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Ramnares P.E.

Load due to deflection

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Simple problem that I can't seem to rationalize.  For a given uniform beam supported on both ends I can calculate the load by rearranging delta = PL/AE.

For the calculated load, is P the load at each end or should I divide by two?

@Krakosky @thekzieg

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Let me see if I understand your question: You have a simply supported beam with a single point load. You know the midpoint deflection and you're trying to figure out what the point load is? 

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It actually is a uniform cross section bar that's supported by two bearings.  I know the deflection, area, material properties, and the distance between bearing centers, L.

As the bar deflects down, it imposes an axial load on the bearings, P.  So I rearranged and solved for P.  

 

Does each bearing see an actual load P or P/2?

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I think I understand. Each reaction would be P/2 (assuming both bearings get the same force applied) and the total force exerted by the bar would be P. If it helps keep things arranged in your head, think of the forces to the bearing as the reaction R, where P=2R, or R=1/2P.

The sum of the forces in any direction always equal 0 in structures world.

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That's exactly what I was thinking.  The total force imposed would be divided between the two bearings.  Thanks for the sanity check!

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Any time!

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You are officially awesome (you were awesome before too).

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Trial by basic statics, I like it.

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We play hardball here in EB.

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Depending on the size of the beam, don't forget to include the distributed load due to beam weight.

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5 hours ago, Audi driver, P.E. said:

Depending on the size of the beam, don't forget to include the distributed load due to beam weight.

Absolutely.  Also. are the bearings on the end restrained from rotation?  If so the midpsan deflection would realistically fall somewhere between  a simple support and a fixed support.  The deflection for a fixed end condition is 1/4 of the deflection of a simple beam for the same concentrated load at the midspan.

 

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