Aykut C. Satici – Generalized Forces

Chapter overview

Lines of action and moments about a point
Resultants and moment transfer
Couples and torque
Equivalent force systems
Virtual work and generalized forces

Goal: replace a distributed set of loads by a consistent force–moment description.

A force has a line of action

A force acting on a rigid body has a magnitude, direction, and line of action.

For F applied at A, its moment about O is

\boxed{M_O=r_{OA}\times F.}

The source calls a vector with a specified line of action a bound vector. A free vector has no specified line of action.

The moment depends on the reference point O.

Sliding along a line of action

Move the application point from A to B along the same line, so r_{AB}=s\widehat f and F=F\widehat f.

r_{OB}\times F=(r_{OA}+s\widehat f)\times F=r_{OA}\times F.

The force and its moment about every point stay unchanged.

This replacement preserves the external rigid-body effect. Internal stresses can depend on where the load enters the actual structure.

Worked example: an offset force

Force of 100 N to the right at A, located 0.3 m right and 0.2 m above O.

r_{OA}=0.3e_x+0.2e_y\ \mathrm m, F=100e_x\ \mathrm N.

M_O=r_{OA}\times F=-20e_z\ \mathrm{N\,m}.

The moment is clockwise. The perpendicular lever arm is 0.2 m.

Planar moments and their signs

With counterclockwise positive,

\boxed{M_{O,z}=xF_y-yF_x.}

Only the perpendicular offset from O to the line of action contributes:

|M_O|=F d_\perp.

A force passing through O has zero moment about O, even if the application point is far away.

The resultant and total moment

For forces F_i and applied free couples T_j,

\boxed{R=\sum_iF_i,\qquad M_O=\sum_i r_{Oi}\times F_i+\sum_jT_j.}

The resultant summarizes the force balance. The total moment contains the effect of the lines of action.

Equal resultants alone do not imply equivalent loading.

Transferring the moment to another point

Let r_{PQ} point from P to Q.

\boxed{M_P=M_Q+r_{PQ}\times R.}

For every force, r_{Pi}=r_{PQ}+r_{Qi}. Therefore,

\sum_i r_{Pi}\times F_i =r_{PQ}\times\sum_iF_i+\sum_i r_{Qi}\times F_i.

The resultant stays the same when the reference point changes.

Worked example: loads on a beam

Horizontal beam with downward loads of 100 N at 0.2 m and 200 N at 0.8 m from O.

Two downward forces act on the beam:

R=-300e_y\ \mathrm N.

About its left end,

M_O=-100(0.2)e_z-200(0.8)e_z, \boxed{M_O=-180e_z\ \mathrm{N\,m}.}

Beam loads: choosing a new reference point

Choose Q at x=0.5 m, so r_{OQ}=0.5e_x m.

M_Q=M_O-r_{OQ}\times R =-180e_z-(0.5e_x)\times(-300e_y)=-30e_z\ \mathrm{N\,m}.

Directly summing moments about Q gives the same result:

(-0.3)(-100)+(0.3)(-200)=-30\ \mathrm{N\,m}.

A couple has zero resultant

Equal and opposite vertical forces separated horizontally by distance d, producing a counterclockwise couple.

A simple couple consists of two equal, opposite, parallel forces with different lines of action.

R=F+(-F)=0, T=(r_A-r_B)\times F.

Its magnitude is Fd, where d is the perpendicular separation.

The torque of a couple is independent of the point

For a couple, R=0. Moment transfer gives

M_P=M_Q+r_{PQ}\times0=M_Q.

Thus its torque T can be treated as a free vector.

For two 50 N forces separated by 0.20 m,

\boxed{|T|=50(0.20)=10\ \mathrm{N\,m}.}

The word couple denotes the force system. Its torque is the moment vector.

Equivalent force systems

Two systems are equivalent for rigid-body analysis if they have

R'=R,\qquad M'_O=M_O

at one chosen point O.

Moment transfer then guarantees equal moments at every other point.

For example, two systems with zero resultant and the same couple torque are equivalent even if their force locations differ.

Replacement by a force and a couple

At any selected point O, replace the entire loading by:

A force R whose line of action passes through O.
A free couple of torque M_O.

\boxed{\text{Equivalent loading at }O:\quad (R,M_O).}

When moving a force to a parallel line through a new point, include the couple required to preserve the moment.

When one force is enough

A single force R at position r can replace the system only if

r\times R=M_O.

For R\ne0, this requires M_O\cdot R=0. Then one solution is

r=\frac{R\times M_O}{\|R\|^2},

and adding any multiple of R gives the same line of action.

For the beam, x(-300)=-180, so the single downward resultant acts at x=0.60 m.

A spatial load may retain a couple

Decompose the moment into parts parallel and perpendicular to R:

M_O=M_\parallel+M_\perp,\qquad M_\parallel=\frac{R\cdot M_O}{\|R\|^2}R.

A change of line of action can absorb M_\perp into r\times R.

The parallel part remains as a couple. This force plus parallel couple is a wrench.

If R=0 and M_O\ne0, the loading is a pure couple.

Distributed loads

For a force density w(x) acting downward on a beam,

R_y=-\int_0^Lw(x)\,dx,\qquad M_{O,z}=-\int_0^Lxw(x)\,dx.

For w(x)=w_0x/L,

R_y=-\frac{w_0L}{2},\qquad M_{O,z}=-\frac{w_0L^2}{3}.

The equivalent resultant acts at x_R=2L/3, toward the heavier end of the triangular load.

Power provides a consistency check

For a rigid body, v_i=v_O+\omega\times r_{Oi}.

The power of all applied forces and couples becomes

\boxed{\mathcal P=\sum_iF_i\cdot v_i+\sum_jT_j\cdot\omega =R\cdot v_O+M_O\cdot\omega.}

Equivalent force systems produce the same instantaneous rigid-body power.

Use the force and moment at the same reference point as the translational velocity.

Generalized forces from virtual work

Let q=(q_1,\ldots,q_n) describe independent configurations and p_i=p_i(q).

With J_i=\partial p_i/\partial q,

\delta W=\sum_iF_i^T\delta p_i =\sum_iF_i^TJ_i\delta q=Q^T\delta q.

\boxed{Q=\sum_iJ_i^TF_i.}

For a rigid-body couple, add J_\omega^TT when \omega=J_\omega\dot q.

Worked example: one revolute joint

A link of length L at angle theta above x, carrying a downward tip force P and a positive joint torque tau.

For q=\theta,

p=L\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix},\quad F=\begin{bmatrix}0\\-P\end{bmatrix}.

With a positive actuator torque \tau,

Q_\theta=\tau+\left(\frac{\partial p}{\partial\theta}\right)^TF =\boxed{\tau-PL\cos\theta}.

Force-analysis bookkeeping

Before writing equilibrium or motion equations:

Identify the body and all external forces and couples.
State the reference point, axes, and positive moment direction.
Reduce the loads to R and M_O if that simplifies the calculation.
Keep every moment with its reference point.

In Chapter 10, these loads determine mass-center acceleration and angular acceleration.

References and next chapter

Primary: Aykut C. Satici, Kinematics and Machine Dynamics, Fall 2020 compiled notes, Chapter 9, pp. 59–61.

Supporting: Kane and Levinson, Dynamics: Theory and Applications (1985), force-system development. Original course Lecture06 covers moments, couples, and replacement.

Virtual work and the numerical examples extend the compiled chapter to connect with equations of motion.

Next: Chapter 10, Formulation of Equations of Motion