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Lecture 7-2: Moment distribution method

SchoolofEngineering ES95D

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Previous lecture

Content

Frames with/without sidesway Moment distribution method for non-sway

frames

Introduction: moment distribution method

Member stiffness, distribution factor and

carry-over factor Procedure

Examples

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Distribution factor

B

FC

DE

AM

AADACAB

A,ofstiffnessTotal

)KK(KM

AK

4

3

MK

KM

MK

KM

M

K

KM

A

ADAD

A

ACAC

A

ABAB

4/3

where terms in brackets are known as: Distribution Factors (DF)

AB

AB

AB L

EIK

)(4

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Frames with/without sidesway

Non-sway frame Note: the rigid joint may rotate andmove as a whole, but members donot rotate against each other. Themoments at the joint should bebalanced in the end.

Sway frame

Sway frames are more difficult becausethe moment is also related to sidesway

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Example 1

DC

A

B

9m

75kN

3m

3m

25kN/m

Step 1: clamp C. Calculate therotational stiffness of CA and CD.

CA (end A fixed): 4EI/6=0.67EIEI constantCD (end D hinged): 3EI/9=0.33EI

Total (rotational) stiffness atpoint C: 0.67EI+0.33EI=EI

Distribution factor:

DFCA : 0.67EI/1EI = 0.67DFCD : 0.33EI/1EI = 0.33

Check: DFC=1, OK!

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Example 1

DC

A

B75kN

25kN/m

Step 2: calculate the fixed endbending moment under loading

EI constant

A

75kN

C

FACM

F

CAM

From the engineeringdata book:

3.566758

1

F

ACM

3.56FCA

M

(anti-clockwise)

(clockwise)

25kN/mFCDM

From moment distributionlecture 1

1.2539258

1 2 FCDM

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C

A

D

0.

67

0.33

56.

3

-253.10

-56.

3After writing down the DFs and FEMs:

The unbalanced moment at Cis: 56.3-253.1= -196.8.

Release C and distribute 196.8according to DFs.

Dont forget the carry-over

from C to A.

66.

0

131.

9

64.9

Example 1

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9.

7

1

88.

2

-188.2 0

9.

7

188.

2

188.2

0

Add up the moment values.

Draw the bending momentdiagram.

From equilibrium of theforces on each individualmember, we can calculate Sforces and reactions(homework).

Example 1

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Example 2

8m

6m 10m 6m

E F

I I

C D50kN/mmBA

3 I3 I 3 I

At point B and C, 3members are joined.Note it is not a constantEI frame.

Rotationalstiffnesses:

BA: 3E(3I)/6=1.5EI

BC: 4E(3I)/10=1.2EI

BE: 4EI/8=0.5EI

Total rotationalstiffness:

(1.5+1.2+0.5)EI=3.2EI

So DFs are:

DFBA

: 1.5EI/3.2EI = 0.47DFBC : 1.2EI/3.2EI = 0.38

DFBE : 1-0.47-0.38=0.15 FEMs can be calculated in thesame way as Example 1.

Similar procedure at point C.

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Example 2, DFs and FEMs

0.380.47 0.38 0.47

0.1

5

0.1

5 CB

225 -417 417 -225

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Example 2, release B

0.380.47 0.38 0.47

0.

15

0.

15 CB

225 -417 417 -225

90 73

29

At point B the unbalanced moment is 225-417=-192, so distribute 192according to DF, carry-over from BC to CB, BE to EB.

E F

37

15

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Example 2, re-clamp B, release C

0.380.47 0.38 0.47

0.

15

0.

15 CB

225 -417 417 -225

90 73

29

At point C the unbalanced moment is 37+417-225=229, so distribute -229according to DF, carry-over from CB to BC, CF to FC.

E F

37

15

-87 -108

-34

-17

-44

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Example 2, iteration

0.380.47 0.38 0.47

0.

15

0.

15 CB

225 -417 417 -225

90 73

29

At point B the unbalanced moment is -44, which was carried over from joint C. Afterdistribution and carry-over at point B, point C is unbalanced because of the carried over 9.

E F

37

1

5

-87 -108

-34

-

17

-44

21 17

7

4

9

-3 -4

-1

-

1

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Example 2: M diagram

88.736

C

E F

D

B

A

254

371

335

18

How would you find shear forces in members? What about reactions?

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What to do next? Example class 5, P1

Solve the tests from 2008, 2009 and 2010