Question

Question:−
piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? Shear modulus of elasticity of copper is 42 x 10⁹ N/m².​

Answer 1

Answer:

Length of the piece of copper, 1 = 19.1mm

Breath of the piece of copper, b = 15.2mm

Area of copper piece: A = lb

$A = 19.1 \times {10}^{ - 3} \times 15.2 \times {10}^{ - 3}$

$2. 9 \times {10}^{ - 4} {m}^{2}$

Tension force applied of the piece of copper,

F = 44500 N

Modulus of elasticity of copper,

$Y = 42 \times {10}^{9} N {m}^{ - 2}$

Modulus of elasticity = Stress/Strain =

(F/A) / Strain

Strain = F/(Y A)

$44500/(2.9 \times {10}^{ - 4 } \times 42 \times {10}^{9} )$

$= 3.65 \times {10}^{ - 3}$

Answer 2

Answer:

Length of the piece of copper, 1 = 19.1mm

Breath of the piece of copper, b = 15.2mm

Area of copper piece: A = lb

Tension force applied of the piece of copper,

F = 44500 N

Modulus of elasticity of copper,

Modulus of elasticity = Stress/Strain =

(F/A) / Strain

Strain = F/(Y A)

Explanation: