1) Calculate the tension stress in a 21.00mm diameter rod subjected to a pull of 30.000kN.
21mm =0.021m
30.000kN = 30000N
A = pi*d^2/4
= 3.1416 * (0.021)^2m^2/ 4
= 3.1416 * 0.000441/ 4
= 0.0013854456/ 4
= 0.0003463614
Stress = Load/ Area
= 30000/ 0.0003463614
= 86614732.47N
= 86614.73247kN
ANS = 86.61473247MPa
2) Consider that the rod was originally 1.000m long, and is stretched 1.090mm by a pulling force. Calculate the strain produced in the rod.
1.090mm = 0.001090m
Strain = 0.00109m/ 1.000m
ANS = 0.00109m
Is these done correctly?
1) Yes, you have correctly calculated the tension stress in the rod. To calculate the stress, you first found the cross-sectional area of the rod using the formula A = πd^2/4, where d is the diameter of the rod. Then you divided the force (load) applied to the rod by the cross-sectional area to find the stress.
2) No, you have made a mistake in calculating the strain. The correct calculation should be:
Strain = change in length / original length
Given that the rod was originally 1.000m long and stretched by 1.090mm, the change in length is 0.00109m. Dividing this by the original length of 1.000m gives:
Strain = 0.00109m / 1.000m
ANS = 0.00109
So, the correct strain produced in the rod is 0.00109.