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  • Trapezoidal loads shear - moment diagram

    Ronan

    Ronan

    @ronan-05ELmM
    Updated: Oct 26, 2024
    Views: 2.0K
    Hi all,

    I'm experiencing a difficulty understanding how the trapezoidal loads are distributed and how to shear moment diagrams are drawn for structural members subjected to trapezoidal loading.

    For example below example is frame member subject to trapeze load, due to the fact that this memeber is extracted from 3D frame building it has initial shear and moment at ends resulted from fixed - end moments(equivalent nodal loads) loading.

    So my hand-calculations is as below(x distance as variabel) and want to compare them with software calulated values.
    Vshear = 150 - q*x2/(2*A);
    Mmoment=150*x -q*x3/(2*3*A);

    A = triangular horizontal leg-length of trapezoidal load. (A=100cm , span = 400cm; q=1kg/cm )

    the above formulas produces the consistent results for boundary conditions and same with software results but in-between values are wrong.

    For example if x = 30cm then Vshear = 145.5 kg where as software finds 142.5 kg.

    [​IMG]

    [​IMG]Any comments will be appreciated.
    Regards,
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  • Capt Spark

    MemberJul 1, 2011

    no idea but are you sure that the SFD from the s/w is right? It should change from parabola to linear where the load becomes uniform from linear. It appears to me to be happening much before at the left end. It may be a graphics problem of the picture though.
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