Physics (Stress & Strain)
posted by Danny .
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?
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