Analysis of semi-rigid frames

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Authors Liu, Soung-Nan, 1936-

Publisher The University of Arizona.

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ANALYSIS OF SEMI-RIGID FRAMES

bySoung-Nan Jan

A Thesis Submitted to the Faculty of theDEPARTMENT OF CIVIL ENGINEERING

In Partial Fulfillment of the Requirements For the Degree ofMASTER OF SCIENCE

In the Graduate CollegeTHE UNIVERSITY^OF ARIZONA

1 9 6 5

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 the rules of the Library„

Brief quotations from this thesis are available without special permission, provided that accurate acknowledgment of the 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;T

APPROVAL BY THESIS DIRECTOR This thesis has been approved on the date shown below:

. . . L . , ______R. M. RICHARD ' V Date

Associate Professor of Civil Engineering

ACKNOWLEDGMENT

The author wishes to express his gratitude to his thesis director. Dr. R. M. Richard, for his advice and valuable assistance.

Special thanks are also expressed to the staff of the Numerical Analysis Laboratories of The University of Arizona.

TABLE OF CONTENTS

ABSTRACT CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 APPENDIX A APPENDIX B APPENDIX C REFERENCES

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INTRODUCTION MATHEMATICAL FORMULATION FLEXIBILITY CONNECTION FACTORS DYNAMIC CHARACTERISTICS APPLICATION CONCLUSIONS

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ABSTRACT

A general formulation for analyzing semi-rigid frames, including dynamic characteristics, is presented in this thesis® The numerical procedure adopted herein is the displacement-method based on Castigliano* s First Theorem®

A linear relationship between moment and rotation is utilized in the analysis = The dynamic response due to a strong ground motion of an earthquake is also investigated for a particular frame®

The solutions generated in this study were achieved utilizing the IBM 7074 digital facilities of the Numerical Analysis Laboratories at The University of Arizona«,

v

CHAPTER 1

INTRODUCTION

In analysis of steel structures, the end connec­tions of a member are usually assumed to be either per­fectly hinged or rigidly fixed0 The former case allows no moment at the ends while the latter case may cause the critical moment (see Fig. 1-1). Neither of the extreme situations is entirely consistent with conditions often encountered in practice.

Extensive experimental research has shown that max­imum economy for beams and connections usually occurs at a

/1 \degree of restraint somewhere between 40 and 75 percent,which implies that both perfectly hinged and rigidly fixedend connections will result in wasteful design. If theeffect of the semi-rigidity of the connection is takeninto consideration in the analysis of a frame, this wouldbe very advantageous.

The semi-rigid connections are now recognized bythe American Institute of Steel Construction and are

(2)classified as type 3. This organization has been

* Numbers in parentheses refer to REFERENCES.

2

wU

(a. ) Hinged Endlb

W ft

12(b) Fixed End

wL12

w-lf

C.cJ Semi-Rigid Connection

Figure I-1 Moment Diagram For Three Different Connections

)

3sponsoring extensive research for the properties of semi­rigid connections which will enable the designer to take advantage of the potential economy to be achieved by par­tial continuity in steel building design.

The relationship between the applied moment and the relative rotation of the beam to the column is found from experimental results to be approximately linear within a specific region, as shown in Fig. 1-2= For prac­tical purposess this relationship will be expressed by M = k8, where k is defined as the semi-rigid connection factor and is the slope of the assumed straight line curve. Its magnitude depends on the type of connection of the member. Formulas for computing the connection factor, and the moment-rotat ion curve for certain types of beam- column connections will be presented herein.)

The method developed in this thesis, through the use of a digital computer,can be easily applied to large structural systems.

Bending

moment M

4

o

Test behavior of semi-rigid connection

•H

( For fixed end G e 0 )Relative angle change

Figure 1-2 Moment-Rotation Curve for Semi- Rigid Connection

CHAPTER 2

MATHEMATICAL FORMULATION

Two basic m e t h o d s a r e available for the analysis of statically indeterminate structural systems. These are the force method (or the flexibility matrix method) and the displacement method (or the stiffness matrix method). For the theoretical basis of the validity of the methods of displacements and forces, Castigliano*s first and second theorems respectively may be used, that is

3U - .

ayi i (2-1)

anddU

i2-2>

in which U denotes the total strain energy of the system, p^ represents the load applied at point i, and y^ is the deflection at point i in the direction of p^.

Since only linear systems are studied in this thesis for which the principle of superposition holds, both the displacement method and the force method are appli­cable in the analysis.

5

When the number of elements is large and the num­ber of redundant forces is relatively small, the flexibil­ity matrix method is preferred. When a large number of redundants is involved, the displacement method will be more applicable. On the whole, both the force and dis­placement method have advantages and disadvantages; there­fore, the choice of one method over the other is not necessarily obvious. In this thesis, only the displacement method will be developed.

In the displacement method, the displacements of the joints are necessary to describe fully the deformed state of the structure. The deformations of the elements can be calculated in terms of these displacements by use of the compatibility relationships, and the internal forces can be determined from stress-strain relationships. Finally, by applying equilibrium criteria at each joint, a set of linear simultaneous equations for the displace­ments can be obtained.

The displacement method can be developed in the f 5)following way:

Letf = column matrix of applied loads A = equilibrium matrix p = column matrix of internal forces v = internal deformations

7K = stress-strain matrixd = column matrix of external displacements

associated with the forces f B = compatibility matrix

At _ transpose of matrix A

Thenv = A^d (2-3)

p = Kv = KA^d (2-4)

f = Ap = AK^^d (2-5)

or

where

f = Kd d (2-6)

Kd = AKA1 (2-7)

Hence

It may be noted that

d "d

whereFd = flexibility matrix.

d - (2-8)

F, = K^'1 (2-9)

8The compatibility matrix, B, which relates the element boundary displacements, y, to the generalized displacements, x, as shown in Fig. 2-1, and which is equal to the transpose of equilibrium matrix, i.e., may be generated as fol­lows .

For a general element ij, let c/ be the angle of rotation, y^ denote the element elongation, yg be the rota­tion of the element at i, y^ be the end rotation of the element at j , x^ denote the generalized horizontal dis­placement at i, Xg denote the generalized vertical dis­placement at i, and let x^ denote the generalized rotation at i . Similarly, x^, x^, and x^ are defined at j. With these definitions, the compatibility relationships for a general element can be expressed in the following form:^^

f V ' COS* sin* 0 COS ( ct + 7T ) sin (* + Tf ) 0 ^ zX 1

x2

ygsin*L

COS<x t“ L 1

sin( * + TT ) cos( * -f TT ) L 0 5

x3

X4

sinoi CO Serf r> sin( * + tt ) COS( o< + TT ) 1 x5I y3j L L U L L / i x.

where L is the length of the element.

9

2

Figure 2 - 1 Compatibility Relationships

10The compatibility matrix as given herein includes

the effect of axial deformation. This effect may be essen­tially removed by setting the axial stiffness term equal to a large number.

To evaluate the flexibility influence coeffi­cients for a large structural system, first assemble the

•j*element stiffness matrix (i.e., B KB) for each member and then transfer it into the master stiffness matrix to get the system stiffness matrix for the unsupported struc­ture.

Due to compatibility, the displacements of the ends of the element must equal the corresponding joint displacements of the structure. Therefore, comparing element (5) in Fig. 2-2b and Fig. 2-3, the following com­parison can be made:

de 1 corresponds to d 4d@ 2 corresponds to d 5dQ 3 corresponds to d 6dQ 4 corresponds to d 10d@ 5 corresponds to d 11dQ 6 corresponds to d 12

The displacement comparison indicates how the element stiffness matrix (i.e., Kg — B KB) is to be placed into the master stiffness matrix, K̂ , i.e., for element (5),

11

(a)

d4

dlOdld!2

(b)d2 dllFigure 2-2

de 2

1

d 5

de 4

Figure 2-3

K@ (1,1) is equivalent to Kd (1,4)Kg (1*2) is equivalent to (1*5)Ke (1,3) is equivalent to (1,6)Ke (1*4) is equivalent to Kd ,(1,10)Ke (1,5) is equivalent to (1*11)Ke (1,6) is equivalent to K^ (1,1̂ )

This procedure is carried out for each element of the entire system. Numbering the joints and members in a spec­ific manner allows the procedure to be 'programmed easily so that the assembly process is carried out by the computer. Since no displacements occur in the direction of reactions, these boundary conditions are accounted for by column and row elimination, and the stiffness matrix for the supported structure may be obtained.

As indicated in eq. (2-9), the inverse of the system stiffness matrix is the flexibility influence coefficient matrix, i.e., K^ = F̂ . It may be that only certain par­ticular elements of the influence coefficients are concerned in the analysis. Using the same techniques for eliminating rows and columns associated with the reactions, the desired flexibility influence coefficient matrix can be easily established. For example, consider a simply supported beam with three intermediate joints (see Fig. 2-4).

13d3 d6 dl2 dl5

dl y- d4 d9_ d? /< dlO dl3

d2 d5B

d8 dllC D

Figure 2-4

dl4E

The system stiffness matrix, Kd, will be a 15 by 15 matrix,

Kd -Hi —1 K1 2 K1 3 K1 13 K1 14 K1 15 '

K2 1 K2 2 K2 3 K2 13 K2 14 K 2 15

K3 1 K3 2 3 3 3 13 K3 14 K3 15I 1 1 1 « t1 f 1 1 » •V « I • 1 «

K13 1 K13 2 K13 3 K13 13 K13 14 K13 15

KH 1 K14 2 KH 3 K14 13 K14 14 K14 15

K15 1 K15 2 K15 3 K15 13 K15 14 K15 15 z

Since there are no vertical displacements at joints A and E, and no horizontal displacements at E, the elements associated with d2, dl3, and d!4 (columns and rows) should be lined out. The stiffness matrix for the reduced system, Kd(R), then becomes a 12 by 12 matrix, i.e.:

14

Kd(R)

K1 1 K1 3— K1 12K

K3 1 K3 3 K3 12K

1 1 t t« f 1 1t 1 V I

K12 1 K12 3"----- K12 12 K

K15 1 K15 3 '" '" K15 12K12 15 15 15

The flexibility influence coefficient for the reduced sys­tem is reached by inverting the above matrix, i.e.:

d(R) K^d(RX

^11 F1 3 ^1 12 *1 15

F* F* ------— F* F*3 1 3 3 3 12 3 15

i i » «• « i •« i i i

F* F* F* F*12 1 12 3 12 12 12 15

F* F* F* F*15 1 15 3 15 12 15 15

/

If only the influence coefficients associated with dis­placements d5, d8, and dll are required for dynamic analysis, the resultant flexibility influence coefficient is found to be

15F* F* F*5 5 5 8 5 11F* F* F*8 5 8 8 8 11

F* F* F*11 5 11 8 11 11

The flexibility influence coefficients denoted byF* are defined as the displacement at i due to a unit d ij

force acting at j. When a force fj is applied at j, the displacement at i becomes d^ = ^jfj. More generally,the displacement at i due to any set of forces at the other points is given by

di = Z Fd ijfj

It may be shown that the influence coefficient is always symmetrical about principal diagonal by the Betti Recip­rocal Theorem.

CHAPTER 3

FLEXIBILITY

The analysis that is developed herein is based on the assumption that the stress-strain relationship obeys Hooke1s law and the principle of superposition holds. Because the relationship between load and deformation is linear, the effect that a given load has on a structure without regard to the presence of other loads can be con­sidered.

The relationship between internal forces and deformations is defined by:

p = Kv or v = Fp where K is the stiffness matrix and F is the flexibilitymatrix.

When an elastic element is subjected to an axialforce (Fig. 3-1), the associated deformation of the elementis simply its elongation. In this case, the flexibility

Longitudinal stiffness AEt — ................. ' p

Figure 3-1

16

17matrix relation is

LV = I 5 P (3-D

and stiffness'matrix relation isAEp = r V. (3-2)

When an element of the system is considered as a simply supported uniform beam of length L and bending stiffness El, and is subjected to end moments (Fig. 3-2),

i

Figure 3-2. Simply Supported BeamSubjected to End Moments

the internal forces and deformations (using the conjugate beam method) can be expressed as follows:

N VM:

or

V ( _L_ L N ✓

►3 El 6EI

4

r CD L L/ 6EI 3 El J

x> ss

' ZfcEIL

2El"L

Z

<

"j.2EIL

iEIL J

M.(3-3)

e.i

0(3-4)

18It can be seen that the stiffness and flexibility

matrices for an elastic structural system are always sym­metric .

Since superposition is applicable for a structure having small displacements, the flexibility and stiffness matrices for an element subjected to both axial and bend­ing forces are found to be

F =

LIE

0

0

0

L 3 El

L6EI

0

L6EI

L 3 El

( 3 - 5 )

K =

AE o

0

0

0

4EI 2EI L L

2EI 4EI

( 3 - 6 )

For a restrained-end element a spring of modulus K is now introduced such that

and

M

M

» . Mii “ M i or = kT

Mjor $4 ™ — J kjj j j

( 3 - 7 )

( 3 - 8 )

Using the conjugate beam method, j- can be taken

equivalent to The diagram is shown in Fig. (3-3).

From Fig. (3~3c) if c^ is taken equal to zero,

^ “ (3EI + k"1 Mi

= - i i Mi

From Fig. (3-3®) with cj taken equal to zero,

^ = - i i mj

^ = ' j l i + ir3] Md

For the total structural system,

9‘ ■ 'jfe * El’ Ml ' 5if "i ,3-91

9J " ‘ jfe "i * ‘jEl * kj’ "i (3"101

The element flexibility matrix will be

20

n M

(b)

MiEl

(c)+

K'i (d)

rij

(e)

Figure 3-3

21From eq. (2-1$) K = F"^

K =

4 ElKjUjL t 3 EfJ 2EI L Kj KjLKi (LKj + 4EI ) t4EI(KjL t 3 E i ; (LKj + 4EI) + 4 E I ( k i L t 3 E i ;

2 EI L Kj Kj-L KI ( L K j t 4 E J ) t + E I ( K j L t 3E I J

_ _ 4 E I K j C K j i t 3E IJ ___L m { L k j + 4 EI ) i 4 E I ( K J L +3EI)

(3-12a)

If the axial deformation is taken into account, the element stiffness coefficient will be

K

AS 0 o

04 ElKj(kiL+3EIJ 2 EI Lkj Kj

LKi ( L K j t 4 E I ) t 4 E I ( K j L t 3 E I ) L K i ( L K j t 4 E I ) t 4 E I ( K j L t S E I )

0 Z E I l K j K j 4 E I k j ( K ; L t 3 E i ;LK1(LKJ t 4 E I ) + 4EI (Kj l+3EIJ L K j ( L K j t 4 E I ) i 4 E l ( K j L t 3 E I )

(3-12b)

CHAPTER 4

CONNECTION FACTORS

Semi-rigid connections are now recognized by the American Institute of Steel Construction and are classi­fied as type 3. The method for determining the elastic properties of the connections is given in References 8 and 9. A brief description of this method follows.

According to the properties of connections, the connection factor, k, will be analyzed for three different types of semi-rigid connections (see Fig. 4-1). The labor­atory results are given in Appendix A.

(a) Type A semi-rigid connection (as shown in Fig.4-la), a web-angle connection without top or seat angles is referred to as standard beam connection. To develop the k - equation for this type of connection, the procedure of the analysis may be divided into the following six steps:

1. Find an expression for the critical moment in the legs of the connection angles due to the outward pull p at the top of the beam, assuming no yield in the rivets (Pig. 4-2).

2. Find an expression for the deflection of the upper ends of the connection angles outward from the face of the column (Fig. 4-2).

22

V L

-i

(•

h<b-

Y

V-

^ J t o<^(

t o

Figure 4-la Type A Semi-Rigid Connection

---- r

I

H

A,

-0 -

Figure 4-lb Type B Semi-Rigid Connection

24

-v*

— <

t /j\ /ki

< ± > - i

- ^ < b -

. . . . . . i . > . . . . . . . . . .

0<b- ■6-<P <p-

6 - -6--0 -- e-

---T/̂~~

Figure 4 1c Type C Semi-Rigid Connection

253. Locate the neutral bending axis of the con­

nection as governed by the elastic behavior of the angles, and find an expression for its distance y down from the top (Fig. 4-4).

4. Note that the angle of strain, (J), equals the deflection, A b, of the heels of the connection angles divided by y, that is.

moment, M, of the connection based on the elastic restraint of the legs of the angles.

(4-D

5. Find an expression for the safe resisting

6. Finally(4-2)

The following is an illustration of the method.

Step 1: The maximum moment in connection angles

t Ma

Figure 4-2 Symbols Figure 4-3 Relative Moment

26

Figure 4-4 Resisting Couple for Each Angle

The angle in this case will be treated as a frame subject to a horizontal load p. Point A is assumed to be fixed so as to prevent rotation or translation while point B allows only horizontal displacement. Hence, the moments at points A, B, and C, by making use of the method of slope-deflection, can be expressed as follows:

Mab “ (26a + % ~ 3Rab) “ ^ (eb ' 3Rab) (4-3a)

Mba = (2Gb + ea ' 3Rba) - 2|I (26b - 3Rba) (4-3b)

Mbc - ^ (26b + 6C - 3Rbc> = (29b) (4-3c)

The joint conditions areMba + Mbc = 0 (4-4a)

andMab + Mba = Pg (4-4b)

Substitute eqs. (4-3a,b,c,d) into eqs. (4-4a) and (4-4b) and solve for 8y and Ray. Hence

0u = ( - ) (-—S— ) (4 - 5 )D 2EI 4g+gi( - f s ! ) ( g+g l ) ( 4 - 6 )3El 4gtg1Rab

and2g+g;4g+g12g

4g+gi

Ma = Pg (4-7a)

Mb = Pg (,_f_ ) (4-7b)

M„ = Pg (— 2— ) (4-7c)4g+g1

The symbols g and g^ in the above equations are the hand­book distances from the center of rivets to the back of the angle minus t, the angle thickness, which can be easily found from the AISC Manual of Steel Construction.

28Step 2. The deflection

For a small deflection, the angle change, Ra ,̂ equals the deflection A ^ divided by length g. That is

Rab = (4-8)

orA b = g Rab (4-9a)

Substitute the expression of in equation (4-8)

A b g ( — ) (£l£l_) - <— ) (££1 ) (4-9b)3El 4g+gi 3El 4g+g1

Step 3. The angle of strain From equation (4-1)

(P£_) (g+gj ) (4-10)y 3Ely 4g+g1

An expression for P may be found from eq. (4-?a). The resisting moment of 1-in. strip at the top of the connec­tion is governed by its strength at point A, i.e., Ma =1 2(-̂ -jst , in which s is the bending stress and t is the thick­ness of the angle.

29

P _ !Ja (4g+gp = (gt2) (4g+gi) (4-11)g 2g+g1 2g+g1

3Substituting eq. (4-11) into eq. (4-10) with I = ~. yields

(j) = (Bgfs) (£^£i_) (4-12)3Ety 2g+g1

Step 4. The resisting moment, M, of the connectionIn Fig. (4-2) let P equal the maximum pull per

angle, in pounds per inch of depth, at the top of the con­nection. The average pull, per inch of depth for each angle, above the neutral axis is ^ Py. From Fig. (4-4)the resisting moment is found to be ~ Py x — = for one2 3 -J""angle. For both angles

M = (4-13a)

Using the expression for P in eq. (4-11), the resisting moment for type A connection is

M = ( h s t ^ (4g1g1) (4-l3b)9g 2g+gi

30

PShear A rea

CompressionArea

/N.A

HbKivet Line A

(a)

Figure 4-5# Neutral Axis and Transformed Section

Step 5# Determination of the neutral axisIn locating the neutral axis of the connection,

the method presented here is analogous to determining the transformed section in reinforced concrete beam analy­sis. Referring to Fig. (4-5), it is clear that the com­pression area (nb(h-y)) below the neutral axis is opposed by a shear area (ty) above the critical section and is the right section of the leg of the angle along rivet line A. Fortunately, the relationship between the shearing stress on this area and the critical bending stress along rivet

31line A can be determined. This relationship gives rise to a stress-ratio factor, n, which is analogous to the

Egn = ~ used in establishing the transformed section in Ecreinforced concrete beam analysis. The value of the pull, P, on the top inch of one angle in terms of bending stress s in equation (4-11) can also be expressed in terms of the unit shearing stress v in the angle as being equal to the total shear on the top inch of the column-connected leg of each angle. Therefore, the unit shear on the top inch is

v = P = st (4s + gj) (4-14)t 6g (2g + g^)

Equation (4-14) shows the desired relationship between the shear v on the shear area(ty) and the critical bending stress s on the section along rivet line A and, accordingly, provides the key to the desired stress-ratio factor, n. Assume that the ratio of bending stress to strain in the column-connected leg of the angle along rivet line A is thesame as that in compression below the neutral axis. Toarrive at the proper shaped transformed area as shown in Fig. (4-5b), either the shear area(ty)must be multiplied by the coefficient of s in eq. (4-14), or the compression area b(h-y), below the neutral axis, must be divided by that coefficient. Then

n = 6g (2g + gj)t (4g + g^)

32

(4-15)

Equating the statical moment of the shear area(ty)above the neutral axis (Fig. 4-5b) to the statical moment of the compression area nb(h-y) below the neutral axis,

ty - 7 * nb(h-y) •2

which gives rise to a quadratic equation in y. But if the plus sign before the radical be discarded, the following results

% -fob - ...^i (4-16)^ nb - t

Step 6. Semi-rigid connection factor, zFrom eq. (4-12), with the expression of (p , M in

eqs. (4-12) and (4-13b ) respectively, the connection factor equation can be easily formed. That is

( S ^ ) (4-17)

in which y can be calculated from eq. (4-16). If the strengthening effect of the fillet be taken into consider­ation, the resisting moment of a pair of connection angles is increased about one-eighth. The resulting equation

for M is approximately

M = ,hst y x ,4g + glN8g V2g + g /

33

(4-18)

(b) Type B semi-rigid connections:As shown in Fig. (4-6a,b), such connections

consist of a top (or clip) and a seat angle extending hori­zontally along the column face. The procedure in finding connection factor, k, will also be based on the method applied to type A semi-rigid connection.

— / —

y,i

i f J L

t, Si ■I

Z

g;

X

HE

ca)

N.A.

(b;Figure 4-6. Type B Semi-Rigid Connection

34

&

t,

1

B'— 63

I j .(a)

/////////

P(b)

Figure 4-7 Symbols and Relative Moment Diagram

1M

Shear AreaR i v e t L i n e

/ / / / / / / / AZZZl

Top Anile

hi y, X-62'

i n n

Seat Angie

Neutral Axis

Figure 4-8 Neutral Axis and Transformed Section

35From Fig. (4-7), it can be seen that the pull, P,

acting on the angle is identical to that of type A connec­tion for convenience. The same notation with different subscripts will be adopted. The procedure will be:

Step 1. Critical moment

Ma *= Pgp (4-19a)2g2 + g34g2 + g3

2g24gg + g3

82>2 o3

Step 2. DeflectionPg?

My ” Pgp (---------) (4-19b)

Mr = Pg2 (---- ) (4-19c )0 4g0 + g.

A b - (-3) (f2 * g3 ) (4-20)b 3 El 4g„ + g.'2 3

Step 3. The angle of strain

Pgp g 4- gf - (rrn (-S--2. ) (4-21)3EI(y1-g2-t1) 4g2 +

2 ,s t twith M = __1, and I = -16 12

2then P = (__1) (̂ 2 _ ^3 ) (4-22)

6g2 2gp + 8,

36Substituting the expression for P in eq. (4-22)

into eq. (4-21) gives

' -

Step 5- The resisting moment from Fig. (4-6a) is

M « pbi(yi + j c) (4-24a)

With the expression of P in eq. (4-22)

M - ( ^ i ) (if2 I ) (y, + f c) (4-24b)g2 2g2 g3 3

The stress-ratio n^ is identical to that for type A connection; thus

6g (2g„ + g )n, = — ---------------- (4-24c)1 tl(8g2 + g 3)3

In determining (the distance from the rivet in the top angle to the neutral axis), there will be two cases in the design problem; that is, the neutral axis falls either above or below the upper rivet line of the seat angle. The former condition will only be dealt with here. Since in the latter case only the elastic affect of the seat angle will affect the computation, this will be ignored.

37From Fig. (4-8), equating the statical moment of

the shear area b-̂ t above the neutral axis to the statical moment of the compression area b^n^h^-y^) below the neutral axis, yields

Step 6. Connection factorFrom eq. (4-2) with the expression for (|> and M in

eqs. (4-23) and (4-24b) respectively yields

(c) Type C semi-rigid connectionsType C connection is the combination of type

A and type B. It was originally treated as a rigid connec­tion consisting of a top and a seat angle extending hori­zontally along the column face and web-angle connection.

b1t1y1 - • I (h1-y1 )

or

With the plus sign before the radical discarded.

k M = a j b ! (y1-g2- t1)(y1 + | c) (4g2 +g3 ) (4-26)

4g| g2 + g3

38The notation used in type A and type B connections will also be adopted here.

t(4g + gi) (4-27)n 6g (2g + g1)

Equation (4-27) applies to the pair of web-angles.

Equation (4-28) applies to the top or clip angle in which the effect of the seat angle on M and Z are negligible.

As shown in Fig. (4-9) (a) and (b), with the same procedures as derived for types A and B connections, we

where H is the distance from the top of the web-angles to the bottom of the column-connected leg of the seat angle. Then

find

2htm (2H-h) (4-29)

K =

39where

y1 = y + d + t1 + g2

04 = (f^ ) 3 (i£_i_El_) (82 + «3) (4-32)1 A.g2 + 63 g +

Moment-rotation curves have been plotted and pub­lished for many semi-rigid connections which were obtained from several research laboratories. To use such curves for finding connection factor k, it is necessary to draw a straight line tangent to the curve at or near its origin, k is equal to the ordinate of this point divided by its abscissa.

Since the moment-rotation curves for different semi-rigid connections are limited, the designer has to confine himself to few joint connections. Equations (4-1 7), (4-26), and (4-3 1) serve useful purposes in find­ing connection factors. The results agreed very nicely with those curves determined experimentally which are given in Appendix A.

CHAPTER 5

DYNAMIC CHARACTERISTICS

In a shear building, the girders of the structures are assumed to be infinitely rigid. The effect of the axial deformations in the columns (and girders) of the structure are not considered. The joints of these struc­tural systems are allowed to move only in the horizontal direction. In the analysis of a structure having semi­rigid connections $ the joint rotations and axial deforma­tions will be taken into account, and a comparison with a shear building analysis will be made. It should be noted, however, that the fundamental formulation for fre­quency a n a l y s i s i s applicable to both systems.

In a dynamic analysis, if a single coordinate or single variable is sufficient to define at any time the position of the structure, the structure is said to have a single degree of freedom. If several coordinates or vari­ables are necessary to define the position of the struc­ture at any time, the structure is said to have multiple degrees of freedom. Such a system has normal modes of vibration, and a general motion of the system may be

40

41represented by a linear combination of these modes. The fundamental frequency of the system corresponds, in gen­eral, to a motion that involves displacement of all of the masses in the same direction. The higher modes correspond to reversals in the directions of motion of the various masses with inflection points in the system between certain of the masses.

By Newton1s second law of motion,

F = ma

where F is the sum of the forces acting on a concentrated mass m and a is the acceleration of mass.

Figure 5-1

For a free undamped system, as shown in Fig. (5-1), the equations of motion will be of the form

mx + Kx = 0 (5-1)

42where x is a n x 1 column matrix of the accelerations of the system, m is a n x n mass matrix, K is a n x n stiff­ness matrix, and x is a n x 1 column matrix of the displace­ments of the system.

Let x = Ae — then

( -m w2 + K) x = 0 (5-2)

where w is the frequency for a normal mode q, i.e., when x = q(r)^ eq. (5-2) becomes

ormq(^) = Kq(^) (5-3a)

wE K"1 mq(^) = q(^) (5-3b)

Define the dynamic matrix

U = K-1 in ( 5 - 4 )

hence(r) „ n(r)

'r

Now the procedure for establishing q^r ̂ and wr by matrix iteration is to assume a mode shape, i.e..

43

(o)

This shape is a linear combination of the unknown(i)modes qv ' with i = 1, 2, 3, ... n, hence

q(o) = x: ck q(k)k=l

orq(o) = q(1) + r: ck qk=2 K

(k)

(5-6a)

(5-6b)

(1)where C^1s are unknown constants. The coefficient of q is arbitrarily set equal to unity because mode shapes are independent of amplitude. The ith mode may be swept out as follows.

i.e.,q l 0 ) = C C q(k) - Ci q(i) (5-?a) k=l K 1

q ,0t - q'0 ’ - C, ql1 ) (5-7b)

Premultiplying eq. (5-6a) by q ^ ^ m and changing k to i for ith mode, gives

q(i)t m q(o) = ^ q(i)t m q(i) = C. m.* (5-8)

wwhere is the transpose of and q ^ ^ m q^^and since

. 5 - 9 1when r = s

Solving eq. (5-8) for C^, gives

. (5-10)-i*

Substituting eq. (5-10) into eq. (5-7b) yields-(o) (o) (i) q ^ ^ m q^0^q = q - q ------

mi* or

q (0 ) - ( I - q 1 q ( l ) ! lS ) q (o) ( 5 - 1 1 )V “i*

Premultiplying by the dynamic matrix gives,

U q (o) - ( U - 0 q(l- l .y i),J2 ) q (o) ( 5 - 12 )

From eq. ( 5 - 5 ) for ith mode,

U q ( i ) = q ( i ) ( 5 - 1 3 )w.2

45So, eq. (5-12) becomes

U q(o) = (U - 3--^ <L.1.).t..fn) q (o) (5-14a)wi2 »i*

Defineq ( i ) _ ( i ) tU S = (U - -____9-----2) (5-Ub)w^2

where U S is the sweeping matrix with the ith mode swept out. In order to use this sweeping procedure, the first mode is obtained first.Iterate, using

u q = q W1

For the second mode, replace U by U S where

a(D d)tU S = (U - --- 3--- 2) (5-15)W.2 ni*

For the third mode, replace U by U S where

„ H = mi - q (1) q (1)t« - q (2) q (2)^ ) (5-16)wl2 ml* «22 m2*

and so on•Such successive approximations have been widely

used in solving engineering problems. This method is com­monly known as the method of matrix iteration.

CHAPTER 6

APPLICATION

Now consider the three-story building frame shown in Fig, 6-1= A detail of the connections is given in Fig= 6-2 and Fig= 6-3= The dynamic characteristics will be analyzed in the following manners

Case I Rigid girder system with rigid connections=

Case II Flexible girder system with rigid connections =

Case III Flexible girder system with semi-rigid connec­tions =

The stiffness matrix, the flexibility influence coefficient matrix and the dynamic characteristics are given in Tables 1, 2, and 3.

46

4720 20 20

M,= 416

12

832

Girders : 18 VAF 96 E = 30 x 10& I = 1675 in4 A - 28.22 in2

Colamns : 14 W 7 48I - 4b5 in4 A = 14.11 in2 14 W 53 1= 542 in4 A = 15.59 in2

lb-secin

lb-secin

lb-secin

Figure 6-1

u8/T-8T

Type A connection 11-3/4"— —i

0 ( > 0O <~ > 0

O < > 6O < > 66 < > 0 -

h 8W

3"

S"S-3"

mini

oo4

48

2 angles4 x 3i x 3 /8 x lf-3w

r 0> O> O> o ) o

18 96

0.512

g = i ( 5 . 5 - 0.512 - 2 x 0 . 3 7 5 ) - 2 . 1 1 9

g1 - 2 . 2 5 - 0 . 375 - 1 . 875n = 6 x 2 . 1 1 9 ( 2 x 2 . 1 1 9 + 1 .8 7 5 ) « 20 04

6 . 3 7 5 (4 x 2 . 1 1 9 4- 1 .8 7 5 )

y = 1 5 ( 2 0 . 0 4 x 4 - 2 0 .0 7 x 4 x 0 . 3 7 5 ) » 3.4.032 0 . 0 4 x 4 - 0 . 37 5

k = 3 0 , 0 0 0 , 0 0 0 X 15 X O.3753 X 1 4 . 042 (4 X 2 . 1 1 9 + 1 .8 7 5 )6 x 2 .1193 (2 .1 1 9 + 1 .8 7 5 )

- 210,800,000

Figure 6-2

Type B c o n n e c t i o n 49

r1, 1<j> i6 :'T 1

cc

— j!

>yk__

—'U '0 |

- ^ - 41- &

- < j > -

L 1•5 K \r 1 --------------- L

■ 1 i.

ro 00in x*-

xh

crO

-̂/L

3̂

-4>— *-

18 VF 9 6

4 — k-

1 T

t1 = 0.5"g 2= 2.25 - 0 .5 = 1.75" , g j = 3 - 0 . 5 = 2.5"

h j= g2 t t , t 18 t 6 = 26.25"

n _ 622(282+83) _ 6x175(2*175+25) _ , 3 2 5

t,(4S2 + i 3) 0.5 (4x1.75 + 2 5;

___ n j h i Y i i - J ( Z n i k i * t i ) i l _ l3.25^26.25to.5-Jc2»<3-25 f Q.5J 0-5 = 24"ni 3.25

K = ---------- 4 *!------- (1 I T g 7 )452c = h i - y j = 2 6 . 2 5 - 24 = 2 .25"

30* )06x 0-534 x f 75K =

- / L' O . L . J ) ~ JL t —

x 0-5̂ (24 -2 25) (24 t -£*2-25) f <4* 1-75 t 2‘5 A — 2 18000000 — 4 * f.753 V 1-75+2-5 V ' ' I"

Figure 6-3

50

Case I: Rigid girders with rigid connections

Stiffnessmatrix

Flexibilityinfluencecoefficientmatrix

Table 1

' 398o4543 -233-8927 0

-233-892? 467-7854 -233-8927

. 0 -233-8927 233-8927

' 0.6075 0.6075 0.6075

0.6075 1.0351 1=0351

v 0.6075 1.0351 1=4626

x 10*lb/in

-5x 10 in/lb

Frequency» w SEC”1

Mode($)2 $3

7-7358 1.0000 1.4966 1.674822.5467 1.0000 -0=0981 -1.018932.2114 1.0000 -1.9797 2=3441

51Case II t Flexible girders with rigid connections

Table 2

Stiffnessmatrix

371=8504 -226.7314 23.8560

-226.7314 377=6432 -174=8741

23=8560 -174=8741 152.9324 y

x 103 lb /in

Flexibilityinflnencecoefficientmatrix

0.5065

0.5745

0.5775

0.5745 0.5775)

1.0581

1.1234

1.1234

1.6756

-5x 10in/lb

Frequency, w SEC™1

Mode

(l)2

6.5546 1.0000 1=6945 2.017019=2668 1.0000 0.1472 -1.239029=2414 1.0000 -1=3587 1.2914

52Case III, Flexible girders with semi-rigid connections

Table 3

'107,5023 -59,3931 7,7247Stiffness matrix = -59.3931 96,6925 -46,5921 x 103

s 7,7247 -46,5921 39,1459 y lb/in

0,8411 0,9360 0,9441 "Flexibility influence = coefficient matrix

0,9360

0,9441

1,7958

1,8918

1,8918

2,7678

x 10”5 in/lb

Frequency, w SEC™1

Mode

*1 *2 03

3,3095 1,0000 1,9730 2,434110,0456 1,0000 0,2392 -1,209415,0886 1,0000 -1,2275 1,1683

53In Case I, it was assumed that the floor girders

were infinitely rigid.A brief study showed that the effects of axial

deformations of the column were extremely small and, there­fore, negligible. This is generally true for short build­ings, but these axial deformations may have a significant effect in the case of tall buildings.

In Case II, the floor girder flexibility is included in the elastic properties rather than assuming infinitely stiff girders. The frequencies, w, for the two systems are as follows:

W1w2w3

Case I Case II7»73585 Sec"1 6.55461 Sec”122.54669 Sec"1 19«26679 Sec"132.21138 Sec"1 29.24138 Sec"1

The change introduced into the frequency of the fundamental mode is about 19% which is due to the flexi­bility of the girders and the corresponding rotation of joints.

It should be noted that Case I and Case II are similar idealizations in that the floor systems move only horizontally. In Case III, the joints are treated as

semi-rigid connections such that relative rotations due to the flexibility of the connections may occur at joints.The frequencies for Case II and Case III are as follows:

Case IIwiw2w~

6.55461 Sec”119.26679 Sec-1

29.24138 Sec”1

Case III 3.30946 Sec”1 10.04557 Sec”1 15.08859 Sec”1

The difference from Case II introduced into the frequencies of the fundamental mode is seen to be 95$.This shows how much the connection factor k in the semi­rigid connections affects the characteristics of the building.

As indicated in Chapter 3» the flexibility matrix for a restrained end element subjected to bending moment is

il 0 0

0 3EI k ±-

0 - — k- L +6EI 3 El

6EIJL kj J

where and kj are the spring constants at i end and j end, respectively. It is apparent that k affects the prin-

55cipal diagonal of the element flexibility matrix directly® For the connection considered in previous application, —

is equal to ° The frequencies for several differ-100ent ratios of ™ to were obtained and are given in

Tables 4 through 9*

(a) k = i n

Table 4

Stiffnessmatrix

Flexibilityinfluencecoefficientmatrix

" 319.3267 -149.1579 14.8472 x

-149.1579 254.2095 -124.5593 x 10314.8472

X -124.5593 110.1556 lb/in

0.6144 0.7174 0.7284 s

0.7174 1.7198 1.8480 x lO’5

0.7284 1.8480 2.8992in/lb

Frequency, w SEC”1

Mode4>i <Pz $3

6.1315 1.0000 2.2047 2.748517.5795 1.0000 0.2972 -1.204425.8470 1.0000 -1.1384 1.0988

<b) k = I§ f3BI>Table 5

56

Stiffnessmatrix

284.9004 -164.7070 15.4004

■164.7070 294.1101 -144.845015.4004 -144.8450 129.8790

x 10"lb/in

Flexibilityinfluencecoefficientmatrix

0.8411

0.9360

0.9441

0.9360

1.7958

1.8918

0.9441

1.8918

2.7678

x 10”5 in/lb

Frequency, wS E C 1

Mode$1 <$2 $3

5.9049 1.0000 1.7461 2.058617.4700 1.0000 0.0839 -1.113924.6477 1.0000 -1.4981 1.5699

k 10 '3EITable 6

Stiffnessmatrix

224=7488 -128,7340 14=1728

•128,7340 223=1896 -109=466914=1728 -109=4669 95=8247

Flexibilityinfluencecoefficientmatrix

0,1068

0,1225

0,1242

0,1225

0,2424

0,2588

0,1242

0,2588

0.3816

Frequencye w SEC”1

Mode%

5=0854 1.0000 1.8180 2.173215=1843 1.0000 0.1310 -1.137422.6343 1.0000 -1.4097 1=4334

x 103 lb/in

x 10"4 in/lb

(d) k - I& (jif)58

Table 7

' 20305187 -116.1163 13.3575Stiffnessmatrix = -116.1163 199.2966 -97.5892

, 13.3575 -97.5892 84.7551 y

" 0.1181 0.1368 0.1390 'Flexibility

0.1368 0.2736influence = 0.2935coefficientmatrix 0.1390 0.2935 0.4340 ,

x 103 lb/in

x 10”4 in/lb

Frequency, w SEC”1

Mode<Pj %

4.7886 1.0000 1.8431 2.213714.3432 1.0000 0.1466 -1.147621.4297 1.0000 -1.3778 -1.3908

( e } k - 10 '3EI‘T& (lEl) 59

Stiffnessmatrix

Flexibilityinfluencecoefficientmatrix

Table 8

186,5095 -105.8913 12.5691

105=8913 180,2058 -88.1133

12,5691 -88,1133 76.0482

0,1288 0.1506 0.1532

0,1506 0.3041 0.3274

0,1532 0.3274 0.4855

x 10^ lb/in

x 10""4. in/lb

Frequency, w SEC”1

Mode$3

4.5470 1.0000 1.8660 2.250213.6468 1.0000 0.1609 -1.155720.4144 1.0000 -1.3524 1.3541

/

tf) k = 30Table 9

60

Stiffnessmatrix

65.1093 -36.4012

-36.4012 55.41085.7734 -26.1840

5.7734

•26.184020.6550

x 103 lb/in

Flexibilityinfluencecoefficientmatrix

0.3933

0.5148

0.5426

0.5148

1.1239

1 .2809

0.5426

1.2809

1.9562

x 10-4- in/lb

Frequency5 w SEC"1

Mode<Pl $ 2 $ 3

0.0676 1.0000 2.1007 2.68330.2546 1.0000 0.3060 -1.22450.3888 1.0000 -1.1516 1.0578

24

18 -

12 r

uifCD

6

20

K -K = 30.2 x 10

10 -6 4

K IOO V3 E I

K — 8 . /8 0 x |0

5 - -

035302510 20155

g . 1 d _

K ( x 10 ̂ 'in

Figure 6 - 1 4 F r e q u e n c i e s F o r T h r e e S t o r / B u i l d i n g With Various Connection Factors

O n

62The curves in Fig. (6-14) are plotted for natural

frequencies, w, vs. connection factors (or spring constant) k. Point a represents the point k == 1.01 x 10'’, i.e., a flexible connection from which the fundamental frequency is found to be w^ ■** 0.06? sec"""*-. Point b is for the true connection factor as used in the example structural system. The connection factor and frequency are 2.180 x 10® lb/in and 3»30946 sec- ,̂ respectively. It is noted that the changes in frequency between points a and c are signifi­cant and may greatly effect the behavior of the struc­tural system. Conversely, the connection factor at point d is twice as much as that of point c, but the change in frequency is only about 3.8% and increases almost with a linear relationship which shows that from point c on, it may be considered a rigidly fixed connection. Consider the ideal shear building, for example. The joint connection is assumed to be infinitely rigid, i.e., k - 00 . The resul- tant frequency is found to be w^ ® 6.5546 sec" . The difference from the frequency at the point where w^ =5.9409 sec” is only 13%. Therefore, a proper selection of joint connection may result in a much more economical design for certain types of dynamic loading.

Next, consider the same building subjected to the N-S component of the ground motion of the El Centro earth­quake* Assume 10$ critical damping and an idealized response spectrum with bounds 1,0D, 1,5V, 2,0 A, i,e,,8,3 in,s 20,6 in/sec., 0,66 g,^^^

Two different cases will be dealt with herein. Case I will deal with the three-story frame having semi­rigid connections. The elastic characteristics are as follows s

Stiffnessmatrix

Flexibilityinfluencecoefficientmatrix

' 107,5023 -59.3931 7.7247

-59,3931 96.6925 -46.5921 x 103

v 7,7247 -46.5921 39,1458 ylb/in

0,2229 0.2713 0.2789 '

0.2713 0,5727 0,6282 x 10-4

0.2789 0,6282 0.9481in/lb

form.The results of this analysis follow in tabular

QuantityMode

1 2 3w sec~^ 3.310 10.046 15.089t sec 1.9 0.625 0.417

Sd in 5.6 1.88 1.083rd floor (roof) 2.434 -1.209 1.1682nd floor 1.975 0.239 -1.2271st floor 1 1 1(<t>)t w 3.484 0.529 0.296m* = 8.282 1.488 2.649

r = 0.421 0.356 0.112X* = r Sd in max 2.36 0.669 0.121

By the absolute value method, one obtains,

X.max3.1504.9586.687

in

100.2J P2 * 80.8k P3 » 73.2JBy the root mean square method, one obtains.

max2.5504.6545.800

> in

64.01 P2 - 52.6 44.7'

where P^ is the lateral force at ith floor.

65Case II considers the three-story frame with hinges

located at the top of the first story columns as shown in Fig. 6-15.

3

Hinges Hinges

Fig. 6-15 Three-Story Frame with Hinges

The elastic characteristics for the case are as follows

Stiffnessmatrix

Flexibility influence = coefficient matrix

' 39.7592 -47.0527 7.2976 "-47.0627 93.4860 -46.4201 x 103v 7.2976 -46 * 4201 39.1214 ; lb/in

' 0.1869 0.1877 0.1877 '0.1877 0.1879 0.1877 x 10"2

x 0.1877 0.1877 0.1880 in/lb

The results of this analysis follow in tabularform.

66

QuantityMode

1 2 3w sec"'̂ 0.379 7.390 14.148T sec 16.6 0.85 0.444Sd in 8.30 2.52 1.23rd floor (roof) 0.989 -1.209 2.5332nd floor 0.996 -0.247 -1.8971st floor 1 1 1(«* M 2.071 0.125 0.305m*= 2.080 1.680 7.941r = 1.000 0.0745 0.0384X* = r Sd in max 8.30 0.188 0.0462

By absolute value method.

'max = (J) X* m

P1 = 17.23max k

f 8.53 8.41 in8.56

P2 - 17.64 P3 - 1 5 , 3 5

By root mean square method.

Xmax8.328.288.23

, in

P-L = 1 1 . 5 0 k P2 = 1 4 . 7 5 k Po - 1 1 . 0 1 k

67The idealized response spectrum for N-S component

of El Centro earthquake was obtained from Reference 11.In the first case, the building is analyzed such

that beam-to-column connections provide partial restraint, while in the second case, hinges are set on the first- story column. It is interesting to note that the resul­tant fundamental frequency and lateral forces due to this earthquake disturbance are extremely different. This shows that if near perfect hinges would fora in a structure dur­ing an earthquake, the dynamic characteristics of the build­ing may change considerably with the result that the ensuing dynamic response will be significantly altered. It should be noted that.the use of such hinges could provide a means for alleviating the earthquake loadings.

CHAPTER 7

CONCLUSIONS

The method of analysis presented herein is devel­oped for the analysis of building frame with semi-rigid connections. Since the operations utilized in this thesis . were much too tedious to attempt manually, a digital com­puter program was written for the analysis. It may be applied to any large structural system.

The procedure developed herein for generating the influence coefficient of the semi-rigid connection is a direct generalization of the procedure used for rigid connection systems. The dynamic characteristics were obtained by matrix iteration techniques.

Since the connection factor, k, at the joints varies extensively, only a brief study was presented for a particular frame.

A preliminary study on dynamic characteristics for a three-story building showed that the effect of axial deformations is extremely small and negligible. This is true for shorter buildings but axial deformations may have significant effects in the analysis of relatively tall buildings.

68

69It was also shown that the frame with semi-rigid

connections had dynamic characteristics which were signif­icantly different from the same frame having rigid connec­tions =

The connection factor frequency curve, as given in Fig. 6-14, shows that the connection factor, up to a certain point, is important; however, beyond this point, the increase of rigidity in a joint connection is of no real significance.

A brief study on a particular building subjected to a strong earthquake ground motion has shown that the resultant lateral forces due to the ground motion for the system with hinges on columns at the first floor level were considerably different from the same system without hinges on the columns at this point.

This indicates that if near perfect hinges should form in a structure during an earthquake, the dynamic characteristics of the building may change considerably with the result that the ensuing dynamic response will be significantly altered. This is especially true for rein­forced concrete buildings where plastic moment reversal may cause the concrete to spall off leaving essentially only the reinforcing steel to act as a tie.

It should be noted that the method of analysis presented in this thesis is also applicable to the analysis of trusses.

APPENDIX A

Typical moraent-rotation curves for different types of connections as developed in reference (3) are presented here*

70

Type A Connection71

Plate 9*5-x jcQ

8 in I* Beam @ 18.4-lb

2‘—ll 2 Ior>&

4.1$ 6*4* 2-x 0—6

6-0

- I n 1

me

ItEUCucHa)"OCftjV)3O

- C

c0)EoY

100

90

80

60

K =40

30 -

20 -

I 0

0.005 0.0150.010 0020

R o t a t i o n In R a d ia n

Mom

ent

In Th

ousa

nds

Inch

P

ound

s

Type A Connection

572t

~T

COml co

6-0

Plate 3"*|x2-0* ^

450

400By El. (4-18)350

300

250K =200

150

100

50

0.004 0.006 0.0080.002 0010

Rotation In Radian

Mom

ent

In Th

ousa

nds

Inch

Fo

unds

Type B Connection

- Plate &* \*2-o'll 1| | 1 2Z/6̂ 4 x^x0Lfe

12 in I-Beam @ 3/.8 l b | 2 ,- n 4 ' “ Jong

x — /2 6x6 * 2 * 0L6 " '

fe­ 0" _____

280.

240..

160

120 . K = —

80.

40 ..

0 0 0 2 0.003 0.004 0.005 0 00 60.001Rotation In Radian

Mom

ent

In Th

ousa

nds

Inch

Po

unds

Type B Connection

t . i f ? P l a t e 14*1* 2 -0

J

II ! ij I 6x4 * | x / '-2*1 /- 5

12-in I-Beam @ 31.8 lb

2- ll̂ -" long

21$ 6 * 6 *. jL * i -2 I_______ 2 - ii t:___

6 - 0 "

640 ■

560

480

400 -

320 . .

240 K= JrL

160

80 -

0.002 0.004 0.006 0.008 0.010 0.012

R otat ion In R a d i a n

Mom

ent

Jn T

hous

ands

In

ch

poun

ds

Type C Connection

12-in I Beam <2 31.8 lb

2Z£ 6*6*̂ "*/-2 /

Plate 9* | x 2 -0*21 5i 2i

75

1 . 2 If 6x4*§ * r-2‘I

2-11$ “ long

4 L § 4 * 3 i x 5 x o ' - 3 "

6 - 0"

9 09 00 00 0

1- 2"

- w“'Im :1 o-H-m

800

700 --

600 --

500 -

400 -- - n

300 -

200 -

100 -

0 002 0.004 0.006 0.008 00/2O.OIO

R ota t ion in Radian

APPENDIX B

Flow Diagrama) Definition of variables

ESM = element stiffness matrix COM = compatibility matrix

EQUIL = equilibrium matrixSM = stiffness matrix of the system

SMIN7 = influence coefficientAE = product of the cross section area and

Young’s modulus El = product of Young’s modulus and moment

inertia Ak = spring constantNZD ” number of zero displacement

(jMass = mass matrixNOM = number of memberNOJ = number of jointNOB = number of reactionMN == number of memberBL = length of member

ALPHA = angle of member in radiansIE = number of first reaction at i end

77JE = number of first reaction at j end Q = assumed mode shape

QQ = generalized mode shape U = dynamic characteristic W = natural frequency

Start

READ, NOW, NOJ NOR, NOZD

TRANSFORM BVSB

FORM ESMFORM COMP

READ, BL, AE, El U , Akl, Ak2, IE JE

INTO SM

( o ver;

* LINE OUT ELEMENTS AT SUPPORTS BY ROWAND COLUMN ELIMINA­TION TO GET SM FORREDUCED SYSTEM

FOR STRUCTURAL SYSTEMINFLUENCE COEFFICIENTINVERT SM TO GET

RESTRAINT JOINT BYROW AND COLUMN ELIM-

SMINV

LINE OUT REACTIONS AT

INATION TO FIND FLEX­IBILITY COEFFICIENT

FORM DYNAMIC CHARAC­TERISTIC U = SMINV *

pMassT(OVER)

END

PRINT RESULTS

NATURAL FREQUENCIES BY ITERATION

FIND MODE SHAPES AND

non

2. Computer Programs 80

* ' COMPILE FORTRAN,EXECUTE FORTRAN,DUMP IF ERRORDISPLACEMENT METHOD FOR OBTAINING INFLUENCE COEFFICIENTMATRIX MULTIPLY AND MATRIX INVERT SUBROUTINES SUBROUTINE MATMU (AA,BB ,RRSS,MS,NS,MMS,NNS) DIMENSION AA(6 ,6 ),BB(6 ,6 ),RRSS(6 ,6 )IF (NS-MMS) 100,102,100

102 DO 3 1=1,MS DO 4 J = 1,NNS RRSS(I»J)=0o DO 5 KK = 1 ,MMS

5 RRSS(I,J)=RRSS(I,J) + AA(I,KK)*BB(KK,J )4 CONTINUE 3 CONTINUE

GO TO 103 100 STOP103 RETURN

END

SUBROUTINE INVER(LL,VAR 1,VAR2)DIMENSION VAR 1(54,54) , VAR2(54,54)DO 31=1,LL DO 3J=1,LL IF(I-J)1 , 2 , 1

1 VAR1(I,J )=0o GO TO 3

2 VAR1(I,J )=1«3 CONTINUE

DO 71=1,LL C=1o/VAR2(1,1)DO 4J = 1 , LLVAR2(I,J)=VAR2(I,J)*C

4 VAR1(I,J )=VAR1(I,J)*CDO 7J=1,LL IF(J-I)5,7,5

5 D=VAR2(J,I)DO 6 JJ = 1 ,LLVAR2(J,JU)=VAR2(J,JJ)-VAR2(I,JJ)*D

6 VAR1(J,JJ)=VARl(J,JJ)-VARKI,JJ)*D7 CONTINUE

RETURN END

SUBROUTINE RED (NORS,NZDS,NNS,ARS,KRRS) DIMENSION NZDS(54),ARS(54,54)DO 160 KS=1,NORS NZS = NZDS(KS)DO 161 IS = 1,NNS IF(IS-NZS) 161,163,160

163 IF(NZS-NNS) 173,166,167

173 KRRS = NNS-KS KRCS = NNS-KS NS = IS-KS+1 MS =NNS—KS+1 DO 164 JS=NS »KRRS DO 164 LS=1,MS ARS(JS,LS) = ARS(JS+1»LS)

164 CONTINUEDO 180 JS=NS » KRRS DO 180 LS=19 KRCS ARS(LS9JS) = ARS(LS,JS+1)

180 CONTINUE 161 CONTINUE160 CONTINUE

GO TO 181167 STOP166 KRRS = NNS-KS

KRCS = NNS-KS181 RETURN

END

SUBROUTINE REDS ( NORSS » NZDSS s. NNSS 9 ARSS » KRRSS ) DIMENSION NZDSS(54)jARSSI54*54)DO 160 KSS = 1 9 NORSS NZSS=NZDSS(KSS)DO 161 ISS=19NNSSIF(ISS-NZSS) 1619 1639 160

163 IF(NZSS-NNSS)17391669167"173 KRRSS=NNSS-KSS

KRCSS=NNSS-KS5NSS=ISS-KSS+1MSS=NNSS-KSS+1DO 164 JSS=NSS9KRRSSDO 164 LSS=1eMSSARSS (JSS 9 LSS ) =ARSS( JSS + ls>LSS)

164 CONTINUEDO 180 JSS=NSS9KRRSSDO 180 LSS = 19KRCSSARSS(LSS9JSS)=ARSS(LSS 9 JSS+1)

180 CONTINUE161 CONTINUE 160 CONTINUE

GO TO 181167 STOP166 KRRSS=NNSS-KSS

KRCSS=NNSS-KSS181 RETURN

END

C START OF MAIN PROGRAMDIMENSION ESM(696)9COMP(6»6)9EQUIL(6 96) DIMENSION BTSB(69 6)»SM{54»54),SMINV(54*54)

82DIMENSION TEMPI(6 9 6 ),NZD(54),NZDD(54)READ 10» NO'M 9 NO J 9 NOR , NOZD 9 NORR 9 NOZDR

10 FORMAT( 8 I 10)READ 82 9 < NZD(I)9 I = 1 9 NOZD)

82 FORMAT(8110)READ 839(NZDD<I)9 1=19 NOZDR)

83 FORMAT(8110)NN = 3*NOJI 1=6 JJ = 3DO 1 KKZ = 1 9NOMREAD 2 9MN 9 BL 9 AE 9 EI 9 ALPHA9AKI9 AK2 9 IE 9J E

2 FORMATfI10*F10*0»2E10o0*F10o0*2E10.0* 110/110)FORMATION OF ELEMENT.STIFFNESS MATRIX DO 80 I = 193 DO 80 J = 193

80 ESM(I 9J ) = OoESM(191) = AE/BLESM(2 92) = 4 o *EI*AK2*(AK1*B L + 3 o *EI)/(BL*AK1*

1(6L*AK2+4o*El)+4o*EI *(AK2*BL + 3«*E1) )ESM(393) = ESM(29 2 )ESM ( 2 93) = 2 o*EI*6L*AKl*AK2/( BL#AK-1* ( BL*AK2 +

14 o * EI)+4o*EI*(AK2*8L + 3 = *EI) )ESM(392) = ESM(293)FORMATION OF ELEMENT COMPATIBILITY MATRIX DO 100 1=193 DO 100 J = 1 9 6

100 COMP(I 9J ) = OoPI = 3ol4l5926 ALFST = ALPHA + PI ONE = COSF(ALPHA)TWO = SINF(ALPHA)

' THREE = COSF(ALFST)FOUR = SINF(ALFST)COMP(1 ,1 ) = ONE COMP(291) = TWO/BL COMP(3,1) = COMP(2 9 1)COMP(1,2) = TWOCOMP(2,2) = -ONE/BL COMP(392) = COMP(2,2)COMP(2,3) = 1 o COMP(3,6 ) = 1o COMP(1,4) = THREE COMP(2,4) = FOUR/BLCOMP(3,4) = COMP(2,4)COMP(1,5) = FOUR COMP(2,5) = -THREE/BL COMP (3,5) = COMP (2 9-5 )

DO 279 J = 1,3 DO 279 J=1 , 6

279 EQUIL(J 9 I) = OoFORMATION OF EQUILIBRIUM MATRIX AND BTSB MATRIX

n n

83DO 101 1=1*3DO 101 J = 1 9 6

101 EQUIL(J 9 I) = COMP(19 J )CALL MATMU (EQUIL9 ESM9 TEMPI 9 I I 9JJ 9JJ 9JJ)CALL MATMU (TEMPI»COMP*BTSB*I I *JJ*JJ» I I)TRANSFER OF ELEMENT BTSB MATRIX TO SYSTEM STIFFNESS MATRIX DO 280 1=1*3I M=IE+I-1 DO 280 J = 1 9 3 JM=IE+J-1

280 SM(JM 9 IM) = SM(JM 9 IM) + BTSB(J 9 1)DO 281 1=1*3IM=IE+I-1 DO 281 J = A 9 6 JM=JE+J-4

281 SM(JM * IM) = SM(JM 9 IM) + BTSB(J*I)DO 282 1=4*6IM=JE+I-4 DO 282 J = 1 9 3 JM=IE+J-l

282 SM(JM 9 IM) = SM(JM 9 IM) + BTSB(J*I)DO 283 1=4*6IM = JE+I-4 DO 283 J = 4 9 6 JM = JE+J-4

283 SM(JM * IM) = SM(JM 9 IM) + BTSB{J 9 I)1 CONTINUE

C SM IS NOW REDUCEDCALL RED(N0R9NZD*NN*SM*NNR)PRINT 305

305 FORMAT(5X970HTHE STIFFNESS MATRIX FOR THE 1REDUCED SYSTEM IS //)PRINT 182 9( (SM(I 9J ) *J = 1gNNR)91=1*NNR)

182 FORMAT(1H 6E15o8)C SM REDUCED IS NOW INVERTED

CALL INVER(NNR»SMINV*SM)PRINT 306

306 FORMAT(1H0)PRINT 307

307 FORMAT(5X*77HTHE INFLUENCE COEFFICIENT MATRIX IFOR THE REDUCED SYSTEM IS //)PRINT 183* ((SMINV(I 9 J) 9J = 1*NNR)9 1 = 1 *NNR)

183 FORMAT(1H 6El5*8)C SMINV IS NOW REDUCED

CALL REDS(NORR 9 NZDD 9 NNR *SMINV9 NNQ)PRINT 405

405 FORMAT(1H0)PRINT 407

407 FORMAT(5X*35HTHE INFLUENCE COEFFICIENT MATRIX IS //) PRINT 193* ((SMINV(I*J)9J=19NNQ )9 I=1 * NNQ)

193 FORMAT(1H 6El5o8)

84C SMINV IS NOW INVERTED

CALL INVER(NNQ,SM,SMINV)PRINT 409

409 FORMAT(1H0)PRINT 411

411 FORMAT(5X,23HTHE STIFFNESS MATRIX IS //) PRINT 1959 ( (SM(I»J ) 9J = 19 NNQ)9 1=19NNQ)END

n n

85

* COMPILE FORTRAN»EXECUTE FORTRAN

MATRIX ITERATION PROGRAM FOR OBTAINING FREQUENCIES,. MODE SHAPES SUBROUTINE INVER(LL,VAR1,VAR2)DIMENSION VAR1(20,20)» VAR2(20,20)DO 31=1,LL DO 3J = 1 ,LL I F ( I - J ) 1 » 2 » 1

1 VARl(I,J)=Oo GO TO 3

2 VAR1(I,J )=1o3 CONTINUE

DO 71=1oLL C=1o/VAR2(1,1)DO 4J=1,LLVAR2(I,J )=VAR2(I,J)*C

4 VAR1 ( I-,J ) =VAR1 ( I 9 J )*C DO 7J=1,LL IF(J—1)5,795

5 D=VAR2(J ,I)DO 6JJ = 1,LLVAR2(J9JJ)=VAR2(J,JJ)-VAR2(I,JJ)*D

6 VAR 1(J,JJ)=VAR1(J9JJ)-VAR1(I,JJ)*D7 CONTINUE

RETURN END

SUBROUTINE MTRAN(LL,MM,NN,VAR 1,VAR2,VAR3) DIMENSION VAR1(20,20),VAR2(20,20),VAR3(20,20) D07001=1»LL D07 00J = 19MM VAR1(I,J )=0 o D0700K = 19 NN

700 VARKI,J)=VAR1(I,J )+VAR2(K,I)*VAR3(K,J)RETURN END

SUBROUTINE MULT(LL,MM,NN»VARl,VAR2,VAR3) DIMENSION VAR 1(20,20),VAR2(20,20),VAR3(20,20) D0700I=1»LL D0700J = 1,MM VARl(I,J )= 0 o D0700K=1,NN

700 VARl(I,J )=VAR1(I,J ) -WAR2(I»K )*VAR3(K ,J )RETURN END

DIMENSION S(20g20),Q(20) ,U(20,20),GMASS(20 , 20) DIMENSION QQ(20),A(20,20)DIMENSION TEMP(20,20) , SIN(20,20) , B(20,20)

86100

405

407

999

3501

4

5005

300

880

980

881981

882

982

DIMENSION TG(20,20)»RT(20,20) FORMAT(8110)READ 100 9 NDO 405 I = 1 , NDO 405 J = 1 9 NS ( I 9 J ) = 0,5(1,1) = 186,50945S(1 ,2) -105=89135(1,3) = 12,569125(2,1 ) = -105=891315(2,2) 180,205845(2,3) = -88=113345(3,1 ) = 12,569125(3,2) = -88, 11,3345(3,3) = 76,04819DO 407 J = 1, NDO 407 I = 1 , NGMASS(I,J ) =0 0GMASS(1 ,1) =0,832 .GMASS(2 92) = 0,832GMASS(3,3) = 0,416 KZY = 0CALL INVER(N,SIN,5)CALL MULT(N,N,N,U,SIN,GMASS)CONTINUEKZY = KZY + 1DO 3 1=1,NQ(I)=1o 0CONTINUECALL MULT(N»l ,N»QQ,U,Q)Z=QQ(1)DO 4 1=1,N QQ(I)=QQ(I)/ZIF(ABSF(00(N )“0(N )) - 0,000001) DO 5 1=1,N Q(I)=00(I)GO TO 501 CONTINUEW2 = Q(N)/(Z*QQ(N))W = SQR T F (W2)PRINT 880FORMAT(////25X11HMODE NUMBER/) PRINT 980, KZY FORMAT(113 0)PRINT 881FORMAT(//27X9HFREQUENCY/)PRINT 981, W FORMAT(1P1E30.8)PRINT 882FORMAT(//26X10HMODE SHAPE/) PRINT 982,(QQ(I),1=1,N)FORMAT(1P4E20,8)

300,300,500

87DO 8 8 8 I=1 ,N

8 8 8 TG( I s-KZY) = QQ( I )IF(KZY-N) 99» 180 » 180

99 CONTINUECALL MTRAN(1sN »N>TEMP»QQ»GMASS)CALL MULT(N,N,1,A,QQ,TEMP)CALL MULT(1*1»N»B»TEMP»QQ)T = 6 (1 ,1 ) * W2 DO 1 1 1=1,N DO 11 J = 1 9 N

1 1 A(I*J) = A (I ,J)/T DO 12 1=1,NDO 12 J=1»N

12 U (I,J ) = U (I 9 J ) - A ( I 9 J )GO TO 999

180 CONTINUECALL MTRAN(N9 N 9N 9 TEMP9TG 9GMASS)CALL MULT(N,N»N,RT*TEMP»TG)PRINT 902

902 FORMAT(1H1)PRINT 903

903 FORMAT(2X19HORTHOGONALITY CHECK/) PRINT 904

904 FORMAT(2X69HTHIS SHOULD BE AN N BY N ID IAGONAL MATRIX/)PRINT 982,((RT(l9J)9J=l,N)9l=l,N)END

APPENDIX C

The following symbols have been adopted for use in this thesis:

(A) = equilibrium matrixA = cross section area of the element a = acceleration of mass

(B) = compatibility matrix(Bq) = equilibrium matrix due to applied loads f(B^) = equilibrium matrix due to redundant forces(Bfc) = transpose of matrix (B)

b ” width of flangec = the distance from bottom of seat angle to

neutral axis d = external displacement E - Young8 s modulus

(F) = flexibility matrix«= stiffness coefficient of the structural

systemg = gage of the rolled steel in handbook H = the distance from the top of web angle to

the bottom of seat angle

88

89

H® = the distance from the ri"ret line of top angle to the bottom of seat angle

h = the length of web connection I = moment inertia

(K) = element stiffness matrix (K^) = influence coefficient of the structural sys­

temL = length of element M - bending moment m = massn = coefficient of transformed section p = internal forces q » mode shape s - bending stress

(S) - sweeping matrixt = thickness of angle or time U = dynamic matrix v = unit shear x - displacement of massy = the distance from the top of angle to neutral

axis= rotation of the member

6 = relative angle change ck — orientation of element w = natural frequency

REFERENCES

1. Johnston, Bruce, and Hechtman, H. A., "Design Economy by Connection Restraint," Engineering News-Record,October 1940.

2. Steel Construction Manual, AISC, 1963«3. Johnston, B. G., and Mount, E. H., "Analysis of Build­

ing Frames with Semi-Rigid Connections," Transaction, ASCE, Vol. 107, 1942.

4. McCracken, D. D., and Dorn, W. S., Numerical Methods and Fortran Programming. John Wiley & Sons, Inc., New York, London, Sydney, 1964.

j5. Pestelj E. C., and Lecki-n, F. A., Matrix Method in Elasto- mechanics, McGraw-Hill Book Company, Inc., New York,San Francisco, Toronto, London, 19o3.

6. Goldberg, J. E., and Richard, R. M., "Analysis of Nonlinear Structures," ASCE Journal of the Structural Division, Vol. 89, Aug. 1963.

7. Maugh, L. C., Statically Indeterminate Structures.John Wiley & Sons, Inc., New York, 1946. .

8. Lothers, J. E., Advanced Design in Structural Steel. Prentice Hall, Inc., Englewood Cliffs, N. J., I960.

9. Lothers, J. E,, "Elastic Restraint Equations for Semi- Rigid Connections," Transaction, ASCE, Vol. 116, 1951.

10. Rogers, G. L., Dynamic of Framed Structures. John Wiley & Sons, Inc., New York, 1959.

11. Blume, J. A., Newmark, N. M., and Corning, L. H.$Design of Multistory Reinforced Concrete Buildings for Earthquake Motion. Portland Cement Association, 1961.

90