these slides are a gentle introduction to mappings in continuum mechanics. goals: introduce students...

A “mapping” is…

1. Input2. Output3. Set of rules giving output from input

COPYRIGHT: [email protected]

Mapping from a scalar to a scalar

( )s f tinput is a scalar

output is a scalar

Example1:

3 which means ( ) 3s t f t t

Example 2:

3 3 which means ( )s t f t t Example 3:

which means ( ) identity operator!s t f t t


Mapping from a vector to a scalar

( )s f v

input is a vector

output is a scalar


Can you think of a function that takes a vector as input and returns a scalar as output?

L x x

EXAMPLE: Length of a vector:

input is a vector x

output is a scalar L( )f x x x

( )f x

Q: What shape is described by = constant? Hint: this says “length is constant”.

A: circle in 2D, sphere in 3D!

Q: Which way does the gradient point? Hint: perpendicular to isosurfaces

A: Radially!

Isosurfaces (contours)are lines of constant f(x)

f


Mapping from a scalar to a vector

( )v sv

input is a scalar

output is a vector

Your physical location in space is a vector(quantified by latitude, longitude and altitude)

When you walk to class, your location (a vector) changes with time (a scalar).

The time rate of your position is velocity, which is tangent to your path.


EXAMPLE 1: MAPPING TO POINTS ON A CIRCLE

scalar vector


cos

sin

x

1-D reference configuration(single scalar ranging from 0 to )

0 2

0

2

Symbolically, we would write this mapping as

Two outputs (components of vector x) determined from just one scalar input (the angle ).

Each maps to a unique x.

Moving from left to right on the domain (reference line segment) moves from right to left, or counterclockwise, on the range (spatial configuration, semicircle).

( )x x

Domain: line segment

Range: semi-circle

Given the velocity of a point in the reference configuration, you can use the mapping to figure out velocity in the spatial configuration.

dd d

dtddt

x x


214

214

cos[ ( 1) ]

sin[ ( 1) ]

t

t

x

1-D reference configuration(single scalar ranging from -1 to )

1t 1t 0t

1t 1t

0t

The “phase plot” is the trace mapped out by the x vectors as the parameter t is varied.

Here, the phase plot is the same as before (a semi-circle).

The distinction is non-uniform mapping of the hash marks.

A finite element code developer might use a mapping like this to generate a finer mesh near =0.

Different domain!

Same range (semicircle), but this is a different mapping

21 12 4

21 12 4

( 1)sin[ ( 1) ]

( 1)cos[ ( 1) ]

t td

dt t t

xλ

is tangent to the curve. Its length is proportional to hash mark spacing.


As always, speed and direction come from the derivative.

EXAMPLE 2: CUBIC SPLINES

(start PowerPoint on your laptop and follow along)


This is a “hands-on” exercise for the students who have brought a laptop to class. Otherwise, download this presentation and play with it on your own.

Powerpoint exercise to show how to playwith a cubic spline…From drawing tools make a straight line


Right-click the line and select “edit points”.The endpoints, initially hollow circles,will now be small squares.


Right-click an endpoint and select “smooth point.”

new “control bar”appears

Do the same with the other endpoint.(if a handle doesn’t appear, turn it into a smooth point as done above)

movethis newhandle

Then move the new handle.


You now have a cubic spline! Try changing the curvature with theleft handle. (Below is a bitmap image – do it on the next page)

drag this point…

down here.

If blue the control bar handle is not visible, right-click the curve and select “edit points” again. The handle re-appears when you single click an endpoint.


Play with it here: Escape out of slideshow mode so that you can see PowerPoint drawing tools. Right-click the curve, select “edit points.” Move the endpoints and the handles to change the shape of the curve.


cont

rol b

ar

The control bar may be regarded as a vector that controls curvature and “stretch.”

control bar


cont

rol v

ecto

rcontrol vector

0c

0p

1c

0t 1t 12t

reference configuration

( )x tx

A curve may be represented parametrically as

, or in slightly different notation, ( ) ( )d

t x tdt

x

λ

Each “t” corresponds to a different x.

12t

0t

1t

0 00 or (0)

t

c λ c

1 11 or (1)

t

c λ c

Then…

1p

The control vectors are tangent to the curve. They point in the direction that x would move if t is increased. A longer control vector corresponds to more distance covered by x for a given increment t, so it makes sense to introduce a “stretch” vector:


A cubic is the highest order polynomial capable of fitting the prescribed data.

1 1 2 3(1) 2 3 c a a a

2 30 1 2 3( )x t t t t a a a a

Input is the scalar parameter t. Output is the x position vector. That’s why the a-coefficients must be vectors.

The a-coefficient vectors are found by enforcing the following requirements:

21 2 3Note: ( ) ( ) 2 3t x t t t a a a

0 0(0)x p a

1 0 1 2 3(1)x p a a a a

0 1(0) c a

This is a system of four equations for the four unknown coefficient vectors! (Method of solution is no different from what it would be if the a’s were scalars).


Mapping from a vector to a vector

( )vv w

input is a vector

output is a vector

Example 1: v=3w (output is 3 times longer than the input)Example 2: rotate w by 30 degrees to get v

Example 3: v=w (identity operator! the output is the same as the input).

1 1

2 2

cos30 sin 30

sin 30 cos30

v w

v w


1 1

2 2

1 0

0 1

v w

v w

Continuum deformation is a mapping from a vector to a vector

( )fx X

input is initial location of a point

output is deformed location of the same point


Everything is Mapping

Suppose that this is a color plot of pressure (= the average of the diagonal components of stress).

Input to the FEM code (impact speeds, mesh parameters, material

properties, etc.) is ultimately… mapped to output from the FEM code (e.g., deformed position vectors in 3D space, stress tensor).

TENSOR (stress)maps to SCALAR (pressure) maps to 1D line segment (the legend)maps to 3D array (RGB color)maps to voltage (pixel illumination)

3D VECTOR (physical position) maps to 2D VECTOR (screen position)

This tells us which pixels need to be lit.

But in what color?

We need to map pressure to color.

sin cos

sin sin

cos

R

G

B

12 1

2

12 1

2

p

1

minp maxp

What additional mappings are needed to make this pressure plot during postprocessing of the FEM output?


TERMINOLOGY

scalar-to-scalar function vector-to-vector function

( )fy x

( )y f x

2y ax bx c

1 11 12 1

2 21 22 2

y m m x

y m m x

1 11 12 1 1

2 21 22 2 2

y m m x b

y m m x b

21

1 111 122 112 11 12 1 122

2 211 222 212 21 22 2 21 2

2

2

xy a a a b b x c

xy a a a b b x c

x x

linear

y mx

y mx b

linear

affine affine

quadratic quadratic

y m x

: y a xx b x c

y m x b


How do “beginners” plot the function

2 3y x x y

0 3

1 5

-1 1

2 7

Let’s plot an affine vector-to-vector mapping the same way

2

1 1

2

1 0.5 3.5

0 1 4

y

y

x

x


-1

0

1

2

3

4

5

-1 0 1 2 3 4 5

1 21

2 2

3.5 0.5

4

X

Xx

Xx

A point originally at1

2

X

X

deforms to

Example:homogeneous deformation:

This mapping is “homogeneous” because each little square deforms the same as all the others

X x

initial location deformed

location

{0,0} {3.5,4}{1,0} {4.5,4}{0,1} {4,5}{1,1} {5,5}

{ 1,1}{ 1, 1} {1, 1}{ 1,0}{0, 1}

{3,5}{2,3}{4,3}

{2.5,4}{3,3}

This is the same as the matrix equation in the previous slide with X and x as the variable names instead of x and y.

Here is how to plot this function using Mathematica. Note that it is a parametric plot. As X varies, the mapping tells how x varies.


The previous slide showed a square grid. This picture shows additional (circle and diagonal) “paint lines” that flow with the material.

Distinctive features of homogeneous mapping: All squares (big or little) deform to self-similar parallelograms, circles deform to ellipses, and straight lines deform to rotated and stretched straight lines.


Here is a quadratic mapping. The little squares don’t all deform in the same way.

Some straight lines deform to straight lines, but others (the diagonals) don’t.

Circles don’t deform to ellipses.

1

2

1

2 1(

3

6)1

44

X

X X

x

x


Here is a another quadratic mapping. 221

22

10.75 1.2X X

X

x

x

Look at the formula for the mapping.

It says that the vertical component remains unchanged after deformation (x2=X2), and the horizontal displacement (u1=x1-X1) increases quadratically with vertical distance from the base.


Here is a generally nonlinear mapping. 11

12 2

3( 2)

sin(3 )

x X

x X X

Each little square deforms differently not homogeneous.

Look at the formula for the mapping. It says that the horizontal displacement (u1=x1-X1=2X1+2) involves doubling the horizontal width and translating horizontally by a distance 2. The vertical displacement in the spatial configuration, u2=x2-X2=sin(3X1), varies sinusoidally as you move horizontally in the reference configuration. COPYRIGHT: [email protected]

A tangent mapping (colored part of the figure) is a homogeneous mapping (self-similar parallelograms) that coincides with a nonlinear mapping at a particular location.

This is like the local straight line that is tangent to a nonlinear curve at a point.

nonlinearmapping

Affine (homogeneous) tangentmapping.


SIMPLE SHEAR PURE SHEAR


ROTATION 11

22

cos 20 sin 20

sin 20 cos 20

x

x

X

X


Vortex:The rotation angle increases with proximity to the origin.

1

2

1

2

cos sin 1 w

sin coshere ~ ,

x Xr

X rx

X

Nonlinear because varies with position


Torsion:The rotation angle increases with distance up the axis, X3.

11

223

cos sin

sin where ~

cos

x

x

XX

X

Nonlinear because varies with position


A positive Jacobian is necessary, but not sufficient, for invertibility of the mapping.

xDeformation gradient: = i

ijj

FX

Jacobian: det[ ]J F

NonphysicalMaterial Interpenetration


This massive bending of a square into a big ring-like shape is locally invertible (positive Jacobian) everywhere on the domain, but material interpenetration makes it not globally invertible.

That’s why FEM codes have contact algorithms!

Deformation of a unit square (or unit cube in 3D)

unitsquare

1

1

3/2

1/4

1/2

2

1

3 / 2 1st column of [ ]

1/ 4F

g

1g

2g

2

1/ 2 2nd column of [ ]

2F

g

3 / 2 1/ 2[ ]

1/ 4 2F

det[ ] 23 / 8 2.875J F deformed size is 2.875 times larger!


deformed parallelogram

QUADRATIC FORM

Visualization of a vector-to-scalar mapping

( ) constantf x

describes an “isosurface”

( ) 1f x x A x

211 1 12 1 2 13 1 3

221 2 1 22 2 23 2 3

231 3 1 32 3 2 33 3

1

A x A x x A x x

A x x A x A x x

A x x A x x A x

2

22 2231 2

2 2 22

1/ 0 0

An ellipsoid, 1, corresponds to [ ] 0 1/ 0

0 0 1/

axx x

A ba b c

c



Quadratic Forms

1/ 2 0 0

[ ] 0 1 0

0 0 1/

A

1 0 0

[ ] 0 1 0

0 0 1/

A

1 x A x

1 2 1 2


• ellipsoid if >0• cylinder as • disk as as • hyperboloid if <0

These plots show the surface changing as changes!

QUADRATIC FORM FOR TENSORS!

Visualization of a tensor-to-scalar mapping

( ) constantf X

describes an “isosurface” in 9D tensor space

( ) : 1f X X A X

EXAMPLE: Plasticity yield criteria say that yield occurs when the stress is on the zero-isosurface of the yield function . The yield surface is embedded in 9D tensor space, and it can be regarded as being in 6D space since stress is symmetric, and it can be visualized in 3D principal space when the yield function depends only on the stress invariants.

( )f σ


these slides are a gentle introduction to mappings in continuum mechanics. goals: introduce students...

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