Analysis of Structural Member Systems

JEROME J. CONNOR, Sc.D., Massachusetts Institute of Technology, is Professor of Civil Engineering at Massachusetts Inst...

Author: Jerome J. Connor

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JEROME J. CONNOR, Sc.D., Massachusetts Institute of Technology, is Professor of Civil Engineering at Massachusetts Institute of Technology. He has been active in teaching and research in structural analysis and mechanics

at the U.S. Army Materials and Mechanics Research Agency and for some years at M.I.T. His primary interest is in computer based analysis methods, and his current research is concerned with the dynamic analysis of prestressed concrete reactor vessels and the development of finite element models for fluid flow problems. Dr. Connor is one of the original developers of ICES-STRUDL, and has published extensively in the structural field.

ANALYSIS OF STRUCTURAL MEMBER

SYSTEMS JEROME J. CONNOR Massachusetts Institute of Technology

THE RONALD

PRESS COMPANY • NEW YORK

Copyright ©

1976 by

Ttrn RONALD PRESS COMPANY

All Rights Reserved

No part of this book may be reproduced in any form without permission in writing from the publisher.

Library of Congress Catalog Card Number: 74—22535 PRINTED IN ThE UNITCD STATES OF AMERICA

Preface

With the development over the past decade of computer-based analysis methods, the teaching of structural analysis subjects has been revolutionized. The traditional division between structural analysis and structural mechanics became no longer necessary, and instead of teaching a preponderance of solution details it is now possible to focus on the underlying theory. What has been done here is to integrate analysis and mechanics in a systematic presentation which includes the mechanics of a member, the matrix formulation of the equations for a system of members, and solution techniques. The three fundamental steps in formulating a problem in solid mechanics—. enforcing equilibrium, relating deformations and displacements, and relating forces and deformations—form the basis of the development, and the central theme is to establish the equations for each step and then discuss how the complete set of equations is solved. In this way, a reader obtains a more unified view of a problem, sees more clearly where the various simplifying assumptions are introduced, and is better prepared to extend the theory. The chapters of Part I contain the relevant topics for an essential background in linear algebra, differential and matrix transformations. Collecting this material in the first part of the book is convenient for the continuity of the mathematics presentation as well as for the continuity in the following development. Part II treats the analysis of an ideal truss. The governing equations for

small strain but arbitrary displacement are established and then cast into matrix form. Next, we deduce the principles of virtual displacements and virtual forces by manipulating the governing equations, introduce a criterion for evaluating the stability of an equilibrium position, and interpret the governing equations as stationary requirements for certain variational principles. These concepts are essential for an appreciation of the solution schemes described in the following two chapters. Part III is concerned with the behavior of an isolated member. For completeness, first are presented the governing equations for a deformable elastic solid allowing for arbitrary displacements, the continuous form of the principles of virtual displacements and virtual forces, and the stability criterion. Unrestrained torsion-flexure of a prismatic member is examined in detail and then an approximate engineering theory is developed. We move on to restrained torsion-flexure of a prismatic member, discussing various approaches for including warping restraint and illustrating its influence for thin-walled iii

PREFACE

and closed sections. The concluding chapters treat the behavior of planar and arbitrary curved members. How one assembles and solves the governing equations for a member sysopen

tern is discussed in Part IV. First, the direct stiffness method is outlined; then a general formulation of the governing equations is described. Geometrically nonlinear behavior is considered in the last chapter, which discusses member force-displacement relations, including torsional-flexural coupling, solution schemes, and linearized stability analysis. The objective has been a text suitable for the teaching of modern structural member system analysis, and what is offered is an outgrowth of lecture notes

developed in recent years at the Massachusetts Institute of Technology. To the many students who have provided the occasion of that development, I am deeply appreciative. Particular thanks go to Mrs. Jane Malinofsky for her patience in typing the manuscript, and to Professor Charles Miller for his encouragement. JEROME J. CONNOR

Cambridge, Mass. January, 1976

Contents

I—MATHEMATICAL PRELIMINARiES 1

Introduction to Matrix Algebra 1—i 1—2 1—3 1—4 1—5 1—6 1—7

1—8

Definition of a Matrix Equality, Addition, and Subtraction of Matrices Matrix Multiplication Transpose of a Matrix Special Square Matrices Operations on Partitioned Matrices Definition and Properties of a Determinant Cofactor Expansion Formula

Cramer's Rule 1—10 Adjoint and Inverse Matrices 1—11 Elementary Operations on a Matrix 1—12 Rank of a Matrix 1—13 Solvability of Linear Algebraic Equations 1—9

2

2—2 2—3 2—4

2—5

5 8 10

12 16 19 21

22 24 27 30

Introduction Second-Order Characteristic-Value Problem Similarity and Orthogonal Transformations The nth-Order Symmetrical Characteristic-Value Problem Quadratic Forms

46 46 48 52 55 57

Relative Extrema for a Function 3—1

3—2

3—3

4

5

Characteristic-Value Problems and Quadratic Forms 2—1

3

3

Relative Extrema for a Function of One Variable Relative Extrema for a Function of n Independent Variables Lagrange Multipliers

66 66 71

75

Differential Geometry of a Member Element 4—1

4—2

Parametric Representation of a Space Curve Arc Length V

81 81

82

CONTENTS 4—3 4—4 4—5 4—6

4—7 4—8

5

Unit Tangent Vector Principal Normal and Binormal Vectors Curvature, Torsion, and the Frenet Equations Summary of the Geometrical Relations for a Space

85 86 88

Curve

91

Local Reference Frame for a Member Element Curvilinear Coordinates for a Member Element

92 94

Matrix Transformations for a Member Element 5—1

5—2 5—3

Rotation Transformation Three-Dimensional Force Transformations Three-Dimensional Displacement Transformations

100 100 103 109

Il—ANALYSIS OF AN IDEAL TRUSS 6

Governing Equations for an Ideal Truss 6—1

General

6—2

Elongation—Joint Displacement Relation for a Bar General Elongation—Joint Displacement Relation Force-Elongation Relation for a Bar General Bar Force—Joint Displacement Relation Joint Force-Equilibrium Equations Introduction of Displacement Restraints; Governing Equations Arbitrary Restraint Direction Initial Instability

6—3

6—4 6—5

6—6 6—7

6—8 6—9

7

132 134 137

Variational Principles for an Ideal Truss 7—1

General

7—2

Principle of Virtual Displacements Principle of Virtual Forces Strain Energy; Principle of Stationary Potential

7—3 7—4

7—5

7—6

8

115 116 120 125 130 130

152 152 153 159

Energy

162

Complementary Energy; Principle of Stationary Complementary Energy Stability Criteria

165 169

Displacement Method—Ideal Truss 8—1

General

8—2

Operation on the Partitioned Equations The Direct Stiffness Method

8—3

178 178 178

180

CONTENTS 8—4

8—5

9

Incremental Formulation; Classical Stability Criterion Linearized Stability Analysis

191

200

Force Method—Ideal Truss 9—1

General

210

9—2

Governing Equations—Algebraic Approach Governing Equations—Variational Approach Comparison of the Force and Mesh Methods

211

9—3

9—4

216 217

Ill—ANALYSIS OF A MEMBER ELEMENT 10

Governing Equations for a Deformable Solid 10—1

General

10—2

Summation Convention; Cartesian Tensors Analysis of Deformation; Cartesian Strains Analysis of Stress Elastic Stress-Strain Relations Principle of Virtual Displacements; Principle of Stationary Potential Energy; Classical Stability Criteria Principle of Virtual Forces; Principle of Stationary Complementary Energy

10—3 10—4

10—5 10—6

10—7

11

229 230 232 240 248

253

257

St. Venant Theory of Torsion-Flexure of Prismatic Members 11—1

11—2 11—3 11—4

11—5 11—6 11—7

12

229

Introduction and Notation The Pure-Torsion Problem Approximate Solution of the Torsion Problem for Thin-Walled Open Cross Sections Approximate Solution of the Torsion Problem for Thin-Walled Closed Cross Sections Torsion-Flexure with Unrestrained Warping Exact Flexural Shear Stress Distribution for a Rectangular Cross Section Engineering Theory of Flexural Shear Stress Distribution in Thin-Walled Cross Sections

271 271

273 281

286 293 303

306

Engineering Theory of Prismatic Members 12—1

12—2

Introduction Force-Equilibrium Equations

330 330 331

CONTENTS 12—3

12—4 12—5 12—6

13

13—2 13—3

13—4 13—5 13—6

13—7 13—8 13—9

371

371 Introduction Displacement Expansions; Equilibrium Equations 372 Force-Displacement Relations—Displacement Model 375 Solution for Restrained Torsion—Displacement Model 379 Force-Displacement Relations—Mixed Formulation 383 Solution for Restrained Torsion—Mixed Formulation 389 Application to Thin-Walled Open Cross. Sections -395 405 Application to Thin-Walled Closed Cross Sections Governing Equations—Geometrically Nonlinear Restrained Torsion 414

Planar Deformation of a Planar Member 14—1

14—2 14—3

14—4

14—5

14—6

14—7 14—8

15

333 339 340 349

Restrained Torsion-Flexure of a Prismatic Member 13—1

14

Force-Displacement Relations; Principle of Virtual Forces Summary of the Governing Equations Displacement Method of Solution—Prismatic Member Force Method of Solution

Introduction; Geometrical Relations Force-Equilibrium Equations Force-Displacement Relations; Principle of Virtual Forces Force-Displacement Relations—Displacement Expansion Approach; Principle of Virtual Displacements Cartesian Formulation Displacement Method of Solution—Circular Member Force Method of Solution Numerical Integration Procedures

425 425 427

429

435 445 449 458 473

Engineering Theory of an Arbitrary Member 15—1

15—2 15—3

15—4 15—5 15—6 15—7

15—8

Introduction; Geometrical Relations Force-Equilibrium Equations Force-Displacement Relations—Negligible Warping Restraint; Principle of Virtual Forces Displacement Method—Circular Planar Member Force Method—Examples Restrained Warping Formulation Member Force-Displacement Relations—Complete End Restraint Generation of Member Matrices

485 485 488 490 .493 499 507 511

517

CONTENTS

Member Matrices—Prismatic Member 15—10 Member Matrices—Thin Planar Circular Member 15—11 Flexibility Matrix—Circular Helix 15—12 Member Force-Displacement Relations—Partial End Restraint 15—9

520 524 531

535

tV—ANALYSIS OF A MEMBER SYSTEM 16

Direct Stiffness Method—Linear System 16—1

16—2 16—3 16—4

17

Introduction Member Force-Displacement Relations System Equilibrium Equations Introduction of Joint Displacement Restraints

545 546 547 548

General Formulation—Linear System

17—4

Introduction Member Equations System Force-Displacement Relations System Equilibrium Equations

17—5

Introduction of Joint Displacement Restraints;

17—1

17—2 17—3

Governing Equations Network Formulation 17—7 Displacement Method 17—8 Force Method 17—9 Variational Principles 17—10 Introduction of Member Deformation Constraints 17—6

18

545

554 554 555 557 559 560 562 565 567 570 573

Analysis of Geometrically Nonlinear Systems 18—1

18—2 18—3 18—4

Index

Introduction Member Equations—Planar Deformation Member Equations—Arbitrary Deformation Solution Techniques; Stability Analysis

585 585 585 591

597

605

Part I MATHEMATICAL PRELIMINARIES

1

Introduction to Matrix Algebra 1—1.

DEFINITION OF A MATRIX

An ordered set of quantities may be a one-dimensional array, such as

a two-dimensional array, such as a11, a12, . . ,a1, a21, a22, . , .

.

.

ami,

a two-dimensional array, the first subscript defines the row location of an element and the second subscript its column location. A two-dimensional array having ,n rows and n columns is called a matrix of order m by n if certain arithmetic operations (addition, subtraction, multiplication) associated with it are defined. The array is usually enclosed in square brackets and written as* a11

a12

a21

a22

a,,,1

Note

am2

-

-

a1,,

-

a2,,

=

=

a

a,,,,,

that the first term in the order pertains to the number of rows and the

second term to the nuiñber of columns. For convenience, we refer to the order of a matrix as simply m x n rather than of order m by n. *

In print, a matrix is represented by a boldfaced letter. 3

INTRODUCTION TO MATRIX ALGEBRA

4

CHAP. 1

A matrix having only one row is called a row matrix. Similarly, a matrix having only one column is called a column matrix or column vector.* Braces

instead ofbrackets are commonly used to denote a column matrix and the column subscript is eliminated. Also, the elements are arranged horizontally instead of vertically, to save space. The various column-matrix notations are: C11

C1

C21

C2

{c1, c2,.

.

.

{c1}

,

=c

If the number of rows and the number of columns are equal, the matrix is said to be square. (Special types of square matrices are discussed in a later section.) Finally, if all the elements are zero, the matrix is called a null matrix, and is represented by 0 (boldface, as in the previous case). Example 3

1—1

x 4 Matrix

2—1

4 3

—7

2

4

1

—3

2

—8 1

1 x 3 Row Matrix [3

2]

4

3 x 1 Column Matrix f3] or

2

2

4Jor{3,4,2}

Square Matrix 5

[2

7

[0 [o

0

2 x 2 Null Matrix o

* This is the mathematical definition of a vector. In mechanics, a vector is defined as a quantity having both magnitude and direction. We will denote a mechanics vector quantity, such as force or moment, by means of an italic letter topped by an arrow, e.g., F. A knowledge of Vector algebra is assumed in this text. For a review, see Ref. 2 (at end of chapter, preceding Problems).

MATRIX MULTIPLICATION

SEC. 1—3. 1—2.

EQUALITY, ADDITION, AND SUBTRACTION OF MATRICES

Two matrices, a and b, are equal if they are of the same order and if corresponding elements are equal:

a=

when

b

If a is of order m x n, the matrix equation

a=b corresponds to mn equations:

= =

=

1,

2,. .

1,

2,.. .

.

,m ,

Addition and subtraction operations are defined only for matrices of the same

order. The sum of two m x n matrices, a and b, is defined to be the m x n matrix + +

=

+

—

=

— bLJ]

Similarly,

For example, if

[1

2

ii —d

then

[1

[0 b=[3

[1

—1

i 0

1

—1

1

and

—1

3

2

—1

—1

It is obvious from the example that addition is commutative and associative:

1—3.

a+b=b+a

(1—6)

a+(b+c)=(a+b)+c

(1—7)


The product of a scalar k and a matrix a is defined to be the matrix in which each element of a is multiplied by k. For example, if

k=5

and

then

ka=[[—10 10

+35 5


6

CHAP. 1

Scalar multiplication is commutative. That is,

ka = ak = {ka11]

To establish the definition of a matrix multiplied by a column matrix, we consider a system of m linear algebraic equations in n unknowns, x1, .x2

+ a12x2 +

+ +

a21x1 + a22x2 + l2miXi

+

am2x2

C1

=

C2

+

+

This set can be written as alkxk

C1

i=

1, 2, .

.

. ,rn

where k is a dummy index. Using column matrix notation, (1—9) takes the form

i= Now, we write (1—9) as a matrix product:

=

i= 1,2,..,,rn

{c1}

(1—11)

1,2

Since (1—10) and (1—Il) must be equivalent, it follows that the definition equation for a matrix multiplied by a column matrix is ax =

ulkxk}

j = 1, 2,. .

.

,m

This product is defined only when the column order of a is equal to the row

order of x. The result is a column matrix, the row order of which is equal to that of a. In general, if a is of order r x s, and x of order s x 1, the product ax is of orderr x 1. Example

1—2 1

a=

11

8

2 x={3}

-4j 1(1)(2) + (—1)(3) 4

+ (3)(3)

9


SEC. 1—3.

We consider next the product of two matrices. This product is associated with a linear transformation of variables. Suppose that the n original variables x1, x2,. . . ,x,, in (1—9) are expressed as a linear combination of s new variables Y1,Y2, . . . ,ys:

k=

Xk =

1,

2,. .

.

i

1,

,n

(1—13)

1=

Substituting for Xk in (1—10),

=

2,. .

.

,m

Interchanging the order of summation, and letting

i = 1,2 j—

k=i

in

(1—14)

the transformed equations take the form

=

1,2,.. .,

i

Noting (1—12), we can write (1—15) as

py =

C

where p is in x .s and y is S x 1. Now, we also express the transformation of variables, in matrix form,

which defines

x = by

where b is n x s. Substituting for x in (1—11),

aby=c and requiring (1—16) and (1—18) to be equivalent, results in the following definition equation for the product, ab:

= ab

=

[bkJ] = [pt,]

k

1,2,.

. .

,n

This product is defined only when the column order of a is equal to the row order of b. In general, if a is of order r x n, and b of order n x q, the product ab is of order r x q. The element at the ith row and jth column of the product is obtained by multiplying corresponding elements in the ith row of the first matrix and the jth column of the second matrix.


8

CHAP. 1

Example 1—3

(1)(1) + (0)(O)

(IXI) + (O)(1)

(1)(O) +

1)

ab = (—l)(1) + (1)(O) (—1)(l) + (1)(l) (—1)(O) + (O)(1) + (2)(O)

(O)(1) + (2)(l)

(0)(0) + (2)(—1)

[+1

+1

0

—l

ab=J_1

0

—1

+4

[

0

(1)(— 1) + (01(3)

(—1)(—1) + (1)(3) (0)(—1)

+ (2)(3)

+2 —2 +6

If the product ab is defined, a and b are said to be confbrmable in the order stated. One should note that a and b will be conformable in either order only when a is in x n and b is n x in. In the previous example, a and b are conformable but b and a are not since the product ha is not defined. When the relevant products are defined, multiplication of matrices is associative,

a(bc) =

(ab)c

(1—20)

and distributive, a(b + c) = ab + ac (b + c)a = ha + Ca

but, in general, not commutative, ab

ba

(1—22)

Therefore, in multiplying b by a, one should distinguish preinultiplication, ab, from postrnultiplication ha. For example, if a and b are square matrices of order 2, the products are [a11 [a21

a121[bij a22j[b21

b121

[b11

b121[aji b22j[a21

aizl

[b21

When ab

1—4.

=

ha,

—

b22j

a22]

—

[aitbji

+ a12b21

a11b12 + a12b22

[a21b11 + a22b21

a21b12 + a22b22

[bjjaj1 + b12a21

b11a12 + b12a22

[b21a11 + b22a21

b21a12 + b22a22

the matrices are said to commute or to be permutable.

TRANSPOSE OF A MATRIX

is defined as the matrix obtained from a by The transpose of a = interchanging rows and columns. We shall indicate the transpose of a by

TRANSPOSE OF A MATRIX

SEC. 1—4

aT

=

9

{a79]:

a

a11

a12

a1,

021

a22

a2, (1—23)

=

=

amj

a,,,

am2

a21 012

= [a79] =

022

am2

a,,, The element, a79, at the ith row and jth column of aT, where now i varies from 1 to n and j from 1 to m, is given by (1—24)

a79 =

where

is the element at the jth row and ith column of a. For example,

[3

2 T

r3

a =[2

1

7

5

1

4

Since the transpose of a column matrix is a row matrix, an alternate notation for a row matrix is

a,] =

[a1, a2

(1—25)

We consider next the transpose matrix associated with the product of two matrices. Let

p==ab (a) where a is m x n and b is n x s. The product, p, is m x s and the element, Pu,

=

m Ilukbkf

(b)

—1 .1 —

The transpose of p will be of order s x m and the typical element is (c)

p79 =

where now I =

1,

s and j = 1, 2,. .

2

. ,m.

Using (1—24) and (b), we can

write (c) as p79 =

k1

It follows from (d) that

=

1,

j =—

k1 =

(ab)T

=

2,.

S

(d)

bTaT

Equation (1—26) states that the transpose of a product is the product of the


10

CHAP. 1

transposed matrices in reversed order. This rule is also applicable to multiple

products. For example, the transpose of abc is (abc)T = cT(ab)T

Example

cTbTaT

(1—27)

1—4

ab =

(ab)T = [4

13

13

6]

6

Alternatively, aT

= [2

—1]

= (ab)T = bTaT = [2

1—5.

= [4

—1]

13

6]

SPECIAL SQUARE MATRICES

If the numbers of rows and of columns are equal, the matrix is said to be square and of order n, where n is the number of rows. The elements (i = 1, 2,. .. , n) lie on the principal diagonal. If all the elements except the principal-diagonal elements are zero, the matrix is called a diagonal matrix. We will use d for diagonal matrices. If the elements of a diagonal matrix are all unity, the diagonal

matrix is referred to as a unit matrix. A unit matrix is usually indicated by where n is the order of the matrix. Example

1—5

Square

Matrix, Order 2 [1

7

[3

2

Diagonal Matrix, Order 3

[2 [o

0

0

5

0

0

3

Unit Matrix, Order 2

12[

LO

0 I

SPECIAL SQUARE MATRICES

SEC. 1—5.

We

introduce the Kronecker delta notation:

oij=0

(1—28)

i—j

+1

With this notation, the unit matrix can be written as

i,j = 1, 2

=

(1—29)

n

Also, the diagonal matrix, d, takes the form

d=

(1—30)

are the principal elements. If the principal diagonal elements . , are all equal to k, the matrix reduces to where d1, d2,. .

=

=

(1—31)

and is called a scalar matrix.

Let a be of order rn x n. One can easily show that multiplication of a by a conformable unit matrix does not change a: a

(1—32)

Ima = a

A unit matrix is commutative with any square matrix of the same order. Similarly, two diagonal niatrices of order n are commutative and the product is a diagonal matrix of order a. Premultiplication of a by a conformable diagonal matrix d multiplies the ith row of a by and postmultiplication multiplies the jth column by Example

1—6

[2 [o

01[2

01[3

01

[6

0

—i][o 5j[O 5j[O _ij[o —5 [2 01[3 'l_[ 6 2 —

ij [2 7] — [—2 —7

[3 11[2 [2

[6

01

7j[0 _1j[4

A square matrix a for which = property that a = If

—' —7

=

is called symmetrical and has the j) and the principal diagonal elements all equal zero, the matrix is said to be skew-symmetrical. In this case, aT = — a. Any square matrix can be reduced to the sum of a symmetrical matrix and a skew-symmetrical matrix: (i

a=b+c = =

+ —

(1-33)

CHAP. 1


12

The product of two symmetrical matrices is symmetrical only when the matrices

are commutative.* Finally, one can easily show that products of the type (aTa)

(aTba)

(aaT)

where a is an arbitrary matrix and b a symmetrical matrix, result in symmetrical matrices.

A square matrix having zero elements to the left (right) of the principal diagonal is called an upper (lower) triangular matrix. Examples are: Upper Triangular Matrix

352 071 004 Lower Triangular Matrix

300 570 214 Triangular matrices are encountered in many of the computational procedures developed for linear systems. Some important properties of triangular matrices are: The transpose of an upper triangular matrix is a lower triangular matrix and vice versa. The product of two triangular matrices of like structure is a triangular matrix of the same structure.

1.

2.

[a11

0 1[b11 I

[a21

1-6.

0

1

I=

b22j

[aijbij

0

[a21b11 + a22b21

a22b22

OPERATIONS ON PARTITIONED MATRICES

Operations on a matrix of high order can be simplified by considering the matrix to be divided into smaller matrices, called .subina.trices or cells. The partitioning is usually indicated by dashed lines. A matrix can be partitioned in a number of ways. For example,

a

a11

012

0131

a21

a22

023

031

a32

a33J

a11

a12

013

032

a33

= a1 031

=

a11

a12

a13

a31

a32

a33

Note that the partition lines are always straight and extend across the entire matrix. To reduce the amount of writing, the submatrices are represented by *

See Prob. 1—7.

SEC. 1—6.

OPERATtONS ON PARTITIONED MATRICES

single symbol. We will use upper case letters to denote the submatrices whenever possible and omit the partition lines. a

Example 1-1 We represent

[au

a12

a13

a=Ia,i

a22

a23

as

[A11 A121 [A21 A22J

a

or

a = [A11

A12

[A21

A22

where Ia11

A11

= [a21

a121

A12 =

I

A21

=

[a31

Ia13 I

La23

A22 = [a33]

a32]

If two matrices of the same order are identically partitioned, the rules of matrix addition are applicable to the submatrices. Let [A11 [A23

[B11

A121 I

A22J

8121 B22j

[823

(134)

where BLJ and A13 are of the same order. The sum is a

+b =

[A11 + 8fl

+ B121 A22 + B22j A12

LA2I + B21

(1-35)

The rules of matrix multiplication are applicable to partitioned matrices provided that the partitioned matrices are conformable for multiplication. In general, two partitioned matrices are conformable for multiplication if the partitioning of the rows of the second matrix is identical to the partitioning of the columns of the first matrix. This restriction allows us to treat the various submatrices as single elements provided that we preserve the order of multiplication. Let a and b be two partitioned matrices: a b

[A131t

= 1, 2,.. I = 1,2

= [B1d

,

M M

(1—36)

k= 1,2,...,S

We can write the product as

C = ab = [CIk]

,,...,

M 1

C when

—

ik

i

.

i1,

—

the row partitions of b are consistent with the column partitions of a.


14

As an

CHAP. 1

illustration, we consider the product

ab =

au

a12

a13

h1

1221

a22

a23

h2

1233

a32

033

b3

Suppose we partition a with a vertical partition between the second and third columns,

a=

1211

1212

a13

a21

a22

a23

a31

a32

a33

= [A11A12]

For the rules of matrix multiplication to be applicable to the submatrices of a, we must partition b with a horizontal partition between the second and third rows. Taking

the product has the form

= [A,1A12]

= A11B11 + A12B21

The conformability of two partitioned matrices does not depend on the horizontal partitioning of the first matrix or the vertical partitioning of the second matrix. To show this, we consider the product ab

a12

a13

£121

1222

a23

1231

a32

1233

b11

b12 1322

b31

b32

Suppose we partition a with a horizontal partition between the second and third rows: a

a11

1212

C1j3

1221

a22

1223

1231

a32

a33

r A11 =

Since the column order of A11 and A21 is equal to the row order of b, no partitioning of b is required. The product is ab =

[A111

[A11b

LA2ijb = [A21b

As an alternative, we partition b with a vertical partition. b12

b=

b21

b22

= [811B12]

b31

In this case, since the row order of B11 and B12 is the same as the column

OPERATIONS ON PARTITIONED MATRICES

SEC. 1—6.

order of a, no partitioning of a is necessary and the product has the form

ab =

a[B11B12]

= [aBj1

aBi2]

To transpose a partitioned matrix, one first interchanges the off-diagonal submatrices and then transposes each submatrix. If A1,,

A11

A12

A21

A22

Arnt

Am2

AT1 AT

AT1 AT

.

.

.

AT

AT

AT

.

.

.

AT

a= then

A particular type of matrix encountered frequently is the quasi-diagonal matrix. This is a partitioned matrix whose diagonal submatrices are square of various orders, and whose off-diagonal submatrices are null matrices. An example is 0

a= 0

0 a22

0

a32

a33

a11

which can be written in partitioned form as a

= [Ai A2]

where

a23]

A2 = [a22

A1 = [a11]

a33

a32

and 0 denotes a null matrix. The product of two quasi-diagonal matrices of like structure (corresponding diagonal submatrices are of the same order) is a quasi-diagonal matrix of the same structure. A1

0

...

0

B1

0

...

0

A1B1

0

...

0

0

A

0

are of the same order. A and We use the term quasi to distinguish between partitioned and unpartitioned matrices having the same form. For example, we call (1—40)

a lower quasi-triangular matrix.


16

1—7.

Cl-lAP. 1

DEFINITION AND PROPERTIES OF A DETERMINANT

The concept of a determinant was originally developed in connection with the solution of square systems of linear algebraic equations. To illustrate how this concept evolved, we consider the simple case of two equations:

a11x1 + a21X1

a12x2

= + a22x2 =

C2

Solving (a) for x3 and x2, we obtain (a11a22 — a12a21)x1

c2a12

=

(a11a22 — a12a21)x2

The scalar quantity, a1 1a22 —

a21 a2

—c1a21

+ c2a11

defined as the determinant of the second-

order square array (i,j 1, 2). The determinant of an array (or matrix) is usually indicated by enclosing the array (or matrix) with vertical lines: a11

a12

a21

a22

= al =

a31a22

—

a12a21

We use the terms array and matrix interchangeably, since they are synonymous. Also, we refer to the determinant of an eth-order array as an nth-order determinant. It shou'd be noted that determinants are associated only with square arrays, that is, with square matrices. The determinant of a third-order array is defined as +a11a22a33

a11

a12

a13

a21

a22

a23 = —a12a21a33

a31

a32

a33

+ a12a23a31

(1—42)

+a13a21a32 — a13a22a31

This number is the coefficient of x1, x2, and x3, obtained when the third-order system ax c is solved successively for x1, x2. and x3. Comparing (l—41) and (1—42), we see that both expansions involve products which have the following properties: 1.

2.

Each product contains only one clement from any row or column and no element occurs twice in the same product. The products differ only in the column subscripts. The sign of a product depends on the order of the column subscripts, e.g., +a11a22a33 and —a11a23a32,

These properties are associated with the arrangement of the column subscripts

and can be conveniently is

described using

the concept of a permutation, which

discussed below.

A set of distinct integers is considered to be in natural order if each integer is followed only by larger integers. A rearrangement of the natural order is called a permutation of the set. For example, (1, 3, 5) is in natural order and

DEFINITION AND PROPERTIES OF A DETERMINANT

SEC. 1—7.

(1,5,3) is a permutation of(1, 3,5). If an integer is followed by a smaller integer, the pair is said to form an inversion. The number of inversions for a set is defined

as the sum of the inversions for each integer. As an illustration, we consider the set (3, 1, 4, 2). Working from left to right, the integer inversions are: Integer

Inversions

Total

3

(3, 1)(3, 2)

2

None (4,2) None

0

1

4 2

1

0 3

This set has three inversions. A permutation is classified as even (odd) if the total number of inversions for the set is an even (odd) integer. According to this convention, (1, 2, 3) and (3, 1, 2) are even permutations and (1, 3, 2) is an odd permutation. Instead of cbunting the inversions, we can determine the number of integer interchanges required to rearrange the set in its natural order since an even (odd) number of interchanges corresponds to an even (odd) number of inversions. For example, (3,2, 1) has three inversions and requires one interchange. Working with interchanges rather than inversions is practical only when the set is small. Referring back to (1—41) and (1—42), we see that each product is a permutation

of the set of column subscripts and the sign is negative the permutation is odd. The number of products is equal to the number of possible permutations of the column subscripts that can be formed. One can easily show that there are possible permutations for a set of n distinct integers. We let , n) and define , ce,,) be a permutation of the set (1, 2,. . .

.

.

.

as

•

•

+

I

when

—

1

when

.

is

. ,

an even permutation (1—43)

.

..

,

a,,)

is an odd permutation

Using (1—43), the definition equation for an ,ith-order determinant can be written as a11

a12

a1,,

a21

a22

a2,,

=

(1—44)

1

where

the summation is taken over all possible permutations of (1, 2,

Factorial n =

= n(n

—

1)(n

—

2)

.

• (2)(1).

.

.

, n).


18

CHAP. 1

Example 1—8 The permutations for n =

3

are

cxi—1

x23

a33 a32

=2

1

=3

a1=1 z1=2

a3=1

a32 a3=1

e123=+1 e132=—1

e231=+1 e312=+1 e321—-—1

Using (1—44), we obtain a11a22a33 — a11a23a32

a11

a12

a13

a21

a22

a23 = —a12a21a33 + a12a23a31

a32

a33

+a13a21a32 — a13a22a31

This result coincides with (1—42).

The following properties of determinants can be established* from (1—44): 1.

2. 3.

4. 5.

6.

7.

If all elements of any row (or column) are zero, the determinant is zero. The value of the determinant is unchanged if the rows and columns are interchanged; that is, aT! = a!. If two successive rows (or two successive columns) are interchanged, the sign of the determinant is changed. If all elements of one row (or one column) are multiplied by a number k, the determinant is multiplied by k. If corresponding elements of two rows (or two columns) are equal or in a constant ratio, then the determinant is zero. If each element in one row (or one column) is expressed as the sum of two terms, then the determinant is equal to the sum of two determinants, in each of which one of the two terms is deleted in each element of that row (or column).

If to the elements of any row (column) are added k times the corresponding elements of any other row (column), the determinant is unchanged.

We demonstrate these properties for the case of a second-order matrix. Let a

=

[a31

[a21

a22

The determinant is a! = a11a22 — a12a21

Properties 1 and 2 are obvious. It follows from property 2 that laTl * See

Probs. 1—17, 1—18, 1—19.

a!. We

COFACTOR EXPANSION FORMULA

SEC. 1—8.

illustrate the third by interchanging the rows of a: [a21

a' =

a22

a12 = a21a12 — a11a22 = —Ia! a'!

Property 4 is also obvious from (b). To demonstrate the fifth, we take

a21 =

a22 = ka12

ka11

Then a! = Next,

a11(kaj2)

a12(ka1j)

=

0

let a12 = b12 + c12

+ c11

a11

According to property 6,

hi + ci

al

where ibi

b11

b12

a21

a22

= ci

a21

a22

result can be obtained by substituting for O.ii and a12 in (b). Finally, to illustrate property 7, we take This

b12 = a12 + ka22 b21 = a21 b22 = a7, Then, ibi

1-8.

=

(a11 + ka21)a22 — (a12 + ka22)a21

=

a!

COFACTOR EXPANSION FORMULA

in the square matrix, a, If the row and column containing an element, are deleted, the determinant of the remaining square array is called the minor of and is denoted by The cofactor of is related to denoted by the minor of by (1—45) = (— As an illustration, we take

a= The values of

and

328 1

7

4

531

associated with a23 and a22 are

M23. =

= —1

A23 = (— 1)5M23 = +

M22 =

= —37

A22 = (—1)4M22 =

1

—37


20

CHAP. 1

Cofactors occur naturally when (.1 —44) is expanded9 in terms of the elements

of a row or column. This leads to the following expansion formula, called Laplace's expansion by cofactors or simply Laplace's expansion: a1kAIk

=

(1 —46)

akJAkJ

Equation (1—46) states that the determinant is equal to the sum of the products of the elements of any single row or column by their cofactors. Since the determinant is zero if two rows or columns are identical, if follows that

=

k1

0

(147) 0

k

s

I

The above identities are used to establish Cramer's rule in the following section.

Example

1—9

We apply (1—46) to a third-order array and expand with respect to the first row:

(1) a11

a12

a13

(121

a23

a23

a31

a32

a33 2

=

a22

023

a33

023

+

a11(a22a33 — a23a32)

+

a31

a52(—a21a33

+ 0j3(—

a22 1) 035

(133

+ a23a31) +

a53(a21a32

a32

— 022035)

To illustrate (1 —47), we take the cofactors for the first row and the elements of the second row:

=

a21(a22a33

—

a23a32)

+ a22(—a21a33 + a23a31) +

a23(a21a32

—

a22a31)

0

(2) Suppose the array is triangular in form, for example, lower triangular. Expanding with respect to the first row, we have

a21 031

0

0

a22

0

a32

= a11

(122

0

032

033

=

(a51)(a22a33)

=

a11a22a33

033

Generalizing this result, we find that the determinant of a triangular matrix is equal to the product of the diagonal elements. This result is quite useful.

* See

Probs. 1—20, 1—21.

f See Ref. 4, sect. 3 15, for a discussion of the general Laplace expansion method. The expansion in terms of cofactors for a iow Or a COlUmn is a special case of the general method.

CRAMER'S RULE

SEC. 1—9.

The evaluation of a determinant, using the definition equation (1—44) or the cofactor expansion formula (1—46) is quite tedious, particularly when the array

is large. A number of alternate and more efficient numerical procedures for evaluating determinants have been developed. These procedures are described in References 9—13.

Suppose a square matrix, say c, is expressed as the product of two square matrices,

c='ab and we want cJ. It can be shown* that the determinant of the product of two square matrices is equal to the product of the determinants:

ci =

(1—48)

a! hi

Whether we use (1—48) or first multiply a and b and then determine lab! depends

on the form and order of a and b. If they are diagonal or triangular, (1—48) is quite efficient. t

Example

1—10

[1

r2

31

5] a! =

hi

3

4

=

Ic!

=

—20

Alternatively,

[[11

c

and

29J [1

a=[0

—20

r2 0 b__[1

31

5] bi = 8

a! = 5

cj =

Ic! = +40

Determining c first, we obtain

rs

121

= [5 20]

1—9.

and

ci = +40

CRAMER'S RULE

We consider next a set of n equations in n unknowns:

= *

j

= 1, 2, .

.

.

, ii

See Ref. 4, section 3—16.

t See Prob. 1 —25 for an important theoretical application of Eq. 1—48.

(a)


22

CHAP. 1

Multiplying both sides of (a) by Air, where r is an arbitrary integer from 1 to n,

and summing with respect to j,

we

obtain (after interchanging the order of

summation)

=

Xk

k1

j=1

k and equals al when r =

Now, the inner sum vanishes when r

k.

This

follows from (1—47). Then, (b) reduces to lalxr

=

The expansion on the right side of (c) differs from the expansion al

=

ajrAj.

only in that the rth column of a is replaced by c. Equation (c) leads to Cramer's rule, which can be stated as follows:

A set of n linear algebraic equations in n unknowns, ax = c, has a n) is unique solution when 0. The expression for Xr (r = 1, 2 the ratio of two determinants; the denominator is al and the numerator

is the determinant of the matrix obtained from a by replacing the rth column by c.

If jaf = 0, a is said to be singular. Whether a solution exists in this ease will depend on c. All we can conclude from Cramer's rule is that the solution, if it exists, will not be unique. Singular matrices and the question of solvability are discussed in Sec. 1 —13. 1—10.

We

ADJOINT AND INVERSE MATRICES

have shown in the previous section that the solution to a system of n

equations in n unknowns,

i,j

1, 2,..., n

can be expressed as 1

(note

that we have taken r =

I

1, 2,. . ., ii

in Eq. c of Sec. 1—9). Using matrix notation,

(b) takes the form [Au]T{cj}

Equation (e) leads naturally to the definition of adjoint and inverse matrices.

ADJOINT AND INVERSE MATRICES

SEC. 1—10.

23

We define the adjoint and inverse matrices for the square matrix a of order n as

adjoint a = Adj a =

a1

inverse a =

(1—49)

Adj a

(1—50)

Note that the inverse matrix is defined only for a nonsingular square matrix. Example

1—11

We determine the adjoint and inverse matrices for

123 412

a= 2

3

1

The matrix of cofactors is 5

0

—10

—1

—10

+7

—7

+5

—1

Also, al = —25. Then —i

5

Adja

—10

+ 1/25 + 2/5

+7/25

0

+2/5

—7/25

+ 1/25

—1/5

=

—-- Adj

a

—7

—10 +5 +7 —1

0

a=

—

1/5

Using the inverse-matrix notation, we can write the solution of (a) as

x= Substituting for x in (a) and c in (d), we see that a1 has the property that

a1a = aa'

= Equation (1—51) is frequently taken as the definition of the inverse matrix instead of (1—50). Applying (1—48) to (i—Si), we obtain

It follows that (1—Si) is valid only when 0. Multiplication by the inverse matrix is analogous to division in ordinary algebra. If a is symmetrical,, then a is also symmetrical. To show this, we take the transpose of (1—5 1), and use the fact that a =. aT: 1

(a_la)T =


24

Premultiplication by a'

1

CHAP. 1

results in —

a"'

and therefore a1 is also symmetrical. One can also show* that, for any nonsingular square matrix, the inverse and transpose operations can be interchanged:

bT,_t =

(1—52)

We consider next the inverse matrix associated with the product of two square matrices. Let

c=

ab

where a and b are both of order n x n and nonsingular. Premultiplication and then b1 results in by

a'c = b

(b'a'')c = It follows from the definition of the inverse matrix that

(ab)1 =

(1—53)

In general, the inverse of a multiple matrix product is equal to the product of the inverse matrices in reverse order. For example,

= The determination of the inverse matrix using the definition equation (1 —50)

is too laborious when the order is large. A number of inversion procedures based on (1—51) have been developed. These methods are described in Ref. 9—13. 1—11.

ELEMENTARY OPERATIONS ON A MATRIX

The elementary operations on a matrix are:' 1.

2. 3.

The interchange of two rows or of two columns. The multiplication of the elements of a row or a column by a number other than zero. The addition, to the elements of a row or column, of k times the corresponding element of another row or column.

These operations can be effected by premultiplying (for row operation) or postmultiplying (for column operation) the matrix by an appropriate matrix, called an elementary operation matrix. We consider a matrix a of order x n. Suppose that we want to interchange rowsj and k. Then, we premultiply a by an rn x in matrix obtained by modifying the mth-order unit matrix, I,,,, in the following way: 1. 2. *

Interchange Interchange

See Prob. 1—28.

and 5k• and

SEC. 1—11.

ELEMENTARY OPERATIONS ON A MATRIX

25

For example, if a is 3 x 4, premultiplication by

001 010 100 interchanges rows 1 and 3 and postmultiplication by

1000 0001 0010 0100 interchanges columns 2 and 4. This simple example shows that to interchange rows, we first interchange the rows of the conformable unit matrix and premultiply. Similarly, to interchange columns, we interchange columns of the conformable unit matrix and postmultiply. The elementary operation matrices for operations (2) and (3) are also obtained by operating on the corresponding conformable unit matrix. The matrix which multiplies row j by is an mth order diagonal matrix having d1 = 1 for i j and = Similarly, postmultiplication by an nth order diagonal matrix having = 1 for i j and = will multiply thejth column by Suppose that we want to add times row jto row k. Then, we insert in the kth row and jth column of and premultiply. To add z times column jto column k, we put in the jth row and kth column of and postmu-ltiply.

We let e denote an elementary operation matrix. Then, ea represents the result of applying a set of elementary operations to the rows of a. Similarly, ac represents the result of applying a set of elementary operations to thc columns of a. In general, we obtain e by applying the same operations to the conformable unit matrix. Since we start with a unit matrix and since the elementary operations, at most, change the value of the determinant by a nonzero scalar factor,* it follows that e will always be nonsingular.

Example We

1—12

illustrate these operations on a third matrix:

a=

1

1/2

1/5

3

7

2

1

5

—2 We

first: 2.

Add (—3) times the first row to the second row. Add (2) times the first row to the third row.

See

properties of determinants (page 18).

1.

*


26

These

CHAP. 1

operations are carried out by premultiplying by

100 0 201

—3

1

and the result is 1

1/2

1/5

0

11/2

7/5

0

2

27/5

Continuing, we multiply the second row by 2/11: 1

0

0

1

1/2

1/5

0

2,'Il

0

0

11/2

7/5

0

0

1

0

2

27/5

=

1

1/2

1/5

0

1

14/55

0

2

27/5

Next, we add (—2) times the second row to the third row: 0

1

0

1/2

1

1

1/5

1

0

1

1

14/55

0

269/55

Finally, we multiply the third row by 55/269. The complete set of operations is

100 010 0

0

55/269

1

0

0

1

0

0110

100

0

11/21/5

012/110 —310 372

—2

=

0

0

1

2

1

1

1/2

1/5

0

1

14/55 =b

00

0

—2

1

1

5

1

This example illustrates the reduction of a square matrix to a matrix using elementary operations on rows, and is the basis for the Gauss elimination solution scheme (Refs. 9, 11, 13). We write the result as

ea =

b

where e is the product of the four operation matrices listed above:

—6/11

e

+ 1870/2959

0

0

2/11

0

—220/2959

55/269

We obtain e by applying successive operations, starting with a unit matrix. This is more convenient than listing and then multiplying the operation matrices for the various steps. The form of e after each step is listed below: Initial

100 0 0 001 1

Step 2

Step 1

100 0 201

—3

1

1

—6/11 2

0

0

2/11

0

0

1

RANK OF A MATRIX

SEC. 1—12. Step 3

0

0

2/11

0

—4/11

0

1

—6/11

27

Step 4

[1

0

0

2/11

0 55/269

—6/11 }

+34/11

L+187o/2959

—220/2959

Two matrices are said to be equivalent if one can be derived from the other by any finite number of elementary operations. Referring to Example 1 —12, the matrices 1

1/2

1/5

3

7

2

—21

5

and

1

1/2

0

1

00

1/5 14/55 1

are equivalent. Tn general, a and b are equivalent if b = paq

(1—54)

where p and q are nonsinqular. This follows from the fact that the elementary operation matrices are nonsingular.

1-12.

RANK OF A MATRIX

The rank, r, of a matrix is defined as the order of the largest square array,

formed by deleting certain rows and columns, which has a nonvanishing deter-

minant. The concept of rank is quite important since, as we shall see in the next section, the solvability of a set of linear algebraic equations is dependent on the rank of certain matrices associated with the set.

Let a be of order in x n. Suppose the rank of a is r. Then a has r rows which are linearly independent, that is, which contain a nonvanishing determinant of order r, and the remaining rn — r rows are linear combinations of these r rows. Also, it has n — r columns which are linear combinations of r linearly independent columns. To establish this result, we suppose the determinant associated with the first r rows and columns does not vanish. If a is of rank r, one can always rearrange the rows and columns such that this condition is satisfied. We consider the (r + 1)th-order determinant associated with the first r rows and columns, row p, and column q where r < p in, r < q n. a11

a12

a21

a22

an

ar2

aIr

01q

azq (1—55)

arr

0rq apq

We multiply the elements in rowj by (j 1, 2,. . . ,r) and subtract the result from the last row. This operation will not change the magnitude of Ar+t (see Sec. 1—7). In particular, we determine the constants such that the first r elements

CHAP. 1


28

in the last row vanish: a11

021

012

022

a1,

a2r

-

=

a,2

0p2

(1—56)

apr

Equation (1—56) has a unique solution since the coefficient matrix is nonsingular. Then (1 —55) reduces to a11

012

a21

022

a1,

Ar+i

(1—57)

0

a,2

0rr

0

0

where = apq —

(1—58)

Orq]

Applying Laplace's expansion formula to (1—57), we see that A,÷1 vanishes when a is of rank r, A,÷1 vanishes for all combinations of p and q. It

follows that apq

= [aiq. 02p

:

.,

m

(1—59)

Apr

Combining (1—56) and (1—59), we have a11

a21

012

022

0r1

0p1

4p1

r

r+2

m (1—60)

Equation (1—60) states that the last m — r rows of a are linear combinations

of the first r rows. One can also show* that the last n — r columns of a are linear combinations of the first r columns.

Example

1—13

Consider the 3 x 4 matrix

1234

a=21 32 5

? See Prob. 1—39.

7

12

14

RANK OF A MATRIX

SEC. 1—12.

29

We see that a is at least of rank 2 since the determinant associated with the first two rows

and columns is finite, Then, the first two rows are linearly independent. We consider the determinant of the third-order array consisting of columns 1, 2, and q: 1

2

ajq

2

1

a2q

5

7

a3q

Solving the system,

+ 223 = 5 22 = 7 we obtain

If a is of rank 2, A3 must vanish. This requires a3q =

2 101q

+

=

22a2q

3ajq

+

a3q

q = 3,4 Since a33 and (134 satisfy this requirement, we conclude that a is of rank 2. The rows are related by (third row) = + 3 (first row) + (second row)

One can show* that the elementary operations do not change the rank of a matrix. This fact can be used to dctcrmine the rank of a matrix. Suppose b defined by (1—61) is obtained by applying elementary operations to a. We know

that band a have the same rank. It follows that a is of rank p. A matrix having the form of b is called an echelon matrix. When a is large, it is more efficient to reduce it to an echelon matrix rather than try to find the largest nonvanishing determinant: (pxpt b12

...

I

a11

0

(121

b2p

II,

B12 (1—61)

0

0

Example

1—14

[i a=)2 [5 First, we eliminate

Prob. 1—40.

3

4

1

3

2

7

12

12

and a31, using the first row: 1

* See

2

2

3

4

0

—3 —3 —6

0

—3 —3

—8


30

CHAP. 1

Next, we eliminate aW, using the second row: —1

2

4

3

—3 —3 —6

0

0

0

0

—2

At this point, we see that r = 3. To obtain b, we multiply the second row by — 1/3, the third row by — 1/2, and interchange the third and fourth columns:

1243 0010

b= 0

1

2

1

Suppose a is expressed as the product of two rectangular matrices: a

=

(rnxn) (nxs) b c

(1—62)

One can show* that the rank of a cannot be greater than the minimum value of r associated with b and C: ruin [r(b), r(c)]

r(a)

(1—63)

As an illustration, consider the product a

[—1/2 — [—1/2

+1/2 +1/2

01

1]

Since each matrix is of rank 2, the rank of a will be we obtain

[0

0

I

Evaluating the product,

0 1

It follows that a is of rank 1. 1—13.

SOLVABILITY OF LINEAR ALGEBRAiC EQUATIONS

We consider first a system of two equations in three unknowns:

[:: ::

(1-64)

Suppose a is of rank 2 and a11

a21 * See

Prob. 1—44.

a22

0

(1—65)

SEC. 1—13.

SOLVABILITY OF LINEAR ALGEBRAIC EQUATIONS

If a is of rank 2, we can always renumber the rows and columns such that (1—65) is satisfied. We partition a and x, [a11 [a21

a

X1

a12

a131

a22

a23j

[A1

A2]

(1—66)

çx1

1x2 and write (1—64) as A1X1 + A2X2 = c. Next, we transfer the term involving

X2 to the right-hand side: A1X1 =

c — A2X2

(1—67)

0, it follows from Cramer's rule that (1—67) has a unique solution Since jA1j for X1. Finally, we can write the solution as

= Aj'(c

—

A2X2)

(1—68)

Since X2 is arbitrary, the system does not have a unique solution for a given c. The order of X2 is generally called the defect of the system. The defect for this system is 1.

If a is of rank 1, the second row is a scalar multiple, say A, of the first row. Multiplying the second equation in (1—64) by 1/A, we have a12x2 + a13x3 = a11x1 + a12x2 + a13x3 =

a11x1 +

C1

c2/A

(1—69)

2cr, the equations are inconsistent and no solution exists. Then, when a is of rank 1, (1—64) has a solution only if the rows of c are related in the same manner as the rows of a. If this condition is satisfied, the two equations in (1—69) are identical and one can be disregarded. Assuming that 0, the solution is Tf c2

x1 =

(1/a11)(c1

—

a12x2

—

a13x3)

(1—70)

The defect of this system is 2. The procedure followed for the simple case of 2 equations in 3 unknowns is also applicable to the general case of in equations in n unknowns: a11

a12

a1,,

x1

C1

(221

a is of rank in, there exists an mth order array which has a nonvanishing

determinant. We rearrange the columns such that the first in columns are


32

CHAP. 1

linearly independent. Partitioning a and x, a12

a11

ai,m+1 a2m÷1

az,,,

am2

{x1

A2

A1 (mxm)

] (1—72)

•.. x4 =

Xm+i

Xm

= [

a,,,,

am,m÷1

amni

X2

az,,

{ X1 (m

1)

X2 }

((n—rn)x 1)

we write (1—71) as

c — A2X2

A3X1

(1—73)

0, (1—73) can be solved for X1 in terms of c and X2. The defect of the set is n — m, that is, the solution involves n — m arbitrary constants represented by X2. Suppose a is of rank r where r < m. Then, a has r rows which contain an rth-order array having a nonvanishing determinant. The remaining m — r rows are linear combinations of these r rows. For (1—71) to be consistent, that is, have a solution, the relations between the rows of c must be the same as those for a. The defect for this case is n — r. Since IA1I

Example

1—15

As an illustration, consider the third-order system

a11x1 + a12x2 + a13x3 = a21x1 + a22x2 + a73x3 = a31x1 + a32x3 + Suppose

C1

(a)

= C3

that r = 2 and the rows of a are related by (third row) =

(first row) +

(second row)

(b)

For (a) to be consistent, the elements of c must satisfy the requirement, A1C1

+ 22C2

To show this, we multiply the first equation in (a) by these equations the third equation. Using (b), we obtain

0=

C3 —

— I12C2

(c)

the second by —A2, and add to

(d)

Unless the right-hand side vanishes, the equations are contradictory or inconsistent and no solution exists. When e 0, (c) is identically satisfied and we see that (a) has a nontrivial 0) only when r < 3. The general case is handled in the same manner.* solution (x

See

Prob. 1—45.

REFERENCES

33

In general, (1 —71) can be solved when r < ,n if the relations between the rows of a and c are identical. We define the augmented matrix, ci, for (1—71) as a11

a12

C1

=

a2,,

a,flfl

afl,2

[a

cJ

(1—74)

Cm

When the rows of a and c are related in the same way, the rank of tz is equal to

the rank of a. It follows that (1—71) has a solution only if the rank of the augmented matrix is equal to the rank of the coefficient matrix:

=

r(a)

(1—75)

Note that (1—75) is always satisfied when r(a) = m for arbitrary c. We can determine r(cz) and i(a) simultaneously using elementary operations on provided that we do not interchange the elements in the last column. The

reduction can be represented as (1-76)

> r(a) and has a nonvanishing element, is of rank r(a). If no solution exists. r(a), (1—71) contains r independent equations involving n unWhen

where

knowns. The remaining m — r equations are linear combinations of these r equations and can be disregarded. Thus, the problem reduces to first finding r@) and then solving a set of r independent equations in n unknowns. The complete problem can be efficiently handled by using the Gauss elimmation procedure (Refs. 9, 11, 13).

REFERENCES 1.

2. 3.

4. 5.

6. 7, 8.

R. A., W. J. DUNCAN and A. R. cOLLAR: Elementary Matrices, cambridge University Press, London, 1963. THOMAS, G. B., JR.: Calculus and Analytical Geometry. Addison-Wesley Publishing Co., Reading, Mass., 1953. BODBWIG, E.: Matrix calculus, Interscience Publishers, New York, 1956. HOUN, F. E.: Elementary Matrix Algebra, Macmillan Co., New York, 1958. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961. HOUSEHOLDER, A. S.: The Theory of Matrices in Numerical Analysis, Blaisdell, Waltham, Mass., 1964. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969. HIL DEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York, 1952.

9.

Faddeeva, V. N.: Computational Methods of Linear Algebra, Dover Publications, New York, 1959.


34

CHAP. 1

RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 1, New York, 1960. 11. RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2, Wiley, New York, 1967. 10.

Wiley,

12.

BEREZIN, I. S. and N. P. ZHIDKOV: Computing Methods, Vols. I and 2, Addison-Wesley

Publishing Co., Reading, Mass., 1965. 13. 14. 15.

FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems, Prentice-Hall, New York, 1967. VARGA, R. S.: Matrix Iterative Analysis, Prentice-Hall, New York, 1962. CONTE, S. D.: Elementary Numerical Analysis, McGraw-Hill, New York, 1965.

PROBLEMS 1—1.

Carry out the indicated operations:

(a) 4

1

0

2

5

.3

321 +713 510 056 (b) [2

1

6j[ 3

1

(c)

—1

2

[1 21 [2 3[3 4j+2[[—i0 ii 3j4[l

[i

(d)

—2152

[—3

4J 55

(e)

[—i

l][4

1

2

—3j [2

3

[

1—2.

[4

11

[—i

I

[2

3j

2

—3

[

Expand the following products: [a1, a2,

.

.

.

,

{b1, b2,

a2, .

.

.

,

[b1, b2

.

(b)

{ai, (c)

[c1

01

[a11

a12

[o

c2j

[a21

a22

.

.

,

b,j

3

PROBLEMS

35

(d)

1—3.

[Cii

a121

0

[021

022j [0

C2

Show that the product of

Sl= a1+ a2 + 03 = S2

+ b2 + b3

=

bk

=

can be written as 3

3

S1S2 I

1

Generalize this result for the sum of n elements. 1—4. Suppose the elements of a and b are functions of y. Let cia —

[dv j

dy

db

—

[dblk]

dy

—

[dy j

Using (1—19), show that if c

= ab

then dc dy 1—5.

db dy

da dy

=a—+—b

Consider the triple product, abc. When is this product defined? Let p

= abc

What is the order of p? Determine Determine an expression for aT. case where c 1—6. Evaluate the following products:

for the

(a) F

[—2

41 [1

21 [5

1

[4

1

ij [2 sj

(b)

1—7.

where a is a square matrix. (a + b)(a + b) Show that the product of two symmetrical matrices is symmetrical

only when they are commutative. 1—8. Show that the following products are symmetrical: (a)

aTa (b)

aTba

where b is symmetrical

bTaTcab

where c is symmetrical

(c)

INTRODUCTION TO MATRtX ALGEBRA

36

CHAP. 1

1-9. Evaluate the following matrix product, using the indicated submatrices: 1

3

3

1

51

4

1—10. Let c = ab. Show that the horizontal partitions of c correspond to those of a and the vertical partitions of c correspond to those of b. Hint: See

Eq. (1—37).

1—11. A matrix is said to be symmetrically partitioned if the locations of the row and column partitions coincide. For example, a11

012

0131

a21

a22 a32

a23 a33

a31

is symmetrically partitioned and

is

a11

a12

a13

a31

a32

033

unsymmetrically partitioned. 1 symmetrical partitions.

Suppose we partition a square matrix with

N—

i,j =

a (a)

ab

If a and b are symmetrically partitioned, show that CJk, AJk, the same order. Illustrate for the case of one partition, e.g., [A11

[A2,

—

1—13.

have

Consider the product of two square nth order matrices.

a (b)

.

Ars =

c= (a)

2,. ., N

Deduce that the diagonal submatrices are square and Ars,

the same order. (b) If a = aT, deduce that 1—12.

1,

are of

A12 A22

Suppose we symmetrically partition c. What restrictions are placed on the partitions of a and b? Does it follow that we must also partition a and b symmetrically? Hint: See Prob. 1—10. Consider the triple product, C=

a symmetric rth-order square matrix and a is of order r x ii. Suppose we symmetrically partition c. The order of the partitioned matrices are indicated in parentheses. (pxp)

(pxq)

(nxn) —

[C11

C12

—

[c21 (q> is the complex conjugate of

= I Finally, the characteristic values and characteristic vectors are 21,2 =

Qt,2 = In general, the characteristic values are complex conjugate quantities when the elements of a are real. Also, the corresponding characteristic vectors are complex conjugates.

2—3.

SIMILARITY AND ORTHOGONAL TRANSFORMATIONS

The characteristic vectors for the relations:

aQ1 = aQ2 =

system satisfy the following (a)

22Q2

We can write (a) as

2j

Q2] = EQ1

(b)

Now, we let

q=[Qj

Q2] (2—17)

= We call q the normalized Column j of q contains the normalized solution for modal matrix* for a. With this notation, (b) takes the form

aq =

(2—18)

* This terminology has developed from dynamics, where the characteristic vectors define the normal modes of vibration for a discrete system.

AND ORTHOGONAL TRANSFORMATIONS

SEC. 2—3.

53

We have shown that the characteristic vectors are always linearly indepen-

dent when a is symmetrical. They are also independent when a is unsymmetrical, provided that 0 except for the case where a is 11.2. Then, unsymmetrical and the characteristic values are equal. If 0, q1 exists and we can express (2—18) as

q'aq =

(2—19)

The matrix operation, p is arbitrary, is called a similarity transformation. Equation (2—19) states that the similarity transformation, q1( )q, reduces a to a diagonal matrix whose elements are the characteristic values of a.

If a is symmetrical, the normalized characteristic vectors are orthogonal, that is, — —

— —

—

—

Also, by definition,

Using these properties, we see that q

[Qfl

q = [QTJ [Qi

Q2]

[1

0

[o

1

and it follows that qT

(2—20)

p'

A square matrix, say p, having the property that is called an = orthogonal matrix and the transformation, pT( )p, is called an orthogonal transformation. Note that an orthogonal transformation is also a similarity transformation. Then, the modal matrix for a symmetrical matrix is orthogonal and we can write qTaq = (2—21) Example

2—3

[2

2 5

+6

Q2{2,-1} —[0

q=[Qi

0

+1]

PROBLEMS

54

We verify that qT

CHAP. 2

= q' and qTaq = 21 [1

1 [1

1 [5

21

—lj[2 [2 21[1 [2

1

aq — —

q

T

aq

[1

I

21

[1

01

1 [6

2

[12

—1

—ij —

0

sj[o i

21[6 01

—ij[o

i[i

21[6

[6

21

_-1][12 —ii = [o

= 5L2

01

ij =

(2)

[1

8 3

= +5

—

01

Lo Since

a is not symmetrical, qT q

-1

(

+1}

Qi =

—ij Actually,

q

[,15/6 — [\/17/6 —

1—

j

[— One can easily verify that

[5

q

[1

01

—2

01

Lo —iJ

[1

1

q involves complex elements. Since the characteristic vectors are complex conjugates, they are linearly independent and q -' exists. We find q — 1, using the definition equation for the inverse (Equation (1—50)):

SEC. 2—4

THE nth-ORDER SYMMETRICAL CHARACTERISTICS

One can easily verify that q

2—4.

[+i

-1

55

01

0

THE nth-ORDER SYMMETRICAL CHARACTERISTIC-VALUE PROBLEM

The nth order symmetrical characteristic-value problem involves determining the characteristic values and corresponding nontrivial solutions for a11x1 + a12x2 + a12x1 + +

+ +

+

+

+

= Ax1 — AX2

(2—22)

We can write (2—22) as

ax = AX (a —

(2—23)

0

In what follows, we suppose a is real. For (2—23) to have a nontrivial solution, the coefficient matrix must be singular. a — AI,4

(2—24)

0

The expansion of the determinant is

=

+

+

0

where

= a11 + a22 ±

+

(2—25) -

is the sum of all the jth order minors that can be formed on the diagonal.* Letting 22,. . , denote the roots, and expressing the characteristic equation in factored form, we see that and

.

=

2522

+ 2523 + ... +

(2-26)

We summarize below the theoretical results for the real symmetrical case. The proofs are too detailed to be included here (see References 1 and 9): 1.

2.

The characteristic values are all real. 22,. . , The normalized characteristic vectors Q1, Q2,. . , Q,,, are orthogonal: .

.

QTQJ =

i,j =

1,2

* Minors having a diagonal pivot (e.g.. delete the kth row and column). They are generally called principal minors.

CHARACTERiSTIC-VALUE PROBLEMS

56 3.

CHAP. 2

a is diagonalized by the orthogonal transformation involving the normalized modal matrix. qTaq

where

= Example 2—4

5—2

0

a= —2

3

—.1

0

—1

1

Since a is symmetrical, its characteristic values are all real. We first determine /3k, $2, f33, using (2—25):

5+ $2 /33

3

+9

+1

+11 + 5 + 2 +18 = 5(2) — (—2)(—2) = +6

The characteristic equation is

182—6=0 and the approximate roots are 22

+0.42 +2.30 +6.28

To determine the characteristic solutions, we expand ax = 2x,

=

(5 — 2)x1

2x2

—x3 = —(3 — 2)x2

(l—,t)x3=

x2

Solving the first and third equations for x1 and x3 in terms of x2, the general solution is

j=l,2,3 Finally, the modal matrix (to 2-place accuracy) is

+0.22

(2)

—0.84

q = [Q1Q2Q3] = +0.50

+0.51 +0.68

+0.85

—0.52

—0.10

120

a= 2

1

0

0 .0

3

+0.54

QUADRATIC FORMS

SEC. 2—5.

The expansion of Ia —

213J

=

57

0 is

and the roots are

23=—i

22—3 Writing out ax = 2x, we have (1—2)x1

=0

+2x2 +(1 — 1)x2

2x1

(3 — 2)x3

When 2 =

3,

= =

0

(a)

0

(a) reduces to

+2x2 =

—2x3

0

—2x2 = 0

(b)

(0)x3=0 We see from (b) that (a —

213)

is of rank 1 when 2 =

xI=c1

x2=cl

3.

The general solution of (b) is

x3=cz

By specializing the constants, we can obtain two linearly independent solutions for the repeated root. Finally, the characteristic vectors for 22 = 3 are

Q2 = (0,01) When 2 = 23 = —1, (a) reduces to 2x3 + 2x2

=

0

2x1 + 2x2

0

4x3

=0

The general solution and characteristic vector for

are

=

and

-

0

o}

This example illustrates the case of a symmetrical matrix having two equal characteristic

values. The characteristic vectors corresponding to the repeated roots are linearly independent. This follows from the fact that a — 213 is of rank I for the repeated roots.

2-5. QUADRATIC FORMS The homogeneous second-degree function

F a quadratic form in

+ 2a12x1x2 +

x2. Using matrix notation, we can express

F as [a11

F=[xix2]t[a12

ajal

T

CHARACTERISTIC-VALUE PROBLEMS

58

CHAP. 2

In general, the function

=

F= where afk =

for]

(2—27)

j=1

k, is said to be a quadratic form in xj, x2, ..

If F =

x

.

, x,,.

x

positive definite a 0, we say that F is positive semidefinite. We define negative definite and negative semidefinite quadratic forms in a similar manner. A quadratic form is negative definite if F 0 for all x and F = 0 only when x = 0. The question as to whether a quadratic form a

we is zero for some x

for

is positive definite is quite important. For example, we will show that an equilibrium position for a discrete system is stable when a certain quadratic form is positive definite.

Consider the quadratic form

F= b1

0

b2

LXIX2 13

x2

(2—28)

1

When F involves only squares of the variables, it is said to be in canonical form.

According to the definition introduced above, F is positive definite when

b1 >0

>0

b2

It is positive semidefinite when b1

0

and at least one of the elements is zero. Now, to establish whether is positive definite, we first reduce a to a diagonal matrix by applying the transformation, q '()q, where q is the orthogonal normalized modal matrix for a. We write xTax = (xTq)(q_taq)(qlx)

= Then, letting

y=

qTx

x

qy

(2—29)

(a) reduces to a canonical form in y:

F = xTax =

(2—30)

It follows that F is positive definite with respect to y when all the characteristic values of a are positive. But y is uniquely related to x and y = 0 only when x = 0. Therefore, F is also positive definite with respect to x. The problem of establishing whether xTax is positive definite consists in determining whether all the characteristic values of a are positive.

QUADRATIC FORMS

SEC. 2—5.

We consider first the second-order symmetric matrix

[a11

cz12

Laiz

a22

Using (2—26), the characteristic values are related by

+

=aii -F a22

,t122 = 132

=

a11a22

=

—

aJ

We see from (a) that the conditions

132>0 are

equivalent to

,t2>0 Suppose we specify that

>

au

at —. a11a22

0

>

—

0

Since a1 > 0, it follows from the second requirement in (d) that a22 > Therefore, (d) is equivalent to (b). We let

= aj1j = A2 =

a11a12

a12a12

0.

a11

=

(2—31)

at

Then, a is positive definite when

132>0 (2—3 2)

A2>0

A1>O The quantities

and are called the invariants and discriminants of a. The above criteria also apply for the case. That is, one can show that a is positive definite when all its invariants are greater than zero.

>

131 > 0

where

is

0

•..

/3,,

>

0

(2—33)

the sum of all the jth-order principal minors. Equivalent conditions

can be expressed in terms of the discriminants. Let represent the determinant of the array consisting of the first j rows and columns.

A=

a11

a12

£112

a22

aU (2—34)

a2J

The conditions, A1 >

0

A2 >

0

...

are sufficient for a to be positive definite.* *

See

Ref. 1 for a detailed proof. Also see Prob. 2—15.

A,,

>

0

(2—35)


60

CHAP. 2

Example 2—5

111

122 123

The discriminants are A1 = +1 = 2 — 1 = +1 = 1(6—4)— 1(3—2) + 1(2—2) = +1 Since all the discriminants are positive, this matrix is positive definite. The corresponding invariants are = 1 + 2 + 3 = +6

$2(2—1)+(3—1)+(64)

Since A2 is negative

+5

= A3 = +1

/13

=

—3),

1

1

1

—2

2

1

2

3

1

this matrix is nor positive definite.

Suppose b is obtained from a by an orthogonal transformation: b = pTap

p1ap

(2—36)

If a is symmetrical, b is also symmetrical:

bT =

pTaTp

pTap

(2—37)

Now, b and a have the same characteristic values.* This follows from —

=

—

a —

(2—38)

Then, if a is positive definite, b is also positive definite. In general, the positive definite character of a matrix is preserved under an orthogonal transformation, REFERENCES 1.

HILDSBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York, 1952.

2. 3.

BODEWIG, E.: Matrix Calculus, Interscience Publishers, New York, 1956. SMiRNOV, V. I.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1964.

4. 5.

TURNBULL, H. W., and A. C. AITKEN: An Introduction to the Theory of Canonical Matrices, Dover Publications, New York. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961.

* See

Prob. 2—5.

PROBLEMS 6.

7.

8. 9.

10.

11. 12.

13.

CRANDALL, S. H.: Engineering Analysis, McGraw-Hill, New York, 1956. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969. FRAZER, R. A., W. J. DUNCAN and A. R. COLLAR: Elementary Matrices, Cambridge University Press, London, 1963. WILKINSON, 3. H.: The Algebraic Eigenvalue Problem, Oxford University Press, London, 1965. FADDEVA, V. N.: Computational Methods of Linear Algebra, Dover Publications, New York, 1953. RALSTON, A., and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2, Wiley, New York, 1967. FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems, Prentice-Hall, New York, 1967. with Band Synimetric PETERS, G., and J. H. WILKINSON: "Eigenvalues of AX = A and B," Comput. J., 12, 398—404, 1969.

PROBLEMS 2—1.

Consider the system

Ay = where A and B arc symmetrical nth-order matrices and is a scalar. Suppose B can be expressed as (see Prob. 1—25)

B = brb where b is nonsingular. Reduce (a) to the form

ax = where x 2—2.

let c1,

c2

by. Determine the expression for a in terms of A and b. Let x1, x2 be two nth-order column matrices or column vectors and be arbitrary scalars. If c1x_I

+ c2x2

0

only when c1 =

= 0, x1 and x2 are said to be linearly independent. It follows that x1 and x2 are linearly dependent when one is a scalar multiple of the other. and Q2 arc linearly independent when Using (2—10) and (2—13), show that 2—3.

Determine the characteristic values and the modal matrix for

(a)

[3 [2 [2 0

3

(b)

to s

0

[3

2

2 7

0

2—4. Following the procedure outlined in Prob. 2—I, determine the characteristic values and modal matrix for

+ 12Y2

l2Y1 +

=


62 .2—5.

Suppose

CHAP. 2

that b is derived from a by a similarity transformation.

b = p'ap Then, lb —

=

t1,4

a

—

and it follows that b and a have the same characteristic equation. (a)

Deduce that 2(b) — '(a) k

k

a(b) — p(a) Yk Pk

=

1,

2,.. ., n

Demonstrate for [1

The fact that (b)

.

, /3,,

.

formation is quite useful. Show that

P[2[1

—21

1

are invariant under a similarity trans-

_. Q U') k

P

—1

(a) k

2—6. When a is symmetrical, we can write qTaq = Use this result to find the inverse of

in terms of q and

Express

[3 a=[2 2—7.

2

Positive integral powers of a square matrix, say a, arc defined as

a2 = a3 =

aa aa2

ar = If al # 0,

exists, and it follows from the definition that

= (a) (b)

2—8.

a''

Show that ar is symmetrical when a is symmetrical. is a characteristic Let 2L be a characteristic value of a. Show that value of ar and is the corresponding characteristic vector.

Hint: Start with = 2,Q, and premultiply by a. A linear combination of nonnegative integral powers of a is called a

polynomial function of a and written as P(a). For example, the third order polynomial has the form

P(a) =

c0a3

+ c1a2 + c2a + c3L

Note that P(a) is symmetrical when a is symmetrical.

PROBLEMS

Let F(1)

0 be the characteristic equation for a. When the characteristic

values of a are distinct, one can show that (see Ref: 1)

F(a) =

0

where 0 is an nth-order null matrix. That is, a satisfies its own characteristic equation. This result is known as the Cayley-Hamilton Theorem. (a) Verify this theorem for [2 1 2

(b)

Note: F(a) = Show that

a2

f31a

= '(a2 (c) 2—9.

+

/3212.

— /31a + /3213)

forn =

3

Establish a general expression for a —' using (2—25). Determine whether the following quadratic forms are positive definite.

(a) (b) 2—10.

F= F = 34 +

+ 4x1x2 +

+

4x1x2 + 6x1x3 — 8x2x3

—

Show that a necessary but not sufficient condition for a to be positive

definite is

a11>O,a22>O (Hint:

= Oforj 1,] = 1,2,.. .,n) ax = 0 has a nontrivial solution, say x1. What is. the

= 0, 2—11. If value of xfax1? Note that 2 0 is a characteristic value of a when a is singular. 2—12. Let C be a square matrix. Show that CTC is positive definite when 0 and positive sernidefinite when CI = 0. IC! (hint: Start with F = xT(CTC)x and let y = Cx. By definition, F can equal zero only when x 0 in order for the form to be positive definite.) 2—13. Consider the product CTaC, where a is positive definite and C is 0 and positive semisquare. Show that C1aC is positive definite when CI definite when CI = 0. Generalize this result for the multiple product, . .

.

.

.

2—14. Let a be an mth-order positive defInite matrix and let C be of order m x n. Consider the product, b = CTaC

Show that b is positive definite only when the rank of C is equal to n. What can we say about b when r(C) < n? 2—15. Consider the quadratic form a11

a12

x1

a12

a22

x2

:

:

a27,

x,,


64

CHAP. 2

We partition a symmetrically, (pXl)

(pxp)

A121 f Xt

NzTvTl [A11

AT (qxp)

where

q=

n — p.

MV ("2

The expansion of F = XTaX has the form

F= Now, we take X2

A

(qxq)

XfA11X1 -i- 2XTA12A2 +

0 and denote the result by XTAIIX1

> 0 for arbitrary X1, A11 must be positive definite. Since 1A111 is equal For to the product of the characteristic values of A11, it follows that Ajj must be positive. (a)

By taking p =

1,

2,.. .

, n,

deduce that

p=1,2,...,n are necessary conditions for a to be positive definite. Note that it remains to show that they are also sufficient conditions. (b)

Discuss the case where

= 0.

2-46. Refer to Prob. t—25. Consider a to be symmetrical. (a) Deduce that one can always express a as the product of nonsingular lower and upper triangular matrices when a is positive definite. (b)

Suppose we take Show that a is positive definite when

j=1,2,...,n and positive semi-definite when

j=1,2,...,n and at least one of the diagonal elements of g is zero. Suppose we take g = hT. Then,

= and A

Show

— Ati.11 —..h2i.2

,..i.2 Upp

that the diagonal elements of b will always be real when a is

positive definite. 2—17. If a quasi-diagonal matrix, say a, is symmetrically partitioned, the submatrix A11 is also a quasi-diagonal matrix. Establish that

a=

i,j =

is positive definite only when A, (i = 1, 2,.

2,. . ., N

1, .

.

,

N) are positive definite.

PROBLEMS

65

Hint: Use the result of Prob, 1—23. Verify for

1100 2300 0021 0052

2—18. Suppose we express a as the product of two quasi-triangular matrices, for example, (pxp)

(nxn)

a

[G11

1)

= [G2,

(qxp)

where p +

q

(pxq)

1 [B11

G22j[O

B12

B22

= n. We take B11

1P

8221q

Show that the diagonal submatrices of g are nonsingular for arbitrary p when a is positive definite.

3

Relative Extrema for a Function 3—1.

RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE

Letf(x) be a function of x which is defined for the interval x1 x x2. If f(x) — f(a) 0 for all values of x in the total interval x1 x x2, except x a, we say the function has an absolute minimum at x a. If f(x) — f(a)> 0 for all values of x except x = a in the subinterval, x containing x = a, we say that f(a) is a relative minimum, that is, it is a minimum with respect to all other values of f(x) for the particular subinterval. Absolute and relative maxima are defined in a similar manner. The relative maximum and minimum values of a function are called relative extrema. One should note thatf(x) may have a number of relative extreme values in the total interval x1 x x2. As an illustration, consider the function shown in Fig. 3—1. The relative extrema are [(a), f(h), f(c), f(d). Using the notation introduced above, we say that f(b) is a relative minimum for the interval x fib. The absolute maximum and minimum values of f occur at x = a and x = d, respectively. f(x)

x1

x a

b

c

Fig. 3—i. Stationary points at points A, 8, C, and 0. 66

X2

SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE

67

In general, values of x at which the slope changes sign correspond to relative extrema. To find the relative extrema for a continuous function, we first determine the points at which the first derivative vanishes. These points are called stationary points. We then test each stationary point to see if the slope changes sign. If the second derivative is positive (negative) the stationary point is a relative minimum (maximum). If the second derivative also vanishes, we must consider higher derivatives at the stationary point in order to determine whether

the slope actually changes sign. In this case, the third derivative must also vanish for the stationary point to be a relative extremum. Example

3—1

Setting the first derivative equal to zero,

x2 + 4x +

dx

I

=

0

2(x

+

2)

and solving for x, we obtain

x1,2 =

—2

±

The second derivative is d2f

=

2x

+4=

Thcn,x = x1 = —2 + corresponds to a relative maximum.

=

J(x) =

(x

—

a)3

x2

= —2—

+c

The first two derivatives are = 3(x

—

a)

Since both derivatives vanish at x =

a,

we must consider the third derivative: d3f dx3

6

The stationary point, x = a, is neither a relative minimum nor a relative maximum since the third derivative is finite. We could have also established this result by considering the expression for the slope. We see from (a) that the slope is positive on both sides of x = a. The general shape of this function is shown in Fig. E3—l.

RELATIVE EXTREMA FOR A FUNCTION

66

CHAP. 3 Fig. E3—1

f(x)

x

I

a

The sufficient condition for a stationary value to be a relative extremum (relative minimum (maximum) when d2f/dx2 > 0 (< 0)) follows from a consideration of the geometry of the f(x) vs. x curve in the vicinity of the stationary point. We can also establish the criteria for a relative extremum from the Taylor series expansion of f(x). Since this approach can he readily extended to functions of more than one independent variable we will describe it in detail. Suppose we know the value of f(x) at x = a and we want f(a + Ax) where Ax is some increment in x. If the first n + 1 derivatives off(x) are continuous in the interval, a x a + Ax, we can express f(a + Ax) as

f(a + Ax)

f(a) Ax +

=

where

remainder

(Ax)2

+

(Ax)" +

denotes the jth derivative of f(x) evaluated at x

(3.1) a, and the

is given by 1

(3-2)

where is an unknown number between a and a + Ax. Equation (3—1) is called

the Taylor series expansion* of f(x) about x =

a.

If f(x) is an eth-degree

polynomial, the (n + 1)th derivative vanishes for all x and the expansion will yield the exact value off(a + Ax) when n terms are retained. In all other cases, there will be some error, represented by due to truncating the series at n terms. Since depends on we can only establish bounds on The following example illustrates this point. See Ref. 1, Article 16—8.

SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE

69

Example 3—2 We expand sin x in a Taylor series about x = 0 taking n = and noting that a = 0, we obtain sin Ax = Ax + R2

2.

Using (3—1) and (3—2),

The bounds on R2j are cos Ax
0 for all points in the neighborhood of Ax e, where a, that is, for all finite values of Ax in some interval, — and e are arbitrary small positive numbers. Considering Ax to be small, the first-order increment dominates and we can write

x

f(a + Ax) — f(a) =

Ax + (second- and higher-order terms) (3—3)

For f(a + Ax) — f(a) to be positive for both positive and negative values of Ax, the first order increment must vanish, that is, df(a)/dx must vanish. Note that this is a necessary but not sufficient condition for a relative minimum, if the first-order increment vanishes, the second-order increment will dominate: f(a + Ax) — f(a) =

(Ax)2

+ (third- and higher-order terms)

(3—4)

It follows from (3—4) that the second-order increment must be positive for > 0 to be satisfied. This requires d2f(a)/dx2 > 0. Finally, f(a + Ax) — the necessary and sufficient conditions for a relative minimum at x = a are df(a) dx

—

0

d2f(a)

dx2

lithe first two derivatives vanish at x = the dominant term in the expansion.

f(a + Ax) + f(a) =

a,

>

3

5

the third-order increment is now

(Ax)3 + (fourth- and higher-order terms)

(3—6)

Since the third-order increment depends on the sign of Ax, it must vanish for


CHAP. 3

f(a) to be a relative extremum. The sufficient conditions for this case are as follows:

Relative Minimum d4f

d3f

dX4> (3—7)

Relative Maximum d4f

d3f

The notation used in the Taylor series expansion off(x) becomes somewhat cumbersome for more than one variable. In what follows, we introduce new notation which can be readily extended to the case of 11 variables. First, we to be the total increment in f(x) due to the increment, Z\x. define

41 = f(x +

Ax)

—

f(x)

(3—8)

This increment depends on Ax as well as x. Next, we define the differential operator, d, (3—9)

The result of operating onf(x) with d is called the first by df:

df=-1Ax =

df(x,Ax)

and is denoted (3—10)

The first differential off(x) is a function of two independent variables, namely, x and Ax. Iff(x) = x, then df/dx = 1 and

c/f = dx = Ax

(3—11)

One can use dx and Ax interchangeably; however, we will use Ax rather than dx.

Higher differentials of f(x) are defined by iteration. For example, the second differential is given by

d2f = d(df)

(3—12)

=

Since Ax is independent of x,

(Ax) =

0

and d2f reduces to

d2f =

(Ax)2

= d2f(x, Ax)

In forming the higher differentials, we take d(Ax) =

0.

(3—13)

SEC. 3—2.

FUNCTION OF n INDEPENDENT VARIABLES

Using differential notation, the Taylor series expansion (3—1) about x can be written as

(3-14)

The first differential represents the first-order increment in f(x) due to the increment, Ax. Similarly, the second differential is a measure of the secondorder increment, and so on. Then,f(x) is a stationary value when df = 0 for all permissible values of Ax. Also, the stationary point is a relative minimum (maximum) when d2f> 0 ( 0 (-—&Xk

k1

OXk

It follows that

df=

j=l

Repeating leads to

d2f

=

Consider the double sum,

f

>

k=t

The first differential (see Prob. 3—9) has the form

df =

(duJwJkvk +

dv1)

dwfkvk +

Introducing matrix notation,

u=

w= =

and letting

du

[w31]

v=

{v1}


74

CHAP. 3

and so forth, we can write df as

df = =

d(urwv) duTwv

+ nTdWv + uTwdv

One operates on matrix products as if they were scalars, but the order must be preserved. As an illustration, consider

f= where

— x7c

a, c are constant and a is symmetrical. Noting that da =

dc

=

0

and dx

Ax,

the first two differentials are AxT(ax — c)

df

d2f =

AxTa Ax

Comparing (g) and (3—24), we see that fO)

ax —

=

c

a

The Euler equations are obtained by setting

equal to 0:

ax

c

The solution of (i) corresponds to a stationary value of (f). If a is positive definite, the stationary point is a relative minimum. One can visualize the problem of solving the system ax = c, where a is symmetrical from the point of view of finding the stationary — XTC. value of a polynomial having the form f =

Suppose

f

u/v. Using the fact that

ldu

3j7u\ \,vJ

UX1

:X3

1(6u

uôv

V

vax1

—

we can write

df =

=

—f dv)

We apply (b) to 1.

=

xx

where a is symmetrical, and obtain (see Prob. 3—5) 2 AXT

= —f---- (ax — 2x) (d)

(122 =

xx

Ax

A

AxT Ax — 2 dA AxTx)

Setting dA = 0 leads to the Euler equations for (c),

ax—Ax=0 which we recognize as the symmetrical

(e)

problem.

LAGRANGE MULTIPLIERS

SEC. 3—3.

75

The quotient xTax/xTx, where x is arbitrary and a is symmetrical, is called Rayleigh's

quotient. We have shown that the characteristic values of a are stationary values of Rayleigh's quotient. This property can he used to improve an initial estimate for a characteristic value. For a more detailed discussion, see Ref. 6 and Prob. 3—11.

3-3.


Up to this point, we have considered only the case where the function is expressed in terms of independent variables. In what follows, we discuss how one can modify the procedure to handle the case where some of the variables are not independent. This modification is conveniently effected using Lagrange multipliers. Suppose f is expressed in terms of n variables, say x1, x2 some of which are not independent. The general stationary requirement is

df= >

(3—28)

j=1

for all arbitrary differentials of the independent variables. We use of

instead

to emphasize that some of the variables are dependent. In order to

establish the Euler equations, we must express df in terms of the differentials of the independent variables. Now, we suppose there are r relations between the variables, of the form =

g5(x1, x2

It =

0

1,

2, .

.

,r

(3—29)

One can consider these relations as constraint conditions on the variables. Actually, there are only n — r independent variables. We obtain r relations between the n differentials by operating on (3—29). Since = 0, it follows that 0. Then,

"a

k=1,2,...,r

j=t

(3—30)

Using (3—30), we can express r differentials in terms of the remaining n — r differentials. Finally, we reduce (3—28) to a sum involving the n — r indepen-

dent differentials. Equating the coefficients to zero leads to a system of n — r

equations which, together with the r constraint equations, are sufficient to determine the stationary points. Example We

3—4

illustrate the procedure for n =

2

and r =

1:

f= g(x1, x2) = 0

The first variation is ox1

Ox2


76

CHAP. 3

Operating on g(x1, x2) we have

ax1

Now, we suppose ag/ax2 0. Solving (b) for dx2 (we replace dx1 by x1 is the independent variable.)

/ag\

dx2= —t—i-----itSx1 8x2/

and substituting in (a), we obtain

df = [PL ax1

—

8x1 ox2

ox2

Finally, the equations defining the stationary points arc

\ox1fOx2/ox2 g(x1, x2) = 0

ax1

To determine whether a stationary point actually corresponds to a relative extremum, we must investigate the behavior of the second differential. The general form of d2f for a function of two variables (which are not necessarily independcnt) is d2f = 2

=

2

2

dx1 +

k1

Of —i—

a quadratic form in the independent differential, d2x1

using (c), and noting

= 0, d2f

+ ax1ax.2

ox2

+

\0x1

+

ax2j Ox2

where

u=

Og Jag

The character of the stationary point is determined from the sign of the bracketed term.

An automatic procedure for handling constraint conditions involves the use of Lagrange multipliers. We first describe this procedure for the case of two variables and then generalize it for n variables and r restraints. The problem consists in determining the stationary values of f(x1, x2) subject to the constraint condition, g(x1, x2) = 0. We introduce the function H, defined by H(x1, x2, A.) = f(x1, x2) + Ag(x1, x2)

(3—31)

where A. is an unknown parameter, referred to as a Lagrange multiplier. We


SEC. 3—3.

77

consider x1, x2 and ,% to be independent variables, and require H to be sta-

tionary. The Euler equations for H are OH

—

Og

+

Ox, Ox2

A

(3-32)

Ox2

Ox2

OH

—

Ox1 —

g(x1, x2) =

0

0. Then, solving the second equation in (3—32) for A, and substituting in the first equation, we obtain

We suppose Og/0x2

A=

(3-33)

0x2/ Ox2

and

=

0

g(x1, x2) =

0

—

Ox1

(3-34)

Equations (3—34) and (e) of the previous example are identical. We see that the Euler equations for II are the stationary conditions for f including the effect of constraints.

Example

3—5

f= g

We form H

f+

+ 2x1 + 7x2

+ =

0

2x1

+

=

— x2

+

+

2g,

H=

7x2

+

A(x1 — x2)

The stationary requirement for H treating x1, x2, and 2 as independent variables is

6x1 + 2 + 2 =

0

2

=

0

x2

=

0

4x2

+7— x1

—

Solving this system for x1, x2 and A we obtain 4x2

A

=

+7 = —9/10

This procedure can be readily generalized to the case of n variables and r constraints. The problem consists of determining the stationary values Of subject to the constraints gk(xl, x2,. . , = 0, where j(x1, x2, . . , , r. There will be r Lagrange multipliers for this case, and H has k = 1, 2, .

.


78

CHAP. 3

the form

H=

f + k1

H(x1, x2

2k9k

.

.

(3—35)

,

The Euler equations for H are

+

0

..,n

(3—36)

= 1, 2,. . ., r

(3—37)

i = 1,2,.

k=1

9k = 0

k

We first solve r equations in (3—36) for the r Lagrange multipliers, and then determine the n coordinates of the stationary points from the remaining n — r equations in (3—36) and the r constraint equations (3—37). The use of Lagrange multipliers to introduce constraint conditions usually reduces the amount of algebra. REFERENCES 1.

THOMAS, G. B., JR., C'alculus and Analytical Geometry, Addison-Wesley Publishing

Co., Reading, Mass., 1953. COURANT; R., Differential and Integral Calculus, Vol. 1, Blackie, London, 1937. COURANT, R., Differential and Integral Calculus, Vol. 2, Interscience Publishers, New York, 1936. 4. HANCOCK, H., Theory of Maxima and Minima, Dover Publications, New York, 1960. 5. APOSTOL, T. M,, Mathematical Analysis, Addison-Wesley Publishing Co., Reading, Mass., 1957. 6. CRANDALL, S. H., Engineering Analysis, McGraw-Hill, New York, 1956. 7. HILDEBRAND, F. B., Methods of Applied Mathematics, Prentice-Hall, New York, 2. 3.

1952.

PROBLEMS 3—1.

Determine the relative extrema for

(a) (b) (c)

f(x) = f(x) = f(x) =

2x2 + —2x2 ax2 +

(e) (f) (g)

f(x) = f(x) =

(x

4x + 5 + 8x + 10 2bx + c

(d) f(x)=x3+2x2+x+10 f(x)=1x3+2x2+4x+15 —

4ax3

+ (x —

a)2

+ 4bx2 + cx + d

3—2. Expand cos x in a Taylor series about x = the upper and lower bounds on R3.

3—3.

n=

3.

Determine

Expand(1 + x)112inaTaylorseriesaboutx = Otakingn =

mine upper and lower bounds on R2. 3—4. Find df and d2f for

(a) f=x2+2x+5

(b) f=3x3+2x2+5x+6 (c) f=x2sinx (d)

0, taking

f=

cosywhcrey = x3

2.

Deter-

PROBLEMS

Let f =

3—5.

79

Show that

u(x)/v(x).

df =

(du

—

d2f =

f dv)

—

fd2v)

—

Let u1, u2, u3 be functions of x and f = f(u1, u2, u3). Determine df. Suppose f = u(x)w(y) where y = y(x). Determine expressions for df and d2f. Apply to 3—6. 3—7.

(a) u=x3—x (b) w=cosy (c) y=x2

Find the first two differentials for the following functions: + + + 5x1 — 4x2 + 6x1x2 + Consider f = uv, where

3—8.

(a) (b)

f= f' =

3—9.

u=

u(y1, .v2)

v

=

v(y1, Y2)

and = y2(x1, x2)

y1(x1, x2)

Yi

Show that

df =

u dv + v du

d(uv)

d2f =

ud2v

+ 2 du dv + vd2u

Note that the rule for forming the differential of a product is independent of whether the terms are functions of the independent variables (x1, x2) or of dependent variables. 3—10. Classify the stationary points for the following functions: (a) — 9x1 + 12x2 — 10 3xl + (b)

f f f f

3—11.

+ 6x1x2 6x1x2 + 2x1 + 6x1x2 + 34 — 3x1 = Consider Rayleigh's quotient, xTax

x is arbitrary. Since a is symmetrical, its characteristic vectors are linearly independent and we can express x as

x= (j = 1, 2,. . Show that

where Q3 (a)

.

,

n)

are the normalized characteristic vectors for a.

= j=j-

CHAP. 3


80

(b) (c)

Suppose x differs only slightly from Qk. Then, ICjI Specialize (a) for this case. Hint: Factor out 2k and Use (b) to obtain an improved estimate for A.

0. Also, if we take y = s, then Example

4—2

Consider the curve defined by (4—3). Using (4—6), the scale factor is

[a2 sin2 y +

cos2 y + c2]'12

a. One can always orient the axes such that this condition is satisfied.

We suppose that b

Then, we express

b2

as

= (b2 + c2)"2 [1 —

k2

sin2 y]"2

where

k2 = b2 — b2 +

2

c2

The arc length is given by s

dy =

(b2

+

[1 — k2 sin2 yJ112 dv

The integral for s is called an elliptic integral of the second kind and denoted by E(k, y). Then,

(b2 +

s

c2)'12

E(k, y)

Tables for E(k, y) as a function of k and y are contained in Ref. 3. When b = is called a circular helix and the relations reduce to

=

(a2

S = ny

+

const.

a,

the curve

SEC. 4—3. 4—3.

UNIT TANGENT VECTOR

85

UNIT TANGENT VECTOR

We consider again the neighboring points, P(y) and Q(y + shown in Figure 4—3. The corresponding position vectors are P(y + ky), and

approaches the tangent to the curve at P. Then, the unit tangent vector at P is given by* As L\y -+ 0,

-

t=

.

Jim

PQ d1 --=—ds

(4—8)

•

Using the chain rule, we can express I as — dP

— —

ds

dy

—

1

dy ds —

—

dF

(4—9)

dy

Since > 0, 1 always points in the positive direction of the curve, that is, in the direction of increasing s (or y). It follows that dP/dy is also a tangent vector and

— dy

/df dP\"2 \dy dy Equation (4—10) reduces to (4—6) when coordinates. +s

is expressed in

Q(y+6.y)

+

r(y) Fig. 4—3. Unit tangent vector at P(y).

* See Ref. 1, p. 401.

(4—10)

cartesian

DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT

86

CHAP. 4

Example 4—3 We determine the tangent vector for the curve defined by (4—3). The position vector is F

= a cos

+ b sin Y12 + cyi3

Differentiating P with respect to y, dP

dy

=

+ h cos Y12 +

—a sin

and using (4—9) and (4—10), we obtain

a = +[a2 sin2 y + b2 cos2 y + c2]"2

=1[—asinyT1 + bcosyi2 + c13] =' coast, and the angle between the t?ngent and the X3 When a b, a [a2 + direction is constant. A space curve having the property that the angle between the tangent and a fixed direction (X3 direction for this example) is constant is called a helix.*

4-4. PRINCIPAL NORMAL AND BINORMAL VECTORS Differentiating

= 1 with respect to y, we have -

dy

=

0

It follows from (a) that di/dy is orthogonal to f. The unit vector pointing in the direction of di/dy is called the principal normal vector and is usually denoted by ii. H=

ldt dy

where

d (1

(4..-lt) dF

The binormal vector, h, is defined by

b='?xh We see that b is also a unit vector and the three vectors.

(4-12) ñ,

b comprise a right-

handed mutually orthogonal system of unit vectors at a point on the curve. Note that the vectors are uniquely defined once y) is specified. The frame associated with b_and ii is called the moving trihedron and the planes determined by (1, ñ), (ii, b) and are referred to as the osculating normal, and rectifying planes (see Fig. 4—4). * See

Ref. 4, Chap.

1.

PRiNCIPAL NORMAL AND BINORMAL VECTORS

SEC. 4—4.

87

Normal plane

Rectifying

plane

Fig. 4—4. Definition of local planes.

Example 4—4 We determine fi and b for the circular helix. We have already found that

a — [a2 + and

=

sin VT1

+ a cos

+ c13]

Differentiating t with respect to y, we obtain

di

a — [cos ytj + Sm

—

Then, i dt fl

dt dy

— C05

— Sm

dy

The principal normal vector is parallel to the plane and points in the inward radial direction. It follows that the rectifying plane is orthogonal to the X1-X2 plane. We can determine b using the expansion for the vector product.

a

—asiny acosy

C

This reduces to

b

C. a

sin

The unit vectors are shown in Fig. E4—4.

C

— — cos

a

+

a £3

a

GEOMETRY OF A MEMBER ELEMENT

88

CHAP. 4 Fig, E4—4

4—5.

CURVATURE, TORSION, AND THE FRENET EQUATIONS

The derivative of the tangent vector with respect to arc length is called the curvature vector, K. K

dt

d2F

c/s

i/s2

ic/i

K

(4—13)

c/s2

Using (4—11), we can write —

ds

— Ku

(4—14)

0. The Note that K points in the same direction as Ft since we have taken K curvature has the dimension L1 and is a measure of the variation of the tangent vector with arc length. We let R be the reciprocal of the curvature:

R=

K1

iS)

In the case of a plane curve, R is the radius of the circle passing through three consecutive points* on the curve, and K = JdO/dsj where 6 is the angle between I and To show this, we express I in terms of 0 and then differentiate with respect to s. From Fig. 4—5, we have cos * See

+ Sifl 012

Ref. 4, p. 14, for a discussion of the terminokgy 'three consecutive points."

SEC. 4—5.

CURVATURE, TORSION, AND THE FRENET EQUATIONS

Then —

K

.

[—sin

+ cos 617]

dO a—

and

K

dO

1

ds

R

dO/ds

+ cos 612]

[— sin

In the case of a space curve, the tangents at two consecutive points, say P and Q, are in the osculating plane at F, that is, the plane determined by and ñ at. P. We can interpret R as the radius of the osculating circle at P. It should be noted

that the osculating plane will generally vary along the curve. x2

\ R

+ R

t i2

it

Fig. 4—S. Radius of curvature for a plane curve.

The binormal vector is normal to both and ñ and therefore is normal to the osculating plane. A measure of the variation of the osculating plane is given by db/ds. Since his a unit vector, db/ds is orthogonal to h. To determine whether db/ds involves we differentiate the orthogonality condition I b 0, with respect to s. -

db

-

ds

dl ds

But dl/ds Kñ and b ii = 0. Then, db/ds is also orthogonal to I and involves only ñ. We express db/ds as

db

= —tn

where r is called the torsion and has the dimension, L

(4—16)

DiFFERENTIAL GEOMETRY OF A MEMBER ELEMENT

90

CHAP. 4

It remains to develop an expression for a. Now, h is defined by

xn Differentiating with respect to s, we have db = di

This reduces to

db

diii

—=t x

ii =

0.

dñ

xn+t

Finally, using (4—16), the torsion is given by

l-dfl —

dñ

—b

——

ds

dy

(4—17)

Note that a can be positive or negative whereas K is always positive, according to our definition. The torsion is zero for a plane curve since the osculating plane coincides with the plane of the curve and b is constant.

Example

4—5

The unit vectors for a circular helix are

=

[—a

sin vij +

—cos

sin

—

b=

cos Yti + cT3]

a

yl' — ccosyi3 + at3]

where

a = (a2 +

c2)112

Then, a

K=-— adv

a

a +c

a

and 1—

a

dñ

c

dy

a

c 2

2—const

We have developed expressions for the rate of change of the tangent and binormal vectors. To complete the discussion, we consider the rate of change of the principal normal vector with respect to arc length. Since fi is a unit vector, dñ/ds is orthogonal to ñ. From (4—17), -

dñ

b— ds

a

SEC. 4—6.

GEOMETRICAL RELATIONS FOR A SPACE CURVE

91

To determine the component of dñ/ds in the I direction, we differentiate the

orthogonality relation, I n = 0. ds

(b)

ds

it follows from (a) and (b) that dñ

-

— = —I(t + tb us

(4—18)

The differentiation formulas for 1, ii, and b are called the Frenet equations. 4—6.

We

SUMMARY OF THE GEOMETRICAL RELATIONS FOR A SPACE CURVE

summarize the geometrical relations for a space curve: Orthogonal Unit Vectors t =

a=

ldi

di

= — = tangent vector exdy

thu

di

1

principal normal vector

—i-- i—

(4—19)

= I x ñ binormal vector di ds — dy

dy

Equations)

Di:fferenriation Formulas

lull

dl

— -— = Kn ds ady db 1db — = —— = —rn ds ctdy dñ -. 1 dñ —Kt ds ady

K= =

+tb

(4—20)

1 di — curvature a dy 1—dñ

—b

a

— = torsion dy

We use the orthogonal unit vectors (I, ñ, b) to define the local reference frame for a member element. This is discussed in the following sections. The Frenet

92


CHAP. 4

equations are utilized to establish the governing differential equations for a

member element.

4—7.

LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT

The reference frame associated with ñ, and b at a point, say P, on a curve is uniquely defined once the curve is specified, that is, it is a property of the

curve. We refer to this frame as the natural frame at P. The components b) are actually the direction cosines for the natural of the unit vectors frame with respect to the basic cartesian frame which is defined by the orthogonal unit vectors (1k, 12, 13). We write the relations between the unit vectors as ft n

£12

=

133

11

t22 £32

(4—21)

12

e33

One can express* the direction cosines in terms of derivatives of the cartesian

coordinates (x1, x2, x3) by expanding (4—19). Since (1, b) are mutually or13) the direction cosines are related by thogonal unit vectors (as well as 1jm6m

=

j,

k

=

1,

2, 3

(4—22)

Equation (4—22) leads to the important result [ljk]T =

(4—23)

and we see that is an orthogonal matrix.f The results presented above arc applicable to an arbitrary continuous curve. Now, we consider the curve to be the reference axis for a member clement and take the positive tangent direction and two orthogonal directions in the normal plane as the directions for the local member frame. We denote the directions of the local frame by (Y1, Y2, 1'3) and the corresponding unit vectors by (t1,

2,

= 1) We will always take the positive tangent direction as the Y1 direction x and we work only with right handed systems t3). This notation is shown in Fig. 4-6. When the centroid of the normal cross-section coincides with the origin of the local frame (point P in Fig. 4—6) at every point, the reference axis is called the centroidal axis for the member. It is convenient, in this case, to take Y2, Y3 as the principal inertia directions for the cross section. In general, we can specify the orientation of the local frame with respect to the natural frame in terms of the angle between the principal normal direction and the I'2 direction. The unit vectors defining the local and natural frames * See

Prob. 4-5.

t See Prob. 4—6.

SEC. 4—7

are

LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT

related by

93

-

tl — t2

=

COS 4)11

+ sin 4)b

(4—24)

çbn + cos 4th

Combining (4—21) and (4—24) and denoting the product of the two direction cosine matrices by the relation between the unit vectors for the local and basic

frames takes the concise form

t=

(4—25)

where

[

€12

€j3

£21cos4)+€31sin4)

€22cos4)+ €32sin4)

€23cosçb+ €33sinqS

[21sin4)+€31cos4)

—€22si+C32cos4)

—€23 sin 41+ £33cos41

Note that the elements of fi are the direction cosines for the local frame with respect to the basic frame. fJjk =

(4—26)

Xk)

'. We will utilize (4—25) in the next Since both frames are orthogonal, J1 chapter to establish the transformation law for the components of a vector. x3 Normat

Y1

Fig. 4—6. Definition of local reference frame for the normal cross section.

CHAP. 4


94

Example 4—6 We determine

for the circular helix. The natural frame is related to the basic frame.by

a.

a

c

a

a

a

sin y

0

c

a

—cosy —

——slay

I

=

— cos

y

—

C. —slay

b

—--cosy a

a

12

=

{Ik}

—

a

Using (4-.25)

a cos y a

a

—sinycosçb ——cosysin4

a

C.

+cosysm4 + —sinycos4

4—8.

a

sin y sin

— — cos

y cos

a

a

I. a

a

CURVILINEAR COORDINATES FOR A MEMBER ELEMENT

We take as curvilinear coordinates (yi, Yz' y3) for a point, say Q, the parameter

of the reference axis and the coordinates (Y2, of Q with respect to the orthogonal directions (Y2, 1'3) in the normal cross section (see Fig. 4—7). Let F(y1) the R(y1, Y2' Y3) be the position vector for Q(Yl, y3) and position vector for the reference axis. They are related by = r + Y2t2 + y3t3 where COS

+ cos4b

=

t3 =

+ Sifl (

We consider 4 to be a function of y1. Y2

y3

—

——

Y2

Fig. 4—7. Curvilinear coordinates for the cross section.

(4—27)

SEC. 4—8.

CURVILft4EAR COORDiNATES FOR A MEMBER ELEMENT

95

The curve through point Q corresponding to increasing Yj with Y2 and y3 held constant is called the parametric curve (or line) for yj. In general, there are three parametric curves through a point. We define as the unit tangent vector for the parametric curve through Q. By definition,

=

Ui

13R (4—28) aIR

= The differential arc length along the

curve is related to

aIR

by (4—29)

=

=

This notation is illustrated in Fig. 4—8. One can consider the vectors (or

to define a local reference frame at Q. x3

y2t2 +y3t3

x2

Fig. 4—8. Vectors defining the curvilinear directions.

Operating on (4—27), the partial derivatives of R are 0R

aR aR

— dy1 =

t2

=

t3

+

dt2 Y2

dy1

+ Y3

dy1

96


CHAP. 4

We see that

t2

g2

ü3=t3

1

(4—30)

g3=1

It remains to determine ü1 and g1. Now, dy1

=

=

Also, d12

(dñ

dy1

\dy1

dt3

=

+ dy1 —bj +

(dii

.

dçb

\dy1

.

j

1db '\dv1

dq5\

—

1db

+ b—)+ dy1j

\dy1

dy1 —11—— dy1

We use the Frenet equations to expand the derivatives of ñ and h. Then, cit2

dyj

dy1

I and finally, = cc(1

—

Ky'2)!1

Y2 COS 4)

+

/

+

d4)'\.

d4)\ _)(Y2t3

y3t2)

(4—31)

J73 sin 4)

We see from Fig. 4—9 that y'2 is the coordinate of the point with respect to the

principal normal direction. y3

\y3

Fig. 4—9.

of y.

Since 13R/ay1 (and therefore ii1) involve the reference frame defined and by iii, u2, will not be orthogonal. However, we can reduce it to an orthogonal

REFERENCES

system

97

by taking dy

=

(4—32)

which requires cer dy

=

(4—33)

150

When (4—32) is satisfied,

aR

=

— Ky'2)t1

and

= =

(4—34) cx(l —

In this case, the local frame at Q coincides with the frame at the centroid. One

should note that this simplification is practical only when ccc can be readily integrated.

Example

4—7

The parameters a and t are constant for a circular helix:

a=

(a2

+ c2)112

C

Then, C

at = — a

and integrating (4—33), we obtain tS

— Yo)

For this curve,

varies linearly with y (or arc length). The parameter g1 follows from (4—34).

hi =

ds1

= a(1

—

Ky2)

/

a

'\

cc-

x(l —

REFERENCES

3.

THOMAS, G. B., JR.: Analytical Geometry and Calculus, Addison-Wesley Publishing Co., Inc., Reading, Mass., 1953. HAY, U. 13.: Vector and Tensor Analysis, Dover Publications, New York, 1953. JA}INKE, E., and F. EMDE: Tables of Functions, Dover Publications, New York, 1943.

4.

STRUm, D. J.: Differential Geometry, Addision-Wesley Publishing Co., Reading,

1.

2.

Mass., 1950.


CHAP. 4

PROBLEMS 4—1.

(a) (b) (c) (d)

Il, b, ; K, x for the following curves: x2 = 3 sin y cos y x3 = 5y x1 3 cos y x2 = 6 sin y x3 = 5y + p313 + = x1 = cos y x2 = sin y

Determine

x1 =

3

x3 = cy where a, /3, c are real constants. 4—2. If 0, the curve lies in the plane. Then,r Oandb ±i3. The sign of b will depend on the relative orientation of ñ with respect to 1.

Suppose the equation defining the curve is expressed in the form

x3=0

x2=J(x1)

Equation (a) corresponds to taking x1 as the parameter for the curve. (a) Determine the expressions for 7, ñ, b, and K corresponding to this representation. Note that Let y and + f(x1)12 +

_Lf' (b)

.f"etc

Apply the results of (a) to

=

4a —

2 x1)

a and b are constants. This is the equation for a parabola symmetrical about x1 = b/2. Let 9 be the angle between and

where (c)

cos0 = I

•11.

= sec 0. Express t, h, !, and K in terms of 0. for the case where 02 is negligible with respect to unity. This approximation leads to Deduce that

(d)

Specialize

sin 0 cos 0 4—3.

tan 0

0

1

A curve is said to be shallow when 02 (e

(6—3 1) —

Loading—Second Segment

F>l) < F F

= 3.

F>1>

,4a12>

F>2>

k'2>(e

Unloading—Second Segment

F

k>1>(e —

One can readily generalize these relations for the nth segment.f *

We

(6—32)

+ (f°> —

are neglecting the Bauschinger etlect. See Ref. 2, Sec. 5.9. or Ref. 3, Art. 74.

t See Prob. 6—8.

(6—33)

GOVERNING EQUATIONS FOR AN IDEAL TRUSS

128

CHAP. 6

Example 6—4 We consider a bilinear approximation, shown in Fig. E6—4. Fig. E6—4

40

41.7 30

(in./in.)

Taking

,4=lin.2

L=lQft=l2Oin we obtain

=

= 83.3 kips/in.

f"> =

1/k"> =

k>2> =

—.— = 41.7 kips/in. L

f"> =

24 x

F"> =

= 3okips

42>

L AE>2>

=

+ ([1)

—

Segment 1 Segment 2

120

12

x

in./kip

in./kip

— 0.36 in.

F

(83.3)(e — 120

F

(41.7)(e —

Suppose a force of 35 kips is applied and the bar is unloaded. The equivalent initial strain is (see Equation 6—33 and Fig. 6—6):

= =

—

+ (f>2) —

=

+ 0.06 in.

The procedure described above utilizes the segment stiffness, which can be interpreted as an average tangent stiffness for the segment. We have to modify the stiffness and equivalent initial elongation only when the limit of the seg-

SEC. 6-4.

FORCE-ELONGATION RELATION FOR A BAR

129

is reached. An alternate procedure is based on using the initial linear stiffness for all the segments. In what follows, we outline the initial st(ffness approach. ment

.4

I''I

/1

-eo,eq.

Fig. 6—7. Notation for the initial stiffness approach.

Consider Fig. 6—7. We write the force-elongation relation F—

= where e0,

eq

is

—

—

kU)(e

A

segment 2 as (6—34)

— eo,eq)

interpreted as the equivalent linear initial strain and is given by eo,eq = A

(6—35)

=

—

—

The equivalent initial strain, eoeq, depends on e, the actual strain. Since e in turn depends on F, one has to iterate on eoeq regardless of whether the segment limit has been exceeded. This disadvantage is offset somewhat by the use for all the segments. The notation introduced for the piecewise linear case is required in order to distinguish between the various segments and the two methods. Rather than continue with this detailed notation, which is too cumbersome, we will drop all the additional superscripts and write the force-deformation relations for bar n in the simple linear form of

= =

+

(6—36)


130

CHAP. 6

where k, f, and e0 are defined by (6—31) through (6—35) for the physically

nonlinear case.

6—5.

GENERAL BAR FORCE—JOINT DISPLACEMENT RELATION

The force-deformation and deformation-displacement relations for bar n are given by (6—22) and (6—36). Combining these two relations leads to an expres-

sion for the bar force in terms of the displacement matrices for the joints at the ends of the bar. The two forms are: F,, =

=

— e0,

F0,,,

+

— k,,y,,u,,

(6—37)

F0,,, = —k,,e0,,,

and —

e0,, + f,,F,,

u,_) = e,,

(6—38)

We can express the force-displacement relations for the "m" bars as a single matrix equation by defining (6 39)

k1

k2

k=

km

and noting (6—24). The generalized forms of (6—37) and (6—38) are:

F=

k(e

—

e0)

=

F0

+

(6—40)

and

d°1I = 6—6.

e0

+ fF

(6—41)

JOINT FORCE-EQUILIBRIUM EQUATIONS

Let F,, be the axial force vector for bar n (see Fig. 6—8). The force vector has the direction of the unit vector, i,,, which defines the orientation of the bar in the deformed state. Now, = fi,,i. Then,

F,, =

(6—42)

When F,, is positive, the sense of F,, is the same as the positive sense for the bar. Continuing, we define F,,,, as the forces exerted by bar n on the joints at the positive and negative ends of the bar. From Fig. 6—8, = — F,, = — F,,fi,,i F,,,,

+F,, =

(6—43

JOINT FORCE-EQUILIBRIUM EQUATIONS

SEC. 6—6.

Joint n_

Fig. 6—8. Notation for barforce.

We consider next joint k. The external joint load vector is Pk, where For equilibrium, the resultant force vector must equal zero. Then, Pk =

- j+=k

Pk

—

The first summation involves the bars which are positive incident on joint k (positive end at joint k) and the second the bars which are negative incident. Using (6--43), the matrix equilibrium equation for joint k takes the form: Pk = Let

(6—44)

—

j+k

be the general external joint load matrix:

=

P2,

(if x

,

(6—45)

1)

We write the complete set of joint force-equilibrium equations as:

=

(6—46)

Note that the rows of pertain to the joints and the columns to the bars. We partition into submatrices of order i x 1. (if x

=

=

1,

2,.. ,j .

m)

k=

and

1,

2.

.

.

,m

(6—47)

Since a bar is incident only on two joints, there will be only two elements in any column of From (6—44), we see that, for column n, = =

=

0

when

(6—48) e

orn_

The matrix can be readily developed using the connectivity table. It will have the same form as dT with y, replaced by n,,. When the geometry is linear,

=

=

and

CHAP. 6


132

Example 6—5 matrix for the truss of Example 6—1 has the following general form:

The

Bar Numbers 1

2

4

3

RT

a

7

8

10

9

11

+llç

1

Iz

6

5

+

nT

T 2

ftT

—plo

OF +p7

DT

-'

nT

P2

A

oT ±1J3

C

oT

OT

nT

C

J3

-I

oT +p4

øT

P6

oT +pU •r

I,

We could have also utilized the connectivity* matrix C to develop ft was pointed out in Example 6—3 that the elements of the kth column of C define the incidence of the bars on joint k. Using this property, we can write the generalized form of (6—44) as

where 0

-.

0

(rn x Lm) o

(6—49)

o

Finally, we have

= 6—7.

(6—50)

INTRODUCTION OF DISPLACEMENT RESTRAINTS; GOVERNING EQUATIONS

We have developed the following equations relating F, e,

and qj,

e = d°l1 = e0 + IF = and are the external joint-displacement and external joint-load matrices arranged in ascending order. Also, in our derivation, we where the elements of

have considered the components to be referred to a basic reference frame. Now, * See Sec. 6—3, Eq. 6—27.

SEC. 6—7.

INTRODUCTtON OF DISPLACEMENT RESTRAINTS

133

joint displacement restraints are imposed, there will be a reduction in the number of joint displacement unknowns and a corresponding increase in d, the number of force unknowns. This will require a rearrangement of and when

Let r be the number of displacement restraints and 11d the number of displace-

ment unknowns. There will be n4 prescribed joint loads and r unknown joint loads (usually called reactions) corresponding to the na unknown joint displacenients and the r known joint displacements. We let U1, U2 be the column matrices of unknown and prescribed joint displacement components and P1, P2 the corresponding prescribed and unknown joint load matrices. The rearranged system joint displacement and joint load matrices are written as U, P: (fld

x

1)

(r x 1) >
2

:13

0

m

0 z C,)

—1

0 C

m

z z

m

0

C)

INITIAL INSTABILITY

SEC. 6—9.

137

Combining (a) and (6—63), we have

U= and it follows that

I) = Since both H and

HPII°'

(6—64)

are orthogonal matrices, D is also an orthogonal matrix.

Using (6—62),

A

dDT

then substituting for d, P.s, and D in terms of the geometrical, connectivity, local rotation matrices lead to and

B

( 6— 65 )

A=

Equation (6—65) is of interest since the various terms are isolated. However, one would not generate A, B with it. 6—9.

INITIAL INSTABILITY

The force equilibrium equations relating the prescribed external joint forces and the (internal) bar forces has been expressed as (see Equation 6—54):

P1=B1F I) and F is (m x 1). When the geometry is nonlinear, B1 depends on the joint displacements as well as on the initial geometry and

where P1 is

neutral

cos3

unstable

cos3 0'

P2

P2

One can show that (n) is equivalent to stable

neutral

dP2

dP2

0

o

dO

unstable

dP2

A transition from stable to unstable equilibrium occurs at point A, the peak of the deflection curve. The solution for 0 is different in that its stable segment is the linear kL) corresponds to a branch point, solution and the neutral equilibrium point (P2 Both the linear and nonlinear branches are unstable. Fig. E7—6B

0

REFERENCES 1

2. 3. 4. 5. 6. 7.

WANG, C. T.: Applied Elasticity, McGraw-Hill, New York, 1953. LANGHAAR, H. L,: Energy Methods in Applied Mechanics, Wiley, New York, 1962. REISSNeR, E.: "On a Variational Theorem in Elasticity," J. Math. Phys., Vol. 29, pages 90-95, 1950. ARGYRIS, i. H., and S. KIsLseY: Energy Theorems and Structural Analysis, Butterworths, London, 1960. CI.IARLTON, T. M.: Energy Principles in Applied Statics, Blackie, London, 1959. HOFF, N. J.: The Anal vsis of Structures, Wiley, & New York, 1956. K.: Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968.

CHAP. 7

VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS

174

PROBLEMS

Consider the two-dimensional symmetrical truss shown. Assume = 03 = 0. (a) Determine the first two differentials of e1 and ez by operating on the

7—1.

expanded expression (equation 6—19) for e. (b) (c)

When a b, we can neglect the nonlinear term involving u12 in the Specialize (a) for this case. expressions for e and When a b, we can neglect the nonlinear term involving u11 in the Specialize (a) for this case. expressions for e and Prob. 7—1

x2

T 21

3

7—2. Refer to the figure of Prob. 7—1. Assume = u3 = 0 and a> b. Using the principle of virtual displacements, determine the scalar force-equilibrium equations for joint 1. 7—3. Suppose a force F is expressed in terms of e,

F=

C1e

+ 4C2e3

where a is related to the independent variable u by a

(a)

u + 1u2

Determine the first two differentials of the work function, W = W(u), defined by W

(b)

F de

=

Suppose (a) applies for increasing e and

F=

C1(e

—

e decreasing from e*. Determine d2 W at a = e*. Refer to Prob. 6—23. The n — 1 independent node equations relating the branch currents are represented by 7—4.

ATI

U

Now, the branch potential differences, e, are related to the n — node potentials, V, by

e=

AV

1

independent

PROBLEMS

175

Deduce that the requirement, 1T de

=

for arbitrary

0

is equivalent to (a). Compare this principle with the principle of virtual displacements for an ideal truss. 7—5. Consider the two-dimensional truss shown. Assume u2 = = 0. (a)

(b)

Using (7—14), obtain a relation between the elongations and ü32. Take the virtual-force system as LxF2 and the necessary bar forces and reactions required to equilibrate AF2. Using (7—12), express u11, u12 in terms of e1, e3. Note that bar 2 is

not needed. One should always work with a statically determinate system when applying (7—12). Prob.

7—5

x2

2

7—6. Refer to Prob. 6—23. One can develop a variational principle similar to the principle of virtual forces by operating on the branch potential difference—node potential relations. Show that -

AiTe=0 for any permissible set of current increments. Note that the currents must satisfy the node equations ATi

0

Deduce Kirchhoff's law (the sum of the voltage drops around a closed loop must equal zero) by suitably specializing Lsi in (a). Illustrate for the circuit shown in Prob. 6—6, using branches 1, 2, 4, and 6. 7—7. By definition, the first differential of the strain-energy function due to an increment in U1 has the form

=

n1

dV.

=

F,, de,,

We work with expressed as a compound function of e = e(U) since it is more convenient than expressing V directly in terms of U1. One can also

VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS

176

write (a)

CHAP 7

as

Using (b), show that the system of if joint force-equilibrium equations expressed in terms of the joint displacements can be written as:

k=

ÔU(k

Equation c is called Castigliano's principle, part I. (b)

Show that an alternate form of (c) is

P(k= Note that (d) is just the expansion of (c). Rework P rob. 7—2, using (d).

7—8. Determine V(e), dv, and d2V for the case where the stress-strain relation has the form (see Prob. 6—10)

a=

—

be3)

7—9. Determine V*(F), dV*, and d2V* for the case where the stress-strain relation has the form

=

(a

+

ca3)

7—10. Show that (7—12) can be written as UkJ

= 0Pkj

is defined by (7—31). This result specialized for U2 = 0 is = called Castigliano's principle, part H. Apply it to Prob. 7—5, part b. Assume linear elastic material and f1 = = = f. 7—11. The current and potential drop for a linear resistance are related by where

ef

Inverting (a), we can express

e0,j

as a function of e1.

= (a)

+

—

which has the property that

Suppose we define a function,

= corresponding to (b).

Determine b

(b)

Let W

where h = total number of branches. Considering

the branch potential drops to be functions of the node potentials, deduce that the actual node potentials V correspond to a stationary value of W. Use the results of Prob. 7—4. The Euler equations for

PROBLEMS

177

W = W(V) are the node current equilibrium equations expressed in (c)

terms of the node potentials. Suppose we define a function

which has the property that

=

(d)

corresponding to (a).

Determine b

(d)

Let W* =

W7. Show that the Euler equations for

H=

iTe

=

—

1T(AV)

—

= H(i,V)

(e)

are the governing equations for a d-c network. Show that the actual currents correspond to a stationary value of One can either introduce the constraint condition, An = 0, in (e) or use the result of Prob. 7—6. 7—12. Investigate the stability of the system shown below. Take k, = aL2k5 (e)

P

Prob.

Linear translational restraint

Rigid rod

ICr

and

consider a to range from 0 to 6.

(Linear rotational restraint)

7—12

8

Displacement Method Ideal Truss 8—1.

GENERAL

The basic equations defining the behavior of an ideal truss consist of forceequilibrium equations and force-displacement relations. One can reduce the system to a set of equations involving only the unknown joint displacements by substituting the force-displacement relations into the force-equilibrium equations. This particular method of solution is called the displacement or method. Alternatively, one can, by eliminating the displacements, reduce the governing equations to a set of equations involving certain bar forces. The latter procedure is referred to as the ,fin'ce or flexibility method. We emphasize that these two methods are just alternate procedures for solving the same basic equations. The displacement method is easier to automate than the force method and has a wider range of application, However, it is a computer-based method, i.e., it is not suited for hand computation. In contrast, the force method is more suited to hand computation than to machine computation. In what follows, we first develop the equations for the displacement method by operating on the governing equations expressed in partitioned form. We then describe a procedure for assembling the necessary system matrices using

only the connectivity table. This procedure follows naturally if one first operates on the unpartitioned equations and then introduces the displacement restraints.

The remaining portion of the chapter is devoted to the treatment of nonlinear behavior. We outline an incremental analysis procedure, apply the classical stability criterion, and finally, discuss linearized stability analysis. 8—2.

The

OPERATION ON THE PARTITIONED EQUATIONS

governing partitioned equations for an ideal truss are developed in

Sec. 6—7. For convenience, we summarize these equations below. 178

SEC. 8—2.

OPERATION ON THE PARTITIONED EQUATIONS

= B1F P2 = B2F

179

eqs.)

P1

(r eqs.) (in eqs.)

F = F, + kA1U1 F, = k(—e0 + A2tJ2)

The unknowns are the in bar forces (F), the r reactions (P2), and the na joint

displacements (U1). One can consider F, to represent the initial bar forces, that is, the bar forces due to the initial elongations and support movements 0. The term kA1U1 represents the bar forces due to U1. When the with U1 material is linear elastic, k and e0 are constant. Also, = BT when the geometry is linear.

We obtain a set of equations relating the flj displacement unknowns, U1, by substituting for F in (a). The resulting matrix equation has the form (B1kA1)U1 =

—

B1F1

We solve (8—i) for U1, determine F from (e), and P2 from (b). The coefficient matrix for U1 is called the system stiffness matrix and written as K11 = B1kA1

(8—2)

as representing the initial joint forces due to the initial One can interpret elongations and support movements with U1 = 0. Then — B1F1 represents the net unbalanced joint forces. When the geometry is linear, K1 reduces to 1

K11 = B1kBT = AfkA1

If the material is linear, k is constant and positive definite for real materials. Then, the stiffness matrix for the linear case is posiLive definite when the system

is initially stable, that is, when r(B1) Conversely, if it is not positive definite, the system is initially unstable. If the material is nonlinear, k and e0 depend on e. We have employed a piecewise linear representation for the force-elongation curve which results in linear relations. However, one has to iterate when the limiting elongation for a segment is exceeded. The geometrically nonlinear case is more difficult since both A and B depend

on U1. One can iterate on (8—1), but this requires solving a nonsymmetrical system of equations. It is more efficient to transform (8—1) to a symmetrical system by transferring some nonlinear terms to the right-hand side. Nonlinear analysis procedures are treated in Sec. 8—4. Even when the behavior is completely linear, the procedure outlined above for generating the system matrices is not efficient for a large structure, since f See Prob. 2—14.

DISPLACEMENT METHOD: IDEAL TRUSS

180

CHAP. 8

it requires the multiplication of large sparse matrices. For example, one obtains the system stiffness matrix by evaluating the triple matrix product,

=AfkA1 One can take account of symmetry and the fact that k is diagonal, but A1 is generally quite sparse. Therefore, what is needed is a method of generating K which does not involve multiplication of large sparse matrices, A method which has proven to be extremely efficient is described in the next section. 8—3.

THE DIRECT STIFFNESS METHOD

We start with (6—37), the force-displacement relation for bar ii:

+

= F0,

= where n4, n. denote the joints at the positive and negative ends of barn. One can consider F0, as the bar force due to the initial elongation with the ends

= 0). Now, we let required to equilibrate the action of

be the external joint force matrices Noting (6—43), we see that

fixed (un,

= =

p,I_

Substituting for

(8—4)

(8—4) expands to

=

+

—

pn_ = One can interpret (b) as end action—joint displacement relations since the elements of ± are the components of the bar force with respect to the

basic frame. Continuing, we let (8—5)

Note that is of order i x i where I = 2 or 3 for a two or three-dimensional truss, respectively. When the geometry is linear, and is sym= y,, metrical. With this notation, (b) takes a more compact form, = =

+

—

—

+

(8—6)

We refer to as the bar stiffness matrix. Equation (8—6) defines the joint forces required for bar n. The total joint forces required are obtained by summing over the bars.

SEC. 8—3.


We have defined p2, .

—

=

.

X 1) (ii x 1) (U

,

{u1, u2

as the general external joint force and joint displacement matrices. Now, we write the complete system of if joint force-equilibrium equations, expressed in terms of the displacements, as

=

+

(8—7)

We refer to if, which is of order if x as the unrestrained system stiffness matrix. The elements of are the required joint forces due to the initial elongations and represents the required joint forces due to the joint displacements.

We assemble if and in partitioned form, working with successive members. The contributions for member n follow directly from (8—6). (Partitioned Form is j x 1) in row

8—8

in row n.

if (Partitioned Form is j x j) +k, —

k,,

—ku

in row column in row column in row n_, column n

Example 8—1

The connectivity table and general form of if and

for the numbering shown in

Fig. E8—l are presented below: Fig. E8—l

0 4

3


182

Bar

k1+k2

4

5

2

2

4

2

1

3

3

4

2

+joint

1

—joint

4

—k1

—k2

k2 + k3 + k5

—k2

—k3

k3 + k4

--k3

Pai ,10

P0.2

—k4

k1 + k4 + k0

—k4

—k5

—k1

U4

U3

U2

U1

P2

3

1

CHAP. 8

=

UT

L'

0.1PI —

L'

pT

0,212 r

L'

UT

t'

Po, 3

—

0.

3P3 —

Po, 4

—

0,

1I'I 1

ftT 0,2P2 nT 0,3P3

øT r0,SPS

UT

0,

4P4

uT 0, 41'4

L'

0. 5P5

Example 8—2 The external force matrix, involves and the displacement matrices for those joints connected to joint j by bars. Now, corresponds to row j and ii,, to column j of ir. By suitably numbering the joints, one can restrict the finite elements of X' to a zone about the diagonal. This is quite desirable from a computational point of view. Fig. E8—2

Sect. 1

0

ft® 2:

(71

fs

(-I -..- —--

/

\

3

%._ —--

L6 ',.

/

;

©

-.

Consider the structure shown. We group the vertical joints into sections. The equilibrium equations for section k involve only the joints in section k and the adjacent sections. For example, the equations for section 3 (which correspond to P6) will involve only the displacement matrices for sections 2, 3, 4. This suggests that we number the joints by section. The unpartitioned stiffness matrix corresponding to the above numbering scheme


SEC. 8—3.

is listed below. Note that has the form of a quasi-tridiagonal band matrix when it is partitioned according to sections rather than individual joints. The submatrices for this truss are of order 4 x 4, U'

U2

k, +k2 —k, P2

U3

U6

U7

U8

I—k2

k,+k3 —k3

—k,

U4

—k4

+k4 —k2

—k3

k2+k3

—k5

—k6

—k7

I

+k6 —k4

—k,

k4+k51

—k8

k6+k9 —k9

—k6

—k10

+k,0 p6

—k7

k8+k7

—k9

—k8

—k1,

—k,2

+k9

I

+k,2 —k,0

JJ7

—k,,

k10+k,, —k,, 3

—k,,

p8

—k,3

k,2+k,,

The introduction of displacement restraints involves first transforming the partitioned elements and to local frames associated with the restraints, permuting the actual rows, and finally partitioning the actual rows. The steps are indicated below.

-+ U -+

We write the system of joint force-equilibrium equations referred to the local

joint frames as

=

+

(8—10)

The transformation la'vs for the submatrices of

= =

€,n= 1,2,...,j

and T

follow from (6—57). (8—11)


184

The step,

—+

P,

CHAP. 8

involves only a rearrangement of the rows of

We

obtain the corresponding stiffness matrix, K, by performing the same operations on both the rows and columns of The rearranged system of equations is written as P = KU + P0 (8-12) Finally, we express (8—12) in partitioned form:

= P2 =

K11U1

+ K1202 + P0,1 + K2202 + P0,2

8

13

The first equation in (8—13) is identical to (8—1).

Example

8—3

It is of interest to express the partitioned elements of K in terms of the geometrical, We start with the general Unconnectivity, and displacement transformation partitioned equations(6—28), (6—40), and (6—44), (6—50):

F0 + kda/1 = F0 + kyC'W

F

Then, substituting for F in (a) and equating the result to (8—7) leads to

= =

The matrix, DTkY, is a quasi-diagonal matrix of order im. The diagonal submatrices arc of order i, and the submatrix at location n has the form, We have defined this product as k,,. Then, if we let

=

[ki I. r,T

2p272

k5

we can express

as

= CTk5c Carrying out (8—9) for n = 1, 2 m is the same as evaluating the triple matrix product. Obviously, (8—9) is more efficient than (f).

The introduction of displacement restraints can be represented as

P= 11

= =

D1dP

(g)

and

= DTU = DfU1 +

(h)


SEC. 8—3.

185

Substituting (g) and (h) in (8—7) and equating the result to (8—13), we obtain

=

K,, =

=

P0.

t—12

DsCTDke0

—

In order to obtain (8—13), we must rearrange the rows and columns of

then partition. This operation is quite time-consuming. Also, it leads to rectangular submatrices. In what follows, we describe a procedure for introducing displacement restraints which avoids these difficulties. We start with the complete system of equations referred to the basic frame, and

(8—14)

We assemble

and

using (8—8) and (8—9). Then, we add to

the

external force matrices for those joints which are unrestrained. It remains to modify the rows and columns corresponding to joints which are either fully or partially restrained. Case A: Fit!! Restraint

is unknown. We replace the equation for Pq by

Suppose uq = Uq. Then

=

Uq

This involves the following operations on the submatrices of X and

On X. Set off diagonal matrix elements in row q and column q equal 1. to 0 and the diagonal matrix element equal to I,.

I=

0

(8—15)

= Ii 2.

On

Add terms in

due to

t

C

X(qUq (8—16)

j ease B: Partial Restraint—Local Frame

We suppose the rth element in

is

prescribed.

= prescribed = = unknown

186


We have to delete the equation corresponding to

CHAP. 8

and replace it with

4= Step I —Assemblage of Basic Matrices

according to the following:

We assemble Eq, Gq, ui', 1.

Eq and Gq. We start with

G=O, and we set G,r

+1

2. us'. We start with an ith-order column vector having zero elements and we set the element in the rth row equal to Note that this matrix involves only the prescribed displacements (local frame) in their natural locations. We start with an ith-order column vector having zero elements and 3. we insert the values of the prescribed joint forces (local frame) in their natural locations. Note that the elements corresponding to the reactions are zero.

When the joint is fully restrained,

E=O, Suppose joint

G=11

5 is partially restrained, The data consist of:

The rotation matrix, R°5, defining the direction of the local frame at 5 with respect to the basic frame. (b) The direction (or directions) of the displacement restraint and the value (or values) of the prescribed displacement. (a)

direction r, (c)

The values of the prescribed joint forces:

j=1,...,i As an illustration, suppose r =

2.

Then, in (b), we read in

In (c), we read in —5

—5

Psi

The four basic matrices are (for r =

E5=

P53 2)

1

0

0

[0

0

0

0

0

0

Gs=IO

1

0

0

0

1

[0 0 0


SEC. 8—3.

187

1°

In

—5

Step 2—Operation on Jr and 1.

Premultiply row q of it" and

by

= " N, q 2.

e

" N, q

'-'q 1%

Postmultiply column q of it" by —

— 1' 2' —

.

T11* and add to PPN.

= 3.

Postmultiply column q of it" by (Eq

irtq = 4.

1,

2, .

.

.

Add Gq to irqq

= it"qq + Gq 5.

and

Add

to

=

P'N,q + U

+

The operation on row q and column q are summarized below.

On Jr = X'qq =

.YV'eq(E9R0")T

+

(Eq R°").Y(qq(Eq

— —

1'

2

I

Oiz 2PN,

q

I?'N,q

=

R0q,

—

1, 2,. . . ,j

(8—18)

+

+0

When ir is symmetrical (this will be the case when the system is geometrically linear), we can work only with the submatrices on and above the diagonal. The contracted operations for, the symmetrical case are threefold:

=

—

it"eq(EqR°")T

€= 1,2,...,q.— 1

(8—19)

CHAP. 8


188

Ttlq*

q—

q

it'qq =

+

+

(8—20)

+ Gq —

*

—

=

(8—21)

The operations outlined above are carried out for each restrained joint. Note that the modifications for joint q involve only row q and column q. We denote the mOdified system of equations by

=

(8—22)

will be Equation (8—22) represents if equations. The coefficient matrix nonsingular when K1 is nonsingular. To show this, we start with the first equation in (8—13) and an additional set of r dummy equations: 1

[K11

Olfuil

-fJ

f—P0,1 —

—

—

-

+

N

Equation (a) represents 1/equations. This system is transformed to (8—22) when to d?tJ, we permute U, They are related by (sec (6—63)) [1°?, J

U

= rVp

where H is a permutation matrix. It follows that

= HT[K11

and, since H is an orthogonal matrix,

=

(8—23)

IK11I

It is more convenient to work with (8—22) rather than (a) since the solution of (8—22) yields the joint displacement matrices listed in their natural order, that is, according to increasing joint number. Once ciii' is known, we convert the joint displacement matrices to the basic frame, using uq =

The bar forces are determined from

F,, = F0,,, + Next, we calculate F,,

and assemble

—

in partitioned form by summing the

THE DiRECT STIFFNESS METHOD

SEC. 8—3.

189

contribution for each member. For member n, we put (see (8—4)) in row n+ in row n_

+ FOIIPf —

Once is known, we convert the force matrix for each partially restrained joint to the local joint reference frame, using

= required to equilibrate the bar forces. This operation The final result is provides a static check on the solution in addition to furnishing the reactions. When the problem is geometrically nonlinear, y,, and depend on the joint displacements. In this case, it is generally more efficient to apply an incremental formulation rather than iterate on (8—22). Example

8—4

We illustrate these operations for the truss shown in Fig. E8—4. Fig. E8—4

/

/50

Ii

1.

Member-Joint Connectivity Table Bar(n)

2.

1

2

3

4

5

+joint(n+)

1

3

1

3

—joint(n...)

2

1

4

2

6

7

8

9

10

11

4

3

5

3

5

6

5

2

4

3

6

4

4

6

Assemblage of

We consider the geometry to be linear. Then, (8—9) results in

listed below.

=

and

=

Applying

DISPLACEMENT METHOD: DEAL TRUSS

190

N

2

1

k1+k2+k3

—k1

—

4

3

—k2

—k3

—k4

—k5

CHAP. 8

51

6

0

k1+k4+k5

2

—k1

3

—k2

---k4

k2+k4+k6 +k7+k8

—k6

—k7

-— k3

4

—k3

—k5

—k6

k3+k5±k6 +k9+k10

—k9

—k10

—k7

—k9

+k7+k9

—k11

5

+k11

0

—

.

—k8

6

—k10

—k11

+k8+k10 +k11

Note that i( 3.

is

symmetrical and quasi-tridiagonal. with submatrices of order 4 x 4.

Introduction of Joint Displacement Restraints

The original equations are =

=

where contains the external joint forces. We start with i?PN If joint q is un— restrained, we put in row q If joint q is fully restrained, we modify and according to (8—15) and (8—16). Finally, if joint q is partially restrained, we use (8—19) through (8—21). Since is symmetrical, we have to list only the submatrices on and above the diagonal. It is convenient to work with successive joint numbers. For this system,

joint 2 is fully restrained and joints 4, 6 are partially restrained. The basic matrices for joints 4,6 and the initial and final forms of.Yt', are listed below. Note that this procedure does not destroy the banding of the stiffness matrix. Joint 4

R°4

(u42 is prescribed)

2

ri

E4=[

[0

01

G4=[0

oj

= {O,ii42j Joint 1

ri E6=[0 — —

6

[

=

(t42 is prescribed) 1

ii ii

01

oJ =

0

INCREMENTAL FORMULATION

SEC. 8—4.

Initial matrices (ir and ("1)

(Us)

(U4)

)r33

ir34

=

ir22 ir4,5

.X44

Sym

Final matrices (ir* and (ui)

(u,)

(U4)

(U3)

o

ir,4E4

1,0

0

.K13

(Us)

U

.K34E4

J35

E4ir46(E6R°6)T

,

—

—

+

+ G4

+

ir56(E6R°6)T

—

(E6R°6)iq66(E6R°6)T

E6R°6(—

--

+ G6

8—4.

INCREMENTAL FORMULATION; CLASSICAL STABILITY CRITERION

Equations (8—13), (8—22) are valid for both linear and nonlinear behavior. However, it is more efficient with respect to computational effort to employ an incremental formulation when the system is nonlinear. With an incremental formulation, one applies the load in increments and determines the corresponding displacements. The total displacement is obtained by summing the displacement increments. An incremental loading procedure can

also be used with (8—13) but, in this case, one is working with total displacement

rather than with incremental displacement. In this section, we develop a set of equations relating the external load and the resulting incremental displacements. These equations are also nonlinear, but if one works with small

load increments, the equations can be linearized. Our approach will be similar to that followed previously. We first establish incremental member force-displacement relations and then apply the direct stiffness method to


192

CHAP. S

the incremental system equations. We complete the section with a discussion of the classical stability criterion. We start with (8—4), which defines the external joint forces required to equilibrate the action of the force for bar n, generate

Pn*

=

=

p,,

Equations (a) are satisfied at an equilibrium position. We suppose an in-

cremental external load AP is applied and define AU as the resulting incremental

displacement for the new equilibrium position. Since F and depend on U, their values will change. Letting AF, AD be the total increments in F, D due to AU, and requiring (a) to be satisfied at both positions, leads to the following incremental force-equilibrium equations: =

Afif +

+

(8—24)

Ap,,.. =

To proceed further, we need to evaluate the increments in e and

D.

The exact

relations are given by (6—22):

= — ci,,

u,,_)

—

+

—

—

—

=

To allow for the possibility of retaining only certain nonlinear terms, we write (a) as fi,, —

= = =

u)Tg,

(u,,÷ —

—

u,,..) +

—

u,,)

(8—25)

Yh(ufl. — u,,_)

If all the nonlinear terms are retained, g,,

=

To neglect a particular displacement component, we delete the corresponding element in For geometrically linear behavior, = 0. Operating on (8—25),

we obtain dv,, =

Au,,)Tg,,

—

(8—26)

and Ae,, =

= d2e,,

+ Au,,)

—

(827)

—

It remains to evaluate AF,,. We allow for a piecewise linear material and employ the relationst developed in Sec. 6—4. For convenience, we drop all the t See (6—31), (6—32), and (6—33).

INCREMENTAL FORMULATION

SEC. 8—4.

193

notation pertaining to a segment and write the "generalized" incremental

expression in the simple form = k(Ae —

(8—28)

where k, are constant for a segment. They have to be changed if the limit is unknown, one of the segment is exceeded or the bar is unloading. Since has to iterate, taking the values of k, Ae0 corresponding to the initial equilibrium position as the first estimate. This is equivalent to using the tangent stiffness. The initial elongation, Ae0, is included to allow for an incremental temperature change. Substituting for (8—28) takes the form

At' _AC' — Finally, we substitute for

+

Q

Lw,, in

LXPn+ =

—

+ if

I

72 Ct

(8—29)

(8—24) and group the terms as follows:

&i,..) +

+

(8—30)

where

= Fag,, + =

(8—31)

=

+

+

-i-

We interpret k7 as the tangent stiffness matrix. The vector, L\p9, contains linear, quadratic, and cubic terms in We have included the subscript g to indicate that it is a nonlinear geometric term. We write the total set of incremental joint equilibrium equations as

+ M'0 +

=

(8-32)

is assembled using (8—9) and MPO + with (8—8). Note that is symmetrical. Finally, we introduce the displacement restraints by ap-

where

plying (8—19)—(8—21). The modified equations are

=

—

(8-33)

—

It is convenient to include the prescribed incremental support displacement terms in involves only the incremental temperature and so that the variable displacement increments. The contracted equations are K1,11 AU1 =

—

AP0,1

—

AP9,1

—

1(1,12

AU2

(8-34)

is symmetrical. We cannot solve (8—33) directly for

where K1,

since contains quadratic and cubic terms in MI. There are a number of techniques for solving nonlinear

algebraic equations. t We describe here the method of successive substitutions, t See Ref. 12.

CHAP. 8


194

which is the easiest to implement, but its convergence rate is slower in com-

parison to most of the other methods. First, we note that and are independent of A'1/1. They depend only on the initial equilibrium position and the incremental loading. We combine and and write (8—33) as X7K MI1 =

(8-35)

—

Now, we let represent the nth estimate for LXa/IJ and determine the (n + 1)th estimate by solving )p* L\cW(n+

=

— &?P

(8—36)

The iteration involves only evaluation of and back-substitution once is transformed to a triangular matrix. The factor method is particularly convenient since X7 is symmetrical. With this method, STS

(8—37)

where S is an upper triangular matrix. We replace (8—36) with

= STQ = A9* S

(8—38) —

In linearized incremental analysis, we delete solution of

in (8—35) and take the

(8-39)

.

as the "actual" displacement increment. One can interpret this scheme as one cycle successive substitution. The solution degenerates when the tangent stiffness matrix becomes singular.

To investigate the behavior in the neighborhood of this point, we apply the classical stability criterion developed in Sec. 7—6. The appropriate form for a truss is given by (7—41):

±

for arbitrary AU1 with AU2 =

den)> 0

0 (a)

We have already evaluated the above terms. Using (8—26), (8—27), and (8—29) with Ae0 = 0,

+

—

—

Au,,)

(b)

and (a) can be written as d2WD

It follows that

ALT ICE,

AU1 > 0

for arbitrary AU1

(8—40)

must be positive definite for a stable equilibrium position.

¶ Iterative techniques are discussed in greater detail in Secs. 18—7, 18—8, 18—9.

_____ iNCREMENTAL FORMULATION

SEC. 8—4.

But K1,

and

are related by HT[Kt.11

(8—41)

where H is a nonsingular permutation matrix, which rearranges the elements of according to (8-42) =H Then, and K1 have the same definiteness Finally, we can classify the stability of an equilibrium position in terms of the determinant of the tangent stiffness matrix: D

Example

= 1K1,

=

D>O

stable neutral unstable

iii

D=0 D< 0

(8—43)

8—5

We illustrate the application of both the total (8—13) and incremental (8—34) formulations

to the truss shown in Fig. E8—5A. To simplify the analysis, we suppose the material is linearly elastic, k1 = k2 = k, and there are no initial elongations or support movement.

b

c•

2.

(b)

Let a, b be square matrices, x a column vector, and f, g scalars defined by

frxTax g = xrbx

(c)

The matrix form of the product, fg, is fg =(xTax)(xTbx) One could expand (d) but it is more convenient to utilize (b) and

write (c)

as

f= g

=

bk(xkxe

fg =

alJbk,xlxfxkx(

= 3.

DIJk(XIXJXkX(

We return to part 1 The inner product of c is a scalar, H,

=

II

xT(aTa)x

Using (b),

H =

c-c1 = 0f50

The outer product is a second-order array, il,

d=

axxraT

ccT

and can be expressed as

=

= alkaf,xkx,.

= AIJk(XkXe

According to the summation convention,

=

d11

+ d22 +

= trace of d

Then, we can write (h) as

H=

d11

=

AIIk(XkX(

4. Let represent square second-order arrays. The inner product is defined as the sum of the products of corresponding elements:

Inner product

ç,) = =

+

+

+

+ g21e21 +

In order to represent this product as a matrix product, we must convert cki,

one-dimensional

(m)

ejj over

to

arrays.

Let represent a one-dimensional set of elements associated with an orthogonal reference frame having directions If the

GOVERNING EQUATIONS FOR A DEFORMABLE SOLID

232

CHAP. 10

corresponding set for a second reference frame is related to the first set by — —

k

=

(10—4)

cos 1, 2, 3

we

say that the elements of b comprise a first-order cartesian tensor. Noting

(5—5), we can write (10—4) as

=

(10—5)

and it follows that the set of orthogonal components of a vector are a first-order cartesian tensor. We know that the magnitude of a vector is invariant. Then, the sum of the squares of the elements of a first-order tensor is invariant. (10—6)

A second-order cartesian tensor is defined as a set of doubly subscripted elements which transform according to =

(10—7)

j. k.

,n.

1.2. 3

An alternate form is

=

(10—8)

The transformation (10—8) is orthogonal and the trace, sum of the principal second-order minors, and the determinant are invariant.t

= fl(2)

where

= = b12 L

021

7

022

+

b22 7

032

b23 7

033

+

b11

b13 1733

In the cases we encounter, b will be symmetrical.

10-3. ANALYSIS OF DEFORMATION; CARTESIAN STRAINS Let P denote an arbitrary point in the undeformed state of a body and the position vector for P with respect to 0, the origin of an orthogonal cartesian reference frame. The corresponding point and position vector in the deformed state are taken as F'; and the movement from P to P' is represented by the

displacement vector, fl. By definition, (10—10)

This notation is shown in Fig. 10—1. f See Prob. 2—5.

SEC. 10—3.

ANALYSIS OF DEFORMATION; CARTESIAN STRAINS

233

Excluding rigid body motion, the displacement from the initial undeformed position will be small for a solid, and it is reasonable to take the initial Cartesian

coordinates (xi) as the independent variables. This is known as the Lagrange

Undeformed

dp

F' (Deformed)

i3 2

112

Fig. 10—1. Geometric notation.

approach. Also, to simplify the derivation, we work with cartesian components for ü. Then, ii

=

We consider a differential line element at P represented by the vector dii. (See Fig. 10—1). The initial length and direction cosines are ds and using the subscript notation for partial differentiation.

=

/

=

We are

(10—12)

Since we are in the deformed state is The corresponding line arid we can write following the Lagrange approach, p =

=

=

(10—13)

The extensional strain, r, is defined as the relative change in length with respect

234


CHAP. 10

to the initial length.t = (1 +

(10— 14)

Using the dot product, (10—14) becomes (1

Finally, we write (a) as

e(1 + 4c) = ap.kejk (10—15) — = One can readily establish that (eJk) is a second-order symmetrical Cartesian tensor4 direction and letting Taking the line element to be initially parallel to the represent the extensional strain, we see that (no sum) = e0 (10—16) -1) = To interpret the off-diagonal terms, we consider 2 initially orthogonal line elements represented by (see Fig. 10—2) and having direction cosines

+

d r',

I

xa

— '/12 p.-

dp'1

x2

Fig. 10—2. Notation for shearing strain. t This is the definition of Lagrangian strain. In the Eulerian approach, the cartesian coordinates for the deformed state are taken as the independent variables, =

and the strain is defined as

=

(1 —

are also called the See Prob. 10—4. It is known as Green's strain tensor. The elements, components of finite strain. They relate the difference between the square of the initial and deformed lengths of the line element, i.e., an alternate definition of Cjk —

ds2 = 2eJkdxJdxk

SEC. 10—3.


235

We define as the angle between the lines in the deformed which is called the shearing strain, follows by state. The expression for taking the dot product of the deformed vectors.

(it

,

COS —

—

=

(1

J=

.

Y12 =

)

Substituting for k)dsf

(sum on k only)

+

and noting that the lines are initially orthogonal,

= (a)

takes the form

(1 +

+

=

(10—17)

shows that

Specializing (10—17) for lines parallel to X,, shearing strain.

(I +

=

+

=

2e13

is related to the (10—18)

Equations (10—15) and (10—17) are actually transformation laws for extensional and shearing strain. The state of strain is completely defined once the strain tensor is specified for a particular set of directions. To generalize these expressions, we consider two orthogonal frames defined by the unit vectors and (see Fig. 10—3), take the initial frame parallel to the global frame = ti), and let = 15 tk. With this notation:

+ (1 +

=

)

(10—19)

=

+

The strain measures (e, y) are small with respect to unity for engineering materials such as metals and concrete. For example, e for steel. Therefore, it is quite reasonable (aside from the fact that it simplifies the expressions) to assume r, y in the strain expressions. The relations for "small" strain are: 1

(10—20)

It remains to expand eJk. Now,

= Differentiating

+ ii

+

u,,Ji,,

with respect to S

OP

=

=

—

+ Un, j)l,n

236

and


substituting into the definition of eJk =

+

k + Uk.

Cl-lAP. 10

(Equation (10—15)) leads to (sum on m only)

4Um, ,u,,,, k

(10—21)

In order to simplify (10—21), we must establish the geometrical significance of the various terms. x3 t3

t2

/

t,3

X2

Fig. 10—3. Unit vectors defining transformation of orthogonal directions.

With this objective, we consider a line element initially parallel to the X1 axis. Figure 10—4 shows the initial and deformed positions, and the angles

which define the rotation of the line toward the X2, K3 directions. The geometrical relations of interest to us are 012,

sin

033 1 +

0j3 =

1421

sine12

(1 + 81)cos 0j3 )2

(1 +

+ uj 1

+

Also, by definition,

+ We solve (a), (b) for u2,

=

=

e11

U1,

1+

i+

i4,

i+

and 03, 03,1 1

(1 + 013

—


SEC. 10—3.

237

and then solve (c) for u1,

= A= 1

(1

+

{1

—

A}112

—

1

(10—23)

sin2 013 + cos2 013 sin2 012

Applying the binomial expansion, (1 — x)"2

= I

—

+

+

(10—24)

we can write (10—23) as

to (1 —

+

+

—

+

—

+

(10—25)

In what follows, we assume small strain and express the derivatives and extensional strain (see Equation (d)) as

1=

u3. 1 = 0(013) U1

a11 =

0(012,

n2

t/12, "13

1—

+

+

u1,

(f)

The various approximate theories are obtained by specializing (f). 'U3

dx1

dx1

1123 dx1

X1,u1

Fig. 10—4. Initial and deformed positions of a line element.

In the linear geometric case, the rotations are neglected with respect to strain. Formally, one sets 012 = 613 = 0 in (f) and the result is a linear relation between

strain and displacement,

-

a11

(g)

Note that, according to this approximation, the deformed orientation coincides

CHAP. 10

GOVERNING EQUATIONS FOR A DEFORMA8LE SOLID

238

with the initial orientation. The general relations for the linear geometric case

(small strain and infinitesimal rotation) are

= =

= =

en

(no sum)

(10—26)

+ ui,,

The next level of approximation is to consider 62 to be of the same order as strain.

02 = sin 0 cos 6

0(s) 0 for arbitrary (E — a°), D and A are positive definite matrices.

There are 21 material constants for a linearly elastic Green-type material. The number of independent constants is reduced if the material structure t See Prob. 10—6, 10—13.


250

CHAP. 10

In what follows, we describe the transition from an anisotropic material to an isotropic material. A material whose structure has three orthogonal axes of symmetry is called orthotropic. The structure of an orthotropic material appears identical after a 1800 rotation about a symmetry axis. To determine the number of independent constants for this case, we suppose X1, X2, X3 are axes of symmetry and consider a 180° rotation about X2. We use a prime superscript to indicate the rotated axes. From Fig. 10—13, exhibits

= —x1

= -x3 = x2 The stress and deformation quantities are related by (we replace 1 by — I and 3 by —3 in the shear terms)

= 1,2.3

= a12 = Y12 =

—a12 Y12

a23

—a23

=

Y23

a13 =

a13

Y13

Y13

Now, the stress-strain relations must be identical in form. We expand e = Acv', and substitute for using (b). Equating the expressions for a'

Fig. 10—13. Rotation of axes for symmetry with respect to the X2-X3 plane.

tA

material whose structure has no symmetry is said to be anisotropic.

ELASTIC STRESS-STRAIN RELATIONS

SEC. 10—5.

and

251

leads to the following relations between the elements of A,

= + a24a12 + a25a23 = a34a12 + a35a23 =

ti15a33

314(T12 —

—a24a12

—

—a34u12

—

a25a23

For (c) to be satisfied, the coefficients must vanish identically. This requires £434 = a15

=

0

a24 =

=

0

£435

a34a350

The symmetry conditions require We consider next the expansions for a46 = a56 = 0. By rotating 1800 about X1, we find = a36 = a45 0 a16 = A rotation about the X3 axis will not result in any additional conditions. Finally, when the strains are referred to the structural symmetry axes, the stress-strain relations for an orthotropic material reduce to a11

a12

0j3

a12

a22

a23

a1,

0 -

— £444 -—

0

0

0

0

Y31

0

0

a12

0

a23

a66

(733

(10-70)

We see that A is quasi-diagonal and involves 9 independent constants. There is no interaction between extension and shear. Also, the shearing effect is uncoupled, i.e., cr12 leads only to An alternate form of the orthotropic stress-strain relations is

AT +

a1 =

—

V32

1

—-----a33

E2

=

/13

1

AT +

(10—71) a33

1

Y12 =

!(733

—

Y23

=

—

— —i—-

1

1

Y31

=

where E4 are extensional moduli,

are shear moduli, Vjk are coupling coefficients, and AT is the temperature increment. The coupling terms are related by E2

E1

E3

E1

E3

E2

(10-72)


252

CHAP. 10

It is relatively straightforward to invert these relations:t One should note that (10—71) apply only when X, coincide with the material symmetry directions4

If the stress-strain relations are invariant for arbitrary directions in a plane, the material is said to be transversely orthotropic or isotropic with respect to the plane. We consider the case where the X1 direction is the preferred direction, i.e., where the material is isotropic with respect to the X2-X3 plane. By definition, A is invariant when we transform from X1-X2-X3 to This c, '—'12'-'31'--'

v32

v23

v

2(1 + v)

1

--

F and

the relations reduce to

AT +

=

(a22

—

= PAT +

+

= pAT +

— va27)

1

2(l+v)

1

Yi2

(10—74)

—

—

y23

Y31

There are five independent constants (F, v, E1, v1, G1).

Lastly, the material is called isotropic when the stress-strain relations are Invariant for arbitrary directions, For this case, A = A' for arbitrary The relations are obtained by specializing (10—74): = p AT +

(at,

+ akk))

—

(10—75)

2(1 + v) F

Note that now there are only two independent constants (F, v). The coupling coefficient, v,is called Poisson's ratio. The inverted form of (10—75) is written as

a= a0

= =

a0

+

+

IO--21.

Prob.

10—22.

+ (10—76)

+ 2G)pAT

t See Prob. 10—19 for the inverted form of (10—7 1). § See

+

SEC. 10—6.

where

PRINCIPLE OF VIRTUAL DISPLACEMENTS

253

G are called Lamé constants and are related to E, v by

G=

shear

modulus =

E 2(1

+ v)

yE A

— (1

(10—77)

+ v)(1 — 2v)

Since D must be positive definite, v is restricted to — 1 < v < 1/2. The limiting case where v = + 1/2 is discussed in Problem 10—24.

10—6.

PRINCIPLE OF VIRTUAL DISPLACEMENTS; PRINCIPLE OF. STATIONARY POTENTIAL ENERGY; CLASSICAL STABILITY CRITERIA

Chapter 7 dealt with variational principles for an ideal truss. For completeness, we derive here the 3-dimensional form of the principle of virtual

displacements, principle of stationary potential energy, and the classical stability criterion. The principle of virtual forces and stationary complementary energy are treated in the next section.

The principle of virtual displacements states that the Iirst-order work done is equal to the first oidcr work done by the internal forces acting on the restraints for an arbitrary virtual displacement of the body from an equilibrium position. f In the continuous case, the external loads and the internal forces are loading consists of body (b) and surface represented by the stress vectors. We follow the Lagrange approach, i.e., we work with Lagrange finite strain components (eJ,j, Kirchhoff stresses and external force measures per unit initial volume or area (b*, p*). This is consistent with our derivation of the equilibrium equations. Let Au denote the virtual displacement. The firstorder external work is by the external forces

= =

dx1 dx2 dx3 + JJj3* Au dx1 dx2 dx3 ± dfI

10—78

where fI is the initial surface area. The total internal deformation work is obtained by summing the first-order work done by the stress vectors acting on a differential volume element. *

= =

dx2dx3 dx1 dx2 dx3

(10—79)

Equating (a) and (b), we obtain the 3-dimensional form of the principle of See Sec. 7—2.

See Fig. 10—12. §

See (10—60).


254

CHAP. 10

virtual displacements, 5WD

=

dx1 dx2 dx3 = fJJh*

dx1 dx2 dx3 +

dx1 dx2 dx3 =

(10—80)

dx1 dx2 dx3 +

Requiring (10—80) to be satisfied for arbitrary (continuous) is equivalent to enforcing the equilibrium equations. To show this, we work with the vector form and utilize the following integration by parts formula: t

=

J

—

dx2dx3

(10-81)

is the direction cosine for the initial outward normal (n) with respect direction. Operating on the left-hand term and equating coefficients

where

to the

in the volume and surface integrals leads directly to (10—50) and (10—54).

The principle of virtual displacements applies for arbitrary loading (static or dynamic) and material behavior. When the behavior is elastic and the loading

is independent of time, it can be interpreted as a variational principle for the displacements. The essential steps required for the truss formulation are described in Sec. 7—4. Their extension to a continuous body is straightforward. When the behavior is elastic, = Letting

VT

denote the total strain energy, the left-hand side of (10—80) reduces to

fJJ öVdx1 dx2 dx3 =

dx1 dx2 dx3

We consider the surface area to consist of 2 zones as shown in Fig. 10—14.

+ where displacements are prescribed on on cd

U1

(10—82)

and surface force intensities arc prescribed on pni

Pni

on

The displacement variation, L\u1, is admissible if it is continuous and satisfies

=

0

on

(10—83)

We also consider the surface and body forces to be independent of the displacements. With these definitions, the principle of virtual displacements is transt See Prob. 10—25.


SEC. 10—6.

255

formed to

= fl,, =

for arbitrary admissible

0

cIx1 dx2 dx3

VT

(1084)

—

is the total potential energy functional. According to (10—84), the displacements defining an equilibrium position correspond to a stationary value of the total potential energy functional. Note that this result applies for arbitrary strain and finite rotations. The only restrictions are elastic behavior where

and static loading.

PH

Fig. 10—14. Classification of boundary zones.

Example 10—2 Direct methods of variational calculus such as Rayleigh-Ritz, Galerkin, weighted residuals, and others are applied to fl,, to determine approximate solutions for the displacements. In the Rayleigh-Ritz method, one expresses the displacements in terms of unknown parameters, q, and prescribed functions, x2, x3), U1

+

=

where

=

0

forj =

>

1.

2

The displacement boundary conditions on fd are called "essential" boundary conditions. to a function of the q's. When the material is linearly Substituting for transforms elastic, V is a quadratic function of the strains. Then, V will involve up to fourth-degree reduces terms for the geometrically nonlinear case. If the behavior is completely linear,

H, = q =

Const.

+ qTQ + . .

.

.

.

.

K is symmetrical

(3N x 1)

256


Finally, requiring

CHAP. 10

to be stationary for arbitrary c5q leads (for linear behavior) to

Kq = Q The strains are evaluated by operating on (a) and the stresses are determined from the

stress-strain relations.

Polynomials and trigonometric functions are generally used to construct the spatial distribution functions. The mathematical basis for direct methods is treated in numerous texts (see Refs. 9, 10).

The "classical" stability criterion for a stable equilibrium position ist —

o2WE

>

0

for arbitrary Ad

is the second-order work done by the external forces where = during the incremental displacement, Ad, and WD = ó(ö WD) is the second-

order work done by the internal forces acting on the restraints during the incremental deformation resulting from Ad. The form of the work terms for a continuous body are obtained by operating on (10—78) and = Sfl = = =

Ad dx1 dx2 dx3 + j(

Ad

Au1 dx1 dx2 dx3 + J$ Au, dx2 dx3 ie11 + dx2 dx3

(10—85)

If = Ô2WE for a particular Ad, the equilibrium position is neutral. The position is unstable if ö2 WD < o2 Note that öb, are itull vectors when the forces are prescribed.

For elastic behavior, the incremental deformation work is equal to the increment in strain energy

=

=

and (10—84) can be written as

>

0

for arbitrary Ad

(10—86)

It follows that a stable equilibrium position corresponds to a relative minimum

value of the total potential energy. Bifurcation (neutral equilibrium) occurs when = 0 for some Ad, say Ada. If the loading is prescribed, and ö2VT = 0 at bifurcation. The governing equations for bifurcation can be obtained by expanding This involves transforming the integrand of ö2WD = by applying (10—81). Since bifurcation corresponds to the existence of an alternate equilibrium position, it is more convenient to form the incremental equations directly. The equations for the case of linearly elastic material and prescribed external forces are listed below. f See Sec. 7—6 for a derivation of the classical stability criterion. See Probs. 10—11, 10—18.

PRINCIPLE OF VIRTUAL FORCES

SEC. 10—7.

L

Equilibrium Equation in the Interior

+ 2.

=

+

0

= 1,2,3

Stress-Boundary Force Equations on

+ 3.

257

+

Au1

0

J = 1, 2, 3

(10—87)

Stress-Strain Relations

= D 4.

Strain-Displacement Relations

=

3 + AUJ, +

=

Au1

10—7.

AIIm, j + Urn, j Am,

0

PRINCIPLE OF VIRTUAL FORCES; PRINCIPLE OF STATIONARY COMPLEMENTARY ENERGY

Let u1 be the actual displacements in a body due to some loading and the geometrically linear strain measures corresponding to u1. The strain and displacement measures are related by

=

+

u1=fl Once the strains are known, we can find the displacements by solving (a) and enforcing (b). The principle of virtual forces is basically a procedure for determining displacements without having to operate on (a). It applies only for linear geometry. We developed its form for an ideal truss in Sec. 7—3. We will follow the same approach here to establish the three-dimensional form. The essential step involves selecting a statically permissible force system, i.e., a force system which satisfies the linear equilibrium equations. For the continuous case, the force system consists of stresses, surface forces, on Static permissibility requires and reactions, on Ac31,3 = 0

on on

= =

(10—88)

If we multiply e13 by Ac13, integrate over the volume using (10—81), and note the static relations, we obtain the following identity,t Acr13 dx1 dx2 dx3 f See Prob. 10—26.

=

u1

0,,

+ $ Th

(10—89)

258


CHAP. 10

which is referred to as the principle of virtual forces (or stresses). This result is

applicable for arbitrary material behavior. 1-lowever, the geometry must be linear.

in the direction defined by is Suppose the translation at a point Q on desired (see Fig. 10—15). Let d0 be the displacement. We apply a unit force at Q in the tq direction and generate a statically permissible stress field. (1) 1q at

The integral on

point Q

and

Acr

reduces to (l)dQ, and it follows that dx1 dx2 dx1 —

=

(10—90)

A second application is in the force method, where one reduces the governing

equations (stress equilibrium and stress displacement) to a set of equations

Fig. 10—15. Notation for determination of the translation at point Q.

involving only force unknowns. We start by expressing the stress field in terms

of a prescribed distribution

and a "corrective" field + cit,

(10—91)

is a particular solution of the equilibrium equations which satisfies the boundary conditions on where

+

0

= and

Thu

on

(10-92)

satisfies

= =

0 0

on

(10—93)

on

Stress fields satisfying (10—93) are called seljequilibrating stress fields. For the

ideal truss, a-° corresponds to the forces in the primary structure due to the prescribed loading and ? represents the contribution of the force redundants.


SEC. 10—7.

259

The governing equations for the force redundants were obtained by enforcing

geometric compatibility, i.e., the bar elongations are constrained by the requirement that the deformed bar lengths fit in the assembled structure.

Geometric compatibility for a continuum requires the strains to lead to continuous displacements. One can establish the strain compatibility equations by operating on the strain- displacement relations. This approach is described

in Prob. 10—10. One can also obtain these equations with the principle of virtual forces by taking a self-equilibrating force system. Letting Aox, Apc denote the virtual stress system, (10--89) reduces to

dx1 dx2 dx3 =

(10—94)

The compatibility equations are determined by expressing in terms of stress functions and integrating the left-hand term by parts. We illustrate its application to the plane stress problem.

Example 10—3 If the stress components associated with the normal direction to a plane are zero, the stress state is called planar. We consider the case where

=

= 033 = 0. The

equilibrium equations and stress-boundary force relations reduce to + b1 = 0 2 + b2 = 0

+ 012 5 +

=

+

a,,2o21

+

•z,,2a22

The stress field, oi,, must satisfy (a) with h1 = h2 = 0 and also p,,1 = = 0 on We can satisfy the equilibrium equations by expressing in terms of a function, follows:t 033 = = = The boundary forces corresponding to

are

=

Pa

OS

where s is the arc length on the boundary (sense is from X1 —* X2).

Substituting for crC,

pC

in terms of i/i, (10-94) expands to + a2

if —

f

Ps

CS

There is no loss in generality by taking 22

f See Prob. 10—14.

11 — y52

+ a2,

=

0

— 712,

12)dx1

0 ,i

on S. Then, integrating (e) by parts, dx1 dx2

0

as

260


CHAP. 10

results in the strain compatibility equation,

and requiring (f) to be satisfied for arbitrary

+

0

—

which is actually a continuity requirement U1 122

We express (g) in terms of

+

211 — (u1 212 + 112 112) = 0

by substituting for the strains in terms of the stresses.t

The principle of virtual forces is also employed to generate approximate solutions for the stresses. It is convenient to shift over to matrix notation for this discussion, and we write (10—94) as dx1 dx2 dx3

ApC

if

We express the stress matrix in terms of prescribed stress states and unknown parameters, a1,

where

= =

+ +

+ (12(l)2

satisfies (10—92) and

(1

+

+ 04,,

'

1, 2,...,r) are self-equilibrating stress

states, i.e., They satisfy the homogenous equilibrium equations and boundary conditions on The corresponding surface forces arc p

p°

+

= p° + 0101 + 0209 + = p (i = 1, 2, = 0

(i =

Taking virtual-force systems corresponding to equations for the parameters. dx1 dx2 dx3 =

jjT9.

1

1, 2,

., r) results in r

1, 2,...,r

(10—97)

In order to proceed, we need to introduce the material properties. When the material is linearly elastic,

+ Ai'=

+

+

and the equations expand to

=

d1

f,j = d, =

i,j =

1,

2,....r

dx1 dx2 dx3 —

r$J1T(a°

(10—98)

+ A6°)dx1 dx2 dx3

One should note that (10—97) are weighted compatibility conditions. The true stresses must satisfy both equilibrium and compatibility throughout the t See Prob. 10—27.


SEC. 10—7.

261

the corrective stress field since it is required to correct the compatibility error due to For completeness, we describe here how one establishes a variational principle for Our starting point is (10—94) restricted to elastic behavior. We define = according to = c5V* = (10—99) domain. We call

and call V* the complementary energy density. The form of V* for a linearly elastic material is = (10—100) + By definition, V* complements V, i.e.,

V+

=

(10—101)

Then, letting

=

cjx1 dx2 dx3

(10—102)

we can write (10—94) as 0

(511. *

TiC

—

for arbitrary = $$

(10—103)

This form is called the principle of stationary complementary energy and shows

that the true stresses correspond to a stationary value of Since is linear in the second variation of reduces to (52fl

=

=

dx1 dx2 dx3

(10—104)

We shift over to matrix notation and express öe as

=

A,

(10— 105)

represents the tangent compliance matrix. Now, must be positive definite in order for the material to be stable.t Then, (5211. > 0 for arbitrary and we see that the solution actually corresponds to a relative minimum where

value of The approximate method described earlier can be applied to 11g. Substituting for given by (10—95) converts to a function of the stress parameters When the material is linearly elastic, (a1. a2, . , ar). .

.

H, =

—

ard + const

(10—106)

The equations for the stress parameters follow by requiring H. to be stationary for arbitrary (511, = — ci) = 0

fa=d A,

(10—107)

The classical stability criterion specialized for elastic material and linear geometry requires SCTD, & > 0 for arbitrary Sc which, in turn, requires D, to be positive definite. Since D1', it follows that A must be positive definite for a stable material.

262


CHAP. 10

Operating on

c52fl = LtaTfLui

(10—108)

and noting that ö211. > 0, we conclude that f is positive definite. REFERENCES I.

CRANDALL, S. J., and N. C. DAHL: An Introduction to the Mechanics of Solids, McGraw-Hill, New York, 1959.

2.

BISPLINGHOFF, R. L., MAR., J. W., and T. H. H. PlAN:

3. 4.

5. 6. 7. 8. 9. 10.

of Deformable Solids. Addison-Wesley, Reading, Mass., 1965. WANG, C. T.: Applied Elasticity, McGraw-I-jill, New York, 1953. TIMOSHENKO, S. J., and J. N. GooDiag: Theory of Elasticity, 3d ed., McGraw-Hill, New York, 1970. SOKOLNIKOFF, I. S.: Mathematical Theory of Elasticity, 2d ed. • McGraw-Hill. New York, 1956. FUNG, Y. C.: Foundations of Solid Mechanics, Prentice-Hall, 1965. LEKIINITSKU, S. G. Theory of Elasticity of an Anisotropic Elastic Body, Holden-Day, San Francisco, 1963. WAsmzu, K. Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968. HLDEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, 1965. CRANDALL, S. J.: Engineering Analysis, McGraw-Hill. New York, 1956.

PROBLEMS 10—1.

Write out the expanded form of the following products. Consider

the repeated indices to range from 1 to 2. (a) (b)

+ u1, ± Urn,

10—2.

where

=

+ Urn. k) —

Let f be a continuous function of x1, x2, x3. Establish the trans-

formation laws for and (3Xk. 10—3. Establish the transformation law for tensors. 10—4. Prove that eJk =

Jbk where

are cartesian

— ôJk)

is a second-order cartesian tensor. Hint: Expand (3/3

P.

(If)

P.

10—5. Equations (10—19) are the strain transformation laws. Since is a symmetrical second-order cartesian tensor, there exists a particular set of directions, say Xi', for which is a diagonal array. What are the strain components for the frame? Consider a rectangular parallelepiped having sides dXy in the undeformed state. What is its deformed shape and relative change with respect to its initial volume? Specialize the expression for in volume, for small strain. Then determine for the initial (Xi) directions and small strain. Finally, show that r.., is invariant.

PROBLEMS

263

10—6.

(a)

Specialize (10—19) for small strain and write out the expressions for

(b)

Let

in terms of ei, 62, .

Develop the form of (c)

.

P13•

P12, P23, y31}. We can express the strain trans= formation (small strain) as = using the results of part a.

Evaluate TE in terms of cos 0, sin 0 for the rotation shown below. Comment on the transformation law for the out-of-plane shear strains P32. Prob. 10—6

x2

10—7.

Tn the Eulerian approach, the cartesian coordinates

for the

deformed state are taken to be the independent variables, i.e., =

Ui

Xj(f/k)

Almansi's strain tensor is defined as — (ds)2

=

2Efk

thik

Determine the expression for EJk in terms of the displacements. Compare the result with (10—21). 10—8.

Consider the case of two-dimensional deformation in the X1-X2 be the extensions in the a, b, c direc-

plane (83 = P13 = P23 = 0). Let 6b, tions defined below and let 6N = {8a, 6b,

We can write

= BE C= (a) (b)

Determine the general form of B. Determine for = 0, 9b Determine B1 for Oa = 0, 6h = 60°,

= 90c. = 120°. (d) Extend (a) to the three-dimensional case. Consider six directions having direction cosines GJ2, with respect to X1, X2, X3. Can we select the six directions arbitrarily? (c)


264

CHAP. 10 Prob. 10—8

xz

-a

For small strain, the volumetric strain is

10—9.

= Rather than work with

+

+

C3

= eti + &22 +

one can express it as the sum of two tensors,

=

+

is called the spherical strain tensor,

where

=

is the deviator strain tensor. and (a) Write out the expanded form for (b) Determine the first invariant of of ejj.

and and compare with the invariant

This question concerns strain compatibility equations. Show that

10—10.

(a)

+

=

+

CX,, cXj, — 8X,,,

(?X,,

where eflk =

ek,

=

1

(CII,,

+

—

2 \CXk

and k, in, n range from I to 3. This expression leads to six independent conditions, called geometric compatibility relations, on the strain measures. (b)

Show that for two-dimensional deformation in the X1-X2 plane = 0; this called plane strain) there is only one com= 813 = patibility equation, and it has the following form: 22 + 83 11 = Is

Y12. 12

the following strain state permissible?

=

+

82 = kx2 Y12 = 2kx1x2 k

= constant

PROBLEMS 10—11.

265

Equation (10—21) defines the strain measures due to displacements,

To analyze geometrically nonlinear behavior, one can employ an incremental formulation. Let represent the displacement increment and Ae1k the incremental strain. We write + = where

contains linear terms (Aug) and öeJk involves quadratic terms. The 5-symbol denotes the first-order change in a functional and is called the variational operator (see Ref. 8). We refer to 5e as the first variation of e. Determine the expressions for 10—12.

Let

i,, be the unit vector defining the initial orientation of the

differential line element d1,, at a point.

=

dsi,,

1,

=

The unit vector defining the orientation in the deformed state is = (1 +

= Determine the general expression for Then specialize it for small strain. 10—13. The several parts of this question concerns stress transformation. in terms of (a) Starting with (i0--41). write out the expressions for all, a22, . , = stress matrix. We express the (b) Let a22, a33, ai2, = stress transformation as a matrix product. .

a' = T,a Develop the form of T,, using the results of part a. (c)

Evaluate 1',, in terms of cos (9, sin (9 for the axes shown. Prob. 1O—13

x2

x,t

xl (d)

Plane stress refers to the case where a13 = with reduced stress and strain matrices,

a23

=

a33

=

0.

We work

{a11, a22,

Er

= and write the transformations in the same form as the three-dimensional case:

a' = a'

=

CHAP. 10


266

Evaluate T,. from part c above and T, from

Prob. 10—6. Verify that

13

10—14. This question develops a procedure for generating self-equilibrating stress fields. (a) Expand the linear equilibrium equations, (10—49) and (10—50). (b) Specialize the equilibrium equations for plane stress (a13 = a23 =

= 0). Suppose we express the two-dimensional stress components in terms of a function = as follows: a33

(c)

a11 = t1'.22 a22 = tI'. ii

—

a12

—

b1

dx1

fx7 b2

dx2

= —1//,12

The notation for body and surface forces is defined in the following sketch. Prob.

10—14

x2

x1

Verify that this definition satisfies the equilibrium equations in the interior. Show that the expressions for and P2 ifl terms of derivatives

with respect to x1, x2, and s are Pi

T t/"1

=

P2

10—15.

b1

—

dx1 b2

—

dx2

The mean stress, a,,,, is defined as am

Rather than work with

=

we

+

a22

+ a33)

can express it as the sum of two tensors, — —

L

aU

PROBLEMS

267

is called the spherical stress tensor,

where

=

óijOrn

and

is the deviator stress tensor. Write out the expanded forms for and Determine the first invariant of 10-46. Establish the stress-equilibrium equations for small-finite rotation and small strain. (a) (b)

10—17.

Starting with (10—52), (10—55) specialized for small strain, establish

Au, Ab*, and the incremental equilibrium equations in terms of Group according to linear and quadratic terms. Specialize these equations

for the case where the initial position is geometrically linear, i.e., where approximate with in the incremental equations. 10—18. Prove (10—60). Hint:

= /= 10—19.

can

+ P.k

Verify that the inverted form of(l0—71) is D(e —

where D11 = E1/C3

D12

D13

C4D11

D22 = E2/C1 + (C2/C1)D12 D23 = v32E2/C1 + (C2/C1)D13 D31 = E3 + v31D13 + v32D23

and

C1 = C2 =

1

—

v21

+ v31v32(E2/E3)

E2C1 C4 =

v31

+

"32

= 0) = = Consider 2 sets of orthogonal directions defined by the unit vectors The stress-strain relations for the two frames are

Specialize for plane strain 10—20.

and

=

+ (a°)' + A'&

Express A' in terms of A and Also determine D'. 1O--21. Consider the three-dimensional stress-strain relations defined by (10—71).

(a)

Specialize for plane stress

=

=

= 0).

GOVERMNG EQUATIONS FOR A DEFORMABLE SOUD

268

CHAP. 10

Let

(b)

a22, cri2}

C= C=

62, Y12}

Verify that D has the following form: V2t

0

I

1 G (1 —

n

=

E2

Assuming X1-X2 in the sketch are material symmetry directions, determine D' for the X'1-X'2 frame. Use the results of Prob. 10—13, 10—20. What relations between the properties are required in order for D' to be identical to D? Prob. 10—21

x2

xI 10—22. Verify (10—73). Start by requiring equal properties for the X2 and X3 directions. Then introduce a rotation about the X1 axis and consider the Isotropy in the X2-X3 plane requires expression for Y23

=7

I.

023

10—23. Verify that the directions of principal stress and strain coincide for an isotropic material. Is this also true for an orthotropic material? 10—24.

Equations (10—76) can be written as

+

a11 = a°&1 +

where

2Ge11

is the volumetric strain. Using the notation introduced in Probs. 10—9

and 10—15——

(a)

Show that

=

Ka,,

+ a0

PROBLEMS

where K is the bulk modulus = (E/3(1 (b)

269 —

2v)). Discuss the case where

Show that

= (c)

Verify that the strain-energy density can be written as V

—

= = (d)

—

+

+

for the isotropic case. Determine and When v = We must work with 7 stress measures ('u' Urn) = and the mean stress has to be determined from an equilibrium consideration. Summarize the governing equations for the incompressible case.

Prove (l0--81) for the two-dimensional case. Is this formula restricted to a specific direction of integration on the boundary? Does it apply for a multi-connected region, such as shown in the figure below? .10—25.

Prob.

10—26.

10—25

Verify Equation (10—89).

Refer to Example 10—3. Express (g) in terms of material to be orthotropic. 10—28. Verify that the stationary requirement 10—27.

=0

Consider the

for arbitrary

where

=

—

—

dx2

— —

= Kirchhoff stress = Lagrange strain =

—

+

+

1u,,,,

= complementary energy density (initial volume) = prescribed force measures (initial dimensions) leads to the complete set of, governing equations for an elastic solid, i.e., stress equilibrium equations 1. 2. stress-displacement relations 3. stress boundary conditions on 4. displacement boundary conditions on 5. expressions for the reaction surface forces on


270

CHAP. 10

This variational statement is called Reissner's principle (see Ref. 8).

Transform HR to by requiring the stresses to satisfy the stress displacement relations. Hint: Note (10—101). (b) Transform 11R to — by restricting the geometry to be linear = and (ui, + and requiring the stresses to satisfy the stress equilibrium equations and stress boundary conditions on Hint: Integrate by parts, using (10—8 1). (a)

10—29.

Interpret (10—90) as dQ ==

where PQ is a force applied at Q in the direction of the displacement measure, dQ.

11

St.

Venant Theory of of

Prismatic Members 11—1.

INTRODUCTION AND NOTATION

A body whose cross-sectional dimensions are small in comparison with its axial dimension is called a member. If the centroidal axis is straight and the the member shape and orientation of the normal cross section are is said to be prismatic. We define the member geometry with respect to a global reference frame (X1, X2, X3), as shown in Fig. 11—1. The X1 axis is taken to coincide with the centroidal axis and X2, X3 are taken as the principal inertia directions. We employ the following notation for the cross-sectional properties: A = if dx2 dx3 = dA 12 — Sj(x3)2 dA

13 = fl(x2)2 dA

Since X2, X3 pass through the centroid and are principal inertia directions, the centroidal coordinates and product of inertia vanish:

'23

jJx2x3 dA = 0

One can work with an arbitrary orientation of the reference axes, but this will

complicate the derivation. St. Venant's theory of torsion-flexure is restricted to linear behavior. It is an

exact linear formulation for a prismatic member subjected to a prescribed t The case where the cross-sectiona' shape is constant but the orientation varies along the centroidal axis is treated in Chapter 15. 271

TORSION-FLEXURE OF PRISMATIC MEMBERS

272

CHAP. it

distribution of surface forces applied on the end cross sections. Later, in

Chapter 13, we modify the St. Venant theory to account for displacement restraint at the ends and for geometric nonlinearity. x2

F3

Fig. 11—i. Notation for prismatic member.

The distribution of surface forces on a cross section is specified in terms of its statically equivalent force system at the centroid. Figure 11—1 shows the Stress components on a positive face. We define M.. as the force and moment vectors acting at the centroid which are statically equivalent to the distribution of stresses over the section. The components of F.., M÷ are called stress resultants and stress couples, respectively, and their definition equations are

F1 = ffcrij c/A M1 = M2 = M3 =

F2

c/A

F3 =

JJ(x2cr13 — x3c12)dA

JJx3crj1 dA

JJcc13 c/A

(11—3)

dA

The internal force and moment vectors acting on the negative face are denoted

byF_,M_. Since

F_ =

—F÷

M_ =

(11—4)

it follows that the positive sense of the stress resultants and couples for the negative face is opposite to that shown in Fig. 11 —1. We discuss next the pure-torsion case, i.e., where the end forces are statically equivalent to only M1. We then extend the formulation to account for flexure

SEC. 11—2.

THE PURE-TORSION PROBLEM

273

and treat torsional-flexural coupling. Finally, we describe an approximate procedure for determining the flexural shear stress distribution in thin-walled

sections. 11—2.


Consider the prismatic member shown in Fig. 11—2. There are no boundary forces acting on the cylindrical surface. The boundary forces acting on the end

cross sections arc statically equivalent to just a twisting moment M1. Also, there is no restraint with respect to axial (out-of-plane) displacement at the ends.

The analysis of this member presents the pure-torsion problem. In what follows, we establish the governing equations for pure torsion, using the approach originally suggested by St. Venant.

0

Fig. 11—2. Prismatic member in pure torsion.

Rather than attempt to solve the three-dimensional problem directly, we impose the following conditions on the behavior and then determine what problem these conditions correspond to. 1.

2.

Each cross section is rigid with respect to deformation in its plane, i.e., = 723 = 0. = Each cross section experiences a rotation w5 about the X3 axist and an out-of-plane displacement u1.

These conditions lead to the following expansions for the in-place displacements: 112 = —C01X3 03

(11—5)

+W3.3.2

The corresponding linear strains are 13 = a3 = Li —Ui 1

U12 +

712 713

Y23

=

U5,

+ U3,

0

012 —

=

(11—6)

05,3 + X20)1, 1

t Problem 11—i treats the general case where the cross section rotates about an arbitrary point.

274


CHAP. 11

Now, the strains must be independent of x1 since each cross section subjected to the same moment. This requires = const = k1 = u1(x2, x3)

is

We consider the left end to be fixed with respect to rotation and express co1,u1aS

=

k1x1

(11—8)

U.1 =

x3) defines the out-of-plane displacement (warping) of a cross = section. The strains and stresses corresponding to this postulated displacement behavior are = 0 = C3 Cl =

where

712

=

x3)

+ x2)

713

and

a11 =

a12 =

Gy12

U13 =

Go'13

=

a22

=

a33

O'23 = 0

=

2 — x3)

+ x2)

a12(x2, x3)

(11—10)

a1 3(x2, x3)

We are assuming that the material is and there are no initial strains. One step remains, namely, to satisfy the stress-equilibrium equations and stress boundary conditions on the cylindrical surface. The complete system of linear stress-equilibrium equations, (10—49), reduces to

=

U21,2 +

0

(11—11)

Substituting for the shearing stresses and noting that Gk1 is constant lead to the differential equation (11-12)

which must be satisfied at all points in the cross section. The exterior normal n for the cylindrical surface is perpendicular to the X1 direction. Then 0, and the stress boundary conditions, (10—49), reduce to Using (11—10), the boundary condition for 2 — x3)

=0

+

Pfli =

+

(11—13)

is

+ x2) =

0

(11—14) —

t Problem 11 —3 treats the orthotropic case.

(on S)


SEC. 11—2.

275

The pure-torsion problem involves solving V2q5, = 0 subject to (11—14). Once

çb, is known, we determine the distribution of transverse shearing stresses from (11—10). Note that depends oniy on the shape of the cross section.

The shearing stress distribution must lead to no shearing stress resultants:

dA =

F2 =

F3 = J$a13 dA

This requires

0

0

JJ ('X3

J J OX2

To proceed further, we need certain integration formulas. We start with dA =

if

(IS

dA leads to

which is just a special case of (10—81). Applying (11—15) to

Green's theorem, JJV2VJdA

0X2

.1

If

11—16

ôn

=

is a harmonic function (i.e.,

+

0),

Green's theorem requires

dS =

0

Now, /, is a harmonic function. For the formulation to he consistent, (11—14) must satisfy (c). Usiiig (11—15), (c) transforms to

#(XH2x3

=

—

0

is specified on the boundary, we cannot apply (11—15) directly Since to (b). In this case, we use the fact that = 0 and write cxi

ax2

ax2j

ox3 \

ax31

(j=2,3)

Integrating (e),

(j=2,3) and then substituting for the normal derivative, verifies (h). The constant k1 is determined from the remaining boundary condition,

=

J$(x2c13

—

x3c12)dA

(11—17)


276

CHAP. 11

We substitute for the shearing stresses and write the result as

Gk1J where J is a cross-sectional property,

+

if

— X3

dA

+ = At this point, we summarize the results for the pure-torsion problem. 1.

Displacements

= 02

U3 = W1X2

= k1x1

=

k1

2.

(if

Stresses M3

J

\(;X2

A'11

+ X2

0j3 = —H J 3.

(11—20)

Governing Equations

mA: on S:

—

It is possible to obtain the exact solution for for simple cross sections. The procedure outlined above is basically a displacement method. One can also use a force approach for this problem. We start by expressing the shearing stresses in terms of a stress function so that the stress-equilibrium equation (Equation 11—li) is identically satisfied. An appropriate definition is 012

X3

(11—21)

013

The shearing stresses for the 2,

v

directions, shown in Fig. 11—3, follow directly

from the definition equation 01A

cv

0lv = —

(11—22) CA.

SEC. 11-2.


277

Taking S 900 counterclockwise from the exterior normal direction, and noting

that the stress boundary condition is

=

0,

lead to the boundary condition

fort/i,

= const on S

(11—23)

We establish the differential equation for t/i by requiring the warping function be continuous. First, we equate the expressions for a in terms of t/i and M1

a12 =

—

a13 =

x3)

+ x2)

=

Now, for continuity,

= Operating on (a), we obtain

= It is convenient to express t/.'

as

(11—24)

aret

The governing equations in terms of

a12 = =

M1 dt/ (11--25)

j

a 13

and

=

(mA)

—2

(on boundary S1)

tJi =

(11—26)

Substituting (11—25) in the definition equation forM1 leads to the following expression for J:

cr7 —

JJ \.

CX3J

that

Applying (10—8 1) to (a) and —

dS =

A1

=

area enclosed by the interior boundary curve, S1

=

C1

(b)

= const

t Equations (11—26) can be interpreted as the governing equations for an initially stretched membrane subjected to normal pressure. This interpretation is called the "membrane See Ref. 3.

The S direction is always taken such that n — S has the same sense as X2 — X3. Then, the + S direction for an interior boundary is opposite to the + S direction for an exterior boundary since the direction for n is reversed. This is the reason for the negative sign on the boundary integral.

278

we can write


J=

dA +

CHAP. 11

(11—27)

= 0 on the exterior boundary. To determine the constants C, for the multiply connected case, we use the is continuous. This requires fact that where

Js ('IS

dS

(11—28)

0

for an arbitrary closed curve in the cross section,

x

x

x2

0

Fig. 11—3. Definition of n-s and A.-vdirections. x3

Fig. 11—4. Graphical representation of sector area.


SEC. 11—2.

279

Consider the closed curve shown in Fig. 11 —4. The shearing strain

is

given by

Yis =

ct52y12

+ 0t53y13

Using (11—9), we can write (a) as

2+

Yis =

=

k1

3 — xacls2

+ (11-29)

+

where p is the projection of the radius vector on the outward normal.t The magnitude of p is equal to the perpendicular distance from the origin to the tangent. Integrating between points P, Q, we obtain = where APQ

=

r50 p

J

dS =

+ 2APQ)

—

(11—30)

area enclosed by the arc PQ and the radius vectors to P and Q. sector

Finally, taking P = 5dS = 2k1A5

(11—31)

where A5 denotes the area enclosed by the curve. Since

=

we can

write

(11-32)

2Gk1A5 =

Note that the +S direction for (11—32) is from .X2 toward X3. Also, this result is independent of the location of the origin. Instead of using (11—9), we could have started with the fact that the cross section rotates about the centroid. The displacement in the + S direction follows from Fig. 11—4:1 x

u,5 =

Substituting for

is)

=

w1p

+

s

k1x1p

(11—33)

in

Yss =

Us

(11—34)

and noting that Ut =

lead to (11—29). Using (11—22), we can write

=

M5

=

(11—35)

t This interpretation of p is valid only when S is directed from X2 to X3, i.e., counterclockwise for this case. See Prob. 11—14 for an alternate derivation.

§ This development applies for arbitrary choice of the +S direction. The sign of p is positive if a rotation about X1 produces a translation in the +S direction. Equation (11—29) is used to determine the warping distribution once the shearing stress distribution is known. See Prob. 11—4.


280

Then, substituting for

CHAP. 11

in (11—32), we obtain

=

(11—36)

.3 s

where n is the outward normal, A5 is the area enclosed by S, and the + S sense is from X2 to X3. This result is valid for an arbitrary closed curve in the cross section. We employ (11—36) to determine the values of 17 at the interior boundaries of a multiply connected cross section. It is of interest to determine the energy functions associated with pure torsion. When the material is linearly elastic and there are no initial strains, the

strain and complementary energy densities are equal, i.e., V = We let

V dA

V

strain energy per unit length

(11—37)

The strain energy density is given by

+

V= Substituting for Y12' '/13,

V=

X3)

2

+(

+

x2)j 2

and integrating (b) over the cross section, we obtain

V= Since

=

V,

(11—38)

and M1 = GJk1, it follows that =

+

(11—39)

xl

WI

dx1

Fig. 11—5. Differential element for determination of the rotational work.

Instead of integrating the strain-energy density, we could have determined the work done by the moments acting on a differential element. Consider the element shown in Fig. 11—5. The boundary forces acting on a face are statically equivalent to just a torsional moment. Also, the cross sections are rigid in

THIN-WALLED OPEN CROSS SECTIONS

SEC. 11—S.

281

their plane and rotate about X1. The relative rotation of the faces is

/((01 + dw1 '\ —dx1 — dx1

and the first-order reduces to

=

dx1

,i

workdone by the external forces due to an increment in wj

5WE =

M1 ,Xk1 dx1

Now,

=

=

dx2dx3 = óVdx1

5jJ

for an elastic body. Then, expanding ö V.

and it follows that dk1

= M1 =

GJk1

V=

11—3.

APPROXIMATE SOLUTION OF THE TORSION PROBLEM FOR THIN-WALLED OPEN CROSS SECTIONS

We consider first the rectangular cross section shown in Fig. 11—6. The exact solution for this problem is contained in numerous texts (e.g., see Art. 5—3 of Ref. 1) and thcrefore we will only summarize the results obtained. x3

I

dl 2

d, 2

HFigS 11—6. Notation for rectangular section.

282


CHAP. 11

occurs at x2 = ± t/2, x3 =

When t d, the maximum shearing stress (points 5, 6). The exact expressions are

0

dt3

J = K1— (11-41) =

K2t

where

K1 =

192 (t'\ 1

I

tanh

— =

K2 =

8

1

1

1

(2n+1)2 cosh

—

A,,

2n+1 Id Values of K1, K2 for d/t ranging from 1 to 10 are tabulated below: d/t 1

K1

K2

0.422

0.675 .930 .985 .997 .999 1.000

3

.687 .789

4

.843

2

5

.873

10

0.936

If t d, we say the cross section is thin. The approximate solution for a thin rectangle is J 4dt3 (113

2—-—x2

=

2Gk1x2

(11-42) x2x3

(t)2 take d/t = in the exact solution.) The shearing stress across the thickness and

(We

M1

varies linearly

3M1

A view of the warped cross section is shown in Fig. 11—7.

Since the stress function approach is quite convenient for the analysis of thin-walled cross sections, we illustrate its application to a thin rectangular


SEC. 11—3.

283

section. Later, we shall extend the results obtained for this case to an arbitrary thin walled open cross section. The governing equations for a simply connected cross section are summarized below for convenience (see (11—26), cross

(11—27)):

=

—2

(in A)

0

(on the boundary)

= J =

(1A

where the S direction is 900 counterclockwise from the is direction.t Since t is small and a12, the shearing Stress component in the thickness direction, must

_:k,1,

Fig. 11—7. Warping function for a rectangular cross section.

= 0 at all points independent of x3. The

vanish on the boundary faces, it is reasonable to assume

in the cross section. This corresponds to taking equations reduce to d2

=

—2

Solving (b), we obtain

-

J

=

dx2

=

M1 = ——---—-- = 2——x2

J

t This applies for X3 counterclockwise from X2. The general requirement is the n — S sense must coincide with the X2-X3 sense.

284


The expression for (x3

CHAP. 11

developed above must be corrected near the ends

± d/2) since it does not satisfy the boundary condition, ti

This will lead to a12 0 near the ends, but will have a negligible effect on J Actually, the moment due to the approximate linear expansion for and is equal to only one half the applied moment: I't/2

x2a13 dx, =

d

P4

/1

dt3) =

J

The corrective stress system (a12) carries M1/2. This is reasonable since, even is small in comparison to amax, its moment arm is large. though

-f-s

t(s)

Fig. 11—8. Notation for thin-walled open cross section.

We consider next the arbitrary thin-walled open cross section shown in Fig. 11 —8. The S curve defines the centerline (bisects the thickness) and the n direction is normal to S. We assume = 0 and take = —n2 + t2/4. This corresponds to using the solution for the thin rectangle and is reasonable when S is a smooth curve. The resulting expressions for I and are

J=

4

t3 dS

(11—43)

M1

a15, ma, =

= Gkitrnax


SEC. 11—3.

The

285

results for a single thin rectangle are also applied to a cross section

consisting of thin rectangular elements. Let of element i. We take J as

t1 denote the length and thickness

J=

(11—44)

Asan illustration, consider the symmetrical section shown in Fig. 11—9. Applying (11—44), we obtain 3 1'ff + w4v ..i

The

3

maximum shearing stress in the center zone of an element is taken as M1

= —7t1 = Gk1t1

(11—45)

In general, there is a stress concentration at a reentrant corner (e.g., point A in Fig. 11—9) which depends on the ratio of fillet radius to thickness. For the case bi

+

(Iw

I Fig. 11—9. Symmetrical wide-flange section.

of an angle having equal flange thicknesses, the formulat

=

\

+

(1146)

4rf/

is the fillet radius and 0rn is given by (14—45), gives good results for where rf/t < 0.3. The stress increase can be significant for small values of rf/t. For example, for Tf = 0,lt. Numerical procedures such as finite differences or the finite element must be resorted to in order to obtain exact solutions for irregular sections. -

t See Ref. 2 and Appendix of Ref. 9. See Ref. 4.

CHAP. 11


286

11—4.

APPROXIMATE SOLUTION OF THE TORSION PROBLEM FOR THIN-WALLED CLOSED CROSS SECTIONS

The stress function method is generally used to analyze thin-walled closed cross sections. For convenience, the governing equations are summarized below (see (11—26), (11—27), (11—36)):

(in A)

—2

(on the exterior boundary) (on the interior boundary, S,) = area enclosed by

— ci J

—

dA +

S

and +S sense from X2 toward X3,)

—

£

j

on

dS

=

—2A5

We consider first the single cell shown in Fig. 11—10. The curve defines the centerline. Since there is an interior boundary, we have to add a term n S.

E

Sect. E-E

Fig. 11—10. Single closed cell.

involving C1 to the approximate expression for We take as + tz

where

used for the open section.

2n\

(11—47)

represents the contribution of the interior boundary. This expression

SEC. 11—4.

satisfies

THIN-WALLED CLOSED CROSS SECTIONS

287

the one-dimensional compatibility equation and boundary conditions, 2

=

atn— +t/2 n = — t/2

at

C1

(a)

and is a reasonable approximation when S is a smooth curve. Differentiating b

(fl

and substituting (b) in the expressions for the shearing stress components lead to 0

M1 /

+ C1\ 7)

(11—48)

cr?5 + The tangential shearing stress varies linearly over the thickness and its average We let q be the shear stress resultant per unit length along S, value is positive when pointing in the + S direction, 1/2

q

=

(11—49)

cr15

J —1/2

and call q the shear flow. Substituting for a

we find (11—50)

q

The additional shearing stress due to the interior boundary (i.e., closed cell) corresponds to a constant shear flow around the cell. One can readily verifyt that the distribution, q = const, is statically equivalent to only a torsional moment,

given by

=

(11—51)

The torsional constant is determined from

J Substituting for

dA + 2C1A1

M1/Gk1

(a)

using (11—47), we obtain

j = Jo + 4

(11—52)

=

t3 dS

Equation (a) was established by substituting for the shearing stresses in terms in the definition equation forM1 and then transforming the integrand. We could have arrived at (11—52) by first expressing the total torsional moment as

M1 = See Prob. 11-5.

+

(11—53)

288

TORSION-FLEXURE OF PRfSMATIC MEMBERS

CHAP. 11

is due to the closure. Next,

where MI is the open section contribution and

we write

M1 = GkIJ

it'll =

Gk1J°

J=

+ JC

= Gk1J'

(11—54)

Then, J0

(11—55)

and it follows that Jc

Jo

(11-56)

Finally, using (11—5 1), we can express JC JC

as

= Mu/(M1/J) =

(11—57)

This result shows that we should work with a modified shear flow, C

q/(M1/J)

(11—58)

rather than with the actual shear flow. Note that C C1 for the single cell. It remains to determine C1 by enforcing continuity of the warping function on the centerline curve. Applying (11—32) to

we have

=

(11—59)

Substituting for

q/t

M1C1

=

leads

(1160) One should note that C1 is a property of the cross section. Once C1 is known, we can evaluate .J from (11—52) and the shearing stress from

M1 (

(11—61)

+ -i—,)

Example 11—1 Consider the rectangular section shown. The thickness is constant and a, b are centerline properties are dimensions. The various CdS

Cl =

t See Prob. 11—6.

2(a+h)

SEC. 11-4.


289

We express J as

=

+

For this section,

1 (r'Y /

J°

We consider a >

b.

Then, Jo

J'

(t'Y

= 01— \\h

The section is said to be thin-walled when

c< b.

In this case, it is reasonable to neglect

Jo vs.

Fig. Eli—i

b H r+tb

+s, q

I

-

The strcss follows from (11—61),

M1C1/ t2'\ = —-———ii ± —-I = J C11

t\,

(I

± —s—

where, for this section, t2

/

h'\t

(t

If the section is thin-walled, we can neglect the contribution of = q/t =

We

i.e., we can take

M1

consider next the section shown in Fig. 11—li. Rather than work with

it is more convenient to work with the shear flows for the segments. We

number the closed cells consecutively and take the + S sense to coincide with the X2-X3 sense. The +S sense for the open segments is arbitrary. We define q3 as the shear flow for çellj and write (11—62)

Note that is the value of on the interior boundary of cell j and the shear flow is constant along a segment. The total shear flow distribution is obtained

290


CHAP. 11

q2. S2

x3

Fig. 11—11. Cross section consisting of closed cells and open segments; A, and A,, are centerline areas.

by superimposing the individual cell flows. Then, the shear flow in the segment common to cells i andj is the difference between qj and q1. The sign depends on the sense of S.

q=

— q2

=

— C2)

q=q2—q1

for S1

(11—63)

forS2

The shearing stress is assumed to vary linearly ovcr the thickness. For convenience, we drop the subscripts on and write the limiting values as cr = ±a° + Cr" where M1 /Cnet

cr=—1-t

(11—64)

It remains to determine C1, C2, and J. We have shown (see (11—55)) that

J=

Jo + Jc

(a)

and

=

(h)

We determine J° from (11—65) segments


SEC. 11—4.

291

Substituting for

MI =

2qjA1 + 2q2A2

+ C2A2) in (b) leads to

+ A2C2)

-

(11—66)

The constants are obtained by enforcing continuity of on the centerline of each cell. This can also be interpreted as requiring each cell to have the same twist deformation, k14

I = 1,2

(11-67)

Substituting for q in terms of C and letting

=

C

dS

JS, t

a22

Cads

dS

C

=

=

.J52

=

—I

Jc

t

where a12 involves the segment common to cells 1, 2, the continuity equations

take the following form:

+ a12C2 = a12C1 + a22C2 =

(11—69) 2A2

C2, then determine f with (11—66), We solve this system of equations for and finally evaluate the stresses with (11—64). We can represent the governing equations in compact form by introducing matrix notation. The form of the equations suggests that we define

c

A

=

a

[a11

a121

a22j

(11—70)

With this notation, JC

= 2ATC (11—7.!)

aC

2A

Substituting for A in the expression for JC

and noting that JC is

CTaC

positive, we conclude that a must be positive definite,

The complementary energy per unit length along the centroidal axis is defined by (11—39), 11

We apply (11—51) to each cell. See (11—32).

2


292

Since

CHAP. 11

ais varies linearly over the thickness, the open and closed stress dis-

tributions are uncoupled, i.e., we can write

=

+

toq

where

=

2G

(11-72) aq)

i

—

It is reasonable to neglect the open contribution when the section is thin-walled.

Examp'e 11—2 The open-section torsional constant for the section shown is

=

± 2(b

+ htfl

+d+

Applying (11 —68) to this section, we obtain

=

hd

A2=hb a11 =

1(h + 2d) + 11

t2

6

012

=

a22

= tl

+ 2b) +

t2

and the following equations for C1. C2 and i.

+

+2

C2

=

2

dt1

—

c1 +

±2

+

C2 = 2bt1

—

J=

Jo + Jo

Finally, the shear stress intensities in the various segments are

=

M1 (C1 (k— + M1 /C1 — C2

J

t2

M1 (C2

=7 =

M1

t3

+ t1

+t2

/

(a)

____ _____________ TORSION-FLEXURE WITH UNRESTRAINED WARPING

SEC. 11—5.

293

Fig. E11—2

tl

t3

03

1

Ii I

I

032 A2

h

X3

a'

M1

When

d=

b.

Cl =

C2

=

—s 2bt3

1 + 213

and the section functions as a single cell with respect to shear flow.

11—5.

TORSION-FLEXIJRE WITH UNRESTRAINED WARPING

Consider the prismatic member shown in Fig. 11—12. There are no boundary

forces acting on the cylindrical surface. The distribution of boundary forces x2

x2

—_____

if

xI

I +S

P2

Fig. 11—12. Prismatic member in shear loading.

on the cross section at x1 = L is statically equivalent to a single force P212, acting at the centroid. Also, the end cross sections are not restrained against displacement. In what follows, we describe St. warping, i.e., Venant's torsion-flexure formulation for this problem. Later, in Chapter 13, we shall modify the theory to include restraint against warping.

294


CHAP. il

We start by postulating expansions for the stresses. The stress resultants and couples required for equilibrium at x1 are

=A4,r0 F2 = P2 M3 = P2(L

—

x1)

Introducing (a) in the definition equations for the stress resultants and couples leads to the following conditions on the stresses:

dA = JJx3a13 c/A = dA = P2(L — x1) Jj712 c/A =

0

P2

$fcrj3 c/A

=

0

S$(x2a13 — x3cr12)dA

—,

0

The expansion,

=

M3 13

=

P2 —

13

satisfies the first three conditions (i.e., F1,

M3) identically since

JJx2 c/A = jJx2x3 dA = 0

dA =

13

The last three conditions (i.c., F2, F3, M1) require a12, a13 to be independent of x1. This suggests that we consider the following postulated stress behavior:

cru =

——--—x2 =

a1 2

a1 2(x2, x3)

13

P2 —

x1)x2

13

(11 —73)

a13 = a13(x2, X3) a22

0

a33

Introducing (11—73) in the stress-equilibrium equations and stress boundary conditions for the cylindrical surface leads to

a21,2 + a31,3 + 2a21 +

P2 13

0

(mA)

=0

(on S)

(11—74)

At this point, we can either introduce a stress function or express (11—74) in terms of a warping function. We will describe the latter approach first. The displacements can be found by integrating the stress-displacement relations. We suppose the material is linearly elastic, isotropic with respect to the X2-X3 plane, and orthotropic with respect to the axial direction. This is a convenient way of keeping track of the coupling between axial and in-plane

SEC. 11—5.

TORSION-FLEXIJRE WITH UNRESTRAINED WARPING

295

deformation. Substituting for the stresses in (10—74), we obtain

=

=

u1

I

P2

E1

E113 V1

= u2,2 =

'Y12

Y23

= U2,3 + U3,2

v1P2 —7—(L

LI3

I

=

xj)x2

—.

x1)x2

—

xj)x2

= function ofx2, x3

1

U1 3 + U3, 1 =

Y13

—(L El3

=

= u1, 2 + u2,

—

V1P2

=

v1

63 = u33 =

(L

= function ofx2, x3

0

Integrating the first three equations leads to u1

=

(Lx1

E1 13

v1P2

(L

—

u3 = ——- (L

—

= v1P2

El3

1

+ f1(x2, x3)

—

+ .f2(xj, x3) x1)x2x3

+

(b)

x2)

The functions f1, f2, f3 are determined by substituting (b) in the last three equations. We omit the details and just list the resulting expressions, which involve seven constants:

= f2 =

C1

+ C5x2 + C6x3 +

C2

—

v1P2 —

=

x3)

+ C4x3 — k1x1x3

C5x1

Xi.

+ —--7-(L —

(c)

2E,iJ 3

C3 — C5x1 — C4x2

+ k1x1x2

The constants C1, C2, ..., C6 are associated with rigid body motion and k3 is associated with the twist deformation.t We consider the following displacement boundary conditions: 1.

The origin is fixed:

u1=u2=u3==0 2.

at(0,0,0)

A line element oh the centroidal axis at the origin is fixed:

= fSee Eq. (11—5).

u3,1

=

0

at(0,0,O)

CHAP. Ii


296 3.

A line element on the X2 axis at the origin is fixed with respect to rotation in the X2-X3 plane: at (0, 0, 0) u2,3 = 0

These conditions correspond to the "fixed-end" case and are sufficient to eliminate the rigid body terms. The final displacement expressions are (Lxj —

u1

=

+ 4)(x2, x3)

+

—

—

(11—75)

—

—

Vj

2

El3

-

— x1)x2x3 + k1x1x2

One step remains, namely, to satisfy the equilibrium equation and boundary condition. The transverse shearing stresses are given by 1

41,2 — k1x3

1

(7j3

+

+ k1x2 —

4),

v1P2

2L13 v1P2

LI3

—.

x2) (11—76)

x2x3

Substituting for the stresses in (11—74), we obtain the following differential equation and boundary condition for 4):

l\

P2f'2v1

+

(mA)

-

The form of the above equations suggests that we express 4) 4)

= kjq5t

+

—

+

1

12) +

v1P2 C

+

- 77

as

(11-78)

where

is the warping function for pure torsion and 4)2. and 4)2d are harmonic functions which define the warping due to flexure. Substituting for 4) leads to the following boundary conditions for 4)2. and 2

2

)

2

One can show, by using (11—15), that cn dS

=

0

=

0

+

(11—79)

TORSION-FLEXURE WITH UNRESTRAINED WARPING

SEC. 11—5.

297

therefore the formulation is consistent. Terms involving vj/E are due to in-plane deformation, i.e., deformation in the plane of the cross section, and setting v1/E = 0 corresponds to assuming the cross section is rigid. Then, defines the flexural warping for a rigid cross section and represents the correction due to in-plane deformation. The shearing stress is obtained by substituting for in (11—76). We write the result as (j = 2,3) (11—SO) + 01j,r + 01j = and

where crU, is the pure-torsion distribution and butions corresponding to and 42d:

d are

r,

flexural distri-

=

2

2 2

P2

x2x3)

The pure torsion distribution is statically equivalent to only a torsional mo-

ment,

=

G1k1J.

One can show thatt dA

J$a12,a dA

dA — 0 dA 0

P2

=

0

(11—82)

Note that the shear stress due to in-plane deformation does not contribute to P2. The total torsional moment consists of a pure torsion term and two flexural terms, M1 = G1k1J +

+ X24)2r 3)dA

S2r

2

S2d

+ X242d,3

X34)2a,2)dA

and depend only on the shape of the cross section, it follows that and S2d are properties of the cross section. For convenience, we let

Since

X3

(1184)

1

13\

£.

J

and (11—83) reduces to

= Now, —

—

(11—85)

is the statically equivalent torsional moment at the centroid due

tSeeProb. Il—lO.

298


CHAP. 11

to the fiexural shear stress distribution. Then,

defines the location of the resultant of the flexural shear stress distribution with respect to the centroid. The twist deformation is determined from

+

k1

(11—86)

where M1 is the applied torsional moment with respect to the centroid. If P2 is applied at the centroid, M1 = 0, and k1 =

= 0. Suppose P2 has an eccentricity e3. In this case (see Fig. 11—13), M1 = —e3P2, and

The cross section will twist unless

k1

P2 —

e3)

For Ilexure alone to occur, e3 must equal x3

Fig. 11—13. Notation for eccentric load.

Now, to Whether twist occurs depends on the relative eccentricity, e3 — find x3, one must determine S2. and S2d. This involves solving two secondorder partial differential equations. Exact solutions can be obtained for simple cross sections. In the section following, we present the exact solution for a rectangular cross section. If the section is irregular, one must resort to such numerical procedures as finite differences to solve the equations. In Sec. 11—7, we describe an approximate procedure for determining the flexural shear stress distribution in thin walled cross sections. Suppose the cross section is symmetrical with respect to the X2 axis. Then, is an odd function of x3. The form of the is an even function of x3 and boundary conditions (11—79) requires and to be even functions of x3

SEC. 11—5.


299

this case. Finally, it Ibilows thatt S2,. = 0 and S2d = 0. Generalizing this result, we can state: The resultant of the shear stress distribution due to fiexure in the is an axis of symmetry direction passes through the centroid when for the cross section. for

x3

Shear center

Fig. 11—14. Coordinates of the shear center.

We consider next the case where the member is subjected to P2, P3 and at the right end (see Fig. 11—14). The governing equations for the P3 loading can be obtained by transforming the equations for the P2 case according to >

X3

U2 —* U3 13

(3

ox2

Ox3

.-

U3—> —U2

—---+-—

—--*-----13

(3

Ox3

Ox2

U12

U13

—a12

13

'*

12

Two additional flexural warping functions must be determined. The expres-

sions defining the flexural shear stress distributions due to P3 are P3

413r. 2 -r 12

cr12 r

3

r

d

'2

=

= t

is even in x>,

((/33r, 3

(11—87)

.v1G1 P3 Vj D

x2x31!

2

L.

12

i —i---

+

+ (,b3d, 3]

is odd in x3, and S2r, Ssd involve only integrals of odd functions of

300


where q53r,

are

CHAP. 11

harmonic functionst satisfying the following boundary

conditions:

= 2

(11—88)

(X2 +

=

2

Note that the distribution due to

leads to no shearing stress resultants. Finally, the total normal stress is given by M2

M3

12

13

(P3

P2

'\

13

J

(11—89)

Superimposing the shearing stresses and evaluating the torsional moment,

we obtain

M1 = where

defines

G1k1J

+

—

(11—90)

the location of the resultant of the flexural shear stress distri-

bution due to P3. One can interpret X2, x3 as the coordinates of a point, called the shear center. The required twist follows from (11—90): k1

(M1 +

=

—

Since (see Fig. 11—14)

M1 + P2x3 — P3x2 = the applied moment with respect to the shear center = MT

(11—91)

we can write (a) as

k1 =

(11—92)

To determine the twist deformation (and the resulting torsional stresses), one must work with the torsional moment with respect to the shear center, not the centroid. For no twist, the applied force must pass through the shear center. In general, the shear center lies on an axis of symmetry. lithe cross section is completely symmetrical, the shear center coincides with the centroid. It is of interest to determine the complementary energy associated with Then torsion-flexure. The only finite stress components are 012, aild V* reduces to =

+

+

dA

(a)

follows directly by substituting (11 —89) and using The contribution from the definition equations for 13. t The total flexural warping function for P3 is P3

(

I

—

'\

+

v,P3(

1

+ —4

SEC. 11—5.


=

1

M2

M2

(11-93)

+ Now, the total shearing stress is the sum of three terms: a pure torsional distribution due to MT the flexural distribution due to F2 the fiexural distribution due to F3

1.

2. 3.

Each of the flexural distributions can be further subdivided into— dr.

the

distribution corresponding to a rigid cross section (defined

by 2.

dId,

the distribution associated with in-plane deformation of the cross

section (defined by 4)jd)

We combine the flexural distributions and express the total stress as C33

d12,t + C12,, + C12.d = 013.1 + C13r + 013,d

where the various terms are defined by (11—81) and (11—87). For example,

r=

F2

F3

• 2

13

2

13

The complementary energy due to pure torsion follows from (11—38) and (11—92):

+

a

as

C12r

F3_

F2

—

2

3

r+

JJ =

1 See Prob. 11—il.

r) and integrating over the cross section, we obtaint F2

+

+

Jj

3

+

2FF

+

F2

(1196)

dA

2+

3413r, 3)dA

=

CIA

JJ


302

The

CHAP. 11

coupling term, I/A23, vanishes when the cross section has an axis of

symmetry. We consider next the coupling between or., and JJ(a12,to12,r +

=

±

2

+ x2)

+

=

+

dA

+

MT ""F2 Jj 13

—

2)

+

F3

+

+

—

(11—97)

dA = 0

12

The remaining terms involve a,a, the shearing stress distribution due to inplane deformation of the cross section, We will not attempt to expand these terms since we are interested primarily in the rigid cross section case. Summarizing, the complementary energy for flexure-torsion with unrestrained warping is given by 1

M2

M2

M2

FF

F2

I

F2

(11-98)

+ terms involving v1/E — We introduce the assumption of negligible M1 + in-plane deformation by setting v1/E = 0. Similarly, we introduce the assumption of negligible warping due to flexure by setting 1/A1 = (&3r 1/A2 1/A23 = 0. In Sec. 11 —7, we develop an approximate procedure, called the engineering theory, for determining the flexural shear stress distribution, which is based upon integrating the stress-equilibrium equation directly. This approach is

where

similar to the torsional stress analysis procedure described in the previous section. Since the shear stress distribution is statically indeterminate when the cross section is closed, the force redundants have to he determined by requiring the warping function to be continuous. For pure torsion, continuity requires (see (11—32))

= 2G1k1A5

where the integration is carried out in the X2-X3 sense around S. and

is the

area enclosed by S. To establish the continuity conditions for flexure, we operate on (11 —81) and (11

There are four requirements:

j (aisd)F2 dS dS =

2,3

2v1G1P2

X3

dA

2v1G1P3 rr ii X2 dA El2 ,jj 4,

(11—99)

SEC. 11—6.

EXACT FLEXURAL SHEAR STRESS DISTRIBUTION

303

In the engineering theory of flex ural shear stress distribution, the cross section

is considered to be rigid, i.e., the distribution due to in-plane deformation is neglected. The consistent continuity condition on the flexural shearing stress

4sajsdSO

(11—100)

One can take the + S direction as either clockwise or counterclockwise. By coincides with the + S direction.

definition, the positive sense for 11—6.

EXACT FLEXURAL SHEAR STRESS DiSTRIBUTION FOR A RECTANGULAR CROSS SECTION

We consider the problem of determining the exact shear stress distribution due to F2 for the rectangular cross section shown in Fig. 11—15. For con-

venience, we first list the governing equations: x2

rd3

'2

—

dt3

A = dt

d

r Fig. 11—15. Notation for rectangular cross section. 1.

Warping functions +

= an

= =

1

2

+ 2

)

+


304 2.

CHAP. 11

Shearing stresses 012

-

= F2

3) +

= T13 Determination

+

+ v1G1E2

+ xi)]

x2x3)

of

The boundary conditions for

are

l(d'\2

d

=0

atx3 =

We can take the solution as .i.

> 1 C,

1

for a closed section. We obtained a similar result for using the displacement-model formulation for a solid section. Since is due to the restrained shearing stress (q'), we see that shear deformation due to q' cannot be neglected for a closed cross section. We discuss next the determination of the normal and shearing stresses due to warping. The general expressions are

°isq q' t

tie


SEC. 13—8.

411

The maximum normal stress occurs at point 2 while the maximum shear stress can occur

at either points I or 3. We consider the same problem as was treated in Example 13—4, i.e., a member fully restrained at one end and subjected to a torsional moment M at the other end. We express the stresses in terms of ag,, the maximum shear stress for unrestrained torsion,

M(

+

= J

C

which reduces to

MC = since

7

M =

we are considering the section to be thin-walled. The maximum stresses are 2

0$

,nax,I

=

i

tanh 2.L

= [3C,1112

—

The variation of

and

2

S.C

with height/width is shown below. We are taking Poisson's

ratio equal to 1/3.

=

h/a

c,

(point 2)

1

0

2

1.04

3

—1.51

(point 1)

(point 3)

0 —0.35 —0.46

+0.44 +0.65

0

For large tanh I and we see that both the normal and shear stress are of the order of the unrestrained-torsion stress. In the open section case, we found the restrainedtorsion shear stress to be of the order of (thickness/depth) times the unrestrained shear stress.

To illustrate the procedure for a multicell section. we consider the section shown in Fig. 13—7. The unrestrained-torsion analysis for this section is treated in Sec. 11—4 (see Fig. 11—11). For convenience, we summarize the essential results here.

We nttmber the cells consecutively and take the +S sense from X2 to X3 for the closed segments and inward for the open segments. The total shear flow is obtained by superimposing the individual ccli flows

q' = qU =

0

for an exterior (open) segment constant for an interior segment

We let —

—

(U

WIT

RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13

412 e

"2

+

q,S

q1 ,S1

Fig. 13—7. Notation for mixed cross section.

The constants C1, C2 are determined by requiring each cell to have the same twist deformation, w1,

Enforcing (11—67),t

=

=

for each cell leads to 2A

where a, A are defined as

f

dS

—

,Jsj t

= a21

dS

=

A = {A1, A2}

The warping function is generated by applying (13—6): 4'

=

(13—76)

a

= Psc —

7

We start at point P1 in cell 1 and integrate around the centerline, enforcing continuity of 4, at the junction points b, c, and d. For example, at b, we require

t See also (11—32).

SEC. 13—8.

413


which leads to a relation between

and 4),,,:

Jb = 4)e +

j

Psc

+

dS =

dS —

Repeating for points C and d results in the distribution of 4) expressed in One can easily verify that 4) is continuous, i.e., & determined terms of determined from segment cdcL. Finally, we from segment ca is equal to evaluate

by enforcing

JJ4)dA=J4)tdS=O where the integral extends over the total centerline. Note that

0 if P1

is taken on an axis of symmetry. The shear flow for restrained torsion is obtained with (13—69): a

=

as

The steps are the same as for the flexural shear determination discussed in Sec. 11—7. We take the shear flow at points P1. P2 as the redundants, =

J = 1, 2

(13—77)

and express the shear flow as

+ ii.

(13—78)

is due to The distribution, has the same form as We just have to replace C with C'S. We generate by integrating (i) around the centerline, and enforcing equilibrium at the junction points. For example, at point b (see Fig. 13—7), where Zj0 is the open section distribution and

+ = Note that = 0 at points P1, P2. e andf The redundant shear flows are evaluated by requiring no energy coupling between qU and qr which is equivalent to requiring qr to lead to no twist deformation, j. Noting (c), we can write

= Finally, substituting for

j = 1,2

0

we obtain aCr = B 1'

f See footnote on page 385.

(13—79)

—

dS

(13—80)


414

Once 4) and zir are known, the cross-sectional properties (1 , Ci., X2r, xar) can be evaluated. Also we can readily generalize the above approach for an

n-cell section. 1

3—9.

GOVERNING EQUATIONS—GEOMETRICALLY NONLINEAR RESTRAINED TORSION

In this section, we establish the governing equations for geometrically nonlinear restrained torsion by applying Reissner's principle. This approach is a mixed formulation, i.e., one introduces expansions for both stresses and dis-

placements. The linear case was treated in Sec. 13—5. To extend the formulation

into the geometrically nonlinear realm is straightforward. One has only to introduce the appropriate nonlinear strain-displacement relations. Our starting point is the stationary requirement t —

V*)d(vol.)

d(surface area)]

—

=

0

V*(o.),

a, are independent variables, C and b are where e(u), prescribed. We take the displacement expansions according to (13—3) and use the strain-

displacement relations for small strain and small finite rotations4 U1

= U1 + C02X3 —

U3 =

+ w1(x2

= Yiz = Y13

W3X2

+

+

/4)

x3)

cn1(x3

112

—

(13—81

+

+ U2,1 + U3,3U3,2 l.1j3 + U31 + 1)2

The in-plane strain measures (62, 63, Y23) are of 0(w2), which is negligible

according to the assumption of sinai! finite rotations. Actually we assume O'22 =

1723

0, i.e., plane Stress. Substituting for the displacements and

noting the definition equations for the force parameters, the first term in (a) expands to d(vol.) =

1

1

1)]

— W3 + + F2[u,2. i + F3[u53 S + — t+ + M2{w2, 1 — co1, 1(u52 i + x3w1, + M3[w3, i — w1, i — T2w1. 1)] M0f1 + MRf + 1

+

i+

MQW1W1, 1}dxj

f See Eqs. 13—33 and corresponding footnote. We are working with Kirchhoff Stress and Lagrangian strain here. See Sec. 10—3. Eq. 10—28. Tile displacement expansions assume small-finite rotation, i.e., sin w and cos w 1. To be consistent, we must use (10—28).

GOVERNING EQUATIONS

SEC. 13—9.

415

where the two additional force parameters are = ÷ MQ = $J(x2c12 + x3a13)dA

The terms involving the external forces have the same form as for the linear case, but we list them again here for convenience (see (13—6)):

JJcbTu d(vol.) + jJpTu d(surface area)

+

+ + mrwi + in2w2 + m3oj3 + rn4f)dxi + F3u33 + MTO1 + M2(02 + M3co3 +

(13—83)

+ F1u1 + where the end forces (the barred quantities) are defined as previously, for example,

=

etc.

(5Jp1

It remains to introduce expansions for the stresses in terms of the independent

force parameters and to expand V*. In the linear case, there are 8 force M3, and M4, MR. Two additional force measures (Me, MQ) are present for the nonlinear case but they can be related to the previous force measures. We proceed as follows. We use the stress expansions employed for the linear case with = They are summarized below for convenience measures. F1

(see Sec. 13--5): F1

a11

±

A

M2

M3

—1—-x3 — -T—X2

13

±

+

+ &ij

= —

+ +

MT

Il

_.-

where 4,, f, q, h2 and h3 are functions of x2, x3. Introducing (a) in the definition

equations for

and MQ leads to

=

f11F1

= $2

+ fl2M2 + fl3M3 +

if if

=

$3 =

if

/34, 4,

+

= +

(13-84)

+


416

andt

+ (k =

+ +xlh3k)dA

+

MQ —

$J(x2h2k

2, 3)

(13—85)

=

+ = Certain coefficients vanish if the cross section has an axis of symmetry4 One can readily verify that fi1F1

(13—86)

0

MQ

when the section is doubly symmetric. For generality, we will retain all the terms here.

The complementary energy density function has the same form as for the linear case: —

=

1

—----'

2Ek.,A

+

+ —-——

13)

'2

1 + ——'

+

+

+

+

((Mw +

+

+ X2rF3)

We have shown that it is quite reasonable to neglect transverse shear deformation due to warping (C. X2r X3r = 0) for a thin-walled open section. Substituting Equations (13—82)—(13—87) in Reissner's functional and re-

quiring it to be stationary with respect to the seven displacement and eight force measures leads to the following governing equations: Equilibrium Equations

F1,1 + b1 = 0 +

j+

+ (1 +

+

+

—

1

= 0 + m2 = M3,1 + F2 + m3 =

+

—

See Prob. 13—12.

1M2} + b2

+

M2, 1 — F3

t See Prob. 13—il.

— w1

1 1

1

w1F3

+ F3 + w1F2 — wi,1M3} + b3 =

—

(1 +

—

+

0 0

+

2J32w1,

0 0

SEC. 13—9.

GOVERNING EQUATIONS

417

where

Relations

=

1FF2

1FF2

+

+

G

M

1

X3r

F3

+

1 + Wj,

1+

1+

—

w3

+ wi[u53, 1 — Wj, 1/33]

1 MrTJ =

+

(02

+

1+

+

=

(13—88)

=

(02,1

+ (0l,j(—US2,j + /32(01,1)

+ +

+ /33(01,1) j

[CrM?+ X3rF2 +

= I' +

Conditions (+ for x1 = L,

—

+

for x1 = 0)

+ T3w1,1) + F2 — wjF3 (01,1M2 = ±F2 — x2w1, 1M3 + F3 + w1F2 — ±F3 1 + 7J1w1, 1

+ (1 + + M2(—u52,j + + (02 prescribed or M2 = ±M2 (03 prescribed or M3 = ± M3 prescribed or = ± + (0j(172F2 +

i)

—

= 1.

u1 prescribed or F1 = prescribed or prescribed or wi prescribed or

+

=

=

Boundarp

i — x2uS3,

+ (I + 1+

+

= ±MT

f

These equations simplify considerably when the cross section is symmetric and transverse shear deformation is neglected.1' We discuss the general solution of (13—88) in Chapter 18. The following example treats one of the cases, a member subjected to an axial force and torsional moment. t See Prob. 13—13.

418

RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER

CHAP. 13

Example 13—7 We consider a prismatic member (see Fig. E13—7A) having a doubly symmetric cross section, fully restrained at one end and loaded by an axial force P and torsional moment M. We are interested here in evaluating the influence of axial force on the torsional behavior. The linear solution (with no axial force) was derived in Example 13—i.

Fig. E13—7A

x2

P

M

L

F

Equilibrium Equations (symmetrical cross section and no distributed load) = F1. d dx1

0

i) =

(M1 +

0

Force-Displacement Relations = GJw11

= ErI,ji F1 =

i

+

=

0

Boundary conditions

.xi=O xj = L

F1

P

M1 + J3tF1w1,1 = Al

Integrating the last two equations in (a) and noting the boundary conditions, lead to

F1 = M1 + /31F1w1,

The first equilibrium equation takes the form 1,11

2

const

= corlst

=P =M

GOVERNING EQUATIONS

SEC. 13—9.

419

where P11 7;:ij

GJA

2GJ

i±P I+

+ F)

This expression reduces to Equation (g) of Sec. 13—6 when P = 0. Once f is known, we can determine the rotation by integrating (d), which expands to + F

f

M—

+ when we substitute for M1 using (b). The general solution is,

f= [GJ

(i

C1

M

cosli ,ux + C2 smh

+ +

=

C3

+ Mx {i +

—

{C1

sinh px + C2 cosh

(We drop the subscript on x1 for convenience.) Finally, specializing (g) for these particular boundary conditions result in

f= wi =

{

—1 + cosh

—

sinh

tanh

{sinh jtx + (1 — cosh

— —

These equations reduce to (13-57) when P = 0. A tensile force (P > 0) increases the torsional stiffness whereas a compressive force (P < 0) decreases the stiffness. Equation (h) shows that the limiting value of P is 1. We let F, represent the critical axial force and the corresponding axial stress

11

(;J

to be less than the yield stress, (J/11) must be small with respect to unity. As an illustration, consider the section shown in Fig. E13—7B. The various coefficients (see Example 13—4) are In order for

J=

+


420

Fig. E13—78

x3

X2

and

r,,

(t'\2(

G

REFERENCES 1.

2. 3.

4. 5. 6.

T., and N. B. CHRISTENSEN: "Methods of Analysis of Torsion with Variable Twist," J. Aero. Sci., pp. 110—124, April 1944. TIMOSHENKO, S. J:: "Theory of Bending, Torsion and Buckling of Thin-Walled Members of Open Cross Section," J. Franklin Inst., pp.559—609, 1945. VON KARMAN, T.. and W. C. CHIEN: "Torsion with Variable Twist," J. Aero. Sci., Vol. 13, No. 10, pp. 503—510, October 1946. BENSCOTER, S. U.: "Secondary Stresses in Thin-Walled Beams with Closed Cross Sections," NACA—TN 2529, Washington, D. C., 1951. BENSCOTER, S. U.: "A Theory of Torsion Bending for Multiceil Beams," J. Appi. Mech., Vol 21, No. 1, 1954. VON

VLASOV, V. Z.: Thin Walled Elastic Brains, israel Program for Scientific Translations,

Office of Technical Services, IJ.S. Dept. of Commerce, Washington. D.C. 1961. HEILIG, R.: "Der Schuberverformungseinfiuss auf die Wölbkrafttorsion Von Stilben mit offenern Profil," Der Stahlbau, April 1961. 8. HEILIG, R,: "l3eitrag zur Theorie der Kastentrhger beliehiger Der Stahlbau, December 1961. 9. J. T.: Mechanics of Elastic Structures, McGraw-Hill. New York, 1967. 7.

10.

KOLLORUNNER, C. F., and K. BASLER: Torsion in Structures, Springer-Verlag. Berlin, 1969.

Ii.

K.: Variational Methods in

and Plasticity, Pergarnon Press.

1968. 12,

MAISEL, B. I.: "Review of Literature Related to the Analysis and Design of ThinWalled Beams," Technical Report 440, Cement and Concrete Association, London, July 1970.

PROBLEMS 13.

14. 15. 16. 17.

18.

421

DABROWSKi, R.: "Gekrüinmte dUnnwandige Trager," Springer-Verlag, Berlin, 1968. GALAMBOS, T. V.: Structural Members and Fiames, Prentice-Hall, 1968. BLEICH, F.: Buckling Strength of Metal Structures, McGraw-Hill, New York, 1952. BURGERMEISTER, G., and H. STEin': Srabilitar Theorie, Part 1. Akademie Verlag, Berlin, 1957.

CHILVER, A. H.: Thin- Walled Structures, Chatto and Windus, London, 1967. REISSER, E.: "Note on Torsion with Variable Twist." .J. AppI. Mech., Vol. 23, No. 2, pp. 315—316, June 1956,

PROBLEMS 13—1.

The shear stress distribution due to

= where

is given by (see (11—95)) F2

F2 (733

2

13

13

'

3

are fiexural warping functions which satisfy

=

— x2

(in A)

(onS) This result applies when the cross section is assumed to be rigid with respect to in-plane deformation. The coordinate of the shear center is defined by

=

if

X3

3

X3

where

is the St. Venant torsional warping function. Hint: See Prob. 11—11

and Equation (11—97). 13—2. 13—3.

(a)

Verify (13—40) and (13—44). This problem reviews the subject of the chapter in two aspects.

No coupling between the unrestrained and restrained torsional distribution requires 0 +

The unrestrained torsional shear stress distribution for twist about the shear center (see Sec. 13—3, Equation (b)) is given by ,f U

IVIT O'12

=

=

—

— X3 + X3]

+ x2

—

—

x2]

The restrained torsional shear stress distribution is determined from (13—39). Verify that

=

MR when ç& =

and (a) is enforced.


422

(b)

When the cross section is thin-walled, (a) and (b) take the form

•fquqr_ = o

is the perpendicular distance from the shear center to the tangent at the centerline, Equation (d) follows from (11—29) and where

Prob. 11—4. We determine qf from (13—43). Finally, the force parameters for the thin-walled case are defined as

= MR = Verify that 1.

2. 3.

= MR when Open section Closed section Mixed section

=

dS Jqrc&

dS

Consider the following cases:

I and compare vs. Mu. Also evaluate L and compare with the unrestrained value. 13—5. Refer to Examples 12—2 and 13—2. Discuss how you would modify the member force-displacement relations developed in Example 12—2 to account for restrained torsion. Consider 1, X3r = 0, and—— (a) warping restrained at both ends (b) warping restrained only at x L 13—6. Refer to Example 13—2. Determine the translations of the shear center. Consider the cross section fixed at x 0. Discuss how the solution has to be modified when the cross section at x = L is restrained against translation. 13—7. Starting with the force-deformation relations based on the mixed formulation (13—49), derive the member force-displacement relations (see Example 12—2). Consider no warping at the end sections and take = + 1. Specialize for— (a) symmetrical cross section (b) no shear deformation due to restrained torsion and flexure—arbitrary cross section,

13—4.

Specialize (13—57) for .L >

at x

13—S.

Consider a thin-walled section comprising discrete elements of

material properties (F, G). Discuss how the displacement and mixed formulations haveto be modified to account for variable material properties. Note: The unrestrained torsion and flexural stress distributions are treated in Prob. 11—14 and 12—1.

Determine the distribution of qr, and expressions for Cr, for the cross sections shown in parts a and b and part e—d of the accompanying sketch (four different sets of data). and qr for the section shown. 13—10. Determine 13—11. Using the fiexural shear distributions listed in Prob. 13—1, show that 13—9.

-

'12 =

423

PROBLEMS

Prob. 13—9

I

Ii

T F— 0.75k (b)

I I

I

2k

/z

+

See part c.

(d)

'i—H

(c)

Prob. 13-10 t

0

0 ç1s2

t I

H

a

I

RESTRAINED TOIRSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13

424

Hint: One can write

22

13 •JJ

Also show that

(x2 V q52r +

113 —

13—12. Specialize Equations (13—84) and (13—85) for the case where the cross section is symmetrical with respect to the X2 axis. Utilize

x3)H0(x2, x3)dA =

0

where He is an even function and H,, an odd function of x3. Evaluate the coefficients for the channel section of Example 13—5. Finally, specialize the equations for a doubly symmetric section. 13—13. Specialize (13—88) for a doubly symmetrical cross Section. Then specialize further for negligible transverse shear deformation due to flexure and warping. The symmetry reductions are X2

=0

=

X2r

X3r = 0

i/A23=O 'li 0

= Consider the two following problems involving doubly symmetric cross section. (a) Establish "linearized" incremental equations by operating on (13—88) !72

13—14.

and retaining only linear terms in the displacement increments. (b)

Specialize for a doubly symmetric cross section (see Prob. 13—12). Consider the case where the cross section is doubly symmetric and the initial state is pure compression (F1 —P). Determine the critical

load with respect to torsional buckling for the following boundary conditions: 1.

co1 =

2.

==

f

0

at x =

=0

at x

0,

L

(restrained warping)

(unrestrained warping)

0, L

Neutral equilibrium (buckling) is defined as the existence of a nontrivial solution of the linearized incremental equations for the same external load. One sets

F1 = U2

—P

U3 = W1 = (02 = (03 = f

0

and determines the value of P for which a nontrivial solution which satisfies the boundary conditions is possible. Employ the notation introduced in Example 13—7.

Determine the form of V, the strain energy density function (strain energy per unit length along the centroidal axis), expressed in terms of displace13—15.

ments. Assume no initial strain but allow for geometric nonlinearity. Note that V = V* when there is no initial strain.

14

Planar Deformation of a Planar Member 14—i.

INTRODUCTION: GEOMETRICAL RELATIONS

A member is said to be planar if— 1.

The centroidal axis is a plane curve.

2.

The plane containing the centroidal axis also contains one of the

3.

principal inertia axes for the cross section. The shear center axis coincides with or is parallel to the axis. However, the present discussion will be limited to the case where the shear center axis lies in the plane containing the centroidal axis.

We consider the centroidal axis to he defined with respect to a global reference frame having directions X1 and K2. '[his is shown in Fig. 14—i. The orthogonal unit vectors defining the orientation of the local frame (Y1, Y2) at a point are x 12 = where points in the positive tangent direction and denoted by 13. Item 2 requires Y2 to be a principal inertia axis for the cross section. x2

Yl n r2

tl

S B

A

n i2

x1

ii

Fig. 14—i.

notation for planecurve. 425

426

PLANAR DEFORMATION OF A PLANAR MEMBER

CHAP. 14

By definition, t

=

=

dx1

+

dx2

Since we are taking t2 according to 11 x t2 = -

t2

dx2

+

=

(14-1)

13, it

follows that

dx5

(142)

The differentiation formulas for the unit vectors are dt1

1

(14-3)

where dt1

1

d2x2 dx1

d2x1 dx2

—

According to this definition, R is negative when d11/dS points in the negative t2 direction, e.g., for segment AB in Fig. 14—1. One could take t2 = ii, the unit normal vector defined by -

1

d11

(14-4) ciS

x 12 = 13 but this choice is inconvenient when there is a reversal in curvature. Also, this definition degenerates at an inflection point, i.e., when dt/dS = O. If the sense of the curvature is constant, one can always orient the X1-X2 frame so that coincides with ñ, to avoid working with a negative R. To complete the geometrical treatment, we consider the general parametric representation for the curve defining the centroidal axis,

rather than according to

x1 = x1(y) x2 = x2(y)

(j45

where y is a parameter. The differential arc length is related to dy by dS

+

d. -

2

2

+ (p)]

1/2

dv

=

dy

(14—6)

According to this definition, the +S sense coincides with the direction of t We summarize here for convenience the essential geometric relations for a plane curve which are developed in Chapter 4.

SEC. 14—2.

FORCE-EQUILIBRIUM EQUATIONS

and 1/R in terms of y are

increasing y. Using (14—6), the expressions for

-

t2 = — ( —-

R

1j

if

-

1

dx2

I 7dx1

t1 = — (

427

+— dy

dx2.. ——-—--

dy

+

dx1

dy

(14_

if._ (It1 = -( t2 dy — —

1

( k\

d2x1 dx2 dv2

dy

+

d2x2 dx1

dy2 dy

A planar member subjected to in-plane forces plane for our notation) will experience oniy in-plane deformation. In what follows, we develop the governing equations for planar deformation of an arbitrary planar member. This formulation is restricted to the linear geometric case. The two basic solution procedures, namely, the displacement and force methods, are described and applied to a circular member.

We also present a simplified formulation (Marguerre's equations) which is valid for a shallow member. Finally, we include a discussion of numerical integration techniques, since one must resort to numerical integration when the cross section is not constant. 14—2.

FORCE-EQUILIBRIUM EQUATIONS

The notation associated with a positive normal cross section, i.e., a cross

section whose outward normal points in the + S direction, is shown in Fig. 14—2.

We use the same notation as for the prismatic case, except that now the vector - I'3

dA

012

Centroidal axis

Fig. 14—2. Force and moment components acting on a positive cross section.


428

CHAP. 14

components are with respect to the local frame (Y1, Y2, Y3) rather than the

basic frame (X1, X2, X3). The cross-sectional properties are defined by

Since

if dy2 dy2

A

=

13

= JJ(y2)2 dii

=

if dii

(14—8)

'2 = .iJ(Y3)2

Y2, Y3 pass through the centroid and are principal directions, it follows

that

dA = flY3 dA =

dii =

SSY2Y3

0

(14—9)

When the member is planar (X1-X2 plane) and is subjected to a planar loading, F3

0

M2

M1

(14—10)

in this case, we work with reduced expressions for F÷ and M÷ (see Fig. 14—3) and drop the subscript on M3:

=

M+ =

M3t3

+ F212' = Mt3

(14-11)

Note that 13 is constant for a planar member. x2

) = t1 x t2

x1

Fig. 14—3. Force and moment components in planar behavior,

To establish the force-equilibrium equations, we consider the differential volume element shown in Fig. 14—4. We define b and as the statically equivalent external force and moment vectors per unit arc length acting at the centroid.

For equilibrium, the resultant force and moment vectors must vanish. These conditions lead to the following vector differential equilibrium equations: dS

—

dM÷

+

,_.

+ —

=

in + r1 x F+ =

o

-

0

(14—12)

SEC. 14—3.

We

expand b and


429

in terms of the unit vectors for the local frame:

b= + = mt3

b212

(14—13)

Introducing the component expansions in (14—12), and using the differentiation formulas for the unit vectors (14—3), lead to the following scalar differential equilibrium equations: dF1

—

F2

+ b1 = 0

(14-14)

dM

+

+m

0

that the force-equilibrium equations are coupled due to the curvature. The moment equilibrium equation has the same form as for the prismatic case. dS

r(S)

Fig. 14—4. Differential element for equilibrium analysis.

The positive sense of the end forces is shown in Fig. 14—5. We work with components referred to the local frame at each end. The end forces are related to the stress resultants and stress couples by

= = Mj52 = MA= —MISA

14—3.

(14-15)

j=1,2

FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUAL FORCES

We establish the force-displacement relations by applying the principal of virtual forces to a differential element. The procedure is the same as for the

CHAP. 14


430

prismatic case described in Sec. 12—3, except that now we work with displacement components referred to the local frame at each point. We define ü and as

=

=

=

centroid. = equivalent rigid-body rotation vector

rigid-body translation vector at the (14—16)

For planar deformation, only u1, u2 and 0J3 are finite, and the terms involving u3, co1, and w2 can be deleted: u1t1 + U2T2 — C03t3

(14—17)

Wt3

The positive sense of the displacement components is shown in Fig. 14—6.

F41

Fig. 14—5. Convention for end forces.

x2

x1

FIg. 14—6. Definition of displacement measures. t "Equivalence" refers to work. See (12—8).


SEC. 14—3.

431

as the complementary energy per unit arc length. For planar

We define

deformation, = (F1, F2, M). One determines by taking expansions for the stresses in terms of F1, F2, M, substituting in the complementary energy density, and integrating with respect to the cross-sectional coordinates Y2, y3. We will discuss the determination of later.

Specializing the three-dimensional principle of virtual forces for the onedimensional elastic case, and writing

=

cF1

=

e1

AF1

+

0F2

AF2

AF1 + e2 AF2 +

k

+

cM

AM (14—18)

AM

lead to the one-dimensional form Ss(ei AF1

+ e2 AF2 +

k

AM)dS

AP1

(14—19)

where is a displacement measure and is the force measure corresponding to d1. The virtual-force system (AF1, AF2, AM, AP1) must be statically permis-

sible, i.e., it must satisfy the one-dimensional equilibrium equations.

(

Fig. 14—7. Virtual force system

We apply (14—19) to the differential element shown in Fig. 14—7. The virtual

force system must satisfy the force-equilibrium equations (14—17), dS

Evaluating

AF÷ =

0

(a)

AP1,

=

+AM÷

+ (b)

+ AF2

= {AF1 —

+

—

+

dS

432


CHAP. 14

and then substituting in (14-49) results in the following relations between the

force and displacement parameters: cj

eV*

du1

U2

du2

u1

(14-20)

dw dS

k

We interpret e1 as an average extension, e, as an average transverse shear deformation, and k as a bending deformation. Actually, k is the relative rotation of adjacent cross sections. In what follows, we discuss the determination of Consider the differential volume element shown in Fig. 14—8. The vector defining the arc QQ1 is QQ1

=

ar2

dy =

+

di2

+

dt-\

dv

Noting that —

dy (112

dv

—7k-ti

=

o

for a planar member, (a) can be written as dS2

By definition,

=

=

—

— = is the complementary energy per unit length along the

centroidal axis. Substituting for dS2 in the general definition, we obtain dS

dS2 dv2 dv Y2,Y3

(14—2 1)

if

.-

In general, V* = V" (ô11, We select suitable expansions for the stress components in terms of F1, F2, M, expand V*, and integrate over the cross section. The only restriction on the stress expansions is that they satisfy the definition equations for the stress resultants and couples identically:

= F2 $5c12 dA SSa13 dA = JJy3aii dA = 0 —ify2ci1 dA = M

dA

J$(y2a13 — y3a12)dA = 0

0

433

PRINCIPLE OF VURTUAL FORCES

SEC. 14—3.

The most convenient choice for iH is the linear expansion,t M

(14—22)

—

where I 13. A logical choice for (when the cross section is thin-walled) is the distribution predicted by the engineering theory of flexural shear stress distribution described in Sec. 11—7: a11 = 1q(F2)

q=

F2t/i

(14—23)

where t denotes the local thickness, and q is the flexural shear flow due to F2. Both expansions satisfy (a). x2

r +1)212 +Y33

r2

r1ty +dy)

Y2

it

Fig. 14—8. Differential volume element.

In what follows, we consider the material to be linearly elastic. The complementary energy density is given by 11*.....

—

0

2

a12

2

where c? is the initial extensional strain. Substituting (a) in (14—21) and taking

the stresses according to (14—22), (14—23) results in the following expression f This applies for a homogeneous beam. Composite beams are more conveniently treated with the approach described in the next section.


434

CHAP. 14

for V*:

= e?Fi + k°M +

+

+

+

2GA2*

(14-24)

where

=

(i

if

dA

-

55

I

—

\\

—

R}

1 and If the section is symmetrical with respect to the 1'3 axis, 1* The deformation-force relations correspoiiding to this choice for —ei e,

M

F1

F2

du2

=

=

+

-w

M

F1

A2.

u2

dr,1

U1

= are

(14-25)

dw

Note that the axial force and moment are coupled, due to the curvature. Inverting (14—25) leads to expressions for the forces in terms of the deformations:

F1 = M— —

LA

e1)

—

—

k

R(l

)

EI*

+

Ô)(el —

R(1

/

k°

(14-26)

F2

We observe that

I —

where p is the radius of gyration and d is the depth of the cross section, For example, 1

d2

AR2 = i2R2 for a rectangular cross section. Then, is of the order of (d/R2) and can be neglected when (dIR)2 1. A curved member is said to be thin when O(d/R) 1.

We set ö =

0

1

and thick when O(d/R)2

for a thick member. The thinness assumption is introduced


SEC. 14—4.

435

neglecting y2/R with respect to unity in the expression for the differential arc length, i.e., by taking by

dS

14 27

-

Assuming a curved member to be thin is equivalent to using the expression

for V* developed for a prismatic member. The approximate form of (14—25) for a thin member is F1

dii1

Li2

(14—28)

i—k° To complete the treatment of the linear elastic case, we list the expanded forms of the principle of virtual forces for thick and thin members. Note that these expressions are based on a linear variation in normal stress over the cross

section. Thick Member

Cit0

+

F1

+

M'\ AF1

+

F2

/XF2

(14—29)

+ (ko +

+

AM} dS —

Thin

J

14—4.

+

+

/ M'\ + (\kO + h-i)

1

dS =

(14—30) d1 AP1

FORCE-DISPLACEMENT RELATIONS—DISPLACEMENT EXPANSION APPROACH; PRINCIPLE OF VIRTUAL DISPLACEMENTS

In the variational procedure for establishing one-dimensional force-displacement relations, it is not necessary to analyze the deformation, i.e., to determine

the strains at a point. One has only to introduce suitable expansions for the stress components in terms of the one-dimensional force parameters. Ndw, we• can also establish force-displacement relations by starting with expansions for the displacement components in terms of one-dimensional displacement parameters and determining the corresponding strain distribution. We express the

PLANAR OEFORMA11ON OF A PLANAR MEMBER

436

CHAP. 14

stresses in

terms of the displacement parameters using the stress-strain relations, and then substitute the stress expansions in the definition equations for F1, F2, and M. The effect of transverse shear deformation is usually neglected in this

approach. To determine the strain distribution, we must first analyze the deformation at a point. This step is described in detail below. Figure 14—9 shows the initial position of two orthogonal line elements, QQ1 and QQ2, at a point (y, Y2' y3). The vectors defining these elements are QQ1

dy2

QQ2 =

a2 =

I

dy2 t2

(14—31)

a —

We use a prime superscript to denote qua Iltities associated with the deformed position of the member, which is shown in Fig. 14—10; for example:

?'=

= position vector to point P(y) in the deformed position (point P'). tangent vector to the deformed centroidal axis. = position vector to Q(y, Y2, y3) in the deformed position (point Q'). x2 Q2 Q1

Pj(y +dy) P(y) axis

x1

Fig. 14—9. Initial geometry for orthogonal curvihnear line elements.

SEC. 14—4.


437

From Fig. 14—10, and noting (14—3 1): —

=

P'P'1 =

/

0))

=

+

or2

/

&Y2

\.

C))J c,u2

—

dy

(14—32)

dy2

The analysis of strain consists of determining the extensions and change in

angle between the line elements. We denote the extensional strains by

(j =

1,

2) and the shearing strain by Y12 The general expressions are

'—12 3—

(1-3)

Sin Y12

Now, we restrict this discussion to small strain, Substituting for the deformed

vectors and neglecting strains with respect to unity, (14—33) expands to

Istj,

au2 c'y

+

2(a2)"

—.

ô,V

1 ('U2

2

"

tj -'

Y12

a2

t2

-±

cc2

The nonlinear terms arc associated with the rotation of the tangent vector. Neglecting these terms corresponds to neglecting the difference between the deformed and undeformed geometry, i.e.. to assuming linear geometry. The next step involves introducing an expansion for in terms of y2. We express ü2 as a linear function of ü

wv211

(14—35)

where co = w(y) and U

U1t1 + U2t2 = 1kv)

(14—36)

is the displacement vector for a point on the centroidal axis. Equation (14—35) implies that a normal cross section remains a plane after deformation. One can interpret co as the rotation of the cross section in the direction from toward t2. This notation is illustrated in Fig. 14—1 1. In what follows, we consider only linear geometry. Substituting for ü2, taking y = S, and evaluating the derivatives lead to the following strain expansions:


438

=

—

= du,

y2k)

u2 —

1

e2

=

CHAP. 14

+

= I61IY20 (14—37)

0)

doi

The vanishing of c2 is due to our choice for ü2. One could include an addiThis would give tional linear term, = $ and, additional terms in the x2 Q2

x1

Fig. 14—10. Deformed geometry for orthogonal curvilinear line elements. u2t2

(u1 —Wy2)tl

Centroidal axis

UI tl

Fig. 14—11. Displacement expansion.


SEC. 14—4.

expressions for

439

that the assumption that a normal cross

and Y12• Note

section remains plane does not lead to a linear variation in extensional strain over the depth when the member is curved. We introduce the assumption of negligible transverse deformation by setting e2 = 0. The resulting expressions for (0 and k in terms of u1 and u2 are e2

=

0

du2

u1

+

dS

R

(14—38)

dIui

—

dS

—

dS2

When transverse shear deformation is neglected, one must determine F2 using the moment-equilibrium equation. The next step involves expressing F1, F,, and M in terms of the one-dimensional deformation parameters e1e2 and k. In what follows, we consider the material to be linearly elastic and take the stress-strain relations for c12 as:

=

= Gy12

E(c1

Substituting for r1, Y12, using (14—37),

=

F 1 —y2/R

— y,k) —

———--—-(e1

Fe1

(14—39)

and then evaluating F1, F2, and M, we obtain

F2

d

= Ge2 if =

(14-40)

+ Ekjj

—Fe1

+

The various integrals can be expressed in terms of only one integral by using the identity 1

1 — y2/R

and noting that Y is a

—

f The relation for

1

axis: $5Y2 dA

member.

y2/R

1-F

is exact only when

=

0

= (733

11

We generally neglect

for a

440


CHAP. 14

One can easily show that

ri

I'

c/A

dA

=

L JJ

(14—41)

1

- y2/R

For completeness, we list the inverted form of (14—40),

=

+

M

+

F2

k

= k° +

F1

lvi

+

where

=

+

= A(1 4—42

e? =

if

k° =

(i

c/A —

if

-

dA

The expressions for e1 are identical with the result (see (14—25)) obtained with

the variational approach. However, the result for k differs in the coefficient for M. This difference (1' or F') is due to the nonlinear expansion used for Example 14—1

We determine I' for the rectangular cross section shown in Fig. F.14—1.

I' =

11 1

y2/R

=h

J—a;2 1 — y2/R

=—R2bd+R3bln To obtain a more tractable form, we expand the log terms, using

(1+x'\

I


SEC. 14—4.

441

This series converges for xI < 1. Then

=

In

d

d3

+

3(d\4

3(d\2

I

ii

+

+

+

and

I' =

+

{

3

d

2

3

d

+ ..

+

Fig. E14—1 H

I

Y3

The relations listed above involve exactintegrals. Now, when the member is thick, we neglect (y2/R)2 with respect to unity. This assumption is introduced by taking

1 —y2/R

=

1

+

+ ...

+

+

and I':

in the expansions for

Co

—.--..e2

Y2

- yJR

=i{i

JJ

dA

+

442

PLANAR DEFORMATiON OF A PLANAR MEMBER

CHAP. 14

To be consistent, we must also neglect 1'/AR2 with respect to unity in the expression for A'2 and I". When the member is thin, we neglect y2/R with respect to unity.

1—y2/R —

y2k

(14—44)

—

at2

It is of interest to establish the one-dimensional form of the principle of virtual displacements corresponding to the linear displacement expansion used in this development. The general three-dimensional form for an orthogonal coordinate system is (see Sec. 10—6):

SJJ(aii

+

)d(vol.) =

+ a12 öy12

represents an external force quantity and d1 is the displacement quantity We consider only and Viz to be finitc, and express the corresponding to differential volume in terms of the cross-sectional coordinates Y2' Y3 and arc length along the ccntroidal axes (see Fig. 14—9): where

d(vol.) =

dS2

(i

dy2 dv3

dS

d7 dy3

—

Then (a) reduces to (a11

(i

+ a12

dA] dS =

(14—45)

—

We take (14—45) as the form of the principle of virtual displacements for planar

deformation.

The strains corresponding to a linear expansion for displacements and linear geometry are defined by (14—37), which are listed below for convenience: ci

—

Y12

—

y2/R

du1

= k

du2

e2

U2

+

u1

— U)

do dS

Substituting for e1, Y12 and using the definition equations for F1, F2, and M,

SEC. 14—4.

we


443

obtain

Ad1 (14—46) + F2 + M ök]dS = Js[Fi This result depends only on the strain expansions, i.e., (c). One can apply it for the geometrically nonlinear case, provided that (cS) are taken as defining

the strain distribution over the cross section. We use the principle of virtual displacements to establish consistent forceequilibrium equations. One starts with one-dimensional deformation-displacement relations, substitutes in (14—46), and integrates the left-hand side by parts. Equating coefficients of the displacement parameters leads to a set of force equilibrium equations and boundary conditions that are consistent with the geometrical assumptions introduced in establishing the deformation-displacement relations. The following example illustrates this application.

Example

14—2

The assumption of negligible transverse shear deformation is introduced by setting e2

equal to zero. This leads to an expression for the rotation. w, in terms of the translation components,

=

(In2

+

and the relations for negligible transverse shear deformation reduce to

hEFt ôe1 + M ök]dS

d (du2

1< =

u1

=

+

Substituting for Aw and the strain variations, d

= Au1 —id

=

(5k

Aui

d2

Au2 1

AU2

—

+

Au2

d

I

and integrating by parts, the left- and right-hand sides of (b) expand to

j 54

[F1

+ M (5k]dS

=

/ F1 /

—

I

\

+

F1 + I

+j

Rj M\

Rj r

J

Au1

dM —

dS

dM

Au2

Au1 — —-- Au2

1Aui

[—

dS

dF1

1

—

+M

d uS

An2

d

+M—

dM1

dS

+ Au2

[ [—

F1

dS

+


444

CHAP. 14

and

+

(b1 +

Ad,

=

dS

+ (p22 + ma) Au32 +

+ (PB! +

+ (p41 +

(b2

MA)

+ (r42

A

A

014) AUA2 + M4 A

The consistent equilibrium equations and boundary conditions for negligible transverse shear deformation follow by equating corresponding coefficients of the displacement variations in (e) and (f):

S4<S<S3 +

dF1

+

d2M

F1

+-

1dM

—

+ b1 +

01

0

drn

+

=0

—

s—sn u1

prescribed or

u2

prescribed or

do2

F1

—

p42

prescribed or

M=

—MA

UI

prescribed or

F1 =

F21

U2

prescribed or

=

— m

S=

du2

prescribed or

—F32

—

in

M

One can obtain (g) by solving the last equation in (14—14) for F2 and substituting in the first two equations. Suppose we neglect u1/R in the expression for w: do2 CD

d2u2

k

This assumptiont is generally referred to as Mush tori's approximation. The equilibrium

equations for the tangential direction reduce to dF1

t See Ref. 5.

CARTESIAN FORMULATION

SEC. 14—5.

445

The other equilibrium equation and the boundary conditions are not changed. Using (h) instead of (a) eliminates the shear term, F2/R, in the tangential force-equilibrium

equation.

14—5.


We consider the case where the equation defining the centroidal axis has the form x2 = f(x1). The geometrical relations for this parametric representation are obtained by taking y x1 in (14—7). They are summarized belowt for convenience and the notation is shown in Fig. 14—12:

dS = dx1

r [

7df\21112

ylx1jj. f'df'\ + I—)

= i[ I

-

I

t2

[ / df \

1

cos0 1 j

-

-

(14—47)

+ '2

= t1 X t2 = 13 d2f

I

ci:

In the previous formulation, we worked with displacement components and external force components referred to the local frame. An alternate approach, originally suggested by involves working with components referred to the basic frame rather than the local frame. The resulting expressions differ, and it is therefore of interest to describe this approach in detail. We start with the determination of the force-equilibrium equations. Consider the differential clement shown in Fig. 14—13. The vector, equilibrium

equations are dF+ dx1 See Prob. 14—1. See Ref. 6.

+

x dx1

-

-

ax1

(14—48)

+

=

0


446

CHAP. 14

x2 Y2

YI

X2

dx1

x1 'I

Fig. 14—12. Notation for Cartesian formulation.

dx1

2

pN2 12 1

F1t1

lj Fig. 14—13. Differential element for equilibrium analysis.


SEC. 14—5.

447

h are the external applied force and moment vectors per unit projected length, i.e., per unit x1. They are related to b and (see Fig. 14—4) by

where fl,

dx1 = b dS = (cth)dx1 hdx1 = iñdS = (cthi)dx1

(1449)

Substituting for the force and moment vectors,

= F1t1 + F2t2 = N171 + N212 it—hi3 + P = N1 = F1 cos 0 — F2 sin 0 N2 = F1 sin 0 + F2 cos 0

(14—50)

the equilibrium equations expand to dN1 —dx1

dx1

d = ——(F1 cosP — dx1

=

(F1

dx1

—1(dM

'\

——-(—-- + hJ=' \dx1 ,i We

sin 0) =

sin0 + F2 cos 0)

F2

(14-51)

P2

—N1 sm0 + N2cosO

restrict this treatment to an elastic material and establish the force-

displacement relations, using the principle of virtual forces,

dx1 =

[e1

AF1 + e2 AF2 + k

dx1 =

d1

(a)

where V* V* (F1, F2 M) is the complementary energy per unit arc length. Consider the differential element shown in Fig. 14—14. The virtual-force system is statically permissible, i.e., it satisfies the force-equilibrium equations identically:

=

dx1

+ J1 x

dx1

Expanding

o

=

d1

x

and then substituting for the displacement and rotation vectors, O= (0

v111

+ 1)212

(013 = (0t3

(14—52)


448

we

CHAP. 14

obtain

=

+

+

(d)

dx1; Finally, substituting for N1, N2, in terms of F1, F2 and equating coefficients of the force increments result in dx1

oV

dx1

do1

do2

cos2 0-— + sinOcosO—dx1

.

e2

a v*

=

—sin 0 COS

do1

,

U

--

do2

+cosO——w dx1

(14—53)

dw

k=

= ——— cos 0

dx1

The member is said to be shallow when 02 j1k4 = fjk

=

+

dx1

(14—78)

Jx,

= —

f

+

+ (k0 +

Example 14—8

Consider the two-hinged arch shown in Fig. E14—8A. We work with reaction components referred to the basic frame and take the horizontal reaction at B as the force redundant. Fig. E14—8A

x2,

P

Primary We must carry out two force analyses on the primary structure (Fig. E14—SB), one for the external forces (condition Z1 = 0) and the other for Z1 1. The results are displayed in Figs. E14—8C and D, respectively. Fig. E14—8B V2

= Z1

zI

R242

CHAP. 14


468

Fig. E14—8C N2

KM 0

Fig. E14—8D

H

(+)

'VI,'

ii

(+) N2,,

M,1

Compatibility Equation We suppose the member is not shallow. The compatibility equations for Z1 follow from (14—74):

I J11—I El

dx t cosO

+ f'N2,1) +

= Jo L

(ko

+

FORCE METHOD OF SOLUTION

SEC. 14—7.

469

Using the results listed above, the various terms in (a) expand to

f

— h

= .ft0E1

±

=

s:

+ f'N2

+ k°

cost?

L

+

dx1

JL[(

JLIJLI(h)(h) +

+

+

k0(f_

!L

LIcosO

L

(+P(xi

- a))dxi

Once the integrals are evaluated, we can determine Z1 from (c)

Finally, the total forces are obtained by superposition of the two loadings:

M = M•0 + Z1M1 R. = + Z1R,

=

(d) 1,

2, 3

R4 =

To evaluate the vertical displacement at point Q, we apply a unit vertical load at Q on the prinwry structure and determine the required internal forces and reactions plotted in Fig. E14—8E.

Fig. F—

XQ

1/L

Q

(i_IL) FQ=+l


470

CHAP. 14

Applying (14—73), we obtain VQ2

=

+

VA2

—

(13B2

VA2)

+

—

('XQ

—

I'

M\

CL

dx1

x1Ik°+—1—---—---—x1

I

\ 1L

El)cosO XQ1 M'\ dx1 ( +— x1(k°+-—J-—— EljcosO L jof \ Jo

/

dx1

(k°+—'l—-— El) cosO

I

A numerical procedure for evaluating these integrals is described in the next section.

Example

14—9

The symmetrical nonshallow two-hinged parabolic arch shown in Fig. E14—9A is subjected to a uniform load per unit horizontal length, that is, per unit x1. The equation for the centroidal axis is

4h( where h is the elevation at mid-span

= L/2). We take the horizontal reaction at the

right end as the force redundant and consider only bending deformation. Figures E14—9B and C carry through an analysis parallel to that of the preceding example,

Deternzination of Z1 and Total Forces The equation for Z1 follows from (14—74): 1L

Jo

EJcosO

pL2

ElcosO

Note that this result is valid for an arbitrary variation of El. Finally, the total forces are N1 — N10 + Z1N1 N2

=

= N20 + Z1N21 = —

M=M,0+Z1M,1 =0 Since M =

0, the deformed shape of the arch coincides with the initial shape when axial deformation is neglected. It follows that (c) also apply for the fixed nonshollolv case.

When the arch is shallow, the effect of axial deformation cannot be neglected. The expression for Z1 follows from (14—78): —

+

FORCE METHOD OF SOLUTION

SEC. 14—7.

471

Fig. E14—9A

p = Coflst

B

xi

Pnmary structure

R2,d2

Fig. E14—9B

N2

xI L

N5,00

M0

pL(

Force System Due to p

Xj\ )

pL


472

CHAP. 14

If E is constant, (d) reduces to JL dx1

pL2

1L1

8h

—

+

—

pL2

—

8h

1

I

+

jL1

The parameter ö is a measure of the influence of axial deformation. As an illustration, we

consider A and I to be constant and evaluate ö for this geometry. The result is

I

15(p'\2

— 8 Ah2

—

8

where p is the radius of gyration for the cross section. Fig.

A'2

N11 = +1

M,1 = +f

N2,1 = 0

Force System Due to Z1 = + I One should note that (e) applies only for the shallow case, in, for (f')2 K< 1. Now, 4/1

/

2x1

For the assumption of shallowness to be valid, 16(h/L)2 must he small with respect to unity. The total forces for the shallow case are

=

N1 =

M=

p1?

f(

pL2

ô

1

\=

PL( —

SEC. 14—8.

NUMERICAL INTEGRATION PROCEDURES

473

It is of interest to determine the rotation at B. The "Q" loading consists of a unit moment applied at B to the primary structure (see Fig. E14—9D). Applying (14—77) (note that Fig. E14—9D

PQ

+l

the stretching terms vanish since Nj,Q = 1L

M

0),

we obtain

P/

x1

j

\

/

,

When El is constant, (i) reduces to

pL3(

(5

2

Since

14—8.

0, the results for the fixed end shallow case will differ slightly from (h).


One of the steps in the force method involves evaluating certain integrals which depend on the member geometry and the cross-sectional properties. Closed-form solutions can be obtained for only simple geometries, and one usually must resort to a numerical integration procedure. In what follows, we describe two proceduresi which can be conveniently automated and illustrate their application in deflection computations.

We consider the problem of evaluating

J t See Ref. 8 for a more detailed treatment of numerical integration schemes.

(14—79)

__________________ PLANAR DEFORMATION OF A PLANAR MEMBER

474

where f(x) is a reasonably smooth function in the interval XA divide the total interval into n equal segments, of length h:

CHAP. 14

x

— XA

h=

We

(14—80)

If f(x) is discontinuous, we work with subintervals and use a different spacing x,, represent the for each subinterval. For convenience, we let x0, x1, the coordinates of the equally spaced points on the x axis, and f0, corresponding values of the function. This notation is shown in Figure 14—17. .

12

,

1;,

a

11

X2

X1

XO

.

B X,,

Fig. 14—17. Coordinate discretization for numerical integration.

The simplest approach consists in approximating the actual curve by a set

of straight lines connecting approximation,

etc., as shown in Fig. 14—18. With this ('xk

EXJk1,k =

f(x)dx dx =

=

h

+

(14—81)

+

If only the total integral is desired, we use,

J

f(x)dx

+ L)

=

(14—82)

+

which is called the trapezoidal rule. A more accurate formula is obtained by approximating the curve connecting

three consecutive points with a second-degree polynomial, as shown in Fig. 14—19. This leads to AJk,k+2

=

fdx

h

+ 4fk+1 +

(14—83)

J

Jk÷2= Jk + To apply (14—83), we must take an even number of segments, that is, n must

be an even integer. If the values of J at odd points are also desired, they can


SEC. 14—8.

be

475

determined using h

= Finally, one can express

12

Jo

{Sfk +

(14—84)

— Jk+2]

as

in =

+

+ 4(11 + .13 +

+ (14—85)

+ 2(f2 +f4 + Equation (14—85) is called Simpson's rule.

f

N fk+1

fk

fk—1

h

S

Xk_1

x

Xk

Fig. 14—18. Linear approximation.

I

S Xk

fk÷2

fk+1

fk

S

Xkf.1

Fig. 14—19. Parabolic approximation.

________________M CHAP. 14


476

Example 14—10 Consider the problem of determining the vertical displacement at Q for the straight member of Fig. E14—lO. We suppose shear deformation is negligible. The deflection due

Fig. E14—10

XQ

M

PQ

+1

XQ(1t)

to bending deformation (we consider the material to be linear elastic) is given by dQ

MQ dx

J

where M is the actual moment and M0 is due to the "Q" loading. Substituting for M, expands to

(a)

=

/ 1L M

CXQ

— J

M

'\

M

M

+ J

—


SEC. 14—8.

477

To evaluate (b), we divide the total length into ii equal segments of length h, number the points from 0 to n, and let

M

—dx J0 El Cx M x—dx El Jo I

I

With this notation, (b) takes the form = Xk

+ "k

—

Xkjk

If, in determining we also evaluate the integrals the interior points, then we can readily determine the displacement distribution using cd).

Example

14—11

Consider the simply supported nonshallow arch shown. We suppose there is some distribution of It'! and we want to determine the vertical deflection at Q. Considering Fig. E14—11

A

ill

El L

only bending deformation, dQ is given by dQ

=

M,

1W

ds

1W,

J

dx


478

CHAP. 14

Now, the distribution of M

is the same as for the straight member, Then, the procedure followed in Example 14—10 is also applicable here. We just have to replace M/EI with M/EI cos 0 in Equation (c) of Example 14—10.

REFERENCES 1.

TIMOSHENKO, S. 1.: Advanced Strength of Materials, Van Nostrand, New York, 1941.

2.

Boan, S. F., and J. J. GENNARO: Advanced Structural Analysis, Van Nostrand, New York, 1959.

3.

REISSNER, E.: "Variational Considerations for Elastic Beams and Shells," J. Lag. Mech. Div., A.S.C.E, Vol. 88, No. EM 1, February 1962.

4.

MARTIN, H. C.: introduction to Matrix lvi etliods of Structural Analysis, McGraw—Hill, New York, 1966. MUSRTARI, K, M., and K. Z. (IALIMOV: "Nonlinear Theory of Thin Elastic Shells,"

5. 6. 7.

8.

Israel Program for Scientific Translations. Jerusalem, 1962 MARGUERRE, K.: "Zur Theoric der gekriimmten Platte grosser Formanderung," Proc. 5th mt. Congress App!. Mccli. pp. 93—101. 1938. Onai'i, J. 1.: Mechanics of Elastic Structures. McGraw-Hill, New York, 1967. HILDEBRAND, F. J.: introduction to Numerical Analysis, McGraw-Hill, New York, 1956.

PROBLEMS

Specialize (14—7) for the case where Yi = x1. Let x2 = f(x1) and 14—1. let 9 be the angle from X1 to Y1 as shown below. Evaluate the various terms for a parabola

f=

+

Finally, specialize the relations for a shallow curve, i.e., where Prob.14—1 32

14—2. Evaluate 1* and the sketch. 14—3. Verify (14—34).

14—4. 14—5.

(see Equation 14-24) for the section defined by

Verify (14—41) and (14—42).

Discuss the difference between the deformation-force relations based

on stress and displacement expansions (Equations (14—25) and (14—42)). Illustrate for the rectangular section treated in Example 14—i. Which set of relations would you employ?

PROBLEMS

479

2t

I

Prob.

14—2

Prob.

14—6

T b=O.75d

d

I 14—6.

Evaluate I' and 1" for the symmetrical section shown.

h=O.75d

t=d120

14—7. Consider a circular sandwich member comprised of three layers, as shown. The core layer is soft (E 0). and the face thickness is small in comparison to the depth d). Establish force-deformation relations based

on strain expansions (see (14—37)). Prob. 14—i

I

II I

Starting with (14—34) and (14—35), derive a set of nonlinear strain 14—8. displacement relations for a thin member. Assume small finite rotation, and


480

CHAP. 14

linearize the expressions with respect to Yz' i.e., take

=

e1 —

Y12

Determine the corresponding force-equilibrium equations with the principle of virtual displacements. 14—9. Refer to Fig. 14—10 and Equation (14—31). If we neglect transverse is orthogonal to t'1 and we can write shear deformation,

=

(1 +

—

tj

ldF' -

= fl1t1 + fl2t2

= —/32t1 + f31t2 1 + e1 .-, di'2 I + e1

dt'1

=

R'

= (a)

Verify that

+

= (1 +

can be expressed as 1

(i+e1

c

Cl

I

- y,k}

Also determine e1 and R' for small strain. Express ü in terms of th initial tangent vectors,

ü= and take y (b)

S (i.e.,

U1t1

+ U2t2

1).

Derive the force-equilibrium equations, starting with the vector equations (see (14—12) and Fig. 14—4),

dS

dM÷

+m +

+

=

0

—

x F÷ = 0

and expanding the force vectors in terms of components referred to the deformed frame: = F11'I + = Mi3 (c)

h=

b11'1

+

Assume small strain. Derive the force-equilibrium equations with the principle of virtual displacements. Take the strain distribution according to Equation (b).

________ PROBLEMS

(d)

481

Derive the nonlinear deformation-displacement and equilibrium equations for the cartesian formulation. Refer the translations and loading to the basic frame, i.e., take

= P

+ V212

Pi'i + P2t2

Specialize the equations for the case of a shallow member. 14—10. The accompanying sketch applies to both phases of this

problem.

Prob. 14—10

h2 = const

(a)

(b)

Determine the complete solution for the circular member shown. Utilize symmetry at point A = co = F2 = 0) and work with (14—58), (14—59). Discuss the effect of neglecting extensional and shear deformation, i.e., setting (1/A) (1/42) = 0. Repeat (a), using Mushtari's equations for a thin member with no

transverse shear deformation, which are developed in Example 14—2. Show that Mushtari's approximation (u1

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