Question

Find the diameter of the test cylinder in which 6660 N force is acting on it with a modulus of elasticity 110 x 103 Pa. The initial length of the rod is 380 mm and elongation is 0.50 mm.

Answer 1

Answer:

The diameter of the test cylinder should be 7.65 meters.

Explanation:

The Hooke's law relation between stress and strain is mathematically represented as

$Stress=E\times strain\\\\\sigma =e\times \epsilon$

Where 'E' is modulus of elasticity of the material

Now by definition of strain we have

$\epsilon =\frac{\Delta L}{L_{o}}$

Applying values to obtain strain we get

$\epsilon =\frac{0.5}{380}=0.001316$

Thus the stress developed in the material equals

$\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}$

Now by definition of stress we have

$\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}$

Solving for 'D' we get

$D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters$