Answer to Question #235171 in Mechanical Engineering for mike

Question

Find the young modulus of the rod of diameter 50 mm and of length 400 mm which is subjected to tensile load of 90 kN, and the extension of the rod is equal to .4 mm

Answer 1

given data:

$D=50mm$

$L=400mm$

$P=90KN$

$\Delta L=0.4mm$

The Youngs modulus of the material can be determined as,

$E=\frac {\sigma}{\epsilon}$

$E=\frac {P/A}{\Delta L/L}$

$E=\frac {PL}{\Delta LA}$

$E=\frac {4PL}{\Delta L(\pi D^2)}$

Substitute the given values to determine the value of E.

$E=\frac{4(90\times10^3N) (400mm)}{0.4mm(\pi(50mm)^2)}$

$E=1.6\times10^6 MPa$