# Does Bending Stress Affect Natural Frequency?

I have been researching about the factors that affect natural frequencies, particularly about the effects of stresses.

Most of what I have found only discusses compressive and tensile stresses though, that they decrease and increase the natural frequency respectively

However, I cannot find anything about how bending/flexural stress affects the natural frequency. If I had to assume though, it probably does not affect it, as bending is just a combination of compression and tension occurring at the two sides, and they just cancel out.

The oscillation comes about as the interaction between a position dependent 'restoring force' and an inertia. I put force between brackets as it is a force in a mass-spring system, but can also a torque.

This interaction can be described by the differential equation $$m*a = -k*x$$ so $$m*x"+k*x = 0$$ For a linear spring, $$x = \sin(\sqrt(k/m)*t)$$ is a solution, which is harmonic and with frequency determined by the inertia term 'm' and the spring stiffness term 'k'.

In your bending beam, the 'restoring force' is actually a torque that is the result of the combination of compression on one side of the neutral line and tension on the other side.

Its hard to imagine how to add bending stress without altering either the shape or otherwise the physics of the beam. For strings (under tension) or columns (under compression) this is easily done.

One note (pun somewhat intended): the frequency can change with changing amplitude of oscillation, when the stiffness is not constant. Many materials exhibit spring stiffening. They are thus non-linear springs. This means that the restoring force depends on amplitude (larger deflection, larger-than-proportionally-larger restoring force. Also, the response will not be a simple harmonic anymore.

• Yeah, I was trying not to bring it up, but the object is a column under combined axial and bending. Can it then be simplified that the effect of bending stress on the natural frequency depends on how much of the cross-section of the column is subject to compressive and tensile stresses respectively? Nov 16, 2022 at 7:23
• Depends on the amount of stress you're talking about. Near the buckling load, the column might have very low stiffness, so also low frequency. Example: minusk.com/content/technology/… For small load, where you're assuming
– RJDB
Nov 16, 2022 at 21:28
• Sorry, last comment incomplete. For small loads (linear elastic behaviour), it can still be that the beam deflects due to the bending load that the shape changes, and the eigenmode with it. Picture a slender vertical rod supporting a load against gravity. Adding a bending load causes deflection, causing more bending because the vertical load now has an arm. Depends on your definition of column and loads.
– RJDB
Nov 16, 2022 at 22:20

Bending stress is actually the general term to refer to tensile and compressive stresses that develop in a long member by transverse loading, applying a moment to it, or mechanically bending it.

All these lead to the member having tensile stress on the side far from the curvature and compression stress near the inside curve.

• So do the effect of these two stresses on the natural frequency effectively cancel each other out? Or does it still depend on how much of the cross-section area experiences tensile/compressive stresses, and whichever dominates will cause their effect on the natural frequency? Nov 15, 2022 at 9:49
• only if the effect they have in increasing and reducing natural frequency are equal in magnitude. lts a reasonable assumption, like tuning a guitar by adding tension or relaxing the wire! Nov 15, 2022 at 17:24

Frequency is related to force, that is applied and released quickly to produce a forced vibration (oscillation) of an object. The neutral frequency is the property of the material that the object is made of. If the forced frequency is equal to the natural frequency, the vibrations' amplitude increases manyfold. This phenomenon is known as resonance.

In a mass-spring system, with mass m and spring stiffness k, the natural frequency can be calculated as:

Note that the spring stiffness/constant has a unit of "force/length". In it, the length is the displacement caused by the applied force, not stress.