(II) An iron bolt is used to connect two iron plates together. The bolt must withstand a shear force of up to about 3300 N. Calculate the minimum diameter for the bolt based on a safety factor of 7.0.

You know that stress is the ratio of force and cross-sectional area.

For this problem, first, you have to find the forces acting on the supports. Then, using the formula\({\bf{shear strength = safety factor \times stress}}\), you will get the diameter.

The maximum shear force is 3300 N.

The safety factor is 7.0.

You know that the shear strength of iron is \({\rm{170}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let d be the diameter of the bolt.

You can write

\(\begin{array}{c}{\rm{shear strength}} = {\rm{safety factor}} \times {\rm{stress}}\\\frac{{{\rm{shear strength}}}}{{{\rm{safety factor}}}} = \frac{F}{A}\\\pi \frac{{{d^2}}}{4} = F\left( {\frac{{{\rm{safety factor}}}}{{{\rm{shear strength}}}}} \right)\\d = \sqrt {\frac{{4F}}{\pi }\left( {\frac{{{\rm{safety factor}}}}{{{\rm{shear strength}}}}} \right)} \end{array}\).

Now, substituting the values in the above equation, \(\begin{array}{c}d = \sqrt {\frac{{4 \times \left( {3300\;{\rm{N}}} \right)}}{\pi }\left( {\frac{{{\rm{7}}{\rm{.0}}}}{{{\rm{170}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}} \right)} \\ = 1.3 \times {10^{ - 2}}\;{\rm{m}}\\ = 1.3\;{\rm{cm}}\end{array}\).

Hence, the diameter of the bolt is 1.3 cm.