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

Answer:

$l=222.803mm$

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,$A_o=\frac{\pi d^2}{4}$

substituting the values in the above equation we get

$A_o=\frac{\pi 0.0038^2}{4}$

or

$A_o=1.134\times 10^{-5}$

now

the stress (σ) is given as:

$\sigma=\frac{Force}{Area}$

and$E=\frac{\sigma}{\epsilon}$

where,

$\epsilon =\ Strain$

also,

$\epsilon=\frac{\Delta l}{l}$

where,

$l=initial \ length$

thus,

$E=\frac{\frac{F}{A_o}}{\frac{\Delta l}{l}}$

or on rearranging we get,

$l=\frac{E\times \Delta l\times A}{F}$

substituting the values in the above equation we get

$l=\frac{102\times 10^9\times 0.00047\times 1.134\times 10^{-5}}{2440}$

or

$l=0.222803m$

or

$l=222.803mm$