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When calculating the reactions of a simply supported beam under any type of loading we assume that the moment about the reaction points is zero. What if our assumption is wrong and in actual practice there is a rotational effect about that point? Why do we make such inappropriate assumption?

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  • $\begingroup$ Awareness of this assumption is important. I was analyzing a canoe with three people in it, and ended up with a non-zero moment. I had assumed the canoe was level. Rotating the canoe to obtain a zero moment produced a tilt that was consistent with empirical measurements. $\endgroup$
    – Phil
    Feb 10, 2016 at 21:33
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    $\begingroup$ @Ethan48's answer is quite sufficient, but just as an aside: the assumption in statics of "sum of moments equals zero" applies to any point whatsoever, not just around reaction points. In a simply supported beam, you can check for moments around each of the supports or around the midspan or even around a point in empty space. We almost always choose to calculate the sum of moments around a support for the simple reason that it eliminates variables since all the force reactions from a support generate no moment around the support itself (though moment reactions must still be considered). $\endgroup$
    – Wasabi
    Feb 10, 2016 at 23:42
  • $\begingroup$ Note that we consider equilibrium condition so sum of Moments = 0 $\endgroup$
    – Fennekin
    Feb 14, 2016 at 7:54

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This rule is typically applied when studying statics. Static means that your structure or object does not move. If the moments didn't all add up to zero, that would mean there was a net force action on the object, which would cause it to accelerate and move. Since it is static and not moving, we know that whatever forces are acting on the object, they must add up to zero. If you were studying something that moves like a see-saw then the moments might not add up to zero and there is potential for the object to move, but if you're studying a beam in a building you can be sure that the moments add to zero unless the building is falling down!

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    $\begingroup$ More to the point, a see-saw is not a simply supported beam. $\endgroup$
    – Air
    Feb 10, 2016 at 19:27
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This is something of a nonsensical question, as zero moment is not an assumption here but part of the definition of this class of beam problem. A "simply supported" beam is any beam that can be represented as supported at one end by a hinge/pin and at the other by a roller (or, sometimes another hinge/pin). These are, again, classes of supports that represent real-world objects; in particular, they represent real-world objects that are not capable of transferring a moment to the member they support. By definition.

"Rotational effect" is not a meaningful term. The requirement that moment is zero means that the ends of the beam are free to rotate. However, if you meant what if there is a nonzero moment transferred at the support(s) of an actual beam, the answer is simply that the type of analysis used for the "simply supported" class of problem would no longer apply.

If you chose to represent a real beam as simply supported in your analysis and then found that there was actually a moment being transferred at one or both ends, the only bad assumption would be that this real beam could accurately be represented as a simply supported beam. You would have to classify it as something else. Not every real-world beam can be adequately analyzed using some common framework like "simply supported" or "fixed-fixed" or "cantilever"—some problems are more complex, and don't have a name. Some, we don't have an analytical solution for at all.

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I got it as we can observe there will be no rotation because of the supports so that is the reason we take moment about that point as zero

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  • $\begingroup$ Rohit, welcome to Engineering SE. You are welcome to answer your own question but this comes at the risk that your answer will be down voted if it is incorrect or lacks explanation. If you are just making your best guess, sometimes that works better as part of the main post or as a comment on another post. $\endgroup$
    – Air
    Feb 10, 2016 at 23:33

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