Posted by **Danny** on Friday, June 28, 2013 at 1:35pm.

1) Calculate the tensile stress in a 33.000mm diameter rod subjected to a pull of 30.000kN.

ANS= MPa (Round to 3 decimal places)

Diameter = 33.000mm = 0.033m

30.000kN = 30000N

Area = pi*d^2/4

= 3.1416 * 0.033^2/ 4

= 3.1416 *0.001089/ 4

= 0.0034212024/ 4

= 8.553006*10^-4

= 0.0008553

Stress= Load/Area

= 30000/ 0.0008553

= 35075412.14Pa

= 35075.41214kPa

ANS = 35.07541214MPa

2) Consider that the rod was originally 1.000 meters long, and it was stretched 1.110mm by the pulling force. Calculate the strain produced in the rod.

ANS=(6decimal places)

1.110mm = 0.00111m

= 0.00111m/1m

ANS = 0.001110

Please check. Thank you.

## Answer this Question

## Related Questions

- Pyhsics(Stress&Strain) - 1) Calculate the tensile stress in a 33.000mm diameter ...
- Physics - 1) Calculate the tensile stress in a 33.000mm diameter rod subjected ...
- Physics - 1) Calculate the tensile stress in a 33.000mm diameter rod subjected ...
- Physics (Stress & Strain) - 1) Calculate the tension stress in a 21.00mm ...
- Physic (Stress & Strain) - 1) Calculate the tension stress in a 21.00mm diameter...
- maths - Determine the stress in a solid steel rod of diameter 30 mm when ...
- Physics(Stress&Strain) - A uniform bar 1.000 meters in length and having a cross...
- Physics (Stress&Strain) - A uniform bar 1.000 meters in length and having a ...
- Physics - A uniform bar 1.000 meters in length and having a cross sectional area...
- Physics - A uniform bar 1.000 meters in length and having a cross sectional area...

More Related Questions