Analytical force method

Analytical force method#

The analytical (or vector) method consists of writing the dynamic equilibrium equations of all the links at once and solving the resulting system of equations. It is systematic and ideal for computer solution. We apply it to the inverse dynamics problem: the motion is known and we look for the forces.

Prerequisite: complete kinematic analysis#

Before starting, the kinematics at the position of interest must have been solved. For each moving link we need:

The acceleration of its center of mass, \(\vec{a}_{G_i}\).
Its angular acceleration, \(\vec{\alpha}_i\).

They are obtained with the acceleration analysis methods of lesson 2 (vector equations of relative velocities and accelerations, polygons…).

Free-body diagrams#

Breakdown: the mechanism is split into each of its links, including the ground if the forces on it are of interest.
Forces: on each link we draw all the forces and torques acting on it:
- External forces: weight \(m_i\vec{g}\), actuator forces, useful or resisting loads.
- Input torque \(\vec{M}_0\) (or driving force), on the input link.
- Reaction forces at each joint: the forces that one link exerts on another.
- Inertia loads: \(-m_i\vec{a}_{G_i}\) and \(-I_{G_i}\vec{\alpha}_i\) (unless the analysis is quasi-static).

Notation of the reaction forces#

To avoid mistakes, we are strict with the notation:

\(\vec{F}_{ij}\) = force that link \(i\) exerts on link \(j\).

By Newton’s third law (action and reaction), the force that \(j\) exerts on \(i\) is the opposite one:

\(\vec{F}_{ji} = -\vec{F}_{ij}\)

This rule “connects” the free-body diagrams with each other: if \(\vec{F}_{12}\) (force of the ground 1 on link 2) appears in the diagram of link 2, then \(\vec{F}_{21} = -\vec{F}_{12}\) appears in the diagram of the ground. This way, each joint contributes a single set of unknowns.

The components of the unknown forces are drawn and written all as positive (\(\vec{F}_{12} = F_{12x}\hat{i} + F_{12y}\hat{j}\)): the sign of the result gives their actual sense.

Unknowns introduced by each joint (planar case, frictionless)#

Joint	Reaction	Unknowns
Revolute (pin)	Force of unknown direction through the pin	2 (\(F_x\), \(F_y\))
Prismatic (slider)	Force perpendicular to the guide and a torque (or normal force at an unknown point)	2
Cam or point contact with sliding	Force along the common normal	1
Rolling without slipping	Force of unknown direction at the contact point	2

If the slider is small compared with the rest of the mechanism and all forces go through its center, the torque from the guide is zero and the normal force is enough (1 unknown), as in the examples of this lesson.

Dynamic equilibrium equations#

For each moving link we write the three planar equations:

\[\begin{split} \left\{ \begin{aligned} \sum \vec{F} + \vec{F}_{in} &= \vec{0} \quad (2 \text{ equations}) \\ \sum M_P + \left( \vec{r}_{PG} \times \vec{F}_{in} \right)_z + M_{in} &= 0 \quad (1 \text{ equation}) \end{aligned} \right. \end{split}\]

where \(P\) is a convenient point (usually \(G\), or a joint to eliminate its unknowns). Taking moments about \(G\), the term \(\vec{r}_{PG}\times\vec{F}_{in}\) vanishes and we directly get the Newton-Euler equations:

\[ \sum \vec{F}_{j} = m\,\vec{a}_G, \qquad \sum_j \left( \vec{r}_{Gj} \times \vec{F}_j \right) + \sum M_j = I_G\, \alpha \]

with \(\vec{r}_{Gj}\) the vector from \(G\) to the point of application of each force \(\vec{F}_j\). In the plane, the cross product of \(\vec{r} = (r_x, r_y)\) and \(\vec{F} = (F_x, F_y)\) only has a \(z\) component:

\[ \left( \vec{r} \times \vec{F} \right)_z = r_x F_y - r_y F_x \]

Counting equations and unknowns#

A planar mechanism with \(n\) moving links provides \(3n\) equations. The unknowns are the components of the joint reactions and the driving loads (one torque or driving force per degree of freedom). By Grübler’s formula, in a mechanism with \(g\) degrees of freedom and no redundant constraints both numbers coincide: the inverse dynamics problem is determinate.

If there are more unknowns than equations, the problem is hyperstatic (statically indeterminate): the deformation of the links must be taken into account. A simple example is how the horizontal forces are shared between the wheels of a vehicle (Example 3.3: Vehicle accelerating on level ground).

Solution: a linear system#

The equations are linear in the unknowns (the forces). Stacking all unknowns in a vector \(\mathbf{x}\), the system reads:

\[ \mathbf{A}\, \mathbf{x} = \mathbf{b} \]

where the matrix \(\mathbf{A}\) only depends on the geometry (the position of the mechanism) and the vector \(\mathbf{b}\) holds the known terms: masses times accelerations, moments of inertia times angular accelerations, and external loads. Even if the system is large (14 equations for the six-bar mechanism of Example 3.7: Transfer mechanism (complete analysis)), it is algebraically simple and can be solved with any computing tool:

% Option 1: symbolic computation, with the equations written as they are
syms F12x F12y F32x F32y M0k real
sol = solve([eq1, eq2, eq3, ...], [F12x, F12y, F32x, F32y, M0k]);

% Option 2: matrix form, A*x = b (more efficient)
x = A \ b;

Both options are developed in example Example 3.7: Transfer mechanism (complete analysis). The matrix form is the basis of multibody dynamics simulation software.

Method summary#

Solve the kinematics: \(\vec{a}_{G_i}\), \(\vec{\alpha}_i\) of all links.
Compute the inertia loads of each link.
Draw the free-body diagrams with the \(\vec{F}_{ij}\) notation, using \(\vec{F}_{ji} = -\vec{F}_{ij}\).
Write the 3 equations of each moving link (Newton-Euler, or equilibrium with D’Alembert).
Solve the linear system and check the results (signs, orders of magnitude, power).

Tip

A simple check of the input torque \(M_0\) is the power balance (lesson 4): the power of the input torque plus that of all external and inertia loads is zero, \(M_0\,\omega_2 + \sum_i \left( \vec{F}_i\cdot\vec{v}_i \right) + \sum_i \left( -m_i\vec{a}_{G_i}\cdot\vec{v}_{G_i} - I_{G_i}\alpha_i\,\omega_i \right) = 0\). It gives \(M_0\) without computing the reactions.