Distributed Loads and Their Resultants

A distributed load acts continuously over a finite region of a body. In planar beam problems it is commonly described by an intensity $q(x)$, which represents force per unit length. The SI unit of a line-load intensity is N/m.

Equivalent concentrated force

A distributed load can be replaced by a resultant concentrated force when its magnitude, direction, and moment effect are preserved. For a load $q(x)$ acting from $a$ to $b$:

$$R=\int_a^b q(x)\,dx.$$

Geometrically, the magnitude of the resultant equals the area under the load-intensity diagram.

Uniformly distributed load

For a uniformly distributed load of intensity $q$ acting over a length $L$, the magnitude of the equivalent concentrated force is:

$$R=qL.$$

The line of action of the resultant passes through the midpoint of the loaded interval.

The load diagram is a rectangle, whose area centroid lies at its midpoint. The line of action of the equivalent force therefore passes through the midpoint of the loaded interval.

Location of the resultant for a varying load

The coordinate of the resultant line of action follows from moment equivalence:

$$x_R=\frac{\int_a^b xq(x)\,dx}{\int_a^b q(x)\,dx}.$$

Thus, the resultant passes through the centroid of the area under the $q(x)$ diagram. For standard rectangular, triangular, and trapezoidal diagrams, familiar geometric centroid locations can be used.

Triangular load

If the intensity varies linearly from zero to a maximum value $q_{max}$ over a length $L$, the resultant magnitude equals the triangular area: $R=q_{max}L/2$. Its line of action is located $L/3$ from the side with maximum intensity, or $2L/3$ from the zero-intensity side.

Trapezoidal load

A trapezoidal load diagram can conveniently be decomposed into a rectangular and a triangular part. Determine the resultant and location for each part, then combine the forces while preserving their moments.

Example

A uniformly distributed downward load $q=3$ kN/m acts over 4 m of a beam. Its equivalent force is $R=qL=3\cdot4=12$ kN and acts at the midpoint of the loaded interval, 2 m from its beginning.

Common mistakes

• confusing the load intensity $q$ in kN/m with a force in kN;
• placing the resultant at the midpoint for every load-diagram shape;
• ignoring the actual starting position of the loaded interval on the beam;
• calculating only the resultant magnitude without preserving its moment about a reference point.