Structural Analysis - II Force Method Analysis

Structural Analysis - II
Indeterminate structures;Indeterminate structures;
Force method of analysis
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 IModule I
Statically and kinematically indeterminate structures
• Degree of static indeterminacy, Degree of kinematic
indeterminacy, Force and displacement method of analysis
• Method of consistent deformation-Analysis of fixed and
i b
continuous beams
• Clapeyron’s theorem of three moments-Analysis of fixed and
continuous beams
• Principle of minimum strain energy-Castigliano’s second theorem-
2
p gy g
Analysis of beams, plane trusses and plane frames.

Types of Framed Structures
• a. Beams: may support bending moment, shear force and axial
force
yp
• b. Plane trusses: hinge joints; In addition to axial forces, a
member CAN have bending moments and shear forces if it has
3
loads directly acting on them, in addition to joint loads

• c. Space trusses: hinge joints; any couple acting on a member
should have moment vector perpendicular to the axis of the
member, since a truss member is incapable of supporting a
twisting momentg
• d. Plane frames: Joints are rigid; all forces in the plane of the
frame, all couples normal to the plane of the frame
4

• e Grids: all forces normal to the plane of the grid all couples• e. Grids: all forces normal to the plane of the grid, all couples
in the plane of the grid (includes bending and torsion)
• f. Space frames: most general framed structure; may support
bending moment, shear force, axial force and torsion
5
g , ,

Deformations in Framed Structures
, ,x y zT M MThree couples:, ,x y zN V VThree forces:
•Significant deformations in framed structures:•Significant deformations in framed structures:
Structure Significant deformationsStructure Significant deformations
Beams flexural
Plane trusses axial
Space trusses axial
Plane frames flexural and axial
Grids flexural and torsional
Space frames axial, flexural and torsional
6

Types of deformations in framed structures
b) axial c) shearing d) flexural e) torsional
7
b) axial c) shearing d) flexural e) torsional

Equilibrium
•Resultant of all actions (a force, a couple or both) must vanish for static
equilibrium
q
0F∑ 0F∑ ∑
•Resultant force vector must be zero; resultant moment vector must be
zero
0xF =∑ 0yF =∑ 0zF =∑
0xM =∑ 0yM =∑ 0zM =∑∑ y∑ ∑
•For 2 dimensional problems (forces are in one plane and
0F =∑ 0F =∑ 0M =∑
•For 2-dimensional problems (forces are in one plane and
couples have vectors normal to the plane),
0xF =∑ 0yF =∑ 0zM =∑
8

Compatibility
•Compatibility conditions: Conditions of continuity ofp y y
displacements throughout the structure
•Eg: at a rigid connection between two members, theg g ,
displacements (translations and rotations) of both members must
be the same
9

Indeterminate Structures
Force method and Displacement method
• Force method (Flexibility method)
• Actions are the primary unknowns
St ti i d t i f k ti th• Static indeterminacy: excess of unknown actions than
the available number of equations of static equilibrium
• Displacement method (Stiffness method)
• Displacements of the joints are the primary unknowns
• Kinematic indeterminacy: number of independent
Kinematic indeterminacy: number of independent
translations and rotations

Static indeterminacy
• Beam:
• Static indeterminacy = Reaction components number of eqns• Static indeterminacy = Reaction components - number of eqns
available 3E R= −
• Examples:• Examples:
• Single span beam with both ends hinged with inclined
l dloads
• Continuous beam
• Propped cantilever
• Fixed beam

Degree of StaticStructure Degree of Static
Indeterminacy
Structure
0
1
3
12

Degree of Static
I d t i
Structure
0
Indeterminacy
0
1
55
2

• Rigid frame (Plane):g ( )
• External indeterminacy = Reaction components - number of eqns
available 3E R= −available
• Internal indeterminacy = 3 × closed frames
3E R
3I a=• Internal indeterminacy = 3 × closed frames 3I a=
• Total indeterminacy = External indeterminacy + Internal indeterminacy
( )3 3T E I R a= + = − +
• Note: An internal hinge will provide an additional eqn
14

Example 1 Example 2p p
( )3 3T E I R a= + = − +( )3 3T E I R a= + = − + ( )
( )3 3 3 3 2 12= × − + × =( )2 2 3 3 0 1= × − + × =
Example 3
Example 4
( )
( )
3 3T E I R a= + = − + ( )
( )
3 3T E I R a= + = − +
( )3 2 3 3 3 12= × − + × = ( )4 3 3 3 4 21= × − + × =

Degree of Static
I d t i
Structure
Indeterminacy
3
63
16

Degree of Static
I d t i
Structure
Indeterminacy
2
1

• Rigid frame (Space):g ( p )
• External indeterminacy = Reaction components - number of eqns
available 6E R= −available
• Internal indeterminacy = 6 × closed frames
6E R
Example 1
( )
( )
6 6T E I R a= + = − +
( )4 6 6 6 1 24= × − + × =
If i l d f ti l t d t ti i d t i i tIf axial deformations are neglected, static indeterminacy is not
affected since the same number of actions still exist in the
structure

Plane truss (general):
External indeterminacy = Reaction components - number of eqns available
3E R= −
Minimum 3 members and 3 joints.
A dditi l j i t i 2 dditi l b
( )3 2 3 2 3m j j= + − = −Hence, number of members for stability,
Any additional joint requires 2 additional members.
( )2 3I m j= − −Hence, internal indeterminacy,
Total (Internal and external) indeterminacyTotal (Internal and external) indeterminacy
( )3 2 3
2
T E I R m j
m R j
= + = − + − −
= + − 2m R j+
• m: number of members
• R: number of reaction components
• j : number of joints
• Note: Internal hinge will provide additional eqn

E l 1Example 1
2 9 3 2 6 0T m R j= + − = + − × =
3 3 3 0E R= − = − =
0I T E= − =
Example 2
2 15 4 2 8 3T R
p
2 15 4 2 8 3T m R j= + − = + − × =
3 4 3 1E R= − = − =
2I T E= − =
Example 3 2 6 4 2 5 0T m R j= + − = + − × =
( )3 1 4 4 Hinge at0 AE R= − + = − =
( )
0I T E= − =

Example 4 2 7 3 2 5 0T m R j= + − = + − × =Example 4 2 7 3 2 5 0T m R j+ +
3 3 3 0E R= − = − =
0I T E 0I T E= − =
2 6 4 2 4 2T m R j= + − = + − × =
3 4 3 1E R= = =
Example 5
3 4 3 1E R= − = − =
1I T E= − =
2 11 3 2 6 2T m R j= + − = + − × =Example 6
3 3 3 0E R= − = − =
2I T E= − =
2I T E= =

• Wall or roof attached pin jointed plane truss (Exception to the• Wall or roof attached pin jointed plane truss (Exception to the
above general case):
• Internal indeterminacy 2I m j= −
• External indeterminacy = 0 (Since, once the member forces
are determined, reactions are determinable)
Example 1 Example 3Example 2Example 1 Example 3p
2
6 2 3 0
T I m j= = −
= − × =
2
5 2 1 3
T I m j= = −
= − × =
2
7 2 3 1
T I m j= = −
= × =
22
7 2 3 1= − × =

• Space Truss:
External indeterminacy = Reaction components - number
of eqns available
Minimum 6 members and 4 joints
6E R= −
( )6 3 4 3 6m j j= + − = −Hence, number of members for stability,
Minimum 6 members and 4 joints.
Any additional joint requires 3 additional members.
( )6 3 4 3 6m j j+Hence, number of members for stability,
( )3 6I m j= − −Hence, internal indeterminacy,
Total (Internal and external) indeterminacy
( )6 3 6T E I R m j= + = − + − −
3m R j= + −

Example
3T m R j= +Total (Internal and external) indeterminacy 3T m R j= + −Total (Internal and external) indeterminacy
12 9 3 6 3T∴ = + − × =
6 9 6 3E R= − = − =

Kinematic indeterminacyKinematic indeterminacy
• joints: where members meet, supports, free ends
• joints undergo translations or rotations
•in some cases joint displacements will be known, from the restraint
conditions
•the unknown joint displacements are the kinematically•the unknown joint displacements are the kinematically
indeterminate quantities
odegree of kinematic indeterminacy: number of degrees of
f dfreedom
Two types of DOF
• Nodal type DOF
J i DOF
Two types of DOF
25
• Joint type DOF

• degree of kinematic indeterminacy (degrees of freedom) is
d fi ddefined as:
• the number of independent translations and rotations in a
structure.
DOF = 1
DOF = 2
DOF 2

• in a truss the joint rotation is not regarded as a degree of• in a truss, the joint rotation is not regarded as a degree of
freedom. joint rotations do not have any physical significance as
they have no effects in the members of the truss
• in a frame, degrees of freedom due to axial deformations can be
neglected

Kinematic Degree
f F d
Structure
of Freedom
2
11
22

Kinematic Degree
f F d
Structure
of Freedom
4
If the effect of the cantilever portion is considered as a joint load atIf the effect of the cantilever portion is considered as a joint load at
the roller support on the far right, kinematic indeterminacy can be
taken as 2.

Kinematic Degree
f F d
Structure
of Freedom
2
3

Kinematic Degree
f F d
Structure
of Freedom
6
2
5

Method of Consistent Deformation
Illustration of the method
Problem
Released structure
32
Released structure

D fl ti f l d
4
5wL
Δ
Deflection of released
structure due to actual
loads
384
B
EI
Δ =
C Deflection of released
structure due to
redundant applied as app
load
33

3
BR L
Deflection due o
48
t B
B
R L
I
R
E
=
3
BR L
Δ
48
B
B
EI
Δ =
4 3
5 BwL R L
Compatibility condition (or equation of
superposition or equation of geometry)
5
384 48
BwL R L
EI EI
=
5
B
wL
R∴ =
8
B
34

•A general approach (applying consistent sign convention
for loads and displacements):
RA l it l d di t BR•Apply unit load corresponding to
3
48
B
L
EI
δ =Let the displacement due to unit load be
BR B BR δDisplacement due to is
35

44
5
384
B
wL
EI
Δ = − (Negative, since deflection is downward)
0B B BR δΔ + = (Compatibility condition)
B
B
B
R
δ
Δ
= −
5
8
B
wL
R∴ =
Bδ 8
36

Example 1: Propped cantilever
AM
PP
A
2l 2l2l 2l
A B
AV BV
Choose VB as the redundant
P
AM
2l 2l
AV
37
Released structure

Find deflection at B of the released structure
3
1 2 5PL l l l Pl− −⎛ ⎞
Δ = + =⎜ ⎟Pl− . . .
2 2 2 2 3 2 48
B
EI EI
Δ = + =⎜ ⎟
⎝ ⎠2
Pl
EI
M
diagram
EIEI
Apply unit load on released structure corresponding to VApply unit load on released structure corresponding to VB
and find deflection at B
3
3
B
l
EI
δ =
1
l
3EI
38

0R δΔ
3
5 3 5B Pl EI P
R
−Δ
= = =0B B BR δΔ + = 3
.
48 16
B
B
R
EI lδ
= = =
+ve sign indicates that VB is in the same direction of the unit load.
i.e., in the upward direction.
P
2l 2l
5 3
2 16 16
A
Pl Pl Pl
M = − =Other reactions
2l 2l
5
16
P11
16
P
16
5
32
Pl
Bending moment diagram
3Pl
39
16
−

Example 2: Continuous beam
10kN
A
2m 2m
A B 4m
C
10kN
A B C
V VVAV CVBV
Choose VB as the redundant
40

10kN
2 6
A
BΔ C
2m 6m
B
Released structure
2.5 kN7.5 kN
15
EI
10
EI
1 251 358 6+⎛ ⎞
A B C
EI
EI
4m
Conjugate beam of the released structure (to find ∆ )
( )
1 25
0.5 15 8 35
EI EI
=× × −
1 358 6
80.5 15 8
3EI EI
+⎛ ⎞ ÷ =× × ×⎜ ⎟
⎝ ⎠
Conjugate beam of the released structure (to find ∆B)
1 73.3334
25 4 0 5 10 4
⎛ ⎞∴Δ ⎜ ⎟ (numerically)
41
25 4 0.5 10 4
3
B
EI EI
⎛ ⎞∴Δ = =× − × × ×⎜ ⎟
⎝ ⎠
(numerically)

1kN1kN
4m
A B 4m
C
4m 4m
Released structure with unit load corresponding to V (to find δ )Released structure with unit load corresponding to VB (to find δB)
3 3
8 10.667
B
l
δ = = =
(numerically)
(di i i ∆ i d d )48 48
B
EI EI EI
δ
0R δΔ +
73.333
6 875B EI
R kN
−Δ −
(direction is same as ∆B i.e., downwards)
0B B BR δΔ + = . 6.875
10.667
B
B
B
R kNm
EIδ
= = = −
i i di t th t V i i th it di ti f th it l d-ve sign indicates that VB is in the opposite direction of the unit load.
i.e., in the upward direction.
42

10kN
Other reactions
10kN
A B CB C
4.063AV kN= 0.938CV kN=6.875 kNA
Bending moment diagramg g
A
B
C
8.122 kNm
A C
3.752 kNm−

Example 3A:
2m
3m
CD
30kN
B
EI is constant
2m
D
MD
A
2m
6m
C
D
HD
30kN
B
VD
2m
B
HA
Choose H as the redundant
44
A
Choose HA as the redundant

D CD
To find ∆HA and δHA (using unit load method)
10kN
2m3m
CD
2m6m
B
C
2m
A
B
2m
A
B
1kNE=2x105N/mm2
I= 2x108mm4
Portion Origin Limits M m Mm m2
AB A 0-2 0 x 0 x2AB A 0 2 0 x 0 x
BC A 2-4 -10(x-2) x -10x2+20x x2
CD A 0 3 20 4 80 16
( ) ( )
4 3
21 1
10 20 80
Mmdx
x x dx dxΔ = = + +∫ ∫ ∫
CD A 0-3 -20 4 -80 16
( ) ( )
2 0
10 20 80HA x x dx dx
EI EI EI
Δ = = − + + −∫ ∫ ∫
[ ]
43
321 1x⎡ ⎤ 306 67−
[ ]
32
0
2
1 1
10 10 80
3
HA
x
x x
EI EI
⎡ ⎤
Δ = − + + −⎢ ⎥
⎣ ⎦
306.67
EI
−
=

4 32 2 4 33
69 3316⎡ ⎤ ⎡ ⎤
4 32 2
0 0
16
HA
m dx x
dx dx
EI EI EI
δ = = +∫ ∫ ∫
33
00
69.3316
3
xx
EIEIEI
⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
0HA A HAH δΔ + =
306.67
4.4.
69 3
3
3
2B
A
EI
H
EI
kN
δ
−Δ
= = =
HA A HA 69.33B EIδ
46

Example 3B: 2I
CD
2m
6m
CD
30kN
2m
I
B
M
2m
A
2
6m
2I
C
D
HD
MD
A
30kN
2m
6m
IVD
Choose H as the redundant
2m
B
H
Choose HA as the redundant
A
HA

2I
D
2I
CD
To find ∆HA and δHA (using unit load method)
30kN
2m6m
I
CD
2m6m
I
B
C
2m
A
B
2m
A
B
1kNE=2x105N/mm2
I= 2x108mm4
Portion Origin Limits M m Mm m2
AB A 0-2 0 x 0 x2
BC A 2-4 -30(x-2) x -30x2+60x x2
CD A 0-6 -60 4 -240 16
( ) ( )
4 6
21 1
30 60 240
Mmdx
x x dx dxΔ = = + +∫ ∫ ∫
CD A 0 6 60 4 240 16
( ) ( )
2 0
30 60 240
2
HA x x dx dx
EI EI EI
Δ = = − + + −∫ ∫ ∫
[ ]
4 63 21 1
10 30 240⎡ ⎤Δ 920−
[ ]
63 2
02
1 1
10 30 240
2
HA x x x
EI EI
⎡ ⎤Δ = − + + −⎣ ⎦
920
EI
−
=

4 62 2
16d
4 63
69 3338x⎡ ⎤ ⎡ ⎤2 2
0 0
16
2
HA
m dx x
dx dx
EI EI EI
δ = = +∫ ∫ ∫
3
00
69.3338
3
xx
EIEIEI
⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
0H δΔ + =
920
. 13.269
69 333
HA
A
EI
H kN
EIδ
−Δ
= = =0HA A HAH δΔ + = 69.333
A
HA EIδ

Example 4: Two or more redundants
60kN
A B C
80kN 50kN
D
3m 3m
A B C
3m 3m 3m 3m
60kN 80kN 50kN
3m 3m
A B C D
3m 3m 3m 3m
Choose VB and Vc as the redundant
50

60kN 80kN 50kN
A B C D
BΔ CΔB C
1kN
A B D
δ
C
BBδ CBδ
1kN
A B D
δ
C
BCδ
CCδ
0B B BB C BCV Vδ δΔ + + = 0C B CB C CCV Vδ δΔ + + =

60kN
A B C
80kN 50kN
DA B
3m 3m 3m 3m 3m 3m
A B C D
150kNm
15
120kNm
60kNm
3m 15m
525kN825kN
240kNm 240kNm
A B C D
360kNm
9 9
240kNm 240kNm
9m 9m
1620kN1620kN
A B C D
125kNm
100kNm
50kNm
15m 3m
687.5kN437.5kN

3420 8280 2325 14025BEIΔ = + + = 2790 8280 2850 13920CEIΔ = + + =
D2kNm
4kNm
A
B C
D
12m
6m
16kN20kN
2kNm
16kN20kN
96BBEIδ = 84CBEIδ =
A D2kNm
4kNm
A
B C
D
12m
6m
20kN16kN
2kNm
84BCEIδ =
96CCEIδ =

0V Vδ δΔ + + = 14025 96 84 0B CV V⇒ + + =0B B BB C BCV Vδ δΔ + + =
0C B CB C CCV Vδ δΔ + + =
14025 96 84 0B CV V⇒ + +
13920 84 96 0B CV V⇒ + + =C B CB C CC B C
82BV kN= − 73.25CV kN= −
( )82BV kN= ↑ ( )73.25CV kN= ↑
( )19.25AV kN= ↑ ( )15.5DV kN= ↑
( )B ( )C

Clapeyron’s theorem of three moments
1. Uniform loading
w kN m w kN m1w kN m
1 1l I
A B C
2w kN m
2 2l I
M
1 1,l I 2 2,l I
1
AM
EI
B C
1
BM
EI
2
BM
EI
2
CM
EI
A B C
M/EI diagram for joint moments
+
B C
2
1 1
18
w l
EI
2
2 2
28
w l
EI
A B C
M/EI diagram for simple beam moments

To find slopes at B using Conjugate Beam Method:
2
1 1 2 2M l l M l l w l l⎡ ⎤⎡ ⎤
From span AB:
1 1 1 1 1 1 1
11 1
1 1 1
1 1 2 2
. . . . . . .
2 3 2 3 3 8 2
A B
B
M l l M l l w l l
lV l
EI EI EI
⎡ ⎤⎡ ⎤
+= + ⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦
3
1 1 1 1
1
1 1 16 3 24
A B
B BA
M l M l w l
V
EI EI EI
θ⇒ = + + =
From span BC:
2
2 2 2 2 2 2 2
22 2
2 2 2
1 1 2 2
. . . . . . .
2 3 2 3 3 8 2
B C
B
M l l M l l w l l
lV l
EI EI EI
⎡ ⎤⎡ ⎤
+= + ⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦⎣ ⎦
3
2 2 2 2
2
3 6 24
B C
B BC
M l M l w l
V
EI EI EI
θ⇒ = + + =
2 2 23 6 24EI EI EI

BAθ
A B C
BA
BCθ
Deflected shape
0BA BC BA BCθ θ θ θ+ = ⇒ = −
33
2 2 2 21 1 1 1
3 6 246 3 24
B CA B M l M l w lM l M l w l
EI EI EIEI EI EI
⎛ ⎞
+ +⇒ + + = −⎜ ⎟
⎝ ⎠2 2 21 1 1 3 6 246 3 24 EI EI EIEI EI EI ⎝ ⎠
3 3
1 2l lM l M l w l w l⎛ ⎞1 21 2 1 1 2 2
1 21 2 1 2
2
4 4
A C
B
l lM l M l w l w l
M
EI EIEI EI EI EI
⎛ ⎞
+⇒ + + = − −⎜ ⎟
⎝ ⎠

2. General loadingg
1a
2a
A
B
C
x x1x 2x
1 1 1 1
1 1 1 16 3
A B
BA
M l M l a x
EI EI EI l
θ = + +
2 2 2 2
2 2 2 23 6
B C
BC
M l M l a x
EI EI EI l
θ = + +
2 2 2 2
1 21 2 1 1 2 26 6
2A C
l lM l M l a x a x
M
⎛ ⎞
+⇒ + + = − −⎜ ⎟BA BCθ θ= −
1 21 2 1 1 2 2
2 BM
EI EIEI EI EI l EI l
+⇒ + + = − −⎜ ⎟
⎝ ⎠
BA BCθ θ

2. General loading with support settlement
1a
2a ;A B
B C
δ δ
δ δ
>
<
A
B
C
x x1x 2x
M l M l δ δ1 1 1 1
1 1 1 1 16 3
A B A B
BA
M l M l a x
EI EI EI l l
δ δ
θ
−
= + + +
2 2 2 2
2 2 2 2 23 6
B C C B
BC
M l M l a x
EI EI EI l l
δ δ
θ
−
= + + +
BA BCθ θ= −
( ) ( )1 21 2 1 1 2 2
1 21 2 1 1 2 2 1 2
6 66 6
2 A B C BA C
B
l lM l M l a x a x
M
EI EIEI EI EI l EI l l l
δ δ δ δ− −⎛ ⎞
+⇒ + + = − − − −⎜ ⎟
⎝ ⎠
1 21 2 1 1 2 2 1 2l l l l⎝ ⎠

Example 1:
20kN m 20kN m
4m
A B C
4m
EI is constant
3 3
l lM l M l l l⎛ ⎞
1 2l l=
3 3
1 21 2 1 1 2 2
1 21 2 1 2
2
4 4
A C
B
l lM l M l w l w l
M
EI EIEI EI EI EI
⎛ ⎞
++ + = − −⎜ ⎟
⎝ ⎠
1 2w w=2
2
4
4
B
wl
M = −
0A CM M= = 0A CM M
2
8
B
wl
M∴ = −
2
20 4
40
8
kNm
×
= − = −
8 8

Example 2:
W
l
A
B Cl l D
EI is constant
EI EI
Span ABC
0M =
EI is constant
3 3
l lM l M l l l⎛ ⎞
1 2l l= 1 2EI EI= 0AM =
3 3
1 21 2 1 1 2 2
1 21 2 1 2
2
4 4
A C
B
l lM l M l w l w l
M
EI EIEI EI EI EI
⎛ ⎞
++ + = − −⎜ ⎟
⎝ ⎠
4 0A B CM M M⇒ + + =
4 0M M⇒ + = ( )1
4 0B CM M⇒ + = ( )1

Wl
1 Wl
a l=
l
A
B Cl l D
Span BCD
4
Wl
EI
2
2
. .
2 4
2
a l
l
x
=
=
0M
( ) 1 1 2 26 6
4
a x a x
l M M M+ +
2
20DM =
( )2 1
4 6
lWl
l M M l
⎛ ⎞
⎜ ⎟
( ) 1 1 2 2
4B C Dl M M M
l l
+ + = − −
( )2
4 6 . .
22 4
B Cl M M l
⎛ ⎞+ = − ⎜ ⎟
⎝ ⎠
3
4
Wl
M M ( )24
8
B CM M+ = − ( )2
Wl− Wl
M
10
C
Wl
M =
40
BM =( ) ( )&1 2

40
Wl
5
Wl
A
B C
40
10
Wl−
D
5
10
Bending moment diagram
63

Example 3: 20kN m 20kN mp
6mA B C
8mEI is constant
20kN m 20kN m
6mA B C
8mImaginary
span A’Aspan A A
Span A’AB
1
2
0
20
w
w kN m
=
= ' 0AM = 1 2EI EI=
Span A AB
3 3
1 2' 1 2 1 1 2 2
2A B
A
l lM l M l w l w l
M
EI EI
⎛ ⎞
++ + = − −⎜ ⎟
⎝ ⎠
1 21 2 1 24 4
A
EI EIEI EI EI EI⎜ ⎟
⎝ ⎠

3
20 6×20 6
2 6 6
4
A BM M
×
× + × = −
2 180M M+ ( )12 180A BM M+ = −
Span ABC
( )1
1 2 20w w kN m= =
3 3
1 21 2 1 1 2 2
1 21 2 1 2
2
4 4
A C
B
l lM l M l w l w l
M
EI EIEI EI EI EI
⎛ ⎞
++ + = − −⎜ ⎟
⎝ ⎠
0CM =
1 21 2 1 2⎝ ⎠
3 3
20 6 20 8
6 2 14
4 4
A BM M
× ×
+ × = − −
4 4
3 14 1820A BM M+ = − ( )2
( ) ( )&1 2 28AM kNm= − 124BM kNm= −

Example 4 (Support settlement):
2
26320EI kN 87δ δEI is constant
20kN m 20kN m20kN m
2
26320EI kNm= 87B C mmδ δ= =EI is constant
6mA
D
C 3m
B3m
EI EI EI= =0M
Span ABC
1 2EI EI EI= =0AM =
( ) ( )3 3
1 21 2 1 1 2 2
66 C BA BA C
l lM l M l w l w l δ δδ δ −−⎛ ⎞ ( ) ( )1 21 2 1 1 2 2
1 21 2 1 2 1 2
66
2
4 4
C BA BA C
B
l lM l M l w l w l
M
EI EIEI EI EI EI l l
δ δδ δ⎛ ⎞
++ + = − − − −⎜ ⎟
⎝ ⎠
( )
( )3 3
1 20 3 20 6 6 6 00.087
18 6
26320 4 26320 4 26320 3 6
B CM M
× × ×−
+ = − − − −
× ×
3 560.78B CM M+ = ( )1

Span BCD
0DM =
( ) ( )3 3
1 21 2 1 1 2 2
6 6
2
4 4
B C D CB D
C
l lM l M l w l w l
M
EI EIEI EI EI EI l l
δ δ δ δ− −⎛ ⎞
++ + = − − − −⎜ ⎟
⎝ ⎠
D
1 21 2 1 2 1 24 4EI EIEI EI EI EI l l⎜ ⎟
⎝ ⎠
( )
( )3 3
1 20 6 20 3 6 0 6 0.087
6 18M M
× × × −
+( )6 18
26320 4 26320 4 26320 6 3
B CM M+ = − − − −
× ×
3 560.78B CM M+ = ( )2
140.195BM kNm= 140.195CM kNm=
Alternatively, from symmetry, B CM M=
67
3 560.78B CM M+ = 4 560.78 140.195B BM M kNm⇒ = ⇒ =

Examples 5 (Fixed Beam)xa p es 5 ( xed ea )
P
A
B
C
l
A
P
A
B
C
A′ B′
l
A
A′ B
Imaginary
span A’A
Imaginary
span BB’span A A span BB
1 21 2 1 1 2 26 6
2A C
l lM l M l a x a x
M
⎛ ⎞
++ +⎜ ⎟
68
1 21 2 1 1 2 2
1 21 2 1 1 2 2
2A C
BM
EI EIEI EI EI l EI l
++ + = − −⎜ ⎟
⎝ ⎠

Span A’AB
1 l
' 0AM =
4
Pl
EI
2
1
. .
2 4
Pl
a l
l
=
1 2EI EI=
4EI
2
2
l
x =A B
6Pl6
2
16
A B
Pl
M M⇒ + = −
S ABB’
( )1
' 0BM = 1 2EI EI=
Span ABB’
2
1 1,
8 2
Pl l
a x= =
8 2
6
2
16
A B
Pl
M M⇒ + = − ( )2
16
( ) ( )&1 2 A B
Pl
M M= = −
69
( ) ( )&1 2
8
A BM M

Examples 6 (Fixed Beam)
w
xa p es 6 ( xed ea )
lA
B
CC
70

ENERGY PRINCIPLES BASED ON DISPLACEMENT FIELDENERGY PRINCIPLES BASED ON DISPLACEMENT FIELD
Principle of Minimum
Total Potential
Castigliano’s
Theorem (Part I)Total Potential
Energy (PMTPE)
Theorem (Part I)
alternative forms ofalternative forms of
Principle of Stationary TotalPrinciple of Stationary Total
Potential Energy (PSTPE)
71

Principle of Stationary Total Potential Energy (PSTPE)p y gy ( )
When the displacement field in a loaded elastic structure is given a
ll d bit t b ti i t i i tibilit dsmall and arbitrary perturbation, maintaining compatibility and
without disturbing the associated force field, then the first variation
of the total potential energy is equal to zero, if the forces are in a
state of static equilibrium.
72

Alternative form of Principle of Stationary Total Potential EnergyAlternative form of Principle of Stationary Total Potential Energy
(PSTPE)
The total potential energy π in a loaded elastic structure expressed as a
function of n independent displacements D1, D2, … Dn in a
compatible displacement field must be rendered stationary with thecompatible displacement field must be rendered stationary, with the
partial derivative of π with respect to every Dj being equal to zero, if
the associated force field is to be in a state of static equilibrium.

Principle of Minimum Total Potential Energy (PMTPE)Principle of Minimum Total Potential Energy (PMTPE)
When the displacement field in a loaded linear elastic structure is given
a small and arbitrary perturbation, maintaining compatibility and
without disturbing the associated force field, then the first variation
of the total potential energy is equal to zero, if the forces are in ap gy q ,
state of static equilibrium.
Castigliano’s Theorem (Part I)
If the strain energy, U, in an elastic structure, subject to a system of
external forces in static equilibrium, can be expressed as a functionq , p
of n independent displacements D1, D2, … Dn satisfying
compatibility, then the partial derivative of U with respect to every
Dj will be equal to the value of the conjugate force, Fj.
74
Dj will be equal to the value of the conjugate force, Fj.

ENERGY PRINCIPLES BASED ON FORCE FIELDENERGY PRINCIPLES BASED ON FORCE FIELD
Principle of Minimum Total
Complementary Potential
Castigliano’s
Theorem (Part II)
Complementary Potential
Energy (PMTCPE)
Theorem of Least
Work
alternative forms of
Principle of StationaryPrinciple of Stationary
Complementary Total Potential
Energy (PSCTPE)

Principle of Stationary Total Complementary Potential EnergyPrinciple of Stationary Total Complementary Potential Energy
(PSTCPE)
When the force field in a loaded elastic structure is given a small and
arbitrary perturbation, maintaining equilibrium compatibility and
without disturbing the associated displacement field, then the firstg p ,
variation of the total complementary potential energy is equal to
zero, if the displacements satisfy compatibility.
76

Alternative form of Principle of Stationary Total Complementary
Potential Energy (PSTCPE)Potential Energy (PSTCPE)
The total complementary potential energy π* in a loaded elasticp y p gy
structure expressed as a function of n independent forces F1, F2, …
Fn in a statically admissible force field must be rendered stationary,
with the partial derivative of π* with respect to every Fj being equalw t t e pa t a de vat ve o w t espect to eve y j be g equa
to zero, if the associated displacement field is to satisfy
compatibility.

Principle of Minimum Total Complementary Potential EnergyPrinciple of Minimum Total Complementary Potential Energy
(PMTCPE)
When the force field in a loaded linear elastic structure is given a small
and arbitrary perturbation, maintaining equilibrium compatibility
and without disturbing the associated displacement field, then theg p ,
first variation of the total complementary potential energy is equal
to zero, if the displacement satisfy compatibility.
78

Castigliano’s Theorem (Part II)Castigliano s Theorem (Part II)
If the complementary strain energy, U*, in an elastic structure, with a
kinematically admissible displacement field, is expressed as a
function of n independent external forces F1, F2, … Fn satisfying
equilibrium, then the partial derivative of U* with respect to every Fjq , p p y j
will be equal to the value of the conjugate displacement, Dj.
If the behaviour is linear elastic, U* can be replaced by U.
Thus, Castigliano’s Theorem (Part II) can otherwise be stated as:
“If U is the total strain energy in a linear elastic structure due to
application of external forces F1 F2 F3 F at points A1 A2 A3 Aapplication of external forces F1, F2, F3, … Fn at points A1, A2, A3,…, An
respectively in the directions AX1, AX2, AX3,…, AXn then the
displacements at points A1, A2, A3,…, An respectively in the directions
AX AX AX AX ∂U/∂F ∂U/∂F ∂U/∂F ∂U/∂F
79
AX1, AX2, AX3,…, AXn are ∂U/∂F1, ∂U/∂F2, ∂U/∂F3,…, ∂U/∂Fn
respectively.”

Proof for Castigliano’s theorem (Part II)Proof for Castigliano s theorem (Part II)
Let x1, x2, x3,… xn be deflections at points A1, A2, A3,… An
due to F F F Fdue to F1, F2, F3,… Fn
Total strain energy, 1 1 2 2 3 3
1 1 1 1
2 2 2 2
n nU F x F x F x F x= + + + +… (1)
Let the load F1 alone be increased by 1Fδ
Let 1 2 3, , , nx x x xδ δ δ δ… be the additional deflections at points A1, A2, A3,… An
I i t iIncrease in strain energy,
1 1 1 2 2 3 3
1
2
n nU F F x F x F x F xδ δ δ δ δ δ⎛ ⎞
= + + + + +⎜ ⎟
⎝ ⎠
…
2⎝ ⎠
1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +… , neglecting small quantities.
80
(2)

( )1 1 2 3, , , nF F F F Fδ+ …Let are acting on the original structure
( )( ) ( ) ( ) ( )
1 1 1 1
U U F F F F Fδ δ δ δ δ δ
Total strain energy,
( )( ) ( ) ( ) ( )1 1 1 1 2 2 2 3 3 3
1 1 1 1
2 2 2 2
n n nU+ U F F x x F x x F x x F x xδ δ δ δ δ δ= + + + + + + + + +…
(3)
1 1 1 1
U F F F F+ + + +(1) 1 1 2 2 3 3
2 2 2 2
n nU F x F x F x F x= + + + +…(1)
1 1 1 1
U F F F Fδ δ δ δ δ⎛ ⎞
⎜ ⎟(3) (1) 1 1 1 1 2 2
2 2 2 2
n nU x F F x F x F xδ δ δ δ δ⎛ ⎞
= + + + +⎜ ⎟
⎝ ⎠
…(3)-(1)
( )2 U x F F x F x F xδ δ δ δ δ= + + + + (4)
1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +…(2)
( )1 1 1 1 2 22 n nU x F F x F x F xδ δ δ δ δ= + + + +… (4)
81
1 1U x Fδ δ=(4)-(2)

Uδ U∂
1
1
U
x
F
δ
δ
= 1 1
1
0,
U
When F x
F
δ
∂
→ =
∂
2 2
2 3
, , n
n
U U U
Similarly, x x x
F F F
∂ ∂ ∂
= = =
∂ ∂ ∂
…
For example, in the case of bending,
2
2
M
U dx
EI
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
∫⎝ ⎠
U M M
dx
F EI F
δ
∂ ∂⎛ ⎞
∴ = = ⎜ ⎟
∂ ∂⎝ ⎠
∫F EI F∂ ∂⎝ ⎠
82

Example 1: Using Castigliano’s Theorem, analyse the continuous
w kN/m
p g g , y
beam shown in figure.
A
B
C
/
LLL
w kN/m
A
B
C
/
LL
RB
L
2
BR
wL −
2
BR
wL −
Let RB be the redundant.
83
B

2
wxR⎛ ⎞
22
B
x
wxR
M xwL
⎛ ⎞= −−⎜ ⎟
⎝ ⎠
From A to B,
0Bδ = 0
B B
U M M
dx
R EI R
∂ ∂⎛ ⎞
⇒ = =⎜ ⎟
∂ ∂⎝ ⎠
∫
2
M x
R
∂ −
=
∂ F th ti AC2BR∂
2
2M M∂ ⎛ ⎞⎛ ⎞
For the entire span AC
2
2
0 0
22 2
B
B
M M xR x wx
dx dxwLx
EI R EI
∂ −⎛ ⎞⎛ ⎞
∴ = ⇒ =− −⎜ ⎟ ⎜ ⎟
∂⎝ ⎠ ⎝ ⎠
∫ ∫
5
B
wL
R⇒ =
4
BR⇒

Example 2: Using Castigliano’s Theorem analyse the frame shownExample 2: Using Castigliano s Theorem, analyse the frame shown
in figure.
3 kN/3 kN/m
2mB
C
3 kN/m
2mB
C
2m
1m
D
2m
1m
D
A
D
A
VD
HD
VD
V
HA
VA
Let HA be the redundant.
Let HA be the redundant.

3 kN/m
2m
1
B
C
From A to B,
x AM H x= −
2m
1m
D
AH A
M
x
H
∂
= −
∂
A
3 AH
V = −
AH
From B to C,
2
3⎛ ⎞HA
3
2
DV
3 AH
V +
2
3
23
22
A
x A
xH
M x H⎛ ⎞= − −+⎜ ⎟
⎝ ⎠
3
2
A
AV = +
2
2A
M x
H
∂
= −
∂
F D t C M H x=
M
x
∂
From D to C, x AM H x= −
A
x
H
= −
∂

0
M M
dx
∂⎛ ⎞
∴ ⎜ ⎟∫ 0
A
dx
EI H
∴ =⎜ ⎟
∂⎝ ⎠
∫
2 2 1
⎧ ⎫⎛ ⎞⎛ ⎞
( )( ) ( )( )
2 2 12
0 0 0
1 3
023 2
22 2
A
A AA
xH x x
dx dx dxH x H xx Hx x
EI
⎧ ⎫⎛ ⎞⎛ ⎞⇒ + + =− −−+ − −− −⎨ ⎬⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎩ ⎭
∫ ∫ ∫
2 2 13 3 3 2 4 2 3
2 33
03 4A A A A A
A
H x x H x H x x H x H x
x H x x
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
+ + =+ − − − − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦0 0 03 2 12 2 16 2 3
A⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
8 8 8 4A AA A
H HH H⎡ ⎤8 8 8 4
03 12 2 8 8
3 32 12 2
A AA A
A A
H HH H
H H
⎡ ⎤+ + =+ − − − − + +⎢ ⎥⎣ ⎦
3 3.19
2
A
A
H
V kN= + =0.39AH kN=
H
3 2.8
2
A
D
H
V kN= − =

Example 4: Using Castigliano’s Theorem, analyse the truss shownp g g , y
in figure. AE is constant.
CD CD
10kN
CD
T10kN
CD
3m 3m
T
4m
A B
4m
A B
T
m
. . 0
P L
P
T AE
∂
=
∂
∑ 6.25T kN=

Example 4: Do the above problem using the method of consistentExample 4: Do the above problem using the method of consistent
deformation.
P kL′
10kN 2
P kL
AET
k L
′
= −
∑
∑
CD
3m
k L
AE
∑
A B
4m
B
P P kT′= +

Note 1: If there are two internal redundants in a truss in method ofNote 1: If there are two internal redundants in a truss, in method of
consistent deformation,
2
P k L k L k k L′ 2
1 1 1 2
1 2 0
P k L k L k k L
T T
AE AE AE
′
+ + =∑ ∑ ∑
2
2 1 2 2
1 2 0
P k L k k L k L
T T
AE AE AE
′
+ + =∑ ∑ ∑AE AE AE
Note 2: If there are both internal redundants external redundants in
a truss, in method of consistent deformation,
2
1 1 1
0BP k L k L k k L
T V
′
+ + =∑ ∑ ∑1 0BT V
AE AE AE
+ + =∑ ∑ ∑
2
P k L k k L k L′
2
1
1 0B B B
B
P k L k k L k L
T V
AE AE AE
′
+ + =∑ ∑ ∑

Example 3: Using Castigliano’s Theorem, analyse the pin-jointed
h i fi
3m 4m
truss shown in figure.
A B C
3m 4m
400A =
A
3m 500A = 400A =
D
80kN
Internally indeterminate to degree 1.
Take force in BD as redundant.
y g
0
P L
P
∂
∑
91
Assume force in BD is T . . 0P
T AE
=
∂
∑

cos45 sin 53.13AD CDF F=cos45 sin 53.13AD CDF F
cos45 cos53.13 80AD CDF F T+ + =
T
57.14 0.714T−
64.64 0.808T−
sin 53.13 cos53.13 80CD CDF F T+ + =
80kN
57.14 0.714CDF T= −
i 53 13
( )
sin 53.13
57.14 0.714 64.64 0.808
cos45
ADF T T= − = −
. . 0
P L
P
T AE
∂
=
∂
∑
( )
( )
( )
( )
4.243 57.14 0.714 5 64.64 0.8083
0.714 3 0.808 0
500 400 400
T TT
E E E
− −⎛ ⎞ ⎛ ⎞⎛ ⎞
× − + × + × − =⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
9249.09T kN∴ =

SummarySummary
Statically and kinematically indeterminate structures
• Degree of static indeterminacy, Degree of kinematic
indeterminacy, Force and displacement method of analysis
• Method of consistent deformation-Analysis of fixed and
continuous beams
• Clapeyron’s theorem of three moments-Analysis of fixed and
continuous beams
• Principle of minimum strain energy-Castigliano’s second theorem-
Analysis of beams, plane trusses and plane frames.
93

Structural Analysis - II Force Method Analysis

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