Beam Deflections Jun 2022

Hand calculate beam deflections
I thought I should reassess and resharpen my basic structural engineering skills.

I present a few problems

    1. Given a cantilever subject to three separate loadings:
      (a) a concentrated load P at the free end,
      (b) a moment M at the free end, and
      (c) a uniform load w over the entire beam.
      The span is L.
      For each loading, determine the beam deflection at the free end of the span.
      Answer: (a)

          \[\delta = \frac{PL^3}{3EI}\]

      Answer: (b)

          \[\delta = \frac{ML^2}{2EI}\]

      Answer: (c)

          \[\delta = \frac{wL^4}{8EI}\]

      See calculation of cantilever deflections

    2. Given a simple beam with a P load at midspan.
      The span is L.
      Determine the beam rotation at the right end of the span.
      Answer:

          \[\theta_{B} = \frac{PL^2}{16EI}\]

      See calculation of symmetric rotation

      I solved this problem using the virtual work method.
      hand calcs virtual work

    3. Given a simple beam with a P load at at a distance a from the left support.
      The span is L; therefore, b = L-a.
      Determine the beam rotation at the right end of the span.
      Determine also the deflection for all points x located left of the P loading.
      Answer:

          \[\theta_{B} = \frac{Pab(L+a)}{6EIL}\]

      See calculation of asymmetric rotation

      I solved this problem using the conjugate beam method (*).hand calcs conjugate beam
      Note that if

          \[a = \frac{L}{2}\]

          \[\text{and therefore}\]

          \[b = \frac{L}{2}\]

      then the third equation is equal to the second equation.


    4. Given a beam (or plate) with uniform load, w, subjected to a 2 point symmetric lift
      The center span, L, is to be determined
      The value, a, is a ratio to be determined
      The left overhang = the right overhang = aL
      lift arrangement
      Determine the span and the overhang that minimizes the absolute value of the moment along the span
      Answer: see calculation of lift load points
      Determine the resulting deflection at the right end of the beam, at x = L+2aL = (1+2a)L
      Answer: see calculation of deflection given minimized moment
      Note that my calculation of a agrees with the well-known value of a.

          \[a=0.3536\]

          \[\frac{aL}{L+2aL} = \frac{a}{1+2a} = \frac{\sqrt{2}-1}{2} \approx 0.2071 \approx 0.21\]

I derived two deflection equations:

    \[\Delta_{cantilever}=\frac{a(1-6{a}^2-3{a}^3)}{24}\frac{{w}{L}^4}{{E}{I}}\]

    \[\Delta_{centerspan}=\frac{(6{a}^2-1)}{48}\frac{{w}{L}^4}{{E}{I}}\]

See calculation of centerspan deflection.
See calculation of cantilever deflection.


  1. Given a pin-supported bent with uniform EI:
    lift arrangement
    The beam span is L.
    The column length is μL.
    Determine the horizontal deflection of the frame at point B.
    Answer:

        \[\Delta_{B} = \Big( \frac{\mu^2L^3(1+2\mu)}{12EI}\Big) H\]

    See calculation of frame deflection

Bonus Material

For the first 2 problems, I checked some algebra using Jupyter, Plotly, Python, and Sympy.

http://khoitsmahq.firstcloudit.com/images%2Fbeam_hand_calc_check.html


Additional Bonus Material

Quite a surprise — I discovered a Sympy package named sympy.physics.continuum_mechanics.beam.Beam (**) and checked some of my results using that package too. Note Sympy’s use of singularity functions in cells 4 and 5. I use singularity functions in cells 6 through 10.

http://khoitsmahq.firstcloudit.com/images%2Fshear_and_moment.html


def f_rotation(x, E, I):
    sf = SingularityFunction
    return (-10*sf(x, 0, 2)/3 + 
            5*sf(x, 48, 2) - 
            5*sf(x, 144, 2)/3 + 
            12800)/(E*I)
    
def f_deflection(x, E, I):
    sf = SingularityFunction
    return (12800*x - 
            10*sf(x, 0, 3)/9 + 
            5*sf(x, 48, 3)/3 - 
            5*sf(x, 144, 3)/9)/(E*I)    


References

(*)   Conjugate Beam Method — Wiki

(**)  https://docs.sympy.org/… continuum_mechanics/beam.html

(***) A bonus from Dr. Myosotis


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