Y,y

(1+d ux /dX)dX

(d uy /dX)dX

ds

u(X,Y)

u(X+dX,Y)

dX

X,x

FIGURE 1.9 General deformation of a two-dimensional body.

are very small, so that the initial and ¬nal positions of a given particle are

practically the same. When the displacements are large, however, this is

no longer the case and one must distinguish between initial and ¬nal co-

ordinates of particles. This is typically done by using capital letters X, Y

for the initial positions and lower case x, y for the current coordinates. It

would then be tempting to extend the use of the above equations to the

nonlinear case by simply replacing derivatives with respect to x and y by

their corresponding initial coordinates X, Y . It is easy to show that for large

displacement situations this would result in strains that contradict the phys-

ical reality. Consider for instance a two-dimensional solid undergoing a 90

degree rotation about the origin as shown in Figure 1.8. The corresponding

displacements of any given particle are seen from the ¬gure to be,

ux = ’X ’ Y (1.17a)

uy = X ’ Y (1.17b)

and therefore the application of the above formulas gives,

µxx = µyy = ’1; µxy = 0 (1.18a,b)

These values are clearly incorrect, as the solid experiences no strain during

the rotation.

It is clearly necessary to re-establish the de¬nition of strain for a contin-

uum so that physically correct results are obtained when the body is subject

to a ¬nite motion or deformation process. Although general nonlinear strain

measures will be discussed at length in Chapter 3, we can introduce some

13

1.3 NONLINEAR STRAIN MEASURES

of the basic ideas by trying to extend the de¬nition of Green™s strain given

in Equation (1.8a) to the two-dimensional case. Consider for this purpose a

small elemental segment dX initially parallel to the x axis that is deformed

to a length ds as shown in Figure 1.9. The ¬nal length can be evaluated

from the displacements as,

2 2

‚uy

‚ux

2

ds = dX + dX + dX (1.19)

‚X ‚X

Based on the 1-D Green strain Equation (1.8a), the x component of the 2-D

Green strain can now be de¬ned as,

ds2 ’ dX 2

Exx =

2dX 2

2 2

‚uy

1 ‚ux

= 1+ + ’1

2 ‚X ‚X

2 2

‚uy

‚ux 1 ‚ux

= + + (1.20a)

‚X 2 ‚X ‚X

Using similar arguments equations for Eyy and (with more di¬culty) the

shear strains Exy = Eyx are obtained as,

2 2

‚uy ‚uy

1 ‚ux

Eyy = + + (1.20b)

‚Y 2 ‚Y ‚Y

1 ‚ux ‚uy 1 ‚ux ‚ux ‚uy ‚uy

Exy = + + + (1.20c)

2 ‚Y ‚X 2 ‚X ‚Y ‚X ‚Y

Clearly, if the displacements are small, the quadratic terms in the above

expressions can be ignored and we recover Equations (1.16a,b,c). It is a

simple exercise to show that for the rigid rotation case discussed above,

the Green strain components are Exx = Eyy = Exy = 0, which coincides

with one™s intuitive perception of the lack of strain in this particular type

of motion.

It is clear from Equations (1.20a“c) that nonlinear measures of strain

in terms of displacements can become much more intricate than in the lin-

ear case. In general, it is preferable to restrict the use of displacements

as problem variables to linear situations where they can be assumed to be

in¬nitesimal and deal with fully nonlinear cases using current or ¬nal posi-

tions x(X, Y ) and y(X, Y ) as problem variables. In a fully nonlinear context,

however, linear displacements will arise again during the Newton“Raphson

solution process as iterative increments from the current position of the

14 INTRODUCTION

x1 x2

k k

F

1 2

FIGURE 1.10 Two-degrees-of-freedom linear spring structure.

body until ¬nal equilibrium is reached. This linearization process is one of

the most crucial aspects of nonlinear analysis and will be introduced in the

next section. Finally, it is apparent that a notation more powerful than

the one used above will be needed to deal successfully with more complex

three-dimensional cases. In particular, Cartesian tensor notation has been

chosen in this book as it provides a reasonable balance between clarity and

generality. The basic elements of this type of notation are introduced in

Chapter 2. Indicial tensor notation is used only very sparingly, although

indicial equations can be easily translated into a computer language such as

FORTRAN.

1.4 DIRECTIONAL DERIVATIVE, LINEARIZATION AND

EQUATION SOLUTION

The solution to the nonlinear equilibrium equation, typi¬ed by (1.11a),

amounts to ¬nding the position x for a given load F . This is achieved in

¬nite deformation ¬nite element analysis by using a Newton“Raphson iter-

ation. Generally this involves the linearization of the equilibrium equations,

which requires an understanding of the directional derivative. A directional

derivative is a generalization of a derivative in that it provides the change in

an item due to a small change in something upon which the item depends.

For example the item could be the determinant of a matrix, in which case

the small change would be in the matrix itself.

1.4.1 DIRECTIONAL DERIVATIVE

This topic is discussed in detail in Chapter 2 but will be introduced here

via a tangible example using the simple linear spring structure shown in

Figure 1.10.

The total potential energy (TPE), Π, of the structure is,

Π(x) = 2 kx2 + 1 k(x2 ’ x1 )2 ’ F x2

1

(1.21)

1 2

where x = (x1 , x2 )T and x1 and x2 are the displacements of the Joints 1 and

15

1.4 DIRECTIONAL DERIVATIVE AND LINEARIZATION

2. Now consider the TPE due to a change in displacements given by the

increment vector u = (u1 , u2 )T as,

Π(x + u) = 2 k(x1 + u1 )2 + 1 k(x2 + u2 ’ x1 ’ u1 )2 ’ F (x2 + u2 )

1

(1.22)

2

The directional derivative represents the gradient of Π in the direction u and

gives a linear (or ¬rst order) approximation to the increment in TPE due to

the increment in position u as,

DΠ(x)[u] ≈ Π(x + u) ’ Π(x) (1.23)

where the general notation DΠ(x)[u] indicates directional derivative of Π at

x in the direction of an increment u. The evaluation of this derivative is

illustrated in Figure 1.11 and relies on the introduction of a parameter

that is used to scale the increment u to give new displacements x1 + u1 and

x2 + u2 for which the TPE is,

Π(x + u) = 2 k(x1 + u1 )2 + 2 k(x2 + u2 ’ x1 ’ u1 )2 ’ F (x2 + u2 )

1 1

(1.24)

Observe that for a given x and u the TPE is now a function of the parameter

and a ¬rst-order Taylor series expansion about = 0 gives,

d

Π(x + u) ≈ Π(x) + Π(x + u) (1.25)

d =0

If we take = 1 in this equation and compare it with Equation (1.23), an

equation for the directional derivative emerges as,

d

DΠ(x)[u] = Π(x + u)

d =0

= kx1 u1 + k(x2 ’ x1 )(u2 ’ u1 ) ’ F u2

= uT (Kx ’ F) (1.26)

where,

2k ’k 0

K= ; F= (1.27)

’k k F

It is important to note that although the TPE function Π(x) was nonlinear

in x, the directional derivative DΠ(x)[u] is always linear in u. In this sense

we say that the function has been linearized with respect to the increment

u.

The equilibrium of the structure is enforced by requiring the TPE to

be stationary, which implies that the gradient of Π must vanish for any

16 INTRODUCTION

Π

Π (x + †u)

Π (x + u)

Π (x)

D Π (x)[u]

x2

x

x1

u

†=0

†=1

†

FIGURE 1.11 Directional derivative.

direction u. This is expressed in terms of the directional derivative as,

DΠ(x)[u] = 0; for any u (1.28)

and consequently the equilibrium position x satis¬es,

Kx ’ F = 0 (1.29)

If the direction u in Equation (1.26) or (1.28) is interpreted as a virtual

displacement δu then, clearly, the virtual work expression of equilibrium is

obtained.

The concept of the directional derivative is far more general than this

example implies. For example, we can ¬nd the directional derivative of the

determinant of a 2 — 2 matrix A = [Aij ] in the direction of the change

U = [Uij ], for i, j = 1, 2 as,

d

D det(A)[U] = det(A + U)

d =0

d

= (A11 + U11 )(A22 + U22 )

d =0

’ (A21 + U21 )(A12 + U12 )

= A22 U11 + A11 U22 ’ A21 U12 ’ A12 U21 (1.30)

We will see in Chapter 2 that for general n — n matrices this directional

derivative can be rewritten as,

D det(A)[U] = det A (A’T : U) (1.31)

n

where, generally, the double contraction of two matrices is A : B = i,j=1 Aij Bij .

17

1.4 DIRECTIONAL DERIVATIVE AND LINEARIZATION

1.4.2 LINEARIZATION AND SOLUTION OF NONLINEAR

ALGEBRAIC EQUATIONS

As a prelude to FE work, let us consider the solution of a set of nonlinear

algebraic equations,

R(x) = 0 (1.32)

where, for example, for a simple case with two equations and two unknowns,

R1 (x1 , x2 ) x1

R(x) = ; x= (1.33a,b)

R2 (x1 , x2 ) x2