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7. MOMENT DISTRIBUTION METHOD. 7.1 MOMENT DISTRIBUTION METHOD - AN OVERVIEW. 7.1 MOMENT DISTRIBUTION METHOD - AN OVERVIEW 7.2 INTRODUCTION 7.3 STATEMENT OF BASIC PRINCIPLES 7.4 SOME BASIC DEFINITIONS 7.5 SOLUTION OF PROBLEMS - PowerPoint PPT Presentation
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1
7. MOMENT DISTRIBUTION METHOD
2
7.1 MOMENT DISTRIBUTION METHOD - AN OVERVIEW
• 7.1 MOMENT DISTRIBUTION METHOD - AN OVERVIEW• 7.2 INTRODUCTION• 7.3 STATEMENT OF BASIC PRINCIPLES• 7.4 SOME BASIC DEFINITIONS• 7.5 SOLUTION OF PROBLEMS• 7.6 MOMENT DISTRIBUTION METHOD FOR STRUCTURES
HAVING NONPRISMATIC MEMBERS
3
7.2 MOMENT DISTRIBUTION METHOD - INTRODUCTION AND BASIC PRINCIPLES
7.1 Introduction(Method developed by Prof. Hardy Cross in 1932)The method solves for the joint moments in continuous beams andrigid frames by successive approximation.
7.2 Statement of Basic PrinciplesConsider the continuous beam ABCD, subjected to the given loads,as shown in Figure below. Assume that only rotation of joints occurat B, C and D, and that no support displacements occur at B, C andD. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D.
15 kN/m 10 kN/m150 kN
8 m 6 m 8 m
A B C DI I I
3 m
4
In order to solve the problem in a successively approximating manner, it can be visualized to be made up of a continued two-stage problems viz., that of locking and releasing the joints in a continuous sequence.
7.2.1 Step IThe joints B, C and D are locked in position before any load is applied on the beam ABCD; then given loads are applied on the beam. Since the joints of beam ABCD are locked in position, beams AB, BC and CD acts as individual and separate fixed beams, subjected to the applied loads; these loads develop fixed end moments.
8 m
-80 kN.m -80 kN.m15 kN/m
A B
6 m
-112.5kN.m 112.5 kN.m
B C 8 m
-53.33 kN.m10 kN/m
C D
150 kN53.33 kN.m
3 m
5
In beam ABFixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.mFixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m
In beam BCFixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62
= -112.5 kN.mFixed end moment at C = + (Pa2b)/l2 = + (150)(3)(3)2/62
= + 112.5 kN.m
In beam ABFixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.mFixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m
6
7.2.2 Step II
Since the joints B, C and D were fixed artificially (to compute the the fixed-end moments), now the joints B, C and D are released and allowed to rotate. Due to the joint release, the joints rotate maintaining the continuous nature of the beam. Due to the joint release, the fixed end moments on either side of joints B, C and D act in the opposite direction now, and cause a net unbalanced moment to occur at the joint.
15 kN/m 10 kN/m
8 m 6 m 8 m
A B C DI I I
3 m
150 kN
Released moments -80.0 -112.5 +53.33 -53.33+112.5Net unbalanced moment
+32.5 -59.17 -53.33
7
7.2.3 Step III
These unbalanced moments act at the joints and modify the joint moments at B, C and D, according to their relative stiffnesses at the respective joints. The joint moments are distributed to either side of the joint B, C or D, according to their relative stiffnesses. These distributed moments also modify the moments at the opposite side of the beam span, viz., at joint A in span AB, at joints B and C in span BC and at joints C and D in span CD. This modification is dependent on the carry-over factor (which is equal to 0.5 in this case); when this carry over is made, the joints on opposite side are assumed to be fixed.
7.2.4 Step IVThe carry-over moment becomes the unbalanced moment at the joints to which they are carried over. Steps 3 and 4 are repeated till the carry-over or distributed moment becomes small.
7.2.5 Step VSum up all the moments at each of the joint to obtain the joint moments.
8
7.3 SOME BASIC DEFINITIONSIn order to understand the five steps mentioned in section 7.3, some words need to be defined and relevant derivations made.
7.3.1 Stiffness and Carry-over Factors
Stiffness = Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions
A
MAMB
A BA
RA RB
L
E, I – Member properties
A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Find A and MB.
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Using method of consistent deformations
L
A
A
MA
BL
fAA
A
B
1
EILM A
A 2
2
EILf AA 3
3
Applying the principle of consistent deformation,
L
MRfR A
AAAAA 2
30
EILM
EILR
EILM AAA
A 42
2
LEIM
khenceLEIM
A
AAA
4;4
Stiffness factor = k = 4EI/L
10
Considering moment MB,
MB + MA + RAL = 0MB = MA/2= (1/2)MA
Carry - over Factor = 1/2
7.3.2 Distribution FactorDistribution factor is the ratio according to which an externally applied unbalanced moment M at a joint is apportioned to the various members mating at the joint
+ ve moment M
A CB
D
A
D
B CMBA
MBC
MBD
At joint BM - MBA-MBC-MBD = 0
I1
L1I3
L3
I2
L2
M
11
i.e., M = MBA + MBC + MBD
MFDMK
KM
MFDMK
KM
Similarly
MFDMK
KKM
KM
KKKMKKK
LIE
LIE
LIE
BDBD
BD
BCBC
BC
BABA
BBABA
BDBCBAB
BBDBCBA
B
).(
).(
).(
444
3
33
2
22
1
11
12
7.3.3 Modified Stiffness FactorThe stiffness factor changes when the far end of the beam is simply-supported.
AMA
A B
RA RB
L
As per earlier equations for deformation, given in Mechanics of Solids text-books.
fixedAB
A
AAB
AA
K
LEI
LEIMK
EILM
)(43
4433
3
13
7.4 SOLUTION OF PROBLEMS -
7.4.1 Solve the previously given problem by the moment distribution method
7.4.1.1: Fixed end moments
mkNwlMM
mkNwlMM
mkNwlMM
DCCD
CBBC
BAAB
.333.5312
)8)(10(12
.5.1128
)6)(150(8
.8012
)8)(15(12
22
22
7.4.1.2 Stiffness Factors (Unmodified Stiffness)
EIEIK
EIEIEIK
EIEILEIKK
EIEILEIKK
DC
CD
CBBC
BAAB
5.08
4
5.084
84
667.06
))(4(45.0
8))(4(4
14
7.4.1.3 Distribution Factors
00.1
4284.0500.0667.0
500.0
5716.0500.0667.0
667.0
5716.0667.05.0
667.0
4284.0667.05.0
5.0
0.0)(5.0
5.0
DC
DCDC
CDCB
CDCD
CDCB
CBCB
BCBA
BCBC
BCBA
BABA
wallBA
BAAB
K
KDF
EIEIEI
KK
KDF
EIEIEI
KK
KDF
EIEIEI
KK
KDF
EIEIEI
KK
KDF
stiffnesswallEI
KK
KDF
15
Joint A B C DMember AB BA BC CB CD DC
Distribution Factors 0 0.4284 0.5716 0.5716 0.4284 1
Computed end moments -80 80 -112.5 112.5 -53.33 53.33Cycle 1
Distribution 13.923 18.577 -33.82 -25.35 -53.33
Carry-over moments 6.962 -16.91 9.289 -26.67 -12.35Cycle 2
Distribution 7.244 9.662 9.935 7.446 12.35
Carry-over moments 3.622 4.968 4.831 6.175 3.723Cycle 3
Distribution -2.128 -2.84 -6.129 -4.715 -3.723
Carry-over moments -1.064 -3.146 -1.42 -1.862 -2.358Cycle 4
Distribution 1.348 1.798 1.876 1.406 2.358
Carry-over moments 0.674 0.938 0.9 1.179 0.703Cycle 5
Distribution -0.402 -0.536 -1.187 -0.891 -0.703
Summed up -69.81 99.985 -99.99 96.613 -96.61 0moments
7.4.1.4 Moment Distribution Table
16
7.4.1.5 Computation of Shear Forces
Simply-supported 60 60 75 75 40 40
reaction
End reaction due to left hand FEM 8.726 -8.726 16.665 -16.67 12.079 -12.08
End reaction due to right hand FEM -12.5 12.498 -16.1 16.102 0 0
Summed-up 56.228 63.772 75.563 74.437 53.077 27.923 moments
8 m 3 m 3 m 8 mI I I
15 kN/m 10 kN/m150 kN
AB C
D
17
7.4.1.5 Shear Force and Bending Moment Diagrams
56.23
3.74 m
75.563
63.77
52.077
74.43727.923
2.792 m
-69.80698.297
35.08126.704
-96.613
31.693
Mmax=+38.985 kN.mMax=+ 35.59 kN.m
3.74 m84.92
-99.985
48.307
2.792 m
S. F. D.
B. M. D
18
Simply-supported bending moments at center of span
Mcenter in AB = (15)(8)2/8 = +120 kN.m
Mcenter in BC = (150)(6)/4 = +225 kN.m
Mcenter in AB = (10)(8)2/8 = +80 kN.m
Problem: Draw the shear force, bending moment and elastic diagram for the continuous beam shown in fig. by MDM due to settlement of 12 mm downward and a clockwise rotation of 0.002 rad at support D. Given data I= 200X10-5 m4 and E=200X106KN/m2
19
20
Solution
21
Settlement:
MCD MDC
* ∆ * 0.012 800 KN-m
22
Rotation:MDC *DC
* 0.012 533.33 KN-m
MCDMDC
KN-m
23
FIXED END MOMENT:
MDC =-800 KN-m MDC =+533.33 KN-m MDC=-266.67 KN-m MCD =-800 KN-m MCD =+266.66 KN-m MCD=-533.33 KN-m
24
Stiffness Factors:
KAB = KBA = KBC = KCB = KCD=KDC=
=
= 0.667EI
25
Distribution Factors:
DFAB==
DFBA==
DFBC==
DFCB==
DFCD==
DFDC==
26
Moment Distribution tableJoint A B C DMember AB BA BC CB CD DCDF 1 0.5 0.5 0.5 0.5 0Cycle-1 FEM 0 0 0 0 -533.33 -266.67
Balance 0 0 0 +266.66 +266.66 0Cycle-2 C.O.M 0 0 +133.33 0 0 +133.33
Balance 0 -66.66 -66.66 0 0 0Cycle-3 C.O.M -33.33 0 0 -33.33 0 0
Balance +33.33 0 0 +16.66 +16.66 0Cycle-4 C.O.M 0 +16.66 +8.33 0 0 +8.33
Balance 0 -12.495 -12.495 0 0 0Cycle-5 C.O.M -6.25 0 0 -6.25 0 0
Balance +6.25 0 0 +3.125 +3.125 0Cycle-6 C.O.M 0 +3.125 +1.56 0 0 +1.56
Balance 0 -2.34 -2.34 0 0 0Cycle-7 C.O.M -1.17 0 0 -1.17 0 0
Balance +1.17 0 0 0.585 +0.585 0Cycle-8 C.O.M 0 +0.585 0.3 0 0 +0.3
Balance 0 - 0.44 -0.44 0 0 0∑M 0 -61.57 +61.58 +246.28 -246.3 -123.15
27
Reaction Finding: Consider member ABMA = 0 (+)=> -RB1*6 – 61.583 + 0 = 0=> RB1 = -10.26 KN
RB1 = 10.26 KN(↓)
∑Fy = 0(↑+)=> RA + RB1 = 0 RA = 10.26 KN
28
Consider member BC
∑MB = 0 (+)=> -RC1*6 + 246.28 + 61.562 = 0
RC1 = 51.307 KN ∑Fy = 0(↑+)=> RC1 + RB2 = 0=> RB2 = -51.307 KN
RB2 = 51.307 KN(↓)
29
Consider member CD
∑MC = 0 (+)=> -RD*6 -123.145 -246.28 = 0=> RD = -61.57 KN
RD = 61.57 KN(↓) ∑Fy = 0(↑+)=> RC2 + RD = 0RC2 = 61.57 KN
30
Actual Reaction
RA = 10.26 KNRB = RB1 + RB2 = -10.26 -51.307 = 61.567 KN(↓)RC = RC1 + RC2 = 51.307 + 61.57 = 112.87 KNRD = 61.57 KN(↓)
31
SFD, BMD & Elastic Curve
32
Analyzing the rigid frame from figure using moment distribution method without sidesway.
32
D
10k
CB
I I
A6' 10'
10'
Solution:F.E.M due to the applied load:MBD = +(10×6) = +60 k-ft
33
Stiffness:
KAB = KBA = (4EI/L)AB = (4EI/L)BA
= (4 × E × 3I)/10 = 12EI / 10 = 1.2EIKBC = KCB = (4EI/L)BC = (4EI/L)CB
= (4 × E × I)/10 = 0.4EIKBD = KDB = O
Distribution Factors:
DFAB = KAB/(KAB + Kwall) = 1.2KI/(1.2EI + α) = 0DFBA = KBA/(KBA + KBC + KBD) =1.2EI/(1.2EI + 0.4EI + 0) = 0.75DFCB = KCB/(KCB + Kwall) = 0DFBC = KBC/(KBC + KBA + KBD) = 0.4EI/(0.4EI + 1.2EI + 0) = 0.25DFBD = DFDB = 0
34
Joint A B CMember AB BA BC BD CBStiffness 1.2EI 1.2EI 0.4EI 0 0.4EI
DF 0 0.75 0.25 0 0FEM 0 0 0 +60 0
Balance 0 - 45 - 15 0 0CO - 22.5 0 0 0 - 7.5
Balance 0 0 0 0 0∑ - 22.5 - 45 - 15 + 60 - 7.5
35
Moment Distribution Table
36
10k
60
45
15
22.5
CDB
A
10'
10'6'
7.5
For DB Span: RB1 = 10K
37
10KB
D60
RB1 = 10k6'
38
For AB Span:
∑MB = 0 ( +)Þ RA × 10 – 45 – 22.5 = 0... RA = 6.75 k
∑ Fx = 0 ( +)Þ RB2 – RA = 0... RBh = 6.75 K
10'
45
22.5
RB2 = 6.75k
RA = 6.75k
B
A
39
For span BC:
∑MB = 0 ( +)Þ RC × 10 – 7.5 – 15 = 0... RC = 2.25 k
∑FX = 0 ( +)Þ RB2 – RC = 0... RB2 = 2.25 k
B C15
7.5
10'RB3 = 2.25k RC = 2.25k
RA = 6.75 kRBh = 6.75 kRB = 10k + 2.25 k = 12.25kRC = 2.25 kRD = -10k
40
2.25
BD
106.75
6.75
2.25
C
A
SFD
41
45
6015
22.5
7.5
A
B CD+
+
-
-
BMD
MAB = - 22.5 MBA = - 45MBD = +60MBC = -15MCB = - 7.5
42
Some Definitions of MDM:• Stiffness Factor: is defined as the moment at the near
end to cause a unit rotation at the near end when the far end is fixed. That means how many load required for one unit deflection.
• Carry-over factor: is defined as the ratio of the moment at the fixed far end to the moment at the rotation near end.(the number + 0.5 is the carry over factor)
• Distribution factor: which may be defined as the fractions by which the total unbalanced at the joint is to be distributed to the member ends meeting at the joint.
43
Condition of sidesway:
Unsymmetrical 2-D frame in both direction Unsymmetrical cross sectional properties Unsymmetrical loading condition Unsymmetrical support settlement and rotation Symmetrical frame with symmetrical cross-sectional
properties and symmetrical vertical loading but joint translation due to lateral loading
44
Steps of Solution:
• Step-01:Find fixed end moment due to applied load• Step-02:Find fixed end moment due to sidesway• Step-03:Find stiffness factors • Step-04:Find distribution factors • Step-05: Table of finding moment due to applied load and sidesway • Step-06: Find the support reaction• Step-07: Calculate the total moment using factor (K) • Step-08: Draw SFD,BMD and Elastic Curve
45
PROBLEM: Analyze the rigid frame shown in figure-01 by the Moment Distribution Method(MDM). Draw shear and Moment diagrams. Sketch the deformed structure.
24k
2I
D
B C
E
F
A
20’ 16’
8’2I
2I3I3I
16’8’8’
Fig-01
Step- 01: Fixed End Moment due to Applied load
46
2I
D
B C
E
F
A
20’16’
8’
2I2I
3I3I
16’8’8’
24k
MAB MBA
Fig-02
Step -02: Fixed End Moment due to Sides ways
47
∆
MAD
D
16 ´
B
E
MEB
MBE
8 '
C ∆
MCF
MDA
MFC
∆
F
A
20 ´
2I 2I
2I
Fig-03
48
Step-3: Stiffness Factor
49
Step-4:Distribution Factors
50
Step-05:Moment Distribution Table(Applied Load case)
JOINT D A B C F E
Member DA AD AB BA BE BC CB CF FC EB
D.F 0 0.35 0.65 0.375 0.25 0.375 0.43 0.57 0 0
F.E.M 0 0 -48 48 0 0 0 0 0 0
Balance 0 16.8 31.2 -18 -12 -18 0 0 0 0
C.O 8.4 0 -9 15.6 0 0 -9 0 0 -6
Balance 0 3.15 5.85 -5.85 -3.9 -5.85 3.87 5.13 0 0
C.O 1.57 0 -2.925 2.95 0 1.935 -2.925 0 2.565 -1.95
Balance 0 1.02 1.90 -1.82 -1.215 -1.823 1.258 1.667 0 0
∑M 9.97 20.97 -20.97 40.85 -17.11 -23.738 -6.797 6.797 2.57 -7.95
Step-05:Moment Distribution Table(Sides way case)
51
JOINT D A B C F EMember DA AD AB BA BE BC CB CF FC EB
D.F 0 0.35 0.65 0.375 0.25 0.375 0.43 0.57 0 0
F.E.M -30 -30 0 0 -46.87 0 0 -187.5 -187.5 -46.87
Balance 0 10.5 19.5 17.58 11.72 17.58 80.62 106.87 0 0
C.O 5.25 0 8.79 9.75 0 40.31 8.79 0 53.44 5.86
Balance 0 -3.08 -5.71 -18.77 -12.52 -18.77 -3.78 -5.01 0 0
C.O -1.54 0 -9.38 -2.855 0 -1.89 -9.385 0 -2.5 -6.26
Balance O 3.28 6.1 1.78 1.19 1.78 4.04 5.35 0 0
∑M -26.29 -19.3 19.3 7.485 46.48 39.01 80.29 -80.29 -136.56 -47.27
Step-6: Now, HD+HE+HF=0--------(i)
So, HD=−(MAD+MDA)/20= −[M’AD+KM”AD+M’DA+KM”DA]/20= −[20.97+K(−19.3)+9.975+K(-26.29)]/20=2.28K −1.55HE=−(MEB+MBE)/16= −[M’EB+KM”EB+M’BE+KM”BE]/16= −[− 7.95+K(−47.275)+(−17.115)+K(-46.485)]/16=1.567+5.86K
52
A
DHD
MAD
MDA
20’
B
E
16’
MBE
MEBHE
53
HF=−(MCF+MFC)/8= −[M’CF+KM”CF+M’FC+KM”FC]/20= −[6.797+K(−80.29)+2.57+K(−136.56)]/8=27.11K −1.17
Putting these value in equation (i)2.28K −1.55+1.567+5.86K+27.11K −1.17=0Þ−1.153+35.25K=0ÞK=0.0327
8’
MCF
MFC
HF
F
C
54
Step-07:Moment Calculation
By using value of kMDA=9.975+(0.0327× −26.29)=9.11 K.ftMAD=20.97+(0.0327× −19.3)=20.34 K.ftMAB= − 20.97+(0.0327× 19.3)= − 20.34 K.ftMBA= 40.852+(0.0327× 7.485)= 41.1K.ftMBE= −17.115+(0.0325 ×-46.485)=-18.64 K.ftMEB= − 7.95+(0.0327× -47.275)= − 9.5 K.ftMBC= − 23.738+(0.0327× 39.01)= − 22.46 K.ftMCB= − 6.797+(0.0327× 80.29)= − 4.17 K.ftMCF= 6.797+(0.0327× − 80.29)= 4.17 K.ftMFC= 2.57+(0.0327× − 136.56)= -1.9 K.ft
55
24k
2I
D
B C
E
F
A
20’ 16’8’
2I2I
3I3I
16’8’8’
20.34
41.1
22.46
4.17
4.17
1.9
9.5
18.64
9.11
20.34
Step-08:
56
Reaction Calculation:
For member AD:∑MA=0 +Þ20.34+9.11-HD×20=0ÞHD= 1.47 ( )∑FX =0 ( + ) Þ-HA+HD=0ÞHA=1.47 ( )
HD
20ft
9.11
20.34
HA
D
A
57
Reaction Calculation:
For member BE:∑MB=0 +Þ-18.64-9.5+HE×16=0ÞHE= 1.76 ( )∑FX =0 ( + ) Þ-HE+HB=0ÞHB=1.76 ( )
HD
16ft
9.5
18.64
HA
E
B
58
Reaction Calculation:For member CF:∑MC=0 +Þ4.17-1.9+HF×8=0ÞHF= -0.28 = 0.28 ( )∑FX =0 ( + ) ÞHC-HF=0ÞHC-(-0.28)=0ÞHC=-0.28 SO, HC= 0.28( )
HF
8ft
1.9
4.17
HC
F
C
59
Reaction Calculation:For member AB:∑MA=0 +Þ -20.34+41.1+(24×8)+VB×16=0Þ VB= -13.3 Þ VB1= 13.3 ( )
∑FY =0 ( + ) Þ -22.46-4.17+VA+13.3=0SO,VA = 10.7 ( )
24VB
VA
20.34 41.1
A B
8ft8ft
60
Reaction Calculation:For member BC:∑MB=0 +Þ -22.46 – 4.17+VC×16=0ÞVC= 1.66 ( )
∑FY =0 ( + ) ÞVB2 – VC= 0SO,VB2 = 1.66( )
B
VC
VB2
22.46 4.17C
16ft
61
1.47
−
+
−
+
+
E
CBA
D
F0.28
1.66
13.3
10.7
−
S.F.D
1.76
Step-08:
62
9.11
8`
CBA
D
F1.9
18.6
20.3
20.3
B.M.D
41.1
4.1
22.46
4.14
65.26
+
+
+
+
-
--
-
-
∆
9.5
Step-08:
63
E
CBA
D
F
Elastic Curve
∆∆ ∆
Step-08:
Problem
64
Analyze the rigid frame shown in the figure by the Moment Distribution Method. Draw the shear and moment diagram and also sketch the deform structure.
65
F.E.M. due to the applied load:F.E.M. due to the applied load will be zero because all load will be applied at the joint.F.E.M. due to the sides way:At the direction ABMAC = MCA = MBD = MDB = - * ∆ = - * 25600 = -100 k
Continued…
Continued…
66
At the direction CDMAC = MCA = MBD = MDB = * ∆ = * 25600 = 100 kMCE = MEC = - * ∆ = - * 25600 = -200 kMDF = MFD = = - * ∆ = - * 25600 = -400 k
Stiffness Factor
67
KAC = KCA = KBD = KDB = = = 0.5EI
KAB = KBA = = = 1EI
KCD = KDC= = = 1.67EI
KCE = KEC = = = 1EI
KDF = KFD = = = 1EI
Distribution Factor
68
DFAB = KAB/(KAB + KAC) = = 0.67
DFBA = KBA/(KBA+KBD) = = 0.67
DFCD = KCD/(KCD+KCA+KCE) = = 0.53
DFDC = KDC/(KDC+KDB+KDF) = = 0.53
DFAC = KAC/(KAC+KAB) = = 0.33
DFCA = KCA/(KCA+KCE+KCD) = = 0.16
Continued…
69
DFBD = KBD/(KBD+KBA) = = 0.33
DFDB = KDB/(KDB+KDF+KDC) = = 0.16
DFCE = KCE/(KCE+KCA+KCD) = = 0.32
DFEC = KEC/(KEC+Kwall) = = 0
DFDF = KDF/(KDF+KDB+KDC) = = 0.32
DFFD = KFD/(KFD+Kwall) = = 0
Table - 1
70
Table - 2
71
Continued…
72
Here,∑MA = 0 (+)Þ - HC*16 – MAC - MCA = 0Þ HC = - (MAC + MCA )/16Similarly,HD = - (MBD + MDB )/16
Now, HC + HD = -3 Þ -{(MAC + MCA)/ 16} – {(MBD + MDB)/ 16} = -3 Þ -MAC- MCA- MBD – MDB = -48 ---------------(1)
MAC
A
C
MCA
3 kip
HC
16
73
Here,∑MC = 0 (+)Þ - HE*16 – MCE - MEC = 0Þ HE = - (MCE + MEC )/16Similarly,HF = - (MDF + MFD )/8
Now,
HE + HF = -9
Þ -{(MCE + MEC)/ 16} – {(MDF + MFD)/ 8} = -9
Þ -MCE- MEC- 2MDF – 2MFD = -144 ---------------(2)
MCE
C
E
MEC
6 kip
HE
Continued…
16
74
From eqn (1) we get,
-(MAC + MCA + MBD + MDB) = -48
Þ K1MAC + K2MAC + K1MCA + K2MCA + K1MBD + K2MBD +
K1MDB + K2MDB = 48
Þ K1 (MAC + MCA + MBD + MDB) + K2 (MAC + MCA + MBD +
MDB) = 48
Þ K1 (-70.98-77.4-70.98-77.4) + K2 (78.93+93.4+90.73+133.92) = 48
Þ -296.76 K1+396.98 K2 = 48 ---------------(3)
Continued…
75
From eqn (2) we get,
-(MCE + MEC + 2MDF + 2MFD) = -48
Þ K1MCE + K2MCE + K1MEC + K2MEC + 2K1MDF + 2K2MDF +
2K1MFD + 2K2MFD = 144
Þ K1 (MCE + MEC+ 2MDF + 2MFD) + K2 (MCE + MEC + 2MDF +
2MFD) = 144
Þ K1 (22.7+10.75+45.28+21.4) + K2 (-188.64-194.1-604.8-736) = 144
Þ 100.13 K1 – 1723.54 K2 = 144 ---------------(4)
Continued…
76
Solving this two eqn we get
K1 = - 0.297K2 = - 0.101
Continued…
77
Now,MEC = (-0.297*10.75) + (-0.101* (-194.1)) = 16.41 MCE = (-0.297*22.7) + (-0.101* (-188.64)) = 12.31 MCD = (-0.297*55.49) + (-0.101*95.6) = -26.14 MCA = (-0.297*( -77.4)) + (-0.101*93.4) = 13.55MAC = (-0.297*( -70.98)) + (-0.101* 78.93) = 13.1MAB = (-0.297*71.1) + (-0.101*( -78.9)) = -13.1MBA = (-0.297*71.1) + (-0.101* (-90.73)) = -11.9MBD = (-0.297*(-70.98)) + (-0.101*90.73) = 11.9MDB = (-0.297* (-77.4)) + (-0.101*133.92) = 9.46MDC = (-0.297*55.45) + (-0.101*171.5) = -33.79MDF = (-0.297*22.64) + (-0.101* (-302.4)) = 23.82MFD = (-0.297*10.7) + (-0.101*( -368)) = 33.99
Continued…
Actual Moment Distribution
78
A B
C D
E
F
13.1
11.9
13.113.55
26.14
12.31
16.41
11.99.46
33.79
33.99
23.82
Reaction
We know,HC = - (MAC+MCA)/16 = - (13.1+13.5)/16 = - 1.66 kip = 1.66 kip (←) HD = - (MBD+MDB)/16 = - (11.9+9.46)/16 = - 1.34 kip = 1.34 kip (←)
HE = - (MCE+MEC)/16 = - (12.31+16.41)/16 = - 1.8 kip = 1.8 kip (←) HF = - (MDF+MFD)/8 = - (23.82+33.99)/8 = - 7.23 kip = 7.23 kip (←)
79
Continued…
80
Reaction at AB member∑MA = 0 (+)Þ-RB*12 -13.1 -11.95 = 0ÞRB = -2.1 kip = 2.1 kip (↓) ∑FY = 0 (↑+)RA -2.1 = 0RA = 2.1 kip (↑)
A B13.1
11.95
RA RB
Reaction at CD member∑MC = 0 (+)Þ-RD*12 -26.14 -33.79 = 0RD = -4.99 kip = 4.99 kip (↓) ∑FY = 0 (↑+)ÞRC -4.99 = 0RC = 4.99 kip (↑)
81
C D26.14
33.79
RCRD
Continued…
Actual reaction
82
HC = 1.66 kip (←)
HD = 1.34 kip (←)
HE = 1.8 kip (←)
HF = 7.23 kip (←)
RB = 2.1 kip (↓)
RA = 2.1 kip (↑)
RD = 4.99 kip (↓)
RC = 4.99 kip (↑)
SFD
83
BMD
84
Elastic Curve
85