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Lecture 41: ME 213 HighlightsPowerPoint Presentation

Lecture 41: ME 213 Highlights

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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

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

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.

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

We’ll limit ourselves to two dimensional motion

center of mass position (y, z)

orientation

z

q

y

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

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

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

How many?

one if it rolls without slipping

two if it can slip

three if it can leave the surface

If I let the vertical rotate

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

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

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

Harmonic motion is proportional to sines and cosines

We can write a general harmonic function as

phase

amplitude

frequency

OR

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.

axis

Complex plane

Harmonic motion

velocity

displacement

blue’s the real part

red’s the imaginary

t

real

axis

acceleration

Real and imaginary parts

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

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!

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

Energy and the Euler-Lagrange equations

z

geometric constraint

y

differentiate

q1

m1

(y1, z1)

substitute and simplify

q2

m2

(y2, z2)

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

Energy and the Euler-Lagrange equations

The energies in terms of the generalized coordinates

Note the special form of T

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

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

Energy and the Euler-Lagrange equations

The governing equations are then

Put the physical variables back so it looks more familiar

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

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

Energy and the Euler-Lagrange equations

Convert to an eigenvalue problem

The determinant simplifies to

The nonzero frequency is the square root of

Energy and the Euler-Lagrange equations

We can find the zero eigenvector by inspection

The other is

which we can demonstrate

The first row

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

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

Suppose that yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant

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

multiply top and bottom by the complex conjugate of the denominator

simplify

The sum of the particular and homogeneous solutions add up denominator

to satisfy the initial conditions

The homogeneous contribution denominator

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

Evaluate these at denominatort = 0

The particular solution contribution denominator

(slide 34, equation (10))

displacement denominator

velocity

algebra continues . . . . denominator

OR

We have part from the forcing

and part from the initial conditions

Procedure Overview in Words denominator

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

Procedure Overview in Symbols denominator

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

Combine the two solutions to deal with initial conditions denominator

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

We generally care most about the long term response hand wall?

The homogeneous solution decays for nonzero z

so we care about the particular solution

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

even if limited to the particular solution

We have, in physical variables, the following

in phase

out of phase

The hand wall?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

We can simplify this a lot, hand wall?

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

Rotating imbalance hand wall?

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

Centripetal force hand wall?

Suppose the system to be constrained from side to side motion

We get a standard undamped equation

Make the usual assumption for an undamped 1 DOF system hand wall?

solve for the amplitude

compare to equation (2.26) in Den Hartog

why is this different?

Support Motion hand wall?

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?

force balance hand wall?

rearrange

divide by m

and so we have our standard problem

the motion is forced inertially

Air Bag Sensors hand wall?

MEMS devices: a cantilever beam with a mass

some damping associated with the gas in the device

These are basically accelerometers

Model as a simple mass-spring system hand wall?

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

We can analyze this using the support motion equation hand wall?

we just derived

Consider a constant decelleration

and neglect the damping

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

For the designer hand wall?

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

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

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

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

L

The bump ends for t >L/V

The governing equation hand wall?

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

Suppose that hand wall?yP is proportional to the forcing function

Substitute into the differential equation (3)

Solve for the constant

Then we need to interpret the solution as a real quantity hand wall?

We need the real part of yP NOT just YP!

To help us do this we need to manipulate YP

multiply top and bottom by the complex conjugate of the denominator

simplify

The sum of the particular and homogeneous solutions add up denominator

to satisfy the initial conditions

The homogeneous contribution denominator

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

Its derivative denominator

Evaluate these at t = 0

The particular solution contribution denominator

displacement denominator

velocity

This is remarkably tedious, denominator

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).

You need to know how to differentiate such an integral denominator

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

Of course denominator

The integral vanishes and we have a solution denominator

which establishes my contention

Apply this to the present case denominator

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

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

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

of leaving the bump

We can do the integral using Mathematica denominator

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

I would like you all to think about it, however

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