Lecture 41: ME 213 Highlights
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Lecture 41: ME 213 Highlights. Part I. Dynamical equations of motion: force and moment balances, degrees of freedom. Harmonic motion: natural frequencies, resonance, and damping, complex numbers. Energy and the Euler-Lagrange equations: constraints, generalized forces.

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Lecture 41: ME 213 Highlights

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Lecture 41 me 213 highlights

Lecture 41: ME 213 Highlights

Part I

Dynamical equations of motion: force and moment balances, degrees of freedom

Harmonic motion: natural frequencies, resonance, and damping, complex numbers

Energy and the Euler-Lagrange equations: constraints, generalized forces

Forced motion: homogeneous and particular solutions, ground motion, rotational imbalance


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Part II

State space: conversion of mechanical and electromechanical equations

Linearization: eigenvalues and eigenvectors, stability and poles

Feedback control: stabilization, gains and pole placement

Advanced feedback control: tracking control and observers


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Dynamical equations of motion

These are based on force and moment balances

For simple problems free body diagrams can be used to find the equation(s) of motion

The Euler-Lagrange approach is better for more complicated problems

The spring force is k(y – y0)

The damping force is cv, where v denotes the rate of change of y.


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Dynamical equations of motion

The number of degrees of freedom is the number of independent motions

We had some examples back in Lecture 4, and we can look at those again now


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Dynamical equations of motion

We’ll limit ourselves to two dimensional motion

center of mass position (y, z)

orientation

z

q

y


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Dynamical equations of motion

What happens when we pin a body to the world?

the body has 3 DOF

the pin removes two

+3

-2

so we are left with

= 1


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Dynamical equations of motion

Suppose we pin two bodies together

+3

+3

-2

= 4


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Dynamical equations of motion

General rules for two dimensional systems

Every element has three degrees of freedom to start

Every pin removes 2 of them

You need to think about how other constraints remove degrees of freedom

For example, let’s look at the next slide


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How many?

Dynamical equations of motion

y

the floor removes two:

the cart can’t move up and down

the cart can’t rotate

the pin removes 2

q

TWO

We start with six, three for each element


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Dynamical equations of motion

How many?

one if it rolls without slipping

two if it can slip

three if it can leave the surface


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Dynamical equations of motion

How many?

+3

+3

-2

-2

2


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Dynamical equations of motion

If I let the vertical rotate

I’ll have a 3D picture that’s the base of a robot


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Dynamical equations of motion

The archetype of the one degree of freedom equation of motion is

Undamped systems have harmonic solutions with natural frequencies define by their parameters

Linearized pendulum

Simple mass-spring system


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Harmonic motion

Harmonic motion is sinusoidal motion at a single frequency

It can be written in terms of trigonometric functions or exponential functions

in several equivalent ways

We saw this in Lecture 2


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Harmonic motion

Harmonic motion is proportional to sines and cosines

We can write a general harmonic function as

phase

amplitude

frequency

OR


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Harmonic motion

The multiple angle formulas will be very helpful

You can use these to connect the two forms on the previous slide

We also care about complex notation. The connections are on the next slide.


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Harmonic motion

Recall that

so that

also represents harmonic motion


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imaginary

axis

Complex plane

Harmonic motion

velocity

displacement

blue’s the real part

red’s the imaginary

t

real

axis

acceleration


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Harmonic motion

Phase issues

sint vs. sin(t +)

in the complex world


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Harmonic motion

Imaginary

axis

Complex plane

resultant

1.6 exp(jwt+jf)

exp(jwt)

f

t

Real

axis


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Harmonic motion

Real and imaginary parts


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Energy and the Euler-Lagrange equations

The Lagrangian, L, is the difference between the kinetic and potential energies

Kinetic energy has translational and rotational components

Potential energy in this course is limited to spring energy and gravitational energy

Damping enters the Euler-Lagrange equations through the Rayleigh dissipation function

External forcing enters through the generalized forces


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Energy and the Euler-Lagrange equations

ENERGY

kinetic energy

potential energy

gravity

spring


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Energy and the Euler-Lagrange equations

SIMPLE PENDULUM

q

m


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Energy and the Euler-Lagrange equations

FIGURE 3.1

k1

k3

k2

m1

m2

y1

y2


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Energy and the Euler-Lagrange equations

CONSTRAINED MOTION

We need as many variables as there are degrees of freedom

and NO more

We can often write down the energies without reference to constraints

and then we have to apply the constraints before we can go on to analysis

This is often a very good thing to do!


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OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

but

so

q

m

(y2, z2)


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Energy and the Euler-Lagrange equations

DOUBLE PENDULUM

q1

m1

(y1, z1)

There are six variables in the figure

four variables in the energy expressions

and but two degrees of freedom

q2

m2

(y2, z2)

We probably want to use the angles as our two independent variables


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Energy and the Euler-Lagrange equations

z

geometric constraint

y

differentiate

q1

m1

(y1, z1)

substitute and simplify

q2

m2

(y2, z2)


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Energy and the Euler-Lagrange equations

GENERALIZED COORDINATE FORMALISM

We need as many coordinates as there are degrees of freedom

no more, no fewer

It is traditional to name them q1, q2, etc.

For the overhead crane we can put y1 = q1 and q = q2

For the double pendulum we can put q1 = q1 and q2 = q2


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Energy and the Euler-Lagrange equations

The energies in terms of the generalized coordinates

Note the special form of T


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Energy and the Euler-Lagrange equations

The Lagrangian L = T - V

The homogeneous Lagrange equations


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Energy and the Euler-Lagrange equations

The process

1. Find T and V as easily as you can

2. Apply geometric constraints to get to N coordinates

3. Assign generalized coordinates

4. Define the Lagrangian


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Energy and the Euler-Lagrange equations

5. Differentiate the Lagrangian with respect to the

derivative of the first generalized coordinate

6. Differentiate that result with respect to time

7. Differentiate the Lagrangian

with respect to the same generalized coordinate

8. Subtract that and set the result equal to zero

Repeat until you have done all the coordinates


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OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

Steps 1-4 lead us to

q

m

(y2, z2)


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OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

but

so

q

m

(y2, z2)


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Energy and the Euler-Lagrange equations

For q1


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Energy and the Euler-Lagrange equations

For q2


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Energy and the Euler-Lagrange equations

The governing equations are then

Put the physical variables back so it looks more familiar


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Energy and the Euler-Lagrange equations

Linearize (in two steps)

Step one

Step two: drop squares and higher products of the angles

and their derivatives

Final linear equations


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Energy and the Euler-Lagrange equations

We can find the natural frequencies of this system in the usual way

There’s no damping, so we can seek harmonic solutions

We can see from 8a that one of the frequencies will be zero

and it represents motion of the cart with the pendulum fixed

But let’s look at this more formally


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Energy and the Euler-Lagrange equations

Convert to an eigenvalue problem

The determinant simplifies to

The nonzero frequency is the square root of


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Energy and the Euler-Lagrange equations

We can find the zero eigenvector by inspection

The other is

which we can demonstrate

The first row


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Energy and the Euler-Lagrange equations

The second row

Substitute for w2

and simplify


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Energy and the Euler-Lagrange equations

The second motion has the cart moving in one direction

and the pendulum moving in the opposite direction

and the relative motion depends on the relative masses


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Forced motion

Put in damping and address the particular solution using complex notation

m

c

f

k


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The general equation

Let the forcing go like the sine

Put in the complex representation of the sine

We can find the solution to (3) and take the real part when we are done


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Suppose that yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant


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Then we need to interpret the solution as a real quantity

We need the real part of yP NOT just YP!

To help us do this we need to manipulate YP


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multiply top and bottom by the complex conjugate of the denominator

simplify


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put the exponential back in

expand

take the real part


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The sum of the particular and homogeneous solutions add up

to satisfy the initial conditions


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The homogeneous contribution

We can write the homogeneous solution in real form from last time,

supposing we have an underdamped situation

I put the H in to emphasize that this satisfies the homogeneous equation —

the original equation with the right hand side set equal to zero. It’s the same

solution we had last time, when we were dealing with an unforced system


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Its derivative

which we can simplify

(collect the sine and cosine terms)


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Evaluate these at t = 0


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The particular solution contribution

(slide 34, equation (10))


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evaluate

at

zero


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displacement

velocity


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From equation (1) we have

From equation (2)

and combining these gives


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algebra continues . . . .

OR

We have part from the forcing

and part from the initial conditions


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Procedure Overview in Words

You have an equation (an ode) and a set of initial conditions

Find the homogeneous and particular solutions to the ode

The particular solution takes care of the forcing term

Use the SUM of the particular and homogeneous solution

to satisfy the initial conditions


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Procedure Overview in Symbols

Find yH such that

This gives you two arbitrary constants

Find yP such that

There is no arbitrary constant here — we “use up” the forcing amplitude


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Combine the two solutions to deal with initial conditions

Determines the two arbitrary constants in the homogeneous solution.

These depend on the forcing amplitude through the particular solution

as well as on the specific initial conditions


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To get back to the physics: what’s the force on the left hand wall?

m

c

f

k


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We generally care most about the long term response

The homogeneous solution decays for nonzero z

so we care about the particular solution


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It takes a lot of algebra to get the force on the wall,

even if limited to the particular solution

We have, in physical variables, the following

in phase

out of phase


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The transmissibility is the ratio of

the magnitude of the wall force to the applied force

In physical terms we have

We can simplify this a lot, and we can write it in a dimensionless manner


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We can simplify this a lot,

and we can write it in a dimensionless manner

where r denotes the ratio of the exciting frequency to the natural frequency

r = 1 is the resonant condition for zero damping


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Plot for several values of z over a range of r

ratio = 1

z = 0.1

z = 0.9


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A closer view

ratio = 1

note peak shifts

with increasing z

etc.


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Rotating imbalance

m

w

denote the offset

distance by a0

M

M includes the yellow mass and the blue rotating mass

but not the black imbalance mass


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Centripetal force

Suppose the system to be constrained from side to side motion

We get a standard undamped equation


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Make the usual assumption for an undamped 1 DOF system

solve for the amplitude

compare to equation (2.26) in Den Hartog

why is this different?


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Support Motion

m

c

f

k

We know what to do if f is zero or a harmonic function,

assuming the wall is fixed

What do we do if f = 0,

but the wall is not fixed?


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m

c

k

y+yW

yW

y denotes the relative motion of the mass

with respect to the wall


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force balance

rearrange

divide by m

and so we have our standard problem

the motion is forced inertially


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Air Bag Sensors

MEMS devices: a cantilever beam with a mass

some damping associated with the gas in the device

These are basically accelerometers


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Model as a simple mass-spring system

damping is not particularly important here

The air bag is triggered if the distance between the mass and the far wall

is less than some critical distance, measured capacitively, perhaps


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We can analyze this using the support motion equation

we just derived

Consider a constant decelleration

and neglect the damping

The vehicle is decellerating, so a is actually a negative number


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particular solution

homogeneous solution

initial conditions


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peak response atwt = π

twice

the

amplitude


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For the designer

Ignition takes place at t = π/wn into the crash

Deflection is a/wn2

So, wn has to be large, and the design deflection small

And, the critical acceleration a must be larger than anything

one would expect under normal operation


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Drive your car over a bump

m

y+yW

c

k

yW


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Let the mass be ¼ of a small car, 250 kg

Find the spring constant by supposing the spring to be compressed

a bit over 10 cm under gravity — 25,000 N/m

The natural frequency is 10 Hz

We want the system to be critically damped — z = 1

c = 2wnm = 5000 N-s/m


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V denotes the speed of the car, L the length of the bump and d its height

L

The bump ends for t >L/V


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The governing equation

Put in the complex representation of the sine

High speed and short bumps increase the apparent frequency

We can use w and d to characterize the problem

We can solve as before


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Suppose that yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant


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Then we need to interpret the solution as a real quantity

We need the real part of yP NOT just YP!

To help us do this we need to manipulate YP


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multiply top and bottom by the complex conjugate of the denominator

simplify


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put the exponential back in

expand

take the real part


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The sum of the particular and homogeneous solutions add up

to satisfy the initial conditions


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The homogeneous contribution

We write the homogeneous for the critically damped situation

I put the H in to emphasize that this satisfies the homogeneous equation —

the original equation with the right hand side set equal to zero


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Its derivative

Evaluate these at t = 0


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The particular solution contribution


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evaluate

at

zero


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displacement

velocity


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From equation (1) we have

From equation (2)

and combining these gives


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This is remarkably tedious,

and it may be time to find a better way to do this

Let f(t) denote a solution to the homogeneous solution such that

In the present case that solution is

I will show that

is a particular solution for any forcing a(t).


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You need to know how to differentiate such an integral

derivative of the integrand

derivative of the upper limit times the integrand evaluated at the upper limit

minus derivative of the lower limit times the integrand evaluated at the lower limit


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Of course


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start with

apply the formula

apply the formula again

We add all these guys up


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The integral vanishes and we have a solution

which establishes my contention


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Apply this to the present case

therefore

This function and its first derivative vanish at t = 0

so it is the full solution so long as the car is on the bump


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When the car leaves the bump, we need a homogeneous solution

with its initial values given by yP and its derivative at the moment

of leaving the bump


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We can do the integral using Mathematica


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The initial conditions for the follow on part of the problem


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This is as far as I want to go with this problem today

I would like you all to think about it, however


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