CONTINUUM MECHANICS for ENGINEERS

FIGURE 4.9
Volume of parallelepiped defined by vectors dX (1) , dX (2) , and dX (3) in the reference configuration
deforms into the volume defined by paralellepiped defined by vectors dx (1) , dx (2) , and dx (3) in
the deformed configuration.
and finally
d = (tr L) dS – dS ⋅ L or (4.11-7)
˙ S dS˙ = v dS −dS
v
i k,k i j j,i
which gives the rate of change of the current element of area in terms of the
current area, the trace of the velocity gradient, and of the components of L.
Consider next the volume element defined in the referential configuration
by the box product
dV° = dX (1) ⋅ dX (2) × dX (3) = εABC = [dX (1) , dX (2) , dX (3) ( 1) ( 2) ( 3)
dX A dX B dXC
]
as pictured in Figure 4.9, and let the deformed volume element shown in
Figure 4.9 be given by
dV = dx (1) ⋅ dx (2) × dx (3) = εijk = [dx (1) , dx (2) , dx (3) ( 1) ( 2) ( 3)
dxi dx j dxk
]
For the motion x = x(X,t), dx = F ⋅ dX so the current volume is the box product
dV = [F ⋅ dX (1) , F ⋅ dX (2) , F ⋅ dX (3) ] = εijkxi,A xj,Bxk,C dX dX B dX
( 1) ( 2) ( 3)
A C
= det F [dX (1) , dX (2) , dX (3) ] = JdV° (4.11-8)
which gives the current volume element in terms of its original size. Since
J ≠ 0 (F is invertible), we have either J < 0 or J > 0. Mathematically, J < 0 is
possible, but physically it corresponds to a negative volume, so we reject it.
Henceforth, we assume J > 0. If J = 1, then dV = dV° and the volume magnitude
is preserved. If J is equal to unity for all X, we say the motion is
isochoric.