Module1 flexibility-2-problems- rajesh sir
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Transcript of 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.
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥−⎢ ⎥=−⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
A B CL
1 1QA =
A B C
1
L
L− 0 0 0
A B C2L
2 1QA =1
2L− L L− 0
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎣ ⎦⎣ ⎦ − − − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } { } { } 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 ⎧ ⎫⎨= ⎬
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩
− −−
⎭ ⎩
−
⎭ ⎩ ⎭ ⎩ ⎭−
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
19Released structure
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 6 12L⎢ ⎥⎣ ⎦
To find [BMS] and [BRS] matrices:
[ ] [ ]B b d i d
To find [BMS] and [BRS] matrices:
[ ]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.
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎢ ⎥⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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− −⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫− −= =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎧ ⎫ ⎧ ⎫= − + = + =⎨ ⎬ ⎨ ⎬
⎩ ⎭
⎧ ⎫⎨ ⎬−⎩ ⎩⎭ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎩ ⎭ ⎩ ⎭ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪
−⎪ ⎪⎨ ⎬⎪ ⎪⎪⎩ ⎭ ⎩ ⎭ ⎩⎩ ⎭⎭ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎧ ⎫ ⎧ ⎫ ⎧ ⎫
= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬−⎩ ⎭
⎧ ⎫⎨
⎩ ⎩ ⎭ ⎩⎭⎬⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
120 kN40 kN/m 20 kN/mA
B C D4 m12 m 12 m 12 m
Dept. of CE, GCE Kannur Dr.RajeshKN
33
Released structure
120 kN40 kN/m 20 kN/mA
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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 −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
36
0 0 0 0 1 2−⎣ ⎦
To find [BMS] and [BRS] matrices:
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎡ ⎤⎡ ⎤ ⎢ ⎥⎣ ⎦⎣ ⎦ − − − − − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
− ⎪ ⎪−⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎪ ⎪∴ = − ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
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 − −⎩ ⎭ ⎩ ⎭ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
Member end actions
{ } { } [ ]{ } { }A A B A B A⎡ ⎤= + + ⎣ ⎦
Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
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⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
45
Reactions other than redundants
{ } { } [ ]{ } { }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⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
× ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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− ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
⎢ ⎥−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
50
24 48 156 192 1296 3456⎢ ⎥−⎣ ⎦
•Problem 4:
Static indeterminacy = 1
Dept. of CE, GCE Kannur Dr.RajeshKN
51Choose horizontal reaction at D as redundant
1RA 1R
2RA3RA
Redundants [AQ] and reactions other than redundants [AR]
3R
Dept. of CE, GCE Kannur Dr.RajeshKN
Redundants [AQ] and reactions other than redundants [AR]
427PL27
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
53
Combined (equivalent +actual) joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN
Member flexibility matrix
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 −⎢ ⎥⎢ ⎥−⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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− −⎡ ⎤
⎢ ⎥⎡ ⎤⎡ ⎤= = −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
1 1 2 1 1 1 0L L L L⎢ ⎥− − − −⎣ ⎦
Redundants:
{ }QD∵ is a null matrix{ } { }1
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⎡ ⎤= − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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+ + ⎟⎜
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
Member end actions
{ } { } [ ]{ } { }A A B A B A⎡ ⎤= + + ⎣ ⎦
Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
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
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭⎩ ⎭ ⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎪ ⎪⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
Reactions other than redundants
{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦
Reactions other than redundants
{ } { } [ ]{ } { }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
⎨ ⎬⎪⎩ ⎪⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪= −⎨ ⎬⎪ ⎪⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪⎪ ⎪−⎨ ⎬⎪ ⎪−
=
⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
169−⎪⎩⎪ ⎪
⎪⎭
Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]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
− − − −⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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− −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
•Problem 5: 30 kN/m
50 kN
30 kN/m
4m
B C
42m
4m
D
A
Static indeterminacy = 3
Dept. of CE, GCE Kannur Dr.RajeshKN
70Choose 3 reactions at D as redundants
3QA 2QA
1QAA
2RA
A
3RA
Released structure
1RA
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
72
actual) joint loads
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⎢ ⎥⎣ ⎦
U bl d fl ibilit 4 2 0 0 0 02 4 0 0 0 0
−⎡ ⎤⎢ ⎥
Unassembled flexibility matrix
[ ]
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 ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
75
1 0
1
44−
0
3 1JA =
1− 1 4− 0
3J
00 0
0
0
1A04−
0
0
01− 1 1QA =1−
Dept. of CE, GCE Kannur Dr.RajeshKN
2 2− 1−1
2− 2 11−
2 1QA = 12Q
1−
0
23 1QA =01−
0
120
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎡ ⎤⎢ ⎥⎡ ⎤⎡ ⎤= = − −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
4 1 1 4 2 1⎢ ⎥− − − −⎣ ⎦
Redundants:
{ }QD∵ is a null matrix{ } { }1
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−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎩ ⎭ ⎩ ⎩⎭ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
80
Reactions other than redundants
{ } { } [ ]{ } { }A A B A B A⎡ ⎤= − + + ⎣ ⎦
Reactions other than redundants
{ } { } [ ]{ } { }R RC RJ J RQ QA 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⎨ ⎬⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
81
Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
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
⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎜ ⎟−⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎨ ⎬⎪ ⎪⎩⎠ ⎭⎝
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥−⎢ ⎥⎡ ⎤⎡ ⎤−⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥− ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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×
Static indeterminacy = 3
Dept. of CE, GCE Kannur Dr.RajeshKN
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
−⎡ ⎤⎢ ⎥
Unassembled flexibility matrix
[ ]
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 ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎩⎭ ⎩ ⎭ ⎭⎩
Dept. of CE, GCE Kannur Dr.RajeshKN
91
Member end actions
{ } { } [ ]{ } { }⎡ ⎤
Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
83.755 4
00
⎧ ⎫⎪ ⎪⎪ ⎪
⎧ ⎫⎪ ⎪⎪ ⎪
[ ]{ } { }55.455.440 3
000 MJ J MQ QB A B A
⎪ ⎪⎪ ⎪
⎡ ⎤= + + =⎨ ⎬ ⎣ ⎦⎪
⎪ ⎪−⎪ ⎪⎨ ⎬⎪ ⎪⎪
[ ]{ } { } 40.340.3
0
02525
Q Q⎣ ⎦⎪ −⎪ ⎪
⎪ ⎪⎪
⎪⎪ ⎪⎪ ⎪
⎩⎩ ⎭⎭⎪
025⎪⎪ ⎪⎩⎩ ⎭− ⎭
⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
92
Reactions other than redundantsReactions other than redundants
Dept. of CE, GCE Kannur Dr.RajeshKN
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
−
−
⎧ ⎫− ×⎪ ⎪×⎨ ⎬⎪
=⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
9433.1 10−⎪ ×⎩⎪⎭
Alternatively if the entire [Fs] matrix is assembled at a time
[ ] [ ] [ ][ ]TS MS M MSF B F B=
Alternatively, if the entire [Fs] matrix is assembled at a time,
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⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
96
253
A
B
C D112
84
74 6679.333 50+ 50
74.667 74.667 9.333+
Equivalent joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN
97
L L−⎡ ⎤
[ ] 3 6Mi
L LEI EIFL L
⎡ ⎤⎢ ⎥
= ⎢ ⎥−⎢ ⎥
Member flexibility matrix
6 3L L
EI EI⎢ ⎥⎢ ⎥⎣ ⎦
Unassembled flexibility matrix2 1 0 0 0 01 2 0 0 0 0
−⎡ ⎤⎢ ⎥⎢ ⎥
Unassembled flexibility matrix
[ ]
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 ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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,
Dept. of CE, GCE Kannur Dr.RajeshKN
101Note the difference here.
M b d ti
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
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.
Dept. of CE, GCE Kannur Dr.RajeshKN
102
Reactions other than redundantsReactions other than redundants
Dept. of CE, GCE Kannur Dr.RajeshKN
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−
×⎨ ⎬⎪ ×⎩
=⎪⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
104
•Problem 7:
Static indeterminacy = 2 ( 1 internal + 1 external )
Dept. of CE, GCE Kannur Dr.RajeshKN105
Choose reaction at B and force in AD as redundants
Released structureReleased structure
Dept. of CE, GCE Kannur Dr.RajeshKN
Dept. of CE, GCE Kannur Dr.RajeshKN
Redundants [AQ] and reactions other than redundants [AR]
Member flexibility matrix of truss member
⎡ ⎤[ ]MiLF
EA⎡ ⎤= ⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
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
⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
108
⎣ ⎦
To find [BMS] and [BRS] matrices:
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
Dept. of CE, GCE Kannur Dr.RajeshKN
109Joint displacements
1 1JA = 2 1JA =
1
1 111A = 13 1JA =
4 1JA =1
Dept. of CE, GCE Kannur Dr.RajeshKN
1101
1A =1 1QA =2 1QA =
1
1 11 1
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎡ ⎤⎢ ⎥⎡ ⎤⎡ ⎤= = − − −⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
113
1 0 1 0 1 0⎢ ⎥−⎣ ⎦
Redundants
{ }QD∵ is a null matrix{ } { }1
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
= ⎢ ⎥− − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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.)
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
116
Reactions other than redundants
{ }⎡ ⎤{ } { } [ ]{ } { }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⎨ ⎬
⎪ ⎪⎩ ⎭
=
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪∴ =
⎪−⎪ ⎪⎨ ⎬⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
0.172⎪ ⎪⎪ ⎪⎩ ⎭
Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]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
⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎢ ⎥=⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
120
•Problem 8:
Static indeterminacy = 1
Dept. of CE, GCE Kannur Dr.RajeshKN
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:
Dept. of CE, GCE Kannur Dr.RajeshKN
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 =
Dept. of CE, GCE Kannur Dr.RajeshKN
123
1 1QA =1 1QA
Dept. of CE, GCE Kannur Dr.RajeshKN
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,
Dept. of CE, GCE Kannur Dr.RajeshKN
matrix need not be assembled. [ ]RSB
Redundants:
{ }QD∵ is a null matrix{ } { }1
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⎧ ⎫
= ⎨ ⎬−⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
1{ } { }1
Q QQ QJ JA F F A−
⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦
[ ]151 0 3.906105.906
EAEA
⎧ ⎫−= − ⎨ ⎬
⎩ ⎭⎩ ⎭
6.614 kN=
Dept. of CE, GCE Kannur Dr.RajeshKN
Member forces:
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
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
⎪ ⎪ ⎢ ⎥= + =⎨ ⎬ ⎢ ⎥⎪
⎪ ⎪⎨ ⎬⎪ ⎪−⎩⎪− ⎢ ⎥⎩ ⎣ ⎦ ⎭⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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 ⎧ ⎫⎨ ⎬
⎩ ⎭ ⎩ ⎭⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
13.226EA ⎨⎩
= ⎬⎭
Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]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⎢ ⎥= −⎢ ⎥
−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
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:
Dept. of CE, GCE Kannur Dr.RajeshKN
•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
Dept. of CE, GCE Kannur Dr.RajeshKN132
Choose the force in AB as redundant
Member flexibility matrix of truss member
L⎡ ⎤[ ]MiLF
EA⎡ ⎤= ⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
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
⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
⎣ ⎦
⎡ ⎤ 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
Dept. of CE, GCE Kannur Dr.RajeshKN
1Q
A
1⎧ ⎫
AQ1=1
0⎧ ⎫⎪ ⎪1
11 732
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪ { }
00
⎪ ⎪⎪ ⎪⎪ ⎪1.732
1.732MQB−⎪ ⎪
⎡ ⎤ = ⎨ ⎬⎣ ⎦ −⎪ ⎪⎪ ⎪
{ }000
TD⎪ ⎪
= ⎨ ⎬⎪ ⎪⎪ ⎪1.732
1
⎪ ⎪−⎪ ⎪⎩ ⎭
00.1
⎪ ⎪⎪ ⎪−⎩ ⎭⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪⎪ ⎪⎪ ⎪−⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
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.
Dept. of CE, GCE Kannur Dr.RajeshKN
138