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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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slide1

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

slide2

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

slide3

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.

slide4

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

slide5

Dynamical equations of motion

We’ll limit ourselves to two dimensional motion

center of mass position (y, z)

orientation

z

q

y

slide6

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

slide7

Dynamical equations of motion

Suppose we pin two bodies together

+3

+3

-2

= 4

slide8

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

slide9

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

slide10

Dynamical equations of motion

How many?

one if it rolls without slipping

two if it can slip

three if it can leave the surface

slide12

Dynamical equations of motion

If I let the vertical rotate

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

slide13

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

slide14

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

slide15

Harmonic motion

Harmonic motion is proportional to sines and cosines

We can write a general harmonic function as

phase

amplitude

frequency

OR

slide16

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.

slide17

Harmonic motion

Recall that

so that

also represents harmonic motion

slide18

imaginary

axis

Complex plane

Harmonic motion

velocity

displacement

blue’s the real part

red’s the imaginary

t

real

axis

acceleration

slide19

Harmonic motion

Phase issues

sint vs. sin(t +)

in the complex world

slide20

Harmonic motion

Imaginary

axis

Complex plane

resultant

1.6 exp(jwt+jf)

exp(jwt)

f

t

Real

axis

slide21

Harmonic motion

Real and imaginary parts

slide22

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

slide23

Energy and the Euler-Lagrange equations

ENERGY

kinetic energy

potential energy

gravity

spring

slide26

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!

slide27

OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

but

so

q

m

(y2, z2)

slide28

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

slide29

Energy and the Euler-Lagrange equations

z

geometric constraint

y

differentiate

q1

m1

(y1, z1)

substitute and simplify

q2

m2

(y2, z2)

slide30

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

slide31

Energy and the Euler-Lagrange equations

The energies in terms of the generalized coordinates

Note the special form of T

slide32

Energy and the Euler-Lagrange equations

The Lagrangian L = T - V

The homogeneous Lagrange equations

slide33

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

slide34

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

slide35

OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

Steps 1-4 lead us to

q

m

(y2, z2)

slide36

OVERHEAD CRANE

Energy and the Euler-Lagrange equations

y1, f1

M

but

so

q

m

(y2, z2)

slide39

Energy and the Euler-Lagrange equations

The governing equations are then

Put the physical variables back so it looks more familiar

slide40

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

slide41

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

slide42

Energy and the Euler-Lagrange equations

Convert to an eigenvalue problem

The determinant simplifies to

The nonzero frequency is the square root of

slide43

Energy and the Euler-Lagrange equations

We can find the zero eigenvector by inspection

The other is

which we can demonstrate

The first row

slide44

Energy and the Euler-Lagrange equations

The second row

Substitute for w2

and simplify

slide45

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

slide46

Forced motion

Put in damping and address the particular solution using complex notation

m

c

f

k

slide47

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

slide48

Suppose that yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant

slide49

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

slide51

put the exponential back in

expand

take the real part

slide53

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

slide54

Its derivative

which we can simplify

(collect the sine and cosine terms)

slide56

The particular solution contribution

(slide 34, equation (10))

slide57

evaluate

at

zero

slide58

displacement

velocity

slide59

From equation (1) we have

From equation (2)

and combining these gives

slide60

algebra continues . . . .

OR

We have part from the forcing

and part from the initial conditions

slide61

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

slide62

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

slide63

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

slide65

We generally care most about the long term response

The homogeneous solution decays for nonzero z

so we care about the particular solution

slide66

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

slide67

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

slide68

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

slide70

A closer view

ratio = 1

note peak shifts

with increasing z

etc.

slide71

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

slide72

Centripetal force

Suppose the system to be constrained from side to side motion

We get a standard undamped equation

slide73

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?

slide74

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?

slide75

m

c

k

y+yW

yW

y denotes the relative motion of the mass

with respect to the wall

slide76

force balance

rearrange

divide by m

and so we have our standard problem

the motion is forced inertially

slide77

Air Bag Sensors

MEMS devices: a cantilever beam with a mass

some damping associated with the gas in the device

These are basically accelerometers

slide78

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

slide79

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

slide80

particular solution

homogeneous solution

initial conditions

slide81

peak response atwt = π

twice

the

amplitude

slide82

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

slide84

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

slide86

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

slide87

Suppose that yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant

slide88

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

slide90

put the exponential back in

expand

take the real part

slide92

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

slide93

Its derivative

Evaluate these at t = 0

slide95

evaluate

at

zero

slide96

displacement

velocity

slide97

From equation (1) we have

From equation (2)

and combining these gives

slide98

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).

slide99

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

slide101

start with

apply the formula

apply the formula again

We add all these guys up

slide102

The integral vanishes and we have a solution

which establishes my contention

slide103

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

slide104

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

slide107

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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