CONTINUUM MECHANICS for ENGINEERS

or
But xj,q = δjq and by Eq 3.4-3, σ ρ , so that the latter equation imme-
qk,q k b + =0
diately above reduces to
ε x σ + x σ + ρb
dV
ijk j,q qk j qk,q k
Again, since volume V is arbitrary, the integrand here must vanish, or
(3.4-5)
By a direct expansion of the left-hand side of this equation, we obtain for
the free index i = 1 (omitting zero terms), ε 123σ 23 + ε 132σ 32 = 0, or σ 23 – σ 32 = 0
implying that σ 23 = σ 32. In the same way for i = 2 and i = 3 we find that
σ 13 = σ 31 and σ 12 = σ 21, respectively, so that in general
(3.4-6)
Thus, we conclude from the balance of moments for a body in which concentrated
body moments are absent that the stress tensor is symmetric, and
Eq 3.4-3 may now be written in the form
(3.4-7)
Also, because of this symmetry of the stress tensor, Eq 3.3-8 may now be
expressed in the slightly altered form
(3.4-8)
In the matrix form of Eq 3.4-8 the vectors and nj are represented by
column matrices.
t ( ˆn )
i
3.5 Stress Transformation Laws
∫
V
[ ( ) ] =
∫
V
ε σ
ijk jk
dV = 0
εijkσjk = 0
σjk = σkj
σ , + ρb = 0 or ⋅ + ρb=
0
ij j i
t
( nˆ
) ˆ
n
( n
= σ or t
)
= ⋅nˆ
i
ij j
Let the state of stress at point P be given with respect to Cartesian axes
Px 1x 2x 3 shown in Figure 3.8 by the stress tensor having components σ ij.
We introduce a second set of axes Px , which is obtained from Px 1′ x2′ x3′
1x2x3 by a rotation of axes so that the transformation matrix [aij] relating the two
0