7. MOMENT DISTRIBUTION METHOD

1

7. MOMENT DISTRIBUTION METHOD

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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

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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

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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

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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.

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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.

9

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

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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

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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

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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

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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

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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

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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

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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(↓)

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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

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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