Question

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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:

Given:-

Length of the piece of copper = l = 19.1 × $10^{-3}$

Breadth of the piece of copper = B = 15.2mm = 15.2×$10^{-3}$

According to question:-

Area of the copper piece, A = l × B

= 19.1 × $10^{-3}$ × 15.2 × $10^{-3}$

= 2.9 × $10^{-4}$ m²

__________________

Tension forced applied on the piece of copper, η = 42 × 10⁹ N/m²

Module of elasticity , η = $\frac{stress}{strain}$

= $(\frac{f}{a} )/strain$

= strain = $\frac{F}{Aη}$

= $\dfrac{44500}{(2.9 \times 10^{-4} \times 42 \times 10⁹)}$

= 3.65 × 10$10^{-3}$

Answer 2

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Calculate the resulting strain? (Modulus of elasticity of copper, Y = 42 × 10^9 Nm^-2 ) ... of 15.2mm×19.1mm is pulled in tension with 44,5 00 N force, producing only elastic deformation

hope its help u mate ✔