statics explained

distributed load 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

A real load rarely sits at a single point. Weight, wind, and water press along a whole length as force per unit length, the intensity w. To work with it you replace it by one resultant equal to the area under the load, acting at the centroid of that area. Gathering a spread force into one point is the whole of this lesson.

Drag the intensity and a distributed load shows what it is. It is force spread along a length, measured in newtons per meter as w, a row of arrows rather than a single arrow. The denser the row, the closer it comes to a continuous press.

Changed into a resultant, the whole load gathers into one force equal to the area under it, and that force sits at the centroid of the area. The effect on the body stays the same. Only the picture changes into a point load, and the equilibrium stays the same.

A uniform load is the simplest. The area is a rectangle, so the resultant is W = w L and it sits at the middle, and dragging the length shows the resultant changing with it. When the intensity is even, the place is always the center.

A triangular load grows from 0 up to the peak. The area is half a rectangle, so W = ½ w0 L, and it acts at the point 2/3 of the way toward the peak, and dragging the peak carries that place along with it. The sharp end pulls the resultant toward itself.

Now solve a beam. Once the distributed load is replaced by its resultant it is just a point load, and dragging the intensity makes the support reactions follow from the sum of moments and the sum of vertical forces. That is the work after the spread picture has collapsed to one point.

In PracticeA distributed load presses along a length with intensity w, force per unit length, and the point is to replace it by one equivalent resultant. The size is the area under the load curve and the point of action is the centroid of that area: a uniform load gives W = w L at the middle, and a triangular load gives W = ½ w0 L at the point 2/3 of the way toward the peak. Reduced to that one force, a beam solves as before with ΣFx=0, ΣFy=0, and ΣM=0. The idea that a spread quantity collapses to an area and a centroid leads straight into the next subject, finding centroids.

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