It's been a while since I've done any comprehensive work with structural dynamics, so let me know if I get anything wrong here.

I have a horizontal beam that is clamped on one end and free on the other. It has the following parameters:

  • length $L$
  • elastic modulus $E$
  • moment of inertia $I$
  • mass $M$

There are no loads on this beam except for its own gravity, which can be modeled by a uniformly distributed load:

$$\psi = \frac{Mg}{L}$$

Using standard beam theory, we can calculate the deflection at points $x$ along the beam, as well as the maximum deflection $\delta_{max}$

$$\delta = \frac{\psi x^2}{24EI}(x^2 + 6L^2 - 4Lx)$$ $$\delta_{max} = \frac{\psi L^4}{8EI}$$

I have a physical constraint that $\delta_{max}$ can be no more than $\epsilon$, so I propose the following:

Spin the beam about the wall it's attached to at angular velocity $\omega$, which creates an axial centrifugal force. This should lower the value of $\delta_{max}$, and for all $\epsilon > 0$, I should be able to find $\omega$ meeting this constraint.

What I'm looking to derive is some function $\delta_{max}(\omega; \psi, L, E, I)$

My question is how can I incorporate the axial force? Each slice of beam $dm$ has centrifugal force equal to $\omega^2 x dm$ where $x$ is the distance from the clamp. I don't immediately see where to proceed from here.

Any help is super appreciated! ^^

  • 2
    $\begingroup$ Why do you say that spinning the beam will reduce its deflection? $\endgroup$
    – hazzey
    Mar 8, 2016 at 19:18
  • $\begingroup$ I thought about the problem as analogous to a conical pendulum - there are two orthogonal forces (gravity and the centrifugal force), and as the centrifugal force increases, it'll reduce the "angular dip" (the angle from the $x$-axis to the tangent of the beam tip) of the beam, thus reducing $\delta_{max}$. Is this not the right way to think about the problem? $\endgroup$
    – anonymouse
    Mar 8, 2016 at 19:22
  • $\begingroup$ Should the title be Removing beam deflection through torsion? $\endgroup$
    – grfrazee
    Mar 8, 2016 at 20:15
  • $\begingroup$ I think that what you are proposing would be similar to adding tension to a string to reduce its deflection, only in this case, the tension is added through centrifugal force. $\endgroup$
    – hazzey
    Mar 8, 2016 at 20:59

2 Answers 2


My question is how can I incorporate the axial force?

Doing so would invalidate the assumptions made in the Bernoulli beam theorem, and therefore render your deflection equation invalid. Per the linked Wikipedia article (emphasis mine),

Euler–Bernoulli beam theory (also known as engineer's beam theory or classical beam theory) is a simplification of the linear theory of elasticity which provides a means of calculating the load-carrying and deflection characteristics of beams. It covers the case for small deflections of a beam that is subjected to lateral loads only.

You could model this in a finite element software if you wanted to really know the answer. Unfortunately, the simplified approach you're outlining here won't work if you need to take axial load or torsion into account.


This is well-known theory in finite element analysis. There are two different effects that contribute to the effective stiffness of the structure in addition to the elastic stiffness $K_e$ given by Euler-Bernoulli-Timoshenko beam theory, and you need to include both of them to get the right answer.

(1) Rotating the beam creates axial tension in the beam, which will be zero at the tip and increases to a maximum value at the rotation axis. This tension creates an additional stiffness that opposes lateral bending of the beam, in the same way that the tension in a guitar string affects its vibration frequency and hence the pitch of the note produced. (Of course in a guitar string, $K_e$ is negligible, unlike your beam). This stiffness term was traditionally called the "geometric stiffness" (for reasons which aren't worth explaining) but a more logical name is "stress stiffness" often denoted by $K_\sigma$, since it depends only on the internal stresses in the structure and not on the material properties.

Incidentally, $K_\sigma$ for a compressive stress is what causes Euler buckling in columns. At the "critical load" for buckling, $K_\sigma$ "cancels out" the elastic stiffness and when $K_e + K_\sigma = 0$, the structure can't resist the applied loads. For tension stresses, $K_\sigma$ increases the stiffness of the structure.

(2) If the beam deflects from its original straight configuration, the direction and/or the point of application of the centrifugal loads changes.

If the beam rotates about a vertical axis and the beam deflects vertically, the CF load on each particle of the beam is still in a horizontal direction, and therefore the CF loads at the beam tip will create a bending moment that tends to straighten out the beam.

But if the beam deflects tangentially (for example while you are ramping up the rotation speed from zero) the direction of the CF load is radially away from the axis, and that generates a bending moment that tends to increase the bending of the beam, not reduce it.

The usual name for both these effects is "follower forces", because the direction of the applied loads "follows" the deflection of the structure. The corresponding stiffness terms are usually called the "load stiffness" and denoted by $K_L$.

The original equation for statics using only the elastic stiffness, $$K_e u = F$$ becomes $$[K_e + K_\sigma(\omega) + K_L(\omega)]u = F.$$ In principle you could get expressions for $K_\sigma$ and $K_L$ for a simple situation (e.g. a rectangular straight beam) but in practice it's much easier to use a Finite Element package that includes these effects.


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