Math, asked by francisreddy6521, 1 month ago

Find the required diameter of steel rod that has to carry an axial pull of 40 kn if the permissible stress is 150 mpa ans d 18.43mm

Answers

Answered by yassersayeed
0

Given:- The diameter is 18.43mm.

The permissible stress is 150mpa

Axial pull P= 40kn = 40,000N

Solution :- Area = \mathrm{A}=\frac{\pi}{4}D^2

\text { Stress }=\frac{\text { Load }}{\text { Area }}=\frac{\mathrm{P}}{\mathrm{A}}

\mathrm{D}^{2}=\frac{4000 \times 4}{\pi \times 150}=18.43

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