1 Answer

Jan Gorzny · Answer 1 · 2021-06-24T02:03:35+0000

Answer:

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

Step-by-step explanation:

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

$\sigma = (P)/(A_(c))$

$\sigma = (4\cdot P)/(\pi \cdot D^(2))$

Where:

$\sigma$ - Normal stress, measured in kilopascals.

- Axial load, measured in kilonewtons.

A_(c) - Cross section area, measured in square meters.

- Diameter, measured in meters.

Given that
$\sigma = 75* 10^(3)\,kPa$ and
$P = 30\,kN$ , diameter is now cleared and computed at last:

$D^(2) = (4\cdot P)/(\pi \cdot \sigma)$

$D = 2\sqrt{(P)/(\pi \cdot \sigma) }$

$D = 2 \sqrt{(30\,kN)/(\pi \cdot (75* 10^(3)\,kPa)) }$

$D = 0.0225\,m$

$D = 22.568\,mm$

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

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