Question

A mild steel flange-mounted single shear pin (Ultimate shear strength = 42 MPa) is used in a flange. The perpendicular distance between the axis of driving shaft and the shear pin axis is 100 mm. If the speed of the driving shaft is 300 rpm, the maximum power the shaft could transmit in kW before the failure of the pin is _____. \textit{[Round off to two decimal places.]}

Hint:

[colframe=blue!30!black, colback=yellow!10!white, coltitle=black] Ensure to use consistent units for the calculation of power. Shear stress must be in Pascals (Pa) and area in square meters.

Answer 1

Correct Answer: 10.2

The power transmitted by the shaft is given by the formula: \[ P = \tau \times A \times v \] where:
- \( \tau \) = shear stress (MPa),
- \( A \) = cross-sectional area of the pin (mm\(^2\)),
- \( v \) = velocity of the shear pin (m/s). The velocity of the shear pin is calculated as: \[ v = \text{angular velocity} \times \text{radius} = \frac{2 \pi \times 300}{60} \times 0.1 = 3.1416 \, \text{m/s}. \] Now, calculate the cross-sectional area \( A \) of the pin, assuming it's a circular cross-section: \[ A = \pi \times \left( \frac{d}{2} \right)^2 = \pi \times \left( \frac{10}{2} \right)^2 = 78.54 \, \text{mm}^2. \] The maximum shear stress \( \tau \) is 42 MPa, which is \( 42 \times 10^6 \, \text{Pa} \). Now, substitute these values into the formula for power: \[ P = 42 \times 10^6 \times 78.54 \times 10^{-6} \times 3.1416 = 103.64 \, \text{W}. \] Thus, the maximum power the shaft could transmit is approximately \( \boxed{10.20} \, \text{kW} \) (rounded to two decimal places).