Distributed Loads — Equivalent Resultants

Replace a spread-out load with a single equivalent force acting at exactly the right place.

StaticsMechanical Engineering Year 1Free preview
⏱️ About 16 min

Snow piled on a roof, water pressing against a dam, and your own weight spread across a diving board are all distributed loads — force spread out over a length or area, not piled at a single point. Before you can find the reactions holding that structure up, you have to turn the spread-out load into one equivalent point force, placed exactly where the original load's 'average' acts.

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The big idea: Any distributed load along a beam can be replaced by a single resultant force whose magnitude equals the total area under the load-intensity diagram and whose line of action passes through the centroid of that same area. Once replaced, the beam reacts as if only that single force were applied — a huge simplification.
🎯 By the end, you'll be able to
  • Compute the resultant force of a uniform and a triangular distributed load
  • Locate the line of action of the resultant using the centroid of the load diagram
  • Apply the resultant to find support reactions on a beam carrying distributed loads
📎 Helpful to know first

From spread-out load to a single force

A distributed load is described by its intensity w(x), measured in force per unit length (N/m or kN/m). Plotting w(x) along the beam gives the load diagram. The total force exerted by the distributed load is the area under that diagram — because force equals intensity times length, integrated over the loaded region.

But force alone isn't enough. To compute moments — and therefore support reactions — you need to know where that total force effectively acts. That point is the centroid of the load diagram, the geometric center of the area. For simple shapes, the centroid is a standard formula you can apply by inspection.

🔑 Two load diagrams every engineer memorizes
  • Uniform load: rectangular diagram. Resultant = w · L, acting at the midpoint (L/2 from either end).
  • Triangular load: linearly varying from zero to wmax. Resultant = ½ wmax · L, acting at L/3 measured from the high side (or 2L/3 from the low side).
\[ R = \int_0^L w(x)\,dx \qquad \bar{x} = \frac{\int_0^L x\,w(x)\,dx}{R} \]
General formulas: resultant magnitude = area under load diagram; resultant location = centroid of that area.
🎮 Interactive: uniform distributed load resultant LIVE
Predict first: If you double the span while keeping w the same, what happens to the resultant?

An interactive slider showing the resultant force of a uniform distributed load as a function of load intensity and span length.

The resultant of a uniform load equals the rectangular area w·L, and it acts at L/2 from the end.
📝 Worked example: A 6 m simply supported beam carries a triangular distributed load that varies from zero at the left end (A) to 9 kN/m at the right end (B). Find the magnitude and location of the equivalent resultant force.
  1. The load diagram is a triangle with base L = 6 m and height w_max = 9 kN/m.
  2. Resultant magnitude = area of triangle = ½ · 9 · 6 = 27 kN.
  3. The centroid of a triangle is located at one-third of the base from the high side. Here the high side is at B, so the resultant acts at 6/3 = 2 m from B, or equivalently 4 m from A.
  4. Sanity check: the load is heavier near B, so the resultant should sit closer to B than to A — 4 m from A (and 2 m from B) confirms this.
✓ R = 27 kN, acting 4 m from A (or 2 m from B).
✏️ Practice: A uniform distributed load of 5 kN/m acts over an 8 m span. What is the magnitude of the equivalent resultant force, in kN?
kN
Solution
  1. For a uniform load, resultant = w · L = 5 · 8 = 40 kN.
✏️ Practice: Using the same triangular load as the worked example (6 m span, 0 at A, 9 kN/m at B), how far from support A does the resultant act, in metres?
m
Solution
  1. The centroid of a right-triangle load diagram sits at L/3 from the high side (B).
  2. Distance from A = L − L/3 = 6 − 2 = 4 m.

Check your understanding

1. A uniform distributed load of 4 kN/m acts over 10 m. Where does the resultant act?
A uniform load's resultant acts at the midpoint of the loaded length: 10/2 = 5 m from either end.
2. A triangular load varies from 0 to w_max over length L. What is the resultant magnitude?
The area of a triangle is ½ · base · height, so the resultant is ½ w_max L.
3. Why must the resultant pass through the centroid of the load diagram?
Equivalence requires both the same total force and the same total moment about every point — placing the resultant at the centroid is what makes the moments match.
🔎 Related lessons
Internal Forces at a Section 2D Equilibrium: ΣF=0, ΣM=0
✅ Key takeaways
  • The resultant of a distributed load equals the area under the load-intensity diagram.
  • The resultant acts through the centroid of that area: midpoint for a rectangle, one-third from the high side for a triangle.
  • Once replaced by its resultant, the beam can be analyzed with ordinary point-load equilibrium.
➡️ With distributed loads reduced to resultants, the next step is to look inside the beam — at the internal forces that hold it together when it is cut.
Want to test yourself on this? Try the Mechanical Aptitude test →
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