Question

A circular section of material is tested. The original specimen is 200 mm long and has a diameter of 13 mm. When loaded to its proportional limit, the specimen elongates by 0.3 mm. The total axial load is 20 kN. Determine the modulus of elasticity and the proportional limit.

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Answer 1

Given that,

Diameter = 13 mm

Length = 200 mm

Elongation = 0.3 mm

Load = 20 kN

We need to calculate the stress of proportional limit

Using formula of proportional limit

$\sigma=\dfrac{P}{A}$

$\sigma=\dfrac{P}{\pi(\dfrac{d}{2})^2}$

Put the value into the formula

$\sigma=\dfrac{20\times10^3}{\pi\times\dfrac{(13\times10^{-3})^2}{4}}$

$\sigma=0.150\times10^{9}\ N/m^2$

$\sigma=0.150\ GN/m^2$

We need to calculate the modulus of elasticity

Using relation of proportional limit and modulus of elasticity

$\sigma=E\epsilon$

$E=\dfrac{\sigma}{\epsilon}$

$E=\dfrac{\sigma}{\dfrac{\Delta l}{l}}$

Put the value into the formula

$E=\dfrac{0.150\times10^{9}\times200\times10^{-3}}{0.3\times10^{-3}}$

$E=1\times10^{11}\ N/m^2$

Hence, (a). The stress of proportional limit is $0.150\ GN/m^2$

(b). The modulus of elasticity is $1\times10^{11}\ N/m^2$