This image shows that the base structure is made up of 3 elements and a plate is placed over it. Subsequently, a long structure is attached to it. The overall structure can be considered as a cantilever beam.

I'm trying to solve this composite cantilever beam for deflection analysis. How should I proceed? I'm not from mechanical background.

The connection between the elements is strong enough that the elements do not fall or get disconnected.

Any input or referral to some useful resources to solve this is greatly appreciated.

EDIT: The connections between elements are strong enough to hold the elements together. The base material(3 cylindrical bases) is rubber which is flexible, and the triangular connecting plate is rigid. The cantilever is also flexible material. The weight of the elements is less.

  • $\begingroup$ The answer will depend on the applied load. 1) The solutions provided by the other commenters are for small (visually imperceptible) deflections. For visible deflections, you will have to do a numerical calculation that takes into account large rotations. 2) A typical composite beam solution found in strength of materials books will not work because the base plates cannot be approximated as beams. $\endgroup$ Mar 31, 2022 at 21:41

2 Answers 2


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The equation of deflection for the smaller rod with a concentrated load is $\dfrac{PL^3}{3EI}$.

$E$ - "Elastic/Young's modulus" of the material

$I$ - "Moment of inertia" of the rod

Once you decided on the material for the beam, you can easily find the values of $E$ & $I$ online, or from a textbook.

  • $\begingroup$ Thanks for the quick reply. I see a complexity here because the 3 cylinders connected to the base are kind of made up of rubber material. However, the 3 cylinders, the plate over them, and the long single cylindrical structure are properly attached. $\endgroup$ Mar 31, 2022 at 6:55
  • $\begingroup$ You are correct, it is much more involved than I have provided. I think you need help from a FEM program. $\endgroup$
    – r13
    Mar 31, 2022 at 12:42
  • $\begingroup$ Couldn't you treat it like multiple pile rows and distribute the moment to the rows that way? $\endgroup$
    – Forward Ed
    Mar 31, 2022 at 15:15
  • $\begingroup$ @ForwardEd I don't quite understand your question. But in general, regardless of the material and size variances, we can take moment at any location along the beam, however, it does not hold for calculating the deflections. $\endgroup$
    – r13
    Mar 31, 2022 at 16:58
  • $\begingroup$ I was just thinking that with bridge abutments, we calculate the moment and vertical force at the base of the abutment. That force is then distributed through the pile cap into various rows of piles. Based on the force in each pile you can then calculate elongation/compression resulting in some rotation angle for the abutment base. Based on the base rotation you take that as your initial starting angle and carrying on with the deflection in the abutment, then carrying on through the deck depending on the deck connection. USUALLY for a bridge abut. this is too small and not worth the time. $\endgroup$
    – Forward Ed
    Mar 31, 2022 at 17:30

The attachment to the wall seems much stiffer than the cantilever beam. So ignoring that base and annotating the free length, L, and the load P. We have

  • $Moment = P*L$
  • $Shear, V= P$
  • E = Modulus of elasticy
  • $D =\text{diameter of the ouside or inside cylinder!}$
  • $I= second\ moment\ of\ area=I_x = π\frac{ (Do4 - Di4)}{ 64} $ and
  • $defelction,\delta=\frac{PL^3}{3EI}$


If your base is flexible, you need to have your rubber's modulus of elasticity, E.

If not you can try stimate it by using a rigid lever with mechanical advantage of say 100, loading your rubber disk and measure the deflection$\delta$ and

$ E=(LP/A)/\delta$

Next we find the neutral axis of the base.

assuming a circle plate with rubber pads at angle of 120 degrees and area of pads A and the horizontal axis at the circle center as test centroid, with radius, R. we verify the sum of area moments to be zero.

$ 2 RA sin 30- RA= 2RA*1/2-RA=0$

so I = 2RA

and $\sigma=MR/I=M/2A$

linear deflection of lower rubber pad is

$ \delta =L_{rubber} \sigma/E_{rubber }/ E = L*M/2A/E$

angular tilt of the base

$ a= sin^{-1}\delta/R$

I did this this on my phone. please check my arithmetic.

  • $\begingroup$ Kamran, thanks for the reply. Here the base structure can not be ignored because of its less stiffness. Could you please suggest some method related to this setup of the rubber base material, rigid plate, and nylon cantilever? $\endgroup$ Mar 31, 2022 at 7:43

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