Page 292 - Mechanical Behavior of Materials

Section 7.5 Octahedral Shear Stress Yield Criterion 293

stress criterion from Eq. 7.35 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 = √ (704.3 − 195.7) + (195.7 − (−20)) + ((−20) − 704.3) = 644.1MPa
2
The yield strength for the 18 Ni maraging steel material from Table 4.2 is 1791 MPa, so that the
safety factor against yielding for the inner wall is
X = σ o / ¯σ H = (1791 MPa)/(644.1MPa) = 2.78 Ans.

Comments For the outer wall, revising the preceding calculation with σ r = σ 3 = 0gives
¯ σ H = 629.6 MPa and X = 2.84. The slightly lower value of X = 2.78 is thus the controlling
one. Note that both safety factors are somewhat higher than those for the maximum shear
criterion from Ex. 7.2.
Second Solution For the octahedral shear criterion, the step of determining principal normal
stresses is not necessary, as the solution can proceed directly from the stresses on the original
r-t-x coordinate system with the use of Eq. 7.36.

1
2
2
2
2
2
2
¯ σ H = √ (σ r − σ t ) + (σ t − σ x ) + (σ x − σ r ) + 6(τ + τ + τ )
tx
rt
xr
2
1
2
2
2
2
¯ σ H = √ (−20 − 600) + (600 − 300) + (300 − (−20)) + 6(0 + 205.3 + 0) = 644.1MPa
2
As expected, ¯σ H is the same as for the first solution, and X = 2.78 (Ans.) is similarly obtained
for the inner wall.
Example 7.5
A block of material is subjected to equal compressive stresses in the x- and y-directions, and it is
confined by a rigid die so that it cannot deform in the z-direction, as shown in Fig. E7.5. Assume
that there is no friction against the die and also that the material behaves in an elastic, perfectly
plastic manner, with uniaxial yield strength σ o .
(a) Determine the stress σ x = σ y necessary to cause yielding, expressing this as a function
of σ o and elastic constants of the material.
(b) What is the value of σ y at yielding if the material is an aluminum alloy with uniaxial
yield strength σ o = 300 MPa and elastic constants as in Table 5.2?