Page 314 - Mechanical Behavior of Materials

Section 7.8 Modified Mohr Fracture Criterion 315

Example 7.8
A gray cast iron has a tensile strength of 214 MPa and a compressive strength of 770 MPa, where
these values are averages from three tests of each type on a single batch of material. Also, in the
compression tests, the fracture was observed to occur on a plane inclined to the direction of
◦
loading by an angle averaging θ c = 37 .
(a) Assuming that the modified Mohr criterion applies, calculate m and σ i values for this
material.
(b) If a shaft of diameter 30 mm of this material is subjected to a torque of 500 N·m,
estimate the safety factor against fracture.
(c) What is the safety factor against fracture if a 100 kN compressive force is applied to the
shaft in addition to the torque?

◦
Solution (a) The value of m can be obtained from θ c = 37 and Eqs. 7.43 and 7.47.
◦
◦
φ = 90 − 2θ c = 16 , m = sin φ = 0.2756 Ans.

To calculate σ i ,use σ uc = σ uc =−770 MPa, as well as σ ut = 214 MPa, with m in Eq. 7.62.
1 + m

σ i =−|σ |+ σ ut =−393.1MPa Ans.
uc
1 − m
(b) The shear stress at the shaft surface due to a torque T is obtained from the shaft radius
of r = 15 mm and expressions from Appendix A:

Tr πr 4
τ xy = , J =
J 2
2T 2(500,000 N·mm)
τ xy = = = 94.31 MPa
πr 3 π(15 mm) 3

Noting that there is a state of plane stress with this τ xy and σ x = σ y = 0, the in-plane principal
normal stresses from Eq. 6.7 are

2

σ x + σ y σ x − σ y
2
σ 1 ,σ 2 = ± + τ xy = 94.31, −94.31 MPa
2 2
The third principal normal stress is σ 3 = 0.
We now have all of the quantities needed to obtain C 12 , C 23 , and C 31 from Eq. 7.58:
C 12 = 260.4, C 23 = 94.31, C 31 = 166.09 MPa

These, along with the principal normal stresses, σ 1 , σ 2 , and σ 3 give the effective stresses and
safety factors for the C–M and maximum normal stress components of the M-M failure criterion.