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?
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.
Yes, you have calculated the stress and strain correctly.
The stress produced in the bar is 4954.545 kPa (or 4954545.455 Pa), and the strain produced is 0.0001273.
Your calculation for the stress appears to be incorrect. The formula for stress is force divided by area:
Stress = Load / Area
Given that the load is 5450.000 N and the area is 11.000 cm^2, you first need to convert the area from cm^2 to m^2:
Area = 11.000 cm^2 * (1 m / 100 cm)^2
= 0.0011 m^2
Now you can calculate the stress:
Stress = Load / Area
= 5450.000 N / 0.0011 m^2
= 4,954,545.45 Pa or 4,954.545 kPa
So the correct stress value is 4,954.545 kPa.
Your calculation for the strain, on the other hand, appears to be correct. The strain is equal to the change in length divided by the original length:
Strain = (2.420 mm / 1.9 m)
= 0.0012737
So the correct strain value is 0.0012737.
In summary, the correct stress value is 4,954.545 kPa and the strain value is 0.0012737.