A load distributed continuously over a line cannot usually be replaced by a point force by simply choosing a convenient location. The replacement must preserve both the total force and the total moment.

Let a load act along a curve parameterized by arc length $s$. If

$$\mathbf q(s)$$

is force per unit length, then a small segment $ds$ carries

$$d\mathbf F=\mathbf q(s),ds.$$

Resultant force

The total force is the vector integral

$$\boxed{\mathbf R=\int_C\mathbf q(s),ds}.$$

This is the force that an equivalent concentrated representation must reproduce.

Resultant moment

Choose a reference point $O$. If

$$\mathbf r(s)$$

points from $O$ to the load element, its moment contribution is

$$d\mathbf M_O=\mathbf r(s)\times d\mathbf F.$$

Therefore

$$\boxed{\mathbf M_O=\int_C\mathbf r(s)\times\mathbf q(s),ds}.$$

The pair

$$\boxed{(\mathbf R,\mathbf M_O)}$$

is an equivalent force-couple system for the distributed load about $O$.

Preserving only $\mathbf R$ is not enough: two different load distributions can have the same total force but different rotational effects.

Changing the reference point

Let $A$ be another reference point and let

$$\mathbf r_{A/O}$$

point from $O$ to $A$. Since

$$\mathbf r_{element/A}=\mathbf r_{element/O}-\mathbf r_{A/O},$$

we obtain

$$\boxed{\mathbf M_A =\mathbf M_O-\mathbf r_{A/O}\times\mathbf R}.$$

The resultant force is independent of the reference point, while the reported moment generally is not.

Parallel planar loads

A particularly common case is a scalar distributed load $w(x)$ acting everywhere in the same transverse direction along

$$a\le x\le b.$$

Its resultant magnitude is

$$\boxed{R=\int_a^b w(x),dx}.$$

The magnitude of its moment about $x=0$ is

$$\boxed{M_0=\int_a^b x,w(x),dx}$$

with the sign determined by the chosen moment convention.

If

$$R\ne0,$$

one concentrated force of magnitude $R$ can reproduce the same force and moment when placed at

$$\boxed{x_R=\frac{\int_a^b xw(x),dx}{\int_a^b w(x),dx}}.$$

Thus $x_R$ is the force-weighted location of the distribution.

For a uniform load,

$$w(x)=w_0,$$

this reduces to the midpoint of the loaded interval. For a nonuniform load, the resultant shifts toward the region carrying more load.

Worked example: triangular load

Suppose a beam segment of length $L$ carries the linearly increasing load

$$w(x)=w_0\frac{x}{L},\qquad0\le x\le L.$$

The resultant is

$$R=\int_0^Lw_0\frac{x}{L},dx =\frac{w_0}{L}\frac{L^2}{2} =\boxed{\frac{w_0L}{2}}.$$

Its first moment about $x=0$ is

$$M_0=\int_0^Lx\left(w_0\frac{x}{L}\right)dx =\frac{w_0}{L}\frac{L^3}{3} =\frac{w_0L^2}{3}.$$

Therefore the equivalent force acts at

$$x_R=\frac{M_0}{R} =\frac{w_0L^2/3}{w_0L/2} =\boxed{\frac{2L}{3}}.$$

The load is strongest near $x=L$, so the resultant lies closer to that end than to the midpoint.

When a single force is not enough

For general three-dimensional distributed loading, the force-couple pair

$$\mathbf R,\mathbf M_O$$

is always a valid equivalent representation, but one single force at one point may not reproduce an arbitrary moment vector. A residual couple can remain.

The simple location formula $x_R=M/R$ applies to parallel planar loads for which the moment can be produced entirely by shifting the line of action of the resultant.

The general principle is broader: integrate force first, integrate moment about a stated reference point, and preserve both when replacing a distributed load by an equivalent system.