Question
If the allowable tensile stress for the bar is $\left(\sigma_{t}\right)_{\text {allow }}=21 \mathrm{ksi},$ and the allowable shear stress for the pin is $\tau_{\text {allow }}=12 \mathrm{ksi},$ determine the diameter of the pin so that the load $P$ will be a maximum. What is this load? Assume the hole in the bar has the same diameter $d$ as the pin. Take $t=\frac{1}{4}$ in and $w=2$ in
Step 1
In this case, the area is the cross-sectional area of the pin, which is a circle with diameter $d$. So, the area $A$ is $\pi d^2/4$. Therefore, the shear stress is given by $\tau = P/A = P/(\pi d^2/4)$. Show more…
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The bar is connected to the support using a pin having a diameter of $d=1$ in. If the allowable tensile stress for the bar is $\left(\sigma_{t}\right)_{\text {allow }}=20 \mathrm{ksi},$ and the allowable bearing stress between the pin and the bar is $\left(\sigma_{b}\right)_{\text {allow }}=30 \mathrm{ksi}$ determine the dimensions $w$ and $t$ so that the gross area of the cross section is $w t=2$ in $^{2}$ and the load $P$ is a maximum. What is this maximum load? Assume the hole in the bar has the same diameter as the pin.
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