Question

A cylindrical specimen of some metal alloy having an elastic modulus of 102 GPa and an original cross-sectional diameter of 3.8 mm will experience only elastic deformation when a tensile load of 2440 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.47 mm.

Answer 1

`l=222.803mm`

Step-by-step explanation:

Given:

Elastic modulus, E = 102 GPa

Diameter, d = 3.8mm = 0.0038 m

Applied tensile load = 2440N

Maximum allowable elongation, = 0.47mm = 0.00047

Now,

The cross-sectional area of the specimen,
`A_o=(\pi d^2)/(4)`

substituting the values in the above equation we get

`A_o=(\pi 0.0038^2)/(4)`

or

`A_o=1.134* 10^(-5)`

now

the stress (σ) is given as:

`\sigma=(Force)/(Area)`

and
`E=(\sigma)/(\epsilon)`

where,

`\epsilon =\ Strain`

also,

`\epsilon=(\Delta l)/(l)`

where,

`l=initial \ length`

thus,

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

or on rearranging we get,

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

substituting the values in the above equation we get

`l=(102* 10^9* 0.00047* 1.134* 10^(-5))/(2440)`

or

`l=0.222803m`

or

`l=222.803mm`