Help out a college student.

How do you derive the deflection equation y(x) of the shown stepped cantilever beam in the figure? Each segment has a unique Young's modulus and moment of inertia. The thickness of segment 1 is larger than segment 2. I will be needing the deflection equation to derive the resonant frequency of the beam using Rayleigh method shown:

$$ \omega^2 = \frac{\int_{0}^{L} EI \left( \frac{d^2y(x)}{dx^2} \right)^2 \,dx}{\int_{0}^{L} \rho A (y(x))^2 \,dx} $$

where L = L1 + L2


I have tried using the deflection equation of a uniform cantilever beam with concentrated load F at the free end $y(x) = \frac{Fx^2}{6EI}(3L-x)$, but when I substitute this equation to Rayleigh equation and compare the resonant frequency values to COMSOL (FEA) simulation results, it does not follow the trend of resonant frequencies obtained when I try to vary the length of segment 1. However, it does follow trend of resonant frequencies when the width and thickness of segment 1 is varied.

I hope you can give me some advice. Thanks!

Step Beam with Point Load at the Free End


1 Answer 1


First of all, there will be two deflection equations, because the stiffness is not uniform. I'm getting the impression you somehow tried to use a single equation for the whole beam, which would obviously fail.

The basic equation for cantilever beam you posted seems correct. Now for the left part you can use that as is, substituting the total length (L1+L2) for L and E1I1 for EI, which gives you the deflection equation for x between 0 and L1, I'll call it yL.

The right part is more complicated, though, because aside from the load at free end, there is a deflection and a rotation on the left side. You can take the basic cantilever equation again, this time using E2I2 for the stiffness, but you need to add the end deflection from the left part, i.e. yL(L1), and account for the rotation as well. To get the rotation equation of the left part, take the derivative of the deflection equation. The resulting equation for the right part will look like yR(x) = yR,basic(x) + yL(L1) + yL'(L1)*(x-L1)

This is the best I can provide with limited time and writing on my mobile, hopefully it sets you on the right track. It would help if you posted your attempt, so that we can see what you actually substituted in the Rayleigh equation and possibly point out what exactly went wrong.

  • $\begingroup$ Hello, thank you for your response. :) I actually tried a procedure similar to what you have suggested. The derivation of y(x) actually started off with the equation of moment $$ M = F(L-x) = EI \frac{d^2y}{dx^2} $$ where $$ L = L_1 + L_2$$ such that $$ \frac{dy}{dx} = \frac{F}{EI} \left( Lx - \frac{x^2}{2} + C \right)$$ $$ y(x) = \frac{F}{EI} \left( \frac{Lx^2}{2} - \frac{x^3}{6} + Cx + D \right) $$ Applying the boundary conditions at the fixed end for the first segment $$ y(0) = 0 $$ and $$ \frac{dy(0)}{dx} =0$$ for the left part, C and D are zero. $\endgroup$
    – p.laine
    Mar 22, 2021 at 17:00
  • $\begingroup$ For the right part, however, we have to satisfy the boundary/continuity equation at $x=L_1$. To do that, I equated the equation for $y(x)$ and $\frac{dy(x)}{dx}$ for the two sides at $x=L_1$. $$ \frac{dy}{dx} = \frac{F}{E_1I_1} \left( Lx - \frac{x^2}{2} \right) = \frac{F}{E_2I_2} \left( Lx - \frac{x^2}{2} + C \right)$$ use the above equation to calculate C $$ y(x) = \frac{F}{E_1I_1} \left( \frac{Lx^2}{2} - \frac{x^3}{6} \right) = \frac{F}{E_2I_2} \left( \frac{Lx^2}{2} - \frac{x^3}{6} + Cx + D \right) $$ using the value for C, use the above equation to calculate D $\endgroup$
    – p.laine
    Mar 22, 2021 at 17:12
  • $\begingroup$ Now, we already have equations for $y(x)$ for the left and right part. For the left part, we have $$ y(x) = \frac{F}{EI} \left( \frac{Lx^2}{2} - \frac{x^3}{6} \right) $$ and for the right part, $$ y(x) = \frac{F}{EI} \left( \frac{Lx^2}{2} - \frac{x^3}{6} + Cx + D \right) $$ wherein the derived values for C and D will be used. These equations were substituted to the Rayleigh equation. $\endgroup$
    – p.laine
    Mar 22, 2021 at 17:16
  • $\begingroup$ My problem, however, is that when I use the Rayleigh model to determine the resonant frequency and compare the results to COMSOL simulation, the trend of resonant frequencies when I vary $L_1$ does not follow the COMSOL results. In my design, the thickness of segment 1, t1 (connected to the fixed end), is made larger (around 3x) than t2. $\endgroup$
    – p.laine
    Mar 22, 2021 at 17:27
  • $\begingroup$ In the COMSOL results, even if I change the length of segment 1, the resonant frequency does not change significantly (plot is almost flat). This makes sense because since segment 1 is thick, then the first mode of resonant frequency is mostly due to the thinner segment, segment 2. However, when using the Rayleigh model, there is a slope in the plot or resonant frequency when I vary $L_1$. $\endgroup$
    – p.laine
    Mar 22, 2021 at 17:28

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