A 12ft-long simple beam carries a uniformly distributed load of 2 kips/ft over its entire span and a concentrated load of 8 kips at its midspan, as shown in Figure . Determine the reactions at the supports A and B of the beam.
Solution
Free-body diagram. The free-body diagram of the entire beam is shown in
Computation of reactions. The distributed loading is first replaced with a single resultant force, as seen in Figure . The magnitude of the resultant force is equal to the area of the rectangular loading (distributed force). Thus, P = [(2 k/ft)(12 ft)], and its location is at the centroid of the rectangular loading Since there is a symmetry in loading in this example, the reactions at both ends of the beam are equal, and they could be determined using the equations of static equilibrium and the principle of superposition, as follows:
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