29741728 Matrix Methods of Analysis

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INTRODUCTION TO MATRIXMETHODS

LECTURE No. 1


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MATRIX METHODS OF ANALYSIS :Broadly the methods of analysis are categorised in two ways

1. Force Methods : Methods in which forces are made unknowns i.eMethod of consistant deformation and strain energy method. In both

these methods solution of number of simultaneous equations isinvolved.

2. Displacement Methods in which displacements are made unknownsi.e slope deflection method, Moment distribution method and KanisMethod (In disguise). In slope deflection method also, the solution of

number of simultaneous equations is involved.In both of the above methods, for the solution of simultaneousequations matrix approach can be employed & such Method is calledMatrix method of analysis.


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FORCE METHOD :Method of consistent deformation is the base and forces are madeunknown


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b = Upward Deflection of point B on primary structure due to all causesbo = Upward Deflection of point B on primary structure due to applied

load(Redundant removed i.e condition Xb = 0)bb = Upward Deflection of point B on primary structure due to Xb

( i.e Redundant )

bb = Upward Deflection of point B on primary structure due to Xb = 1bb = bb . Xb

@b = bo + bb Substituting forbb -

b = bo + bb . Xb Called Super position equation

Using the compatibility condition that the net displacement at B = 0 i.e

b = 0 we get Xb = bo / bb

To Conclude we can say, [ ] = [L ] +[R]

@


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DISPLACEMENT METHODThis method is based on slope deflection method and displacementsare made unkowns which are computed by matrix approch instead ofsolving simultaneous equations and finaly unknown forces arecalculated using slpoe deflection equations.

Mab = Mab + 2EI / L ( 2 Ua + Ub + 3H / L)

Mab = Final Moment and may be considered as net force P at the joint

Mab = Fixed end moment i.e Force required for the condition of zerodisplacements & is called locking force. ( i.e. P)

The second term may be considred as the force required to produce therequired displacements at the joints. (i.e Pd )Therefore the aboveequation may be written as [P] = [P] + [Pd]


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Thus, there are Two Methods in matrix methodsMATRIX METHODS

FORCE METHOD DISPLACEMENT METHOD

The force method is also called by the names 1) Flexibility Method2)Static Method 3)Compatibility.

Similarly the displacement method is also called by the names

1)StiffnessM

ethod 2) Kinematic method 3) EquilibriumM

ethod.In both force method & displacement method there are twodifferent approaches 1) System Approach 2) Element Approach.


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To study matrix methods there are some pre-requisites :i) Matrix Algebra - Addition, subtraction ,Multiplication &

inversion of matrices (Adjoint Method )

ii) Methods of finding out Displacements i.e. slope & deflection

at any point in a structure, such as a)Unit load method orStrain energy method b) Moment area method etc.

According to unit load method the displacement at any point jis given by

j= 0 Mmjds / EI Where M B M dueto applied loads & mj B M due to unit load at j

L


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ij = 0 mi.mj.ds / EI When unit load is applied at i and is

called flexibility coefficient.The values of jand ij can be directly read from the tabledepending upon the combinations of B M diagrams & thesetables are called Diagram Multipliers.iii) Study of Indeterminacies Static indeterminacy & kinematic

indeterminacy


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THANK YOU


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INDETERMINATESTRUCTURES

1.Statically Indeterminate Structure

2.Kinematically Indeterminate Structure

LECTURE No. 2


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INDETERMINATE STRUCTURESStatically Indeterminate Structure :Any structure whose reaction components or

internal stresses cannot be determined by usingequations of static equilibrium alone, (i.e.7Fx = 0,7Fy = 0, 7Mz = 0) is a statically IndeterminateStructure.

The additional equations to solve staticallyindeterminate structure come from the conditionsof compatibility or consistent displacements.


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Roller Support : No. of reactions, r = 1

Hinged Support : No. of reactions, r = 2

Fixed Support : No. of reactions, r = 3


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1. Pin Jointed Structures i.e. TrussesInternal static indeterminacy : (Dsi) No. of

members required for stability is given by 3joints 3 members every additional jointrequires two additional members.

@ m = 2 (j 3)+3 j = No. of joints@ m = 2j 3 Stable and statically

determinate

Dsi = m m Where m = No. of members ina structure

Dsi = m (2j 3)


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External static indeterminacy (Dse)

r = No. of reaction components

Equations of static equilibrium = 3 (i.e.7

Fx= 0, 7Fy = 0, 7Mz = 0)

@ Dse = r 3

Total static indeterminacy Ds = Dsi + Dse@ Ds = m (2j 3)+(r 3)

@ Ds = (m+r) 2j


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Rigid Jointed Structures : No. of reactioncomponents over and above the no. ofequations of static equilibrium is called a

degree of static indeterminacy.Ds = r-3

Equations of static equilibrium = 3

(i.e.7

Fx = 0,7

Fy = 0,7M

z = 0)


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No. of reaction components r = 5 (as shown)@ Ds = r 3 = 5 3 = 2 Ds = 2

Example 1


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Introduce cut in the member BC as shown. At the cut the internalstresses are introduced i.e. shear force and bending moment asshown.Left part : No. of unknowns = 5 Equations of equilibrium = 3

@ Ds = 5 3 = 2Right Part : No. of unknowns = 4 Equations of equilibrium = 2

@ Ds = 4 2 = 2@ Ds = Static Indeterminacy = 2


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Example 2

Fig. (A) Fig. (B) Fig. (C)


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Another Approach : For Every member in a rigidjointed structure there will be 3 unknowns i.e. shearforce, bending moment, axial force.Let r be the no. of reaction components and m be theno. of members

Total no. of unknowns = 3m + rAt every joint three equations of static equilibrium areavailable

@ no. of static equations of equilibrium = 3j (where j isno. of joints)

@ Ds = (3m + r) 3j@ In the example r = 6, m = 6, j = 6

@Ds = (3 x 6 + 6) (3 x 6 ) = 6


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A structure is said to be kinematically indeterminate if thedisplacement components of its joints cannot be determined by

compatibility conditions alone. In order to evaluate displacementcomponents at the joints of these structures, it is necessary to consider theequations of static equilibrium. i.e. no. of unknown joint displacements overand above the compatibility conditions will give the degree of kinematicindeterminacy.

Fixed beam : Kinematically determinate :

Simply supported beam Kinematically indeterminate

Kinematic Indeterminacy :LECTURE No. 3


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Any joint Moves in three directions in a plane

structureTwo displacements Hx in x direction, Hy in y direction,U rotation about z axis as shown.

Roller Support :

r = 1, Hy = 0, U & Hx exist DOF = 2 e = 1

Hinged Support :

r = 2, Hx = 0, Hy = 0, U exists DOF = 1 e = 2

Fixed Support :

r = 3, Hx = 0, Hy = 0, U = 0 DOF = 0 e = 3

i.e. reaction components prevent the displacements @ no. of restraints = no. of reaction components.


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Degree of kinematic indeterminacy :Pin jointed structure :Every joint two displacements components and norotation

@ Dk = 2j e where, e = no. of equations of compatibility= no. of reaction components

Rigid Jointed Structure : Every joint will have three displacementcomponents, two displacements and one rotation.Since, axial force is neglected in case of rigid jointed structures, it isassumed that the members are inextensible & the conditions due toinextensibility of members will add to the numbers of restraints. i.e to the e

value.@ Dk = 3j e where, e = no. of equations of compatibility

= no. of reaction components +constraints due to in extensibility


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Example 1 : Find the static and kinematic indeterminacies

r = 4, m = 2, j = 3Ds = (3m + r) 3j

= (3 x 2 + 4) 3 x 3 = 1Dk = 3j e

= 3 x 3 6 = 3i.e. rotations at A,B, & C i.e. Ua, Ub & Uc

are the displacements.(e = reaction components + inextensibility conditions = 4 + 2 = 6)

Example 2 :Ds = (3m+r) 3jm = 3, r = 6, j = 4@ Ds = (3 x 3 + 6) 3 x 4 = 3

Dk = 3j e e = no. of reactioncomponents + conditions of inextensibility

= 6+3 = 9Dk = 3 x 4 9 = 3 i.e. rotationUb, Uc & sway.


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Example 3 :

Ds = (3m + r) 3jr = 6, m = 10, j = 9@Ds = (3 x 10 + 6) 3 x 9 = 9

Conditions of inextensibility :

Joint : B C E F H I1 1 2 2 2 2 Total= 10

Reaction components r = 6

@ e = 10 + 6 = 16

@ Dk = 3j e= 3 x 9 16 = 11


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FORCE METHOD :This method is also known as flexibility methodor compatibility method. In this method the degree of staticindeterminacy of the structure is determined and the redundantsare identified. A coordinate is assigned to each redundant.

Thus,P1, P2 - - - - - -Pn are the redundants at the coordinates 1,2, -- - - - n.If all the redundants are removed , the resulting structureknown as released structure, is statically determinate. Thisreleased structure is also known as basic determinate structure.From the principle of super position the net displacement at any

point in statically indeterminate structure is some of thedisplacements in the basic structure due to the applied loads andthe redundants. This is known as the compatibility condition andmay be expressed by the equation;

LECTURE No. 4


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1 = 1L + 1R Where 1 - - - - n = Displ. At Co-ord.at 1,2 - -n2 = 2L + 2R 1L ---- nL = Displ.At Co-ord.at 1,2 - - - - -n| | | Due to aplied loads| | | 1R ----nR = Displ.At Co-ord.at 1,2 - - - - -n

n = nL + nR Due to Redudants

The above equations may be return as [] = [L] + [R] - - - - (1)1 = 1L + 11 P1 + 12 P2 + - - - - - 1nPn2 = 2L + 21 P1 + 22 P2 + - - - - - 2nPn

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

n = nL + n1 P1 + n2 P2 + - - - - - nnPn


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@ = [ L] + [] [P] - - - - - - (3)

@[P]= []-1 {[] [ L]} - - - - - - (4)

If the net displacements at the redundants are zero then

1, 2 - - - - n =0,

Then @ [P] = - [] -1 [ L] - - - - - -(5)

The redundants P1,P2, - - - - - Pn are Thus determined

DISPLACEMENT METHOD :

This method is also known as stiffness or equilibrium. In thimethod the degree of kinematic indeterminacy (D.O.F) of thstructure is determined and the coordinate is assigned to eacindependent displacement component.


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In general, The displacement components at thesupports and joints are treated as independent displacementcomponents. Let 1,2, - - - - - n be the coordinates assigned tothese independent displacement components 1, 2 - - - - n.

In the first instance lock all the supports and the joints to

obtain the restrained structure in which no displacement ispossible at the coordinates. Let P1, P2 , - - - - Pn be the forcesrequired at the coordinates 1,2, - - - - - n in the restrained structurein which the displacements 1, 2 - - - - n are zero. Next, Let thesupports and joints be unlocked permitting displacements 1, 2 -- - - n at the coordinates. Let these displacements require forcesin P1d, P2d, - - - - Pnd at coordinates 1,2, - - - - n respectively.


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If P1, P2- - - - - - Pn are the external forces at the coordinates 1,2, -- - - n, then the conditions of equilibrium of the structure may beexpressed as:

P1 = P1 +P1dP2 = P2 +P2d - - - - - - - - - - - - -(1)

| | || | |

Pn = Pn + Pnd

or [P ] = [P] + [ Pd] - - - - - - - -- -(2)


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P1 = P1 +K11 1+ K12, 2+ K133 + - - - - -K1n n

P2 = P2 +K21 1+ K22 2+ K233 + - - - - - K2n n

| | | | | |

| | | | | | - - - - - - - - - (3)

Pn = Pn +Kn1 1+ Kn2, 2+ Kn33 + - - - - - - - - -Knn n @

i.e [P ] = [P ] + [ K ] [ ] - - - - - - - - - - (4)@ [ ]= [ K ] 1 {[P] [P]} - - - - - - - - - (5)

Where P = External forces

P = Locking forcesPd=Forces due to displacements

If the external forces act only at the coordinates the terms P1, P2 ,- - - - Pn vanish. i. e the Locking forces are zero,then


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[ ]= [ K ]1

[ P ] - - - - - - - - - - (6)On the other hand if there are no external forces at thecoordinates then [P]=0 then

[ ]= [ K ] 1 [ P ] - - - - - - - - - - -(7)Thus the displacements can be found out. Knowing the

displacements the forces are computed using slope deflectionequations:

Mab= Mab+ 2EI / L (2Ua+ Ub+3H / L)Mba=Mba+ 2EI / L (Ua+ 2Ub+3H / L)

Where Mab& Mba are the fixed end moments for the memberAB due to external loading


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FLEXIBILITYAND STIFFNESS MATRICES : SINGLE CO-ORD.P=K x D

P=K x PL3 / 3EI

K=3EI / L3K=Stiffness Coeff.

D= x P

D=PL3/3EI = x P

@ =L3 /3EI=Flexibility Coeff.

D= ML / EI

D= x M=ML/EI@ =L / EI

=Flexibility Coeff.

M=K x D =K x ML/EI

@ K=EI / LK=Stiffness Coeff.

X K= 1

LECTURE No. 5


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TWO CO-ORDINATE SYSTEM

Unit Force At Co-ord.(1)


11=L3 / 3EI 21=L2 / 2EI

12=21 =L2 / 2EI 22=L / EI

=

D1= 11P1 + 12 X P2 & D2= 21P1+ 22P2@ D1 11 12 P1

D2 21 22 P2

L3

/ 3EIL2 / 2EIL

2

/ 2EIL / EI[]=

FLEXIBILITY MATRIX

Unit Force At Co-ord.(1)Unit Force At Co-ord.(1)Unit Force At Co-ord.(1)Unit Force At Co-ord.(1)


Unit Force At Co-ord.(1)Unit Force At Co-ord.(1)





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STIFFNESS MATRIX

Unit Displacement at (1)

K11=12EI / L3

K21= 6EI / L2Forces at Co-ord.(1) & (2)

Unit Displacement at (2) Forces at Co-ord.(1) & (2)

K12= 6EI / L2

K22=4EI / L

=@P1=K11D1+K12D2

P2=K21D1+K22D2

P1

P2

K11 K12

K21 K22

D1

D2 K =

12EI / L3 6EI / L2

6EI / L2 4EI / L


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Develop the Flexibility and stiffness matrices for frameABCD with reference to Coordinates shown

The Flexibility matrix can be developed byapplying unit force successively atcoordinates (1),(2) &(3) and evaluating thedisplacements at all the coordinates

ij = mi mj / EI x ds ij =displacement at I due to unit load at j


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Unit Load at (1) Unit Load at (2) Unit Load at (3)

Portion DC CB BA

I I 4I 4I

Origin D C B

Limits 0 - 5 0 - 10 0 - 10

m1 x 5 5 - xm2 0 x 10

m3 - 1 -1 -1


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11 = m1.m1dx / EI = 125 / EI21 = 12 = m1.m2dx / EI = 125 / 2EI

31 = 13 = m1.m3dx / EI = -25 / EI

22 = m2.m2dx / EI = 1000 / 3EI23 = 32 = m2.m3dx / EI = -75 / 2EI33 = m3.m3dx / EI = 10 / EI

@ = 1 / 6EI

750 375 -150

375 2000 -225-150 -225 60


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INVERSING THE FLEXIBILITY MATRIX [ ]

THE STIFENESS MATRIX [ K ] CAN BE OBTAINED

K = EI

0.0174 0.0028 0.0541

0.0028 0.0056 0.0282

0.0541 0.0282 0.3412


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STIFFNESS MATRIX

Given Co-ordinates

Unit Displacement at Co-ordinate (1)

K11=(12EI / L3)AB + (12EI / L3)CD = 0.144EI

K21= (6EI / L2)AB = 0.24EI

K31= (6EI / L2)CD = 0.24EI

LECTURE No. 6


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Unit Dspl. at Co-ord.(2)

K12=K21= (6EI / L2)AB = 0.24EIK22=(4EI / L)AB +(4EI / L)BC= 3.2EI

K32=(2EI / L)BC= 0.82EI

Unit Dspl. at Co-ord.(3)

K13=K31= (6EI / L2)CD = 0.24EIK23= K32=(2EI / L)BC= 0.82EI

K33= (4EI / L)BC+(4EI/L)CD=2.4EI

K=EI0.144 0.24 0.24

0.24 3.2 0.8 0.24 0.8 2.4


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1. Determine the degree of static indeterminacy (degree of

redundancy), n

1. Determine the degree of kinematic indeterminacy ,

(degree of freedom), n

2. Choose the redundants. 2. Identify the independentdisplacement components

3. Assign coordinates 1, 2, ,n to the redundants

3. Assign coordinates 1, 2, , nto the independentdisplacement components.

Force Method ( Flexibility orcompatibility method )

Displacement method(Stiffness or equilibrium method)


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4. Remove all the redundants to

obtain to obtain the releasedstructure.

4.Prevent all the independent

displacement components toobtain the restrainedstructure.

5.Determine [L], the

displacements at the coordinatesdue to applied loads acting onthe released structure

5. Determine [P1] the forces

at the coordinates in therestrained structure due tothe loads other than thoseacting at the coordinates

6. Determine [R], the

displacements at the coordinatesdue to the redundants acting onthe released structure

6. Determine [P] the forces

required at the coordinates inthe unrestrained structure tocause the independentdisplacement components[]


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7. Compute the netdisplacement at the coordinates[] = [L ] + [R]

7. Compute the forces atthe coordinates

[P] = [ P1] + [P]

8. Use the conditions of

compatibility of displacement sto compute the redundants[P] = [] -1 {[] [L]}

8. Use the conditions of

equilibrium of forces tocompute the displacements[] = [k] -1 { [P] [P1] }

9. Knowing the redundants,compute the internal member forces by using equations of statics.

9. Knowing thedisplacements, compute theinternal member forces byusing slope deflectionequation.


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LECTURE No. 7


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6LECTURE No. 8


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LECTURE No. 9


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LECTURE No. 10


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29741728 Matrix Methods of Analysis