# Understanding Compression in Eccentrically Loaded Columns and Beams

• curiousnoncat
In summary, the two points of load in Column B will cancel each other out, resulting in a uniform compression of a single value along the beam's cross-section. However, an element in a state of plane stress (e.g. Element C in the figure) is not considered to be in compression in the y direction.
curiousnoncat
I'm a student taking a non-calculus strength of materials course, and I believe I have what is probably a very simple misconception.

I'm wondering why in the "eccentrically loaded column" b), which has two point loads placed at distances of equal magnitude from a centroidal axis, the moments will cancel each other out, and the cross section will be under uniform compression of a single value, but an element on beam a), which is in a state of plane stress, is not considered to be in compression in the y direction (at least in my textbook).

These seem completely analogous to me, besides the material's dimensions. I've got this sneaking suspicion that compression must exist in the y direction but is "negligible" (would it be negligible in an I-beam?), and also not uniformly distributed.

What is the y direction?

curiousnoncat said:
I'm wondering why in the "eccentrically loaded column" b), which has two point loads placed at distances of equal magnitude from a centroidal axis, the moments will cancel each other out, and the cross section will be under uniform compression of a single value
That is an approximation for beams that are much taller than wide (using the orientation of (b) for those words).

mfb said:
What is the y direction?

That is an approximation for beams that are much taller than wide (using the orientation of (b) for those words).
By the y direction, I meant the vertical direction on element c, which has no stress indicated.

And what I'm really interested in hearing is a response to that part of the question, about the state of stress in the horizontal beam.

Last edited:
This is again an approximation. Close to the contact points of the forces there will be compression, but you do not have an opposing force on the other side of the beam that would give the usual way of compression. You get shear stress instead.

curiousnoncat said:
And what I'm really interested in hearing is a response to that part of the question, about the state of stress in the horizontal beam.
I think I see where your misconception is. You're thinking of the beam as a static whole.
curiousnoncat said:
I've got this sneaking suspicion that compression must exist in the y direction but is "negligible" (would it be negligible in an I-beam?), and also not uniformly distributed.
Notice how there is particular attention paid to element c. This is an attempt to show you that stress/strain is not necessarily uniform throughout an object. Precisely as you surmised.

jackwhirl said:
I think I see where your misconception is. You're thinking of the beam as a static whole.

Notice how there is particular attention paid to element c. This is an attempt to show you that stress/strain is not necessarily uniform throughout an object. Precisely as you surmised.

Thanks Jack, but I am aware of the derivation for bending stress and the distribution of shear stress in vertical and longitudinal planes, so what I want to know is how you would calculate vertical compression or tension in a beam at an arbitrary point. For example, surely as mfb said, right above actual beam supports there would be compressive stresses approximately equal to the reaction force divided by the area of application.

curiousnoncat said:
Thanks Jack, but I am aware of the derivation for bending stress and the distribution of shear stress in vertical and longitudinal planes, so what I want to know is how you would calculate vertical compression or tension in a beam at an arbitrary point.
Well, I would model the system in a 3D CAD tool and run it through finite element analysis...
If you're looking for something more sophisticated, I must leave that to our more erudite fellows.

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