Page 272 - Mechanical Behavior of Materials

262 Chapter 6 Review of Complex and Principal States of Stress and Strain

Substitution of the general form of the invariants and manipulation then gives

σ x + σ y + σ z
σ h = (6.35)
3
1
2
2
2
2
2
τ h = (σ x − σ y ) + (σ y − σ z ) + (σ z − σ x ) + 6(τ xy + τ 2 yz + τ ) (6.36)
zx
3
These more general expressions may be used to compute σ h and τ h for stresses described with
respect to any coordinate system, so that it is not necessary to first determine the principal stresses.
Since the octahedral stresses σ h and τ h are functions of the stress invariants, these quantities are
themselves invariant. Hence, any equivalent representation of a given state of stress will give the
same values of σ h and τ h .
The octahedral shear stress is an important quantity, as it is used as a basis for predicting
yielding and other types of material behavior under complex states of stress. This is considered
starting in the next chapter, as is the similar use of the maximum shear stress. Since τ max occurs on
only two planes, τ h occurs twice as frequently as does τ max . (Compare Figs. 6.8 and 6.15.) Also, for
all possible states of stress, it can be shown that τ h is always similar in magnitude to τ max , with the
√
ratio τ h /τ max being confined to the range 0.866 to unity, or more precisely, 3/2tounity.

6.6 COMPLEX STATES OF STRAIN

In the discussion of complex states of stress, it was noted that equilibrium of forces leads to
transformation equations for obtaining an equivalent representation of a given state of stress on
a new set of coordinate axes. Of particular interest are two sets of axes, one containing the principal
normal stresses, and the other the principal shear stresses. The mathematics involved is common
to all physical quantities classed as symmetric second-order tensors, as distinguished from vectors,
which are first-order tensors, or scalars, which are zero-order tensors.
Strain is also a symmetric second-order tensor and so is governed by similar equations. In this
case, the basis of the equations is simply the geometry of deformation. Detailed analysis (see the
References) gives equations that are identical to those for stress, except that shear strains are divided
by two. Hence, the various equations developed for stress can be used for strain by changing the
variables as follows:
γ xy γ yz γ zx
σ x ,σ y ,σ z → ε x ,ε y ,ε z , τ xy ,τ yz ,τ zx → , , (6.37)
2 2 2

These apply in general and also to the special case where the x-y-z axes are axes of principal strain,
1-2-3.
Advanced textbooks on continuum mechanics, theory of elasticity, and similar subjects often
redefine shear strains as being half as large as the usual engineering shear strains used here, calling
these tensor shear strains, so that the equations become identical to those for stress. However, we
will continue to use engineering shear strains.