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q
`




vp
Q
p


time = t
X1, x 1 P
X2, x 2


time = 0

FIGURE 3.9 Particle velocity.


3.11.2 MATERIAL TIME DERIVATIVE
Given a general scalar or tensor quantity g, expressed in terms of the ma-
terial coordinates X, the time derivative of g(X, t) denoted henceforth by

g(X, t) or dg(X, t)/dt is de¬ned as,
dg ‚g(X, t)

g= = (3.82)
dt ‚t
This expression measures the change in g associated with a speci¬c particle
initially located at X, and it is known as the material time derivative of
g. Invariably, however, spatial quantities are expressed as functions of the
spatial position x, in which case the material derivative is more complicated
to establish. The complication arises because as time progresses the speci¬c
particle being considered changes spatial position. Consequently, the mate-
rial time derivative in this case is obtained from a careful consideration of
the motion of the particle as,
g(φ(X, t + ∆t), t + ∆t) ’ g(φ(X, t), t)

g(x, t) = lim (3.83)
∆t
∆t’0

This equation clearly illustrates that g changes in time, (i) as a result of a
change in time but with the particle remaining in the same spatial position
and (ii) because of the change in spatial position of the speci¬c particle.
Using the chain rule Equation (3.83) gives the material derivative of g(x, t)
28 KINEMATICS



as,
‚g(x, t) ‚g(x, t) ‚φ(X, t) ‚g(x, t)

g(x, t) = + = + ( g)v (3.84)
‚t ‚x ‚t ‚t
The second term, involving the particle velocity in Equation (3.84) is often
referred to as the convective derivative.

EXAMPLE 3.9: Material time derivative
Here Example 3.1 is revisited to illustrate the calculation of a material
time derivative based on either a material or spatial description. The
material description of the temperature distribution along the rod is
™ ™
T = Xt2 , yielding T directly as T = 2Xt. From the description of
motion, x = (1+t)X, the velocity is expressed as v = X or v = x/(1+t)
in the material and spatial descriptions respectively. Using the spatial
description, T = xt2 /(1 + t) gives,
(2t + t2 )x t2
‚T (x, t) ‚T (x, t)
= ; T= =
(1 + t)2
‚t ‚x (1 + t)

Hence from (3.84), T = 2xt/(1 + t) = 2Xt.



3.11.3 DIRECTIONAL DERIVATIVE AND TIME RATES
Traditionally, linearization has been implemented in terms of an arti¬cial
time and associated rates. This procedure, however, leads to confusion
when real rates are involved in the problem. It transpires that linearization
as de¬ned in Chapter 2, Equation (2.101), avoids this confusion and leads
to a much clearer ¬nite element formulation. Nevertheless it is valuable
to appreciate the relationship between linearization and the material time
derivative. For this purpose consider a general operator F that applies to
the motion φ(X, t). The directional derivative of F in the direction of v
coincides with the time derivative of F , that is,
d
DF [v] = F (φ(X, t)) (3.85)
dt
To prove this, let φX (t) denote the motion of a given particle and F (t) the
function of time obtained by applying the operator F on this motion as,

F (t) = F (φX (t)); φX (t) = φ(X, t) (3.86)

Note ¬rst that the derivative with respect to time of a function f (t) is related
to the directional derivative of this function in the direction of an increment
29
3.11 VELOCITY AND MATERIAL TIME DERIVATIVES



in time ∆t as,
d df
Df [∆t] = f (t + ∆t) = ∆t (3.87)
d dt
=0

Using this equation with ∆t = 1 for the functions F (t) and φX (t) and
recalling the chain rule for directional derivatives given by Equation (2.105c)
gives,
d dF
F (φ(X, t)) =
dt dt
= DF [1]
= DF (φX (t))[1]
= DF [DφX [1]]
= DF [v] (3.88)

A simple illustration of Equation (3.85) emerges from the time derivative
of the deformation gradient tensor F which can be easily obtained from
Equations (3.6) and (3.80) as,
d ‚φ ‚ ‚φ

F= = = (3.89)
v
dt ‚X ‚X ‚t
Alternatively, recalling Equation (3.70) for the linearized deformation gra-
dient DF gives,

DF [v] = v=F (3.90)


3.11.4 VELOCITY GRADIENT
We have de¬ned velocity as a spatial vector. Consequently, velocity was ex-
pressed in Equation (3.81) as a function of the spatial coordinates as v(x, t).
The derivative of this expression with respect to the spatial coordinates de-
¬nes the velocity gradient tensor l as,
‚v(x, t)
l= = (3.91)
v
‚x
This is clearly a spatial tensor, which, as Figure 3.10 shows, gives the relative
velocity of a particle currently at point q with respect to a particle currently
at p as dv = ldx. The tensor l enables the time derivative of the deformation
gradient given by Equation (3.89) to be more usefully written as,
‚v ‚v ‚φ

F= = = lF (3.92)
‚X ‚x ‚X
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X3, x 3 vq = vp + d v



q
`


dx dv

Q vp
p

time = t
X1, x 1 P
X2, x 2


time = 0

FIGURE 3.10 Velocity gradient.



from which an alternative expression for l emerges as,

l = F F ’1 (3.93)



3.12 RATE OF DEFORMATION

Consider again the initial elemental vectors dX 1 and dX 2 introduced in
Section 3.4 and their corresponding pushed forward spatial counterparts
dx1 and dx2 given as (see Figure 3.11),

dx1 = F dX 1 ; dx2 = F dX 2 (3.94a,b)

In Section 3.5 strain was de¬ned and measured as the change in the scalar
product of two arbitrary vectors. Similarly, strain rate can now be de¬ned
as the rate of change of the scalar product of any pair of vectors. For the
purpose of measuring this rate of change recall from Section 3.5 that the
current scalar product could be expressed in terms of the material vectors
dX 1 and dX 2 (which are not functions of time) and the time-dependent
right Cauchy“Green tensor C as,

dx1 · dx2 = dX 1 · C dX 2 (3.95)
31
3.12 RATE OF DEFORMATION



X3, x 3
time = t

v dt
`
d x2


d x1


p
d X2
d X1
time = t + dt

X1, x 1 P
X2, x 2

time = 0

FIGURE 3.11 Rate of deformation.



Di¬erentiating this expression with respect to time and recalling the relation-
ship between the Lagrangian strain tensor E and the right Cauchy“Green
tensor as 2E = (C ’I) gives the current rate of change of the scalar product
in terms of the initial elemental vectors as,

d ™ ™
(dx1 · dx2 ) = dX 1 · C dX 2 = 2 dX 1 · E dX 2 (3.96)
dt

where E, the derivative with respect to time of the Lagrangian strain tensor,
is known as the material strain rate tensor and can be easily obtained in

terms of F as,

1 ™T
™ ™ ™
E = 1 C = 2 (F F + F T F ) (3.97)
2


The material strain rate tensor, E, gives the current rate of change of
the scalar product in terms of the initial elemental vectors. Alternatively,
it is often convenient to express the same rate of change in terms of the
current spatial vectors. For this purpose, recall ¬rst from Section 3.4 that
Equations (3.94a,b) can be inverted as,

dX 1 = F ’1 dx1 ; dX 2 = F ’1 dx2 (3.98a,b)

Introducing these expressions into Equation (3.96) gives the rate of change
32 KINEMATICS



X3, x 3

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