FEA Based Level 3 Assessment of Deformed Tanks with Fluid Induced Loads
Module2 rajesh sir
1. Structural Analysis - II
Slope deflection methodSlope deflection method
Moment distribution method
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKNDept. of CE, GCE Kannur Dr.RajeshKN
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2. Module IIModule II
Displacement method of analysis
• Slope deflection method-Analysis of continuous beams and
Displacement method of analysis
Slope deflection method Analysis of continuous beams and
frames (with and without sway)
• Moment distribution method- Analysis of continuous beams
and frames (with and without sway)and frames (with and without sway).
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3. Displacement method
Example 1: Propped cantilever (Kinematically indeterminate to first
degree)
p
degree)
• Required to get Bθ
•degrees of freedom: one
• Kinematically determinate structure is obtained by restraining allKinematically determinate structure is obtained by restraining all
displacements (all displacement components made zero - restrained
structure)
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4. Restraint at B causes a reaction of MB as shown.
2
12
B
wL
M =
12
The actual rotation at B is Bθ
To induce a rotation of at B, it is required to
apply a moment of MB anticlockwise.
Bθ
Bθ
4
B B
EI
M θ=
pp y
B BM
L
θ
2
4 J i ilib i i
3
wL
2
4
12
B
wL EI
L
θ=
Joint equilibrium equation
(or equation of action superposition)
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448
B
wL
EI
θ∴ =
5. •A general approach (applying consistent sign convention
for loads and displacements):
• Restrained structure: Restraint at B causes a reaction of MB
2
L (N t th i ti
Restrained structure: Restraint at B causes a reaction of MB.
2
12
B
wL
M =
(Note the sign convention:
clockwise positive)
Bθ•Apply unit rotation corresponding to
4EI
BmLet the moment required for this unit rotation be
4
B
EI
m
L
= −
anticlockwise
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Bθ B Bm θ• Moment required to induce a rotation of is
6. 0B B BM m θ+ = (Joint equilibrium equation)
2
4
0
wL EI
θ
3
B
B
wLM
θ∴ = − =
B B B (Joint equilibrium equation)
0
12
B
L
θ− = 48
B
B EIm
θ∴
m (Moment required for unit rotation) is the stiffness coefficient hereBm (Moment required for unit rotation) is the stiffness coefficient here.
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66
7. Sign convention for momentsSign convention for moments
(for Slope Deflection and Moment Distribution methods)
• A support moment acting in the anticlockwise direction will
be taken as positive (reactive moment is clockwise)
• A support moment acting in the clockwise direction will be
taken as negative (reactive moment is anticlockwise)
+ve
A B
ve+ve−
A B
support moments
reactive
moment
reactive
moment
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support moments
8. anticlockwise support
moment (reactive
moment is clockwise)
anticlockwise support
moment (reactive
moment is clockwise)moment is clockwise) moment is clockwise)
ve+ ve+
A B
Moment distribution
distribution/slope
deflection signdeflection sign
convention
veve+
A B
ve−ve+
Usual sign convention
for drawing BMD
A B
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9. clockwise support
moment (reactive
moment is
anticlockwise support
moment (reactive
moment is clockwise)moment is
anticlockwise)
moment is clockwise)
ve+ve−
A B
Moment
distribution/slope
deflection signdeflection sign
convention
ve− ve−
A B
ve ve
Usual sign convention
for drawing BMD
A B
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10. clockwise support
moment (reactive
moment is
clockwise support
moment (reactive
moment ismoment is
anticlockwise)
moment is
anticlockwise)
ve− ve−
A B
Moment distribution
distribution/slope
deflection signdeflection sign
convention
ve− ve+
A B
ve ve+
Usual sign convention
for drawing BMD
A B
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11. Sign convention for slopes and deflections
A l k i t ti i t k iti d ti l k i
g p
(for Slope Deflection and Moment Distribution methods)
• A clockwise rotation is taken as positive and anticlockwise
rotation as negative
ve−
Aθ
ve+
Bθ
•If one end of a beam settles, the rotation at both ends are taken as
positive if the beam as a whole rotates clockwise, and negative if the
beam as a whole rotates anticlockwise
Bθ
beam as a whole rotates anticlockwise
Aθ
ve+
δ ve+
B
ve−δ ve−
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Bθ
ve+ Aθ ve−
12. Sl d fl ti th dSlope deflection method
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13. Introduction
• This method is based on the relationships of end moments with
slopes and deflections (called slope-deflection equations) for eachp ( p q )
member.
Approach to solve problems
• The slope-deflection equations are written for each member.
pp p
• Joint equilibrium conditions are written.
• Solving the joint equilibrium conditions, unknown displacements
are found out.
• Substituting these unknown displacements back in the slope-
deflection equations, we get the unknown end moments.
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14. Derivation of fundamental equations
BAM
M
=BA
ABM
Aθ Bθ
AFEM( )1
+
ABFEM BAFEM( )1
( )2
ABM′
Aθ′ Bθ′
+
( )2
BAM′
A
M ′′ M ′′
+
( )3
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ABM BAM
15. ( ) Ends assumed as fixed (zero rotation) This requires restraining( )1 Ends assumed as fixed (zero rotation). This requires restraining
moments (fixed end moments) FEMAB and FEMBA. External
loads are acting.
( )2 Rotations are forced at ends. This requires moments M’AB and
M’BA
( )3 If there is a support settlement, moments M’’AB and M’’BA will
be induced.
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16. ( )2
M′ BAM′
Aθ′ Bθ′
ABM BA
=
BAM′2Bθ′2Aθ′
M′
1Aθ′ 1Bθ′
+
BA
ABM′
BAM′
+ABM′
+
BAM l
θ
′−
′
EI
+AB
EI
1
3
AB
A
M l
EI
θ
′
′ =
2
6
BA
A
EI
θ′ =
l′ M l′
Conjugate beams
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1
6
AB
B
M l
EI
θ
′−
′ = 2
3
BA
B
M l
EI
θ′ =
17. AB BAM l M l
θ θ θ
′ ′
′ ′ ′ AB BAM l M l
θ θ θ
′ ′
′ ′ ′1 2
3 6
AB BA
A A A
EI EI
θ θ θ′ ′ ′= + = − 1 2
6 3
AB BA
B B B
EI EI
θ θ θ′ ′ ′= + = − +
( )3 Rotation at the end A due to
support settlement
δABM′′
BAM′′
support settlement
= Rotation at the end B due to
support settlement
BA
=
l
δl
Total rotations at the ends are:
3 6
AB BA
A A
M l M l
l EI EI l
δ δ
θ θ
′ ′
′= + = − +
3 6
BA AB
B B
M l M l
l EI EI l
δ δ
θ θ
′ ′
′= + = − +
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3 6l EI EI l 3 6l EI EI l
18. 2 3
2
EI
M
δ
θ θ
⎛ ⎞′ ⎜ ⎟
Solving the above two equations, we get:
2 3
2
EI
M
δ
θ θ
⎛ ⎞′ = +⎜ ⎟2AB A BM
l l
θ θ
⎛ ⎞′ = + −⎜ ⎟
⎝ ⎠
2BA B AM
l l
θ θ= + −⎜ ⎟
⎝ ⎠
Hence the final moments at the supports are:
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
= + +⎜ ⎟
Hence the final moments at the supports are:
2AB A B ABM FEM
l l
θ θ= + − +⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
These are the slope deflection equations
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21. Illustration of the method
B3
5kN 8kN
2 5
Example 1
A
B
C
3m 2.5m
5m 5m
Problem structure
A B CB
Problem structure
A B
2 4kN 3 6kN
C
5kNm 5kNm
B
2.4kNm 3.6kNm 5kNm 5kNm
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Fixed end moments (reactive)
22. Fixed end moments
2 2
2 2
5 3 2
2.4
5
AB
Pab
FEM kNm
l
− − × ×
= = = −
2 2
2 2
5 3 2
3.6
5
BA
Pa b
FEM kNm
l
× ×
= = =
8 5
5
8 8
BC
Pl
FEM kNm
− − ×
= = = −
8 5
5
8 8
CB
Pl
FEM kNm
×
= = =
8 8
Known displacements
0A Cθ θ= = 0A B Cδ δ δ= = =
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23. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2.4
5
AB B
EI
M θ⇒ = −
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
= + +⎜ ⎟ ( )
2
2 3 6
EI
M θ⇒ +2BA B A BAM FEM
l l
θ θ= + − +⎜ ⎟
⎝ ⎠
( )2 3.6
5
BA BM θ⇒ = +
( )
2
2 5
5
BC B
EI
M θ⇒ = −
2 3
2BC B C BC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠ 5l l⎝ ⎠
2 3
2CB C B CB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
5
5
CB B
EI
M θ⇒ = +
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24. Joint equilibrium condition
0BA BCM M+ =
( ) ( )
2 2
2 3.6 2 5 0B B
EI EI
θ θ
⎛ ⎞ ⎛ ⎞
⇒ + + − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( ) ( )2 3.6 2 5 0
5 5
B Bθ θ⇒ + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1.6 1.4 0BEIθ − =
0.875
B
EI
θ⇒ =
( )
2 2 0.875
2.4 2.4 2.05
5 5
AB B
EI EI
M kNm
EI
θ
⎛ ⎞
= − = − = −⎜ ⎟
⎝ ⎠
( )
5 5
AB B
EI
⎜ ⎟
⎝ ⎠
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25. ⎛ ⎞2 0.875
2 3.6 4.3
5
BA
EI
M kNm
EI
⎛ ⎞
= × + =⎜ ⎟
⎝ ⎠
2 0.875
2 5 4.3BC
EI
M kNm
⎛ ⎞
= × − = −⎜ ⎟
⎝ ⎠
2 5 4.3
5
BCM kNm
EI
×⎜ ⎟
⎝ ⎠
2 0.875
5 5.35
5
CB
EI
M kNm
EI
⎛ ⎞
= + =⎜ ⎟
⎝ ⎠
A B C4.3kNm2.05kNm 4.3kNm 5.35kNm
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26. A B C4 3kNm2 05kNm 4 3kNm 5.35kNm
A B C4.3kNm2.05kNm 4.3kNm
A
B
C
2 05
2.6 5.175
4.3
2.05
5.35
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27. Procedure to solve problems
• The slope-deflection equations are written for each member.
l b d• Joint equilibrium conditions are written.
• Solving the joint equilibrium conditions, unknown displacementsg j
are found out.
• Substituting these unknown displacements back in the slope-g p p
deflection equations, we get the unknown end moments.
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29. Fixed end moments
2 2
20 6 40 6
90
12 8 12 8
AB BA
wl Pl
FEM FEM kNm
× ×
− = = + = + =
2
20 6 60 6
105
12 8
BC CBFEM FEM kNm
× ×
− = = + =
12 8
80 6
60FEM FEM kNm
×
= = =
K di l t
60
8
CD DCFEM FEM kNm− = = =
Known displacements
0A B C Dδ δ δ δ= = = =A B C D
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30. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2 90
6
AB A B
EI
M θ θ⇒ = + −
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2 90
6
BA B A
EI
M θ θ⇒ = + +
( )
2
2 105
6
BC B C
EI
M θ θ⇒ = + −
2 3
2BC B C BC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
6
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
⎜ ⎟
BC B C BC
l l
⎜ ⎟
⎝ ⎠
( )
2
2 105
EI
M θ θ
3
2CB C B CBM FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )2 105
6
CB C BM θ θ⇒ = + +
2EI2 3EI δ⎛ ⎞
( )
2
2 60
6
CD C D
EI
M θ θ⇒ = + −
2 3EI δ⎛ ⎞
2 3
2CD C D CD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2EI
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2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2 60
6
DC D C
EI
M θ θ⇒ = + +
31. Joint equilibrium conditions
0ABM = ( )
2
2 90 0
6
A B
EI
θ θ⇒ + − =
6
0.667 0.333 90A BEI EIθ θ+ =
( )1
90 0.333
135 0.5
0.667
B
A B
EI
EI EI
θ
θ θ
−
= = −
( )
2
2 60 0
EI
θ θ0DCM = ( )2 60 0
6
D Cθ θ⇒ + + =
0 667 0 333 60EI EIθ θ
( )2
0.667 0.333 60D CEI EIθ θ+ = −
60 0.333
90 0 5CEI
EI EI
θ
θ θ
− −
= =
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31
( )290 0.5
0.667
D CEI EIθ θ= = − −
37. Fixed end moments
10 8
10
8 8
AB BA
Pl
FEM FEM kNm
×
− = = = =
2 2
5 8
26.667
12 12
BC CB CD DC
wl
FEM FEM FEM FEM kNm
×
− = = − = = = =
12 12
Known displacements
0A B C Dδ δ δ δ= = = =
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38. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
2 10
8
AB A B
EI
M θ θ⇒ = + −
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
2 10
8
BA B A
EI
M θ θ⇒ = + +
( )
6
2 26 667
EI
M θ θ⇒ = + −
2 3
2BC B C BC
EI
M FEM
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟ ( )2 26.667
8
BC B CM θ θ⇒ = + −
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
= + +⎜ ⎟
2BC B C BCM FEM
l l
θ θ+ +⎜ ⎟
⎝ ⎠
( )
6
2 26 667
EI
M θ θ2CB C B CBM FEM
l l
θ θ= + − +⎜ ⎟
⎝ ⎠
( )2 26.667
8
CB C BM θ θ⇒ = + +
6EI2 3EI δ⎛ ⎞
( )
6
2 26.667
8
CD C D
EI
M θ θ⇒ = + −
2 3EI δ⎛ ⎞
2 3
2CD C D CD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
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38
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
6
2 26.667
8
DC D C
EI
M θ θ⇒ = + +
39. Joint equilibrium conditions
0ABM = ( )
4
2 10 0
8
A B
EI
θ θ⇒ + − = 10 0.5A BEI EIθ θ⇒ = −
0DCM = 17.778 0.5D CEI EIθ θ⇒ = − −( )
6
2 26.667 0
8
D C
EI
θ θ⇒ + + =
50 0M M+ − = ( ) ( )
4 6
2 10 2 26.667 50B A B C
EI EI
θ θ θ θ
⎛ ⎞ ⎛ ⎞
⇒ + + + + − =⎜ ⎟ ⎜ ⎟50 0BA BCM M+ = ( ) ( )2 10 2 26.667 50
8 8
B A B Cθ θ θ θ⇒ + + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2 5 0 5 0 75 66 667EI EI EIθ θ θ⇒ + +2.5 0.5 0.75 66.667B A CEI EI EIθ θ θ⇒ + + =
( )2.5 0.5 0.75 66.66710 0.5B CBEI EIEIθ θθ⇒ + + =−
( )1
( )2.5 0.5 0.75 66.66710 0.5B CBEI EIEIθ θθ⇒ + +
2.25 0.75 61.667B CEI EIθ θ⇒ + =
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2.25 0.75 61.667B CEI EIθ θ⇒ +
43. Example 4
90kN
20kN30kN m
90kN
A B C D
2.5m 2.5m
E2I I 2.4I
7.5m 5m 5m 3m
20 3 60DEM kNm= − × = −
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44. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
150
7.5
AB B
EI
M θ⇒ = −
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
2 150
7 5
BA B
EI
M θ⇒ = +
l l⎝ ⎠ 7.5
( )
2
2 62.5BC B C
EI
M θ θ⇒ = + −
2 3
2BC B C BC
EI
M FEM
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟ ( )
5
BC B C
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
⎜ ⎟
2BC B C BCM FEM
l l
θ θ+ +⎜ ⎟
⎝ ⎠
( )
2
2 62 5
EI
M θ θ
3
2CB C B CBM FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )2 62.5
5
CB B CM θ θ⇒ = + +
( )
4.8
2
EI
M θ θ
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
⎜ ⎟
2 3EI δ
θ θ⎛ ⎞
( )
4.8
2
EI
M θ θ
( )2
5
CD C DM θ θ⇒ = +
2 3
2CD C D CD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
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44
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )2
5
DC D CM θ θ⇒ = +
45. Fixed end moments
2 2 2 2
2 2 2 2
90 2.5 5 90 2.5 5
150
7.5 7.5
AB BA
Pab Pa b
FEM FEM kNm
l l
× × × ×
− = = + = + =
2 2
30 5
62.5BC CB
wl
FEM FEM kNm
×
− = = = =
12 12
BC CB
0FEM FEM= =
Known displacements
0CD DCFEM FEM= =
0A B C Dδ δ δ δ= = = =
Known displacements
0Aθ =
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46. Joint equilibrium conditions
0BA BCM M+ =
( ) ( )
4 2
2 150 2 62.5 0
7.5 5
B B C
EI EI
θ θ θ
⎛ ⎞ ⎛ ⎞
⇒ + + + − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
1.867 0.4 87.5B CEI EIθ θ⇒ + = − ( )1
0CB CDM M+ =
2 4 8EI EI⎛ ⎞ ⎛ ⎞
( ) ( )
2 4.8
2 62.5 2 0
5 5
B C C D
EI EI
θ θ θ θ
⎛ ⎞ ⎛ ⎞
⇒ + + + + =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0.4 2.72 0.96 62.5B C DEI EI EIθ θ θ⇒ + + = − ( )2
Dept. of CE, GCE Kannur Dr.RajeshKN
47. ( )
4.8
2 60 0
EI
θ θ⎛ ⎞
⇒ + − =⎜ ⎟0M M+ = ( )2 60 0
5
D Cθ θ⇒ + − =⎜ ⎟
⎝ ⎠
0DC DEM M+ =
1 92 0 96 60EI EIθ θ⇒ + =1.92 0.96 60D CEI EIθ θ⇒ + =
31.25 0.5D CEI EIθ θ⇒ = −
( ) ( )0.4 2.72 0.96 31.25 0.5 62.52 B C CEI EI EIθ θ θ⇒ + + − = −
( )30.4 2.24 92.5B CEI EIθ θ⇒ + = −
39.532BEIθ = − 34.235CEIθ = −
( )31.25 0.5 31.25 0.5 48.36834.235D CEI EIθ θ= − = − =−
Dept. of CE, GCE Kannur Dr.RajeshKN
48. 4 4EI
( ) ( )
4 4
150 39.532 150 171.084
7.5 7.5
AB B
EI
M kNmθ∴ = − = − − = −
( ) ( )
4 4
2 150 2 39.532 150 107.833
7.5 7.5
BA B
EI
M kNmθ= + = ×− + =
( ) ( )
2 2
2 62.5 2 39.532 34.235 62.5 107.82
5 5
BC B C
EI
M kNmθ θ= + − = ×− − − =
5 5
( ) ( )
2 2
2 62 5 39 532 2 34 235 62 5 19 3
EI
M kNmθ θ= + + = − + ×− + =( ) ( )2 62.5 39.532 2 34.235 62.5 19.3
5 5
CB B CM kNmθ θ= + + = − + ×− + =
( ) ( )
4.8 4.8
2 2 34 235 48 368 19 3
EI
M kNθ θ
4 8 4 8EI
( ) ( )2 2 34.235 48.368 19.3
5 5
CD C DM kNmθ θ= + = ×− + = −
Dept. of CE, GCE Kannur Dr.RajeshKN
( ) ( )
4.8 4.8
2 2 48.368 34.235 60
5 5
DC D C
EI
M kNmθ θ= + = × − =
49. Example 5
Support B settles by 10 mm.
6 4
200 , 50 10E GPa I mm= = ×
kN6 6 4 4 2
2
200 10 50 10 10
kN
EI m kNm
m
−
= × × × =
Dept. of CE, GCE Kannur Dr.RajeshKN
50. Fixed end moments
20
8
AB BA
Pl
FEM FEM kNm− = = =
2
16
12
BC CB
wl
FEM FEM kNm− = = =
12
C C
K di l t
0A Cδ δ= =
Known displacements
10B mmδ =
0Cθ =
Dept. of CE, GCE Kannur Dr.RajeshKN
51. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2
4 3 10
2 20
8 8
AB A B
EI
M θ θ
−
⎛ ⎞×
⇒ = + − −⎜ ⎟
⎝ ⎠
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2
4 3 10
2 20
8 8
BA B A
EI
M θ θ
−
⎛ ⎞×
⇒ = + − +⎜ ⎟
⎝ ⎠l l⎝ ⎠ 8 8⎝ ⎠
2
6 3 10
2 16
EI
M θ
−
⎛ ⎞×
⇒ = + −⎜ ⎟
2 3
2BC B C BC
EI
M FEM
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟ 2 16
8 8
BC BM θ⇒ = +⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
2BC B C BCM FEM
l l
θ θ+ +⎜ ⎟
⎝ ⎠
2
6 3 10EI −
⎛ ⎞×2 3
2CB C B CB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
6 3 10
16
8 8
CB B
EI
M θ
⎛ ⎞×
⇒ = + +⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
51
52. Joint equilibrium conditions
0ABM =
2
4 3 10
2 20 0
8 8
A B
EI
θ θ
−
⎛ ⎞×
⇒ + − − =⎜ ⎟
⎝ ⎠⎝ ⎠
2
6 10
0.5 20 0
32
A B
EI
EI EIθ θ
−
×
⇒ + − − =
32
2
3
4
20 6 10
0.5 3.875 10
10 32
A Bθ θ
−
−×
⇒ + = + = × ( )1
0BA BCM M+ =
2 2
4 3 10 6 3 10
2 20 2 16 0
8 8 8 8
B A B
EI EI
θ θ θ
− −
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞× ×
⇒ + − + + + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
420 16
9.375 100.5 1.5B A B
EI EI
θ θ θ −⎛ ⎞ ⎛ ⎞⇒ + = − ×+ + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
B A B
EI EI
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
56. Example 6Example 6
B
10kN 2kN m
A
B C
2
2I I
332m
3m I
3m3m
D
I
D
Dept. of CE, GCE Kannur Dr.RajeshKN
57. Fixed end moments
2 2
2 2
10 2 3
7.2
5
AB
Pab
FEM kNm
l
− − × ×
= = = −
2 2
2 2
10 2 3
4.8
5
BA
Pa b
FEM kNm
l
× ×
= = =
2 2
2 3
1 5
wl
FEM FEM kN
×
2 2
5l
3
1.5
12 12
BC CB
wl
FEM FEM kNm− = = = = 0BD DBFEM FEM= =
Known displacements
0A B C Dδ δ δ δ= = = =
0A Dθ θ= =
Dept. of CE, GCE Kannur Dr.RajeshKN
58. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
7.2
5
AB B
EI
M θ⇒ = −
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
2 4.8
5
BA B
EI
M θ⇒ = +
l l⎝ ⎠ 5
( )
2
2 1 5
EI
M θ θ⇒ = + −
2 3
2BC B C BC
EI
M FEM
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟ ( )2 1.5
3
BC B CM θ θ⇒ = + −
2 3EI δ⎛ ⎞
2BC B C BCM FEM
l l
θ θ+ +⎜ ⎟
⎝ ⎠
2EI2 3
2CB C B CB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2 1.5
3
CB C B
EI
M θ θ⇒ = + +
⎛ ⎞2 3
2BD B D BD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2
3
BD B
EI
M θ⇒ =
Dept. of CE, GCE Kannur Dr.RajeshKN
58
2 3
2DB D B DB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
3
DB B
EI
M θ⇒ =
59. Joint equilibrium conditions
0CBM = ( )
2
2 1.5 0
3
C B
EI
θ θ⇒ + + =
1.333 0.667 1.5C BEI EIθ θ⇒ + = − ( )1
0BA BC BDM M M+ + =
( ) ( ) ( )
4 2 2
02 4.8 2 1.5 2
5 3 3
B B C B
EI EI EI
θ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ + + =+ + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠5 3 3⎝ ⎠ ⎝ ⎠ ⎝ ⎠
4.267 0.667 3.3B CEI EIθ θ⇒ + = − ( )2
0.6481BEIθ = − 0.801CEIθ = −
Dept. of CE, GCE Kannur Dr.RajeshKN
60. ( )
4EI
( )
4
( )
4
7.2
5
AB B
EI
M θ= − ( )
4
0.6481 7.2 7.718
5
kNm= − − = −
( )
4
2 4.8
5
BA B
EI
M θ= + ( )
4
2 0.6481 4.8 3.763
5
kNm= ×− + =
( )
2
2 1.5
3
BC B C
EI
M θ θ= + − ( )
2
2 0.6481 0.801 1.5 2.898
3
kNm= ×− − − = −
( )
2
2
3
BD B
EI
M θ= ( )
2
2 0.6481 0.864
3
kNm= ×− = −( )
3
BD B
2EI
( )
3
2
Dept. of CE, GCE Kannur Dr.RajeshKN
60
( )
2
3
DB B
EI
M θ= ( )
2
0.6481 0.432
3
kNm= − = −
62. P BAM CDMP
B
C
BA CD
2l
DH D1l
DCM
AH
A
ABMAB
AB BA
A
M M
H
+
= CD DCM M
H
+
= 0H H P
Dept. of CE, GCE Kannur Dr.RajeshKN
1
AH
l
2
DH
l
= 0A DH H P+ + =
63. M CDM
B
CBAM CDM
a
P
l
2l
DH D
1l
DCM
AH
A
ABMAB
CD DCM M
H
+
= 0H H P
AB BA
A
M M Pa
H
+ −
=
Dept. of CE, GCE Kannur Dr.RajeshKN
63
2
DH
l
= 0A DH H P+ + =
1
AH
l
66. Fixed end moments
2 2
16 1 4
10.24BC
Pab
FEM kNm
− − × ×
= = = −
Fixed end moments
2 2
10.24
5
BCFEM kNm
l
2 2
16 1 4
2 56
Pa b
FEM kN
× ×
2 2
16 1 4
2.56
5
CB
Pa b
FEM kNm
l
= = =
Known displacements
0A Dθ θ= =
δ=Let horizontal movement (sidesway)
Dept. of CE, GCE Kannur Dr.RajeshKN
66
67. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
5 5
B
EI δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
2
5 5
B
EI δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
2 3
2BC B C BC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
2 10.24B C
EI
θ θ= + −
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
⎜ ⎟
BC B C BC
l l
⎜ ⎟
⎝ ⎠
( )
2EI
( )
5
B C
3
2CB C B CBM FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
( )
2
2 2.56
5
C B
EI
θ θ= + +
2 3EI δ⎛ ⎞
2 3EI δ⎛ ⎞
2 3
2CD C D CD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
2 3
2
5 5
C
EI δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
67
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
5 5
C
EI δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
71. Example 8
10 kN B C
p
4m
4m
4m
EI
A D
4m 4m
EI EI
• No fixed end moments
• Known displacements: 0θ =• Known displacements: 0Aθ =
δ=• Let horizontal movement (sidesway)
Dept. of CE, GCE Kannur Dr.RajeshKN
0DCM =
72. Slope deflection equations
2 3
2AB A B AB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
4 4
AB B
EI
M
δ
θ
⎛ ⎞
⇒ = −⎜ ⎟
⎝ ⎠
2 3
2BA B A BA
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
2
4 4
BA B
EI
M
δ
θ
⎛ ⎞
⇒ = −⎜ ⎟
⎝ ⎠
( )
2
2
4
BC B C
EI
M θ θ⇒ = +
2 3
2BC B C BC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
4
2 3
2
EI
M FEM
δ
θ θ
⎛ ⎞
⎜ ⎟
BC B C BC
l l
⎜ ⎟
⎝ ⎠
( )
2
2
EI
M θ θ
3
2CB C B CBM FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )2
4
CB C BM θ θ⇒ = +
2 3EI δ⎛ ⎞2 3EI δ⎛ ⎞ 2 3
2
4 4
CD C D
EI
M
δ
θ θ
⎛ ⎞
⇒ = + −⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
2 3
2CD C D CD
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
Dept. of CE, GCE Kannur Dr.RajeshKN
72
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
2
4 4
DC D C
EI
M
δ
θ θ
⎛ ⎞
⇒ = + −⎜ ⎟
⎝ ⎠
73. Joint equilibrium conditions
0DCM = 2 3
2 0
4 4
D C
EI δ
θ θ
⎛ ⎞
⇒ + − =⎜ ⎟
⎝ ⎠
3
2
4
D C
δ
θ θ⇒ + =
4 4⎝ ⎠
( )1
4
3
8 2
C
D
δ θ
θ⇒ = −
0BA BCM M+ = ( )
2 3 2
2 02
4 4 4
B B C
EI EIδ
θ θ θ
⎡ ⎤⎛ ⎞ ⎡ ⎤⇒ − + =+⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦4 4 4⎣ ⎦⎝ ⎠⎣ ⎦
( )2
3
4
4
B C
δ
θ θ⇒ + =
3
16 4
C
B
δ θ
θ⇒ = −
0CB CDM M+ = ( )
2 32
2 02
4 4
C DC B
EIEI δ
θ θθ θ
⎡ ⎤⎛ ⎞⎡ ⎤⇒ + + − =+ ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦
4 16 4
CB CD ( )
4 44
C DC B ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦
3 3 3
4 C Cδ θ δ θ δ
θ
⎡ ⎤
⇒ + + =⎢ ⎥ 3 25 0 1875 0θ δ =
Dept. of CE, GCE Kannur Dr.RajeshKN
( )3
4
16 4 8 2 4
Cθ⇒ + − + − =⎢ ⎥⎣ ⎦
3.25 0.1875 0Cθ δ− =
74. 0H H P 0AB BA CD DCM M M M
P
+ +
⇒ + +0A DH H P+ + =
1 2
0AB BA CD DC
P
l l
⇒ + + =
M M M
10 0
4 4
AB BA CDM M M+
⇒ + + = 40AB BA CDM M M⇒ + + = −
2 3 2 3 2 3
2 2 40
4 4 4 4 4 4
B B C D
EI EI EIδ δ δ
θ θ θ θ
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⇒ − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦
80 9
3 2
4
B C D
EI
δ
θ θ θ
−
⇒ + + = +
4EI
3 3 80 9
3 2
16 4 8 2 4
C C
C
EI
δ θ δ θ δ
θ
−⎡ ⎤ ⎡ ⎤
⇒ − + + − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
( )4
16 4 8 2 4
C
EI⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
80
0.75 1.3125Cθ δ
−
− =
Dept. of CE, GCE Kannur Dr.RajeshKN
74
( )40.75 1.3125C
EI
θ δ
75. ( )3.25 0.1875 0CEI θ δ− =( )3 ( )C
( )0.75 1.3125 80CEI θ δ− = −
3.636
63.03
CEI
EI
θ
δ
=
=
( )
( )4
3 C
DEI EI
δ θ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
3 63.03 3.636
21 818EIθ
×
= − =
8 2
DEI EIθ ⎜ ⎟
⎝ ⎠
3 C
EI EI
δ θ
θ
⎛ ⎞
= ⎜ ⎟
21.818
8 2
DEIθ = − =
3 63.03 3.636
10 909EIθ
×
16 4
BEI EIθ = −⎜ ⎟
⎝ ⎠
10.909
16 4
BEIθ = − =
( )
2 3
0.5 10.909 0.75 63.03 18.182
4 4
AB B
EI
M kNm
δ
θ
⎛ ⎞
= − = − × = −⎜ ⎟
⎝ ⎠
( )
2 3
2 0.5 2 10.909 0.75 63.03 12.727BA B
EI
M kNm
δ
θ
⎛ ⎞
= − = × − × = −⎜ ⎟
Dept. of CE, GCE Kannur Dr.RajeshKN
75
( )2 0.5 2 10.909 0.75 63.03 12.727
4 4
BA BM kNmθ × ×⎜ ⎟
⎝ ⎠
76. ( ) ( )
2EI
( ) ( )
2
2 0.5 2 10.909 3.636 12.727
4
BC B C
EI
M kNmθ θ= + = × + =
( ) ( )
2
2 0.5 2 3.636 10.909 9.091
4
CB C B
EI
M kNmθ θ= + = × + =
( )
2 3
2 0.5 2 3.636 21.818 0.75 63.03 9.091CD C D
EI
M kNm
δ
θ θ
⎛ ⎞
= + − = × + − × = −⎜ ⎟ ( )2 0.5 2 3.636 21.818 0.75 63.03 9.091
4 4
CD C DM kNmθ θ+ × + ×⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
76
77. M t di t ib ti th dMoment distribution method
Dept. of CE, GCE Kannur Dr.RajeshKN
78. Stiffness, Carry-over factor and Distribution factor
Beam hinged at both ends
A
B
M
ABθ
BAθ
(Applied
moment)
M
( )
2
2
EI
M θ θ
( )
2
2 0BA BA AB
EI
M
L
θ θ= + =
)
( )2AB AB BAM
L
θ θ= +
( )
L
1
θ θ⇒ = −
2
BA ABθ θ⇒ = −
2 1 3
2
EI EI
M θ θ θ
⎛ ⎞
⎜ ⎟
Dept. of CE, GCE Kannur Dr.RajeshKN
78
2
2
AB AB AB ABM
L L
θ θ θ
⎛ ⎞
∴ = − =⎜ ⎟
⎝ ⎠
79. 3M EI3AB
AB
M EI
Lθ
=
3EI
i.e., the moment required at A to induce a unit rotation at A is
(when the far end B is free to rotate)
3EI
L
This moment, i.e., moment required to induce a unit rotation,
is called stiffness (denoted by k).is called stiffness (denoted by k).
Dept. of CE, GCE Kannur Dr.RajeshKN
80. Beam hinged at near end and fixed at far end
A B
ABθ
0BAθ =
A B
M
(Applied
2EI
moment)
4M EI
( )
2
2 0AB AB
EI
M
L
θ= +
4AB
AB
M EI
Lθ
⇒ =
i.e., the moment required at A to induce a unit rotation at A is
(when the far end B is fixed against rotation)
4EI
L
(when the far end B is fixed against rotation)
Dept. of CE, GCE Kannur Dr.RajeshKN
80
81. ( )
2EI 2 AB ABEI M M
M L
⎛ ⎞
⎜ ⎟( )
2
0BA AB
EI
M
L
θ= +
2
4 2
AB AB
BA
EI M M
M L
L EI
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
A moment applied at the near end induces at a fixed far end a
moment equal to half its magnitude, in the same direction.
Half of moment applied at the near end is carried over to the fixed
far end.
Carry over factor is 1/2.
Dept. of CE, GCE Kannur Dr.RajeshKN
82. Several members meeting at a joint
1 13E I
M kθ θ= =1 1
1
M k
L
θ θ= =
2 24E I2 2
2 2
2
4E I
M k
L
θ θ= =
3 3
3 3
3
3E I
M k
L
θ θ= =
4 4
4 4
4E I
M k
L
θ θ= =
4L
1 2 3 4 1 2 3 4: : : :: : : :M M M M k k k k
Dept. of CE, GCE Kannur Dr.RajeshKN
1 2 3 4 1 2 3 4: : : :: : : :k k k k
83. 1
1
1 2 3 4
k
M M
k k k k
=
+ + +
1k
M
k
=
∑
i
i
k
M M
k
=
∑
A moment applied at a joint, where several members meet, will be
distributed amongst the members in proportion to their stiffnessdistributed amongst the members in proportion to their stiffness.
i
i
k
M M
k
=
∑
distribution factordistribution factor
Dept. of CE, GCE Kannur Dr.RajeshKN
84. Illustration of the method
B3
5kN 8kN
2 5
Example 1
A
B
C
3m 2.5m
5m 5m
Problem structure
A B CB
Problem structure
A B C
5kNm 5kNm2.4kNm 3.6kNm 5kNm 5kNm
Dept. of CE, GCE Kannur Dr.RajeshKN
Fixed end moments (reactive)
86. A B C
0 5 0 5 di t ib ti f t0.5 0.5
-2.4 +3.6 -5.0 +5.0 Fixed End Moments
0 0 distribution factors
+0.7 +0.7
+0 35 +0 35
Distribution
Carry over+0.35 +0.35 Carry over
Distribution
-2.05 +4.3 -4.3 +5.35 Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
87. 5 35kN
A B C4.3kNm2.05kNm 4.3kNm 5.35kNm
A
B
CA C
2.05
5 35
4.3
5.35
Dept. of CE, GCE Kannur Dr.RajeshKN
88. Example 2
40kN
80kN60kN
20kN m
A B C
D
3m 3m 3m 3m 3m 3m
Fi d d t
2
wl Pl
Fixed end moments
2
20 6 40 6× ×
12 8
AB BA
wl Pl
FEM FEM− = = +
2
wl Pl
20 6 40 6
60 30 90
12 8
kNm
× ×
= + = + =
2
20 6 60 6× ×
80 6Pl ×
12 8
BC CB
wl Pl
FEM FEM− = = +
20 6 60 6
60 45 105
12 8
kNm
× ×
= + = + =
Dept. of CE, GCE Kannur Dr.RajeshKN
80 6
60
8 8
CD DC
Pl
FEM FEM kNm
×
− = = = =
89. A B C D
0 5 0 5 0 5 0 51 1 Di t ib ti f t0.5 0.5
-90 +90 -105 +105 -60 +60 Fixed End Moments
0.5 0.51 1 Distribution factors
+90 +7.5 +7.5 -22.5 -22.5 -60
+3.5 +45 -11.25 +3.5 -30 -11.25
Distribution
Carry over
-3.5 -16.875 -16.875 +13.25 +13.25 +11.25
8 434 1 75 +6 625 8 434 +5 625 +6 625
Distribution
Carry over-8.434 -1.75 +6.625 -8.434 +5.625 +6.625
+8.434 -2.438 -2.438 +1.405 +1.405 -6.625
Carry over
Distribution
0 +121.48 -121.44 +92.221 - 92.22 0 Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
90. A B C D
-90 +90 -105 +105 -60 +60 Fixed End Moments
0.571 0.429 Distribution factors0.429 0.571
90 +90 105 +105 60 +60
+90 +45 -30 -60
Release A& D,
and carry over
0 +135 -105 +105 -90 0
-12.87 -17.13 -8.565 -6.435
Initial moments
Distribution
-4.283 -8.565
+1 837 +2 445 4 89 3 674
Carry over
Di t ib ti+1.837 +2.445 4.89 3.674
+2.445 1.223
Distribution
Carry over
-1.049 -1.396 -0.698 -0.524
0 +122.92 -122.92 +93.29 - 93.29 0 Final Moments
Distribution
Dept. of CE, GCE Kannur Dr.RajeshKN
103. A B C
0 333 0 6670.333 0.667
-38.75 +1.25 12.125 44.125 Fixed End Moments
1 0
Release A and
38.75 19.375
0 0 20 625 12 125 44 125
Release A, and
carry over
Initial Moments0.0 20.625 12.125 44.125
-10.906 -21.844 0 Distribution
-10.922 Carry over
Distribution
0.0 + 9.719 -9.719 +33.203 Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
104. Example 6
Support B settles by 10 mm.
6 4
200 , 50 10E GPa I mm= = ×
Dept. of CE, GCE Kannur Dr.RajeshKN
107. A B C
0.333 0.667
12 -5.0 -10
1 0
Joint couple dist.
6 -5
-38 75 +1 25 12 125 44 125 Fixed End Moments
Carry over
-38.75 +1.25 12.125 44.125
38.75 19.375
Fixed End Moments
Release A, and
carry over
-0.0 26.625 12.125 39.125
-12.904 -25.834 0
Initial Moments
Distribution
-12.917 Carry over
Di t ib ti
12.0 + 8.721 -23.709 +26.208
Distribution
Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
110. Example 7Example 7
B
10kN 2kN m
A
B C
2
2I I
332m
3m I
3m3m
D
I
D
Dept. of CE, GCE Kannur Dr.RajeshKN
111. Fixed end moments
2 2
2 2
10 2 3
7.2AB
Pab
FEM kNm
− − × ×
= = = −2 2
7.
5
AB kNm
l
2 2
10 2 3
4 8
Pa b
FEM kN
× ×
2 2
4.8
5
BAFEM kNm
l
= = =
2 2
2 3
1.5
12 12
BC CB
wl
FEM FEM kNm
×
− = = = = 0BD DBFEM FEM= =
Dept. of CE, GCE Kannur Dr.RajeshKN
112. Distribution factors
( )
( ) ( ) ( )
4 2 5
4 3 42 5 3 3
BA
EI
DF
EI EI EI
=
+ +
( )
( ) ( ) ( )
3 3
4 3 42 5 3 3
BC
EI
DF
EI EI EI
=
+ +
Distribution factors
( ) ( ) ( )4 3 42 5 3 3EI EI EI+ +
( ) ( ) ( )
0.407= 0.254=
( )
( ) ( ) ( )
4 3
4 3 42 5 3 3
BD
EI
DF
EI EI EI
=
+ +( ) ( ) ( )4 3 42 5 3 3EI EI EI+ +
0.339=
Dept. of CE, GCE Kannur Dr.RajeshKN
113. AB BA BD BC CB
0.407 0.339 0.254
-7.2 4.8 0.0 -1.5 1.5 Fixed End Moments
0 1 Distribution factors
-0.75 -1.5
-7.2 4.8 0.0 -2.25 0.0
Release C, and
carry over
Initial moments
-1.04 -0.864 -0.648
-0.52 Carry over
Distribution
Initial moments
-7.72 3.72 -0.864 -2.9 0.0
y
Final Moments
Distribution
0.864
0.432
2
DBM kNm
−
= = −
Dept. of CE, GCE Kannur Dr.RajeshKN
114. Moment Distribution for frames:Moment Distribution for frames:
sidesway
Dept. of CE, GCE Kannur Dr.RajeshKN
115. • Assume sway is prevented by giving a support at CAssume sway is prevented by giving a support at C.
• This causes a reaction R at C, along with end-moments M.
• This reaction R has to be cancelled out, since actually sway is not
prevented.
M
RR
Dept. of CE, GCE Kannur Dr.RajeshKN
116. • It is required to find out what member-end-moments cause this
reaction R, in the absence of actual external loads (Let this unknown
moment be MBA)
MBA M’BA
R’R’
• Let us assume that a moment of M’ ( = 100kNm say) is causing a
Dept. of CE, GCE Kannur Dr.RajeshKN
• Let us assume that a moment of M BA ( = 100kNm, say) is causing a
reaction of R’.
117. • If M’BA= MBA R+R’ must be zeroIf M BA MBA , R+R must be zero.
And the total moment will be M + M’BA .
• But since M’BA≠ MBA , R+ R’ ×(MBA /M’BA) =0
• Therefore, MBA /M’BA= – R / R’ = C1 i.e., MBA = C1 × M’BA
• Hence the total moment is
M M M C M’ M M’ R / R’M + MBA = M+ C1 ×M’BA = M – M’BA × R / R’
Dept. of CE, GCE Kannur Dr.RajeshKN
118. R
R’
1 0R C R′+ =
Dept. of CE, GCE Kannur Dr.RajeshKN
119. • When a moment of M’BA ( = 100kNm, say) is applied, what will be theBA ( , y) pp ,
value of M’CD?
Ratio of sway moments at column heads
2
1 1
2
2 2
BA
CD
M I L
M I L
′
=
′
Both column bases hinged:
2
1 1
2
2 2
BA
CD
M I L
M I L
′
=
′
Both column bases fixed:
2
1 1
2
2BAM I L
M I L
′
=
′One column base hinged and the other fixed:
2 2CDM I Lg
Dept. of CE, GCE Kannur Dr.RajeshKN
120. Both column bases hinged
3 3
1 1 2 2
3 3
P P
EI EI
δ = =
δδP
C
g
1 23 3EI EIB C
3 3EI EIδ δ
1 2 1 2
1 23 3
1 2
3 3
,
EI EI
P P
δ δ
= =
1 1 2 2,BA CDM P M P′ ′= =A
D
2P
2
M P I′
1P
1 1 1 1
2
2 2 2 2
BA
CD
M P I
M P I
= =
′
Dept. of CE, GCE Kannur Dr.RajeshKN
, 0, 0AB DCAlso M M′ ′= =
121. Both column bases fixed
P
C
δδ
1 2
2 2
1 2
6 6
,BA CD
EI EI
M M
δ δ
′ ′= =
B C
1 2
1 2
2
1 1
2
2 2
BA
CD
M I
M I
′
=
′A
D
2P
1P
, ,BA AB CD DCAlso M M M M′ ′ ′ ′= =
Dept. of CE, GCE Kannur Dr.RajeshKN
122. One column base fixed and the other hinged
3
2 2
3
P
EI
δ =
g
δδP
C 23EI
23EI δ
B C
2
2 3
2
3EI
P
δ
=1 2
1 2
2 22 2
1 2
6 3
,BA CD
EI EI
M M P
δ δ
′ ′= = =A
D
2P
1 2
2
1 1
2
2BAM I′
=
1P
2
2 2CDM I′
0Al M M M′ ′ ′
Dept. of CE, GCE Kannur Dr.RajeshKN
, , 0BA AB DCAlso M M M′ ′ ′= =
123. Example 8
1 0R C R′+ =
R R’
Dept. of CE, GCE Kannur Dr.RajeshKN
124. ( )4 5EI( )
( ) ( )
4 5
4 45 5
BA BC CB CD
EI
DF DF DF DF
EI EI
= = = =
+
0.5=
2 2
2 2
16 1 4
10.24
5
BC
Pab
FEM kNm
l
− − × ×
= = = −
5l
2 2
16 1 4
2 56
Pa b
FEM kN
× ×
2 2
2.56
5
CBFEM kNm
l
= = =
Dept. of CE, GCE Kannur Dr.RajeshKN
125. To find M values
A B C D
0 5 0 5 0 5 0 50 DF00.5 0.5
-10.24 2.56
5.12 5.12 -1.28 -1.28
FEM
Di t
0.5 0.50 DFs0
2.56 -0.64 2.56 -0.64
0.32 0.32 -1.28 -1.28
CO
Dist
Dist
0.16 -0.64 0.16 -0.64
0.32 0.32 -0.08 -0.08 Dist
CO
Dist
0.16 -0.04 0.16 -0.04
0.02 0.02 -0.08 -0.08
2.88 5.78 -5.78 +2.72 -2.72 -1.32 Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
126. 2.88 5 5.78xA− × = −
( )1.73xA⇒ = →
R
( )x →
1.32 5 2.72xD− + × =
( )0.81xD⇒ = ←
1.73 0.81 0R− − =
( )0 92R ( )0.92R = ←
Assume M’BA= -100 kNm
2
1 1
2
BAM I L
M I L
′
=
′
Ratio of sway moments at column heads for both
column bases fixed:
Dept. of CE, GCE Kannur Dr.RajeshKN
2 2CDM I L
Hence M’CD= -100 kNm
Also, M’AB= M’DC= -100 kNm
127. To find M’BA values
A B C DA B C D
0.5 0.5
-100.0 -100.0 -100.0 -100.0 FEM
0.5 0.50 DFs0
50.0 50.0 50.0 50.0
25.0 25.0 25.0 25.0
FEM
CO
Dist
-12.5 -12.5 -12.5 -12.5
-6.25 -6.25 -6.25 -6.25
CO
CO
Dist
6.25 6.25 6.25 6.25
3.125 3.125 3.125 3.125
1.563 1.563 1.563 1.563
Dist
CO
CO
-0.781 -0.781 -0.781 -0.781
-79.687 -60.156 60.157 60.156 -60.157 -79.687 Final Moments
Dist
Dept. of CE, GCE Kannur Dr.RajeshKN
129. All the end-moments are to be found as:
1FINAL BAM M C M′= +
All the end moments are to be found as:
R=0.92 R’=56R =56
Dept. of CE, GCE Kannur Dr.RajeshKN
131. Example 9 Δ Δ
20 kN B C
4m
B C
4m
4m
4m
EI
A D
EI EI
DA
( )4 4EI
DF DF= =
( )4 4EI
DF =
( )
( ) ( )4 44 4
BA BCDF DF
EI EI
= =
+
0.5=
( )
( ) ( )4 34 4
CBDF
EI EI
=
+
0.571= 0.57
( )
( ) ( )
3 4
4 34 4
CD
EI
DF
EI EI
=
+
Dept. of CE, GCE Kannur Dr.RajeshKN
( ) ( )
0.429=
132. To find M values
No fixed end moment since there is no member force
(only a joint force of 20 kN).
To find M values
(o y a jo t o ce o 0 N).
R= -20 kN
Assume fixed end moment due to sway M’BA as -10 kNm.
Ratio of sway moments at column heads for One column base hinged
and the other fixed: 2
1 12BAM I L′ 1 1
2
2 2
BA
CDM I L
=
′
2=
5CDM kNm′∴ = −
Dept. of CE, GCE Kannur Dr.RajeshKN
Also, M’AB= -10 kNm, M’DC= 0
133. To find M’BA values
A B C D
0.5 0.5 0.571 0.4290 DFs0.5 0.5
-10 -10 -5
5 5 2.86 2.14
FEM
Dist
0.571 0.429 DFs
2.5 1.43 2.5
-0.72 -0.71 -1.43 -1.07
CO
Dist
Dist
-0.36 -0.72 -0.36
0.36 0.36 0.21 0.15 Dist
CO
0.18 0.1 0.18
-0.05 -0.05 -0.1 -0.08 Dist
CO
-7.68 -5.41 5.41 3.86 -3.86 Final Moments
Dept. of CE, GCE Kannur Dr.RajeshKN
134. ( )7.68 4 5.41xA− + × = ( )3.27xA⇒ = ←
4 3 86D × = ( )0 965D⇒ = ←4 3.86xD × = ( )0.965xD⇒ = ←
3 27 0 965 0R′+ ( )4 235R′⇒ = →3.27 0.965 0R− − + = ( )4.235R⇒ = →
20
1 0R C R′+ = 120 4.235 0C⇒− + = 1
20
4.723
4.235
C∴ = =
All the end-moments are to be found as:
1 1FINAL BA BAM M C M C M′ ′= + =
Dept. of CE, GCE Kannur Dr.RajeshKN
136. SummarySummary
Displacement method of analysis
• Slope deflection method-Analysis of continuous beams and
Displacement method of analysis
Slope deflection method Analysis of continuous beams and
frames (with and without sway)
• Moment distribution method- Analysis of continuous beams
and frames (with and without sway)and frames (with and without sway).
Dept. of CE, GCE Kannur Dr.RajeshKN
136
