Page 293 - Mechanical Behavior of Materials

294 Chapter 7 Yielding and Fracture under Combined Stresses

Solution (a) Apply Hooke’s law for the z-direction, Eq. 5.26(c), letting σ x = σ y , and noting
that preventing deformation in the z-direction requires that the strain in that direction be zero
(ε z = 0):

1
1
ε z = σ z − ν(σ x + σ y ) , 0 = σ z − ν(σ y + σ y ) , σ z = 2νσ y
E E
Here, solving the second expression for σ z gives the third. Since there are no shear stresses, the
x-y-z axes are also the principal axes, 1-2-3, and the principal normal stresses are

σ 1 = σ x = σ y , σ 2 = σ y , σ 3 = σ z = 2νσ y

The effective stress for the octahedral shear criterion is
1
2
2
¯ σ H = √ (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 2
2
1
2
2
2
¯ σ H = √ (σ y − σ y ) + (σ y − 2νσ y ) + (2νσ y − σ y ) = σ y (1 − 2ν)
2
Since ¯σ H = σ o at the point of yielding, the desired result is
σ o
σ y = Ans.
1 − 2ν
(b) For the aluminum alloy with uniaxial yield strength σ o = 300 MPa, assume that σ o =
−300 MPa applies for uniaxial compression. Substituting this and ν = 0.345 from Table 5.2,

Figure E7.5 Block of material stressed equally in two directions, with rigid walls
preventing deformation in the third direction.