Question 3.138: An aluminum alloy cylindrical roller with diameter 1.25 in a...

An aluminum alloy cylindrical roller with diameter 1.25 in and length 2 in rolls on the inside of a cast-iron ring having an inside radius of 6 in, which is 2 in thick. Find the maximum contact force F that can be used if the shear stress is not to exceed 4000 psi.

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v_{1}=0.333, E_{1}=10.4  Mpsi , l=2 \text { in, } d_{1}=1.25 \text { in, } v_{2}=0.211, E_{2}=14.5  Mpsi , d_{2}=-12  in.

with b=K_{c} F^{1 / 2}

\begin{aligned}K_{c} &=\left(\frac{2}{\pi(2)} \frac{\left(1-0.333^{2}\right) /\left[10.4\left(10^{6}\right)\right]+\left(1-0.211^{2}\right) /\left[14.5\left(10^{6}\right)\right]}{1 / 1.25+1 / 12}\right)^{1 / 2} \\&=2.336\left(10^{-4}\right)\end{aligned}

By examination of Eqs. (3-75), (3-76), and (3-77, it can be seen that the only difference in the maximum shear stress for the two materials will be due to poisson’s ratio in Eq. (3-75). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum roller will be the critical element in this interface. Instead of applying these equations, we will assume the poisson’s ratio for aluminum of 0.333 is close enough to 0.3 to make Fig. 3-39 applicable.

\begin{aligned}&\tau_{\max }=0.3  p_{\max } \\&p_{\max }=\frac{4000}{0.3}=13300  psi\end{aligned}

From Eq. (3-74), p_{\max }=2 F /(\pi b l) , so we have

p_{\max }=\frac{2 F}{\pi l K_{c} F^{1 / 2}}=\frac{2 F^{1 / 2}}{\pi l K_{c}}

So

\begin{aligned}F &=\left(\frac{\pi l K_{c} p_{\max }}{2}\right)^{2} \\&=\left(\frac{\pi(2)(2.336)\left(10^{-4}\right)(13300)}{2}\right)^{2} \\&=95.3  lbf\end{aligned}

 

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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}}}

 

 

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