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.

a) 0.00996 m

b) 109090909 Pa

Explanation:

Unit conversions:

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

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

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

And its corresponding diameter is

c. newton's third law of motion