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Tapered Cantilever Beam

  1. Sep 4, 2014 #1
    Hello all,

    I need to calculate the bending stress of a tapered cantilever beam. However, I would like to know how to calculate the moment of inertia.

    For a uniform shape beam, I know the Ix=(bh3)/12, but since h changes throughout the length of the beam how would inertia, I, be calculated?

    I was able to figure out the height of the beam as a function of x, H(x).

    Equation for H(x)= [(a-b)/L](x)+a
    a= height of beam at fix wall
    b= height of beam at end
    L= total length of beam
    x= point along length of beam.

    Any help would be greatly appreciated!!!

    Attached Files:

  2. jcsd
  3. Sep 4, 2014 #2


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    I is still calculated in the usual manner, but instead of being a constant w.r.t. the length of the beam, I is now a function of the length coordinate.

    [itex]I(x) = \frac{b(x)h(x)^{3}}{12}[/itex]

    and I(x) will still be about the centroid of the section.

    To calculate bending stress, you can calculate the section modulus of the beam at the point or points where you want to determine the bending stress and use the formula σ = M / SM as before.

    All you have to do to find the height is plug in a value for 'x'.
  4. Sep 4, 2014 #3
    Hi SteamKing,

    Thank you for your help. Taking the cross-section at the wall I can calculate the bending stress at that point.

    However, this led to more questions.

    1) What would be the equation to determine the deflection at the very tip? I don't believe
    δ=(FL3)/(3EI) since beam is tapered..
    2) If I know the deflection needed, can I manipulate the deflection equation to obtain the force?
    3) Would the moment of inertia change since taken at the very tip?

    Again, thank you for your help..
  5. Sep 4, 2014 #4


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    Now you have found where things get more complicated with varying cross-section beams.

    When I is no longer constant w.r.t. the length of the beam, the deflection equation must be re-derived from first principles, knowing the bending moment in the beam as a function of the beam loading and the position of the loads. In other words,

    [itex]θ(x) = \frac{1}{E}\int^{L}_{0}\frac{M(x)}{I(x)}dx + C_{1}[/itex]

    [itex]δ(x) = \int^{L}_{0}θ(x)dx + C_{1}x + C_{2}[/itex]

    assuming that E, Young's Modulus, is still constant, and M(x) represents the bending moment as a function of the length coordinate. The two unknown constants of integration, [itex]C_{1}[/itex] and [itex]C_{2}[/itex] are determined by applying the boundary conditions of the cantilever known at the fixed end, namely the slope and deflection there are both equal to zero.

    Now, you can assume a unit force acting at the tip of the beam and work out the deflection at the tip based on that force. Once you have calculated the deflection due to a unit force, you can determine the force needed to produce a given deflection at the tip using a simple ratio.

    The two integrations need not be carried out symbolically unless you are interested; numerical integration can be used if you are only interested in what happens say, at the the tip of the beam.
  6. Sep 4, 2014 #5
    OK, I see. I will work it out and see what I can come up with.. Thanks!!
  7. Mar 28, 2015 #6
    Hi d.saldana,
    Greetings to you. I have the same query as yours. I tried a lot, but could not get the defection and bending stress equations.
    I would like to know whether you got the solution.
  8. Jan 5, 2017 #7
    I need help with this also, I have a tapered beam, and need to work out the theoretical deflection along intervals of x across the beam. I have the beam length, the butt diameter, the tip diameter, and the beam thickness. After that I have not a clue where to start, this forum seems to make sense to me but then i get lost when trying to implement what has been said into Excel
  9. Jan 5, 2017 #8


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    Please start a new thread for this . If this is homework then start your new thread in homework section .

    Anyway we need to know more about the problem . Try sketching the beam so that we can see what shape it is and where the loads act .
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