ME451 Kinematics and Dynamics

of Machine Systems

Numerical Integration

October 28, 2013

Radu SerbanUniversity of Wisconsin-Madison

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Before we get started… Last Time:

Specifying position and velocity initial conditions for dynamics Recovering constraint reaction forces

Today Wrap-up discussion on recovering constraint reaction forces Begin section on numerical integration

Assignments: 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

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Final Project I encourage you to come up with a problem on your own

Something related to your research Something related to a personal interest of yours Something that you are curious about

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 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

Constraint Reaction Forces6.6

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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

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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

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Reaction Forces: Important Observation

What we obtain by multiplying the transposed Jacobian of a constraint, , with the computed corresponding Lagrange multiplier(s), , 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 , , and a torque due to the constraint We would like to report these quantities as acting at some point on a body

The strategy:Look for a real force which, when acting on the body at the point , would lead to a generalized force equal to

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Reaction Forces: Framework

Assume that the -th joint in the system constrains points on body and on body

We are interested in finding the reaction forces and torques and acting on body at point , as well as and acting on body at point

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 and , 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.

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Reaction Forces: Main Result

Let the constraint equations defining the -th joint be

Let the Lagrange multipliers associated with this joint be

Then, the presence of the -th joint leads to the following reaction force and torque at point on body

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Reaction Forces: Comments

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 and , there will also be a , pair associated with such constraints, representing the constraint reaction forces on body According to Newton’s third law, they oppose and , 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

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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

An alternative expression for the reaction torque is

Note: The expression of for all the usual joints is known, so a boiler plate approach can be used to obtain the reaction force in all these joints

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Reaction force in a Revolute Joint[Example 6.6.1]

Numerical Integration

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Numerical Methods

Numerical Analysis is the study of quantitative approximations to the solution of problems of mathematical analysis (i.e., calculus) Study of algorithms

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.

Algorithm(arithmetic model)

Program(set of procedures)

Code(computer implementations)

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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

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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

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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 Methodwith constant step-size

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FE: Geometrical Interpretation

IVP

Forward Euler integration formula

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FE: Example

Solve the IVP

using Forward Euler (FE) with a step-size

Compare to the exact solution

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Forward Euler: Effect of Step-Size

0 5 10 15 20 25 30 35 40 45 50-0.06

-0.04

-0.02

0

0.02

0.04

0.06

h = 0.001h = 0.01h = 0.1

% IVP (RHS + IC)f = @(t,y) -0.1*y + sin(t);y0 = 0;tend = 50; % Analytical solutiony_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 errorscolors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];Figure, hold on, box onh = [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,:));endlegend('h = 0.001', 'h = 0.01', 'h = 0.1');

FE errors for different values of the step-size

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0 5 10 15 20 25 30 35 40 45 50-6

-4

-2

0

2

4

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h = 0.1h = 1h = 5

FE: Effect of Step-Size

% IVP (RHS + IC)f = @(t,y) -0.1*y + sin(t);y0 = 0;tend = 50; % Loop over the various step-size values and plot errorscolors = [[0, 0.4, 0]; [1, 0.5, 0]; [0.6, 0, 0]];Figure, hold on, box onh = [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,:))endlegend('h = 0.1', 'h = 1', 'h = 5');

FE accuracy at different values of the step-size