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We know this that if the a cantilever beam (of any cross section) is fixed at one end and a transverse force is applied on the other, then it will bend. However, we just casually assume that the moment only depends on the type of transverse loading applied on it and the distance from the loading to the support, while drawing a bending moment diagram.

I couldn't understand if this is actually the case in reality or not? I mean shouldn't the bending moment have some dependence on the cross sectional area of the beam as well? Since the change in cross sectional area would mean that the moment of inertia has changed, and hence the stiffness of the beam has changed. Should the bending moment at the same distance from the fixed support end also change or not?

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    $\begingroup$ Are you referring to the resulting deformation affecting the geometry? Because there is the concept of actual stress and engineering stress. If not, then I I think you're confusing bending moment with stress. $\endgroup$
    – DKNguyen
    Oct 13, 2021 at 19:40
  • $\begingroup$ No. I am just talking about a simple beam subjected to load (of any type) at one end and fixed on the other. I can calculate the bending moment at any cross section, right? And draw a bending moment diagram. Then by using that moment at a specific cross section, I can find the stresses (which should change linearly over that cross section under linear elastic range of the material) on that cross section. Now, the moment picked from the bending moment diagram has no dependence on the cross section. Does it in reality, or not. Thats my question. $\endgroup$ Oct 13, 2021 at 19:49
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    $\begingroup$ Yes, so the linear analog of your question is "If I am pushing a block, does the force I am pushing a block with depend on the geometry of the block (i.e. area across which I am exerting the force against)?" The answer is a lot more obvious in the linear case which is why I said it sounds like you're confusing bending moment with stress. $\endgroup$
    – DKNguyen
    Oct 13, 2021 at 21:15
  • $\begingroup$ This is a common question among undergraduates. To understand the reason, you have to go to the definition of the bending moment. en.wikipedia.org/wiki/… gives you an explanation that shows that the area is used for integration and hence does not appear again when using bending moments. $\endgroup$ Oct 14, 2021 at 3:20

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You seem to be mixing up a few concepts.

As others have mentioned, bending moment is independent of a structure's cross-sectional dimensions. After all, bending moment is simply the sum of $F_i\ell_i$, where $F_i$ represents the different forces applied on a structure and $\ell_i$ is the perpendicular distance from the force to the point of study. That is it, there's no room for a structure's cross-section to interfere.

Or is there?

If you're dealing with a hyperstatic (statically-indeterminate) structure, where you have more unknowns (support reactions) than static equilibrium equations, then you can't trivially determine the support reactions and therefore can't determine the bending moment along the beam.

In these cases, one must also use compatibility equations, which basically ensure that the beam behaves appropriately (no deflection or rotation discontinuities, etc). However, these equations are a function of the structure's stiffness $EI$ (or $EA$ for axial loads), with stiffer elements "pulling" more of the load.

So if you're dealing with a hyperstatic structure with beams of different stiffness, you'll get different support reactions than if the stiffnesses were all the same (in the case of different stiffnesses, the reactions close to the stiffer element will be greater and those further away will be lower).

And if the reactions are a function of the beams' stiffnesses, then the bending moments are also affected by them. So yes, in this case, bending moment is indirectly affected by a beam's cross-section.


However, you were thinking about the effect of a beam's deflection on the bending moment.

For this, there is a simple answer and a complicated one:

Let's start with the simple answer: no, deflections and rotations don't really influence bending moments. Structural engineering works almost exclusively under the hypothesis of small deflections and rotations. That is, that the loads applied to the structure will cause deflections so small, we can basically ignore whatever effect they'll have on the structure.

So the simple answer is: who cares? Even if there's an effect, it's so insignificant we can just ignore it.

Now for the more complicated answer: yes, deflections and rotations most certainly do influence bending moments. Indeed, as stated above, that effect is usually insignificant, but "usually" just ain't good enough for safety.

For a classical example, just look at column buckling. That is, the fact that very slender columns under axial load will "collapse" under loads far below their crushing load. For a visual cue, think of holding a straw or piece of spaghetti between your fingers and pressing down. Under barely any load the straw bends in the middle. That's buckling.

Why does buckling occur? Because deflections and rotations impact bending moments.

Basically, imagine you have a perfectly straight column with a compressive load that's perfectly vertical and perfectly centered along the column's axis. If you were to do a bending moment calculation, you'd see that the column is under zero bending moment, as expected.

However, real life ain't perfect. Columns aren't perfectly straight, loads aren't perfectly vertical nor are they perfectly aligned with the column's longitudinal axis.

But let's keep this simple and assume that the column is still perfectly straight and the load is still perfectly vertical, but it's not perfectly aligned with the axis. It is just an infinitesimal distance $\delta$ off the axis. A value so small it's almost non-existent. Almost, but not quite.

That tiny $\delta$ means the compressive load now causes a minuscule bending moment along the column's length. That bending moment causes the column to deflect ever so slightly. That increases the $\delta$, increasing the bending moment, increasing the deflection, increasing $\delta$...

For properly designed columns, this process will eventually converge to a stable value for $\delta$ and therefore for the bending moment, and everything is fine. If the column is too slender, however, the process diverges and $\delta$ quickly goes to infinity. Or, in real-world terms, the column buckles.

This is a classic example where yes, the beam's stiffness and deflection have a significant impact on the bending moment. But the simple answer of "they don't" will serve you well 99% of the time.

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  • $\begingroup$ so even if the compressive force on the column is perfectly aligned with the column axis, so no matter how much force we apply (ofcourse, less than the proof load at which the column yields), we won't see any buckling? Bending moment is always required for buckling? This is what you meant? Assume I am conducting a FEA analysis and applying a compressive load at the column's axis, so no matter how much I increase the load, I won't be seeing any buckling? $\endgroup$ Oct 14, 2021 at 17:53
  • $\begingroup$ @RameezUlHaq I'm not an expert in FEA, so take this with a grain of salt, but yes, if your FEA is perfectly meshed in such a way that you don't even have floating-point errors, a perfectly aligned force will not cause buckling (unless the FEA algorithm chosen has that built-in, of course). However, the real world isn't perfect, so you can always expect that any real column will never be perfectly straight or perfectly vertical, nor will forces be perfectly vertical and aligned with the column's axis. $\endgroup$
    – Wasabi
    Oct 14, 2021 at 18:11
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A bending moment is an external moment that does not have anything to do with the section or material properties.

Let's say we have a concentrated load applied at the center of a simply supported beam.

The moment is the same for both.

$$M=\frac{PL}{4}$$

Regardless of if you have for the beam a big W 12 x 12 x 120 or a small 2 by 4-inch lumber.

Of course, the same moment will create drastically different stresses in that two beams. That is where the engineer calculates the stresses and deflections to verify the section is suitable.

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Yes, the load is independent of the cross-section of a structure in the example you are quoting and for most cases that an engineer would meet ( wasabi had the patience to write a detailed and more complete answer than my dumbed down version- hopefully his effort and thoroughness will be appreciated) .

If the cross-section is not enough the the structure will fail. It is the engineer's job to select an adequate and economic structure.


In the example of the cantilever beam loaded at the free end), then if the structure doesn't fail and it remains in the small displacement then internal forces and moments develop to resist the load.

Those internal loads are independent of the cross-section.

On the other hand (and that is probably confusing the author of the OP), is that the magnitude of those internal loads (bending moments, shear forces, axial forces) AND the cross-section determines the stresses inside the structure.

The end result is that when you apply a load on a structure you either need:

  • a smaller cross-section with a stronger material (i.e. one that can withstand higher stresses)
  • a larger cross-section with a weaker material (i.e. one that can withstand less stresses)
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  • $\begingroup$ So basically you are saying that the moment generated at a cross section due to the load (in order to ensure equilibrium) is independent of the cross sectional moment of inertia, and hence the stiffness, right? $\endgroup$ Oct 13, 2021 at 19:59
  • $\begingroup$ No I am saying that the beam will fail if you load it too much. $\endgroup$
    – NMech
    Oct 13, 2021 at 20:01
  • $\begingroup$ Well, you said that force is independent of the cross sectional area so I was assuming that you were also asserting that moment is independent of it. My bad. $\endgroup$ Oct 13, 2021 at 20:17
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A moment can be thought as the work energy produced by a force - $F*L$. The energy can be stored on any structural element of any size but, due to the requirement of structural equilibrium, it will cause larger internal stress on a smaller element, and smaller stress on a larger element (see case 1 on the sketch below), so $\sum M$ always equals to zero for any case.

However, If we hold the stress constant, then the larger element will then experience a larger moment due to the fact that $M = f_b*S_X$, and is said to have a larger "Moment Capacity".

enter image description here

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