Chapter 9, Problem 54P

To determine

Minimum diameter of the bolt.

Answer to Problem 54P

Solution: $1.3cm$

Explanation of Solution

Given: Shear force, $F=3300\text{ N}$

Factor of safety, $FOS=7$

Shear strength of the iron is $\tau =170\times {10}^6{\text{N/m}}^{\text{2}}$

Formula Used:

Shear stress is given by

  $\begin{array}{l}\tau =\frac{F}{A}\\ A=\frac{F}{\tau }\end{array}$

Where, A is the area of cross section, F is the applied force, $\tau$ is the shear stress.

Calculation:

Total load acting with factor of safety is

  $\begin{array}{l}{F}_{total}=F\times FOS\\ F_{total}=3300\times 7\\ F_{total}=2.31\times 10^{4}N\end{array}$

From the stress equation,

  $\begin{array}{l}\tau =\frac{F_{total}}{A}\\ A=\frac{F_{total}}{\tau }\\ A=\frac{2.31\times 10^4}{170\times 10^6}\\ A=1.358\times 10^{-4}{m}^2\end{array}$

The required cross section is $1.358\times 10^{-4}{m}^2$ .

  $\begin{array}{l}A=\frac{\pi d^2}{4}\\ d=\sqrt{\frac{4A}{\pi }}\\ d=\sqrt{\frac{4\times 1.358\times 10^{-4}}{\pi }}\\ d=0.0130\text{mm}\\ d=1.3cm\end{array}$

Conclusion:Minimum diameter of the bolt is $1.3cm$ .