part i - basics flexibility

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Matrix Methods for Structural Analysis

Part I Fundamentals & The Flexibility Method

October 2014Bedilu HabteCivil Engineering, AAU

Matrix Methods for Structural Analysis, AAiT Civil - Bedilu Habte

Course Outline 1

1. Fundamental Principles of Structural Analysis

Deformations in framed structures

Equilibrium and Compatibility

Principle of Superposition

Flexibility and Stiffness Matrices

2


Course Outline 2

2. The Flexibility Method Basic Concepts

Flexibility of Prismatic Members

Action Transformation and System Flexibility Equation

Solution Procedures

3Matrix Methods for Structural Analysis, AAiT Civil - Bedilu Habte

Course Outline 33. The Stiffness Method

Basic concepts

Stiffness of Prismatic Members

Axis Transformation and Master Stiffness Equation

Direct Stiffness Method & Solution Procedure

4. Additional Topics for the Stiffness Method Curved Members

Non Prismatic Members

Oblique Supports

Elastic Supports

Discontinuities in Members

4


References1. Matrix Analysis of Framed Structures

by Weaver & Gere

2. Matrix Structural Analysis by McGuire & Gallagher

3. Any other Matrix Structural Analysis books


Revise1. Matrix Manipulation

Addition, subtraction, multiplication

Determinant, inverse,

Solution of simultaneous equations

Sub-matrix

2. Determinate Structures

Condition for stability & determinacy of structures

Analysis of determinate structures

Forces in truss, beam, plane frame

Deflection of determinate structures

Moment-area, Conjugate-beam, Virtual Work6

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Software for the coursePractice how to use:

SciLabA free software for scientific computation and

visualzation.

http://www.scilab.org

http://www.scilab.org/products/scilab/downloads


Topics

Introduction

Deformations in Framed Structures


Static & Kinematic Determinacy

Structural Analysis Methods


Equivalent Joint Load


Energy Concepts

8


Introduction

The objectives of structural analysis are: to determine the reactions and the displacement

at various points of interest, and to evaluate stresses and strains within the elements present in the structure. All structural forms used for load transfer

from one point to another are 3-D in

nature. Using the fact that one or two dimensions are smaller than the other(s),

analytical models of the structure are adopted for simplicity.


Introduction

Classification of structures (Analytical models)

1. Skeletal structures: consist of line elements

2. Plated structures: plates & shells3. Solid structures: have all three dimensionsActions/stresses in structures may be:

Axial: in cables, truss elements, arches, membranes

Flexural: in beams, frames, grids, plates

Torsional: in grids, 3D frames

Shear: Frames, grids, shear walls

2D Models 3D Models

Plane Trusses Grids

Beam Space Trusses

Plane Frames Space Frames

10



Deformations in framed structures:

Axial:

Torsional:



Bending:

Shear:

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Equilibrium equations set the externally

applied loads to be equal to the sum of the

internal forces at all joints or node points of a structure.

In simple form, the equilibrium equations in

three dimensional coordinate system are:



Compatibility condition

refers to the continuity of

displacements and must also be satisfied. In the analysis of a structural system of discrete elements, all

elements connected to a

joint or node must have the same absolute displacement

at that node.

14


Static & Kinematic Indeterminacy

Overall statical classification of plane structures

Derive similar relationships for the statical classification of space (3D) truss and frame structures.

15

Statically



Static indeterminacy may be internal or external (or both), depending on the

redundancy. The total number of releases required to make a structure statically

determinate is called the degree of statical indeterminacy.

This truss system is

statically determinate

16


The degree of kinematic indeterminacy refers to the number of independent joint displacements that are

unknown and are needed to describe the displaced shape of the structure. It is also known as the

number of degrees of freedom.

This truss system is

kinematically

Indeterminate to

the 2nd degree



Structural Analysis Methods

The force method:

the redundant(s) of a statically indeterminate structure are removed and solutions are obtained by setting the relative displacements at the redundant(s) to zero.

The displacement method:

constraints are added to the structure, and subsequently equations are written satisfying

the equilibrium conditions, the solution of which is the constrained displacements.

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Where:

A is the action, D is the displacement, f is the flexibility coefficient and s is the stiffness coefficient of the spring.

DsA

AfD

]][[][2221212

2121111

AfD

AfAfD

AfAfD

]][[][2221212

2121111

DsA

DsDsA

DsDsA


Equivalent Joint Load

In matrix structural analysis, loads must be

placed at the joints.

Equivalent joint load

20


Principles of Superposition

The forces acting on a structure may be separated or divided into any convenient fashion and the structure

analyzed for the separate cases. Then the final results can be obtained by adding up the individual results.

Provided:

1. the geometry of the structure is not appreciably

altered under load, and

2. the structure is composed of a material in which the stress is linearly related to the strain.


Contents


Flexibility & Stiffness

Energy Methods

Strain Energy

Castiglianos Theorem

Virtual Work

22



The principle of superposition

the deflection at a given point in a structure

produced by several loads acting

simultaneously on the structure can be found

by superposing deflections at the same point

produced by the loads acting individually.


A, D

Where:A is the action, D is the displacement, f is the flexibility coefficient and s is the stiffness coefficient of the spring.

DsA

AfD


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A1,D1 A2,D2

]][[][

2221212

2121111

AfD

AfAfD

AfAfD

]][[][

2221212

2121111

DsA

DsDsA

DsDsA


Flexibility equation Stiffness equation

25 Matrix Methods for Structural Analysis, AAiT Civil - Bedilu Habte

Energy Methods

Are techniques to study the consequences of

deformation in structures .

They are useful for the formulation of the

stiffness and flexibility matrix of an element

in a structure and also for the analysis of

indeterminate structures as a whole.

26


Strain Energy

Defined as:

the work stored within the structure due to

the deformation it undergoes.

x

xy

y

z

zyzxxz

yzxy


Stresses Strains Displacements

Strain Energy

(u,v,w) are the x, y and z

components of displacement

x xy

y xz

z yz

u u v

x y x

v u w

y z x

w w v

z y z

zx

yz

xy

z

y

x

28


Evaluated as the internal work of

stresses acting through incremental

strains, integrated over the volume.

Strain Energy

V

TdVU

2

1


Evaluated as the internal work of strains

multiplied by incremental stresses,

integrated over the volume.

Complementary Strain Energy

V

TdVU

2

1*

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Evaluated similarly as the strain energy:

External Work of Loads

A

D

DAW T2

1


Evaluated similarly as the complementary

strain energy:

Complementary Work of Loads

ADW T2

1

A

D

32


Conservation of Energy:

The work of loads W equals the strain energy U stored in the structure

Conservation of Energy

DSDUW T2

1

Conservation of Complementary Energy:

Leads to W* equals U*

AFAUW T2

1**


1st Theorem:

If the strain energy of an elastic structure can be expressed as a function of a set of

displacement , the first partial derivative of that function, with respect to a particular

displacement equals the corresponding

action.

Castiglianos Theorems

j

j

AD

U

34


2nd Theorem:

If the complementary strain energy is expressed as a function of a set of applied

actions, the first partial derivative of that function, with respect to a particular action

equals the corresponding displacement.

Castiglianos Theorems

*

j

j

UD

A


The external virtual work of the real actions A

multiplied by virtual displacements D is

equal to the internal virtual work of the real stresses multiplied by the virtual strains

, integrated over the volume.

Virtual Work Principle

V

T

T

dVU

DAW

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The external work of the virtual actions A

multiplied by real displacements D is equal

to the internal work of the virtual stresses multiplied by the real strains ,

integrated over the volume.

Complementary Virtual Work

V

T

T

dVU

DAW

*

*


Unit Displacement Method:

to obtain stiffness coefficients

Unit Load/ Unit Displacement

V

jT

j dVA 1

Unit Load Method:

to obtain flexibility coefficients

V

Tjj dVD 1

38


Contents

Matrix Methods Truss Analysis1

The Flexibility Method Basic Approach2

The Flexibility Method - Beam3

The Flexibility Method - Truss4 The Flexibility Method - Truss4

Other Effects; Fixed End Actions5


Matrix Methods

Use matrix equations to analyze the truss and loading shown below.

40


External & internal unknowns (3+5 = 8) 4*2 = 8 equilibrium equations available Assume all

member forcesare tension (+ve) Y

X

Matrix Methods


Matrix Methods

Equilibrium equations are:

Equation Joint Axis Equation

1 1 X R1 + F14 + 0.6F13 = 0

2 1 Y R2 + F12 + 0.8F13 = 0

3 2 X 10 + F23 = 0

4 2 Y -5 F12 = 0

5 3 X -F23 0.6F13 + 0.6F34 = 0

6 3 Y -5 0.8F13 0.8F34 = 0

7 4 X -F14 0.6F34 = 0

8 4 Y 0.8F34 + R3 = 0

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Matrix Methods

Matrix form of equilibrium equations:

Eqn. R1 R2 R3 F12 F13 F14 F23 F34 RHS

1 1 0.6 1 0

2 1 1 0.8 0

3 1 -10

4 -1 5

5 -0.6 -1 0.6 0

6 -0.8 -0.8 5

7 -1 -0.6 0

8 1 0.8 0


A =

1. 0. 0. 0. 0.6 1. 0. 0. 0. 1. 0. 1. 0.8 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. - 1. 0. 0. 0. 0. 0. 0. 0. 0. - 0.6 0. - 1. 0.6 0. 0. 0. 0. - 0.8 0. 0. - 0.8 0. 0. 0. 0. 0. - 1. 0. - 0.6 0. 0. 1. 0. 0. 0. 0. 0.8

b = [0 0 -10 5 0 5 0 0]'

-10. 0.8339.167

- 5. 5.20836.875

- 10.- 11.4583

R1

R2

R3

F12

F13

F14

F23

F34

Solution is:

Solving the equilibrium equations (SciLab):

X = A \ b

Matrix Methods

44


Matrix Methods

The Flexibility method is also called 1)Force Method 2)Static Method 3)Compatibility Method.

Similarly the Stiffness method is also called 1)Displacement Method 2) Kinematic Method 3)Equilibrium Method.

Both force & displacement methods are used for the analysis of indeterminate structures.


In this method the degree of static indeterminacy of thestructure is determined and the redundants are identified.A coordinate is assigned to each redundant.

Thus, Aq1, Aq2 , Aqn are the redundants at thecoordinates 1,2, n. If all the redundants are removed,the resulting structure known as primary-structure, isstatically determinate.

From the principle of superposition the total displacementat any point in statically indeterminate structure is the sumof the displacements in the basic structure due to theapplied loads and the redundants. This is known as thecompatibility condition and may be expressed by thefollowing equations.

The Flexibility Method - General

46


D1 = D1(AL)+ D1(Aq) Where D1 Dn = Displ. at co-ord. 1,2 nD2 = D2(AL)+ D2(Aq) D1(AL) Dn(AL) = displ. at coord. 1,2 n | | | due to aplied loads| D1(Aq) Dn(Aq) = displ. at coord. 1,2 n Dn = Dn(AL)+ Dn(Aq) due to redudants

D1 = D1(AL)+ f11 Aq1 + f12 Aq2 + - - - - - f1nAqnD2 = D2(AL)+ f21 Aq1 + f22 Aq2 + - - - - - f2nAqn

| | | | || | | | | (2)

Dn = Dn(AL)+ fn1 Aq1 + fn2 Aq2 + - - - - - fnnAqn

The Flexibility Method - General


The above equations may be written in compact form as

[D] = [DL] + [DAq] - - - - - - (2)

[D] = [DL] + [F] [Aq] - - - - - - (3)

where [F] is the array of flexibility coefficients

[Aq]= [F]-1 {[D] [DL]} - - - - - - (4)

If the net (total) displacements at the redundants are zero then

D1, D2 Dn = 0,

Then [Aq] = - [F] -1 [DL] - - - - - - (5)

The redundants Aq1, Aq2, Aqn are thus determined.

The Flexibility Method Basic Approach

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Method of consistent deformation is the basis of the method

Redundant forces are made the primary unknowns

Beam is indeterminate to 1st degree

A stable and determinate structure (primary structure) is obtained by removing the vertical reaction at A (AqA) as redundant.

The Flexibility Method - Beam


Da = Upward Deflection of A on primary structure due to all causesDao = Upward Deflection of A on primary structure due to applied load (Redundant removed i.e condition Aqa = 0)Daa = Upward Deflection of A on primary structure due to Aqa( i.e Redundant )faa = Upward Deflection of A on primary structure due to Aqa = 1;

this is the flexibility coefficient (faa).

Daa = faa . AqaDa = Dao + Daa Substituting for Daa

Da = Dao + faa . Aqa Superposition equation

Due to compatibility, the net displacement at A = 0 i.e

Da = 0 we get Aqa = Dao / faa


50



8

3

0

8

34

3

wL

f

DAq

D

fAqDD

EI

wLD

EI

Lf

aa

aLa

a

aaaaLa

ao

aa


Flexibility Method Beam

Analyze the beam shown below using the basic flexibility method.

52

W P M

L1 L2 L3

Given: W = 5(n+1)P = 10(n+1)

M = 5n+1L1=L2-1.5=L3-1=2.5+0.2nwhere n is your roll number


Flexibility method Truss


Flexibility method Truss

Member A L P p pPL/A p2L/A

AB 2A/sqrt(3) 2L/sqrt(3) -P -1/sqrt(3) PL/(A.sqrt(3)) L/(3A)

AC A L 0 1 0 L/A

AD 4A/sqrt(3) 2L/sqrt(3) P -1/sqrt(3) -PL/(2A.sqrt(3)) L/(6A)

SUM PL/(2A.sqrt(3)) 3L/(2A)

Using the virtual load method (Virtual work):

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{DQ} = {DQL} + {DQT} + {DQP} + {DQR} + [F] {Aq} - - - - - - (7)

{DQC}

Let {DQC} = {DQL} + {DQT} + {DQP} + {DQR}

[Q]= [F]-1 ({DQ} {DQC}) - - - - - - (8)

where

{DQL} displacement due to applied loads on released structure

{DQT} displacement due to temperature change on released structure

{DQP} displacement due prestrain (initial displacement from any cause)

{DQR} displacement due restraint (support) settlement not selected as redundant supports

Refere Text book Section 2.4

Effect of temperature, prestrains & support settlement


Using principle of superposition

{DJ} = {DJL} + [DJR]{Aq}

{AM} = {AML} + [AMR]{Aq}

{AR} = {ARL} + [ARR]{Aq}where DJL = joint disp in primary structure due to loads

DJR = joint disp in primary structure due to unit value of redundant

AML = member end action in primary structure due to loads

AMR = in primary structure due to unit value of redundant

ARL = support reaction in primary structure due to loads

ARR = in primary structure due to unit value of redundant

Effect of temperature change, prestrains, and support settlement are accommodated in the first equation by replacing {DJL} by {DJC} but not in the remaining two.

where {DJC} = {DJL} + {DJT} + {DJP} + {DJR}

56

The Flexibility Method Other Unknowns


Flexibility Matrix Formalized Approach

Basic Approach Recap

Flexibilities of Prismatic Members

Flexibility for a Truss Member

Flexibility for a Beam Member

Flexibility for a Plane Frame Member

Flexibility for a Grid Member

Flexibility for a Space Frame Member

Transformation from member flexibility matrix to assembled system flexibility matrix

Solution of System flexibility equations

Determination of member forces, joint displacements and support reactions


Flexibility Method Basic Approach Procedure

Flexibility equation: Compatibility is written as

RfDD L where: D array of the total displacement on the structure (known) DL array of displacements on the released structure due to the applied actionR array of redundant forces to be solvedf matrix of flexibility coefficients

LDDfR 1

LDfR1

If D is zero.58


Effects other than loading: if a structure is subjected to the effect of temperature, pre-strain or support settlement,

RfDDDDD SPTL where: DT array of displacements on the released structure due to temperature DP ditto due to pre-strainDS ditto due to support settlement

If D is [0],

SPTL DDDDfR 1

Flexibility Method Basic Approach Procedure


Flexibility for Truss Member

EA

LFMi

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Flexibility for Beam Member

EI

L

EI

LEI

L

EI

L

FMi

2

232

23


Flexibility for Plane Frame Member

EI

L

EI

LEI

L

EI

LEA

L

FMi

20

230

00

2

23

62


Flexibility for Grid Member

EI

L

EI

LGJ

LEI

L

EI

L

FMi

02

00

20

3

2

23


Flexibility for Space Frame Member

zz

yy

yy

zZ

Mi

EI

L

EI

L

EI

L

EI

LGJ

L

EI

L

EI

L

EI

L

EI

LEA

L

F

0002

0

002

00

00000

02

03

00

2000

30

00000

2

2

23

23

64


Formalized Flexibility MethodFlexibility matrix of a space frame element

65

EIzL

EIzL

EIyL

EIyL

GJL

EIyL

EIyL

EIzL

EIzL

EAL

f

0000

0000

00000

0000

0000

00000

2

2

23

23

2

2

23

23


Recall Energy MethodsPrinciples of Conservation of Energy: the work of loads equals the strain energy stored in the structure. For linearly elastic structures, U = W and U* = W*, which could be expressed as:

AfAWU

DsDWUT

T

21**

21

Principles of virtual work: the external virtual work of the real action equals the work of the real stresses times the virtual strains.

V

T

T

dvU

DAW

DsA

AfD

using

using

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Formalized Flexibility Method

Flexibility equation

for an individual element

for the whole structure with n elements (arranged in matrix form)

MiMiMi AfD

Mn

Mi

M

M

Mn

Mi

M

M

Mn

Mi

M

M

A

A

A

A

f

f

f

f

D

D

D

D

...

...

00000

0...0000

00000

000...00

00000

00000

...

...

2

1

2

1

2

1


Structure Flexibility equation in compact form

called unassembled flexibility equation.

Relationship between Am and As (member vs. structure actions related through equilibrium equations)

where:BMS : action transformation matrixAS, AJ, and AQ : structure-, joint- as well as redundant-actions

MMM AfD

Q

J

MQMJ

SMSM

A

ABB

ABA


68


Arbitrary virtual load: AS = [AJ AQ] would cause a member force of:

External complementary virtual work: is given by

Internal complementary virtual work: is given by

Equating complementary virtual works: W *=U*

Q

J

MQMJSMSM A

ABBABA

Q

JT

Q

T

JS

T

S D

DAADAW *

M

T

M DAU *

SMSMTMSTSM

T

MS

T

S

ABfBA

DADA



Structure displacement:

the structure flexibility equation is obtained as

In terms of joint and redundant displacements:

the redundant action is given by:

the member forces is given by:

MSM

T

MSS

SSS

SMSM

T

MSS

BfBf

where

AfDei

ABfBD

,.

MQM

T

MQQQMJM

T

MQQJ

MQM

T

MJJQMJM

T

MJJJ

Q

J

QQQJ

JQJJ

Q

J

S

BfBfBfBf

BfBfBfBf

whereA

A

ff

ff

D

DD

,

JQJQQQQ AfDfA 1

QMQJMJMfM ABABAA


70


Formalized Method Solution Procedure Steps in applying the formalized flexibility method

1. Establish the idealized structure and identify the nodal points (load points, support points & member connection points)

2. Identify the internal member forces and specify the redundants

3. Select system coordinates where external forces are applied and where displacement measurements are desired

4. Select element coordinates so that system coordinates occur only at their ends


Formalized of the Flexibility Method 5. Form the member flexibility matrix (unassembled)

[FM] from the individual members

6. Calculate [BMJ] as the matrix of internal forces for unit values of Applied forces {AJ} (only one force acts at a time)

}{][}{ MMM AFD

}{

...

}{

...

}{

}{

][00000

0...0000

00][000

000...00

0000][0

00000][

`

}{

...

}{

...

}{

}{

2

1

2

1

2

1

Mn

Mi

M

M

Mn

Mi

M

M

Mn

Mi

M

M

A

A

A

A

F

F

F

F

D

D

D

D

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Formalized Flexibility Method7. Calculate [BMQ] as the matrix of internal forces for

unit values of Redundant forces {AQ} (only one force acts at a time)

Action transformation matrix [BMS] which relates {AM} and {AS} (member vs. structure actions related through equilibrium equations) is formed from both [BMJ] & [BMJ] considered as partitions

[BMS] = [BMJ | BMQ]

8. Calculate the assembled system flexibility matrix [FS][FS] = [BMS]

T[FM] [[BMS] or [FS] = [BMJ|BMQ]

T[FM] [BMJ|BMQ]


Formalized Flexibility Method 9. Partition the system flexibility matrix as

10. Calculate the redundant forces {AQ} from the partitioned equation

]][[][][]][[][][

]][[][][]][[][][

,

MQMT

MQQQMJMT

MQQJ

MQMT

MJJQMJMT

MJJJ

Q

J

QQQJ

JQJJ

Q

J

S

BFBFBFBF

BFBFBFBF

whereA

A

FF

FF

D

DD

}]{[}{][}{ 1 JQJQQQQ AFDFA

74



11. Calculate the internal forces and displacements

where {AMF} represents fixed end action

12. Compute the support reaction

where {ARC} denotes the combined loads (actual & equivalent) applied at supports

}]{[}]{[}{

}]{[}]{[}{}{

QJQJJJJ

QMQJMJMFM

AFAFD

ABABAA

Q

J

RQRJRCSRSRCR A

ABBAABAA ][}{}]{[}{}{



Example Analyze the truss shown below using the

formalized method.

Use P1=10kN, P2=5kN

EA=constant for all members

76


Beam Example

Analyse the beam shown below using the generalized flexibility method


Beam has two DOF (rotation both at the roller-supports Dj1 and Dj2); it is indeterminate to the 2nd degree, hence

Step 1. To create a primary structure, the moment reaction at the fixed-support is removed and also a hinge is inserted just to the right of the middle support (Aq1 and Aq2).

Beam Example - continued

Step 2. Flexibility matrix of a beam element and the unassembled structure flexibility matrix are given by:

EI

L

EI

LEI

L

EI

L

f Mi

2

232

23

2

1

0

0

M

M

mf

fF

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Step 2. Unassembled structure flexibility matrix :

5125200

2525400

00354

00549

1

..

..

.

.

EIFM

Step 3. Action Transformation Matrix Bms:

0010

3/103/10

1001

3/13/103/1

Bms



Step 4. Assembled structure flexibility matrix Fs:

Step 5. The redundant actions are obtained as:

5.15.025.01

5.0105.0

25.005.00

15.001

ms

T

ms BFmBFs

kNmAj

5.7

0 TAmf 5.7105.715

kNmAjfqjfqqAq

5.1

75.0\

80



Step 6. Member end forces

0

7

9

75.15

AsBmsAmfAm

Step 7. Structure displacement :

0

0

375.3

125.1

1;

EIFsAqAj

AsFsDq

DjDs


Plane Frame Example

Analyze the plane frame shown below using the formalized flexibility method

For both members use:

E = 200GPa

I = 200E6 mm4

A = 5000 mm2

82


Plane Frame Example (continued)

The frame has four DOFs and it is statically indeterminate to the 2nd degree

Primary structure & redundants

DOFs


Plane Frame Example - continued

Flexibility matrix of a plane frame element and the unassembled structure flexibility matrix are given by:

EI

L

EI

LEI

L

EI

LEA

L

f Mi

20

230

00

2

23

2

1

0

0

M

M

mf

fF

3

1

10015.00225.00

0225.0450

006.01

xE

fM

3

2

1002.004.00

04.0667.1060

008.01

xE

fM

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Unassembled structure flexibility matrix :

7

8

1010002.00000

0002.0533333.00000

00002.0000

000105.70001125.00

0000001125.0225.00

000000015.0

x

xFM



300001001

010010

100100

300000001

100000

010000

Bms

Action Transformation Matrix Bms:

86



Assembled structure flexibility matrix Fs:

msT

ms BFmBFs

0.0000002 - 0.0002 0. 1.000E-07 -0.0002 0.0001120

- 0.0002 0.5333333 0. -0.0002 0.5333333 0.0006

0. 0. 0.002 0. 0. 0.002

1.000E-07 - 0.0002 0. 1.000E-07 -0.0002 -0.0000003

- 0.0002 0.5333333 0. -0.0002 0.5348333 0.0006

0.0001120 0.0006 0.002 -0.0000003 0.0006 0.2263266



The redundant actions are obtained as:

TxAj 31045151005.6715000 TAmf 450005.67015000300

kNcheckAjfqjfqqAq \

88



Structure displacement :

T

FsAqAj

FsAsDqDjDs

0000286.002969.003644.000344.0

;

;

Redundant forces:


Contents

Basics1

Vectors and Matrices2

More Matrix Operations3

Plotting Graphs4

Further Topics5

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Basics

SciLab window


Basics

Scilab is a software for computation and visualization

is developed in 1990 at INRIA (France)

is open source & free of charge

has all matrix operations as built in functions

excellent tool for subjects involving matrices

has a built-in programming language

has a number of toolboxes:

2-D and 3-D graphics, animation, linear algebra, sparse matrices, polynomials and rational functions,

differential equation solvers, etc

92


Basics

Simple calculations

>> (-1+2+3)*5 - 2/3 // simple arithmetic

>> 2^3 // 2 to the power 3

>> exp (2) // e to the power of 2

>> %e // constant e

>> %pi // constant

>> cos( %pi/3 ) // trig functions.

>> sin( %pi/4 )

>> 22/7 4*atan(1.0)

>> 6*(5/3) - 10 // round-off error


Basics

Variables

>> 10-3 // values are stores in ans

ans =

7.

>> a=2; b=3; // declaring variables

>> c=b-2*a

c =

-1.

>> y= sqrt(-4); // y = 2.i (imaginary)

>> Y= acos(-0.5); // case sensitive, yY

94


Vectors and Matrices

Vectors

>> A = [1 2 3]; B = [4,3,2] // row vector

>> C = [0.5; 1; 1.5] // column vector

>> f=2*A; D=A+B; e=B-A; At = A //algebra

>> s=A*B; sc=sum(A .* B) // dot product

What about cross product??

>> a=[1:3] // means [1, 2, 3], range

>> b=[1:0.5:3] // means [1,1.5,2,2.5,3]

>> v = linspace(0,9,5) //[0, 2.25, 4.5, 6.75, 9]


Vectors and Matrices

Matrices

Separate the elements of a row with blanks or commas.

Use a semicolon ; or a new line at the end of each row.

Surround the entire list of elements with square brackets, [ ].

A = [1 2 3; 4,3,2; -1 3, 2] // 3x3 matrix

p=A(1:2,2:3), A(1,:), A(:,2) //sub-matrix

s=size(p) // 2x2

B = [-1 2.5 3; 1 1, 1]

c=B*A; d=c-2*B; b=A/A, S=A^2

e = A .* A // element-wise product

D = [A;B] // Juxtapose

z=zeros(3,2),o=ones(2,3),y=eye(4,3)96

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17


More Matrix Operations

Matrices

The element in row i and column j of A is denoted by A(i , j)

A = [-1 3 1; 1,3,2; 3 0, 1] // 3x3 matrix

d=det(A), inv(A), diag(A), diag([1 2 -1])

Solution of simultaneous equations

Let b = [-1; 2.4; 3]

To solve for A*x = b, use

x = A \ b //

R = rand(3,3)

D = R*inv(R)

v = clean(R*inv(R))97 Matrix Methods for Structural Analysis, AAiT Civil - Bedilu Habte

Plotting Graphs

x = 0:%pi/100:2*%pi;

y = sin(x);

plot(x,y)

Label the axes and add a title.

xlabel('x = 0:2\pi')

ylabel('Sine of x')

title('Plot of the Sine Function','FontSize',12)

Multiple graphs

y2 = sin(x-0.25); plot(x,y,x,y2)

legend('sin(x)','sin(x-.25)');

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Further Topics

Working with Polynomials

Matrix of polynomials

Calculus, DEs

3D graphs

Functions

Programming


Graph Example

function [x,y,z] = f3(alpha,theta)

x = cos(alpha).*cos(theta);

y = cos(alpha).*sin(theta);

z = sinh(alpha);

endfunction

alphagrid = linspace(-%pi/2,%pi/2,40);

thetagrid = linspace(0,2*%pi,20);

[x1,y1,z1] = eval3dp(f3, alphagrid, thetagrid);

plot3d1(x1,y1,z1);

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part i - basics flexibility

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