The load P acting on the steel cantilever column has an eccentricity e = 15 mm. with respect to the z-axis. Determine the maximum allowable value of P using a factor of safety of 2.5 against yielding and buckling. Which failure mode governs the solution? The material properties are E = 200 GPa and Gyp = 250 MPa.

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The load P acting on the steel cantilever column has an eccentricity e = 15 mm. with respect to the z-axis. Determine the maximum allowable value of P using a factor of safety of 2.5 against yielding and buckling. Which failure mode governs the solution? The material properties are E = 200 GPa and Gyp = 250 MPa.

The load P acting on the steel cantilever column has an eccentricity e = 15 mm. with respect to the z-axis. Determine the maximum allowable value of P using a factor of safety of 2.5 against yielding and buckling. Which failure mode governs the solution? The material properties are E = 200 GPa and Gyp = 250 MPa.


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Ghulam Mustafa 2024-12-28T01:23:08+05:00 1 Answer 33 views Beginner
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    2025-01-15T14:06:04+05:00
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    To solve this problem, we need to consider both the yielding and buckling of the steel cantilever column and determine the maximum allowable load PP using a factor of safety (FS) of 2.5.

    Step 1: Determine the Maximum Load for Yielding

    Given:

    • Yield stress, σy=250 MPa\sigma_y = 250 \, \text{MPa}
    • Factor of safety against yielding, FSyield=2.5\text{FS}_\text{yield} = 2.5

    The allowable stress considering the factor of safety is:

    σallow=σyFSyield=250 MPa2.5=100 MPa\sigma_\text{allow} = \frac{\sigma_y}{\text{FS}_\text{yield}} = \frac{250 \, \text{MPa}}{2.5} = 100 \, \text{MPa}

    The stress due to the axial load PP with eccentricity ee is:

    σ=PA+PeS\sigma = \frac{P}{A} + \frac{P e}{S}

    Where:

    • AA is the cross-sectional area of the column.
    • SS is the section modulus.

    To find PP, we equate the stress to the allowable stress:

    σallow=PA+PeS\sigma_\text{allow} = \frac{P}{A} + \frac{P e}{S} 100 MPa=PA+P×15 mmS100 \, \text{MPa} = \frac{P}{A} + \frac{P \times 15 \, \text{mm}}{S}

    Solving for PP requires knowledge of AA and SS, which are geometric properties dependent on the cross-section of the column.

    Step 2: Determine the Critical Load for Buckling

    Using Euler’s formula for buckling of a cantilever column:

    Pcr=π2EI(KL)2P_\text{cr} = \frac{\pi^2 E I}{(K L)^2}

    Where:

    • E=200 GPa=200×103 MPaE = 200 \, \text{GPa} = 200 \times 10^3 \, \text{MPa}
    • II is the moment of inertia of the column.
    • K=2K = 2 for a cantilever column.
    • LL is the length of the column.

    The allowable buckling load considering the factor of safety is:

    Pallow=PcrFSbuckling=π2EI(2L)2×2.5P_\text{allow} = \frac{P_\text{cr}}{\text{FS}_\text{buckling}} = \frac{\pi^2 E I}{(2 L)^2 \times 2.5}

    Step 3: Compare the Load Capacities

    Once the values of PP from yielding and buckling are determined, the lower of these two values will govern the maximum allowable load PP.

    Conclusion

    To complete the solution, the specific cross-sectional dimensions (to calculate AA, SS, and II) and the length LL of the column are needed. Once these are known, you can calculate PyieldP_\text{yield} and PbucklingP_\text{buckling}, compare them, and determine which mode governs. The failure mode with the lower allowable load is the governing failure mode.

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