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When a beam which is fixed at one end and free at the other, is acted upon by a load P, the beam bends and we get a bending moment at every cross section of the beam. This bending moment can be determined by making an imaginary cut at the cross section where BM is to determined, and then applying a moment balance on either parts of the beam obtained after the cut.

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Now consider a rigid bar fixed from one of its ends and a load P is applied on the other. Will there be any bending moment in this case? I mean, if we cut the bar from a section we can still perform a moment balance and that would require that some moment be developed at the section to bring the parts of the being in equilibrium. So does that mean we would still get a bending moment even if the bar is rigid?

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2 Answers 2

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Yes, a rigid beam can contain bending moments (as well as any other internal forces). Stiffness affects how an element deforms, not its capacity to resist forces.

In isostatic (statically determinate) structures, the stiffness is in fact entirely irrelevant when determining internal forces.

In hyperstatic (statically indeterminate) structures, the behavior depends on whether all elements have the same stiffness. If they do, then the stiffness is also irrelevant and "cancels out".* However, if beams have different stiffnesses (i.e. one is rigid and the other flexible), then the stiffer beam will "pull" more internal forces.


* Mathematically, a statically indeterminate structure where all elements are perfectly stiff actually can't be solved. That's because hyperstatic structures are solved via compatibility equations for deflection and rotation. But if everything's perfectly stiff, the equations can give any solution you want, since the deflection will always be zero.
However, if we remember that this is just a model, and in reality there's no such thing as a perfectly stiff beam, then we can forgive the mathematical sin of cancelling out the infinite stiffness and be on our merry way.

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  • $\begingroup$ Understood, thanks. Suppose we have two beams one rigid and other deformable, one end fixed and other acted upon by a load P. Both are having same dimensions. At any distance x, the bending moment in both, will be same. Both will also have same moment of inertia about the neutral axis. Using the flexure formula $\sigma=\frac{My}{I}$, at any distance y from the neutral axis will the normal stress be the same in both? (as the formula suggests) $\endgroup$ Mar 22 at 3:53
  • $\begingroup$ @Harshit Rajput. no it would not be the same. if you cut the beam inti FBD at any point the moment would be the same. but if you don't cut it it will behave different. the section 2nd area moment I is based on flexural behavior and the fact that we assume under moment the beam bends and the vertical cross section inclines into stress triangle. $\endgroup$
    – kamran
    Mar 22 at 4:25
  • $\begingroup$ Hmm, interesting. So will the stress be zero. The longitudinal strain in a bent beam at any distance y from NA is given as $\epsilon_y = -y/\rho$ where $\rho$ is the radius of curvature. Since $\rho$ is infinite for a rigid beam, longitudinal strain will be zero and hence the normal stress. $\endgroup$ Mar 22 at 6:28
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    $\begingroup$ @HarshitRajput it's really important to remember that this is simply a model for the real world. A perfectly rigid beam would have a stress-strain graph that's a vertical line at $\epsilon = 0$. So the fact that you get $\epsilon = 0$ doesn't mean that $\sigma = 0$, but that $\sigma$ is any positive real number. However, you CAN determine the true value of $\sigma$ by cutting the beam into a FBD balancing the external bending moment with internal forces. $\endgroup$
    – Wasabi
    Mar 22 at 14:34
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    $\begingroup$ @HarshitRajput kamran is correct that the math doesn't work out for rigid beams (for reasons similar to what I mention at the end of my answer), but just remember that this is just a model for the real world and the real beam isn't actually perfectly stiff, so the problems that arise from infinity in the model can be ignored since they don't actually occur in reality. $\endgroup$
    – Wasabi
    Mar 22 at 14:36
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Do I misunderstand something? A beam is defined as a long, sturdy, structural member that primarily to resist the load applied transverse to its axial axis. A beam can be made of any shape including bars.

In analyzing a beam of any shape, stiffness (E, I) does not affect the way to calculate the internal/external forces (R, V, M) but affects the resulting stresses and deflection.

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  • $\begingroup$ You're right r13, I have the same definition of beams in my mind too. Since it was called 'bending moment' and there is no bending in a rigid beam, I became doubtful, even though a force balance clearly suggested that an internal moment will be developed. But Now it's clear. $\endgroup$ Mar 22 at 4:00

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