Slope Deflection Method
All structures must satisfy:
Load-displacement relationship
 Equilibrium of forces
 Compatibility of displacements

Using the principle of superposition by
considering separately the moments
developed at each support of a typical
prismatic beam (AB) shown in Fig. 1(a) of a
continuous beam, due to each of the
displacements ,
, and the
applied loads. Assume clockwise moments
are +ive.
1.
Assume ends A and B are fixed, i., e., the
rotations
. This means that we
have to apply counterclockwise moment at
end A
and clockwise moment at end B
due to the applied loads to cause zero
rotation at each of ends A and B. Table (1)
gives
for different loading conditions.
Table (1)
2. Release end A against rotation at end A (rotates to
its final position
) by applying clockwise
moment
while far end node B is held fixed as
shown in Fig. 1.
3. Now, the clockwise moment
- rotation
relationship is:
4. The carry over moment at end B is:
5. In a similar manner, if end B of the beam rotates to
its final position
, while end A is held fixed. The
clockwise moment
– rotation
relationship
is:
6. The carry over moment at end A is:
7.
If node B is displaced relative to as shown in Fig.
(1), so that the cord of the member rotates
clockwise i., e., positive displacement and yet both
ends do not rotate, then equal but anticlockwise
moments are developed in the member as shown
in the figure.
Slope-Deflection Equation
Load-displacement relationship
If the end moments due to each displacement and
the loading are added together, the resultant
moments at the ends may then be written as:
For prismatic beam element, equation (1) may be
written as:
The slope deflection equations (1 or 2) relate
the unknown moments applied to the nodes
to the displacements of the nodes for
any span of the structure.
To summarize application of the slopedeflection equations, consider the
continuous beam shown in Fig. (2) which has
four degrees of freedom.
Now equation (2) can be applied to each of
the three spans.
Fig. (2)
From Fig.(2):
Equilibrium conditions
Compatibility conditions



These equations would involve the four unknown
rotations  A ,  B ,  C ,  D .
Solving for these four unknown rotations. It may be
noted that there is no relative deflection between the
supports, so that
The values of the obtained
rotations may then be substituted in to the slope
deflection equations to determine the internal
moments at the ends of each member.
If any of the results are negative, they indicate
counterclockwise rotation.
Example (1)
Solution
1.
Draw the shear and moment diagrams for the beam
shown in Fig.(3). EI is constant.
Using the formulas for the
tabulated in Table
(1) for the given loadings:
Fig. (3)
2. There are two slopes at B and C, i., e.,
are unknowns. Since end A is fixed,
Also,
since the supports do not settle, nor are they
displaced up or down
Now, by applying the equilibrium conditions:
Substituting the computed values in to moment
equations (a), (b), (c), and (d):
By considering the values of support moments and
the applied loads, the support reactions may then be
determined:
RA = 8.3625 kN
RB = 10.2042 kN
RC = 1.8333 kN
Shearing force and bending moment digrams are
shown in Fig. (4).
Fig. (4) Shearing Force & Bending Moment Digrams
Example (2)
Determine the internal moments at the supports of the
beam shown in Fig. (5). The support at B is displaced
(settles) 12 mm.
Solution
1.
Two spans must be considered. FEMs are determined
using Table (1).
 AB
 BC
0.012  0

 6.667 x10 4
18
0  0.012

 0.1x10 4
12
2. Using equation 2:
2 EI
M AB  (
) AB ( B  3 x6.667 x10 4 )
18
2 EI
M BA  (
) AB (2 B  3 x6.667 x10  4 )
18
2 EI
M BC  (
) BC (2 B   C  3 x0.1x10 4 )
12
2 EI
M CB  (
) BC (2 C   B  3 x0.1x10 4 )  0
12
(i )
( j)
(k )
(l )
3. Equilibrium condition:
M
B
 0 and M C  0
2 EI
M AB  (
) AB ( B  3x6.667 x10  4 )
( m)
18
2 EI
2 EI
(
) AB (2 B  3x6.667 x10  4 )  (
) BC (2 B   C  3x0.1x10  4 )  0 (n)
18
12
2 EI
(
) BC (2 C   B  3x0.1x10  4 )  0
( p)
12

In order to obtained the rotations  B and C equations
(n) & (p) may then be solved simultaneously, it may be
noted that  A  0 since A is fixed support. Thus,
 B  4.65294x104 rad. and C  2.47647x104 rad.
Substituting these values into equations (i to l) yields
M AB   621556 k N.mm
anticlock wise
M BA   433112 k N.mm
anticlock wise
M BC   433112 k N.mm
clock wise
M CB  0
Example (3)
If end A in example (1) is simply supported, and by
applying the compatibility condition, their will be three
unknown rotations, ( A , B , C )
Now,
2 EI
M AB  (
) AB (2 A   B )  4.5
(a)
2. 4
2 EI
M BA  (
) AB (2 B )  4.5
(b)
2.4
2 EI
M BC  (
) BC (2 B   C )  1.62
(c )
3 .6
2 EI
M CB  (
) BC (2 C   B )  1.62
(d )
3 .6
Applying the equilibrium conditions:
2 EI
M AB  (
) AB (2 A   B )  4.5  0
(1)
2 .4
2 EI
2 EI
(
) AB (2 B )  4.5  (
) BC (2 B   C )  1.62  0
2 .4
3 .6
2 EI
M CB  (
) BC (2 C   B )  1.62  0
(3)
3. 6
( 2)
By solving equations (1, 2 & 3) for  A ,  B , C and
substitute the values into equations (a, b, c, d):
M AB  0
M BA  4.158 kN.m clockwise
M BC  4.158 kN.m anticlockwise
M CB  0
Shearing force and bending moment diagrams are shown
in the following figure.
Example (4)
Determine the moments at each joint of the frame shown in
Fig.(7). EI is constant.
Fig. (7)
( FEM ) BC
5(24 )(8) 2

 80 kN.M
96
( FEM ) CB
5(24 )(8) 2

 80 kN.M
96
Because ends A and D are fixed supports.
and  AB   BC   CD  0 since no sidesway will occur.
2 EI
) AB ( B )  0.1667 EI B
12
2 EI
) AB ( 2 B )  0.3334 EI B
M BA  (
12
2 EI
) BC (2 B   C )  80  0.5 EI B  0.25 EI C  80
M BC  (
8
2 EI
) BC (2 C   B )  80  0.5 EI C  0.25 EI B  80
M CB  (
8
2 EI
) CD ( 2 C )  0.3334 EI C
M CD  (
12
2 EI
) CD ( C )  0.1667 EI C
M DC  (
l
M AB  (
Equilibrium conditions:
M
BA
 M BC  0
0.3334 EI B  0.5 EI B  0.25 EI C  80  0
Or
0.8334 EI B  0.25 EI C  80  0
M
CB
(1)
 M CD  0
0.5 EI C  0.25 EI B  0.3334 EI C  80  0
Or
0.8334 EI C  0.25 B  80  0
( 2)
Solving simultaneously yields
137 .1
137 .1
B 
and C  
EI
EI
Therefore,
M AB  22 .9 kN.m
clockwise
M BA  45 .7 kN.m
clockwise
M BC  45 .7 kN.m anticlockwise
M CB  45 .7 kN.m
clockwise
M CD  45 .7 kN.m anticlockwise
M DC  22 .9 kN.m anticlokwise
The bending moment diagram is shown in Fig.(8).
Fig.(8)
Example (5)
Determine the internal moments at each of the frame
shown in Fig.(9).
Solution
Fig.(9)
1.
Fixed end moments:
Span AB : ( FEM ) AB
Span BC : ( FEM ) BC
w(1.5L) 2
w(1.5L) 2

, ( FEM ) BA  
12
12
wL2
wL2

, ( FEM ) CB  
12
12
Span BD : ( FEM ) DB
Pl
Pl

, ( FEM ) BD  
8
8
2. Joint moments:
M AB
2 EI
w(1.5L) 2
(
) AB (2 A   B ) 
1.5L
12
(i )
M BA
2 EI
w(1.5L) 2
(
) AB (2 B   A ) 
1.5L
12
( j)
2 EI
wL2
M BC  (
) BC (2 B   C ) 
L
12
2 EI
wL2
M CB  (
) BC (2 C   B ) 
L
12
2 EI
PL
M BD  (
) BD (2 B   D ) 
L
8
2 EI
PL
M DB  (
) BD (2 D   B ) 
L
8
(k )
(l )
( m)
( n)
3. Equilibrium conditions:
M AB
M
w(1.5L) 2
2 EI
(
) AB (2 A   B ) 
0
1.5L
12
B
(1)
0
w(1.5L) 2
2 EI
2 EI
wL2
(
) AB (2 B   A ) 
(
) BC (2 B   C ) 

1.5L
12
L
12
2 EI
PL
(
) BD (2 B   D ) 
0
 (2)
L
8
2 EI
wL2
M CB  (
) BC (2 C   B ) 
0
L
12
2 EI
PL
M DC  (
) BD ( B ) 
L
8
(3)
(4)
Solving equations (1,2,3) simultaneously yields  A ,  B ,  C
Substituting the rotation values into equations (i to n) to
determine the joint moments.
Example (6)
Determine the joint internal moments of the frame shown in
Fig.(10), both ends A and D are fixed.
Assume
(
EI
EI
EI
) AB  ( ) BC  1 and ( ) CD  1.5
L
L
L
Fig.(10)
Solution
1. Fixed end moments:
Span AB : ( FEM ) AB  
( FEM ) BA  
10 (1.8)(3.6) 2
(5.4)
Span BC : ( FEM ) BC
( FEM ) CB
10 (3.6)(1.8) 2
2
(5.4)
2
 4.0 kN.m
 8.0 kN.m
3(7.2) 2
 12 .96 kN.m

12
3(7.2) 2
 12 .96 kN.m

12
It is assumed that the axial deformation is neglected so that
BBO  CCO  
as shown in the following figure.
It may be noted that
 CD  1.5 AB and  BC   A   D  0
2. Joint moments:
M AB  ( B  3 AB )  4
2 EI
) AB (2 B   AB )  8
1 .5 L
 (2 B   C )  12 .96
(i )
M BA  (
( j)
M BC
(k )
M CB  1.5(2 C   B )  12 .96
(l )
M CD  (2 C  4.5 AB )
 3 C  6.75 AB
( m)
M DC  1.5( C  4.5 AB )
 1.5 C  6.75 AB
( n)
3. Equilibrium conditions:
Jo int B :
M
BA
 M BC  0
Or
4 B   C  3 AB  4.96
Jo int C :
M
CB
(1)
 M CD  0
Or
 B  5 C  6.75 AB  12 .96
( 2)
Since a horizontal displacement  occurs, the summing forces on the
entire frame in the x-direction. This yields

F
X
 0 :  10  H A  H D  0
In which :
HA 
10 M AB  M BA

3
5.4
and
M CD  M DC
3.6
10 M AB  M BA M CD  M DC
  10  

0
3
5 .4
3.6
Or
HD 
 B  2.25 C  4.75 AB  10 .667
(3)
Solving equations (1, 2, 3) yields
 B  2.8208 ,  C  0.565 ,  AB  1.9194
By substituting these values into moment equations (i to n):
M AB  6.9374 kN.m
M BA  7.8833 kN.m
M BC  7.8833 kN.m
M CB  14 .6509 k .N .m
M CD  14 .6509 kN.m
M DC  13 .8035 kN.m
Bending moment diagram is plotted in the following figure.