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A pseudo dynamic method for structural analysis
Item Type text; Thesis-Reproduction (electronic)
Authors Zimmerman, Eugene George, 1945-
Publisher The University of Arizona.
Rights Copyright © is held by the author. Digital access to this materialis made possible by the University Libraries, University of Arizona.Further transmission, reproduction or presentation (such aspublic display or performance) of protected items is prohibitedexcept with permission of the author.
Download date 19/06/2018 06:59:06
Link to Item http://hdl.handle.net/10150/318174
STATEMENT BY'AUTHOR
This thesis has been submitted in partial fulfillment of requirements for an advanced degree at The University of Arizona and is deposited in the University Library to be made available to borrowers under rules of the Library*
Brief quotations from this thesis are allowable without special permission, provided that accurate acknowledgment of source is made, Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or the Dean of the Graduate College when in his judgment the proposed use of the material is in the interests of scholarship. In all other instances, however, permission must be obtained from the author*
SIGNED:
APPROVAL BY THESIS DIRECTOR .
This thesis has been approved on the date shown below:
ACKNOWLEDGMENT
The author wishes to express his gratitude to Dr *
Ho M 0 Richard for his guidance and suggestions which greatly
aided in the development and presentation of this thesis*
The author also wishes to thank his wife, Jan, for
her patient work in typing the manuscript*
iii
TABLE OF CONTENTS
Page
LIST OF ILLUSTRATIONS . , v
LIST OF TABLFS . . . . . . . . . . . . . . . . . . . . vxi
A B S T RAC T ■ . . . . . . . . # @ . . . . . . . . . . . . . xz i x i
CHARTER 1 - INTRODUCTION . . . . . . . . . . . . . . 1
CHAPTER 2 - THEORETICAL DEVELOPMENT AND .,PRESENTATION OF THE METHOD . . . . . . . . 5
Stiffness Formulation by theDirect Stiffness Method . . . . . . . . ., . . 5
Dynamic Formulation Using LinearAcceleration Model . . . . . . . « . .. . . . / 8
Application of the Pseudo Dynamic Analysis . . 13 ,Content of the Computer Program . . . . 15
CHAPTER 3 - APPLiCATION TO A BUILT-IN BEAM . . ... . . 18,Effect of Damping Coefficients . . . . .. . « . 21Effect of Time Increment Size 22
CHAPTER 4 - APPLICATION TO A PORTAL FRAME . . . . . . . 26Solution for Laterally Loaded Frame . . . . . . 26Adjustment of the Axial Frequencies . . . . . . 35
Elimination of the Axial Frequencyof the Beam . . . . . . . . © . . . . . . 35
Adjustment of the Axial Frequencyof the Columns . . . . . . . . . . . . . 36
Solution for Frame with Verticaland Lateral Load . . . ... . . . . . . . . . 3 7
CHAPTER 5 - APPLICATION TO: A BUILDING FRAME . © . . . . 46
CHAPTER 6 - SUMMARY AND CONCLUSIONS . . . . . . . . . . 56 .
APPENDIX . . . O . O . . . o O* . 0 . 0 . 0 O . O O O O 5 9Flow Chart for Computer Analysis . . . . . . . 59Fortran Program . . . . . . . . . . o . . . . . 6T
REFERENCES . . . . . . . . . . . . . . . . . . . . . . 67
- ■ . , , . . ' ■ . . iv . :
LIST OF ILLUSTRATIONS
Figure Page
1 „ Axial Force, Element . . 6
2 o Flexural Element . . . » « <, „ » . . <. » » « . 6
3 e Compatibility Relations ............ . . . 6
4 0 Single Degree of Freedom System » . » = <, . = . <> 9
5,- Linear Approximation of the Acceleration <= » « « 9
6 c Solution of a Single Degree of FreedomSystem Using Heavy Damping „ « „ ., . , • - 14 ■
7o Formulation of the System.Stiffness Matrix « » = 14
Be Built-in Beam Example . » . * . « „ • . • •« * • . « 19
9, .Relationship.Between Damping and Damped Period• « 19
1Oo Plots of Undamped, 10% Critically Damped, and30% Critically Damped System. «. 23
11. Plots of 50%, 7 0 % , and 90% Critical Damping . . . 24
12. Effect of Time Increment Size for 70%Critical Damping . . . . .. . . . . . . . . o 25
13. Portal Frame Example . . . . . . . . . . . . . . 27
14. Obtaining the Element Masses . . . . . . . . . . 27
15. Plot of Number of Increments Required forSolution vs. Size of Time Increment . . . . . 30
16. Horizontal Displacement of Node 2 . . . . . . . . 32
17. Vertical Displacement of Node 2 . . . . . . . . . 33
18. Rotational Displacement of Node 2 . . . . . . . . 34
19. Laterally and Vertically Loaded Frame . « . . . . 40
.:.v ■ V ■■ v • ■.■■■■■/; -■
yi
LIST OF ILLUSTRATIONS— Continued
Figure Page ■
20. Horizontal and Vertical Displacements Underthe 40 Kip Load (Signs Neglected) e e « . . « 43
21 „ Four Story Building Frame » . . . . . - . . •. « . . 47
22* Idealization f o r .Rayleigh Method , , . « « « . , 49
23c Sidesway Displacements of Four Story Frame . . . 52
. " LIST OF TABLES '
Table .. . Page
1® Solution to Built-in Beam e o © © @ © o © © 0 O O 21
2© Displacements for Portal Frame © © © . O . © o © © 31
3© Element Forces for Portal F r a m e © © © © © e 0.0 o 31
4© Displacements Obtained in 35 Increments byReducing the Axial Frequencies ® « » 38
5 e . Element Forces Obtained in 35 Incrementsby Reducing the Axial Frequencies . „ e . . . 38
6 c, Displacements for Laterally and VerticallyLoaded Portal Frame « e „ e « „ « •« e = . , •" 42
7, Member Forces for Laterally and VerticallyLoaded Frame @ © © «. © « © © © © © © © © « © 42
8© Sidesway Accelerations of Nodes 2, 3, and 4 © © © 45
. 9® Properties of Four Story Frame © « © © © © © . . 48
10. Displacements of Four Story Frame . © , . © © © © 53
11© Element Forces for Four Story Frame © © © © © = ;© 54
vii
ABSTRACT
.A method is presented herein for the analysis of
frame structures under static loading,' assuming that the .
structure is initially.unloaded and then determining the
displacements as the loading is applied. This is done by
formulating the stiffness matrix by the Direct Stiffness
Method and then generating the dynamic equations of the
system. Damping is included in the dynamic equations so
that the displacements converge to the static values.
The method is applied first to a built-in beam
and the effects of damping coefficients are studied, as
well as the effect of time increment size. Next it is
applied to a portal frame and procedures are developed to
decrease the axial frequencies of the members so as to pro
duce, a good solution in fewer time increments. Finally a
tall building frame is analyzed with the method.
The. method gives very accurate solutions for both
laterally and vertically loaded frames and includes the
effect of axial deformations. However more computer time
and storage space are required than for the more conventional
finite element methods.
Solutions were obtained using a CDC 6400 computer.
. CHAPTER 1 ;:; v
INTRODUCTION
. The advent of the modern-day computer has radically
changed the analytical approach to the solution of large
statically indeterminate structures such as building frames.
In the past approximate methods of analysis, such as Moment-
Distribution,: 51 ope-Deflection, and Newmark's Method, which
are suitable to manual calculations, were used extensively6 :
The capabilities of stbring large masses of information and
quickly performing large numbers of arithmetic operations
have been made possible by the computere" This has given
rise to more fundamental methods of analysis„ The stress- .
strain relationships and the compatibility requirements for
each individual structural member can be generalized in matrix
form to obtain "exact" equations relating the displacements
and forces of the entire systemB The computer is ideally
suited to perform ' the. matrix operations needed to solve these
equations to obtain the element forces and displacements«
In the Direct Stiffness Method, (Turner et al, 1956)
as in most other finite-element m e t h o d s i t is assumed that
the structure has already been loaded and that the displace
ments have reached their final equilibrium position, the
internal forces also having reached their final values«.
Assuming a linear, stress-strain relationship, the member
distortions v , can be then related to the member forces p,
such that
p=kv, (1-1)
where k is the element stiff ness matrix»
Next, a compatibility matrix B, can be formed which
•relates the member distortions to .the system nodal displace
ments d , so that
; v=Bd, ' (1-2)
Then element stiffness coefficients can be calculated for
each element by obtaining the product B^kB. These coeffi
cients for each member are then added in the appropriate
manner to obtain the system stiffness array, K« The applied
forces f, are then related to the system displacements by
f=Kd (1-3)
so that the displacements can now. be found by pre-multi
ply in g f by the inverse of K.. Finally the element forces /
can be found by the equation
. p=kBd o (1-4)
A fundamentally different approach has been
formulated by A. So Day (1965)* His "Dynamic Relaxation"
Method assumes that the structure is initially unloaded and
follows the development of the internal forces with each
increment of time* Day uses the. classical slope-deflection
equations of equilibrium and converts them to dynamic
equations« These are then replaced:by finite difference
equations which can be solved and integrated by an incre
mental process, to obtain the final member forces« In his:
paper Day treated a simple portal frame.with equal moments
applied at the two ends of the top member, thus avoiding
problems of sidesway. (He also deals with some plate and
shell problems.)
' A hew approach for structural analysis, using D a y ’s
basic concepts that the structure is initially unloaded and
that the internal forces are gradually developed throughput
the structure, is presented in this thesis* This method,
which is termed the "Pseudo Dynamic" Method for static
structural analysis, utilizes the basic equilibrium and
compatibility equations of the Direct Stiffness Method*
Thus this method takes into account axial deformations*
In the Pseudo Dynamic Method, the system stiffness
array K is formed as before* However, this large array is
not inverted to solve for the displacements* Instead the
structure is idealized to a system of discrete masses and
continuous flexibility* The equations of motion for this
system under the constant external forces are written and
integrated assuming that the acceleration is linear through
out each time interval* Heavy damping is applied to the
system so that the "response" of the structure quickly
converges to the static displacements. These static
' ' ' ■ . ■ ' ' : 4 displacements are then related to the member forces as
indicated bef ore s
P=kSd '
and the problem is solved.
The purpose of this thesis is to present the -
Pseudo Dynamic Method and to apply it to built-in beams,
simple portal frames with sidesway, and large building
frames® The proper choice of masses, damping coefficients,
and time increments is investigated, as well as methods for
decreasing the number of time increments necessary for a
solution. The accuracy of this method is compared to the
"exact" solutions obtained by the Direct, Stiffness Method• * •
and the feasibility of the method with regard to computer
storage space and computer time is analyzed, The method
is used only for linear s y s t e m s b u t could also be extended
to certain nonlinear systems.
CHAPTER 2
THEORETICAL DEVELOPMENT AND PRESENTATION
OF THE METHOD
This method of structural analysis combines portions
of the Direct Stiffness approach with the dynamic equations
of the system. Heavy damping is applied to reach a solution
for the static displacements and forces.
Stiffness Formulation by the Direct Stiffness Method
The element stiffness matrix k for a general frame
element can be derived using an axial force element and a
flexural element.
For the axial force element, Fig. 1, assuming that
the stress in the bar is proportional to the strain,
Pi v iX = E f . (2-1)
Rearranging this equation gives
Pf = ££ Vi (2-2)1 1
where AE_ represents the axial stiffness of the member.1For the flexural element, Fig. 2, using the usual
slope-deflection relationships, it can be shown that
P j " 1 r v j + V k - ( 2- 3 )
5
FIGURE 1. AXIAL FORCE ELEMENT
FIGURE 2. FLEXURAL ELEMENT
J END
I END
FIGURE 3. COMPATIBILITY RELATIONS
8
~v i COSOC.£ s in # ^ 0 C O S +Tr) sin(o<i + ir) 0
Vj
= s i n c X ^ -cosou 1 s in (c x i +ir) - c o s (** +TT) 0
1 1 1 1
v k sinc*^ cosoc 0 sin(<x^+Tij - c o s (o l +Tf) 1
1 1 1 1
6_ (2-7)
If a coordinate system is defined for the structure,
as shown in Fig. 3, and we define 1 as the projection of the
element on the x-axis and 1 as the projection on the y-axis,
we can reformulate the compatibility matrix:
B =
R x 0 ix cF
1 i 1 1
{»~^x12 -
1- h1̂
Xxl z
0
1 y ~ 1X 0 >s| 1—1 1 lx 1
i z ,2I2 r
(2-8)
Dynamic Formulation Using Linear Acceleration Model
For a single degree of freedom system with viscous
damping, as shown in Fig. 4, the equation of motion is
mx+cx+kx=P(t )• (2-9)
Here m is the mass of the system, c is the damping coeffi
cient, k is the spring constant or stiffness coefficient,
and P(t) is the dynamic load. The displacement, measured
from the initial equilibrium position of the mass, is de
noted by x and the velocity and acceleration are denoted by
x and x, respectively. The coefficients c and k are constants
for linear systems.
Equation (2-9) can be solved by several numerical
•procedures such as Taylor Series Solution or Runge Kutta.
9
FIGURE 4. SINGLE DEGREE OF FREEDOM SYSTEM
Actual Accel
^Linear Approxim
:x
— t — >
t-2T t-T t t+T
Time, T*
FIGURE 5. LINEAR APPROXIMATION OF THE ACCELERATION
However the method used for this thesis assumes that the
acceleration varies linearly with time over each time
increment. This is the so-called Linear Acceleration
Method (Norris et al, 1959).
the curve obtained with the linear acceleration approxima
tion. This method will provide a very good approximation to
the actual acceleration if a small time increment T is used.
From Fig• 5, the acceleration at time T* = t may be written
as
where a is the slope of the chord of the acceleration curve
in this time interval. Integrating equation (2-10) with
respect to time over the interval T yields
*t = y t-TT + ^ + *t-T. (2-11
Integrating again with respect to time yields
Figure 5 shows the actual acceleration curve and
(2-1 0)
2 (2-1 2)Solving Equation (2-10) for the slope a yields
a x t x t-T.T (2-13)
Substituting this expression for a into Equation (2-11) gives
11Now, define the quantity X such that
X E x t_TT ♦ x T2 (2-15)
and substitute this into E quation(2-14) to obtain
x. = x. T + X 1 2 (2-16)
Similarly, substituting Equation (2-13) into Equation (2-12)
yields
x t = V T + X t-IT + T x t-T + X t-T. 6 3 (2-17)
Defining the quantity & such that
£ ~ Xt-TT + X t-TT + X t-T3 (2-18)
and substituting into Equation (2-17) gives
*x -2 t “ tx i. = x *.T + g .
& (2-19)Substitute Equations (2-16) and (2-19) into Equation (2-9)
to obtain
= P(t) - c/x\T +)f\ - k/xfT 2 ++ X) - kf X tT + i)(2-20)
or
x.fm + cT + kT^) = P(t) - c% - k& .1 T T (2-20a)
Define the equivalent mass as,2m* = m + T_ + T .
2 6 k ( 2 - 2 1 )
so that
x t = 1, (p(t) - cS - kS).m* (2-2 2)
12For the analysis technique developed in this thesis,
the initial accelerations, velocities, and displacements may
be taken to be zero. The loads on the structure will be con
stant instead of time-varying and so the loads will be re
ferred to as the vector P rather than P(t).
To extend the method to structural systems with a
finite number of degrees of freedom, the equations derived
may be put into matrix form:
t-T t-T3
^ = T *xt_T +
(Y1* = |Y| + T r + T 2 .,2 C ~ K
-1 x (2-23)x. = (Yl* (P - CX - K6)
x i. = if + T ..2 x t
xt = V l 2x to u
These are the recurrence formulas for the system, where
x is the vector of the accelerations of the masses at L — Itime t-T, x̂ _.-p is a vector of the velocities of the masses,
x k_,rp is a vector of the displacements of the masses, M is a
diagonal matrix of the system masses, C is a square matrix
of damping coefficients, and K is the square matrix of
stiffness coefficients.
13• Application of the Pseudo Dynamic Analysis
The above dynamic equations (2-23) are now applied
to the static structure. The damping coefficients are picked
so that the displacements will quickly damp out to the static
solution as shown in Fig. 6. Thus the method is termed the
Pseudo Dynamic Method for Structural Analysis.
To idealize the structure, the system is replaced by
an equivalent mass-1inkage model with a finite number of
degrees of freedom. The massless segments joining the masses
at the nodes are assumed to possess the same properties as
the corresponding portions of the real structure. All loads
are applied at the node points. The members are assumed to
be uniform and the effects of geometry changes are not con
sidered.
Two important assumptions are made in this method.
First, the matrix of damping coefficients C, is assumed to
be a diagonal matrix: that is no off-diagonal terms which
would be present in a coupled system are included. This is
a good approximation since the stiffness matrix tends to
couple the damping effects. Second, since M is a diagonal
matrix, the off-diagonal terms of the M* matrix are neglected
and the inverse of M* is obtained by merely inverting each
element on the diagonal of M*. This is a good approximation
if T is small since
m* = M + T +2 6 (2-24)
14
4->
Q.
Q
o T ime
FIGURE 6. SOLUTION OF A SINGLE DEGREE OF FREEDOM SYSTEM USING HEAVY DAMPING.
3
J END
FIGURE 7. FORMULATION OF THE SYSTEM STIFFNESS MATRIX
/
15and therefore
m # ii = m ii + | c + ^ K > m *ij = h + r - K (i#J)-(2-25)
There are three degrees of freedom at each nodepoint:
lateral displacement, vertical displacement, and angular
rotation. Therefore the order of the system stiffness array
is 3n x 3n where n is the number of nodepoints. Accordingly,
M and C are also of order 3n x 3n. Since M and C are diagonal
matrices, these arrays may be formed from column matrices M-j
and . These vectors will consist of horizontal, vertical,
and rotational components of the mass or damping coefficients
at each node point.
Content of the Computer Program
The application of this method involves three main
calculations. first, the Direct Stiffness Method is used to
create the system stiffness array K . Then the damping matrix
and the equivalent mass matrix are formed and the solution for
the displacements is obtained. finally these displacements
are multiplied by the stiffness and compatibility matrices of
each element to give the member forces acting on each element.
A flow chart and a copy of the computer program are given in
the Appendix. Solutions were obtained on a CDC 5400.
The program is written so as to conserve storage space
by considering each element individually. The element com
patibility matrix, B, and the element stiffness matrix, k ,
are formed and the product B^kB is found. This new matrix,
of order 6 x 6 , is then packed into the proper position of the
stiffness array, K . . To illustrate the procedure' consider the
system shown in Fig. 7. From Fig. 7 it is apparent that the
generalized displacements of member 4, i.e. x ^ , x^j x ^ , x4 >
X5 S correspond respectively to the generalized system
displacements d^, ^5 » ^6 » ̂ 10 $ ^ 11> ^12' The same relation
ship would also exist for any loads on the system. Therefore $
for element 4,
B kB(i , j) wouId be placed into K (i>3#' j +’3) i = 1,2,3 j=1, 2 , 3
B^k8 (i,j) would be placed into K(i+3, j+6 ) 1=1,2,3 j=4,5 ,6
B^kB(i,j) would be placed into K (i+6 , j+3) i=4,5,6 j=1,2,3; t : ■ ■' ■ .. ■ ■ • /B kB(i ,j) would be placed into K (1 +6 , j+6 ) i=4,5,6 j=4,5,6.
The elements of the system stiffness matrix are additive and
the system stiffness coefficients for a node at which several
elements are connected will be the sum of the element stiff
ness coefficients of each of these elements.
After the system stiffness matrix has been completely
populated, the support conditions are applied. For a support
restraining the i ^ degree of freedom, the i^h r0uu 0f the K.
matrix is changed so that all elements in the row are zero
except, that unity is placed on the diagonal. Thus the K
matrix now represents the actual structure.
Next the damping matrix C is formed from the damp-
ing coefficients read in as a vector. Then the equivalent
mass matrix is formed from the mass vector and inverted by
taking the reciprocal of each member on the diagonal.
17Gamma and delta are then calculated from the initial
conditions by Equations (2-23). Using these values plus the
time increment T , the accelerations, velocities, and displace
ments corresponding to each degree of freedom are computed
from Equations (2-23). Using these values a new gamma and
delta are calculated and the procedure is repeated until the
desired values for the displacements are obtained. To deter
mine this point, the sum of the absolute values of the dif
ferences between the present and previous values of the dis
placements is compared with a norm. If
where n equals the number of degrees of freedom, C is a pre
selected norm, and the subscript represents the increment
number, then the present displacements are considered the
static displacements of the system.
stiffness matrix k and the compatibility matrix B, element by
element, and takes the product times the appropriate displace
ments to obtain the final end moments and axial force for each
element in the system.
do not converge after a specified number of intervals (due,
for example, to an unstable system), the program will ter
minate.
n(2-26)
i = 1
At this point the computer regenerates the element
The program is written so that if the displacements
CHAPTER 3
APPLICATION TO A BUILT-IN BEAM
To assess the accuracy of the Pseudo Dynamic Method,
a typical built-in beam was analyzed. Using this simple
problem, the effects of different damping coefficients and
time increments could be investigated.
As shown in Fig. 8, the 2 7WF94 is loaded at midspan
with a load of 40 kips. Since all loads must be applied at
node points, the idealized system is divided into two ele
ments, each 20 feet long. There are 9 degrees of freedom,
but 6 of these are constrained by the support conditions.
Since the only deflection of this system will be in
the vertical direction at node 2, all of the mass is concen
trated at this point. This mass is equal to
and will be applied in the vertical direction at node two.
Therefore the resulting mass vector is:
94 lbs (40 ft), ft
sec^ ft 9.74 lb-sec^in (3-1)32.2 ft (12 in)
0.0.0. 0.
M 1 = 9.74 (3-2)0.0.0. 0.
18
Actual System(a)
P = 40,000 pounds
1 = 4 0 feet = 480 in
E = 29,000,000 psi
I = 3266.7 in4
A = 27.65 in^
FIGURE 8. BUILT-IN BEAM EXAMPLE
Idealized System(b)
3
Sign Convention
D
1
or
1n/n cr
= 1
FIGURE 9. RELATIONSHIP BETWEEN DAMPING AND DAMPED PERIOD
20
To determine the frequency of the system, the stiff
ness coefficient is calculated, knowing that for a built-in
beam loaded at midspan with a load P the midspan deflection
i 5
A m = EL3
192EI . (3-3)
Hence,
K = 192EI = 164,000. • (3-4)L3
Now the frequency of a single degree of freedom system,
where m is the mass is
60 = \i~ErN m . (3-5)
For this example oj = 130 radians/second. The natural period,
I, is then given by ,
T = 2TT = 0.0483 seconds. (3-6)to
For a critically damped system, the damping coeffi
cient, c, is
c = 2m \| K ' = 1770.N m (3-7)
The norm, which is used to determine the convergence
of the solution (see Equation (2-26)), was arbitrarily set as
.0001 and should therefore give displacements accurate to
approximately three significant figures.
The initial conditions were all set equal to zero.
Using critical damping and a time increment of one-
fifth the natural period, the solution was determined in
ten increments• The displacements and moments obtained
are listed in Table 1 along with the exact values. As can
be seen, the displacement is accurate to three decimal
places and the moments are accurate to four.
TABLE 1. SOLUTION TO BUILT-IN BEAM
DISPLACEMENT (in)
x(5) =
MOMENTS (in -lb)
MEMBER ZEND JEND
1 -2,399,637 -2,399,637CALCULATED -0.24317 2 +2,399,637 +2,399,637
EXACT -0.2440 1 -2,400,000 -2,400,0002 +2,400,000 +2,400,000
Effect of Damping Coefficients
Changing the damping coefficients changes the time
required for the solution as well as the shape of the dis
placement-time curve.
The damped period also changes as the damping co
efficient changes. As can be seen in Fig. 9, there is a
definite relationship between the ratio of the damped period
to the natural period, Tq /T, and the ratio of the applied
damping to the critical damping, n/ncr i
Solving the above for Tp, we find thatT'
The solution for the system with n o ■damping applied
and a time increment equal to one fifth the natural period
was obtained by the computer0 Damping equal to 10% of the
critical damping was also applied with a time increment of
one "fif th: the damped period as calculated by Equation (3-9) <,
Similarly 30%, 50^, 70%., and 90% critical damping were also
applied. The results are plotted in Figures 1.0 and 11* As
can be seen, the number of oscillations above the static
solution decreases as the per cent critical damping increases.
F or 90% critical damp ing, the solution converges directly to
the static solution. The solution converges in 14 increments.
for 50%, 70%, and 90% critical damping, while convergence
takes T8 .increments for. 30% and 33 increments for 10%. f
Effect of Time Increment Size
As the time; increment size is increased, the number
of increments required for convergence decreases. For example :
the number of time increments required for convergence at 70%
damping reduces from 14 when T = 1/5Tq to 8 when T = 2Tq•
However, if the time increment is made too large the solution
becomes unstable and diverges rapidly, as when T = 1,OOT^
(as shown in Fig. 12), This fact becomes important in trying
to decrease the number of increments necessary for the solu
tion of frames and is discussed in the next chapter.
Disp
lace
ment
(n
egat
ive)
in in
ches
23
Undamped Response
10% Critical Damping
* 30% Critical Damping
Number of Time Increments
FIGURE 10. PLOTS OF UNDAMPED, 10% CRITICALLY DAMPED, AND 30% CRITICALLY DAMPED SYSTEM
Disp
lace
ment
(n
egat
ive)
in in
ches
2 4
50% Critical Damping
70% Critical Damping
x static
90% Critical Damping
0
Number of Time Increments
FIGURE 11. PLOTS OF 50%, 70%, AND 90% CRITICAL DAMPING
Disp
lace
ment
in
inch
es
25
-.5
0
70% Critical Damping'for ■"Both Curves ------------
2 4 6 8Number of Time Increments
FIGURE 12. EFFECT OF TIME INCREMENT SIZE FOR 70% CRITICAL DAMPING
CHAPTER 4
APPLICATION TO A PORTAL FRAME
The method was next applied to the solution of por
tal frames* This introduced several problems which had to
be overcome to obtain solutions in a reasonable number of :.
time .increments<,
Solution for Laterally Loaded Frame
The frame that was analyzed is shown in Fig„ 13« As
can be seen, the frame is loaded with a lateral load of 20
kips to cause sidesway. The system has 12 degrees of free
dom , ■ 6 of which are constrained by the supports*
The masses for the system were determined as shown
in Fig., 14* Half of the column weights was applied at the
base nodes in the vertical direction .and the other half was .
applied at nodes 2 and 3. Half of the weight of the beam was
applied in the vertical direction at node .2, the other half
at node 3* The weights were converted to masses by dividing
by 386 in/sec^, The resulting mass vector is s
IY11=' (0V 0.648 0. 1 .22 0.648 0. 1 .22 0.648 0. 0. 0.648 0. ).. . (4-1)
To determine an approximate value of K , relating
the lateral load, to the sidesway deflection, slope deflection
relations were used. .
• ' ■ ' . ' • : ' 26 : ■
27
El El
— ~ 10 ' — ^ Actual System
P = 20,000 lbs
L = 120 inches
E = 29,000,000 psi
Ay(120,120)P
4EI
El
7777
Idealized SystemIdealized(b)
Members 1 & 3: 2
Member 2:
A = 28 in2
I = 2000 in4
Weight = 50 lb/ft Weight = 94 lb/ft
FIGURE 13. PORTAL FRAME EXAMPLE
A = 15 in'
I = 500 in'
470 lb 470 lb
F : j250 lb
^{250 lb7" 77
470 lb = 1.22 lb-sec'250 lb 396 in
<
250 lb ’ = 0.648 lb-sec2250 lb 386 in
FIGURE 14. OBTAINING THE ELEMENT MASSES
' ' - ' . ' . 20
1t can be shown that for a portal frame having all members
of equal length L .in which the beam moment of inertia equals
41, where I is the moment of inertia of each of the columns,
. K = 21.4EI
- ; L ‘ \ (4-2)
Using this, a value of 180,000 lb/in was obtained
for K . To compute the frequency, the total mass above the
dashed line in Fige 14 was used as the system mass, m » This
seemed reasonable, Since the lower portion of the frame goes
through little vibration, A frequency of 219 radians per
second was obtained from Equation (3-5)„
Seventy per cent critical damping was used and a
horizontal damping coefficient of 1150 iJlllJL£L£ was appliedin
at nodes 2 and 3, The damping vector was theref ore i : ,
Ci= (oV 0. 0. 1150. 0 6 0. 1150. 0. 0 . 0 . 0. 0.)(4-3)
The initial conditions were set equal to zero and the
norm was arbitrarily chosen as 0.0001.
The damped period was calculated as 0.020.5 seconds,
Applying a time increment T of one-tenth of this, the solution
was unstable and diverged. Obviously, the time increment was
too large and so was decreased drastically«
Using a time increment of 0.0001, the solution con
verged in 368 increments. The increment was then increased
until a solution was obtained in 83 increments with a time
increment of .0007225. The number of time increments required
29
for convergence decreases as the size of the time.increment
increases (see Fig, 15), •
The displacements and forces obtained in 83 incre
ments were accurate as shown in Tables 2 and 3 where they
are listed with the "exact" values obtained by the Direct
Stiffness Method, Figures 15, 17, and 18 show how the
solution for the horizontal, vertical, and rotational dis
placements at node 2 converge smoothly to their static
values«
It was then found that by decreasing the damping
coefficients to 1000, a solution for T = ,0007225 was ob
tained in 72 increments and that these answers were slightly
improved.
Up to this point, solutions had been, obtained with
only horizontal damping applied. Next vertical damping was
applied at nodes 2 and 3 to determine if this had any effect,
on the solution. One hundred per cent critical damping
calculated on the basis of the axial frequency of the col
umns was applied and the solution was obtained in 72 incre
ments again, the answers not being quite as accurate as
before. Therefore it was decided that application of
vertical damping was unnecessary.
Next, the time increment was increased to 0.000820
and the solution became unstable. The reason for this was
: thoroughly investigated and the results are explained be-
1 ow, -
X
31TABLE- 2. DISPLACEMENTS FOR PORTAL FRAME
NODE DIRECTION PSEUDO DY.NAMIC SOLUTION (in)
"EXACT"SOLUTION (in)
1 Hor 0.
1 Vei't 0. 0.
1 Rot 0. : D-
2 H p r V11 377054 .-114482 35 .
2 Vert .00260765 .00262494
2 Rot .00024219 .00024369
3 Hor .11230091 .11301272
3 Vert .00260765 - .00262494
3 7 Rot .00023607 .00023756
. 4 Hor 0. . o.
4 Vert 0. ■ 0.
4 Rot 0. 0.
TABLE 3. ELEMENT FORCES,FOR PORTAL FRAME
MEMBER AXIAlC. FORCE Lb)
I END,MOMENT (in-lb)
J END MOMENT (in-lb)
EXACT PROGRAM EXACT ’ PROGRAM EXACT PROGRAM
12
V 3
9515
-9944
-9515
9452 -
-9944
^9452
-632,774
573,885
-567,965
-628,834
570,402
-564,385
-573,985
567,965
-625,375
-570,304
564,483
-621,434
Hori
zont
al
Disp
lace
ment
in
inch
es
32
12
10
08
04
02
0
Number of Time Increments
FIGURE 16. HORIZONTAL DISPLACEMENT OF NODE 2
Vert
ical
Di
spla
ceme
nt
in in
ches33
003
0025
002
0015
001
.0005
20 8040 60Number of Time Increments
FIGURE 17. VERTICAL DISPLACEMENT OF NODE 2
Rota
tion
in
radi
ans
34
0003
.00025 9 static
0002
00015
0001
00005
0 20 40 50 80Number of Time Increments
FIGURE 18. ROTATIONAL DISPLACEMENT OF NODE 2
35Adjustment of the Axial Frequencies
The sidesway frequency of the system had been cal
culated as 219 radians per second. However the columns and
the beam vibrate axially also, and at a much higher fre
quency. For example, the axial stiffness K for each column
is
and the resulting frequency, using the total mass of the
column, is
period of sidesway motion. The same is true for the beam.
These smaller periods caused the solution to become unstable
when using a larger time increment. Therefore, in order to
increase the time increment further so as to obtain a faster
solution, the axial frequencies had to be decreased. Several
approaches were tried.
Elimination of the Axial Frequency of the Beam
To decrease the axial frequency of the beam, K could
be decreased or m could be increased. However to do so would
also effect the sidesway frequency and displacements--so this
was not used. Also the axial frequency could be eliminated
by equating the sidesway displacements, equating the side
sway velocities and equating the sidesway accelerations at
K = AE = 3,620,000 lb/in L
(4-4)
u> = ^ = 1670 rad/sec.(4-5)
This means that the axial period of the columns is 1/8 the
36nodes 2 and 3« This would in effect eliminate axial effects
in the top member. However, this would give sidesway dis
placements equal to twice the desired values0 To circumvent
this, half of the load was applied at node 2 and half at node
3 0 Using a time increment of .000820 a solution was found in
-64 increments which was quite good, except that the axial
force in the top member was, of course, calculated to be zero0
Adjustment of the Axial Frequency of the Columns
An increase in the time increment to 0.00091 caused:
the solution to become unstable again, this time due to the
high axial frequency of the columns. Three methods might be
considered: t'cr overcome this. ■
. First, twice the critical vertical damping could be ,
applied at nodes 2 and 3. Overdamping the axial motion
would not eliminate the problem, however, and the solution
would again be unstable»
Next, it could be reasoned that decreasing the cross-
sectional area of the columns would reduce their axial stiff
ness and thus the axial frequency. Reducing the column
cross-sectional areas to one fifth their actual values a
solution was obtained for T = .00091 in 65 increments. This
method did indeed decrease the axial frequency; however, the
displacements and forces were accurate to only one signifi
cant figure.
Therefore the second method was not used; instead it
■>■■■ ' . • ; : ' ■ : . - " ; ' 37 .
was reasoned that increasing the column masses would decrease
their axial Frequencies* The vertical masses applied at nodes
2 and 3 were therefore doubled and a solution was obtained
for T = .0 0091- in 58 increments* The displacements and forces
were quite accurate * The final values, of course, are in
sensitive to the s izes of the masses *
Using this procedure, a T of 0*00205 was used to ob
tain a good solution in 35 increments. This program used
vertical masses approximately 6 times their actual value and
eliminated the axial frequency of the beam as outlined above.
The displacements and forces calculated are listed in Tables
4 and 5 and compared with the Direct Stiffness solution for
the portal frame with half the load applied at node 2 and
half at node 3, The results compare favorably with the
"exact" values,
. The time required for the solution in 35 increments
was about twice, the time required for the Direct Stiff ness
* Method. ' .
Increasing the vertical masses still more was fppnd
not to be useful since the much larger masses tended to give
less accurate, displacements and the number of increments re
quired for solution began to increase due to making the axial
frequencies smaller than the sidesway frequencies.
Solution for Frame with Vertical and Lateral Load
Most frames are loaded both laterally and vertically
and so it was decided to apply a vertical load at the center
38TABLE.: 4, DISPLACEmENJS OBTAINED IN 35 INCREMENTS
BY REDUCING THE AXIAL FREQUENCIES
NODE DIRECTION PROGRAM (in) 1 "EXACT" SOLUTION (in)
1 Hor o. 0.
1 Vert Oc o.1 Rot 0. 0.
: 2 . Hor .11353338 =11374753
\ 2 Vert ,00261995 : .00262494
2 . Rot .00024012 =00024062
3 Hor .1135 3338 .11374753
/3 : - . V e r t 00261995 -.00262494
3 Rot .00024012 .00024062
4 Hor 0 o : o.
4 Vert 0. 0.
4 Rot 0. 0.
TABLE 5> ELEMENT FORCES OBTAINED IN 35 INCREMENTS
' BY REDUCING THE AXIAL FREQUENCIES
MEMBER AXIAL FORCE (lb)
I END MOMENT (in-lb) .
J END MOMENT (in-lb)
EXACT PROGRAM EXACT PROGRAM EXACT PROGRAM.
- I. - : . 9515 9497 -629,075 -627,902 -570,925 -569,872
2 0 O' ' 570,925 569,719 570,925 569,719
3 -9515 -9497 -570,925 -569,872 -629,075 -627,902
39of the top member of the portal frame. This would extend the
Pseudo Dynamic Method to a more typical frame problem. Since
all loads must be applied at node points, the top member was
divided into two elements, extending the number of degrees of
freedom to 15. The resulting structure is shown in Fig. 19.
The masses were applied in a similar matter as before,
except that the horizontal masses at nodes 2, 3, and 4 were
each given as a third of the total mass of the beam (members
2 and 3). The vertical mass of node 3 was given a value
equal to the total mass of the beam. The resulting mass
vector was therefore:
-o..648
0..91 .649
0..91
2.43 = 0.
.81
.648 0.0..648
0. (4-6)
The stiffness of the beam was assumed to be the same
as for a built-in beam, that is
K = 192EI = 6,450,000 L3
and the frequency would then be
uo =y|lp = 1 ,630 rad/sec.
(4-7)
(4-9)
z
40
20 kiPS @ j
©
VTV
40 kips
10
( D4
© 10
5 -
Properties of the
members are the
same as given in
Fig. 13.
FIGURE 19. LATERALLY AND VERTICALLY LOADED FRAME
41Then applying 50% critical vertical damping the damping co
efficient would be approximately 4000 Ib-sec/in. The re
sulting vector of damping coefficients is
0 0 01000 0 01000
C = 400001000 0 0 0 00 . (4-9)
Setting the initial conditions equal to zero, a norm
of 0,0001 was used, The time increment was set as 0.00025
and a good solution was obtained in 262 increments. The
resulting displacements and forces are listed in Tables 6
and 7 along with the "exact* values obtained by the Direct
Stiffness Method. Although most of the displacements are
accurate only to two significant figures, the vertical
displacement under the 40 kip load is accurate to 8 signi
ficant figures, since this displacement converges much
faster than the others. This is illustrated in Fig. 20
which is a plot of the horizontal and vertical displace
ments at node 3 (signs neglected) versus time. Of course,
this means that the vertical period is much shorter than
the sidesway period and using a time increment of just
0.00041 will cause the solution to become unstable.
TABLE 6, DISPLACEMENTS F O R 'LATERALLY AND VERTICALLY /
LOADED PORTAL FRAME
NODE DIRECTION PROGRAM (in) "EXACT" SOLUTION (in)
1 Ho.r 0. 0.
1 Vert 0, ' . o. •
1 - Rot 0 e - 0.
2 Hor „ 11258264 11 466605
2 : Vert -.00294120 -.002892 30
2 Rot .00065379 .00065824
3 Hor .11165965 .11374753
3 Vert -.0024252746 -.0024252746
3 . Rot >.00005365 -.00005469
4 Hor . .11,074560 .11282901
4 Vert -o00809328 -.00814218
4 Rot -.00018145 -o00017700
5 Hor 0. 0.
5 Vert : 0 » ■ o. .
■ 5 Rot 0. D.
TABLE 7» MEMBER FORCES FOR LATERALLY AND
VERTICALLY LOADED FRAME
MEMBER AXIAL FORCE I END MOMENT J END MOMENT(]-b) (in-lb) (in-lb)
EXACT PROGRAM EXACT PROGRAM EXACT. PROGRAM
1 -10,485 -10,662 -5 33,699 -522,187 -374,625 -364,1872 -12,431 -12,491 374,625 364,157 -1,003,670 -1,003,5 653 -12,431 -12,370 1,003,700 1,003,835 767,225 756,7574 -29,515 -29,338 -767,225 -756,788 -724,450 -712,938
Disp
lace
ment
s in
inch
es
z43
Horizontal Displacement08
04
02
Vertical Displacement
25020015010050Number of Time Increments
FIGURE 20. HORIZONTAL AND VERTICAL DISPLACEMENTS UNDER THE 40 KIP LOAD (SIGNS NEGLECTED)
. To obtain a solution in fewer increments„ a much
larger vertical mass was applied at node 3 and larger verti
cal masses were applied at nodes 2 and 4. The lateral load
was divided up with half the load at node 2 and half at n o d e .
4 and the horizontal displacements (as well as velocities
and accelerations) of nodes 2, 3» and 4 were equated. Using
a time increment of 0.00205 a solution was obtained in 62
increments®. However, the values obtained for the sidesway
displacements were quite inaccurate: instead of the sidesway
displacement being .114 inches as before, it had increased
to .142 inches--an error of approximately 25%! Taking the
vertical load off, however, gave the correct sidesway dis
placement. . . ... ■
The reason for this is as follows. Equating the
sidesway accelerations is a good approximation for a frame
with lateral loads only. However it is not a good approxi
mation for the frame with lateral and vertical loads. Table
8 gives the values of the horizontal accelerations at nodes
2, 3, and 4 obtained for the previous Solution. Notice that
the accelerations are not even approximately equal until
after 50 increments. Equating them, therefore, introduces
the resulting error in the sidesway displacements. If, how
ever, the accelerations and velocities are not equated, the
axial motion of the beam will not be eliminated and the
solution will become unstable for the larger time increment.
45So to obtain accurate solutions., a small time increment
must be used,
In summary, the Pseudo Dynamic Method gives very
good solutions for laterally and vertically loaded portal
frames» If the frame is not loaded vertically, certain
approximations can be introduced to reduce the axial fre
quencies so- that good.solutions pan be obtained in a small
number of increments„
The method was then extended to larger frames as .
discussed in the following chapter0
TABLE S'. SIDESWAY ACCELERATIONS OF NODES 2, 3, and 4
INCREMENT NO. NODE 2 in/sec2 NODE 3 in/sec2 NODE 4 in/sec21 +18,569.7 ;. o; 0
4 •—107.4 -4373.5 +13,635.9
' 8 +3901.4 .-1763.6" +586.2
12 +625.3 +782.0 -1319.5
20 . +41.0 . +622.2 -1623.1
28 -261.3 -545.3 -41.9
36 “ 321.7 -358.9 -117.2
50 -210.4 -235.1 -207.0
100 -100.3 -100.4 -100.2
286 -8 .1 ^3 ©1 -8.1
CHAPTER 5
APPLICATION TO A BUILDING FRAME
Finally the method was used to analyze a typical
building frame„ It was decided to use a four story, single
bay frame ^ith lateral loading* The problem is shown in
Fig. 21 and the member properties are listed in Table 9„
There are 30 degrees of freedom, 6 of which are constrained
by the supportso
The vertical mass at each node was obtained by
taking half of the weight of the upper column, (if existing)
connected at that node and half that of the lower column
(if existing) connected at the node„ The horizontal mass
at each node was obtained by taking half of the weight of
the beam (if any) connected at that node. . The weights,
were divided by 386 as before to convert them to Tb-sec^/ine
The resulting mass vector wasi
|V11 = (.Do .558 0. 0. .558 0. .493 1 .01 0. .493
1 .01 0. .493 .8,45 0. .49 3 .845 0. .493
.718 0. .493 .718 0. ,357 .319 0. .357
■ . .319 0.). , (5-1)
To obtain an approximation for the sidesway fre
quency , the frame was idealized to the simple structure
shown in Fig. 22a. A deformed shape V was assumed as
; : ■ : - \ ■
CL|C
N
47
P = 700 lb E = 29,000,000 psi
P
P CD
P
7T715'
0
©
( 0,0) 7/4 -̂----------- x
Actual Structure (a)
Idealization(b)
FIGURE 21. FOUR STORY BUILDING FRAME
48
TABLE 9-. PROPERTIES OF FOUR STORY FRAME
MEmBER - U E m i ON AREA (in2 ) MOMENT OF INERTIA (in4 )
1 8WF35 10.30 126.5
2 8WF35 10.30 126.5
3 1 0 125 o 4 7.38 122.1
4 8 M2 B 8.23 90.1 .
5 8M28 8.23 90.1
6 10125®4 7.38 122.1
7 6M25 7.35 47.0
8 • 6 M2 5 7.35 47.0 .
9 10125.4 7.33 122.1
10 6M20 ■ 5.88 38.3
11 6M20 . ■ 5.88 38.8
12 .i- - '■
.8 M B . 4 5 © 3 4 56.9
49
Assumed Shape
k
k
k
k
(b)
P = Holding Force
P1 = k-| (1 )
= -k1(1)
= 0(1) 31st column k1
2
P3of =
Stiffness-k10
array0
P4 = 0(1 )
FIGURE 22. IDEALIZATION FOR RAYLEIGH METHOD
50shown and a diagonal mass matrix was formed with m ^ being
the total mass at nodes 9 and 10, m^^ being the total mass
at nodes 7 and 8, and so on. Then a stiffness K was cal
culated for each story using
K = 24EIu3 (5-2)
and a 4 x 4 system stiffness array was computed. A unit
lateral displacement was given to each node, one by one, with
the other nodes held in place, and the resulting holding
forces were computed: these corresponded to the columns of
the stiffness array. For example, the computation of the
first column of the stiffness matrix is shown in Fig. 22b.
The resulting matrices were:
0m = 1.34 0 0 00 2.39 0 00 0 2.63 00 0 0 2.94
K = 2.33(10 ) 3.88 -3.88 0 0-3.88 8.58 -4.70 00 -4.70 13.71 -9.010 0 -9.01 21.66
(5-3)
Then, according to the Rayleigh Method for obtaining
the fundamental frequency of a system,
T* = VTmV
V* = V KV (5-4)
and CO = V* T*.
From these, the fundamental frequency was estimated
as 35.6 radians per second. The fundamental natural period
would therefore be 0.176 seconds.
51
The damping coefficients were calculated using
C = • 575m (01 27.40 0 0 00 49.00 0 00 0 53.90 00 0 0 50.20 (5-5)
Thus half of 27.40 was applied in the lateral direction at
node 9 and half at node 10, and so on for the entire system.
Using a norm of 0.00005 with a time increment of
0.000755, and setting the initial conditions equal to zero,
an excellent solution was obtained in 333 increments. Figure
23 is a graph of the sidesway displacement of each story. The
building displaces as a unit, instead of one story moving at
a time. The final displacements and forces are listed in
Tables 10 and 11, along with the "exact" solution obtained by
the Direct Stiffness analysis of the idealized structure of
Fig. 21,
Equating the horizontal displacements, as well as
velocities and accelerations, at each story and placing all
the load on one side of the building gave displacements and
forces approximately the same as those obtained above.
Placement of the upper three loads on the right side of the
frame and the load on the first story on the left side, again
gave approximately the same results. Placing half of the
load on each story on the left side and half on the right
side also gave similar results except that the axial forces
in the beams were calculated as approximately zero, as would
be expected. Therefore the final solution does not seem to
vary significantly with different placements of the loads.
Disp
lace
ment
in
inch
es52
4th Story
•3rd Story
-2nd Story
1st Story
12060 180 240 300Number of Time Increments
FIGURE 23. SIDESWAY DISPLACEMENTS OF FOUR STORY FRAME
TABLE 1 0 o DiSPLACEjYlENTS OF FOUR STORY FRAME
NODE DIRECTION COMPUTER (in) "EXACT" (in)
1 Hor 0 01 Vert .0 0T R o t 0 02 Hor 0 02 Vert 0 02 Rot 0 03 . Hor •.15756800 o158578693 Vert .00153023 .001544523 . Rot .00103988 .001047174 Hor .15727498 .158285834 Vert -.00153023 ; -.001544524 Rot .00103336 .00104565'•S- Hor .37768359 . .38044749
; 5 Vert .00263690 .002663125 Rot .00086971 .000877586 Hor .37738837 .380132686 Vert -.00263690 -.002663126 Rot .00087005 .000877937 Hor .56932926 .574154587 Vert .0031 3161 .003164187 Rot .00047552 .000481388 Hor .56903443 .57386038
. 8 Vert -.00313161 -.003164188 Rot , .00047654 .000481519 Hor .66230604 . .668536259 Vert .00326170 .003296269 Rot .00028777 .0002922610 Hor .66210054 .6683327310 Vert -.00326170 -.0032962610. Rot . .00028831 .00029281
54
TABLE 11. ELEMENT FORCES FOR FOUR STORY FRAME
v; MEMBER I AXIAL FORCE (lb)
I END.MOMENT (in-lb)
J END (i
MOMENT, n-lb)
EXACT COMPUTER EXACT COMPUTER EXACT C o m p u t e r .'-
1 3204 31 74 -114,975 -114,273 -61,620 -61,2 90
2 -3204 -3174 -114,741 -114,039 -61,464 -61,113
3 ' -348 . -348 121,512 120,670 121,452 120,611
-4 1854 18 34 -59,892 -59,380 -66,046 -65,556
5 -1854 -1834 -59,988 -59,477 -66,075 -65,585
: ■ e •-351 -351 100,102 99,208 100,116 99,221
L %? - / 742 732 -34,056 -33,652 -41,557 — 41 ,114
8 -742 -732 -34,041 -33,637 -41,545 -41,103
9 -350 • -351 52,673 52,024 52,678 52,029
: 10 156 154 -11,116 -10,911 -14,071 -13,846
11 -156 -154 -11,132 -10,928 -14,081 -13,856
:C. 12 -175 -177 14,071 13,845 : 14,081 13,855
. 55The example problem showed that the Pseudo Dynamic
Method,, is ^applicable to tall building frames and gives very
good results for sidesway loading, the solution converging to
the first modee The number of time increments required for
solution could be decreased by increasing the time increment
size and reducing the axial frequencies as discussed for the
Portal frame in the previous chapter;
CHAPTER 6
SUMMARY AND CONCLUSIONS
A method has been presented for the analysis of
structural .frames with static loading, assuming that the
structure is initially unloaded and that the internal dis
placements and forces are gradually developed throughout the:
structureo The method takes into account the effects of
axial deformations and sidesway«.
As has been demonstrated, the Pseudo Dynamic Method
gives accurate results for both displacements and forces of
built-in beams, portal frames, and building frames,
- Generally, the procedure for analyzing a frame can
be summarized as presented below. Idealize the structure as
a system of nodes and massless segments with'masses applied
at the nodes,. Each node will have three components of mass.
The rotatipnal component may be taken to be zero. The ‘hopi-
zdntai component should be taken as half the mass of each
beam connected at that pointj the vertical component may be
taken as half the mass of each column connected at that point<
To reach a solution quickly, at least 50% critical
damping should be applied to the system» Damping need only
be applied.in the dirOotion of the loads* That is, for
lateral loading, the only non-zero damping coefficient will
57be applied In the horizontal direction at the non-supported
nodes®* For vertical loads simply apply vertical damping co
efficients at the nodes which are vertically loaded. For
sidesway loading the sidesway frequency may be estimated
using siope-def1action equations or the Rayleigh Method dis
cussed in the previous chapter® The frequency of vertically
loaded beams can be calculated, assuming a fixed-ended beam®
These frequencies can then be used to calculate the approxi
mate damping coefficients®
In order to obtain the quickest possible solution,
use the largest possible time increment, realizing that
using an increment that is much larger than half of the
'Smallest axial period of the system will cause the solutionI
to become unstable. One noteworthy item is that the accuracy
of the solution does not decrease when using a larger time
increment, up to the limit previously mentioned®
The axial period of the columns can be increased by
increasing the column masses to several times their actual
values, without appreclab1y affeating the accuracy of the
results® The period of vertically loaded horizontal members
can also be increased by increasing their vertical masses®
For lateral load only, the axial period of horizontal mem
bers may be1: eliminated by equating the lateral displacements,
velocities, and accelerations at the ends of the member®
This * cannot be done for vertically loaded members without
introducing large errors into the final results®
, 58Initial velocitiesaccelerations, and. displacements
may all be 'taken as zero. The value of the norm may be cho
sen so as to obtain any degree.of accuracy desired. A value
of 0.0001 seems to be adequate for most frames.
The method seems to work quite well for tall build
ing frames and the solution converges in the first mode.
More work could be done to investigate the possibil
ity of having the computer generate the damping coefficients,.,
the time increment size, and even the masses. Some study was
made in this direction, but the task proved to be quite com
plex for large systems with many different frequencies present.
The basic equations of the method could be applied to
frames under dynamic loading and the effect of damping coef
ficients could be studied. The.knowledge thus obtained
cou l d .possibly be.applied to actual structures under earth
quake or blast loading.
Although the method is quite accurate, it does not
compare favorably with the Direct Stiffness Method in either
computer time or storage space .required. The computer time-
required for solution varied from twice to ten times the time
required for the Direct Stiffness Method. However the Pseudo
Dynamic Method does give.one a better feel for the way the
initially unloaded .structure deforms after the load is ap~.
plied. And it gives one a better idea as to how damping
would affect a structure under dynamic loading.
APPENDIX
Flow Chart for Computer Analysis
Start
Do For Each Element
Read Data
Print Data
Obtain the Product B 1*ES*B
Generate Compatibility Matrix (B )
Form the Equivalent Mass Matrix (MSTAR)
Zero System Stiffness Matrix (SS)
Generate Element Stiffness Matrix (ES)
Apply Support Conditions to the SS Matrix
Form System Damping Matrix (SDAMP) From Vector Read In (CDAMP)
Place the Elements of B ESB in the Proper Locations of the SS Matrix
Invert MSTAR to Obtain SMI byAssuming it to be a Diagonal Matrix &— —
Fortran Program 61
COMPUTER PROGRAM FOR PSEUDO DYNAMIC METHOD NUMEL=NO. OF ELEMENTS,NSUPS =NO». OF SUPPORTS NL=NO. OF LOADSv NODES= NO» OF NODES, NP=NODE POINT NO
- - ■ ■ - ■ ■X=X COORDINATE, Y=Y COORDINATE■ ,NUM=ELEMENT NO., IE=I END, JE=J END, A=AREA EMOD=MODULUS OF ELASTICITY, SECM =MOMENT OF INERTIA ELS=ELEMENT LENGTH, LP= LOAD POINT NO*PTL-APPLIED LOAD, NS=SUPPORT NO*, TAU=TIME INCREMENTCONSt=NORM FOR TESTING CONVERGENCESUBROUTINE BES (ELX,ELY,A,EMOD,SECM,B,ES,ELS,ZZ)DI MENS ION A(100),EMOD(100),B {3,6),£SI 3,31rSECM i lOOI ■ 1 = ZZ 'B M , 1)=-ELX/ELS B(It2)=-ELY/ELS B(l:,3)=0o B ( 1, A }=— B ( i , 1)
1,2)8 (1 , B ) T— 6 aB(2,il=ELY/ELS**2 -B(2,21=-ELX/ELS**2B(2,3) = l.B ( 2, 4 )=— B ( 2,1)B(2,5)=-B(2*2)B(2$65-0o B 13,1 } = B{ 2? 1)B ( 3, 2 )-B{ 2, 2 i B {3,3)=0»B(3,4)=B(2,4)B ( 3 , 5 )— 8 ( 2, 5 )B ( 3, 6 ) = 1»ESC 1,1 ) = A( I ) *EMOD{ I )./ELS ESC 1,21=0.ESCT , 3>-0.ESC2,1)=0.ES(2,2) = 4.*EM0D( I >*SECM.U )/ELSES C 2 »3) = o 5 &E S C 2,2) 'TESI3,T1-0.B (3V21 = ES( 2, 3 )ES C3,31= £S (2,2)RETURN ’ :• END 'SUBROUTINE MUTRA (A,B,C)DI MENS ION AC 3,6),B (3, 6),C C 6,6)NCA=6 NCB=6 'NRB=3.DO-1 N-1,NCA DO 1 M=T,NCB C C N, M ) = 0 .DO 1 L=1,NRBC C N,M) = C(N,M)+A(L»N)*B{L,M)
62RETURNENDSUBROUTINE MATMU (A*6,C*NRAsNRBfN C B )DIMENSION A'(3,i3),8C3,6).,C(3,-6)- DO 1 N=1» NRA DO 1 M=1,NCB
. CI Nv MI = OT .
DO 1 L-IrNRB 1 C ( NVM,) = C ( N yM) -̂ A ( N f L ) ̂ B ( L p M )
RETURNEND; . . - :MAIN PROGRAM
.DIMENSION BTESBI 6,6)COMMON SSI75$75)?NP(25)$X( 25 3 $Yi 25),B(3»6)»E S (3»3 3 $NSC 603DIMENSION NUM( 1003 $ IE( 100) ,06( 1003 ,A( 100) , SEC M (10,03 , E M 00(1003DIMENSION LP(60),PTL(60 3DIMENSION ESB(3,6 3,P(75),0(6,1),FORCE(3,13DIMENS ION CMASS (75 3 , STARM ( 75 3 , SDAMP ( 75 $ 75.)D I-MENS I ON C D AMP { 7 5 3 , SM T( 7 5, 75 3DIMENSION D EL T A C 7 5 3,G A M (7 5 3 ,T E M P (7 5,2 3 ,X X ( 75 3 ,X D (75 3 ,XDD(75 3DIMENSION XX0LD(75),DIF(75) .
. READ- 600'vN.UMEL.,.NODES,NSUPSjNL600 FORMAT (8110).
PRINT 700700 FORMATt IHOyAX,5HNUMEL,5X,5HN0DES,5X,5HNSUPS,8X,2HNL/ 3
PRINT 600$NUMEL,NODES,NSUPS,NLPRINT 701
701 FORMAT ( 1H0, 2X, 7HN0DE PT, 1 3X,7HX COORD, .13X, 7HY COORD/3 DO 6 1=1,-NODESREAD 601,N P ( 13,X ( 13,Y (I 3
6 PRINT 601? NP( I ) $ X ( 13 $ Y (I 3 ..601 FORMAT (I10,2F20.5)
PRINT 7027 0 2 FORM A T (1HO,9 HNUM 8 ER E L ,5 X ,5 HI END$5X,5HJ END,6X,4MAREA
,6Xi4HSECM, 'T'feX $■ 4HEM 00/ 3
. DO 7 1=1, NUMEIRE AD 602 , NUM ( I 3 , I E ( I 3, J E ( I 3 , A ( I ) $ SECMI- 13 , EMOD ( 13
' 7 PRINT 602 , NUM ( I 3, IE( 13, J E ( I 3, A (I 3 , S£CM,( I 3 ,-E MODI T)602 FORMAT (3110,2F10e3,E10o33
K0RDS=3*N0DESR E AD 603,(L P ( I 3,I = 1,NL 3
603 FORMAT ( 8110 3, PRINT 703 .
703 FORMAT;( IHOtlBHLOAD POINT NUMBERS/) - PRINT 60 3,(LP(I 3,1 = 1,NL 3PRINT 704
704 FORMAT (1H0,8HPT LOADS/3 READ 604,(PTL(I 3,1 = 1,NL 3
z . ■
604 FORMAT(8F10„2) .• PR INT 604s {P T L (I )s I = ls N L )
PRINT 769 7.69 FORM AT ( 1H0» 1 5HS UP PO R T MUM BE R S / )
READ 605»(MS( I ? »I-1,NSUPS)605 -FORMAT (1615) ;
PRINT 605,(M S (I ),1=1,N S UPS)PRINT 706
706 FORMAT (1H0,9H EL MASS)DO 300 I=1,KORDS READ 606,CMASS( I)
300 PRINT 606,CMASS(I>606 FORMAT (FI0.3)
READ 607,CONST607 FORMAT (F10.6) -
PRINT 609609, FORMAT (1H0, 5HC0MST)
PRINT 607,CONSTREAD 608,(XX(I),1=1,KURDS)PRINT 612
612 FORMAT (IHOsROHINITlAL DISPLACEMENT) PRINT 608$ ( X X U ) , I = 1,K0RDS )READ 608,(XD(I),1=1,K0RDS)PRINT 611
611 FORMAT (1HO,16HINITIAL VELOCITY) PRINT 608, (XD( I ) , I = 1,.K0RDS)READ 608,(XDD(I),1=1,KORDS)PRINT 610
610 FORMAT (1H0,20HINITIAL ACCELERATION) "PRINT 608,(XDD(I),1=1,KORDS) '
608 FORMAT (8F10o2 5 PRINT 492
492:FORMAT (1H0,5HCDAMP)READ 490, t CD AM PI 1 ) $ 1= 1, KORDS )
490 FORMAT. (F10»3)PRINT 490, (COAMP(I),1=1,KORDS)PRINT 493
493 FORM.AT (1H0, 3HTAU ). READ 491, TAU
491.FORMAT (F10.8)PRINT 49.1, TAU • - •DO. 911. MMM=1,KORDS DO 911 LLL=1,KORDS
9 11 SS (MMM, LLL ). = 0 oDO 777 1=1,KORDS
777 P(I)=0«: D O 3 1 = 1, NL J=LP(I)
3 P (J 5 = PTL(I )C GENERATE MATRICES
DO 999 1=1,NUMEL KKK=JE(I)
' z ' - - ; ' '' ■; ■ ' • .
- • . ' ' :. .64.J<3 J= I E( I )ELX=X { i<KK)-X{ JJJ )EL Y= Y { KKK)^ Y ( J J J )
..ELS=SQRT(ELX**2+ELY**2) - 11=1
: CALL BES ( ELXiE.LYf A', E M O D » SECM»B,ES.»ELS, ZZ-i / NRA=3 .
NR 8=3 NCB=6CALL MATMU ( ES».B, ES'B.,NRA» NRB,NC8 )CALL MUTRA(8,ESB?BTESB)DO 10 1 1 = U 3 DO 10 JJ = 1» 3 • IS*JJJ#3 + 11-3 JS=JJJ*3+JJ-3 ISS=KKK*3+II-3
- JSS-KKK#3>JJ“ 3 : S S (ISfJS)=SS( IS?JS)+BTES8{IlvJJ)
S S (IS? 3 S S )= S S (ISf J S S )+ B T E S B (11» JJ + 31 S S (1S S $ J S ) = S S ( IS S> J S ) + BIE S B ( 11 + 3»J J )
10 SS ( ISS» JSS ) = SS( ISS» J S S H B T E S B { I 14-3 $ J J + 3 )999 CONTINUE
C APPLY BoC. AND SOLVEDO 20 1 = 1» NSUPS J = N S (I)DO 21 K=lf KORDS
21 SS<J?K)-0.20 SS(JeJ)=1.
C FORM SYSTEM DAMPING MATRIX (SDAMP)DO 220 l = lLKOR.DS DO 2 21 J = 1.»K0RDS
221 SDAMP( IfJ ) = 0o 220 SDAMPT I® I )=CDAMP{ I )
C FORM M STAR.(STARM)• DO 214 t=I,KORDS
ST ARM (I ) = CMASS ( I 3 + C TAU/2 ® ) # SDAMP (I »I ) +'< T AU##2o / 6® M S S ( I , I )
214 CONTINUE •C INVERT MSTAR (SMI) ' .
DO 270 1 = 1? KORDSDO 270 J=l?KORDSIF (I-J) 305,306,305
305 S M I (I,3)=0,•GO TO 270 . . - . -
3.06 SMIT I, J )= 1 o /STARM ( I )2 7 0 % 0 N T 1 N U E ' '
■ ITER =. 1 C .CALCULATE DELTA AND GAMMA
501 DO 2.22 1 = 1,KORDSIF (C M A S S ( I )) 350,350,351
350 D E L T A (I3 = XX{ I )G A M ( I )=(TAU/2,)*XDD(I)+XD(1)
/
65GO TO 222
351 DELTA(I) = ((T AU** 2 » )/3 «)* X D D (I )+TAU*XDIi) + XXtI)GAM( I ) = tTAU/2o )*XDDC I )-fXD( I )
- 222 CONTINUEDO 230 1=1,KURDS ,
2 30 XXOLDI lS = XX{ I) .PRINT 502,ITER
502 FORMAT { 1H0>13HDISPLACEMENTS»5 X s13HINCREMENT N0oT4J C CALCULATE XX,XD, AND XDD
DO 223 I-1,K0RDS TEMPI I, I )**£>.XDD ( I ) = 0 o DO 217 J=1,K0RDS
217 T EMP { X, 1 j ?=TEMP ( 1,1) TSOAMP (I , J ) *GAM( J >>SS.( i ,J i *DELTA( 3 ) 2 23 TEMP ( I, 2 1~P ( I )“ TEMP ( 1,1)
: ■Cl=TAU/2o C2=(TAU**2e )/6 o DO 275 1=1,KORDS
. DO 276 J= I, KORDS -276 XDD( n=XDD( 1 }-$-SMI ( I, J )*TEMP( J,2) "
X'D(I)-GAM( I )+Cl*XDDi I )XXI I ) = D£LTA( I ) ■f.C2*XDD ( I )
215 CONTINUE : : '800 PRINT 801,(XXt1),1=1,KORDS)•801 FORMAT ( 8E15.1)
C CHECK CONVERGENCE OF XXI I )IF (ITER-1) 555,555,559
550 SUM =0«DO 250 1=1,KORDS D I F (I) = ABS(X X {I)-XXOLD(I) )
2 50 SUM=SUM-5-D IF { I )IF (SUM-CONST) 556,556,555
555 ITER = ITER > 1C CHECK NUMBER OF INCREMENTS (ITER)
IF (TTER-500) 554,554,900 ':v. >5 54 PR INT 570, SUM
570 FORMAT {THO,4HSUM=F10®7)' ; 571 GO TO 501; 556 PRINT 910,SUM
910 FORMAT {1H0,4HSUM=F10«7)557 PRINT 905905 FORMAT .£ 1H0,28HTHESE ARE THE ELEMENT FORCES/) J
PRINT 906906 FORMAT(1HQ, 10H EL EM NUM,9 X,11HA XIA L F ORC E ,8 X ,12 HI END
MOMENT,8X,1■ 22HJ END ̂ MOMENT/) ' .
- ;C' CALCULATE ELEMENT FORCES . ■DO: 998 I = 1,NUMEL KKK=JE(I)j j j = i e ( n ■ ■ '■ • :ELX=X{KKK)-X(JJJ)ELY=Y(KKK)-Y(JJJ)
E L S= S Q RT ( E L X * *2+E L Y* * 2)ZZ=ICALL BES (ELXpELY,A»EMOD$ SECM»B„ES»ELSPZZ) N:RA=.3 NR 6=3
. NC6=6CALL MATMy (.ES*BrESBfNRAtNR8vNCBL:,;.DO 997 11=1,3IA=3*JJJ-34II IBr=3*KKK-3>ri D( II>=XX(1A)
997 D (II>3}=X X (IB1 .:■ N*R A= 3 - : ' ■■■"'
NRB=6 ■ ' - " V:N C B= 1. DO 1 N= 1, NRA •:DO 1 M=1,NCB FORCE(N,M)=0c DO 1 L=1.,NRB
1 F O R C E (N,M) = FORCEi N $ N )+ E S B ( N ,L )* D (L ,M)PR I NT 904, 11 ( ( FORCE ( IK, J K ) , IK = 1, 3) , JK= 1, 1')
904 FORMAT (I10,3E20o7)998 CONTINUE 900 STOP
END '
Day s, A®
Norris,
Turner,
REFERENCES
S® "An Introduction to Dynamic Relaxation,"The Engineer, Uol® 219, No® 5688$ pp® 218- 221, January. 29, 1965®
C® H®, R* J® Hansen, M. J.® Hollery Jr®, J® M® Biggs, S„ Namyet, and J® K« Minami® Structural Design for Dynamic■Loads, McGraw-Hill, New York, 1959®
M® Je, R® W® Clough, H® C® Martin, and L® J®Topp. "Stiffness and Deflection Analysis of Complex Structures," Journal of Aeronautical Sciences® Uol® 23, No® 9, pp® 805-824, September, 1956®
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