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Failures Resulting from Static Loading 237
8
which can more closely represent some of the data. However, this introduces a nonlin-
ear equation for the sake of a minor correction, and will not be presented here.
EXAMPLE 5–5 Consider the wrench in Ex. 5–3, Fig. 5–16, as made of cast iron, machined to dimen-
sion. The force F required to fracture this part can be regarded as the strength of the
component part. If the material is ASTM grade 30 cast iron, find the force F with
(a) Coulomb-Mohr failure model.
(b) Modified Mohr failure model.
Solution We assume that the lever DC is strong enough, and not part of the problem. Since grade
30 cast iron is a brittle material and cast iron, the stress-concentration factors K t and K ts
are set to unity. From Table A–24, the tensile ultimate strength is 31 kpsi and the com-
pressive ultimate strength is 109 kpsi. The stress element at A on the top surface will be
subjected to a tensile bending stress and a torsional stress. This location, on the 1-in-
diameter section fillet, is the weakest location, and it governs the strength of the assem-
bly. The normal stress σ x and the shear stress at A are given by
M 32M 32(14F)
σ x = K t = K t = (1) = 142.6F
I/c πd 3 π(1) 3
Tr 16T 16(15F)
τ xy = K ts = K ts = (1) = 76.4F
J πd 3 π(1) 3
From Eq. (3–13) the nonzero principal stresses σ A and σ B are
142.6F + 0 142.6F − 0
2
2
σ A ,σ B = ± + (76.4F) = 175.8F, −33.2F
2 2
This puts us in the fourth-quadrant of the σ A ,σ B plane.
(a) For BCM, Eq. (5–31b) applies with n = 1 for failure.
175.8F (−33.2F)
σ A σ B
− = − = 1
3
3
S ut S uc 31(10 ) 109(10 )
Solving for F yields
Answer F = 167 lbf
(b) For MM, the slope of the load line is |σ B /σ A |= 33.2/175.8 = 0.189 < 1.
Obviously, Eq. (5–32a) applies.
175.8F
σ A
= = 1
3
S ut 31(10 )
Answer F = 176 lbf
As one would expect from inspection of Fig. 5–19, Coulomb-Mohr is more conservative.
8 See J. E. Shigley, C. R. Mischke, R. G. Budynas, Mechanical Engineering Design, 7th ed., McGraw-Hill,
New York, 2004, p. 275.