Question 3.137: An AISI 1018 steel ball with 1-in diameter is used as a roll...

Aluminum Plate-Ball interface: From Eq. (3-68),

From Eq. (3-69),

By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference in the maximum shear stress for the plate and the ball will be due to poisson’s ratio in Eq. (3-70). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum plate will be the critical element in this interface. Applying the equations for the aluminum plate,

From Eq. (3-72),

Comparing this stress to the allowable stress, and solving for F,

Table-Ball interface: From Eq. (3-68),

From Eq. (3-69),

The steel ball has a higher poisson’s ratio than the cast iron table, so it will dominate.

From Eq. (3-72),

Comparing this stress to the allowable stress, and solving for F,

The steel ball is critical, with F = 3.16 lbf.

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Eq. 3-68 : a=\sqrt[3]{\frac{3 F}{8} \frac{\left(1-v_{1}^{2}\right) / E_{1}+\left(1-v_{2}^{2}\right) / E_{2}}{1 / d_{1}+1 / d_{2}}}

Eq. 3-69 : p_{\max }=\frac{3 F}{2 \pi a^{2}}

Eq. 3-70 : \sigma_{1}=\sigma_{2}=\sigma_{x}=\sigma_{y}=-p_{\max }\left[\left(1-\left|\frac{z}{a}\right| \tan ^{-1} \frac{1}{|z / a|}\right)(1+v)-\frac{1}{2\left(1+\frac{z^{2}}{a^{2}}\right)}\right]

Eq. 3-71 : \sigma_{3}=\sigma_{z}=\frac{-p_{\max }}{1+\frac{z^{2}}{a^{2}}}

Eq. 3-72 : \tau_{\max }=\tau_{1 / 3}=\tau_{2 / 3}=\frac{\sigma_{1}-\sigma_{3}}{2}=\frac{\sigma_{2}-\sigma_{3}}{2}