Module2 rajesh sir

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
1

Module IIModule II
Displacement method of analysis
• Slope deflection method-Analysis of continuous beams and
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
2

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)
3

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)
448
B
wL
EI
θ∴ =

•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
Bθ B Bm θ• Moment required to induce a rotation of is

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.
66

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
support moments

anticlockwise support
moment (reactive
moment is clockwise)
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

clockwise support
moment (reactive
moment is
moment (reactive
moment is clockwise)moment is
anticlockwise)
moment is clockwise)
ve+ve−
A B
Moment
distribution/slope
convention
ve− ve−
A B
ve ve
for drawing BMD
A B

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
convention
ve− ve+
A B
ve ve+
for drawing BMD
A B

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−
Bθ
ve+ Aθ ve−

Sl d fl ti th dSlope deflection method

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.
13

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
ABM BAM

( ) 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.

( )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
1
6
AB
B
M l
EI
θ
′−
′ = 2
3
BA
B
M l
EI
θ′ =

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
′ ′
′= + = − +
17
3 6l EI EI l 3 6l EI EI l

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
18

Fixed end moments
8
PL−
8
PL
+
2L 2L2L 2L
19

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
Fixed end moments (reactive)

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= = =
22

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⇒ = +
23

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
⎜ ⎟
⎝ ⎠
24

⎛ ⎞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
25

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
26

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.
27

Example 2
40kN
80kN60kN
20kN m
A B C
D
3m 3m 3m 3m 3m 3m

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

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
30
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⇒ = + +

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
− −
= =
31
( )290 0.5
0.667
D CEI EIθ θ= = − −

0BA BCM M+ = 0BA BCM M+
( ) ( )
22
2 105 02 90
EIEI
θ θθ θ
⎛ ⎞⎛ ⎞⇒ + + =+ +⎜ ⎟ ⎜ ⎟( ) ( )2 105 02 90
66
B CB A
θ θθ θ⇒ + + − =+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0 333 1 333 0 333 15EI EI EIθ θ θ+ +0.333 1.333 0.333 15A B CEI EI EIθ θ θ+ + =
( )0 333 1 333 0 333 15135 0 5 EI EIEI θ θθ + + =( )1
( )
( )0.333 1.333 0.333 15135 0.5 B CB EI EIEI θ θθ + + =−( )1
( )31.166 0.333 30B CEI EIθ θ+ = −
32

0CB CDM M+ = 0CB CDM M+
( ) ( )
22
2 60 02 105
EIEI
θ θθ θ
⎛ ⎞⎛ ⎞
⎜ ⎟ ⎜ ⎟( ) ( )2 60 02 105
66
C DC B
θ θθ θ
⎛ ⎞⎛ ⎞⇒ + + − =+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
0.333 1.333 0.333 45B C DEI EI EIθ θ θ+ + = −
( )0 333 1 333 0 333 90 0 5 45EI EI EIθ θ θ+ + − − = −( )2
( )40.333 1.167 15B CEI EIθ θ+ = −
( )0.333 1.333 0.333 90 0.5 45B C CEI EI EIθ θ θ+ + − − = −( )2
( )4B C
( ) 0 333 0 388 0 111 103 EI EIθ θ ( )5( ) 0.333 0.388 0.111 103 B CEI EIθ θ× ⇒ + = − ( )5
( ) 1.167 0.388 1.362 17.54 B CEI EIθ θ× ⇒ + = − ( )6
C

7 5
( ) ( ) 1.251 7.56 5 CEIθ− ⇒ = −
7.5
6.0
1.251
CEIθ
−
⇒ = = −
0.333 1.167 6.0 15 24B BEI EIθ θ+ ×− = − ⇒ = −( )4
( )135 0.5 14724AEIθ = − =−
90 0.5 6.0 87DEIθ = − − ×− = −D
4 2
90EI EIθ θ
⎛ ⎞
⎜ ⎟( )
2
2 90
EI
M θ θ 90
6 6
B AEI EIθ θ
⎛ ⎞
= + +⎜ ⎟
⎝ ⎠
( )2 90
6
BA B AM θ θ⇒ = + +
⎛ ⎞4 2
24 147 90
6 6
⎛ ⎞
= ×− + × +⎜ ⎟
⎝ ⎠
123=

( )
2EI 4 2⎛ ⎞
( )
2
2 105
6
BC B C
EI
M θ θ⇒ = + −
4 2
105 123
6 6
B CEI EIθ θ
⎛ ⎞
= + − = −⎜ ⎟
⎝ ⎠
( )
2
2 105
6
CB C B
EI
M θ θ⇒ = + +
4 2
105 93
6 6
C BEI EIθ θ
⎛ ⎞
= + + =⎜ ⎟
⎝ ⎠
( )
2
2 60
6
CD C D
EI
M θ θ⇒ = + −
4 2
60 93
6 6
C DEI EIθ θ
⎛ ⎞
= + − = −⎜ ⎟
⎝ ⎠

Example 3

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= = = =
37

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
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
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⇒ = + +

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⇒ + =
2.25 0.75 61.667B CEI EIθ θ⇒ +

0CB CDM M+ = ( ) ( )
6 6
2 26 667 2 26 667 0
EI EI
θ θ θ θ⎛ ⎞ ⎛ ⎞
⇒ + + + + − =⎜ ⎟ ⎜ ⎟0CB CDM M+ ( ) ( )2 26.667 2 26.667 0
8 8
C B C Dθ θ θ θ⇒ + + + + − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3 0 75 0 75 0EI EI EIθ θ θ⇒ + + =3 0.75 0.75 0C B DEI EI EIθ θ θ⇒ + + =
( )3 0.75 0.75 17.778 0.5 0C B CEI EIθ θ θ⇒ + + − − =
2.625 0.75 13.333C BEIθ θ⇒ + = ( )2
( )12.25 0.75 61.667B CEI EIθ θ⇒ + =
28.421BEIθ =
3 041EIθ = −
( )10 0.5 10 0.5 28.421A BEI EIθ θ= − = −
3.041CEIθ = −
4.211AEIθ⇒ = −
17.778 0.5 17.778 0.5 3.041D CEI EIθ θ= − − = − − ×−
C
16.258DEIθ⇒ = −

( )
4
2 10 0 5 10
EI
M EI EIθ θ θ θ+ + + +( )2 10 0.5 10
8
BA B A B AM EI EIθ θ θ θ= + + = + +
28.421 0.5 4.211 10 36.32kNm= + ×− + =
( )
6
2 26.667 1.5 0.75 26.667
8
BC B C B C
EI
M EI EIθ θ θ θ= + − = + −
8
1.5 28.421 0.75 3.041 26.667 13.68kNm= × + ×− − =
( )
6
2 26.667 1.5 0.75 26.667
8
CB C B C B
EI
M EI EIθ θ θ θ= + + = + +
( )
6
2 26 667 1 5 0 75 26 667
EI
M EI EIθ θ θ θ
1.5 3.041 0.75 28.421 26.667 43.42kNm= ×− + × + =
( )2 26.667 1.5 0.75 26.667
8
CD C D C DM EI EIθ θ θ θ= + − = + −
1.5 3.041 0.75 16.258 26.667 43.42kNm= ×− + ×− − = −
41

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= − × = −

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
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
44
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )2
5
DC D CM θ θ⇒ = +

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 =
45

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

( )
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= − = − =−

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= + = ×− + = −
( ) ( )
4.8 4.8
2 2 48.368 34.235 60
5 5
DC D C
EI
M kNmθ θ= + = × − =

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
−
= × × × =

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 =

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
⎛ ⎞×
⇒ = + +⎜ ⎟
⎝ ⎠
51

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 −⎛ ⎞ ⎛ ⎞⇒ + = − ×+ + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
B A B
EI EI
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

20 16 4
4 4
20 16
2.5 0.5 9.375 10
10 10
B Aθ θ −
⇒ + + − = − ×
4 44 4
2.5 0.5 9.375 10 4 10B Aθ θ − −
⇒ + = − × − ×
( )24
2.5 0.5 13.375 10B Aθ θ −
+ = − ×
3
0.5 3.875 10A Bθ θ −
+ = × ( )1
3
1.456 10Bθ −
= − ×3
4.603 10Aθ −
= ×

2
4 3 10EI −
⎛ ⎞×4 3 10
2 20
8 8
BA B A
EI
M θ θ
⎛ ⎞×
= + − +⎜ ⎟
⎝ ⎠
4 2
4 10 3 10⎛ ⎞
( )
4 2
3 34 10 3 10
2 1.456 10 4.603 10 20
8 8
−
− −⎛ ⎞× ×
= − × + × − +⎜ ⎟
⎝ ⎠
9.705 kNm=
2
6 3 10
2 16
8 8
BC B
EI
M θ
−
⎛ ⎞×−
= − −⎜ ⎟
⎝ ⎠
4 2
36 10 3 10
2 1.456 10 16
8 8
−
−⎛ ⎞× ×−
= ×− × − −⎜ ⎟
⎝ ⎠
9.705 kNm= −
2
6 3 10
16
8 8
CB B
EI
M θ
−
⎛ ⎞×−
= − +⎜ ⎟
⎝ ⎠
⎝ ⎠
8 8
⎜ ⎟
⎝ ⎠
4 2
36 10 3 10
1 456 10 16
−
−⎛ ⎞× ×−
= − × − +⎜ ⎟ 33 21 kNm=
54
1.456 10 16
8 8
= − × − +⎜ ⎟
⎝ ⎠
33.21 kNm=

Slope Deflection for frames:
No sideswayy

Example 6Example 6
B
10kN 2kN m
A
B C
2
2I I
332m
3m I
3m3m
D
I
D

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= =

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⇒ =
58
2 3
2DB D B DB
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
( )
2
3
DB B
EI
M θ⇒ =

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 = −

( )
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
60
( )
2
3
DB B
EI
M θ= ( )
2
0.6481 0.432
3
kNm= − = −

Slope Deflection for frames:
S dSidesway

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
1
AH
l
2
DH
l
= 0A DH H P+ + =

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
+ −
=
63
2
DH
l
= 0A DH H P+ + =
1
AH
l

CCDM
B
C
BAM
CDM
l
2l
w
DH
D
1l
D
DCM
AH
A
ABM
2
2 2CD DC
D
M M wl
H
+ +
= 2 0A DH H wl+ − =AB BA
A
M M
H
l
+
=
ABM
64
2
D
l1
A
l

Example 7

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)
66

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
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
67
2 3
2DC D C DC
EI
M FEM
l l
δ
θ θ
⎛ ⎞
= + − +⎜ ⎟
⎝ ⎠
2 3
5 5
C
EI δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠

0BA BCM M+ = ( )
2 3 2
2 02 10.24B B C
EI EIδ
θ θ θ
⎡ ⎤⎛ ⎞ ⎡ ⎤⇒ − + =+ −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦
BA BC ( )5 5 5
B B C⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦
( )1( )1.6 0.4 0.24 10.24B CEI θ θ δ+ − = ( )
0M M+ ( )
2 32
2 02 2 56
EIEI δ
θθ θ
⎡ ⎤⎛ ⎞⎡ ⎤⇒ ++ + ⎜ ⎟⎢ ⎥⎢ ⎥
( )1.6 0.4 0.24 10.24B CEI θ θ δ+
0CB CDM M+ = ( ) 2 02 2.56
5 55
CC B
θθ θ
⎡ ⎤⇒ + − =+ + ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦
( )2( )1.6 0.4 0.24 2.56C BEI θ θ δ+ − = −

0H H+ = 0AB BA CD DCM M M M+ +
⇒ + 5l l0A DH H+ =
1 2
0AB BA CD DC
l l
⇒ + = 1 2 5l l= =
2 3 2 3 2 3 2 3
2 2 0
5 5 5 5 5 5 5 5
B B C C
EI EI EI EIδ δ δ δ
θ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⇒ − + − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
3 3 3 3
2 2 0
5 5 5 5
B B C C
δ δ δ δ
θ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⇒ − + − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0.8 0B Cθ θ δ⇒ + − = ( )3
845
BEIθ =
272
CEIθ
−
=
48
7
EIδ =
105
BEIθ
105
CEIθ 7
69

2 3EI
M
δ
θ
⎛ ⎞
= ⎜ ⎟
2 845 3 48
1 5733 kNm
⎛ ⎞
= − =⎜ ⎟
5 5
AB BM θ= −⎜ ⎟
⎝ ⎠
2 3EI δ⎛ ⎞
1.5733
5 105 5 7
kNm= − =⎜ ⎟
⎝ ⎠
2 845 3 48⎛ ⎞2 3
2
5 5
BA B
EI
M
δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
2 845 3 48
2 4.838
5 105 5 7
kNm
⎛ ⎞
= × − =⎜ ⎟
⎝ ⎠
( )
2
2 10.24
5
BC B C
EI
M θ θ= + −
2 845 272
2 10.24 4.838
5 105 105
kNm
⎛ ⎞
= × − − = −⎜ ⎟
⎝ ⎠
( )
2
2 2.56
5
CB C B
EI
M θ θ= + +
2 272 845
2 2.56 3.707
5 105 105
kNm
−⎛ ⎞
= × + + =⎜ ⎟
⎝ ⎠
( )
5
2 3
2
5 5
CD C
EI
M
δ
θ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
5 105 105⎝ ⎠
2 272 3 48
2 3.707
5 105 5 7
kNm
−⎛ ⎞
= × − × = −⎜ ⎟
⎝ ⎠
2 3EI
M
δ
θ
⎛ ⎞
= −⎜ ⎟
5 5
CD C⎜ ⎟
⎝ ⎠ 5 105 5 7
⎜ ⎟
⎝ ⎠
2 272 3 48
2.682 kNm
−⎛ ⎞
= − × = −⎜ ⎟
5 5
DC CM θ= ⎜ ⎟
⎝ ⎠
2.682
5 105 5 7
kNm×⎜ ⎟
⎝ ⎠

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)
0DCM =

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⎛ ⎞
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
⎛ ⎞
⇒ = + −⎜ ⎟
⎝ ⎠

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 =
( )3
4
16 4 8 2 4
Cθ⇒ + − + − =⎢ ⎥⎣ ⎦
3.25 0.1875 0Cθ δ− =

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
−
− =
74
( )40.75 1.3125C
EI
θ δ

( )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
⎛ ⎞
= − = × − × = −⎜ ⎟
75
( )2 0.5 2 10.909 0.75 63.03 12.727
4 4
BA BM kNmθ × ×⎜ ⎟
⎝ ⎠

( ) ( )
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+ × + ×⎜ ⎟
⎝ ⎠
76

M t di t ib ti th dMoment distribution method

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
⎛ ⎞
⎜ ⎟
78
2
2
AB AB AB ABM
L L
θ θ θ
⎛ ⎞
∴ = − =⎜ ⎟
⎝ ⎠

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).

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)
80

( )
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.

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
1 2 3 4 1 2 3 4: : : :: : : :k k k k

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

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
Fixed end moments (reactive)

A
B
C1.4kNm
Unbalanced moment
A
B
C0.7kNm0.7kNm Unbalanced moment
distributed amongst
members
B 0 35kNm
A
B
C
0.35kNm0.35kNm
Distributed moments
carried over to far
ends of members

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

5 35kN
A B C4.3kNm2.05kNm 4.3kNm 5.35kNm
A
B
CA C
2.05
5 35
4.3
5.35

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
× ×
= + = + =
80 6
60
8 8
CD DC
Pl
FEM FEM kNm
×
− = = = =

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

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

Example 3
91

A B C DA B C D
Distrib.
factors
1 0.2727 0.7273 0.6667 0.3333 0
Fixed-end
moments
-14.700 +6.300 -8.333 +8.333 -12.500 +12.500
Release A andRelease A and
Carry over
+14.700 +7.350
Initial
moments
0.00 13.65 -8.333 +8.333 -12.500 +12.500
moments
Dist 1 -1.450 -3.867 +2.779 +1.388
CO 1 1.39 -1.934 +0.694
Dist 2 -0.379 -1.011 1.29 0.644
CO 2 0.645 -0.506 0.322
Dist 3 -0.176 -0.469 0.338 0.168
Final
moments
0 +11.645 -11.645 +10.3 -10.3 +13.516
92

93

Example 3
2 2
5 8l2 2
5 8
26.667
12 12
wl
kNm
×
= =
10 8
10
8 8
Pl
kNm
×
= =

Distribution factors
( )
( ) ( )
1
3 2 8
3 42 8 3 8
BA
EIK
DF
K K EI EI
= =
+ +
0.333=
( ) ( )1 2 3 42 8 3 8K K EI EI+ +
( )2
4 3 8
0 667
EIK
DF = = =
( )
( ) ( )1 2
0.667
3 42 8 3 8
BCDF
K K EI EI
= = =
+ +
( )
( ) ( )
4 3 8
3 43 8 3 8
CB
EI
DF
EI EI
=
+
0.571=
0.429=
( )
( ) ( )
3 3 8
3 43 8 3 8
CD
EI
DF
EI EI
=
+( ) ( )

A B C D
0.333 0.667 0.571 0.4291 1 Distribution factors
Joint couple dist.16.65 33.35
Fixed End Moments
Carry over16.675
-10 +10 -26.667 +26.667 -26.667 +26.667
Release A& D,
and carry over
Initial moments
+10 +5.0 -13.333 -26.667
0.0 +15.0 -26.667 +43.342 -40.0 0.0
Carry over
Distribution3.885 7.782 -1.905 -1.437
-0.953 +3.891
Carry over
Distribution0.317 0.636 -2.218 -1.673
-1.109 0.318
Distribution
Final Moments
0.369 0.74 -0.181 -0.137
0 +36.22 13.78 +43.28 -43.25 0

90kN
20kN30kN m
90kN
Example 4
A B C D
2.5m 2.5m
E2I I 2.4I
7.5m 5m 5m 3m
2 2
30 5
62.5
12 12
wl
kNm
×
= =
2 2
2 2
90 2.5 5
100
7.5
Pab
kNm
l
× ×
= =
2 2
2 2
90 2.5 5
50
7.5
Pa b
kNm
l
× ×
= =
20 3 60kNm× =
( )
( ) ( )
4 2 7.5
4 42 7 5 5
BA
EI
DF
EI EI
=
+
( )
( ) ( )
4 5
4 42 7.5 5
BC
EI
DF
EI EI
=
+0.571=
0.429=
7.5l
( ) ( )4 42 7.5 5EI EI+
( ) ( )
( )4 5EI
DF 0 357 ( )3 2.4 5EI
( )
( ) ( )4 35 2.4 5
CBDF
EI EI
=
+ 0.643=0.357= ( )
( ) ( )
3 2.4 5
4 35 2.4 5
CD
EI
DF
EI EI
=
+

A B C D
0.571 0.429
-150.0 150.0 -62.5 62.5 0.0 0.0 -60.0 FEM
0.357 0.6430 DFs1 0
30.0 60.0
-150.0 150.0 -62.5 62.5 30.0 60.0 -60.0
Balance D, and
carry over
Initial moments
-49.96 -37.54 -33.02 -59.48
-24.98 -16.51 -18.77 CO
Dist
9.43 7.08 6.7 12.07
4.72 3.35 3.54 CO
Dist
-1.91 -1.44 -1.26 -2.28
-0.96 -0.63 -0.72
Dist
CO
0.36 0.27 0.26 0.46
-171.2 107.92 -107.92 +19.23 -19.23 60.0 -60.0 Final Moments
Dist

Example 5
6 4
200 , 50 10E GPa I mm= = ×
( )3 2 8EI
DF
( )4 3 8EI
DF0 333 0 667
( )
( ) ( )3 42 8 3 8
BADF
EI EI
=
+
( )
( ) ( )3 42 8 3 8
BCDF
EI EI
=
+
0.333= 0.667=

2 2
3 8l20 8Pl × 2 2
3 8
16
12 12
wl
kNm
×
= =
20 8
20
8 8
Pl
kNm
×
= =
2
6
8
AB
Pl EI
FEM
L
δ−
= −
8 L
6 6 12 3
2
6 2 200 10 50 10 10 10 10
20
− −
× × × × × × × ×
= − − 2
20
8
20 18.75 38.75 kNm= − − = −
2
6
8
BA
Pl EI
FEM
L
δ
= −
8 L
6 6 12 3
2
6 2 200 10 50 10 10 10 10
20
8
− −
× × × × × × × ×
= −
2
8
20 18.75 1.25 kNm= − =

22
2
6
12
BC
wl EI
FEM
L
δ−
= +
6 6 12 3
2
6 3 200 10 50 10 10 10 10
16
8
− −
× × × × × × × ×
= − +
8
16 28.125 12.125 kNm= − + =
2
2
6
12
CB
wl EI
FEM
L
δ
= + 2
12
CB
L
6 6 12 3
6 3 200 10 50 10 10 10 10
16
− −
× × × × × × × ×
= + 2
16
8
= +
16 28 125 44 125 kNm= + =
16 28.125 44.125 kNm= + =

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

Example 6
6 4
200 , 50 10E GPa I mm= = ×

( )3 2 8EI
DF =
( )4 3 8EI
DF =0 333= 0 667=
( ) ( )3 42 8 3 8
BADF
EI EI
=
+ ( ) ( )3 42 8 3 8
BCDF
EI EI
=
+
0.333 0.667=
2 2
3 8
16
wl
kN
×20 8
20
Pl
kN
×
16
12 12
kNm= =20
8 8
kNm= =
6Pl EIδ
2
6
8
AB
Pl EI
FEM
L
δ
= − −
6 6 12 3
6 2 200 10 50 10 10 10 106 6 12 3
2
6 2 200 10 50 10 10 10 10
20
8
− −
× × × × × × × ×
= − −
20 18 75 38 75 kN20 18.75 38.75 kNm= − − = −
6Pl EI
FEM
δ
= − 2
8
BAFEM
L
6 6 12 3
6 2 200 10 50 10 10 10 10
20
− −
× × × × × × × ×
=
2
20
8
= −
20 18.75 1.25 kNm= − =

2
2
6
12
BC
wl EI
FEM
L
δ−
= + 2
12 L
6 6 12 3
2
6 3 200 10 50 10 10 10 10
16
− −
× × × × × × × ×
= − + 2
8
16 28.125 12.125 kNm= − + =
2
2
6
12
CB
wl EI
FEM
L
δ
= + 2
12
CB
L
6 6 12 3
6 3 200 10 50 10 10 10 10
16
− −
× × × × × × × ×
= + 2
16
8
= +
16 28 125 44 125 kNm= + =
16 28.125 44.125 kNm= + =

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

Moment Distribution for frames:
No sideswayy

Example 7Example 7
B
10kN 2kN m
A
B C
2
2I I
332m
3m I
3m3m
D
I
D

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= =

( )
( ) ( ) ( )
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
=
+ +
( ) ( ) ( )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=

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
−
= = −

Moment Distribution for frames:Moment Distribution for frames:
sidesway

• 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

• 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
• Let us assume that a moment of M BA ( = 100kNm, say) is causing a
reaction of R’.

• 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’

R
R’
1 0R C R′+ =

• 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

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
= =
′
, 0, 0AB DCAlso M M′ ′= =

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′ ′ ′ ′= =

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′ ′ ′
, , 0BA AB DCAlso M M M′ ′ ′= =

Example 8
1 0R C R′+ =
R R’

( )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
= = =

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

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:
2 2CDM I L
Hence M’CD= -100 kNm
Also, M’AB= M’DC= -100 kNm

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

80 5 60xA− + × =
( )28xA⇒ = ←
R′
80 5 60xD− + × =
( )28D⇒ = ←( )28xD⇒ = ←
28 28 0R− − + =
( )56R′ = →
0 92
1 0R C R′+ = 10.92 56 0C⇒− + = 1
0.92
0.0164
56
C∴ = =

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

13.01 2.99

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
=
+
( ) ( )
0.429=

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′∴ = −
Also, M’AB= -10 kNm, M’DC= 0

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

( )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′ ′= + =

4 723 7 68M × 36 273 kNm=4.723 7.68ABM = ×−
4 723 5 41M = ×−
36.273 kNm= −
25 551 kNm= −4.723 5.41BAM = ×
4 723 5 41M = ×
25.551 kNm
25 551 kNm=4.723 5.41BCM = ×
4.723 3.86CBM = ×
25.551 kNm=
18.231 kNm=.7 3 3.86CB
4.723 3.86CDM = ×− 18.231 kNm= −4.723 3.86CDM
0DCM = 0DCM

SummarySummary
• Slope deflection method-Analysis of continuous beams and
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).
136

Module2 rajesh sir

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