1. 7. MOMENT DISTRIBUTION METHOD

2. 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
(Method developed by Prof. Hardy Cross in 1932)
The method solves for the joint moments in continuous beams and
rigid frames by successive approximation.
7.2 Statement of Basic Principles
Consider the continuous beam ABCD, subjected to the given loads,
as shown in Figure below. Assume that only rotation of joints occur
at B, C and D, and that no support displacements occur at B, C and
D. Due to the applied loads in spans AB, BC and CD, rotations occur at B, C and D.
150 kN
15 kN/m
10 kN/m
3 m A D B C I I I
8 m
6 m
8 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 I
The 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.
-80 kN.m
15 kN/m
10 kN/m
-80 kN.m
-112.5kN.m
kN.m
112.5 kN.m
53.33 kN.m
150 kN
3 m A B C B D
8 m C
8 m
6 m

5. In beam AB
Fixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.m
Fixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m
In beam BC
Fixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62
= kN.m
Fixed end moment at C = + (Pa2b)/l2 = + (150)(3)(3)2/62
= kN.m
In beam AB
Fixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = kN.m
Fixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = kN.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.
150 kN
15 kN/m
10 kN/m
3 m A D B C I I I
8 m
6 m
8 m
-53.33
Released moments
+112.5
-112.5
+53.33
Net unbalanced moment
-53.33
-59.17
+32.5

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 IV
The 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 V
Sum up all the moments at each of the joint to obtain the joint moments.

8. 7.3 SOME BASIC DEFINITIONS
In 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 MB
A clockwise moment MA is applied at A to produce a +ve bending in beam AB. Find A and MB. MA A B A
A RA RB L
E, I – Member properties

9. Stiffness factor = k = 4EI/L
Using method of consistent deformations MA A
fAA B B L L A A 1
Applying the principle of consistent deformation,
Stiffness factor = k = 4EI/L

10. A C B D Considering moment MB, 7.3.2 Distribution Factor
MB + MA + RAL = 0
MB = MA/2= (1/2)MA
Carry - over Factor = 1/2
7.3.2 Distribution Factor
Distribution 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 M B A
MBC C A
MBA C B I2 L2 I1 L1
MBD I3 L3
At joint B
M - MBA-MBC-MBD = 0 D D

11. i.e., M = MBA + MBC + MBD

12. 7.3.3 Modified Stiffness Factor
The stiffness factor changes when the far end of the beam is simply-supported. MA A A B L RA RB
As per earlier equations for deformation, given in Mechanics of Solids text-books.

13. 7.4 SOLUTION OF PROBLEMS - 7.4.1.1: Fixed end moments
7.4.1 Solve the previously given problem by the moment distribution method
: Fixed end moments
Stiffness Factors (Unmodified Stiffness)

14. Distribution Factors

15. 7.4.1.4 Moment Distribution Table

16. 7.4.1.5 Computation of Shear Forces
10 kN/m
15 kN/m
150 kN B C A D I I I
8 m
3 m
3 m
8 m

17. S. F. D. B. M. D 7.4.1.5 Shear Force and Bending Moment Diagrams
52.077
75.563
2.792 m
56.23
27.923
3.74 m
74.437
63.77
S. F. D.
Mmax= kN.m
Max= kN.m
31.693
35.08
48.307
3.74 m
84.92
98.297
2.792 m
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

19. 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 rad at support D. Given data I= 200X10-5 m4 and E=200X106KN/m2

20. Solution

21. Settlement:
MCD= MDC = −𝟔𝐄𝐈 𝐋 𝟐 * ∆ = −(𝟔∗𝟐𝟎𝟎∗ 𝟏𝟎 𝟔 ∗𝟐𝟎𝟎∗ 𝟏𝟎 −𝟓 ) 𝟔 𝟐 * =−800 KN-m

22. Rotation:
MDC = 𝟒𝐄𝐈 𝐋 *𝛉DC = (𝟒∗𝟐𝟎𝟎∗ 𝟏𝟎 𝟔 ∗𝟐𝟎𝟎∗ 𝟏𝟎 −𝟓 ) 𝟔 * = KN-m MCD = 𝟏 𝟐 ∗MDC =𝟎.𝟓∗𝟓𝟑𝟑.𝟑𝟑=𝟐𝟔𝟔.𝟔𝟔KN-m

23. FIXED END MOMENT:
MDC =-800 KN-m MDC = KN-m  MDC= KN-m MCD =-800 KN-m MCD = KN-m  MCD= KN-m

24. Stiffness Factors:
KAB = KBA = KBC = KCB = KCD=KDC= 𝟒𝑬𝑰 𝑳 = 𝟒𝐄𝐈 𝟔 = 0.667EI

25. Distribution Factors:
DFAB= 𝑲 𝑨𝑩 𝑲 𝑨𝑩 = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝑬𝑰 =𝟏 DFBA= 𝑲 𝑩𝑨 𝑲 𝑩𝑨 +𝑲 𝑩𝑪 = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝐄𝐈+𝟎.𝟔𝟔𝟕𝑬𝑰 =𝟎.𝟓 DFBC= 𝑲 𝑩𝑪 𝑲 𝑩𝑪 +𝑲 𝑩𝑨 = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝐄𝐈+𝟎.𝟔𝟔𝟕𝑬𝑰 =𝟎.𝟓 DFCB= 𝑲 𝑪𝑩 𝑲 𝑪𝑩 +𝑲 𝑪𝑫 = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝐄𝐈+𝟎.𝟔𝟔𝟕𝑬𝑰 =𝟎.𝟓 DFCD= 𝑲 𝑪𝑫 𝑲 𝑪𝑫 +𝑲 𝑪𝑩 = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝐄𝐈+𝟎.𝟔𝟔𝟕𝑬𝑰 =𝟎.𝟓 DFDC= 𝑲 𝑫𝑪 𝑲 𝑫𝑪 +𝐖𝐀𝐋𝐋(𝛂) = 𝟎.𝟔𝟔𝟕𝑬𝑰 𝟎.𝟔𝟔𝟕𝐄𝐈+∞ =𝟎

26. Moment Distribution table
Joint A B C D
Member AB BA BC CB CD DC DF 1
0.5
Cycle-1
FEM
Balance
Cycle-2
C.O.M
-66.66
Cycle-3
-33.33
+33.33
+16.66
Cycle-4
+8.33
Cycle-5
-6.25
+6.25
+3.125
Cycle-6
+1.56
-2.34
Cycle-7
-1.17
+1.17
0.585
+0.585
Cycle-8
0.3
+0.3
- 0.44
-0.44 ∑M
-61.57
+61.58
-246.3

27. Reaction Finding: Consider member AB
MA = 0 (+) => -RB1*6 – = 0
=> RB1 = KN
 RB1 = KN(↓)   ∑Fy = 0(↑+) => RA + RB1 = 0  RA = KN  

28. Consider member BC  RB2 = 51.307 KN(↓)
∑MB = 0 (+) => -RC1* = 0 RC1 = KN   ∑Fy = 0(↑+) => RC1 + RB2 = 0 => RB2 = KN
 RB2 = KN(↓)

29. Consider member CD
∑MC = 0 (+) => -RD* = 0 => RD = KN
RD = KN(↓)   ∑Fy = 0(↑+) => RC2 + RD = 0 RC2 = KN

30. Actual Reaction
RA = KN RB = RB1 + RB2 = = KN(↓) RC = RC1 + RC2 = = KN RD = KN(↓)

31. SFD, BMD & Elastic Curve

32. Analyzing the rigid frame from figure using moment distribution method without sidesway.
10k B D C I I
10' A 6'
10'
Solution:
F.E.M due to the applied load:
MBD = +(10×6) = +60 k-ft 32

33. Stiffness:
KAB = KBA = (4EI/L)AB = (4EI/L)BA
= (4 × E × 3I)/10
= 12EI / 10
= 1.2EI
KBC = KCB = (4EI/L)BC = (4EI/L)CB
= (4 × E × I)/10
= 0.4EI
KBD = KDB = O

34. Distribution Factors: DFAB = KAB/(KAB + Kwall) = 1. 2KI/(1
Distribution Factors: DFAB = KAB/(KAB + Kwall) = 1.2KI/(1.2EI + α) = 0 DFBA = KBA/(KBA + KBC + KBD) =1.2EI/(1.2EI + 0.4EI + 0) = 0.75 DFCB = KCB/(KCB + Kwall) DFBC = KBC/(KBC + KBA + KBD) = 0.4EI/(0.4EI + 1.2EI + 0) = 0.25 DFBD = DFDB = 0

35. Joint A B C Member AB BA BC BD CB Stiffness 1.2EI 0.4EI DF 0.75 0.25
Moment Distribution Table
Joint A B C
Member AB BA BC BD CB
Stiffness
1.2EI
0.4EI DF
0.75
0.25
FEM
+60
Balance
- 45
- 15 CO
- 22.5
- 7.5 ∑
+ 60

36. 10k B
7.5 D C 15 60 45
10'
22.5 A 6'
10'

37. 10K B D 60
RB1 = 10k 6'
For DB Span: RB1 = 10K

38. For AB Span: ∑MB = 0 ( +) RA × 10 – 45 – 22.5 = 0 ... RA = 6.75 k
∑ Fx = 0 ( )
RB2 – RA = 0
... RBh = 6.75 K
RB2 = 6.75k B 45
10'
22.5 A
RA = 6.75k

39. For span BC: ∑MB = 0 ( +) RC × 10 – 7.5 – 15 = 0 ... RC = 2.25 k
∑FX = 0 ( )
RB2 – RC = 0
... RB2 = 2.25 k
7.5 B C 15
RC = 2.25k
RB3 = 2.25k
10'

40. RA = 6. 75 k RBh = 6. 75 k RB = 10k + 2. 25 k = 12. 25k RC = 2
RA = 6.75 k RBh = 6.75 k RB = 10k k = 12.25k RC = 2.25 k RD = -10k
2.25
2.25 B D C
6.75 10
6.75 A
SFD

41. - - MAB = - 22.5 MBA = - 45 MBD = +60 MBC = -15 MCB = - 7.5 7.5 + 45 B
BMD

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 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 B C A 3I 3I 8’ 2I 2I F
16’
20’ 2I E D 8’ 8’
16’
Fig-01

46. Step- 01: Fixed End Moment due to Applied load
24k B C A 3I 3I
MAB
MBA 8’ 2I 2I
16’
20’ 2I F E D 8’ 8’
16’
Fig-02

47. Step -02: Fixed End Moment due to Sides ways∆ ∆ B A
MAD
MBE ∆ C 2I
MCF 2I
16 ´
20 ´
8 ' 2I
MFC
MEB F E
MDA D
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
0.35
0.65
0.375
0.25
0.43
0.57
F.E.M
-48 48
Balance
16.8
31.2
-18
-12
C.O
8.4 -9
15.6 -6
3.15
5.85
-5.85
-3.9
3.87
5.13
1.57
-2.925
2.95
1.935
2.565
-1.95
1.02
1.90
-1.82
-1.215
-1.823
1.258
1.667 ∑M
9.97
20.97
-20.97
40.85
-17.11
-6.797
6.797
2.57
-7.95

51. Step-05:Moment Distribution Table(Sides way case)
JOINT D A B C F E
Member DA AD AB BA BE BC CB CF FC EB
D.F
0.35
0.65
0.375
0.25
0.43
0.57
F.E.M
-30
-46.87
-187.5
Balance
10.5
19.5
17.58
11.72
80.62
106.87
C.O
5.25
8.79
9.75
40.31
53.44
5.86
-3.08
-5.71
-18.77
-12.52
-3.78
-5.01
-1.54
-9.38
-2.855
-1.89
-9.385
-2.5
-6.26 O
3.28
6.1
1.78
1.19
4.04
5.35 ∑M
-26.29
-19.3
19.3
7.485
46.48
39.01
80.29
-80.29
-47.27

52. Step-6: Now, HD+HE+HF=0--------(i)A
MAD
20’
So, HD=−(MAD+MDA)/20 = −[M’AD+KM”AD+M’DA+KM”DA]/20 = −[20.97+K(−19.3) K(-26.29)]/20 =2.28K −1.55 HE=−(MEB+MBE)/16 = −[M’EB+KM”EB+M’BE+KM”BE]/16 = −[− 7.95+K(−47.275)+(−17.115)+K( )]/16 = K
MDA D HD B
MBE
16’ HE
MEB E

53. = −[M’CF+KM”CF+M’FC+KM”FC]/20 = −[6.797+K(−80.29)+2.57+K(−136.56)]/8
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 − K+27.11K −1.17=0
− K=0
K=0.0327 C
MCF F 8’
MFC HF

54. Step-07:Moment Calculation
By using value of k MDA=9.975+(0.0327× −26.29)=9.11 K.ft MAD=20.97+(0.0327× −19.3)=20.34 K.ft MAB= − (0.0327× 19.3)= − K.ft MBA= (0.0327× 7.485)= 41.1K.ft MBE= − ( × )= K.ft MEB= − 7.95+(0.0327× )= − 9.5 K.ft MBC= − (0.0327× 39.01)= − K.ft MCB= − (0.0327× 80.29)= − 4.17 K.ft MCF= (0.0327× − 80.29)= 4.17 K.ft MFC= 2.57+(0.0327× − )= -1.9 K.ft

55. Step-08:
24k
22.46
4.17
20.34
41.1 B C
4.17 A 3I 3I
20.34
18.64 8’ 2I 2I
16’
20’
1.9 2I
9.5 F
9.11 E 8’ 8’
16’ D

56. Reaction Calculation:
For member AD:
∑MA=0 +
HD×20=0
HD= ( )
∑FX =0 ( )
-HA+HD=0
HA=1.47 ( ) HA
20.34
20ft
9.11 HD D

57. Reaction Calculation:
For member BE:
∑MB=0 +
HE×16=0
HE= ( )
∑FX =0 ( )
-HE+HB=0
HB=1.76 ( ) B HA
18.64
16ft
9.5 HD E

58. Reaction Calculation:
For member CF:
∑MC=0 +
HF×8=0
HF= = ( )
∑FX =0 ( )
HC-HF=0
HC-(-0.28)=0
HC=-0.28
SO, HC= 0.28( ) HC C
4.17
8ft
1.9 HF F

59. Reaction Calculation:VB
For member AB:
∑MA=0 +
(24×8)+VB×16=0
VB=
VB1= ( )
∑FY =0 ( )
VA+13.3=0
SO,VA = ( ) 24
41.1
20.34 B A VA
8ft
8ft

60. Reaction Calculation:
For member BC:
∑MB=0 +
– 4.17+VC×16=0
VC= ( )
∑FY =0 ( )
VB2 – VC= 0
SO,VB2 = 1.66( ) VC
4.17
22.46 C B
16ft
VB2

61. Step-08:
10.7 +
1.66 C B + A −
13.3 − +
0.28 F −
1.76 E
1.47
S.F.D D

62. Step-08:
65.26
4.1 ∆ +
18.6 C B + A
4.14 - - -
20.3
22.46 -
20.3 8` -
41.1 +
1.9 F
9.5 +
B.M.D
9.11 D

63. Step-08: ∆ ∆ ∆ C B A F E
Elastic Curve D

64. Problem
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. Continued…
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 AB MAC = MCA = MBD = MDB = - 6EI L 2 * ∆ = - 6E∗2I L 2 * = -100 k

66. Continued…
At the direction CD MAC = MCA = MBD = MDB = 6EI L 2 * ∆ = 6E∗2I L 2 * = 100 k MCE = MEC = - 6EI L 2 * ∆ = - 6E∗4I L 2 * = -200 k MDF = MFD = = - 6EI L 2 * ∆ = - 6E∗2I L 2 * = -400 k

67. Stiffness Factor
KAC = KCA = KBD = KDB = 4EI L = 4∗E∗2I 16 = 0.5EI KAB = KBA = 4EI L = 4∗E∗3I 12 = 1EI KCD = KDC= 4EI L = 4∗E∗5I 12 = 1.67EI KCE = KEC = 4EI L = 4∗E∗4I 16 = 1EI KDF = KFD = 4EI L = 4∗E∗2I 8 = 1EI

68. Distribution Factor
DFAB = KAB/(KAB + KAC) = 1EI 1EI+0.5EI = 0.67 DFBA = KBA/(KBA+KBD) = 1EI 1EI+0.5EI = 0.67 DFCD = KCD/(KCD+KCA+KCE) = 1.67EI 1.67EI+0.5EI+1EI = 0.53 DFDC = KDC/(KDC+KDB+KDF) = 1.67EI 1.67EI+0.5EI+1EI = 0.53 DFAC = KAC/(KAC+KAB) = 0.5EI 0.5EI+1EI = 0.33 DFCA = KCA/(KCA+KCE+KCD) = 0.5EI 0.5EI+1EI+1.67EI = 0.16

69. Continued…
DFBD = KBD/(KBD+KBA) = 0.5EI 0.5EI+1EI = 0.33 DFDB = KDB/(KDB+KDF+KDC) = 0.5EI 0.5EI+1EI+1.67EI = 0.16 DFCE = KCE/(KCE+KCA+KCD) = 1EI 1EI+0.5EI+1.67EI = 0.32 DFEC = KEC/(KEC+Kwall) = 1EI 1EI+∞ = 0 DFDF = KDF/(KDF+KDB+KDC) = 1EI 1EI+0.5EI+1.67EI = 0.32 DFFD = KFD/(KFD+Kwall) = 1EI 1EI+∞ = 0

70. Table - 1

71. Table - 2

72. Continued…
Here, ∑MA = 0 (+)  - HC*16 – MAC - MCA = 0  HC = - (MAC + MCA )/16 Similarly, HD = - (MBD + MDB )/16 Now, HC + HD = -3  -{(MAC + MCA)/ 16} – {(MBD + MDB)/ 16} = -3  -MAC- MCA- MBD – MDB = (1)
3 kip A
MAC
16
MCA HC C

73. Continued…
Here, ∑MC = 0 (+)  - HE*16 – MCE - MEC = 0  HE = - (MCE + MEC )/16 Similarly, HF = - (MDF + MFD )/8 Now, HE + HF = -9  -{(MCE + MEC)/ 16} – {(MDF + MFD)/ 8} = -9  -MCE- MEC- 2MDF – 2MFD = (2)
6 kip C
MCE
16
MEC HE E

74. Continued…
From eqn (1) we get, -(MAC + MCA + MBD + MDB) = -48  K1MAC + K2MAC + K1MCA + K2MCA + K1MBD + K2MBD + K1MDB + K2MDB = 48  K1 (MAC + MCA + MBD + MDB) + K2 (MAC + MCA + MBD + MDB) = 48  K1 ( ) + K2 ( ) = 48  K K2 = (3)

75. Continued…
From eqn (2) we get, -(MCE + MEC + 2MDF + 2MFD) = -48  K1MCE + K2MCE + K1MEC + K2MEC + 2K1MDF + 2K2MDF + 2K1MFD + 2K2MFD = 144  K1 (MCE + MEC+ 2MDF + 2MFD) + K2 (MCE + MEC + 2MDF + 2MFD) = 144  K1 ( ) + K2 ( ) = 144  K1 – K2 = (4)

76. Solving this two eqn we get K1 = - 0.297 K2 = - 0.101
Continued…
Solving this two eqn we get K1 = K2 =

77. Continued…
Now, MEC = (-0.297*10.75) + (-0.101* (-194.1)) = MCE = (-0.297*22.7) + (-0.101* ( )) = MCD = (-0.297*55.49) + (-0.101*95.6) = MCA = (-0.297*( -77.4)) + (-0.101*93.4) = MAC = (-0.297*( )) + (-0.101* 78.93) = 13.1 MAB = (-0.297*71.1) + (-0.101*( -78.9)) = MBA = (-0.297*71.1) + (-0.101* (-90.73)) = MBD = (-0.297*(-70.98)) + (-0.101*90.73) = 11.9 MDB = (-0.297* (-77.4)) + (-0.101*133.92) = 9.46 MDC = (-0.297*55.45) + (-0.101*171.5) = MDF = (-0.297*22.64) + (-0.101* (-302.4)) = MFD = (-0.297*10.7) + (-0.101*( -368)) = 33.99

78. Actual Moment Distribution
11.9 B A
13.1
11.9
13.1
13.55
9.46
33.79 C D
26.14
23.82
12.31
33.99
16.41 F E

79. Reaction
We know, HC = - (MAC+MCA)/16 = - ( )/16 = kip = 1.66 kip (←) HD = - (MBD+MDB)/16 = - ( )/16 = kip = 1.34 kip (←)
HE = - (MCE+MEC)/16 = - ( )/16 = kip = 1.8 kip (←) HF = - (MDF+MFD)/8 = - ( )/8 = kip = 7.23 kip (←)

80. Continued…
Reaction at AB member ∑MA = 0 (+) -RB* = 0 RB = -2.1 kip = 2.1 kip (↓) ∑FY = 0 (↑+) RA -2.1 = 0 RA = 2.1 kip (↑)
11.95 A
13.1 B RA RB

81. Continued… Reaction at CD member ∑MC = 0 (+)
-RD* = 0
RD = kip
= 4.99 kip (↓)
∑FY = 0 (↑+)
RC = 0
RC = 4.99 kip (↑)
33.79 C
26.14 D RD RC

82. Actual reaction
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 (↑)

83. SFD

84. BMD

85. Elastic Curve