Couple, wrench, moment of P about points and lines

Couple

A couple is a pair of equal, opposite and parallel forces whose lines of action do not coincide. It produces pure rotation, no translation, because the resultant force is zero. Its moment is M = F × d, where d is the perpendicular distance between the forces (the arm of the couple). It is a vector perpendicular to the plane of the couple.

A couple is a free vector

Let forces F at A and −F at B form a couple. Take any point O, with position vectors r_A and r_B.

Moment about O = r_A × F + r_B × (−F)
               = (r_A − r_B) × F
               = r_AB × F

The result contains only the vector r_AB between the two points of application. It does not contain the position of O. So the moment of a couple is the same about every point. The couple can be moved anywhere in its plane, to a parallel plane, or turned in its plane, without a change in its effect on a rigid body, provided magnitude and direction of the moment vector remain the same. Hence its moment is a free vector.

Reduction of a force system to a wrench

1. Reduce the system to a resultant force R = ΣF at a chosen point O and a resultant couple M_O = Σ(r × F).
2. Resolve M_O into M₁, parallel to R, and M₂, perpendicular to R: M₁ = (R·M_O / R²) R.
3. The pair R and M₂ can be replaced by R alone, moved to a new line of action displaced from O by r = (R × M_O)/R² (distance d = M₂/R).
4. What remains is R along its line of action and the couple M₁ parallel to it. This is the wrench; the ratio p = M₁/R is its pitch. A wrench is the simplest form to which a general 3D system can be reduced.

Moments of force P on the cube

Reading of the figure: cube side a = 4 m, O at the origin, x right, y up, z toward the viewer. E(0,0,4), F(4,0,4), G(4,0,0), D(0,4,0), A(0,4,4), B(4,4,4), C(4,4,0). P = 20 kN acts along F→C, i.e. direction (0, 1, −1)/√2.

P = 20/√2 × (0, 1, −1) = (0, 14.14, −14.14) kN, applied at F(4, 0, 4).

a) About A r = F − A = (4, −4, 0) m M_A = r × P = (4, −4, 0) × (0, 14.14, −14.14) = ((−4)(−14.14) − 0, 0 − 4(−14.14), 4(14.14) − 0) = (56.57, 56.57, 56.57) kN·m |M_A| = 98.0 kN·m

b) About edge AB (A to B is along +x, unit vector i) M_AB = M_A · i = 56.57 kN·m

c) About the diagonal AG A(0,4,4) to G(4,0,0): vector (4, −4, −4), unit vector λ = (0.577, −0.577, −0.577) M_AG = M_A · λ = 0.577 (56.57 − 56.57 − 56.57) = −32.66 kN·m The magnitude is 32.66 kN·m; the negative sign means it acts opposite to the direction A→G.