Question

Calculate the force ‘F’ needed to punch a 1.46 cm diameter hole in a steel plate 1.27cm thick. The ultimate shear strength of steel is 345$N/ {m^2}$ (approx)

Answer 1

Hint: According to the question there is a steel plate 1.27cm thick. The shear strength of the steel is $345N/{m^2}$, or$3.45 \times {10^8}KN$. Shearing force is defined as the force transverse to the beam at a given section tending to cause it to shear at that section.

Complete answer:
The force of the punch is the shear stress multiplied by area
F= Shear strength × Area
$F = 3.45 \times {10^8} \times 2\pi rl$
$ \Rightarrow F = 3.45 \times {10^8} \times 2 \times 3.14 \times 0.73 \times {10^{ - 2}} \times 1.27 \times {10^{ - 2}}$
$\therefore 200KN $
So, the total force needed to punch the plate in 200KN

Note:
In engineering, shear strength is the strength of a material or component against the type of yield or structural failure when the material or component fails in shear. A shear load is a force that tends to produce a sliding failure on a material along a plane that is parallel to the direction of the force.
Now there will be a question: what is the difference between shear stress and shear strength.
Well, Shear stress is relative and it changes in relation of shear load applied to a material per unit area. And ,on the other hand, shear strength is a fixed and definite value in the general nature of a material.