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7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
http://slidepdf.com/reader/full/chapter-4-systems-of-non-linear-equationspptx 1/21
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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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.
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
][
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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 equa- tions. 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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
7/27/2019 Chapter 4 - Systems of Non-Linear Equations.pptx
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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
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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
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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
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Assignment No. 7
Solve the following equations using the Newton-Raphson Method:
equation 1: x2 + 2y 2 = 22equation 2: 2x2 – xy + 3y = 12