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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.

User Ladookie
by
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1 Answer

4 votes

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.

User Paras Santoki
by
8.1k points
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