Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm. A tensile force of 1500 N produces an elastic reduction in diameter of 6.7 × 10–…

Question

asked Jan 10, 2020 179k views

1 Answer

Adam Shamsudeen · Answer 1 · 2020-01-17T06:27:51+0000

Answer:

The elastic modulus is 99.7 GPa

Solution:

As per the question:

Diameter of the metal alloy, D = 10.0 mm = 0.01 m

Tensile force, F = 1500 N

Elastic reduction in the diameter of the metal alloy,
$\Delat D = 6.7* 10^(- 4) mm = 6.7* 10^(- 7) m$

Poisson's ratio,
$\mu = 0.35$

Now,

Lateral strain is given by:

$\varepsilon_(D) = (\Delta D)/(D) = (6.7* 10^(- 4))/(10) = 6.7* 10^(- 5)$

Longitudinal strain,
$\varepsilon_(l)$ :

$\varepsilon_(l) = (\varepsilon_(D))/(\mu) = (6.7* 10^(- 5))/(0.35) = 1.914* 10^(- 4)$

Now, stress is given by:

$\sigma = (F)/(Area,\ A)$

where

$A = \pi ((d)/(2))^(2) = \pi ((0.01)/(2))^(2) = 7.854* 10^(- 5) m^(2)$

Now,

$\sigma = (1500)/(7.854* 10^(- 5)) = 19.098* 10^(6) = 19.098 MPa$

Now, the elastic modulus is given by:

$E = (\sigma)/(\varepsilon_(l))$

E = (19.098* 10^(6))/(1.914* 10^(- 4)) = 99.784* 10^(9)Pa = 99.7 GPa