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2 years ago

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An aluminium alloy bar of diameter 12.5 mm and length 27 m loaded in uniaxial tension to a force of 3 kN. Determine the length o

f the bar when the force is applied. The elastic modulus of the aluminium alloy is 69 GPa.

1 answer:

KonstantinChe [14]2 years ago

7 0

Answer:

27009.56 mm

Explanation:

Given:

Diameter of the aluminium alloy bar, d = 12.5 mm

Length of the bar, L = 27 m = 27 × 10³ mm

Tensile force, P = 3 KN = 3 × 10³ N

Elastic modulus of the bar, E = 69 GPa = 69 × 10³ N/mm²

Now,

for the uniaxial loading, the elongation or the change in length (δ) due to the applied load is given as:

$\delta=\frac{PL}{AE}$

where, A is the area of the cross-section

$A=\frac{\pi d^2}{4}$

or

$A=\frac{\pi\times12.5^2}{4}$

or

A = 122.718 mm²

on substituting the respective values in the formula, we get

$\delta=\frac{3\times10^3\times27\times10^3}{122.718\times69\times10^3}$

or

δ = 9.56 mm

Hence, the length after the force is applied = L + δ = 27000 + 9.56

= 27009.56 mm

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Explanation:

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