Question

E MCH 213- Spring 2019 (Figure 1)

Answer 1

Area of shear plane of the plate

$A=\pi DL=\pi \times 0.08\times 0.03=7.54\times 10^{-3}m^{2}$

Allowable shear stress

$\tau _{allow}=\frac{\tau _{fail}}{FS}=\frac{150}{2.5}=60 Mpa$

Maximum allowable force P is

$P=\tau _{allow}A=(60\times 10^{6})(7.54\times 10^{-3})=452400 N$

or

P=452.4 KN