Module1 flexibility-1- rajesh sir

Structural Analysis - III

Fl ibilit M th d 1Flexibility Method - 1

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

Dept. of CE, GCE Kannur Dr.RajeshKN

1

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 &

Matrix analysis of structures

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.


2

FUNDAMENTALS OF FLEXIBILITY METHOD

Introduction

•This method is a generalization of the Maxwell-Mohrmethod(1874)method(1874)

•Not conducive to computer programming, because the choice f d d t i t iof redundants is not unique

•Unknowns are the redundant actions, which are arbitrarily, ychosen


3

Flexibility method (Explanation using principle of superposition)

Example 1: Single redundant - Continuous 2-span beam

Flexibility method (Explanation using principle of superposition)

Released structure

45384B

wLEI

−Δ =

Deflection of released structure due to actual loads

Dept. of CE, GCE Kannur Dr.RajeshKN4

(Negative, since deflection is downward)

BRApply unit load corresponding to

C

3

BL

=δDisplacement due to unit load 48B EI

δDisplacement due to unit load,

BR B BR δDisplacement due to is


Deflection of released B BR δstructure due to redundant applied as a load

C

0R δΔ + = Compatibility condition (or equation of 0B B BR δΔ + = superposition or equation of geometry)

58

BB

B

wLR −Δ∴ = =

δ

δ(Displacement due to unit load corresponding to ) BR

Bδ flexibility coefficient


6

Example 2 – More than one redundant

Ch i f d d t


Choice of redundants

Q Q b th d d tL t 1 2,Q Q be the redundantsLet

Deflections corresponding to redundantsredundants

&D D Displacements in the released 1 2&QL QLD D Displacements in the released

structure corresponding to redundants, due to external loads


8

•To get flexibility coefficientsTo get flexibility coefficients

1 2&Q Q•Apply unit loads corresponding to

Fl ibilit ffi i t Flexibility coefficients

Net deflection is zero at B and C

1 11 1 12 2 0QLD F Q F Q+ + =

0D F Q F Q1 11 12 1 0

0QLD F F Q

D F F Q⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎡ ⎤

+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎩ ⎭⎩ ⎭⎣ ⎦⎩ ⎭


2 21 1 22 2 0QLD F Q F Q+ + = 2 21 22 2 0QLD F F Q⎢ ⎥⎩ ⎭⎩ ⎭⎣ ⎦⎩ ⎭

•Generally, net deflection need not be zero

1 1 11 1 12 2Q QLD D F Q F Q= + +

D D F Q F Q+ +2 2 21 1 22 2Q QLD D F Q F Q= + +

Q Qh di l di 1 2,Q QD D 1 2,Q Q•Where :support displacements corresponding to

{ } { } [ ]{ }D D F Q{ } { } [ ]{ }Q QLD D F Q= +

D⎧ ⎫ D⎧ ⎫ F F⎡ ⎤ Q⎧ ⎫{ } 1

2

QQ

Q

DD

D⎧ ⎫

= ⎨ ⎬⎩ ⎭

{ } 1

2

QLQL

QL

DD

D⎧ ⎫

= ⎨ ⎬⎩ ⎭

[ ] 11 12

21 22

F FF

F F⎡ ⎤

= ⎢ ⎥⎣ ⎦

{ } 1

2

QQ

Q⎧ ⎫

= ⎨ ⎬⎩ ⎭

F QQDFlexibility coefficient is sometimes denoted as


10

F QQDFlexibility coefficient is sometimes denoted as

{ } [ ] { } { }( )1Q QLQ F D D−= −

{ } { }0D•If there are no support displacements

{ } [ ] { }1Q F D−∴

{ } { }0QD =•If there are no support displacements,

{ } [ ] { }QLQ F D∴ = −


11

• Example: To find out redundants

: 2P P M PL PGiv n P Pe P= = = =1 2 3: 2P P M PL PGiv n P Pe P= = = =


12

313PL397PL3

11324QL

PLDEI

= 29748QL

PLDEI

=

⎡ ⎤3 269748QL

PLDEI

⎡ ⎤⎡ ⎤ = ⎢ ⎥⎣ ⎦

⎣ ⎦∴


13

⎣ ⎦

33

11 3LFEI

=

35L21

56

LFEI

=

35LF12 6F

EI=

3

2283

LFEI

= [ ]3 2 5

5 166LFEI

⎡=∴

⎤⎢ ⎥⎣ ⎦


14

22 3EI 5 166EI ⎣ ⎦

⎡ ⎤[ ] 13

16 565 27

EIFL

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦⎣ ⎦

Q⎡ ⎤[ ] [ ] 11

2QL

QQ F D

Q−⎡ ⎤⎡ ⎤= = −⎢ ⎥ ⎣ ⎦

⎣ ⎦

316 5 26⎡ ⎤ ⎡ ⎤ ⎡ ⎤3

3

16 5 2665 2 977 48

EI PLL EI

−⎡ ⎤ ⎡ ⎤−= ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

696456

P ⎡=

⎤⎢ ⎥−⎣ ⎦⎣ ⎦

1 269 8, ,P Pi e Q Q −

= =


15

1 2. ., ,56 7

i e Q Q

Temperature changes, pre-strains and p g , psupport displacements not corresponding to redundants

L t

Displacements corresponding to redundants due to { }D

Let:

Displacements corresponding to redundants due to temperature changes, in the released structure

{ }QTD

Displacements corresponding to redundants due to pre strains in the released structure

{ }QPDpre-strains, in the released structure

Di l di d d d { } Displacements corresponding to redundants due to support displacements not corresponding to redundants in the released structure

{ }QRD


16

redundants, in the released structure

{ } { } { } { } { } [ ]{ }{ } { } { } { } { } [ ]{ }Q QL QT QP QRD D D D D F Q= + + + +

{ } { } { } { } { }QC QL QT QP QRD D D D D= + + +•Let { } { } { } { } { }

{ } { } [ ]{ }Q QCD D F Q= +•Hence, and

{ } [ ] { } { }( )1Q QCQ F D D−= −


17

Member end actions

• Member end actions are the couples and forces that act at the ends of a member when it is considered to be isolated the ends of a member when it is considered to be isolated from the remainder of the structure

• In the above case, member end actions are the SFs and BMs at the ends of members AB and BC


18

• In the above figure are the member end actions considered (upward forces and

l k )

1 2 3 4, , ,M M M MA A A A

anticlockwise moments are +ve).

•The first two are just to the left of B, and the last two are just to the right of B

A A+ gives the reaction at B and 1 3M MA A+ gives the reaction at B, and

A A i th b di t t B2 4M MA A+ gives the bending moment at B


19

Joint displacements, member end actions, and

•Once the redundants are found all the joint

support reactions

Once the redundants are found, all the joint displacements, member end actions, and support reactions can be found subsequentlyq y

•But it is easier to incorporate such calculations into the b i t ti i t d f t i th t basic computations, instead of postponing them as separate calculations


20

• Example: To find outj i di l • joint displacements,

• member end actions and • reactions other than redundants• reactions other than redundants

Actual structure


1 2 3: 2Given P P M PL P P P P= = = =

{ } 1JJ

DD

D⎧ ⎫

= ⎨ ⎬⎩ ⎭

Joint displacements in the actual structured l d2JD⎩ ⎭ due to loads

Q⎧ ⎫{ } 1

2

QQ

Q⎧ ⎫

= ⎨ ⎬⎩ ⎭

Redundants


22

Joint displacements in the released structure due to loads{ } 1

2

JLJL

JL

DD

D⎧ ⎫

= ⎨ ⎬⎩ ⎭2JLD⎩ ⎭

{ } 1RLA⎧ ⎫⎨ ⎬ Reactions in the released structure due to { } 1

2

RLRL

RL

AA

⎧ ⎫= ⎨ ⎬⎩ ⎭

Reactions in the released structure due to loads


23

Joint displacements in the released structure due to i l f d dJQijD

unit values of redundantsJQij

RQijA Reactions in the released structure due to unit


24

RQijAvalues of redundants

D D D Q D Q= + +Joint displacements 1 1 11 1 12 2

2 2 21 1 22 2

J JL JQ JQ

J JL JQ JQ

D D D Q D Q

D D D Q D Q

= + +

= + +

J p

{ } { } { }D D D Q⎡ ⎤

oIn matrix form,

{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦

D D⎧ ⎫ ⎧ ⎫{ } { }1 1

2 2

, ,, J JLJ JL

J JL

D DD D

D Dw ereh

⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭

{ }11 12 1

21 22 2

,JQ JQJQ

JQ JQ

D D QD Q

D QD⎡ ⎤ ⎧ ⎫

⎡ ⎤ = = ⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭⎣ ⎦21 22 2JQ JQD QD ⎩ ⎭⎣ ⎦

D Joint displacement i in the released structure


25JQijD J p

due to unit value of redundant j

• If there are j joint displacements to be obtained,

{ } { } { }D D D Q⎡ ⎤

and there are q redundants,

{ } { } { }11 1

J JL JQqj j j q

D D D Q×× × ×

⎡ ⎤= + ⎣ ⎦


26

Reactions (other than redundants)

1 1 11 1 12 2R RL RQ RQA A A Q A QA A A Q A Q

= + +

= + +

A AA A Q⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧ ⎫

2 2 21 1 22 2R RL RQ RQA A A Q A Q= + +

11 121 1 1

21 222 2 2

RQ RQR RL

RQ RQR RL

A AA A QA AA A Q⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧ ⎫

= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎣ ⎦

{ } { } { }R RL RQA A A Q⎡ ⎤= + ⎣ ⎦{ } { } { }Q⎣ ⎦

• If there are r reactions to be obtained (other than

{ } { } { }A A A Q⎡ ⎤= + ⎣ ⎦

redundants) and q redundants,


27

{ } { } { }11 1

R RL RQqr r r q

A A A Q×× × ×

⎡ ⎤= + ⎣ ⎦

Member end actions

• Member end actions { } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦

{ }A Member end actions in the released structure

{ } { } { }M ML MQ Q⎣ ⎦

{ }MLA Member end actions in the released structure due to loads

MQA⎡ ⎤⎣ ⎦Member end actions in the released structure due to unit values of redundants


If th b d ti d d d t

{ } { } { }A A A Q⎡ ⎤

If there are m member end actions and q redundants,

{ } { } { }11 1

M ML MQqm m m q

A A A Q×× × ×

⎡ ⎤= + ⎣ ⎦


29

• In the given example,In the given example,

1 2 32P P M PL P P P P= = = =

[ ] 6964

PQ ⎡ ⎤= ⎢ ⎥

⎣ ⎦As found out earlier,


[ ]6456

Q ⎢ ⎥−⎣ ⎦,

{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦

{ } { } { }A A A Q⎡ ⎤+ ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦

{ } { } { }R RL RQA A A Q⎡ ⎤= + ⎣ ⎦

To be found out


[ ]DTo get [ ]JLDTo get

2 10PL ⎡ ⎤2

154JLPLDEI

=2

2138JLPLDEI

= [ ]2 10

138JLPLDEI

∴⎡ ⎤

= ⎢ ⎥⎣ ⎦


32

D⎡ ⎤⎣ ⎦To get JQD⎡ ⎤⎣ ⎦To get

2 23L L⎡ ⎤2

11 12

2 221 22

31 32 21 424

JQ JQJQ

JQ JQ

L LD D LEI EIDD D EIL L

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤

⎡ ⎤ = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎣ ⎦⎣ ⎦


21 22

2 2JQ JQ

EI EI⎢ ⎥ ⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦

[ ] 69PQ ⎡ ⎤= ⎢ ⎥Already we know [ ]

6456Q = ⎢ ⎥−⎣ ⎦

Already we know,

Joint displacements

⎡ ⎤ ⎡ ⎤ ⎡ ⎤{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦2 210 1 3 69

13 1 4 648 2 56PL L PEI EI

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 17PL ⎡ ⎤⎢ ⎥=

5112EI ⎢ ⎥−⎣ ⎦=


{ }RLATo get { }RL

M AM

Released structure with loads d i

AVand reactions

1 23

3 22 2 2A

PL P L PLM M P L −= − + − =

1 2 3 2AV P P P P= + − =

{ } 12

RLRL

PAA PL

⎧ ⎫⎧ ⎫ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬


{ }2 2

RLRL

PLA⎨ ⎬ ⎨ ⎬−⎩ ⎭ ⎪ ⎪⎩ ⎭

RQA⎡ ⎤⎣ ⎦To get

LL L L

R l d t t ith it l 1 Released structure with unit value of redundant Q1

2L

1L L

R l d t t ith it l

1 1A

− −⎡ ⎤⎡ ⎤ = ⎢ ⎥⎣ ⎦

Released structure with unit value of redundant Q2


2RQAL L

⎡ ⎤ = ⎢ ⎥⎣ ⎦ − −⎣ ⎦

{ } { } { }⎡ ⎤

Reactions (other than redundants)

{ } { } { }R RL RQA A A Q⎡ ⎤= + ⎣ ⎦

2 691 1PP− −⎡ ⎤⎧ ⎫ ⎡ ⎤⎪ ⎪ 69

64562

1 12L

PPL L

⎡ ⎤⎢

⎡ ⎤⎪ ⎪= +⎨ ⎬ ⎢ ⎥−− ⎣ ⎦⎪ ⎪⎩ ⎭⎥− −⎣ ⎦

107P ⎡ ⎤

2⎩ ⎭

1073156

PL

⎡=

⎤⎢ ⎥⎣ ⎦


37

Choice of member end actions

AM BMA

V VBA

AV BV

•Any number among the 4 member end actions can be

chosen for analysis

•Usually two among the 4 are chosen•Usually two among the 4 are chosen

•Any two of the 4 member end actions can be chosen for

analysis

•Usually the moments at both ends are chosen


Usually the moments at both ends are chosen

Member end actions

{ }MLATo get1MA A

L BA1M 2MA

Member end actions considered

are the reactive moments at the end of members in the actual structure

1 2,M MA A

1 2,ML MLA A are the reactive moments at the end of 1 2ML MLmembers in the released structure


PLM PL=

2P P= 3P P=1 2P P=

2PL Released

structure

2 3

2P

2P L

32PL

L

2P

B

2 P 0

L BA

1MLA = reactive moment just to the right of A2PL−

=

2MLA = reactive moment just to the left of B 32PL

=


40

2

P2

PL

L

P P

0

3MLA = reactive moment just to the right of BPL−

=0

L

B C0

3MLA reactive moment just to the right of B2

A = reactive moment just to the left of C 0=4MLA = reactive moment just to the left of C 0=


41

A B CPL 2P P PPL

A B C2

L L2P

P L3

2PL2P

2PL P P

2 2

0

L BA2

0L

B C0

⎧ ⎫ ⎧ ⎫

2 P 0 B C

{ }

1

2

23 2

ML

ML

A PLA PL

A

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬{ }

3

4

20

MLML

ML

AA PLA

= =⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭


42

4ML ⎩ ⎭⎩ ⎭

A B CLMQA⎡ ⎤⎣ ⎦To get

11L L MQA⎡ ⎤⎣ ⎦

1 21 1

2Q Q

L L= =

⎡ ⎤L 0 0

0 20L L

L− −⎡ ⎤⎢ ⎥⎢ ⎥

0

A B C2 LL L

00 0

L⎢ ⎥=

−⎢ ⎥⎢ ⎥⎣ ⎦1

1

L L 0 0⎣ ⎦

2L L L−

0


43

H b d tiHence, member end actions

{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤+ ⎣ ⎦

2 2PL L L− − −⎧ ⎫ ⎡ ⎤ 0 554PL⎧ ⎫2 23 2 0 69

PL L LPL L P

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎡ ⎤⎪ ⎪ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥⎢ ⎥ ⎣ ⎦

0.5540.357

PLL

⎧ ⎫⎪ ⎪⎪= ⎪⎨ ⎬2 0 6456

0 0 0PL L⎨ ⎬ ⎢ ⎥− − −⎢ ⎥ ⎣ ⎦⎪ ⎪ ⎢ ⎥⎪ ⎪⎩ ⎭ ⎣ ⎦

0.6430

L= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭0⎩ ⎭


Member end actions (with a different choice of member end actions for analysis)

Member end actions considered

{ }MLATo get

A A

In the released structure,

are SF and BM (equal to reactions) just to the left of B, and

1 2,ML MLA A


3 4,ML MLA A are SF and BM just to the right of B

PLM PL=

2P P= 3P P=1 2P P=

2PL Released

structure

2 3

2P3PL

2P L

32PL

L

2P

BA

2 P 0A

1MLA = Shear force just to the left of B 3 2 0P P= − =

2MLA = Bending moment just to the left of B

2 3P L PLP L M


46

23 2 2

P L M= − + =

2PL P P2

0L

B CB C

A = Shear force just to the right of B3MLA = Shear force just to the right of B

12 2 2 0P P P P= − = − =

4MLA = Bending moment just to the right of B4ML g j g

122 2PL PLPL M−

= − + +2 2

22PL PL PLPL PL− −+ +


47

22 2 2

PL PL= − + + =

A B CPL 2P P PPL

A B C2

L L

P L 3P L2P PL P

2P

2P L

2

0L

2P2

PL

L

P P

0⎧ ⎫

2 P0 0

{ }

1

2

03

2ML

ML

A PLA

A

⎧ ⎫⎪ ⎪⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪

⎨ ⎬ ⎨ ⎬{ } 2

3

4

20

MLML

ML

ML

AAA PL

⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎪ ⎪


48

4

2ML⎩ ⎭ −⎪ ⎪⎩ ⎭

A B CLMQA⎡ ⎤⎣ ⎦To get

11L L

MQA⎡ ⎤⎣ ⎦

1 21 1

1 1Q Q= =

⎡ ⎤L 0 0

1 10 L⎡ ⎤⎢ ⎥⎢ ⎥1 1

L0

L

A B C2 LL L

0 10 L

⎢ ⎥=−⎢ ⎥

⎢ ⎥⎣ ⎦1

1

L L 0 L−⎣ ⎦

2L L

L

L−

1L

1Dept. of CE, GCE Kannur Dr.RajeshKN

491 111−

Hence, member end actions

{ } { } { }A A A Q⎡ ⎤+ ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦

5⎡ ⎤

01 13PL

⎧ ⎫⎪ ⎪ ⎡ ⎤⎪ ⎪ 5

20LP⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

=

30 692

0 0 1 6456

PLL P

⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎡ ⎤⎪ ⎪ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎪ ⎪ 6456

36L⎢ ⎥⎢

⎦⎥

⎣

0 0 1 64560PL L

⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎪ ⎪ ⎢ ⎥⎪ ⎪ −⎣ ⎦−⎪ ⎪⎩ ⎭2⎪ ⎪⎩ ⎭


Flexibilities of prismatic membersp

•Flexibility coefficients of a structure are calculated from the contributions of individual members

•Hence it is worthwhile to construct member flexibility matrices for various types of actionsyp

•Member oriented axes (local coordinates) and structure oriented axes (global coordinates)


51

Member flexibility matrices for prismatic members with

B b

one end fixed and the other free3 2L L⎡ ⎤•Beam member

[ ] 11 122

21 22

3 2M MMi

M M

L LF F EI EIFF F L L

⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦21 22

2M MF F L L

EI EI⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦


[ ] LF =•Truss member [ ]MiFEA

=•Truss member


•Plane frame member Plane frame member


1

FM33

11 12 13 3 2

0 0M M M

LEAF F F

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎢ ⎥[ ]

11 12 13 3 2

21 22 23

231 32 33

03 2

M M M

Mi M M M

M M M

L LF F F FEI EI

F F FL L

⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦


231 32 33

02

M M M L LEI EI

⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

G id b •Grid member


FM31

FM33

3 2L L⎡ ⎤

[ ]11 12 13

03 2

M M M

L LEI EIF F F

L

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥[ ] 21 22 23

231 32 33

0 0Mi M M M

M M M

LF F F FGJ

F F FL L

⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥


02L LEI EI

⎢ ⎥⎢ ⎥⎣ ⎦

•Space frame member •Space frame member


0 0 0 0 0LE A

⎡ ⎤⎢ ⎥⎢ ⎥3 2

0 0 0 03 2Z Z

E AL LE I E I

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

[ ]

3 2

0 0 0 03 2

Z Z

Y Y

L LE I E I

F

⎢ ⎥−⎢ ⎥

⎢ ⎥⎢ ⎥[ ]

0 0 0 0 0M iF

LG J

⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥2

0 0 0 02 Y Y

L LE I E I

⎢ ⎥−⎢ ⎥

⎢ ⎥⎢ ⎥

2

0 0 0 02 Z Z

L LE I E I

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


Formalization of the Flexibility method(Explanation using principle of complimentary virtual work)

{ } [ ]{ }D F A

For each member,

{ } [ ]{ }Mi Mi MiD F A=

contains relative displacements of the k end { }DH contains relative displacements of the k end with respect to j end of the i-th member{ }MiDHere


•If there are m members in the structure,

{ } [ ] [ ] [ ] [ ] [ ] { }11 10 0 0 0MM MFD A⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥{ }{ }

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

{ }{ }

22 2

33 3

0 0 0 00 0 0 0

MM M

MM M

FD AFD A

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

{ } [ ] [ ] [ ] [ ] [ ] { }0 0 0 0MiMi MiFD A

⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

{ } [ ] [ ] [ ] [ ] [ ] { }0 0 0 0 MmMm MmFD A

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦

{ } [ ]{ }D F A


{ } [ ]{ }M M MD F A=

{ } [ ]{ }M M MD F A={ } [ ]{ }M M M

[ ]MF is the unassembled flexibility matrix of the entire structure

{ }MA{ }A

•Member end actions in will be related to the structure actions applied to the released structure

{ }SA { }JA { }QAconsists of joint loads and redundant actions

{ }SAstructure actions applied to the released structure.

{ }S { }J { }Qj

{ } [ ]{ }M MS SA B A=Hence,A i f i i Action transformation matrix (equilibrium matrix)

{ } [ ]{ }{ }

JM MJ MQ

AA B B

⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤= ⎨ ⎬⎣ ⎦⎣ ⎦i.e.,


{ } [ ] { }M MJ MQQA⎨ ⎬⎣ ⎦⎣ ⎦

⎪ ⎪⎩ ⎭

[ ]B { }A { }A andl t to[ ]MJB

MQB⎡ ⎤⎣ ⎦

{ }MA { }JA

{ }QA

andrelate to

relate to{ }MA

•Each column in the submatrix consists of [ ]MJB

MQB⎡ ⎤⎣ ⎦ { }Qrelate { }M

•Each column in the submatrix consists of member end actions caused by a unit value of a joint load applied to the released structure.

[ ]MJB

pp

•Each column in the submatrix consists of member end actions caused by a unit value of a

MQB⎡ ⎤⎣ ⎦member end actions caused by a unit value of a redundant applied to the released structure.


{ }SAδ•Suppose an arbitrary set of virtual actions { }SSuppose an arbitrary set of virtual actions is applied on the structure.

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

JM MS S MJ MQ

AA B A B B

A

δδ δ

δ

⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤= = ⎨ ⎬⎣ ⎦⎣ ⎦⎪ ⎪⎩ ⎭

{ } [ ]{ } [ ] { }QQAδ⎣ ⎦⎣ ⎦

⎪ ⎪⎩ ⎭

External complimentary virtual work produced by the { }SAδ { }SD

External complimentary virtual work produced by the virtual loads and actual displacements is

{ } { } { } { }TT T* JDW A D A Aδ δ δ δ

⎧ ⎫⎡ ⎤ ⎨ ⎬{ } { } { } { }T T JS S J Q

QW A D A A

Dδ δ δ δ⎡ ⎤= = ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭


Internal complimentary virtual work produced by the virtual member end actions and actual (relative) { }MAδend displacements is

{ }M

{ }MD

{ } { }T*M MU A Dδ δ=


•Equating the above two (principle of complimentary i t l k)

{ } { } { } { }T TS S M MA D A Dδ δ=

virtual work),

{ } { } { } { }S S M MA D A Dδ δ

{ } [ ]{ }{ } [ ]{ }A A { } [ ]{ }M M MD F A={ } [ ]{ }M MS SA B A=But and

{ } [ ]{ }M MS SA B Aδ δ=Also,

{ } { } { } [ ] [ ][ ]{ }TT TS S S MS M MS SA D A B F B Aδ δ=Hence, { } { } { } [ ] [ ][ ]{ }S S S MS M MS S


{ } [ ] [ ][ ]{ }TD B F B A{ } [ ] [ ][ ]{ }S MS M MS SD B F B A=

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

[ ] [ ] [ ][ ]TF B F B

Where,

the assembled flexibility matrix [ ] [ ] [ ][ ]TS MS M MSF B F B= , the assembled flexibility matrix

for the entire structure.


[ ]SF is partitioned into submatrices related to:[ ]S

{ }JA

p

joint loads

{ }QAand redundant actions

{ } [ ]{ }{ } [ ] { }JJ JQJ JF FD A

A⎡ ⎤⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎣ ⎦ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬{ } [ ]{ }

{ }{ }

[ ] { }{ }

QJ J

Q QQJ

S S S

QQ

DD AF F

F A ⎪ ⎪ ⎣ ⎦ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

=⎣ ⎦⎣ ⎦

⇒

T T⎡ ⎤ ⎡ ⎤

Where,

[ ] [ ] [ ][ ]TJJ MJ M MJF B F B= [ ] [ ]T

JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

T⎡ ⎤ ⎡ ⎤

T⎡ ⎤ ⎡ ⎤ ⎡ ⎤


[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ [ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦

{ } [ ]{ } { }D F A F A⎡ ⎤= + ⎣ ⎦{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦

{ } { } { }D F A F A⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦{ } { } { }Q QJ J QQ QD F A F A⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

1{ } { } { }1

Q QQ Q QJ JA F D F A−⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣⇒ ⎦


In the subsequent calculations, the above {AQ} should be d

However the final values of redundants are obtained

used.

However, the final values of redundants are obtained by including actual or equivalent joint loads applied directly to the supports.y pp

{ } { } { }Q QC QFINALA A A= − +Thus, { } { } { }Q QC QFINAL,

{ }QCA represents actual and equivalent joint loads applied directly to the supports, corresponding to redundants


70

redundants.

{ }{ }QA

{ }JD• Once redundants are found,

can be found out from,

{ } [ ]{ } { }D F A F A⎡ ⎤+ ⎣ ⎦

{ }J ,

{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦


71

• Similarly, support reactions caused by joint loads and redundant can be obtained with an action transformation matrix [ ]RSB[ ]RS

{ } [ ]{ } [ ]{ }JA⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤{ } [ ]{ } [ ]{ }{ }

JR RS S RJ RQ

Q

A B A B BA

⎪ ⎪⎡ ⎤⎡ ⎤= = ⎨ ⎬⎣ ⎦⎣ ⎦⎪ ⎪⎩ ⎭

•Each column in the submatrix consists of support i d b i l f j i l d li d

[ ]RJBreactions caused by a unit value of a joint load applied to the released structure.

B⎡ ⎤RQB⎡ ⎤⎣ ⎦•Each column in the submatrix consists of support reactions caused by a unit value of a


pp yredundant applied to the released structure.

• If actual or equivalent joint loads are applied • If actual or equivalent joint loads are applied directly to the supports,

{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦

t bi d j i t l d ( t l d { }RCA represents combined joint loads (actual and equivalent) applied directly to the supports.


•As seen earlier, member end actions due to actual loads ,are obtained by superimposing member end actions due to restraint actions and combined joint loads

{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦{ } { } [ ]{ } { }M MF MJ J MQ Q⎣ ⎦

{ }Awhere represents fixed end actions{ }MFAwhere represents fixed end actions


Comparison of the procedures explained with principle of superposition and principle of complimentary virtual work•For calculating redundants,

{ } [ ] { } { }( )1Q F D D−P i i l f iti

Principle of complimentary virtual work

{ } [ ] { } { }( )Q QLQ F D D= −Principle of superposition

{ } { } { }1

Q QQ Q QJ JA F D F A−⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣ ⎦


{ } { } { }Q QQ Q QJ J⎣ ⎦ ⎣ ⎦⎣ ⎦

[ ]F F⎡ ⎤⎣ ⎦H [ ] QQF F⎡ ⎤= ⎣ ⎦Hence,

{ } { }D F A⎡ ⎤d{ } { }Q A


{ } { }QL QJ JD F A⎡ ⎤= ⎣ ⎦and{ } { }QQ A=

•For calculating joint displacements,g j p

{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦Principle of superposition


{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦

{ } [ ]{ }JL JJ JD F A=Hence and{ } [ ]{ }JL JJ JD F A

JQ JQD F⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

Hence, and

Q Q⎣ ⎦ ⎣ ⎦


•For calculating member end actions,g ,

{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦Principle of superposition


{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦

{ } { } [ ]{ }ML MF MJ JA A B A= +Hence, and

MQ MQA B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦


•For calculating support reactions•For calculating support reactions,

{ } { } { }R RL RQA A A Q⎡ ⎤= + ⎣ ⎦Principle of superposition


{ } { } { }R RL RQ Q⎣ ⎦p p p

{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦

{ } { } [ ]{ }RL RC RJ JA A B A= − +Hence, and

RQ RQA B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦


Member flexibility matrix for a beam member with t t th d b d timoments at the ends as member end actions

Required to find out rotations at the ends due to unit h d l

12x d yM EI−

= = −

moments at each end separately

1

1 1x

2xM EIL dx

= =

2d1L

1L

x 2

12dy xEI Cdx L

= +

3

1 26xEIy C x CL

= + +


79

1 26L

0 Ly C −= ⇒ =0 0y C= ⇒ = 10

6x Ly C== ⇒ =2

00 0

xy C== ⇒ =

212 6

dy x Ldx EI L

⎛ ⎞∴ = −⎜ ⎟

⎝ ⎠⎝ ⎠

L Ly y−′ ′0

,6 3x x L

y yEI EI= =

= =

1 1 3MLFE I

= 21 6MLF

EI−

=6

1


1

Similarly,

LFLF −

111 3MF

EI=12 6MF

EI=

11 12 3 6L L

F F EI EI−⎡ ⎤

⎢ ⎥⎡ ⎤[ ] 11 12

21 22

3 6

6 3

M MMi

M M

F F EI EIFF F L L

EI EI

⎢ ⎥⎡ ⎤∴ = = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦


81

6 3EI EI⎢ ⎥⎣ ⎦

Member flexibility matrix for a beam member with t d h t d b d timoment and shear at one end as member end actions

3 2L LF F

⎡ ⎤⎢ ⎥⎡ ⎤[ ] 11 12

221 22

3 2M MMi

M M

F F EI EIFF F L L

⎢ ⎥⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥


2EI EI⎢ ⎥⎣ ⎦

SummarySummary

Flexibility method

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


83

James Clerk Maxwell (13 June 1831 – 5 November 1879) was a Scottish theoretical physicist and 1879) was a Scottish theoretical physicist and mathematician. His most important achievement was classical electromagnetic theory. Maxwell also developed the Maxwell–Boltzmann distribution, a p ,statistical means of describing aspects of the kinetic theory of gases. These two discoveries helped usher in the era of modern physics, laying the foundation for p y y gsuch fields as special relativity and quantum mechanics.Maxwell is also known for creating the first true colour photograph in 1861 and for his foundational work on the rigidity of rod-and-joint frameworks like those in many bridges.

Maxwell is considered by many physicists to be the 19th-century scientist with the greatest influence on 20th-century physics. His contributions to the science are considered by many to be of the same magnitude as those of Isaac Newton y y gand Albert Einstein. Einstein himself described Maxwell's work as the "most profound and the most fruitful that physics has experienced since the time of Newton." Einstein kept a photograph of Maxwell on his study wall, alongside


84

pictures of Michael Faraday and Newton.

Christian Otto Mohr (October 8, 1835 – October 2, 1918) was a German civil engineer one of the most 1918) was a German civil engineer, one of the most celebrated of the nineteenth century.Starting in 1855, his early working life was spent in railroad engineering for the Hanover and Oldenburg railroad engineering for the Hanover and Oldenburg state railways, designing some famous bridges and making some of the earliest uses of steel trusses.Even during his early railway years, Mohr had g y y y ,developed an interest in the theories of mechanics and the strength of materials. In 1867, he became professor of mechanics at Stuttgart Polytechnic, and p g yin 1873 at Dresden Polytechnic in 1873. In 1874, Mohr formalised the idea of a statically determinate structure.

In 1882, he famously developed the graphical method for analysing stress known as Mohr's circle and used it to propose an early theory of strength based on shear stress He also developed the Williot-Mohr diagram for truss on shear stress. He also developed the Williot Mohr diagram for truss displacements and the Maxwell-Mohr method for analysing statically indeterminate structures, it can also be used to determine the displacement of truss nodes and forces acting on each member. The Maxwell-Mohr method is


85

galso referred to as the virtual force method for redundant trusses.