)(.\ .179+1)2 ,1)1)1)+1)0 ,1)1)1)+1)1) ,l)t)i)+1)() .276+0 2 .000+00 ,(1(1(1+1}(.\ ,0(.\(1+00 .1?9+02 ,1)1)(1+1)0 ,1)1)1)+1)0 ,l)t)(J+00 .1281)6 ELEI'IEMT FOP.CES AMD I'IOI'IBITS FOP. LOAl' CASE I ? EL P1J1 1!2Jl V)Jl T1J1 l'I2JI M3JI P1J2 \12J2 .720+00 .252+(10 .00(1+00 ,01)(1+(1(1 ,(1(1(1+1)(1 .109(15 .72(1+0(1 . 252E+t)(i 'l .391)+1)0 â€¢ 642+(11) .000+00 â€¢ i)t)l)+1)!) . . 504+1)2 .390+00 .642+01) J. T .390+00 .642+00 ,(1(1(1+00 .000+(1(1 .0(10 + 0 0 ,78(1+02 .390+(1(1 â€¢ b42+(10 J 4 ,721)+1)1) .252+1)1) â€¢ 001 )+0 0 â€¢ i)l)1 )+1)f) ,i)t)fl+1)0 .504+02 .?20+1)0 . 252+00 V3J2 T1J2 112J2 !'nJ2 .OOOE+t)Q .000+(10 â€¢ OOOE +(I (I . 5 04 +(12 â€¢ (JI)I)+1)1) ,1)1)1)+1)1) â€¢ (100+1)1) â€¢ ?80+02 . ON!+(1 0 .000+00 .000+(10 .504+02 ,1)1)0+1)1) J)l)l)+1)1) .000+1)1) . 3 4?1)6
PAGE 102
APPENDIX C COMPUTER PROGRAM FOR THE NONLINEAR SYSTEM
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CCCC94 PROGRAM THESIS LINEAR AND NOLINEAR THIS SUBROUTINE IS TO FIND THE SOLUTION OF LINEAR EQUATIONS. THEN, USE THEM TO FIND NON LINEAR SOLUTIO NS. C************** DIMENSION INTEGER NTIME REAL PAR(4) REAL A(4, 5) REAL L(4),S(4) REAL K(4),P(4),KK(4) REAL QD(4),QW(4),QV(4),QH(4) REAL LAMDA(4) ,DLAMDA(4) REAL CP(4),DP(4) REAL LP(4),GUESS(4) REAL SETA INTEGER NSIG,NOFE,ITMAX REAL WK(lOO) REAL ZK(2) REAL FNORM EXTERNAL FCN C======================================================l C L :THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM FEMUCD I C K :THE CONSTANT RELATIVE TO AXIS FORCE I C S :THE CONSTANT IS EQAL TO EI/L I C QV:THE CONCENTRATE LOAD VERTICAL TO XAXIS AT I C EACH JOINT. I C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO EACH I c BAR I C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO I c QD*L**3/ (E*I) I C QH: THE LOADS PARAL TO AAXIX, HORIZONTAL LOADS, I C AT EACH JOINT. I c KK: KK=K*K I C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED I c I C======================================================l C************** COMMON COMMON/LS/L,S
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CCCCCCCOMMON/LOADS/QD,QW,QV,QH COMMON/LAMDA/LAMDA,DLAMDA COMMON/ANGLE/SETA COMMON/LP/LP 95 OPEN ALL OF THE INPUT AND OUTPUT FILES IN THIS PROGRAM CALL OPEN INPUT ALL OF INPUT DATA WHICH ARE KNOWN OR INITIAL GUESS CALL INPUT (K) CALL SUBROUTINE LINEAR TO SOLVE LINEAR EQATIONS CALL LINEAR( A , K ,l) NSIG= 4 NOFE= 4 ITMAX=200 CALL ZSPOW(FCN,NSIG,NOFE,ITMAX,PAR, K ,FNORM & ,WK,NO_IER) CCCCCCCALL LINEAR( A , K , 2) CALCULATE AND OUTPUT THE FINAL RESULTS OF THIS SYSTEM CALL RESULT( A , K , 3) PLOTE THE FRAME AFTER THE DISPLACEMENTS ARE CALCULATED CALL PLOTFRAME(A, K ) THE ORTHER SUBROUTINES CAN BE USED I N THE SEQUENCE STOP END
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CC96 SUBROUTINE OPEN OPEN ALL OF INPUT AND OUTPUT DATA FILES IN PROGRAM THESIS OPEN(l,FILE='PROPER . INP' ,STATUS='OLD') OPEN(3,FILE='LOADS.INP' ,STATUS='OLD') OPEN(4,FILE='ANGLE.INP',STATUS='OLD') OPEN(S,FILE='FORCES . INP' ,STATUS='OLD') OPEN(lO, FILE='SOLUTION .DAT' ,STATUS='NEW') OPEN(13,FILE='INPUT .DAT' ,STATUS='NEW') OPEN(15,FILE='WXl .OUT' ,STATUS='NEW') OPEN(16,FILE='WX4 .0UT' ,STATUS='NEW') C=====================================================l C 1 :INPUT FOR E,A,I,L I C 3 : INPUT FOR QD, Q, QH I C 4 :INPUT FOR ANGLE, SETA. I C 5 :INPUT INITIAL GUESS AXIAL FORCE, P . I C 11 : OUTPUT MATRIX A I C 10:0UTPUT THE INTERMINATE AND FINAL SOLUTIONS OF I C NOLINEAR I C 12 : OUTPUT MATRIX B I C 13 : OUTPUT COMBINE ALL OF THE INPUT DATA FILES I C=====================================================l RETURN END
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97 C=============================== ======================i c I C PROPER THE IDETITY OF PROPERTY OF MATERIALS I C LOAD THE IDETITY OF LOADS I C FORCE THE IDETITY OF FORCES I C SETAA THE IDETITY OF ANGLE, SETA. I c I C TITLEP THE TITLE OF PROPERTY OF MATERIALS I C TITLEL THE TITLE OF LENGTH, L. I C TITLEA THE TITLE OF AREA, AREA. I C TITLEF THE TITLE OF AXIAL FORCES I C=================================================== ==l CREAD INPUT DATA FILES c**** ********** c ************** READ TITLE LINE OF EACH INPUT DATA FILE DO N=1,2 READ(1,100)TITLEP(N) READ(3,100)TITLEL(N) READ(4,100)TITLEA(N) READ(5,100)TITLEF(N) END DO C ************** READ PROPERTIES, LOADS, AND INITIAL C**** ********** GUESS (P) . DO 1=1,4 READ(1, 200 ) PROPER(J),E(J),A(J),I(J),L(J) READ(3,300)LOAD(J) ,QD(J),QV(J),QH(J) READ(5,400)FORCE(J) ,P(J) END DO C ************** READ INPUT DATA FILE ANGLE, SETA. CCCCREAD(4,400)SETAA,SETA CHANGE ANGLE, SETA, FROM DEGREE TO RADIUM PI=4.0*ATAN(1. 0) SETA=PI*SETA/ 180 . 0 CALCULATE CONSTANT S AND QW, IN WHICH S=E*I/L, QW=QD*L**3/(E*I) CALCULATE LAMDA, LAMDA=L/(SQRT(I/A)),
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CCCC100 200 300 400 DLAMDA=LAMDA*LAMDA DO 1 = 1 ,4 S(J)=E(J)*I(J)/L(J) QW(J)=QD(J) *L(J)*L(J)/S(J) LAMDA(J)=L(J)/(SQRT(I(J)/A(J))) DLAMDA(J)=LAMDA(J)*LAMDA(J) END DO CALCULATE K, KK=K*K DO 1 = 1 , 4 K(J)=SQRT(P(J)*L(J)/S(J)) KK(J)=P(J)*L(J)/S(J) END DO OUTPUT ALL OF THE INITIAL INPUT DATA CALL OUTPUTDAT (K) FORMAT FORMAT(A80) FORMAT(A20, 4F15 . 8) FORMAT(A20,3F15.10) FORMAT(A20,F15.10) RETURN END 98
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CC99 SUBROUTINE OUTPUTDAT (X) THIS SUBROUTINE IS WRITE ALL OF THE INPUT AND CALCULATED DATA C************** DIMENSION CHARACTER*80 PROPER(4),LOAD(4),FORCE(4) CHARACTER*80 TITLEL(2) , TITLEF(2),TITLEP(2) & , TITLEA(2) CHARACTER*80 SETAA REAL E(4),A(4),I(4) REAL K(4),P(4),KK(4) REAL L(4),S(4) REAL QD(4),QW(4),QV(4),QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA C======================================================l C L :THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM FEMUCD I C K :THE CONSTANT RELATIVE TO AXIS FORCE I C S :THE CONSTANT IS EQAL TO EI/L I C QV:THE CONCENTRATE LOAD VERTICAL TO XAXIS AT I C EACH JOINT. I C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO EACH I C BAR I C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO I c QD*L** 3 / (E*I) I C QH: THE LOADS PARAL TO A AXIX, HORIZONTAL LOADS, I C AT EACH JOINT. I c KK : KK=K*K I C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED I c I C======================================================l C ************** COMMON COMMON/EAI/E,A,I COMMON/LS/L,S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA COMMON/CHARACTER/TITLEL,TITLEF,TITLEP,TITLEA, & PROPER,LOAD,FORCE,SETAA
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100 C=====================================================l c I C PROPER THE IDETITY OF PROPERTY OF MATERIALS I C LOAD THE IDETITY OF LOADS I C FORCE THE IDETITY OF FORCES I C SETAA THE IDETITY OF ANGLE, SETA. I c I C TITLEP THE TITLE OF PROPERTY OF MATERIALS I C TITLEL THE TITLE OF LENGTH, L. I C TITLEA THE TITLE OF AREA, AREA. I C TITLEF THE TITLE OF AXIAL FORCES I C=================================================== = =l CCALCULATE P DO N=1,4 P(N)=X(N) *X(N)* S(N)/L(N) END DO C** * ****** ***** WRIRE PROPERTIES DO J=1,2 WRITE(13,100)TITLEP(J) END DO DO J =1,4 WRITE(13,200)PROPER(J),E(J),A(J),I(J),L(J) END DO c ************** WRITE LOADS DO J=1,2 WRITE(13,100)TITLEL(J) END DO DO J=1,4 WRITE(13,300)LOAD(J),QD(J),QV(J) ,QH(J) END DO c *****"'******* WRITE FORCES DO J=1,2 WRITE(13,100)TITLEF(J) END DO DO J=1,4 WRITE(13,400)FORCE(J),P(J) END DO
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101 c********** **** OUTPUT LAMDA, AND DLAMDA WRITE(13,500) DO J=l ,4 WRITE(13,600)J,LAMDA(J),DLAMDA(J) END DO C********** **** WRITE INPUT DATA FILE ANGLE, C************** SETA IN RADIUM. C100 200 300 400 500 600 & & & DO J=1,2 WRITE(13,100)TITLEA(J) END DO WRITE(13,400)SETAA,SETA FORMAT FORMAT(III,5X,A80,II) FORMAT(5X,A20,4F15.6,1) FORMAT(5X,A20,3F15 .6,1) FORMAT(5X,A20,F15.6,1) FORMAT (I I I, ' THIS IS LAMDA AND LAMDA*LAMDA=DLAMDA', I,36X, 'LAMDA' , 15X, 'DLAMDA' ,II) FORMAT(' ***** ELEMENT ',Il,' ***** = 2(5X,F15 . 6) ) RETURN END
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CCCC102 SUBROUTINE LINEAR(A,X ,ISW) THIS SUBROUTINE IS TO FIND THE SOLUTION OF LINEAR EQUATIONS, A(l,l),A(1,2), ... , TO A(4 , 5) DIMENSION INTEGER ISW REAL A(4,5) REAL L(4) ,S(4) REAL P(4),X(4) REAL QD(4),QW(4),QV(4) ,QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA C L :THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM P.C . FRAME! C K :THE CONSTANT RELATIVE TO AXIS FORCE I C S :THE CONSTANT IS EQAL TO EI/L I C QV:THE CONCENTRATE LOAD VERTICAL TO XAXIS AT EACH! c JOINT. I C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO I C EACH BAR I C QW:THESE LOADS ARE MODIFIED WHIR ARE EQUAL TO I c QD* L**3 / ( E *I) I C QH:THE LOADS PARAL TO A AXIX, HORIZONTAL LOADS, ATI C EACH JOINT. I c KK : KK =K*K I C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED I C AS RADIUM I C ISW :THE SWITCH TO DEFINE LINEAR OR NOLINEAR I C SOLUTION I C IF ISW= l IS TO FIND LINEAR SOLUTION I C ORTHERWISE NOLINEAR SOLU. I C============================== ========= ===============l C ************** COMMON COMMON/LS/L, S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA
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C CC CCCCCCCCCCCCC Cgg 103 CALCULATE A(1 ,3),A(2,3),A(3,3), AND A(4,3) CALL MATRIXl(A,X) USE CONSTANTS A(1,3), A(2,3), A(3,3), A(4,3) TO CALCULATE A(l,l),A(2,1),A(2,2),A(3,1), A(3,2),A(4, 1) CALL MATRIX2(A,X) USE CONSTANTS A(1,3),A(2,3),A(3,3),A(4, 3) AND A(l,l),A(2,1),A(2,2),A(3,1),A(3,2), A(4,1) TO CALCULATE A(2,4),A(2,5)1A (3,4), A(3,5) CALL MATRIX3(A,X) USE ISW TO OPTION WHEATER OR NOT OUTPUT A , AND X IF ISW= l OUTPUT THE LINEAR SOLUTIONS, IF ISW IS NOT EQAL TO 1 THEN THE SOLUTION IS NOLINEAR WHICH WON'T BE OUTPUTED. IF(ISW.GT.l)GO TO gg OUTPUT THE INITIAL SOLUTION OF LINEAR EQUATIONS CALL OUTPUTA( A ,X) CALCULATE AND OUTPUT THE INITIAL RESULT CALL RESULT( A , X ,l) THE ORTHER SUBROUTINES CAN BE PUT HERE IF ANY MORE. RETURN END
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104 SUBROUTINE MATRIX1(A,X) CCCCTHIS SUBROUTINE USES EQUATION 1, 3 , 5, AND 7 TO EVALUATE CMATRIX A1 AND B1, AND TO FIND A(1,3),A(2,3), A(3,3),A(4,3). DIMENSION INTEGER IJOB,NM REAL A(4,5) REAL A1(4,4) REAL B1 (4, 1) REAL L(4),S(4) REAL P(4),X(4),XX(4) REAL QD(4),QW(4) ,QV( 4) ,QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA REAL WK(24) C============ = = ======== = ========= = ================== ===l C A THE SOLUTION MATRIX OF A OF FULL SYSTEM I C A1 THE COEFFICIENT 4X4 MATRIX OF EQ. 1, 3 , 5, I c AND 7 I C B1 THE CONSTANT VECTOR OF EQUATION 1 , 3, 5, I c AND 7 I C L : THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM P.C . FRAMEI C K :THE CONSTANT RELATIVE TO AXIS FORCE I C S :THE CONSTANT IS EQAL TO EI/ L I C QV:THE CONCENTRATE LOAD VERTICAL TO X AXIS AT EACH C JOINT. C QD:DISTRIBUTION LOADS WHICH IS VERTICAL C TO EACH BAR C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO C QD* L**3/(E*I) C QH: THE LOADS PARAL TO A AXIX, HORIZONTAL LOADS, AT C EACH JOINT. C KK:KK= K * K C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED C AS RADIUM C====================================================== C ************* COMMON COMMON/LS/L,S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA
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C. CC INITIALIZE MATRIX A DO N=1,4 DO M=1,5 A(N,M)=O.O END DO END DO CALCULATE XX DO N=1, 4 XX(N)=X(N)*X(N) P(N)=XX(N) * S(N)/L(N) END DO EVALUATE MATRIX Al c ************* INITIALIZE MATRIX Al DO N =1,4 DO M =1,4 Al(N ,M)=O.O END DO END DO c ************* EVALUATE MATRIX Al Al(l,l)=P(l) A1(1,2)=P(2)* SIN(SETA) A1(2,2)=P(2)*COS(SETA) A1(2,3)=P(3)*COS(SETA) A1(3,2)=P(2)* SIN(SETA) A1(3,3)=P(3)* SIN(SETA) A1(4,3)=P(3)* SIN(SETA) Al(4,4)=P(4) c ************* MATRIX IS DONE CEVALUATE MATRIX Bl c ************* INITIALIZE MATRIX Bl 105
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DO N=1,4 Bl(N,l)=O.O END DO 106 C************* EVALUATE MATRIX Bl Bl(l,l)=QH(2)P(2)*COS(SETA) B1(2,1)=P(2)*SIN(SETA)+QV(3)+P(3)*SIN(SETA) & QW(2)*S(2)/L(2)*COS(SETA)+QW(3) & *S(3)/L(3)*COS(SETA) B1(3,1)=QW(2) *S(2)/L(2)* SIN(SETA)+QH(3) & +QW(3)*S(3)/L(3)*SIN(SETA) & +P(2)*COS(SETA)P(3)*COS(SETA) B1(4,1)=QH(4)+P(3) *COS(SETA) c ************* MATRIX Bl IS DONE CCCCCCFIND THE SOLUTION OF AlX=Bl IN WHICH THE SOLUTION IS X BY CALL THE SUBROUTINE LINV3F OF IMSL LIBRARY IN VAX/VMS IJOB=3 NM=4 Dl=1.0 CALL LINV3F(Al,Bl,IJOB ,NM,NM,Dl,D2,WK,IER1) PUT THE SOLUTION OF AlX= Bl INTO THE FULL SOLUTION A A(1 , 3) = A(2 , 3) = A(3,3)= A(4,3)= RETURN END Bl(l,l) B1(2,1) B1(3,1) B1(4,1)
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CCCCCC107 SUBROUTINE MATRIX2(A,X) THIS SUBROUTINE USES EQUATION 2, 4, 6, 8, 9 AND 10 TO EVALUATE MATRIX A2 AND B2, AND TO FIND A(1,1),A(2,1) ,A(2,2),A(3,1),A(3,2) , AND A(4,1). DIMENSION INTEGER IJOB,NM REAL A(4,5) REAL A2 (6, 6) REAL B2(6,1) REAL L(4),S(4) REAL P(4),X(4) ,XX(4) REAL QD(4) ,QW(4),QV(4) ,QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA REAL WK(24) C======================================================l C A THE COEFFICIENT MATRIX OF FULL SYSTEM I C A2 THE COEFFICIENT 4X4 MATRIX OF EQ. 2,4,6,8,9, I c AND 10 I C B2 THE CONSTANT VECTOR OF EQUATION 2,4,6,8,9, I c AND 10 I C L : THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM P.C. FRAME! C K :THE CONSTANT RELATIVE TO AXIS FORCE I C S :THE CONSTANT IS EQAL TO EI/L I C QV:THE CONCENTRATE LOAD VERTICAL TO XAXIS AT EACH! c JOINT. I C QD:DISTRIBUTION LOADS WHICH IS VERTICAL I C TO EACH BAR I C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO I c QD*L**3/ (E*I) I C QH: THE LOADS PARAL TO AAXIX, HORIZONTAL LOADS, AT I C EACH JOINT. I c KK:KK=K* K I C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED I C AS RADIUM I C======================================================l C ************** COMMON COMMON/LS/L,S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA
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CCCOMMON/LAMDA/LAMDA,DLAMDA CALCULATE XX, AND P DO N=1,4 XX(N)=X(N)*X(N) P(N)=XX(N)*S(N)/L(N) END DO EVALUATE MATRIX A2 c************** INITIALIZE MATRIX A2 DO N=1,6 DO M=1,6 A2(N,M)=O . O END DO END DO c************** EVALUATE MATRIX A2 A2(1,1)=P(l)*L(l)*SIN(X(l)) A2(1,3)=P(2)*L(2) A2(2,2)=P(2)*L(2)*SIN(X(2)) A2(2,3)=P(2)*L(2)* COS(X(2)) A2(2,5)=P(3)xL(3)xCOS(X(3)) A2(3 , 5) =P(3)*L(3) A2(3,6)=P(4)*L(4)* SIN (X(4)) A2(4 ,1)=X(l)* COS(X(l) ) A2(4,2)=X(2) A2(5,2)= X(2) * COS(X(2)) A2(5,3)=X(2) â€¢SIN(X(2)) A2(5,4)=X(3) * COS(X(3)) A2(5,5)= X(3) *SIN(X(3)) A2(6,4)= X(3) A2(6,6)=X(4) *COS(X(4)) c************** MATRIX A2 IS DONE CEVALUATE MATRIX B2 c************** INITIALIZE MATRIX B2 108
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DO N =1,6 B2(N,l)= O . O END DO 109 C ****** ******* * EVALUATE MATRIX B2 B2(1 ,1)=QW(2)*S(2)/XX(2) B2(2,1)=QW(2)* S(2)/XX(2)QW(3) *S(3)/XX(3) B2(3 , 1)=QW(3) *S(3)/XX(3) B2(4,1)=A(1,3)+A(2,3) B2(5,1)=A(2,3)+A(3, 3) QW(2)/XX(2)+QW(3) & /XX(3) B2(6,1)=A(3 , 3) + A(4 , 3) c ************** MATRIX B2 IS DONE CCCC& C& FIND THE SOLUTION OF A2X=B2 IN WHICH THE SOLUTION IS X BY CALL THE SUBROUTINE LINV3F OF IMSL LIBRARY IN VAX IJOB = 3 NM= 6 Dl=1.0 CALL LINV3F(A2,B2,IJOB,NM,NM,Dl,D2,WK, IER2) PUT THE SOLUTION OF A2X=B2 INTO THE FULL SOLUTION A A(l, l)= B2( 1 , 1 ) A(2 , 1)= B2( 2 , 1 ) A ( 2,2)= B2( 3 , 1 ) A (3,1)= B2( 4 , 1) A(3,2)= B2(5 , 1) A(4,1)= B2(6 , 1) RETURN END
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llO SUBROUTINE MATRIX3(A,X) C CCC THIS SUBROUTINE USES EQUATION 11, 12, 13 , AND 14 TO. EVALUATE CMATRIX A3 AND B3, AND TO FIND A(2,4),A(2,4) ,A(3,4),A(3,5). DIMENSION REAL A(4,5) REAL A3(4,4) REAL B3 ( 4 , 1) REAL L(4),S(4) REAL P(4), X(4),XX(4) REAL QD(4),QW(4),QV(4),QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA REAL WK(24) C=================== = === ========== ====== ===============l c I C A THE COEFFICIENT MATRIX OF FULL SYSTEM I C A3 THE COEFFICIENT 4X4 MATRIX OF EQ.11 , 12 , 13, I c AND 14 I C B3 THE CONSTANT VECTOR OF EQUATION 11,12, 13, I C AND 14 C L :THE LENGTH OF EACH BAR C P :AXIAL FORCE WHICH BE CALCULATE FROM P . C . FRAME C K :THE CONSTANT RELATIVE TO AXIS FORCE C S :THE CONSTANT IS EQAL TO EI/ L C QV:THE CONCENTRATE LOAD VERTICAL TO X AXIS AT EACH C JOINT. C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO EACH C BAR C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO C QD* L**3 /(E*I) C QH:THE LOADS PARAL TO A AXIX, HORIZONTAL LOADS, AT C EACH JOINT. C KK:KK= K * K C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED C AS RADIUM C=============================== ======================= C * * ********* *** COMMON COMMON/LS/L, S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA
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CCCALCULATE XX, AND P DO N=1,4 XX(N)=X(N)*X(N) P(N)=XX(N) *S(N)/L(N) END DO EVALUATE MATRIX A3 c **** *** ****** * INITIALIZE MATRIX A3 DO N=1, 4 DO M=1,4 A3(N,M)=O.O END DO END DO c ************** EVALUATE MATRIX A3 A3(1 , 1) =SIN(SETA) A3(1,2)=COS(SETA) A3(2,1)=COS(SETA) A3( 2 , 2) = SIN (SETA) A3(3,1)=COS(SETA) A3(3,2)= SIN (SETA) A3(3,3)=COS(SETA) A3( 3 , 4)=SIN(SETA) A3(4,3)= SIN(SETA) A3(4 , 4) =COS(SETA) c ************** MATRI X IS DONE CEVALUATE MATRIX B3 c ************** INITIALIZE MATRIX B3 DO N=1,4 B3( N,l)= O . O END DO 111
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112 C *********** *** EVALUATE MATRIX B3 C++++++++++++++++++++++ EVALUATE VECTOR B3 B3(1 ,1)=W(X,A,l,l.O)+A(2,2)* SIN(SETA) B3(2,1)=U(X,A ,l,l. O)A(2 , 2) *COS(SETA) B3(3,1)=U(X,A,3,l.O)* SIN(SETA) W(X,A,3,1.0) & *COS(SETA) & U(X ,A,2,l.O)* SIN(SETA) & W(X,A,2,l.O)*COS(SETA) B3(4 , 1) =W(X,A,4,1.0)A(3 , 2) * SIN(SETA) c * * ************ MATRIX B3 IS DONE C CCC& CCFIND THE SOLUTION OF A3X=B3 IN WHICH THE SOLUTION IS X BY CALL THE SUBROUTINE LINV3F OF IMSL LIBRARY IN VAX IJOB = 3 NM=4 Dl=1.0 CALL LINV3F(A3,B3,IJOB ,NM,NM,Dl,D2,WK, IER3) PUT THE SOLUTION OF A3X=B3 I NTO THE FULL SOLUTION A A(2,4)= B3(1 , 1) A(2 , 5) = B3(2 , 1) A(3,4)= B3(3 , 1) A(3,5)= B3(4 , 1) RETURN END
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CCCCCC100 & & 200 & 300 & & 400 SUBROUTINE OUTPUTA(A,K) THIS SUBROUTINE IS TO OUTPUT THE SOLUTION OF LINEAR EQUATIONS DIMENSION REAL A(4,5) REAL K(4) OUTPUT K WRITE(10,300) DO I=1,4 WRITE(10,400 )I,K(I) END DO OUTPUT A WRITE (10, 100) DO I =1,4 WRITE(10,200 ) I , (A(I,J) ,1=1,5) END DO FORMAT 113 FORMAT(///,' THIS IS INITIAL SOLUTION OF LINEAR EQUATIONS,' I I' 30X, , A1, '20X,) A2, '20X, 'A3' 20X 'A4' 20X 'AS' //) ' , ' ' ' FORMAT(' ***** ELEMENT ' ,I1,' *****' ,5(5 F15 . 6)) FORMAT(///,' THIS IS INITIAL SOLUTION OF K ', 5X, 'FROM K1 TO K4', / / ) FORMAT( /,' ***** K(' ,I1, ') ***** = ' ,F20.5) RETURN END
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CC114 SUBROUTINE FCN(X,F,N,PAR) THIS SUBROUTINE IS USED IN SUBROUTINE ZSPOW OF IMSL LIBRARY c************** DIMENSION INTEGER N REAL X(N),F(N) ,PAR(N) REAL LAMDA(4),DLAMDA(4) REAL QV(4) ,QD(4),QH(4),QW(4) REAL L(4),P(4) ,S(4),XX(4) REAL A(4,5) REAL LP(4) REAL SETA EXTERNAL U,W C ************* COMMON CCC& & & & COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA COMMON/LOADS/QD,QW,QV,QH COMMON/LS/L, S COMMON/LP/LP REDO THE LINEAR SOLUTION FOR NEW X CALL LINEAR(A,X,2) SET UP PAR PAR(l) =QV(2) PAR(2)=QV(4) PAR(3) =0.0 PAR( 4) = 0 . 0 FUNCTION F F(l)=(X(l))*(X(l))*S(l)/L(l) A(2,3)*(X(2))*(X(2))*S(2)/L(2)*COS(SETA) (X(2)) *(X(2))*S(2)/L(2)* SIN(SETA)PAR(l) F(2)=(X(3))*(X(3))*S(3)/L(3)* SIN(SETA) +A(3,3)*(X(3))*(X(3))*S(3)/L(3)*COS(SETA) +(X(4))* (X(4))*S(4)/ L(4)PAR(2)
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& & & F(3)=U(X,A,3,l.O)*COS(SETA)+W(X,A,3,1.0) *SIN(SETA)+U(X,A,2,l.O)*COS(SETA) . W(X,A,2,l.O)*SIN(SETA)PAR(3) F(4)=U(X,A,4,1.0)U(X,A,3,0.0)*SIN(SETA) +W(X,A,3,0 . 0)*COS(SETA)PAR(4) RETURN 115
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SUBROUTINE RESULT(A,X,ISW) CTHIS FUNCTION IS USED IN SUBROUTINE FCN c********* ***** DIMENSION INTEGER ISW,NTIME REAL TOTALP, WX1 REAL QV(4),QD(4),QH(4),QW(4) REAL X(4) REAL LAMDA(4),DLAMDA(4) REAL A(4,5) REAL L(4),S(4) REAL P(4) REAL LP(4) REAL PV,PH REAL SETA C ************** COMMON COMMON/LAMDA/LAMDA,DLAMDA COMMON/LOADS/QD,QW,QV,QH COMMON/LS/L,S COMMON/ANGLE/SETA . CEVALUATE ALL OF U AT ANY JOINT CCUlO=U(X,A,l,O .O)*L(l) Ull=U(X,A ,l,l.O)*L(l) U20=U(X,A,2 , 0 . 0) *L(2) U21=U(X,A ,2,1.0)*L(2) U30=U(X , A,3 , 0 . 0) *L(3) U31=U(X,A,3,l.O)*L(3) U40=U(X,A,4,0.0)*L(4) U41=U(X,A,4,l .O)*L(4) EVALUATE W WlO= W(X,A,l , O .O)*L(l) Wll =W(X,A,l,l.O)*L(l) W20=W(X,A,2, 0 . 0) *L(2) W21=W(X,A,2,l.O)*L(2) W30=W(X,A,3, 0 . 0) *L(3) W31=W(X,A,3,l .O)*L(3) W40=W(X,A,4,0.0)*L(4) W41=W(X,A,4,l.O)*L(4) EVALUATE WX 116
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C10 20 & WX10=WX(X,A,1,0.0) WX11=WX(X,A,1,1 . 0) WX20=WX(X,A,2 , 0 . 0) WX21=WX(X,A,2 ,1.0) WX30=WX(X,A,3,0.0) WX31=WX(X,A,3 , 1 .0) WX40=WX(X,A,4,0.0) WX41=WX(X,A,4,1 .0) OUTPUT U, W , AND WX IF(ISW.EQ.1)GO TO 10 IF(ISW.EQ . 3)GO TO 30 WRITE(10,700) GO TO 20 WRITE(10,800) WRITE (10, 100) WRITE(10,200)U10 ,U1 1 ,U 20 ,U21, U30,U31, U40,U41 WRITE(10,300) WRITE(10,400)W10,W11 ,W20,W21, W30,W31, 117 & W40,W41 WRITE(10,500 ) WRITE(10,600)WX10 ,WX11,WX20,WX21,WX30, & WX3l,WX40,WX41 30 & C100 200 & & GO TO 99 TOTALP=QV(2)+ QV(3)+QV(4) + QD(2) *L(2) *COS(SETA)+ QD(3) *L(3)*COS(SETA) WXl=WX11 WX4=WX41 WRITE(15,900)WX1,TOTALP WRITE(16,900)WX4,TOTALP FORMAT FORMAT (I I I' 24X, 'UlO, 'llX, 'U11 , '11X, 'U20,' 11X , 'U21', llX,' U30' , llX, 'U31', llX, ' U40, 'llX, 'U41, 'I I) FORMAT(' ***** U ***** ' , 8 (2X,Fl2.6))
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300 400 500 600 700 800 900 99 & & & & & & & & 118 FORMAT(/ I I' 24X, 'WlO' 'llX, 'Wll' 'llX, 'W20'' llX, 'W21', llX, 'W30', llX, 'W31' ,llX, 'W40' ,llX, 'W41' ,//) FORMAT(' ***** W ***** ',8(2X,Fl2.6)) FORMAT(/ I I' 24X, 'WXlO' 'lOX, 'WXll' 'lOX, 'WX20' ' lOX, 'WX21', lOX, 'WX30', lOX 'WX31' lOX 'WX40' lOX 'WX41' //) ' ' ' ' ' ' FORMAT(' ***** WX ***** ',8(2X,Fl2.6)) FORMAT(///,' THESE SOLUTIONS DEVELOPED BY A, AND K ARE THE NO LINEAR RESULTS OF THIS SYSTEM.',///) FORMAT(///,' THESE SOLUTIONS DEVELOPED BY A, AND K ARE THE LINEAR RESULTS OF THIS SYSTEM.',///) FORMAT(2(5X,F20.10)) RETURN END
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FUNCTION U(X,A,N,DX) CTHIS FUNCTION IS USED IN SUBROUTINE FCN C*************** DIMENSION INTEGER N REAL DX,Ul,U2,U3,U4,U5,U6,U7,U8,U9,Ul0, & Ull,U12,U REAL QV(4),QD(4),QH(4),QW(4) REAL X(4) REAL LAMDA(4),DLAMDA(4) REAL A(4,5) C********* ****** COMMON CCOMMON/LAMDA/LAMDA,DLAMDA COMMON/LOADS/QD,QW,QV,QH FUNCTION U IF(DX.EQ .O.O)GO TO 10 IF(DX .EQ.l.O)GO TO 20 Ul=A(N,5)X(N) *X(N)*DX/DLAMDA(N) .U2=UlX(N)/4.*A(N,l)*A(N,l)*(X(N)*DX+ & SIN(X(N) *DX)*COS(X(N)*DX)) U3=U2X(N)/4.* A(N,2) * A(N,2) * (X(N) *DX& SIN(X(N) *DX)*COS(X(N)*DX)) U4=U3QW(N)*QW(N)*DX*DX* DX/6./X(N) & /X(N)/X(N)/X(N) U5=U4A(N,3) * A(N,3) * DX/2. U6=U5+X(N)/4.*A(N,l)* A(N,2)*(1.COS(2. & * (X(N) * DX))) U7=U6+A(N,l) â€¢QW(N)/ X(N)/X(N) /X(N)A(N,l) & *QW(N)*COS(X(N)*DX) & /X(N)/X(N) / X(N) U8=U7A(N,l)*QW(N)*DX* SIN(X(N) *DX) & /X(N)/X(N) U9=U8A(N,l) * A(N,3)*SIN(X(N) *DX) U10=U9+A(N,2)*QW(N)*SIN(X(N) *DX) & /X(N)/X(N)/X(N) Ull=U10A(N,2) *QW(N)*DX*COS(X(N)*DX) & /X(N)/X(N) U12=Ull+A(N,2) * A(N,3) * (1COS(X(N)*DX)) U=U120.5*A(N,3)*QW(N)*DX* DX/X(N)/X(N) 119
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10 20 gg & & & & & & & & & & GO TO 99 U=A(N,S) GO TO 99 Ul=A(N,S)X(N)*X(N)/DLAMDA(N) U2=UlX(N)/4.*A(N,l)* A(N,l)*(X(N) +SIN(X(N)) * COS(X(N))) U3=U2X(N)*A(N,2)*A(N,2)/4.*(X(N) SIN(X(N)) * COS(X(N))) QW(N)* QW(N)/(6.*X(N)*X(N)*X(N)*X(N)) A(N,3)*A(N,3)/2. 120 U4=U3+X(N)/4 .*A(N,l)* A(N,2) *(1.COS(2* X(N))) +A(N,l)* QW(N)/(X(N)*X(N) * X(N)) A(N,l)*QW(N)* COS(X(N))/(X(N) *X(N)*X(N)) A(N ,l)*QW(N)*SIN(X(N))/(X(N ) *X(N)) U5=U4A(N,l)*A(N, 3) * SIN(X(N)) + A(N,2) *QW(N) . *SIN(X(N))/(X(N)*X(N) * X(N)) A(N,2) *QW(N)* COS(X(N))/(X(N) *X(N)) U=U5+A(N,2)* A(N,3) *(1.COS(X(N))) A(N,3) * QW(N)/2./(X(N) *X(N)) RETURN
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CCCC121 FUNCTION W(X,A,N,DX) THIS FUNCTION IS TO FIND SIDEWAY DISPLACEMENT DIMENSION INTEGER N REAL DX,Wl,W2,W3,W REAL A(4,5) REAL X(4) REAL QV(4) ,QD(4),QW(4),QH(4) COMMON COMMON/LOADS/QD,QW,QV,QH EVALUATE W Wl=A(N,l) * SIN(X(N) *DX) W2=Wl+ A(N,2) *COS(X(N)*DX) W3=W2+A(N,3)*DX+A(N, 4) W=W3+QW(N)*DX* DX/(2 *X(N)* X(N)) RETURN END
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FUNCTION WX(X,A,N,DX) C cTHIS SUBROUTINE IS TO EVALUATE RATATION OF EACH JOINT C********** ***** DIMENSION INTEGER N REAL DX, WXl , WX2 , WX REAL X(4) REAL A(4,5) REAL QD(4),QW(4),QV(4),QH(4) C *************** COMMON CCOMMON/LOADS/QD,QW,QV,QH EVALUATE WX WXl=A(N,l)*X(N) *COS(X(N)*DX) WX2=WX1A(N, 2) *X(N)* SIN(X(N) *DX) WX=WX2+ A(N,3) +QW(N)* DX/X(N)/X(N) RETURN END 122
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SUBROUTINE PLOTFRAME(A,X) C+++++++++++++THESE DIMENSION ARE FOR CALCOM PLOTER PARAMETER IN=lOOl REAL Xl(IN),Yl(IN) REAL X01(2),X02(2),X03(2),X04(2) REAL Y01(2),Y02(2),Y03(2),Y04(2) REAL XF1(101),XF2(101),XF3(101),XF4(101) REAL YF1(101),YF2(101),YF3(101),YF4(101) REAL X(4) 123 CC LX: LENGTH OF XAXIS(INCH) (MAX.=8 INCH) C LY: LENGTH OF YAXIS(INCH) (MAX.=S INCH) C (XO,YO): LEFTLOWER ORIGIN COORDINATE OF PICTURE C MEASURED C FROM LEFTLOWER CORNER (0,0) OF THE C PAPER C XO: XCOORDINATE DATA VALUE AT THE ORGIN OF THE C PICTURE C YO: YCOORDINATE DATA VALUE AT THE ORGIN OF THE C PICTURE C DX: XDATA INCREAMENT BETWEEN TWO NEIGHBOR C TICKS(l INCH) C DY: YDATA INCREAMENT BETWEEN TWO NEIGHBOR C TICKS(l INCH) C++++++++++++++ C++++++++++++++ THIS DIMENSION ARE FOR ALL SYSTEM DIMENSION REAL A(4,5) REAL L(4),S(4) REAL P(4) REAL QD(4),QW(4) ,QV(4),QH(4) REAL LAMDA(4),DLAMDA(4) REAL SETA C====================================================l C L : THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM I C FEMUC I C K :THE CONSTANT RELATIVE TO AXIS FORCE I
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C S :THE CONSTANT IS EQAL TO EI/L C QV:THE CONCENTRATE LOAD VERTICAL TO X AXIS AT C EACH JOINT. C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO EACH C BAR C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO C QD*L** 3 /(E*I) C QH:THE LOADS PARAL TO AAXIX, HORIZONTAL LOADS, C AT EACH JOINT. C KK:KK=K* K C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE C MODIFIED AS RADIUM C==================================================== c ************** COMMON CCCC++++++ & C++++++ COMMON/LS/ L , S COMMON/LOADS/QD,QW,QV,QH COMMON/ANGLE/SETA COMMON/LAMDA/LAMDA,DLAMDA DATA FOR PLOTETER SYSTEM DATA LX,LY/ 8 , 6 / DATA XO,Y0/ 3 .0,2. 0 / DATA XO,Y0/ 0 . 0 , 0 . 0 / DATA DX,DY/ 0 .5,0.5/ HERE IS FOR INPUT DATA HERE IS FOR ORIGINAL FRAME CALL ORIFRAME(XOl,YOl, X02,Y02 , X03,Y03, X04,Y04,SETA) HERE IS FOR ROTATION OF ELEMEN T 2 , AND 3 CALL PLOTRESULT(A, X , XFl,YFl ,XF2,YF2,XF3, YF3,XF4,YF4,SETA) HERE IS FOR ORIGINAL FRAME FIGURE 124
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125 CHERE IS FOR DEFORMATION FRAME C=====================================================l c I C THOSE SUBROUTINE ARE REQUIRED FOR CALCOM I c I C=====================================================l Cc Cc CCCALL PLOTS(O. O,O.O,O) CALL LETTER CALL LANDS CALL PLOT(XO,Y0,3) CALL NEWPEN (7) HERE IS FOR AXIS IF ANY HERE IS FOR SYMBOL IF ANY HERE IS FOR CALCOM CALL CALCOM(X01,Y01,2,XO,DX,YO,DY,l) CALL CALCOM(X02,Y02,2,XO,DX,YO,DY,l) CALL CALCOM(X03,Y03 , 2,XO,DX , YO,DY,l ) CALL CALCOM(X04,Y04,2,XO,DX,YO,DY,l) CALL CALCOM(XFl,YFl,lOl,XO,DX,YO,DY,2) CALL CALCOM(XF2,YF2,101,XO,DX,YO,DY, 2) CALL CALCOM(XF3,YF3, 101,XO ,DX,YO,DY,2) CALL CALCOM(XF4 ,YF4, 101,XO ,DX,YO,DY, 2) HERE IS FOR PLOTE CALL PLOT(0. 0 , 0 . 0 , 999 ) RETURN END
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C===================================================l c I C SUBROUTINE CALCOM I c I C===================================================l SUBROUTINE CALCOM(X,Y,K1,STARTX,DELTAX,STARTY, DELTAY,MK) DIMENSION XX(1001) ,YY(1001),X(K1),Y(K1) DO 100 I=1,K1 XX(I)=X(I) YY(I)=Y(I) 100 CONTINUE XX(K1+1)=STARTX XX(K1+2)=DELTAX YY(K1+1)=STARTY YY(K1+2)=DELTAY CALL LINE(XX,YY,K1,1,0,MK) RETURN END 126
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& CSUBROUTINE ORIFRAME(XOl,YOl, X02,Y02 , X03, Y03, X04,Y04,SETA) DIMENSION REAL SETA,ALPH2,ALPH3,L REAL X01(2),X02(2),X03(2),X04(2) REAL Y01(2),Y02(2),Y03(2),Y04(2) 127 REAL ELEM2(2,2),ELEM3(2,2),R2(2,2),R3(2, 2) REAL C2(2,2),C3(2,2) CCCSPECIFY L, AND PI L = l . O PI=4 . 0 *ATAN( l .O) SPECIFY THE ORIGINAL COORDINATE OF ORIGINAL FRAME C ************** ELEMENT 1 XOl( l)=O.O X01(2)=0 . 0 YOl(l)=O . O Y01( 2)=L c *""************ ELEMEN T 4 DO 1 = 1 , 2 X04(I)=X01( I )+ 2 * L *COS(SETA) Y04(I)=Y01( I ) END DO c ************** ELEMENT 2, AND ELEMENT 3 C+++++ + C++++++ DO I = 1 , 2 X02( I)=X01( I ) Y02( I)=Y01( I ) X03(I)=X01( I ) Y03(I)=Y01(I) END DO CHANGE THE COORDINATE TO THE FORM O F MATRIX TO FIND ROTATION
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128 c *** * ********** ELEMENT 2 ELEM2(1,1)=X02(1) ELEM2(1,2)=Y02(1) ELEM2(2,1)=X02(2) ELEM2(2,2)=Y02(2) c***** **** * **** ELEMENT 3 C++++++ C+++ + + + ELEM3(1,l)=X03(1) ELEM3(1,2)=Y03(1) ELEM3(2,1) = X03(2) ELEM3(2,2) = Y03(2) ROTATION MATRIX ALPH2= 3 *PI/ 2 +SETA ALPH3=PI/2.SETA R2(1,1)=COS(ALPH2) R2(1,2)=SIN(ALPH2) R2(2,1)= SIN(ALPH2 ) R2(2 ,2)=COS(ALPH2) R3( 1 , 1 )=COS(ALPH3) R3(1,2)=SIN(ALPH3) R3( 2 , 1)=SIN(ALPH3) R3( 2 , 2) =COS(ALPH3) CALCULATE ROTATION COORDINATE CALL VMULFF(ELEM2,R2,2,2,2,2,2,C2,2, IE) CALL VMULFF(ELEM3,R3,2,2, 2 , 2 ,2,C3,2, IER3) X02( 1)=C2( 1,1) Y02( 1)=C2( 1 , 2 )+ L X02( 2) =C2( 2,1) Y02( 2) =C2( 2 , 2 )+ L X03(1) = C3(1 , 1) + 2 * L *COS(SETA) Y03(1) = C3(1 , 2) + L X03(2) = C3(2 , 1 )+ 2 * L *COS(SETA) Y03(2) = C3(2 , 2) + L RETURN END
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129 & SUBROUTINE PLOTRESULT(A,X,XFl,YFl,XF2,YF2 ,XF3,YF3,XF4,YF4,SETA) CDIMENSION REAL XF1(101),XF2(101),XF3(101),XF4(101) REAL YF1(101),YF2(101),YF3(101),YF4(101) REAL U1(101),U2(101),U3(101),U4(101) REAL W1(101),W2(101),W3(101),W4(101) REAL R2(2,2),R3(2,2) REAL C2(101 ,2),C3(101,2) REAL ELEM2(101,2),ELEM3(101,2) REAL DY(lOl) REAL X(4) C++++++++++++++ THE DIMENSION IS FOR SYSTEM REAL A(4,5) REAL L(4),S(4) REAL P(4) REAL QD(4),QW(4),QV(4) ,QH(4) LAMDA(4),DLAMDA( 4) REAL SETA,ALPH2,ALPH3,LL,DL C======================================================l C L : THE LENGTH OF EACH BAR I C P :AXIAL FORCE WHICH BE CALCULATE FROM FEMUCD I C K :THE CONSTANT RELATIVE TO AXIS FORCE C S :THE CONSTANT IS EQAL TO EI/L C QV:THE CONCENTRATE LOAD VERTICAL TO X AXIS AT EACH C JOINT. C QD:DISTRIBUTION LOADS WHICH IS VERTICAL TO C EACH BAR C QW:THESE LOADS ARE MODIFIED WHIH ARE EQUAL TO C QD*L**3 /(E*I) C QH:THE LOADS PARAL TO A AXIX, HORIZONTAL LOADS, AT C EACH JOINT. C KK:KK=K*K C SETA:ANGLE WHICH WHEN INPUT IS IN DEGREE MODIFIED C AS RADIUM C====================================================== c ************** COMMON COMMON/LS/L,S
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COMMON/LOADS/QD,QW,QY,QH COMMON/LAMDA/LAMDA,DLAMDA 130 CCSPECFY LL, THE UNIT LENGTH IN THIS GRAPH, AND PI PI=4.0*ATAN(1.0) LL=1. 0 DL=LL/100 . DO I=1, 101 DY(I) =(I1)*DL END DO CDO I=1,101 c ************** u U1(I)=U(X,A,1,DY(I)) U2(I)=U(X,A,2,DY(I)) U3(I)=U(X,A,3,DY(I)) U4(I)=U(X,A,4,DY(I)) c ************** w W1(I) = W(X,A,1,DY(I)) W2(I) =W(X,A, 2 , DY(I) ) W3(I)=W(X,A, 3 , DY(I)) W4(I)=W(X,A,4,DY(I)) c ************* wx WRITE(20, *)DY(I),U1(I),U2(I),U3(I),U4(I), & W2(I) , W3(I) END DO C ************** ELEMENT 1 DO I=1,101 XF1(I) =O.OW1(I) YF1(I)=DY(I) +U1(I) END DO c ************** ELEMENT 4 DO I=1,101 XF4(I)=0.0+2.*LL* COS(SETA)W4(I) YF4(I) = DY(I) +U4(I) . END DO c ************** ELEMENT 2 , AND ELEMENT 3 DO I=1,101
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XF2(I)=O . OW2(I) YF2(I)=U2(1)+DY(I)(I1)*U2(101)/100. c************** ELEMENT 3 C++++++ C++++++ XF3(I)=O.OW3(I) YF3(I)=U3(1)+DY(I)(I1)*U3(101)/100. END DO CHANGE THE COORDINATE TO THE FORM OF MATRIX TO FIND ROTATION c ************** ELEMENT 2 DO I=1,101 ELEM2(I, 1)=XF2(I) ELEM2(I,2)=YF2(I) c************** ELEMENT 3 C++++++ C++++++ & & ELEM3(I,1)=XF3(I) ELEM3(I,2)=YF3(I) END DO ROTATION MATRIX ALPH2=3.*PI/2.+SETA ALPH3=PI/2.SETA R2(1,l)=COS(ALPH2) R2(1 , 2 )=SIN(ALPH2) R2(2,1)=SIN(ALPH2 ) R2(2,2)=COS(ALPH2) R3(1,1)=COS(ALPH3) R3(1,2)=SIN(ALPH3) R3(2,1)=SIN(ALPH3) R3(2,2)=COS(ALPH3) CALCULATE ROTATION COORDINATE CALL VMULFF(ELEM2,R2,101,2,2,101,2,C2, 101 , IER2) CALL VMULFF(ELEM3,R3,101,2,2,101,2 ,C3, 101 , IER3) C++++++++++++++ ELEMENT 2 131
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DO 1=1, 101 XF2(I)=C2(I,1) YF2(I)=C2(I,2)+LL C++++++++++++++ ELEMENT 3 XF3(I)=C3(I,1)+2.0*LL*COS(SETA) YF3(I)=C3(I,2)+LL END DO RETURN END 132
