Answer to Question #159977 in Mechanical Engineering for Sandy

Question

Find the Young's modulus of a rod of diameter 30mm and length 300 mm which is subjected to a tensile load of 60 kN and the extension of the rod is equal to 0.4mm

Answer 1

Solution:

As given that,

Diameter of Rod($D$) = 30mm

Length of Rod($L$) = 300mm

Tensile load($T$) = 60KN

Extension($\delta l$) Occurred in rod due to Tensile load = 0.4mm

In general, The Extension of Rod which is subjected to tensile load is given by the following formula,

$\delta l=\frac {TL} {AE}$

Where,

$A=\frac {\pi} {4} D^2$ = Cross sectional area of the rod

E = Young's Modulus of Rod

So that from above Equation, Young's modulus of the rod is given by

$E = \frac {TL} {T \delta l}$

$= \frac {60 \times 10^3 \times 3 } {\frac {\pi} {4} \times0.03^2 \times 0.4 \times 10^{-3}} \\ = \frac {18 \times 10^{3}} {0.0002826 \times 10^{-3}} \\ = 63.694 \times 10^9 N/m^2 \\ =63.6 \times 10^{9}N/m^2 \\ =63.6 GN/m^2$

Young's modulus of the rod = 63.6 GN/m²