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Slab hanging:
Three similar uniform slabs each having length '2a' are resting on edge of table as shown in the figure given below. If each slab is overhung by the maximum possible amount, find out amount by which bottom slab is overhanging. As the bottom beam overhang by a/3 amount
Sol.: The maximum overhang of the top beam is 'a',
Now taking moment about the point A2, considering all the load acting on middle beam.
-W(a - X) + W.X = 0
on solving X = a/2 ...(i)
Now taking the moment about point A3
-W(a - Y) + W[Y - (a - X)] + W(X + Y) = 0
-Wa + WY + WY - Wa + WX + WX + WY = 0
3Y - 2a + 2X = 0
Y = a/3 .......ANS
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