Question 5.5: Consider the wrench in Ex. 5–3, Fig. 5–16, as made of cast i...

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 compressive 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-indiameter section fillet, is the weakest location, and it governs the strength of the assembly. The normal stress $σ_{x}$ and the shear stress at A are given by

Table A–24  Mechanical Properties of Three Non-Steel Metals (a) Typical Properties of Gray Cast Iron [The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.]

*Polished or machined specimens.
†The modulus of elasticity of cast iron in compression corresponds closely to the upper value in the range given for tension and is a more constant value than that for tension.

$σ_{x} =K_{t} \frac {M}{I/c} = K_{t} \frac {32M}{πd^{3}} = (1) \frac {32(14F)}{π(1)^{3}} = 142.6F$
$τ_{xy} = K_{ts} \frac {Tr}{J} = K_{ts} \frac {16T}{πd^{3}} = (1) \frac {16(15F)}{π(1)^{3}} = 76.4F$

From Eq. (3–13) the nonzero principal stresses $σ_{A}$ and $σ_{B}$ are

$σ_{1}, σ_{2} =\frac {σ_{x} + σ_{y}}{2}± \sqrt { \left(\frac {σ_{x} – σ_{y}}{2} \right)+τ^{2}_{xy}}$                   (3-13)

$σ_{A}, σ_{B} =\frac {142.6F+ 0}{2}± \sqrt { \left(\frac {142.6F- 0}{2} \right)+(76.4F)^{2}} = 175.8F,−33.2F$

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.

$\frac {σ_{A}}{S_{ut}} −\frac {σ_{B}}{S_{uc}} =\frac {1}{n}              σ_{A} ≥ 0 ≥ σ_{B}$                         (5–31b)

$\frac {σ_{A}}{S_{ut}} −\frac {σ_{B}}{S_{uc}}=\frac {175.8F}{31(10^{3})} − \frac {(−33.2F)}{109(10^{3})} = 1$

Solving for F yields

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.

$σ_{A} ≥ 0 ≥ σ_{B}   and  | \frac {σ_{B}}{σ_{A}} | ≤ 1$             (5–32a)

$\frac {σ_{A}}{S_{ut}}=\frac {175.8F}{31(10^{3})}=1$

F = 176 lbf

As one would expect frominspection of Fig. 5–19, Coulomb-Mohr ismore conservative.