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.
If you apply the force at the center of the shaft ( LBC=LAC ), what safety factor SFmid would you have against failure? Provide a numerical (dimensionless) value:
If you want to apply the force FC as close as possible to the wall at B, but you also want to maintain a minimum safety factor SFmin=2 against failure, what is the minimum (LBC/LAC) that you can have? Provide a numerical (dimensionless) value:
To find the safety factor against failure when the force is applied at the center of the shaft, we can use the formula:
SF = σf / σ
Where SF is the safety factor, σf is the allowable stress (226 MPa), and σ is the stress in the material.
The formula for stress in a cylindrical bar under axial load is:
σ = FC / (π/4 * d^2)
Where FC is the applied force, π is the constant pi, and d is the diameter of the bar.
Given:
FC = 30 kN = 30,000 N
d = 15 mm = 0.015 m
σf = 226 MPa = 226 * 10^6 Pa
We can substitute these values into the formula and calculate σ:
σ = (30,000 N) / (π/4 * (0.015 m)^2)
= 8,638,171.67 Pa
Now we can calculate the safety factor:
SF = 226 * 10^6 Pa / 8,638,171.67 Pa
= 26.16
Therefore, the safety factor when the force is applied at the center of the shaft is 26.16.
To find the minimum (LBC/LAC) that will maintain a safety factor of at least 2, we can use the following formula:
SF = σf / σ
Given:
SFmin = 2
Using the same formula for stress in a cylindrical bar, we can set up the equation:
SFmin = σf / σ = 2
Solving for (LBC/LAC), we find:
(LBC/LAC) = (FC / (σf * π/4 * d^2)) * SFmin
Substituting the given values:
(LBC/LAC) = (30,000 N / (226 * 10^6 Pa * π/4 * (0.015 m)^2)) * 2
= 2.49
Therefore, the minimum (LBC/LAC) value to maintain a safety factor of at least 2 is approximately 2.49.