Page 288 - Mechanical Behavior of Materials

Section 7.5 Octahedral Shear Stress Yield Criterion 289

so that the failure criterion is
1
2
2
τ ho = (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 2 (at yielding) (7.30)
3
As was done for the maximum shear stress criterion, it is useful to express the critical value in terms
of the yield strength from a tension test. Substitution of the uniaxial stress state with σ 1 = σ o and
σ 2 = σ 3 = 0 into the octahedral shear criterion gives
√
2
τ ho = σ o (7.31)
3
From the three-dimensional geometry of the octahedral planes, as described in the previous
chapter, it can be shown that the plane on which the uniaxial stress acts is related to the octahedral
plane by a rotation through the angle α of Fig. 6.15, where

1
−1
α = cos √ = 54.7 ◦ (7.32)
3
The same result can also be obtained from Mohr’s circle by noting that, in uniaxial tension, the
normal stress on the octahedral plane is
σ 1 + σ 2 + σ 3 σ 1
σ h = = (7.33)
3 3
Locating the point that satisfies this on Mohr’s circle leads to the aforementioned values of α and
τ ho , as shown in Fig. 7.7.
σ
1
τ
σ
1
3
σ
1 (σ , τ )
h h
2α

σ 0 σ
1
(σ , 0)
1
α = 54.7 o
σ
σ = 1
h 3
τ = 2 σ 1
h 3

Figure 7.7 The plane of octahedral shear in a uniaxial tension test.