Moment-Distribution Method

Structural Analysis By

Aslam Kassimali

Theory of Structures-II

M Shahid MehmoodDepartment of Civil Engineering

Swedish College of Engineering & Technology, Wah Cantt

FRAMES

Analysis of Frames Without Sidesway

• The procedure for the analysis of frames without sidesway issimilar to that for the analysis of continuous beam.

• Unlike the continuous beams, more than two members may beconnected to a joint of a frame.

2

Example 1

• Determine the member end moments and reactions for the frameshown by the moment-distribution method.

3

A B

DC E

2 k/ft

40 k

10 ft

10 ft

I = 8

00 in

4

I = 8

00 in

4

I = 1,600 in4 I = 1,600 in4

30 ft 30 ft

E = 29,000 ksi

Solution

1.Distribution Factors

• Distribution Factors at Joint C,

4

571.0

30

1600

20

800

30

1600

429.0

30

1600

20

800

20

800

CD

CA

DF

DF

Checks 1571.0429.0 CDCA DFDF

• Distribution Factors at Joint D,

5

3.0

30

1600

4

3

30

1600

20

800

30

1600

4

3

4.0

30

1600

4

3

30

1600

20

800

30

1600

3.0

30

1600

4

3

30

1600

20

800

20

800

DE

DC

DB

DF

DF

DF

Checks 14.03.02 DCDEDB DFDFDF

• Distribution Factors at Joint E,

2.Fixed-End Moments (FEMs)

3.Moment Distribution

4.Final Moments

6

1EDDF

ft-k 150

ft-k 150

0

ft-k 100

ft-k 100

EDDC

DECD

DBBD

CA

AC

FEMFEM

FEMFEM

FEMFEM

FEM

FEM

7

Distribution Factors

1.Fixed-end Moments

ACMember Ends

2.Balance Joints

3.Carryover

4.Balance Joints

5.Carryover

7.Carryover

8.Balance Joints

9.Carryover

-10.7

10.Balance Joints

13.Final Moments +92.1 -115.9 +115.9 -186.4 -19.4 +205.6

CA CD DC DB DE ED BD

0.429 0.571 0.4 0.3 0.3 1

+100 -100 +150 -150 +150 -150

-21.4 -28.6 +150

-14.3 +75

-24.3 -18.2 -18.2

-12.2

6.Balance Joints +5.2 +7

+2.6 +3.5

-1.4 -1.1 -1.1

-0.7

+0.4+0.3

11. Carryover +0.2 +0.2

-0.1 -0.1 -0.112.Balance Joints

0

-9.1

-0.6

-9.7

Carryover

8

A

C

2 k/ft

40 k

92.1

115.9

115.9C

186.4D

B

D

9.7

19.4

2 k/ft

D E205.6

Analysis of Frames With Sidesway

Consider the rectangular frame shown in Figure.

A qualitative deflected shape of the frame for an arbitrary loadingis shown in the figure using an exaggerated scale.

9

A B

DC

w

P

EI = constant

Δ Δ

While the fixed joints A and B of the frame are completelyrestrained against rotation as well as translation, the joints C and Dare free to rotate and translate.

Since the members of the frame are assumed to be inextensibleand the deformations are assumed to be small, the joints C and Ddisplace by the same amount Δ, in the horizontal direction only.

10

A B

DC

w

P

Actual Frame

Δ Δ

The MD analysis of such a frame, with sidesway, is carried out intwo parts.

In the first part, the sidesway of the frame is prevented by addingan imaginary roller to the structure.

External loads are then applied to this frame, and MEM arecomputed by applying the MD process in the usual manner.

11

A B

DC

w

PFrame with sidesway prevented

With the MEM known, the restraining force (reaction) R thatdevelops at the imaginary support is evaluated by applying theequations of equilibrium.

In the second part of the analysis, the frame is subjected to theforce R, which is applied in the opposite direction, as shown in thenext slide.

12

A B

DC

w

P

R

Imaginary roller

Frame with sidesway prevented

The moments that develop at the member ends are determinedand superimposed on the moments computed in the first part toobtain the member end moments in the actual frame.

If M, MO, and MR denote, respectively, the MEM in the actualframe, the frame with sidesway prevented, and the framesubjected to R, then we can write

13

A B

D

C R

Δ Δ

Frame subjected to R

RO MMM

14

A B

DC

w

P

Δ Δ

A B

DC

w

P

R

A B

D

C R

Δ Δ

+

=

M Moments

MO Moments MR Moments

RO MMM

An important question that arises in the second part is “how todetermine the member end moments MR that develop when theframe undergoes sidesway under the action of R”.

The MDM cannot be used directly to compute the moments due tothe known lateral load R, we employ an indirect approach in whichthe frame is subjected to an arbitrary known joint translation Δ’caused by an unload load Q acting at the location and in thedirection of R.

From the known joint translation, Δ’, we determine the relativetranslation between the ends of each member, and we calculatethe member FEMs in the same manner as done previously in thecase of support settlements.

15

16

A B

D

C Q

Δ Δ

Frame subjected to an arbitrary Translation Δ’MQ Moments

The FEMs thus obtained are distributed by the MD process todetermine the MEMs MQ caused by the yet-unknown load Q.

Once the MEMs MQ have been determined, the magnitude of Qcan be evaluated by the application of equilibrium equations.

With the load Q and the corresponding moments MQ known, thedesired moments MR due to the lateral load R can now bedetermined easily by multiplying MQ by the ratio R/Q; that is

By substituting this Equation into the last Equation, w can expressthe MEMs in the actual frame as

17

QR MQ

RM

QO MQ

RMM

Example 2

• Determine the member end moments and reactions for the frameshown by the moment-distribution method.

18

A

B

DC

40 kN

7 m

EI = constant

4 m3 m

5 m

Solution

• Distribution Factors

At joint C

At joint D

19

A

B

DC

40 kN

7 m

EI = constant

4 m3 m

5 m

5.0

72

7 I

I

DFDF CDCA

583.0

57

5

417.0

57

7

II

I

DF

II

I

DF

DB

DC

Checks 1583.0417.0 DBDC DFDF

• Part 1: Sidesway Prevented

The sidesway of frame is prevented by adding an imaginary rollerat joint C.

Assuming that joint C and D of this frame are clamped againstrotation, we calculate the FEMs due to the external loads to be

20

A

B

DC

40 kN

Imaginary RollerFrame with sidesway prevented

0

m . kN 4.29

m . kN 2.39

DBBDCAAC

DC

CD

FEMFEMFEMFEM

FEM

FEM

R

The MD of these FEMs is then performed, as shown on the MDTable to determine the MEMs “MO” in the frame with sideswayprevented.

21

AC

-9.8

-12 -24 +23.9 -24 +24

CA CD DC DB

0.5 0.5 0.417 0.583

+39.2 -29.4

-19.6 -19.6

-9.8 +8.6

+4.1 +5.7

+2.1

-1.1 -1.1

-1.6

-0.6 +0.4 -0.6 +0.5

+0.3-0.2

+12

BD

+12.3 +17.1

+6.2

-3.1 -3.1

-1.6 +2.9

+0.7 +0.9

-0.2 +0.3

Member End Moments for Frame with Sidesway Prevented - MO

To evaluate the restraining force R that develops at the imaginaryroller support, we first calculate the shears at the lower ends of thecolumns AC and BD by considering the moment equilibrium of thefree bodies of the columns.

22

A

B

DC

5.14

12

7.2

12

7.2

24

5.14

24

7 m

5 m

Next, by considering the equilibrium of the horizontal forces actingon the entire frame, we determine the restraining force R to be

Restraining force acts to the right, indicating that if the roller wouldnot have been in place, the frame would have swayed to the left.

23

kNRRFx 06.2 02.714.5 0

A

B

DC

R = 2.06

5.14

7.2

• Part 2: Sidesway Permitted

Since the actual frame is not supported by a roller at joint C, weneutralize the effect of the restraining force by applying a lateralload R = 2.06 kN in the opposite direction to the frame.

MD method cannot be used directly to compute MEMs MR due to thelateral load R = 2.06 kN, we use an indirect approach in which theframe is subjected to an arbitrary known joint translation Δ’ causedby an unknown load Q acting at the location in the direction of R.

24

A

B

DC

R = 2.06 kN

Frame subjected to R = 2.06 kNMR Moments

Assuming that the joints C and D of the frame are clamped againstrotation as shown in figure on next slide, FEMs due to thetranslation Δ’ are given by

25

A

B

DCQ

Frame subjected to an ArbitraryTranslation Δ’MQ Moments

Δ’ Δ’

0

25

'6

5

'6

49

'6

7

'6

2

2

DCCD

DBBD

CAAC

FEMFEM

EIEIFEMFEM

EIEIFEMFEM

In which negative sign have been assign to the FEMs for thecolumns, because these moments must act in the clockwisedirection, as shown.

Instead of arbitrarily assuming a numerical value for Δ’ to computethe FEMs, it is usually more convenient to assume a numericalvalue for one of the FEMs, evaluate Δ’ from the expression of thatFEM, and use the value of Δ’ to compute the remaining FEMs.

26

A

B

DC

FEMs due to Known Translation Δ’

Δ’ Δ’

Thus, we arbitrarily assume the FEMAC to be -50 kN.m

by solving for Δ’, we obtain

by substituting this value of Δ’ into the expressions for FEMBD andFEMDB, we determine the consistent values of these moments tobe

These FEMs are then distributed by the usual MD process, todetermine the MEMs MQ caused by the yet unknown load Q.

27

kN.m 5049

'6

EIFEMFEM CAAC

EI

33.408'

kN.m98

25

33.4086

DBBD FEMFEM

28

AC

+12.5

-42.3 -34.5 +34.3 +45.4 -45.4

CA CD DC DB

0.5 0.5 0.417 0.583

+25 +25

+12.5 +28.6

-5.2 -7.3

-2.6

+1.3 +1.3

-5.2

+0.7 +1.1 +0.7 +1.5

-0.3-0.6

-71.8

BD

+40.9 +57.1

+20.5

-10.3 -10.3

-5.2 -3.7

+2.2 +3

-0.6 -0.4

Member End Moments Due to Known Translation Δ’- MQ

-50 -50 -98 -98

-0.3 -0.2 -0.3 -0.2

+0.1+0.1 +0.1 +0.2

To evaluate the magnitude of Q that corresponds to these MEMs,we first calculate shears at the lower ends of the columns byconsidering their moment equilibrium and then apply the equationof equilibrium in the horizontal direction to the entire frame

29 kNQQFx 41.34 044.2397.10 0

A

B

DC

10.97

42.3

23.44

71.8

23.44

45.4

10.97

34.5

7 m

5 m

A

B

DC

Q = 34.41

10.97

23.44

which indicates that the moments MQ computed are caused by alateral load Q = 34.41 kN.

Since the moments are linearly proportional to the magnitude ofthe load, the desired moment MR due to the lateral load R = 2.06kN must be equal to the moment MQ multiplied by the ratio R/Q =2.06/34.41.

• Actual Member End Moments

The actual MEMs , M, can now be determined by algebraicallysumming the MEMs MO and 2.06/34.41 times the MEMs MQ.

30

31

NS m . kN 7.78.7141.34

06.212

NS m . kN 3.214.4541.34

06.224

NS m . kN 3.214.4541.34

06.224

NS m . kN 263.3441.34

06.29.23

NS m . kN 1.265.3441.34

06.224

NS m . kN 5.143.4241.34

06.212

AM

AM

AM

AM

AM

AM

BD

DB

DC

CD

CA

AC

32

A

C

14.5

26.1

C21.3

D

B

D

7.7

21.3

40 kN

26.1

Actual Member End Moments (kN . m)

Example 3

• Determine the member end moments and reactions for the frameshown by the moment-distribution method.

33

A

B

DC30 k

20 ft12 ft

16 ft

EI = constant

12 ft

8 ft

Solution

• Distribution Factors

At joint C

34

A

B

DC30 k

20 ft12 ft

16 ft

EI = constant

12 ft

8 ft

5.0

202

20

I

I

DFDF CDCA

Solution

• Distribution Factors

At joint D

35

A

B

DC30 k

20 ft12 ft

16 ft

EI = constant

12 ft

8 ft

49.0

42.144

3

20

20

II

I

DFDC51.0

42.144

3

20

42.144

3

II

I

DFDB

MEMs due to an Arbitrary Sidesway Δ’

Since no external loads are applied to the members of the frame,the MEMs MO in the frame restrained against sidesway will bezero.

To determine the MEMs M due to the 30-k lateral load, we subjectthe frame to an arbitrary known horizontal translation Δ’ at joint C.

Figure on the next slide shows a qualitative deflected shape of theframe with all joints clamped against rotation and subjected to thehorizontal displacement Δ’ at joint C.

36

37

A

B

DC30 k

Δ’

4

3

5

5

4

3

3

2

13C’

D’

D1

3

213

D

D1

D’

Δ

32

13

5

4

3

'3

2

'4

3

FEMs due to an Arbitrary Translation Δ’

MEMs due to an Arbitrary Sidesway Δ’

Note that, since the frame members are assumed to beinextensible and deformations are assumed to be small, an end ofa member can translate only in a direction perpendicular to themember.

From this figure, we can see that the relative translation ΔAC

between the ends of members AC in the direction perpendicular tothe member can be expressed in terms of the joint translation Δ’ as

38

'25.1'4

5' CCAC

MEMs due to an Arbitrary Sidesway Δ’

Similarly, the relative translation for members CD and BD are givenby

The FEMs due to the relative translation are

39

'202.1'3

13'

'417.1'4

3'

3

2'1

DD

DD

BD

CD

2

2

2

42.14

'202.16

20

'417.16

20

'25.16

EIFEMFEM

EIFEMFEM

EIFEMFEM

DBBD

DCCD

CAAC

MEMs due to an Arbitrary Sidesway Δ’

in which the FEMs for members AC and BD are CCW (positive),whereas those for member CD are CW (negative).

If we arbitrarily assume that

then

and therefore

40

ft-k 10042.14

'202.162

EI

FEMFEM DBBD

2.2883'EI

ft-k 3.61

ft-k 1.54

DCCD

CAAC

FEMFEM

FEMFEM

The FEMs are distributed by the MD process to determine theMEMs MQ.

To determine the magnitude of the load Q that corresponds to theMEMs MQ we first calculate the shears at the ends of the girder CDby considering the moment equilibrium of the free body of thegirder as shown.

41

A

C

55.3

56.5

C55.2

D

B

D

55.2

56.4

11.17

11.17

5.58 5.58

5.58

5.585.58

5.58

8.32

8.32

Evaluation of Q

42

AC

+1.8

+55.3 +56.5 -56.4 -55.2 +55.2

CA CD DC DB

0.5 0.5 0.49 0.51

+3.6 +3.6

+1.8

+23.6 +24.6

+11.8

-5.9 -5.9

+2.4

-3 -0.6 -3

+1.5+0.3

0

BD

-19 -19.7

-9.5

+4.8 +4.8

+2.4

-1.2 -1.2

+0.3 +1.5

Member End Moments Due to Known Translation Δ’- MQ

+54.1 +54.1 +100 +100

+0.2 +0.8 +0.2

-0.1-0.4 -0.4 -0.1

1

-0.2

-61.3 -61.3

-0.2

-50

+0.1

-100

+0.1

MD TABLE

The girder shears (5.58 k) thus obtained are then applied to thefree bodies of the inclined members AC and BD.

Next, we apply the equations of moment equilibrium to membersAC and BD to calculate the horizontal forces at the lower ends ofthese members.

43

AB

DC

11.17

8.32

Evaluation of Q

The magnitude of Q can now be determined by considering theequilibrium of horizontal forces acting on the entire frame asshown below

44

AB

DCQ = 19.49 k

11.17

8.32

Evaluation of Q

k 49.19 032.817.11 0 QQFx

Actual MEMs

The actual MEMs, M, due to the 30-k lateral load can now beevaluated by multiplying the moments MQ computed in Table bythe ratio 30/Q=30/19.49:

45

NS 0

NS ft -k 852.5549.19

30

NS ft -k 852.5549.19

30

NS ft -k 8.864.5649.19

30

NS ft -k 875.5649.19

30

NS ft -k 1.853.5549.19

30

AM

AM

AM

AM

AM

AM

BD

DB

DC

CD

CA

AC

Member End Forces

46

A

C

85.1

87

C85

D

B

D

85

86.8

17.2

17.2

8.59 8.59

8.59

8.59

8.59

8.59

12.8

12.8

Evaluation of Q

12.8 12.8C

8.59

8.59

17.2

30 kD

8.59

12.8

8.59

Support Reactions

47

A

B

DC30 k

85.1 k-ft

17.2 k

8.59 k

8.59 k

12.8 k