Page 294 - Mechanical Behavior of Materials

Section 7.6 Discussion of the Basic Failure Criteria 295

the stress to cause yielding is

−300 MPa
σ y = =−967.7MPa Ans.
1 − 2(0.345)
Discussion If the same block of material is not confined in the z-direction, the stress in that
direction is zero, and an analysis similar to the previous one gives simply σ y = σ o . However,
preventing deformation in the z-direction is seen to cause a stress σ z = 2νσ y to develop, which
in turn causes the value of σ x = σ y at yielding to substantially exceed the uniaxial yield strength.
Hence, constraining the deformation makes it more difficult to yield the material. This occurs
because the stresses σ x , σ y , and σ z all have the same sign and so combine to create significant
hydrostatic stress, which in effect subtracts from the ability of the applied stresses to cause
yielding. For this particular situation, the maximum shear stress criterion gives an identical result.

7.6 DISCUSSION OF THE BASIC FAILURE CRITERIA

The three failure criteria discussed so far, namely, the maximum normal stress, maximum shear
stress, and octahedral shear stress criteria, may be considered to be the basic ones among a larger
number that are available. It is useful at this point to discuss these basic approaches. We will
also consider some design issues and some additional failure criteria that are modifications or
combinations of the basic ones.

7.6.1 Comparison of Failure Criteria
Both the maximum shear stress and the octahedral shear stress criteria are widely used to predict
yielding of ductile materials, especially metals. Recall that both of these indicate that hydrostatic
stress does not affect yielding, and also that the hexagonal-tube yield surface of the maximum
shear criterion is inscribed within the circular-cylinder surface of the octahedral shear criterion.
Hence, these two criteria never give dramatically different predictions of the yield behavior under
combined stress, there being no state of stress where the difference exceeds approximately 15%.
This can be seen in Fig. 7.10, where the distance from the cylinder axis to the two yield surfaces
differs by a maximum amount at the various points where the circle is farthest from the hexagon.
√
From geometry, the distances at these points have the ratio 2/ 3 = 1.155. Hence, safety factors and
effective stresses for a given state of stress cannot differ by more than this. For plane stress, σ 3 = 0,
such a maximum deviation occurs for pure shear, where σ 1 =−σ 2 =|τ|, and also for σ 1 = 2σ 2 ,as
in pressure loading of a thin-walled tube with closed ends.
However, note that, in some situations, the maximum shear and octahedral shear yield criteria
do give dramatically different predictions than a maximum normal stress criterion. Compare the
tubular yield surfaces of either with the cube of Fig. 7.3, and consider states of stress near the tube
axis (σ 1 = σ 2 = σ 3 ), but well beyond the boundaries of the cube. For plane stress, the three failure
criteria compare as shown in Fig. 7.11. Where both principal stresses have the same sign, the maxi-
mum shear stress criterion is equivalent to a maximum normal stress criterion. However, if the prin-
cipal stresses are opposite in sign, the normal stress criterion differs considerably from the other two.