Moment Area Theorems:
Theorem 1:
When a beam is subjected to
external loading, it under goes
deformation. Then the intersection
angle between tangents drawn at
any two points on the elastic curve
is given by the area of bending
moment diagram divided by its
flexural rigidity.
1
Moment Area Theorems:
Theorem 2:
The vertical distance between any
point on the elastic curve and
intersection of a vertical line
through that point and tangent
drawn at some other point on the
elastic curve is given by the
moment of area of bending
moment diagram between two
points taken about first point
divided by flexural rigidity.
2
Fixed end moment due to a point load at the mid span:
 1 WL
  M M 
 L    A B L
 
2 4
  2  0
AB  
EI
WL
MA MB  - - - -  (1)
4
1 
1
 MA L  L  
2
3 
AA1  
3
MA 2MB   WL
8
2   1 WL  L
1
L
 MB L  L    
2
3
2
4

 
2
EI
     ( 2)
3
From (1) and (2) we get
WL
MB  
8
WL
WL  WL 
WL
 MA  
 MB  
 



4
4
8 
8

Both moments are negative and
hence they produce hogging
bending moment.
4
Stiffness coefficients
a) When far end is simply supported
BB ' 
Moment of area of BMD between A & B about B
EI
1
2
  M L  L
2
3

EI
2
ML

3EI
ML2
 BB  AL 
3EI
I
M 
3EI
A
L
5
b) When far end is fixed
1
1

 MAL - MBL 
area of BMD  2
2

A 

EI
EI

M A - M B  L
2 EI
- - - - - - - -  1
Moment of the Area of BMD between A & B about A
EI
1
L
1
2
 MAL    MB L 
2
3
2
 3  0

EI
MA   2MB
- - - - - - - --  2
AA 1 
6
Substituting in (1)
MA  MB  L  
A
2EI
 2MB - MB  L  
A
2EI
2EI
 MB 
A
L
4EI
From (2) MA  2 MB 
A
L
7
Fixed end moments due to yielding of support.
area of BMD between A and B
0
EI
 MA MB 

L
2
ie.  (MA  MB )  0
 AB 
Hence
MA   MB
Moment of area of BMD b/n A and B about B
EI
L
1
2 
1
 M B  L   M AL  L 
3
2
3 
 2
EI






 M  2M A  2 
- B
L
6
EI


   BB 1 
8
  M A  2M A  2 
-
L  Since M A   M B
6 EI


M A L2

6 EI
M A 
Now
6 EI

2
L
MB   MA  
Hence hogging BM
6 EI

2
L
Hence sagging BM
9
Fixed end moment for various types of loading
10
11
Assumptions made in slope deflection method:
1) All joints of the frame are rigid
2) Distortions due to axial loads, shear stresses being small
are neglected.
3) When beams or frames are deflected the rigid joints are
considered to rotate as a whole.
12
Sign conventions:
Moments: All the clockwise moments at the ends of members
are taken as positive.
Rotations: Clockwise rotations of a tangent drawn on to an
elastic curve at any joint is taken as positive.
Sinking of support: When right support sinks with respect to
left support, the end moments will be anticlockwise and are
taken as negative.
13
Development of Slope Deflection Equation
Span AB after deformation
Effect of loading
Effect of rotation at A
14
Effect of rotation at B
Effect of yielding of support B
Hence
M AB  FAB 
4 EI
2 EI
6 EI
2 EI
A 
B 
  FAB 
L
L
L
L2
3 

2




B
 A
L 

4 EI
2 EI
6 EI
2 EI
B 
 A  2  FBA 
L
L
L
L
3 

2




B
 A
L 
Similarly M BA  FBA 
15
Slope Deflection Equations
MAB  FAB 
2EI 
3 
2




A
B
L 
L 
MBA  FBA 
2EI 
3 
2




A
B
L 
L 
16
17
Example: Analyze the propped cantilever shown by using slope
deflection method. Then draw Bending moment and shear force
diagram.
Solution:
FAB
wL2
wL2

, FBA  
12
12
18
Slope deflection equations
2EI
2A  B 
L
 wL2 2EI


B
12
L
MAB  FAB 
2EI
2B  A 
MBA  FBA 
L
wL2 4EI


B
12
L
 (1)
 ( 2)
19
Boundary condition at B
MBA=0
MBA
wL2 4EI


B  0
12
L
wL3
 EIB  
48
Substituting in equations (1) and (2)
MAB
wL2 2  wL3 
wL2

 

12 L  48 
8
wL2 4  wL3 
MBA  
 
0
12 L  48 
20
Free body diagram
 MB  0
V  0
wL2
L
RA  L 
 wL 
8
2
5
R A  wL
8
5
RB  wL  R A  wL  wL
8
RB 
3
wL
8
21
9
wL2
128
3
L
8
3
L
8
SX  
3
wL  wX  0
8
3
X  L
8
 Mmax 
9
wL2
128
22
Example: Analyze two span continuous beam ABC by
slope deflection method. Then draw Bending moment &
Shear force diagram. Take EI constant
23
Solution:
FAB
 Wab 2
100  4  22


 44.44KNM
2
2
L
6
FBA
Wa 2b
100  42  2


 88.89KNM
2
2
L
6
 wL2
20  52
FBC 

 41.67KNM
12
12
FCB
 wL2 20  52


 41.67KNM
12
12
24
Slope deflection equations
2EI
2A  B 
MAB  FAB 
L
2EI
 44.44 
B
6
1
 44.44  EI B    (1)
3
MBA  FBA 
2EI
2B  A 
L
 88.89 
2EI  2B
6
2
 88.89  EIB      (2)
3
2EI
2B  C 
L
2EI
2B  C 
 41.67 
5
4
2
 41.67  EIB  EIC  (3)
5
5
MBC  FBC 
MCB
2EI
2C  B 
 FCB 
L
2EI
2C  B 
 41.67 
5
 41.67 
4EI
2
C  EIB  ( 4)
5
5
25
Boundary conditions
i. -MBA-MBC=0
MBA+MBC=0
ii. MCB=0
Now
2
4
2
EIB 41.67  EIB  EIC
3
5
5
22
2
 47.22 
EIB  EIC  0      (5)
15
5
2
4
MCB  41.67  EIB  EIC  0      (6)
5
5
MBA  MBC  88.89 
Solving
EIB  20.83
EIC  41.67
26
1
MAB  – 44.44   20.83  51.38 KNM
3
2
MBA   88.89   20.83  75.00 KNM
3
4
2
MBC  – 41.67   20.83   41.67  75.00 KNM
5
5
MCB   41.67 
2
 20.83  4  41.67  0
5
5
27
Free body diagram
Span AB:
MA = 0
V = 0
Span BC: M C = 0
V = 0
RB ×6 = 100 × 4 + 75 - 51.38
RB = 70.60 KN
R A +RB = 100KN
 RA = 100 - 70.60 = 29.40 KN
5
RB × 5 = 20 × 5 × + 75
2
 RB = 65 KN
RB + RC = 20 × 5 = 100KN
RC = 100 - 65 = 35 KN
28
BM and SF diagram
29
Example: Analyze continuous beam ABCD by slope
deflection method and then draw bending moment diagram.
Take EI constant.
Solution:
Wab 2
100  4  22
FAB  

 - 44.44 KN M
2
2
L
6
Wa 2b
100  42  2
FBA  

  88.88 KNM
2
2
L
6
30
wL2
20  52
FBC  

 - 41.67 KNM
12
12
wL2
20  52
FCB  

  41.67 KNM
12
12
FCD  20  1.5  - 30 KN M
31
Slope deflection equations:
2EI
2A B   44.44  1 EIB
L
3
- - - - - - - --  1
2EI
2
2B A   88.89  EIB
MBA FBA 
L
3
- - - - - - - --  2
MAB FAB 
2EI
2B C   41.67  4 EIB  2 EIC
L
5
5
- - - - - - - -  3
2EI
4
2
2C B   41.67  EIC  EIB
MCB FCB 
L
5
5
- - - - - - - -  4
MBC FBC 
MCD  30 KNM
32
Boundary conditions
MBA MBC  0
MCB MCD  0
Now, MBA MBC  88.89 
 47.22 
2
4
2
EIB 41.67  EIB  EIC
3
5
5
22
2
EIB  EIC  0
15
5
4
2
EIC  EIB 30
5
5
2
4
 11.67  EIB  EIC
5
5
- - - - - - - -  5 
And, MCB MCD  41.67 
Solving
EIB  32.67
       6 
EIC  1.75
33
Substituting
MAB  44.44 
1
 32.67  61.00 KNM
2
MBA  88.89 
2
 32.67  67.11 KNM
3
MBC  41.67 
4
 32.67   2 1.75  67.11 KNM
5
5
4
2
MCB  41.67  1.75   32.67  30.00 KNM
5
5
MCD  30 KNM
34
35
Example: Analyse the continuous beam ABCD shown in figure
by slope deflection method. The support B sinks by 15mm.
Take E  200  105 KN / m2 and I  120  106 m4
Solution:
Wab2
FAB   2  44.44 KNM
L
FBA  
2
Wa b
 88.89 KNM
2
L
FBC
wL2

 41.67 KNM
8
FCB
wL2

 41.67 KNM
8
FCD  20  1.5  30 KNM
36
FEM due to yielding of support B
For span AB:
mab  mba  
6  200
15
6EI
5
6



10

120

10

 6 KNM

2
2
6
1000
L
For span BC:
6EI
6  200
15
5
6



10

120

10

 8.64KNM
mbc  mcb   2 
2
5
1000
L
37
Slope deflection equation
EI
2 A B   6EI2   - 44.44  1 EIB 6
L
L
3
1
 50.44  EIB
- - - - - - - - - -  1
3
MAB  F AB 
2EI
6EI
2
(2B  A )  2   88.89  EIB 6
L
L
3
2
 82.89  EIB
- - - - - - - - - - - -  2
3
MBA  FBA 
2EI
6EI
2
(2B C )  2  - 41.67  EI2B C   8.64
L
L
5
4
2
 33.03  EIB  EIC
- - - - - - - --  3 
5
5
MBC  FBC 
38
2EI
6EI
2
(2C B )  2   41.67  EI2C B   8.64
L
L
5
4
2
 50.31  EIC  EIB
- - - - - - - --  4 
5
5
MCD  30 KNM
- - - - - - - --  5
MCB  FCB 
Boundary conditions M M  0
BA
BC
MCB MCD  0
Now
Solving
22
2
MBA MBC  49.86 
EIB  EIC  0
15
5
2
4
MCB MCD  20.31  EIB  EIC  0
5
5
EIB  31.35
EIC  9.71
39
Final moments
1
 31.35  60.89 KNM
3
2
MBA  82.89   31.35  61.99 KNM
3
4
2
MBC  33.03   31.35   9.71  61.99 KNM
5
5
4
2
MCB  50.31   9.71   31.35  30.00 KNM
5
5
MAB  50.44 
MCD  30 KNM
40
41