Page 225 - Intro to Tensor Calculus

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Figure 2.3-6. Shearing parallel to the y axis

which can be rewritten in the form

∂σ pk dv k
e ijk σ jk + x j ( + %b k − % ) − %v j v k =0. (2.3.21)
∂x p dt
In the equation (2.3.21) the middle term is zero because of the equation (2.3.17). Also the last term in
(2.3.21) is zero because e ijk v j v k represents the cross product of a vector with itself. The equation (2.3.21)
therefore reduces to
e ijk σ jk =0, (2.3.22)

which implies (see exercise 1.1, problem 22) that σ ij = σ ji for all i and j. Thus, the conservation of angular
momentum requires that the stress tensor be symmetric. Consequently, there are only 6 independent stress
components to be determined. This is another fundamental law for a continuum.

Strain in Two Dimensions

Consider the matrix equation

x 1 0 x
= (2.3.23)
y β 1 y
which can be used to transform points (x, y)topoints (x, y). When this transformation is applied to the
unit square illustrated in the figure 2.3-6(a) we obtain the geometry illustrated in the figure 2.3-6(b) which
represents a shearing parallel to the y axis. If β is very small, we can use the approximation tan β ≈ β and
then this transformation can be thought of as a rotation of the element P 1 P 2 through an angle β to the
0
0
position P P when the barred axes are placed atop the unbarred axes.
1 2
Similarly, the matrix equation

x 1 α x
= (2.3.24)
y 0 1 y
can be used to represent a shearing of the unit square parallel to the x axis as illustrated in the figure
2.3-7(b).