This document discusses the flexibility method for analyzing structures. It provides the definitions of flexibility and stiffness influence coefficients and describes how to develop flexibility matrices for truss, beam, and frame elements using the physical and energy approaches. It then shows how to assemble the total flexibility matrix of a structure and use it to analyze simple structures like plane trusses, continuous beams, and plane frames. The document includes an example problem of a two-member structure to illustrate the flexibility method steps, such as determining static indeterminacy, developing member and system flexibility matrices, evaluating joint displacements and member end actions.

1. Structural Analysis IIIStructural Analysis - III
Fl ibilit M th d 2Flexibility Method -2
ProblemsProblems
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1

2. 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 dplane 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

3. •Problem 1:
Static indeterminacy = 2
Choose reactions at B and C as redundants
Static indeterminacy = 2
Dept. of CE, GCE Kannur Dr.RajeshKN
3
Released structure

4. 1JA 2JA
1QA 2QA1QD 2QD
Joint actions & corresponding displacementsJoint actions & corresponding displacements
Redundants & corresponding displacements
Reactions other than redundants
Dept. of CE, GCE Kannur Dr.RajeshKN

5. Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
5
Equivalent joint loads

6. Member end actions consideredMember end actions considered
2MA 4MA
1MA L
A B
3MA L
B C
Hence member flexibility matrix,
L L−⎡ ⎤
[ ] 11 12
21 22
3 6M M
Mi
L L
F F EI EI
F
F F L L
⎡ ⎤
⎢ ⎥⎡ ⎤
= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥21 22
6 3
M MF F L L
EI EI
⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
6

7. L L−⎡ ⎤
[ ]1
6 12
;M
EI EI
F
L L
⎡ ⎤
⎢ ⎥
= ⎢ ⎥
−⎢ ⎥
⎢ ⎥
Member1:
12 6EI EI
⎢ ⎥
⎢ ⎥⎣ ⎦
L L⎡ ⎤
[ ]2
3 6
M
L L
EI EI
F
L L
−⎡ ⎤
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
Member2: [ ]
6 3
L L
EI EI
−⎢ ⎥
⎢ ⎥⎣ ⎦
Unassembled flexibility matrix 2 1 0 0−⎡ ⎤
⎢ ⎥
[ ]
1 2 0 0
0 0 4 212
M
L
F
EI
⎢ ⎥−
⎢ ⎥=
−⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 2 4
⎢ ⎥
−⎣ ⎦

8. 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.RajeshKN
2 1QA =

9. 1 1JA =
A B C
1
1− 1 0 0
A B C
0
1A 2 1JA =
1
1− 1 1− 1
A B C
0
1− 1 1− 1
Dept. of CE, GCE Kannur Dr.RajeshKN

10. 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

11. AJ1 AJ2 AQ1 AQ2
1 1 2L L− − − −⎡ ⎤
⎢ ⎥
J1 J2 Q1 Q2
=1 =1 =1 =1
[ ] [ ]
1 1 0
0 1 0MS MJ MQ
L
B B B
L
⎢ ⎥
⎢ ⎥⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦ ⎢ ⎥− −
⎢ ⎥
0 1 0 0
⎢ ⎥
⎣ ⎦
AJ1 AJ2 AQ1 AQ2
=1 =1 =1 =1
[ ] [ ]
0 0 1 1
1 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

12. [ ] [ ] [ ][ ]
T
S 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 2
1 1 1 1 1 2 0 0 1 1 0
L L
LL
− − − − − −⎡ ⎤⎡ ⎤ ⎡ ⎤
⎢ ⎥⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥⎢ ⎥ ⎢ ⎥=
⎢ ⎥⎢ ⎥ ⎢ ⎥0 0 0 0 0 4 2 0 1 012
2 0 0 0 2 4 0 1 0 0
L LEI
L L L
⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN

13. 1 1 0 0 3 3 2 5
1 1 1 1 3 3 4
L L
L LL
− − − − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥
0 0 0 0 6 0 412
2 0 0 6 0 2
L LEI
L L L L
⎢ ⎥ ⎢ ⎥=
− − −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
− −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
6 6 3 9L L⎡ ⎤
[ ]
6 6 3 9
6 18 3 15 JJ JQ
L L
F FL LL
⎡ ⎤
⎢ ⎥ ⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥ ⎢ ⎥2 2
2 2
3 3 2 512
9 15 5 18
QJ QQ
L L L LEI F F
L L L L
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦⎢ ⎥
⎣ ⎦9 15 5 18L L L L⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
13

14. 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 2
2 5 3 3
112 12 12 12
L L L L
PLEI EI EI EI
A
−
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎧ ⎫
⎢ ⎥ ⎢ ⎥ ⎨ ⎬{ } 3 3 2 2
295 18 9 15
12 12 12 12
QA
L L L L
EI EI EI EI
⎢ ⎥ ⎢ ⎥= − ⎨ ⎬
⎢ ⎥ ⎢ ⎥ ⎩ ⎭
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦12 12 12 12EI EI EI EI⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
14

15. 18 5 1 1 1PL ⎡ ⎤ ⎡ ⎤ ⎧ ⎫ 18 5 3P −⎡ ⎤ ⎧ ⎫
{ }
18 5 1 1 1
5 2 3 5 233
Q
PL
A
L
−⎡ ⎤ ⎡ ⎤ ⎧ ⎫−
= ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎩ ⎭
18 5 3
5 2 1333
P ⎡ ⎤ ⎧ ⎫−
= ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎩ ⎭
3
3
P
P
⎧
=
⎫
⎨ ⎬
⎩ ⎭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 P
A A A
P−⎧ ⎫ ⎧ ⎫
+ +⎨ ⎬ ⎨ ⎬
⎧ ⎫
⎨ ⎬Thus
Dept. of CE, GCE Kannur Dr.RajeshKN
{ } { } { } 3 2 3Q QC QFINAL
A A A
P P P
= − + = − + =⎨ ⎬ ⎨ ⎬
−⎩ ⎭
⎨
⎩ ⎭ ⎩
⎬
− ⎭
Thus,

16. 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 9
6 18 2 3 1512 9 1
3
2 3
L LL PL L
L LEI EI
P
P
⎧ ⎫⎡ ⎤ ⎡ ⎤
= +⎨ ⎬⎢ ⎥ ⎢ ⎥
⎩ ⎭⎣ ⎦ ⎣ −⎦
⎧ ⎫
⎨ ⎬
⎩ ⎭
2 2
3 2PL PL⎧ ⎫ ⎧ ⎫3 2
7 418 12
PL PL
EI EI
⎧ ⎫ ⎧ ⎫
= −⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
2
0PL ⎧ ⎫
⎨= ⎬
Dept. of CE, GCE Kannur Dr.RajeshKN
118EI
⎨
⎩
= ⎬
⎭

17. 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 2
1 1 1 0
0 1 2 0
3
3
9
3
2 9 3M
PL
PL P
PL
L L
LPL
A
PL
− − − −⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= + +⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪
− ⎧ ⎫
⎨ ⎬
⎩ ⎭0 1 2 09
0 1 0 0
2 9 3
2 9
PL P
P
L
L
− −⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪
⎢ ⎥ ⎢ ⎥⎪ ⎪⎩ ⎭ ⎣ ⎦ ⎣ ⎦
−⎩ ⎭
−
3 3 3
3 3 3 3
3PLPL PL PL
PL PL PL PL
⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
−
− − −3 3 3
2 9 2 9 3
2 9 9 0
3
2
3
0
PL PL PL
PL PL PL
PL PL
PL
PL
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩
− −
−
⎭ ⎩
−
⎭ ⎩ ⎭ ⎩ ⎭−
Dept. of CE, GCE Kannur Dr.RajeshKN
2 9 9 02 0PL PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭−

18. 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 29
3
3
R
P
PL
A PL
L L
P
P
−⎧ ⎫
− −⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪
= − + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
⎩ ⎭⎣ ⎦ ⎣ ⎦⎪
⎧ ⎫
⎨
⎪
⎬
⎩ ⎭
{ }
1 1 2 29
3
3L L P⎢ ⎥ ⎢ ⎥− − − −− ⎩ ⎭⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭
−⎩ ⎭
22 0 0P
PL PL PL
P
PL
⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎧ ⎫
⎪ ⎪
⎨ ⎬
⎪ ⎪⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
18
3 3 3 3
−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩⎪ ⎪⎩ ⎩ ⎭ ⎩ ⎭ ⎭⎭

19. •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

20. Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
20
Equivalent joint loads

21. 2MA
A B
4MA
L
B C
1MA
L
B
3MA
L
Member end actions considered
Hence member flexibility matrix,
3 2
L L
F F
⎡ ⎤
⎢ ⎥⎡ ⎤
[ ] 11 12
2
21 22
3 2M M
Mi
M M
F F EI EI
F
F F L L
⎢ ⎥⎡ ⎤
⎢ ⎥= =⎢ ⎥
⎢ ⎥⎣ ⎦
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
2EI EI⎢ ⎥⎣ ⎦

22. 3 2
L L⎡ ⎤
[ ]1 2
6 4
;M
L L
EI EI
F
L L
⎡ ⎤
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
⎢ ⎥
Member1:
4 2EI EI⎢ ⎥⎣ ⎦
⎡ ⎤
[ ]
3 2
2 2
3 2
M
L L
EI EI
F
L L
⎡ ⎤
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
Member2: [ ]2 2
2
M
L L
EI EI
⎢ ⎥
⎢ ⎥⎣ ⎦
Unassembled flexibility matrix 2
2 3 0 0L L⎡ ⎤
⎢ ⎥
[ ] 2
3 6 0 0
12 0 0 4 6
M
LL
F
EI L L
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 6 12L
⎢ ⎥
⎣ ⎦

23. 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 redundants
separately.
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

24. 1 1JA =
A B C
1
A B C
0
2 1MLA =1
4 0MLA =
1
0 1 0MLA = 3 0MLA =
2 1JA =
1
1ML
A B C
1
0
1 1
0
1
Dept. of CE, GCE Kannur Dr.RajeshKN
0
00

25. L A B C
1
L
L L
1 1QA =
A B C
0 0
L
1 01
A B C2L
2 1QA =
1
L 0L 0
1
2L
Dept. of CE, GCE Kannur Dr.RajeshKN
1 11

26. AJ1 AJ2 AQ1 AQ2
=1 =1 =1 =1
0 0 1 1
1 1 0 L
⎡ ⎤
⎢ ⎥
⎢ ⎥
=1 =1 =1 =1
[ ] [ ]
1 1 0
0 0 0 1
0 1 0 0
MS MJ MQ
L
B B B ⎢ ⎥⎡ ⎤⎡ ⎤= =⎣ ⎦⎣ ⎦ ⎢ ⎥
⎢ ⎥
⎣ ⎦0 1 0 0⎣ ⎦
[ ] [ ]
0 0 1 1
B B B
− −⎡ ⎤
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ] 1 1 2RS RJ RQB B B
L L
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦ − − − −⎣ ⎦
[ ] [ ] [ ][ ]
T
S MS M MSF B F B=[ ] [ ] [ ][ ]S MS M MS
2
0 1 0 0 0 0 1 12 3 0 0L L⎡ ⎤⎡ ⎤ ⎡ ⎤
⎢ ⎥⎢ ⎥ ⎢ ⎥
2
0 1 0 1 1 1 03 6 0 0
1 0 0 0 0 0 0 112 0 0 4 6
LLL
EI L L
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥=
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
26
1 1 0 0 1 0 00 0 6 12L L
⎢ ⎥⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎣ ⎦

27. 2 2
0 1 0 0 3 3 2 5
0 1 0 1 6 6 3 9
L L L L
L LL
⎡ ⎤⎡ ⎤
⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥
2
1 0 0 0 12 0 6 0 4
1 1 0 0 12 0 6
EI L L
L L
⎢ ⎥⎢ ⎥=
⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
6 6 3 9L L⎡ ⎤
⎢ ⎥ [ ]F F⎡ ⎤⎡ ⎤⎣ ⎦
2 2
6 18 3 15
3 3 2 512
L LL
L L L LEI
⎢ ⎥
⎢ ⎥=
⎢ ⎥
⎢ ⎥
[ ]JJ JQ
QJ QQ
F F
F F
⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=
⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦
2 2
9 15 5 18L L L L
⎢ ⎥
⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN

28. 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 3
112 12 12 12
29
L L L L
PLEI EI EI EI
−
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎧ ⎫
⎢ ⎥ ⎢ ⎥= − ⎨ ⎬
⎢ ⎥ ⎢ ⎥ ⎩ ⎭
3 3 2 2
295 18 9 15
12 12 12 12
L L L L
EI EI EI EI
⎨ ⎬
⎢ ⎥ ⎢ ⎥ ⎩ ⎭
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
18 5 1 1 1 18 5 3PL P− −⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫− −
= =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
28
5 2 3 5 2 5 2 1333 33L
⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

29. 3P⎧ ⎫
{ }
3
3Q
P
P
A
⎧
−
=
⎫
⎨ ⎬
⎩ ⎭
The final values of redundants are obtained by includingy g
actual or equivalent joint loads applied directly to the
supports.
Th { } { } { }
3 10 33P P P
A A A
⎧ ⎫ ⎧ ⎫
⎨ ⎬ ⎨ ⎬
⎧ ⎫
⎨ ⎬Thus,{ } { } { } 3 2 3Q QC QFINAL
A A A
P P P
⎧ ⎫ ⎧ ⎫
= − + = + =⎨ ⎬ ⎨ ⎬
⎩ ⎭
⎧ ⎫
⎨ ⎬
−⎩ ⎩⎭ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
29

30. Joint displacements
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
Joint displacements
{ } [ ]{ } { }J JJ J JQ Q⎣ ⎦
{ }
6 6 1 3 9
6 18 2 3 1512 9 12
3
3J
L LL PL L
D
L LEI E
P
I P
⎡ ⎤ ⎧ ⎫ ⎡ ⎤
= +⎨
⎧ ⎫
⎨ ⎬
−
⎬⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎩ ⎦ ⎩⎭ ⎣ ⎭⎣ ⎦ ⎩ ⎦ ⎩⎭ ⎣ ⎭
{ }
2 2 2
0
11
3 2
7 41 12 88
J
PL PL
D
EI EI
PL
EI
⎧ ⎫ ⎧ ⎫
= − =⎨ ⎬
⎧
⎨ ⎬
⎩ ⎭ ⎩ ⎭
⎫
⎨ ⎬
⎩ ⎭117 41 12 88EI EI EI⎩ ⎭ ⎩ ⎭ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
30

31. Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦{ }⎣ ⎦
2 0 0 1 1P⎧ ⎫ ⎡ ⎤ ⎡ ⎤
{ }
2 0 0 1 1
3 1 1 1 0
0 0 2 0 1
3
39
M
P
PL LPL
A
P
P
P
⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= + +⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪
⎧ ⎫
⎨ ⎬
−⎩ ⎭0 0 2 0 1 39
2 9 0 1 0 0
P P
PL
⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪
⎢ ⎥ ⎢ ⎥⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎣ ⎦
⎩ ⎭
2 0 0
3 3 3
2
3
P
PL PL PL
P
PL
⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎧ ⎫
⎪ ⎪
⎪ ⎪3 3 3
0 3
3
2 3
2 9 2 9 0 0
PL PL PL
P P
PL PL
PL
P
− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪
= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
−⎪ ⎪
⎨ ⎬
⎪ ⎪
⎪⎩ ⎭ ⎩ ⎭ ⎩⎩ ⎭⎭ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
31
2 9 2 9 0 0PL PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪⎩ ⎭ ⎩ ⎭ ⎩⎩ ⎭⎭ ⎪

32. 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 PL
A
P− − −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤
= − + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
⎧ ⎫
⎨ ⎬{ }
3 1 1 2 29 3RA
PL L L P
= + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− − − − −⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ −⎦
⎨ ⎬
⎩ ⎭
{ }
2
3
2 0 0
3 3 3R
P
P
P
A
PL PL PL L
⎧ ⎫ ⎧ ⎫ ⎧ ⎫
= + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬
−⎩ ⎭
⎧ ⎫
⎨
⎩ ⎩ ⎭ ⎩⎭
⎬
⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
32
33 3 3 PPL PL PL L⎩ ⎭ ⎩ ⎩ ⎭ ⎩⎭ ⎭

33. •Problem 3:
A D
120 kN40 kN/m 20 kN/m
A
B C D4 m
12 m 12 m 12 m
Static indeterminacy = 2
Choose reactions at C and D as redundants
A D
120 kN40 kN/m 20 kN/m
A
B C D4 m
12 m 12 m 12 m
Dept. of CE, GCE Kannur Dr.RajeshKN
33
Released structure

34. 120 kN40 kN/m 20 kN/m
A
B C D
/ 20 kN/m
4 m
12 m 12 m 12 m
2
wl
480
12
wl
= 480 240 240
wl 240 120 120
240
2
wl
= 240 120 120
213.33 106 673.33 106.67
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
3488.89 31.11
Fixed end actions

35. 480 266 67 133.33 240
A B C D
480 266.67
240
240 88.89
328 89
+
=
120 31.11
151 11
+
120
328.89= 151.11=
Equivalent joint loads
A
1JA 2JA
3JA
4JA
A
B C
D
A AA A
Structure with redundants and other reactions
1QA 2QA1RA 2RA
Dept. of CE, GCE Kannur Dr.RajeshKN
35
and joint actions

36. Member flexibility matrix
11 12 3 6M M
L L
F F EI EI
−⎡ ⎤
⎢ ⎥⎡ ⎤
[ ] 11 12
21 22
3 6
6 3
M M
Mi
M M
F F EI EI
F
F F L L
EI EI
⎢ ⎥⎡ ⎤
= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
2 1 0 0 0 0
1 2 0 0 0 0
−⎡ ⎤
⎢ ⎥
⎢ ⎥
[ ]
1 2 0 0 0 0
0 0 2 1 0 02
0 0 1 2 0 0MF
EI
−
⎢ ⎥
−⎢ ⎥
= ⎢ ⎥
⎢ ⎥0 0 1 2 0 0
0 0 0 0 2 1
0 0 0 0 1 2
EI −⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
36
0 0 0 0 1 2−⎣ ⎦

37. To find [BMS] and [BRS] matrices:
are found from the released structure when[ ]MSB [ ]RSB
1 1 1 1 1 1A A A A A A
and
it 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

38. 1 1JA =
1
A B
C
D
1
12
1 0 0 0 0 0
1212
2 1JA =
A B
C
D
1
12
1
12
0 1 0 0 0 0
Dept. of CE, GCE Kannur Dr.RajeshKN
38

39. 3 1JA =
A B
C
D
1
12
1
12
0 1 1− 1 0 0
1212
1A 4 1JA =
A B DA B
C
D
1
12
1
12
0 1 1− 1 1− 1
Dept. of CE, GCE Kannur Dr.RajeshKN

40. 1 1QA =
A B C D
21
0 12 12− 0 0 0
1 1QA
A B C D
2 32
2 1QA =
0 24 24− 12 12− 0
Dept. of CE, GCE Kannur Dr.RajeshKN

41. Hence action transformation matrixHence, action transformation matrix,
AJ1 AJ2 AJ3 AJ4 AQ1 AQ2
1 1 1 1 1 1
1 0 0 0 0 0⎡ ⎤
⎢ ⎥
=1 =1 =1 =1 =1 =1
[ ] [ ]
0 1 1 1 12 24
0 0 1 1 12 24
B B B
⎢ ⎥
⎢ ⎥
⎢ ⎥− − − −
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ] 0 0 1 1 0 12
0 0 0 1 0 12
MS MJ MQB B B⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦
⎢ ⎥
⎢ ⎥− −
⎢ ⎥
0 0 0 1 0 0
⎢ ⎥
⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN

42. A
B C
D
B C
1RA 2RA
AJ1 AJ2 AJ3 AJ4 AQ1 AQ2
1 1 1 1 1 1
[ ] [ ]
1 1 1 1 12 241
RS RJ RQB B B
⎡ ⎤
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦
=1 =1 =1 =1 =1 =1
[ ] [ ]
1 1 1 1 24 3612
RS RJ RQB B B⎡ ⎤⎡ ⎤ ⎢ ⎥⎣ ⎦⎣ ⎦ − − − − − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
42

43. 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 60 602
24 48 156 192EI
−⎡ ⎤
= ⎢ ⎥−⎣ ⎦
⎣ ⎦ ⎣ ⎦ 24 48 156 192EI ⎣ ⎦
480−⎧ ⎫
⎪ ⎪
{ }
1
576 1296 12 24 60 60 266.672 2
1296 3456 24 48 156 192 133.33QA
EI EI
− ⎪ ⎪−⎛ ⎞⎡ ⎤ ⎡ ⎤⎪ ⎪
∴ = − ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
240
⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎪ ⎪
⎪ ⎪⎩ ⎭

44. { }
3456 1296 18560.281
1296 576 49600 68311040
QA
−⎡ ⎤ ⎧ ⎫−
= ⎨ ⎬⎢ ⎥
⎣ ⎦ ⎩ ⎭
{ } 1296 576 49600.68311040 −⎣ ⎦ ⎩ ⎭
{ }
138153.61
4515868 8
0.4
31104
442
0 14 5186QA
⎧ ⎫
⎨ ⎬
−
−⎧ ⎫−
= =⎨ ⎬
⎩⎩ ⎭ ⎭4515868.8311040 14.5186⎩⎩ ⎭ ⎭
{ } { } { }
151.11 0.4442 151.55
120 14 5186 105 48Q QC QFINAL
A A A
−⎧ ⎫ ⎧ ⎫ ⎧ ⎫
∴ = − + = − + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬
− −⎩ ⎭ ⎩ ⎭ ⎩ ⎭120 14.5186 105.48FINAL − −⎩ ⎭ ⎩ ⎭ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

45. 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 0
480 0 1 1 1 480 12 24
⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎪ ⎪ ⎢ ⎥ ⎢ ⎥− −⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥
0
449
⎧ ⎫
⎪ ⎪−
⎪ ⎪480 0 1 1 1 480 12 24
213.33 0 0 1 1 266.67 12 24 0.4442
10667 0 0 1 1 13333 0 12 145186
⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎪ ⎪− − − −⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎧ ⎫⎪ ⎪
= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
⎩ ⎭⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
449
449
174
⎪ ⎪
⎪ ⎪
⎨ ⎬
⎪ ⎪
=
106.67 0 0 1 1 133.33 0 12 14.5186
240 0 0 0 1 240 0 12
240 0 0 0 1 0 0
− − −⎩ ⎭⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎩ ⎭
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎩ ⎭ ⎣ ⎦ ⎣ ⎦
174
174
0
−⎪ ⎪
⎪ ⎪
⎪
⎩ ⎭
⎪
240 0 0 0 1 0 0
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 0⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
45

46. 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 12
240
⎩ ⎭ ⎣ ⎦ ⎣ ⎦⎩ ⎭⎪ ⎪
⎪ ⎪⎩ ⎭
202.518
380 447
⎧
=
⎫
⎨ ⎬
⎩ ⎭380.447⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
46

47. Joint displacements
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
J p
2 1 1 1− − −⎡ ⎤
⎢ ⎥
[ ] [ ] [ ][ ]T
JJ MJ M MJF B F B=
1 2 2 22
1 2 8 8EI
⎢ ⎥−
⎢ ⎥=
−⎢ ⎥
⎢ ⎥
1 2 8 14
⎢ ⎥
−⎣ ⎦
[ ] [ ]T
F B F B⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
12 24
24 482
− −⎡ ⎤
⎢ ⎥
⎢ ⎥[ ] [ ]JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
60 156
60 192
EI
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
47
60 192⎣ ⎦

48. ⎧ ⎫⎡ ⎤ ⎡ ⎤
{ }
2 1 1 1 480 12 24
1 2 2 2 266.67 24 48 0.44422 2
JD
− − − − − −⎧ ⎫⎡ ⎤ ⎡ ⎤
⎪ ⎪⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎢ ⎥
{ }
1 2 8 8 133.33 60 156 14.5186
1 2 8 14 240 60 192
JD
EI EI
+⎨ ⎬ ⎨ ⎬
− − −⎢ ⎥ ⎢ ⎥⎩ ⎭⎪ ⎪
⎢ ⎥ ⎢ ⎥⎪ ⎪− ⎩ ⎭⎣ ⎦ ⎣ ⎦
990−⎧ ⎫
⎪ ⎪5402
371EI
⎪ ⎪
⎪ ⎪
⎨ ⎬
−⎪
⎪
=
⎪
⎪545.378⎪ ⎪⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

49. Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]
T
S MS M MSF B F B=
⎡ ⎤1 0 0 0 0 0
0 1 0 0 0 0
0 1 1 1 0 0
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥0 1 1 1 0 0
0 1 1 1 1 1
0 12 12 0 0 0
−⎢ ⎥
= ⎢ ⎥
− −⎢ ⎥
⎢ ⎥−0 12 12 0 0 0
0 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 0
1 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 12
EI
× ⎢ ⎥⎢ ⎥
− ⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥− − −
⎢ ⎥⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 0 0 1 2 0 0 0 1 0 0
⎢ ⎥⎢ ⎥
− ⎢ ⎥⎣ ⎦ ⎣ ⎦

50. 1 0 0 0 0 0 2 1 1 1 12 24
0 1 0 0 0 0 1 2 2 2 24 48
0 1 1 1 0 0 0 0 3 3 24 602
− − − − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥0 1 1 1 0 0 0 0 3 3 24 602
0 1 1 1 1 1 0 0 3 3 12 48
0 12 12 0 0 0 0 0 0 3 0 24
E I
− − − − −⎢ ⎥ ⎢ ⎥
= ⎢ ⎥ ⎢ ⎥
− −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥− − −0 12 12 0 0 0 0 0 0 3 0 24
0 24 24 12 12 0 0 0 0 3 0 12
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
− −⎣ ⎦ ⎣ ⎦
2 1 1 1 12 24
1 2 2 2 24 48
− − − − −⎡ ⎤
⎢ ⎥−
⎢ ⎥
[ ]
1 2 2 2 24 48
1 2 8 8 60 1562
1 2 8 14 60 192
JJ JQF F
EI F F
⎢ ⎥
⎡ ⎤⎡ ⎤−⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥
⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥1 2 8 14 60 192
12 24 60 60 576 1296
24 48 156 192 1296 3456
QJ QQ
EI F F− ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
⎢ ⎥−
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
50
24 48 156 192 1296 3456
⎢ ⎥
−⎣ ⎦

51. •Problem 4:
Static indeterminacy = 1
Dept. of CE, GCE Kannur Dr.RajeshKN
51
Choose horizontal reaction at D as redundant

52. 1RA 1R
2RA 3RA
Redundants [AQ] and reactions other than redundants [AR]
3R
Dept. of CE, GCE Kannur Dr.RajeshKN
Redundants [AQ] and reactions other than redundants [AR]

53. 4
27
PL
27
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
53

54. Combined (equivalent +actual) joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN

55. Member flexibility matrix
3 6
L L
F F EI EI
−⎡ ⎤
⎢ ⎥⎡ ⎤
[ ] 11 12
21 22
3 6
6 3
M M
Mi
M M
F F EI EI
F
F F L L
EI EI
⎢ ⎥⎡ ⎤
= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
2 1 0 0 0 0
1 2 0 0 0 0
−⎡ ⎤
⎢ ⎥−
⎢ ⎥
[ ]
1 2 0 0 0 0
0 0 4 2 0 0
0 0 2 4 0 012
M
L
F
EI
⎢ ⎥
−⎢ ⎥
= ⎢ ⎥
⎢ ⎥
[ ]
0 0 2 4 0 012
0 0 0 0 2 1
EI −⎢ ⎥
⎢ ⎥−
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
55
0 0 0 0 1 2
⎢ ⎥
−⎣ ⎦

56. Joint displacements
R d d tRedundants
Reactions 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.

57. L/2 0
1A
L/2
0
2 1JA =
1
0 0
1/2
1/2
Dept. of CE, GCE Kannur Dr.RajeshKN
57

58. 0 00
1
3 1JA = 0
0
0
1
0
4 1JA =
0
1
0 0 4 1JA
0
1/L 1/L
0
0
/
1/L
0
0
Dept. of CE, GCE Kannur Dr.RajeshKN
58
1/L
1/L

59. 00
0
1
1
L/2 L/2
L/2
L/2
0
1 1 1QA =
1A
0 1
1 1Q
5 1JA =1/L
1/L
0 0
00
Dept. of CE, GCE Kannur Dr.RajeshKN
59

60. AJ1 AJ2 AJ3 AJ4 AJ5 AQ1
=1 =1 =1 =1 =1 =1
1 0 0 0 0 0
1 2 0 0 0 2L L
⎡ ⎤
⎢ ⎥
⎢ ⎥
=1 =1 =1 =1 =1 =1
[ ] [ ]
1 2 0 0 0 2
1 2 1 0 0 2
L L
L L
B B B
⎢ ⎥
−⎢ ⎥
⎢ ⎥− −
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]
0 0 0 1 1 2
0 0 0 0 1 2
MS MJ MQB B B
L
L
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦
⎢ ⎥
⎢ ⎥− −
⎢ ⎥
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⎢ ⎥− − − −⎣ ⎦

61. Redundants:
{ }QD∵ is a null matrix{ } { }
1
Q QQ QJ JA F F A
−
⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦
T
3
L
[ ]
T
QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3
L
E I
=
[ ][ ]
T
QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
2
9 2 2 3 3 9 2
L
L L L L L⎡ ⎤= − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
9 2 2 3 3 9 2
12
L L L L L
EI
⎡ ⎤⎣ ⎦

62. 0⎧ ⎫
{ } 2
0
2
3
9 2 2 3 3 9 2 4 27
P
EI L
A L L L L L PL
⎧ ⎫
⎪ ⎪
⎪ ⎪− ⎪ ⎪
⎡ ⎤∴ = ⎨ ⎬⎣ ⎦{ } 3
9 2 2 3 3 9 2 4 27
12
2 27
0
QA L L L L L PL
L EI
PL
⎡ ⎤∴ = − − −⎨ ⎬⎣ ⎦
⎪ ⎪
⎪ ⎪
⎪ ⎪⎩ ⎭0⎪ ⎪⎩ ⎭
5
12
P−
=
12
{ } { } { } 5 5
0Q QC Q
P
A A A
P⎛ ⎞
= − + = + =⎟
−
−⎜
⎝ ⎠
{ } { } { } 12
0
12
Q QC QFINAL
A A A+ + ⎟⎜
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN

63. 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 0
0
0 1 2 0 0 0 2
2
L L
P
⎧ ⎫ ⎡ ⎤ ⎡ ⎤
⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎪ ⎪
2
4 27 1 2 1 0 0 2 5
4 27
2 27 0 0 0 1 1 2 12
2 27
P
PL L L P
PL
PL L
PL
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎪ ⎪− −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎛ ⎞
= + +−⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟− ⎝ ⎠⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎪ ⎪
2 27
0 0 0 0 0 1 2
0
0 0 0 0 0 1 0
PL
L
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭
⎩ ⎭ ⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
63

64. 0 0 0⎧ ⎫ ⎧ ⎫ ⎡ ⎤
0⎧ ⎫
⎪ ⎪0 0 0
0 4 1
4 27 43 108 1 5
PL
PL PL PL
⎧ ⎫ ⎧ ⎫ ⎡ ⎤
⎪ ⎪ ⎪ ⎪ ⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎛ ⎞
24
24
PL
PL
⎪ ⎪
⎪ ⎪
−⎪ ⎪4 27 43 108 1 5
2 27 2 27 1 24
0 0 1
PL PL PL
PL PL
− −⎪ ⎪ ⎪ ⎪ ⎢ ⎥ −⎛ ⎞
= + +⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎜ ⎟
− ⎝ ⎠⎪ ⎪ ⎪ ⎪ ⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎢ ⎥
24
5 24
5 24
PL
PL
PL
⎪ ⎪
⎨ ⎬
−⎪ ⎪
⎪ ⎪
=
0 0 1
0 0 0
⎪ ⎪ ⎪ ⎪ ⎢ ⎥−
⎪ ⎪ ⎪ ⎪ ⎢ ⎥
⎩ ⎭ ⎩ ⎭ ⎣ ⎦
5 24
0
PL⎪ ⎪
⎪ ⎪
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

65. 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
5
20 1 1 2 1 1 1 04 27
27 12
7 1 1 2 1 1 1 02 27
P
P P
L L L L PL
L L L L PL
⎪ ⎪− −⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎛ ⎞⎢ ⎥= − + − +−⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎝ ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦7 1 1 2 1 1 1 02 27
0
L L L L PL⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − − −⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦⎪ ⎪
⎪ ⎪⎩ ⎭
1
5
12
P
−⎧ ⎫
⎪ ⎪
⎨ ⎬=
12
7
⎨ ⎬
⎪
⎩
⎪
⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
65

66. { } [ ]{ } { }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− − −⎡ ⎤
[ ] [ ] [ ][ ]T
JJ MJ M MJF B F B=
2
10 7 2 4 2 2
7 2 3 2 2
4 2 4 2 2
L
L L L L L
L
L
⎡ ⎤
⎢ ⎥− −
⎢ ⎥
= − − −⎢ ⎥[ ] [ ] [ ][ ]JJ MJ M MJF B F B 4 2 4 2 2
12
2 2 4 4
2 2 4 10
L
EI
L
L
= ⎢ ⎥
⎢ ⎥
− −⎢ ⎥
⎢ ⎥⎣ ⎦2 2 4 10L⎢ ⎥− −⎣ ⎦
9 2L⎧ ⎫
T
⎡ ⎤ ⎡ ⎤
2
9 2
2
L
L
L
−⎧ ⎫
⎪ ⎪
⎪ ⎪⎪ ⎪
[ ] [ ]T
JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ 3
12
3
L
L
EI
L
⎪ ⎪
= −⎨ ⎬
⎪ ⎪
⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
66
9 2L
⎪ ⎪
⎪ ⎪⎩ ⎭

67. { } [ ]{ } { }D F A F A⎡ ⎤∴ = + ⎣ ⎦{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤∴ = + ⎣ ⎦
2 2
10 7 2 4 2 2 0 9 2
7 2 3 2 2 2 2
5
L L
L L L L L P L
L L P
− − − −⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥− −
⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎛ ⎞5
4 2 4 2 2 4 27 3
12 12 12
2 2 4 4 2 27 3
L L P
L PL L
EI EI
L PL L
⎪ ⎪ ⎪ ⎪⎢ ⎥ −⎪ ⎪ ⎪ ⎪⎛ ⎞
= +− − − − −⎢ ⎥⎨ ⎬ ⎨ ⎬⎜ ⎟
⎝ ⎠⎢ ⎥⎪ ⎪ ⎪ ⎪− −⎢ ⎥⎪ ⎪ ⎪ ⎪
2 2 4 10 0 9 2L L
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥− − ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦
2
133
62L
PL
−⎧ ⎫
⎪ ⎪
⎪ ⎪⎪ ⎪
{ }
2
106
2592
34
JD
PL
EI
⎪ ⎪⎪ ⎪
−⎨ ⎬
⎪ ⎪−
=
⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
169−⎪⎩
⎪ ⎪
⎪⎭

68. Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]
T
S MS M MSF B F B=
y [ ]
1 0 0 0 0 0 2 1 0 0 0 0 1 0 0 0 0 0
1 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 2
1 2 1 0 0 2 0 0 4 2 0 0 1 2 1 0 0 2
0 0 0 1 1 2 0 0 2 4 0 0 0 0 0 1 1 212
0 0 0 0 1 2 0 0 0 0 2 1 0 0 0 0 1 2
L L L L
L L L LL
L LEI
L 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 0
L L⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥
−⎢ ⎥ ⎢⎣ ⎦⎣ ⎦ ⎣ ⎦⎥
1 1 1 0 0 0 3 2 0 0 0 2
0 2 2 0 0 0 3 0 0 0
L L
L L L L
− − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥0 0 1 0 0 0 4 2 4 2 2 3
0 0 0 1 0 0 2 2 4 4 312
0 0 0 1 1 1 0 0 0 0 3
L LL
L 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
⎢ ⎥ ⎢ ⎥
− −⎣ ⎦ ⎣ ⎦

69. 2 2
10 7 2 4 2 2 9 2
7 2 3 2 2 2
L L
L L L L L L
− − − −⎡ ⎤
⎢ ⎥− −
⎢ ⎥
[ ]
7 2 3 2 2 2
4 2 4 2 2 3
2 2 4 4 312
JJ JQ
L L L L L L
F FL LL
L LEI F F
⎢ ⎥
⎡ ⎤⎡ ⎤− − − −⎢ ⎥ ⎣ ⎦⎢ ⎥= =⎢ ⎥
− − ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦
2 2
2 2 4 4 312
2 2 4 10 9 2
9 2 2 3 3 9 2 4
QJ QQ
L LEI F F
L 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

70. •Problem 5: 30 kN/m
50 kN
30 kN/m
4m
B C
4
2m
4m
D
A
Static indeterminacy = 3
Dept. of CE, GCE Kannur Dr.RajeshKN
70
Choose 3 reactions at D as redundants

71. 3QA 2QA
1QA
A
2RA
A
3RA
Released structure
1RA
Dept. of CE, GCE Kannur Dr.RajeshKN
with redundants and other reactions

72. 60 kN 60 kN
40 kNm 40 kNm
60 kN 60 kN
40 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

73. Member flexibility matrix
11 12 3 6M M
L L
F F EI EI
−⎡ ⎤
⎢ ⎥⎡ ⎤
[ ] 11 12
21 22
3 6
6 3
M M
Mi
M M
F F EI EI
F
F F L L
EI EI
⎢ ⎥⎡ ⎤
= = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦6 3EI EI⎢ ⎥⎣ ⎦
U bl d fl ibilit
4 2 0 0 0 0
2 4 0 0 0 0
−⎡ ⎤
⎢ ⎥
Unassembled flexibility
matrix
[ ]
2 4 0 0 0 0
0 0 4 2 0 02
0 0 2 4 0 06
MF
EI
⎢ ⎥−
⎢ ⎥
−⎢ ⎥
= ⎢ ⎥[ ]
0 0 2 4 0 06
0 0 0 0 2 1
0 0 0 0 1 2
M
EI
⎢ ⎥
−⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
73
0 0 0 0 1 2
⎢ ⎥
−⎣ ⎦

74. 40A = 40A =2 40JA = 3 40JA =
1 50JA =
2JD 3JD
1 50JA
1JD
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.

75. 0 0 1 0
1A =
0
0
0
0
1 0
0 01 1JA =
2 1JA =
0 0
4 0 0
1−4 01−
0
0 0
Dept. of CE, GCE Kannur Dr.RajeshKN
75

76. 1 0
1
4
4−
0
3 1JA =
1− 1 4− 0
3J
00 0
0
0
1A
0
4−
0
0
01− 1 1QA =
1−
Dept. of CE, GCE Kannur Dr.RajeshKN

77. 2 2− 1−1
2− 2 1
1−
2 1QA = 12Q
1−
0
2
3 1QA =01−
0
1
2
0
Dept. of CE, GCE Kannur Dr.RajeshKN

78. AJ1 AJ2 AJ3 AQ1 AQ2 AQ3
=1 =1 =1 =1 =1 =1
4 1 1 4 2 1
0 1 1 4 2 1
− − − −⎡ ⎤
⎢ ⎥
⎢ ⎥
[ ] [ ]
0 1 1 4 2 1
0 0 1 4 2 1
0 0 1 0 2 1
MS MJ MQB B B
⎢ ⎥
⎢ ⎥− − − −
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦[ ] [ ]
0 0 1 0 2 1
0 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 0
1 0 0 0 1 0
4 1 1 4 2 1
RS RJ RQB B B
⎡ ⎤
⎢ ⎥⎡ ⎤⎡ ⎤= = − −⎣ ⎦⎣ ⎦ ⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
4 1 1 4 2 1⎢ ⎥− − − −⎣ ⎦

79. Redundants:
{ }QD∵ is a null matrix{ } { }
1
Q QQ QJ JA F F A
−
⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦
T
256 48 72
2
⎡ ⎤
⎢ ⎥[ ]
T
QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2
48 72 30
6
72 30 30
EI
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦⎣ ⎦
96 48 72−⎡ ⎤
[ ][ ]
T
QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
96 48 72
2
16 0 24
6
24 12 24
EI
⎡ ⎤
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦24 12 24−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
79

80. { }
1
⎡ ⎤ ⎡ ⎤{ } { }
1
Q QQ QJ JA F F A
−
⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦
1260 720 3774 3840
1
720 2496 4224 1760
− −⎡ ⎤⎧ ⎫
− ⎪ ⎪⎢ ⎥
⎨ ⎬
10
53 333
⎧ ⎫
⎪ ⎪
⎨ ⎬720 2496 4224 1760
87552
3774 4224 16128 720
⎢ ⎥= − ⎨ ⎬⎢ ⎥
⎪ ⎪− − −⎢ ⎥⎣ ⎦⎩ ⎭
53.333
53.333
−⎨ ⎬
⎪ ⎪
⎩ ⎭
=
60 10 70−⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎧ ⎫
⎪ ⎪
{ } { } { } 0 53.333
0 5
53.333
53.3333.333
Q QC QFINAL
A A A
⎪ ⎪ ⎪ ⎪
= − + = − + − =⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎩ ⎭
⎪ ⎪
−⎨ ⎬
⎪
⎩ ⎩⎭
⎪
⎭0 5 53.3333.333⎩ ⎭ ⎩ ⎩⎭ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
80

81. 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 10
0 1 0 0 40 0 1 0 53.333
0 4 1 1 40 4 2 1 53 333
RA
− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= − + − − + − −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭0 4 1 1 40 4 2 1 53.333⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − −⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭
50
3.333
⎧ ⎫
=
⎪ ⎪
⎨ ⎬3.333
0
⎨ ⎬
⎪ ⎪
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
81

82. 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 10
40 0 0 1 4 2 1
40 53 333
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎧ ⎫ ⎧ ⎫
− − − −⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪
+ +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
15.636
15.636
⎧ ⎫
⎪ ⎪−
⎪ ⎪
⎪ ⎪
⎨ ⎬40 53.333
40 0 0 1 0 2 1
40 53.333
0 0 0 0 0 2 1
= + − + −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
−⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
⎩ ⎭ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥− −
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
54.648
54.648
⎨ ⎬
−⎪ ⎪
⎪ ⎪
⎪ ⎪
=
0 0 0 0 0 0 1
⎪ ⎪ ⎢ ⎥ ⎢ ⎥
⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 52.018
⎪ ⎪
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
82

83. Joint displacements
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
Joint displacements
{ } [ ]{ } { }J JJ J JQ Q⎣ ⎦
[ ] [ ] [ ][ ]T
64 24 24
2
− −⎡ ⎤
⎢ ⎥[ ] [ ] [ ][ ]T
JJ MJ M MJF B F B=
2
24 12 12
6
24 12 24
EI
⎢ ⎥= −
⎢ ⎥
−⎢ ⎥⎣ ⎦
T
⎡ ⎤ ⎡ ⎤
96 16 24
2
− −⎡ ⎤
⎢ ⎥
[ ] [ ]T
JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
2
48 0 12
6
72 24 24
EI
⎢ ⎥=
⎢ ⎥
⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
83

84. { }
64 24 24 50 96 16 24 10
2 2
− − − −⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥
{ }
2 2
24 12 12 40 48 0 12 53.333
6 6
24 12 24 40 72 24 24 53.333
JD
EI EI
⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥∴ = − − + −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎢ ⎥⎣ ⎦⎩ ⎭ ⎣ ⎦⎩ ⎭
3200 3093.3 35.56
1
26.67
2 106.68
2 1
1200 1120 80.00
6 3E II E EI
−⎛ ⎞⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎜ ⎟
= − + = − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟
⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎧ ⎫
⎪ ⎪
−⎨ ⎬
⎪ ⎪0
6 3
720 720 0
E II E EI
⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟
⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎜ ⎟−⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎨ ⎬
⎪ ⎪
⎩⎠ ⎭⎝
Dept. of CE, GCE Kannur Dr.RajeshKN
84

85. [ ] [ ] [ ][ ]
T
S 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 1
0 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 12
0 0 1 0 2 1 0 0 2 4 0 0 0 0 1 0 2 16
0 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 16 6 6 24 4 6
1 1 0 0 0 0 8 6 6 24 4 6
− − − −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
1 1 0 0 0 0 8 6 6 24 4 6
1 1 1 1 0 0 0 0 6 16 12 62
4 4 4 0 0 0 0 0 6 8 12 66
2 2 2 2 2 0 0 0 0 0 4 3
E 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 2
2 4 1 2 2 4 7 2 2 4 2 42
9 6 4 8 7 2 2 5 6 4 8 7 26
1 6 0 2 4 4 8 7 2 3 0
JJ JQ
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 0
2 4 1 2 2 4 7 2 3 0 3 0
⎢ ⎥
⎢ ⎥
−⎣ ⎦

86. •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

87. [ ] 3 6
Mi
L L
EI EI
F
−⎡ ⎤
⎢ ⎥
= ⎢ ⎥Member flexibility matrix [ ]
6 3
MiF
L L
EI EI
⎢ ⎥
−⎢ ⎥
⎢ ⎥⎣ ⎦
2 1 0 0 0 0
1 2 0 0 0 0
−⎡ ⎤
⎢ ⎥
Unassembled flexibility matrix
[ ]
1 2 0 0 0 0
0 0 4 2 0 03
0 0 2 4 0 06
MF
EI
⎢ ⎥−
⎢ ⎥
−⎢ ⎥
= ⎢ ⎥[ ]
0 0 2 4 0 06
0 0 0 0 2 1
0 0 0 0 1 2
M
EI
⎢ ⎥
−⎢ ⎥
⎢ ⎥−
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
87
0 0 0 0 1 2−⎣ ⎦

88. 2525
Fixed end actions 50 50
2525 2525
Equivalent joint loads 50 50
1JD 2JD
3JD
Joint displacements{ }JD
A C D30mm 1QD
Dept. of CE, GCE Kannur Dr.RajeshKN
Support settlements { }QD
B

89. AJ1 AJ2 AJ3 AQ1 AQ2 AQ3
1 1 1 1 1 1
1 1 1 3 9 12
1 1 1 0 6 9
− − − − − −⎡ ⎤
⎢ ⎥
⎢ ⎥
=1 =1 =1 =1 =1 =1
[ ] [ ]
1 1 1 0 6 9
0 1 1 0 6 9
0 1 1 0 0 3
MS MJ MQB B B
⎢ ⎥
⎢ ⎥− − − −
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦
⎢ ⎥0 1 1 0 0 3
0 0 1 0 0 3
0 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

90. Redundants:
1
{ }
{ } { } { }
1
Q QQ Q QJ JA F D F A
−
⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣ ⎦
{ }QD is NOT a null matrix
[ ]
T
F 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

91. { } { } { }
1
A F D F A
−
⎡ ⎤⎡ ⎤ ⎡ ⎤{ } { } { }Q QQ Q QJ JA F D F A⎡ ⎤⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦⎣ ⎦
1
0.03 0
0 25F F
−
−⎛ ⎞⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪⎜ ⎟
⎡ ⎤ ⎡ ⎤= − −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟
62.3
29 4
−⎧ ⎫
⎪ ⎪
⎨ ⎬0 25
0 25
QQ QJF F⎡ ⎤ ⎡ ⎤= − −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟
⎪ ⎪ ⎪ ⎪⎜ ⎟
⎩ ⎭ ⎩ ⎭⎝ ⎠
29.4
13.4
= ⎨ ⎬
⎪ ⎪−⎩ ⎭
0 62.3 62.3−⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
−⎧ ⎫
⎪ ⎪
{ } { } { } 50 29.4
50 13 4
79.4
36 6
Q QC QFINAL
A A A
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
= − + = − − + =⎨ ⎬ ⎨ ⎬
⎪
⎧ ⎫
⎪ ⎪
⎨ ⎬
⎪ ⎪
⎩
⎪ ⎪ ⎪− −⎭ ⎩ ⎭ ⎭⎩
Thus,
50 13.4 36.6⎩⎭ ⎩ ⎭ ⎭⎩
Dept. of CE, GCE Kannur Dr.RajeshKN
91

92. Member end actions
{ } { } [ ]{ } { }⎡ ⎤
Member end actions
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
83.7
55 4
0
0
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎧ ⎫
⎪ ⎪
⎪ ⎪
[ ]{ } { }
55.4
55.4
40 3
0
0
0 MJ J MQ QB A B A
⎪ ⎪
⎪ ⎪
⎡ ⎤= + + =⎨ ⎬ ⎣ ⎦
⎪
⎪ ⎪
−⎪ ⎪
⎨ ⎬
⎪ ⎪⎪
[ ]{ } { } 40.3
40.3
0
0
25
25
Q Q⎣ ⎦
⎪ −⎪ ⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎪ ⎪
⎩⎩ ⎭⎭
⎪
025
⎪⎪ ⎪
⎩⎩ ⎭− ⎭
⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
92

93. Reactions other than redundantsReactions other than redundants
Dept. of CE, GCE Kannur Dr.RajeshKN
93

94. 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=
[ ] [ ]T
JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
[ ]
62.30
29.425JJ JQF F
−⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎡ ⎤= +−⎨ ⎬ ⎨ ⎬⎣ ⎦[ ]
13.425
JJ JQ⎨ ⎬ ⎨ ⎬⎣ ⎦
⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
⎧ ⎫3
3
7.5 10
0.5 10
−
−
⎧ ⎫− ×
⎪ ⎪
×⎨ ⎬
⎪
=
⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
94
3
3.1 10−⎪ ×⎩
⎪
⎭

95. Alternatively if the entire [Fs] matrix is assembled at a time
[ ] [ ] [ ][ ]
T
S 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 12
1 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 93
0 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
⎢ ⎥
⎢ ⎥
⎣ ⎦

96. •Problem 6a ( Same problem as above, with a differentProblem 6a ( Same problem as above, with a different
approach)
25 28
A D
25 25
6 03EI
28
B
C2
6 .03
112
3
EI ×
=
2
6 .03
28
6
EI ×
=
112
12 .03EI ×
74.667
3
12 .03
6
EI ×
9.333 50 50
3
3
74.667=
6
9.333=
Fixed end actions
Dept. of CE, GCE Kannur Dr.RajeshKN
96

97. 253
A
B
C D
112
84
74 667
9.333 50+ 50
74.667
74.667 9.333+
Equivalent joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN
97

98. L L−⎡ ⎤
[ ] 3 6
Mi
L L
EI EI
F
L L
⎡ ⎤
⎢ ⎥
= ⎢ ⎥
−⎢ ⎥
Member flexibility matrix
6 3
L L
EI EI
⎢ ⎥
⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
2 1 0 0 0 0
1 2 0 0 0 0
−⎡ ⎤
⎢ ⎥
⎢ ⎥
Unassembled flexibility matrix
[ ]
1 2 0 0 0 0
0 0 4 2 0 03
0 0 2 4 0 06
MF
EI
⎢ ⎥−
⎢ ⎥
−⎢ ⎥
= ⎢ ⎥[ ]
0 0 2 4 0 06
0 0 0 0 2 1
M
EI
⎢ ⎥
−⎢ ⎥
⎢ ⎥−
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
98
0 0 0 0 1 2
⎢ ⎥
−⎣ ⎦

99. AJ1 AJ2 AJ3 AQ1 AQ2 AQ3
=1 =1 =1 =1 =1 =1
1 1 1 3 9 12
1 1 1 0 6 9
− − − − − −⎡ ⎤
⎢ ⎥
⎢ ⎥
[ ] [ ]
0 1 1 0 6 9
0 1 1 0 0 3
MS MJ MQB B B
⎢ ⎥
⎢ ⎥− − − −
⎡ ⎤⎡ ⎤= = ⎢ ⎥⎣ ⎦⎣ ⎦
⎢ ⎥0 1 1 0 0 3
0 0 1 0 0 3
0 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

100. Redundants:
{ } { } { }
1
A F D F A
−
⎡ ⎤⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦{ } { } { }Q QQ Q QJ JA F D F A⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣ ⎦
[ ]
T
F 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

101. 0 84−⎛ ⎞⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪⎜ ⎟
21.7⎧ ⎫
⎪ ⎪
{ }
1
0 3
0 25
Q QQ QJA F F
− ⎪ ⎪ ⎪ ⎪⎜ ⎟
⎡ ⎤ ⎡ ⎤= −⎨ ⎬ ⎨ ⎬⎣ ⎦ ⎣ ⎦⎜ ⎟
⎪ ⎪ ⎪ ⎪⎜ ⎟
⎩ ⎭⎩ ⎭⎝ ⎠
20
13.4
⎪ ⎪
⎨ ⎬
⎪−⎩
=
⎪
⎭⎩ ⎭⎩ ⎭⎝ ⎠
Note the differences hereNote the differences here.
{ } { } { }
84 21.7 62.3⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
−⎧ ⎫
⎪ ⎪
{ } { } { } 59.4 20
50
79.4
36.613.4
Q QC QFINAL
A A A
⎪ ⎪ ⎪ ⎪
=− + =− − + =⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪− −
⎪ ⎪
⎨ ⎬
⎪
⎩⎩ ⎭ ⎩ ⎭
⎪
⎭
Thus,
Dept. of CE, GCE Kannur Dr.RajeshKN
101
Note the difference here.

102. 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
⎧ ⎫
⎪ ⎪
⎪ ⎪
−⎪ ⎪
⎡ ⎤⎨ ⎬
55.4
55.4
⎧ ⎫
⎪ ⎪
⎪ ⎪
−⎪ ⎪
⎨ ⎬{ } [ ]{ } { }28
25
M MJ J MQ QA B A B A
⎪ ⎪
⎡ ⎤= + +⎨ ⎬ ⎣ ⎦−⎪ ⎪
⎪ ⎪
40.3
40 3
⎪ ⎪
⎨ ⎬
−⎪
⎪
=
⎪
⎪25
25
⎪ ⎪
⎪ ⎪
−⎩ ⎭
40.3
0
⎪
⎪
⎩
⎪
⎪
⎭
Note the difference here.
Dept. of CE, GCE Kannur Dr.RajeshKN
102

103. Reactions other than redundantsReactions other than redundants
Dept. of CE, GCE Kannur Dr.RajeshKN
103

104. { }⎡ ⎤
Joint displacements
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
[ ]
21.784
203JJ JQF F
−⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎡ ⎤= +⎨ ⎬ ⎨ ⎬⎣ ⎦[ ] 203
13.425
JJ JQF F⎡ ⎤+⎨ ⎬ ⎨ ⎬⎣ ⎦
⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
3
3
7.5 10
0 5 10
−
−
⎧ ⎫− ×
⎪ ⎪
×⎨ ⎬=
3
0.5 10
3.1 10−
×⎨ ⎬
⎪ ×⎩
=
⎪
⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
104

105. •Problem 7:
Static indeterminacy = 2 ( 1 internal + 1 external )
Dept. of CE, GCE Kannur Dr.RajeshKN
105
Choose reaction at B and force in AD as redundants

106. Released structureReleased structure
Dept. of CE, GCE Kannur Dr.RajeshKN

107. Dept. of CE, GCE Kannur Dr.RajeshKN
Redundants [AQ] and reactions other than redundants [AR]

108. Member flexibility matrix of truss member
⎡ ⎤
[ ]Mi
L
F
EA
⎡ ⎤
= ⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
1 0 0 0 0 0
0 1.414 0 0 0 0
⎡ ⎤
⎢ ⎥
⎢ ⎥
[ ]
0 0 1.414 0 0 0
0 0 0 1 0 0M
L
F
EA
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
0 0 0 0 1 0
0 0 0 0 0 1
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
108
⎣ ⎦

109. 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 3JD
2JD
1JD 5
2 3
D
1 4
4JD
6
Dept. of CE, GCE Kannur Dr.RajeshKN
109Joint displacements

110. 1 1JA =
2 1JA =
1
1 1
11A = 13 1JA =
4 1JA =1
Dept. of CE, GCE Kannur Dr.RajeshKN
110
1

111. 1A =1 1QA =
2 1QA =
1
1 11 1
Dept. of CE, GCE Kannur Dr.RajeshKN
111

112. •Each column in the submatrix consists of member[ ]MJB
forces 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 0.707
0 0 0 0 0 1
−⎡ ⎤
⎢ ⎥
⎢ ⎥
=1 =1 =1 =1 =1 =1
0 0 0 0 0 1
1.414 0 0 0 1.414 1
⎢ ⎥
⎢ ⎥
⎢ ⎥
= ⎢ ⎥[ ] [ ]MS MJ MQB B B⎡ ⎤⎡ ⎤= ⎣ ⎦⎣ ⎦
1 0 1 0 1 0.707
1 0 0 0 0 0.707
= ⎢ ⎥
− − −⎢ ⎥
⎢ ⎥− −
⎢ ⎥
[ ] [ ]MS MJ MQ⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 0 1 0 0.707
⎢ ⎥
−⎢ ⎥⎣ ⎦

113. •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 ofQ⎣ ⎦
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 0
1 1 0 0 1 0
1 0 1 0 1 0
RS RJ RQB B B
⎡ ⎤
⎢ ⎥⎡ ⎤⎡ ⎤= = − − −⎣ ⎦⎣ ⎦ ⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
113
1 0 1 0 1 0⎢ ⎥−⎣ ⎦

114. 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 2.707
2.707 4.828
L
EA
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
[ ][ ]
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

115. 1−⎧ ⎫
⎪ ⎪
{ }
1
3.828 2.707 3.828 0 1 0 2
2.707 4.828 3.414 0.707 0.707 0.707 0
Q
EA PL
A
L EA
− ⎪ ⎪− −⎡ ⎤ ⎡ ⎤− ⎪ ⎪
∴ = ⎨ ⎬⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎪ ⎪
0
⎪ ⎪
⎪ ⎪⎩ ⎭
⎧ ⎫1.172
0.243
P
⎧
=
⎫
⎨ ⎬
−⎩ ⎭
(There are no loads applied directly to the
supports.)
Dept. of CE, GCE Kannur Dr.RajeshKN

116. 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 0.707
0 0 0 0 1 0 1
1 414 0 0 0 2 1 414 1 1 172
−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥−⎧ ⎫⎢ ⎥ ⎢ ⎥
⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎪ ⎪
1.828
0.243
0
⎧ ⎫
⎪ ⎪−
⎪ ⎪
⎪ ⎪1.414 0 0 0 2 1.414 1 1.172
1 0 1 0 0 1 0.707 0.243
1 0 0 0 0 0 0 707
P P
⎪ ⎪−⎢ ⎥ ⎢ ⎥ ⎧ ⎫⎪ ⎪
= +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥
− − − −⎩ ⎭⎪ ⎪⎢ ⎥ ⎢ ⎥
⎪ ⎪⎢ ⎥ ⎢ ⎥⎩ ⎭
0
0
1 172
P
⎪ ⎪
⎨ ⎬
⎪ ⎪
⎪ ⎪
=
1 0 0 0 0 0 0.707
0 0 0 1 0 0.707
⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎩ ⎭
⎢ ⎥ ⎢ ⎥
−⎣ ⎦ ⎣ ⎦
1.172
0.172
⎪ ⎪
⎪ ⎪
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
116

117. Reactions other than redundants
{ }⎡ ⎤{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦
1⎧ ⎫1
1 0 0 1 1 0
2 1.172
1 1 0 0 1 0
0 0 243
P P
−⎧ ⎫
− − −⎡ ⎤ ⎡ ⎤⎪ ⎪− ⎧ ⎫⎪ ⎪⎢ ⎥ ⎢ ⎥= − − + −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎩ ⎭0 0.243
1 0 1 0 1 0
0
⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ −⎩ ⎭⎪ ⎪−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎪ ⎪⎩ ⎭
0.172
1 828 P
−⎧ ⎫
⎪ ⎪
⎨ ⎬= 1.828
0.172
P⎨ ⎬
⎪ ⎪
⎩ ⎭
=
Dept. of CE, GCE Kannur Dr.RajeshKN
117

118. { } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦Joint displacements { }⎣ ⎦J p
4.828 0 1 0
0 1 0 0
−⎡ ⎤
⎢ ⎥
[ ] [ ] [ ][ ]T
JJ MJ M MJF B F B=
0 1 0 0
1 0 1 0
L
EA
⎢ ⎥
⎢ ⎥=
−⎢ ⎥
⎢ ⎥
0 0 0 1
⎢ ⎥
⎣ ⎦
3.828 3.414⎡ ⎤
[ ] [ ]T
JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
3.828 3.414
0 0.707
1 0 707
L
EA
⎡ ⎤
⎢ ⎥−
⎢ ⎥=
− −⎢ ⎥1 0.707
0 0.707
EA − −⎢ ⎥
⎢ ⎥
−⎣ ⎦1.172−⎧ ⎫
⎪ ⎪
{ }
1.828
0
J
P
A
D
L
E
⎪
∴ =
⎪−⎪ ⎪
⎨ ⎬
⎪ ⎪
Dept. of CE, GCE Kannur Dr.RajeshKN
0.172
⎪ ⎪
⎪ ⎪⎩ ⎭

119. Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]
T
S MS M MSF B F B=
T
0 1 0 0 0 0 707⎡ ⎤0 1 0 0 0 0.707
0 0 0 0 0 1
1 414 0 0 01 414 1
−⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥1.414 0 0 01.414 1
1 0 1 0 1 0.707
1 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 0.707
0 1.414 0 0 0 0 0 0 0 0 0 1
0 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 0
0 0 0 0 1 0 1 0 0 0
L
EA
⎢ ⎥
× ⎢ ⎥
−⎢ ⎥
⎢ ⎥
1 0.707
0 0 707
⎢ ⎥
⎢ ⎥
− −⎢ ⎥
⎢ ⎥
Dept. of CE, GCE Kannur Dr.RajeshKN
0 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 1
⎢ ⎥ −
⎢ ⎥
⎣ ⎦
0 0.707
0 0.707
⎢ ⎥−
⎢ ⎥
−⎢ ⎥⎣ ⎦

120. 4 828 0 1 0 3 828 3 414−⎡ ⎤4.828 0 1 0 3.828 3.414
0 1 0 0 0 0.707
1 0 1 0 1 0 707
−⎡ ⎤
⎢ ⎥−
⎢ ⎥
⎢ ⎥
[ ]
1 0 1 0 1 0.707
0 0 0 1 0 0.707S
L
F
EA
− − −⎢ ⎥
= ⎢ ⎥
−⎢ ⎥
3.828 0 1 0 3.828 2.707
3.414 0.707 0.707 0.707 2.707 4.828
⎢ ⎥
⎢ ⎥−
⎢ ⎥
− − −⎣ ⎦⎣ ⎦
[ ]JJ JQF F⎡ ⎤⎡ ⎤⎣ ⎦⎢ ⎥=
QJ QQF F
⎢ ⎥=
⎢ ⎥⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
120

121. •Problem 8:
Static indeterminacy = 1
Dept. of CE, GCE Kannur Dr.RajeshKN
121
Choose the force in CD as redundant

122. 1JA =
1JD
2JA =
2JD
2JA
DJ1 and DJ2
: Joint displacements
M b fl ibilit
Released structure with loads
Member flexibility
matrix of truss member: [ ]Mi
L
F
EA
=
5 0 0⎡ ⎤
[ ]
5 0 0
1
0 2 0MF
EA
⎡ ⎤
⎢ ⎥=
⎢ ⎥
Unassembled flexibility matrix:
Dept. of CE, GCE Kannur Dr.RajeshKN
122
0 0 5
EA ⎢ ⎥
⎢ ⎥⎣ ⎦

123. 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

124. 1 1QA =1 1QA
Dept. of CE, GCE Kannur Dr.RajeshKN
124

125. •Each column in the submatrix consists of member[ ]MJB
forces 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

126. Redundants:
{ }QD∵ is a null matrix{ } { }
1
Q QQ QJ JA F F A
−
⎡ ⎤ ⎡ ⎤= − ⎣ ⎦ ⎣ ⎦
[ ]
T
F 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 0 625
EA
⎡ ⎤⎡ ⎤
⎢ ⎥⎢ ⎥=
⎢ ⎥⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
5.906
EA
=
0 0 5 0.625⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦
0⎧ ⎫
[ ][ ]
T
QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
01
3.906EA
⎧ ⎫
= ⎨ ⎬
−⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

127. 1
{ } { }
1
Q QQ QJ JA F F A
−
⎡ ⎤ ⎡ ⎤∴ = −⎣ ⎦ ⎣ ⎦
[ ]
151
0 3.906
105.906
EA
EA
⎧ ⎫−
= − ⎨ ⎬
⎩ ⎭⎩ ⎭
6.614 kN=
Dept. of CE, GCE Kannur Dr.RajeshKN

128. 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 0.625
15
0 0 1 6.614
10
−⎡ ⎤ ⎡ ⎤
⎧ ⎫⎢ ⎥ ⎢ ⎥= +⎨ ⎬⎢ ⎥ ⎢ ⎥⎩ ⎭10
0.833 0.625 0.625
⎢ ⎥ ⎢ ⎥⎩ ⎭
− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
6.245 4.134 10.379⎧ ⎫ ⎡ ⎤
⎪ ⎪ ⎢ ⎥
⎧ ⎫
⎪ ⎪
0 6.614
18.745 4.134
6.614
14.611
⎪ ⎪ ⎢ ⎥= + =⎨ ⎬ ⎢ ⎥
⎪
⎪ ⎪
⎨ ⎬
⎪ ⎪−⎩
⎪− ⎢ ⎥⎩ ⎣ ⎦ ⎭⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
128
⎩⎩ ⎣ ⎦ ⎭⎭

129. { } [ ]{ } { }D F A F A⎡ ⎤= + ⎣ ⎦Joint displacements: { } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦Joint displacements:
6 939 01 ⎡ ⎤
[ ] [ ] [ ][ ]T
JJ MJ M MJF B F B=
6.939 01
0 3.906EA
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
[ ] [ ]T
F 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 3.906
JD
EA EA
⎧ ⎫ ⎧ ⎫⎡ ⎤
∴ = +⎨ ⎬ ⎨ ⎬⎢ ⎥ −⎩ ⎭ ⎩ ⎭⎣ ⎦
104.0851 ⎧ ⎫
⎨ ⎬
⎩ ⎭ ⎩ ⎭⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
13.226EA
⎨
⎩
= ⎬
⎭

130. Alternatively, if the entire [Fs] matrix is assembled at a time,
[ ] [ ] [ ][ ]
T
S MS M MSF B F B=
T
0833 0625 0625 5 0 0 0833 0625 0625⎡ ⎤ ⎡ ⎤⎡ ⎤0.833 0.625 0.625 5 0 0 0.833 0.625 0.625
1
0 0 1 0 2 0 0 0 1
0833 0625 0625 0 0 5 0833 0625 0625
EA
− −⎡ ⎤ ⎡ ⎤⎡ ⎤
⎢ ⎥ ⎢ ⎥⎢ ⎥=
⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦0.833 0.625 0.625 0 0 5 0.833 0.625 0.625− − − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
T
0.833 0.625 0.625 4.165 3.125 3.125− −⎡ ⎤ ⎡ ⎤0.833 0.625 0.625 4.165 3.125 3.125
1
0 0 1 0 0 2
0833 0625 0625 4165 3125 3125
EA
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥=
⎢ ⎥ ⎢ ⎥
− − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
[ ]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 3.906
0 3.906 5.906
EA
⎢ ⎥= −
⎢ ⎥
−⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
QJ QQ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

131. 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 appliedMQB⎡ ⎤⎣ ⎦ 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

132. •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 lmembers 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.RajeshKN
132
Choose the force in AB as redundant

133. Member flexibility matrix of truss member
L⎡ ⎤
[ ]Mi
L
F
EA
⎡ ⎤
= ⎢ ⎥⎣ ⎦
Unassembled flexibility matrix
1000 0 0 0 0 0
0 1000 0 0 0 0
⎡ ⎤
⎢ ⎥
⎢ ⎥
[ ]
0 0 577.36 0 0 01
0 0 0 577.36 0 0
MF
EA
⎢ ⎥
⎢ ⎥
= ⎢ ⎥
⎢ ⎥
0 0 0 0 577.36 0
0 0 0 0 0 1000
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN
⎣ ⎦

134. ⎡ ⎤ 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 2
4
3 5
D
A B
1 1QA =
6
Dept. of CE, GCE Kannur Dr.RajeshKN
1Q

135. A
1⎧ ⎫
AQ1
=1
0⎧ ⎫
⎪ ⎪1
1
1 732
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪ { }
0
0
⎪ ⎪
⎪ ⎪
⎪ ⎪1.732
1.732
MQB
−⎪ ⎪
⎡ ⎤ = ⎨ ⎬⎣ ⎦ −⎪ ⎪
⎪ ⎪
{ }
0
0
0
TD
⎪ ⎪
= ⎨ ⎬
⎪ ⎪
⎪ ⎪
1.732
1
⎪ ⎪−
⎪ ⎪
⎩ ⎭
0
0.1
⎪ ⎪
⎪ ⎪
−⎩ ⎭⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

136. 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⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦
8196
AE
=
0⎧ ⎫0
0
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪
{ } [ ]
0
1 1 1.732 1.732 1.732 1
08196
QT
AE
A
⎪ ⎪−
∴ = − − − ⎨ ⎬
⎪ ⎪
0
0 1
⎪ ⎪
⎪ ⎪
⎪ ⎪
−⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN
0.1−⎩ ⎭0.89677 tons=

137. Member forces (due to lack of fit):
{ } { }⎡ ⎤
Member forces (due to lack of fit):
{ } { }MT MQ QTA B A⎡ ⎤= ⎣ ⎦
1
1
⎧ ⎫
⎪ ⎪
⎪ ⎪
0.89677
0 89677
⎧ ⎫
⎪ ⎪
⎪ ⎪
{ } { }
1
1.732
0.89677MTA
⎪ ⎪
−⎪ ⎪
= ⎨ ⎬
0.89677
1.553
tons
⎪ ⎪
−⎪ ⎪
= ⎨ ⎬{ } { }
1.732
1.732
MT ⎨ ⎬
−⎪ ⎪
⎪ ⎪−
⎪ ⎪
1.553
1.553
⎨ ⎬
−⎪ ⎪
⎪ ⎪−
⎪ ⎪
1
⎪ ⎪
⎩ ⎭ 0.89677
⎪ ⎪
⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN

138. 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