Resultant of a General Distributed Force System | Engineering Mechanics - Civil Engineering (CE) PDF Download

Replacing distributed loads by a resultant load

Many practical problems involve loads spread over a length, area or surface rather than concentrated forces. Such loads are called distributed loads. A distributed load acting along a line (for example, on a beam) with intensity w(x) (force per unit length) can be replaced by a single resultant force and, if required, an associated resultant couple about a chosen reference point O. The equivalent single force and couple produce the same net force and the same moment about O as the original distributed loading.

Definitions and expressions for a linearly distributed load w(x)

Consider a one-dimensional distributed load w(x) acting along the x-axis from x = 0 to x = L, referenced to a point O at x = 0.

• Resultant force F is the integral of the load intensity over the length:
  F = ∫₀^L w(x) dx
• Resultant moment about O M_O is the integral of the moment of the elemental load about O:
  M_O = ∫₀^L x·w(x) dx
• Location of resultant (distance d from O along x) is obtained by taking the moment of the resultant about O and equating it to M_O:
  d = M_O / F provided F ≠ 0
• If the resultant is applied at the point whose coordinate is d found above, the resultant couple about that point is zero. If the resultant is applied at any other point, an equivalent couple (moment) equal to M_O - F·x_app must be carried to preserve equilibrium about O.

Physical interpretation

The formula for d is identical to the formula for the centroid of the area under the curve y = w(x) between x = 0 and x = L. Therefore, the resultant force of the distributed load acts through the centroid of the load area (the area obtained by plotting w(x) vs x and integrating between the limits of the load).

Worked examples

Example 1 - Uniformly distributed load

A beam of length L carries a uniform distributed load w₀ (force per unit length) along its length. Find the resultant force and its line of action measured from the left end (x = 0).

Sol.

F = ∫₀^L w₀ dx

F = w₀ L

M_O = ∫₀^L x·w₀ dx

M_O = w₀ ∫₀^L x dx

M_O = w₀ (L² / 2)

d = M_O / F

d = [w₀ (L² / 2)] / [w₀ L]

d = L / 2

Ans. The resultant force is F = w₀L acting at the mid-span, distance L/2 from the left end.

Example 2 - Triangular distributed load (zero at left, w₀ at right)

A triangular load varies linearly from w = 0 at x = 0 to w = w₀ at x = L. Determine the resultant force and its position measured from x = 0.

Sol.

Express w(x) as a linear function: w(x) = (w₀/L) x

F = ∫₀^L (w₀/L) x dx

F = (w₀/L) ∫₀^L x dx

F = (w₀/L) (L² / 2)

F = (1/2) w₀ L

M_O = ∫₀^L x · (w₀/L) x dx

M_O = (w₀/L) ∫₀^L x² dx

M_O = (w₀/L) (L³ / 3)

M_O = (1/3) w₀ L²

d = M_O / F

d = [(1/3) w₀ L²] / [(1/2) w₀ L]

d = (2/3) L

Ans. The resultant force is F = (1/2) w₀ L acting at a distance 2L/3 from the left end (measured from the smaller end of the triangle toward the larger).

Equivalent force-couple system and choice of reference point

When replacing a distributed load by an equivalent concentrated force, the choice of the point at which that resultant is applied affects whether an additional couple (moment) is required to represent the original loading. If the resultant is applied at its centroid, no additional couple is needed. If the resultant is moved to a different point, a couple equal to the force times the distance between the points must be included to preserve moments.

• General procedure:
  1. Compute the resultant force F by integrating the load intensity.
  2. Compute the moment M_O of the load about the chosen reference point O by integrating x·w(x).
  3. Compute the position d of the resultant from O as d = M_O/F.
  4. If the resultant is not applied at O, include a couple M_O about O (or equivalently, if applied at a different point x_app, include couple M = M_O - F·x_app).
• Sign convention: Take care with sign conventions for w(x), moments and distances. Loads acting downward are usually taken as negative in vertical equilibrium, and moments may be taken positive in a chosen sense (e.g., anticlockwise).

Applications and remarks

• The method applies to line loads, surface loads (replace area integrals for force and first moments for centroids) and volume loads in three dimensions.
• For area or surface loads, replace w(x) by pressure p(x,y) and use double integrals to obtain the resultant and first moment(s) to find centroidal coordinates.
• Many structural analyses and design problems (beam shear and moment diagrams, reaction calculations, stability checks) begin by replacing distributed loads with their resultant equivalent to simplify free-body diagrams.
• The centroidal property is useful: determining the centroid of the load area is equivalent to locating the line of action of the resultant force for line loads.

Summary

To replace a distributed load by a single resultant force and a possible resultant couple about a reference point:

• Compute the resultant force: F = ∫ w(x) dx.
• Compute the moment about a point O: M_O = ∫ x·w(x) dx.
• Locate the resultant from O by d = M_O / F (if F ≠ 0). The resultant acts through the centroid of the w(x) vs x area.
• Include the appropriate couple if the resultant is not applied at the reference point.