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My professor gave us yesterday a problem to calculate the equivalent spring of the system shown in the attached figure. The problem: a cantilever beam supported by two springs $k_1$ and $k_2$ located at distances $a$ and $b$, respectively, from the left end. The beam has length $l$ and flexural rigidity $EI$. The idea is to approximate the motion of the mass in the vertical direction by a simple one degree of freedom system, so the first step on it is to calculate the equivalent stiffness coefficient of the system.

enter image description here

I am having some trouble to understand his solutions, because I don't agree entirely with what he said. His approach to the problem:

  • Given the deflection equation of a cantilever beam with a load $P$ applied at the tip, $$ v=\frac{-Px^2}{6EI}(3l-x) $$

he says an equivalent spring $k_{eq12}$ substitutes the springs $k_1$ and $k_2$ at the right-end of the beam. To calculate $k_{eq12}$, the total potential energy is $$\begin{align} U&=\frac{1}{2}k_{eq_{12}}y_l^2=\frac{1}{2}k_{1}y_a^2+\frac{1}{2}k_{2}y_b^2\\ k_{eq_{12}}&=k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2 \end{align}$$ where $y_a$, $y_b$ and $y_l$ are the deflections of the beam for a load located at $a$, $b$ and $l$. He claims $y_a$ and $y_b$ are the deflection of the springs.

Therefore, for a unit load, $$\begin{align} y_a&=\frac{a^2}{6EI}(3l-a) \tag1\\ y_b&=\frac{b^2}{6EI}(3l-b) \tag2\\ y_l&=\frac{l^3}{3EI} \tag3 \end{align}$$

Now, if $k_{eq_{12}}$ is an equivalent spring at the right-end of the beam, then the equivalent spring of the beam is in series with this equivalent spring, so the equivalent spring of the entire system supporting the mass is, $$ \frac{1}{k_{eq}}=\frac{1}{k_{eq_{12}}}+\frac{1}{k_{beam}} $$

Substituting the deflections $y_a$, $y_b$, $y_l$, and the equivalent stiffness of the beam being $k_{beam}=3EI/l^3$, we get $$ k_{eq}=\frac{\left( k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2 \right)k_{beam}}{k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2+k_{beam}} $$ which can be simplified, substituting $(1)$, $(2)$, $(3)$ and the $k_{beam}$, to $$ k_{eq}=\frac{3EIk_1a^4(3l-a)^2+3EIk_2b^4(3l-b)^2}{l^3\left[k_1a^4(3l-a)^2+k_2b^4(3l-b)^2+12l^3EI\right]} $$

That is the answer he gave us. Particularly, I don't agree with a few steps along the way, but the key point here for me is that we cannot use the deflection equation for a cantilever beam a tip load at the end, because this beam is not exactly a cantilever beam, but it has two extra supports, the springs.

My approach to this problem would be to, given an applied load $P$ at the left end, find the deflection equation for the system with the springs and the approximate the equivalent stiffness with the new equation. The deflection equation now would have to be calculated from the $EI\frac{d^4v}{dx^4}=w(x)$ elastic curve. A much harder work, because it is a cantilever beam supported by two springs, and it is a statically indeterminate beam. Is this line of though correct or was his approach correct?

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  • $\begingroup$ Are we really sure that they are in series? if you find an equivalent stiffness for the springs at the end of the beam, the resulting springs will work in parallel I believe since they will deflect with a common displacement at the tip of the beam... $\endgroup$ – Umut Tabak Feb 22 at 11:11
  • $\begingroup$ Yeah, I just don't agree with everything he said along the explanation of this particular exercise. If his approach is correct, then the equivalent spring at the end should work in parallel, exactly as you said. $\endgroup$ – Thales Feb 22 at 13:58
  • $\begingroup$ Well since this is a 1-dof system, it should be easy to write the equations of motion and find the equivalent stiffness, I can give an answer in the weekend I believe... $\endgroup$ – Umut Tabak Feb 22 at 16:53

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