7 Distributed Loads and Centroids

Learn how to replace distributed loads with equivalent forces, locate centroids of load diagrams and areas, and use these results in rigid-body equilibrium.

Distributed loads and equivalent forces

A acts across a length, area, or volume rather than at one point. For a beam, its intensity is commonly written as $w(x)$ , with units such as $\text{N}/\text{m}$ or $\text{kN}/\text{m}$ .

The equivalent is the signed area under the intensity curve:

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

For a load acting in a constant direction, the equivalent force acts in that direction. Its position is found by matching the moment of the original distribution:

\bar{x} = \frac{\int_a^b x\,w(x)\,dx}{\int_a^b w(x)\,dx}, \qquad R \ne 0.

The resultant preserves the total force and moment of the original distribution, so it can replace that distribution when analyzing a rigid body. Its magnitude is the area under the load diagram, and its line of action passes through the diagram’s .

Common load shapes

For a of intensity $w_0$ over length $L$ , the load diagram is a rectangle. Its resultant and location are

R = w_0L, \qquad \bar{x} = \frac{L}{2}.

For a that rises from zero to peak intensity $w_{\max}$ over length $L$ , the resultant is

R = \frac{1}{2}w_{\max}L.

It acts $L/3$ from the high-intensity end, equivalently $2L/3$ from the zero-load end. A trapezoidal load can be treated as a rectangle plus a triangle, or evaluated by integration.

For example, a downward load increasing linearly from zero to $8\,\text{kN}/\text{m}$ over a $3\,\text{m}$ beam segment has resultant

R = \frac{1}{2}(8)(3) = 12\,\text{kN}.

It acts downward $1\,\text{m}$ from the high-intensity end, or $2\,\text{m}$ from the zero-load end.

Centroids of areas and load diagrams

A is the geometric balance point of a shape. For a plane area $A$ , its coordinates are

\bar{x} = \frac{1}{A}\int_A x\,dA, \qquad \bar{y} = \frac{1}{A}\int_A y\,dA.

Useful locations include:

Rectangle: the intersection of its diagonals, halfway along each side.
Triangle: the intersection of its medians, one-third of the altitude from the base, or two-thirds from the opposite vertex.
Symmetric shape: on each axis of symmetry.

For a , find the area $A_i$ and coordinates $(x_i,y_i)$ of each piece. Combine them by area-weighted averages:

\bar{x} = \frac{\sum_i A_i x_i}{\sum_i A_i}, \qquad \bar{y} = \frac{\sum_i A_i y_i}{\sum_i A_i}.

Treat a hole as a negative area. The same idea applies to load diagrams: the load intensity acts as the diagram’s height, so the load’s equivalent point of application is at the diagram’s area .

Combining load shapes

When a load diagram is easier to divide into familiar shapes, find the resultant $R_i$ and location $x_i$ of each piece. Then combine the pieces using force and moment sums:

R = \sum_i R_i, \qquad \bar{x} = \frac{\sum_i R_i x_i}{\sum_i R_i}, \qquad R \ne 0.

Use signed forces when some loads act in opposite directions.

For example, consider a $3\,\text{m}$ beam segment with a downward of $2\,\text{kN}/\text{m}$ and a downward increasing from zero at the left end to $4\,\text{kN}/\text{m}$ at the right. The rectangle has a resultant of $6\,\text{kN}$ at $1.5\,\text{m}$ ; the triangle has a resultant of $6\,\text{kN}$ at $2\,\text{m}$ . Thus,

R = 12\,\text{kN}, \qquad \bar{x} = \frac{(6)(1.5)+(6)(2)}{12} = 1.75\,\text{m}

from the left end, downward.

Loads over areas

A $p(x,y)$ over a surface has resultant

R = \int_A p\,dA.

For pressure acting normal to a flat surface in one direction, the resultant’s location is the pressure-weighted :

\bar{x} = \frac{\int_A x p\,dA}{R}, \qquad \bar{y} = \frac{\int_A y p\,dA}{R}.

Uniform pressure acts through the area . When pressure varies, its resultant generally does not act through that . The same integration approach applies to distributed body forces, using the appropriate force per unit volume.

Using resultants in equilibrium

On a , represent a either by its original arrows or by its equivalent resultant, not by both. Replacing the distribution with a point force preserves the external force and moment for rigid-body equilibrium calculations.

Check that the resultant’s units are force, its direction matches the loading, and its line of action is located using the load diagram’s . If the signed total load is zero, the -location formula is undefined; opposing loads may instead produce a pure couple, which must be represented by a moment.