2019 CIVE1143 Tut 8 Moment Distribution Method PDF

Preview

Summary

Tutorial 8 (week 8)CIVE1143 Analysis of ComplexStructures2 May 2019: Moment Distribution Method By Dr. Jianhu Shen (Chris) Office: 10.12. Email: [email protected] Consultation (29 April -5 May): Tuesday 10:30pm-12:30pm. Thursday 11:00am-12:00pm 1:30pm-4:30pmMoment Distribution MethodDefinition: Using ...

Description

Tutorial 8 (week 8)

CIVE1143 Analysis of Complex Structures 2 May 2019: Moment Distribution Method By Dr. Jianhu Shen (Chris) Office: 10.12.15.2 Email: [email protected] Consultation (29 April -5 May): Tuesday 10:30pm-12:30pm. Thursday 11:00am-12:00pm 1:30pm-4:30pm

Moment Distribution Method

Definition: Using moment at nodes as variables and distribute total fixed end moments at each joint to near end of the beam members connected to it and then transfer the newly distributed moment from near end to the far end of a beam member; repeating this Dist&CO iteration until the residual moment is small enough (1% of its original value). General Procedure: (1) Calculating distribution factor at joints (2) Calculating FEM for all spans (3) Form a table to do Dist&CO iterations until residual moment is small enough (4) Sum all moments at each iterations to get internal moments for each end in each span (5) Solve other items as requested in the problem 2

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

The bending stiffness of a span (beam member)

The bending stiffness of an span with pin or free end With pin end:

With free end: 3

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)

∑ ∑

4

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

Carry-over (CO) factor for beam member without end support 

Recall the Case 1 in derivation of SDEs:

   -

Solving for

A

  

  

  

and equating these eqn,

CO factor The CO factor is always +0.5 for beam member cut from continuous beam or frame - The plus sign indicates both moments act in the same direction

-

(6) 5

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

Carry-over (CO) factor for beam member with end support (or special span with far end pinned or free)

With pin end:

With free end:

(6) 6

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

How to solve Fix End Moment (FEM) Analytically by force method By using table for common loads Note the direction of the moment at two ends which are different from the sign conventions for SDEs.

Real life examples to use displacement method Simple bridges Open garage design One-level factory building

P1-Tutors demonstrate by hand calculation

(1) Divide structures to beam members or end span with support, mark the positive direction of moments and slope.

A

B

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

C

D

2. Distribution factors For Joint A and D are included in the end spans, so it will be disregarded: For Joint B:

A

B

C

Bending stiffness of members connecting to B, i.e., BA and BC:

Distributing factor of members connecting to B:

2. Distribution factors B

C

For Joint C: Bending stiffness of members connecting to B, i.e., BA and BC:

Distributing factor of members connecting to C:

D

3. FEM

4. Iteration with table

Joint

A

Member

AB

DF FEM

0

B

C

D

BA

BC

CB

CD

0.652

0.348

0.348

0.652

24

-100

100

-24

Dis-01

-(-(-26.45 100+2 100+2 4)×0.6 4) 52=49. ×0.348 55 =26.45

Co-01

-13.22

13.22

-49.55

DC

0

4-Con

Joint

A

B

Member

AB

BA

BC

CB

CD

DC

DF

1

0.652

0.348

0.348

0.652

1

FEM

0

24

-100

100

-24

0

49.55

26.45

-26.45

-49.55

-13.22

13.22

4.6

-4.6

Dis-01 Co-01 Dis-02

=-(13.22)×0.6 52=8.62

C

D

-8.62

4-Con-

Joint

A

B

Member

AB

BA

BC

CB

CD

DC

DF

1

0.652

0.348

0.348

0.652

1

FEM

0

24

-100

100

-24

0

49.55

26.45

-26.45

-49.55

-13.22

13.22

4.6

-4.6

-2.3

2.3

Dis-01 Co-01 Dis-02 Co-02

8.62

C

D

-8.62

4-Con

Joint

A

B

Member

AB

BA

BC

CB

CD

DC

DF

1

0.652

0.348

0.348

0.652

1

FEM

0

24

-100

100

-24

0

49.55

26.45

-26.45

-49.55

-13.22

13.22

4.6

-4.6

-2.3

2.3

0.8

-0.8

-0.4

0.4

0.14

-0.14

-0.07

0.07

0.026

-0.026

Dis-01 Co-01 Dis-02

8.62

Co-02 Dis-03

1.5

Co-03 Dis-04

0.26

Co-04 Dis-05

0.044

C

D

-8.62

-1.5

-0.26

-0.044

5. Sum up to get unknown internal moments Joint

A

B

Member

AB

BA

BC

CB

CD

DC

DF

1

0.652

0.348

0.348

0.652

1

FEM

0

24

-100

100

-24

0

49.55

26.45

-26.45

-49.55

-13.22

13.22

4.6

-4.6

-2.3

2.3

0.8

-0.8

-0.4

0.4

0.14

-0.14

-0.07

0.07 -0.026

Dis-01 Co-01 Dis-02

=-(13.22)×0.652 =8.62

Co-02 Dis-03

1.5

Co-03 Dis-04

0.26

Co-04

C

Dis-05

0.044

0.026

Su m

84

-84 84

D

-8.62

-1.5

-0.26

-0.044

-84

6. Sum of results according to requirements

Alternative solution for P1 using a different approach by cutting all supports out all spans A brief alternative solution will be provided for this problem

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

(1) Divide structures to beam members or end span with support, mark the positive direction of moments and slope.

A

B

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

C

D

2. Distribution factors For Joint A and D are included in the end spans, so it will be disregarded: For Joint A:

A

B

Bending stiffness of members connecting to A, i.e., A pin (-) and AB:

Distributing factor of members connecting to A:

For Joint A and D are included in the end spans, so it will be disregarded: For Joint B:

A

B

C

Bending stiffness of members connecting to B, i.e., BA and BC:

Distributing factor of members connecting to B:

B

C

For Joint C: Bending stiffness of members connecting to B, i.e., BA and BC:

Distributing factor of members connecting to C:

For Joint D is similar to Joint A:

D

3. FEM

Due to symmetry:

(4) Table to Do Iteration Joints Member s

A -

DF FEM Dist00 Co01 Dist01 Co02 Dist02 Co03 Dist03 Co04 Dist04 Co05 Dist05 Co06 Dist06 Sum

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

AB

B

C

D

BABC CBCD DC0.714280.28571 0.28571 0.71428 1 64 4 6 1 -16 16-100 100 -16 16 16 60 24 -24 -60 -16 30 8 -12 12 -8 -30 2.85714 1.14285 -30 3 7 1.14286 2.85714 30 1.42857 - 0.57142 1 -15 0.57143 9 15 1.42857 - 11.1224 - 1.42857 1.42857 5 4.44898 4.44898 11.1224 1 5.56122 0.71428 4 0.71429 2.22449 2.22449 6 5.56122 - 2.09912 - 5.56122 5.56122 5 0.83965 0.83965 2.09913 4 1.04956 - 0.41982 2.78061 3 2.78061 0.41983 5 2 1.04956 - 2.28602 0.91441 - 1.04956 1.04956 7 1 0.91441 2.28603 3 1.14301 - 0.45720 0.52478 3 0.52478 0.45721 5 1 1.14301 - 0.70141 0.28056 - 1.14301 1.14301 9 8 0.28057 0.70142 3 - 0.14028 0.57150 0.35071 0.57151 0.14028 4 7 0.35071 - 0.50842 0.20336 0.35071 2 9 0.20337 0.50842 0.35071 0 84 -84 84 -84 0

P2.-Student to do it in class

1. Stiffness factor and distribution factors Stiffness factor for members connected to Joints (not to ends)

Distribution factors for all members: A is a fixed end, so Member BA and BC are connected to B C is a pin end, so

2. Fixed End Moment

3 Form the table and do iteration in the table

Joint

A

B

C

Member

AB

BA

BC

CB

DF

0

0.4

0.6

1

FEM

-25

25

0

0

Dist Co ∑M

3 Iteration in Table

Joint

A

B

C

Member

AB

BA

BC

CB

DF

0

0.4

0.6

1

FEM

-25

25

0

0

-25×0.4=10

-25×0.6=15

Dist Co ∑M

3 Iteration in Table

Joint

A

B

C

Member

AB

BA

BC

CB

DF

0

0.4

0.6

1

FEM

-25

25

0

0

-10

-15

Dist Co ∑M

=-10×0.5=5

3 Sum all iteration to get internal moment

Joint

A

B

C

Member

AB

BA

BC

CB

DF

0

0.4

0.6

1

FEM

-25

25

0

0

-10

-15

=25+(10)=15

=0+(-15)=- 0 15

Dist Co

-5

∑M

=-25+(5)=-30

Please consider why we stop the iteration? Answer: Next step for the iteration is to distribute the carry-over moments at Joints. Only Joint A has a carryover moment and the DF for AB is 0, so all value in the next line will be zeros.

P3. Student to complete after class

1. Stiffness factors and Distribution Factors

Stiffness factors

Distribution factors

2.

3 Iteration in Table

4. Sum up to get unknown internal moments

Please consider why we stop the iteration? Answer: Next step for the iteration is to carry-over the distributed load from near end to far end. However, A and C are pinned end, so the Co=0 from BA to A and from BC to C. So, similar all value in the next line will be zeros.

7. Solve other unknown forces and plot SFD and BMD

plot SFD and BMD on tension side

P4.-Supplimentary for Bridge Project-Excel file will be available on canvas Determine the moments at B and C. EI is constant. Assume B and C are rollers and A and D are fixed.

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

Solution Guide

Joint Member DF FEM Dist00 Co01 Dist01 Co02 Dist02 Co03 Dist03 Co04 Dist04 Sum

-

A AB

B

C D BABC CB CD DC 0.714280.28571 0.28571 0.71428 4 6 0 0 64 -16 16-100 100 -16 16 0 60 24 -24 -60 0 30 0 -12 12 0 -30 8.57142 3.42857 0 9 1 3.42857 8.57143 0 4.28571 - 1.71428 4 0 1.71429 6 0 4.28571 0.48979 0 1.22449 6 -0.4898 1.22449 0 0.61224 0.24489 5 0 -0.2449 8 0 0.61224 0.17492 0.06997 0 7 1 0.06997 0.17493 0 0.08746 - 0.03498 4 0 0.03499 5 0 0.08746 0.00999 0 0.02499 6 -0.01 0.02499 0 19.0 86.0 -86.0 86.0 -86.0 -19.0

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