Answer
$$P_{MAX} = 67.86 \space kN$$
Work Step by Step
Solution:
1) Shear Force Distribution Areas:
For Shearing areas: $A=\pi * D * t_{PLATE}$
$\bullet Steel Rod:$
$A_{STEEL} = \pi * 12 *10^{-3} * 10 * 10^{-3} = 3.770 *10 ^{-4} \space m^2$
$\bullet Aluminium Plate:$
$A_{ALUMINIUM} = \pi * 40 *10^{-3} * 8 * 10^{-3} = 1.005 *10 ^{-3} \space m^2$
2) Max Allowable Force:
Max force can be obtained from the shear stress equation as follows:
$\tau_{MAX} = \frac{P_{MAX}}{A} \space \Rightarrow \space P_{MAX} = \tau_{MAX} * A$
Now Calculate the max force needed to shear aluminium and steel:
$\bullet Steel Rod:$
$P_{STEEL} = 180*10^6 *3.770 *10 ^{-4} = 67860 \space N$
$P_{STEEL} = 67.860 \space kN$
$\bullet Aluminium Rod:$
$P_{ALUMINIUM} = 70*10^6 * 1.005 *10 ^{-3} = 70350 \space N$
$P_{ALUMINIUM} = 70.350 \space kN$
3)Taking the smallest of the above values as the limiting one we get the final answer:
$\boxed{P_{MAX} = 67.86 \space kN} \leftarrow \space\space ANS$
