A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in
tension with a load of 89,000 N. Under this load it experiences an elongation of 0.10 mm.
Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.

Respuesta :

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, [tex]\sigma=\frac {F}{A}[/tex]

where F is applied force and A is original area

Also, engineering strain, [tex]\epsilon=\frac {\triangle l}{l}[/tex] where l is original area and [tex]\triangle l[/tex] is elongation

We also know that Hooke's law states that [tex]E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}[/tex]

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

[tex]\triangle l= 0.1 mm= 0.1\times 10^{-3} m[/tex]

By substitution we obtain

[tex]E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa[/tex]

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