Page 281 - Mechanical Behavior of Materials

282 Chapter 7 Yielding and Fracture under Combined Stresses

Example 7.1
A sample of gray cast iron is subjected to the state of generalized plane stress of Ex. 6.3. Gray
cast iron normally behaves in a brittle manner, and this particular material has ultimate strengths
in tension and compression of σ ut = 214 and |σ uc |= 770 MPa, respectively. What is the safety
factor against fracture?
Solution In Ex. 6.3, the given state of stress is σ x = 100, σ y =−60,σ z = 40,τ xy = 80,
and τ yz = τ zx = 0 MPa. From these, principal normal stresses σ 1 = 133.1, σ 2 =−93.1, and
σ 3 = 40 MPa are calculated. Equation 7.13 thus gives

¯ σ NT = MAX(σ 1 ,σ 2 ,σ 3 ) = MAX(133.1, −93.1, 40.0) = 133.1MPa
X = σ ut / ¯σ NT = (214 MPa)/(133.1MPa) = 1.61 Ans.

The limitations accompanying Eq. 7.13 need to be checked. Clearly, ¯σ NT > 0. Also, |σ max | =
133.1 , and |σ min | = 93.1MPa, so that |σ max | > |σ min |, and the answer above is valid as to this
being a tension-dominated case.

7.4 MAXIMUM SHEAR STRESS YIELD CRITERION

Yielding of ductile materials is often predicted to occur when the maximum shear stress on any
plane reaches a critical value τ o , which is a material property:

(at yielding) (7.14)
τ o = τ max
This is the basis of the maximum shear stress yield criterion, also often called the Tresca criterion.
For metals, such an approach is logical because the mechanism of yielding on a microscopic size
scale is the slip of crystal planes, which is a shear deformation that is expected to be controlled by
a shear stress. (See Chapter 2.)

7.4.1 Development of the Maximum Shear Stress Criterion
From the previous chapter, recall that the maximum shear stress is the largest of the three principal
◦
shear stresses, which act on planes oriented at 45 relative to the principal normal stress axes, as
illustrated in Fig. 6.8. These principal shear stresses may be obtained from the principal normal
stresses by Eq. 6.18, which is repeated here for convenience:

|σ 2 − σ 3 | |σ 1 − σ 3 | |σ 1 − σ 2 |
τ 1 = , τ 2 = , τ 3 = (7.15)
2 2 2
Hence, this yield criterion can be stated as follows:

|σ 1 − σ 2 | |σ 2 − σ 3 | |σ 3 − σ 1 |
τ o = MAX , , (at yielding) (7.16)
2 2 2