Equivalent Force Systems

NCEES FE Civil spec: Statics, B. Equivalent force systems. FE Reference Handbook: Statics chapter, moments (couples) and systems of forces.

Two force systems are equivalent when they produce the same resultant force and the same resultant moment about every point. On the FE Civil exam this shows up as moving a force to a new point, replacing a distributed load with a single force, or reducing a group of loads to one force and one couple.

What the NCEES spec covers

Statics section B, "Equivalent force systems", covers moments of forces, couples, force-couple systems and the replacement of distributed loads. It is the bridge between simple vector addition (section A) and equilibrium (section C): you cannot find beam reactions quickly unless you can replace a distributed load with its resultant at the right location.

Key concepts

Moment of a force about a point is force times the perpendicular distance to its line of action. Counterclockwise is usually taken as positive in 2D.
A couple is two equal, opposite, non-collinear forces. Its moment is the same about every point, so a couple is a "free vector" that can be moved anywhere on the body.
Force-couple system. A force F at point A can be moved to point O if you add a couple equal to the moment of F about O.
Distributed loads are replaced by a force equal to the area under the load diagram, acting through the centroid of that area.

Core formulas

M_{O} = F d_{⊥} M_{O} = x F_{y} - y F_{x} (2D, counterclockwise positive)

M_{O} = r \times F (3D)

R = \sum F_{i} M_{O}^{R} = \sum (r_{i} \times F_{i}) + \sum C_{j}

Resultants of common distributed loads over a length L:

Uniform w : R = w L at \frac{L}{2} Triangular (0 \to w) : R = \frac{1}{2} w L at \frac{2 L}{3} from the zero end

Where it is in the FE Reference Handbook

The Statics chapter covers moments (couples) and systems of forces. The vector cross product is in the Mathematics chapter under vectors. Centroid locations for the triangle and other shapes you need for distributed loads are in the Statics tables of plane-area properties.

Worked example

A horizontal bracket is fixed at O and extends 1.2 m to point A. A 600 N force acts at A, pointing down and to the right at 60° below the horizontal. Replace it with an equivalent force-couple system at O. Then, separately, find the resultant of a triangular load that rises from 0 to 6 kN/m over a 3 m beam.

Part 1: components of the force.

F_{x} = 600 cos 6 0^{\circ} = 300 N F_{y} = - 600 sin 6 0^{\circ} = - 519.6 N

Moment about O (A is at x = 1.2 m, y = 0):

M_{O} = x F_{y} - y F_{x} = (1.2) (- 519.6) - (0) (300) = - 623.5 N \cdotp m

The equivalent system at O is the same 600 N force (300 N right, 519.6 N down) plus a 623.5 N·m clockwise couple. The horizontal component passes through O, so it adds no moment.

Part 2: triangular load.

R = \frac{1}{2} (6) (3) = 9 kN \overset{x}{ˉ} = \frac{2}{3} (3) = 2 m from the zero end

A single 9 kN force located 2 m from the zero end (1 m from the high end) is equivalent to the triangular load.

Common mistakes

Moving a force to a new point without adding the couple.
Placing a triangular load resultant at L/3 from the wrong end. It sits closer to the high end of the load.
Using the full distance between points instead of the perpendicular distance to the line of action.
Sign errors: pick counterclockwise positive (or clockwise) once and keep it for every term.
Replacing a distributed load before cutting a section, then using the whole-load resultant for a partial segment.

See the equivalent force systems example for a second walk-through.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A 200 N force is applied at the end of a 0.75 m bar that pivots at O. The force acts at 60° to the bar. What is the magnitude of the moment about O? See the worked solution or practice the Statics & Equilibrium deck
A 10 N force has a perpendicular distance of 0.4 m from its line of action to a point. What is its moment magnitude about that point? See the worked solution or practice the Statics & Equilibrium deck
Two parallel downward forces act along a horizontal beam: 10 kN at x = 0 and 20 kN at x = 3 m. Where does their resultant act, measured from x = 0? See the worked solution or practice the Statics & Equilibrium deck
A 6 m simply supported beam carries a uniformly distributed load of 4 kN/m over its full length. What is the reaction at each support? See the worked solution or practice the Statics & Equilibrium deck

Related Calculators and Examples

Frequently Asked Questions

How do you move a force to a different point?

Move the force unchanged to the new point and add a couple equal to the moment of the original force about that new point. The force plus the couple is an equivalent force-couple system.

Where does the resultant of a triangular distributed load act?

Its magnitude is half the base times the peak intensity, and it acts through the centroid of the triangle, which is two thirds of the length from the zero end (one third from the high end).

Why can a couple be moved anywhere on a rigid body?

A couple is two equal and opposite forces, so its net force is zero and its moment is the same about every point. Moving it does not change the external effect on the body.

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