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Deflection in a Simply supported beam

  1. Oct 23, 2014 #1
    Hi All

    As shown in the attached image, I have a simply supported beam with a load of 150kN acting at the center of the beam span.
    E = 210 GPa
    rho = 7800 kgm-3
    span = 250 mm

    After solving this in FE solver, I got maximum displacement to be ~4mm (at node 52)
    But, on using the theoretical relation of delta = P* l^3 / (48*E*I), it is about ~11mm.

    Can somebody please help me on this ? how to co-relate this.

    Some observations:
    1. There is no hourglassing.
    2. Beam looks compressed by about 0.8mm at the center at the end of solution.

    Attached Files:

  2. jcsd
  3. Oct 23, 2014 #2


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    Staff Emeritus
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    Just eyeballing your problem, it appears you have a short, deep beam with a central load applied. The standard beam formulas for deflection are generally applicable to long slender beams where shear deflection doesn't need to be taken into account. The standard formulas are also approximations which assume that the slope of the beam in the deflected condition is very small. Without knowing more about the cross section of you beam, I can't provide any more insight into your dilemma.
  4. Oct 23, 2014 #3
    Cross section:
    Thickness = 2mm
    height = 50 mm
  5. Oct 23, 2014 #4


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    Staff Emeritus
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    Doing some simple calculations, your simply supported beam is wildly overloaded, with the calculated bending stress lying far outside the elastic range for the material used to construct it. Therefore, any deflections you are calculated using the deflection formula for a simply supported beam are certainly incorrect. The deflection your FE solver is giving you is suspect as well: this beam should probably have snapped in two.

    Code (Text):

    Beam test:

    L = 250 mm

    depth = 50 mm
    width =   2 mm

    E = 210 GPa

    P = 150 kN @ L/2

    Simple supports

    M = PL/4 = 150 kN * 0.25/4 = 9375 N-m

    d = PL^3/(48EI)

    I = (1/12)*0.002*0.05^3
    I = 2.083E-8 m^4

    d = 150000(0.25)^3/(48*210*10^9*2.083*10^-8)

    d = 0.0111 m = 11.1 mm deflection

    sigma = My/I = 9375 * 0.025 / 2.083*10^-8

    sigma = 1.125*10^10 Pa = 11.25 GPa bending stress

    yield strength = 186-758 MPa for steel

    1 MPa = 145 psi

    yield strength = 26970 - 109910 psi for carbon steels
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