Questions: Calculate the force required to punch a circular hole, 65 mm in diameter in a steel plate 10 mm thick. The ultimate shear stress of the material is 200 MN / m^2.

Solution Steps

The shear area is the surface area that will be sheared when the hole is punched. For a circular hole, the shear area is the circumference of the hole multiplied by the thickness of the plate.

• Diameter of the hole, \( d = 65 \) mm
• Thickness of the plate, \( t = 10 \) mm

First, calculate the circumference of the hole: \[ \text{Circumference} = \pi \times d = \pi \times 65 \text{ mm} \]

\[ \text{Shear Area} = \text{Circumference} \times \text{Thickness} \] \[ \text{Shear Area} = \pi \times 65 \text{ mm} \times 10 \text{ mm} \] \[ \text{Shear Area} = 650\pi \text{ mm}^2 \]

The force required to punch the hole is given by the product of the shear area and the ultimate shear stress of the material.

• Ultimate shear stress, \( \tau = 200 \text{ MN/m}^2 = 200 \times 10^6 \text{ N/m}^2 \)

Convert the shear area to square meters: \[ \text{Shear Area} = 650\pi \text{ mm}^2 = 650\pi \times 10^{-6} \text{ m}^2 \]

Calculate the force: \[ F = \text{Shear Area} \times \tau \] \[ F = 650\pi \times 10^{-6} \text{ m}^2 \times 200 \times 10^6 \text{ N/m}^2 \] \[ F = 650\pi \times 200 \text{ N} \] \[ F = 130000\pi \text{ N} \] \[ F \approx 408407 \text{ N} \]

Final Answer