Physic (Stress & Strain)
A uniform bar 1.900m in length and having a cross sectional area of 11.000cm^2 is subject to a loading of 5450.000N. this load causes an extension to the bar length of 2.420mm. Calculate the stress and the strain produced.
Area = 11.000cm^2 = 11*10^3m
Load = 5450.000N
Stress = Load/ Area
= 5450.000N/(11*10^3)
= 4954545.455Pa
ANS = 4954.545kPa
Strain:
2.420*10^4m/ 1.9m
ANS =0.0001273
Have I done this correctly?
asked by
Danny

L=1.9 m, A= 11 cm²= 11•10⁻⁴m²
F=5450 N
ΔL = 2.42 mm =2.42•10⁻³ m
Stress
σ= F/A = 5450/11•10⁻⁴=4.95•10⁶ N/m²=
=4.95 MPa
Strain
ε = ΔL/L= 2.42•10⁻³/1.9 =1.27•10⁻³.
posted by Elena
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