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Mathematically, we can see that with downward loads only (only upward loads at the ends), the maximum bending moment for a fixed end beam of span $L$ either occurs at $x=0$, $x=L$, or where the shear force is zero (or discontinuous, or otherwise undefined).

Is is the case that the maximum (absolute value) bending moment is in fact always found at one of the ends?

The most extreme case I can envision is a point load at the midpoint of the beam, this gives the (absolute value) bending moment equal, and maximised, at the ends and the midpoint: $$M(0)=M(L/2)=M(L).$$

I have asked this in the mathematical context here. I have a semi-rigorous proof there.

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  • $\begingroup$ I think I have an argument using the fact that the average bending moment is zero. $\endgroup$ Commented Nov 30, 2022 at 10:04

2 Answers 2

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the maximum bending moment for a fixed end beam of span L either occurs at x=0 , x=L , or where the shear force is zero (or discontinuous, or otherwise undefined).

The most extreme case I can envision is a point load at the midpoint of the beam, this gives the (absolute value) bending moment equal, and maximised, at the ends and the midpoint: M(0)=M(L/2)=M(L).

If you place a torque in the midspan, the maximum moment will occur in the midspan.

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  • $\begingroup$ Yes, I included this in the question. $\endgroup$ Commented Apr 30, 2023 at 7:20
  • $\begingroup$ Double-check your statements. $\endgroup$
    – r13
    Commented Apr 30, 2023 at 15:43
  • $\begingroup$ My apologies, I had never heard of such a torque and assumed you were talking about a point load. $\endgroup$ Commented May 1, 2023 at 10:07
  • $\begingroup$ In the mathematical context I had assumed $M$ continuous but I can see this is not generally the case. I am not sure how the torque fits into Euler-Bernoulli. Apologies again. $\endgroup$ Commented May 1, 2023 at 10:13
  • $\begingroup$ The Bernoulli-Euler beam theory (Euler pronounced 'oiler') is a model of how beams behave under axial forces and bending. Torque applied in the xy plane = Moment = Force x arm. $\endgroup$
    – r13
    Commented May 1, 2023 at 15:21
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We know two things about the bending moment $M(x)=-EI\,w''(x)$ for this kind of boundary conditions and load: it is has zero average and is concave. Zero average because of $w'(0)=w'(L)=0$ and concavity because of $q(x)=-EI\,w''''(x)\le 0$.

Now starting with the worst case of a point load in the middle, the bending moments at the ends are of same absolute value as the one at the center. Any deviation from this load leads to an increase of the absolute value of the bending moment at one end relative to the maximum in between. For an asymmetric point load the piecewise linear function $M(x)$ is easily found. For a set of point loads and/or a distributed load we note that increasing concavity shifts both, the maximum (center) and the minima (ends), of an average-free function to lower values (increasing absolute value of minima and decreasing absolute value of maximum), which becomes apparent from a sketch by adding some distributed load (concavity) to the case of a symmetric point load.

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  • $\begingroup$ This is just a verbal description of the answer I gave on MSE... $\endgroup$ Commented Dec 1, 2022 at 3:44
  • $\begingroup$ I am afraid to have noticed your answer too late, but it confirms your idea. I was also thinking of a more engineering approach, e.g. dividing the clamped-clamped beam into cantilevers by the zero-crossings of the bending moment, but then to find their length ratios you need again to use concavity and it is basically the same. $\endgroup$ Commented Dec 1, 2022 at 9:09
  • $\begingroup$ The engineering approach goes like this, downward loads push the beam down, by BCs and minimum, there are two inflection points corresponding to two cantilevers per inflection point with transversal force of same value but opposite sign. At the point of maximal displacement (minimum when pushed down) is an imaginary clamp. Any asymmetric point load leads to maximum absolute value of bending moment at an end. Distributed (downward) load increases this effect by increasing the clamping moment at the ends and decreasing at the imaginary clamping in the middle (equilibrium at each cantilever). $\endgroup$ Commented Dec 1, 2022 at 9:46

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