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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