ME451
Kinematics and Dynamics
of Machine Systems
Dynamics of Planar Systems
November 17, 2011
Finish 6.3
© Dan Negrut, 2011
ME451, UW-Madison
“To improve is to change; to be perfect is to change often.”
Winston Churchill
Before we get started…
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Last Time

Looked at the contribution to the EOM of an external force or torque that is applied at a
point P (how to get a generalized force QA)
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

Started discussion about how to get the EOM for a set of bodies (a mechanism)
Today:
 Wrap up EOM discussion after we introduce the Lagrange Multiplier Theorem


Example: Pendulum, setting a set of ICs
HW due Tuesday, Nov 29:



Example: EOM for a slider-crank mechanism
Discuss why and how to specify Initial Conditions (ICs)


Discussions of TSDA & RSDA
6.1.1 thru 6.1.4, and 6.2.1
MATLAB component will be emailed to you
NOTE: Syllabus has been updated, we’ll discuss on Tu
2
Getting Rid of the Internal Forces: Summary

Our Goal: Get rid of the constraint forces QC since we don’t
know them

For this, we had to compromise…


We gave up our requirement that
holds for any
arbitrary virtual displacement
Instead, by only selecting virtual displacements ±q that are
consistent with the set of constraints present in the system, we
saw that we can get rid of QC :
…whenever…
NOMENCLATURE: Constrained
Variational Equations of Motion
This is the condition that it takes for
a virtual displacement q to be
consistent with the set of constraints
3
Short Detour
~ Lagrange Multiplier Theorem ~

Theorem:
4
[Ex. 6.3.3]
Example: Lagrange Multipliers


First show that any for any x=[x1 x2 x3]T, one has
that xTb=0 as soon as Ax=0
Show that there is indeed a vector  such that
AT + b = 0
5
Going back to EOMs

Variational Form of the EOMs:
…whenever…
NOMENCLATURE: Constrained
Variational Equations of Motion
This is the conditions that it takes for
a virtual displacement q to be
consistent with the set of constraints
Use this notation and apply the
Lagrange Multiplier Theorem
introduced two slides ago
Lagrange Multiplier Form of the
Equations of Motion
}
6
Summary of the Lagrange form of the
Constrained Equations of Motion

Equations of Motion:

Position Constraint Equations:
The Most
Important Slide
of ME451

Velocity Constraint Equations:

Acceleration Constraint Equations:
7
Some Practical Issues…

One relevant question: How do you know that you can actually compute the
acceleration and the Lagrange Multipliers?

By coupling the Equations of Motion with the Acceleration Constraint
Equations, one ends up with this linear system:

Assuming that your Jacobian q is healthy (that is, has full row rank), it can be
proved that because the kinetic energy of a system is always positive, the
coefficient matrix of the linear system above is nonsingular

This means that a solution exists, and not only that, but it is also unique
8
Putting Things in Perspective…

Suppose you solve the linear system and retrieve the
accelerations are Lagrange Multipliers

What are they good for?

Since you have the accelerations, you can integrate them twice and
find the velocity and position of each part in your system

As for the Lagrange Multipliers, in a next lecture we’ll see that they
can be used to compute the reaction force in each joint in the system.


These are the “internal” forces that keep the system together
They are the forces produced by joints that we just eliminated before
9
Example 6.3.5: Slider Crank Mechanism

Derive the equations of motion (EOM) for the slider
crank model in the figure
10
[New Topic]
The Need for Initial Conditions (ICs)

When dealing with a differential
equation, one needs a set of initial
conditions to uniquely find the
solution of the “problem”

In layman’s words, the “problem”
here can be formulated as follows:



I give you at each point in plane the
direction (slope) of an unknown
function
Can you find me the unknown
function if additionally I also give you
the value of this function at time t=0?
A note on notation

t0 or t=0 is the same in ME451
Picture shows the slopes
of the unknown function
everywhere in plane.
You simply have to start
somewhere and follow
the direction of the slope
11
ODE: Infinite Number of Solutions


12
0.1t
ODE Problem: y  0.1y 100e
Initial Condition: y0 =[-1000:50:1000]
The Need for Initial Conditions (ICs)

Suppose that you just computed the accelerations, and are ready
to apply some numerical formula to integrate the acceleration

What you need though is a set of initial conditions (just like in
ME340, if you recall)

How many ICs can/should you specify at time t0?

Recall that you have a set of n generalized coordinates, but they
are not totally arbitrary, in that they need to satisfy the set of m
constraints present in the system:
13
simEngine2D Comment
14
Specifying Initial Conditions (ICs)

So you have m equations that must be satisfied by n generalized coordinates.
You have m constraints
(kinematic and driving, that is)

You need to know the initial configuration of the mechanism (positions and
velocities) to be able to start the numerical integration procedure

What does it take to uniquely define the initial configuration of the mechanism?


You can specify an additional set of ndof conditions that your generalized
coordinates must satisfy
Recall how we chose the driving constraints…
15
Specifying Position ICs

Important: since you have ndof generalized coordinates that you can choose, it
will be up to you to decide the configuration in which the mechanism starts

For a simple pendulum: should it start in a vertical position or in a horizontal position?


You decide based on the problem you solve…
How do you exactly specify the conditions


You do exactly what you did for the kinematic analysis of a mechanism: you had
some excess DOFs and prescribed motions (drivers) D(q,t) to “occupy” them
For IC in dynamics you’ll specify ndof initial conditions IC(q,t0) that will implicitly
determine the configuration of the mechanism at time t0
Only at the beginning of the simulation,
you specify more IC constraints to
determine configuration of mechanism
16
Specifying Position ICs



How do you know whether the set of conditions you specified to determine the
initial conditions are “healthy”?
Healthy is going to be that IC(q,t0) that leads to a nonsingular constraint
Jacobian:
If this is the case, one can apply Newton’s method to solve the system of
nonlinear equations below to get the initial configuration of the mechanism q0
17
Specifying Velocity ICs

Specifying a set of *position* ICs is not enough

We are dealing with a second order differential problem: We need two sets of ICs



Position ICs (we’ve just done this)
Velocity ICs
Specifying Velocity ICs:

You can play it safe and take the same generalized coordinates that you assigned
initial positions and also specify for them some initial speeds (done most often)

You can decide to use a completely new set of generalized coordinates for which you
prescribe the value of the velocity at time t0 (strange, but not inconceivable)
18
Specifying Velocity ICs


Playing it safe, you find the velocities at time t0 as the
solution of the linear system
If you want to be fancy, you replace the matrix qIC with a
different matrix D, chosen such that the coefficient matrix
continues to be nonsingular (not going to elaborate on this)
19
ICs: Concluding Remarks

This IC issue is actually simple if you remember what we did back
when we dealt with the Kinematics problem

You only do this at time t=0 (provide a starting configuration for your
mechanism). Otherwise, your dynamics problem is underdefined

You take care of this IC issue by

Specifying in the adm file a good (consistent) initial configuration for your mechanism


This is what you should do for simEngine2D; i.e., have consistent guesses for positions and velocities
Adding some extra constraints and relying on the Kinematic solver to get the good (consistent)
configuration at time t=0

This is what a general purpose solver might do, such as ADAMS
20
Example 6.3.6: Simple Pendulum

How do you go about specifying ICs?

I’d like the pendulum to start from a
vertical configuration (hanging
down), and angular velocity to be 2
rad/s.

Assume that L=0.2 m
Pin joint at point P
(between pendulum and ground)
21
End ICs
Beginning Constraint Reaction Forces (6.6)
22
Reaction Forces: The Framework

Remember that we jumped through some hoops to get rid
of the reaction forces that would show up in joints

I’d like to go back and recover them, since they are
important





Durability analysis
Stress/Strain analysis
Selecting bearings in a mechanism
Etc.
It turns out that the key ingredient needed to compute the
reaction forces in all joints is the set of Lagrange multipliers 
23
Reaction Forces: The Basic Idea

Recall the partitioning of the total force acting on our mechanical system

Applying a variational approach (principle of virtual work) we ended up
with this equation of motion

After jumping through hoops, we ended up with this:

It’s easy to see that
24
The Important Observation

IMPORTANT OBSERVATION:

Actually, you don’t care for the “generalized” QC flavor of the reaction force,
but rather you want the actual force represented in the Cartesian global
reference frame



You’d like to have Fx, Fy, and a torque T that is due to the constraint
You report these quantities as they would act at a point P
The strategy:

Look for a force (the classical, non-generalized flavor), that when acting on
the body would lead to a generalized force equal to QC
25
The Nuts and Bolts



There is a joint acting
between Pi and Pj and we
are after finding the reaction
forces/torques Fi and Ti, as
well as Fj and Tj
Figure is similar to Figure
6.6.1 out of the textbook
Textbook covers topic well (pp. 234), I’m only modifying one thing:


The book expresses the reaction force/torque Fi and Ti in a body-fixed
reference frame
attached at point Pi
I didn’t see a good reason to do it that way

Instead, start by deriving in global reference frame OXY and then
multiply by AT to take it to the body-fixed reference frame
26
The Main Result
(Expression of reaction force/torque in a joint)

Suppose that two bodies i and j
are connected by a joint, and that
the equation that describes that
joint, which depends on the
position and orientation of the two
bodies, is

Suppose that the Lagrange multiplier associated with this joint is

Then, the presence of this joint in the mechanism will lead at point P on
body i to the presence of the following reaction force and torque:
27
Comments
(Expression of reaction force/torque in a joint)

Note that there is a Lagrange multiplier associated with each constraint equation

Number of Lagrange multipliers in mechanism is equal to number of constraints

Each Lagrange multiplier produces (leads to) a reaction force/torque combo

Therefore, to each constraint equation corresponds a reaction force/torque
combo that throughout the time evolution of the mechanism “enforces” the
satisfaction of the constraint that it is associated with


Since each constraint equation acts between two bodies i and j, there will also
be a Fj/Tj combo associated with each constraint, acting on body j


Example: the revolute joint brings along a set of two kinematic constraints and
therefore there will be two Lagrange multipliers associated with this joint
According to Newton’s third law, they oppose Fi and Ti, respectively
Note that you apply the same approach when you are dealing with driving
constraints (instead of kinematic constraints)

You will get the force and/or torque required to impose that driving constraint
28
Reaction Forces
~ Remember This ~


As soon as you have a joint
(constraint), you have a Lagrange
multiplier 

As soon as you have a Lagrange
multiplier you have a reaction
force/torque:
The expression of  for all the usual
joints is known, so a boiler plate
approach renders the value of the
reaction force in all these joints
Just in case you want another form for the torque T above, note that
29
Example 6.6.1: Reaction force in
Revolute Joint of a Simple Pendulum
Pendulum driven by motion:
1) Find the reaction force in the
revolute joint that connects
pendulum to ground at point O
2) Express the reaction force in
the
reference frame
30
End Constraint Reaction Forces
31