What is the required load on a punch to create a hole 3 cm in diameter in a steel plate that is 8 cm thick, given a shear stress of 670 GPa?

To determine the required load on a punch to create a hole in a steel plate, one must consider the shear stress and the area being sheared. The shear stress is given as 670 GPa, which is a measure of the force per unit area that causes the material to fail in shear.

The area of the hole that needs to be created can be calculated using the formula for the area of a circle, which is A = πr². Given that the diameter of the hole is 3 cm, the radius (r) would be 1.5 cm or 0.015 m when converted to meters:

[ A = \pi (0.015 \text{ m})^2 \approx 7.0686 \times 10^{-4} \text{ m}^2. ]

Knowing the area, the next step is to calculate the load required to produce shear failure across this area. The shear load (P), using the shear stress (τ), can be calculated with the formula:

[ P = τ \cdot A. ]

Before applying this formula, it's important to ensure that the shear stress is converted into consistent units. Since the shear stress is given in GPa, we need to