Page 285 - Mechanical Behavior of Materials

286 Chapter 7 Yielding and Fracture under Combined Stresses

Hence, the maximum shear stress criterion predicts that hydrostatic stress alone does not cause
yielding. This seems surprising, but is in fact in agreement with experimental results for metals
under hydrostatic compression. Testing in hydrostatic tension is essentially impossible, but it is
likely that brittle fracture without yielding would occur at a high stress level even in normally
ductile materials.
Interpretation of the safety factor in terms of lengths of lines from the origin in principal stress
space, as discussed earlier, is also valid for the maximum shear stress criterion. Since stresses are
expected to affect yielding only to the extent that they deviate from the axis of the hexagonal tube,
the projections of lengths normal to this axis can also be used:

D S
X = (7.27)
D

Here, D is the projected distance corresponding to yielding, and D to the applied stress, as shown

S
in Fig. 7.6(b).
Example 7.2
Consider the pipe with closed ends of Ex. 6.5, with wall thickness 10 mm and inner diameter
0.60 m, subjected to 20 MPa internal pressure and a torque of 1200 kN·m. What is the safety
factor against yielding at the inner wall if the pipe is made of the 18 Ni maraging steel of
Table 4.2?
Solution The hoop, longitudinal, and radial stresses due to pressure, and the shear stress
due to torsion, are calculated in Ex. 6.5 as σ t = 600,σ x = 300,σ r =−20 (inside), and τ tx =
205.3 MPa. These give principal normal stresses at the inner wall of

σ 1 = 704.3, σ 2 = 195.7, σ 3 =−20 MPa

The effective stress for the maximum shear stress criterion from Eq. 7.21 is

¯ σ S = MAX (|σ 1 − σ 2 | , |σ 2 − σ 3 | , |σ 3 − σ 1 |)

¯ σ S = MAX (|704.3 − 195.7| , |195.7 − (−20)| , |(−20) − 704.3|) = 724.3MPa
The yield strength for this material from Table 4.2 is 1791 MPa, so the safety factor against
yielding for the inner wall is

X = σ o / ¯σ S = (1791 MPa)/(724.3MPa) = 2.47 Ans.

Comment For the outer wall, revising the preceding calculation with σ r = σ 3 = 0gives
¯ σ S = 704.3 MPa and X = 2.54. The slightly lower value of X = 2.47 is thus the control-
ling one.