Question

A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (30 * 106 psi) and an original diameter of 10.2 mm (0.40 in.) experiences only elastic deformation when a tensile load of 8900 N (2000 lbf ) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in.).

Answer 1

The maximum length of specimen is 0.475 m

Step-by-step explanation:

Given :

Maximum elongation
`\Delta l = 0.25 * 10^(-3)` m

Tensile force
`F = 8900` N

Diameter
`d =10.2 * 10^(-3)` m

Elastic modulus
`E = 207 * 10^(9)` Pa

So area of cylindrical specimen is given by,

`A = \pi (d^(2) )/(4)`

From the formula of elongation,

`(F)/(A) = (\Delta l E)/(l)`

Where
`l =` maximum length of specimen,

`l = (\Delta l EA)/(F)`

`l = (\Delta l \pi E d^(2) )/(4F)`

Put the value and find maximum length,

`l = (0.25 * 10^(-3) * 207 * 10^(9) * \pi (10.2 * 10^(-3)) ^(2) ) )/(4 * 8900)`

`l = 0.475` m