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(1+d ux /dX)dX

(d uy /dX)dX



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

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
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
1 ‚ux
= 1+ + ’1
2 ‚X ‚X

2 2
‚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

x1 x2
k k
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


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.

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 2

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

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 )

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

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,

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,

DΠ(x)[u] = Π(x + u)
d =0

= kx1 u1 + k(x2 ’ x1 )(u2 ’ u1 ) ’ F u2
= uT (Kx ’ F) (1.26)

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
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

Π (x + †u)
Π (x + u)

Π (x)
D Π (x)[u]

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
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 det(A)[U] = det(A + U)
d =0

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

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


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