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 = Load/ Area
= 5750.000N/ 0.10m
= 575000Pa
ANS = 575.000kPa
Strain = Change in Length/ Length
= 0.002740m/ 1m
ANS = 0.002740
Please check. Thank you.
cross sectional area of 10.000cm^2 =>
Area =10 cm²=10•10⁻⁴ m²=0.001 m² !!!!!
Stress =Load/Area =
= 5750/0.001 =5750000 Pa =
= 5750 kPa
Strain = ΔL/L = 2.74•10⁻³/1 = 2.74•10⁻³=0.00274
The Cross Sectional Area is 10cm^2.
Does that mean that the bar would be square in shape?
10cm = 0.10m
10cm^2 = 0.10m * 0.10m * 0.10m * 0.10m
= 10 * 10^-4
= 0.001