Question

Assume if the shear stress in steel exceeds about $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$, the steel ruptures. Determine the shearing force necessary to (a) shear a steel bolt 1.00cm in diameter and (b) punch a 1.00cm diameter hole in a steel plate thick.

Short answer

$\begin{array}{l}\left(a\right)F=3.14\times {10}^{4}N\\ \left(b\right)F=6.28\times {10}^{4}N\end{array}$

Step-by-step solution

To find the shear a steel and in diameter.

Shear stress $=4\times {10}^{8}N/m^2$d( diameter of steel bolt )=1 cm=0.01 m

$t\left(thicknessofthehole\right)=0.5cm=0.005m$

(a)

The shear stress is defined as the ratio of the shearing force to the area:

Shear stress$=\frac{F}{A}$

Solve for (F) and

F= Shear stress

substitute$\pi \frac{d}{2}$ for (A) :

$F=Shearstress\times \pi \frac{d^2}{2}$

Substitute numerical values:

$F=(4\times {10}^8)\times \pi \frac{0.01}{2}=3.14\times {10}^{4}N$

To find the total surface area.

(b)

We need to find the total surface area that the perimeter of a 1 cm-diameter hole would have which is the surface area of a cylinder:

$A=2\pi rt=\left(2r\right)\pi t=d\pi t$

Substitute numerical values:

$A=(0.01)\left(\pi \right)(0.005)=1.57\times {10}^{-4}{m}^2$

Therefore, the shearing force is given by Equation as:

$F=(4\times {10}^8)(1.57\times {10}^{-4})=6.28\times {10}^{4}N$