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Consider a uniform beam that is supported by 3 pivots that are the same distance from each other (shown below). What will the normal force exerted by each pivot be?

The situation seemed rather paradoxical to me because depending on the order you place the pivots in the normal force exerted by each pivot will be different.

For instance, in (1), I first place in the 2 pivots on the sides, and then the one in the centre. Logically, the final weight will be borne entirely by the 2 pivots on the side... right? Since the 2 systems (the rod and 2 pivots, and the 3rd pivot) are entirely stable on their own, and there is no need for the weight to redistribute.

But I can also place in the centre pivot first, and then the side pivots (2) and then I get a different distribution of weight!

In fact I think the solution can be anything that looks like the situation in (3).

But of course, we can do the experiment in real life, and swap the pivots with mass balances, and each time we perform the experiment we can expect to get the same reading. So what is going on? Is there some missing information in the reasoning above that leads to the solution being indeterminate?

enter image description here

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    $\begingroup$ Appears to be a statically indeterminate beam $\endgroup$ – Bob D Oct 10 '20 at 7:19
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Is there some missing information in the reasoning above that leads to the solution being indeterminate?

The beam is statically indeterminate. It is not "missing information" that makes it so. It is because there is a redundant vertical reaction. If you remove any one of the supports the beam remains stable and the reactions can be determined solely by the equations for equilibrium, i.e., the sum of the moments equals zero and the sum of the forces equals zero.

When the structure is statically indeterminate you need to use additional equations with respect to the constraints in deformation, such as equations of deflection.

For an introduction to the method of analysis, check out the following.

http://ocw.nthu.edu.tw/ocw/upload/8/259/Chapter_10-98.pdf

By the way, you didn't indicate what type of supports are involved (pins and/or rollers). More than one pin would mean the beam is statically indeterminate with respect to horizontal reactions as well a vertical reactions.

Hope this helps.

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The solution to your apparent paradox is that when the beam is supported at each end it deflects and sags slightly in the middle. When you introduce a third support in the middle you change this deflection and so redistribute the weight of the beam between the supports. This is why the load on the central support is not zero.

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