Question

A 2.2-m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 8.5-kN tension force. Knowing that E = 200 GPa, determine (a) the smallest diameter rod that should be used, (b) the corresponding normal stress in the rod.

Answer 1

Answer:

a) 0.00996 m

b) 109090909 Pa

Explanation:

Unit conversions:

$E = 200GPa = 200\times10^9 Pa$

1.2 mm = 0.0012 m

8.5 kN = 8500 N

If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is

$\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455$

With elastic modulus being E = 200 GPa, then its maximum stress must be

$\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa$

Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

$A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2$

And its corresponding diameter is

$A = \pi d^2/4$

$7.79\times10^{-5} = \pi d^2/4$

$d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}$

$d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm$