module1 1 introduction-tomatrixms - rajesh sir

Structural Analysis - III

Introduction to M t i M th dMatrix Methods

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

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Module I

Matrix analysis of structures

Module I

• Definition of flexibility and stiffness influence coefficients –d l t f fl ibilit t i b h i l h &


development of flexibility matrices by physical approach & energy principle.

Flexibility method

• Flexibility matrices for truss beam and frame elements –• Flexibility matrices for truss, beam and frame elements –load transformation matrix-development of total flexibility matrix of the structure –analysis of simple structures –

l t ti b d l f d l l d plane truss, continuous beam and plane frame- nodal loads and element loads – lack of fit and temperature effects.


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Force method and Displacement method p

•These methods are applicable to discretized structures •These methods are applicable to discretized structures of all types

• Force method (Flexibility method)

• Actions are the primary unknowns

• Static indeterminacy: excess of unknown actions than the available number of equations qof static equilibrium


• Displacement method (Stiffness method)

• Displacements of the joints are the primary unknowns

• Kinematic indeterminacy: number of d d l d ( hindependent translations and rotations (the

unknown joint displacements)

• More suitable for computer programming


Types of Framed Structures• a. Beams: may support bending moment, shear force and

Types of Framed Structures

axial force

• b. Plane trusses: hinge joints; In addition to axial forces, a member CAN have bending moments and shear forces if it


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ghas loads directly acting on them, in addition to joint loads

• c. Space trusses: hinge joints; any couple acting on a c. Space trusses: hinge joints; any couple acting on a member should have moment vector perpendicular to the axis of the member, since a truss member is incapable of

ti t i ti tsupporting a twisting moment

• d. Plane frames: Joints are rigid; all forces in the plane of the frame all couples normal to the plane of the frame


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the frame, all couples normal to the plane of the frame

G id ll f l t th l f th id ll • e. Grids: all forces normal to the plane of the grid, all couples in the plane of the grid (includes bending and torsion))

• f. Space frames: most general framed structure; may support bending moment shear force axial force and


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support bending moment, shear force, axial force and torsion

Deformations in Framed Structures

, ,x y zT M MThree couples: , ,x y zN V VThree forces:

•Significant deformations in framed structures:•Significant deformations in framed structures:

Structure Significant deformationsStructure Significant deformationsBeams flexuralPlane trusses axialPlane trusses axial

Space trusses axialPlane frames flexural and axialPlane frames flexural and axialGrids flexural and torsionalSpace frames axial flexural and torsional


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Space frames axial, flexural and torsional

Types of deformations in framed structures


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b) axial c) shearing d) flexural e) torsional

Static indeterminacy

B

Static indeterminacy

• Beam:

• Static indeterminacy = Reaction components - number of y peqns available

3E R= −

• Examples:

• Single span beam with both ends hinged with inclined loads

C ti b• Continuous beam• Propped cantilever• Fixed beam



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• Rigid frame (Plane): g d a e ( a e):

• External indeterminacy = Reaction components - number of il bl eqns available 3E R= −

• Internal indeterminacy = 3 × closed frames 3I a=

• Total indeterminacy = External indeterminacy + Internal indeterminacy

( )3 3T E I R a= + = − +

• Note: An internal hinge will provide an additional eqn


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Example 1 Example 2Example 1 Example 2

( )3 3T E I R a= + = − +( )3 3T E I R a= + = − + ( )( )3 3 3 3 2 12= × − + × =( )2 2 3 3 0 1= × − + × =

Example 3 Example 4

( )( )

3 3T E I R a= + = − + ( )( )

3 3T E I R a= + = − +


( )3 2 3 3 3 12= × − + × = ( )4 3 3 3 4 21= × − + × =


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• Rigid frame (Space): g d a e (Space):

• External indeterminacy = Reaction components - number of il bl eqns available

6E R= −

• Internal indeterminacy = 6 × closed frames

Example 1

( )6 6T E I R a= + = − +

Example 1

( )4 6 6 6 1 24= × − + × =

If axial deformations are neglected, static indeterminacy is not affected since the same number of actions still exist in


not affected since the same number of actions still exist in the structure

Plane truss (general):

External indeterminacy = Reaction components - number of eqns available

3E R= −

Minimum 3 members and 3 joints. A dditi l j i t i 2 dditi l b

Hence, number of members for stability,

Any additional joint requires 2 additional members.

( )3 2 3 2 3m j j= + − = −


( )2 3I m j= − −Hence, internal indeterminacy,

l ( l d l) dTotal (Internal and external) indeterminacy

( )3 2 3T E I R m j= + = − + − −

2m R j= + −

b f b• m : number of members• R : number of reaction components• j : number of jointsj j

• Note: Internal hinge will provide additional eqn


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Example 1p

2 9 3 2 6 0T m R j= + − = + − × =

3 3 3 0E R= − = − =

0I T E= − =

Example 22 15 4 2 8 3T m R j+ + ×2 15 4 2 8 3T m R j= + − = + − × =

3 4 3 1E R= − = − =

2I T E= − =Example 3

2 6 4 2 5 0T m R j= + − = + − × =

( ) 3 1 4 4 Hinge at 0 AE R= − + = − =


0I T E= − =

Example 4Example 42 7 3 2 5 0T m R j= + − = + − × =

3 3 3 0E R= − = − =3 3 3 0E R= = =

0I T E= − =

2 6 4 2 4 2T m R j= + − = + − × =Example 5

2 6 4 2 4 2T m R j+ +

3 4 3 1E R= − = − =

1I T E 1I T E= − =

Example 62 11 3 2 6 2T m R j= + − = + − × =

3 3 3 0E R= − = − =

p


2I T E= − =

• Wall or roof attached pin jointed plane truss (Exception • Wall or roof attached pin jointed plane truss (Exception to the above general case):

• Internal indeterminacy 2I m j= −

• External indeterminacy = 0 (Since, once the member forces are determined, reactions are determinable)forces are determined, reactions are determinable)

Example 1 Example 3Example 2Example 1 Example 3Example 2

26 2 3 0

T I m j= = −= − × =

25 2 1 3

T I m j= = −= − × =

27 2 3 1

T I m j= = −


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6 3 0 7 2 3 1= − × =

• Space Truss:• External indeterminacy = Reaction components -number of equations available 6E R= −

E l 1

3T m R j= + −Total (Internal and external) indeterminacy

Example 1

Total (Internal and external) indeterminacy3T m R j= + −

( ) y

12 9 3 6 3T∴ = + − × =


6 9 6 3E R= − = − =

Actions and displacements

•Actions:

p

• External actions (Force or couple or combinations) and

• Internal actions (Internal stress resultants – BM, SF, axial forces, twisting moments)


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•Displacements: A translation or rotation at some pointsp ace e ts: t a s at o o otat o at so e po t

•Displacement corresponding to an action: Need not be causedb th t tiby that action


•Notations for actions and displacements:•Notations for actions and displacements:

D32

D33


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Equilibrium•Resultant of all actions (a force, a couple or both) must vanish for static equilibrium

q

q

•Resultant force vector must be zero; resultant moment vector must be zero

0xF =∑ 0yF =∑ 0zF =∑∑ ∑ ∑

must be zero

0xM =∑ 0yM =∑ 0zM =∑

F 2 di i l bl (f i l •For 2-dimensional problems (forces are in one plane and couples have vectors normal to the plane),

0xF =∑ 0yF =∑ 0zM =∑•In stiffness method, the basic equations to be solved


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In stiffness method, the basic equations to be solved are the equilibrium conditions at the joints

Compatibility•Compatibility conditions: Conditions of continuity of displacements throughout the structure

•Eg: at a rigid connection between two members, the displacements (translations and rotations) of both members displacements (translations and rotations) of both members must be the same

I fl ibili h d h b i i b l d •In flexibility method, the basic equations to be solved are the compatibility conditions


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Action and displacement equations

A SD=D FA=•Spring:

1S F −=•Stiffness

1F S −=•Flexibility:

•The above equations apply to any linearly elastic structure


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•Example 1:

Flexibility and stiffness of a beam subjected to a single load

3

48LFEI

=Flexibility 3

48EISL

=Stiffness


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48EI L

•Example 2:

Flexibility coefficients of a beam subjected to several loads

A ti th bActions on the beam

Deformations corresponding to actions


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30Unit load applied corresponding to each action, separately

1 11 12 13D D D D= + + 2 21 22 23D D D D= + + 3 31 32 33D D D D= + +

1 11 1 12 2 13 3D F A F A F A= + +

2 21 1 22 2 23 3D F A F A F A= + +

3 31 1 32 2 33 3D F A F A F A= + +

11 12 13, ,F F F Flexibility coefficients etc.

• Flexibility coefficient F12: Displacement corresponding to A1caused by a unit value of A2.caused by a unit value of A2.

• In general, flexibility coefficient Fij is the displacement

Dept. of CE, GCE Kannur Dr.RajeshKN31

In general, flexibility coefficient Fij is the displacement corresponding to Ai caused by a unit value of Aj.

•Example 3: Stiffness coefficients of a beam subjected to several loads

Actions on the beam



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Unit displacement applied corresponding to each DOF separately


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Unit displacement applied corresponding to each DOF, separately, keeping all other displacements zero

A A A A= + +1 11 12 13A A A A= + +

A S D S D S D+ +1 11 1 12 2 13 3A S D S D S D= + +

2 21 1 22 2 23 3A S D S D S D= + +

3 31 1 32 2 33 3A S D S D S D= + +

11 12 13, ,S S S Stiffness coefficients: etc.

• Stiffness coefficient S12: Action corresponding to D1 caused by a unit value of D2.a unit value of D2.

• In general, stiffness coefficient Sij is the action corresponding


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In general, stiffness coefficient Sij is the action corresponding to Di caused by a unit value of Dj.

•Example 4: Flexibility and stiffness coefficients of a beam Example 4: Flexibility and stiffness coefficients of a beam

Flexibility coefficients:y


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Stiffness coefficients:


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•Example 5: Flexibility and stiffness coefficients of a trussExample 5: Flexibility and stiffness coefficients of a truss


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Flexibility and stiffness matrices1. Flexibility matrix

Th tibilit ti

1 11 1 12 2 13 3 1... n nD F A F A F A F A= + + + +

oThe compatibility equations are:

2 21 1 22 2 23 3 2...............................................................

n nD F A F A F A F A= + + + +

1 1 2 2 3 3 ...n n n n nn nD F A F A F A F A= + + + +

oIn matrix form,

1 11 12 1 1

2 21 22 2 2

...

...n

n

D F F F AD F F F A

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬

1 2

... ... ... ... .........n n nn nn F F F AD

⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭


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1 211

n n nn nnn n nn × ××

⎣ ⎦ ⎩ ⎭⎩ ⎭

D = FA {D} Displacement matrix (vector), {D} Displacement matrix (vector),

[F] Flexibility matrix,

{A} A ti t i ( t ){ } [ ]{ }D F A= {A} Action matrix (vector){ } [ ]{ }D F A

F are the flexibility coefficientsijF are the flexibility coefficients


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2. Stiffness matrix2. Stiffness matrix

oThe equilibrium equations are:

1 11 1 12 2 13 3 1... n nA S D S D S D S DA S D S D S D S D= + + + += + + + +2 21 1 22 2 23 3 2...

............................................................n nA S D S D S D S D

A S D S D S D S D

= + + + +

+ + + +1 1 2 2 3 3 ...n n n n nn nA S D S D S D S D= + + + +

oIn matrix form,

1 11 12 1 1

2 21 22 2 2

...

...n

n

A S S S DA S S S D

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬

1 2

... ... ... ... .........n n nn nn S S S DA

⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭


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1 211

n n nn nnn n nn × ××

⎣ ⎦ ⎩ ⎭⎩ ⎭

A = SD {A} Action matrix (vector) ,

{ } [ ]{ }A S D=[S] Stiffness matrix,

{D} Displacement matrix (vector) p ( )

ijS are the stiffness coefficientsijS are the stiffness coefficients

• Relationship between flexibility and stiffness matrices

{ } [ ]{ } [ ][ ]{ }A S D S F A

Relationship between flexibility and stiffness matrices

[ ] [ ] 1F S −∴

{ } [ ]{ } [ ][ ]{ }A S D S F A= =


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[ ] [ ]F S∴ =

•Example: Cantilever element3L 2L L

11 ;3LFEI

= 21 12 ;2LF FEI

= = 22LFEI

=

3 2L L3 2

1 1 23 2L LD A AEI EI

= +

2

2 1 22L LD A AEI EI

= +

3 2L L⎡ ⎤3 2

1 12

3 2L L

D AEI EID AL L

⎡ ⎤⎢ ⎥⎧ ⎫ ⎧ ⎫⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎩ ⎭ ⎩ ⎭2 2

2D AL L

EI EI⎢ ⎥⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦


42[ ]F

12EI 6EI− 4EI11 3

12 ;EISL

= 21 12 2

6 ;EIS SL

= = 224EISL

=

1 1 23 2

12 6EI EIA D DL L

= −

2 1 22

6 4EI EIA D DL L

−= +

3 2

12 6EI EIA DL L

−⎡ ⎤⎢ ⎥⎧ ⎫ ⎧ ⎫3 2

1 1

2 22

6 4A DL LA DEI EI

L L

⎢ ⎥⎧ ⎫ ⎧ ⎫= ⎢ ⎥⎨ ⎬ ⎨ ⎬−⎩ ⎭ ⎩ ⎭⎢ ⎥⎢ ⎥⎣ ⎦L L⎢ ⎥⎣ ⎦

[ ][ ] [ ][ ] [ ]1 0⎡ ⎤ [ ]S


43[ ][ ] [ ][ ] [ ]

1 00 1

F S S F I⎡ ⎤= = =⎢ ⎥

⎣ ⎦[ ]S

•Flexibility matrix and stiffness matrix are relating actions and corresponding displacements

The flexibility matrix obtained for a structure analysed by [ ]Fy y y

flexibility method may not be the inverse of the stiffness matrix

obtained for the same structure analysed by stiffness method

[ ]S

obtained for the same structure analysed by stiffness method

because different sets of actions and corresponding displacements

may be utilized in the two methods.


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Equivalent joint loads

•Analysis by flexibility and stiffness methods requires that loads must act only at joints

Equivalent joint loads

must act only at joints.

•Thus, loads acting on the members (i.e., loads that are not acting at the joints) must be replaced by equivalent loads acting at the joints.

•The loads that are determined from loads on the members are called equivalent joint loads.


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•Equivalent joint loads are added to the actual joint loads to get combined joint loads.

•Analysis carried out for combined joint loads

•Combined joint loads can be evaluated in such a manner that the resulting displacements of the structure are same as the di l t d d b th t l l ddisplacements produced by the actual loads

•This is achieved thru the use of fixed end actions to get gequivalent joint loads


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•Example 1•Example 1

Beam with actual applied loads

Applied joint loads


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Applied joint loads

Applied loads other than joint loads (To be converted to equivalent joint loads) ( o be co e ted to equ a e t jo t oads)


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Member fixed end actions (Due to applied loads other than joint loads )


49Fixed end actions for the entire beam

Equivalent joint loads (Negative of fixed end actions)Equivalent joint loads (Negative of fixed end actions)

Combined joint loads (A li d j i t l d + E i l t j i t l d )


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(Applied joint loads + Equivalent joint loads)

Fixed end actionsM

M( )2 2Mb a b

l− a b ( )2 2Ma b a

l−

3

6Mabl 3

6Mabl

M

M

4M

2l 4

M

2l

32Ml

32Ml


2l2l

120 kN40 kN/m 20 kN/m•Example 2

AB C D

/ 20 kN/m

4 m12 m 12 m 12 m

2l 2

48012wl

= 480 240 240

wl 240 120 120240

2wl

= 240 120 120

213 33 106 67213.33 106.67

Fi d d ti


5288.89 31.11

Fixed end actions

480 266 67 133.33 240

A B C D

480 266.67

B C

240240 88.89

328 89+

=120 31.11

151 11+ 120328.89= 151.11=

Equivalent joint loads(Opposite of fixed end actions)

Combined joint loads are same as equivalent joint loads here, i th l d li d t j i t di tl since there are no loads applied to joints directly


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•Superposition of combined joint loads and restraint actions gives the actual loads.the actual loads.

•Superposition of joint displacements due to the combined j i l d d i i i h di l joint loads and restraint actions gives the displacements produced by the actual loads.

•But joint displacements due to restraint actions are zero. Thus, joint displacements due to the combined joint loads give th di l t d d b th t l l dthe displacements produced by the actual loads

•But member end actions due to actual loads are obtained by ysuperimposing member end actions due to restraint actions and combined joint loads


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SummarySummary


• Definition of flexibility and stiffness influence coefficients –d l t f fl ibilit t i b h i l h &


development of flexibility matrices by physical approach & energy principle.


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module1 1 introduction-tomatrixms - rajesh sir

Engineering

gce kannur dr

t e i r

number of d d

internal indeterminacy

external indeterminacy

axial deformations

t ti b d

axial forces