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

Step-by-step explanation:

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

`A_(o) = (pi)/(4) d^(2)`

=
`(3.14)/(4) * ((4.4)/(1000))^(2)`

=
`1.5197 * 10^(-5) m^(2)`

Equation of stress is as follows.

`\sigma = (F)/(A_(o))`

And, the equation of strain is as follows.

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

Hence, the Hook's law is as follows.

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

E =
`((F)/(A_(o)))/((\Delta l)/(l_(o)))`

=
`(F * l_(o))/(A_(o) * \Delta l)`

or,
`l_(o) = (E * \Delta l * A_(o))/(F)`

=
`(129 * 10^(9) * (0.48)/(1000) * 1.662 * 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.