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 divide the load applied to the bar by the cross-sectional area of the bar. In this case, the load is 5750.000 Newtons and the area is 0.10 square meters. So, Stress = 5750.000 N / 0.10 m^2 = 575,000 Pascal (Pa). Since kPa = Pa / 1000, you convert Pa to kPa by dividing by 1000, resulting in 575.000 kPa (rounded to 3 decimal places).
To calculate the strain, you divide the change in length of the bar by the original length of the bar. In this case, the change in length is 2.740mm, which is equivalent to 0.002740 meters, and the original length is 1 meter. So, Strain = 0.002740 m / 1 m = 0.002740 (rounded to 6 decimal places).