A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of the rod be under this load?

Question

asked Jan 2, 2021 184k views

1 Answer

Willem De Jong · Answer 1 · 2021-01-08T02:19:48+0000

Answer:

0.29 m

Step-by-step explanation:

9 mm = 0.009 m in diameter

Cross-sectional area
$A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36* 10^(-5) m^2$

Let the tensile modulus of Nickel
E = 170 * 10^9Pa .

The elongation of the rod can be calculated using the following formula:

$\Delta L = (F L)/(A E) = (6283*50)/(6.36* 10^(-5) * 170 * 10^9) = (314150)/(1081200) = 0.29 m$