Question

A specimen of some metal having a rectangular cross section 9.61 mm x 11.5 mm is pulled in tension with a force of 8850 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.

Answer 1

Strain = =0.00102

Step-by-step explanation:

Hooke's law says Youngs modulus (E) = Stress(σ) / Strain(ε)

Also Stress = Load/ Area

So, the dimensions of the metal are 9.61 mm x 11.5 mm

Let's convert to m.

Will be; 9.61 x 10^(-3)m and 11.5 x 10^(-3)m

Area of a rectangle is length x breadth

So since load = 8850N

Stress = 8850/ (9.61 x 10^(-3) x 11.5 x 10^(-3) = 8850/0.00011= 8.045 x 10^(7) N/m2

Now youngs modulus from the question = 79GPa = 79 x 10^(9) Pa or 79 x 10^(9) N/m2

Thus putting the value of Youngs modulus and the stress in the first equation, we have;

79 x 10^(9) = 8.045 x 10^(7)/Strain(ε)

Thus strain = 8.045 x 10^(7)/ (79x 10^(9)) =0.00102

Answer 2

Step-by-step explanation:

In elastic region

• Strain = stress / elastic modulus

• Stress = load applied / Cross section area

• Cross section area = 0.00961 x 0.0115 = 0.000110515m²

• Stress = 8850N / 0.000110515m² = 80079627.2N/m² = 80.0796272MPa

• Strain = 80.0796272 / 79000 = 0.001014 or 1.014 x10-3