Analysis of statically indeterminate structures

Analysis of Statically
Indeterminate Structures
Clasical Methods
Prof. Dr. Ahmed H. Zubydan

Analysis of Statically Indeterminate
Structures – Clasical Methods
Prof. Dr. Ahmed H. Zubydan
Faculty of Engineering
Port Said University
Egypt

Contents
Chapter 1 Method of Consistent Deformations
Chapter 2 Three Moment Equation
2.1 Derivation of Three Moment Equation
2.2 Application of Three Moment Equation
Chapter 3 Slope Deflection Equation
3.1 Slope-Deflection Equation
Chapter 4 Moment Distribution Method
4.1 Definitions and Terminology
4.2 Basic Concept of the moment-Distribution method
4.3. Analysis of frames with Side-sway
1
31
31
35
53
53
77
77
83
84

1
Method of Consistent Deformations
This method, known also as a force method, can be applied to any intermediate
structure under the effects of applied loads, support settlement, temperature changes,
or any other effect. In applying the method to analyze an indeterminate structure, as
shown in Fig. 1.1a, or the number of redundant more than the minimum necessary
for static equilibrium of the structure. The outline of the method is as follows:
1. Reduce the structure to a condition of determinacy and stability by removing the
redundant. The resulting structure is called the primary structure (Fig. 1.1b, where
R1 is removed).
2. Determine the resulting errors in geometry incurred by subjecting the primary
structure to loads on the original indeterminate structure (Fig. 1.1b; the
displacement Δ10 is the error in geometry).
(a) Statically indeterminate structure (b) Statically determinate primary
structure
(c) Unit value of R1
Fig. 1.1 Determination of a redundant reaction by the method of consistent
deformation.
3. The primary structure must be altered to meet the boundary conditions. This
alteration is accomplished by introducing a unit value of the redundant reaction
(R11 = 1) on the primary structure, as is shown in Fig 1.1c. The displacement
corresponding to the released redundant in this case is identified as Δ11, which
expresses the deflection at the point and in the direction of R1 that is caused by
a unit value of R1.
4. Since Δ11 is the displacement that results from a unit value of R1, then the
displacement for the value of R1 is Δ11 R1. So, the desired solution of Fig. 1.1a,
the solutions of Figs. 1.1b and 1.1c must be superimposed. Specifically, the
11
R31
R41
R21
a
b
R11 =1
R1
1=0
R3
R4
R2
a b
R30
R40
R20
a b
R10 =0
10

2 Analysis of Statically Indeterminate Structures, by Prof. Dr. Ahmed Zubydan
displacements Δ10 and Δ11 R1 combined to give the final displacement, Δ11 R1.
Thus, we have
10 11 1 1R + = …………………………………….(1.1)
Solving for R1, we have
( )1 10
1
11
R
 − 
=

…………………………...………….(1.2)
In this case, all displacements are considered to be positive when upward. Thus, Δ10
is actually negative, as shown in Fig. 1.1b. Also, Δ1 is zero, as shown in Fig. 1.1a,
because of the constraint associated with the redundant reaction.
5. Once R1 has been determined, statics can be applied to determine any reaction
force or any internal force component, then
0 1 1S S S R= + ………..…………………..……….(1.3)
where S0 is the value of S on the primary structure when the actual loading of the
given structure is applied, and S1 is the value of S on the primary structure when a
unit load value of R1 is applied.
Consider next the continuous beam structure of Fig 1.2a which is twice statically
indeterminate. One way to reduce the given structure to a statically determinate
primary structure is to remove the two interior reactions, as shown in Fig. 1.2b.
These redundant reaction components are identified as R1 and R2.
The primary structure can now be analyzed by the method of statics and then
displacements Δ10 and Δ20, on the lines of action of the redundant reactions, can be
evaluated. The required solution is accomplished by introducing unit values of the
redundant reactions on the primary structure and determining the effects that these
individual loading cases have on the displacements where compatibility is to be
restored. These unit load cases are shown in Figs. 1.2c and 1.2d. The displacements
Δ1 and Δ2 on the line of actions of the redundant reaction for the original structure
are determined as follows:
10 11 1 12 2 1R R + + = ……………………………..….(1.4a)
and,
20 21 1 22 2 2R R + + = ……………………………….(1.4b)
Equations (1.4a) and (1.4b) can be solved for R1 and R2. For instance, if S is taken
as such a response quantity, then
0 1 1 2 2S S S R S R= + +
…………………………………….…….(1.5)
where S0 is the value of S on the primary structure when the actual loading of the
given structure is applied, and Si is the value of S on the primary structure when a
unit load value of Ri is applied.

3Method of consistent deformations
(a) Statically indeterminate beam
(b) Statically determinate primary structure
(c) Unit value of R1
(d) Unit value of R2
Fig. 1.2 Statically indeterminate continuous beam.
EXAMPLE 1.1
Construct the moment diagram for the
structure shown. EI is the same for each
member.
Solution
Primary structure and loading Bending moment of primary structure
(M0)
R2
R5
R4
R3
R1
b
a
c
d
2 =01 =0
R20 = 0
R50
R40
R30
R10 = 0
10 20
R21 = 1
R51
R41
R31
R11 = 1
11 21
R22 = 1R52
R42
R32
R12 = 0
2212
a b c
8 m 5 m
120 kN
20 kN/m
3 m
a b c
120 kN
20 kN/m
600
160

Unit value of X1 (M1)
The required displacements are
calculated based on the virtual work method
0 1
10
L
M M
dx
EI
 = 
10
1 2 1
160 8 4 600 8 5.33
3 2EI
 
 =    −    
 
( )
1 5
2 600 8 600 3
6EI
 
+ −    +  
 
18887
EI
−
=
1 1
11
L
M M
dx
EI
 = 
1 1 341.3
8 8 5.33 2
2EI EI
 
=     = 
 
Then,
10 11 1 0x +  =
or,
118887 341.3 0x− + = → 1 55.33 .x kN m=
The final moments are calculated as follows:
0 1 1M M M x= + 
or,
0 0 0
600 8 55.33 157.3 .
0 0 0
a
b
c
M
M kN m
M
       
       
= − +  = −       
       
       
Final bending moment
EXAMPLE 1.2
Construct the moment diagram for the structure
shown. EI is the same for each member.
Solution
8
a b c
X11= kN
157.3
166
a b
c
5 m 5 m
15 kN/m

(M0)
The required displacements
are calculated using the
virtual work method as follows:
0 1
10
L
M M
dx
EI
 = 
( )10
1 2 5
46.875 5 0.75 2 156.3 0.5 156.3 1
2 3 6EI
 
 = −    −    +  
 
3
1 15 5 8 1
0.5 156.3 5 0.33
24 15 2EI
   
+ −   −     
  
=
340
EI
−
( )11
1 5
2 1 1 2 0.5 0.5 1 0.5 0.5 1
2 6EI
 
 =    +   +  +  
 
 
1
0.5 0.5 5 0.33
EI
+   
1.875
EI
=
Then, 10 11 1 0x +  =
or,
1
340 1.875
0x
EI EI
− +  = → 3.181x1 =
0 1 1M M M x= + 
or,
0 1 181.3
181.3 .
156.3 0.5 65.6
a
b
M
kN m
M
− −       
= +  =       
−      
a
b
c
15 kN/m
156.346.875
1
0.5
b
a
X11=1 kN.m
c
181.3
65.6

EXAMPLE 1.3
Solution
(M0)
The required displacements are
calculated using the virtual work method
10
1 2 1
33.75 3 4.5 705 3 6
2 3 2EI
 
 = −    −    
 
1 1
720 2 4
2EI
 
+ −    
 
1 2 1
180 6 3 720 6 4
3 2EI
 
+    −    
 
=
12684
EI
−
11
1 1
9 3 6
2 2EI
 
 =    
 
1 1
6 2 4
2EI
 
+    
 
1 1
6 6 4
2EI
 
+    
 
136.5
EI
=
Then,
10 11 1 0x +  =
or,
− + =1126844 136.5 0x → =1 92.93x kN
The final moments are determined using the following relationship:
0 1 1M M M x= + 
or,
705 9 131.3
92.93 .
720 6 162.4
a
c
M
kN m
M
− −       
= +  =       
− −      
a b
c d
3 m 2 m 6 m
80 kN
30 kN/m 40 kN/m
a b
c
d
80 kN
30 kN/m 40 kN/m
705
720
180
33.75
a
c
9
6
b d
X11 =1.0 kN

EXAMPLE 1.4
member.
Solution
Since rotation at point c is equal zero, then the structure can be simplified as follows:
Simplified structure Primary structure and loading
Bending moment of primary structure (M0) Unit value of X1 (M1)
Using the virtual work method, the following displacements are determined:
10
1 1 2
250 8 0.667 160 8 0.5
2 3EI
 
 =    −    
 
1 1
250 5 1
2EI
 
+    
 
=
865
EI
 11
1 1 1 7.67
1 8 0.667 1 5 1
2EI EI EI
 
 =    +   = 
 
Then,
10 11 1 0x +  =
or,
+ =1865 7.67 0x or, =1 112.8 .x kN m
To obtain the final moment, the following relation is deduced:
0 1 1M M M x= + 
131.3
162.4
a b
c
d e
8 m 5 m 5 m 8 m
100 kN
20 kN/m 20 kN/m
a b
c
50 kN
20 kN/m
8 m 5 m
a b c
50 kN
20 kN/m
250
160
a
b
1.0 1.0
X11 = 1 kN.m

or,
( )
250 1 137.2
112.8 .
0 1 112.8
b
c
M
kN m
M
− − −       
= +  − =       
−      
EXAMPLE 1.5
Solution
(M0)
Unit value of X1
(M1)
Using the virtual work method, the following displacements are calculated:
137.2
112.8
137.2
a b
c
d
e
4 m 2 m 6 m
5 m
4 m
15 kN
10 kN
10 kN/m
a b
c
de
15 kN
10 kN
10 kN/m
30.7
50.7
20
54.08
a b
X11=10 kN
9.0
5.0
50
9.0

10
1 1
50.7 10.817 6
2EI
 
 = −    
 
( )
1 2 7.21
54.08 7.21 2 30.7 9 30.7 5
3 6EI
 
+ −   −    +  
 
=
4313.4
EI
−
( )11
1 1 7.21
9 10.817 6 2 9 9 2 5 5 9 5 5 9
2 6EI
 
 =    +    +   +  +  
 
1 1
5 5 3.33
2EI
 
+    
 
696.6
EI
=
Then,
10 11 1 0x +  =
or,
14313.4 696.6 0x− + = 1 6.192x kN=
The final bending moment is determined as follow:
0 1 1M M M x= + 
or,
30.7 9 25
50.7 9 6.192 5 .
0 5 30.9
da
dc
c
M
M kN m
M
− −       
       
= + −  = −       
       − −       
EXAMPLE 1.6
Solution
5
30.9
2520
a b
c
d
e
f
g
4 m 3 m 3 m 4 m
4 m
50 kN
50 kN
50 kN

Primary structure and loading Bending moment of primary structure(M0)
Unit value of X1
(M1)
The following displacements are determined using the virtual work method:
 10
1
75 4 2 2
EI
 = −     
1
375 6 4
2EI
+ −   =
5700
EI
−
11
1 1
4 4 2.67 2
2 2EI
 
 =     
 
 
1
4 6 4
2EI
+  
90.67
EI
=
Then,
10 11 1 0x +  =
or,
15700 90.67 0x− + = → 1 62.87x kN=
0 1 1M M M x= + 
or,
75 0 75
75 4 176.4
62.87 .
375 4 123.6
300 0 300
c
bc
be
ba
M
M
kN m
M
M
− −       
       −       
= +  =       
−       
             
a
b
c
d
e
f
g
h
50 kN
25 kN
50 kN
25 kN
300
75
75
375 375
75
75
300
a g
4
4
4
X11 =1.0 kN
300
176.4
123.6 123.6
75 75
176.4
300

EXAMPLE 1.7
member.
ks = 6800 kN/m, EI = 8105
kN.m2
Solution
Since the displacements and the rotation at b is equal zero, then this node can be
replaced to a fixed end. So, the beam can be simplified to the following system:
Simplified beam
he concepts of consistent deformation method are applied to the simplified system
as follows:
Primary structure and loading Bending moment of primary structure (M0)
Unit value of X1
(M1)
The following displacements are determined using the virtual work method:
1 2 2 1
10 2 3 3 2
1 20480
1280 8 8 320 8 8
EI EI
 =     −     =  
1 2
11 2 3
1 170.67
8 8 8
EI EI
 =     =  
Then,
10 11 1
1
0
s
x
k
 
 +  +  = 
 
or,
8 m 8 m
40 kN/m
EI EI
a
b
c
ks ks
40 kN/m
ba
x11=1.0 kN
8
b c
-
8 m
40 kN/m
EI
a b
ks
1280
320
b c
+
-

16 6
20480 170.67 1
0
200 10 0.004 200 10 0.004 6800
x
 
+ +  =     
→ 1 71x kN=−
0 1 1M M M x= + 
or,
1280 8 ( 71) 711.7 .bM kN m=− −  − =−
EXAMPLE 1.8
member.
Solution
711.7
a
b
c
6 m
2 m
2 m
80 kN
30 kN/m
a
b
c d
e
f
80 kN
a
c
b
d
e
f
30 kN/m
160
160
a
b
c d
e
f
+
-
-
e
-

2
10 3
1
135 6 1
EI
 = −     
1 2
2 3
1
160 2 1
EI
+ −      
1
2
1 1126.7
160 6 1
EI EI
−
+ −    =  
 11
2
1 6 1
EI
 =   1 2
2 3
4 14.7
1 2 1
EI EI
+     =  
Then,
10 11 1 0x +  =
or,
1
1126.7 14.7
0x
EI EI
− + = → 1 76.8 .x kN m=
0 1 1M M M x= + 
or,
160 1 82.3
0 1 76.8
76.8
0 1 76.8
0 1 76.8
a
c
d
f
M
M
M
M
  − −     
       − −       
= +  =       
− −       
             
a
b
c d
f
+
e
_
_ _X11 = 1 kN.m
83.2
76.8 76.8
76.8
76.8
a
b
c d
e
f

EXAMPLE 1.9
Solution
Unit load of X1 (M1) Unit value of X2 (M2)
The required displacements are calculated using the virtual work method as follows:
( )10
1 2 6
45 6 3 2 340 6 40 6
3 6EI
 
 =    −    +  
 
( )
1 4
2 340 6 2 660 10 340 10 660 6
6EI
 
+ −    +   +  +  
 
( )
1 4
2 660 10 2 1220 14 660 14 660 14 1220 10
6EI
 
+ −    +   +  +  +  
 
=
66073
EI
−
( )20
1 4
2 660 4 340 4
6EI
 
 = −    +  
 
( )
1 4
2 660 4 2 1220 8 660 8 1220 4
6EI
 
+ −    +   +  +  
 
27733
EI
−
=
11
1 1 914.7
14 14 9.33
2EI EI
 
 =    = 
 
22
1 1 170.7
8 8 5.33
2EI EI
 
 =    = 
 
a
b c d
2 m 6 m 4 m
20 kN 60 kN
10 kN/m
4 m
a b c
d
20 kN 60 kN
10 kN/m
40
340
660
1220
45
8
a b c
X22=1 kN
d
6
14
a b
X11 =1 kN
c d

( )12
1 8 362.7
2 14 8 6 8
6EI EI
 
 =    +  = 
 
21 12
362.7
EI
 =  =
Then,
10 11 1 12 2 0x x +  +  =
or,
1 266073 914.7 362.7 0x x− + + = …………………………………..……….(a)
and,
20 21 1 22 2 0x x +  +  =
or,
1 227733 362.7 170.7 0x x− + + = …………..………………………..……….(b)
Solving (a) and (b) for x1 and x2 we get
1 49.6x = and 2 57.1x =
The final moments are determined as follows:
0 1 1 2 2M M M x M x= +  + 
or,
40 0 0 40
340 6 49.6 0 57.1 42.5 .
1220 14 8 68.8
b
c
d
M
M kN m
M
− −         
         
= − +  +  = −         
         − −         
EXAMPLE 1.10
shown. EI/EA = 0.1 m2
Solution
40 42.5
64.4
68.8
a b
c
d
e
5 m 5 m
5 m
3 m
40 kN
20 kN/m20 kN/m
EI EI
EI EI
EA

Unit value of X1 (M1) Unit value of X2 (M2)
10
1 1
200 5 3.33
2EI
 
 = −    
 
( )
1 2 5.831
72.9 5.831 6.5 2 200 5 2 391.5 8 200 8 391.5 5
3 6EI
 
+ −    −    +   +  +  
 
=
24830
EI
−
( )20
1 2 5.831
72.9 5.831 1.5 2 391.5 3 200 3
3 6EI
 
 = −    −    +  
 
1 2 1 5999
72.9 5.831 1.5 391.5 5.831 2
3 2EI EI
 
+ −    −    = 
 
11
1 1
5 5 3.33 2
2EI
 
 =     
 
( )
1 5.831
2 5 5 2 8 8 8 5 5 8 2
6EI
 
+    +   +  +   
 
584.8
EI
=
a b
c
d
e
40 kN
20 kN/m 20 kN/m
40 kN
96.6 kN 136.6 kN
200
200
391.5
72.9 72.9
a b
X11 =1.0 kN
5
8
5
8
c d
e
1.0 kN
a b
c d
e
X22=1.0 kN X22=1.0 kN
3.0 3.0

22
1 1
3 5.831 2 2
2EI
 
 =     
 
35
EI
=
( )12
1 5.831
2 8 3 3 5 2
6EI
 
 =    +   
 
122.5
EI
=
21 12 = 122.5
EI
=
Then,
10 11 1 12 2 0x x +  +  =
or,
1 2
24830 584.8 122.5
0x x
EI EI EI
−
+ + = ……………………………..……….(a)
and,
20 21 1 22 2 0
cd
L
x x
EA
  
 +   +  +  =  
  
or,
1 2
5999 122.5 35 10
0x x
EI EI EI EA
−  
+ + + = 
 
……………………………….(b)
Solving (a) and (b) for x1 and x2, we obtain
1 26.3x kN= and 2 77.2x kN=
The final moments are calculated using the following relationship:
0 1 1 2 2M M M x M x= +  + 
or,
200 5 0 68.6
391.57 8 26.3 3 77.2 50.5 .
0 5 0 131.4
c
e
d
M
M kN m
M
−         
         
= + −  + −  = −         
         − −         
68.6
68.6
50.5
131.4
50.5

EXAMPLE 1.12
Construct the moment diagram for shown
continuous beam due the following support
settlement: Δa = -13 mm; Δb = -29 mm; Δc
= -50 mm; Δd = -36 mm. EI = 5105
kN.m2
for all members
Solution
Primary structure M0 = 0 Deflected shape of the structure
Unit load of X1 (M1) Unit value of X2 (M2)
The following displacements are calculated using virtual work method:
11
1 1
5.54 8 3.69
2EI
 
 =    
 
1 1
5.54 18 3.69
2EI
 
+    
 
265.9
EI
=
22 11
265.9
EI
 =  =
12
1 1
5.54 8 1.64
2EI
 
 =    
 
( )
1 10
2 5.54 2.46 2 2.46 5.54 5.54 5.54 2.46 2.46
6EI
 
+    +   +  +  
 
1 1
2.46 3.69
2EI
 
+   
 
224.8
EI
=
a b c d
8 m 10 m 8 m
a b c d a
c
db
13 mm 20.1 mm
8.9 mm
21.1 mm
36 mm28.9 mm
5.538
2.462
X11 =1
a
b c
d
5.538
2.462
X22=1
a
b c
d

21 12 = 224.8
EI
=
Then,
10 11 1 12 2 1x x +  +  =
or,
1 2
265.9 224.8 8.9
0
1000
x x
EI EI
+ + = − …………………………………………... (a)
and,
20 21 1 22 2 2x x +  +  =
or,
1 2
224.8 265.9 21.1
0
1000
x x
EI EI
+ + = − …………………………………………... (b)
Solving (a) and (b) for x1 and x2, we get
1 58.65x kN= and 2 89.2x kN=−
0 1 1 2 2M M M x M x= +  + 
or,
( )
0 5.54 2.46 105.3
58.65 89.2 .
0 2.46 5.54 349.7
b
c
M
kN m
M
− − −         
= +  +  − =         
− −        
EXAMPLE 1.13
Calculate the normal force in all members of the truss
shown. EA is the same for each member.
Solution
105.3
349.7
a b
c d
a b
c
d
e
f
4 m 4 m
3 m
3 m
10 kN
40 kN

Normal forces of primary structure (F0) Normal forces due to value of X1 (F1)
Member L (m) F0 (kN) F1 (kN) F0 F1L F1 F1L
ac 5 12.5 -1.25 -78.125 7.8125
ad 6 -20 0.75 -90 3.375
bc 5 0 -1.25 0 7.8125
bf 6 -27.5 0.75 -123.75 3.375
cd 5 33.3 -1.25 -208.125 7.8125
ce 3 -40 0 0 0
cf 5 45.8 -1.25 -286.25 7.8125
de 4 -36.7 1.0 -146.8 4
ef 4 -36.7 1.0 -146.8 4
-1080 -29.3
From table, the required displacements are given as follows:
0 1
10
1080F F L
EA EA
−
 = =
1 1
11
29.3F F L
EA EA
−
 = =
Then, 10 11 1 0x +  =
or,
1
1080 29.3
0x
EA EA
−
− = → 1 23.475x kN=
The final member forces are calculated as follows:
0 1 1F F F x= +
Member F0 (kN) F1 (kN) F(kN)
ac 12.5 -1.25 -16.8
ad -20 0.75 -2.4
bc 0 -1.25 -29.3
bf -27.5 0.75 -9.9
a b
c
d e
f
10 kN
40 kN
-20
12.5 0
-27.5
33.3
-40
45.8
-36.7 -36.7
10 kN
12.5 kN 27.5 kN
a b
c
d
e
f
X11=1 kN
0.75
-1.25 -1.25
0.75
-1.25
0
-1.25
1 1
1 kN

cd 33.3 -1.25 4.0
ce -40 0 -4.0
cf 45.8 -1.25 16.5
de -36.7 1.0 -13.2
ef -36.7 1.0 -13.2
Final member forces
EXAMPLE 1.14
Calculate the normal force in all members of the truss
shown. EA is the same for each member.
Solution
Normal forces of primary structure (F0) Normal forces due to value of X1 (F1)
Member L (m) F0 (kN) F1 (kN) F0 F1L F1 F1L
ac 5 35.4 1.67 295.6 13.9
a b
c
d
e
f
10 kN
40 kN
-2.4 -16.8 -29.3
-9.9
4
-40
16.5
-13.2 -13.2
13.5 kN
12.5 kN
23.5 kN
27.5 kN
a b
c
d
e
f
4 m 4 m
3 m
3 m
3 m
10 kN
10 kN
10 kN
20 kN
a b
c
d e
f
10 kN 10 kN
10 kN
20 kN
-5
35.4 -2.1
-35
-2.1 35.4
-10.4
-22.9
28.3 kN
16.3 kN
1.7 kN
36.3 kN
a b
c
d e
f
X11 =1 kN
X11 =1 kN
-1
1.67 1.67
-1
0.83 0.83
-0.83 -0.83
1.33 kN 1.33 kN

ad 6 -5 -1.0 30 6
bc 5 -2.1 1.67 -17.5 13.9
be 6 -35 -1.0 210 6
cd 5 -2.1 0.83 -8.7 3.44
ce 5 35.4 0.83 146.9 3.44
df 5 -10.4 -0.83 43.16 3.44
ef 5 -22.9 -0.83 95 3.44
796 53.7
The required displacement are calculated based on the virtual work method as
follows:
0 1
10
796F F L
EA EA
 = =
1 1
11
53.7F F L
EA EA
 = =
Then, 10 11 1 0
cf
L
x
EA
  
 +  + =     
or,
1
796 53.7 6
0x
EA EA EA
 
+ + = 
 
→ 1 13.33x kN=−
The final forces are calculated using the following relationship:
0 1 1F F F x= +
Member F0 (kN) F1 (kN) F(kN)
ac 35.4 1.67 13.2
ad -5 -1.0 8.3
bc -2.1 1.67 -24.3
be -35 -1.0 -21.7
cd -2.1 0.83 -13.2
ce 35.4 0.83 24.3
cf 0 1.0 -13.33
df -10.4 -0.83 0.7
ef -22.9 -0.83 -11.8

Final member forces
EXAMPLE 1.16
Calculate the normal force in all members of
the truss shown. EA is the same for each
member.
Solution
Normal forces of primary structure (F0)
Normal forces due to value of X1 (F1) Normal forces due to value of X2 (F2)
Mem. L (m) A F0 (kN) F1 (kN) F2 (kN) F0 F1L F0 F2L F1 F1L F2 F2L F1 F2L
ac 5 2A -300 3.75 0 -2812.5 0 35.2 0 0
bc 6.4 2A 256.1 -1.6 0 -1311.3 0 8.2 0 0
bd 5 A 100 -2.5 0 -1250 0 31.25 0 0
cd 4 A 0 0 -0.62 0 0 0 1.54 0
ce 5 A 0 1.25 -0.78 0 0 7.8 3.04 -4.88
cf 6.4 A -128.1 1.6 1 -1311.75 -819.8 16.4 6.4 10.24
df 5 A 100 -2.5 -0.78 -1250 -390 31.25 3.04 9.75
ef 4 A 80 0 -0.62 0 -198.4 0 1.54 0
eg 5 2A 0 1.25 0 0 0 3.91 0 0
fg 6.4 2A 0 -1.6 0 0 0 8.2 0 0
a b
c
d e
f
10 kN 10 kN
10 kN
20 kN
8.3
13.2 -24.3
-21.7
-13.2
24.3
-13.3
.7
-11.8
10.6 kN
16.3 kN
19.4 kN
36.3 kN
a
b
c
d
e
f
g
5 m 5 m 5 m
4 m
80 kN 80 kN
2A
2A
A
A
A
A
A
A
A
2A
2A
-128.1
a
b
c
d
e
f
g
80 kN 80 kN
-300
256.1
100
0
0
100
80
0
0
300 kN
300 kN
160 kN
a
b
c
d
e
f
g
X11 =1kN
3.75
-1.6
-2.5
0
1.25
1.6
-2.5
0
1.25
-1.6
3.75 kN
3.75 kN
1 kN
a
b
c
d
e
f
g
X
22 =1
kN
0
0
0
-0.62
-0.78
1.0
-0.78
-0.62
0
0

-7938 1408 142.2 15.6 15.1
The required displacements are given as follows:
0 1
10
7938F F L
EA EA
−
 = = and 0 2
20
1408F F L
EA EA
 = =
1 1
11
142.2F F L
EA EA
 = = and 2 2
22
15.6F F L
EA EA
 = =
1 2
12
15.1F F L
EA EA
 = = and 21 12
15.1
EA
 =  =
Then, 10 11 1 12 2 0x x + + =
or,
1 2
7938 142.2 15.1
0x x
EA EA EA
−
+ + = …………………………...…………….. (a)
and, 20 21 1 22 2 0
de
L
x x
EA
  
 +  +  + =  
  
or,
1 2
1408 15.1 15.6 6.4
0x x
EA EA EA EA
−  
+ + + = 
 
…………………………...…….. (b)
Solving (a) and (b) for x1 and x2, we get
1 52.9x kN= and 2 27.7x kN=
The final member forces are calculated as follows:
0 1 1 2 2F F F x F x= + +
Member F0 (kN) F1 (kN) F2 (kN) F (kN)
ac -300 3.75 0 -101.7
bc 256.1 -1.6 0 171.5
bd 100 -2.5 0 -32.2
cd 0 0 -0.62 -17.3
ce 0 1.25 -0.78 44.5
cf -128.1 1.6 1 -128.1
df 100 -2.5 -0.78 -53.8
ef 80 0 -0.62 62.7
eg 0 1.25 0 66.1
fg 0 -1.6 0 -84.6

Final member forces
EXAMPLE 1.17
Calculate the normal force in all members of the
truss shown due to
a) Lake of fit of 3 cm in member cd
b) Temperature change of +120 C0
in members
bd and de. α = 1210-6
per C0
, EA = 104
for each
member.
Solution
Primary structure (N0 = 0) Normal forces due to value of X1 (N1)
a) Lake of fit of 3 cm in member cd
Member Δ 0
(m)
F1
(kN)
Δ 0 F1 F1 F1L
ab 0 0 0 0
ac 0 2.67 28.44
bc 0 -1.67 0 13.89
bd 0 -1.33 0 7.11
cd -0.03 1.0 -0.03 3
ce 0 1.33 0 7.11
de 0 -1.67 0 13.89
-0.03 73.44
The required displacement are calculated as follows:
27.7171.5
a
b
c
d
e
f
g
80 kN 80 kN
-101.7
-32.2
-17.3
44.5
-15.7
-53.8
62.7
66.1
-84.6
101.7 kN
101.7 kN
107.1 kN
52.9 kN
a
b
c
d
e
4 m 4 m
3 m
a
b
c
d
e
a
b
c
d
e
X11 =1 kN
0
2.67
-1.67
-1.33
1
1.33
-1.67
2.67 kN
2.67 kN
1 kN

10 10
0.03F =  = −
1 1
11
73.44F F L
EA EA
 = =
Then, 10 11 1 0x + =
or,
1
73.44
0.03 0x
EA
− + = → 1 4.085x kN=
The final forces are calculated as follows:
0 1 1F F F x= +
Member F0 (kN) F1 (kN) F (kN)
ab 0 0 0
ac 0 2.67 10.9
bc 0 -1.67 -6.8
bd 0 -1.33 -5.4
cd 0 1.0 4.1
ce 0 1.33 5.4
de 0 -1.67 -6.8
b) Temperature change of +120 C0 in members bd and de.
0 tL =
Member L
(m)
Δ 0
(m)
F1 (kN) Δ 0 F1 F1 F1L
ab 3 0 0 0 0
ac 4 0 2.67 28.44
bc 5 0 -1.67 0 13.89
bd 4 0.00576 -1.33 -0.00768 7.11
cd 3 1.0 0 3
ce 4 0 1.33 0 7.11
de 5 0.0072 -1.67 -0.012 13.89
-0.0197 73.44
The required displacements are calculated as follows:
10 0 1 0.03F =  = −
1 1
11
73.44F F L
EA EA
 = =

Then, 10 11 1 0x + =
or,
1
73.44
0.0197 0x
EA
− + = → 1 2.68x kN=
The final forces are calculated as follows:
0 1 1F F F x= +
Member F0 (kN) F1 (kN) F (kN)
ab 0 0 0
ac 0 2.67 7.1
bc 0 -1.67 -4.5
bd 0 -1.33 -3.6
cd 0 1.0 2.7
ce 0 1.33 3.6
de 0 -1.67 -4.5

Problems
(a) Draw the bending moment diagram for the shown structures due to applied loads:
(1) (2)
(3) (4)
(5) (6)
(7) (8)
4 m 3 m
40 kN
20 kN/m
a b
c
2 m 2 m 4 m 2 m
100 kN
a b
c
d e
EI2EI 2EI
2 m 4 m 4 m 2 m 6 m
40 kN 80 kN
12 kN/m
a b
c
d e
f2EI 2EI EI
2 m 4 m
5 m
50 kN 150 kN
a
b
c
d
2EI
EIEI
3 m
8 m
5 m
60 kN
50 kN/m
25kN/m
2EI
EI EI
a
b
d
c
8 m 1.5
4 m
1.5 m
40 kN
20 kN/m
a
b
c
d
1.5
60 kN
4 m 4 m
4 m
40 kN
150 kN
EI
2EI 2EI
EI
a
b
c
d
e
4 m 4 m 4 m 4 m
3 m
100 kN 100 kN
EA EA
EA
EA EA
EI

(9) (10)
(11) (12)
(b) Draw the bending moment diagram for the beam in (3) due to support settlements
as follows:
Δb = 13 mm and Δd = 23 mm.
(c) Calculate the member forces for the shown trusses due to the applied loads:
(13) (14)
6 m 3 m 3 m
4 m
100 kN
a
b
c
d
EI EI
EA EA
2 m 6 m 2 m
4 m
20 kN
15 kN
30 kN
a
b
c d e
f g
a
b c
d
e
f
2 m 6 m 2 m
4 m
20 kN/m
a b
c d
e
f g
h
i
3 m 3 m
3 m
2 m
2 m
20 kN
15 kN
40 kN
15 kN/m
a b
c
d e f
2.5 m 2.5 m 2.5 m 2.5 m
2.5 m
2.5 m
40 kN
100 kN
c
4 m 4 m 3 m
4 m
2 m
80 kN
20 kN 40 kN
f
g
a
b
d
e

(15) (16)
(17)
(d) For the truss shown in (15), calculate the member forces due to support
settlement of 12 mm at g.
(e) For the truss shown in (14), calculate the member forces due to 50 Co
increase in
temperature of members ac, cd, df and fg.
b d f h j
c g
5 at 4 m = 20 m
3 m
a
e i
k
l
40 kN 80 kN
40 kN
a
b c d
e
f g h
4 m 4 m 4 m 4 m
3 m
20 kN 60 kN 20 kN
2A2A
2A
A
A
A
2A
A
2A
A
A
A
2AA
A
2 m 2 m
1.5 m
1.5 m
1.5 m
10 kN
10 kN
5 kN
f
g
h
i
a
d
c
e
j
k

2
Three-MomentEquation
The three-moment equation represents, in a general form, the compatibility
condition that the slope of the elastic curve be continuous at an interior support of
the continuous beam. Since the equation involves three moments, the bending
moments at the support under consideration and at the two adjacent supports, it
commonly is referred to as the three-moment equation. When using this method, the
bending moments at the interior (and any fixed) supports of the continuous beam are
treated as the redundant. The three-moment equation is then applied at the location
of each redundant to obtain a set of compatibility equations which can be solved for
the unknown redundant moments.
2.1. Derivation of Three-Moment Equation
Consider an arbitrary continuous beam subjected to external loads and support
settlements as shown in Fig. 2.1(a). This beam can be analyzed by the method of
consistent deformations by treating the bending moments at the interior supports
to be the redundant. From Fig. 2.1(a), we can see that the slope of the elastic curve
of the indeterminate beam is continuous at the interior supports. When the restraints
corresponding to the redundant bending moments are removed by inserting internal
hinges at the interior support points, the primary structure thus obtained consists of
a series of simply supported beams. As shown in Figs. 2.1(b) and (c), respectively,
when this primary structure is subjected to the known external loading and support
settlements, discontinuities develop in the slope of the elastic curve at the locations
of the interior supports. Since the redundant bending moments provide continuity of
the slope of the elastic curve, these unknown moments are applied as loads on the
primary structure as shown in Fig. 2.1(d), and their magnitudes are determined by
solving the compatibility equations based on the condition that, at each interior
support of the primary structure, the slope of the elastic curve, due to the combined
effect of the external loading, support settlements, and unknown redundant, must be
continuous.
The three-moment equation uses the foregoing compatibility condition of slope
continuity at an interior support to provide a general relationship between the
unknown bending moments at the support where compatibility is being considered

and at the adjacent supports to the left and to the right, in terms of the loads on the
intermediate spans and any settlements of the three supports.
To derive the three-moment equation, we focus our attention on the compatibility
equation at an interior support c of the continuous beam, with prismatic spans and a
constant modulus of elasticity, shown in Fig. 2.1(a). As indicated in this figure, the
adjacent supports to the left and to the right of c are identified as l and r, respectively;
the subscripts l and r are used to refer to the loads and properties of the left span, lc,
and the right span, cr, respectively; and the settlements of supports l; c, and r are
denoted by Δl; Δc, and Δr, respectively. The support settlements are considered
positive when in the downward direction, as shown in the figure.
(a) Deformed shape
(b) Primary beam subject to external loading
(c) Primary beam subject to support settlement
(d) Primary beam subject to redundant bending moments
Fig. 2.1 Continuous beam.
Ll Deformed shape
EIl
Lr
EIr
c
l
r
c
rl
l1
r1
1c
l r
l2
r2
c
l
r2
l3
r3
3
Mc
Ml Mr

Three-Moment Equation 33
From Fig. 2.1(a), we can see that the slope of the elastic curve of the indeterminate
beam is continuous at c. In other words, there is no change of slope of the tangents
to the elastic curve at just to the left of c and just to the right of c; that is, the angle
between the tangents is zero. However, when the primary structure, obtained by
inserting internal hinges at the interior support points, is subjected to external loads,
as shown in Fig. 2.1(b), a discontinuity develops in the slope of the elastic curve at
c, in the sense that the tangent to the elastic curve at just to the left of c rotates relative
to the tangent at just to the right of c.
The change of slope (or the angle) between the two tangents due to external loads is
denoted by ϴ1 and can be expressed as (see Fig. 2.1(b))
𝜃1 = 𝜃 𝑟1 + 𝜃𝑙1 (2.1)
in which ϴl1 and ϴr1 denote, respectively, the slopes at the ends c of the
spans to the left and to the right of the support c, due to external loads.
Similarly, the slope discontinuity at c in the primary structure, due to support
settlements (Fig. 2.1(c)), can be written as
𝜃2 = 𝜃𝑟2 + 𝜃𝑙2 (2.2)
in which ϴl2 and ϴr2 represent, respectively, the slopes of the spans to the left and
to the right of c, due to support settlements. Finally, when the primary structure is
loaded with the redundant support bending moments, as shown in Fig. 2.1(d), the
slope discontinuity at c can be expressed as
𝜃3 = 𝜃𝑟3 + 𝜃𝑙3 (2.3)
in which ϴl3 and ϴr3 denote, respectively, the slopes at end c of the spans to the left
and to the right of the support c, due to unknown redundant moments.
The compatibility equation is based on the requirement that the slope of the elastic
curve of the actual indeterminate beam is continuous at c; that is, there is no change
of slope from just to the left of c to just to the right of c. Therefore, the algebraic
sum of the angles between the tangents at just to the left and at just to the right of c
due to the external loading, support settlements and the redundant bending moments
must be zero. Thus,
𝜃1 + 𝜃2 + 𝜃3 = 0 (2.4)
By substituting Eqs. (2.1) through (2.3) into Eq. (2.4), we obtain

(𝜃 𝑟1 + θ𝑙1) + (𝜃 𝑟2 + 𝜃𝑙2) + (𝜃 𝑟3 + 𝜃𝑙3) = 0 (2.5)
Since each span of the primary structure can be treated as a simply supported beam,
the slopes at the ends c of the left and the right spans, due to the external loads (Fig.
2.1(b)), which can be determined by any conventional method, is equal to the elastic
reactions at c, so,
𝜃𝑙1 = (
𝑟
𝐸𝐼
)
𝑙
(2.6. b)
𝜃𝑟1 = (
𝑟
𝐸𝐼
)
𝑟
(2.6. b)
The slopes ϴl2 and ϴr2, of the left and the right spans, respectively, due to support
settlements, can be obtained directly from the deformed positions of the spans
depicted in Fig. 2.1(c). Since the settlements are assumed to be small, the slopes can
be expressed as
𝜃𝑙2 =
∆𝑙 − ∆ 𝑐
𝐿𝑙
(7. a)
𝜃𝑟2 =
∆ 𝑟 − ∆ 𝑐
𝐿 𝑟
(7. b)
The slopes at ends c of the left and the right spans, due to redundant support bending
moments, (Fig. 2.1(d)), can be determined conveniently by using the beam-
deflection formulas. Thus,
𝜃𝑙3 =
𝑀𝑙
6
(
𝐿
𝐸𝐼
)
𝑙
+
𝑀𝑐
3
(
𝐿
𝐸𝐼
)
𝑙
(8. a)
𝜃𝑟3 =
𝑀𝑐
3
(
𝐿
𝐸𝐼
)
𝑟
+
𝑀𝑟
6
(
𝐿
𝐸𝐼
)
𝑟
(8. b)
in which Ml, Mc and Mr denote the bending moments at supports l, c and r,
respectively. As shown in Fig. 2.1(d), these redundant bending moments are
considered to be positive in accordance with the beam convention that is, when
causing compression in the upper fibers and tension in the lower fibers of the beam.
By substituting Eqs. (2.6) through (2.8) into Eq. (2.5), we write the compatibility
equation as

(
𝑟
𝐸𝐼
)
𝑙
+ (
𝑟
𝐸𝐼
)
𝑟
+
∆𝑙 − ∆ 𝑐
𝐿𝑙
+
∆ 𝑟 − ∆ 𝑐
𝐿 𝑟
𝑀𝑙
6
+ (
𝐿
𝐸𝐼
)
𝑙
+
𝑀𝑐
3
(
𝐿
𝐸𝐼
)
𝑙
+
𝑀𝑐
3
(
𝐿
𝐸𝐼
)
𝑟
+
𝑀𝑟
6
(
𝐿
𝐸𝐼
)
𝑟
= 0
By simplifying the foregoing equation and rearranging it to separate the terms
containing redundant moments from those involving loads and support settlements,
we obtain the general form of the three-moment equation:
𝑀𝑙 × (
𝐿
𝐸𝐼
)
𝑙
+ 2𝑀𝑐 [(
𝐿
𝐸𝐼
)
𝑙
+ (
𝐿
𝐸𝐼
)
𝑟
] + 𝑀𝑟 × (
𝐿
𝐸𝐼
)
𝑟
= −6 [(
𝑟
𝐸𝐼
)
𝑙
+ (
𝑟
𝐸𝐼
)
𝑟
] − 6 (
∆𝑙 − ∆ 𝑐
𝐿𝑙
+
∆ 𝑟 − ∆ 𝑐
𝐿 𝑟
) (2.9)
in which Mc is the bending moment at support c where the compatibility is being
considered; Ml, Mr are the bending moments at the adjacent supports to the left and
to the right of c, respectively; EI is the modulus of elasticity; Ll; Lr are the lengths of
the spans to the left and to the right of c, respectively; Il; Ir are the moments of inertia
of the spans to the left and to the right of c, respectively; rl; rr are the elastic reactions
of the applied loads.
If the moments of inertia of two adjacent spans of a continuous beam are equal (i.e.,
Il = Ir = I), then the three-moment equation simplifies to
𝑀𝑙 × 𝐿𝑙 + 2𝑀𝑐(𝐿𝑙 + 𝐿 𝑟) + 𝑀𝑟 × 𝐿 𝑟
= −6(𝑟𝑙 + 𝑟𝑟) − 6𝐸𝐼 (
∆𝑙 − ∆ 𝑐
𝐿𝑙
+
∆ 𝑟 − ∆ 𝑐
𝐿 𝑟
) (2.10)
The foregoing three-moment equations are applicable to any three consecutive
supports, l, c and r, of a continuous beam, provided that there are no discontinuities,
such as internal hinges, in the beam between the left support l and the right support
r.
2.2. Application of Three-Moment Equation
The following step-by-step procedure can be used for analyzing continuous beams
by the three-moment equation.
1. Select the unknown bending moments at all interior supports of the beam as
the redundant.

2. By treating each interior support successively as the intermediate support c,
write a three-moment equation. When writing these equations, it should be
realized that bending moments at the simple end supports are known. For
such a support with a cantilever overhang, the bending moment equals that
due to the external loads acting on the cantilever portion about the end
support. The total number of three-moment equations thus obtained must be
equal to the number of redundant support bending moments, which must be
the only unknowns in these equations.
3. Solve the system of three-moment equations for the unknown support
bending moments.
4. Compute the span end shears. For each span of the beam, (a) draw a free-
body diagram showing the external loads and end moments and (b) apply the
equations of equilibrium to calculate the shear forces at the ends of the span.
5. Determine support reactions by considering the equilibrium of the support
joints of the beam.
6. If so desired, draw shear and bending
Example 2.1
Calculate and draw the bending moment
diagram of the shown beam due to the
applied loads using the three moment
equation method.
Solution
• Calculation of elastic reaction
Considering the bending moment diagram for the simply supported beams shown,
the elastic reactions are calculated as follows:
Bending moment diagram for simply supported beams
𝑟𝑏𝑎 = 100 × 1 +
1
2
× 100 × 2 = 200
50kN 50kN 15kNm
2m 2m 2m 8m
a b
c
EI 2EI
50kN 50kN 15kNm
a bb c
120100
a b c
rba
rbc
6m 8m

𝑟𝑏𝑐 =
2
3
× 120 × 4 = 320
Applying the three moment equation at b leads to
0 + 2𝑀 𝑏 (
6
1
+
8
2
) + 0 = −6 (
200
1
+
320
2
)
Solving the equation for Mb,
𝑀 𝑏 = −108 𝑘𝑁. 𝑚
Bending moment diagram
Example 2.2
Calculate and draw the bending moment diagram of the shown beam due to the
applied loads using the three moment equation method.
Solution
The fixed support is replaced by an imaginary interior roller support
with an adjoining end span of zero length simply supported at its outer
end, as shown in figure. The reaction moment at the actual fixed
support is now treated as the redundant bending moment at the
imaginary interior support, and the three-moment equation when
applied to this imaginary support satisfies the compatibility condition
of zero slope of the elastic curve at the actual fixed support.
Calculation of elastic reaction
120100 108
100kN 20kNm
a
b c
4m 4m 10m
d
2m
40kN
0m
d
d
d’

Bending moment diagram for
simply supported beams
𝑟𝑏𝑎 =
1
2
× 200 × 4 = 400
𝑟𝑏𝑐 =
2
3
× 250 × 5 = 833.3
𝑟𝑐𝑏 =
2
3
× 250 × 5 = 833.3
Applying the three-moment equation at c putting Mb = -80 kN.m
−80 × 8 + 2𝑀𝑐(8 + 10) + 𝑀 𝑑 × 10 = −6(400 + 833.3) (a)
𝑀𝑐 × 10 + 2𝑀 𝑑(10 + 0) + 0 = −6(833.3 + 0) (b)
By solving Eqs. (a) and (b), Mc and Md are given as
𝑀𝑐 = −137.4 𝑘𝑁. 𝑚 , 𝑀 𝑑 = −181.3 𝑘𝑁. 𝑚
Example 2.3
Calculate and draw the bending moment diagram of the shown symmetric beam
due to the applied loads using the three moment equation method.
Solution
200 250rba rbc
rcb
db c d’
0m8m 10m
0
80
137.4
181.3
200
a b c d
20kNm20kNm 100 kN
6m 3m 3m 6m
a
b c
d

Bending moment diagram for
simply supported beams
𝑟𝑎𝑏 = 𝑟𝑏𝑎 =
2
3
× 90 × 3 = 180
𝑟𝑏𝑐 =
1
2
× 150 × 3 = 225
Applying the three-moment equation at a
0 + 2𝑀 𝑎(0 + 6) + 𝑀 𝑏 × 6 = −6(0 + 180) (a)
Applying the three-moment equation at b
𝑀 𝑎 × 6 + 2𝑀 𝑏(6 + 6) + 𝑀𝑐 × 6 = −6(180 + 225) (b)
For the shown symmetric beam, 𝑀 𝑏 = 𝑀𝑐, then Eq. (b) can be written as
𝑀 𝑎 × 6 + 2𝑀 𝑏(6 + 6) + 𝑀 𝑏 × 6 = −6(180 + 225) (c)
By solving Eqs. (a) and (c), Ma and Mb are given as
𝑀 𝑎 = −55 𝑘𝑁. 𝑚 , 𝑀 𝑏 = −70 𝑘𝑁. 𝑚
Example 2.4
diagram of the shown symmetric frame due to
the applied loads using the three moment
equation method.
Solution
90 150a’ a b
c0
rab
rba
rbc
6m 6m
55 70 5570
90 150 90
a
b c
d
ef
50kN 50kN
4m
2m6m2m
50 50

For the shown frame, 𝑀𝑓 = 𝑀 𝑎 = 𝑀𝑒 = 𝑀 𝑑
The given frame can also be treated as a sum of simply supported beams, and then
the three-moment equation is applied at the nodes at which the unknown moments
are required
𝑟𝑓 𝑒 =
1
2
× 100 × 2 + 100 × 3 = 400
𝑟𝑓 𝑎 = 0
Applying the three-moment equation at f
𝑀 𝑎 × 4 + 2𝑀𝑓(4 + 10) + 𝑀𝑒 × 10 = −6(0 + 400)
Putting 𝑀𝑓 = 𝑀 𝑎 = 𝑀𝑒 = 𝑀 𝑑, leads to
𝑀𝑓 × 4 + 2𝑀𝑓(4 + 10) + 𝑀𝑓 × 10 = −6(0 + 400)
By solving the equation, 𝑀𝑓 = 𝑀 𝑎 = 𝑀𝑒 = 𝑀 𝑑 = −57.1 kN. m
a ef
100 100
rfa
rfe
4m 10m
57.1
57.1 57.1
57.1
a
b c
d
ef
100 100
100 100

Example 2.5
diagram of the shown symmetric frame due to
the applied loads using the three moment
equation method.
Solution
For the shown frame, member fe can be extracted and solved individually as
follows:
The frame is now solved without member fe
knowing that 𝑀 𝑎 = 𝑀 𝑑
Considering the bending moment diagram for the
members individually
𝑟𝑎𝑑 =
1
2
× 100 × 2 + 100 × 3 = 400
0 + 2𝑀 𝑎(4 + 10) + 𝑀 𝑑 × 10 = −6(−400 + 0)
Putting 𝑀 𝑎 = 𝑀 𝑑, leads to
a b c
d
ef
50kN 50kN
4m
2m6m2m
50 50
a b c
ef
50kN 50kN
50 50
50
5050
50
f
d
a b c
100 100
50 50
a
a
a d
f
f
rad
raf
10m
4m
d

0 + 2𝑀 𝑎(4 + 10) + 𝑀 𝑎 × 10 = −6(−400 + 0)
The equation is now solved for 𝑀 𝑎, so
𝑀 𝑎 = 𝑀 𝑑 = 63.1 kN.m
Example 2.6
diagram of the shown frame due to the applied
loads using the three moment equation method.
Solution
For the shown frame, members cd and de are statically determinate, so, they are
solved individually as follows:
The frame abcf can now be solved using the three-moment equation
a
b c
d
ef
63.1 63.1
100 100
100
40kN/m60kN 40kN
a b c d
e
f
3m 3m 6m 2m
3m
3m
40kNn60kN
a b c
e
f
40
26.7
26.7
40kN
c d
40
26.7
26.7

The elastic load for member bc which is subjected to
triangle load can be calculated as shown the figure
Considering the bending moment diagram for the members individually, the elastic
reactions are calculated as follows:
𝑟𝑏𝑎 =
1
2
× 90 × 3 = 135
𝑟𝑏𝑐 =
𝑤𝐿3
45
=
40 × 63
45
= 192
𝑟𝑐𝑏 =
7𝑤𝐿3
360
=
7 × 40 × 63
360
= 168
𝑟𝑐𝑓 =
1
2
× 40 × 3 = 60
Applying the three-moment equation at b,
0 + 2𝑀 𝑏(6 + 6) + 𝑀𝑐 × 6 = −6(135 + 192) (a)
And applying the three-moment equation at c
𝑀 𝑏 × 6 + 2𝑀𝑐(6 + 6) + 0 = −6(168 + 60) (b)
Solving Eqs. (a) and (b) leads to
𝑀𝑐 = −39 𝑘𝑁. 𝑚, 𝑀 𝑏 = −72 𝑘𝑁. 𝑚
b c
L
b c
wL3/45
w
wL3/24
7wL3/360
8L/15 7L15
40
90
6m 6m 6m
a
b c
f
rba rbc rcb
rcf

Example 2.7
Calculate and draw the bending moment diagram of
the shown symmetric frame due to the applied loads
using the three moment equation method.
Solution
The shown frame is symmetric in which 𝑀𝑐 = 𝑀 𝑑 and 𝑀𝑒 = 𝑀𝑓
For the given frame, member cd is solved
individually as shown in figure
72
39
40
90
80 kN/m
50kN
2m 2m
3m
3m
a b
c
d
e f
40kN 40
e’ f’
1m 1m
80 kN/m
50kN
a b
c
d
e f
c
d
40kN 40kN
e’ f’

The elastic load caused by cantilevers ee’ and ff’ is applied to member ef as shown
in figure
Considering the bending moment diagram for the
members individually, the elastic reaction are calculated as follows:
𝑟𝑒𝑓 = 133.33 and 𝑟𝑒 𝑐 = 𝑟𝑐 𝑒 = 𝑟𝑐 𝑎 = 0
Applying the three-moment equation at c,
0 + 2𝑀𝑐(3 + 3) + 𝑀𝑒 × 3 = 0 (a)
And applying the three-moment equation at e,
𝑀𝑐 × 3 + 2𝑀𝑒(3 + 4) + 𝑀𝑓 × 3 = −6(0 + 133.3) (b)
Putting Mf = Me lead to
𝑀𝑐 × 3 + 2𝑀𝑒(3 + 4) + 𝑀𝑒 × 3 = −6(0 + 133.3) (c)
Solving Eqs. (a) and (c), then
𝑀𝑐 = 11.6 𝑘𝑁. 𝑚 , 𝑀𝑒 = −46.4 𝑘𝑁. 𝑚
The bending moment diagram is obtained by adding the simply supported bending
moments to the obtained values as follows:
50kN
e f
40kN 40
e’ f’
a
c e f
160
rca rec
rce ref
40 403m 3m
4m

Example 2.8
Calculate and draw the bending moment, shear
force and normal force diagrams of the shown
frame due to the applied loads in addition to a
downward settlement of 10 mm at support a
using the three moment equation method. Use EI
= 48,000 kN.m2
Solution
The elastic load caused by cantilevers ec is
applied to member cd as shown in figure
Considering the bending moment diagram for the members individually, the elastic
reactions are calculated as follows:
86.4 86.4
4040
11.6 11.6
50
160
46.4 46.4
20 kN 80 kN
40 kN
2m 1m 2m 4m
2m
2m
a
b
c d
e
f
20 kN
c
d
e
40
40
137.14
20
20
274.3205.7
140
20
20
0
a c
d
rcdrca
rac
a’

𝑟𝑎 𝑐 =
1
4
[20 × (2 +
2
3
) − 20 ×
2
3
× 2] = 6.67
𝑟𝑐 𝑎 = 6.67
𝑟𝑐𝑑 =
1
7
[20.5.7 × (4 + 1) + 274.3 ×
2
3
× 4 − 140 ×
2
3
× 7] = 158.1
0 + 2𝑀 𝑎(0 + 4) + 𝑀𝑐 × 4 = −6(0 − 6.67) (a)
The settlement at support a causes a sway in
member cd as shown in figure
Remember the three-moment equation
considering the member sway which can be
applied at c
𝑀 𝑎 × (𝐿)𝑙 + 2𝑀𝑐[(𝐿)𝑙 + (𝐿) 𝑟] + 𝑀 𝑑 × (𝐿) 𝑟
= −6[(𝑟)𝑙 + (𝑟) 𝑟] − 6𝐸𝐼 (
∆ 𝑎 − ∆ 𝑐
𝐿𝑙
+
∆ 𝑑 − ∆ 𝑐
𝐿 𝑟
)
Then,
𝑀 𝑎 × 4 + 2𝑀𝑐[4 + 7] + 0
= −6[6.67 + 158.1] − 6 × 48,000 (0 +
0 − 0.01
7
) (b)
Solving Eqs. (a) and (b) leads to
𝑀 𝑎 = 19.9 𝑘𝑁. 𝑚 , 𝑀𝑐 = −29.8 𝑘𝑁. 𝑚
The bending moment diagram is obtained by adding the simply supported bending
moments to the obtained values as follows:
10 mm
10 mm
a
b
c de
f
40
69.8
137.14
19.9
25
15
29.8
40

To draw the shearing force and the normal force diagrams, the end forces at each
member free body diagram should be evaluated due to the applied loads and due to
the end bending moments
Shear end forces due to applied loads Shear end forces due to end moment
Total shear end forces.
Considering the equilibrium of each member and at the frame joints, the normal
and the shearing force diagrams can be evaluated as follows:
80 kNc d
45.7 34.3
40 kN
a
b
c
f
10
10
c d
69.8
69.8/7=9.97
a
b
c
f
(19.9+29.8)/4=12.43
19.9
29.8
80 kNc d
55.7 24.3
40 kN
a
b
c
f
22.4
22.4

Normal force diagram Shear force diagram
Example 2.9
diagram for the shown frame due to the applied
loads.
Solution
The frame should be divided into two
statically indeterminate structures by
disconnect then at intermediate hinge at
node c as shown in figure. Each part is
individually solved using the three-
moment equation.
Column ce can be easily solved, so the final results for this part is presented. On
the other hand, the other part of the frame includes three unknowns. These
unknowns are the bending moments about joint b. The solution of this part is
illustrated as follows:
-20
55.7
-24.3
-22.4
40
-22.4
-115.7
-75.7
30kN 30kN
14kN/m
40kN 60kN
a
b
c
d e
2m 2m 6m
3m
3m
2m
60kN
c
e
30kN 30kN
14kN/m
40kN
a
b
d

The equilibrium of bending moments at
joint b for the shown frame should be
evaluated as follows
𝑀 𝑏𝑎 − 𝑀 𝑏𝑐 − 𝑀 𝑏𝑑 = 0 (a)
Considering the bending moment
diagrams for members ab and bc as
simply supported beams as shown in
figure and applying the three-moment
equation at b, the following equation is
obtained
0 + 2(𝑀 𝑏𝑎 × 6 + 𝑀 𝑏𝑐 × 6) + 0 = −6(120 + 126) (b)
Now, considering the bending moment
diagrams for members ab and bd as simply
supported beams as shown in figure and
applying the three-moment equation at b,
the following equation is obtained
0 + 2(𝑀 𝑏𝑎 × 6 + 𝑀 𝑏𝑑 × 6) + 0 = −6(120 + 90) (c)
Solving equations (a), (b) and (c) leads to
𝑀 𝑏𝑎 = −76 kN. m, 𝑀 𝑏𝑐 = −47 𝑘𝑁. 𝑚 , 𝑀 𝑏𝑑 = −29 𝑘𝑁. 𝑚
30kN 30kN 40kN
a b
d
60 60
6m
120 120
6m
60
90 90
30kN 30kN
14kN/m
a b
c
60 60
63
6m 6m
120 120 126 126
a b
c
d
e
Mbd
Mbc
Mba
0 0
0
29
76 47
67.5

Problems
(1) Calculate and draw the bending moment diagram for the following structure
due to the applied loads using the 3-moment equation method.
(a) (b)
(c) (d)
(e) (f)
120kN
a
b c
4m 4m 10m
d
2m
50kN 40kNn
2EI EI
a
b
cd
50kN 50kN
80kN 4m
2m 2m 2m 2m
3m 3m 2 m 2m 3m 3m
6m
a
b
c d e
f
g
2EI
EI
2EI 2EI
EI
100kN 100kN40kN
a
b c
d
80kN
4m
4m 4m
e
f
20kN/m
60kN20kN
40kN
40kNa
b c d
e f
3m 3m 6m 2m2m
2m 3m
3m
20kN/m
60kN20kN
40kN
40kNa
b c d
e f
3m 3m 6m 2m2m
2m 3m
3m

(g) (h)
(i)
(2) Calculate and draw the shearing force diagrams for problems (a), (b), (c), (d)
and (e) due to the applied loads.
(3) Recalculate and draw the bending moment diagram for problem (a) due to
support settlement 10mm at c and 4 mm. Use EI = 60,000 kN.m2
.
(4) Recalculate and draw the bending moment diagram for problem (e) due to
support settlement 12mm at e. Use EI = 50,000 kN.m2
.
20kN/m
60kN
a
b
c d
e f
3m3m 6m2m
3m
3m
80 kN/m
50kN
2m 2m
3m
3m
a b
c d
e f
40kN 40
e’ f’
1m 1m
120kN
2m 2m3m 3m
6m
a b c d
e
2EI 2EI2EI
EI EI

3
Slope-Deflection Method
The slope-deflection method for the analysis of indeterminate beams and frames is
considered a classical formulation of the displacement method. The method takes
into account only the bending deformations of structures. Although the slope-
deflection method is itself considered to be a useful tool for analyzing indeterminate
beams and frames, an understanding the fundamentals of this method provides a
valuable introduction to the matrix stiffness method, which forms the basis of most
computer software currently used for structural analysis. We first derive the
fundamental relationships necessary for the application of the slope-deflection
method and then develop the basic concept of the slope-deflection method. We
consider the application of the method to the analysis of continuous beams and
present the analysis of the frames in which joint translations are prevented. Finally,
we consider the analysis of frames with joint translations.
3.1 Slope-Deflection Equations
When a continuous beam or a frame is subjected to external loads, internal moments
generally develop at the ends of its individual members. The slope-deflection
equations relate the moments at the ends of a member to the rotations and
displacements of its ends and the external loads applied to the member.
To derive the slope-deflection equations, let us focus our attention on an arbitrary
member AB of the continuous beam shown in Fig. 3.1(a). When the beam is
subjected to external loads and support settlements, member AB deforms, as shown
in the figure, and internal moments are induced at its ends. The free-body diagram
and the elastic curve for member AB are shown using an exaggerated scale in Fig.
3.1(b). As indicated in this figure, double-subscript notation is used for member end
moments, with the first subscript identifying the member end at which the moment
acts and the second subscript indicating the other end of the member. Thus, MAB
denotes the moment at end A of member AB, whereas MBA represents the moment
at end B of member AB. Also, as shown in Fig. 3.1(b), θA and θB denote,
respectively, the rotations of ends A and B of the member with respect to the
undeformed (horizontal) position of the member; Δ denotes the relative translation
between the two ends of the member in the direction perpendicular to the
undeformed axis of the member; and the angle c denotes the rotation of the member’s
chord (i.e., the straight line connecting the deformed positions of the member ends)

due to the relative translation D. Since the deformations are assumed to be small, the
chord rotation can be expressed as
𝜓 =
Δ
𝐿
(3.1)
• The sign convention used in this chapter is as follows:
The member end moments, end rotations, and chord rotation are positive when
counterclockwise.
(a) Loaded continuous beam and the deformed shape
(b) Deformed shape of member AB
(c) Bending moment diagram
L
Deformed shape
B
A
MAB
MBA
ML
B

BA
A


B
AB
B’
A
Tangent at B
Elastic curve
Tangent at A
L
EI
MAB
MBA
BA

Slope deflection 55
Tangential deviations due to MAB Tangential deviations due to MBA
(d) Tangential deviations due to external loads (e) Fixed-end moments
Fig. 3.1
Note that all the moments and rotations are shown in the positive sense in Fig. 3.1(b).
The slope-deflection equations can be derived by relating the member end moments
to the end rotations and chord rotation by applying the second moment-area theorem.
From Fig. 3.1(b), we can see that
𝜃𝐴 =
Δ 𝐵𝐴 + Δ
𝐿
(3.2a)
𝜃 𝐵 =
Δ 𝐴𝐵 + Δ
𝐿
(3.2b)
By substituting Δ 𝐿⁄ = 𝜓 into the preceding equations, we write
𝜃𝐴 − 𝜓 =
Δ 𝐵𝐴
𝐿
(3.3a)
𝜃 𝐵 − 𝜓 =
Δ 𝐴𝐵
𝐿
(3.3b)
in which, as shown in Fig. 3.1(b), ΔBA is the tangential deviation of end B from the
tangent to the elastic curve at end A and ΔAB is the tangential deviation of end A
from the tangent to the elastic curve at end B. According to the second moment-area
theorem, the expressions for the tangential deviations ΔBA and ΔAB can be obtained
by summing the moments about the ends B and A, respectively, of the area under
the M/EI diagram between the two ends. The bending moment diagram for the
member is constructed in parts by applying MAB; MBA, and the external loading
separately on the member with simply supported ends. The three simple-beam
MAB
BA
MBA
BA
B
A
FEMAB FEMBA
B
   = A B
= =
A
𝑀𝐴𝐵 𝐿2
6𝐸𝐼
𝑀𝐴𝐵 𝐿2
3𝐸𝐼
𝑔 𝐵
𝐸𝐼
𝑔 𝐴
𝐸𝐼
𝑀 𝐵𝐴 𝐿2
6𝐸𝐼
𝑀 𝐵𝐴 𝐿2
3𝐸𝐼

bending moment diagrams thus obtained are shown in Fig. 3.1(c). Assuming that the
member is prismatic, that is, EI is constant along the length of the member, we sum
the moments of the area under the M/EI diagram about the ends B and A,
respectively, to determine the tangential deviations:
Δ 𝐵𝐴 =
1
𝐸𝐼
[(
𝑀𝐴𝐵 𝐿
2
) (
2𝐿
3
) − (
𝑀 𝐵𝐴 𝐿
2
) (
𝐿
3
) − 𝑔 𝐵]
or
Δ 𝐵𝐴 =
𝑀𝐴𝐵 𝐿2
3𝐸𝐼
−
𝑀 𝐵𝐴 𝐿2
6𝐸𝐼
−
𝑔 𝐵
𝐸𝐼
(3.4a)
and
Δ 𝐴𝐵 =
1
𝐸𝐼
[− (
𝑀𝐴𝐵 𝐿
2
) (
𝐿
3
) + (
𝑀 𝐵𝐴 𝐿
2
) (
2𝐿
3
) − 𝑔 𝐴]
or
Δ 𝐴𝐵 = −
𝑀𝐴𝐵 𝐿2
6𝐸𝐼
+
𝑀 𝐵𝐴 𝐿2
3𝐸𝐼
+
𝑔 𝐴
𝐸𝐼
(3.4b)
in which gB and gA are the moments about the ends B and A, respectively, of the
area under the simple-beam bending moment diagram due to external loading (ML
diagram in Fig. 3.1(c)). The three terms in Eqs. (3.4a) and (3.4b) represent the
tangential deviations due to MAB; MBA, and the external loading, acting separately
on the member (Fig. 3.1(d)), with a negative term indicating that the corresponding
tangential deviation is in the direction opposite to that shown on the elastic curve of
the member in Fig. 3.1(b). By substituting the expressions for ΔBA and ΔAB (Eqs.
(3.4)) into Eq. (3.3), we write
θ 𝐴 − 𝜓 =
𝑀𝐴𝐵 𝐿
3𝐸𝐼
−
𝑀 𝐵𝐴 𝐿
6𝐸𝐼
−
𝑔 𝐵
𝐸𝐼 𝐿
(3.5a)
θ 𝐵 − 𝜓 = −
𝑀𝐴𝐵 𝐿
6𝐸𝐼
+
𝑀 𝐵𝐴 𝐿
3𝐸𝐼
+
𝑔 𝐴
𝐸𝐼 𝐿
(3.5b)
To express the member end moments in terms of the end rotations, the chord
rotation, and the external loading, we solve Eqs. (3.5a) and (3.5b) simultaneously
for MAB and MBA. Rewriting Eq. (3.5a) as
𝑀𝐴𝐵 𝐿
3𝐸𝐼
=
2𝑀𝐴𝐵 𝐿
3𝐸𝐼
−
2𝑔 𝐵
𝐸𝐼 𝐿
− 2(θ 𝐴 − 𝜓)
By substituting this equation into Eq. (5b) and solving the resulting equation for
MAB, we obtain
𝑀𝐴𝐵 =
2𝐸𝐼
𝐿
(2θ 𝐴 + θ 𝐵 − 3𝜓) +
2
𝐿2
(2𝑔 𝐵 − 𝑔 𝐴) (3.6a)
and by substituting Eq. (3.6a) into either Eq. (3.5a) or Eq. (3.5b), we obtain the
expression for MBA:

Slope deflection 57
𝑀 𝐵𝐴 =
2𝐸𝐼
𝐿
(θ 𝐴 + 2θ 𝐵 − 3𝜓) +
2
𝐿2
(𝑔 𝐵 − 2𝑔 𝐴) (3.6b)
As Eqs. (3.6) indicate, the moments that develop at the ends of a member depend on
the rotations and translations of the member’s ends as well as on the external loading
applied between the ends. Now, suppose that the member under consideration,
instead of being a part of a larger structure, was an isolated beam with both its ends
completely fixed against rotations and translations, as shown in Fig. 3.1(e). The
moments that would develop at the ends of such a fixed beam are referred to as
fixed-end moments, and their expressions can be obtained from Eqs. (3.6) by setting
θA = θB = ψ = 0; that is,
𝐹𝐸𝑀𝐴𝐵 =
2
𝐿2
(2𝑔 𝐵 − 𝑔 𝐴) (3.7a)
𝐹𝐸𝑀 𝐵𝐴 =
2
𝐿2
(𝑔 𝐵 − 2𝑔 𝐴) (3.7b)
Equations (3.8), which express the moments at the ends of a member in terms of its
end rotations and translations for a specified external loading, are called the slope-
deflection equations. These equations are valid only for prismatic members
composed of linearly elastic material and subjected to small deformations. Also,
although the equations take into account the bending deformations of members, the
deformations due to axial forces and shears are neglected. From Eqs. (3.8), we
observe that the two slope-deflection equations have the same form and that either
one of the equations can be obtained from the other simply by switching the
subscripts A and B. Thus it is usually convenient to express these equations by the
following single slope-deflection equation:
𝑀𝐴𝐵 =
𝐸𝐼
𝐿
(4𝜃𝐴 + 2𝜃 𝐵) −
6𝐸𝐼
𝐿2
∆ + 𝐹𝐸𝑀𝐴𝐵 (3.8a)
𝑀 𝐵𝐴 =
𝐸𝐼
𝐿
(4𝜃 𝐵 + 2𝜃𝐴) −
6𝐸𝐼
𝐿2
∆ + 𝐹𝐸𝑀 𝐵𝐴 (3.8b)
Members with One End Hinged
The slope-deflection equations derived previously (Eqs. (3.8)) are based on the
condition that the member is rigidly connected to joints at both ends, so that the
member end rotations θA and θB are equal to the rotations of the adjacent joints.
When one of the member’s ends is connected to the adjacent joint by a hinged
connection, the moment at the hinged end must be zero. The slope-deflection
equations can be easily modified to reflect this condition. With reference to Fig.
3.1(b), if the end B of member AB is hinged, then the moment at B must be zero. By
substituting MBA = 0 into Eqs. (3.8), we write
𝑀𝐴𝐵 =
𝐸𝐼
𝐿
(4𝜃𝐴 + 2𝜃 𝐵) −
6𝐸𝐼
𝐿2
∆ + 𝐹𝐸𝑀𝐴𝐵 (3.9a)

𝑀 𝐵𝐴 = 0 =
𝐸𝐼
𝐿
(4𝜃 𝐵 + 2𝜃𝐴) −
6𝐸𝐼
𝐿2
∆ + 𝐹𝐸𝑀 𝐵𝐴 (3.9b)
Solving Eq. (3.9b) for θB, we obtain
𝜃 𝐵 = −
𝜃𝐴
2
+
3∆
2𝐿 𝐵𝐴
−
𝐿
4𝐸𝐼
𝐹𝐸𝑀 𝐵𝐴 (3.10)
To eliminate θB from the slope-deflection equations, we substitute Eq. (3.10) into
Eq. (3.9a), thus obtaining the modified slope-deflection equations for member AB
with a hinge at end B:
𝑀𝐴𝐵 =
𝐸𝐼
𝐿
(3𝜃𝐴) −
3𝐸𝐼
𝐿2
∆ + (𝐹𝐸𝑀𝐴𝐵 −
𝐹𝐸𝑀 𝐵𝐴
2
) (3.11a)
𝑀 𝐵𝐴 = 0 (3.11b)
Similarly, it can be shown that for a member AB with a hinge at end A, the modified
slope-deflection equations for member AB with a hinge at end A:
𝑀 𝐵𝐴 =
𝐸𝐼
𝐿
(3𝜃 𝐵) −
3𝐸𝐼
𝐿2
∆ + (𝐹𝐸𝑀 𝐵𝐴 −
𝐹𝐸𝑀𝐴𝐵
2
) (3.12a)
𝑀𝐴𝐵 = 0 (3.12b)
Example 3.1
diagram for the shown loaded beam using
the slope deflection method. Consider EI is
the same for all members
Solution
For simplicity, the stiffness ratios are used instead of the real values. For this
reason, the obtained values of rotations are not real. The beams is solved in the
following steps:
• Stiffness ratio (EI/L)
ab : bc
𝐸𝐼
4
:
𝐸𝐼
6
1.5 : 1
• Fixed-end moments
40kN/m 120kNa
b c
4m 3m 3m

Slope deflection 59
• End moments
Using the member stiffness ratios instead the actual values, the end moments are
evaluated as follows:
𝑀 𝑎𝑏 = 1.5(4𝜃 𝑎 + 2𝜃 𝑏) + 53.3
= 3𝜃 𝑏 + 53.3 ………………………………… (a)
𝑀 𝑏𝑎 = 1.5(4𝜃 𝑏 + 2𝜃 𝑎) − 53.3
= 6𝜃 𝑏 − 53.3 ………………………………… (b)
𝑀 𝑏𝑐 = 1.0(4𝜃 𝑏 + 2𝜃𝑐) + 90
= 4𝜃 𝑏 + 90 ………………………………… (c)
𝑀𝑐𝑏 = 1.0(4𝜃𝑐 + 2𝜃 𝑏) − 90
= 2𝜃 𝑏 − 90 ………………………………… (d)
• Equilibrium conditions
The valid equilibrium conditions is the summation of bending moments about joint
b is equal zero as follows
∑ 𝑀 𝑏 = 0; 𝑀 𝑏𝑎 + 𝑀 𝑏𝑐 = 0
Substituting from Eqs (b) and (c)
6𝜃 𝑏 − 53.3 + 4𝜃 𝑏 + 90 = 0
Then,
𝜃 𝑏 = −3.66
Since only a stiffness ratio are used, the obtained rotation is not the actual.
• Final end moments
𝑀 𝑎𝑏 = 3𝜃 𝑏 + 53.3 = 3(−3.66) + 53.3 = 42.3
𝑀 𝑏𝑎 = 6𝜃 𝑏 − 53.3 = 6(−3.66) − 53.3 = −75.3
𝑀 𝑏𝑐 = 4𝜃 𝑏 + 90 = 4(−3.66) + 90 = 75.3
𝑀𝑐𝑏 = 2𝜃 𝑏 − 90 = 2(−3.66) − 90 = −97.3
+ve -ve
40kN/m
a
b +ve -ve
120kN
b c

The final end bending moment
values are applied to the beam as
shown in figure
The final bending moment diagram is plotted at the beam as follows
Example 3.2
Calculate and draw the bending
moment diagram for the shown loaded
beam using the slope deflection
method.
Solution
42.3
75.3 97.3
42.3 -75.3 75.3 -97.3
a
b
c
42.3 75.3 75.3 97.3
40kN/m 120kNa
b
b
c
63.756.388.371.8
71.8 56.3
-63.7
-88.3
a
b c
30 kN 100 kN 20 kN/m
a
b c d
1.5 4m 4m 10m

Slope deflection 61
bc : cd
𝐸𝐼
8
:
𝐸𝐼
10
1.25 : 1
Fixed-end moment of member cd
The fixed –end moment of member bc is calculated due to mid-span loading and
the cantilever load
Fixed-end moment of member bc
due to entire span load
Fixed-end moment at c for member bc
due to cantilever load noting that the
moment at c is equal to the half value of
the moment at the other end b
The total fixed-end moment at c can then be evaluated as follows
𝐹𝐸𝑀𝑐𝑏 = −100 − 100/2 + 22.5
= −127.5
• End moments
+ve -ve
20 kN/m
c d
45/2
= 22.5
4530 kN
a
b
c
100 kN
b
c
100 kN30 kN
b
c
-ve
30(1.5)
=45
202.5

The member bc is treated as a hinged-end member
𝑀𝑐𝑏 = 1.25(3𝜃𝑐) − 127.5
= 3.75𝜃𝑐 − 127.5 ………………………… (a)
𝑀𝑐𝑑 = 1.0(4𝜃𝑐 + 2𝜃 𝑑) + 166.67
= 4𝜃𝑐 + 166.67 ………………………… (b)
𝑀 𝑑𝑐 = 1.0(4𝜃 𝑑 + 2𝜃𝑐) − 166.67
= 2𝜃𝑐 − 166.67 ………………………… (c)
c is equal zero
∑ 𝑀𝑐 = 0; 𝑀𝑐𝑏 + 𝑀𝑐𝑑 = 0
Substituting from Eqs (a) and (b), then
3.75𝜃𝑐 − 127.5 + 4𝜃𝑐 + 166.67 = 0
The equation is solved to give 𝜃𝑐 = −5.054
Using the obtained rotation, the final-end moments are evaluated by substituting in
Eqs.(a) to (c) as follows:
𝑀𝑐𝑏 = 3.75𝜃𝑐 − 127.5 = 3.75(−5.054) − 202.5 = −146.5
𝑀𝑐𝑑 = 4𝜃𝑐 + 166.67 = 4(−5.054) + 166.67 = 146.5
𝑀 𝑑𝑐 = 2𝜃𝑐 − 166.67 = 2(−5.054) − 166.67 = −176.8
values are applied to the beam as
shown in figure
45 -146.5 146.5 -176.8
a
b c d
45
146.5 176.8

Slope deflection 63
Example 3.3
diagram for the shown loaded frame using the
slope deflection method.
Solution
The member ef is solved individually, so the
other part of the frame which will be solved
using the slope deflection method is shown
in figure
eb : bc
𝐸𝐼
3
:
𝐸𝐼
4
1.0 : 0.75
The fixed-end moment at b for member be is
given as shown in figure
• End moments
𝑀 𝑏𝑒 = 1.0(3𝜃 𝑏) + 22.5
= 3𝜃 𝑏 + 22.5 ………………………… (a)
𝑀 𝑏𝑐 = 0.75(4𝜃 𝑏 + 2𝜃𝑐)
60kN
40 kN40 kN
a b c d
e f
1m 2m 2m 1m
1.5m
1.5m
30 30
+ve
+
40 kN
b
e
30kN 30kN
60kN
e f
30kN 30kN
40 kN40 kN
a b c d
e f
30 30

From symmetry, 𝜃 𝑏 = −𝜃𝑐 , then
𝑀 𝑏𝑐 = 1.5𝜃 𝑏 ………………………… (b)
b is equal zero
∑ 𝑀 𝑏 = 0; 𝑀 𝑏𝑒 + 𝑀 𝑏𝑐 + 𝑀 𝑏𝑎 = 0
Substitution from Eqs (a) and (b), noting that the
cantilever moment Mba is given as shown in figure
3𝜃 𝑏 + 22.5 + 1.5𝜃 𝑏 + 30 = 0
The equation is solved to give 𝜃 𝑏 = −11.667
Substituting the rotation in Eqs (a) and (b) to gives
𝑀 𝑏𝑒 = 3𝜃 𝑏 + 22.5 = 3(−11.667) + 22.5 = −12.5
𝑀 𝑏𝑐 = 1.5𝜃 𝑏 = 1.5(−11.667) = −17.5
The final end bending moment values are
applied to the beam as shown in figure
-12.5
-17.530
a
b c
d
e f
30 3017.5
12.5 12.5
30(1)
=30
+ve
a
b
30

Slope deflection 65
Example 3.4
Calculate and draw the bending
moment diagram for the shown loaded
frame using the slope deflection
method.
Solution
bc : cd : ce : eg
2𝐼
6
:
𝐼
4
:
2𝐼
6
:
𝐼
4
1.0 : 0.75 : 1.0 : 0.75
due to the entire uniform loads
due to the entire cantilever load
The total fixed-end moment is then
calculated as shown in figure
Fixed-end moments for member ce
20 kN/m30 kN
60 kN 30 kN
20 kN
a
b c
d
e
f
g
h
2I 2I
I I
1.5 6m 3m 3m 1m
3m
1m
-ve
20 kN/m
30 kN
a
b
c
20 kN/m
a
b
c
+
30 kN
a
b
c
60 kN
c e -ve+ve

𝐹𝐸𝑀𝑐𝑑 = 11.25 + 3.75 2⁄ = 13.125 𝑘𝑁. 𝑚
𝐹𝐸𝑀𝑒𝑔 = −(−5.625) + 9.375/2 2⁄
= 10.31 𝑘𝑁. 𝑚
• End moments
Using the member stiffness ratios, the end moments are evaluated as follows:
𝑀𝑐𝑏 = 1.0(3𝜃𝑐) − 67.5
= 3𝜃𝑐 − 67.5 ………………………… (a)
𝑀𝑐𝑒 = 1.0(4𝜃𝑐 + 2𝜃𝑒) + 45
= 4𝜃𝑐 + 2𝜃𝑒 + 45 ………………………… (b)
𝑀𝑒𝑐 = 1.0(4𝜃𝑒 + 2𝜃𝑐) − 45
= 2𝜃𝑐 + 4𝜃𝑒 − 45 ………………………… (c)
𝑀𝑐𝑑 = 0.75(3𝜃𝑐) + 13.125
= 2.25𝜃𝑐 + 13.125 ………………………… (d)
𝑀𝑒𝑔 = 0.75(3𝜃𝑒) + 10.31
= 2.25𝜃𝑒 + 10.31 ………………………… (e)
The valid equilibrium conditions are considered as follows:
20 kN
c
d
f
g
30 kN.m
e

Slope deflection 67
∑ 𝑀𝑐 = 0; 𝑀𝑐𝑏 + 𝑀𝑐𝑒+𝑀𝑐𝑑 = 0
Substituting from Eqs (a), (b) and (d), then
3𝜃𝑐 − 67.5 + 4𝜃𝑐 + 2𝜃𝑒 + 45 + 2.25𝜃𝑐 + 13.125=0
Which may simplified as
9.25𝜃𝑐 + 2𝜃𝑒 = 9.375 ………………………… (i)
∑ 𝑀𝑒 = 0; 𝑀𝑒𝑐 + 𝑀𝑒𝑔 = 0
Substituting from Eqs (c) and (e), then
2𝜃𝑐 + 4𝜃𝑒 − 45 + 2.25𝜃𝑒 + 10.31 = 0
Or
2𝜃𝑐 + 6.25𝜃𝑒 = 34.69 ………………………… (ii)
Solving Eqs (i) and (ii) yields 𝜃𝑐 = −0.167 and 𝜃𝑒 = 5.58
Substituting into Eqs (a) to (e), the final end-moments are calculated as follows:
𝑀𝑐𝑏 = 3𝜃𝑐 − 67.5
= 3(−0.167) − 67.5 = −68
𝑀𝑐𝑒 = 4𝜃𝑐 + 2𝜃𝑒 + 45
= 4(−0.167) + 2(5.64) + 45 = 55.4
𝑀𝑒𝑐 = 2𝜃𝑐 + 4𝜃𝑒 − 45
= 2(−0.167) + 4(5.64) − 45 = −23
𝑀𝑐𝑑 = 2.25𝜃𝑐 + 13.125
= 2.25(−0.167) + 13.125 = 12.7
𝑀𝑒𝑔 = 2.25𝜃𝑒 + 10.31
= 2.25(5.64) + 10.31 = 23
The final bending moment diagram is plotted at the beam as follows:
45 -68 55.4
12.7
-23
23a
b c
d
e
f
g
h
45
68 55.4
12.7
23
23
5.28
24.7
30

Example 3.4
diagram for the shown loaded beam using
the slope deflection method.
Solution
The present example is almost as the previous example. The only change is the
replacement the support at b to roller instead of the hinge. This roller allows the
girder abce to move horizontally. This type of frames is call sway permitted frame.
This frame contains the same unknowns as the frame in the previous example in
addition to the sway displacement Δ. The stiffness ratios and the fixed-end
moments are the same as the previous example.
Assuming the that points a,
b, c and e moves Δ as shown
in figure, the sway moment
ratios are calculated as
shown in figure. Only
columns cd and eg are
subjected to sway
• Sway moment ratio
cd : eg
Δ : Δ
• End moments
Using the member stiffness ratios and the sway moment ratios, the end moments
are evaluated as follows:
𝑀𝑐𝑏 = 1.0(3𝜃𝑐) − 67.5
= 3𝜃𝑐 − 67.5 ………………………… (a)
𝑀𝑐𝑒 = 1.0(4𝜃𝑐 + 2𝜃𝑒) + 60
= 4𝜃𝑐 + 2𝜃𝑒 + 45 ………………………… (b)
𝑀𝑒𝑐 = 1.0(4𝜃𝑒 + 2𝜃𝑐) − 60
= 2𝜃𝑐 + 4𝜃𝑒 − 45 ………………………… (c)
𝑀𝑐𝑑 = 0.75(3𝜃𝑐) − ∆ + 13.125
= 2.25𝜃𝑐 − ∆ + 13.125 ………………………… (d)
𝑀𝑒𝑔 = 0.75(3𝜃𝑒) − ∆ + 10.31
20 kN/m
30 kN
60 kN 30 kN
20 kN
a
b c
d
e
f
g
h
1.5 6m 3m
3m 1m
3m
1m2I 2I
I I
-ve
-vea
b c
d
e
f
g
h
 

Slope deflection 69
= 2.25𝜃𝑒 − ∆ + 10.31 ………………………… (e)
The previous equilibrium conditions are applied as follows:
∑ 𝑀𝑐 = 0; 𝑀𝑐𝑏 + 𝑀𝑐𝑒+𝑀𝑐𝑑 = 0
Substituting from Eqs (a), (b) and (d), then
3𝜃𝑐 − 67.5 + 4𝜃𝑐 + 2𝜃𝑒 + 45 + 2.25𝜃𝑐 − ∆ + 13.125=0
or
9.25𝜃𝑐 + 2𝜃𝑒 − ∆= 9.375 …………………..…. (i)
and
∑ 𝑀𝑒 = 0; 𝑀𝑒𝑐 + 𝑀𝑒𝑔 = 0
Substituting from Eqs (c) and (e), then
2𝜃𝑐 + 4𝜃𝑒 − 45 + 2.25𝜃𝑒 − ∆ + 10.31 = 0
or
2𝜃𝑐 + 6.25𝜃𝑒 − ∆ = 34.69 …………………..…. (ii)
Only one equilibrium equation is required. This equation is formed by considering
summation the horizontal equilibrium of horizontal forces in the frame is equal zero
as follows:
∑ 𝐹𝑥 = 0;
or
𝑋 𝑑 + 𝑋 𝑔 + 20 = 0 ………. (f)
To find Xd and Xg as a function of the unknowns, each column is studied as
follows:
For column cd
∑ 𝑀𝑐 = 0; 𝑋 𝑑 = 𝑀𝑐𝑑 4⁄ − 5
20 kN/m
30 kN
60 kN 30 kN
20 kN
a
b c
d
e
f
g
h
2I 2I
I
I
Xd Xg
20 kN
c
d
Xd

For column eg
∑ 𝑀𝑒 = 0; 𝑋 𝑔 = 𝑀𝑒𝑔 4⁄ − 7.5
Substituting of Xd and Xg in Eq. (f) leads to
𝑀𝑐𝑑 4⁄ − 5 +𝑀𝑒𝑔 4⁄ − 7.5 + 20 = 0
or
𝑀𝑐𝑑 + 𝑀𝑒𝑔 = −30
Substituting from Eqs (d) and (e), then
2.25𝜃𝑐 − ∆ + 13.125 + 2.25𝜃𝑒 − ∆ + 10.31 = −30
or
2.25𝜃𝑐 + 2.25𝜃𝑒 − 2∆= −53.44 …………………..…. (iii)
Solving Eqs (i), (ii) and (iii) gives 𝜃𝑐 = 3.21 , 𝜃𝑒 = 11.43 and ∆= 43.2
Once again, the obtained deformation values are not the actual values since a
stiffness ratios and a sway moment ratio are used.
The final end moments are obtained by substituting the deformation values into
Eqs (a) to (e) as follows:
𝑀𝑐𝑏 = 3𝜃𝑐 − 67.5
= 3(3.21) − 67.5 = −57.8
𝑀𝑐𝑒 = 4𝜃𝑐 + 2𝜃𝑒 + 45
= 4(3.21) + 2(11.43) + 45 = 80.7
𝑀𝑒𝑐 = 2𝜃𝑐 + 4𝜃𝑒 − 45
= 2(3.21) + 4(11.43) − 45 = 7.14
𝑀𝑐𝑑 = 2.25𝜃𝑐 − ∆ + 13.125
= 2.25(3.21) − 43.2 + 13.125 = −22.9
𝑀𝑒𝑔 = 2.25𝜃𝑒 − 43.2 + 10.31
= 2.25(11.43) − 43.2 + 10.31 = −7.14
30 kN
e
f
g
h
Xg

Slope deflection 71
values are applied to the beam
as shown in figure
Example 3.4
Calculate and draw the bending moment diagram
for the shown loaded beam using the slope
deflection method.
Solution
ab : bc : cd
𝐸𝐼
6.71
:
𝐸𝐼
6
:
𝐸𝐼
6
0.894 1.0 : 1.0
There is no member loads, therefore no fixed end-moments are calculated
To calculate the sway moment ratios, the frame deformed shape should be evaluated.
Assuming that point c moves horizontally by Δ, the shown deformed shape of the
frame is considered
45 -57.8 80.7
-
22.9
7.14
-7.14a
b c
d
e
f
g
h
45
57.8 80.7
22.9
7.14
30
80 kN
a
b
c
d
3m 6m
6m

So, the shown sway moment ratios may be
considered
• End moments
Using the member stiffness ratios and the sway moment ratios, the end moments
are evaluated as follows:
𝑀 𝑏𝑎 = 0.894(3𝜃 𝑏) + 0.896∆
= 2.682𝜃 𝑏 + 0.896∆ …………………..…. (a)
𝑀 𝑏𝑐 = 1.0(4𝜃 𝑏 + 2𝜃𝑐) − ∆
= 4𝜃 𝑏 + 2𝜃𝑐 − ∆ …………………..…. (b)
𝑀𝑐𝑏 = 1.0(4𝜃𝑐 + 2𝜃 𝑏) − ∆
= 4𝜃𝑐 + 2𝜃 𝑏 − ∆ …………………..…. (c)
𝑀𝑐𝑑 = 1.0(4𝜃𝑐 + 2𝜃 𝑑) + ∆
= 4𝜃𝑐 + 0.1667∆ …………………..…. (d)
𝑀 𝑑𝑐 = 1.0(4𝜃 𝑑 + 2𝜃𝑐) + ∆
= 2𝜃𝑐 + ∆ …………………..…. (e)
Three equilibrium equations are required to solve the given frame as follows:
∑ 𝑀 𝑏 = 0; 𝑀 𝑏𝑎 + 𝑀 𝑏𝑐 = 0
Substituting from Eqs (a) and (b)
2.682𝜃 𝑏 + 0.896∆ + 4𝜃 𝑏 + 2𝜃𝑐 − ∆=0
or
6.682𝜃 𝑏 + 2𝜃𝑐 − 0.104∆= 0 …………………..…. (i)
and


a
b c
d
a
b c
d

 -ve
+ve
+ve

Slope deflection 73
∑ 𝑀𝑐 = 0; 𝑀𝑐𝑏 + 𝑀𝑐𝑑 = 0
Substituting from Eqs (c) and (d)
4𝜃𝑐 + 2𝜃 𝑏 − ∆ + 4𝜃𝑐 + 2∆= 0
or
2𝜃 𝑏 + 8𝜃𝑐 + ∆= 0 …………………..…. (ii)
And finally, the horizontal forces equilibrium is considered as follows
∑ 𝐹𝑥 = 0; 𝑋 𝑎 + 𝑋 𝑑 = 0 ……..…. (f)
To find Xa and Xd as a function of the unknowns, the free body diagram of the
frame members is studied as follows:
For member cd
∑ 𝑀 𝑏 = 0;
then,
𝑋 𝑎 = 𝑀 𝑏𝑎 6⁄ −𝑀 𝑏𝑐 12⁄ −𝑀𝑐𝑏 12⁄ − 40
For member ab
80 kN
a
b
c
dXa Xd
c
dXd
80 kN
80 kN
a
b
Xa
b c

∑ 𝑀𝑐 = 0;
then
𝑋 𝑑 = 𝑀𝑐𝑑 6⁄ + 𝑀 𝑑𝑐 6⁄
Substituting in Eq. (f), we get
𝑀 𝑏𝑎 6⁄ −𝑀 𝑏𝑐 12⁄ −𝑀𝑐𝑏 12⁄ − 40 + 𝑀𝑐𝑑 6⁄ + 𝑀 𝑑𝑐 6⁄ = 0
or
2𝑀 𝑏𝑎−𝑀 𝑏𝑐−𝑀𝑐𝑏 + 2𝑀𝑐𝑑 + 2𝑀 𝑑𝑐 = 480
Substituting from Eqs (a) to (e) we can find
2(2.682𝜃 𝑏 + 0.896∆ ) − (4𝜃 𝑏 + 2𝜃𝑐 − ∆ )
−(4𝜃𝑐 + 2𝜃 𝑏 − ∆ ) + 2(4𝜃𝑐 + 2∆ ) + 2(2𝜃𝑐 + 2∆ ) = 480
or
−0.636𝜃 𝑏 + 6𝜃𝑐 + 11.79∆= 480 …………………..…. (iii)
By solving Eqs (i), (ii) and (iii) we get
𝜃 𝑏 = 2.52, 𝜃𝑐 = −6.12, ∆= 43.97
Substituting into Eqs (a) to (e) we can find
𝑀 𝑏𝑎 = 2.682𝜃 𝑏 + 0.896∆= 2.682(2.52) + 0.896(43.97 ) = 46.1
𝑀 𝑏𝑐 = 4𝜃 𝑏 + 2𝜃𝑐 − ∆= 4(2.52) + 2(−6.12) − 43.97 = −46.1
𝑀𝑐𝑏 = 4𝜃𝑐 + 2𝜃 𝑏 − ∆= 4(−6.12) + 2(2.52) − 43.97 = −63.4
𝑀𝑐𝑑 = 4𝜃𝑐 + 2∆= 4(−6.12) + 2(43.97 ) = 63.4
𝑀 𝑑𝑐 = 2𝜃𝑐 + 2∆= 2(−6.12) + 2(43.97 ) = 75.7
46.1
-46.1 -63.4
75.7
a
b c
d
75.7
46.1
63.4

Slope deflection 75
Problems
(a) Calculate and draw the bending moment diagram for the following structure due
to the applied loads using the 3-moment equation method.
(1) (2)
(3) (4)
(5) (6)
(7) (8)
50kN/m
100kN
a
b c
6m 4m 4m
2EIEI
20 kN 120 kN 16 kN/m
a
b c d
1.5 3m 3m 8m
60kN
a b c d
e f
1m 2m 2m 1m
1.5m
1.5m
a
b c d
80kN 80kN
4m
3m
e
f g
h
3m3m
a b
c d
e
4m 4m
4m
4m
80 kN
10 kN/m
20 kN
80 kN 20 kN
30 kN
a
b c
d
e
f
g
h
2I 2I
I I
1.5 6m 3m 3m 1m
3m
1m
6m 3m 3m
2m
1.5m 1m
2I 2I
15kN/m
I I
30kN
40 kN
90 kN
15 kN
2m
a b
c
d
e f
g
60kN
a
b c d
4m 4m
6m

(9) (10)
(b) Calculate and draw the shearing force diagrams for problems (2), (4) and (6) due
to the applied loads .
(c) Recalculate and draw the bending moment diagram for problem (6) due to
support settlement 10mm at d. Use EI = 60,000 kN.m2
.
3m 6m 3m
2I
I I
6m
40 kN
a
b c
d
80 kN
a
b c
d
3m 6m
6m2EI
EI
2EI

4
Moment-Distribution Method
The moment-distribution method can be used only for the analysis of continuous
beams and frames, taking into account their bending deformations only. The
moment-distribution method is classified as a displacement method, and from a
theoretical viewpoint, it is very similar to the slope-deflection method. However,
unlike the slope-deflection method in which all the structure’s equilibrium equations
are satisfied simultaneously, in the moment distribution method the moment
equilibrium equations of the joints are solved iteratively by successively considering
the moment equilibrium at one joint at a time, while the remaining joints of the
structure are assumed to be restrained against displacement. We first derive the
fundamental relations necessary for the application of the moment-distribution
method and then develop the basic concept of the method. We next consider the
application of the method to the analysis of continuous beams and frames without
side-sway and, finally, discuss the analysis of frames with side-sway.
4.1. Definitions and Terminology
Before we can develop the moment-distribution method, it is necessary to adopt a
sign convention and define the various terms used in the analysis.
Sign Convention
In applying the moment-distribution method, we will adopt the same sign
convention as used previously for the slope-deflection method: Counterclockwise
member end moments are considered positive. Since a counterclockwise moment
at an end of a member must act in a clockwise direction on the adjacent joint, the
foregoing sign convention implies that clockwise moments on joints are considered
positive.
Member Stiffness
Consider a prismatic beam AB, which is hinged at end A and fixed at end B, as
shown in Fig. 4.1(a). If we apply a moment M at the end A, the beam rotates by an
angle y at the hinged end A and develops a moment MBA at the fixed end B, as shown
in the figure. The relationship between the applied moment M and the rotation θ can
be established by using the following slope-deflection equation:

𝑀 𝑛𝑓 =
𝐸𝐼
𝐿
(4𝜃 𝑛 + 2𝜃𝑓) −
6𝐸𝐼
𝐿2
∆ + 𝐹𝐸𝑀 𝑛𝑓
By substituting 𝑀 𝑛𝑓 = 𝑀 , 𝜃𝐴 = 𝜃 and 𝜃𝑓 = Δ = 𝐹𝐸𝑀 𝑛𝑓 = 0 into the slope
deflection equation, we obtain
𝑀 = (
4𝐸𝐼
𝐿
) 𝜃 (4.1)
(a) Beam with far end fixed (b) Beam with far end hinged
Fig. 4.1
The bending stiffness, K, of a member is defined as the moment that must be applied
at an end of the member to cause a unit rotation of that end. Thus, by setting θ = 1
rad in Eq. (4.1), we obtain the expression for the bending stifness of the beam of Fig.
4.1(a) to be
𝐾̅ =
4𝐸𝐼
𝐿
(4.2)
When the modulus of elasticity for all the members of a structure is the same (i.e., E
= constant), it is usually convenient to work with the relative bending stiffnesses of
members in the analysis. The relative bending stiffness, K, of a member is obtained
by dividing its bending stiffness, K, by 4E. Thus, the relative bending stiffness of
the beam of Fig. 4.1(a) is given by
𝐾 =
𝐾̅
4𝐸
=
𝐼
𝐿
(4.3)
Now, suppose that the far end B of the beam of Fig. 4.1(a) is hinged, as shown in
Fig. 4.1(b). The relationship between the applied moment M and the rotation θ of
the end A of the beam can now be determined by using the following modified slope
deflection equation:
𝑀𝑟ℎ =
𝐸𝐼
𝐿
(3𝜃𝑟) −
3𝐸𝐼
𝐿2
∆ + (𝐹𝐸𝑀𝑟ℎ −
𝐹𝐸𝑀ℎ𝑟
2
)
By substituting 𝑀𝑟ℎ = 𝑀 , 𝜃𝑟 = 𝜃 and Δ = 𝐹𝐸𝑀𝑟ℎ = 𝐹𝐸𝑀ℎ𝑟 = 0 into the slope
deflection equation, we obtain
𝑀 = (
3𝐸𝐼
𝐿
) 𝜃 (4.4)
By setting θ = 1 rad, we obtain the expression for the bending stiffness of the beam
of Fig. 4.1(b) to be

M=applied
moment
M =BA carryover
moment
L
EI = const
A B

M=applied
moment
L
EI = const
B
A

Moment distribution 79
𝐾̅ =
3𝐸𝐼
𝐿
(4.5)
A comparison of Eqs. (4.2) and (4.5) indicates that the stiffness of the beam is
reduced by 25 percent when the fixed support at B is replaced by a hinged support.
The relative bending stiffness of the beam can now be obtained by dividing its
bending stiffness by 4E:
𝐾 =
3𝐼
4𝐿
(4.6)
From Eqs. (4.1) and (4.4), we can see that the relationship between the applied end
moment M and the rotation θ of the corresponding end of a member can be
summarized as follows: From Eqs. (4.1) and (4.4), we can see that the relationship
between the applied end moment M and the rotation θ of the corresponding end of a
member can be summarized as follows:
𝑀 = {
(
4𝐸𝐼
𝐿
) 𝜃 if far end of member is fixed
(
3𝐸𝐼
𝐿
) 𝜃 if far end of member is hinged
(4.7)
Similarly, based on Eqs. (4.2) and (4.5), the bending stiffness of a member is given
by
𝐾̅ = {
4𝐸𝐼
𝐿
if far end of member is fixed
3𝐸𝐼
𝐿
if far end of member is hinged
(4.8)
and the relative bending stiffness of a member can be expressed as (see Eqs. (4.3)
and (4.6))
𝐾 = {
𝐼
𝐿
3
4
(
𝐼
𝐿
) if far end of member is hinged
(4.9)
Carryover Moment
Let us consider again the hinged-fixed beam of Fig. 4.1(a). When a moment M is
applied at the hinged end A of the beam, a moment MBA develops at the fixed end B,
as shown in the figure. The moment MBA is termed the carryover moment. To
establish the relationship between the applied moment M and the carryover moment
MBA, we write the slope-deflection equation for MBA by substituting 𝑀 𝑛𝑓 = 𝑀 𝐵𝐴 ,
𝜃𝑓 = 𝜃 and 𝜃 𝑛 = 𝜓 = 𝐸𝐸𝑀 𝑛𝑓 = 0 into Eq. (4.9):
𝑀 𝐵𝐴 = (
2𝐸𝐼
𝐿
) 𝜃 (4.10)

By substituting 𝜃 = 𝑀𝐿/4𝐸𝐼 from Eq. (4.1) into Eq. (4.10), we obtain
𝑀 𝐵𝐴 =
𝑀
2
(4.11)
As Eq. (4.11) indicates, when a moment of magnitude M is applied at the hinged end
of a beam, one-half of the applied moment is carried over to the far end, provided
that the far end is fixed. Note that the direction of the carryover moment, MBA, is the
same as that of the applied moment, M. When the far end of the beam is hinged, as
shown in Fig. 4.1(b), the carryover moment MBA is zero. Thus, we can express the
carryover moment as
𝑀 𝐵𝐴 = {
𝑀
2
0 if far end of member is hinged
(4.12)
The ratio of the carryover moment to the applied moment (MBA/M) is called the
carryover factor of the member. It represents the fraction of the applied moment M
that is carried over to the far end of the member. By dividing Eq. (4.12) by M, we
can express the carryover factor (COF) as
𝐶𝑂𝐹 = {
1
2
0 if far end of member is hinged
(4.13)
Distribution Factors
When analyzing a structure by the moment-distribution method, an important
question that arises is how to distribute a moment applied at a joint among the
various members connected to that joint. Consider the three-member frame shown
in Fig. 4.3(a), and suppose that a moment M is applied to the joint B, causing it to
rotate by an angle y, as shown in the figure. To determine what fraction of the applied
moment M is resisted by each of the three members connected to the joint, we draw
free-body diagrams of joint B and of the three members AB; BC, and
BD, as shown in Fig. 4.3(b). By considering the moment equilibrium of the free
body of joint B (i.e., ∑ 𝑀 𝐵 = 0 ), we write
𝑀 + 𝑀 𝐵𝐴 + 𝑀 𝐵𝐶 + 𝑀 𝐵𝐷 = 0
or
𝑀 = −(𝑀 𝐵𝐴 + 𝑀 𝐵𝐶 + 𝑀 𝐵𝐷) (14)
Since members AB; BC, and BD are rigidly connected to joint B, the rotations of
the ends B of these members are the same as that of the joint. The moments at the
ends B of the members can be expressed in terms of the joint rotation y by applying
Eq. (4.7). Noting that the far ends A and C, respectively, of members AB and BC
are fixed, whereas the far end D of member BD is hinged, we apply Eqs. (4.7)
through (4.9) to each member to obtain

𝑀 𝐵𝐴 = (
4𝐸𝐼1
𝐿1
) 𝜃 = 𝐾̅ 𝐵𝐴 𝜃 = 4𝐸𝐾 𝐵𝐴 (4.15)
𝑀 𝐵𝐶 = (
4𝐸𝐼2
𝐿2
) 𝜃 = 𝐾̅ 𝐵𝐶 𝜃 = 4𝐸𝐾 𝐵𝐶 𝜃 (4.16)
(a) Frame geometry and applied moment
(b) Frame joint equilibrium
Fig. 4.2
𝑀 𝐵𝐷 = (
3𝐸𝐼3
𝐿3
) 𝜃 = 𝐾̅ 𝐵𝐷 𝜃 = 4𝐸𝐾 𝐵𝐷 𝜃 (4.17)
Substitution of Eqs. (4.15) through (4.17) into the equilibrium equation (Eq. (4.14))
yields
𝑀 = − (
4𝐸𝐼1
𝐿1
+
4𝐸𝐼2
𝐿2
+
3𝐸𝐼3
𝐿3
) 𝜃
= −(𝐾̅ 𝐵𝐴 + 𝐾̅ 𝐵𝐶 + 𝐾̅ 𝐵𝐷)𝜃 = −(∑ 𝐾̅ 𝐵 )𝜃 (4.18)
in which Σ𝐾̅ 𝐵 represents the sum of the bending stiffnesses of all the members
connected to joint B. The rotational stiffness of a joint is defined as the moment
required to cause a unit rotation of the joint. From Eq. (4.18), we can see that the
rotational stiffness of a joint is equal to the sum of the bending stiffnesses of all the
members rigidly connected to the joint. The negative sign in Eq. (4.18) appears



M
L , I2 2
L , I3 3L , I1 1
EI = const
A
B
C
D
MBD
MBD
MBC
MBC
MBA
MBA
M
A
B
B
B B
C
D

because of the sign convention we have adopted, according to which the member
end moments are considered positive when in the counterclockwise direction,
whereas the moments acting on the joints are considered positive when they act in
the clockwise direction. To express member end moments in terms of the applied
moment M, we first rewrite Eq. (4.18) in terms of the relative bending stiffnesses of
members as
𝑀 = −4𝐸(𝐾 𝐵𝐴 + 𝐾 𝐵𝐶 + 𝐾 𝐵𝐷)𝜃 = −4𝐸 (∑ 𝐾̅ 𝐵 ) 𝜃
from which
𝜃 = −
𝑀
4𝐸 ∑ 𝐾 𝐵
(4.19)
By substituting Eq. (4.19) into Eqs. (4.15) through (4.17), we obtain
𝑀 𝐵𝐴 = − (
𝐾 𝐵𝐴
∑ 𝐾 𝐵
) 𝑀 (4.20)
𝑀 𝐵𝐶 = − (
𝐾 𝐵𝐶
∑ 𝐾 𝐵
) 𝑀 (4.21)
𝑀 𝐵𝐷 = − (
𝐾 𝐵𝐷
∑ 𝐾 𝐵
) 𝑀 (4.22)
From Eqs. (4.20) through (4.22), we can see that the applied moment M is distributed
to the three members in proportion to their relative bending stiffnesses. The ratio
𝐾 = ∑ 𝐾 𝐵 for a member is termed the distribution factor of that member for end B,
and it represents the fraction of the applied moment M that is distributed to end B of
the member. Thus Eqs. (4.20) through (4.22) can be expressed as
𝑀 𝐵𝐴 = −𝐷𝐹𝐵𝐴 𝑀 (4.23)
𝑀 𝐵𝐶 = −𝐷𝐹𝐵𝐶 𝑀 (4.24)
𝑀 𝐵𝐷 = −𝐷𝐹𝐵𝐷 𝑀 (4.25)
In which 𝐷𝐹𝐵𝐴 = 𝐾 𝐵𝐴 ∑ 𝐾 𝐵⁄ , 𝐷𝐹𝐵𝐶 = 𝐾 𝐵𝐶 ∑ 𝐾 𝐵⁄ , and 𝐷𝐹𝐵𝐷 = 𝐾 𝐵𝐷 ∑ 𝐾 𝐵⁄ are the
distribution factors for ends B of members AB, BC and BD, respectively.
Fixed-End Moments
The fixed-end moment expressions for some common types of loading conditions
as well as for relative displacements of member ends are given inside the back cover
of the book for convenient reference. In the moment-distribution method, the effects
of joint translations due to support settlements and side-sway are also taken into
account by means of fixed-end moments. Consider the fixed beam of Fig. 4.3(a). As
shown in this figure, a small settlement Δ of the left end A of the beam with respect
to the right end B causes the beam’s chord to rotate counterclockwise by an angle
𝜓 = Δ 𝐿⁄ . By writing the slope-deflection equations for the two end moments
with 𝜓 = Δ 𝐿⁄ and by setting θA; θB, and fixed-end moments FEMAB and FEMBA due
to external loading, equal to zero, we obtain

𝐹𝐸𝑀𝐴𝐵 = 𝐹𝐸𝑀 𝐵𝐴 = −
6𝐸𝐼Δ
𝐿2
(4.26)
(a) (b)
Fig. 3
in which FEMAB and FEMBA now denote the fixed-end moments due to the relative
translation D between the two ends of the beam. Note that the magnitudes as well as
the directions of the two fixed-end moments are the same. It can be seen from Fig.
4(a) that when a relative displacement causes a chord rotation in the
counterclockwise direction, then the two fixed-end moments act in the clockwise
(negative) direction to maintain zero slopes at the two ends of the beam. Conversely,
if the chord rotation due to a relative displacement is clockwise, as shown in Fig.
4.3(b), then both fixed-end moments act in the counterclockwise (positive) direction
to prevent the ends of the beam from rotating.
4.2. Basic Concept of the moment-Distribution method
The moment-distribution method is an iterative procedure, in which it is initially
assumed that all the joints of the structure that are free to rotate are temporarily
restrained against rotation by imaginary clamps applied to them. External loads and
joint translations (if any) are applied to this hypothetical fixed structure, and fixed-
end moments at the ends of its members are computed. These fixed-end moments
generally are not in equilibrium at those joints of the structure that are actually free
to rotate. The conditions of equilibrium at such joints are then satisfied iteratively
by releasing one joint at a time, with the remaining joints assumed to remain
clamped. A joint at which the moments are not in balance is selected, and its
unbalanced moment is evaluated. The joint is then released by removing the clamp,
thereby allowing it to rotate under the unbalanced moment until the equilibrium state
is reached. The rotation of the joint induces moments at the ends of the members
connected to it. Such member end moments are referred to as distributed moments,
and their values are determined by multiplying the negative of the unbalanced joint
moment by the distribution factors for the member ends connected to the joint. The
bending of these members due to the distributed moments causes carryover moments
6𝐸𝐼Δ
𝐿2 6𝐸𝐼Δ
𝐿2


L
EI = const
A
B 

L
EI = const
A B
6𝐸𝐼Δ
𝐿2
6𝐸𝐼Δ
𝐿2

to develop at the far ends of the members, which can easily be evaluated by using
the member carryover factors. The joint, which is now in equilibrium, is reclamped
in its rotated position. Next, another joint with an unbalanced moment is selected
and is released, balanced, and reclamped in the same manner. The procedure is
repeated until the unbalanced moments at all the joints of the structure are negligibly
small. The final member end moments are obtained by algebraically summing the
fixed-end moment and all the distributed and carryover moments at each member
end. This iterative process of determining member end moments by successively
distributing the unbalanced moment at each joint is called the moment-distribution
process.
4.3. Analysis of frames with Side-sway
Thus far, we have considered the analysis of structures in which the translations of
the joints were either zero or known (as in the case of support settlements). In this
section, we apply the moment-distribution method to analyze frames whose joints
may undergo both rotations and translations that have not been prescribed. Such
frames are commonly referred to as frames with side-sway. Consider, for example,
the rectangular frame shown in Fig. 4.4(a). A qualitative deflected shape of the frame
for an arbitrary loading is also shown in the figure using an exaggerated scale. While
the fixed joints A and B of the frame are completely restrained against rotation as
well as translation, the joints C and D are free to rotate and translate. However, since
the members of the frame are assumed to be inextensible and the deformations are
assumed to be small, the joints C and D displace by the same amount, D, in the
horizontal direction only, as shown in the figure.
(a) Actual frame (M) (b) Frame with side sway prevented (MO)
 
A
C
D
B A
R
C
D
B

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