A uniform bar 1.900 meters in length and having a cross sectional area of 11.000cm^2 is subject to a loading of 5450.000 Newtons. this load causes an extension to the bar length of 2.420mm. Calculate the stress and strain produced.
11cm = .11m
2.420mm = .02420m
length = 1.9m
Area = L * W
= 1.9m * .11m
= .209m
Stress
= load/ Area
= 5450.000N/.209m
= 26076.555Pa
ANS = 26.076555kPa
Strain
= .02420m/1.9m
ANS = .0127368
Is this done correctly? Thank you.
Yes, your calculations are correct.
Well, your calculations seem to be on point, but let me tell you one thing - stress and strain can cause more than just mathematical calculations. They can also make us question our sanity! So, let's not stress too much about it and just hope for a strain-free existence. After all, life is all about finding the right balance between stress and strain, right?
Yes, your calculations for stress and strain are correct. The stress produced is 26.076555 kPa and the strain produced is 0.0127368.
Yes, your calculations for stress and strain appear to be correct.
To calculate stress, you first need to find the area of the cross section of the bar. In this case, the given cross sectional area is 11 cm^2, which is equivalent to 0.11 m^2.
Next, you divide the applied load (in this case, 5450 N) by the cross sectional area (0.11 m^2) to get the stress. The calculated stress is 26076.555 Pa. Since the value is quite large, it is often expressed in kilopascals (kPa), so the final answer is 26.076555 kPa.
To calculate strain, you divide the change in length (2.420 mm = 0.02420 m) by the original length (1.9 m). The calculated strain is approximately 0.0127368.
Overall, your calculations for stress and strain are correct.