Question

Assume if the shear stress in steel exceeds about 4.00 108 N/m2 the steel ruptures.
(a) Determine the shearing force necessary to shear a steel bolt 0.75 cm in diameter.

N

(b) Determine the shearing force necessary to punch a 2.60-cm-diameter hole in a steel plate 0.550 cm thick.

Answer 1

Data Given

Stress = 4.00 * 10⁸ N/m²

Part A) We know that

$Stress = \frac{Force}{Area}$

$Force = Stress \times Area$

Here Radius r = 0.75/2 = 0.375 cm

So

$Force = Stress \times \pi r^{2}$

$Force = 4.00\times 10^{8}N/m^{2} \times \pi \times (0.375\times 10^{-2}m)^{2}= 1.77\times 10^{4}N$

Part B) In this case we we consider surface area

$Area = Perameter \times thickness$

$Area = 2\pi r \times t = 2 \pi \times \frac{2.60 \times 10^{-2}m}{2}\times 0.55\times 10^{-2}m= 4.5\times 10^{-4}m^{2}$

$Force = 4.00\times 10^{8}N/m^{2} \times 4.5\times 10^{-4}m^{2}= 18 \times 10^{4}N$