Question

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

Answer 1

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.