Question
A hard steel ball ( $E=200 \mathrm{GPa}$ and $v=0.29$ ) of diameter $50 \mathrm{~mm}$ is pressed against a thick aluminum plate $(E=$ $72.0 \mathrm{GPa}, v=0.33$, and $Y=450 \mathrm{MPa}$ ). Determine the magnitude of load $P_Y$ required to initiate yield in the aluminum plate according to the octahedral shear-stress criterion of failure.
Step 1
The diameter of the steel ball is given as \(50 \, \text{mm}\). Therefore, the radius \(R\) is: \[ R = \frac{50 \, \text{mm}}{2} = 25 \, \text{mm} = 0.025 \, \text{m} \] Show more…
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The 10 -mm-diameter shank of the steel bolt has a bronze sleeve bonded to it. The outer diameter of this sleeve is $20 \mathrm{mm}$. If the yield stress for the steel is $\left(\sigma_{Y}\right)_{\text {de }}=640 \mathrm{MPa}$, and for the bronze $\left(\sigma_{y}\right)_{h r}=520 \mathrm{MPa}$, determine the largest possible value of $P$ that can be applied to the bolt. Assume the materials to be clastic perfectly plastic. $E_{\mathrm{st}}=200 \mathrm{GPa}, E_{\mathrm{br}}=100 \mathrm{GPa}$.
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