Fig.1.1. Shows a simply supported beam AB of span l carrying a uniformly distributed load w per unit run over the whole span . Let Va and Vb be the vertical reaction at the supports A and B respectively.
Since the loading is symmetrical on the span, each vertical reaction equals half the total load on the span.
The S.F. diagram is a straight line. The S.F. uniformly changes from + wl/2 at A to – wl/2 at B and obviously the S.F. at mid span is zero.
The B.M. diagram is a parabola. The B.M. increases according to a parabolic law from zero at A to + wl²/8 at the mid span C and from this value the B.M. decreases to zero at B following the parabolic law .
(v) Simply Supported Beam Carrying a uniformly distributed load over part of its span .
(a) When the beam carries a uniformly distributed load for a certain distance from one end.
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