A piece of copper having a rectangular cross-section of $15.2 m m \times 19.1 m m$ is pulled in tension with $44, 500 N$ force, producing only elastic deformation.Calculate the resulting strain? Shear modulus of elasticity of copper is $42 \times 10^{9} N / m^{2}$ .

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Solution

$A = 15.2 \times 10^{- 3} \times 19.1 \times 10^{3}$
$= 290.32 \times 10^{- 6} m^{2}$
$F = 44500 N$
$\Rightarrow \frac{F / A}{S t r a i n} =$ Shear modulus of elasticity.
$\Rightarrow$ Strain $= \frac{44500}{290.32 \times 10^{- 6} \times 42 \times 10^{9}}$
Strain $= 3.65 \times 10^{- 3}$

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

A=15.2×10−3×19.1×103=290.32×10−6m2F=44500N⇒F/AStrain= Shear modulus of elasticity.⇒ Strain =44500290.32×10−6×42×109Strain =3.65×10−3

A piece of copper having a rectangular cross-section of $15.2 m m \times 19.1 m m$ is pulled in tension with $44, 500 N$ force, producing only elastic deformation.Calculate the resulting strain? Shear modulus of elasticity of copper is $42 \times 10^{9} N / m^{2}$ .

$A = 15.2 \times 10^{- 3} \times 19.1 \times 10^{3}$
$= 290.32 \times 10^{- 6} m^{2}$
$F = 44500 N$
$\Rightarrow \frac{F / A}{S t r a i n} =$ Shear modulus of elasticity.
$\Rightarrow$ Strain $= \frac{44500}{290.32 \times 10^{- 6} \times 42 \times 10^{9}}$
Strain $= 3.65 \times 10^{- 3}$