Question

A cylindrical specimen of some metal alloy having an elastic modulus of 129 GPa and an original cross-sectional diameter of 4.4 mm will experience only elastic deformation when a tensile load of 1570 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm.

Answer 1

Explanation:

The cross-sectional area of the specimen is calculated as follows.

         $A_{o} = \frac{pi}{4} d^{2}$

                     = $\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}$

                     = $1.5197 \times 10^{-5} m^{2}$

Equation of stress is as follows.

              $\sigma = \frac{F}{A_{o}}$

And, the equation of strain is as follows.

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

Hence, the Hook's law is as follows.

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

       E = $\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}$

          = $\frac{F \times l_{o}}{A_{o} \times \Delta l}$

or,    $l_{o} = \frac{E \times \Delta l \times A_{o}}{F}$          

                   = $\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}$

                  = 0.6554 m

or,         $l_{o}$ = 655.4 mm

Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.