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.
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(2000)}{\pi(40)} \frac{\left(1-0.292^{2}\right) /\left[207\left(10^{3}\right)\right]+\left(1-0.211^{2}\right) /\left[100\left(10^{3}\right)\right]}{1 / 150+1 / \infty}\right)^{1 / 2} \\b &=0.2583 mm\end{aligned}p_{\max }=\frac{2 F}{\pi b l}=\frac{2(2000)}{\pi(0.2583)(40)}=123.2 MPa
\begin{aligned}\sigma_{x} &=-2 v p_{\max }\left(\sqrt{1+\frac{z^{2}}{b^{2}}}-\left|\frac{z}{b}\right|\right)=-2(0.292)(123.2)\left(\sqrt{1+0.786^{2}}-0.786\right) \\&=-35.0 MPa \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)=-123.2\left(\frac{1+2\left(0.786^{2}\right)}{\sqrt{1+\left(0.786^{2}\right)}}-2(0.786)\right) \\&=-22.9 MPa \end{aligned}
\sigma_{z}=\frac{-p_{\max }}{\sqrt{1+\frac{z^{2}}{b^{2}}}}=\frac{-123.2}{\sqrt{1+0.786^{2}}}=-96.9 MPa
\tau_{\max }=\frac{\sigma_{y}-\sigma_{z}}{2}=\frac{-22.9-(-96.9)}{2}=37.0 MPa
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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}}}