CONTINUUM MECHANICS for ENGINEERS

(
cosθ12 =
ˆ ˆ I )/ ˆ
1+ I2 2 ⋅C⋅I2 2/ 2
=
2 3
6
and θ 12 = 54.7°. Thus the original 45° angle is enlarged by 9.7°.
It is evident from Eq 4.8-14 that if the coordinate axes are chosen in the
principal directions of C, the deformed angle θ12 is a right angle (C12 = 0 in
this case) and there has been no change in the angle between elements in
the X1 and X2 directions. By the same argument, any three mutually perpendicular
principal axes of C at P are deformed into three mutually perpendicular
axes at p. Consider, therefore, the volume element of a rectangular
parallelepiped whose edges are in the principal directions of C (and thus
also of E). Since there is no shear strain between any two of these edges, the
new volume is still a rectangular parallelopiped, and in the edge directions
(i = 1,2,3) the unit strains are
ˆN i
so that now
e ( i i
ˆ ) ( ˆ N N )
= Λ −1
(i = 1,2,3) (4.8-16)
dx (i) = dX (i) + dX (i) [ ] = dX (i) − 1
, (i = 1,2,3) (4.8.17)
Λ
( Ni ˆ )
and the ratio of the deformed volume to the original becomes
which, when Eq 4.8-4 is used, becomes
Λ ( ˆ Ni )
dV dx dx dx
o
dV dX dX dX
= = ( 1) ( 2) ( 3)
Λ Λ Λ
( 1) ( 2) ( 3)
ˆ ˆ ˆ
( N1) ( N2) ( N3)
dV
= C C C o ( 1) ( 2) ( 3) = IIIC
dV
(4.8-18a)
(4.8-18b)
The importance of the second form of Eq 4.8-18b is that it is an invariant
expression and can be calculated without reference to principal axes of C.
Example 4.8-2
Determine the volume ratio dV/dV° for the deformation of Example 4.8-1
using Eq 4.8-18a, and verify using Eq 4.8-18b.
Solution
As the student should show, a set of principal axes for the C tensor of Example
4.8-1 are Nˆ = Iˆ + Iˆ + Iˆ
/ , N = I −I
, and N = I + I −2I
6.
ˆ ˆ ˆ ( ) 2 ( 1 2) / 2 ˆ ˆ ˆ ˆ
3 ( 1 2 3)
/
1 1 2 3 3