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slavikrds [6]
3 years ago
14

Find the minimum diameter of an alloy, tensile strength 75 MPa, needed to support a 30 kN load.

Engineering
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

The minimum diameter to withstand such tensile strength is 22.568 mm.

Explanation:

The allow is experimenting an axial load, so that stress formula for cylidrical sample is:

\sigma = \frac{P}{A_{c}}

\sigma = \frac{4\cdot P}{\pi \cdot D^{2}}

Where:

\sigma - Normal stress, measured in kilopascals.

P - Axial load, measured in kilonewtons.

A_{c} - Cross section area, measured in square meters.

D - Diameter, measured in meters.

Given that \sigma = 75\times 10^{3}\,kPa and P = 30\,kN, diameter is now cleared and computed at last:

D^{2} = \frac{4\cdot P}{\pi \cdot \sigma}

D = 2\sqrt{\frac{P}{\pi \cdot \sigma} }

D = 2 \sqrt{\frac{30\,kN}{\pi \cdot (75\times 10^{3}\,kPa)} }

D = 0.0225\,m

D = 22.568\,mm

The minimum diameter to withstand such tensile strength is 22.568 mm.

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