A STUDY OF MATRIX STRUCTURAL ANALYSIS

by

MORRIS RAY SCALES, B.S, in C.E.

A THESIS

IN

CIVIL ENGINEERING

Submitted to the Graduate Faculty of Texas Technological College in Partial Fulfillment of the Requirements for

the Degree of

MASTER OF SCIENCE IN CIVIL ENGINEERING

Approved

Accepted

<9 Dfe' an of the Gracfuafiir Sc/iool

August, 1964

Ht Kl-Mt5.^{

yM L

5 19^

Acknowledgments

I am indebted to Professor Albert J, Sanger for his direction of

this report and to the other members of my committee, Professors

Keith R, Marmion and George A, Whetstone, for their helpful criticism.

This report is dedicated to my wife, Barbara, whose encouragement

and understanding made this work possible.

ii

, e s » s , o , o , o o a , , s 9 a o « » a ; ' 0

, 0 0 0 0 0 9 0 0 , o o e e a a a , 0 , 0

CONTENTS

Page

l i 1 ^ I yJr 1 / \ 0 1 J C 0 O O O O O O O O C I O O O O O O O O O O O C V C C i V

LIST OF FIGURES

I , INTRODUCTION,

Ila INFLUENCE COEFFICIENTS AND ENERGY THEOREMS, . , , = . 5

Influence Coefficients , , , , , * » , , , , o , 5

Ovram energy, o o o o a a o o o j a o o o i o , **'

V X r b U a J L TV o r i s . a a o o a o o a o a e a s a a e a

1 X X a rV /KV^C I M C i n v J L / a a a o a a o a a a o o o t a o a a a o e

I V , S T I F F N E S S M E T H O D . , 8 0 o a o o e o o e e « a o o e a * ^

I n t r o d u c t i o n a « a a o o a o a « o e a e o 9 g o 2 8

Pin-Ended Plane Frames , , , , , , , , a , , , , 29

Rigid-Jointed Plane Frames , , , o , , * , , . . 37

Three-Dimensional Analysis a , = , , . , , , , , 39

V 0 VJ vJINvrflilJO X vJINO 0 0 0 0 0 0 0 0 0 0 > c Q 3 0 f t o 5 o a o *^\)

LJJL^I \Jr i \ C 1 C t x C P I v i C d o o o o o o o o o o o o o o o o f t o o o o o ^ X

i\r tClHU l A o o o o o o o o Q o o o o o o o o o o Q o o o o o o o A ^

'V, Nomenclature , , , . o , a , , , . , , , , , , , , , 43

B, Fundamentals of Matrix Algebra, , a , , , , » , , , , 44

i i i

LIST OF TABLES

Table Page

I \7 * o o , , , a , o o o e , o o o o , o o o o o o c o v o o c * ' —

iv

LIST OF FIGURES

Figure Page

X, v sn Li jLever DO am , 0 0 , 0 0 9 0 0 0 0 0 0 0 0 0 0 , 0 0 ^

2 , Appl icat ion of Unit Loads, 0 0 0 0 0 0 0 , 0 0 0 0 0 0 7

3, Application of Unit Displacements, 0 , 0 , 0 0 = 0 0 , 9

4, Cantilever Beam Example of Force Method, , , , , , , 0 11

5, Propped Beam Example of Force Method , 0 0 0 , , , , , 20

60 Truss Example of Force Method, , , , , 0 , , , , , 0 , 23

7, Rigid Frame Example of Force Method, , , , , , , , , , 25

8a Typical Pin-Ended Truss Member , , , , , , , 0 , 0 , , 29

9o Truss Example of Stiffness Method, , , , , , , , , , , ^

10, Derivation of Transformation Matrix, , , , , , , , , , ^^

11, Typical Rigid Frame Member , , , , , , , , , , , , , , ^^

CHAPTER I

J INTRODUCTION

The development of new techniques of analysis to meet the in­

creasing requirements of greater accuracy, and the revolutionary

development of the high speed electronic digital computer have

signaled a new era in structural analysis. The necessity for the

calculation of stresses and strains in complex elastic structures

has motivated the aeronautical industry to develop new and better

methods of analysis. Since the analysis of even a relatively

simple structure composed of linear elastic members undergoing

small deformations involves the solution of a number of simultaneous

linear equations in the static case and a corresponding number of

linear differential equations in the dynamic case, the advantages of

matrix formulation have been widely recognized. The purpose of this

paper is to present the basic principles governing the application of

matrix theory to structural analysis.

Matrix formulation provides a means of systematizing generalized

procedures for the machine computation of structural problems. This

is especially important in highly redundant structures where the use

of matrix language offers the following advantages in structural

analysis; (1) simplicity of presentation and calculation through

its concise mathematical language which is perfectly adaptable to the

electronic digital computer, and (2) division of labor into formula­

tion and subsequent machine operation by someone who may be unfamiliar

with formulation procedures. This paper will deal only with the

formulation portion of matrix structural analysis.

Structural analysis consists of the computation of external

reactions, internal forces and accompanying stresses, strains, and

deflections of a structure. The two conditions which must be satisfied

in the analysis are: (1) the forces developed in the members must be

in equilibrium, and (2) the deformations of the members must be com­

patible, or consistent with each other and with the boundary conditions

The forces and deflections must also be consistent with the stress-

strain relationship for the material used,

rThe analysis may be approached from two different points of view,

I-^ the^analysis of statically determinate structures, the external

reactions and internal forces and stresses are computed first. The

strains are then determined from the stresses, and subsequently the

deflections are computed. This same procedure may be followed in

analyzing indeterminate structures by a number of the classic methods

of analysis, such as the superposition-equation or dummy load method,

Castigliano*s second theorem, and the use of the three-moment equation.

In this approach, the forces acting on or in the members of the

structure are considered as unknown quantities. Since there are an

infinite number of such force systems satisfying the equations of

equilibrium, the correct one must also satisfy the conditions of

compatibility. Thus, the redundant forces and/or couples are first

computed by solving an equal number of simultaneous equations, each

of which expresses a known condition of the primary or statically

determinate cutback structure in terms of the redundants. This

approach is referred to as the force or flexibility method of

structural analysis. It is especially appropriate for structures

which are not highly redundant.

The other approach to the analysis of indeterminate structures

considers the displacements of the joints in the structure as unknown

quantities. Since an infinite number of systems of mutually compatible

deformations in the members are possible, the correct system is the

one for which the equations of eqilibrium are satisfied. Thus, the

key displacement components are expressed in terms of internal forces

and/or couples and substituted into the equations of equilibrium.

The solution of this simultaneous set of equations yields the values

of the displacements and the analysis is completed as if dealing with

a statically determinant structure. This approach is known as the

direct stiffness or displacement method of structural analysis and

has the advantage of being completely independent of the concept of

static determinateness or indeterminateness, It is the basis for

slope-deflection and the relaxation types of analysis, such as moment

distribution, (1)

The degree of redundancy determines the most appropriate method

of analysis for a particular structure. The basic principles in the

formulation of the flexibility and stiffness coefficient matrices

will be discussed in Chapter 11^ and example problems covering the

force and stiffness methods will be presented in Chapters III and IV,

respectively, A summary of matrix notation and algebra is presented

in the Appendix, The primary reference for this paper is Matrices

for Structural Analysis by Sydney John McMinn (2).

CHAPTER II

INFLUENCE COEFFICIENTS AND ENERGY THEOREMS

Influence Coefficients

If an elast ic spring deflects an amount A under the action of

an axial load W, and i f Hooke»s Law applies, the f l ex ib i l i ty f of

the spring is defined by the quotient,

f « ^ (2,1)

Thus,vthe f l ex ib i l i ty f is the displacement produced by a unit load

W»l, The deflection L may be writen as

A • f"W (2.2)

The reciprocal of the f l ex ib i l i ty is called the stiffness and is

defined by k « 1/f. The load is thus related to the displacement

by the equation

W . k»A (2,3)

where k i s the force required to produce a unit deflection.

The above relations can be extended to the two-dimensional body

by considering a load W. applied to a structure at point j and produc^

ing a directly proportional deflection Ai at point i . Thus Ai«fij*W.

where f i j i s a constant. Under the principle of superposition, the

action of a number of loads on the structure would result in the

equation Ai« f i i W • fi2'W2 • fi3''W3 • <""> * fin'Wn (2,4)

where the deflection Ai i s in the same direction as the load Wi,

This same procedure for displacements at other points or at i in

another direction will result in a set of simultaneous equations.

In matrix notation, these equations which replace equations (2.2)

and (2o3), respectively, may be written

A » F^W, and (2.5)

W « F"^»A « K-A (2.6)

The flexibility influence matrix F has as its general element f|^,

which is the displacement produced at point i, in the direction of

Wi, by a unit load at j, in the direction of Wj. The stiffness in­

fluence matrix K has as its general element kij, which is the force

required at i, in the direction of Wi, to produce a unit displacement

at j in the direction of Wj. The elements must obey Maxwell's Law

of Reciprocity, resulting in symmetic influence coefficient matrices

as long as they are referred to orthogonal coordinate systems=

These influence coefficient matrices derive their names from

theii similarity to sets of influence lines. In order to obtain the

deflection Ai, the products of the applied loads times the correspond­

ing ordinates of the influence line are summed. Here, the elements

of the ith row represent the ordinates of the influence line for Aj

at the individual load points

9* I

Figure 1, Cantilever beam (2,p,58)

The derivation of the influence coefficient matrices for the

simple cantilever of Figure 1 is presented as an example. The

canti lever has uniform cross-section and loads applied as shown To

obtain the f l ex ib i l i t y matrix, equation (2,5) becomes

P2 V2 -62

5ll ^21 ^31

! l 2 *22 '32

fl3 ^23

^33

:x2 •v2 (2,7) M; '2

where the subscripts of the loads and displacements denote the node

numbers. Unit loads are now applied to node 2 as shown in Figure 2,

and deformations due to shear are considered negligible.

Q)

(3)

1

Figure 2, Application of Unit Loads (2,po59)

The displacements are computed using the M/g^ diagrams of Figure 2

and the moment-area method.

For unit P2 applied alone.

U, * AaE ? V2 • 0 » ©2

8

For unit S2 applied alone.

u.

^ - 1»L 2 " E«I

62 «ilL '• E,I

For unit M2 applied alone.

k 2 L I

2L * L" 3 3EoI

. , L2

2E'I

u.

V2 •

® 2 - F T T

L

L

L 1

a „ L ZE«I

L

The signs are recorded according to pos i t ive and negative deflec-

t ion re la t ive to the coordinate system, and clockwise f») or counter­

clockwise (••') rotation in order to conform with normal pos i t ive shear

and bending moment conventions. Tension on a member i s considered

p o s i t i v e . From these elements, the f l e x i b i l i t y influence matrix i s

P2 S2 M2

u-L3

SToT 2EcI -L2 J ^

IfoT E I_ The underlined terms are for guidance purposes only.

0.

L AoE 0

0 0

=L2 (2,8)

To obtain the stiffness matrix, unit displacements are applied

singly as indicated in Figure 3,

r ®miHi Rt

ra)

Figure 3 . Application of Unit Displacements (2,p,60)

Putting U2 « 1 with V2 « ©2 ' "

A'-E'l , Sj • 0 , and M2 « 0

For V2 • 1 with U2 • ©2 " ^

Fx

2E«I V2 Ml • M2 « i | - - (202 • ©1 -3 j ^ ) »

S2 « (M2+Mi)/L - 12E»I

6E'I

For ©2 « 1 with U2 • V2 • 0?

M, = 2Ea (^2) . .^i-L 2 L L

M « 2E-I ,.,«. . 2E»I

Sj • (Ml •Mj) /L . ^

Pj = 0

10

Thus, the stiffness matrix is

K =

AoE L

0

0

12E°I

6E'I

11 0

6E'I ^ ^

4Ea L J

(2.9)

Obviously, the calculation of the flexibility matrix will become

prohibitive as the structure increases in size and complexity. It

will therefore be beneficial to offer an alternate procedure based

upon the principles of strain energy and its accompanying theorems.

This approach will also offer increased versatility through the easy

calculation of stresses and moments in addition to deflections. The

following sections contain derivations available in many references,

but offered here in matrix notation in order to facilitate an under­

standing of the examples to follow, (2)

Strain Energy

Since we are dealing with elastic structures which obey Hooke's

Law, both strain energy and complementary energy are expressed in

terms of the loads and deflections by the same formula

In matrix notation

V » I E Wi Ai

V « i W* A, 2 .. -, (2.10)

where W and A are column matrices. Using equations (2c5) and (2.6),

V = i w* 1 F«W = J A*(F-^)*»F»K'A = * K'A, (2,11)

11

with F and K being symmetrical.

The strain energy for a member can also be expressed in terms of

the internal stresses o and strains e as

V • Z O'E'dxcdy^dz,

or for the whole structure,

V = i///ECToEcdx^dycdz

Preferring to work with large portions of the structure and

utilize superposition, we will refer to the cantilever shown in

Figure 4 and divide it into the lengths AB, BC and CD, The forces

acting on the elements of the stmcture are bending moment, shear

forces, and axial forces, which will be designated stress resultants.

The strain energy for the cantilever will be calculated as though

each length were a simple cantilever

I" I We iwc I

*1 ftM cU P i S4 B "^ -^c ^^Mo

/EOtns. fiOir^J^aOm >md

Figure 4, Cantilever Beam Example of Force Method (2, po61)

Let the column matrix S ^ represent the stress resultants

acting on member AB and let the column matrix © g represent the

corresponding displacements of the end B relative to end A. There­

fore, the strain energy in AB is

V.„ « i S.o* AB ' J ^AB • ©AB (2.12)

Since 0^g - ^AB'^AB *"** ^AB'^AB'^AB* ^« ?«*

^AB = K B ' ^ A B ' S ^ B " 7 ®AB'^AB'©AB

The total internal strain energy is therefore

or in matrix notation

V « |s*o0.

where

' ^ A B '

^BC

^CD B 1

, and 0

e AB

01 BC

0 CD

in the form of partitioned matrices. Since

0 • f"S,

the diagonal partitioned matrix f is

0

0

^AB 0

BC

0 CD

Then the total strain energy for the structure is

V a i S*»f^S « l0*cko0.

where k is the inverse of f,

k » f"^ «

\i' ' 0

0 0

0

0

-1 'CD

AB

0

0

0

BC

0

0

CD

12

(2ol3)

(2J4)

(2,15)

(2,16)

(2,17)

(2,18)

These matrices f and k are the flexibility and stiffness matrices of

the unassembled structure,

13

We will neglect shear and axial deflections in this example,

with bending energy being the only strain energy considered. Then,

from (2.8) the f l ex ib i l i ty equation for each length is

0

3E^I

-L'

IE'I (2.19)

LM. 2E»I E°I

where S and M are the shear force and bending moment at the right=

hand end. It is fundamental to matrix analysis that structures

have loads applied at nodes only. The procedure to be followed in

the event of intermediately applied loads will be discussed subsequently

Using (2,19) to form the flexibility matrix f of (2,16) and

nuttinp in numerical values results in

5,760 -72 0

.10^ £oI

72

0

1.2 0

720 -18

0

0

0

0

0

0

0

0

18

0

0

0,6 0

0 720

0 -18

18

0,6

(2,20)

The strain energy due to external or internal conditions is the

same, or

W*. A = S*^ 0 (2.21)

The internal stress resultants are produced by the applied loads and

are related to them by a set of simultaneous equations

S = B«W (2,22)

14

For the example problem,

(2-23) Sg « Wg Wc Wijj

Mg » -60 Wc-120Wjj*Wg*M +V ;

Sc " Wc Wpj

Mc « -60Wp+lCf *Wo;

SD • %'^

MD - ^ ^

where M i s the bending moment at B, and F? i s the externally applied

moment at B, Placing equations (2,23) in matrix form (2.22) gives

hi MB

S^

M C

D

L"D.

1 -60 1 -120

o l

1

0

1

0

1

(2.24)

A W W

From equation (2^22) we have S * W "B . Substi tution of th is into

equation (2o21) results in

W*°A = W*''B*'0

Then

A = B ' 0

(2o2S)

(2.26)

The relationship between equations (2,22) and (2^26) is important,(2)

Virtual Work

If a structure which is in equilibrium under a system of external

loads is given a virtual deformation by the application of additional

15

loading, then the strain energy becomes

Expanding,

V+6V « i(W4.«W)* (A*6A ) (2.27) 2

V+5V • i[W*»A*6W*<.A-»'W*«6A*6W**6Al 2

Disregarding the higher order terms and subtracting V«iw*''A

results in

6V - i[6W*»A* W*'6A] (2,28)

Since 6W 'A* 6W •F-'W, and using the rule that the transpose of the

product equals the product of the transposes in reverse order, we have

6W*»F«W « [(FoW)*''6W]* « [W*-F*''6W]*

« [W*»F*6W]*, (2.29)

where F is symmetric. Prom A«F»W follows 6A« F-'fiW and

6W*»F''W • [W*''6A]* or

6W*»A « [W »6A] (2,30)

Since V is a scalar quantity, both 6W*»A and W*«6A are also

scalar quantities and equal to their own transposes. Therefore, from

equations (2,28) and (2,30),

6V « W*''6A (2.31)

5V « 6W -A (2.32)

Similarly, i t can be found that

6V « S*'60, and (2.33)

(5V « 6S* 0 (2.34)

Upon combining

6W*<'A - 6S*'0 (2.35)

16

The load term W includes the redundants as well as the applied

loads in the case of indeterminate stxuctures. However, in equation

(2.35), ^ is selected so that the redundants do not change. Thus,

the structure is reduced to a statically determinate structure by

disregarding the redundants. Under the principle of superposition,

the effects of the redundants are subsequently added. Therefore,

6W of equation (2,35) is applied to a statically determinate system

resulting in the relationship for 6S of •••(• T • ., ; ' L '

63 • B « 6W (2.36)

In the whole.structure, we have seen that

S - B«W

In matrix notation 6S includes as many elements as S, but those

corresponding to the redundant members are zero. Similarly, B^ is

the same size as B, but contains rows of zeros corresponding to the

zeros of 6S, But equation (2,35) can be written as

6W*»F"W - 6S*»f»S - 6W*'>Bj<.f<.B.W

from equations (2,36) and (2,22), Therefore, for an arbitrary 6W,

F » B*»f»B, or (2.37) < f •

F - B*-f-B (2.38) J ' - • :

where BQ applies to the statically determinate cutback structure of

an indeterminate system, and B applies to the determinate system.

In the example cantilever, matrix B is as indicated in equation

(2,24) and f in equation (2.20), Therefore, from equation (2,38),

the product f,B is

10 2 FT

5760

-72

0

0

0

0

72

1,2

0

0

0

080

144

720

-18

0

0

-72

1,2

-18

0,6

0

0

14,400

-216

1800

-54

720

= 18

.72

l c 2

18

0,6

18

0.6

17

(2,39)

and then F » B of .B or

ri2i E o i

576

-7,2

»7,2 1,008 -7.2 1,440

0,12 -14,4 Ool2 -21.6

1,008 -14,4 1,944 -16,2 2,916

-7,2 0,12 -16,2 0,18 -27

1,440 -21,6 2,916 -27 4,608

L -7,2 0,12 -16,2 0,18 -28,8

=7.2

0.12

-16,2

0.18

-28.8

0.24

(2.40)

The f l ex ib i l i ty matrix could have been calculated directly but

without the inherent simplicity and versat i l i ty afforded by the use

of strain energy and virtual work.

Since 6S « Bo«6W(2,36) and 6W*»A-6S*»0(2,35), then

6W*oA « 6W*«Bo*»0 , and (2.41)

A - Bo*-0 (2,42)

Obviously, the displacement Ai i s equal to the product of the ith

row of BQ and the column matrix 0, The ith row of B© , which is

the ith column of BQ, is the l i s t of stress resultants produced in

the s tat ica l ly determinate basic structure by a unit load at W.(2)

CHAPTER III

FORCE METHOD

The force method of analysis of statically indeterminate struc­

tures is simply the adaptation of the dummy load-superposition method

of analysis familiar to all structural engineers to matrix notation

and subsequent computer solution. The flexibility matrix, F, of the

whole structure is obtained from the flexibility matrix, f, of the

unassembled structure through the formula F • B^ "f^B, equation (2,38),

The stress resultants are determined from S • B<>W, ectuation (2,22),

However, in the case of indeterminate structures, the matrix B can

not be calculated directly, so the structure is reduced to a

statically determinate structure by removing selected redundants.

Thus S • Bj,»W*BioR (3,1)

where S is the column matrix of the stress resultants, W is the

column matrix of the loads, and R is the column matrix of the re­

dundants. The matrix B^ is derived from the statically determinate

structure, and B^ is derived from primary structures where only

dummy forces are acting at the point of application of the redundants.

If Si represents the stress-resultants resulting from the

redundants acting alone on the structure, then

Si • B^oR (3,2)

From equation (2,26), the displacements of the points of

application of the redundants are given by

Ar • Bi*»©,

18

19

where 0 is the column,of actual distortions. If we consider L^" 0,

where there i s no sett l ing of the supports, e t c , then

Ar - 0 « Bi*of,s (3,3)

Then, from equation (3 ,1 ) ,

0 « Bi*°f'[Bjj»W*BioR] or

Bj*«f<>BQ»W • - B *'f°Bi»R

Let

Thus

Dp - Bi*''foBo, and (3,4)

Dj • B2*»f«Bi

Djj'W - - Dj'R, or (3,5)

R • - Dj' oDooW (3,6)

Therefore, from equation (3 ,1 ) ,

S - BQOW - Bi»Di"^oDQ«W, and (3.7)

S - [Bo-BjoDf ^oDo] W, or

S « BoW

where

B « [Bo-Bi-Pi'l-Do] (3,8)

Uti l iz ing the equations above, any structure can be fully

analyzed as the following examples wil l indicate.

Example Problem 3 .1 .

The cantilever with the redundant at node 3 in Figure 5 is

analyzed. Shear deformation will be ignored.

20

^?\ m lO" " /30K" Ji5<"

"•4*°

Figure 5, Propped Beam Example of Force Method (2,p.77)

Let S - col(S2,M2,S3,M3,S4,M4) and W - col iyi2$^2»^S»^Af>^i^ ^

The applicable equations linking S and W are:

S2 .

M2 '

S3 '

Ms •

S4 .

M4 •

• " 2 * V 3 !

• -2loyl/^•^2•^i*^^* ^^o R3;

• V3: • . 60W4- f f3*R4j

. W 4 i

• " 4 .

Thus, the matrix B^ is derived by applying unit loads to the cutback

structure at the points of application of external loads in the

appropriate sense. The matrix Bj, is derived by applying unit forces

individually to the primary structure upon which only the dummy load

is acting.

(3.9)

B<

£1 M2

S3

! ^

£i M^

W2

" 1

0

0

0

0

0

A 0

-1

0

0

0

0

J5i 0

-1

0

-1

0

0

^

1

-210

1

-60

1

0

h 0

1

0

1

0

1

21

Bi - c o l ( - l , 150, - 1 , 0, 0, 0 )

The basic f l ex ib i l i ty matrix for each length, from equation (2 ,8 ) , is

0

' L3

L2 I FT M

Thus the flexibility matrix, f, of the unassembled structure is

576 -7,2 0 0 0 0

f • 10^ E«I

-7,2 0,12

1125 -11,25

-11,25 0,15

72

-1,8

-1,8

0,06 0 0 0 0 -J

Matrix multiplication leads to

Bi**f « 10"^ (-1656, 25,2, -1125, 11,25, 0, 0 )

Di « B,*»f«B2 • ,6561

D

1 - "1

D.

1.524158

.-4

Bj=Dj-^»Do«10-2

Bj*-f«BQ - 10"^ (-1656, -25,2, -36,45, -8748, 36,45)

DJ"^«DQ - 10"^ (-2524, -38,4, -55.56, -13,333.3, 55,56)

25.2401 0.3841 0,5556 133.3333 -0.5556

-3786.0090 -57.6135 -83.3340 -20,000.0010 83,3340

25,2401 0.3841 0,5556 133.3333 -0,5556

0 0 0 0 0

0

0

0

0

0

0

0

0

0

0

22

B-[Bo-Bi 'Dr l .D^]

0,7475 -0,0038 -0,0056 -0.3333 0.0056

37,8601 -0.4239 -0,1667 -10.0000 0,1667

i ' . . !• r,

foB • 10 -4

-0,2524

0

0

0

158.025

-0,840

283,951

2,840

0

L 0

-0.0038 -0,0056 -0.3333 0.0056

-60.0000 1.0000

1.0000 0

0,840

-0.023

-4,321

0,043

0

1.0000.

=2,000 -120,000 2,000

0,020 1,200 -0,020

5c000 299,999 -5,000

-0,087 -5.250 .0,087

72,000 - 1 . 8

-1,800 0,060.

158.025 0.840 -2,000 -120,000

F«BQ*'fB • 10"^

.

0,840 0.023 -0.020

-2,000 -0,020 0,067

120,000 -1,200 4,050

2,000 0,020 -0.067

The loads are expressed as

W - col ( 7 , 10, 30, 6 , 15 )

Therefore, from equation ( 3 , 6 ) ,

Rj • - D J - I ' D Q ' W • 9,89^

From equation (2,22)

S - B»W

-1,200

4,050

314.998

-5.850

2.000

-0.020

-0.067

-5.850

0.127

S - col ( 3 , 1 1 , 198,282, - 3 . 8 8 9 , -375 , 6 , 15 )

23

From equation (2,5),

F,W

A « col (,0365, ,0002, ,0011, .1072, .0021 )

Example Problem 3,2,

The truss, illustrated in Figure 6, is indeterminate to the

second degree. The members selected as redundants are 2-4 and 3-6

E'A*ccnsfenf

Figure 6. Truss Example of Force Method

The flexibility matrix for each member is

AoE so the flexibility matrix, f, for the unassembled stxucture is

1-2 1-4 1-3 2-4 2-3 4-3 4-6 4-5 3-5 5-6 3-6

A'E

144 0 0 0 0 0 0 0 0 0 0

0 192 0 0 0 0 0 0 0 0 0

0 0

240 0 0 0 0 0 0 0 0

0 0 0

240 0 0 0 0 0 0 0

0 0 0 0

192 0 0 0 0 0 0

0 0 0 0 0

144 0 0 0 0 0

0 0 0 0 0 0

192 0 0 0 0

0 0 0 0 0 0 0

240 0 0 0

0 0 0 0 0 0 0 0

192 0 0

0 0 0 0 0 0 0 0 0

144 0

0 0 0 0 0 0 0 0 0 0

240

24

Since there are only two external loads, B^ and Bj are given by

W w.

1-2 0 0

Bo «

1-4

1-3

2-4

2-3

4-3 -.1.

4-6

4-5

3-5

5-6

3-6

0,667

-0,833

0

0

0,500

0

0,833

-0,667

-0,500

0

0.500

0.625

0

- l o O O O

-0,375

0

0,625

-0,500

-0,375

0

B, «

2-4

-0 .600

-0c800

1,000

1.000

=0,800

-0 ,600

0

3-6

0

0

0

0

0

-0 ,600

-0 .800

0

J Lo

1.000

-0,800

-0,600

1.000J

Using matrix multiplication and the applicable equations,

* 1 r-86. ^1 'f- F T L 0

4 -153,6 240 240 -153,6 -86,^ 0 0 0 0 0 0 0 0 -86,4 -153.6 240 -153.6 -86.4 24 a

n * r n 1 r 829,44 51,84 Dj . Bj 'f .Bj • /[Tf L 51.84 829.44.

Dj"^ - A»E 0,001210 -0,0000761

-0,000076 0,001210J

The procedure from this point is identical with that of the

previous example problem, and the structure wil l be completely

analyzed.

25

Example Problem 3,3

The rigid frame indicated in Figure 7 is analyzed by the same

procedures. However, the computations will include the effect of

axial forces and deformations. The cutback structure is formed by

releasing the fixed reaction at node 5, and thus reducing the third

to H

1*< ®

30o1

T J @

A E - cor^sfartf

• — * r v s

Figure 7. Rigid Frame Example of Force Method

degree indeterminate structure to a statically determinate one. The

structure can now be treated as a cantilever fixed at node 1. Unit

loads are applied in place of the externally applied loads to obtain

BQ and unit M, V, and H at node 5 to obtain B^. The sign convention

is the same one mentioned in Chapter II, Therefore,

26

B

1-2

2-3

3-4

4-5

{ {

P 5"

f

M

0 1 0 0 0 0 0 0 0 0 0 0

w,

-1 0

240 0 1 0 0 0 0 0 0 0

ili h

B

L

0 1

-300 -1

0 -300

-1 0

-300 0 1 0

1 0

420 0

-1 280

0 -1

0 -1

0 0

Mt

0 0 1 0 0 1 0 0 1 0 0 1

The f matrix for each element or span of the frame, from equation

(2,8), is

f «

L A»E

0

0 0 L3 L2

3FT "21^1

The f matrix for the complete unassembled structure is

f «

^1-2 0

0

0

0

f

0

0

0 0

2-3 0 0

^3-4 0

0 f 4-5

The procedure from this point of the analysis is identical with that

of example problem 1. This solution will give results which differ

from those obtained by moment distribution due to consideration of

axial deformation. This variance can be quite significant

If a pinned reaction were present at node 5 in place of the

fixed one, BQ and B^ would be reduced by the row that corresponds to

27

the moment at node 5, Likewise, the M column of matrix Bj wculd

not ex i s t , and the f matrix of the element 4»5 wil l be reduced by

the row and column corresponding to the moment at node 5 which no

longer ex i s t s , (2)

CHAPTER IV

STIFFNESS METHOD

Introduction

In the direct stiffness method of analysis, the displacements of

the joints in a structure are considered to be the unknown quantities.

The procedure to be followed in the analysis is similar to that of the

Force Method with the complex structure replaced by an equivalent

idealized structure consisting of basic structural parts that are

connected to each other at selected node points. Stiffness matrices

are needed for each basic structural unit appearing in the idealized

structure. While all other nodes are held fixed, a given node is

disnlaced in a particular coordinate direction, The forces required

for this displacement and the resulting reactions generated at connected

nodes are determined by the stiffness matrix of each individual member.

These forces and reactions thus form one column of the composite stiff­

ness matrix. Repeating this process until all possible freedoms at a

node have been considered will result in the complete development of

the stiffness matrix. In the general case, this matrix will be of

order 3n by 3n, where n equals the number of nodes. This singular

matrix so developed will be rendered nonsingular by imposing the

desired support conditions and striking out columns and corresponding

rows for which zero displacements have been specified. With given

external forces at the nodes, matrix calculations applied to the

stiffness matrix will yield all components of node displacement plus

28

29

external reactions. Internal forces can then be found by applying

the appropriate force^deflection relations. The sign convention used

in the Stiffness Method will disagree somewhat with those of the Force

Method of the previous chapter, since the sign convention of the

Stiffness Method agrees with the usual positive directions of the

coordinate system and clockwise rotation will be taken as posit ive,

Pin-Ended Plane Truss

Figure 8 shows a typical pin-ended truss member. In a pin-ended

truss, the only forces that can be applied to a member are equal and

opposite axial forces acting at the nodes. Each node has two possible

degrees of freedoms, translations in the x and y directions. There­

fore, prior to introduction of boundary or support conditions, the

st i f fness matrix of this member will be of order 4 by 4,

£>„^iu->a'

cos Ousya

I L^'^^ (b)

Figure 8, Typical Pin-Ended Truss Member

30

To develop one column of the member K, subject the member to

"2 ^ 0,u,«v,sv2«0, as indicated in Figure 8 (a) . Then

AL = U2eosajj « U2X ,

where the change in the angle is neglected since i t is a virtual dis

placement. The axial force necessary to produce AL is

Therefore, the components of P at node 2 are

Px2 - Pc°s ©x - ( ^ ) U 2 ^ ^

V2 p cos e y -{^|U2^^ Under the conditions of equilibrium, the forces at node 1 are

^xi • -Px2

Pyl " -Py2

This gives, from equation (2,6),

F, xl

>1

x2

_Py2_

AE r

L -Xy

X2

Xy

"1

^1

u^

JL^^J To develop another column, subject the member to V2 f* 0,Ui«U2«vi"0,

as indicated in Figure 8 (b), Then

AL • V2 cos © • V2P

P . AE ^ IT

Fx2» P cos ©x« _

Fy2" P COS ©y" —-

Therefore ^^ AL - V2y

A^ V2yX

P cos ©v" i^ V2p2

Pxl" -Px2» and Fy^ - -F y2

31

Repeating this process gives

K = AE L

"1

X2

Xy

-X^

= Xy

l l Xy

2 y

-Xy

. y 2

^

-x2

-Xy

X2

Xy

IL -Xy

2

- y

Xy 2 y

(4,1)

This matrix is singular, since its determinant vanishes and its in­

verse does not exist, but imposing boundary conditions will render it

nonsingular. Thus, boundary conditions will prevent the member from

moving as a rigid body. The stiffness matrix of the truss indicated

in Figure 9 will be developed. The length of the members, since

variable, will remain inside the brackets, giving X«X ,i/«y , etc.

The data for the truss is as indicated

AE conshan+

Figure 9. Truss Example of Stiffness Method

in Table 1

32

Member

1-2

1-3

2-3

2-4

3-4

X

0

L

L

L

0

y

L

L

0

=L

-L

Length

L

Vzl

L

|2L

L

TABLE 1

X

0

l / f 2

1

1/P 0

y

1

i / R

0

X2

0

1/2

1

- l / f 2 1/2

-1 0

2 y

1

1/2

0

/ l / 2

1

Xy

0

1/2

0

-1 /2

0

P 0

1/2 KJL

1/L

1/2f2L

0

y

l/L

1/2(2L

0

fl /2f2L

1/L

XIT

0

1/2y2L

0

- l /2 ) f2L

0

AE

1 The composite singular matrix for the structure is (4,2)

l /2 f iL l/2ldL 0 0 -l/2|f2L -l/2)f5L 0 0

l/2lf2L l/L*l/2lf2L 0 -1/L ^l/2flL -l/2!f2L 0 0

0 0 l/L+l/2lf2L -l/2y2L -1/L 0 .1/2K2L l/2f2L

0 -1/L .1/2/21 l/L.l/2f2L 0 0 l/2lf2L l/2|flL

0 l/L*l/2f2L 1/2 RL 0 0

0 1/2RL l/2+l/2(2L 0 -1/L

l/2lf2L 0 0 l/2f2L -l/2ir2L

l/2f2L 0 -1/L -l/2)f2L 1/L-1/2(2L

Imposing the boundary conditions u^-vj « u -v ^O by striking out their

rows and corresponding columns gives the nonsingular composite s t i f f -

-l/2ir5L -1/212L -1/L

-l/2f2L -l/2r2L 0

0 0 -l/2y2L

0 0 1/2|(2L

ness matrix (4,3^

Px2

Fy2

''x3

L'y3j

AE IT

1 1+IT7

1 -7T7

- 1

0

1 -Tf7

1 l-TTT

0

0

- 1

0 1

i*ir? 1

Tf?

0

0 1

IT? 1

i*ifr

"2

^2

u. 3

L^3 J

33

Inverting this 4 by 4 matrix, and using equation (2.5) will give the

values of the displacements at nodes 1 and 2, Considering the

partitioned matrix (4,2), it can be represented as

2^2. D

B 2x4 2x2

4x2 4x4 H

4x2

2x2 I 2x4 2x2

Expanding equation (4,3) gives the following additional sets of

equations?

(4,4) "F^ "" XI

L^ij • [B]

"u," V "3 V3

F X4

L^J - [H]

Uo 2

^2 "3

(4.5)

Equations (4,4) and (4,5) will give the reactions after the joint

deflections at nodes 2 and 3 have been solved. In the matrix (4.2),

the 2 by 2 submatrices along the diagonal represent self terms, or

deflections at a node due to two forces. The elements along the

diagonal are known as self-self terms and are displacements at a node

due to forces at the same node in the same direction. The submatrices

off the diagonal are connecting terms between two nodes. The self

terms in matrix (4,2) consisted of the sum of the matrix self terms

for all members entering the particular node. The connecting terms

are simply the connecting terms for the particular member between the

two nodes in question. The terms all come fro» the basic member

34

matrix (4,1) with the appropriate sign. (1)

Another approach offered is a consideration of the basic stiff­

ness matrix (2.9) derived for the cantilever in Chapter II. This

matrix was derived based on each load being applied as an axial force

and a shear or transverse force. In order to bring all joint displace­

ments into the same reference system, a rotation or transformation

matrix T is utilized. If we consider the member 1-2 in

p. /\—4 (a) (b)

Figure 10, Derivation of Transformation Matrix (2,p.112)

Figure 10 (a ) , and displace a joint 6^ only, with 6y«0, then the

displacement in the P direction is 6x°cosa and in the S direction

= 6x'sin a. Repeating with 6yf«0, and 6jj"0» *"^ adding the two

sets gives

6p « 6«ocos o ••• 6v''Sin a

fie ~ -6v»sin a ••• fiv'cosa

35

In matrix notation this is

cos a sin a

-sina cos a

Calling the coefficient matrix T, where 0j»T»Aj, and replacing its

elements with their equivalent from Figure 10 (a) gives

6 . y .

X r Y

T

Y r X r

(4.6)

When dealing with a pin-ended truss member, the shear forces are zero

and the axial forces are in equilibrium. The stress resultants are

12 • • 1

Si '21 • . e i

fip 1

>Si , and ©2

With node 2 fixed, Sj2*l^i2*®i» where

12

E«A

0

(4.7)

Repeating with node 1 fixed gives S22"ki2'©2'' '^^ reaction produced

at node 1 with this translation is Si2«-S2i"-J^i2°®2» Superposing

these gives

Si2 "1^12 (®i-62) • "^21-

Utilizing T, the components of the applied forces Sj2 in any direction

are related to the components Sj2* in the x and y directions by

Si2 « TcSj2'

The transformation matrix T is orthogonal, or T'^ * T'

36

Therefore

» - 1

^12 -"T 12

= T-i°k.(02-02)

= T-l-kj2 T(Ai-A2)

The joints are in equilibrium, so

Wi-J Sj2' • 0

Therefore

W = Z T-^°kj2°'^(^r^2^ ^^"^^

= (E T-lckoT)Ai-J:(T-i»k°T«A2)

where the summations are taken over all joints 2 which are directly

connected to node 1.

Wi);h equations (3,6) and (3.7)

(T-lck«T)j2 = H ^

V Y Since ~ • cos a and — » sin a,

L L (T-l»k»T) E o A

L

X^

L 2

XY

J7

'7 Xy

XY 1 7

Y2

Xy

"2 y

12

with the notation previously used. This will be recognized as the

individual self-self term contained in matrix (4.1), Thus the member

stiffness matrix can be derived as shown in Chapter II and rotated

to its proper orientation relative to the x and y axes by

M - [ "'] hp] b] C2)

37

Rigid-Jointed Plane Frames

The member 1-2 indicated in Figure 11 has three possible

degrees of freedom at each end.

Figure 11, Typical Rigid Frame Member

With node 2 fixed, slope deflection gives the following relationships:

M 12

M

2EI L

2EI 21

. . ,6Si 4EI01 6El5s (201-3-^) - -=-1; 1 ^

' - 6Elgsi L

(Mi2*M2l) . 6Eiei 12EI«sl

F'A'fipl L

In matrix form, this is

p.

Ml

E'A L

0

0

0 0

12E'I -6E.I

•^i? [7 -6E«I 4E'I

°P1

«si

©1

or Si"ki2*©i« The reactions produced by these displacements at

node 2 are related to St2 by

38

S2 - -Sj

M2 s -Mj-Sj^L

or in matrix form.

• P 2 "

^2

. " 2 .

B

.1 0

0 -1

0 =L

0

0

• p l "

Si

_ " l . If A is f ixed.

^2

^2

_ " 2 .

S

E,A L

0

0

0 0

12E-I 6E »I

U ^ L2 6E»I 4Ea

T? L

«P2

«S2

3. The reactions at node 1 are

a 0

0 -1

0 L

"Pl"

h

_"l_

s

0

0

1

" P 2 "

^2

3. As with the truss members, the displacements in the x and y

directions are related to the displacements in the P and S directions

by the transformation matrix

cos ot

- s i r a

sm a

cos a

0 0

39

L

Y L

Y

r

L

0

0

Using the transformation matrix and the proper member K matrix, the

absolute s t i f fness matrix can be obtained, and then the structure

may be analyzed as shown in the pin-ended truss example, (2)

Three-Dimensional Analysis

The procedure followed in the case of three-dimensional s t ructures

i s but an expansion in scope and size of the appropriate two-dimensional

case by the addition of the th i rd dimension. I t i s obvious that the

composite s t i f fness matrix wil l become quite large and the analysis

more complex

CHAPTER V

CONCLUSIONS

The complexity of modern structures is increasing the use of

computers for analysis which was formerly prdiibitive without the

use of simplifying assumptions which were self-defeating and

uneconomical.

The choice of which method of analysis to use is complex. The

force method requires the inversion of a matrix of the order of n by

n, where n is the number of redundants. The stiffness method requires

the inversion of an n by n matrix, where n is the number of possible

freedoms of the joints. The complexity of aircraft structures has

resulted in the aeronautical industry advocating the use of the direct

stiffness method, (1) Therefore, it appears that the redundancy of

the structure will determine the method of analysis to be used.

Because higher orders of redundancy should prove to be most economi­

cally feasible for computer use, and matrix notation is compatible

to computer use only, the direct stiffness method of analysis should

offer the most promising approach to future development and study.

40

LIST OF REFERENCES

1, Turner, M. J., et al. "Stiffness and Deflection Analysis of Complex Structures", Journal of the Aeronautical Sciences, Vol. 25, No. 9, September, 1956, p. 805.

2, McMinn, Sidney John, Matrices for Structural Analysis, London, E, § F. No Spon, Ltd., 1962. — = = ===«,.==

3, Pipes, Louis A. Matrix Methods for Engineering, Prentice-Hall, Inc., Englewood ciitts, N. J., 1963. '^

4, Morice, P, B, Linear Structural Analysis, The Ronald Press Company, New YorkT*^

41

APPENDIX

A, Nomenclature

B, Fundamentals of Matrix Algebra

42

43

APPENDIX A: NOMENCLATURE

A = cross sectional area of member

B - coefficient matrix for complete structure

BQ - coefficient matrix for primary structure

Bj - coefficient matrix for reactions

E - modulus of elasticity

f - flexibility matrix for unassembled structure

F - flexibility influence matrix

Fy. - force at node 1 in the y direction

I - moment of inertia of member

k - stiffness matrix of a member

K - stiffness influence matrix

M - moment

P - axially applied load

R - reaction

S - transverse load

S,2 - internal stress resultants

T - transformation matrix

u - translation in the x direction

V - translation in the y direction

V - strain energy

W - load column matrix

A - deflection column matrix

0 - rotation angle

X - angle between x - axis and member

p - angle between y - axis and member

44

APPENDIX B: FUNDAMENTALS OF MATRIX ALGEBRA

I, Definition of a Matrix

A matrix is a rectangular array of elements defined by

certain rules of operation that prescribe the manner in which

these arrays are to be manipulated, A matrix of order n by s has

n rows and s columns and is denoted in the following manner:

A « [a] «

a n ^12

*21 ^22

a

a

13

23

^15

^25

/ n l *n2 an3 • ^ ^5

The elements a.^^ of the matrix [a] may be real or complex

numbers. The general element a^^ is that element in the ith row

and jth column where i » 1, 2,°"'», n and j e l , 2, 3,'°°°s, A

square matrix is one where n • s, or the number of rows equals the

number of columns, A matrix must be square in order to form a

determinant from its elements. The matrix with an asterisk, [aj*,

is obtained by a complete interchange of the rows and columns, so

that a^j - a^^, and is called the transposed matrix of [a], A

symmetric matrix is a square matrix whose elements are symmetric

about its principal diagonal so that a j « a... It is obvious

that this implies [a] * [a] . A matrix may be a column matrix,

designated by A * col.(a2j[, a22» ^is) » or a row matrix, designated

as A s (an, ai2» ^i$) >

45

II, Matrix Algebra

Matrices are equal if and only if their corresponding

elements are identical. If two matrices are of the same order,

then their sum is c^j « a^j -»• b^^ and their difference is

Cij » a.j -b...

The product of a scalar term and a matrix is effected by

multiplying each element of [a] by the scalar and obtaining a

new matrix whose elements are ka^j. The product of a matrix [a]

of order (n,s) by a matrix [b] of order (s,m) is defined to be

the matrix [c] given by

^ij • ^.1 ^ij * ^Jk

where the product is [a] [b] • [c], The number of columns of [a]

must equal the number of rows of [b] for multiplication to be

possible, and the matrices are then said to be conformable*

For example, if

i-s Z &ii • b,

Then C «

2

3

2

3

1

4

r 4

, and B «

~ 6 "

3

6

3

"2x6+

3x 6*

Obviously, the order of [cj is 2 by 1, or the number of rows of

[ a ] , the pre-mul t ip l ie r , by the number of columns of [b] , the

pos t -mul t ip l ie r . Matrix mult ipl icat ion is not normally commuta­

t i v e , since [a] [bji^fb] [ a ] , except in special cases.

46

The determinant of a square matrix is equal to the determinant

of its elements. The cofactor matrix of any square matrix [a] with

n rows and columns, denoted by cof A, is the matrix obtained by

replacing each element of A by its cofactor. The cof actor of [a]

is the product of the determinant of the matrix with n-1 rows and

columns of [a], by (-l)i^J.

The inverse of the matrix A, denoted by A"^ is a matrix

such that

AA"^ « I

where I is the identity matrix. A"^ is given by

1. (cof A)* '^ det A

where (cof A)* is called the adjoint matrix. The identity matrix

contains only unit elements along the leading diagonal, with the

other elements all zero. This is but one means of determining

the inverse of a matrix. The determinant of a matrix must not

vanish for the inverse to exist. If the inverse of a matrix

exists, the matrix is said to be a nonsingular. (3)

An important rule is that * * *

(AB) = B - A

or the transpose of the product equals the product of the transposes

in reverse order.

Many normal rules of algebra are applicable. For example,

W s K'A

K-i«W = loA A = K-1»W ' F»W

• t X A S TFCHNnLDGICAi: CniJlXGfS LIBRARY