Question

Assume that if the shear stress in steel exceeds about \[4.00 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\] the steel raptures. Determine, the shearing force necessary to:
(a) shear a steel bolt \[1.00\,{\text{cm}}\] in diameter and
(b) punch a \[1.00\,{\text{cm}}\] diameter hole in a steel plate \[0.500\,{\text{cm}}\] thick.

Answer 1

Hint:Use the formula for stress on an object. This formula gives the relation between the stress acting on the object, shearing force on the object and area on which the force is applied. For the first part, determine the area of the steel bolt and determine the shearing force on the steel bolt. For the second part, determine the area on which the force is applied by taking the multiplication of circumference of hole and thickness of plate and then calculate the required shearing force.

Formulae used:
The stress on an object is given by
\[{\text{Stress}} = \dfrac{F}{A}\] …… (1)
Here, \[F\] is a force acting on the object and \[A\] is the area on which the force is applied.
The area \[A\] of a circle is
\[A = \pi {r^2}\] …… (2)
Here, \[r\] is the radius of the circle.

Complete step by step answer:
(a) We have given that the stress acting on the steel is \[4.00 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\].
\[{\text{Stress}} = 4.00 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\]
We have given that the diameter of the steel bolt is \[1.00\,{\text{cm}}\].
\[d = 1.00\,{\text{cm}}\]
We have asked to determine the shearing force on the steel bolt.
Hence, the radius of the steel bolt becomes
\[r = \dfrac{{1.00\,{\text{cm}}}}{2}\]
\[ \Rightarrow r = 5 \times {10^{ - 3}}\,{\text{m}}\]
Hence, the radius of the steel bolt is \[5 \times {10^{ - 3}}\,{\text{m}}\].

Substitute \[\pi {r^2}\] for \[A\] in equation (1).
\[{\text{Stress}} = \dfrac{F}{{\pi {r^2}}}\]
Rearrange the above equation for the sharing force \[F\] on the steel bolt.
\[ \Rightarrow F = \left( {{\text{Stress}}} \right)\pi {r^2}\]
Substitute \[4.00 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\] for \[{\text{Stress}}\], \[3.14\] for \[\pi \] and \[5 \times {10^{ - 3}}\,{\text{m}}\] for \[r\] in the above equation.
\[ \Rightarrow F = \left( {4.00 \times {{10}^8}\,{\text{N/}}{{\text{m}}^2}} \right)\left( {3.14} \right){\left( {5 \times {{10}^{ - 3}}\,{\text{m}}} \right)^2}\]
\[ \therefore F = 3.14 \times {10^4}\,{\text{N}}\]

Hence, the shearing force on the steel bolt is \[3.14 \times {10^4}\,{\text{N}}\].

(b) We have given that the diameter of the hole is \[1.00\,{\text{cm}}\] and the thickness of the steel plate is \[0.500\,{\text{cm}}\].
\[d = 1.00\,{\text{cm}}\]
\[ \Rightarrow t = 0.500\,{\text{cm}}\]
Let us convert the unit of thickness of the steel plate in the SI system of units.
\[t = \left( {0.500\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)\]
\[ \Rightarrow t = 5 \times {10^{ - 3}}\,{\text{m}}\]
Hence, the thickness of the steel plate is \[5 \times {10^{ - 3}}\,{\text{m}}\].

The radius of the hole becomes
\[r = \dfrac{{1.00\,{\text{cm}}}}{2}\]
\[ \Rightarrow r = 5 \times {10^{ - 3}}\,{\text{m}}\]
The area on which the shearing force is acting is the multiplication of the circumference of the hole and thickness of the steel plate.
\[A = 2\pi rt\]
Substitute \[2\pi rt\] for \[A\] in equation (1).
\[{\text{Stress}} = \dfrac{F}{{2\pi rt}}\]
Rearrange the above equation for the sharing force \[F\] to punch the hole in steel plate.
\[ \Rightarrow F = \left( {{\text{Stress}}} \right)2\pi rt\]

Substitute \[4.00 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\] for \[{\text{Stress}}\], \[3.14\] for \[\pi \], \[5 \times {10^{ - 3}}\,{\text{m}}\] for \[r\] and \[5 \times {10^{ - 3}}\,{\text{m}}\] for \[t\] in the above equation.
\[ \Rightarrow F = \left( {4.00 \times {{10}^8}\,{\text{N/}}{{\text{m}}^2}} \right)2\left( {3.14} \right)\left( {5 \times {{10}^{ - 3}}\,{\text{m}}} \right)\left( {5 \times {{10}^{ - 3}}\,{\text{m}}} \right)\]
\[ \therefore F = 6.28 \times {10^4}\,{\text{N}}\]

Hence, the shearing force to punch the hole in the steel plate is \[6.28 \times {10^4}\,{\text{N}}\].

Note:The students should not forget to consider the thickness of the steel plate while determining the area on which the shearing force is applied for punching the hole in the steel plate because this force is applicable to the whole thickness and circumference of the hole in the steel plate. If this area is not determined correctly then the value of the shearing force will also be incorrect.