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View Full Version : Beam Bending....



mvalenti
23rd Nov '09, 20:22
Ok, lets say I have (2) strips of sheet metal, 0.125in thick 5052AL, x 0.63in wide, 10 inches long. (shush you metric folk, its just a fad and you know it) Sandwiched between is a piece of rubber, size and shape arnt important, for the discussion, it is simply applying a load in the middle of the 10 inch span. The sandwich is clamped together at each end. If the total resistance of the load of the rubber is 10 lbs, am I correct that only 5lbs is acting on each piece of sheetmetal? Neither side of the sandwich is supported on its length.
Thanks!
-Mark

keuninkske
4th Jan '10, 04:05
according to the basic force calculations the total weight is distributing itself trought the construction to the ground or etc.

so, when you start calculating from the lowest plate, there is 10 lbs on the lowest AND 10lbs on the upperest plate.

when you calculate out of the gravity center of the rubber piece (what is very unusual) there is 5lbs on each side of the rubber piece.

tip: start your calculation always from a fitted (not variable) point and search then your bending amount and shape changes due to the forces

svenny
16th Feb '10, 17:31
I wasn't sure if this question was resolved. Something I always think of to try to help myself visualize the answers for these types of situations is to go back to a basic thought. i.e. If we imagined the two steel plates were people, and the rubber in between was a bathroom weigh-scale, then if I wanted the scale to read 10lbs I would have to push with 10 lbs force, but so would the person on the opposite side. If they didn't push back equally there would be no force on the back of the scale. In effect the opposite person acts as a floor or fixed location.
So my answer to this would be 10lbs either side...but maybe I'm wrong?!