johnsoninsaidoo johnsoninsaidoo

31-03-2021
Engineering

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The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform

Respuesta :

AFOKE88 AFOKE88

07-04-2021

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

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