The cylindrical bar in the figure has a total length L=1.2 m, a diameter d=15 mm, and is confined by fixed supports at the walls A and B. A concentrated axial load FC=30 kN is applied to the bar at point C as indicated in the figure, with LAC>LBC. The bar is homogenous with Young’s modulus E. The material will fail if the magnitude of axial stress in the material exceeds σf=226 MPa.
I do not see a question here. Whether or not the bar fails will depend upon where the axial load is placed.
This may represebnt your loading situation:
http://www.engineersedge.com/beam_bending/beam_bending19.htm
To determine whether the material will fail or not, we need to calculate the axial stress in the material and compare it with the failure criterion.
To find the axial stress, we can use the formula:
σ = F / A
where σ is the axial stress, F is the axial load applied to the bar, and A is the cross-sectional area of the bar.
To calculate the cross-sectional area, we need to find the radius (r) of the cylindrical bar first. The diameter (d) of the bar is given as 15 mm, so the radius can be found by dividing the diameter by 2:
r = d / 2 = 15 mm / 2 = 7.5 mm = 0.0075 m
Now we can calculate the cross-sectional area:
A = π * r^2
Substituting the value of r into the equation:
A = π * (0.0075 m)^2
Once we have the cross-sectional area, we can calculate the axial stress:
σ = FC / A
Substituting the given value of axial load (FC = 30 kN) and the calculated value of cross-sectional area (A), we can calculate the axial stress (σ).
Finally, we compare the calculated axial stress (σ) with the failure criterion (σf = 226 MPa). If the axial stress is greater than the failure criterion, the material will fail.
Please note that the figure mentioned in the question is not available, so you may need to refer to the original question for any specific information related to the figure.