0

ˆ

ˆ

n3 = F N 3 = ° 0 »

0.7138

(continued)

22 KINEMATICS

EXAMPLE 3.7 (cont.)

ˆ

Since N ± are principal directions, n± are orthogonal vectors and the

corresponding elemental spatial volume is conveniently,

ˆ ˆ ˆ

dv = n1 n3 = 1

n2

ˆ

thus demonstrating the isochoric nature of F .

EXAMPLE 3.8: Simple shear

x2

X2

1

X1, x1

Sometimes the motion of a body

is isochoric and the distortional component of F coincides with F . A

well-known example is the simple shear of a 2-D block as de¬ned by

the motion,

x1 = X1 + γX2

x2 = X2

for any arbitrary value of γ. A simple derivation gives the deformation

gradient and its Jacobean J as,

1γ

F= ; J = det F = 1

01

and the Lagrangian and Eulerian deformation tensors are,

10 10

γ γ

E= ; e=

γ2 2 γ ’γ 2

2γ

3.9 AREA CHANGE

Consider an element of area in the initial con¬guration dA = dA N which

after deformation becomes da = da n as shown in Figure 3.7. For the

purpose of obtaining a relationship between these two vectors, consider an

23

3.10 LINEARIZED KINEMATICS

X 3, x 3

` n

dl

da

N da

p

dL

dA

dA

time = t

P

X1 , x1

X 2 ,x2

time = 0

FIGURE 3.7 Area change.

arbitrary material vector dL, which after deformation pushes forward to dl.

The corresponding initial and current volume elements are,

dV = dL · dA (3.66a)

dv = dl · da (3.66b)

Relating the current and initial volumes in terms of the Jacobian J and

recalling that dl = F dL gives,

JdL · dA = (F dL) · da (3.67)

The fact that the above expression is valid for any vector dL enables the

elements of area to be related as,

da = JF ’T dA (3.68)

3.10 LINEARIZED KINEMATICS

The strain quantities de¬ned in the previous section are nonlinear expres-

sions in terms of the motion φ and will lead to nonlinear governing equa-

tions. These governing equations will need to be linearized in order to enable

a Newton“Raphson solution process. It is therefore essential to derive equa-

24 KINEMATICS

tions for the linearization of the above strain quantities with respect to small

changes in the motion.

3.10.1 LINEARIZED DEFORMATION GRADIENT

Consider a small displacement u(x) from the current con¬guration x =

φt (X) = φ(X, t) as shown in Figure 3.8. The deformation gradient F can

be linearized in the direction of u at this position as,

d

DF (φt )[u] = F (φt + u)

d =0

d ‚(φt + u)

=

d ‚X

=0

d ‚φt ‚u

= +

d ‚X ‚X

=0

‚u

=

‚X

= ( u)F (3.69)

Note that if u is given as a function of the initial position of the body

particles X (the material description) then,

‚u(X)

DF [u] = =u (3.70)

‚X

where indicates the gradient with respect to the coordinates at the initial

con¬guration.

3.10.2 LINEARIZED STRAIN

Using Equation (3.69) and the product rule seen in Section 2.3.3, the La-

grangian strain can be linearized at the current con¬guration in the direction

u as,

DE[u] = 1 (F T DF [u] + DF T [u]F )

2

= 1 [F T uF + F T ( u)T F ]

2

= 1 F T [ u + ( u)T ]F (3.71)

2

Note that half the tensor inside [ ] is the small strain tensor µ, and, therefore,

DE[u] can be interpreted as the pull back of the small strain tensor µ as,

DE[u] = φ’1 [µ] = F T µF (3.72)

—

25

3.10 LINEARIZED KINEMATICS

X 3, x 3

u

`t

u

u (xp )

p

P time = t

X1 , x1

X 2 ,x2

time = 0

FIGURE 3.8 Linearized kinematics.

In particular, if the linearization of E is performed at the initial material

con¬guration, that is, when x = X and therefore F = I, then,

DE 0 [u] = µ (3.73)

Similarly, the right and left Cauchy“Green deformation tensors de¬ned

in Equations (3.15,17) can be linearized to give,

DC[u] = 2F T µF (3.74a)

Db[u] = ( u)b + b( u)T (3.74b)

3.10.3 LINEARIZED VOLUME CHANGE

The volume change has been shown earlier to be given by the Jacobian J =

det F . Using the chain rule given in Section 2.3.3, the directional derivative

of J with respect to an increment u in the spatial con¬guration is,

DJ[u] = D det(F )[DF [u]] (3.75)

26 KINEMATICS

Recalling the directional derivative of the determinant from (2.119) and the

linearization of F from (3.69) gives,

‚u

DJ[u] = J tr F ’1

‚X

= J tr u

=J ·u (3.76)

Alternatively, the above equation can be expressed in terms of the linear

strain tensor µ as,

DJ[u] = J trµ (3.77)

Finally, the directional derivative of the volume element in the direction

of u emerges from Equation (3.57) as,

D(dv)[u] = trµ dv (3.78)

3.11 VELOCITY AND MATERIAL TIME DERIVATIVES

3.11.1 VELOCITY

Obviously, many nonlinear processes are time-dependant; consequently, it is

necessary to consider velocity and material time derivatives of various quan-

tities. However, even if the process is not rate-dependant it is nevertheless

convenient to establish the equilibrium equations in terms of virtual ve-

locities and associated virtual time-dependant quantities. For this purpose

consider the usual motion of the body given by Equation (3.1) as,

x = φ(X, t) (3.79)

from which the velocity of a particle is de¬ned as the time derivative of φ

as (see Figure 3.9),

‚φ(X, t)

v(X, t) = (3.80)

‚t

Observe that the velocity is a spatial vector despite the fact that the equation

has been expressed in terms of the material coordinates of the particle X.

In fact, by inverting Equation (3.79) the velocity can be more consistently

expressed as a function of the spatial position x and time as,

v(x, t) = v(φ’1 (x, t), t) (3.81)

27

3.11 VELOCITY AND MATERIAL TIME DERIVATIVES

X3, x 3 vq