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Sergeeva-Olga [200]
3 years ago
12

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

ed in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Physics
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

strain = 1.4 \times 10^{-3}

Explanation:

As we know by the formula of elasticity that

E = \frac{stress}{strain}

now we have

E = 110 GPA

E = 110 \times 10^9 Pa

Area = 15.2 mm x 19.1 mm

A = 290.3 \times 10^{-6}

now we also know that force is given as

F = 44500 N

here we have

stress = Force / Area

stress = \frac{44500}{290.3 \times 10^{-6}}

stress = 1.53 \times 10^8 N/m^2

now from above formula we have

strain = \frac{stress}{E}

strain = \frac{1.53 \times 10^8}{110 \times 10^9}

strain = 1.4 \times 10^{-3}

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