Module1 flexibility-2-problems- rajesh sir

Structural Analysis IIIStructural Analysis - III

Fl ibilit M th d 2Flexibility Method -2ProblemsProblems

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

•Problem 1:

Static indeterminacy = 2Choose reactions at B and C as redundantsStatic indeterminacy = 2

Dept. of CE, GCE Kannur Dr.RajeshKN3

Released structure

1JA 2JA

1QA 2QA1QD 2QD

Joint actions & corresponding displacements Joint actions & corresponding displacements Redundants & corresponding displacements Reactions other than redundants


Fixed end actions


5Equivalent joint loads

Member end actions consideredMember end actions considered

2MA 4MA

1MA LA B

3MA LB C

Hence member flexibility matrix,

L L−⎡ ⎤

[ ] 11 12

21 22

3 6M MMi

L LF F EI EIFF F L L

⎡ ⎤⎢ ⎥⎡ ⎤

= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥21 22

6 3M MF F L L

EI EI⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦


6

L L−⎡ ⎤

[ ]16 12 ;MEI EIFL L

⎡ ⎤⎢ ⎥

= ⎢ ⎥−⎢ ⎥

⎢ ⎥

Member1:

12 6EI EI⎢ ⎥⎢ ⎥⎣ ⎦

L L⎡ ⎤

[ ]23 6

M

L LEI EIFL L

−⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥

Member2: [ ]

6 3L L

EI EI−⎢ ⎥

⎢ ⎥⎣ ⎦

Unassembled flexibility matrix 2 1 0 0−⎡ ⎤⎢ ⎥

[ ] 1 2 0 00 0 4 212M

LFEI

⎢ ⎥−⎢ ⎥=−⎢ ⎥

⎢ ⎥


0 0 2 4⎢ ⎥−⎣ ⎦

To find [BMS] and [BRS] matrices:

[BMS] and [BRS] are found from the released structure when it is subjected to 1 2 1 21, 1, 1, 1J J Q QA A A A= = = = separately.

1 1JA =

Q Q

2 1JA =2J

1 1QA =

Dept. of CE, GCE Kannur Dr.RajeshKN2 1QA =

1 1JA =

A B C1

1− 1 0 0

A B C0

1A 2 1JA =

1

1− 1 1− 1

A B C0

1− 1 1− 1


A B CL

1 1QA =

A B C

1

L

L− 0 0 0

A B C2L

2 1QA =1

2L− L L− 0


AJ1 AJ2 AQ1 AQ2

1 1 2L L− − − −⎡ ⎤⎢ ⎥

J1 J2 Q1 Q2=1 =1 =1 =1

[ ] [ ] 1 1 00 1 0MS MJ MQ

LB B B

L

⎢ ⎥⎢ ⎥⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦ ⎢ ⎥− −⎢ ⎥0 1 0 0⎢ ⎥⎣ ⎦

AJ1 AJ2 AQ1 AQ2=1 =1 =1 =1

[ ] [ ] 0 0 1 11 1 2RS RJ RQB B B

L L− −⎡ ⎤

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦⎣ ⎦

=1 =1 =1 =1

[ ] [ ]1 1 2Q L L⎣ ⎦⎣ ⎦ − − − −⎣ ⎦


11

[ ] [ ] [ ][ ]TS MS M MSF B F B=[ ] [ ] [ ][ ]S MS M MSF B F B

1 1 0 0 2 1 0 0 1 1 2L L⎡ ⎤⎡ ⎤ ⎡ ⎤1 1 0 0 2 1 0 0 1 1 21 1 1 1 1 2 0 0 1 1 0

L LLL

− − − − − −⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 0 0 0 4 2 0 1 012

2 0 0 0 2 4 0 1 0 0L LEIL L L

⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦


1 1 0 0 3 3 2 51 1 1 1 3 3 4

L LL LL

− − − − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥

0 0 0 0 6 0 4122 0 0 6 0 2L LEIL L L L

⎢ ⎥ ⎢ ⎥=− − −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

6 6 3 9L L⎡ ⎤[ ]

6 6 3 96 18 3 15 JJ JQ

L LF FL LL

⎡ ⎤⎢ ⎥ ⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥ ⎢ ⎥

2 2

2 2

3 3 2 5129 15 5 18

QJ QQL L L LEI F FL L L L

⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦9 15 5 18L L L L⎣ ⎦


13

R d d t

{ } { }1−⎡ ⎤⎡ ⎤ ⎡ ⎤

Redundants:

{ } { } { }1

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

is a null matrix1

1

is a null matrix{ } { }1

Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤∴ = − ⎣ ⎦ ⎣ ⎦

{ }

13 3 2 22 5 3 3112 12 12 12

L L L LPLEI EI EI EIA

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎢ ⎥ ⎢ ⎥ ⎨ ⎬{ } 3 3 2 2 295 18 9 1512 12 12 12

QAL L L LEI EI EI EI

⎢ ⎥ ⎢ ⎥= − ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦12 12 12 12EI EI EI EI⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


14

18 5 1 1 1PL ⎡ ⎤ ⎡ ⎤ ⎧ ⎫ 18 5 3P −⎡ ⎤ ⎧ ⎫{ } 18 5 1 1 1

5 2 3 5 233QPLAL

−⎡ ⎤ ⎡ ⎤ ⎧ ⎫−= ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎩ ⎭

18 5 35 2 1333

P ⎡ ⎤ ⎧ ⎫−= ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎩ ⎭

33

PP

⎧=

⎫⎨ ⎬⎩ ⎭3P−⎩ ⎭

In the subsequent calculations, the above values of {AQ}

H h fi l l f d d b i d b

q , { Q}should be used.

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

{ } { } { } 33 10 3P PA A A

P−⎧ ⎫ ⎧ ⎫+ +⎨ ⎬ ⎨ ⎬

⎧ ⎫⎨ ⎬Thus


{ } { } { } 3 2 3Q QC QFINALA A A

P P P= − + = − + =⎨ ⎬ ⎨ ⎬−⎩ ⎭

⎨⎩ ⎭ ⎩

⎬− ⎭Thus,

Joint displacements:

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

Joint displacements:

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

6 6 1 3 9 3L L P⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎧ ⎫6 6 1 3 96 18 2 3 1512 9 1

32 3

L LL PL LL LEI EI

PP

⎧ ⎫⎡ ⎤ ⎡ ⎤= +⎨ ⎬⎢ ⎥ ⎢ ⎥

⎩ ⎭⎣ ⎦ ⎣ −⎦

⎧ ⎫⎨ ⎬⎩ ⎭

2 23 2PL PL⎧ ⎫ ⎧ ⎫3 27 418 12

PL PLEI EI

⎧ ⎫ ⎧ ⎫= −⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎩ ⎭

2 0PL ⎧ ⎫⎨= ⎬


118EI ⎨⎩= ⎬

⎭

Member end actions:

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

Member end actions:

{ } { } [ ]{ } { }M MF MJ J MQ Q⎡ ⎤⎣ ⎦

1 1 23PL L L− − − −⎧ ⎫ ⎡ ⎤ ⎡ ⎤

{ }

1 1 21 1 1 00 1 2 0

33

93

2 9 3M

PLPL PPL

L LLPLA

PL

− − − −⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪

− ⎧ ⎫⎨ ⎬⎩ ⎭0 1 2 09

0 1 0 02 9 32 9PL PP

LL

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

−⎩ ⎭−

3 3 33 3 3 3

3PLPL PL PLPL PL PL PL

⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

−− − −3 3 32 9 2 9 32 9 9 0

3

23

0

PL PL PLPL PL PLPL PL

PLPL

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩

− −−

⎭ ⎩

−

⎭ ⎩ ⎭ ⎩ ⎭−


2 9 9 02 0PL PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭−

Reactions other than redundants:

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

Reactions other than redundants:

{ }A represents combined joint loads (actual and { }RCA represents combined joint loads (actual and equivalent) applied directly to the supports.

{ }2 0 0 1 1 1

1 1 2 2933R

PPLA PL L L

PP

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

⎧ ⎫⎨

⎪⎬

⎩ ⎭{ }

1 1 2 293

3L L P⎢ ⎥ ⎢ ⎥− − − −− ⎩ ⎭⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭−⎩ ⎭

22 0 0PPL PL PL

PPL

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪


183 3 3 3

−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩⎪ ⎪⎩ ⎩ ⎭ ⎩ ⎭ ⎭⎭

•Problem 2: (Same problem as above, with a different choice of member end actions)

Choose reactions at B and C as redundants

Static indeterminacy = 2

Choose reactions at B and C as redundants


19Released structure

Fixed end actions


20Equivalent joint loads

2MA

A B

4MA

LB C

1MAL B

3MAL

Member end actions considered

Hence member flexibility matrix,

3 2L LF F

⎡ ⎤⎢ ⎥⎡ ⎤[ ] 11 12

221 22

3 2M MMi

M M

F F EI EIFF F L L

⎢ ⎥⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥


2EI EI⎢ ⎥⎣ ⎦

3 2L L⎡ ⎤

[ ]1 2

6 4 ;M

L LEI EIFL L

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥

Member1:

4 2EI EI⎢ ⎥⎣ ⎦

⎡ ⎤

[ ]

3 2

2 2

3 2M

L LEI EIFL L

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥

Member2: [ ]2 2

2

M L LEI EI

⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled flexibility matrix 22 3 0 0L L⎡ ⎤⎢ ⎥

[ ] 2

3 6 0 012 0 0 4 6M

LLFEI L L

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥


0 0 6 12L⎢ ⎥⎣ ⎦


[ ] [ ]B b d i d


[ ]MSB [ ]RSBand are member end actions and support reactions in the released structure when it is subjected to unit loads corresponding to joint actions and redundantsunit loads corresponding to joint actions and redundantsseparately.

are found from the released structure

when it is subjected to [ ]MSB [ ]RSB

1 2 1 21, 1, 1, 1J J Q QA A A A= = = =i.e., and

when it is subjected to 1 2 1 2, , ,J J Q Q

separately.


23

1 1JA =

A B C1

A B C0

2 1MLA =14 0MLA =

1

01 0MLA = 3 0MLA =

2 1JA =

1

1ML

A B C1

01 1

0

1


000

L A B C

1

LL L

1 1QA =

A B C

0 0L

1 01

A B C2L

2 1QA =1L 0L 0

1

2L


1 11

AJ1 AJ2 AQ1 AQ2=1 =1 =1 =1

0 0 1 11 1 0 L⎡ ⎤⎢ ⎥⎢ ⎥

=1 =1 =1 =1

[ ] [ ] 1 1 00 0 0 10 1 0 0

M S M J M Q

LB B B ⎢ ⎥⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦0 1 0 0⎣ ⎦

[ ] [ ] 0 0 1 1B B B

− −⎡ ⎤⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]

1 1 2RS RJ RQB B BL L

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦ − − − −⎣ ⎦

[ ] [ ] [ ][ ]TS MS M MSF B F B=[ ] [ ] [ ][ ]S MS M MS

20 1 0 0 0 0 1 12 3 0 0L L⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥

2

0 1 0 1 1 1 03 6 0 01 0 0 0 0 0 0 112 0 0 4 6

LLLEI L L

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥


261 1 0 0 1 0 00 0 6 12L L⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2 20 1 0 0 3 3 2 50 1 0 1 6 6 3 9

L L L LL LL

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

21 0 0 0 12 0 6 0 41 1 0 0 12 0 6

EI L LL L

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

6 6 3 9L L⎡ ⎤⎢ ⎥ [ ]F F⎡ ⎤⎡ ⎤⎣ ⎦

2 2

6 18 3 153 3 2 512

L LLL L L LEI

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥

[ ]JJ JQ

QJ QQ

F F

F F

⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦

2 29 15 5 18L L L L⎢ ⎥⎣ ⎦

⎣ ⎦ ⎣ ⎦⎣ ⎦


Redundants

{ } { } { }1

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

{ }QD is a null matrix{ } { }1−

⎡ ⎤ ⎡ ⎤{ }

1−⎡ ⎤ ⎡ ⎤

{ } { }Q QQ QJ JA F F A⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦

13 3 2 2

3 3 2 2

2 5 3 3112 12 12 1229

L L L LPLEI EI EI EI

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎢ ⎥ ⎢ ⎥= − ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭3 3 2 2 295 18 9 1512 12 12 12

L L L LEI EI EI EI

⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

18 5 1 1 1 18 5 3PL P− −⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫− −= =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥


285 2 3 5 2 5 2 1333 33L ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

3P⎧ ⎫{ } 33Q

PP

A ⎧−

=⎫

⎨ ⎬⎩ ⎭

The final values of redundants are obtained by including y gactual or equivalent joint loads applied directly to the supports.

Th { } { } { } 3 10 33P P PA A A ⎧ ⎫ ⎧ ⎫

⎨ ⎬ ⎨ ⎬⎧ ⎫⎨ ⎬Thus, { } { } { } 3 2 3Q QC QFINAL

A A AP P P

⎧ ⎫ ⎧ ⎫= − + = + =⎨ ⎬ ⎨ ⎬

⎩ ⎭

⎧ ⎫⎨ ⎬−⎩ ⎩⎭ ⎭


29

Joint displacements

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

Joint displacements

{ } [ ]{ } { }J JJ J JQ Q⎣ ⎦

{ }6 6 1 3 96 18 2 3 1512 9 12

33J

L LL PL LDL LEI E

PI P

⎡ ⎤ ⎧ ⎫ ⎡ ⎤= +⎨

⎧ ⎫⎨ ⎬−⎬⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎩ ⎦ ⎩⎭ ⎣ ⎭⎣ ⎦ ⎩ ⎦ ⎩⎭ ⎣ ⎭

{ }2 2 2 0

113 27 41 12 88J

PL PLDEI EI

PLEI

⎧ ⎫ ⎧ ⎫= − =⎨ ⎬

⎧⎨ ⎬

⎩ ⎭ ⎩ ⎭

⎫⎨ ⎬⎩ ⎭117 41 12 88EI EI EI⎩ ⎭ ⎩ ⎭ ⎩ ⎭


30

Member end actions

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

2 0 0 1 1P⎧ ⎫ ⎡ ⎤ ⎡ ⎤

{ }

2 0 0 1 13 1 1 1 0

0 0 2 0 1339M

PPL LPLAP

PP

⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪

⎧ ⎫⎨ ⎬−⎩ ⎭0 0 2 0 1 39

2 9 0 1 0 0P PPL

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

2 0 03 3 3

23

PPL PL PL

PPL

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎧ ⎫⎪ ⎪⎪ ⎪3 3 3

0 33

2 32 9 2 9 0 0

PL PL PLP PPL PL

PLP

− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪

−⎪ ⎪⎨ ⎬⎪ ⎪⎪⎩ ⎭ ⎩ ⎭ ⎩⎩ ⎭⎭ ⎪


31

2 9 2 9 0 0PL PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪⎩ ⎭ ⎩ ⎭ ⎩⎩ ⎭⎭ ⎪

Reactions other than redundants

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

{ }A represents combined joint loads (actual and { }RCA represents combined joint loads (actual and equivalent) applied directly to the supports.

{ }2 0 0 1 1 1 3P PLA

P− − −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤= − + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

⎧ ⎫⎨ ⎬{ }

3 1 1 2 29 3RAPL L L P

= + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− − − − −⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ −⎦⎨ ⎬⎩ ⎭

{ }2

32 0 0

3 3 3R

PP

PA

PL PL PL L⎧ ⎫ ⎧ ⎫ ⎧ ⎫

= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬−⎩ ⎭

⎧ ⎫⎨

⎩ ⎩ ⎭ ⎩⎭⎬⎭


32

33 3 3 PPL PL PL L⎩ ⎭ ⎩ ⎩ ⎭ ⎩⎭ ⎭

•Problem 3:

A D

120 kN40 kN/m 20 kN/mA

B C D4 m12 m 12 m 12 m

Static indeterminacy = 2Choose reactions at C and D as redundants

A D


B C D4 m12 m 12 m 12 m


33

Released structure


B C D

/ 20 kN/m

4 m12 m 12 m 12 m

2wl 48012wl

= 480 240 240

wl 240 120 120240

2wl

= 240 120 120

213.33 106 673.33 106.67

Fixed end actions


3488.89 31.11Fixed end actions

480 266 67 133.33 240

A B C D

480 266.67

240240 88.89

328 89+

=120 31.11

151 11+ 120328.89= 151.11=

Equivalent joint loads

A

1JA 2JA3JA

4JA

AB C D

A AA A

Structure with redundants and other reactions

1QA 2QA1RA 2RA


35and joint actions

Member flexibility matrix

11 12 3 6M M

L LF F EI EI

−⎡ ⎤⎢ ⎥⎡ ⎤[ ] 11 12

21 22

3 6

6 3

M MMi

M M

F F EI EIFF F L L

EI EI

⎢ ⎥⎡ ⎤= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦

Unassembled flexibility matrix

2 1 0 0 0 01 2 0 0 0 0

−⎡ ⎤⎢ ⎥⎢ ⎥

[ ]

1 2 0 0 0 00 0 2 1 0 020 0 1 2 0 0MF

EI

−⎢ ⎥−⎢ ⎥

= ⎢ ⎥⎢ ⎥0 0 1 2 0 0

0 0 0 0 2 10 0 0 0 1 2

EI −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦


36

0 0 0 0 1 2−⎣ ⎦


are found from the released structure when [ ]MSB [ ]RSB1 1 1 1 1 1A A A A A A

andit is subjected to 1 2 3 4 1 21, 1, 1, 1, 1, 1J J J J Q QA A A A A A= = = = = =

separately. p y


37

1 1JA =

1A B C D112

1 0 0 0 0 0

1212

2 1JA =

A B C D1

12

112

0 1 0 0 0 0


38

3 1JA =

A B C D1

12

112

0 1 1− 1 0 0

1212

1A 4 1JA =

A B DA B C D1

12

112

0 1 1− 1 1− 1


1 1QA =A B C D

21

0 12 12− 0 0 0

1 1QA

A B C D2 32

2 1QA =

0 24 24− 12 12− 0


Hence action transformation matrix Hence, action transformation matrix,

AJ1 AJ2 AJ3 AJ4 AQ1 AQ21 1 1 1 1 1

1 0 0 0 0 0⎡ ⎤⎢ ⎥

=1 =1 =1 =1 =1 =1

[ ] [ ]

0 1 1 1 12 240 0 1 1 12 24

B B B

⎢ ⎥⎢ ⎥⎢ ⎥− − − −

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]0 0 1 1 0 120 0 0 1 0 12

MS MJ MQB B B⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥− −⎢ ⎥0 0 0 1 0 0⎢ ⎥⎢ ⎥⎣ ⎦


AB C DB C

1RA 2RA

AJ1 AJ2 AJ3 AJ4 AQ1 AQ21 1 1 1 1 1

[ ] [ ] 1 1 1 1 12 241RS RJ RQB B B

⎡ ⎤⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦

=1 =1 =1 =1 =1 =1

[ ] [ ]1 1 1 1 24 3612RS RJ RQB B B⎡ ⎤⎡ ⎤ ⎢ ⎥⎣ ⎦⎣ ⎦ − − − − − −⎣ ⎦


42

Redundants

{ }QD∵ is a null matrix{ } { }1

Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦

T⎡ ⎤ ⎡ ⎤ ⎡ ⎤

576 12962 ⎡ ⎤

{ }Q QQ Q⎣ ⎦ ⎣ ⎦

[ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦576 12962

1296 3456EI⎡ ⎤

= ⎢ ⎥⎣ ⎦

[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦12 24 60 60224 48 156 192EI

−⎡ ⎤= ⎢ ⎥−⎣ ⎦⎣ ⎦ ⎣ ⎦ 24 48 156 192EI ⎣ ⎦

480−⎧ ⎫⎪ ⎪

{ }1

576 1296 12 24 60 60 266.672 21296 3456 24 48 156 192 133.33QA

EI EI

− ⎪ ⎪−⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎪ ⎪∴ = − ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎪ ⎪


240⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎪ ⎪

⎪ ⎪⎩ ⎭

{ } 3456 1296 18560.2811296 576 49600 68311040QA

−⎡ ⎤ ⎧ ⎫−= ⎨ ⎬⎢ ⎥

⎣ ⎦ ⎩ ⎭{ } 1296 576 49600.68311040 −⎣ ⎦ ⎩ ⎭

{ } 138153.614515868 8

0.431104

4420 14 5186QA ⎧ ⎫

⎨ ⎬−−⎧ ⎫−

= =⎨ ⎬⎩⎩ ⎭ ⎭4515868.8311040 14.5186⎩⎩ ⎭ ⎭

{ } { } { } 151.11 0.4442 151.55120 14 5186 105 48Q QC QFINAL

A A A−⎧ ⎫ ⎧ ⎫ ⎧ ⎫

∴ = − + = − + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭ ⎩ ⎭120 14.5186 105.48FINAL − −⎩ ⎭ ⎩ ⎭ ⎩ ⎭


Member end actions

{ } { } [ ]{ } { }A A B A B A⎡ ⎤= + + ⎣ ⎦

Member end actions


480 1 0 0 0 0 0480 0 1 1 1 480 12 24

⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥− −⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥

0449

⎧ ⎫⎪ ⎪−⎪ ⎪480 0 1 1 1 480 12 24

213.33 0 0 1 1 266.67 12 24 0.4442106 67 0 0 1 1 133 33 0 12 14 5186

⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪− − − −⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎧ ⎫⎪ ⎪= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥

449449174

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪

=106.67 0 0 1 1 133.33 0 12 14.5186240 0 0 0 1 240 0 12240 0 0 0 1 0 0

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

174174

0

−⎪ ⎪⎪ ⎪⎪⎩ ⎭

⎪240 0 0 0 1 0 0

⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 0⎩ ⎭


45


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

480−⎧ ⎫⎪ ⎪240 1 1 1 1 266.67 12 24 0.44421 1

328.89 1 1 1 1 133.33 24 36 14.518612 12

⎪ ⎪−⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎧ ⎫⎪ ⎪=− + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− − − − − − − − −⎩ ⎭ ⎣ ⎦ ⎣ ⎦⎩ ⎭⎪ ⎪3 8.89 33.33 36 .5 8612 12240

⎩ ⎭ ⎣ ⎦ ⎣ ⎦⎩ ⎭⎪ ⎪⎪ ⎪⎩ ⎭

202.518380 447⎧

=⎫

⎨ ⎬⎩ ⎭380.447⎩ ⎭


46

Joint displacements

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

J p

2 1 1 1− − −⎡ ⎤⎢ ⎥

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

1 2 2 221 2 8 8EI

⎢ ⎥−⎢ ⎥=−⎢ ⎥⎢ ⎥1 2 8 14⎢ ⎥−⎣ ⎦

[ ] [ ]TF B F B⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

12 2424 482− −⎡ ⎤⎢ ⎥⎢ ⎥[ ] [ ]JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ 60 156

60 192EI

⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦


47

60 192⎣ ⎦

⎧ ⎫⎡ ⎤ ⎡ ⎤

{ }

2 1 1 1 480 12 241 2 2 2 266.67 24 48 0.44422 2

JD

− − − − − −⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

{ }1 2 8 8 133.33 60 156 14.51861 2 8 14 240 60 192

JDEI EI

+⎨ ⎬ ⎨ ⎬− − −⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪− ⎩ ⎭⎣ ⎦ ⎣ ⎦

990−⎧ ⎫⎪ ⎪5402

371EI

⎪ ⎪⎪ ⎪⎨ ⎬−⎪⎪

=⎪⎪545.378⎪ ⎪⎩ ⎭


Alternatively, if the entire [Fs] matrix is assembled at a time,

[ ] [ ] [ ][ ]TS MS M MSF B F B=

⎡ ⎤1 0 0 0 0 00 1 0 0 0 00 1 1 1 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥0 1 1 1 0 00 1 1 1 1 10 12 12 0 0 0

−⎢ ⎥= ⎢ ⎥− −⎢ ⎥⎢ ⎥−0 12 12 0 0 00 24 24 12 12 0

2 1 0 0 0 0 1 0 0 0 0 0

⎢ ⎥⎢ ⎥− −⎣ ⎦

− ⎡ ⎤⎡ ⎤2 1 0 0 0 0 1 0 0 0 0 01 2 0 0 0 0 0 1 1 1 12 24

0 0 2 1 0 0 0 0 1 1 12 242

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎢ ⎥− − − − −⎢ ⎥⎢ ⎥⎢ ⎥0 0 1 2 0 0 0 0 1 1 0 12

0 0 0 0 2 1 0 0 0 1 0 12EI

× ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎢ ⎥


0 0 0 0 1 2 0 0 0 1 0 0⎢ ⎥⎢ ⎥− ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 0 0 0 0 0 2 1 1 1 1 2 2 40 1 0 0 0 0 1 2 2 2 2 4 4 80 1 1 1 0 0 0 0 3 3 2 4 6 02

− − − − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥0 1 1 1 0 0 0 0 3 3 2 4 6 02

0 1 1 1 1 1 0 0 3 3 1 2 4 80 1 2 1 2 0 0 0 0 0 0 3 0 2 4

E I− − − − −⎢ ⎥ ⎢ ⎥

= ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −0 1 2 1 2 0 0 0 0 0 0 3 0 2 4

0 2 4 2 4 1 2 1 2 0 0 0 0 3 0 1 2⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

2 1 1 1 12 241 2 2 2 24 48

− − − − −⎡ ⎤⎢ ⎥−⎢ ⎥ [ ]

1 2 2 2 24 481 2 8 8 60 15621 2 8 14 60 192

JJ JQF F

EI F F

⎢ ⎥⎡ ⎤⎡ ⎤−⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥1 2 8 14 60 192

12 24 60 60 576 129624 48 156 192 1296 3456

QJ QQEI F F− ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

⎢ ⎥−⎢ ⎥⎣ ⎦


50

24 48 156 192 1296 3456⎢ ⎥−⎣ ⎦

•Problem 4:



51Choose horizontal reaction at D as redundant

1RA 1R

2RA3RA

Redundants [AQ] and reactions other than redundants [AR]

3R



427PL27

Fixed end actions


53

Combined (equivalent +actual) joint loads



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

⎢ ⎥⎡ ⎤= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦

Unassembled flexibility matrix 2 1 0 0 0 01 2 0 0 0 0

−⎡ ⎤⎢ ⎥−⎢ ⎥

[ ]

1 2 0 0 0 00 0 4 2 0 00 0 2 4 0 012M

LFEI

⎢ ⎥−⎢ ⎥

= ⎢ ⎥⎢ ⎥

[ ]0 0 2 4 0 0120 0 0 0 2 1

EI −⎢ ⎥⎢ ⎥−⎢ ⎥


550 0 0 0 1 2⎢ ⎥−⎣ ⎦

Joint displacements R d d tRedundantsReactions other than redundants

are member end actions found from the released structure when it is subjected to[ ]MSB [ ]RSBand


56

structure when it is subjected to

1 2 3 4 5 11, 1, 1, 1, 1, 1J J J J J QA A A A A A= = = = = = separately.

L/2 0

1A

L/20

2 1JA =

1

0 0

1/21/2


57

0 00

1

3 1JA = 00

0

10

4 1JA =0

1

0 0 4 1JA

01/L 1/L

0

0

/ 1/L

00


58

1/L 1/L

00

0

1

1

L/2 L/2

L/2

L/2

0

1 1 1QA =

1A

0 11 1Q

5 1JA =1/L

1/L

0 0

00


59

AJ1 AJ2 AJ3 AJ4 AJ5 AQ1=1 =1 =1 =1 =1 =1

1 0 0 0 0 01 2 0 0 0 2L L

⎡ ⎤⎢ ⎥⎢ ⎥

=1 =1 =1 =1 =1 =1

[ ] [ ]

1 2 0 0 0 21 2 1 0 0 2

L LL L

B B B

⎢ ⎥−⎢ ⎥⎢ ⎥− −

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]0 0 0 1 1 20 0 0 0 1 2

MS MJ MQB B BLL

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥− −⎢ ⎥

0 0 0 0 1 0⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤

AJ1 AJ2 AJ3 AJ4 AJ5 AQ1=1 =1 =1 =1 =1 =1

[ ] [ ]0 1 0 0 0 1

1 1 2 1 1 1 0RS RJ RQB B B L L L L− −⎡ ⎤

⎢ ⎥⎡ ⎤⎡ ⎤= = −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥


1 1 2 1 1 1 0L L L L⎢ ⎥− − − −⎣ ⎦

Redundants:


Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦

T3L[ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3LE I

=

[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

29 2 2 3 3 9 2L L L L L L⎡ ⎤= − −⎣ ⎦


9 2 2 3 3 9 212

L L L L LEI

⎡ ⎤⎣ ⎦

0⎧ ⎫

{ } 2

02

3 9 2 2 3 3 9 2 4 27P

EI LA L L L L L PL

⎧ ⎫⎪ ⎪⎪ ⎪− ⎪ ⎪⎡ ⎤∴ = ⎨ ⎬⎣ ⎦{ } 3 9 2 2 3 3 9 2 4 27

122 27

0

QA L L L L L PLL EI

PL⎡ ⎤∴ = − − −⎨ ⎬⎣ ⎦

⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭0⎪ ⎪⎩ ⎭

512

P−=

12

{ } { } { } 5 50Q QC QPA A A P⎛ ⎞= − + = + =⎟

−−⎜⎝ ⎠

{ } { } { } 120

12Q QC QFINALA A A+ + ⎟⎜

⎝ ⎠


Member end actions

{ } { } [ ]{ } { }A A B A B A⎡ ⎤= + + ⎣ ⎦

Member end actions


0 1 0 0 0 0 00

0 1 2 0 0 0 22

L LP

⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪2

4 27 1 2 1 0 0 2 54 272 27 0 0 0 1 1 2 12

2 27

PPL L L PPLPL L

PL

⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪− −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎛ ⎞= + +−⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎜ ⎟− ⎝ ⎠⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥

⎪ ⎪ ⎪ ⎪2 270 0 0 0 0 1 2

00 0 0 0 0 1 0

PLL

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


63

0 0 0⎧ ⎫ ⎧ ⎫ ⎡ ⎤0⎧ ⎫

⎪ ⎪0 0 00 4 1

4 27 43 108 1 5PL

PL PL PL

⎧ ⎫ ⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎛ ⎞

2424

PLPL

⎪ ⎪⎪ ⎪−⎪ ⎪4 27 43 108 1 5

2 27 2 27 1 240 0 1

PL PL PLPL PL

− −⎪ ⎪ ⎪ ⎪ ⎢ ⎥ −⎛ ⎞= + +⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎜ ⎟− ⎝ ⎠⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥

245 24

5 24

PLPL

PL

⎪ ⎪⎨ ⎬−⎪ ⎪⎪ ⎪

=

0 0 10 0 0

⎪ ⎪ ⎪ ⎪ ⎢ ⎥−⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎩ ⎭ ⎩ ⎭ ⎣ ⎦

5 240

PL⎪ ⎪⎪ ⎪⎩ ⎭





{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤+ + ⎣ ⎦0

0 0 1 0 0 0 12P⎧ ⎫⎪ ⎪⎧ ⎫ ⎧ ⎫⎡ ⎤0 0 1 0 0 0 12

520 1 1 2 1 1 1 04 2727 12

7 1 1 2 1 1 1 02 27

PP PL L L L PL

L L L L PL

⎪ ⎪− −⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎛ ⎞⎢ ⎥= − + − +−⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎝ ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦7 1 1 2 1 1 1 02 270

L L L L PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − − −⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦⎪ ⎪⎪ ⎪⎩ ⎭

15

12P

−⎧ ⎫⎪ ⎪⎨ ⎬=

127

⎨ ⎬⎪⎩ ⎪⎭


65

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

10 7 2 4 2 2L− − −⎡ ⎤

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

2

10 7 2 4 2 27 2 3 2 24 2 4 2 2

LL L L L L

L L

⎡ ⎤⎢ ⎥− −⎢ ⎥

= − − −⎢ ⎥[ ] [ ] [ ][ ]JJ MJ M MJF B F B 4 2 4 2 212

2 2 4 42 2 4 10

LEI

LL

= ⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦2 2 4 10L⎢ ⎥− −⎣ ⎦

9 2L⎧ ⎫

T⎡ ⎤ ⎡ ⎤

2

9 22LL

L

−⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪[ ] [ ]T

JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ 312

3

L LEI

L

⎪ ⎪= −⎨ ⎬⎪ ⎪⎪ ⎪


669 2L

⎪ ⎪⎪ ⎪⎩ ⎭

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

2 2

10 7 2 4 2 2 0 9 27 2 3 2 2 2 2

5

L LL L L L L P L

L L P

− − − −⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎛ ⎞54 2 4 2 2 4 27 3

12 12 122 2 4 4 2 27 3

L L PL PL LEI EI

L PL L

⎪ ⎪ ⎪ ⎪⎢ ⎥ −⎪ ⎪ ⎪ ⎪⎛ ⎞= +− − − − −⎢ ⎥⎨ ⎬ ⎨ ⎬⎜ ⎟⎝ ⎠⎢ ⎥⎪ ⎪ ⎪ ⎪− −⎢ ⎥⎪ ⎪ ⎪ ⎪

2 2 4 10 0 9 2L L⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

2

13362L

PL

−⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪{ }

2

1062592

34JD PL

EI

⎪ ⎪⎪ ⎪−⎨ ⎬⎪ ⎪−

=

⎪ ⎪


169−⎪⎩⎪ ⎪

⎪⎭


[ ] [ ] [ ][ ]TS MS M MSF B F B=

y [ ]

1 0 0 0 0 0 2 1 0 0 0 0 1 0 0 0 0 01 2 0 0 0 2 1 2 0 0 0 0 1 2 0 0 0 2

T

L L L L−⎡ ⎤ ⎡ ⎤⎡ ⎤

⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥1 2 0 0 0 2 1 2 0 0 0 0 1 2 0 0 0 21 2 1 0 0 2 0 0 4 2 0 0 1 2 1 0 0 20 0 0 1 1 2 0 0 2 4 0 0 0 0 0 1 1 2120 0 0 0 1 2 0 0 0 0 2 1 0 0 0 0 1 2

L L L LL L L LL

L LEIL L

⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − − −⎢ ⎥

= ⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − − − −0 0 0 0 1 2 0 0 0 0 2 1 0 0 0 0 1 2

0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 1 0L L⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢⎣ ⎦⎣ ⎦ ⎣ ⎦⎥

1 1 1 0 0 0 3 2 0 0 0 20 2 2 0 0 0 3 0 0 0

L LL L L L− − −⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 1 0 0 0 4 2 4 2 2 3

0 0 0 1 0 0 2 2 4 4 31 20 0 0 1 1 1 0 0 0 0 3

L LLL LE I

L

− − − −⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥


0 2 2 2 2 0 0 0 0 0 3 2L L L L L⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

2 2

10 7 2 4 2 2 9 27 2 3 2 2 2

L LL L L L L L

− − − −⎡ ⎤⎢ ⎥− −⎢ ⎥ [ ]

7 2 3 2 2 24 2 4 2 2 32 2 4 4 312

JJ JQ

L L L L L LF FL LL

L LEI F F

⎢ ⎥⎡ ⎤⎡ ⎤− − − −⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥− − ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

2 2

2 2 4 4 3122 2 4 10 9 2

9 2 2 3 3 9 2 4

QJ QQL LEI F FL L

L L L L L L

− − ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥− −⎢ ⎥− −⎣ ⎦9 2 2 3 3 9 2 4L L L L L L− −⎣ ⎦


•Problem 5: 30 kN/m

50 kN

30 kN/m

4m

B C

42m

4m

D

A



70Choose 3 reactions at D as redundants

3QA 2QA

1QAA

2RA

A

3RA

Released structure

1RA


with redundants and other reactions

60 kN 60 kN40 kNm 40 kNm

60 kN 60 kN40 kNm 40 kNm

0 k50 kN

Fixed end actions

Combined (equivalent +actual) joint loads


72

actual) joint loads


11 12 3 6M M

L LF F EI EI

−⎡ ⎤⎢ ⎥⎡ ⎤[ ] 11 12

21 22

3 6

6 3

M MMi

M M

F F EI EIFF F L L

EI EI

⎢ ⎥⎡ ⎤= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦

U bl d fl ibilit 4 2 0 0 0 02 4 0 0 0 0

−⎡ ⎤⎢ ⎥


[ ]

2 4 0 0 0 00 0 4 2 0 020 0 2 4 0 06MF

EI

⎢ ⎥−⎢ ⎥−⎢ ⎥

= ⎢ ⎥[ ]0 0 2 4 0 060 0 0 0 2 10 0 0 0 1 2

M EI ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦


73

0 0 0 0 1 2⎢ ⎥−⎣ ⎦

40A = 40A =2 40JA = 3 40JA =

1 50JA =2JD 3JD

1 50JA1JD

Joint displacements and di j i t l dcorresponding joint loads

are found from the released structure when it is subjected to 1 2 3 1 2 31, 1, 1, 1, 1, 1J J J Q Q QA A A A A A= = = = = =

t l

[ ]MSB [ ]RSBand


74

separately.

0 0 1 0

1A =

00

00

1 0

0 01 1JA =

2 1JA =

0 0

4 0 01−4 01−

0

0 0


75

1 0

1

44−

0

3 1JA =

1− 1 4− 0

3J

00 0

0

0

1A04−

0

0

01− 1 1QA =1−


2 2− 1−1

2− 2 11−

2 1QA = 12Q

1−

0

23 1QA =01−

0

120


AJ1 AJ2 AJ3 AQ1 AQ2 AQ3=1 =1 =1 =1 =1 =1 4 1 1 4 2 10 1 1 4 2 1

− − − −⎡ ⎤⎢ ⎥⎢ ⎥

[ ] [ ]

0 1 1 4 2 10 0 1 4 2 10 0 1 0 2 1MS MJ MQB B B

⎢ ⎥⎢ ⎥− − − −

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]0 0 1 0 2 10 0 0 0 2 1

MS MJ MQ ⎢ ⎥⎣ ⎦⎣ ⎦⎢ ⎥⎢ ⎥− −⎢ ⎥0 0 0 0 0 1⎢ ⎥⎢ ⎥⎣ ⎦

A A A A A A

0 0 0 1 0 0−⎡ ⎤

AJ1 AJ2 AJ3 AQ1 AQ2 AQ3=1 =1 =1 =1 =1 =1

[ ] [ ]0 0 0 1 0 01 0 0 0 1 0

4 1 1 4 2 1RS RJ RQB B B

⎡ ⎤⎢ ⎥⎡ ⎤⎡ ⎤= = − −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥


4 1 1 4 2 1⎢ ⎥− − − −⎣ ⎦

Redundants:


Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦

T256 48 72

2⎡ ⎤⎢ ⎥[ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦2 48 72 30

672 30 30

EI⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

96 48 72−⎡ ⎤[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

96 48 722 16 0 24

624 12 24

EI

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦24 12 24−⎢ ⎥⎣ ⎦


79

{ } 1⎡ ⎤ ⎡ ⎤{ } { }1

Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦

1260 720 3774 38401 720 2496 4224 1760

− −⎡ ⎤⎧ ⎫− ⎪ ⎪⎢ ⎥⎨ ⎬

1053 333

⎧ ⎫⎪ ⎪⎨ ⎬720 2496 4224 1760

875523774 4224 16128 720

⎢ ⎥= − ⎨ ⎬⎢ ⎥⎪ ⎪− − −⎢ ⎥⎣ ⎦⎩ ⎭

53.33353.333−⎨ ⎬⎪ ⎪⎩ ⎭

=

60 10 70−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪

⎧ ⎫⎪ ⎪{ } { } { } 0 53.333

0 553.333

53.3333.333Q QC QFINAL

A A A ⎪ ⎪ ⎪ ⎪= − + = − + − =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭

⎪ ⎪−⎨ ⎬⎪⎩ ⎩⎭ ⎪⎭0 5 53.3333.333⎩ ⎭ ⎩ ⎩⎭ ⎭


80


{ } { } [ ]{ } { }A A B A B A⎡ ⎤= − + + ⎣ ⎦



60 0 0 0 50 1 0 0 10⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫{ }

60 0 0 0 50 1 0 0 100 1 0 0 40 0 1 0 53.3330 4 1 1 40 4 2 1 53 333

RA− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫

⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= − + − − + − −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭0 4 1 1 40 4 2 1 53.333⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − −⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

503.333⎧ ⎫

= ⎪ ⎪⎨ ⎬3.333

0⎨ ⎬⎪ ⎪⎩ ⎭


81

Member end actions


Member end actions

{ } { } [ ]{ } { }M MF MJ J MQ Q⎣ ⎦

0 4 1 1 4 2 1− − − −⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥

2.304⎧ ⎫0 0 1 1 4 2 1

50 1040 0 0 1 4 2 1

40 53 333

⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎧ ⎫ ⎧ ⎫

− − − −⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪+ +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

15.63615.636

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪⎨ ⎬40 53.333

40 0 0 1 0 2 140 53.333

0 0 0 0 0 2 1

= + − + −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎢ ⎥ ⎢ ⎥

54.64854.648

⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪

=

0 0 0 0 0 0 1⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 52.018

⎪ ⎪⎩ ⎭


82

Joint displacements

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

Joint displacements

{ } [ ]{ } { }J JJ J JQ Q⎣ ⎦

[ ] [ ] [ ][ ]T64 24 24

2− −⎡ ⎤

⎢ ⎥[ ] [ ] [ ][ ]TJJ MJ M MJF B F B= 2 24 12 12

624 12 24

EI⎢ ⎥= −⎢ ⎥−⎢ ⎥⎣ ⎦

T⎡ ⎤ ⎡ ⎤96 16 24

2− −⎡ ⎤⎢ ⎥[ ] [ ]T

JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦2 48 0 12

672 24 24

EI⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


83

{ }64 24 24 50 96 16 24 10

2 2− − − −⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫

⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥{ } 2 224 12 12 40 48 0 12 53.3336 6

24 12 24 40 72 24 24 53.333JD

EI EI⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥∴ = − − + −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎢ ⎥⎣ ⎦⎩ ⎭ ⎣ ⎦⎩ ⎭

3200 3093.3 35.561 26.67

2 106.682 11200 1120 80.00

6 3E II E EI

−⎛ ⎞⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎜ ⎟= − + = − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎧ ⎫⎪ ⎪−⎨ ⎬⎪ ⎪0

6 3720 720 0

E II E EI⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎜ ⎟−⎩ ⎭ ⎩ ⎭ ⎩ ⎭

⎨ ⎬⎪ ⎪⎩⎠ ⎭⎝


84

[ ] [ ] [ ][ ]TS MS M MSF B F B=

Alternatively, if the entire [Fs] matrix is assembled at a time,[ ] [ ] [ ][ ]S MS M MSF B F B

4 1 1 4 2 1 4 2 0 0 0 0 4 1 1 4 2 10 1 1 4 2 1 2 4 0 0 0 0 0 1 1 4 2 1

T− − − − − − − − −⎡ ⎤ ⎡ ⎤⎡ ⎤

⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 1 4 2 1 0 0 4 2 0 0 0 0 1 4 2 120 0 1 0 2 1 0 0 2 4 0 0 0 0 1 0 2 160 0 0 0 2 1 0 0 0 0 2 1 0 0 0 0 2 1

EI

⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − − − − − − −⎢ ⎥

= ⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − − − −⎢ ⎥ ⎢ ⎥⎢ ⎥0 0 0 0 0 1 0 0 0 0 1 2 0 0 0 0 0 1⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

4 0 0 0 0 0 1 6 6 6 2 4 4 61 1 0 0 0 0 8 6 6 2 4 4 6

− − − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥1 1 0 0 0 0 8 6 6 2 4 4 6

1 1 1 1 0 0 0 0 6 1 6 1 2 624 4 4 0 0 0 0 0 6 8 1 2 66

2 2 2 2 2 0 0 0 0 0 4 3E I

− −⎢ ⎥ ⎢ ⎥⎢ ⎥− − − − − −⎢ ⎥

= ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥2 2 2 2 2 0 0 0 0 0 4 3

1 1 1 1 1 1 0 0 0 0 2 3

⎢ ⎥ ⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎣ ⎦⎣ ⎦

6 4 2 4 2 4 9 6 1 6 2 4− − − −⎡ ⎤⎢ ⎥

[ ]2 4 1 2 1 2 4 8 0 1 22 4 1 2 2 4 7 2 2 4 2 429 6 4 8 7 2 2 5 6 4 8 7 26

1 6 0 2 4 4 8 7 2 3 0

J J J Q

Q J Q Q

F F

E I F F

⎢ ⎥−⎢ ⎥⎡ ⎤⎡ ⎤−⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥− ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

⎢ ⎥


1 6 0 2 4 4 8 7 2 3 02 4 1 2 2 4 7 2 3 0 3 0

⎢ ⎥⎢ ⎥−⎣ ⎦

•Problem 6 (Support settlement)

Analyse the beam. Support B has a downward settlement of 30mm. EI=5.6×103 kNm2×



Choose reactions at B,C,D as redundants

[ ] 3 6Mi

L LEI EIF

−⎡ ⎤⎢ ⎥

= ⎢ ⎥Member flexibility matrix [ ]

6 3

MiFL L

EI EI

⎢ ⎥−⎢ ⎥

⎢ ⎥⎣ ⎦

2 1 0 0 0 01 2 0 0 0 0

−⎡ ⎤⎢ ⎥


[ ]

1 2 0 0 0 00 0 4 2 0 030 0 2 4 0 06MF

EI

⎢ ⎥−⎢ ⎥−⎢ ⎥

= ⎢ ⎥[ ]0 0 2 4 0 060 0 0 0 2 10 0 0 0 1 2

M EI ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦


87

0 0 0 0 1 2−⎣ ⎦

2525

Fixed end actions 50 502525 2525

Equivalent joint loads 50 50

1JD 2JD3JD

Joint displacements { }JD

A C D30mm 1QD

Dept. of CE, GCE Kannur Dr.RajeshKNSupport settlements { }QD

B

AJ1 AJ2 AJ3 AQ1 AQ2 AQ31 1 1 1 1 1

1 1 1 3 9 121 1 1 0 6 9− − − − − −⎡ ⎤⎢ ⎥⎢ ⎥

=1 =1 =1 =1 =1 =1

[ ] [ ]

1 1 1 0 6 90 1 1 0 6 90 1 1 0 0 3MS MJ MQB B B

⎢ ⎥⎢ ⎥− − − −

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦⎢ ⎥0 1 1 0 0 3

0 0 1 0 0 30 0 1 0 0 0

⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦0 0 1 0 0 0⎢ ⎥⎣ ⎦

[ ] [ ] [ ]RS RJ RQB B B⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦[ ] [ ] [ ]RS RJ RQ⎣ ⎦⎣ ⎦


89

Redundants:1

{ }

{ } { } { }1

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

{ }QD is NOT a null matrix

[ ]TF B F B

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤

0.00883900.00642900.0016070

[ ]QQ MQ M MQF B F B ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦0 10290000 06509000 0088390

0.06509000.04339000.0064290

⎣ ⎦

⎡ ⎤⎢ ⎥

0.10290000.06509000.0088390

0 00080360 00080360 0008036

[ ][ ]T

QJ MQ M MJF B F B⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ ⎢ ⎥

0.00723200.00723200.0040180

0.00080360.00080360.0008036


90

⎢ ⎥⎢ ⎥⎣ ⎦

0.01286000.01205000.0056250

{ } { } { }1A F D F A

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

10.03 00 25F F

−−⎛ ⎞⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎜ ⎟⎡ ⎤ ⎡ ⎤= − −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟

62.329 4−⎧ ⎫⎪ ⎪⎨ ⎬0 25

0 25QQ QJF F⎡ ⎤ ⎡ ⎤= − −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟

⎪ ⎪ ⎪ ⎪⎜ ⎟⎩ ⎭ ⎩ ⎭⎝ ⎠

29.413.4

= ⎨ ⎬⎪ ⎪−⎩ ⎭

0 62.3 62.3−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪

−⎧ ⎫⎪ ⎪{ } { } { } 50 29.4

50 13 479.436 6

Q QC QFINALA A A

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= − + = − − + =⎨ ⎬ ⎨ ⎬⎪

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩⎪ ⎪ ⎪− −⎭ ⎩ ⎭ ⎭⎩

Thus, 50 13.4 36.6⎩⎭ ⎩ ⎭ ⎭⎩


91

Member end actions

{ } { } [ ]{ } { }⎡ ⎤

Member end actions


83.755 4

00

⎧ ⎫⎪ ⎪⎪ ⎪

⎧ ⎫⎪ ⎪⎪ ⎪

[ ]{ } { }55.455.440 3

000 MJ J MQ QB A B A

⎪ ⎪⎪ ⎪

⎡ ⎤= + + =⎨ ⎬ ⎣ ⎦⎪

⎪ ⎪−⎪ ⎪⎨ ⎬⎪ ⎪⎪

[ ]{ } { } 40.340.3

0

02525

Q Q⎣ ⎦⎪ −⎪ ⎪

⎪ ⎪⎪

⎪⎪ ⎪⎪ ⎪

⎩⎩ ⎭⎭⎪

025⎪⎪ ⎪⎩⎩ ⎭− ⎭

⎪


92

Reactions other than redundantsReactions other than redundants


93

Joint displacements

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

Joint displacements

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

JJ MJ M MJF B F B=

[ ] [ ]TJQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

[ ]62.30

29.425JJ JQF F−⎧ ⎫ ⎧ ⎫

⎪ ⎪ ⎪ ⎪⎡ ⎤= +−⎨ ⎬ ⎨ ⎬⎣ ⎦[ ]13.425

JJ JQ⎨ ⎬ ⎨ ⎬⎣ ⎦⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

⎧ ⎫3

3

7.5 100.5 10

−

−

⎧ ⎫− ×⎪ ⎪×⎨ ⎬⎪

=⎪


9433.1 10−⎪ ×⎩⎪⎭

Alternatively if the entire [Fs] matrix is assembled at a time

[ ] [ ] [ ][ ]TS MS M MSF B F B=


1 1 1 3 9 12 2 1 0 0 0 0 1 1 1 3 9 121 1 1 0 6 9 1 2 0 0 0 0 1 1 1 0 6 9

T− − − − − − − − − − − − −⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥1 1 1 0 6 9 1 2 0 0 0 0 1 1 1 0 6 9

0 1 1 0 6 9 0 0 4 2 0 0 0 1 1 0 6 930 1 1 0 0 3 0 0 2 4 0 0 0 1 1 0 0 36EI

⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − − − − − − −⎢ ⎥

= ⎢ ⎥ ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢⎢ ⎥

⎥⎥0 0 1 0 0 3 0 0 0 0 2 1 0 0 1 0 0 3

0 0 1 0 0 0 0 0 0 0 1 2 0 0 1 0 0 0

⎢ ⎥ ⎢⎢ ⎥− − − − −⎢ ⎥ ⎢⎢ ⎥−⎢ ⎥ ⎢⎣ ⎦⎣ ⎦ ⎣ ⎦

⎥⎥⎥

⎡ ⎤⎡ ⎤0.01205000.00723200.00080360.00160700.00160700.0005357

0.00562500.00401800.00080360.00053570.00053570.0005357⎡ ⎤⎢ ⎥⎢ ⎥

[ ]JJ JQ

QJ QQ

F F

F F

⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦0.06509000.04339000.00642900.00723200.00723200.0040180

0.00883900.00642900.00160700.00080360.00080360.0008036

0.01286000.00723200.00080360.00214300.00160700.0005357⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥


95

0.10290000.06509000.00883900.01286000.01205000.0056250⎢ ⎥⎢ ⎥⎣ ⎦

•Problem 6a ( Same problem as above, with a different Problem 6a ( Same problem as above, with a different approach)

25 28

A D

25 25

6 03EI

28

BC

2

6 .03 1123

EI ×=

2

6 .03 286

EI ×=

112

12 .03EI ×74.667

3

12 .036

EI ×9.333 50 50

3374.667=

69.333=

Fixed end actions


96

253

A

B

C D112

84

74 6679.333 50+ 50

74.667 74.667 9.333+

Equivalent joint loads


97

L L−⎡ ⎤

[ ] 3 6Mi

L LEI EIFL L

⎡ ⎤⎢ ⎥

= ⎢ ⎥−⎢ ⎥


6 3L L

EI EI⎢ ⎥⎢ ⎥⎣ ⎦

Unassembled flexibility matrix2 1 0 0 0 01 2 0 0 0 0

−⎡ ⎤⎢ ⎥⎢ ⎥


[ ]

1 2 0 0 0 00 0 4 2 0 030 0 2 4 0 06MF

EI

⎢ ⎥−⎢ ⎥−⎢ ⎥

= ⎢ ⎥[ ]0 0 2 4 0 060 0 0 0 2 1

M EI ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥


98

0 0 0 0 1 2⎢ ⎥−⎣ ⎦

AJ1 AJ2 AJ3 AQ1 AQ2 AQ3=1 =1 =1 =1 =1 =1

1 1 1 3 9 121 1 1 0 6 9− − − − − −⎡ ⎤⎢ ⎥⎢ ⎥

[ ] [ ] 0 1 1 0 6 90 1 1 0 0 3MS MJ MQB B B

⎢ ⎥⎢ ⎥− − − −

⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦⎢ ⎥0 1 1 0 0 3

0 0 1 0 0 30 0 1 0 0 0

⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦

(Same as in the previous approach) 0 0 1 0 0 0⎢ ⎥⎣ ⎦


99

Redundants:

{ } { } { }1A F D F A

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

[ ]TF B F B

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤

0.00883900.00642900.0016070

[ ]QQ MQ M MQF B F B ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦0.10290000.06509000.0088390

0.06509000.04339000.0064290

⎣ ⎦

⎡ ⎤⎢ ⎥0 00080360 00080360 0008036

[ ][ ]T

QJ MQ M MJF B F B⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ ⎢ ⎥

0.00723200.00723200.0040180

0.00080360.00080360.0008036


100

⎢ ⎥⎢ ⎥⎣ ⎦

0.01286000.01205000.0056250

0 84−⎛ ⎞⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎜ ⎟

21.7⎧ ⎫⎪ ⎪

{ } 10 30 25

Q QQ QJA F F− ⎪ ⎪ ⎪ ⎪⎜ ⎟⎡ ⎤ ⎡ ⎤= −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟

⎪ ⎪ ⎪ ⎪⎜ ⎟⎩ ⎭⎩ ⎭⎝ ⎠

2013.4

⎪ ⎪⎨ ⎬⎪−⎩

=⎪⎭⎩ ⎭⎩ ⎭⎝ ⎠

Note the differences hereNote the differences here.

{ } { } { }84 21.7 62.3⎧ ⎫ ⎧ ⎫

⎪ ⎪ ⎪ ⎪−⎧ ⎫⎪ ⎪{ } { } { } 59.4 20

5079.436.613.4

Q QC QFINALA A A ⎪ ⎪ ⎪ ⎪=− + =− − + =⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪− −

⎪ ⎪⎨ ⎬⎪⎩⎩ ⎭ ⎩ ⎭ ⎪⎭

Thus,


101Note the difference here.

M b d ti


Member end actions

{ } { } [ ]{ } { }Q Q⎣ ⎦

112⎧ ⎫⎪ ⎪

83.7⎧ ⎫⎪ ⎪

{ } [ ]{ } { }112

28

⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪

⎡ ⎤⎨ ⎬

55.455.4

⎧ ⎫⎪ ⎪⎪ ⎪−⎪ ⎪⎨ ⎬{ } [ ]{ } { }28

25

M MJ J MQ QA B A B A⎪ ⎪

⎡ ⎤= + +⎨ ⎬ ⎣ ⎦−⎪ ⎪⎪ ⎪

40.340 3

⎪ ⎪⎨ ⎬−⎪⎪

=⎪⎪25

25⎪ ⎪⎪ ⎪−⎩ ⎭

40.30

⎪⎪⎩

⎪⎪⎭

Note the difference here.


102

Reactions other than redundantsReactions other than redundants


103

{ }⎡ ⎤

Joint displacements

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

[ ]21.784203JJ JQF F

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎡ ⎤= +⎨ ⎬ ⎨ ⎬⎣ ⎦[ ] 203

13.425JJ JQF F⎡ ⎤+⎨ ⎬ ⎨ ⎬⎣ ⎦

⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

3

3

7.5 100 5 10

−

−

⎧ ⎫− ×⎪ ⎪×⎨ ⎬=

3

0.5 103.1 10−

×⎨ ⎬⎪ ×⎩

=⎪⎭


104

•Problem 7:

Static indeterminacy = 2 ( 1 internal + 1 external )


Choose reaction at B and force in AD as redundants

Released structureReleased structure




Member flexibility matrix of truss member

⎡ ⎤[ ]MiLF

EA⎡ ⎤= ⎢ ⎥⎣ ⎦


1 0 0 0 0 00 1.414 0 0 0 0⎡ ⎤⎢ ⎥⎢ ⎥

[ ] 0 0 1.414 0 0 00 0 0 1 0 0M

LFEA

⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥0 0 0 0 1 00 0 0 0 0 1

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


108

⎣ ⎦


are found from the released structure when it [ ]MSB [ ]RSBand

is subjected to 1 2 3 4 1 21, 1, 1, 1, 1, 1J J J J Q QA A A A A A= = = = = =separately.

3JDD 3JD2JD

1JD 5

2 3

D

1 4

4JD6


109Joint displacements

1 1JA = 2 1JA =

1

1 111A = 13 1JA =

4 1JA =1


1101

1A =1 1QA =2 1QA =

1

1 11 1


111

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

[ ]

•Each column in the submatrix consists of member forces caused by a unit value of a redundant applied to the

MQB⎡ ⎤⎣ ⎦

released structure.AJ1 AJ2 AJ3 AJ4 AQ1 AQ2=1 =1 =1 =1 =1 =1

0 1 0 0 0 0.7070 0 0 0 0 1

−⎡ ⎤⎢ ⎥⎢ ⎥

=1 =1 =1 =1 =1 =1

0 0 0 0 0 11.414 0 0 0 1.414 1

⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥[ ] [ ]MS MJ MQB B B⎡ ⎤⎡ ⎤= ⎣ ⎦⎣ ⎦ 1 0 1 0 1 0.7071 0 0 0 0 0.707

= ⎢ ⎥− − −⎢ ⎥⎢ ⎥− −⎢ ⎥

[ ] [ ]MS MJ MQ⎣ ⎦⎣ ⎦


0 0 0 1 0 0.707⎢ ⎥

−⎢ ⎥⎣ ⎦

•Each column in the submatrix consists of support [ ]RJB•Each column in the submatrix consists of support reactions caused by a unit value of a joint load applied to the released structure.

[ ]RJ

RQB⎡ ⎤⎣ ⎦•Each column in the submatrix consists of Q⎣ ⎦support reactions caused by a unit value of a redundant applied to the released structure.

AJ1 AJ2 AJ3 AJ4 AQ1 AQ2

1 0 0 1 1 0− − −⎡ ⎤

AJ1 AJ2 AJ3 AJ4 AQ1 AQ2=1 =1 =1 =1 =1 =1

[ ] [ ]1 0 0 1 1 01 1 0 0 1 0

1 0 1 0 1 0RS RJ RQB B B

⎡ ⎤⎢ ⎥⎡ ⎤⎡ ⎤= = − − −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥


113

1 0 1 0 1 0⎢ ⎥−⎣ ⎦

Redundants


Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦ { }Q{ } { }Q QQ QJ J⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

3 828 2 707⎡ ⎤[ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦3.828 2.7072.707 4.828

LEA

⎡ ⎤= ⎢ ⎥

⎣ ⎦

[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

3.828 0 1 0L −⎡ ⎤⎢ ⎥3.414 0.707 0.707 0.707EA

= ⎢ ⎥− − −⎣ ⎦


1−⎧ ⎫⎪ ⎪

{ }13.828 2.707 3.828 0 1 0 2

2.707 4.828 3.414 0.707 0.707 0.707 0QEA PLAL EA

− ⎪ ⎪− −⎡ ⎤ ⎡ ⎤− ⎪ ⎪∴ = ⎨ ⎬⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎪ ⎪0

⎪ ⎪⎪ ⎪⎩ ⎭

⎧ ⎫1.1720.243

P ⎧=

⎫⎨ ⎬−⎩ ⎭

(There are no loads applied directly to the supports.)


Member forces

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

0 1 0 0 0 0 707⎡ ⎤ ⎡ ⎤ 1.828−⎧ ⎫0 1 0 0 0 0.7070 0 0 0 1 0 1

1 414 0 0 0 2 1 414 1 1 172

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎧ ⎫⎢ ⎥ ⎢ ⎥⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎪ ⎪

1.8280.243

0

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪1.414 0 0 0 2 1.414 1 1.172

1 0 1 0 0 1 0.707 0.2431 0 0 0 0 0 0 707

P P⎪ ⎪−⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎪ ⎪= +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− − − −⎩ ⎭⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭

00

1 172

P⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪

=

1 0 0 0 0 0 0.7070 0 0 1 0 0.707

⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎩ ⎭⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

1.1720.172

⎪ ⎪⎪ ⎪⎩ ⎭


116


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

1⎧ ⎫11 0 0 1 1 0

2 1.1721 1 0 0 1 0

0 0 243P P

−⎧ ⎫− − −⎡ ⎤ ⎡ ⎤⎪ ⎪− ⎧ ⎫⎪ ⎪⎢ ⎥ ⎢ ⎥= − − + −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭0 0.2431 0 1 0 1 0

0

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

0.1721 828 P−⎧ ⎫

⎪ ⎪⎨ ⎬= 1.828

0.172P⎨ ⎬

⎪ ⎪⎩ ⎭

=


117

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

4.828 0 1 00 1 0 0

−⎡ ⎤⎢ ⎥

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

0 1 0 01 0 1 0

LEA

⎢ ⎥⎢ ⎥=−⎢ ⎥

⎢ ⎥0 0 0 1⎢ ⎥⎣ ⎦

3.828 3.414⎡ ⎤

[ ] [ ]TJQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

3.828 3.4140 0.7071 0 707

LEA

⎡ ⎤⎢ ⎥−⎢ ⎥=− −⎢ ⎥1 0.7070 0.707

EA − −⎢ ⎥⎢ ⎥−⎣ ⎦1.172−⎧ ⎫

⎪ ⎪{ }

1.8280J

PA

D LE

⎪∴ =

⎪−⎪ ⎪⎨ ⎬⎪ ⎪


0.172⎪ ⎪⎪ ⎪⎩ ⎭


[ ] [ ] [ ][ ]TS MS M MSF B F B=

T0 1 0 0 0 0 707⎡ ⎤0 1 0 0 0 0.7070 0 0 0 0 1

1 414 0 0 01 414 1

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥1.414 0 0 01.414 1

1 0 1 0 1 0.7071 0 0 0 0 0 707

⎢ ⎥= ⎢ ⎥− − −⎢ ⎥⎢ ⎥1 0 0 0 0 0.707

0 0 0 1 0 0.707

⎢ ⎥− −⎢ ⎥

−⎢ ⎥⎣ ⎦

⎡ ⎤⎡ ⎤1 0 0 0 0 0 0 1 0 0 0 0.7070 1.414 0 0 0 0 0 0 0 0 0 10 0 1 414 0 0 0 1 414 0 0 01 414 1

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥0 0 1.414 0 0 0 1.414 0 0 01.414 1

0 0 0 1 0 0 1 0 1 00 0 0 0 1 0 1 0 0 0

LEA

⎢ ⎥× ⎢ ⎥ −⎢ ⎥

⎢ ⎥1 0.707

0 0 707

⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥


0 0 0 0 1 0 1 0 0 00 0 0 0 0 1 0 0 0 1⎢ ⎥ −⎢ ⎥⎣ ⎦

0 0.7070 0.707

⎢ ⎥−⎢ ⎥

−⎢ ⎥⎣ ⎦

4 828 0 1 0 3 828 3 414−⎡ ⎤4.828 0 1 0 3.828 3.4140 1 0 0 0 0.7071 0 1 0 1 0 707

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥[ ] 1 0 1 0 1 0.707

0 0 0 1 0 0.707SLF

EA− − −⎢ ⎥

= ⎢ ⎥−⎢ ⎥3.828 0 1 0 3.828 2.7073.414 0.707 0.707 0.707 2.707 4.828

⎢ ⎥⎢ ⎥−⎢ ⎥− − −⎣ ⎦⎣ ⎦

[ ]JJ JQF F⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=QJ QQF F

⎢ ⎥=⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦


120

•Problem 8:



121Choose the force in CD as redundant

1JA =

1JD

2JA =2JD

2JADJ1 and DJ2

: Joint displacements

M b fl ibilit

Released structure with loads

Member flexibility matrix of truss member: [ ]Mi

LFEA

=

5 0 0⎡ ⎤[ ]

5 0 01 0 2 0MF

EA

⎡ ⎤⎢ ⎥= ⎢ ⎥Unassembled flexibility matrix:


1220 0 5

EA ⎢ ⎥⎢ ⎥⎣ ⎦

are found from the released structure when [ ]MSB [ ]RSBand are found from the released structure when it is subjected to separately. 1 2 11, 1, 1J J QA A A= = =[ ]MSB [ ]RSBand

1 1JA =

2 1JA =


123

1 1QA =1 1QA


124

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

[ ]MJ

•Each column in the submatrix consists of member forces caused by a unit value of a redundant applied to

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

[ ] [ ]0.833 0.625 0.625

0 0 1MS MJ MQB B B−⎡ ⎤

⎢ ⎥⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦ ⎢ ⎥[ ] [ ]0.833 0.625 0.625

MS MJ MQ⎣ ⎦⎣ ⎦ ⎢ ⎥− −⎢ ⎥⎣ ⎦

•In this problem, since support reactions can be easily found out once the forces in members are obtained,


matrix need not be assembled. [ ]RSB

Redundants:


Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦

[ ]TF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦[ ]QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦

5 0 0 0.625⎡ ⎤ ⎡ ⎤[ ] 1 0 2 00.625 1 0.625 1

0 0 5 0 625EA

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

5.906EA

=

0 0 5 0.625⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

0⎧ ⎫[ ][ ]T

QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦01

3.906EA⎧ ⎫

= ⎨ ⎬−⎩ ⎭


1{ } { }1

Q QQ QJ JA F F A−

⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦

[ ]151 0 3.906105.906

EAEA

⎧ ⎫−= − ⎨ ⎬

⎩ ⎭⎩ ⎭

6.614 kN=


Member forces:


Member forces:

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

0.833 0.625 0.62515

0 0 1 6.61410

−⎡ ⎤ ⎡ ⎤⎧ ⎫⎢ ⎥ ⎢ ⎥= +⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭10

0.833 0.625 0.625⎢ ⎥ ⎢ ⎥⎩ ⎭− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

6.245 4.134 10.379⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥

⎧ ⎫⎪ ⎪0 6.614

18.745 4.1346.61414.611

⎪ ⎪ ⎢ ⎥= + =⎨ ⎬ ⎢ ⎥⎪

⎪ ⎪⎨ ⎬⎪ ⎪−⎩⎪− ⎢ ⎥⎩ ⎣ ⎦ ⎭⎭


128

⎩⎩ ⎣ ⎦ ⎭⎭

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

6 939 01 ⎡ ⎤[ ] [ ] [ ][ ]TJJ MJ M MJF B F B=

6.939 010 3.906EA

⎡ ⎤= ⎢ ⎥

⎣ ⎦

[ ] [ ]TF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦01 ⎧ ⎫

⎨ ⎬[ ] [ ]JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ 3.906EA= ⎨ ⎬−⎩ ⎭

{ }6.939 0 15 01 1 6.614

0 3.906 10 3.906JDEA EA

⎧ ⎫ ⎧ ⎫⎡ ⎤∴ = +⎨ ⎬ ⎨ ⎬⎢ ⎥ −⎩ ⎭ ⎩ ⎭⎣ ⎦

104.0851 ⎧ ⎫⎨ ⎬

⎩ ⎭ ⎩ ⎭⎣ ⎦


13.226EA ⎨⎩

= ⎬⎭


[ ] [ ] [ ][ ]TS MS M MSF B F B=

T0 833 0 625 0 625 5 0 0 0 833 0 625 0 625⎡ ⎤ ⎡ ⎤⎡ ⎤0.833 0.625 0.625 5 0 0 0.833 0.625 0.62510 0 1 0 2 0 0 0 1

0 833 0 625 0 625 0 0 5 0 833 0 625 0 625EA

− −⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦0.833 0.625 0.625 0 0 5 0.833 0.625 0.625− − − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

T0.833 0.625 0.625 4.165 3.125 3.125− −⎡ ⎤ ⎡ ⎤0.833 0.625 0.625 4.165 3.125 3.12510 0 1 0 0 2

0 833 0 625 0 625 4 165 3 125 3 125EA

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[ ]F F⎡ ⎤⎡ ⎤⎣ ⎦6.939 0 0

1⎡ ⎤⎢ ⎥

0.833 0.625 0.625 4.165 3.125 3.125⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[ ]JJ JQ

QJ QQ

F F

F F

⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦

1 0 3.906 3.9060 3.906 5.906

EA⎢ ⎥= −⎢ ⎥

−⎢ ⎥⎣ ⎦


QJ QQ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

Lack of fit and temperature change problems:

{ } { }1 T−⎡ ⎤ ⎡ ⎤

ac o t a d te pe atu e c a ge p ob e s:

{ } { }1 T

QT QQ MQ TA F B D⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦Redundants:

{ }TD are the displacements (change in length in the case of trusses) due to lack of fit or temperature changes.) p g

MQB⎡ ⎤⎣ ⎦ are the member forces when unit load is applied MQB⎡ ⎤⎣ ⎦ are the member forces when unit load is applied corresponding to the redundants separately.

{ } { }MT MQ QTA B A⎡ ⎤= ⎣ ⎦Member forces:


•Problem 9: Member AB is too short by 1 mm. (i.e., AB is 1 mm shorter than C( ,required, hence it has to be pulled to fit in the frame). All

b h i l members have cross sectional areas 35 cm2 and E=2.1x103 t/cm2

D

A B

D

300

300

300

300

A B

10m

300 300

Internal indeterminacy = 1


Choose the force in AB as redundant

Member flexibility matrix of truss member

L⎡ ⎤[ ]MiLF

EA⎡ ⎤= ⎢ ⎥⎣ ⎦


1000 0 0 0 0 00 1000 0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥

[ ] 0 0 577.36 0 0 010 0 0 577.36 0 0MF

EA

⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥

0 0 0 0 577.36 00 0 0 0 0 1000

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


⎣ ⎦

⎡ ⎤ are the member forces when a unit load is MQB⎡ ⎤⎣ ⎦

are the member forces when a unit load is applied corresponding to the redundant.

CC

1 24

3 5D

A B

1 1QA =6


1Q

A

1⎧ ⎫

AQ1=1

0⎧ ⎫⎪ ⎪1

11 732

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪ { }

00

⎪ ⎪⎪ ⎪⎪ ⎪1.732

1.732MQB−⎪ ⎪

⎡ ⎤ = ⎨ ⎬⎣ ⎦ −⎪ ⎪⎪ ⎪

{ }000

TD⎪ ⎪

= ⎨ ⎬⎪ ⎪⎪ ⎪1.732

1

⎪ ⎪−⎪ ⎪⎩ ⎭

00.1

⎪ ⎪⎪ ⎪−⎩ ⎭⎩ ⎭


Redundants (due to lack of fit):

{ } { }1 T

QT QQ MQ TA F B D−

⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦

T⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 8196

{ } { }Q QQ Q⎣ ⎦ ⎣ ⎦

[ ]T

QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦8196AE

=

0⎧ ⎫00

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪{ } [ ]

01 1 1.732 1.732 1.732 1

08196QTAEA

⎪ ⎪−∴ = − − − ⎨ ⎬

⎪ ⎪00 1

⎪ ⎪⎪ ⎪⎪ ⎪−⎩ ⎭


0.1−⎩ ⎭0.89677 tons=

Member forces (due to lack of fit):

{ } { }⎡ ⎤

Member forces (due to lack of fit):

{ } { }MT MQ QTA B A⎡ ⎤= ⎣ ⎦

11

⎧ ⎫⎪ ⎪⎪ ⎪

0.896770 89677⎧ ⎫⎪ ⎪⎪ ⎪

{ } { }

11.732

0.89677MTA

⎪ ⎪−⎪ ⎪

= ⎨ ⎬

0.896771.553

tons

⎪ ⎪−⎪ ⎪

= ⎨ ⎬{ } { }1.7321.732

MT ⎨ ⎬−⎪ ⎪⎪ ⎪−⎪ ⎪

1.5531.553

⎨ ⎬−⎪ ⎪⎪ ⎪−⎪ ⎪

1⎪ ⎪⎩ ⎭ 0.89677

⎪ ⎪⎩ ⎭


SummarySummary

Fl ibilit th dFlexibility method

• Analysis of simple structures – plane truss, continuous beam and plane frame- nodal loads and element loads –lack of fit and temperature effectslack of fit and temperature effects.


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