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# Answer to Question #159977 in Mechanical Engineering for Sandy

Question #159977

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

1
2021-02-02T01:14:53-0500

Solution:

As given that,

Diameter of Rod("D") = 30mm

Length of Rod("L") = 300mm

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}} \\\\\n= \\frac {18 \\times 10^{3}} {0.0002826 \\times 10^{-3}} \\\\\n= 63.694 \\times 10^9 N\/m^2 \\\\\n=63.6 \\times 10^{9}N\/m^2 \\\\\n=63.6 GN\/m^2"

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

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