This document discusses the moment distribution method for analyzing statically indeterminate structures. It begins with definitions of key terms used in the method like fixed end moments, distribution factors, carryover factors, and flexural stiffness. It then outlines the steps of the method, which involve calculating fixed end moments, distribution factors based on member stiffness, distributing moments at joints iteratively until equilibrium is reached, and calculating shear and bending moment diagrams. An example problem is then presented and solved using the moment distribution method.

1. COURSE NO-416
( PRE-STRESSED CONCRETE LAB)
Presented By
Sharmeen Mahmud
Id no:10.01.03.078
Sec: B
Year & sem : 4-2

2. Solving Statically
Indeterminate
Structure: Moment
Coefficient Method.

3. Introduction
 In the moment distribution method, every joint of the structure
to be analysed is fixed so as to develop the fixed-end moments.
Then each fixed joint is sequentially released and the fixed-end
moments (which by the time of release are not in equilibrium)
are distributed to adjacent members until equilibrium is
achieved. The moment distribution method in mathematical
terms can be demonstrated as the process of solving a set
of simultaneous equations by means of iteration.
 The moment distribution method falls into the category
of displacement method of structural analysis.

4. Indeterminate structure
 The structures which cannot be solved by the
equilibrium equation are known as
indeterminate structure.

5. Moment distribution method
 The moment distribution method is
a structural analysis method for statically
indeterminate beams and frames developed
by Hardy Cross. It was published in 1930 in
an ASCE journal. The method only accounts for
flexural effects and ignores axial and shear
effects. From the 1930s until computers began to
be widely used in the design and analysis of
structures, the moment distribution method was
the most widely practiced method.

6. Some Basic Definition

7. Flexural stiffness
 The flexural stiffness (EI/L) of a member is
represented as the product of the modulus of
elasticity (E) and the second moment of area (I)
divided by the length (L) of the member. What
is needed in the moment distribution method is
not the exact value but the ratio of flexural
stiffness of all members.

8. Fixed end moments
 Fixed end moments are the moments produced
at member ends by external loads when the
joints are fixed.

10. Distribution factors
 When a joint is released and begins to rotate under the
unbalanced moment, resisting forces develop at each
member framed together at the joint. Although the
total resistance is equal to the unbalanced moment, the
magnitudes of resisting forces developed at each
member differ by the members' flexural stiffness.
Distribution factors can be defined as the proportions
of the unbalanced moments carried by each of the
members.

11. Carryover factors
 When a joint is released, balancing moment
occurs to counterbalance the unbalanced
moment which is initially the same as the
fixed-end moment. This balancing moment is
then carried over to the member's other end.
The ratio of the carried-over moment at the
other end to the fixed-end moment of the initial
end is the carryover factor.

13. Steps
The steps involved in
moment distribution method
are described below :

14.  1.Calculate the fixed end moments for all the spans of
the beams by considering all the joints as fixed.
 2. Calculate the stiffness coefficients for all the
members. The stiffness coefficient ( kAB ) for a
member AB is calculated as follows:
 kAB = 4EI/L if the far end B is fixed .
 kAB = 3EI/L if the far end B is pinned or on a roller.
 E is modulus of elasticity, I is moment of inertia of
section and L is span AB. While calculating the
stiffness coefficients all the intermediate joints are
taken as fixed but the joints at end of beam are kept as
they are.

15.  3.Calculate the distribution factors based on the stiffness





coefficient of the member.
Distribution factor can be defined as the ratio of stiffness
coefficient of a member to the sum of the stiffness
coefficient of all the members meeting at that joint. If
BA, BC and BD are connected at joint B, then the
Distribution Factor (D) can be easily calculated as
follows;
DBA= kBA/(kBA+kBC+kBD)
DBC=kBC/(kBA+kBC+kBD)
DBD=kBD/(kBA+kBC+kBD)
Distribution factor for a pinned support or roller at the of
the beam is taken as 1 whereas for a fixed support at the
end of beam the distribution factor are taken as zero.

16.  4. Balance all the joints by applying the balancing
moments in the proportions of distribution factors.
 5.Carry over half of the balancing moments to the
opposite ends of the span. If the opposite end is
pinned there should be no carry over moment to
that end (as in the case of pinned support at the
ends of the beams ).

17.  6. Continue these cycles of balancing and carry
over till the joints reach equilibrium state when
the unbalanced moment is negligible (based on
desired accuracy).
 7. Take the sum of all the moments (fixed end
moment, balancing moment, carry-over
moment) at each end.

18. Sign convention
 Once a sign convention has been chosen, it has to be
maintained for the whole structure. The traditional
engineer's sign convention is not used in the
calculations of the moment distribution method
although the results can be expressed in the
conventional way. In the BMD case, the left side
moment is clockwise direction and other is
anticlockwise direction so the bending is positive.

19. Framed Structures
 Framed structures with or without sidesway
can be analysed using the moment distribution
method.

20. Sides way Frame

21. Problem

22. 150 kN
15 kN/m
10 kN/m
3m
A
I
8m
B
I
6m
C
I
8m
D

23. SOLUTION OF PROBLEMS -
Solve the previously given problem by the moment
distribution method
Fixed end moments
M AB
M BA
M BC
M CB
M CD
M DC
wl 2
12
wl
8
wl 2
12
(15)(8) 2
12
(150)(6)
8
(10)(8) 2
12
80 kN.m
112.5 kN.m
53.333 kN.m
Stiffness Factors
K AB
K BA
K BC
K CB
K CD
K
DC
4EI
(4)(EI )
L
8
4EI
(4)(EI )
L
6
4
EI 0.5EI
8
4EI
8
4EI
0.5EI
8
0.5EI
0.667EI

24. Distribution Factors
K
DFAB
DF
BA
DF
BC
K B A K wall
K
BA
K BA K BC
K BC
K
BA
K
DFCB
K
CB
K
DF
CD
DF
DC
BA
K CB
K DC
K
DC
K
BC
CB
K
CD
CD
K CD
1.00
0 .5
0.5EI
( wall stiffness
)
0.5EI
0.5EI 0.667EI
0.4284
0.667EI
0.5EI 0.667EI
0.0
0.5716
0.667EI
0.667EI 0.500EI
0.5716
0.500EI
0.667EI 0.500EI
0.4284

25. Moment Distribution Table
Joint
A
Member
AB
Distribution Factors
Computed end moments
0
-80
B
BA
C
BC
CB
D
CD
0.4284 0.5716 0.5716 0.4284
80
-112.5
112.5
DC
1
-53.33
53.33
-33.82 -25.35
-53.33
9.289
-26.67
-12.35
9.662
9.935 7.446
12.35
4.968
4.831 6.175
3.723
-2.84
-6.129 -4.715
-3.723
-3.146
-1.42 -1.862
-2.358
1.798
1.876 1.406
2.358
0.938
0.9 1.179
0.703
-0.402
-0.536
-1.187 -0.891
-0.703
99.985
-99.99
Cycle 1
Distribution
Carry-over moments
13.923 18.577
6.962
-16.91
Cycle 2
Distribution
Carry-over moments
7.244
3.622
Cycle 3
Distribution
Carry-over moments
-2.128
-1.064
Cycle 4
Distribution
Carry-over moments
1.348
0.674
Cycle 5
Distribution
Summed up
moments
-69.81
96.613
-96.61
0

26. Computation of Shear Forces
15 kN/m
10 kN/m
150 kN
B
C
A
D
I
I
3m
8m
Simply-supported
75
I
3m
8m
60
60
75
40
8.726
-8.726
16.665
-16.67
12.079
-12.5
12.498
-16.1
16.102
0
56.228
63.772
75.563
74.437
53.077
40
reaction
End reaction
due to left hand FEM
-12.08
End reaction
due to right hand FEM
Summed-up
moments
0
27.923

27. Shear Force and Bending Moment Diagrams
52.077
75.563
2.792 m
56.23
27.923
74.437
3.74 m
63.77
S. F. D.
Mmax=+38.985 kN.m
Max=+ 35.59 kN.m
126.704
31.693
35.08
-69.806
3.74 m
48.307
84.92
98.297
-99.985
2.792 m
-96.613
B. M. D