A wheel of diameter d and width w carrying a load F rolls on a flat rail.
Assume that Fig. 3–39, which is based on a Poisson’s ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. At this critical depth, calculate the Hertzian stresses \sigma_{x}, \sigma_{y}, \sigma_{z}, and τmax for the wheel.
Note to the Instructor: The first printing incorrectly had a width w = 1.25 mm instead of w = 1.25 in. The solution presented here reflects the correction which will be made in subsequent printings.
Use Eqs. (3-73) through (3-77).
\begin{aligned}b &=\left(\frac{2 F}{\pi l} \frac{\left(1-v_{1}^{2}\right) / E_{1}+\left(1-v_{2}^{2}\right) / E_{2}}{1 / d_{1}+1 / d_{2}}\right)^{1 / 2} \\&=\left(\frac{2(250)}{\pi(1.25)} \frac{\left(1-0.211^{2}\right) /\left[14.5\left(10^{6}\right)\right]+\left(1-0.211^{2}\right) /\left[14.5\left(10^{6}\right)\right]}{1 / 3+1 / \infty}\right)^{1 / 2} \\b &=0.007095 \text { in }\end{aligned}p_{\max }=\frac{2 F}{\pi b l}=\frac{2(250)}{\pi(0.007095)(1.25)}=17946 psi
\begin{aligned}\sigma_{x} &=-2 v p_{\max }\left(\sqrt{1+\frac{z^{2}}{b^{2}}}-\left|\frac{z}{b}\right|\right)=-2(0.211)(17946)\left(\sqrt{1+0.786^{2}}-0.786\right) \\&=-3680 psi =-3.68 kpsi \end{aligned}
\begin{aligned}\sigma_{y} &=-p_{\max }\left(\frac{1+2 \frac{z^{2}}{b^{2}}}{\sqrt{1+\frac{z^{2}}{b^{2}}}}-2\left|\frac{z}{b}\right|\right)=-17946\left(\frac{1+2\left(0.786^{2}\right)}{\sqrt{1+\left(0.786^{2}\right)}}-2(0.786)\right) \\&=-3332 psi =-3.33 kpsi \end{aligned}
\sigma_{z}=\frac{-p_{\max }}{\sqrt{1+\frac{z^{2}}{b^{2}}}}=\frac{-17946}{\sqrt{1+0.786^{2}}}=-14109 psi =-14.1 kpsi
\tau_{\max }=\frac{\sigma_{y}-\sigma_{z}}{2}=\frac{-3332-(-14109)}{2}=5389 psi =5.39 kpsi
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Eq. (3-73): b=\sqrt{\frac{2 F}{\pi l} \frac{\left(1-v_{1}^{2}\right) / E_{1}+\left(1-v_{2}^{2}\right) / E_{2}}{1 / d_{1}+1 / d_{2}}}
Eq. (3-74): p_{\max }=\frac{2 F}{\pi b l}
Eq. (3-75): \sigma_{x}=-2 v p_{\max }\left(\sqrt{1+\frac{z^{2}}{b^{2}}}-\left|\frac{z}{b}\right|\right)
Eq. (3-76): \sigma_{y}=-p_{\max }\left(\frac{1+2 \frac{z^{2}}{b^{2}}}{\sqrt{1+\frac{z^{2}}{b^{2}}}}-2\left|\frac{z}{b}\right|\right)
Eq. (3-77): \sigma_{3}=\sigma_{z}=\frac{-p_{\max }}{\sqrt{1+z^{2} / b^{2}}}