[tex] \huge\boxed{ \underline{ \underline{ \sf \purple{question}}}}[/tex]
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.
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Answer:
Given:-
Length of the piece of copper = l = 19.1 × [tex]10^{-3}[/tex]
Breadth of the piece of copper = B = 15.2mm = 15.2×[tex]10^{-3}[/tex]
According to question:-
Area of the copper piece, A = l × B
= 19.1 × [tex]10^{-3}[/tex] × 15.2 × [tex]10^{-3}[/tex]
= 2.9 × [tex]10^{-4}[/tex] m²
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Tension forced applied on the piece of copper, η = 42 × 10⁹ N/m²
Module of elasticity , η = [tex]\frac{stress}{strain}[/tex]
= [tex](\frac{f}{a} )/strain[/tex]
= strain = [tex]\frac{F}{Aη}[/tex]
= [tex]\dfrac{44500}{(2.9 \times 10^{-4} \times 42 \times 10⁹)}[/tex]
= 3.65 × 10[tex]10^{-3}[/tex]
[tex] \huge\boxed{ \underline{ \underline{ \sf \purple{Answer}}}}[/tex]
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
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