Distributed Loads & Resultants

A distributed load is described by its intensity w(x), the force carried per metre of beam at position x, in N m⁻¹. An intensity is not a force: it cannot enter a force balance or a torque equation until it has been multiplied by a length. Intensities come from physical data. A steel beam of 25 kg m⁻¹ loads itself with w = 25 × 9.81 = 245 N m⁻¹. Snow 0.40 m deep, of density 200 kg m⁻³, gathered from a 1.5 m wide strip of roof that bears on the beam, adds w = ρ g d b = 200 × 9.81 × 0.40 × 1.5 = 1.18 kN m⁻¹. Water behind a vertical gate of width b presses with w = ρ g h b, growing linearly with depth h. Plot w against position above the beam, a load diagram, and every later question becomes a question about that graph.

Cut the loaded length into strips of width Δx. The strip at xᵢ carries w(xᵢ)Δx, the total is Σ w(xᵢ)Δx, and as the strips shrink that sum becomes R = ∫w dx. Geometrically R is the area under the load diagram, and the units agree: N m⁻¹ × m = N. On a 4.0 m span the steel beam above carries 245 N m⁻¹ × 4.0 m = 0.98 kN of its own weight and 1.18 kN m⁻¹ × 4.0 m = 4.7 kN of snow, 5.7 kN in all. Rectangles and triangles need only geometry; a curved profile needs the integral. A smooth mound of snow modelled as w = w₀ sin(πx/L), with w₀ = 1.5 kN m⁻¹ over L = 4.0 m, carries R = 2w₀L/π = 3.82 kN, 27% more than a triangle with the same peak. Portions that push up, such as wind suction on a roof, enter with w < 0 and subtract.

Knowing how much is half the job, because equilibrium also needs the load's moment. The strip at x has lever arm x about the origin, so the load's moment is ∫x w dx. A single force R at x̄ has moment R x̄, and demanding that the two agree gives x̄ = ∫x w dx / ∫w dx: the average position of the load, with each strip weighted by how much it carries. That is the centroid of the load diagram, not of the beam. For a triangle rising from 0 to w₀, w = w₀x/L, so ∫x w dx = w₀L²/3 while R = w₀L/2, giving x̄ = 2L/3. The weighting pulls the resultant toward the heavy end, where the strips carry most. For w₀(x/L)ⁿ the same two integrals give x̄ = (n + 1)L/(n + 2), which starts at L/2 for a uniform load and creeps toward L as the load piles up.

The origin for x was a choice, so the line of action had better not depend on it. Take moments about a pivot at x = p instead. Each strip now has lever arm x − p, and ∫(x − p) w dx = ∫x w dx − p∫w dx = R(x̄ − p), which is exactly the moment of R acting at x̄. Force and moment that match about one point therefore match about every point, and that is what licenses the single force in a torque equation about any pivot you like. Use the freedom. Measure s from the heavy end of a triangle instead: w = w₀(1 − s/L), ∫s w ds = w₀L²/6, and s̄ = L/3, the same line as 2L/3 from the light end. On a 4.5 m span rising to 2.0 kN m⁻¹ that is R = 4.5 kN acting 1.5 m from the heavy end and 3.0 m from the light end.

Integrals add, so loads superpose. Split the load diagram into rectangles and triangles, find each piece's resultant Rᵢ and centroid x̄ᵢ from one origin, and combine: R = Σ Rᵢ and x̄ = Σ Rᵢx̄ᵢ / R. A 6.0 m platform carries a uniform 0.50 kN m⁻¹ along its length, plus a 1.0 m stack of pallets adding 6.0 kN m⁻¹ at the far end. The pieces are R₁ = 3.0 kN at 3.0 m and R₂ = 6.0 kN at 5.5 m, so R = 9.0 kN at x̄ = (3.0 × 3.0 + 6.0 × 5.5)/9.0 = 4.67 m. The plain average of the two centroids, 4.25 m, falls 0.42 m short, because the pallets carry twice the load spread along the whole platform. A missing piece enters as negative area: a uniform load interrupted by an unloaded gap is the full rectangle minus the gap's rectangle, each at its own centroid.

The resultant reproduces the net force and moment on the whole rigid body, so it gives the right support reactions. It does not reproduce how the load is shared inside. A 6.0 m beam resting on supports at its ends under 2.0 kN m⁻¹ has reactions of 6.0 kN each, whether the load is spread or replaced by 12 kN at midspan. Now isolate the left half and ask what moment the right half must exert on it across the midspan cut. With the real load, the reaction contributes 6.0 kN × 3.0 m = 18 kN m, and the 6.0 kN resting on the left half, acting 1.5 m from the cut, takes back 9.0 kN m, leaving 9.0 kN m. With the point load sitting at the cut, nothing takes anything back and the answer doubles to 18 kN m. Any part of a body needs the resultant of its own share of the load, found after cutting, and sag and stress inherit the same limit.