This example illustrates the use of a failure theory to determine the strength of a mechanical element or component. The example may also clear up any confusion existing between the phrases strength of a machine part, strength of a material, and strength of a part at a point.
A certain force F applied at D near the end of the 15-in lever shown in Fig. 5–16, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would
be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force.
Question 5.3: This example illustrates the use of a failure theory to dete...
We will assume that lever DC is strong enough and hence not a part of the problem. A 1035 steel, heat-treated, will have a reduction in area of 50 percent or more and hence is a ductile material at normal temperatures. This also means that stress concentration at shoulder A need not be considered. A stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This point, on the 1-in- diameter section, is the weakest section, and governs the strength of the assembly. The two stresses are
σ_{x} =\frac {M}{I/c} =\frac {32M}{πd^{3}} =\frac {32(14F)}{π(1^{3}) }= 142.6F
τ_{zx} =\frac {Tr}{J} =\frac {16T}{πd^{3}} =\frac {16(15F)}{π(1^{3})} = 76.4F
Employing the distortion-energy theory, we find, from Eq. (5–15), that
σ′ =\left (σ^{2}_{x} − σ_{x}σ_{y} + σ^{2}_{y} + 3τ^{2}_{xy} \right)^{1/2} (5–15)
σ′ =\left (σ^{2}_{x} + 3τ^{2}_{xy} \right)^{1/2}=[(142.6F)^{2} + 3(76.4F)^{2}]^{1/2}=194.5F
Equating the von Mises stress to S_{y} , we solve for F and get
F =\frac {S_{y}}{194.5} =\frac {81 000}{194.5 }= 416 lbf
In this example the strength of the material at point A is S_{y}= 81 kpsi. The strength of the assembly or component is F = 416 lbf.
Let us apply the MSS theory for comparison. For a point undergoing plane stress with only one nonzero normal stress and one shear stress, the two nonzero principal stresses will have opposite signs, and hence the maximum shear stress is obtained from the Mohr’s circle between them. From Equation (3-14)
\tau_{\max }=\sqrt{\left(\frac{\sigma_{x}}{2}\right)^{2}+\tau_{j x}^{2}}=\sqrt{\left(\frac{142.6 F}{2}\right)^{2}+(76.4 F)^{2}}=104.5 FSetting this equal to S_{y} / 2, from Equation (5-3) with $n=1$, and solving for F, we get
F=\frac{81000 / 2}{104.5}=388 \mathrm{lbf}which is about 7 percent less than found for the DE theory. As stated earlier, the MSS theory is more conservative than the DE theory.
(5.3) :
\tau_{\max }=\frac{S_{y}}{2 n} \quad \text { or } \quad \sigma_{1}-\sigma_{3}=\frac{S_{y}}{n}(3.14) :
\tau_{1}, \tau_{2}=\pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}}