Displacement Method of Analysis: Moment Distribution...

Displacement Method of Analysis Moment Distribution Method

Chapter 12

– In this method, joint rotations & displacements are used as unknowns in carrying out the analysis.

– Unlike the slope deflection method, the moment-distribution method does not require the solution of simultaneous equations, instead answers are obtained by a procedure of successive approximations (iteration technique).

– From the slope deflection equations it is evident that the moments acting on the ends of a member in a frame consist of several distinct parts;• The fixed-end moments, which are the moments caused by the

load acting on the member, with the ends of the member assumed to be fixed.

• The moment due to the rotations that actually take at the ends of the member.

• The moments caused by the translation of one end of the member relative to the other

Introduction

General Principles

• Sign convention– Same sign as slope deflection equations; clockwise moments are

considered positive & counterclockwise moments are negative.

• Member stiffness factor– If the beam is pinned at one end & fixed at the other

2 2 3 ( )AB A B AB

IM E FEM

L L

2

2 0 0 0AB A

EIM

L

4AB A

EIM

L

For unit A = 1

4AB

EIK

L

It is referred to as the stiffness factor at A and can be

defined as as the amount of moment M required to rotate

the end A of the beam A = 1 radian

General PrinciplesJoint Stiffness Factor

The total stiffness factor for joint A is

equal to the summation of the stiffness

factor of each member.

A AB AD ACK K K K

1000 4000 5000 10000AK

That means a moment of 10000 needed to

rotate joint A (1 radian)

Distribution Factor (DF)

If a moment M is applied to joint A, that will rotate joint A an amount of

then each member connected to that joint will rotate by this same amount

M

1 i nM M M M 1 i n iK K K K

The distribution factor for ith member is

i ii

i

M KDF

M K

K

DFK

General Principles• Example; Fined the moment at each side of joint A (MAB, MAD & MAC).

M=2000N.m

The total stiffness factor of joint A

1000 4000 5000 10000K The distribution factors for each members

at joint A

40000.4

10000ABDF

50000.5

10000ACDF

10000.1

10000ADDF

The moment of each member at the joint

0.4 2000 800 .ABM N m

0.5 2000 1000 .ACM N m

0.1 2000 200 .ADM N m

General Principles

• Member relative-stiffness

– As in the most cases, the frame or the beam, will be made from the same material, same modulus of elasticity, then the 4E is same for all members.

1 1

1 1

1 1

1 1

4

4 4 4i n i n

i n i n

EI I

K L LDF

EI EI EI I I IK

L L L L L L

R

IK

L

General Principles

• Carry-Over Factor

– The applied moment M at A cause A to rotate A.

– Applying slope deflection;

2 42 3AB A B AB A

EI EIM FEM

L L L

2 22 3BA B A BA A

EI EIM FEM

L L L

1

2BA ABM M

That means that, the moment M at the pin induces a moment 0.5 M at the

fixed end. In this case the carry over factor is +0.5.

The carry-over represent the fraction of M that is carried over from the

pin to the wall.

Problem 1• Determine the internal moment at each support of the beam

1. Calculate the distribution factors at all

joints that are free to rotate 4 (300)

4 (20)15

BA

EK E

4 (600)4 (30)

20BC

EK E

Joint B 4 (20)

0.44 (20) 4 (30)

BA

EDF

E E

4 (30)0.6

4 (20) 4 (30)BC

EDF

E E

For joint A & C the distribution factors = 0. 4 (20)

0.4 (20)

AB

EDF

E

4 (30)0.

4 (30)CB

EDF

E

2. Calculate the fixed end moment

assuming all joints to be locked

0.AB BAFEM FEM

2240(20)8000 .

12BCFEM lb ft 8000 .CBFEM lb ft

3. Begin the actual moment distribution process

0.60.40. 0.D.F.

80000.0. 8000F.E.M.

480032000. 0.Balancing Joint B

1600 2400Carry over moment

Final Ms 320032001600 10400

Problem 1

Example 12-1• Draw the bending moment diagram.

1. The distribution factors

4

12AB

EIK

4

12BC

EIK

Joint B 4 /12

0.54 /12 4 /12

BA

EIDF

EI EI

4 /120.5

4 /12 4 /12BC

EIDF

EI EI

Joint C 4 /12

0.44 / 8 4 /12

CB

EIDF

EI EI

4 / 80.6

4 / 8 4 /12CD

EIDF

EI EI

4

8CD

EIK

2. Fixed end moment

0.AB BAFEM FEM

220(12)240 .

12BCFEM kN m 240 .CBFEM kN m

250(8)250 .

8CDFEM kN m 250 .DCFEM kN m

1 36

6 0.3

0.05 1.8

6120

Example 12-1

D.F.

2400.0. 250F.E.M.

120Balancing

60 3C.O.M

Final Ms 125.3125.362.5 234.3

0.50.50. 0.0.60.4

250240

4

281.5281.5

2 60

1Balancing

0.5 18C.O.M

24

12 0.5

6Balancing

3 0.15C.O.M

0.2

0.1 3

0.05Balancing

0.02 0.9C.O.M

1.2

0.6 0.02

Balancing 0.30.3 0.010.01

3. Begin the actual moment distribution process

Example 12-1

Problem 2• Determine the internal moment at each support.

1. The distribution factors

4

6AB

EIK

8

6BC

EIK

Joint B 4 / 6 1

4 / 6 8 / 6 3BA

EIDF

EI EI

8 / 6 2

4 / 6 8 / 6 3BC

EIDF

EI EI

Joint C 8 / 6

18 / 6

CB

EIDF

EI

2. Fixed end moment23(6)

9 .12

BCFEM t m 9 .CBFEM t m

8(6)6 .

8ABFEM t m 6 .BAFEM t m

A3t/m8t

B C2II

3m 3m 6m

3. Begin the actual moment distribution process

9

1

1.5

0.17

3

0.34

0.5

2

D.F.

966 9F.E.M.

1Balancing

0.5 1C.O.M

Final Ms 8.938.934.54 0.

2 / 31/ 30. 1.

4.5

1.5Balancing

0.75 1.5C.O.M 0.5

0.16Balancing

0.08 0.17C.O.M 0.75

0.25Balancing

0.12 0.25C.O.M 0.08Balancing 0.060.02 0.25

0.

0.

0.

0.

0.

Problem 2

Example 12-2

Example 12-2

Example 12-2

Stiffness-Factor Modification• Member pin supported at far end.

– Using slop-deflection equations

22 3AB A B AB

EIM FEM

L L

22 3BA B A BA

EIM FEM

L L

0.BAM

2 32 0 0

2

AAB A A

EI EIM

L L

The stiffness factor for this beam is;

3EIK

L

2

0. 2 0 0B A

EI

L

2

AB

To model the case of having the far end pinned or

roller supported

Problem 3• Back to Problem 3- Solution A

1. The distribution factors

4

6AB

EIK

6

6BC

EIK

Joint B 4

0.44 6

BADF

60.6

4 6BCDF

Joint C 8 / 6

18 / 6

CB

EIDF

EI

2. Fixed end moment23(6)

13.5 .8

BCFEM t m

8(6)6 .

8ABFEM t m 6 .BAFEM t m

A3t/m8t

B C2II

3m 3m 6m

4.5

D.F.

13.566F.E.M.

3Balancing

1.5C.O.M

Final Ms 994.5 0.

0.60.40.

0.

1.

Balancing 0. 0. 0.

• Determine the internal moment at each support.1t/m9t.m

A

B C

12m4m2m4m 2m

D9t.m1. The distribution factors

4

6AB

EIK

4

12BC

EIK

Joint B 4 / 6 2

4 / 6 4 /12 3BA

EIDF

EI EI

4 /12 1

4 / 6 4 /12 3BC

EIDF

EI EI

2. Fixed end moment21(12)

12 .12

BCFEM t m 12 .CBFEM t m

2

9 2(2 4 2)3 .

6ABFEM t m

4

6CD

EIK

Joint C 4 / 6 2

4 / 6 4 /12 3CD

EIDF

EI EI

4 /12 1

4 / 6 4 /12 3CB

EIDF

EI EI

2

9 4(2 2 4)0. .

6BAFEM t m

2

9 2(2 4 2)3 .

6DCFEM t m

2

9 4(2 2 4)0. .

6CDFEM t m

Problem 4

Problem 41t/m9t.m

A

B C

D9t.m

3. Begin the actual moment distribution process

0.67 1.33

0.11 0.22

0.02 0.04

84

D.F.

120.3 3F.E.M.

8Balancing

4 4C.O.M

Final Ms 9.599.591.8 1.8

1/ 32 / 30. 0.2 / 31/ 3

0.12

4

9.599.59

2 2

1.33Balancing

0.67 0.67C.O.M

0.67

0.33 0.33

0.22Balancing

0.11 0.11C.O.M

0.11

0.06 0.06

0.04Balancing

0.02 0.02C.O.M

0.02

Stiffness-Factor Modification

22 3BC B C BC

EIM FEM

L L

2AB B

EIM

L

The stiffness factor is;

2EIK

L

2

2 0 0BC B B

EIM

L

For the center span of Symmetric beam & load

From symmetry of load C = -B

Symmetric beam & loading.– Using slop-deflection equations

Stiffness-Factor Modification

22 3BC B C BC

EIM FEM

L L

6AB B

EIM

L

The stiffness factor is;

6EIK

L

2

2 0 0BC B B

EIM

L

For the center span of Symmetric beam & load

From symmetry of load C = B

Symmetric beam with Anti-symmetric Loading.– Using slop-deflection equations

Problem 5• Back to Problem 4

1t/m9t.mA

B C

12m4m2m4m 2m

D9t.m

1. The distribution factors

4

6AB

EIK

2

12BC

EIK

Joint B 4 / 6

0.84 / 6 2 /12

BA

EIDF

EI EI

2 /120.2

4 / 6 2 /12BC

EIDF

EI EI

2. Fixed end moment

12 .

12 .

3 .

0 .

BC

CB

AB

BA

FEM t m

FEM t m

FEM t m

FEM t m

2.4

D.F.

120.3F.E.M.

9.6Balancing

4.8C.O.M

Final Ms 9.69.61.8 1.8

0.20.80.

9.69.6

Example 12-3

Determine the internal moment at the supports for the beam shown. EI is constant

Example 12-4

Determine the internal moment at the supports for the beam shown. E is constant

Example 12-4

• Determine the internal moment at each support when support B sinks by 10mm. EI = 200t.m2

1. The distribution factors4

3AB

EIK

3

6BC

EIK

Joint B 4 / 3

0.734 / 3 3 / 6

BA

EIDF

EI EI

3 / 60.27

4 / 3 3 / 6BC

EIDF

EI EI

Joint C 4 / 6

14 / 6

CB

EIDF

EI

2. Fixed end moment

22(6) /8 9 .BCFEM t m

6(3) / 8 2.25 .ABFEM t m

2.25 .BAFEM t m

1.5m

A2t/m6t

B C

3m 6m

2t

Moment developed by the settlements Member AB

2 2

6 6 200 0.01

3AB BA

EIFEM FEM

L

1.33 .AB BAFEM FEM t m

Member BC

2 2

3 3 200 0.01

6BC

EIFEM

L

0.165 .BCFEM t m

Cantilevers moment2 1.5 3 .CM t m

Problem 6

3. Begin the actual moment distribution process

7.34

1.74

0.165

D.F.

92.252.25F.E.M.

1.33FEM due to

Final Ms 5.65.61.24 3

0.270.730. 1.

0.92

4.68Balancing

2.34C.O.M

1.33

3.58

0.

3

A

B C

3

Total Modified F.E.M.

Problem 6

Balancing joint C 3

1.5C.O.M

Balancing 0.0.

0.

0

Problem 7: Frame with No Sway

• Determine the internal moment at each support. EI constant

4t

C

3t/m

8m3m

3m

A

B

C

A

B 0.5 0.

16 16

5.75

10.25

2.875

18.875

1. The distribution factors

3

6AB

EIK

4

8BC

EIK

Joint B 1/ 2

0.51/ 2 1/ 2

BADF

1/ 20.5

1/ 2 1/ 2BCDF

2. Fixed end moment

23(8) /12 16 .BCFEM t m

16 .CBFEM t m

3

(4) 6 4.5 .16

BAFEM t m

38.4 13.3326.6757.6

0.

0.

0 013.147.89

-9.9

0

• Determine the internal moment at each support. EI constant

1. The distribution factors

4

5AB

EIK

4

3BC

EIK

Joint B 4 / 5

0.2554 / 5 4 / 3 1

BADF

4 / 30.425

4 / 5 4 / 3 1BCDF

2. Fixed end moment

26.67 .BCFEM kN m

13.33 .CBFEM kN m

38.4 .ABFEM kN m

57.6 .BAFEM kN m

1m 2m

C

60kN

4m

A B

3m 2m

D

80kN

4

4BD

EIK

10.32

4 / 5 4 / 3 1BDDF

0BD DBFEM FEM

0.255 0.425

0.3

2

0. 0.

0.

-4.9

5

3.945 6.57

-4.9

5

42.345 6.7639.8149.71

-9.9

Problem 8

Example 12-5

Determine the internal moment at the joints for the frame shown.

There is a pin support at E and D and a fixed support at A. EI is constant

Example 12-5

Frames with Sway

Example 12-6• Determine the moment at each joint. EI is constant

(C) (B)(A)

Moment at A is equal to the moment at B plus the moment at C

0.32 0.08

0.3

20

.

0.

-0.0

8

Example 12-6• Structure B

DFBA= 0.5

Distribution factors

Joint B

Joint C

Fixed end moment

DFBC= 0.5

DFCB= 0.5

DFCD= 0.5

FEMBC= -10.24 kN.m

FEMCB= 2.56 kN.m

10.24 2.56

0.

0.

0.

0.0.5

0.5

0.5

0.

0.

0.5

5.12 1.28

5.1

20

.

0.

-1.2

8

2.8

8

5.76 2.64

5.7

6

-2.6

4-1

.322

.56

0.

0.64 2.56

-0.6

4

0.

0.32 1.28

0.3

20

.

0.

-1.2

80.64 0.16

0.1

6

-0.6

4

0.

0.

0.1

6

-0.0

4

Example 12-6

R = HA + HD

Determine the joint resistance RFx = 0.

By taking the moment at C we get

HD = -0.79

B

HA

5.76

2.88A

C

HD

-2.64

-1.32D

R

By taking the moment at B we get

HD = 1.73

R = 1.73 – 0.79 = 0.94 kN

• Structure C0.94

kN

2

6EIModified Fixed End Moment FEM

L

6

25AB BA CD DC

EIMFEM MFEM MFEM MFEM

Assuming all the joints are locked, the

produced by the load R caused a moment M

As is equal in column AB & CD & FEM = 0.

0.BC CBMFEM MFEM

0.19 0.19

0.1

90.

0.

0.1

9

Example 12-6The final moment produced by the R = 0.94 kN can be determined by;

Take EI = 25 and then fined out the moment on the frame

0. 0.

-6-6

-6-60.5

0.5

0.5

0.

0.

0.5

3 3

30.

0.

3

-4.8

3.6 3.6

-3.6

-3.6

-4.8

1.5

0.

1.5 1.5

1.5

0.

0.75 0.75

-0.7

50.

0.

-0.7

5

0.375 0.375

-0.3

75 -0

.375

0.

0.

0.1

0

0.1

06AB BA CD DCMFEM MFEM MFEM MFEM

0.BC CBMFEM MFEM

Example 12-6

R’ = H’A + H’D

Determine the joint resistance R’ that produced EI = 25Fx = 0.

By taking the moment at C we get

H’D = 1.68

H’A

B

-3.6

-4.8AH’

D

C

-3.6

-4.8D

R’

By taking the moment at B we get

HA = 1.68

R’ = 1.68 + 1.68 = 3.36 kN

0.94 25

0.94

3.36R EIM M

The moment produced by the reaction R = 0.94 kN would be

1.01 1.01

-1.0

1-1

.34 -1

.34

-1.0

1

Example 12-6

The final moment would be the moment of B plus the moment of C

5.76 1.01 4.75 .BCM kN m

2.64 1.01 3.65 .CBM kN m

2.88 1.34 1.54 .ABM kN m

5.76 1.01 4.75 .BAM kN m

2.64 1.01 3.65 .CDM kN m

1.32 1.34 2.66 .DCM kN m

3.125 3.125

3.1

25

0.

0.

3.1

25

Example 12-6 Solution B

0. 0.

-10

0-1

00

-100

-10

0

0.5

0.5

0.5

0.

0.

0.5

50 50

50

0.

0.

50

-80

1.56

0.195

-60

0.1

95

-80

25

0.

25 25

25

0.

12.5 12.5

-12.5

0.

0.

-12.5

6.25 6.25

-6.5

-6.2

50.

0.

1.5

6

1.5

6

100AB BA CD DCMFEM MFEM MFEM MFEM

0.BC CBMFEM MFEM

1.56

-0.7

8-0

.39

0.

0. 0.

0.780.78

0.39

60

0.39

0.195

60

0.

-60

0.

-0.7

80.

0.1

95

-0.3

9

Example 12-6 Solution B

R’ = H’A + H’D

Determine the joint resistance R’ that produced MFEM=100

Fx = 0.

By taking the moment at C we get

H’D = 28 kN

H’A

B

-60

-80AH’

D

C

-60

-80D

R’

By taking the moment at B we get

HB = 28 kN

R’ = 28 + 28 = 56.0 kN

0.94

0.94100

56.0RM MFEM

The moment produced by the reaction R = 0.94 kN would be

1.01 1.01

-1.0

1-1

.34 -1

.34

-1.0

1

Example 12-6

The final moment would be the moment of B plus the moment of C

5.76 1.01 4.75 .BCM kN m

2.64 1.01 3.65 .CBM kN m

2.88 1.34 1.54 .ABM kN m

5.76 1.01 4.75 .BAM kN m

2.64 1.01 3.65 .CDM kN m

1.32 1.34 2.66 .DCM kN m

Example 12-7Determine the moment at each joint of the frame shown. E is constant

Example 12-7

Example 12-7

Example 12-7

Problem 9• Determine the moment at each joint. EI is constant

(C)(B)(A)

• Structure BDistribution factors

Joint B (symmetry)

Fixed end momentFEMBC = -16.33 kN.m

FEMCB = 16.33 kN.m

4

8,6BA

EIK

2

7BC

EIK

4 / 8.60.62

4 / 8.6 2 / 7BA

EIDF

EI EI

2 / 70.38

4 / 8.6 2 / 7BC

EIDF

EI EI

0.38

16.33

6.21

10.12

0.

Determine the joint resistance R

R = 6 kN (symmetry HA = HD)

Problem 9

• Structure C

2

6

8.6AB BA CD DC

EIMFEM MFEM MFEM MFEM

Assuming all the joints are locked, the

produced by the load R caused a moment M

2

6 6.98( 1.163 )

7 49BC CB

EI EIMFEM MFEM

6 kN

A

B

D

C

AB

BC

CD

B1

2

2

6EIMFEM FEM

L

0.

take AB =

BC = 2(AB cos) = 2 5/8.6 = 1.163

CD = AB =

Take EI = 8.62 and then fined out the moment on the frame

6AB BA CD DCMFEM MFEM MFEM MFEM

10.53BC CBMFEM MFEM

Problem 9

0.190.19

10.5310.532.492.49

7.67.6

1.251.25

0.690.69

0.350.35

0.550.55

The joint resistance R’ that produced EI = 8.62

R’ = H’A + H’D

Fx = 0.

Problem 9

Free body diagram for column AB & taking the moment at B = 0.

A

B

H’A

6.8

7.6

5m

7m

VA

' 7 6.8 7.6 5 0A AH V

Free body diagram for Beam BC& taking the moment at C = 0.

7 7.6 7.6 0BV

' 3.6AH

2.17A BV V

VB

C

7.1B

7.1

VC

C

H’D

6.6

6.8

5m

7m

D

VD

Free body diagram for column CD & taking the moment at C = 0.

' 7 6.8 7.6 5 0D DH V

7 7.6 7.6 0CV

' 3.6DH

2.17D CV V

Free body diagram for Beam BC& taking the moment at B = 0.

Problem 9

R’ = 3.6 + 3.6= 7.2 kN

26 8.6

6

7.2R EI

M M

The moment produced by the reaction R = 6 kN would be

The final moment would be the moment of B plus the moment of C

10.12 6.26 3.79 .BCM kN m

10.12 6.33 16.45 .CBM kN m

5.06 5.67 0.61 .ABM kN m

10.12 6.33 3.79 .BAM kN m

10.12 6.33 16.45 .CDM kN m

5.06 5.67 10.73 .DCM kN m

6.336.33

Problem 9