huangyhg

shear in thin metal sheets
hello, i have two questions
a.where are the greatest amount of stress in a thin panel under a sideway load?
b.what formula is used to calculate that stress?
here is a rough sketch i have prepared of what i think is happening.
shear is not the problem in what you have sketched.
draw a mohr's stress circle for what you have sketched and examine the stresses on an element turned 45 degrees. you will notice that it is in tension on one face and compression on the other. since the element you are considering is very thin it has a very low buckling stress. with a little bit of ingenuity you should be able to test this with a piece of aluminum foil
don,
in practice, the thin sheet shown in the sketch would probably just buckle and collapse. but assuming that it is thick enough to prevent that, here is what i suggest.
isolate the sheet as a free body floating in space. in the first instance, the 100 pound load is pushing the sheet to the right. this is prevented by the fixity along the bottom edge, which results in shear flow, q, of 100 / 12 = 8.333 pounds per inch. shear stress = q/t = 8.333/.032 = 260.4 psi. represent this force along the bottom edge as an arrow along the length of the plate from right to left. now the sheet is balanced against translation, but will spin clockwise unless we balance the moment caused by 100 pounds acting over a distance, i. e., moment arm, of 12 inches.
from the center of the bottom edge of the sheet, draw a straight line downward and to the right, extending to a point 3 inches below the lower right corner of the sheet. draw a vertical line connecting the right end of this line to the lower right corner of the sheet. this creates a triangle 3 inches high and 6 inches wide below the sheet. draw the mirror image of this triangle along the left side of the bottom edge of the sheet. within the left-hand triangle, draw several parallel arrows pointing downward. within the right-hand triangle, draw a similar group of arrows pointing upward. these arrows represent distributed loads, having their maximum values at the right and left bottom corners of the sheet, and diminishing linearly to zero at the center of the bottom edge. these are the equal and opposite loads which prevent the sheet from rotating. the resultant loads of the triangular distributed loads are 2 inches from the left and right edges of the sheet. the moment to be balanced is 1200 inch-pounds, and the distance between the resultants is 8 inches, so the loads must be 150 pounds downward on the left side and 150 pounds upward on the right side. the shear load and the trianglarly distributed loads are the reactions at the base of the sheet due to the applied load of 100 lbs. turn your sketch 90 degrees and you can see that it is a cantilever beam, subject to the same bending stress of my/i as the tube.
a shear panel behaves a bit more like your sketch. such a panel must be supported on all sides, and the support
thanks, i'm working on a reply and a new sketch to see if i understand completely what has been said. don

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