A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced.
ANS 1 = kPa (Round to 3 decimal places)
ANS 2 = (Round to 6 decimal places)
Bar length = 1.000m
Area = 10cm = 0.10m
Load = 5754.000N
Bar extension = 2.740mm = 0.002740m
Stress = Load/ Area
Strain = Change in length/ length
Stress = 5750.000N/ 0.10m
= 575000Pa
ANS = 575.000kPa
Strain = 0.002740/ 1
ANS = 0.002740
Please check. Thank you.
Your calculations are correct.
To calculate the stress, you use the formula: Stress = Load/Area.
In this case, the load is 5750.000 N and the area is 10.000 cm^2, which is equal to 0.10 m^2.
So, Stress = 5750.000 N / 0.10 m^2 = 575000 Pa.
To calculate the strain, you use the formula: Strain = Change in length / Length.
In this case, the change in length is 2.740 mm, which is equal to 0.002740 m, and the original length is 1.000 m.
So, Strain = 0.002740 m / 1.000 m = 0.002740.
Therefore, the stress is 575.000 kPa (rounded to 3 decimal places) and the strain is 0.002740 (rounded to 6 decimal places).