1.5 Couples and force-couple systems
After this lesson you should be able to:
- Define a couple and show that its moment is the same about every point.
- Add couples and recognise when a system reduces to a pure couple.
- Replace a force by an equal force at another point plus a couple, and reverse the process.
Theory
A couple is two forces of equal magnitude, opposite sense and parallel, non-collinear lines of action. Their vector sum is zero, so a couple cannot translate a body, but it does turn it. Its only effect is a moment, M = F d, where d is the perpendicular distance between the two lines of action.
Remarkably, the moment of a couple is the same about every point in its plane. This makes a couple a free vector: it can be moved anywhere in its plane, or to any parallel plane, without changing its effect on a rigid body. A couple is completely described by its moment and sense; the individual forces and their spacing do not matter, so 100 N at 0.6 m and 300 N at 0.2 m are the same couple.
A couple cannot be balanced by a single force, only by another couple of equal and opposite moment. This is why a steering wheel turned with two hands produces pure turning, and why a beam end built into a wall develops a fixing moment as well as a reaction force.
The force-couple transformation is one of the most useful tools in statics. A force F acting at A has the same external effect on a rigid body as the same force F acting at another point B together with a couple equal to the moment of the original force about B. Engineers use it constantly: an eccentric load on a column becomes an axial load plus a bending moment at the column axis, and a load on a bracket becomes a force plus a moment at the bolt group.
Derivation
- Take a couple with forces +F and −F, separated by perpendicular distance d. Choose any point O on the perpendicular, at distance a from the line of −F, on the side away from +F.
- Moment about O: +F acts at distance (a + d), −F at distance a, with opposite senses; M_O = F (a + d) − F a = F d.
- The result does not contain a, so it is the same for every point: the moment of a couple is independent of the moment centre.
- Force-couple: add a pair of equal and opposite forces +F and −F at B. The original F at A and the −F at B form a couple of moment r_BA × F (in 2D, M_B = (x_A − x_B) Fy − (y_A − y_B) Fx). What remains is +F at B plus that couple.
Couple moment C = F d (N·m or kN·m), independent of the moment centreResultant of several couples: C = Σ Ci (signed)Force F at A moved to B: F at B plus M_B = (x_A − x_B) Fy − (y_A − y_B) FxSingle force equivalent to F at O plus couple M: same F at perpendicular distance d = |M| / F, placed so its moment about O equals M
- The body is rigid; moving a force with a compensating couple preserves external effects only.
- Plane problems: couples are perpendicular to the plane and add as signed scalars.
- Anticlockwise positive.
Codes and standards
Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.
- No design code governs this lesson. Eccentric loads treated with the force-couple idea appear in column design to IS 456:2000 (minimum eccentricity, cl. 25.4) and in eccentric bolted connections designed to IS 800:2007; those clauses govern design, not the mechanics.
Worked examples
Problem. Two parallel 150 N forces of opposite sense act 0.4 m apart. Find the couple moment, and verify that its moment about a point 1.0 m from the nearer force (outside the pair) is the same.
- C = F d = 150 × 0.4 = 60 N·m.
- About the outside point: the far force acts at 1.4 m and the near force at 1.0 m with opposite senses.
- M = 150 × 1.4 − 150 × 1.0 = 210 − 150 = 60 N·m, the same as the couple moment.
Answer. 60 N·m about every point
Problem. A bracket carries a force F = (3, −4) kN (3 kN horizontal, 4 kN downward) at point A(2 m, 1 m). Replace it by an equivalent force-couple system at the origin O, which is the bolt-group centre.
- The force is unchanged: (3, −4) kN at O, magnitude 5 kN.
- Couple at O = moment of F about O: M_O = x Fy − y Fx = 2 × (−4) − 1 × 3 = −8 − 3 = −11 kN·m.
- The equivalent system is 5 kN with components (3, −4) kN at O plus a clockwise couple of 11 kN·m.
Answer. F = (3, −4) kN at O plus 11 kN·m clockwise
Step-by-step procedure
- Identify couples: equal, opposite, parallel, non-collinear forces.
- Compute each couple moment as F d with its sense; add couples algebraically.
- To move a force to point B, keep the force and add M_B equal to its moment about B.
- To reduce a force plus couple to a single force, shift the force sideways by d = |M| / F on the side that reproduces the sense of M.
- Check by taking moments of the original and equivalent systems about a third point.
Site notes
- Eccentric loads, such as a crane rail on a corbel or a beam bearing on one side of a column, produce moments in the supporting member equal to load times eccentricity.
- Base plates under moment are designed as an axial force plus a couple at the column centreline, which is the force-couple idea in practice.
- Pairs of bolts or anchor rods resisting overturning act as a couple: a larger spacing gives the same resisting moment with smaller bolt forces.
- Trying to balance a couple with a single force.
- Forgetting to add the couple when moving a force to a new point.
- Getting the sign of the transfer couple wrong; always compute it as the moment of the original force about the new point.
- Thinking a couple's effect depends on where it is applied on a rigid body.
- Adding couple moments to forces in the same sum.
When a force of F kN is moved sideways by e metres, the couple to add is F × e kN·m. A 100 kN column load with 50 mm eccentricity gives 5 kN·m: small eccentricities matter for heavily loaded columns.
Summary
- A couple is two equal, opposite, parallel forces; its moment is C = F d.
- A couple's moment is the same about every point: it is a free vector.
- Only a couple can balance a couple.
- Force at A = same force at B + couple equal to its moment about B.
Practice questions
Tap a question to see the answer.
1A couple of 25 N·m is made by two forces 0.5 m apart. Find the force.
F = 25 / 0.5 = 50 N.
2A body carries couples of +40 kN·m and −15 kN·m. What single couple is equivalent?
+25 kN·m, i.e. 25 kN·m anticlockwise.
3A 12 kN column load acts 0.25 m off the column axis. What moment does it produce about the axis?
M = 12 × 0.25 = 3 kN·m.
4Can a single force balance a couple? Explain.
No. A force always has a resultant force; a couple has zero resultant force. Only an equal and opposite couple can balance it.
5A downward 10 kN force at O and a 30 kN·m clockwise couple act on a beam along the x-axis. Replace them by a single force.
A 10 kN downward force at x = +3 m: its moment about O is x Fy = 3 × (−10) = −30 kN·m, the same clockwise couple.
References
Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.
- F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
- J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
- S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.