SECOND YEAR CIVIL

SECOND TERM

CHAPTER 1

Moment Distribution Method

Prepared
By

Dr. Fahmy F. El-Bab

&

Dr. Gamal A. Al-Saadi

CHAPTER 1
Moment Distribution Method
Moment Distribution is a numerical method of analysis of statically indeterminate structures in which the
joints are temporarily held against rotation and translation, and then through a relaxation process, rotations
of the joints are allowed. Corrections for joint translations (Side-Sway) are made in general through a
separate consideration, or by another process of relaxation. The method is very convenient for analysis of
highly statically indeterminate structures because it avoids the solution of simultaneous equations which
result in the analysis by other methods.

1- Definitions
A. Fixed End Moments, abbreviated by (F.E.M), are the moments at the fixed ends of a beam. Figure
1 shows the F.E.M for prismatic beams subjected to several cases of loading.
B. Sign convention: The sign of the end moments of a beam is positive, when anticlockwise and
negative when clockwise, see Figure 2.
C. Stiffness (S) and Carry-Over-Factor (C.O.F), see Figure 3.
a. Rotation
For fixed-fixed prismatic beam element (AB), the end moments due to rotation a are:
4 EI
2 EI
Ma =
a , M b =
a
L
L

C.O.F =

Mb
= 0.5
Ma

if a = 1 , then M a =

4 EI
= S ab
L

For fixed-hinged prismatic beam element (AB), the end moments due to rotation a are:
3EI
Ma =
a , M b = 0
L
C.O.F =

Mb
=0
Ma

if a = 1 , then M a =

3EI
= S ab
L

b. Translation
For fixed-fixed prismatic beam element (AB), the end moments due to translation ;
located at A or B; are:
6 EI
6 EI
Ma = 2 , Mb = 2
L
L

C.O.F =

Mb
=1
Ma

6 EI
= S ab
L2
For fixed-hinged prismatic beam element (AB), the end moments due to translation ;
located at A or B; are:
3EI
Ma = 2 , Mb = 0
L
if = 1 , then M a =

C.O.F =

Mb
=0
Ma

3EI
= S ab
L2
To prove the above relations, the following methods may be used; (as yielding of supports)
1- Conjugate beam (Elastic weight) method,
2- Virtual work method,
3- Three-moments equation method,
4- Column analogy method, and
5- Slope deflection method.
if = 1 , then M a =

2- Distribution of moment (M) at a joint among the members meeting at this joint
Figure 4, shows a three members; (a, b, and c); with stiffness S12 , S13 , and S14 , and subjected to
moment (M) at joint 1. This moment is distributed among these three members as follows:
M = M 12 + M 13 + M 14 = S12 + S13 + S14
4

or, M = (S12 + S13 + S14 ) = S1i

i =2

or, =

, then

1i

i =2

M 12 = S12

1i

i =2

M 13 = S13

S13

M = D13 M
1i

i =2

i =2

1i

1i

i =2

M 14 = S14

M = D12 M

i =2

S12
4

1i

S14

M = D14 M

1i

i =2

Where D1i is the distribution factor for members 1 i, i=2, 3, and 4.

3- Distribution of moment due to span loads

Figure 5, shows a three members; (a, b, and c); with stiffnesses S12 , S13 , and S14 , and subjected to
span loads. The moments among these three members due to span loads are given below:
Form equilibrium at joint 1
M 12 + M 13 + M 14 = 0 1

if is the resulting rotation at jo int 1 , then

M 12 = S12 + M 12F

M 13 = S13 + M 13F 2

M 14 = S14 + M 14F
Substitute from equation 2 in equation 1

(S12 + S13 + S14 ) + (M 12F + M 13F + M 14F ) = 0

or , (S12 + S13 + S14 ) = (M 12F + M 13F + M 14F ) = M u
Where Mu = the unbalanced moment at joint 1
M
then = 4 u
S1i
i =2

Substitute in equation 2
M
S
M 12 = S12 4 u + M 12F = 4 12 M u + M 12F = D12 M u + M 12F
S1i
S1i
i =2

M 13 = S13

i =2

Mu

+ M 13F =

1i

i =2

M 14 = S14

Mu
i =2

M u + M 13F = D13 M u + M 13F

1i

i =2

+ M 14F =

S13
4

1i

S14

M u + M 14F = D14 M u + M 14F

1i

i =2

Where D1i is the distribution factor for members 1i, i=2, 3, and 4.

4- Moment distribution procedure

a- Relative joint stiffnesses and distribution factors,
b- Fixed end moments (joint locked),
c- Moment distribution table, and
d- Joints are allowed to rotate, or released. To do this, an external moment of the same magnitude
as the artificial moment, but opposite in direction, is applied at each joint. This moment is
called the unbalanced moment (Mu). It is equal to (-F.E.M.) at each joint in the first cycle,
and equal to (-C.O.M.) at each joint in the subsequent cycles. This unbalanced moment, in
each joint is distributed among the members meeting at this joint in proportion to the
distribution factors. These distributed moments will balance the moments at the joint, and, at
the same time, releasing of the joint is effected. The distributed moments are carried-over from
every joint to the adjacent joints. These carry-over-moments will again unbalance the moments

at the joints. This cycle of moment distribution and carry-over-moment is repeated a number
of times until the C.O.M. become negligible at all the joints. The final moments are obtained
by adding up the accumulated moments at every joint.

Symmetrical structures subjected to symmetrical or anti-symmetrical loads and the

equivalent structures
It is of advantage to produce rotations at the two ends of the beam element at the same time. For this
reason, consider the following two cases:
1. When both ends A and B are given symmetrical unit rotations, the stiffness (S) is equal to 2EI/L.
2. When both ends A and B are given anti-symmetrical unit rotations, the stiffness (S) is equal to
6EI/L.
In the above two cases, the bending moment is symmetrical and anti-symmetrical, and the carry-overfactor becomes meaningless. In general, for each case of the above two cases, there are two sub cases for
each one as follows:
1- Symmetrical structures subjected two symmetrical loads
a. Axis of symmetry passes through the middle of a member, in this case, the half structure is
considered, and the stiffness of the middle member is taken equal to 2EI/L.
b. Axis of symmetry passes through a support or a member, in this case, the equivalent
structure is considered. This equivalent structure is the half of the real structure, and the
middle support or member is replaced by totally fixed support.
2- Symmetrical structures subjected two anti-symmetrical loads
a. Axis of symmetry passes through the middle of a member, in this case, the half structure is
considered, and the stiffness of the middle member is taken equal to 6EI/L, or the
equivalent structure is considered. This equivalent structure is the half of the real structure
with putting a roller support at the middle point of each middle member.
b. Axis of symmetry passes through a support, in this case, the equivalent structure is
considered. This equivalent structure is the half of the real structure, and the same middle.
c. Axis of symmetry passes through a member, in this case, the equivalent structure is
considered. This equivalent structure is the half of the real structure with half of the middle
member inertia and loading.

The structures with joints translations (side-sways)

The number of possible independent displacements of the joints indicates the number of degrees of
freedom of the structure. Side-sway induces additional bending moments in comparison with the moments
in the same structure if side-sway is somehow prevented.
1. Frames with one degree of freedom of side-sway
a. Case 0 (prevent side-sway)
In this case, the joints are prevented from translation by introducing a Holding Force F10
(or a support) at one of the joints. In this case, the bending moment diagram which is
denoted by M0-Diagram can be readily determined by the moment distribution as usual.
The magnitude of the necessary holding force F10 is then found by satisfying the statical
condition x = 0, or y = 0 according to the direction of side-sway.
b. Case 1 (allow side-sway)

In this case, the joints are given an arbitrary side-sway without rotations. The resulting
fixed end moments (F.E.M.) are related to . The arbitrary side-sway is chosen with
suitable value. The joints are now allowed to rotate. The unbalanced moments are
distributed in a table and the bending moment diagram, for case 1, which is denoted by
M1-Diagram, is obtained. The necessary holding force F11 is then calculated by satisfying
the statical condition x = 0, or y = 0 according to the direction of side-sway.
To satisfy the statical condition, the following equation is applied:
F10 + F11 = 0, or = - F10 / F11, and the final bending moments (Mf) are given by
Mf = M0 + M1
2. Frames with two degrees of freedom of side-sway
a. Case 0 (prevent side-sway)
In this case, the joints are prevented from translation by introducing a Holding Forces F10
and F20 (or a supports) at two of the joints. In this case, the bending moment diagram which
is denoted by M0-Diagram can be readily determined by the moment distribution as usual.
The magnitude of the necessary holding forces F10 and F20 are then found by satisfying the
statical condition x = 0, or y = 0 according to the direction of side-sway.
b. Case 1 (allow side-sway in level 1, and prevent side-sway in level 2)
In this case, the joints in level 1 are given an arbitrary side-sway without rotations. The
resulting fixed end moments (F.E.M.) are related to . The arbitrary side-sway is
chosen with suitable value. The joints are now allowed to rotate. The unbalanced moments
are distributed in a table and the bending moment diagram, for case 1, which is denoted
by M1-Diagram, is obtained. The necessary holding forces F11 and F21 are then calculated
by satisfying the statical condition x = 0, or y = 0 according to the direction of sidesway.
c. Case 2 (allow side-sway in level 2, and prevent side-sway in level 1)
In this case, the joints in level 2 are given an arbitrary side-sway without rotations. The
resulting fixed end moments (F.E.M.) are related to . The arbitrary side-sway is
chosen with suitable value. The joints are now allowed to rotate. The unbalanced moments
are distributed in a table and the bending moment diagram, for case 2, which is denoted
by M2-Diagram, is obtained. The necessary holding forces F12 and F22 are then calculated
by satisfying the statical condition x = 0, or y = 0 according to the direction of sidesway.
To satisfy the statical condition, the following equations are applied:
F10 + 1 F11 + 2 F 12 = 0
F20 + 1 F21 + 2 F 22 = 0
By solving the above two equations, one can get the correction factors 1 and 2, and the final
bending moments (Mf) are given by:
Mf = M0 + 1 M1 + 2 M2
3. General case, Frames with n degrees of freedom of side-sway
a. Case 0 (prevent side-sway)
In this case, the joints are prevented from translation by introducing a Holding Forces F10 ,
F20 , ., and Fn0 (or a supports) at n joints. In this case, the bending moment diagram
which is denoted by M0-Diagram can be readily determined by the moment distribution as
usual. The magnitude of the necessary holding forces F10 , F20 , ., and Fn0 are then

found by satisfying the statical condition x = 0, or y = 0 according to the direction of

side-sway.
b. Case 1 (allow side-sway in level 1, and prevent side-sway in the rest levels)
In this case, the joints in level 1 are given an arbitrary side-sway without rotations. The
resulting fixed end moments (F.E.M.) are related to . The arbitrary side-sway is
chosen with suitable value. The joints are now allowed to rotate. The unbalanced moments
are distributed in a table and the bending moment diagram, for case 1, which is denoted
by M1-Diagram, is obtained. The necessary holding forces F11 , F21 , ., and Fn1 are then
calculated by satisfying the statical condition x = 0, or y = 0 according to the direction
of side-sway.
c. Case 2 (allow side-sway in level 2, and prevent side-sway in rest levels)
In this case, the joints in level 2 are given an arbitrary side-sway without rotations. The
resulting fixed end moments (F.E.M.) are related to . The arbitrary side-sway is
chosen with suitable value. The joints are now allowed to rotate. The unbalanced moments
are distributed in a table and the bending moment diagram, for case 2, which is denoted
by M2-Diagram, is obtained. The necessary holding forces F12 , F22 , ., and Fn2 are then
calculated by satisfying the statical condition x = 0, or y = 0 according to the direction
of side-sway.
d. Case n (allow side-sway in level n, and prevent side-sway in rest levels)
In this case, the joints in level n are given an arbitrary side-sway without rotations. The
resulting fixed end moments (F.E.M.) are related to . The arbitrary side-sway is
chosen with suitable value. The joints are now allowed to rotate. The unbalanced moments
are distributed in a table and the bending moment diagram, for case n, which is denoted
by Mn-Diagram, is obtained. The necessary holding forces F1n , F2n , ., and Fnn are then
calculated by satisfying the statical condition x = 0, or y = 0 according to the direction
of side-sway.
To satisfy the statical condition, the following equations are applied:
F10
F20
M
M
Fn 0

+ 1 F11
+ 1 F21
M
M
+ 1 Fn1

+ 2 F12
+ 2 F22
M
M
+ 2 Fn 2

+ L L + n F1n
+ L L + n F2 n
O L
M
L O
M
+ L L + n Fnn

= 0 1
= 0 2
M
M
=0 n

By solving the above equations, one can get the correction factors 1, 2, ..,n and the final
bending moments (Mf) are given by:
Mf = M0 + 1 M1 + 2 M2 ++ nM n

pL2 /12

-pL2 /12
p

pL/2

pL/2

Pab /L2

-Pa2 b/L2

pL/4

pL2 /20

-pL2 /30

L
Pa 2 (a+3b)/L3

Pb (3a+b)/L
pL2 /32

-pL2 /32

7pL/20

3pL/20
-M2

M1
p
d/2

R1

pL/4

d/2
L

M.a.(2b-a)/L2

R2

pL/4

M.b.(2a-b) /L2

-5pL2 /96

pL/4

a
2

5pL2 /96

M1= pd [ ab + (a-2b)d /12 ] / L

2

M2= pd [ a 2 b + (b-2a)d /12 ] / L

R1= pd [ (2a + L) b + (a-b)d 2 /4 ] / L

a

R2= pd [ (2b + L) a2 - (a-b)d 2 /4 ] / L

L
6Mab/L3

6Mab/L
M

P1

P2

M1

P1

a
a

P2

R2

R2
P1

P2

P1

M'2

P3

M2

P1

R1

R1
P1

M'1

P3

P2

A m0

L
M = M'1 + M'2 / 2
We can calculate R1 and R2
from equilibrium equations

Figure 1

M1= A m 0 / L = - M2
R1= P1 + 0.5 P2 + pc = R2
Where:
A m 0 = area of moment diagram of simple beam

-pLL'/12
or
p

p(

pLL'/12

L'

h.p

pL /12

.)

2 2
' /1
L
p

pL/2

L'

pL'/2

' /2
pL

L'
' /2
pL

pL'/2

2 2
' /1
-p L

-pL2 /12

pL/2

Figure 1 ( continued )

( +ve )

P1

P2

( -ve )

P3

Sign Convention
Figure 2
Ma = 4EI/L = S

Mb = 2EI/L

a =1

Ma = 3EI/L = S

a =1

L , EI

B
L , EI

Ma = 3EI/L/L

Ma = 6EI/L/L
A

Mb = 2EI/L/L

B
L , EI

=1
A

=1

L , EI
Ma = 3EI/L/L

Figure 3

=1

B
L , EI

a
b

Figure 4

a
b

Figure 5

Examples
Example 1
For the shown beam, Determine the Internal Straining Actions ( N.F.D, S.F.D, & B.M.D ) ,
Using the Method of Moment Distribution.
Solution
Step 1: Stiffnesses and Distribution Factors
Joint B

Stiffnesse s
4 EI
S ba =
= 0 .5 EI
8
4 EI
2 EI
S bc =
=
6
3
Sb

7 EI
6

D .F .

3
0 .43
7
4
M
0 .57
7

Step 2: Fixed End Moments

Member AB

M ab = M ba

2 (8) 2
=
= 10 23 10.667 t.m
12

Member BC

bc

= M

cb

8 6
= 6 t .m
8

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
B.M.

Joint A
AB
0
10.667
0.000
1.003
0.000
11.670

Joint B
BA
BC
0.43
0.57
-10.667 6.000
2.007
2.660
0.000
0.000
0.000
0.000
-8.660
8.660

Joint C
CB
0
-6.000
Cycle 1
0.000
1.330
Cycle 2
0.000
-4.670

Example 1
8t
2 t/m
A

B
8.0 m

3.0 m

3.0 m

Constant EI

10.667 t.m

-10.667 t.m

2 t/m

8t

6 t.m

-6 t.m
C

8.0 m

3.0 m

3.0 m

Fixed End Moments ( F.E.M )

16 t

11.67 t.m

8.66 t.m
B

8.376 t

8t

8.66 t.m

7.624 t

4.67 t.m

4.665 t

3.335 t

Free-body-diagram
8.376
4.605
+

S.F.D.

3.335

Units in ton

7.624
11.67
8.66
4.67
16
+

12

5.335

B.M.D.
Units in t.m

Example 2
For the sketched continuous beam with variable moment of inertia,
Determine the Internal Straining Actions , Using the Method of Moment Distribution.
Solution
Step 1: Stiffnesses and Distribution Factors
Joint B

D. F .

Stiffnesses
4 (5 EI )
S ba =
= 2 EI
10
3 ( 4 EI )
S bc =
= EI
12

2
0 .667
3
1
M 0.333
3

Sb

= 3 EI

Step 2: Fixed End Moments

Member AB

ab

= M

ba

2 . 4 (10 ) 2
=
= 20 t .m
12

Member BC

bc

1 . 5 8 12
= 18 t . m
8

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
B.M.

Joint A
AB
0
20.000
0.000
0.667
0.000
20.667

Joint B
BA
BC
0.667
0.333
-20.000 18.000
1.334
0.666
0.000
0.000
0.000
0.000
-18.666 18.666

Cycle 1
Cycle 2

Example 2 ; Another solution

In this solution, one consider the joint C is temporary fixed and then calculate
the stiffnesses and F.E.M. To correct the joint C, put the D.F. at C = 1
Solution
Step 1: Stiffnesses and Distribution Factors
Joint B

D .F .

Stiffnesse s
4 ( 5 EI )
= 2 EI
10
4 ( 4 EI )
4 EI
=
=
12
3

S ba =
S bc

3
= 0 .6
5
2
M
= 0 .4
5
M

10 EI
3

2 . 4 (10 ) 2
= 20 t .m
12

Sb

Step 2: Fixed End Moments

Member AB

ab

= M

ba

Member BC

bc

= Mcb =

8 12
= 12 t .m
8

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
B.M.

Joint A
AB
0
20.000
0.000
2.400
0.000
-1.800
0.000
0.240
0.000
-0.180
0.000
20.660

Joint B
BA
BC
0.6
0.4
-20.000 12.000
4.800
3.200
0.000
6.000
-3.600 -2.400
0.000
-0.800
0.480
0.320
0.000
0.600
-0.360 -0.240
0.000
-0.080
0.048
0.032
-18.632 18.632

Joint C
CB
1
-12.000
12.000
1.600
-1.600
-1.200
1.200
0.160
-0.160
-0.120
0.120
0.000

Cycle 1
Cycle 2
Cycle3
Cycle 4
Cycle 5

Example 2
8t
2.4 t/m
A

B
10 m

6m

6m

5EI

4EI

8t

12 t.m

-12 t.m
C

20 t.m

6m

-20 t.m

2.4 t/m

6m

8t

18 t.m
B

A
10 m

B
6m

6m

Fixed End Moments ( F.E.M )

24 t

20.667 t.m

18.666 t.m
B

12.2 t

8t

18.666 t.m

11.8 t

5.556 t

2.444 t

Free-body-diagram
12.2
5.556
+

S.F.D.

2.444

Units in ton

11.8
20.667

M = 30 t.m
18.666

M = 24 t.m
4.67

B.M.D.
+
14.667

Units in t.m

Example 3
For the sketched continuous beam with constant moment of inertia,
Determine the Internal Straining Actions , Using the Method of Moment Distribution.
Solution
Step 1: Stiffnesses and Distribution Factors
Joint B

Stiffnesse s
3 ( EI )
S ba =
= 0.5 EI
6
4 ( EI )
S bc =
= 0 .5 EI
8
Sb
= EI

D .F .
M

0 .5

0.5

Step 2: Fixed End Moments

Member DA

Mad = 2 1.5 = 3 t.m

Member AB

M ba

1.5 5 4.8 (6) 2

=
= 13.5 t.m
96

Member BC

3.2 3 (5) 2 3.2 5 (3) 2

M bc = M cb =
+
= 3.75 + 2.25 = 6 t.m
(8) 2
(8) 2
Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
B.M.

Joint A
AD
AB
0
1
-3.000
0.000
0.000
3.000

-3.000

3.000

Joint B
BA
BC
0.5
0.5
-13.500 6.000
3.750
3.750
1.500
0.000
-0.750 -0.750
0.000
0.000
0.000
0.000
-9.000
9.000

Joint C
CB
0
-6.000
0.000
1.875
0.000
-0.375
0.000
-4.500

Cycle 1
Cycle 2
Cycle 3

Example 3
2t
A

1.5 m

9 t.m

3m

3m

3m

2m

A
3m

3m

3.2 t

6 t.m
1.5 m

3.2 t

-6 t.m

-13.5 t.m

4.8 t/m

3m

-9 t.m

4.8 t/m

2 t -3 t.m

3.2 t

3.2 t

4.8 t/m

A
B
3m

3m

2m

3m

3m

Fixed End Moments ( F.E.M )

2 t 3 t.m
A

2t

14.4 t

3 t.m

9 t.m
B

3.2 t

4.5 t.m
C

8.2 t

6.2 t

3.2 t

9 t.m

3.7625 t

2.6375 t

Free-body-diagram
6.2
3.7625

S.F.D.
Units in ton 2

0.5625

2.6375
8.2

M = 14.4 t.m

9
4.5

B.M.D.
Units in t.m

9.6
6.7875

9.6
+
7.9125

Example 4
The sketched Frame with variable moment of inertia, carries loads as shown;
Determine the Internal Straining Actions , Using the Method of Moment Distribution.
Solution
Step 1: Relative Stiffnesses and Distribution Factors
Assume EI = 1
Joint D

Joint E

Stiffnesses
S dc
=0
4(1)
2
S da =
=
6
3
4( 2) 2
S de =
=
12
3
4
Sd =
3

D.F .
0
0.5
0.5
1

Stiffnesse s
4(2)
2
S ed =
=
12
3
3 (1 )
1
S eb =
=
9
3
4(2)
2
S ef =
=
12
3
5
Se =
3

D .F .
0 .4
0 .2
0 .4
1

Step 2: Fixed End Moments

Member DC

M dc = 5 ( 2) = 10
Member DE

M de = M ed

5 4 (12 ) 2
=
= 30 t .m
96

Member EF

M ef = M fe

9 4 (8) 2 9 8 (4) 2
=
+
= 16 + 8 = 24 t.m
(12) 2
(12) 2

Member EB

2.4 3 (6) 2 0.5 2.4 6 (3) 2

= (3.2 + 0.8) = 4 t.m
M eb =
+
2
2
(
9
)
(
9
)

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
B.M.

Joint A
AD
0
0.000
0.000
-5.000
0.000
-0.500
0.000
-0.250
0.000
-0.025
0.000
-5.775

Joint D
DA
DC
DE
0.5
0
0.5
0.000 -10.000 30.000
-10.000 0.000 -10.000
0.000
0.000
2.000
-1.000
0.000
-1.000
0.000
0.000
1.000
-0.500
0.000
-0.500
0.000
0.000
0.100
-0.050
0.000
-0.050
0.000
0.000
0.050
-0.025
0.000
-0.025
-11.575 -10.000 21.575

Joint E
ED
EB
0.4
0.2
-30.000 -4.000
4.000
2.000
-5.000
0.000
2.000
1.000
-0.500
0.000
0.200
0.100
-0.250
0.000
0.100
0.050
-0.025
0.000
0.010
0.005
-29.465 -0.845

EF
0.4
24.000
4.000
0.000
2.000
0.000
0.200
0.000
0.100
0.000
0.010
30.310

Joint F
FE
0
-24.000
0.000
2.000
0.000
1.000
0.000
0.100
0.000
0.050
0.000
-20.850

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Example 4
5t

9t

4 t/m
D

9t

F
( 2EI )

6m

3m

( 2EI )

2.4 t

( EI )

( EI )

6m

6m

2m

30 t.m

6m

4m

4m

9t

-30 t.m 24 t.m

4 t/m

4m

9t

-24 t.m

E
D

E
( 2EI )

( 2EI )

5t

6m

-10 t.m

-4 t.m

Fixed End Moments ( F.E.M )

21.575 t.m

24 t

B
B

16.3425 t
11.3425
5t

11.575 t.m

30.31 t.m 9 t

29.465 t.m

2.892 t

6m

( EI )

6m

( EI )

1.6 t.m

4.586 t
2.892 t

12.6575 t
2.892 t

10 t.m

4m

2.4 t

2.4 t

2m

4m

3m

-3.2 t.m

4m

3m

6m

9t

20.85 t.m

4.586 t

22.4455 t
9.788 t

0.845 t.m
2.4 t

8.212 t

1.694 t

2.892 t
5t

5.775 t.m
16.3425 t

Free-body-diagram

0.706 t
22.4455 t

Example 4 ( Continued )
2.892

4.586

N.F.D.
Units in ton

16.3425

22.4455
11.3425

9.788

1.694

0.788

8.212
2.892

S.F.D.

12.6575

Units in ton
0.706

M = 48 t.m
21.575
10
-

30.31
20.85

29.465
-

0.845

11.575

36

36

8.846
4.8

11.998

4.236

+ 5.775

B.M.D.
Units in t.m

Example 5
The sketched Symmetrical Frame with variable moment of inertia,
carries loads as shown; Determine the Internal Straining Actions,
Using the Moment Distribution Method.
Solution
Use the equivalent Structure; as shown in figure
Step 1: Relative Stiffnesses and Distribution Factors
Joint C

D .F .

Stiffnesse s
3 (2 )
=1
6
=
0

S ca =

S ce

S cd =

0 . 375
0

4 (5)
5
=
12
3

0 . 625

8
3

Sb

Step 2: Fixed End Moments

Member CE

Mce = 4 1.5 = 6 t.m

Member CD

cd

= M

dc

2 (12 ) 2
=
= 24 t .m
12

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
B.M.

CE
0
-6.000
0.000
0.000
0.000
-6.000

Joint C
CA
CD
0.375
0.625
0.000 24.000
-6.750 -11.250
0.000
0.000
0.000
0.000
-6.750 12.750

Joint D
DC
0
-24.000
0.000
-5.625
0.000
-29.625

Cycle 1
Cycle 2

4t

Example 5

4t
2 t/m

6m

( EI )

6m

A'

12 m

4 t -6 t.m

2 t/m
D

( 5EI )

E'
( 2EI )

12 m

4t

C'

( 5EI )

( 2EI )

1.5

( 5EI )

24 t.m

1.5

-24 t.m

2 t/m

1.5

12 m

( 2EI )

Fixed End Moments ( F.E.M )

1.5

12 m

The equivalent structure

C.L.

4 t 6 t.m

24 t

12.75 t.m

29.625 t.m
1.125 t

1.125 t
E

4t

1.125

10.59375 t

6.75 t.m

1.125 t

13.40625 t

14.59375

N.F.D.
Units in ton

1.125 t

Free-body-diagram

14.59375 t

C.L.
M = 36 t.m 29.625

1.125

13.40625

S.F.D.

10.59375

Units in ton

1.125

6
6.75
-

4
+

12.75

13.40625

10.59375

Units in t.m

B.M.D.

1 - Symmetrical Structures Subjected to Symmetrical loads

a- Axis of Symmetry Passes Through the middle of a Member

C.L.
F

p t/m

p t/m

B
L1

C'

L3

L2

A'

B'
L2

L1

C.L.

Deformed shape due to Applied Loads

C.L.

2EI/L

2EI/L
a

p t/m

B
a=-

b
b=1

L , EI

B
L1

L2

0.5L3

Considered Structure

C.L.
F1

F1
P1 t/m

C'
D'

F2
H2

P2 t/m
B

L1

L2

A'

H2

F2

H1

H3

H1

H3

P2 t/m

B'

L3

L2

L1

C.L.

Deformed shape due to Applied Loads

C.L.

F1
P1 t/m
C

H1

H3

F2

Considered Structure

H2

P2 t/m
B

L1

0.5L3

L2

b- Axis of Symmetry Passes Through a Support or a Member

C.L.
F

p t/m
A

p t/m
C

B
L1

L3

L2

L3

L2

C.L.

Deformed shape due to Applied Loads

F
p t/m
A

B
L1

L2

Equivalent Structure

L3

A'

B'
L1

C.L.
F1

F1
P1 t/m

P1 t/m

C'

F2
H2

P2 t/m
B

L1

L2

L3

L3

L2

Deformed shape due to Applied Loads

F1
P1 t/m
C

H1

H3

F2

Equivalent Structure

H2

P2 t/m
B

L1

L2

L3

P2 t/m

B'

C.L.

A'

H2

F2

H1

H1

H3

H3

D'

L1

2 - Symmetrical Structures Subjected to Anti-Symmetrical loads

a- Axis of Anti-Symmetry Passes Through the middle of a Member

C.L.

F
p t/m

B'

C'

A'

B
L1

p t/m

L3

L2

L2

L1

C.L.

6EI/L

6EI/L
a
b

B
a=

b=1

L , EI

OR

3EI/(L/2) = 6EI/L
a

Equivalent Beam Element

a=1

L/2 , EI

Deformed shape due to Applied Loads

C.L.

p t/m

p t/m

B
L1

L2

0.5L3

Considered Structure

L1

0.5L3

L2

Equivalent Structure

OR
C.L.

F1

P1 t/m
C

C'

H3

H3

H1

F2
H2

P2 t/m

H2

F2

L1

L2

H1

D'

F1
A'

P2 t/m

B'

L3

L2

L1

C.L.

Deformed shape due to Applied Loads

C.L.

F1

F1

P1 t/m

P1 t/m

F2

H2

P2 t/m

F2
A

H2

H3

H1

H1

H3

P2 t/m

L1

L2

0.5L3

L1

OR

Considered Structure

L2

0.5L3

Equivalent Structure

b- Axis of Anti-Symmetry Passes Through a Support or a Member

C.L.
F
p t/m
A

B'

L1

A'

p t/m
L3

L2

L3

L2

C.L.

Deformed shape due to Applied Loads

F
p t/m
A

B
L1

L2

Equivalent Structure

L3

L1

C.L.
F1
P1 t/m
C

C'
D'

F3

F2

H2

( EI )

P2 t/m

H2

F2

H3

H1

H3

P1 t/m

L1

L2

L3

L3

L2

Deformed shape due to Applied Loads

F1
P1 t/m
C

E
H3

H1

F2

0.5F3

( 0.5EI )

H2

P2 t/m
B

L1

L2

L3

Equivalent Structure

A'

P2 t/m

B'

C.L.

F1

H1

L1

Example 6
The sketched Symmetrical Frame with variable moment of inertia,
carries loads as shown; Determine the Internal Straining Actions,
Using the Moment Distribution Method.
Solution
The load may be divided into two cases as follows:
Case 1; is a symmetrical case, and case 2 is an antisymmetrical case.
By solving these two cases, and using superposition of two cases, one can obtain
the final solution.
Case 1: Symmetrical Case
Step 1: Relative Stiffnesses and Distribution Factors
Joint B

Stiffnesse

D .F .

3 (1 )
1
=
9
3
2 (2)
1
=
=
12
3
4 (1 )
2
=
=
6
3

S ba =
S bc
S be

Sb

4
3

0 . 25

0 . 25

0 .5

Step 2: Fixed End Moments

Member BA

Mba =

1.569
= 10..125 t.m
8

Step 3: Moment Distribution Table

Joint B
BA
BC
D.F.
0.25
0.25
F.E.M. -10.125 0.000
D.M.
2.531
2.531
C.O.M. 0.000
0.000
D.M.
0.000
0.000
B.M1. -7.594 2.531

BE
0.5
0.000
5.063
0.000
0.000
5.063

Joint E
EB
0
0.000
0.000
2.531
0.000
2.531

Cycle 1
Cycle 2

Case 2: Antisymmetrical Case

Step 1: Relative Stiffnesses and Distribution Factors
Joint B

D.F .

Stiffnesse s
3 (1) 1
=
9
3
3 ( 2)
=
=1
6
4 (1) 2
=
=
6
3

S ba =

S bc

S be

Sb

=2

M
M

1
0.167
6
0.5
1
0.333
3
1

Step 2: Fixed End Moments

Member BA

Mba =

1.569
= 10..125 t.m
8

Step 3: Moment Distribution Table

Joint B
BA
BC
D.F.
0.167
0.5
F.E.M. -10.125 0.0000
D.M. 1.6909 5.0625
C.O.M. 0.0000 0.0000
D.M. 0.0000 0.0000
B.M2. -8.4341 5.0625

BE
0.333
0.0000
3.3716
0.0000
0.0000
3.3716

Joint E
EB
0
0.0000
0.0000
1.6858
0.0000
1.686

Cycle 1
Cycle 2

The final solution is given by using superposition as follows:

For left hand side of the frame ( L.H.S. ), F.B.M = Case 1 + Case 2
For right hand side of the frame ( R.H.S. ), F.B.M = - Case 1 + Case 2
Joint B
BA
BC
F.B.M -16.028 7.594

F.B.M

CB
2.531

Joint C
CD
-0.840

BE
8.434

Joint E
EB
4.217

L.H.S.

CF
-1.691

Joint F
FC
-0.845

R.H.S.

( EI )

6m
4.5 m

( EI )

( EI )

4.5 m

6t

12 m

9m

6t

( EI )

D
( EI )

( 2EI )

6m
4.5 m

D
( EI )

( 2EI )

Example 6

12 t

( EI )

( EI )

4.5 m

12 m

4.5 m

4.5 m

Symmetrical Case

6t
B

A
( EI )

( 2EI )

6m
4.5 m

( EI )

4.5 m

( EI )

( EI )

6t

4.5 m

12 m

4.5 m

Anti-symmetrical Case

C.L.

6t

6t

A
( EI )

( EI )

( 2EI )

6m

6m

( 2EI )
( EI )

( EI )

4.5 m

4.5 m

6m

Symmetrical Case

4.5 m

4.5 m

6m

The equivalent structure for

Anti-symmetrical Case

Example 6 ( Continued )
Symmetrical Case
6t

Anti-symmetrical Case
6t

-10.125 t.m
B

4.5 m

-10.125 t.m
B

4.5 m

4.5 m

4.5 m

Fixed End Moments ( F.E.M )

12 t
1.687 t
A

4.2187 t

16.032 t.m 7.595 t.m

1.687 t

0.421 t

0.421 t

0.8435 t

8.6248 t
B

2.108 t

4.212 t.m

8.437 t.m

2.108 t

0.8435 t

7.7813 t

0.842 t.m

2.527 t.m

0.0936 t

0.9371 t
1.685 t.m

0.421 t

0.421 t

8.6248 t

Free-body-diagram

0.842 t.m

0.9371 t

0.0936 t

Example 6 ( Continued )

0.421
1.687

0.9371

8.6248

N.F.D.
Units in ton

4.2187
0.8435

0.0936
-

0.421

7.7813

2.108

S.F.D.
Units in ton

16.032

M = 27 t.m

7.595
-

1.685
+
2.527 0.842
+

+
-

18.984

+ 8.437

4.212

0.842

B.M.D.
Units in t.m

Example 6; Another Solution

In this solution, the entire structure was taken into consideration.
Solution
Step 1: Relative Stiffnesses and Distribution Factors
Assume EI = 1
Joint B

Joint C

Stiffnesse s
3 (1 )
S ba =
=
9
4 (1)
S be =
=
6
4 (2)
S bc =
=
12
5
Sd =
3

Stiffnesse s
3 (1)
1
S cd =
=
9
3
4 (1)
2
=
S cf =
6
3
4 (2)
2
=
S cb =
12
3
5
Sd =
3

D .F .
1
3
2
3
2
3

0 .2
0 .4
0 .4
1

D .F .
0.2
0.4
0.4
1

Step 2: Fixed End Moments

Member BA

M ba =

1 .5 12 (9 )
= 20 .25 t .m
8

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
B.M.

Joint E
EB
0
0.000
0.000
4.050
0.000
0.000
0.000
0.162
0.000
0.000
0.000
4.212

BE
0.4
0.000
8.100
0.000
0.000
0.000
0.324
0.000
0.000
0.000
0.013
8.437

Joint B
BA
0.2
-20.250
4.050
0.000
0.000
0.000
0.162
0.000
0.000
0.000
0.006
-16.032

BC
0.4
0.000
8.100
0.000
0.000
-0.810
0.324
0.000
0.000
-0.032
0.013
7.595

CB
0.4
0.000
0.000
4.050
-1.620
0.000
0.000
0.162
-0.065
0.000
0.000
2.527

Joint C
CD
0.2
0.000
0.000
0.000
-0.810
0.000
0.000
0.000
-0.032
0.000
0.000
-0.842

CF
0.4
0.000
0.000
0.000
-1.620
0.000
0.000
0.000
-0.065
0.000
0.000
-1.685

Joint F
FC
0
0.000
0.000
0.000
0.000
-0.810
0.000
0.000
0.000
-0.032
0.000
-0.842

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Structures with Side-Sways

Deformed Shape
One Degree of Side-Sway (Horizontally)

Deformed Shape
Two Degrees of Side-Sway (Horizontally)

Two Degrees of Side-Sway

(Horizontally and Vertically)

Deformed Shape

Example 7
For the shown frame, determine the Internal Straining Actions ;
Using the moment distribution method, taking into consideration joint translation ( sidesway )
Solution
Step 1: Relative Stiffnesses and Distribution Factor ( General Calculation )
Assume EI = 1
Joint
B

Stiffnesses
Sce =
Sca = 3(3) / 9 =
Scd = 4(5) / 10 =
Sb =
Sdf =
Sdc = 4(5) / 10 =
Sdb = 4(1.5) / 6 =
Sb =

0
1
2
3

0
2
1
3

D.F.
0
0.333
0.667
1
0
0.667
0.333
1

A: Case of preventing joint translation ( Case 0 )

Step 2: the fixed end moment
Member CE

M ce = 5 ( 2) = 10 t.m

Member CD

M cd = M dc =

Member DF

Member DB

M db = M bd =

df

3 (10) 2
= 25 t.m
12

= 2 ( 2 ) = 4 t .m

4 (6)
= 3 t .m
8

Step 3: Moment Distribution Table

CE
D.F.
0
F.E.M. -10.000
D.M.
0.000
C.O.M. 0.000
D.M.
0.000
C.O.M. 0.000
D.M.
0.000
C.O.M. 0.000
D.M.
0.000
C.O.M. 0.000
D.M.
0.000
M0
-10.000

Joint C
CA
0.333
0.000
-4.995
0.000
-1.999
0.000
-0.556
0.000
-0.222
0.000
-0.062
-7.834

CD
0.667
25.000
-10.005
6.003
-4.004
1.668
-1.113
0.668
-0.445
0.186
-0.124
17.834

Joint D
DC
DF
0.667
0
-25.000 4.000
12.006 0.000
-5.003
0.000
3.337
0.000
-2.002
0.000
1.335
0.000
-0.556
0.000
0.371
0.000
-0.223
0.000
0.149
0.000
-15.586 4.000

DB
0.333
3.000
5.994
0.000
1.666
0.000
0.667
0.000
0.185
0.000
0.074
11.586

Joint B
BD
0
-3.000
0.000
2.997
0.000
0.833
0.000
0.333
0.000
0.093
0.000
1.256

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

F10 =

3.2699 Which prevent joints translation

B: Case of allowing joint translation ( Case 1 )

Step 1: the relative fixed end moment due to sidesway
Assume EI = 1

ca

db

= M

bd

3(3)
1
6 (1 . 5 )
1
M
=
=
2
2
9
4
(9 )
(6 )
M ca
4
400
or
=
=
M db
9
900
Mca = -400
Mdb = Mbd = -900
Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1
F11 =
=

M0
M1
F.B.M

CE
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00

Joint C
CA
0.333
-400.00
133.20
0.00
-99.95
0.00
14.81
0.00
-11.12
0.00
1.65
-361.40

CD
0.667
0.00
266.80
300.15
-200.20
-44.49
29.67
33.38
-22.27
-4.95
3.30
361.40

Joint D
DC
DF
0.667
0
0.00
0.00
600.30
0.00
133.40
0.00
-88.98
0.00
-100.10
0.00
66.77
0.00
14.84
0.00
-9.90
0.00
-11.13
0.00
7.43
0.00
612.62
0.00

DB
0.333
-900.00
299.70
0.00
-44.42
0.00
33.33
0.00
-4.94
0.00
3.71
-612.62

Joint B
BD
0
-900.00
0.00
149.85
0.00
-22.21
0.00
16.67
0.00
-2.47
0.00
-758.16

DB
11.586
-7.457
4.129

Joint B
BD
1.256
-9.229
-7.97318

-268.62
0.0122
(-F10/F11)=
In which is the correction factor
The final moment is equal to ( M0 + 1 )

CE
-10
0
-10

Joint C
CA
-7.834
-4.399
-12.233

CD
17.834
4.3993
22.233

DC
-15.586
7.4574
-8.129

Joint D
DF
4
0
4

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Example 7
5t

2t
3 t/m

3m

( 5EI )

4t
3m

9m

( 1.5EI )

( 3EI )
B

10 m

2m

2m

Case 0 ( Prevent Sidesway )

5t

2t
3 t/m

F10
C

( 5EI )

3m

4t
3m

9m

( 1.5EI )

( 3EI )
B

2m

5t

-10 t.m
C

2m

10 m

25 t.m

3 t/m

2m

-25 t.m

4 t.m

10 m

2m
D

3m

3 t.m

4t
3m

Fixed End Moments ( F.E.M )

-3 t.m
B

2t

Example 7 ( Continued )
5t

11.586 t.m
C

0.87044 t

4.14033 t

F10 =3.26989 t

4t

4t

( 3EI )

3m

9m

( 1.5EI )

0.14033 t

0.14033 t

0.87044 t

( 5EI )

3m

7.834 t.m

2t
3 t/m

1.256 t.m
0.87044 t

2m

10 m

2m

Case 1 ( Allow Sidesway )

F

F11

-400
9m

6m

-900

Deformed shape due to sidesway

-900

without joint rotation and corresponding

relative fixed end moments

10 m

2m

361.4

2m

612.62

40.156

228.463

F11 = 268.62

228.463

228.463

40.156

758.16
A

40.156

758.16
A

Example 7 ( Continued )
10 t.m

30 t

22.233 t.m

1.359 t

4 t.m

8.129 t.m
1.359 t

2t

21.4104 t

13.5896 t

16.4104 t

15.5896 t

5t
12.233 t.m

2t

1.359 t

1.359 t

4.129 t.m
4t

1.359 t

2.641

Free-body-diagram

21.4104 t

7.97318 t.m

15.5896 t
16.4104

2
+

1.359

1.359

13.5896

S.F.D.

15.5896

N.F.D.

2.641

Units in ton

Units in ton
22.233

M = 37.5 t.m
8.129
4
4.129

10
12.233

21.4104

1.359

0.0511 6
-

5t

7.97318

B.M.D.
Units in t.m

Example 8
For the shown frame, determine the Internal Straining Actions ;
Using the moment distribution method, taking into consideration joint translation ( sidesway )
Solution
Step 1: Relative Stiffnesses and Distribution Factor ( General Calculation )
Assume EI = 1
Joint
D

Stiffnesses
Sda = 4(3) / 6 =
Sde = 4(4) / 8 =
Sd =
Sed = 4(4) / 8 =
Seb = 4(3) / 6 =
Sef = 4(3) / 3 =
Se =
Sfe = 4(3) / 3 =
Sfg = 4(6) / 12 =
Sf =
Sgf = 4(6) / 12 =
Sgc = 4(4.5) / 9 =
Sg =

D.F.
0.5
0.5
1
0.25
0.25
0.5
1
0.667
0.333
1
0.5
0.5
1

2
2
4

2
2
4
8

4
2
6

2
2
4

A: Case of preventing joint translation ( Case 0 )

Step 2: the fixed end moment

Member FG

M fg = M gf =

Member DE

M de = M ed =

4 (12) 2
= 48 t.m
12

12 (8)
= 12 t.m
8

Step 3: Moment Distribution Table ( Table 0 )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0

Joint D
DA
DE
0.5
0.5
0.000 12.000
-6.000 -6.000
0.000
1.500
-0.750 -0.750
0.000
2.376
-1.188 -1.188
0.000
0.672
-0.336 -0.336
0.000
0.356
-0.178 -0.178
-8.452 8.452

D.F.
F.E.M.
D.M.
C.O.M.

Joint A
AD
0
0.000
0.000
-3.000

Joint E
ED
EB
0.25
0.25
-12.000 0.000
3.000
3.000
-3.000 0.000
4.752
4.752
-0.375 0.000
1.344
1.344
-0.594 0.000
0.711
0.711
-0.168 0.000
0.206
0.206
-6.123 10.014
Joint B
BE
0
0.000
0.000
1.500

EF
0.5
0.000
6.000
-16.008
9.504
-5.003
2.689
-2.251
1.423
-0.657
0.412
-3.891

Joint F
FE
FG
0.667
0.333
0.000
48.000
-32.016 -15.984
3.000
12.000
-10.005 -4.995
4.752
1.998
-4.502
-2.248
1.344
0.624
-1.313
-0.656
0.711
0.281
-0.662
-0.330
-38.691 38.691

Joint G
GF
GC
0.5
0.5
-48.000 0.000
24.000 24.000
-7.992 0.000
3.996
3.996
-2.498 0.000
1.249
1.249
-1.124 0.000
0.562
0.562
-0.328 0.000
0.164
0.164
-29.971 29.971

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Joint C
CG
0
0.000
Cycle 1
0.000
12.000
Cycle 2

D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0
F10 =
F20 =

0.000
-0.375
0.000
-0.594
0.000
-0.168
0.000
-4.137

Cycle 2
0.000
1.998
Cycle 3
0.000
0.624
Cycle 4
0.000
0.281
Cycle 5
0.000
14.903

0.000
2.376
0.000
0.672
0.000
0.356
0.000
4.904

14.582 Which prevent joints translation at level DE

-9.2077 Which prevent joints translation at level FG

B: Cases of allowing joint translation.

Case 1:
Step 1: the relative fixed end moment due to sidesway
Mda = Mad
6(3)/36
0.5
50

Meb = Mbe
6(3)/36
0.5
50

Mda = Mad = -50

Mef = Mfe
6(3)/9
2
200
Meb = Mbe = -50

Mef = Mfe = 200

Step 2: Moment Distribution Table ( Table 1 )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1

Joint D
DA
DE
0.5
0.5
-50.000 0.000
25.000 25.000
0.000 -18.750
9.375
9.375
0.000
6.775
-3.388 -3.388
0.000 -2.149
1.075
1.075
0.000
1.124
-0.562 -0.562
-18.500 18.500

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1

Joint A
AD
0
-50.000
0.000
12.500
0.000
4.688
0.000
-1.694
0.000
0.537
0.000
-33.969

F11 =
F21 =

ED
0.25
0.000
-37.500
12.500
13.550
4.688
-4.298
-1.694
2.247
0.537
-0.623
-10.593

Joint E
EB
EF
0.25
0.5
-50.000 200.000
-37.500 -75.000
0.000 -66.700
13.550 27.100
0.000
12.506
-4.298
-8.597
0.000
-7.295
2.247
4.495
0.000
1.954
-0.623
-1.246
-76.624 87.217
Joint B
BE
0
-50.000
0.000
-18.750
0.000
6.775
0.000
-2.149
0.000
1.124
0.000
-63.001

-78.695 Which allow joints translation at level DE

49.184 Which prevent joints translation at level FG

Case 2:
Step 1: the relative fixed end moment due to sidesway

Joint F
FE
FG
0.667
0.333
200.000 0.000
-133.40 -66.600
-37.500 0.000
25.013 12.488
13.550
8.325
-14.591 -7.284
-4.298
-1.561
3.908
1.951
2.247
0.911
-2.106
-1.052
52.823 -52.823

Joint G
GF
GC
0.5
0.5
0.000
0.000
0.000
0.000
-33.300 0.000
16.650 16.650
6.244
0.000
-3.122 -3.122
-3.642 0.000
1.821
1.821
0.976
0.000
-0.488 -0.488
-14.861 14.861
Joint C
CG
0
0.000
0.000
0.000
0.000
8.325
0.000
-1.561
0.000
0.911
0.000
7.675

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Mef = Mfe
6(3)/9
2
180

Mgc = Mcg
6(1.5)/81
0.111
10

Mef = Mfe = -180 Mgc = Mcg = -10

Step 2: Moment Distribution Table ( Table 2 )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M2

Joint D
DA
DE
0.5
0.5
0.000
0.000
0.000
0.000
0.000 22.500
-11.250 -11.250
0.000 -7.504
3.752
3.752
0.000
2.683
-1.342 -1.342
0.000 -1.172
0.586
0.586
-8.254 8.254

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M2

Joint A
AD
0
0.000
0.000
0.000
0.000
-5.625
0.000
1.876
0.000
-0.671
0.000
-4.420

F12 =
F22 =

Joint E
ED
EB
EF
0.25
0.25
0.5
0.000
0.000 -180.000
45.000 45.000 90.000
0.000
0.000
60.030
-15.008 -15.008 -30.015
-5.625 0.000 -15.841
5.367
5.367
10.733
1.876
0.000
7.504
-2.345 -2.345
-4.690
-0.671 0.000
-2.449
0.780
0.780
1.560
29.374 33.794 -63.168

Joint F
FE
FG
0.667
0.333
-180.000 0.000
120.06 59.940
45.000
2.500
-31.683 -15.818
-15.008 -7.493
15.008
7.493
5.367
1.977
-4.898
-2.445
-2.345
-0.937
2.189
1.093
-46.310 46.310

Joint G
GF
GC
0.5
0.5
0.000 -10.000
5.000
5.000
29.970 0.000
-14.985 -14.985
-7.909 0.000
3.954
3.954
3.746
0.000
-1.873 -1.873
-1.223 0.000
0.611
0.611
17.292 -17.292

Joint B
BE
0
0.000
0.000
22.500
0.000
-7.504
0.000
2.683
0.000
-1.172
0.000
16.507

Joint C
CG
0
-10.000
0.000
2.500
0.000
-7.493
0.000
1.977
0.000
-0.937
0.000
-13.952

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

42.764 Which prevent joints translation at level DE

-39.965 Which allow joints translation at level FG

F10 + 1 F11 + 2 F12 = 0

F10 + 1 F11 + 2 F12 = 0
By solving the above two equations, one can obtain
1 =
2 =

0.181427
-0.007118

M f = M 0 + 1M 1 + 2 M 2

M0
1.1
2.2
F.B.M.

Joint D
DA
DE
-8.452 8.452

ED
-6.123

Joint E
EB
10.014

EF
-3.891

Joint F
FE
FG
-38.691 38.691

Joint G
GF
GC
-29.971 29.971

-3.356

3.356

-1.922

-13.902

15.824

9.583

-9.583

-2.696

2.696

0.059

-0.059

-0.209

-0.241

0.450

0.330

-0.330

-0.123

0.123

-11.749

11.749

-8.254

-4.128

12.383

-28.777

28.777

-32.790

32.790

Joint A
AD

Joint B
BE

Joint C
CG

M0
1.1
2.2

-4.137
-6.163

4.904
-11.430

14.903
1.392

0.031

-0.118

0.099

F.B.M.

-10.268

-6.644

16.395

4 t/m

Example 8
3m

12 t

F
E

G
( 6EI )

( 3EI )

6m

( 4EI )

( 4.5EI )

( 3EI )

( 3EI )

4m

4m

12 m

Case 0 ( Prevent Sidesway )

4 t/m

3m

12 t
F10

6m

12 t

12 t.m

( 4.5EI )

( 3EI )

( 3EI )

-12 t.m

4m

12 m

48 t.m

-48 t.m

4 t/m
G

F
D

4m

( 6EI )

( 3EI )

( 4EI )

4m

F20

12 m

4m

Fixed End Moments ( F.E.M )

38.691 t.m

3m

F10

14.194 t
3.891 t.m

8.452 t.m
2.098 t

4m
A

4.137 t.m

F10 = 14.582 t

4m
B

10.014 t.m
2.486 t

6m

6m

3m

12 t

4 t/m

14.194 t

F20
G

38.691 t.m
29.971 t.m

3.891 t.m
758.16

4.986 t

14.903 t.m

4.904 t.m

12 m

F20 = -9.208 t

Example 8 ( Continued )
Case 1 ( Allow Sidesway ) at Level DE

Deformed shape due to sidesway at level DE

without joint rotation and corresponding

3m

relative fixed end moments

F11

F21

200
200

G
E

-50
6m

-50

-50

4m

-50

4m

12 m

F21

46.68

87.217

F11

3m

18.5

76.624

8.7448

33.969

23.2708

52.823

6m

6m

3m

52.823
G

46.68

87.217

2.504

7.675

63.001

F11 = -78.6956

14.861

12 m

F21 = 49.184

Example 8 ( Continued )
Case 2 ( Allow Sidesway ) at Level FG

Deformed shape due to sidesway at level FG

without joint rotation and corresponding
relative fixed end moments

F22

-180

3m

-10
F12

-180

6m

4m

-10

4m

12 m

F12

36.493
63.168

D
E

8.254
2.1123
4.42

33.794
A

F22

46.31

36.493

17.292

63.168

8.3835

16.507

F12 = 42.7642

6m

6m

3m

3m

46.31

3.472

13.952
12 m

F22 = -39.965

Example 8 ( Continued )

48 t

28.777 t.m
23.666 t
28.777 t.m

5.465 t

12 t

11.749 t.m

8.254 t.m
E

3.67 t

3.67 t

5.465 t

24.334 t

12.383 t.m

32.79 t.m
6.437 t

23.666 t
5.563 t
29.229 t
4.128 t.m

6.437 t
11.749 t.m

5.465 t

3.67 t

1.795 t
16.395 t.m

3.67 t

24.334 t

23.666 t

5.465 t

5.465 t

32.79 t.m

1.795 t

10.268 t.m

5.465 t

6.644 t.m
24.334 t

6.437 t

29.229 t

Free-body-diagram
23.666

29.229

N.F.D.

1.795

24.334

Units in ton

S.F.D.

Units in ton

M = 72 t.m
32.79

28.777

+
10.268

32.79

12.383

28.777 M = 24 t.m
11.749
8.254
4.128
+
13.9985
+

6.644

24.334

3.67
6.437

5.465

5.563

3.67

6.437
+

23.666

5.465
-

+
16.395

B.M.D.
Units in t.m

5.465

Structures Resting on Elastic Supports

Sketched Example
F

P1 t/m

P2 t/m

S1

S2
b

a
0.5L1

0.5L1
EI1

L2

L3

EI2

EI3

Case 0 ( Prevent Sidesway )

F

P1 t/m

P2 t/m
B

F20

F10

M ab

M ba

P1 t/m

B
0.5L1

0.5L1

M bc

M cb

P2 t/m
C

B
L2

M dc
D

C
b

a
L3

Fixed End Moments ( F.E.M )

M cd

Sketched Example ( Continued )

Case 1 ( Allow Sidesway ) at B
Deformed shape due to sidesway at B
without joint rotation and corresponding
relative fixed end moments

=1
B
A

F21

F11

F11 = Summation of Vertical Forces at B + S 1

S1

Case 2 ( Allow Sidesway ) at C

Deformed shape due to sidesway at C
without joint rotation and corresponding
relative fixed end moments
=1

F12
1

F22 = Summation of Vertical Forces at C + S 2

F22

S2

Example 9
For the shown beam, Determine the Internal Straining Actions ,
Using the Method of Moment Distribution.
Solution
Step 1: Relative Stiffnesses and Distribution Factor ( General Calculation )
Assume EI = 1
Joint
B

D.F.
0.5
0.5
1

Stiffnesses
Sba=4(1.5)/6=
1
1
Sde=2(4)/8=
Sd =
2
Member AB

ab

= M

ba

12 6
= 9 t .m
8

Member BC

bc

= M

cb

3 (8 ) 2
=
= 16 t . m
12

Step 3: Moment Distribution Table

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
M0

Joint A
AB
0
9.000
0.000
-1.750
0.000
7.250

F10 =

18.875

Joint B
BA
0.5
-9.000
-3.500
0.000
0.000
-12.500

BC
0.5
16.000
-3.500
0.000
0.000
12.500

Cycle 1
Cycle 2

Which prevent joints translation at Joint B

B: Case of allowing joint translation ( Case 1 )

Step 1: the fixed end moment due to sidesway ( unit dispalcement )
=1
Mab = Mba = 6(1.5)(10800)/36 = 2700

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
M1

Joint A
AB
0
2700.00
0.00
-675.00
0.00
2025.00

Joint B
BA
0.5
2700.00
-1350.00
0.00
0.00
1350.00

BC
0.5
0.00
-1350.00
0.00
0.00
-1350.00

Cycle 1
Cycle 2

F11 =
=

-2562.5 Which allow joints translation at Joint B

0.007366

(-F10/F11)=

The final moment is equal to ( M0 + 1 )

M0
M1
F.B.M.

Joint A
AD
7.250
14.916
22.166

Joint B
DA
-12.500
9.944
-2.556

DE
12.500
-9.944
2.556

Example 10
For the shown frame, determine the Internal Straining Actions ;
Using the moment distribution method,
Taking into consideration joint translation ( sidesway )
Solution
Step 1: Relative Stiffnesses and Distribution Factor ( General Calculation )
Assume EI = 1
Joint
D

Stiffnesses
Sda=4(1.5)/6=
1
Sde=4(3)/12=
1
Sd =
2
Sed=4(3)/12=
1
Seb=3(2)/6=
1
Sec=3(8/3)/4=
2
Se =
4

D.F.
0.5
0.5
1
0.25
0.25
0.5
1

A: Case of preventing joint translation ( Case 0 )

Step 2: the fixed end moment

5 6 (12) 2
=
= 45 t.m
96

Member DE

M de = M ed

Member DA

M ad = M da =

8 ( 6)
= 6 t .m
8

Step 3: Moment Distribution Table ( Table 0 )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0
F10 =

Joint A
AD
0
6.000
0.000
-9.750
0.000
-1.406
0.000
-0.305
0.000
-0.044
0.000
-5.505

Joint D
DA
0.5
-6.000
-19.500
0.000
-2.813
0.000
-0.609
0.000
-0.088
0.000
-0.019
-29.029

DE
0.5
45.000
-19.500
5.625
-2.813
1.219
-0.609
0.176
-0.088
0.038
-0.019
29.029

ED
0.25
-45.000
11.250
-9.750
2.438
-1.406
0.352
-0.305
0.076
-0.044
0.011
-42.379

Joint E
EB
0.25
0.000
11.250
0.000
2.438
0.000
0.352
0.000
0.076
0.000
0.011
14.126

EC
0.5
0.000
22.500
0.000
4.875
0.000
0.703
0.000
0.152
0.000
0.022
28.252

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

-14.4644 Which prevent joints translation at level DE

B: Case of allowing joint translation ( Case 1 ).

Step 1: the fixed end moment due to sidesway ( unit dispalcement )
=1
Meb
Mec
Mda = Mad
6(1.5)(24000)/36
3(2)(24000)/36
3(8/3)(24000)/16
6000
4000
12000
Step 2: Moment Distribution Table ( Table 1 )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1

Joint A
AD
0
6000.00
0.00
-1500.00
0.00
-250.00
0.00
-46.88
0.00
-7.81
0.00
4195.31

Joint D
DA
0.5
6000.00
-3000.00
0.00
-500.00
0.00
-93.75
0.00
-15.63
0.00
-2.93
2387.70

DE
0.5
0.00
-3000.00
1000.00
-500.00
187.50
-93.75
31.25
-15.63
5.86
-2.93
-2387.70

ED
0.25
0.00
2000.00
-1500.00
375.00
-250.00
62.50
-46.88
11.72
-7.81
1.95
646.48

Joint E
EB
0.25
4000.00
2000.00
0.00
375.00
0.00
62.50
0.00
11.72
0.00
1.95
6451.17

EC
0.5
-12000.00
4000.00
0.00
750.00
0.00
125.00
0.00
23.44
0.00
3.91
-7097.66

F11 =
=

5146.777 Which allow joints translation at level DE

(-F10/F11)=
0.00281
The final moment is equal to ( M0 + 1 )
Joint D
Joint A
Joint E
AD
DA
DE
ED
EB
M0
-5.505
-29.029
29.029
-42.379
14.126
M1
11.790
6.710
-6.710
1.817
18.130
F.B.M.
6.286
-22.318
22.318
-40.562
32.256

EC
28.252
-19.947
8.305

Cycle 1
Cycle 2
Cycle 3
Cycle 4
Cycle 5

Example 9

12 t

12 t
3 t/m
A

S = 20 t/cm = 2000 t/m

3m

3m

8m

1.5EI

3m

3m
1.5EI

4EI

EI = 10800 t.m2

Case 0 ( Prevent Sidesway )

C.L.
12 t
3 t/m
A

F10
12 t

9 t.m

-9 t.m

16 t.m

A
3.0 m

3 t/m

3.0 m

-16 t.m

8.0 m

Fixed End Moments ( F.E.M )

7.25 t.m
A

5.125 t

12 t

12.5 t.m

24 t

12.5 t.m

12 t

6.875 t
F10 = 18.875 t

12.5 t.m
C

12 t

Example 9 ( Continued )
Case 1 ( Allow Sidesway ) at B
Deformed shape due to sidesway at B
without joint rotation and corresponding
relative fixed end moments

C.L.
=1
2700

=1

-2700

-2700
F11
1

2700 F11

F11 = Summation of Vertical Forces at B + S 1

2025 t.m
A

562.5 t

1350 t.m

1350 t.m

562.5 t

F11

F11 = 2562.5 t

1350 t.m
C

Example 9 ( Continued )
22.166 t.m

12 t

2.556 t.m
B

9.268 t

24 t

2.556 t.m

2.556 t.m
C

12 t

2.732 t

12 t

Free-body-diagram

12
9.268
2.732

2.732

S.F.D.

Units in ton

M = 18 t.m

22.166

+
5.639

C.L.
2.556
-

M = 24 t.m

+
21.444

B.M.D.

Units in t.m

12

9.268

Example 10
4m

C
( 8/3 EI )

6 t/m

S
D

8t

( 1.5EI )

6m

3m

( 3EI )

( 2EI )

3m

EI = 24000 t.m
S = 12 t/cm = 1200 t/m
6m

6m

Case 0 ( Prevent Sidesway )

-45 t.m
E

6m

6 t/m

F10
D

8t

( 1.5EI )

( 2EI )

3m

3m

( 3EI )

3m

8t

6 t.m

( 8/3 EI )

6m

3m

-6 t.m

4m

6 t/m

6m

45 t.m

6m

6m

Fixed End Moments ( F.E.M )

7.063

F10 = 14.4644 t

29.029

28.252
F10

42.379

29.029

14.126

M0 and Reactions

8t
5.505
1.7557

Units in ton-meter
A

2.3543

Example 10 ( Continued )
Case 1 ( Allow Sidesway ) at Level DE
Deformed shape due to sidesway at level DE
without joint rotation and corresponding
relative fixed end moments

-12000
F11
D

6000

6000

4000

1774.415
C

7097.66
2387.7

F11 = 5146.778

646.48

F11
E

2387.7
6451.17

1097.1683

4195.31
A

M1 and Reactions

1075.195

2.076 t

Example 10 ( Continued )

22.318 t.m

36 t

8.305 t.m

40.562 t.m

6.672 t
16.48 t

3.372 t

6.672 t

16.48 t

2.076 t

19.52 t

3.372 t

19.52 t
32.256 t.m

22.318 t.m

6.672 t

5.376 t

8t

6.286 t.m

1.328 t

5.376 t

Free-body-diagram
19.52 t

16.48 t

2.076
16.48
+

+
-

6.672

19.52

16.48

1.328

N.F.D.

19.52

S.F.D.
Units in ton

5.376
M = 72 t.m

Units in ton

40.562
-

22.318
22.318
-

+
2.302

M = 12 t.m

8.305

6.286

B.M.D.
Units in t.m

32.256
-

6.672

Structures with Inclined Members ( Non-Rectangular Panels )

Sketched Example

4 t/m

4.5EI

3.0 m

2.5EI
6t

4.0 m

12 t

2EI

3.0 m

1.5EI
A

2.0 m

2.0 m

6.0 m

Case 0 ( Prevent Sidesway )

4 t/m

12 t

6t

F10

F20
B

-6 t.m
12 t.m

12 t
D

6 t.m

6.0 m

2.0 m

-12 t.m

4 t/m

2.0 m

Fixed End Moments ( F.E.M )

Case 1 ( Allow Sidesway ) at Level DE

Deformed shape due to sidesway at level DE
without joint rotation and corresponding
relative fixed end moments
1

/tan

/sin

E
D

1 = 1.333
2

30

80

= 1.667

-80

80

F21
2

Case 2 ( Allow Sidesway ) at C

Deformed shape due to sidesway at C
without joint rotation and corresponding
relative fixed end moments

/tan

/sin

= 1.333

-80 F11

100
-100

2 = 1.667

F12

E
D

-100

F22

100

100

100

Example 11
For the shown frame, determine the Internal Straining Actions ;
Using the moment distribution method, taking into consideration joint translation ( sidesway )
Solution

Assume EI = 1

8
4
8
20

4
5 1/3
9 1/3

D.F.
0.333
0.667
0
1
0.4
0.2
0.4
1
0.6
0.4
1

Step 1: Relative Stiffnesses and Distribution Factor ( General Calculation )

Joint
D

Sed =
Seb =
Sef =
Se =

Stiffnesses
Sda =
4
Sde =
8
Sdg =
0
Sd =
12

Sfe =
Sfg =
Sf =

Mdg = -6 t.m

3 (8) 2
= 16 t .m
12

A: Case of preventing joint translation ( Case 0 )

Member DG

Step 2: the fixed end moment

Member DE

M de = M ed =

Member EF

Mef = M fe =

5 4.5 (8)2
= 15 t.m
96

DE
0.667
16.000
-6.670
0.200
-0.133
-0.233
0.155
0.025
-0.017
-0.030
0.020
9.318

Step 3: Moment Distribution Table ( Table 0 )

Joint D
DG
0
-6.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
-6.000
ED
0.4
-16.000
0.400
-3.335
-0.466
-0.067
0.051
0.078
-0.059
-0.008
0.006
-19.400

Joint A
AD
0
0.000
0.000
-1.665
0.000
-0.033
0.000
0.039
0.000
-0.004
0.000
-1.664
DA
0.333
0.000
-3.330
0.000
-0.067
0.000
0.078
0.000
-0.008
0.000
0.010
-3.318

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0
-13.361 Which prevent joints translation at level GDEF

Meb
3(8.944)(1.118)/45
2/3
200

Joint E
EB
0.2
0.000
0.200
0.000
-0.233
0.000
0.025
0.000
-0.030
0.000
0.003
-0.034

EF
0.4
15.000
0.400
4.500
-0.466
-0.060
0.051
0.070
-0.059
-0.008
0.006
19.434

Mef = Mfe
6(16)(5/6)/64
1.25
375

3 = 0.333

F10 =

B: Case of allowing joint translation ( Case 1 ).

Mde = Med
6(16)(0.5)/64
0.75
225

2 = 1.118

Step 1: the relative fixed end moment due to sidesway

1 = 0.5
Mda = Mad
6(6)/36
1
300

Joint F
FE
FC
0.6
0.4
-15.000
0.000
9.000
6.000
0.200
0.000
-0.120
-0.080
-0.233
0.000
0.140
0.093
0.025
0.000
-0.015
-0.010
-0.030
0.000
0.018
0.012
-6.015
6.015

4 = 1.054

Mfc = Mcf
6(8.433)(1.054)/40
1 1/3
400

Joint C
CF
0
0.000
0.000
3.000
0.000
-0.040
0.000
0.047
0.000
-0.005
0.000
3.002

Cycle 1

Cycle 2

Cycle 3

Cycle 4

Cycle 5

5 = 1+3 = 5/6

x300

Mda = Mad = 300

Mde = Med = 225

DE
0.667
225.00
-350.18
-10.00
6.67
36.52
-24.36
-1.27
0.85
4.63
-3.09
-115.23

Step 2: Moment Distribution Table ( Table 1 )

Joint D
DG
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00

EF
0.4
-375.00
-20.00
-7.50
73.04
3.00
-2.53
-10.96
9.25
0.38
-0.32
-330.64

Joint F
FE
FC
0.6
0.4
-6.015
6.015
-17.231 17.231
-23.246 23.246

Joint F
FE
FC
0.6
0.4
-375.00 400.00
-15.00
-10.00
-10.00
0.00
6.00
4.00
36.52
0.00
-21.91
-14.61
-1.27
0.00
0.76
0.51
4.63
0.00
-2.78
-1.85
-378.05 378.05

Joint C
CF
0
3.002
17.773
20.775

Joint C
CF
0
400.00
0.00
-5.00
0.00
2.00
0.00
-7.30
0.00
0.25
0.00
389.95

Mfc = Mcf = 400

Joint E
EB
0.2
200.00
-10.00
0.00
36.52
0.00
-1.27
0.00
4.63
0.00
-0.16
229.72

EF
0.4
19.434
-15.070
4.364

Mef = Mfe = 375

Joint E
EB
0.2
-0.034
10.470
10.436

Meb = 200

ED
0.4
225.00
-20.00
-175.09
73.04
3.34
-2.53
-12.18
9.25
0.42
-0.32
100.92

Joint A
AD
0
300.00
0.00
-87.41
0.00
1.67
0.00
-6.08
0.00
0.21
0.00
208.38
DA
0.333
300.00
-174.83
0.00
3.33
0.00
-12.16
0.00
0.42
0.00
-1.54
115.23

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1

ED
0.4
-19.400
4.600
-14.800

293.149 Which allow joints translation at level GDEF

DE
0.667
9.318
-5.252
4.066

F11 =
0.04558

DA
0.333
-3.318
5.252
1.934

Joint D
DG
0
-6.000
0.000
-6.000

(-F10/F11)=
Joint A
AD
0
-1.664
9.498
7.834

M0
M1
F.B.M.

Cycle 1

Cycle 2

Cycle 3

Cycle 4

Cycle 5

Example 11
12 t

4.5 t/m

3 t/m
D

6m

( 6 EI )

( 16 EI )

( 8.944 EI )

( 8.433 EI )

2m

( 16 EI )

5m

3m

8m

2m

Case 0 ( Prevent Sidesway )

12 t

4.5 t/m

3 t/m

F10

6m

( 6 EI )

-6 t.m

3 t/m
G

2m

16 t.m

( 16 EI )

( 8.944 EI )

( 8.433 EI )

2m

( 16 EI )

5m

3 t/m

3m

-16 t.m

8m

15 t.m

8m

Fixed End Moments ( F.E.M )

2m

4.5 t/m

-15 t.m
F

8m

Example 11 ( Continued )
9t
9.318 t.m

19.434 t.m

19.4 t.m

3 t/m

6.015 t.m
3.9138 t

4.5 t/m

12.5604 t

12.5604 t

3.9138 t

13.26 t
10.74 t
6 t.m
3 t/m

10.677 t

7.323 t

13.3907 t

F10 = 13.3907 t
6t
16.74 t

3.318 t.m

32.937 t
0.034 t.m

7.323 t
F

E
D

0.8303 t

16.4742 t

16.4742 t

0.8303 t
1.664 t.m

3.002 t.m

3.9138 t

M0 and Reactions

32.937 t

16.74 t

6.015 t.m

3.9138 t

7.233 t

Units in ton-meter

Case 1 ( Allow Sidesway ) at Level DE

Deformed shape due to sidesway at level CDEF
without joint rotation and corresponding
relative fixed end moments

F11
G

225

300

225

-375

4
1

300
A

400

-375
200
2

400
2

Example 11 ( Continued )
115.23

100.92

330.64

378.05
157.529

239.214

239.214

157.529

88.586

88.586

1.789
1.789
293.149
G

F11 = 293.149 t
1.789
115.23

389.95

24 t

M1 and Reactions
9t

14.8 t.m

4.364 t.m

11.12 t

10.66 t

18 t

23.246 t.m
11.12 t

6.64 t

11.36 t

13.34 t

6 t 6 t.m
D

6t

6t

1.63 t
A

16.66 t

1 .5

12.75 t

11.12 t
95 t
14.2
F 23.246 t.m
t
6.96

12.75 t
B

28.98 t

20.775 t.m

6t

7.834 t.m

Free-body-diagram

11.12 t

1 .5

1.63 t

11.36 t
F
23.246 t.m
6t

6t

1.934 t.m

28.98 t
10.436 t.m
10.436 t.m

31.

16.66 t

31.

88.586

1.63 t

1.63 t

157.529

86.797

1.789

378.05

157.529

81.685

4.066 t.m

81.685

53.935
208.38

88.586

53.935

86.797

229.72

20.775 t.m

6.96

11.36 t

95 t
14.2

Example 11 ( Continued )
1.63
+

11.12

31.
6

95
14.2

16.66

N.F.D.
Units in ton
10.66

6.64
+

6.286

+
-

13.34

11.36

1 .5
6

S.F.D.

1.63

6.96

Units in ton

23.246

M = 24 t.m
4.066

- 1.934

4.364
- +

46
23.2

7.834

14.8

10.
436

M = 24 t.m

B.M.D.
Units in t.m

75
20.7

Example 12
For the shown frame, and using the moment distribution method;
Determine the internal straining actions, due to:
1- an outwared sliding of 1 cm of support B,
2- a downwared settlement of 1.5 cm of support A, and
3- a clockwise rotation of 0.003 radians of support B.
Given: E=2000000 t/sq.m.
N.B. the C.S. of the frame elements is rectangular.
Solution
Step 1: Stiffnesses and Distribution Factors
Joint C

D.F .

Stiffnesses
0.3 (0.6) 3
E
12
= 0.0036 E
S ca = S db =
6
0.3 (1) 3
E
4
12
S cd = S dc =
= 0.01E
10
S c = S d
= 0.0136 E
4

A: Case of preventing joint translation ( Case 0 )

Step 2: the fixed end moment
Member CD

M cd = M dc = 45 t.m
Member BD

M db = 28.8 t.m

M bd = 39.6 t.m

M 0.265

M 0.735
M 1

Step 3: Moment Distribution Table ( Table "0" )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0
F10 =

Joint A
AC
0
0.000
0.000
5.963
0.000
-3.594
0.000
0.805
0.000
-0.485
0.000
2.689

Joint C
CA
0.265
0.000
11.925
0.000
-7.187
0.000
1.611
0.000
-0.971
0.000
0.218
5.595

CD
0.735
-45.000
33.075
27.122
-19.934
-6.078
4.467
3.663
-2.692
-0.821
0.603
-5.595

Joint D
DC
DB
0.735
0.265
-45.000 -28.800
54.243 19.557
16.538
0.000
-12.155 -4.382
-9.967
0.000
7.326
2.641
2.233
0.000
-1.642
-0.592
-1.346
0.000
0.989
0.357
11.219 -11.219

Joint B
BD
0
-39.600
0.000
9.779
0.000
-2.191
0.000
1.321
0.000
-0.296
0.000
-30.988

Cycle 1
Cycle 2
Cycle3
Cycle 4
Cycle 5

-5.65388 Which prevent joints translation at level CD

B: Case of allowing joint translation ( Case 1 ).

Step 1: the relative fixed end moment due to sidesway
Mac = Mca = Mdb = Mbd = 100
Step 2: Moment Distribution Table ( Table "1" )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1
F11 =
=

M0
M1
F.B.M.

Joint A
AC
0
100.00
0.000
-13.250
0.000
4.869
0.000
-1.789
0.000
0.658
0.000
90.488

Joint C
CA
0.265
100.00
-26.500
0.000
9.739
0.000
-3.579
0.000
1.315
0.000
-0.483
80.492

CD
0.735
0.000
-73.500
-36.750
27.011
13.506
-9.927
-4.963
3.648
1.824
-1.341
-80.492

Joint D
DC
DB
0.735
0.265
0.000
100.00
-73.500 -26.500
-36.750 0.000
27.011
9.739
13.506
0.000
-9.927
-3.579
-4.963
0.000
3.648
1.315
1.824
0.000
-1.341
-0.483
-80.492 80.492

Joint B
BD
0
100.00
0.000
-13.250
0.000
4.869
0.000
-1.789
0.000
0.658
0.000
90.488

56.99306 Which allow joints translation at level CD

(-F10/F11)=
Joint A
AC
2.689
8.977
11.665

0.099203
Joint C
CA
5.595
7.985
13.580

CD
-5.595
-7.985
-13.580

Joint D
DC
DB
11.219 -11.219
-7.985
7.985
3.234
-3.234

Joint B
BD
-30.988
8.977
-22.011

Cycle 1
Cycle 2
Cycle3
Cycle 4
Cycle 5

30 x 60 cm

30 x 60 cm

6.0 m

Example 12 [ Yielding of Supports ], for the shown frame, and using the Moment Distribution Method;
Determine the internal straining actions, due to :
1- an outwared sliding of 1 cm of support B,
2- a downwared settlement of 1.5 cm of support A,and
3- a clockwise rotation of 0.003 radians of support B.
Given E=2000000.0 t/m 2
C
N.B. the C.S. of the frame elements is rectangular.
30 x 100 cm

B
10.0 m

Case 0 ( Prevent Sidesway )

M1=-6EI ( 0.015 )/L/L =-6(50000)(0.015)/10/10= -45 t.m

M1=-45 t.m
1

F10

M2=6EI ( 0.01 )/L/L = -6(10800)/6/6 = -18 t.m

M1=-45 t.m

M2=-18 t.m

0.015 m

A
M2=-18 t.m

0.01 m
2

F10
M3/2 = -10.8 t.m

B
= 0.003 rad.

M3=-4EI(0.003)/L=-4(10800)(0.003)/6=-21.6 t.m

M3 = -21.6 t.m t.m

Example 12 ( Continued )
- 45 t.m

-45 t.m

F10
-28.8 t.m

A+B
-39.6 t.m

-28.8 t.m

- 45 t.m

-45 t.m

-39.6 t.m

The Fixed End Moments Due to Yielding of Supports

5.595 t.m

11.219 t.m

F10 = 5.6539

F10
11.219 t.m
5.595 t.m

2.689 t.m

1.3806

30.988 t.m

7.0345

Example 12 ( Continued )

Case 1 ( Allow Sidesway ) at Level CD

D

F11

C
100
6.0 m

100

100

100
A

B
10.0 m

Deformed shape due to sidesway

without joint rotation and corresponding
relative fixed end moments

F11
80.492

80.492

F11 = 56.993
90.488

90.488

28.4967

28.4967

13.58

3.234

4.21

4.21

1.035

1.035
1.035

4.21

1.035

4.21

13.58

3.234

11.665
4.21
1.035

22.011

( Units in ton - meter )

Free-body-diagram

4.21

1.035

Example 12 ( Continued )

4.21

1.035

1.035

N.F.D
( Units in ton )

4.21

4.21

1.035

S.F.D
( Units in ton )

13.58

3.234
3.234

13.58

11.665

B.M.D
( Units in t.m )

22.011

Example 13
For the shown frame, and using the moment distribution method;
Determine the internal straining actions, due to:
a rise of temperature of 30C of the outer side, and 10C of the inner side.
Given: E=2000000 t/sq.m., = 0.00005
N.B. the C.S. of the frame elements is rectangular.
Solution
Step 1: Stiffnesses and Distribution Factors
Joint C

D. F .

Stiffnesse s
0 .3 ( 0 .6 ) 3
E
3
12
S ca =
= 0 . 00324 E
5
0 .3 (1) 3
E
4
12
S cd =
= 0 .01666 E
6
Sc =
= 0 . 01990 E

M 0 .163

M 0 .837
M 1

Joint D

D. F .

Stiffnesse s
3

0 .3 ( 0 .8 )
E
12
S db =
= 0 .008533 E
6
0 . 3 (1) 3
4
E
12
S dc =
= 0 .01666 E
6
Sd =
= 0 .0252 E
4

A: Case of preventing joint translation ( Case 0 )

Step 2: the fixed end moment
Mca
2.484

Mcd
-2.0833

Mdc
18

Mdb
-6.4

Mbd
6.4

M 0 .339

M 0 .661
M 1

Step 3: Moment Distribution Table ( Table "0" )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M0
F10 =

Joint C
CA
CD
0.163
0.837
2.484
-2.083
-0.065
-0.335
0.000
-3.806
0.620
3.186
0.000
0.055
-0.009
-0.046
0.000
-0.526
0.086
0.441
0.000
0.008
-0.001
-0.006
3.115
-3.115

Joint D
DC
0.661
17.917
-7.613
-0.168
0.111
1.593
-1.053
-0.023
0.015
0.220
-0.146
10.854

DB
0.339
-6.400
-3.904
0.000
0.057
0.000
-0.540
0.000
0.008
0.000
-0.075
-10.854

-1.2961 Which prevent joints translation at level CD

B: Case of allowing joint translation ( Case 1 ).

Step 1: the relative fixed end moment due to sidesway
0.3 (0.6) 3
= 10800 t.m 2
12
0.3 (1) 3
= 2000000
= 50000 t.m 2
12
0.3 (0.8) 3
= 2000000
= 25600 t.m 2
12

EI ac = 2000000
EI cd
EI bd

3 10800
(1.25) = 1620
52
6 50000
= M dc =
(0.75 ) = 6250
62
6 25600
= M bd =
= 4266.667
62

M ca =
M cd
M db

Mca
Mcd=Mdc
Mcd=Mdc

486
-1875
1280

Joint B
BD
0
6.400
0.000
-1.952
0.000
0.028
0.000
-0.270
0.000
0.004
0.000
4.210

Cycle 1
Cycle 2
Cycle3
Cycle 4
Cycle 5

Step 2: Moment Distribution Table ( Table "1" )

D.F.
F.E.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
C.O.M.
D.M.
M1
F11 =
=

M0
M1
F.B.M.

Joint C
CA
CD
0.163
0.837
486.000 -1875.000
226.407 1162.593
0.000
196.648
-32.054 -164.594
0.000
-192.118
31.315
160.803
0.000
27.199
-4.433
-22.766
0.000
-26.573
4.331
22.241
711.567 -711.567

Joint D
DC
DB
0.661
0.339
-1875.000 1280.000
393.295
201.705
581.297
0.000
-384.237 -197.060
-82.297
0.000
54.398
27.899
80.402
0.000
-53.145
-27.256
-11.383
0.000
7.524
3.859
-1289.147 1289.147

Joint B
BD
0
1280.000
0.000
100.853
0.000
-98.530
0.000
13.949
0.000
-13.628
0.000
1282.644

856.613 Which allow joints translation at level CD

0.001513

(-F10/F11)=
Joint C
CA
3.115
1.077
4.191

CD
-3.115
-1.077
-4.191

Joint D
DC
10.854
-1.951
8.904

DB
-10.854
1.951
-8.904

Joint B
BD
4.210
1.941
6.151

Cycle 1
Cycle 2
Cycle3
Cycle 4
Cycle 5

30 x 80 cm

30

4.0 m

x6
0c
m

Example 13 [ Temprature changes ], for the shown frame, and using the Moment Distribution Method;
Determine the internal straining actions,
o
due to a rise of temprature of 30 C of the outer side,
o
and 10 C of the inner side.
C
30 x 100 cm
2
Given Alpha ( ) = 0.00005 , E = 2000000.0 t/m
N.B. the C.S. of the frame elements is rectangular.

2.0 m

3.0 m

Solution

T2 = 10 c

30 c

6.0 m

T1 = 20 c

10 c

T2 = 10 c

T1 = 20 c

Cross section

Case 0 ( Prevent Sidesway )

A: Case of uniform temperature change ( T1 )
8

T1 ( L )

= 0.00001(20)(5) = 0.001 m

= 0.00001(20)(6) = 0.0012 m

= 0.00001(20)(6) = 0.0012 m

= 0.00001(20)(5)(0.6) = 0.0006 m

= 0.00001(20)(5)(0.8) = 0.0008 m

= 0.0012 - 0.0008 = 0.0004 m

= (0.0012+0.0006)(0.75)-0.0004 = 0.00095 m

= ( 0.0012 + 0.0006 ) / 0.8 = 0.00225 m

M = 6EI (

M1

M
2

)/L/L

M1 = 6(2000000)(0.025)(0.00095)/6/6 = 7.917 t.m

M2 = 3(2000000)(0.0054)(0.00225)/5/5 = 2.916 t.m

M1

cos = 0.6
sin = 0.8

F10

6t
.m

Example 13 ( Continued )

7.917 t.m

- 2.

91

7.917 t.m

Fixed End Moments ( F.E.M )

A: Uniform temperature ( T1 )

B: Case of linear temperature change ( T2 )

M
M=2

T2( EI ) / h

M
+
-

M3 = -10 t.m
+
-

+
-

+
-

+
-

+
-

+
-

M3 = 10 t.m
+
-

+
-

+
-

+
-

1=

5. 4

+
+
-

+
-

- 3.
6t
.m

+
+
-

+
-

1=

+
-

t.m

+
-

1=

3. 6

t.m

M2 = -6.4 t.m

Fixed End Moments ( F.E.M )

-2.0833 t.m

17.9167 t.m
-6.4 t.m

2.4

84

t.m

A+B

M2 = 6.4 t.m

Fixed End Moments ( F.E.M )

6.4 t.m

+
-

Example 13 ( Continued )

1.2898

3.115 t.m

F10
1.2898

0.1886
D

t.m

1.2898
15
3.1

1.1073

10.854 t.m

1.2898
C

10.854 t.m

0.1886

0.1886

0.1886

F10 = 1.296

1.1073

M0 and Reactions

1.2898

4.21 t.m

1.2898

Case 1 ( Allow Sidesway ) at Level CD

Deformed shape due to sidesway at level CD
without joint rotation and corresponding
relative fixed end moments
-1875

1280

-1875
486

/tan = /1.333 = 0.75

/sin = /0.8 = 1.25
A
2

1280
B

711.567

1289.147

333.4523

333.4523

427.981
D

333.4523

428.6318

67

1.5

427.981
C

427.981
A

333.4523

F11 = 856.613

M1 and Reactions

F11

1289.147

427.981

333.4523

71

=
2 =
1

F11

cos = 0.6
sin = 0.8
tan = 1.333

428.6318

1282.644
333.4523

Example 13 ( Continued )

0.7855

4.191

0.459

0.459
0.7855

0.7855
8.904
53

0.7855

0.3

0.459

4.1

91

0. 8

38
2

0.459

4.1

91

8.904

0.459

( Units in ton - meter )

Free-body-diagram

6.151

38
2

0.459

0. 8

0.3

53

0.7855

0.7855

0.7855
0.459

0. 8

38
2

0.3

53

S.F.D

N.F.D
( Units in ton )

( Units in ton )

0.7855

8.904
8.904

4.1

91

4.191

B.M.D
( Units in t.m )

6.151

0.459