e

σ = »(trµ)I + 2µµ

This equation can be rewritten in terms of the fourth-order elasticity

tensor C as,

σ = C : µ; C = »I — I + 2µI; Cijkl = »δij δkl + 2µδik δjl

Alternatively the above relationship can be inverted to give the strain

expressed in terms of the stress tensor. To achieve this, note ¬rst that

taking the trace of the above stress“strain equation gives,

trσ = (3» + 2µ)trµ

and consequently µ can be written as,

1 »trσ

µ= σ’ I

2µ 2µ(3» + 2µ)

or in terms of the Young™s modulus E and Poisson™s ratio ν as,

1 µ(3» + 2µ) »

µ= [(1 + ν)σ ’ ν(trσ)I]; E= ; µ=

E »+µ 2» + 2µ

Hence the inverse elasticity tensor can be de¬ned as,

ν 1+ν

µ = C ’1 : σ; C ’1 = ’ I — I + I

E E

25

2.3 LINEARIZATION AND THE DIRECTIONAL DERIVATIVE

f

f(x0)

x

x0 x1 x2

FIGURE 2.3 One-degree-of-freedom nonlinear problem f (x) = 0.

2.3 LINEARIZATION AND THE DIRECTIONAL

DERIVATIVE

Nonlinear problems in continuum mechanics are invariably solved by lin-

earizing the nonlinear equations and iteratively solving the resulting linear

equations until a solution to the nonlinear problem is found. The Newton“

Raphson method is the most popular example of such a technique. Correct

linearization of the nonlinear equations is fundamental for the success of

such techniques. In this section we will consolidate the concept of the direc-

tional derivative introduced in Chapter 1. The familiar Newton“Raphson

scheme will be used as the initial vehicle for exploring the ideas that will

eventually be generalized.

2.3.1 ONE DEGREE OF FREEDOM

Consider the one-degree-of-freedom nonlinear equation shown in Figure 2.3,

f (x) = 0 (2.87)

Given an initial guess of the solution, x0 , the function f (x) can be ex-

pressed in the neighborhood of x0 using a Taylor™s series as,

1 d2 f

df

(x ’ x0 )2 + · · ·

f (x) = f (x0 ) + (x ’ x0 ) + (2.88)

2 dx2

dx x0 x0

If the increment in x is expressed as u = (x’x0 ) then (2.88) can be rewritten

as,

1 d2 f

df

u2 + · · ·

f (x0 + u) = f (x0 ) + u+ (2.89)

2 dx2

dx x0 x0

26 MATHEMATICAL PRELIMINARIES

To establish the Newton“Raphson procedure for this single-degree-of-freedom

case, (2.89) is linearized by truncating the Taylor™s expression to give,

df

f (x0 + u) ≈ f (x0 ) + u (2.90)

dx x0

This is clearly a linear function in u, and the term u(df /dx)|x0 is called

the linearized increment in f (x) at x0 with respect to u. This is generally

expressed as,

df

Df (x0 )[u] = u ≈ f (x0 + u) ’ f (x0 ) (2.91)

dx x0

The symbol Df (x0 )[u] denotes a derivative, formed at x0 , that operates in

some linear manner (not necessarily multiplicative as here) on u.

Using Equation (2.90) the Newton“Raphson iterative procedure is set up

by requiring the function f (xk + u) to vanish, thus giving a linear equation

in u as,

f (xk ) + Df (xk )[u] = 0 (2.92)

from which the new iterative value xk+1 , illustrated in Figure 2.3, is obtained

as,

’1

df

u= ’ f (xk ); xk+1 = xk + u (2.93)

dx xk

This simple one-degree-of-freedom case will now be generalized in order

to further develop the concept of the directional derivative.

2.3.2 GENERAL SOLUTION TO A NONLINEAR PROBLEM

Consider a set of general nonlinear equations given as,

F (x) = 0 (2.94)

where the function F (x) can represent anything from a system of nonlinear

algebraic equations to more complex cases such as nonlinear di¬erential

equations where the unknowns x could be sets of functions. Consequently x

represents a list of unknown variables or functions.

Consider an initial guess x0 and a general change or increment u that, it

is hoped, will generate x = x0 + u closer to the solution of Equation (2.94).

In order to replicate the example given in Section 2.3.1 and because, in

general, it is not immediately obvious how to express the derivative of a

complicated function F with respect to what could also be a function x, a

single arti¬cial parameter is introduced that enables a nonlinear function

27

2.3 LINEARIZATION AND THE DIRECTIONAL DERIVATIVE

F(†)

f

F (0)

f(x0)

x

†

u

x0 0 1 2

†

0 1 2 3

FIGURE 2.4 Single DOF nonlinear problem f (x) = 0 and F ( ) = 0.

F in , (not equal to F ), to be established as,

F( ) = F (x0 + u) (2.95)

For example, in the one-degree-of-freedom case, discussed in Section 2.3.1,

F( ) becomes,

F ( ) = f (x0 + u) (2.96)

This is illustrated for the one-degree-of-freedom case in Figure 2.4.

A more general case, illustrating (2.95), involving two unknown variables

x1 and x2 is shown in Figure 2.5. Observe how changes the function F in

the direction u and that clearly F( ) = F (x).

In order to develop the Newton“Raphson method together with the as-

sociated linearized equations, a Taylor™s series expansion of the nonlinear

function F( ) about = 0, corresponding to x = x0 , gives,

1 d2 F

dF 2

F( ) = F(0) + + + ··· (2.97)

2d 2

d =0 =0

Introducing the de¬nition of F given in Equation (2.95) into the above Tay-

lor™s series yields,

2

d2

d

F (x0 + u) = F (x0 ) + F (x0 + u) + F (x0 + u) + · · ·

2d2

d =0 =0

(2.98)

Truncating this Taylor™s series gives the change, or increment, in the non-

28 MATHEMATICAL PRELIMINARIES

f

F( †)

f( x1 , x 2 )

x2

†

x1

0 1 2

x0

u

†

FIGURE 2.5 Two-degrees-of-freedom nonlinear problem f (x1 , x2 ) = 0 and F ( ) = 0.

linear function F (x) as,

d

F (x0 + u) ’ F (x0 ) ≈ F (x0 + u) (2.99)

d =0

Note that in this equation is an arti¬cial parameter that is simply being

used as a vehicle to perform the derivative. In order to eliminate from the

left-hand side of this equation, let = 1, thereby giving a linear approxima-

tion to the increment of F (x) as

d

F (x0 + u) ’ F (x0 ) ≈ 1 F (x0 + u) (2.100)

d =0

where the term on the right-hand side of the above equation is the directional

derivative of F (x) at x0 in the direction of u and is written as,

d

DF (x0 )[u] = F (x0 + u) (2.101)

d =0

Note that u could be a list of variables or functions, hence the term “in the

direction” is, at the moment, extremely general in its interpretation. With

the help of the directional derivative the value of F (x0 + u) can now be

linearized or linearly approximated as,

F (x0 + u) ≈ F (x0 ) + DF (x0 )[u] (2.102)

Returning to the nonlinear Equation (2.94), setting F(x0 + u) = 0 in

(2.102) gives,

F (x0 ) + DF (x0 )[u] = 0 (2.103)

29

2.3 LINEARIZATION AND THE DIRECTIONAL DERIVATIVE

which is a linear equation with respect to u* . Assuming that (2.103) can

be solved for u, then a general Newton“Raphson procedure can be re-

established as,

DF (xk )[u] = ’F (xk ) ; xk+1 = xk + u (2.104)

2.3.3 PROPERTIES OF THE DIRECTIONAL DERIVATIVE

The directional derivative de¬ned above satis¬es the usual properties of the

derivative. These are listed for completeness below,

(a) If F (x) = F 1 (x) + F 2 (x) then,

DF (x0 )[u] = DF 1 (x0 )[u] + DF 2 (x0 )[u] (2.105a)

(b) The product rule: if F (x) = F 1 (x) · F 2 (x), where “ · ” means any type

of product, then,

DF (x0 )[u] = DF 1 (x0 )[u] · F 2 (x0 ) + F 1 (x0 ) · DF 2 (x0 )[u]

(c) The chain rule: if F (x) = F 1 (F 2 (x)) then,