ME451
Kinematics and Dynamics
of Machine Systems
Numerical Integration
October 28, 2013
Radu Serban
University of Wisconsin-Madison
Before we get started…

Last Time:



Today



Wrap-up discussion on recovering constraint reaction forces
Begin section on numerical integration
Assignments:




Specifying position and velocity initial conditions for dynamics
Recovering constraint reaction forces
Matlab 7 – due Wednesday, October 30, Learn@UW (11:59pm)
Your simEngine2D can now perform complete Kinematic analysis!
Test the provided visualization function (available on the SBEL course webpage)
Miscellaneous



Draft proposals for the Final Project due on Friday, November 1
Midterm 2 – Wednesday, November 6, 12:00pm in ME 1143
Review session – Monday, November 4, 6:30pm in ME 1143
2
3
Final Project

I encourage you to come up with a problem on your own




Some suggestions




The Reuleaux collection of kinematic mechanisms at Cornell University http://kmoddl.library.cornell.edu/
Mechanical computers http://www.youtube.com/watch?v=s1i-dnAH9Y4
Extend simEngine2D to include contact dynamics (DEM – penalty method)
Default project




Something related to your research
Something related to a personal interest of yours
Something that you are curious about
simEngine2D kinematic and dynamic analysis of the web cutting mechanism
Problems 5.2 and 8.2 in the textbook
You decide what the interesting/relevant questions are
Rules of Engagement



The final project can use either simEngine2D or ADAMS (or both)
You can work individually or in groups of two
For the proposal, send (email) a PDF (at most one page, typeset) with a description of the problem, how you
propose to tackle it, and what you expect to be able to answer about it



Draft proposal due on Friday, November 1; Final proposal (if needed) due on Friday, November 8
If you decide to work on the default problem, simply send an email indicating that
Final report





Due on Wednesday, December 11
Use Word or LaTeX and generate a PDF
Describe the problem, the analysis procedure, and the results
Include all necessary figures, derivations, equations
Include your simEngine2D (if applicable) and/or ADAMS models
6.6
Constraint Reaction Forces
Reaction Forces

Remember that we jumped through some hoops to get rid of the reaction
forces that develop in joints

Now, we want to go back and recover them, since they are important:
 Durability analysis
 Stress/Strain analysis
 Selecting bearings in a mechanism
 Etc.

The key ingredient needed to compute the reaction forces in all joints is
the set of Lagrange multipliers 
5
6
Reaction Forces: The Basic Idea

Recall the partitioning of the total force acting on the 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
Reaction Forces: Important Observation

What we obtain by multiplying the transposed Jacobian of a constraint,
𝚽𝐪𝑇 , with the computed corresponding Lagrange multiplier(s), 𝛌, is the
constraint reaction force expressed as a generalized force:

Important Observation:
What we want is the real reaction force, expressed in the GRF:


We would like to find F𝑥 , F𝑦 , and a torque T due to the constraint

We would like to report these quantities as acting at some point 𝑃 on a body
7
The strategy:
Look for a real force which, when acting on the body at the point 𝑃, would
lead to a generalized force equal to 𝐐𝐶
Reaction Forces: Framework

Assume that the 𝑘-th joint in the
system constrains points 𝑃𝑖 on body 𝑖
and 𝑃𝑗 on body 𝑗

We are interested in finding the
(𝑘)
reaction forces and torques 𝐅𝑖 and
(𝑘)
𝑇𝑖
acting on body 𝑖 at point 𝑃𝑖 , as
(𝑘)
(𝑘)
well as 𝐅𝑗 and 𝑇𝑗
𝑗 at point 𝑃𝑗
acting on body

The book complicates the formulation for no good reason by expressing these
reaction forces with respect to some arbitrary body-fixed RFs attached at the
points 𝑃𝑖 and 𝑃𝑗 , respectively.

It is much easier to derive the reaction forces and torques in the GRF and, if
desired, re-express them in any other frame by using the appropriate rotation
matrices.
8
Reaction Forces: Main Result

Let the 𝑚𝑘 constraint equations defining
the 𝑘-th joint be

Let the 𝑚𝑘 Lagrange multipliers associated
with this joint be

Then, the presence of the 𝑘-th joint leads to the following reaction force and
torque at point 𝑃𝑖 on body 𝑖
9
Reaction Forces: Comments

10
Note that there is one Lagrange multiplier associated with each constraint equation


Number of Lagrange multipliers in mechanism is equal to number of constraints
Example: the revolute joint brings along a set of two kinematic constraints and therefore
there will be two Lagrange multipliers associated with this joint

Each Lagrange multiplier produces (leads to) a reaction force/torque combo

Therefore, to each constraint equation corresponds a reaction force/torque pair that
“enforces” the satisfaction of the constraint, throughout the time evolution of the
mechanism

For constraint equations that act between two bodies 𝑖 and 𝑗, there will also be a 𝐅𝑗 ,
𝑇𝑗 pair associated with such constraints, representing the constraint reaction forces
on body 𝑗

According to Newton’s third law, they oppose 𝐅𝑖 and 𝑇𝑖 , respectively

If the system is kinematically driven (meaning there are driver constraints), the
same approach is applied to obtain reaction forces associated with such constraints

In this case, we obtain the force/torque required to impose that driving constraint
11
Reaction Forces: Summary

A joint (constraint) in the system requires a (set of) Lagrange multiplier(s)

The Lagrange multiplier(s) result in the following reaction force and torque
Note: The expression of 𝚽 for all the
usual joints is known, so a boiler plate
approach can be used to obtain the
reaction force in all these joints

An alternative expression for the reaction torque is
Reaction force in a Revolute Joint
[Example 6.6.1]
12
Numerical Integration
14
Numerical Methods

Numerical Analysis is the study of quantitative approximations to the solution of
problems of mathematical analysis (i.e., calculus)


Study of algorithms
Algorithm
Program
Code
(arithmetic model)
(set of procedures)
(computer implementations)

Study of approximation errors
 Truncation (discretization)
 Round-off

Study of numerical stability
Examples: interpolation and extrapolation, solving linear system of equations,
eigenanalysis, solving nonlinear systems of equations, optimization (linear and
nonlinear), numerical quadrature, numerical differential equations, etc. ,etc.
Numerical Integration

The particular numerical methods used to solve differential equations are typically
called “numerical integrators”, or “integration formulas”

A numerical integrator generates an approximate solution at discrete time
points (also called grid points, station points, nodes)
 This is in fact just like in Kinematics, where the solution is computed on a time
grid

Different numerical integrators generate different solutions, but these solutions
are typically very close together, and (hopefully) close to the actual solution of the
problem

Putting things in perspective: In 99% of the cases, the use of numerical
integrators is the only alternative for solving complicated systems described by
non-linear differential equations
15
Basic Concept

IVP

In general, all we can hope for is approximating the solution at a sequence of discrete
points in time


Uniform grid (constant step integration)
Adaptive grid (variable step integration)

Basic idea: somehow turn the differential problem into an algebraic problem (approximate
the derivatives)

IVP in dynamics:

What we calculate are the accelerations
Oversimplifying, we get something like

This is a second-order DE which needs to be integrated to obtain velocities and positions

16
17
Simplest method: Forward Euler

Starting from the IVP

Use the simplest approximation to the derivative

Rewrite the above as
and use ODE to obtain
Forward Euler Method
with constant step-size ℎ
FE: Geometrical Interpretation

IVP

Forward Euler integration formula
18
FE: Example

Solve the IVP
using Forward Euler (FE) with a step-size ℎ = 0.01

Compare to the exact solution
19
20
Forward Euler: Effect of Step-Size
% IVP (RHS + IC)
f = @(t,y) -0.1*y + sin(t);
y0 = 0;
tend = 50;
% Analytical solution
y_an = @(t) (0.1*sin(t) - cos(t) + exp(-0.1*t)) / (1+0.1^2);
% Loop over the various step-size values and plot errors
colors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];
Figure, hold on, box on
h = [0.001 0.01 0.1];
for ih = 1:length(h)
tspan = 0:h(ih):tend;
y = zeros(size(tspan)); err = zeros(size(tspan));
y(1) = y0;
err(1) = 0;
for i = 2:length(tspan)
y(i) = y(i-1) + h(ih) * f(tspan(i-1), y(i-1));
err(i) = y(i) - y_an(tspan(i));
end
plot(tspan, err, 'color', colors(ih,:));
end
legend('h = 0.001', 'h = 0.01', 'h = 0.1');
FE errors for different values of the step-size
ℎ = 0.001, 0.01, 0.1
0.06
h = 0.001
h = 0.01
h = 0.1
0.04
0.02
0
-0.02
-0.04
-0.06
0
5
10
15
20
25
30
35
40
45
50
21
FE: Effect of Step-Size
% IVP (RHS + IC)
f = @(t,y) -0.1*y + sin(t);
y0 = 0;
tend = 50;
FE accuracy at different values of the step-size
ℎ = 0.1, 1.0, 5.0
6
% Loop over the various step-size values and plot errors
colors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];
Figure, hold on, box on
h = [0.1 1.0 5.0];
for ih = 1:length(h)
tspan = 0:h(ih):tend;
y = zeros(size(tspan));
y(1) = y0;
for i = 2:length(tspan)
y(i) = y(i-1) + h(ih) * f(tspan(i-1), y(i-1));
end
plot(tspan,y, 'color', colors(ih,:))
end
legend('h = 0.1', 'h = 1', 'h = 5');
h = 0.1
h=1
h=5
4
2
0
-2
-4
-6
0
5
10
15
20
25
30
35
40
45
50