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A specimen of some metal having a rectangular cross section 10.5 mm x 13.7 mm is pulled in tension with a force of 2650 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.

Respuesta :

Answer:

resulting strain will be 0.00322

Explanation:

given data

cross section = 10.5 mm × 13.7 mm

force = 2650 N

elastic modulus = 79 GPa

solution

we apply here elastic modulus that is express as

elastic modulus = [tex]\frac{stress}{strain}[/tex]    ....................1

E = [tex]\frac{\sigma }{\epsilon }[/tex]  

and here [tex]\sigma[/tex] = [tex]\frac{force}{area}[/tex]    

[tex]\epsilon = \frac{\sigma }{E}[/tex]  

[tex]\epsilon = \frac{F }{AE}[/tex]

[tex]\epsilon = \frac{34300}{10.8 \times 12.5 \times 10^{-6} \times 79 \times 10^9}[/tex]  

[tex]\epsilon[/tex] = 0.00322

so resulting strain will be 0.00322