Chapter 4 - Systems of Non-Linear Equations.pptx

7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx

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P R E P A R E D B Y :

E N G R . R O M A N O A . G A B R I L L O

MENG’G - M F G . E

Chapter 4Systems of Non-linear Equations



Introduction

In this chapter, a solution technique for solving asystem of non-linear equations will be discussednamely the Newton-Raphson Method.

In the real world no system behaves in a linearmanner. There is no ideal material, ideal supportcondition and a perfect structure. Due to

imperfections the behavior can always be described by a set of non-linear equations only.



Newton-Raphson Method

Suppose there are n unknowns:

{x}T = x1 x2 … … xn

is a solution for non-linear equationsf 1 (x1, x2, … ... xn)=0

f 2 (x1, x2, … ... xn)=0

… … … … f n (x1, x2, … ... xn)=0



If x approximates x bar, then the increment from x tox bar be denoted by:

Where ∆x j is the increment from x j to x j for j=1, 2,-n

It is now required to find vector ∆x, and to find thedirection and distance to more from x (in n – space), to get the desired point: x bar = x + ∆x

nnn x

x

x

x

x x

x x

x x

x x

x x x

.........

}{}{ 3

2

1

33

22

11



It is necessary to seek the exact increment ∆x

that satisfies:

Expanding this in the Taylor’s series and neglectinghigher order terms, we get:

.........,2,10)( ni for x x f i

0...)()(

0...)()(

0...)()(

2

2

1

1

2

2

1

1

22

2

2

1

1

11

n

n

nn

n

n

n

n

dx x

f dx

x

f dx

x

f x f x x f

dx x

f dx

x

f dx

x

f x f x x f

dx x f dx

x f dx

x f x f x x f



Where partial derivatives are evaluated at the current values of x=x bar.This linear system can be expressed in matrix form as:

k

n

k

n

k k

k k

n

nnn

n

x x

x x

x x

x

x f

x

x f

x

x f

x

x f

x

x f

x

x f

k x

...

............

...

)(...

)()(........................

)(...

)()(

1

2

1

2

1

1

1

1

21

1

2

1

1

1

k xn

x f

x f

x f

)(

)(

)(

2

1



or

k

k x

xn

k

n

k

n x f

x f

x f

k

x

x

x

x

x

x

)(

)(

)(

][ 2

1

1

2

1

1

2

1

k xn

nnn

n

x

f

x

f

x

f

x

f

x

f

x

f

k where

21

1

2

1

1

1

][



This method based on this formula is calledNewton-Raphson Method

We have seen this method for single degree of freedom system as:

There is a number but a non-linear set of equations. It is a matrix [k]-1 . The above

equation is written for non-linear system of equations as:

)('

)(

1k

k

k k x f

x f

x x

)('

1

k x f

k x

k k x x f k x xk

)}({][}{}{ 11



Example No. 1

Consider a non-linear spring problem as shown below. The springs used are non-linear whose forcedeflection characteristics are given by:

Spring 1 F1 = 598x1 + 6060x13

Spring 2 F2 = 657x2 + 919x23

Spring 3 F3

= 69x3

+ 196x3

3

The loads are P1 = -120, P2 = 398



It is required to find the deflection of the spring

x1 = deflection in spring 1 = U1

x2 = deflection in spring 2 = (U3 – U1)

x3 = deflection in spring 3 = U3

The equilibrium equations are

F1 – F2 = P1

F2 + F3 = P2

F1 = 598 U1 + 6060 U13

F2 = 657 (U3 – U1) + 919 (U3 – U1)3

F3 = 69 U3 + 196 U33



Substituting:

1255 U1 + 6060 U13 – 657 U3 – 919 (U3 – U1)3 = -120

-657 U1 + 919 (U3 – U1)3 + 726U3 + 196U3

3 = 398

If U1 and U3 are not exact we will get residue as:f 1(x) = r1 = 120 + 1255U1 + 6060U13 - 657U3 - 919(U3 – U1)

3

f 2(x) = r2 = -398 - 657U1 + 919(U3 – U1)3 + 726U3 + 196U3

3

The matrix [k] =

DC

B A

U

f

U

f

U

f

U

f

3

2

1

2

3

1

1

1



where A = 1255 + 18180U12 + 2757(U3 – U1)

2 B = -657 – 2757(U3 – U1)

2 = CD = 2757(U3 – U1) + 726 + 588U3

2

First assume U1 = U3 = 0.3, k=0

k

k k

x f

x f k

U

U

U

U

)(

)(][

2

11

3

1

1

3

1

0

1

0

01

3

1

372

463

92.778657

6572.891.2

3.0

3.0

U

U



Take the Inverse of matrix [k]

10

01

92.778657

6572.2891

92.778657

6572.891.21

0

10

01

92.778657

6572.28910

2

01

E

E

10227241189.0

0000345877.0

62.6290

227241283.01

657

2.2891/1

1

0

2

1

2

0

1

1

1

E E E

E E



At k=0, and substituting the [k]-1

001588259.0000360917.0

00036119.0000345877.0

10

01

62.629/

227.01

2

2

2

22

11

21

E E

E E E

001588259.0000360917.0

00036119.0000345877.0][ 1 k

0

1

0

01

3

1

372

463

001588259.0000360917.0

00036119.0000345877.0

3.0

3.0

U

U



Multiplying, and checking the ∊=0.01

0

01

3

1

42372.0

0637.0

3.0

3.0

U

U

723.0

273.01

3

1

U

U

01.0265.0237.0

3.0237.0

01.0585.0

723.0

3.0723.0



Substitute this again to find [k] & we get

)(

)(

2

1

x f

x f

1

1

1

12

3

1

67.150

46.82

92.163893.1307

92.13072924

723.0

237.0

U

U

1

1

10

01

92.163893.1307

92.13072924

][

k

1

0

2

0

1

10

01

92.163893.1307

92.13072924

E

E

1

1

1

0

2

1

2

0

1

1

1

144730716.0

0000341997.0

82119.10980

447305061.01

92.1307

2924/

E E E

E E



At k = 1

1

1

1

0

2

1

2

11

11

21

000910066.0000407079.0

000407077.0000524085.0

10

01

82.1098/

44.0

E E E

E E E

1

1

1

12

3

1

67.150

46.82

000910066.0000407079.0

000407077.0000524085.0

723.0

237.0

U

U

1

12

3

1

103551909.0

018118242.0

723.0

237.0

U

U

6196.0

2192.02

3

1

U

U

0812.02192.0

237.02192.0

16688.0

6196.0

723.06196.0



At k = 2

2

1

2

23

3

1

43.13

14.7

74.13931099

109952.2570

6196.0

2192.0

U

U

2

1

2

1

10

01

74.13931099

109952.2570

][

k

2

1

2

0

2

0

1

10

01

74.13931099

109952.2570

E

E

2

1

2

1

1

0

2

1

2

0

1

1

1

1427539574.0

0000389026.0

8735917.9230

427539953.01

1099

52.2570/

E E E

E E



At k=3

2

1

2

1

2

2

2

2

2

1

1

2

1

001082399.0000462768.0

000462768.0000586877.0

10

01

87.923/

427.0

E E

E E E

2

23

3

1

01123455.0

002024672.0

6196.0

2192.0

U

U

60857.0

2174.03

3

1

U

U

008.02174.0

2192.02174.0



Final Answer

Finally we get,

U1 = 0.2174

U3 = 0.6087

U2 = U3 – U1 = 0.6087 – 0.2174=0.3913

Deflection in spring 1 = 0.2174

Deflection in spring 2 = 0.3913 Deflection in spring 3 = 0.6087



Assignment No. 7

Solve the following equations using the Newton-Raphson Method:

equation 1: x2 + 2y 2 = 22equation 2: 2x2 – xy + 3y = 12

Chapter 4 - Systems of Non-Linear Equations.pptx

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Transcript of Chapter 4 - Systems of Non-Linear Equations.pptx