Distributed loads
Nothing real is loaded at a point. A floor presses on a beam along its whole length, and the single arrow you would rather draw instead has both a size and a place — the area under the loading curve, and the centroid of that area.
Every force so far has been an arrow through a point, and no force in a real structure is anything like that. A concrete slab bears on a beam along its entire length. Snow lies over a roof. Water pushes on the whole face of a dam. The point force is a convenience, and it has been earning its keep quietly since the first free-body diagram — but when the load is genuinely spread out, the convenience has to be paid for, and the price is knowing exactly where to put the arrow.
Load per unit length
A load spread along a line is described by its intensity: how much force there is per metre of length. Write it \(w(x)\), measured in newtons per metre, and plot it above the beam so that the height of the plot at any station is the intensity there.
That plot is a graph, not a picture of a force. Its height is measured in N/m and its base in metres, and neither axis is newtons. The single most common misreading of a loading diagram is to see \(w = 5\ \mathrm{kN/m}\) written beside it and carry 5 kN into the equilibrium equations. Nothing on that diagram is a force until you multiply by a length.
Take a slice of the beam between x and x + dx. The load on it is
which has units of N/m times m, so newtons, as required. This is the only genuine force in the picture, and everything that follows is built by adding these up.
Line loads usually arrive as surface loads first. A pressure \(p\) in N/m² acting over a strip of width \(b\) gives
so a floor carrying 4 kN/m² and framing into beams at 3 m centres puts roughly 12 kN/m along each beam. The width you multiply by is the width of the strip that beam is responsible for, not the width of the beam itself — an error that is easy to make and expensive to keep, because it is silently wrong by a factor of ten rather than loudly wrong by a sign.
Something presses on the beam everywhere at once.
The equivalent force
The slices are a system of parallel forces, all pointing the same way, spread along a line. You already know what happens to such a system: it reduces to a single force, and to no couple at all provided you put that force on the right line. That was the whole business of force system resultants, and a distributed load is simply the case where there are infinitely many forces and they are infinitely small.
The magnitude comes from summing them, which is an integral:
and the right-hand side is nothing more exotic than the area under the loading curve. That is the fact worth carrying: the equivalent force equals the area of the load diagram. Not its height, not its average height, not its peak. If the diagram is a triangle you take half base times height and stop; there is no integration to do because the area of a triangle is already known.
The units confirm it. An area on that diagram is a height in N/m times a base in m, and the result is newtons. It is one of the few places in statics where an area is a force, and treating the load diagram as an area rather than a shape is most of the skill.
When the loading is made of several pieces — a uniform stretch, then a ramp, then nothing — you do not need a single expression for \(w(x)\). Compute the area of each piece separately and add the pieces. The integral is additive over the span, so chopping the diagram into shapes whose areas you already know is always legitimate.
Every slice contributes its own small force.
Where it acts
A magnitude on its own is half an answer. Two systems are equivalent only when their force sums agree and their moment sums agree, so the single force has to be put on the one line that reproduces the original moment. Take moments about the left end. Each slice contributes its own force times its own arm,
and the equivalent force, placed at some unknown \(\bar{x}\), must contribute the same total:
That ratio has a name. It is the centroid of the area under the loading curve — the geometric balance point of the shape you have drawn. So the rule is entirely visual: the equivalent force has the size of the load diagram’s area and sits on the vertical line through that area’s centroid. The machinery for computing such ratios for any shape at all is the subject of centroids and centres of gravity; here you need only the two or three shapes a load diagram is normally made of.
Now the caution, and it matters more than anything else in this topic. The substitution is exact for external effects and only for those. Because it preserves \(\sum \mathbf{F}\) and \(\sum \mathbf{M}_O\) for the body as a whole, every support reaction comes out identical. It does not preserve what happens inside the loaded region, because there the arrangement of the load is precisely what you are asking about.
The numbers make the point better than the principle does. A simply supported beam of span \(L\) under a uniform load \(w\) has reactions \(wL/2\) at each end and a largest bending moment of \(wL^{2}/8\) at midspan. Replace the load by its resultant \(wL\) at midspan and the reactions do not budge — they are still \(wL/2\) — but the bending moment becomes \(wL^{2}/4\), twice as large. Design to that and you have doubled the beam for nothing; design a real point load as if it were spread and you have halved the beam, which is worse.
The working rule follows. Reduce the load when you are solving for reactions on the whole body. Keep \(w(x)\) intact the moment you cut the beam somewhere inside the loaded stretch, and reduce only the part of the load that lies on the piece you kept.
A load growing from nothing at A to its peak at B.
Uniform and triangular loads
Two shapes cover most of what you will meet, and both are worth knowing without thinking. A uniform load of intensity \(w\) over a length \(L\) draws a rectangle, so
which is the one case where instinct is right. A load ramping linearly from zero to a peak \(w_0\) draws a triangle, so
and this is where instinct is not. The centroid of a triangle sits a third of the way in from its base, meaning the deep end where the load is heaviest, and two thirds of the way from the point. Half the errors in this topic are that one third measured from the wrong end; the sanity check is that the force must be closer to the heavy side, because that is where the load is.
Anything trapezoidal — load \(w_1\) at one end growing to \(w_2\) at the other — is best not memorised at all. Split it into a rectangle of intensity \(w_1\) across the whole span with a triangle of peak \(w_2 - w_1\) sitting on top. Each piece has a resultant you already know, and the two combine the way any two parallel forces do:
Curved diagrams turn up when the loading follows a physical law rather than a straight line. The one worth having is the parabolic spandrel, \(w = kx^{2}\), rising from nothing at one end to \(w_0\) at the other: its area is \(w_0L/3\) and its centroid is at \(3L/4\) from the pointed end. Anything stranger than that, integrate.
One more thing the splitting method quietly handles: loads that change sign. A beam with uplift over part of its span has a diagram partly above the axis and partly below, and the areas are then signed. Add them with their signs and the moment sum with its signs, and the formula above still gives the right line of action — though when the two areas nearly cancel, that line runs a long way off the beam, which is the correct answer to a nearly-couple.
Uniform: a rectangle, so the force sits at midspan.
Fluid pressure
The best reason to care about triangular loads is that fluids make them. In a fluid at rest the pressure at a point is the same in every direction and depends on nothing but depth:
with \(z\) measured downwards from the free surface. Atmospheric pressure presses equally on both sides of a gate or a wall and cancels out of the balance, so \(p\) here is a gauge pressure and the diagram starts from zero at the top.
On a vertical face of width \(b\), the pressure turns into a line load by multiplying by that width, \(w(z) = \rho g z\, b\), and since \(w\) is proportional to \(z\) the load diagram is a triangle: nothing at the surface, largest at the bottom. Everything from the last section now applies without change. The area of that triangle is
and its centroid is one third of the depth up from the base, so the resultant acts at \(h/3\) above the bottom of the face and not at mid-depth.
Both of those results are worth staring at. The force goes as the square of the depth, so a tank filled to twice the height holds four times the push on its wall. And because the resultant also sits lower in proportion, the overturning moment about the base,
goes as the cube. This is why retaining walls and dams are thick at the bottom and thin at the top, and why an extra metre of water is never a small change.
Two variations cover most problems. If the panel is submerged with its top edge below the free surface, the pressure at that edge is not zero and the diagram is a trapezoid — split it into a rectangle and a triangle exactly as before. And if the face is inclined, the pressure still depends on the vertical depth, but the length you integrate over is the slant length of the face, so the diagram is drawn against the slope and the resultant comes out normal to the surface.