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.
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