chapter 5a. simultaneous linear equations

1

� A. J. Clark School of Engineering �Department of Civil and Environmental Engineering

by

Dr. Ibrahim A. AssakkafSpring 2001

ENCE 203 - Computation Methods in Civil Engineering IIDepartment of Civil and Environmental Engineering

University of Maryland, College Park

CHAPTER 5a.SIMULTANEOUS LINEAR EQUATIONS

© Assakkaf

Slide No. 1

� A. J. Clark School of Engineering � Department of Civil and Environmental Engineering

ENCE 203 � CHAPTER 5a. SIMULTANEOUS LINEAR EQUATIONS

Introduction

� Systems of simultaneous equations can be found in many engineering applications and problems.

� Systems that consist of small number of equations can be solved analytically using standard methods from algebra.

� Systems of large number of equations require the use of numerical methods and computers.

2

© Assakkaf

Slide No. 2



Introduction

� Simultaneous Linear Equations and Engineering� The system of simultaneous equations is

probably one of the most important topics in modern engineering computations.

� This is not an exaggeration if one considers that recent technological advances were made possible by the ability of solving larger and larger systems of equations.

© Assakkaf

Slide No. 3



Introduction

� Simultaneous Linear Equations and Engineering

39743223421523

321

321

321

=++=++=++

XXXXXXXXX

7.71102.03.02.193.071.0

85.72.01.03

=+−−=−+

=−−

ZYXZYXZYX

113 828

5.68 472 22 3 2

1.201042

20321

20321

20321

20321

−=+++−

−=−+−−=++−+−=++−+

XXXX

XXXXXXXXXXXX

L

M

L

L

L

Small Systems Larger System

3

© Assakkaf

Slide No. 4



Introduction

� Simultaneous Linear Equations and Engineering� Technological advances in engineering

that involve large number of simultaneous equations include:

� The finite element method� The finite difference method� The analysis of structural, mechanical, and

electrical systems.

© Assakkaf

Slide No. 5



Introduction

� Simultaneous Linear Equations and Engineering� In the past, the numerical methods (such

as the finite element and finite difference methods) that were used to solve systems of simultaneous equations were not attractive owing to the tremendous amount of calculations involved.

� However, computers have changed that and altered our approach to engineering problem solving.

4

© Assakkaf

Slide No. 6



Introduction

� Engineering Examples� Electrical Circuit

I2I3I1 R2R3

R1

V2V1

I = currentV = voltageR = resistance

© Assakkaf

Slide No. 7



Introduction


� Kirchhoff�s first law states that the algebraic sum of current flowing into a junction of a circuit must equal zero

� Kirchhoff�s second law states that the algebraic sum of the electromotive forces around a closed circuit must equal the sum of voltage drops around the circuit.

I2I3I1 R2R3R1

V2V1

5

© Assakkaf

Slide No. 8



Introduction


I2I3I1 R2R3

R1

V2V1c e

fdb

a

© Assakkaf

Slide No. 9



Introduction


� Applying Kirchhoff�s first law at junction c, yields the following linear equation:

� Applying Kirchhoff�s second law to loop acdbyields the following linear equation:

0321 =−+ III

33111 IRIRV +=

6

© Assakkaf

Slide No. 10



Introduction


� Applying Kirchhoff�s second law to loop aefbyields the following linear equation:

221121 IRIRVV −=−

I2I3I1 R2R3R1

V2V1

© Assakkaf

Slide No. 11



Introduction


� If R1 = 2, R2 = 4, R3 = 5, V1 = 6, and V2 = 2, then

The solution to these three equations produces the current flows in the network.

4 4265 20

21

31

321

=−=+=−+

IIIIIII

I2

I3

I1 R2R3R1

V2V1

7

© Assakkaf

Slide No. 12



Introduction

� Engineering Examples� Analysis of Statically Determinant Truss

F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

© Assakkaf

Slide No. 13



Introduction


F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

∑∑

=−+=

==

01000

0

21

2

VVF

HF

V

H

8

© Assakkaf

Slide No. 14



Introduction


F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

1000 lb

F1 F3

∑∑

=−−−=

=+−=

060sin30sin1000

060cos30cos

31

31

FFF

FFF

V

H

6030

(1)

© Assakkaf

Slide No. 15



Introduction

� Engineering Examples

F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

∑∑

=+=

=++=

030sin

030cos

12

212

FVF

FFHF

V

H

H2

V2

F2

F1

30

(2)

0

9

© Assakkaf

Slide No. 16



Introduction


F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

60

V3

F2

F3

∑∑

=+=

=−−=

060sin

060cos

33

32

FVF

FFF

V

H (3)

© Assakkaf

Slide No. 17



Introduction


� Combining the three systems of equations (systems 1, 2, and 3), we obtain one system of simultaneous equation as shown in the next viewgraph:

∑∑

=−−−=

=+−=

060sin30sin1000

060cos30cos

31

31

FFF

FFF

V

H

∑∑

=+=

=++=

030sin

030cos

12

212

FVF

FFHF

V

H

∑∑

=+=

=−−=

060sin

060cos

33

32

FVF

FFF

V

H

10

© Assakkaf

Slide No. 18



Introduction


0 866.0 0 5.0 0 5.0 0 866.0 1000 866.0 5.0 0 5.0 866.0

33

32

21

221

31

31

=+=−−=+=++=−−=+−

VFFF

VFHFF

FFFF

F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

© Assakkaf

Slide No. 19



Introduction


� The solution to the previous system of equations provides the following force values for the truss:

lb 750lb 250

expected) (as 0lb 866

lb 433lb 500

3

2

2

3

2

1

===−=

=−=

VVHFFF

F1 F3

F2

1000 lb

90o

60o30o

1

32H2

V2 V3

11

© Assakkaf

Slide No. 20



Introduction

� Engineering Examples� Engineering Dynamics Example:

� Suppose that a team of three parachutists is connected by a weightless cord while free-falling at a velocity of 5 m/s as shown in the figure. Compute the tension in each section of the cord and the acceleration of the team, given the masses of each parachutist and the dreg coefficients as provided in the table.

© Assakkaf

Slide No. 21



Introduction

� Engineering Examples� Engineering Dynamics Example

R

T

a

3

2

1

18503

15702

11801

Drag Coefficient

(kg/s)Mass (kg)Parachutist

12

© Assakkaf

Slide No. 22



Introduction


Free-body diagrams are needed for each of the parachutists as shown in the next viewgraph.

Summing forces in the vertical direction and using Newton�s second law of motion, gives:

amRvcgmamRvcTgmamvcTgm

333

222

111

=+−=−−+=−−

(4)

© Assakkaf

Slide No. 23



Introduction


3 2 1

c3v

m3g R

c2v R

m2g T

c1v T

m1g

13

© Assakkaf

Slide No. 24



Introduction


3 2 1

c3v

m3g R

c2v R

m2g T

c1v T

m1g

3 2 1

c3v

m3g R

c2v R

m2g T

c1v T

m1g

amTvcgm 111 =−−amRvcTgm 222 =−−+amvcRgm 313 =−+

© Assakkaf

Slide No. 25



Introduction


Substituting the values for parachutists masses and drag coefficients, the system of equations provided by Eq. 4, gives

aRaRTaT

50)5(18 )8.9(5070)5(15)8.9(7080 )5(11)8.9(80

=+−=−−+=−−

14

© Assakkaf

Slide No. 26



Introduction


After rearranging and simplifying,

The solution to the above system of equations gives the following values:

400 5061170729 80

=−=+−=+

RaRTa

Ta

N 35 and N, 33 ,m/s 7.8 2 === RTa

© Assakkaf

Slide No. 27



General Form for a System of Equations

� DefinitionA linear equation is one in which a variable only appears to the first power in every term of a given equation.Thus, a system of m linear equations in nunknowns Xj, j = 1, 2, �,n, can be represented as

∑=

==n

jijij miCXa

1,,2,1 , L

15

© Assakkaf

Slide No. 28




� Expanded Form

mnmnmm

nn

nn

CXaXaXa

CXaXaXaCXaXaXa

=+++

=+++=+++

L

M

L

L

2211

22222121

11212111

aij = known coefficients of the equationsXj = unknown variablesCi = known constants

(5a)

© Assakkaf

Slide No. 29




� Expanded Form (m = n)

nnnnnn

nn

nn

CXaXaXa

CXaXaXaCXaXaXa

=+++

=+++=+++

L

M

L

L

2211

22222121

11212111

aij = known coefficients of the equationsXj = unknown variablesCi = known constants

(5b)

16

© Assakkaf

Slide No. 30




� Classification of Systems of Equations1. A set of equations in which the number of

unknowns is equal to the number of equations (n = m)

2. A set of equations in which the number of unknowns is less than the number of equations (n < m)

3. A set of equations in which the number of unknowns is greater than the number of equations (n > m)

© Assakkaf

Slide No. 31



Solution of a System of Two Equations

� The solution of two equation gives insight to understand the classification of systems of equations based on graphical interpretation in two-dimensional space.� If n equals 2, then Eq. 5 reduces to

2222121

1212111

CXaXaCXaXa

=+=+ (6a)

(6b)

17

© Assakkaf

Slide No. 32




� A solution for simple system can be obtained by substitution.

� Solving Eq. 6a for X1 gives

� The expression for X1 in the above equation can be substituted into Eq. 6b:

11

21211 a

XaCX −=

222211

212121 CXa

aXaCa =+−

(7a)

(7b)

© Assakkaf

Slide No. 33




� Equation 7b is a single equation with one unknown, X2.

� This equation can be solved for X2 to give

� Eq. 8a can be substituted back into Eq. 7a to give

12212211

1212112 aaaa

CaCaX−−= (8a)

12212211

2121221 aaaa

CaCaX−−= (8b)

18

© Assakkaf

Slide No. 34




� It seems that the solution procedure for this set of equations is simple to apply.

� One should imagine the effort that would be required to solve 15 or 20 simultaneous equations using the substitution procedure.

� Many complex engineering problems involve hundreds or even thousands of simultaneous equations.

� Hence, the need for alternative solution procedures is justified.

© Assakkaf

Slide No. 35



Types of Numerical Procedures

� Because of the wide-spread use of computers nowadays, numerical solution methods are widely used. There are three general types:

1. Elimination methods,2. Iteration methods, and 3. Method of determinants.

19

© Assakkaf

Slide No. 36



Classification of Systems of Equations Based on Graphical Interpretation

� Systems of equations can be classified based on their solutions to the following types:

1. Systems that have solutions,2. Systems without solution, and3. Systems with an infinite number of

solutions.

chapter 5a. simultaneous linear equations

Documents