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

The diameter of the test cylinder should be 7.65 meters.

Step-by-step explanation:

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

`Stress=E* strain\\\\\sigma =e* \epsilon`

Where 'E' is modulus of elasticity of the material

Now by definition of strain we have

`\epsilon =(\Delta L)/(L_(o))`

Applying values to obtain strain we get

`\epsilon =(0.5)/(380)=0.001316`

Thus the stress developed in the material equals

`\sigma = 110* 10^(3)* 0.001316=144.76N/m^(2)`

Now by definition of stress we have

`\sigma =(Force)/(Area)\\\\\therefore Area=(Force)/(\sigma )\\\\(\pi D^(2))/(4)=(6660N)/(144.76)=46m^(2)`

Solving for 'D' we get

`D=\sqrt{(4* 46)/(\pi )}=7.653meters`