The stiffness of a uniformly loaded cantilever beam is given as follows:

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I am trying to derive the stiffness, but I am facing a problem. I would be glad, if someone could help me out.

What I have done so far:

From my understanding, the stiffness $k$ is defined as: $$k = \frac{F}{\delta}$$ where $F$ is the force that moves the beam at a certain point by $delta$

Hence, I calculate the tip deflection and the equivalent point load at the tip and use those two values to get the stiffness.

I know that the deflection curve is:

$$y(x) = \frac{p}{24EI}(6L^2x^2-4Lx^3+x^4)$$

The tip deflection becomes: $$\delta_{tip} = y(L) = \frac{p}{24EI}(6L^4-4L^4+L^4) = \frac{pL^4}{8EI}$$

For the equivalent force at the tip, I do the following:

  1. I calculate the equivalent point load

  2. Then I calculate the equivalent tip load from the point load (moment = 0 around fixed end)

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But then I end up with the following stiffness:

$$k=\frac{\frac{1}{2}pL}{\frac{pL^4}{8EI}} = \frac{4EI}{L^3}$$

which is wrong...

  • $\begingroup$ @PhilSweet maybe, i don't quite understand your comment, but to me it seems that the problem is not to get the deflection, but to get the correct expression for the force. $\endgroup$
    – james
    Mar 25, 2020 at 13:46

2 Answers 2


You came so close to the answer, you ran right past it!

As you've said, the stiffness coefficient is defined as force over distance: how much force you need to apply to deflect an object (a beam or a spring, for example) in a given way a certain distance.

As you correctly derived, the deflection at the point of a beam under a distributed load is

$$\delta = \dfrac{pL^4}{8EI}$$

Now, that's actually basically the answer right there, actually. It's saying that if you apply a distributed load $p$ over a beam, it will deflect in a given way a certain distance $\delta$. That's really close to our definition of a stiffness coefficient above, except that here we're talking about a distributed load instead of a force.

So, how can we convert this distributed load to a force? You actually tried too hard. All we have to do is calculate the total load $P$ created by the distributed load, which you correctly derived as

$$P = pL$$

And we're done. There's no need to perform any sort of transformation to the force's location or anything. If we just plug that into the deflection equation, we get

$$\begin{align} \delta &= \dfrac{PL^3}{8EI} \\ \therefore \dfrac{P}{\delta} &\equiv k = \dfrac{8EI}{L^3} \end{align}$$


The thing to remember here is that stiffness coefficients are unique to the exact loading and boundary configuration given, and to the specific point they are calculated for. If you convert a distributed load into a concentrated one and calculate the resulting deflection, you'll get a different stiffness coefficient (after all, a beam under distributed loading deflects differently from one under a concentrated load). If you use that same beam with the same loading, but calculate the deflection at midspan, you'll get another stiffness coefficient. None of these coefficients are wrong, they're just correct for different cases.

So calculating the stiffness coefficient is the easy step: calculate the deflection at the desired point, sum up the total force applied throughout the entire structure, and then divide the deflection by that.

  • $\begingroup$ Great answer ! ...Just one think is still not very clear to me: I thought that the force (in the definition of k) had to act at the position of the deformation. Is my definition wrong ? Because, if it is not wrong, then the force that you get from the distributed load is acting at the center of the beam and not at the tip ? Thanks for your clarification . $\endgroup$
    – james
    Mar 26, 2020 at 15:32
  • 1
    $\begingroup$ @James. Your definition is wrong. The stiffness is based on whatever force is applied and whatever point of displacement is desired. $\endgroup$
    – JohnHoltz
    Mar 28, 2020 at 4:41
  • 1
    $\begingroup$ @james as John said and I try to explain in my answer, the stiffness is for a given point in a structure given the specific loading applied to the whole structure. Don't get confused with the concept of resultant force (conversion of a distributed load into a concentrated force), which is only useful when calculating support reactions. You'll convert loads into forces when calculating reactions in a simply supported beam, but the shear, bending and deflection diagrams you'd get from a distributed load are different than from its equivalent resultant force. The same applies to stiffness. $\endgroup$
    – Wasabi
    Mar 28, 2020 at 4:54
  • $\begingroup$ I see, thank you very much for your answer ! $\endgroup$
    – james
    Mar 28, 2020 at 8:24
  • $\begingroup$ @JohnHoltz Thank you for your answer. This helped me a lot. Do you know of a reference that states the definition of stiffness as you described it ? $\endgroup$
    – james
    Mar 28, 2020 at 8:26

you may not substitute distributed load with a concentrated load for deflection. in your case the deflection of the original loading if we call $ \ \omega L=P \ $ is:

$\delta_{max}=\frac{\omega L^4}{8EI}=\frac{P L^3}{8EI} $

but if we substitute and place P at the center of the beam

$ \delta_{max}=\frac{5P L^3}{48EI}=5/6 \frac{P L^3}{8EI} $

As we see the substitution has lead to 1/6 reduction in the reflection and that's is how it should be because of the fact the parts of the distributed load past the center of the beam are more effective in bending it than those on nearer to the support with less moment.

  • $\begingroup$ Thanks for your answer! $\endgroup$
    – james
    Mar 26, 2020 at 15:33

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