WEEK 8 - Moment Distribution Method Lecture PDF

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CIVE 1143 – ANALYSIS OF COMPLEX STRUCTURESDr. Jianhu Shen(Chris)Office : 10.Email : [email protected] times:Tuesday 10:30AM-12:30PM in 10.12-Other time will be arranged from time to timeplease check canvas announcement for detailsToday 20 April 2018 2:00pm-4:00pm in 10.Analysis of Complex ...

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CIVE 1143 – ANALYSIS OF COMPLEX STRUCTURES

Dr. Jianhu Shen(Chris) Office : 10.12.15 Email : [email protected] Consultation times: Tuesday 10:30AM-12:30PM in 10.12.15-2 Other time will be arranged from time to time please check canvas announcement for details Today 20 April 2018 2:00pm-4:00pm in 10.12.15 1

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Analysis of Complex Structure

Lecture 7-Moment Distribution Method

Reference: Structural Analysis EIGHTH EDITION IN SI UNITS  R. C. HIBBELER 

Chapter 12 Moment distribution

Detailed references 12.1 12.2 12.3 12.4

3

General Principles and Definitions Moment Distribution for Beams Stiffness-Factor Modifications Moment Distribution for Frames: No Sidesway

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

General Guidance for Moment Distribution Method By answering the following questions relating to the “Displacement Method” in analysis of complex structure:

(3) (1)

(8) (5)

Moment Distribution Method (6)

(2) (4)

4

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

(7)

(1)Why do we call this method “Moment Distribution Method”? (2)What are the assumptions (3)What is the key features? (4)What is the unknown variables used in this method? (5)What is the direct objective of this method? (6)What is the general procedure to use this method (7)Is there an easier way to solve a specific problem using this method? (8)What is the application of this method

An example of load path -gravity load pass

How is the gravity load on buildings transferred from all levels to the ground?

(1)

5

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Load transfer on beams through effect of bending

5kN

275mm

A

B

600mm

A question, will the internal moment on span AB is zero?

C

E

D

550mm

From deflected shape using qualitative analysis, the answer is no

600mm

How the bending is transferred from BD to Other spans AB and DE? (1)

6

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Load transfer on beams through effect of bending To consider it, let see what will happen if we fix point B and D by constraining the rotation at Point B and D.

275mm

D

MAB =0 A

5kN B

600mm

FEMAB C

550mm

MDE =0

FEMBA

E

600mm

Note: the moments at B and C will be the fixed end moment FEMAB and FEMBA 7

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

(1) (2)

Q1: How is the moment carried by the fixed support at Joint D (FEMDB) distributed to MDB and MDE?

Load transfer on beams through effect of bending Now we release the fixed support at D.

Distribute 275mm

D

MAB =0 A

5kN B

FEMBD C

FEMDB MDB

FEMBD MDE

E

Carry over 600mm

550mm

MED

600mm

Q2: How is this newly distributed moment carried over from D to E as MED ? (3) 8

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

For Q1: How is the moment carried by fixed support distributed to MDB and MDE at connection D? It will be determined the by the distribution factor, i.e.,

:

1

The Distribution Factor will be determined the by the bending stiffness of each beam connecting to Point D, i.e.,

4

4

For a beam span with fixed end

4

, , (6)

9

Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

The bending stiffness of a beam connecting to others Recall the derivation of slope-deflection equation for superposition Case No. 1:

4

1

KAB Another example:

3

3 10 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

More example for stiffness factor:



Member stiffness factor

4 

Total stiffness for connection A

∑ 

4000 5000 1000 10000

The distribution factors all beams connected to A:

4000 10000

1000 10000

0.4

5000 10000 11 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

0.5

0.1

In summary for Distribution Factor:



Distribution Factor (DF) That fraction of the total resisting moment supplied by the member is called the distribution factor (DF)

∑ ∑

12 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Some Special Cases for stiffness factor

∑ 

Member relative stiffness factor - Quite often a continuous beam or a frame will be made from the same material

 E will therefore be constant 

Let’s check further: - Quite often a continuous beam or a frame will be made from the same material and all cross-sections are the same  EI will therefore be constant

1

This is the end for moment distribution at connection, now consider carry over factor 13 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Q2: How is this newly distributed moment carried over from D to E as MEC ? 

It is done be using a Carry-over (CO) factor



Similarly, we recall the Case 1 in derivation of SDEs:

4   -

Solving for

A

and equating these eqn,

  

2  

A;

  

0.5 CO factor

In the case of a beam with the far end fixed, the CO factor is +0.5 - The plus sign indicates both moments act in the same direction (6) -

14 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

A

The assumptions for Moment Distribution Method

This method is based on the Slope and Deflection Equations in Displacement method Thus Similar assumption for Displacement method will be applied to this method as well.

• • • •

Linear Elastic Small Deformation Only consider bending effect (neglect shear deformation)

15 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

(2)

DF & CO for Normal Spans (a span cut from continuous beam or frame) Lengths of all spans are L. Section properties are all EI except that span AE is 2EI. I.E. we cut all supports out of the spans The distribution factor at Joint A:

4 4

4

4

4 4 4 4 8 4 4 4 4 8

0.2 0.2

8

4 4 4 4 8 8 4 4 4 8

16 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

0.2 0.4

DF & CO for Normal Spans (a span cut from continuous beam or frame) Lengths of all spans are L. Section properties are all EI except that span AE is 2EI.

E-

Consider other joints: EA

Note: the bending stiffness for fix end is infinite, for free end and pin end is 0. For B,

4

0 0

For C,

1 4 1

For D,

DA

BA

0 CA

0

C-

0

4 0

D-

1

17 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

For E, it is the same as D.

B-

DF & CO for Several Special End Spans with Supports Lengths of all spans are L. Section properties are all EI except that span AE is 2EI.

The carry over factors for all spans are 0.5

18 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

DF & CO for Several Special Spans Lengths of all spans are L. Section properties are all EI except that span AE is 2EI. If we consider two special end spans with free end and pin end. The distribution factor at Joint A: For normal span Pin end span Free end span

4

3 3

0 0 0 3 4 8 4 0 3 4 8

0 4

0 4 15

8 3

0 3 4 8 8 0 3 4 8

19 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

0.2 8 15

(7)

DF & CO for Several Special Spans Lengths of all spans are L. Section properties are all EI except that span AE is 2EI.

The carry over factors for spans AD and AE are COAD=0.5 and COAE=0.5. The carry over factor for End span AB is COAB=0 The carry over factor for End span AC is COAC=0

20 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

(7)

Applications Similar to Displacement method, can deal with more DOFs

Bridges

Portal Frames

(8) 21 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Let’s use a simple example to practise how to use this method to determine the internal moments

(6) 22 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

12.2 Moment Distribution for Beams

(1) Calculating distribution factor at connection point

4 (120)(10 6 ) 4 (40)(106 ) mm 4 /m 3 4 (240)(106 ) 4 (60)(106 ) mm4 /m 4

4 (40) 4 (40) 4 (60) 4 (60)

0 0

23 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

4 (40) 0.4 4 (40) 4 (60) 4 (60) 0.6 4 (40) 4 (60)

Moment Distribution for Beams

(2) Calculating Fixed End Moments for all spans with inter-span loads

2

(

)

12

8000 N • m

2

(

24 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

)

12

8000 N • m

Moment Distribution for Beams (3) Distribute fixed end moment at joints which can rotate and carry over the distributed moment to far end We begin by assuming joint B is fixed or locked  The fixed end moment at B then holds span BC in this fixed or locked position  To remove this constrain, we will apply an equal but opposite moment of 8000 N•m to the joint B and allow the joint to rotate freely 

25 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams • As a result, portions of this moment are distributed in spans BC and BA in accordance with the DFs of these spans at the joint • Moment in BA is 0.4(8000) = 3200 N•m • Moment in BC is 0.6(8000) = 4800 N•m • These moment must be carried over since moments are developed at the far ends of the span • Using the carry-over factor of +0.5, the results are shown • Sum all moments together to get M at each end of each span 26 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams • • •

•

•

(4)

(5)

The steps are usually presented in tabular form with DFs The Dist and CO will be in the same line representing one cycle/iteration CO is indicated by an arrow where the distributed moments are carried over from near end to the far end. The carry-over factor is +0.5. The wall supports at A and C “absorb” the moments and no further joints have to be balanced to satisfy joint equilibrium The sum of moment at all ends will get the moment at those ends

Note: In this particular case only one cycle of moment distribution is enough 27 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams

(4) Mark the final solution on spans with real direction and work out shear forces following the similar procedure as in Displacement Method

This is the end of this problem. Any questions? 28 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Puzzle for Moment Distribution Method By answering the following questions relating to the puzzle of “Displacement Method” in analysis of complex structure:

(3) (1)

(8) (5)

(2) (4)

(6)

(7)

Moment Distribution Method 29 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

(1)Why do we call this method “Moment Distribution Method”? (2)What are the assumptions (3)What is the key features? (4)What is the unknown variables used in this method? (5)What is the direct objective of this method? (6)What is the general procedure to use this method (7)Is there an easier way to solve a specific problem using this method? (8)What is the application of this method

Moment Distribution for Beams Another example with more cycles Example 12.2 Determine the internal moment at each support of the beam. The moment of inertia of each span is indicated.

30 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams Example 12.2 (Solution)-Clockwise as positive (1)

Calculating distribution factor at connection point

The moment does not get distributed in the overhanging span AB So the distribution factor (DF)BA =0 Span BC is based on 4EI/L since the pin rocker is not at the far end of the beam

4 (300)(10 6 ) 4 4 ( 240)(10 6 ) 3

6

300 300 320

0.484

6

320 300 320

0.516

300(10 )

320(10 )

Note: The overhanging span requires the internal moment to the left of B to be +4000 N•m. 31 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams

(2) Calculating Fixed End Moments for all spans with inter-span loads and moment from span with free end 2

(

)

12

1500 4 2 12

2000 N • m

2

(

)

∑M

12

B

0

2000 N • m

+

2000 2 0

MBA

4000

32 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

VBA

Moment Distribution for Beams (3) Distribute fixed end moment at joints which can rotate and carry over the distributed moment to far end using a table Note: all carry over factor will be 0.5 for all normal spans.

I will do this by hand so that it will be more clear then directly give you the table,

33 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Beams Example 12.2 (Solution)

Mark the value and real direction of the internal moments and forces on each span

Plot Bending Moment Diagram

This is the end of this problem. Any questions? 34 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Frames: No Sidesway

Example 12.5 Determine the internal moments at the joints of the frame as shown. There is a pin at E and D and a fixed support at A. EI is constant.

35 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Frames: No Sidesway Example 12.5 (Solution) (1) Calculating distribution factor at connection point By inspection, the pin at E will prevent the frame from sidesway. The stiffness factors of CD and CE can be computed using K = 3EI/L since far ends are pinned. The 60-kN load does not contribute a FEM since it is applied at joint B.

4 5

;

4 6

;

3 5

0

4

4 /5 /5 4 /6

0.545

1 0.545 0.455 36 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

;

3 4

Moment Distribution for Frames: No Sidesway Example 12.5 (Solution)

4

4 /6 3

/6 /5 3

4

3 /6 3

/5 /5 3

/4

0.330

/4

1 0.330 0.298 0.372 1;

1

37 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

0.298

Moment Distribution for Frames: No Sidesway

(2) Calculating Fixed End Moments for all spans with inter-span loads

2

(

)

12

135 kN • m

2

(

)

12

135 kN • m

38 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Frames: No Sidesway Example 12.5 (Solution) (3) Distribute fixed end moment at joints which can rotate and carry over the distributed moment to far end using a table The data are shown in table. The distribution of moments successively goes to joints B & C. The final moments are shown on the last line.

39 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd

Moment Distribution for Frames: No Sidesway

Example 12.5 (Solution) Summarize the solution as required in the question:

44.5 89 .1 89 .1 115

51.2 64 .1

40 Structural Analysis Eighth Edition l © 2012 Pearson Education South Asia Pte Ltd...