It seems obvious that when a beam deforms under a point load, it deforms into a U-shape and when a cable deforms, it deforms into a V-shape.
Is there any graphical method of understanding this beyond common sense?
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$\begingroup$ Define common sense. It's generally not so common, certainly not in the sense that common means shared. $\endgroup$ – JMLCarter Oct 24 '17 at 15:16
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$\begingroup$ suggest "draw a force diagram". $\endgroup$ – JMLCarter Oct 24 '17 at 15:17
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$\begingroup$ @JMLCarter Could you clarify? $\endgroup$ – Michael Oct 25 '17 at 6:17
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$\begingroup$ en.wikipedia.org/wiki/Free_body_diagram $\endgroup$ – JMLCarter Oct 25 '17 at 6:45
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$\begingroup$ @JMLCarter Yes, I understand what is a free body diagram. Which is why I asked for a graphical method. What is the point of me drawing a free diagram if it is only based on my assumption that tension is parallel to the cable? So what? It does not give any reason why it should v-shaped. $\endgroup$ – Michael Oct 25 '17 at 7:15
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When you say "$\mathsf U$-shape," you're talking about a large radius of curvature $\rho$. When you say "$\mathsf V$-shape," you're talking about a small (zero for a perfect $\mathsf V$) radius of curvature.
But beam theory tells us that the radius of curvature is proportional to the stiffness of the material (as determined by the Young's modulus $E$) and the moment of inertia (more precisely, the second moment of area $I$) (and inversely proportional to the bending moment $M$ due to the load, which we'll assume is constant to compare the two cases of a beam and a cable): $$\rho=\frac{EI}{M}$$
(The product $EI$ is known as the flexural rigidity.)
Let's assume the beam and cable are made of the same material. The only thing left that can change is the moment of inertia, which is essentially a measure of how far the material is positioned from the center of its cross section. And this parameter differs substantially between a beam and a cable. An I beam is specifically designed so that its horizontal flanges are far from the center of the cross section in the intended direction of bending. In contrast, a cable strand has a relatively small cross-section area, and the strands are designed to easily slip past each other. As a result, cables have negligible flexural rigidity and can't bear a bending moment, therefore exhibiting a negligible radius of curvature (corresponding to a $\mathsf V$ shape) when exposed to a point load.
For the same reason, a cable can't be used as a cantilever and instead hangs limp unless attached at both ends. Here, the bending moment is caused by gravity.
Therefore, the graphical method you're looking for is the bending moment diagram. For a beam, an applied lateral load will generally induce a bending moment that propagates down the beam, transmitted by its nonzero flexural rigidity. In contrast, the bending moment for a cable under the same load is just a single spike at the load point. The cable simply can't transmit that bending moment down its length because its flexural rigidity is so small.
answered Oct 26 '17 at 22:17
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$\begingroup$ Can you please clarify the last two sentences? In a bending moment diagram, the rigidity of the beam does not affect the moment. $\endgroup$ – Michael Oct 27 '17 at 4:27
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$\begingroup$ That’s only because of the small-deflection assumption used for beams, which is exactly because their flexural rigidity is large. If you drop this assumption, the bending moment becomes geometry dependent and, for a cable (just as with a hinge), disappears away from the load point. $\endgroup$ – Chemomechanics Oct 27 '17 at 15:45
