# Can a beam be too short to experience bending stress?

The question in the title says it all. I've got a a configuration where I'm using a bolt to support a load. The way its set up, the bolt will have a very short cantilever on it. When I calculate the bending stress in the bolt, it is over twice the yield stress for the material. The calculated shear is quite low. A similar configuration has been done at our plant in the past with no issues. So this got me wondering if there is a point at which bending stress is not applicable. If it was applicable, clearly the bolt would have failed the last time this operation was completed.

*edit: I have attached a sketch. It Though there might be a generic rule about length of beams when bending might not apply.

*There is a machine called a tugger which is an electric motor with a reel. It is used to pull long heavy lengths of cables out through conduit. It will be placed on grating that is supported by c-channel. A threaded rod will be attached to the tugger and one of the c-channel will be used to hold the tugger in place as pulls the cable. The second threaded rod (on the left in the image) along with the lower support piece are merely to hold the rod in place that will take the load. My assumption in that all the load will go to the cantilever shown in detail A. Again this has already been done with no issues. I'm new at the job and have some free time so I wanted to actually try the calcs. Current engineer just "knows" these things.

**I mistyped when I said the bending stress is twice yield. I meant to say it is twice the allowable.

• If there is no lever then there is no bending stress. Commented Jun 28, 2021 at 11:39
• When you calculated "twice the yield stress", did you include the pre-load in the bolt caused the torque when you tightened it? Note, if there are no redundant load paths, the structure is statically determinate so the load of the bolt does not depend on the flexibility of the beam. Commented Jun 28, 2021 at 13:03
• Could you provide a sketch and you calculations?
– NMech
Commented Jun 28, 2021 at 13:07
• It does not sound right if your shear stress is low. Provide the configuration and show some calculations, we might find where is the problem.
– r13
Commented Jun 28, 2021 at 13:29
• @NMech added a sketch Commented Jun 28, 2021 at 14:48

This sketch is traced off from yours, you may update any missing/incorrect details. However, my impression is you were calculating the forces on plane a-a with the force applied as shown. For such a case, the bolt acts as a shear lug/pin and subjects to shear force only, no bending will occur or only with a negligible amount of moment resulted from the force times half of the top plate/element thickness, which is usually ignored unless the force is applied directly to the embedded part of the bolt.

Note, the bolt will fail in bending if the force is applied at 2" below plane a-a, provided that the bolt is rigidly attached to the top plate/element, and the top plate/element is fixed against translation and rotation.

• OK thanks for the response. The grating is wide enough that it will not support the threaded rod. All the force will be applied at 2" below the a-a plane (it will be pressing against the c-channel), effectively creating a 2" cantilever on the rod, with the 500lb pulling at the end. So in that case, you are saying my analysis is correct? Commented Jun 29, 2021 at 12:38
• If the force is at 2" below a-a plane, provides the element holding the bolt/threaded rod is fixed against translation and rotation so the bolt/threaded rod acting as a fixed end cantilever beam, then your calculation was correct. But, it looks like the assembly is a hanger that consists of a rod holding the channel below the grate, then using a plate washer on top of the grate to hold the rod and channel. If this is the case, the fixed end condition may not exist, thus, no cantilever action and the moment could be zero. You need to figure out the fixity of the support for the rod.
– r13
Commented Jun 29, 2021 at 13:08

The short answer, is that there are bending stresses in the configuration you are presenting. The big difference is that for such small aspect ratio beams the shear stress are comparable to the bending stresses. Because, the Euler-Bernoulli beam theory (which is a cornerstone of structural engineering) ignores shear stresses, the equations are not applicable. However, you can use the Timoshenko beam which accounts for the shear stresses.

The equation for simple bending (I assume that is the one you used) has certain assumptions in order to be valid.

The following is an excerpt from learn about structures

Bernoulli-Euler Assumptions The two primary assumptions made by the Bernoulli-Euler beam theory are that 'plane sections remain plane' and that deformed beam angles (slopes) are small.

The plane sections remain plane assumption is illustrated in Figure 5.1. It assumes that any section of a beam (i.e. a cut through the beam at some point along its length) that was a flat plane before the beam deforms will remain a flat plane after the beam deforms (i.e. it will not curve out-of-its-own-plane as shown in the lower right image within Figure 5.1). This assumption is generally relatively valid for bending beams unless the beam experiences significant shear or torsional stresses relative to the bending (axial) stresses. Shear stresses in beams may become large relative to the bending stresses in cases where a beam is very deep and short in length.

The beam theory is generally applicable for beams of aspect ratio greater than around 5–10.

In your case because the diameter is 0.4'' and the span is about 2'', the aspect ratio is about 5, which the limit at which you can safely use Euler-Bernoulli beam theory (to be perfectly honest I wouldn't use it)

• Ok thanks, the comment that beam theory is only applicable for aspect ratios of 5-10 is the sort of guideline I was looking for. So if euler-bernoulli beam theory does not apply to my problem, would I then assume that the bending stress of ~140ksi is not accurate? @NMech Commented Jun 29, 2021 at 13:01
• The stress value is very close to the "real" value. you will find the principal stress from the values of the normal (i.e. bending) and the shear stress (however, keep in mind that both of them have a variation through the thickness). Even in the worst case scenario (if your calculations for bending and shear are correct - I've not crosschecked them) the max principal stress should be less than $\sqrt{140.8^2 + 5.2^2)}\approx 140.9$, so the difference is pretty small (less than 1%)
– NMech
Commented Jun 30, 2021 at 4:17