Statics

Distributed Loads: Replace Them with One Resultant

Distributed-load intensity w, resultant=area at the centroid, uniform W=wL at mid, triangular W=1/2 w0L at 2/3, beam reactions

Real loads are rarely a single point - weight, wind, and water press along a whole length as an intensity w in force per length. To handle one, replace it with a single resultant equal to the area under the load, acting at that area's centroid.

Drag the intensity. A distributed load is a force spread along a length, measured as w in newtons per meter - not a single arrow but a whole row of them.

Toggle to the resultant. The whole load collapses to one force equal to the area under it, placed at the centroid of that area - same effect on the body.

A uniform load is the simplest: its area is a rectangle, so the resultant is W = w L, sitting at the midpoint. Drag the length and watch W follow.

A triangular load grows from zero to a peak. Its area is half the box, so W = ½ w0 L, and it acts two-thirds of the way toward the peak. Drag the peak.

Now solve a beam. Replace the distributed load by its resultant, then it is just a point load - drag w and the support reactions follow from ΣM and ΣFy.

In PracticeA distributed load presses along a length with intensity w (force per length), and the key move is to replace it with one equivalent resultant: its magnitude is the area under the load curve, and it acts at the centroid of that area. A uniform load gives W = w L at the midpoint; a triangular load gives W = ½ w0 L at two-thirds toward the peak. Once reduced to that single force, a beam is solved exactly as before, with ΣFx=0, ΣFy=0, ΣM=0. The idea that a spread-out quantity reduces to an area and a centroid leads straight into the next topic - finding centroids themselves.

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