VTU EDUSAT PROGRAMME 7

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VTU EDUSAT PROGRAMME 7 B.E. V SEM (CIVIL
ENGG) SUB STRUCTURAL ANALYSIS II (CV51) MOMENT
DISTRIBUTION METHOD FOR ANALYSIS OF INDETERMINATE
STRUCTURES BY A.B. HARWALKAR SELECTION GRADE
LECTURER PDA COLLEGE OF ENGG, GULBARGA
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SIGNIFICANCE OF STRUCTURAL ANALYSIS
SUBJECT AND METHODS OF ANALYSIS OF INDETERMINATE
STRUCTURES
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• FOR DESIGN OF STRUCTURES
• (BOTH ADAPTIVE AND PASSIVE DESIGN METHODOLOGY)

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REQUIREMENTS TO BE SATISFIED FOR ANALYSIS

• 1. EQILIBRIUM EQUATIONS2. LOAD DISPLACMENT
  RELATIONS3. COMPATIBILITY CONDITIONS

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• METHOD OF ANALYSIS OF INDETERMINATE STRUCTURES
• FORCE METHOD
• DISPLACEMENT METHOD

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FORCE METHOD
RB
Redundant RB applied ?? BB RB ? fBB
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MA
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DISPLACEMENT METHOD
1) Determine the degrees of freedom after
identifying the nodes hence the unknown
displacements.
2)Then these unknown displacement are written in
terms of loads by using load displacement
relations. 3)To then write an equilibrium
equation for each unknown degree of freedom for
the structure.
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4)Then these equations are solved for the
displacements. 5) Once the displacements are
obtained, the unknown loads are determined from
the compatibility equations, using
load-displacement relations. Slope deflection
method of analysis is one of the example for
displacement method of analysis
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• MOMENT DISTRIBUTION METHOD
• Sign Convention
• Clockwise ve Anticlockwise ve for
  moment distribution.
• Sagging ve Hogging ve for BMD

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ii) FEM
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• Member Stiffness Factor
• k4EI/L far end fixed
• k3EI/L far end hinged
• ii) Distribution Factor
• DF k/?k
• iii) Carry over factor ½ far end fixed

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Steps involved in Moment Distribution Method
(i)Fixed end moments for each loaded span are
determined assuming both ends fixed. (ii)The
distribution factors can be determined from
equation DF k/?k for a fixed end DF 0 and DF
1 for an end pin or roller support (iii)Assume
that all joints at which the moments in the
connecting spans must be determined are initially
locked
Then determine the moment that is needed to
put each joint in equilibrium.
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Release or unlock the joints and distribute
the counter balancing moments into connecting
span at each joint using distribution factors
Carry these moments in each span over to its
other end by multiplying each moment by carry
over factor. By repeating this cycle of
locking and unlocking the joints, it will be
found that the moment corrections will diminish
since the beam tends to achieve its final or
nearly final deflected shape.Each column of FEMs,
distributed moments and carry over moment should
then be added to get the final moments at the
joints.
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Example1 Analyze the beam shown in figure 6 (a)
by moment distribution method and draw the BMD.
Assume EI is constant
250kN
Solution  (i) FEM calculation MFAB MFBA
0 MFBC MFCB 240 kNm MFCB - 250
kNm FFDC 250 kNm
20kN/m
Fig. 6 (a)
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Calculation of distribution factors
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(i)                   The moment distribution is
carried out in table below  
 
    After writing FEMs we can see that there is
a
Balance
Balance
Balance
Balance
The moment distribution is carried out in table
below
BMD is shown in figure 6 (b).
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
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Fig 6 (b)
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VTU EDUSAT Programme 7 Class B.E. V Sem
(Civil Engineering) Sub Structural Analysis II
(CV51) Session on 25.09.2007   Moment
distribution method for analysis of indeterminate
structures (contd.)     Ex 2 Analyse the
continuous beam shown in fig 7 (a) by moment
distribution method and draw BMD SFD (VTU July
2006 exam)           Fig. 7 a     Fig. 7
(a)   Solution FEM MFAB - MFBC - kNm
MFAB MFDC 3 kNm MDE -10 x 1
?10kNm MFCD -9 kNm   DF
 
 
Fig. 7 (b)
Fig. 7 (c)
Fig. 7 (d)
                  Ex 3 Analyse the continuos
beam as shown in figure 8 (a) by moment
distribution method and draw the B.M. diagrams
(VTU January 2005 exam)               Fig.
8 (a)   Support B sinks by 10mm E 2 x 105
N/mm², I 1.2 x 10-4 m4   Solution Fixed End
Moments MFAB FEM due to load FEM due to
sinking ?60 40 MFAB -100
kNm MFBA FEM due to load FEM due to sinking
60 ?40 MFBA 20 kNm   MFBC FEM due to
loading FEM due to sinking ?24
57.6 MFBA 33.6 kNm MFCB MFCB
93.6kNm   MFCD due to load only ( C D are at
some level)   MFCD MFDC 26.67 kNm      
  Note Since support A is simply supported end
the relative stiffness value of has been taken
and also since D can be considered as simply
supported with a definite moment relative
stiffness of CD has also been calculated using
the formula .   Moment distribution table
 
 
    FBD of various spans is shown in fig. 7 (b)
and 7 (c) and BMD, SFD have been shown in fig. 7
(d)
     
Release of joint A and adjusting moment at D
              Bending moment diagram is shown in
fig. 8 (b)                                   Fig
. 8 (b)  
Balance
Example2Analyse the continuous beam shown in fig
7 (a) by moment distribution method and draw BMD
SFD (VTU July 2006 exam)
 
Fig. 7 a
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
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VTU EDUSAT Programme 7 Class B.E. V Sem
(Civil Engineering) Sub Structural Analysis II
(CV51) Session on 25.09.2007   Moment
distribution method for analysis of indeterminate
structures (contd.)     Ex 2 Analyse the
continuous beam shown in fig 7 (a) by moment
distribution method and draw BMD SFD (VTU July
2006 exam)           Fig. 7 a     Fig. 7
(a)   Solution FEM MFAB - MFBC - kNm
MFAB MFDC 3 kNm MDE -10 x 1
?10kNm MFCD -9 kNm   DF
 
 
Fig. 7 (b)
Fig. 7 (c)
Fig. 7 (d)
                  Ex 3 Analyse the continuos
beam as shown in figure 8 (a) by moment
distribution method and draw the B.M. diagrams
(VTU January 2005 exam)               Fig.
8 (a)   Support B sinks by 10mm E 2 x 105
N/mm², I 1.2 x 10-4 m4   Solution Fixed End
Moments MFAB FEM due to load FEM due to
sinking ?60 40 MFAB -100
kNm MFBA FEM due to load FEM due to sinking
60 ?40 MFBA 20 kNm   MFBC FEM due to
loading FEM due to sinking ?24
57.6 MFBA 33.6 kNm MFCB MFCB
93.6kNm   MFCD due to load only ( C D are at
some level)   MFCD MFDC 26.67 kNm      
  Note Since support A is simply supported end
the relative stiffness value of has been taken
and also since D can be considered as simply
supported with a definite moment relative
stiffness of CD has also been calculated using
the formula .   Moment distribution table
 
 
    FBD of various spans is shown in fig. 7 (b)
and 7 (c) and BMD, SFD have been shown in fig. 7
(d)
     
Release of joint A and adjusting moment at D
              Bending moment diagram is shown in
fig. 8 (b)                                   Fig
. 8 (b)  
Balance
Calculation of Distribution Factors
 
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
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Release of joint A and adjusting moment at D
Release of joint A and adjusting moment at D
Balance
Balance
Note Since support A is simply supported end
and also since D can be considered as simply
supported with a definite moment value relative
stiffness of BA CD has been calculated using
the formula
 
K
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
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Release of joint A and adjusting moment at D
Release of joint A and adjusting moment at D
Balance
Balance
Moment distribution table
 
Balance
Balance
Balance
Balance
Balance
Balance
Balance
Balance
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FBD of various spans is shown in fig. 7 (b) and 7
(c) and BMD, SFD have been shown in fig. 7 (d)
Fig. 7 (b)
Fig. 7 (c)
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Fig. 7 (d)
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Ex 3 Analyze the continuous beam as shown in
figure 8 (a) by moment distribution method and
draw the B.M. diagrams (VTU January 2005 exam)
Fig. 8 (a)
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Fig. 8 (a)
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Calculation of Distribution Factors
DF
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Moment Distribution Table
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Bending Moment Diagram
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Moment Distribution Method for Analysis of Frames
(Non Sway Type)
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Ex 4 Analysis the frame shown in figure 9 (a) by
moment distribution method and draw BMD assume EI
is constant
Fig. 9 (a)
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Calculation of distribution factors
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The moment distribution shown below
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By inspection the hinge at E will prevent the
frame from side sway. The BMD is shown below in
fig. 9 (b)
Fig. 9 (b)
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Ex 5 Analyse the frame shown in fig.10 (a) by
moment distribution method and draw BMD and SFD
(VTU Jan/Feb 2005 exam)
Fig. 10 (a)
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Moment Distribution table
Fig. 10 b
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FBD of various spans are shown in fig. 10 (b) BMD
is shown in fig. 10 (c) SFD is shown in Fig. 10
(d)
Fig. 10 b
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Fig. 10 c
Fig.10(c )
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Fig. 10 c
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Moment Distribution Method for Analysis of Frames
(Sway Type)
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Analysis Steps for Sway Types of Frames
Step1 First do the non-sway analysis considering
frame is prevented from sway as shown in
fig.11(b) Step2 After determining sway force
apply the sway force on the frame do the sway
analysis as shown in fig.11( c ) .
Addition of moments for both cases fig 11 (b)
fig 11(c) will yield the actual moments in the
frame.
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Ex 6 Analyse the frame shown in figure 12 (a) by
moment distribution method. Assume EI is constant
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Solution i) Non Sway Analysis First
consider the frame held from side sway
as shown in fig 12 (b) FEM MFAB MFBA
MFCD MFDC 0 MFBC ? -10.24
kNm MFCB 2.56 kNm
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Calculation of distribution
factors
 
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FBD of Columns
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Figure 12 (d)
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FBD of columns
Using for the entire frame R?
28 28 56 KN (?) Hence R? 56KN creates the
sway moments shown in above moment distribution
table. Corresponding moments caused by R
0.91KN can be determined by proportion. Thus
final moments are calculated by adding non sway
moments and sway moments calculated for R
0.91KN, as shown below
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3.7 kNm
Fig 12(e)
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Ex 7 Analysis the rigid frame shown in figure 13
(a) by moment distribution method and draw BMD
(VTU July 2006 exam).
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Non Sway Analysis First consider the frame
held from side sway as shown in figure 13 (b).
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Fig 13 (d) Fig 13 (e)
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Fig. 13 (f)
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Since ends A D are hinged and columns AB CD
are of different lengths
- 3EI?/L²1 M?CD -
3EI?/L²2
?
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FBD of columns AB CD for sway analysis moments
is shown in fig. 13 (g)
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Using ?fx 0 for the entire frame R? 11.12 KN
(?) Hence R? 11.12 KN creates the sway moments
shown in the above moment distribution table.
Corresponding moments caused by R 5.34KN can be
determined by proportion. Thus final moments are
calculated by adding non sway moments and sway
moments determined for R 5.34 KN as shown
below.
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Fig. 13 (h)