Module1 flexibility-1- rajesh sir

Structural Analysis - III
Fl ibilit M th d 1Flexibility Method - 1
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. 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 dplane truss, continuous beam and plane frame- nodal loads
and element loads – lack of fit and temperature effects.
2

FUNDAMENTALS OF FLEXIBILITY METHOD
Introduction
•This method is a generalization of the Maxwell-Mohr
method(1874)method(1874)
•Not conducive to computer programming, because the choice
f d d t i t iof redundants is not unique
•Unknowns are the redundant actions, which are arbitrarily, y
chosen
3

Flexibility method (Explanation using principle of superposition)
Example 1: Single redundant - Continuous 2-span beam
Flexibility method (Explanation using principle of superposition)
Released structure
4
5
384
B
wL
EI
−
Δ =
Deflection of released
structure due to actual loads
4
(Negative, since deflection is downward)

BRApply unit load corresponding to
C
3
B
L
=δDisplacement due to unit load
48
B
EI
δDisplacement due to unit load,
BR B BR δDisplacement due to is
5

Deflection of releasedB BR δ
structure due to redundant
applied as a load
C
0R δΔ + =
Compatibility condition (or equation of
0B B BR δΔ + = superposition or equation of geometry)
5
8
B
B
B
wL
R
−Δ
∴ = =
δ
δ
(Displacement due to unit load corresponding to )BR
Bδ flexibility coefficient
6

Example 2 – More than one redundant
Ch i f d d t
Choice of redundants

Q Q b th d d tL t 1 2,Q Q be the redundantsLet
Deflections corresponding to
redundantsredundants
&D D Displacements in the released
1 2&QL QLD D Displacements in the released
structure corresponding to redundants, due to
external loads
8

•To get flexibility coefficientsTo get flexibility coefficients
1 2&Q Q•Apply unit loads corresponding to
Fl ibilit ffi i tFlexibility coefficients
Net deflection is zero at B and C
1 11 1 12 2 0QLD F Q F Q+ + =
0D F Q F Q
1 11 12 1 0
0
QLD F F Q
D F F Q
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎡ ⎤
+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥
⎩ ⎭⎩ ⎭⎣ ⎦⎩ ⎭
9
2 21 1 22 2 0QLD F Q F Q+ + = 2 21 22 2 0QLD F F Q⎢ ⎥
⎩ ⎭⎩ ⎭⎣ ⎦⎩ ⎭

•Generally, net deflection need not be zero
1 1 11 1 12 2Q QLD D F Q F Q= + +
D D F Q F Q+ +2 2 21 1 22 2Q QLD D F Q F Q= + +
Q Qh di l di1 2,Q QD D 1 2,Q Q•Where :support displacements corresponding to
{ } { } [ ]{ }D D F Q{ } { } [ ]{ }Q QLD D F Q= +
D⎧ ⎫ D⎧ ⎫ F F⎡ ⎤ Q⎧ ⎫
{ } 1
2
Q
Q
Q
D
D
D
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
{ } 1
2
QL
QL
QL
D
D
D
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
[ ] 11 12
21 22
F F
F
F F
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
{ } 1
2
Q
Q
Q
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
F QQDFlexibility coefficient is sometimes denoted as
10
F QQDFlexibility coefficient is sometimes denoted as

{ } [ ] { } { }( )1
Q QLQ F D D
−
= −
{ } { }0D•If there are no support displacements
{ } [ ] { }1
Q F D
−
∴
{ } { }0QD =•If there are no support displacements,
{ } [ ] { }QLQ F D∴ = −
11

• Example: To find out redundants
: 2P P M PL PGiv n P Pe P= = = =1 2 3: 2P P M PL PGiv n P Pe P= = = =
12

3
13PL
3
97PL3
1
13
24
QL
PL
D
EI
= 2
97
48
QL
PL
D
EI
=
⎡ ⎤3
26
9748
QL
PL
D
EI
⎡ ⎤
⎡ ⎤ = ⎢ ⎥⎣ ⎦
⎣ ⎦
∴
13
⎣ ⎦

33
11
3
L
F
EI
=
3
5L
21
5
6
L
F
EI
=
3
5L
F12
6
F
EI
=
3
22
8
3
L
F
EI
= [ ]
3
2 5
5 166
L
F
EI
⎡
=∴
⎤
⎢ ⎥
⎣ ⎦
14
22
3EI 5 166EI ⎣ ⎦

⎡ ⎤
[ ]
1
3
16 56
5 27
EI
F
L
− −⎡ ⎤
= ⎢ ⎥−⎣ ⎦⎣ ⎦
Q⎡ ⎤
[ ] [ ]
11
2
QL
Q
Q F D
Q
−⎡ ⎤
⎡ ⎤= = −⎢ ⎥ ⎣ ⎦
⎣ ⎦
3
16 5 26⎡ ⎤ ⎡ ⎤ ⎡ ⎤
3
3
16 5 266
5 2 977 48
EI PL
L EI
−⎡ ⎤ ⎡ ⎤−
= ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
69
6456
P ⎡
=
⎤
⎢ ⎥−⎣ ⎦⎣ ⎦
1 2
69 8
, ,
P P
i e Q Q
−
= =
15
1 2. ., ,
56 7
i e Q Q

Temperature changes, pre-strains andp g , p
support displacements not corresponding to redundants
L t
Displacements corresponding to redundants due to{ }D
Let:
Displacements corresponding to redundants due to
temperature changes, in the released structure
{ }QTD
Displacements corresponding to redundants due to
pre strains in the released structure
{ }QPD
pre-strains, in the released structure
Di l di d d d{ } Displacements corresponding to redundants due to
support displacements not corresponding to
redundants in the released structure
{ }QRD
16
redundants, in the released structure

{ } { } { } { } { } [ ]{ }{ } { } { } { } { } [ ]{ }Q QL QT QP QRD D D D D F Q= + + + +
{ } { } { } { } { }QC QL QT QP QRD D D D D= + + +•Let { } { } { } { } { }
{ } { } [ ]{ }Q QCD D F Q= +•Hence, and
{ } [ ] { } { }( )1
Q QCQ F D D
−
= −
17

Member end actions
• Member end actions are the couples and forces that act at
the ends of a member when it is considered to be isolatedthe ends of a member when it is considered to be isolated
from the remainder of the structure
• In the above case, member end actions are the SFs and
BMs at the ends of members AB and BC
18

• In the above figure are the
member end actions considered (upward forces and
l k )
1 2 3 4, , ,M M M MA A A A
anticlockwise moments are +ve).
•The first two are just to the left of B, and the last two are
just to the right of B
A A+ gives the reaction at B and1 3M MA A+ gives the reaction at B, and
A A i th b di t t B2 4M MA A+ gives the bending moment at B
19

Joint displacements, member end actions, and
•Once the redundants are found all the joint
support reactions
Once the redundants are found, all the joint
displacements, member end actions, and support reactions
can be found subsequentlyq y
•But it is easier to incorporate such calculations into the
b i t ti i t d f t i th tbasic computations, instead of postponing them as separate
calculations
20

• Example: To find out
j i di l• joint displacements,
• member end actions and
• reactions other than redundants• reactions other than redundants
Actual structure
1 2 3: 2Given P P M PL P P P P= = = =

{ } 1J
J
D
D
D
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
Joint displacements in the actual structure
d l d2JD⎩ ⎭ due to loads
Q⎧ ⎫
{ } 1
2
Q
Q
Q
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
Redundants
22

Joint displacements in the released
structure due to loads{ } 1
2
JL
JL
JL
D
D
D
⎧ ⎫
= ⎨ ⎬
⎩ ⎭2JLD⎩ ⎭
{ } 1RLA⎧ ⎫
⎨ ⎬
Reactions in the released structure due to
{ } 1
2
RL
RL
RL
A
A
⎧ ⎫
= ⎨ ⎬
⎩ ⎭
Reactions in the released structure due to
loads
23

Joint displacements in the released structure due to
i l f d d
JQijD
unit values of redundants
JQij
RQijA Reactions in the released structure due to unit
24
RQijA
values of redundants

D D D Q D Q= + +Joint displacements
1 1 11 1 12 2
2 2 21 1 22 2
J JL JQ JQ
J JL JQ JQ
D D D Q D Q
D D D Q D Q
= + +
= + +
J p
{ } { } { }D D D Q⎡ ⎤
oIn matrix form,
{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦
D D⎧ ⎫ ⎧ ⎫
{ } { }1 1
2 2
, ,,
J JL
J JL
J JL
D D
D D
D D
w ereh
⎧ ⎫ ⎧ ⎫
= =⎨ ⎬ ⎨ ⎬
⎩ ⎭ ⎩ ⎭
{ }11 12 1
21 22 2
,
JQ JQ
JQ
JQ JQ
D D Q
D Q
D QD
⎡ ⎤ ⎧ ⎫
⎡ ⎤ = = ⎨ ⎬⎢ ⎥⎣ ⎦
⎩ ⎭⎣ ⎦21 22 2JQ JQD QD ⎩ ⎭⎣ ⎦
D Joint displacement i in the released structure
25
JQijD J p
due to unit value of redundant j

• If there are j joint displacements to be obtained,
{ } { } { }D D D Q⎡ ⎤
and there are q redundants,
{ } { } { }
11 1
J JL JQ
qj j j q
D D D Q
×× × ×
⎡ ⎤= +⎣ ⎦
26

Reactions (other than redundants)
1 1 11 1 12 2R RL RQ RQA A A Q A Q
A A A Q A Q
= + +
= + +
A AA A Q⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧ ⎫
2 2 21 1 22 2R RL RQ RQA A A Q A Q= + +
11 121 1 1
21 222 2 2
RQ RQR RL
RQ RQR RL
A AA A Q
A AA A Q
⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎧ ⎫
= +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥
⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎣ ⎦
{ } { } { }R RL RQA A A Q⎡ ⎤= +⎣ ⎦{ } { } { }Q⎣ ⎦
• If there are r reactions to be obtained (other than
{ } { } { }A A A Q⎡ ⎤= + ⎣ ⎦
redundants) and q redundants,
27
{ } { } { }
11 1
R RL RQ
qr r r q
A A A Q
×× × ×
⎡ ⎤= + ⎣ ⎦

Member end actions
•Member end actions { } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦
{ }A Member end actions in the released structure
{ } { } { }M ML MQ Q⎣ ⎦
{ }MLA Member end actions in the released structure
due to loads
MQA⎡ ⎤⎣ ⎦
Member end actions in the released structure
due to unit values of redundants

If th b d ti d d d t
{ } { } { }A A A Q⎡ ⎤
If there are m member end actions and q redundants,
{ } { } { }
11 1
M ML MQ
qm m m q
A A A Q
×× × ×
⎡ ⎤= + ⎣ ⎦
29

• In the given example,In the given example,
1 2 32P P M PL P P P P= = = =
[ ]
69
64
P
Q
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
As found out earlier,
[ ]
6456
Q ⎢ ⎥−⎣ ⎦
,

{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦
{ } { } { }A A A Q⎡ ⎤+ ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦
{ } { } { }R RL RQA A A Q⎡ ⎤= +⎣ ⎦
To be found out

[ ]DTo get [ ]JLDTo get
2
10PL ⎡ ⎤2
1
5
4
JL
PL
D
EI
=
2
2
13
8
JL
PL
D
EI
= [ ]
2
10
138
JL
PL
D
EI
∴
⎡ ⎤
= ⎢ ⎥
⎣ ⎦
32

D⎡ ⎤⎣ ⎦To get JQD⎡ ⎤⎣ ⎦To get
2 2
3L L⎡ ⎤
2
11 12
2 2
21 22
3
1 32 2
1 424
JQ JQ
JQ
JQ JQ
L L
D D LEI EID
D D EIL L
⎡ ⎤
⎢ ⎥⎡ ⎤ ⎡ ⎤
⎡ ⎤ = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦
⎢ ⎥ ⎣ ⎦⎣ ⎦
21 22
2 2
JQ JQ
EI EI
⎢ ⎥ ⎣ ⎦⎣ ⎦
⎢ ⎥⎣ ⎦

[ ]
69P
Q
⎡ ⎤
= ⎢ ⎥Already we know [ ]
6456
Q = ⎢ ⎥−⎣ ⎦
Already we know,
Joint displacements
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦
2 2
10 1 3 69
13 1 4 648 2 56
PL L P
EI EI
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2
17PL ⎡ ⎤
⎢ ⎥=
5112EI ⎢ ⎥−⎣ ⎦
=

{ }RLATo get { }RL
MAM
Released structure with loads
d i
AV
and reactions
1 2
3
3
2
2 2 2
A
PL P L PL
M M P L
−
= − + − =1 2 3 2AV P P P P= + − =
{ } 1
2
RL
RL
P
A
A PL
⎧ ⎫
⎧ ⎫ ⎪ ⎪
= =⎨ ⎬ ⎨ ⎬
{ }
2
2
RL
RL
PL
A
⎨ ⎬ ⎨ ⎬
−⎩ ⎭ ⎪ ⎪⎩ ⎭

RQA⎡ ⎤⎣ ⎦To get
LL L L
R l d t t ith it l1 Released structure with unit value
of redundant Q1
2L
1
L L
R l d t t ith it l
1 1
A
− −⎡ ⎤
⎡ ⎤ = ⎢ ⎥⎣ ⎦
Released structure with unit value
of redundant Q2
2RQA
L L
⎡ ⎤ = ⎢ ⎥⎣ ⎦ − −⎣ ⎦

{ } { } { }⎡ ⎤
Reactions (other than redundants)
{ } { } { }R RL RQA A A Q⎡ ⎤= + ⎣ ⎦
2
691 1
P
P− −⎡ ⎤
⎧ ⎫
⎡ ⎤⎪ ⎪ 69
6456
2
1 1
2L
P
PL
L
⎡ ⎤
⎢
⎡ ⎤⎪ ⎪
= +⎨ ⎬ ⎢ ⎥−− ⎣ ⎦⎪ ⎪⎩ ⎭
⎥− −⎣ ⎦
107P ⎡ ⎤
2⎩ ⎭
107
3156
P
L
⎡
=
⎤
⎢ ⎥
⎣ ⎦
37

Choice of member end actions
AM BM
A
V V
BA
AV BV
•Any number among the 4 member end actions can be
chosen for analysis
•Usually two among the 4 are chosen•Usually two among the 4 are chosen
•Any two of the 4 member end actions can be chosen for
analysis
•Usually the moments at both ends are chosen
Usually the moments at both ends are chosen

Member end actions
{ }MLATo get
1MA A
L BA
1M 2MA
Member end actions considered
are the reactive moments at the end of
members in the actual structure
1 2,M MA A
1 2,ML MLA A are the reactive moments at the end of1 2ML ML
members in the released structure

PL
M PL= 2P P= 3P P=1 2P P=
2
PL
Released
structure
2 3
2P
2
P L
3
2
PL
L
2P
B
2 P 0
L
BA
1MLA = reactive moment just to the right of A
2
PL−
=
2MLA = reactive moment just to the left of B
3
2
PL
=
40
2

P
2
PL
L
P P
0
3MLA = reactive moment just to the right of B
PL−
=
0
L
B C
0
3MLA reactive moment just to the right of B
2
A = reactive moment just to the left of C 0=4MLA = reactive moment just to the left of C 0=
41

A B C
PL 2P P PPL
A B C
2
L L
2P
P L
3
2
PL2P
2
PL P P
2
2
0
L BA
2
0
L
B C
0
⎧ ⎫ ⎧ ⎫
2 P 0 B C
{ }
1
2
2
3 2
ML
ML
A PL
A PL
A
−⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪
= =⎨ ⎬ ⎨ ⎬{ }
3
4
2
0
ML
ML
ML
A
A PL
A
= =⎨ ⎬ ⎨ ⎬
−⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭
42
4ML ⎩ ⎭⎩ ⎭

A B CL
MQA⎡ ⎤⎣ ⎦To get
11
L L MQA⎡ ⎤⎣ ⎦
1 21 1
2
Q Q
L L
= =
⎡ ⎤
L 0 0
0 2
0
L L
L
− −⎡ ⎤
⎢ ⎥
⎢ ⎥
0
A B C2L
L L
0
0 0
L
⎢ ⎥=
−⎢ ⎥
⎢ ⎥
⎣ ⎦1
1
L L 0 0⎣ ⎦
2L L L−
0
43

H b d tiHence, member end actions
{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤+ ⎣ ⎦
2 2PL L L− − −⎧ ⎫ ⎡ ⎤ 0 554PL⎧ ⎫2 2
3 2 0 69
PL L L
PL L P
⎧ ⎫ ⎡ ⎤
⎪ ⎪ ⎢ ⎥
⎡ ⎤⎪ ⎪ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥⎢ ⎥ ⎣ ⎦
0.554
0.357
PL
L
⎧ ⎫
⎪ ⎪
⎪
=
⎪
⎨ ⎬2 0 6456
0 0 0
PL L
⎨ ⎬ ⎢ ⎥− − −⎢ ⎥ ⎣ ⎦⎪ ⎪
⎢ ⎥⎪ ⎪⎩ ⎭ ⎣ ⎦
0.643
0
L
= ⎨ ⎬
⎪ ⎪
⎪ ⎪⎩ ⎭0⎩ ⎭
44

Member end actions (with a different choice of member
end actions for analysis)
Member end actions considered
{ }MLATo get
A A
In the released structure,
are SF and BM (equal to reactions) just to
the left of B, and
1 2,ML MLA A
3 4,ML MLA A are SF and BM just to the right of B

PL
M PL= 2P P= 3P P=1 2P P=
2
PL
Released
structure
2 3
2P
3PL
2
P L
3
2
PL
L
2P
BA
2 P 0
A
1MLA = Shear force just to the left of B 3 2 0P P= − =
2MLA = Bending moment just to the left of B
2 3P L PL
P L M
46
2
3
2 2
P L M= − + =

2
PL
P P
2
0
L
B CB C
A = Shear force just to the right of B3MLA = Shear force just to the right of B
12 2 2 0P P P P= − = − =
4MLA = Bending moment just to the right of B4ML g j g
1
2
2 2
PL PL
PL M
−
= − + +
2 2
2
2
PL PL PL
PL PL
− −
+ +
47
2
2 2 2
PL PL= − + + =

A B C
PL 2P P PPL
A B C
2
L L
P L
3PL2P PL P
2P
2
P L
2
0
L
2P
2
PL
L
P P
0⎧ ⎫
2 P
0 0
{ }
1
2
0
3
2
ML
ML
A PL
A
A
⎧ ⎫
⎪ ⎪⎧ ⎫
⎪ ⎪⎪ ⎪
⎪ ⎪ ⎪ ⎪
⎨ ⎬ ⎨ ⎬{ } 2
3
4
2
0
ML
ML
ML
ML
A
A
A PL
⎪ ⎪ ⎪ ⎪
= =⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎩ ⎭
⎪ ⎪
48
4
2
ML⎩ ⎭ −⎪ ⎪
⎩ ⎭

A B CL
MQA⎡ ⎤⎣ ⎦To get
11
L L MQA⎡ ⎤⎣ ⎦
1 21 1
1 1
Q Q= =
⎡ ⎤
L 0 0
1 1
0 L
⎡ ⎤
⎢ ⎥
⎢ ⎥1 1
L
0
L
A B C2L
L L
0 1
0 L
⎢ ⎥=
−⎢ ⎥
⎢ ⎥
⎣ ⎦1
1
L L 0 L−⎣ ⎦
2L L
L
L−
1
L
1
49
1
1
11−

Hence, member end actions
{ } { } { }A A A Q⎡ ⎤+ ⎣ ⎦{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦
5⎡ ⎤
0
1 13PL
⎧ ⎫
⎪ ⎪ ⎡ ⎤
⎪ ⎪ 5
20LP
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
=
3
0 692
0 0 1 6456
PL
L P
⎡ ⎤
⎪ ⎪ ⎢ ⎥
⎡ ⎤⎪ ⎪ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎪ ⎪ 6456
36L
⎢ ⎥
⎢
⎦
⎥
⎣
0 0 1 6456
0PL L
⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎪ ⎪
⎢ ⎥⎪ ⎪ −⎣ ⎦−⎪ ⎪
⎩ ⎭2
⎪ ⎪
⎩ ⎭
50

Flexibilities of prismatic membersp
•Flexibility coefficients of a structure are calculated from
the contributions of individual members
•Hence it is worthwhile to construct member flexibility
matrices for various types of actionsyp
•Member oriented axes (local coordinates) and structure
oriented axes (global coordinates)
51

Member flexibility matrices for prismatic members with
B b
one end fixed and the other free
3 2
L L⎡ ⎤
•Beam member
[ ] 11 12
2
21 22
3 2M M
Mi
M M
L L
F F EI EI
F
F F L L
⎡ ⎤
⎢ ⎥⎡ ⎤
⎢ ⎥= =⎢ ⎥
⎢ ⎥⎣ ⎦21 22
2
M MF F L L
EI EI
⎢ ⎥⎣ ⎦
⎢ ⎥⎣ ⎦

[ ]
L
F =•Truss member [ ]MiF
EA
=•Truss member

•Plane frame memberPlane frame member

1
FM33
11 12 13 3 2
0 0
M M M
L
EA
F F F
⎡ ⎤
⎢ ⎥
⎡ ⎤ ⎢ ⎥
[ ]
11 12 13 3 2
21 22 23
231 32 33
0
3 2
M M M
Mi M M M
M M M
L L
F F F F
EI EI
F F F
L L
⎡ ⎤ ⎢ ⎥
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦
231 32 33
0
2
M M M
L L
EI EI
⎢ ⎥ ⎢ ⎥⎣ ⎦
⎢ ⎥
⎢ ⎥⎣ ⎦

G id b•Grid member

FM31
FM33
3 2
L L⎡ ⎤
[ ]
11 12 13
0
3 2
M M M
L L
EI EIF F F
L
⎡ ⎤
⎢ ⎥
⎡ ⎤ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
[ ] 21 22 23
231 32 33
0 0Mi M M M
M M M
L
F F F F
GJ
F F F
L L
⎢ ⎥ ⎢ ⎥= =
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦
⎢ ⎥
0
2
L L
EI EI
⎢ ⎥
⎢ ⎥⎣ ⎦

•Space frame member•Space frame member

0 0 0 0 0
L
E A
⎡ ⎤
⎢ ⎥
⎢ ⎥3 2
0 0 0 0
3 2Z Z
E A
L L
E I E I
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
[ ]
3 2
0 0 0 0
3 2
Z Z
Y Y
L L
E I E I
F
⎢ ⎥
−⎢ ⎥
⎢ ⎥
⎢ ⎥[ ]
0 0 0 0 0
M iF
L
G J
⎢ ⎥=
⎢ ⎥
⎢ ⎥
⎢ ⎥2
0 0 0 0
2 Y Y
L L
E I E I
⎢ ⎥
−⎢ ⎥
⎢ ⎥
⎢ ⎥2
0 0 0 0
2 Z Z
L L
E I E I
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦

Formalization of the Flexibility method
(Explanation using principle of complimentary virtual work)
{ } [ ]{ }D F A
For each member,
{ } [ ]{ }Mi Mi MiD F A=
contains relative displacements of the k end{ }DH contains relative displacements of the k end
with respect to j end of the i-th member
{ }MiDHere

•If there are m members in the structure,
{ } [ ] [ ] [ ] [ ] [ ] { }11 10 0 0 0MM MFD A⎧ ⎫ ⎧ ⎫⎡ ⎤
⎪ ⎪ ⎪ ⎪⎢ ⎥
{ }
{ }
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
{ }
{ }
22 2
33 3
0 0 0 0
0 0 0 0
MM M
MM M
FD A
FD A
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
{ } [ ] [ ] [ ] [ ] [ ] { }0 0 0 0MiMi MiFD A
⎪ ⎪ ⎪ ⎪⎢ ⎥
=⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
{ } [ ] [ ] [ ] [ ] [ ] { }0 0 0 0 MmMm MmFD A
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥
⎩ ⎭ ⎩ ⎭⎣ ⎦
{ } [ ]{ }D F A
{ } [ ]{ }M M MD F A=

{ } [ ]{ }M M MD F A={ } [ ]{ }M M M
[ ]MF is the unassembled flexibility matrix of the entire structure
{ }MA
{ }A
•Member end actions in will be related to the
structure actions applied to the released structure
{ }SA { }JA { }QAconsists of joint loads and redundant actions
{ }SAstructure actions applied to the released structure.
{ }S { }J { }Qj
{ } [ ]{ }M MS SA B A=Hence,
A i f i iAction transformation matrix
(equilibrium matrix)
{ } [ ]
{ }
{ }
J
M MJ MQ
A
A B B
⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤= ⎨ ⎬⎣ ⎦⎣ ⎦
i.e.,
{ } [ ]
{ }M MJ MQ
QA
⎨ ⎬⎣ ⎦⎣ ⎦
⎪ ⎪⎩ ⎭

[ ]B { }A { }A andl t to[ ]MJB
MQB⎡ ⎤⎣ ⎦
{ }MA { }JA
{ }QA
andrelate to
relate to{ }MA
•Each column in the submatrix consists of[ ]MJB
MQB⎡ ⎤⎣ ⎦ { }Qrelate { }M
•Each column in the submatrix consists of
member end actions caused by a unit value of a joint
load applied to the released structure.
[ ]MJB
pp
•Each column in the submatrix consists of
member end actions caused by a unit value of a
MQB⎡ ⎤⎣ ⎦
member end actions caused by a unit value of a
redundant applied to the released structure.

{ }SAδ•Suppose an arbitrary set of virtual actions { }SSuppose an arbitrary set of virtual actions
is applied on the structure.
{ } [ ]{ } [ ]
{ }
{ }
J
M MS S MJ MQ
A
A B A B B
A
δ
δ δ
δ
⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤= = ⎨ ⎬⎣ ⎦⎣ ⎦
⎪ ⎪⎩ ⎭
{ } [ ]{ } [ ]
{ }Q
QAδ⎣ ⎦⎣ ⎦
⎪ ⎪⎩ ⎭
External complimentary virtual work produced by the
{ }SAδ { }SD
External complimentary virtual work produced by the
virtual loads and actual displacements is
{ } { } { } { }
TT T* JD
W A D A Aδ δ δ δ
⎧ ⎫⎡ ⎤ ⎨ ⎬{ } { } { } { }T T J
S S J Q
Q
W A D A A
D
δ δ δ δ⎡ ⎤= = ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭

Internal complimentary virtual work produced by the
virtual member end actions and actual (relative){ }MAδ
end displacements is
{ }M
{ }MD
{ } { }
T*
M MU A Dδ δ=

•Equating the above two (principle of complimentary
i t l k)
{ } { } { } { }
T T
S S M MA D A Dδ δ=
virtual work),
{ } { } { } { }S S M MA D A Dδ δ
{ } [ ]{ }{ } [ ]{ }A A { } [ ]{ }M M MD F A={ } [ ]{ }M MS SA B A=But and
{ } [ ]{ }M MS SA B Aδ δ=Also,
{ } { } { } [ ] [ ][ ]{ }
TT T
S S S MS M MS SA D A B F B Aδ δ=Hence, { } { } { } [ ] [ ][ ]{ }S S S MS M MS S

{ } [ ] [ ][ ]{ }
T
D B F B A{ } [ ] [ ][ ]{ }S MS M MS SD B F B A=
{ } [ ]{ }S S SD F A=
[ ] [ ] [ ][ ]
T
F B F B
Where,
the assembled flexibility matrix
[ ] [ ] [ ][ ]
T
S MS M MSF B F B= , the assembled flexibility matrix
for the entire structure.

[ ]SF is partitioned into submatrices related to:[ ]S
{ }JA
p
joint loads
{ }QAand redundant actions
{ } [ ]{ }
{ } [ ] { }JJ JQJ J
F FD A
A
⎡ ⎤⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎣ ⎦ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬{ } [ ]{ }
{ }
{ }
[ ] { }
{ }
QJ J
Q QQJ
S S S
QQ
D
D AF F
F A
⎪ ⎪ ⎣ ⎦ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦
=
⎣ ⎦⎣ ⎦
⇒
T T
⎡ ⎤ ⎡ ⎤
Where,
[ ] [ ] [ ][ ]
T
JJ MJ M MJF B F B= [ ] [ ]
T
JQ MJ M MQF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
T
⎡ ⎤ ⎡ ⎤
T
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
[ ][ ]
T
QJ MQ M MJF B F B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ [ ]
T
QQ MQ M MQF B F B⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ ⎣ ⎦

{ } [ ]{ } { }D F A F A⎡ ⎤= + ⎣ ⎦{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
{ } { } { }D F A F A⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦{ } { } { }Q QJ J QQ QD F A F A⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦
1
{ } { } { }
1
Q QQ Q QJ JA F D F A
−
⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣⇒ ⎦

In the subsequent calculations, the above {AQ} should be
d
However the final values of redundants are obtained
used.
However, the final values of redundants are obtained
by including actual or equivalent joint loads applied
directly to the supports.y pp
{ } { } { }Q QC QFINAL
A A A= − +Thus, { } { } { }Q QC QFINAL
,
{ }QCA
represents actual and equivalent joint loads applied
directly to the supports, corresponding to
redundants
70
redundants.

{ }{ }QA
{ }JD
• Once redundants are found,
can be found out from,
{ } [ ]{ } { }D F A F A⎡ ⎤+ ⎣ ⎦
{ }J ,
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
71

• Similarly, support reactions caused by joint loads and
redundant can be obtained with an action transformation
matrix [ ]RSB[ ]RS
{ } [ ]{ } [ ]
{ }JA⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤{ } [ ]{ } [ ]
{ }
{ }
J
R RS S RJ RQ
Q
A B A B B
A
⎪ ⎪⎡ ⎤⎡ ⎤= = ⎨ ⎬⎣ ⎦⎣ ⎦
⎪ ⎪⎩ ⎭
•Each column in the submatrix consists of support
i d b i l f j i l d li d
[ ]RJB
reactions caused by a unit value of a joint load applied to
the released structure.
B⎡ ⎤RQB⎡ ⎤⎣ ⎦•Each column in the submatrix consists of
support reactions caused by a unit value of a
pp y
redundant applied to the released structure.

• If actual or equivalent joint loads are applied• If actual or equivalent joint loads are applied
directly to the supports,
{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦
t bi d j i t l d ( t l d
{ }RCA
represents combined joint loads (actual and
equivalent) applied directly to the supports.

•As seen earlier, member end actions due to actual loads,
are obtained by superimposing member end actions due
to restraint actions and combined joint loads
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦{ } { } [ ]{ } { }M MF MJ J MQ Q⎣ ⎦
{ }Awhere represents fixed end actions{ }MFAwhere represents fixed end actions

Comparison of the procedures explained with principle
of superposition and principle of complimentary virtual
work
•For calculating redundants,
{ } [ ] { } { }( )1
Q F D D
−
P i i l f iti
Principle of complimentary virtual work
{ } [ ] { } { }( )Q QLQ F D D= −Principle of superposition
{ } { } { }
1
Q QQ Q QJ JA F D F A
−
⎡ ⎤⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦⎣ ⎦
{ } { } { }Q QQ Q QJ J⎣ ⎦ ⎣ ⎦⎣ ⎦
[ ]F F⎡ ⎤⎣ ⎦H [ ] QQF F⎡ ⎤= ⎣ ⎦Hence,
{ } { }D F A⎡ ⎤d{ } { }Q A
{ } { }QL QJ JD F A⎡ ⎤= ⎣ ⎦and{ } { }QQ A=

•For calculating joint displacements,g j p
{ } { } { }J JL JQD D D Q⎡ ⎤= + ⎣ ⎦
Principle of superposition
{ } [ ]{ } { }J JJ J JQ QD F A F A⎡ ⎤= + ⎣ ⎦
{ } [ ]{ }JL JJ JD F A=Hence and{ } [ ]{ }JL JJ JD F A
JQ JQD F⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦
Hence, and
Q Q⎣ ⎦ ⎣ ⎦

•For calculating member end actions,g ,
{ } { } { }M ML MQA A A Q⎡ ⎤= + ⎣ ⎦Principle of superposition
{ } { } [ ]{ } { }M MF MJ J MQ QA A B A B A⎡ ⎤= + + ⎣ ⎦
{ } { } [ ]{ }ML MF MJ JA A B A= +Hence, and
MQ MQA B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

•For calculating support reactions•For calculating support reactions,
{ } { } { }R RL RQA A A Q⎡ ⎤= +⎣ ⎦Principle of superposition
{ } { } { }R RL RQ Q⎣ ⎦p p p
{ } { } [ ]{ } { }R RC RJ J RQ QA A B A B A⎡ ⎤= − + + ⎣ ⎦
{ } { } [ ]{ }RL RC RJ JA A B A= − +Hence, and
RQ RQA B⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦

Member flexibility matrix for a beam member with
t t th d b d timoments at the ends as member end actions
Required to find out rotations at the ends due to unit
h d l
1
2
x d y
M EI
−
= = −
moments at each end separately
1
1 1x
2xM EI
L dx
= =
2
d1
L
1
L
x 2
1
2
dy x
EI C
dx L
= +
3
1 2
6
x
EIy C x C
L
= + +
79
1 2
6L

0
L
y C
−
= ⇒ =0 0y C= ⇒ = 10
6x L
y C
=
= ⇒ =2
0
0 0
x
y C
=
= ⇒ =
2
1
2 6
dy x L
dx EI L
⎛ ⎞
∴ = −⎜ ⎟
⎝ ⎠⎝ ⎠
L L
y y
−
′ ′
0
,
6 3x x L
y y
EI EI= =
= =
1 1
3
M
L
F
E I
= 21
6
M
L
F
EI
−
=
6
1
1

Similarly,
L
F
L
F
−
1
11
3
MF
EI
=12
6
MF
EI
=
11 12 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
⎢ ⎥⎡ ⎤
∴ = = ⎢ ⎥⎢ ⎥ −⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦
81
6 3EI EI⎢ ⎥⎣ ⎦

Member flexibility matrix for a beam member with
t d h t d b d timoment and shear at one end as member end actions
3 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
⎢ ⎥⎡ ⎤
⎢ ⎥= =⎢ ⎥
⎢ ⎥⎣ ⎦
⎢ ⎥
2EI EI⎢ ⎥⎣ ⎦

SummarySummary
Flexibility method
• Flexibility matrices for truss beam and frame elements –
load transformation matrix-development of total flexibility
matrix of the structure .
83

James Clerk Maxwell (13 June 1831 – 5 November
1879) was a Scottish theoretical physicist and1879) was a Scottish theoretical physicist and
mathematician. His most important achievement was
classical electromagnetic theory. Maxwell also
developed the Maxwell–Boltzmann distribution, ap ,
statistical means of describing aspects of the kinetic
theory of gases. These two discoveries helped usher in
the era of modern physics, laying the foundation forp y y g
such fields as special relativity and quantum
mechanics.
Maxwell is also known for creating the first true colour
photograph in 1861 and for his foundational work on
the rigidity of rod-and-joint frameworks like those in
many bridges.
Maxwell is considered by many physicists to be the 19th-century scientist with
the greatest influence on 20th-century physics. His contributions to the science
are considered by many to be of the same magnitude as those of Isaac Newtony y g
and Albert Einstein. Einstein himself described Maxwell's work as the "most
profound and the most fruitful that physics has experienced since the time of
Newton." Einstein kept a photograph of Maxwell on his study wall, alongside
84
pictures of Michael Faraday and Newton.

Christian Otto Mohr (October 8, 1835 – October 2,
1918) was a German civil engineer one of the most1918) was a German civil engineer, one of the most
celebrated of the nineteenth century.
Starting in 1855, his early working life was spent in
railroad engineering for the Hanover and Oldenburgrailroad engineering for the Hanover and Oldenburg
state railways, designing some famous bridges and
making some of the earliest uses of steel trusses.
Even during his early railway years, Mohr hadg y y y ,
developed an interest in the theories of mechanics
and the strength of materials. In 1867, he became
professor of mechanics at Stuttgart Polytechnic, andp g y
in 1873 at Dresden Polytechnic in 1873. In 1874, Mohr
formalised the idea of a statically determinate
structure.
In 1882, he famously developed the graphical method for analysing stress
known as Mohr's circle and used it to propose an early theory of strength based
on shear stress He also developed the Williot-Mohr diagram for trusson shear stress. He also developed the Williot Mohr diagram for truss
displacements and the Maxwell-Mohr method for analysing statically
indeterminate structures, it can also be used to determine the displacement of
truss nodes and forces acting on each member. The Maxwell-Mohr method is
85
g
also referred to as the virtual force method for redundant trusses.

Module1 flexibility-1- rajesh sir

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