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March 28, 2015

March 28, 2015

Posted by **Danny** on Sunday, June 23, 2013 at 9:42pm.

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?

- Physic (Stress & Strain) -
**MathMate**, Monday, June 24, 2013 at 6:05amBoth are numerically correct.

In #2, you divide m by m, hence no units.

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