Notes from April 30 – Thursday

Many engineering problems can be solved using a matrix approach. Although the Cartesian Coordinate System is the most common for the simpler problems, there are many more suitable systems for more advanced problems.

Most unusually, many physical problems have a “Natural Coordinate System.” A complicated formulation in the Cartesian system can be much more simplified using the Natural Coordinate System. For example, 200 coupled simultaneous equations in the Cartesian System can be simplified into 200 independent equations in the Natural system. In terms of matrix algebra, a fully coupled full matrix can be simplified into a diagonal matrix.

In the loose set of examples which follow, some important applications are touched upon. (1) The concept of principle stresses. (2) Orthogonal Transformation made possible by an eigenvalue problem. (3) Transforming coupled ordinary differential equations of a structural analysis into independent single degree of freedom differential equations. Other applications: (4) Space curves, the tangent-normal-binormal coordinate system. (5) How to deal with slowly convergent series. (6) Finite Difference formulation for Partial Differential Equations.

A = 4 5 3 5 6 5 3 5 1 octave:> [Q,D]=eig(A) Q = -0.025121 0.836295 0.547704 0.536662 -0.450956 0.713185 -0.843423 -0.311848 0.437479 D = Diagonal Matrix -2.09210 0 0 0 0.18517 0 0 0 12.90693 octave:> Q*Q' ans = 1.0000e+000 0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000 0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000 octave:13> Q'*A*Q ans = -2.0921e+000 0.0000e+000 0.0000e+000 0.0000e+000 1.8517e-001 0.0000e+000 0.0000e+000 0.0000e+000 1.2907e+001

We started with a symmetric matrix A and obtained its eigenvectors as the columns of matrix Q. It can be shown Q*Q’ or (Q’)*Q is an identity matrix. That means Q is an orthogonal transformation. The transformation (Q’)*A*Q would transform A in to a diagonal matrix D which contains its eigenvalues.

Here is a strategy often used to solve complicated problems. Consider this simple matrix example first: Ax=b. “x” is in the xyz system, so the matrix A is full. Transform x to a new system using Qx=y. Then equation system becomes, Ax=AQy=b. Now pre-multiply the entire system by Q’, then (Q’)*A*Qy=(Q’)y, or Dy=(Q’)b. Normally, a Greek letter, like eta, is used for y. In this transformed coordinate system, the matrix equation is diagonal. It would not be worth the effort to use this method for such a simple problem, but for complicated structural response calculation, this is the most common method. Having a 3000 degrees of freedom system, we can obtain a small sample of eigenvalues and eigenvectors, perhaps 100 or less, the results would be good enough. There is no need to do direct inversion of the 3000 by 3000 matrix.

These are the methods used by engineers worldwide. Create “stick” models of complicated structures, such as a nuclear reactor containment building, and use modal (eigenvalue) analysis to calculate the responses.

A = 35 -15 0 -15 25 -10 0 -10 10 B = 3 0 0 0 2.5 0 0 0 2 octave:> [VB,DB]=eig(B) VB = 0 -0 1 0 1 0 1 0 0 DB = 2 0 0 0 2.5 0 0 0 3 octave:> R=inv(sqrt(DB)) R = 0.70711 0 0 0 0.63246 0 0 0 0.57735 octave:> PBP=VB*R PBP = 0 -0 0.57735 0 0.63246 0 0.70711 0 0 octave:> AP=PBP'*A*PBP AP = 5 -4.4721 0 -4.4721 10 -5.4772 0 -5.4772 11.667 octave:> [VAP,DA]=eig(AP) VAP = 0.74243 -0.62072 0.25198 0.59051 0.42875 -0.68371 0.31636 0.65641 0.68487 DA = 1.443 0 0 0 8.089 0 0 0 17.135 octave:> phi=PBP*VAP phi = 0.18265 0.37898 0.39541 0.37347 0.27117 -0.43242 0.52498 -0.43892 0.17817 octave:> phi'*A*phi 1.443 0 0 0 8.089 0 0 0 17.135 octave:> phi'*B*phi 1 0 0 0 1 0 0 0 1

Space Curves

There are many cases in transportation engineering which require a special coordinate system, the tnb system. It has three unit vectors in the tangent, normal and binormal directions. These unit vectors change as a vehicle traverses the path. Examples are (1) Roller Coasters. The new ones are totally complicated and the vehicle path goes up and down with many curves. (2) Fighter Jet airplanes travel in erratic paths. (3) On and off ramps from highway or freeways. Very exotic designs of highway bridges, clover leave highway interchanges.

The vectors have a very simple form, the velocity vector has only one component, the tangent component. It means the path is design for the vehicle to pass through it in the tangent direction. The acceleration has two components, the tangential change of speed and the centripetal acceleration calculated as v^2/rho, in which rho is the radius of curvature at that location. The binormal unit vector forms the complete set for force related calculations.

We are lucky in ce108 to have introduced the variable spacing interpolation formulas. We would need those delta functions. One parameter describes the path and the distance “s” marks the location of a point along the path. That point is prescribed with x(s), y(s) and z(s). The derivatives are not simple but we can calculate all of them using divided differences. We have all the tools, we can say with a Jamaican accent, “no problem, mon.”

Oh, have you notice about the “superelevation” design put on the race tracks or on the ramps to fast freeways? To counteract the centripetal acceleration using the weight of the vehicle, a slight slope (or large slope for Olympic Bicycle Racing Tracks) were put in to make driving more pleasant and to save some tire wear. The weight of the vehicle in the negative z direction, but the normal force holding the weight of the vehicle is in the binormal direction, we need orthogonal transformation matrices along the path too. With computer programming, all of these things can be accomplished routinely. It is fun and exciting to code a computer. But seriously, DO NOT WAIT FOR Computer Science graduates to do this. They don’t know any Physics or math higher than Calculus III.

Numerical Determination of Unit Vectors

Thex, y, andz coordinates of a space curve can be defined as a function of parameter,ξ, as

x = x(ξ) , y = y(ξ) , z = z(ξ) .

The tangential unit vector,et, can be obtained as

et =d~r

ds=

d~r

dξ

dξ

ds=

dξ

ds

dx

dξ

dy

dξ

dz

dξ

,

in whichds =

√(dx)2 + (dy)2 + (dz)2 ,

and

ds

dξ=

√(dx

dξ

)2

+(

dy

dξ

)2

+(

dz

dξ

)2

=√

∆ ordξ

ds=

1√∆

The above formulation guarantees that|et| = 1. To find the normal unit vector, perform anotherderivative with respect tos,

1ρen =

det

ds=

d

ds

(d~r

ds

)=

d

ds

(d~r

dξ

dξ

ds

)=

d2~r

dξ2

(dξ

ds

)2

+d~r

dξ

(d2ξ

ds2

)

Withdξ

ds=

1√∆

,

it follows that

d2ξ

ds2 = −12

1∆

√∆

d∆ds

= −12

1∆

√∆

d∆dξ

dξ

ds= −1

21

∆2

d∆dξ

= − 1∆2

[(dx

dξ

) (d2x

dξ2

)+

(dy

dξ

) (d2y

dξ2

)+

(dz

dξ

) (d2z

dξ2

)]

Therefore,

1ρen =

det

ds=

(dξ

ds

)2d2~r

dξ2 −(

dξ

ds

)4 [(dx

dξ

) (d2x

dξ2

)+

(dy

dξ

) (d2y

dξ2

)+

(dz

dξ

) (d2z

dξ2

)]d~r

dξ

et et

A

B

C

D E F GH

(et)D =

0.8206

−0.5601−0.1137

(en)D =

−0.2953

−0.2452−0.9234

ρ = 0.6164 ρ = 2.1708

(en)F =

−0.9122

−0.0643−0.4047

(et)F =

0.3557

−0.6148−0.7039

enen

x

y

z

Node x y z

C 3.7610 3.2590 3.07D 4.2240 3.0329 3.20E 4.5298 2.7604 3.02F 4.7798 2.4358 2.67G 4.9269 1.9745 2.10

Non-Equally Spaced LaGrange Interpolation

Using LaGrange’s non-equally spaced interpolation formula, thex, y andz coordinates of a spacecurve can be approximated by

x(ξ) = x0δ0(ξ) + x1δ1(ξ) + x2δ2(ξ)

y(ξ) = y0δ0(ξ) + y1δ1(ξ) + y2δ2(ξ)

z(ξ) = z0δ0(ξ) + z1δ1(ξ) + z2δ2(ξ)

in which

δ0(ξ) =(ξ − ξ1)(ξ − ξ2)

(ξ0 − ξ1)(ξ0 − ξ2); δ1(ξ) =

(ξ − ξ0)(ξ − ξ2)(ξ1 − ξ0)(ξ1 − ξ2)

; δ2(ξ) =(ξ − ξ0)(ξ − ξ1)

(ξ2 − ξ0)(ξ2 − ξ1).

The derivatives can be evaulated accurately only at the center point,ξ1 and those expressions are:

δ′0(ξ1) =

ξ1 − ξ2

(ξ0 − ξ1)(ξ0 − ξ2); δ′

1(ξ1) =2ξ1 − ξ0 − ξ2

(ξ1 − ξ0)(ξ1 − ξ2); δ′

2(ξ1) =ξ1 − ξ0

(ξ2 − ξ0)(ξ2 − ξ1).

δ′′0 (ξ1) =

2(ξ0 − ξ1)(ξ0 − ξ2)

; δ′′1 (ξ1) =

2(ξ1 − ξ0)(ξ1 − ξ2)

; δ′′2 (ξ1) =

2(ξ2 − ξ0)(ξ2 − ξ1)

.

The parameterξ can be the accumulated chord lengths calculated between adjacent points as

Si =√

(xi − xi−1)2 + (yi − yi−1)2 + (zi − zi−1)2 .

Rotational Information for tnb system

In the tnb system

~v =

v

00

tnb

; ~a =d~v

dt=

v

00

tnb

+

0

0v/ρ

tnb

× v

00

tnb

=

v

v2/ρ0

tnb

The angular velocity of thetnb system is defined here as,

~Ω =v

ρeb .

In some applications, the angular acceleration of the frame is also needed.

~Ω =d

dt~Ω =

(v

ρ+ v

d

dt

(1ρ

))eb +

v

ρ

deb

dt.

The last term can be set to zero because

deb

dt=

v

ρeb × eb = ~0 .

Recall the definition ofet,

et =

dx

ds

dy

ds

dz

ds

;det

ds=

d2x

ds2

d2y

ds2

d2z

ds2

,

and sincedet

ds=

1ρen

anden is a unit vector, the magnitude ofρ can be obtained as

1ρ

=

√(d2x

ds2

)2

+(

d2y

ds2

)2

+(

d2z

ds2

)2

,

and its time derivative can be written as

d

dt

(1ρ

)= v

[12ρ

[2

(d2x

ds2

)d3x

ds3 + 2(

d2y

ds2

)d3y

ds3 + 2(

d2z

ds2

)d3z

ds3

]],

and the resulting angular acceleration of thetnb frame is

~Ω =[v

ρ+ ρv2

(d2x

ds2

d3x

ds3 +d2y

ds2

d3y

ds3 +d2z

ds2

d3z

ds3

)]eb .

Slowly Convergent Alternating Series

Given a slowly convergent alternating series of the form

S = V0 − V1 + V2 − V3 + V4 − . . . ,

and with the conditions that Vn > 0 and Vn+1 < Vn, then the series S can be rewritten in terms ofits forward differences as

S =∞∑0

(−1)nVn =12V0 − 1

4∆V0 +

18∆2V0 − 1

16∆3V0 + . . .

in which ∆Vn = Vn+1 − Vn and ∆r+1Vn = ∆rVn+1 − ∆rVn.

Example: Perform summation of the series

S = 1 − 13

+15

− 17

+19

− 111

+ . . . =π

4,

the sum after 6 terms is 0.744011, compared to the exact value of 0.785398. Perform now theforward differences as shown below:

Vi ∆Vi ∆2Vi ∆3Vi ∆4Vi ∆5Vi Sum

1 12 = 0.500000

13 − 2

323 = 0.666667

15 − 2

15815

1115 = 0.733333

17 − 2

358

105 − 1635

1621 = 0.761905

19 − 2

638

315 − 16315

128315

244315 = 0.774603

111 − 2

998

693 − 161155

1283465 − 256

69327043465 = 0.780375

The sum can be transformed as

S =12(1) − 1

4(−2

3) +

18(

815

) − 116

(−1635

) +132

(128315

) − 164

(−256693

) + . . .

the sum after 6 terms is 0.780375, an improvement. The intermediate sums are shown in the tableabove.

Example: Perform summation of the slowest convergent series

S = 1 − 12

+13

− 14

+15

− 16

+ . . . = log 2 ,

the sum after 6 terms is 0.616667, compared to the exact value of 0.693147. Perform now theforward differences as shown below:

Vi ∆Vi ∆2Vi ∆3Vi ∆4Vi ∆5Vi Sum

1 12 = 0.500000

12 − 1

258 = 0.625000

13 − 1

613

23 = 0.666667

14 − 1

12112 − 1

4131192 = 0.682292

15 − 1

20130 − 1

2015

661960 = 0.688542

16 − 1

30160 − 1

60130 − 1

613271920 = 0.691146

The sum can be transformed as

S =12(1) − 1

4(−1

2) +

18(13) − 1

16(−1

4) +

132

(15) − 1

64(−1

6) + . . .

the sum after 6 terms is 0.691146, 0.28% from the exact value, an improvement. The intermediatesums are shown in the table above.

Example: Perform summation of the highly oscillatory Sine Integral

∫ ∞

0

sin x

xdx =

∫ π

0

sin x

xdx +

∫ 2π

π

sin x

xdx +

∫ 3π

2π

sin x

xdx +

∫ 4π

3π

sin x

xdx + . . .

=1.85194 − 0.43379 + 0.25661 − 0.18260 + 0.14180 − 0.11533 + . . . =π

2,

the sum after 100 terms is 1.56761, and after 1000 terms is 1.57048, compared to the exact value of1.5708. Needless to say, the series is very slowly convergent. Perform now the forward differencesas shown below:

Vi ∆Vi ∆2Vi ∆3Vi ∆4Vi ∆5Vi Sum

1.85194 0.92597

0.43379 -1.41815 1.28051

0.25661 -1.17718 1.24097 1.43563

0.18260 -0.07401 0.10317 -1.13780 1.50674

0.14180 -0.04080 0.03321 -0.06996 1.06784 1.54011

0.11533 -0.02647 0.01433 -0.01888 0.05108 -1.01676 1.55600

The sum can be transformed as∫ ∞

0

sin x

xdx =

12(1, 85194) − 1

4(−1.41815) +

18(1.24097)

− 116

(−1.13780) +132

(1.06784) − 164

(−1.01676) + . . .

the sum after 6 terms is 1.55600, less than 1% from the exact value, an amazing improvement. Theintermediate sums are shown in the table above.