CH3 Moment Distribution Beam PDF

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Chapter 3 1 HD in Civil Engineering│ CHAPTER 3 │Moment Distribution Method Learning Objectivesz Acquire the knowledge of determining the degree of kinematic indeterminacy of beams.z Analyse indeterminate beams by using Moment Distribution Method.z Understand the advantages and conditions of using m...

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CBE3027 STRUCTURAL ANALYSIS II

│CHAPTER 3│ │ Moment Distribution Method  z

z

z

z

Learning Objectives Acquire the knowledge of determining the degree of kinematic indeterminacy of beams. Analyse indeterminate beams by using Moment Distribution Method. Understand the advantages and conditions of using modified stiffness in Moment Distribution Method. Acquire the skills and knowledge in drawing free-body diagrams, shear force diagram and bending moment diagram.

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1.

Moment Distribution Method

Moment Distribution Method is a method for analyzing indeterminate beams and frames. The method was published in 1932 by Hardy Cross, a professor of civil engineering at the University of Illinois. From the 1930s through the 1960s, moment distribution was the most widely used method for analysis of structures. Since the 1970s, as computers became increasingly available, the use of the method has gradually been replaced by computer-oriented matrix methods of structural analysis. The development of the moment distribution method has been one of the most notable advances in structural analysis during the 20th century. The main reason for its popularity, especially in the pre-computer era, was due to the fact that it does not require the solution of many simultaneous equations. Moment distribution is still preferred by some engineers for analyzing smaller structures, since it provides a better insight into the behaviour of structures. The method is still useful for checking the results of the computer analysis.

1.1

Sign Convention and Terminology

Sign Convention Clockwise member end moments are considered positive. Counter-clockwise joint moments are considered positive. Since a moment at the end of a member must act in the opposite direction on the adjacent joint, counter-clockwise moments at joints are positive. (Note: This convention is a sign convention for analysis. It is different from the sign convention for moment diagrams, which is a sign convention for design.)

Positive moment: Clockwise on member ends Counter-clockwise on joints

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Fixed End Moments (FEMs) - Extracted from Ref. 1

PL 8

FEM AB =

FEM AB

Pb2 a = 2 L

FEM BA =

FEM BA =

PL 8

Pa2 b L2

FEM AB =

FEM AB =

2PL 9

FEM AB =

15PL

FEM AB =

wL 12

2

11wL2 192

FEM AB =

FEM AB =

FEM BA =

48

FEM AB =

FEM AB =

FEM BA =

wL2

FEM BA =

P L2

FEM AB =

FEM AB =

FEM AB =

5 wL2 192

FEM AB =

wL2 30

⎛ 2 a 2b ⎞ ⎜b a + ⎟ ⎜ 2 ⎟⎠ ⎝

PL 3

45PL 96

wL2 12

FEM BA =

20

2PL 9

15PL 48

FEM BA =

3PL 16

FEM AB =

wL2 8

9 wL2 128

FEM AB =

wL2 15

5 wL2 64

5wL 2 96

FEM BA =

5 wL2 96

6 EI∆

FEM BA =

3 EI∆ 6 EI∆ FEM AB = 2 2 L L

2

L

FEM AB =

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Basic Assumptions Used z

z

z

All the members of the structure are prismatic (EI does not vary within a member, different members could have different EI). All members are primarily subjected to bending. Axial deformations are ignored. The structural behaviour is linear elastic (i.e. obey Hooke’s Law)

Member Stiffness Factor Consider the prismatic beam shown, which is free to rotate at end A and fixed at end B. The relationship between the applied moment M and the rotation θ at A ⎛ 4EI ⎞ is given by M = ⎜ ⎟θ . ⎝ L ⎠

θ

M

B A L EI = Constant The bending stiffness or stiffness factor K of a member is defined as the moment that must be applied at an end of the member to cause a unit rotation of that end. By setting θ = 1 rad., we obtain the member stiffness factor. Stiffness factor K = 4EI/L Distribution Factors (DFs)

When analyzing a structure by the moment distribution method, an important question that arises is how to distribute a moment applied at a joint among the various members connected to that joint. Consider the frame shown on the next page, where an unbalanced moment occurs when joint E is theoretically unlocked.

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B

B

P A

P C

E

D

E A

C

FEM EC

FEM CE

D

(a)

(b) Joints A and E assumed fixed

B

θ EB θEA

E

A

θ

C

EC

θED D Unbalanced moment = - FEM EC (c)

θ EA = θEB = θ EC = θ ED Σ ME = 0 = FEMEC – MEA – MEB – MEC - MED Moment equilibrium at E requires that ΣME = 0 = MEA + MEB + MEC + MED + Σ (FEM)E where Σ(FEM)E represents the algebraic sum of the FEMs at joint E.

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The member moments at joint E are in direct proportion to each other by their stiffness values because all the members undergo the same joint rotation at E, i.e. MEA MEB M EC M ED = = = K EA K EB K EC K ED Substituting the above into the moment equilibrium equation give the following results: M EA = [KEA / Σ KE]* ΣFEME = DFEA* ΣFEME M EB = [KEB / Σ KE]* ΣFEME = DFEB* ΣFEME M EC = [KEC / Σ KE]* ΣFEME = DFEC* ΣFEME M ED = [KED / ΣKE]* ΣFEME = DFED* Σ FEME

where and

ΣKE = KEA + KEB + KEC + KED DFEA = KEA / ΣKE DF EB = KEB / ΣKE DF EC = KEC / ΣKE DF ED = KED / ΣKE

The stiffness ratios given above are referred to as distribution factors. The resisting moments listed above are also known as distribution moments. In general, the distribution factor (DF) for a member is equal to the stiffness factor of the member divided by the total stiffness of the joint. DF =

K ΣK

The sum of the DFs at a joint is always equal to 1.0

It is also useful to note that distribution factors can be computed from the relative stiffnesses(I/L or 1/L) of the members. It is not necessary to use the absolute stiffness values.

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Carry-over Moment and Carry-over Factor The moment MBA induced at the fixed end B of member AB by a moment M acting at A is called a carry-over moment.

M BA= M/2

θ

M

B A L

EI = Constant The carry-over moment is half the magnitude of the acting moment and has the same sign as the acting moment.

The ratio of the induced moment to the applied moment is called the carry-over 1 factor (COF). For prismatic members, the carry-over factor is equal to . 2 1 ) (i.e. COF = 2 1.2

Basic Concepts of the Moment Distribution Method

The moment distribution method is a step-by-step procedure for calculating the end moments in members of indeterminate beams and frames. The method is based on the idea that the sum of moments in members framing into a joint must equal zero because the joint is in equilibrium. First, all the joints in the structure are assumed to be fixed and the external loads are applied – given rise to fixed-end moments. Next, those joints that can actually rotate are allowed to do so. This adds moments to the fixed-end moments according to the relative stiffnesses of the members. Thirdly, the carry-over effects of the added moments on the opposite end of each member are considered. The final moments will be the sum of all the above moments. Carrying out moment distribution procedure requires the use of (a) Fixed-end moments (FEMs) (b) Distribution factors (DFs) (c) Carry-over moments or carry-over factor (COs or COF) (d) A sign convention must be adopted.

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1.3

Introduction to the Moment Distribution Method

Consider the two-span continuous beam in the figure below. The beam will deflect as shown by the dotted line and develop the end moments MAB, MBA, MBC and MCB, which are to be determined.

8 kN A

C

B 10m

10m

10m

A

B

B

M BA

M AB

C

M BC

M CB

(a) Actual Structure

8 kN

Locking moment

A

C

B 10m

10m

10m 20

A 20

B

B 20

C

B 20

(b) Structure with all joints locked

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unlocking moment A

C

B 10m

10m

10m 20 13.3

A 3.3

B 6.7

6.7

B 13.3

B

C 6.7

(c) Unlocking of joint B Fixed end moment = PL/8 = 8*20/8 = 20 kNm Locking moment applied at joint B = 20 kNm (clockwise) Unlocking moment applied at joint B = 20 kNm (counter-clockwise) First of all, we assume the beam to be fixed at all joints as in figure (b). This gives rise to fixed end moment of 20 kNm at both ends of member AB and makes it necessary to apply an artificial moment of 20 kNm to joint B. This external moment, called a locking moment, is required to prevent joint B from rotating. As joint B is free to rotate in the actual structure, we have to apply an unlocking moment to remove the external moment. The unlocking moment is equal and opposite to the locking moment. For equilibrium at joint B, the unlocking moment must be balanced by moments applied to the joint by members AB and BC. (in other words, when joint B rotates, the rotation is resisted by the members framing into the joint). The magnitudes of these resisting moments are proportional to the stiffnesses (4EI/L) of the members. Therefore the moment in member BC (13.33 kNm) is twice that of the moment in member AB (6.67 kNm). The rotation of joint B induces moments NOT ONLY at B BUT ALSO at A and C. The moment induced at A due to the rotation at B is called a carry-over moment (3.33 kNm), it is half that of the member end moment at B (6.67 kNm) and has the same direction.

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The final moments in the actual structure are obtained by adding the member moments produced by the unlocking process to the fixed end moments. M AB = 20 + 3.3 = 23.3 kNm M BA = 20 – 6.7 = 13.3 kNm M BC = 0 + 13.3 = 13.3 kNm M CB = 0 + 6.7 = 6.7 kNm

In carrying out the moment distribution process for an actual structure, it is not necessary to consider both the member moments and the joint moments as in this example. It is adequate to calculate only the moments that act on the ends of the members.

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2.

ANALYSIS OF CONTINUOUS BEAMS (BY MOMENT DISTRIBUTION METHOD)

1.

Calculate the relative stiffness (1/L or I/L or EI/L) of each member.

2.

Calculate the distribution factors at each joint that can rotate. (a) The distribution factor for a member end is computed by dividing the relative stiffness of the member by the sum of the relative stiffnesses of all the members connected to the joint. (b) The sum of all distribution factors at a joint must equal 1.

3.

Find the fixed end moments for the member ends (clockwise fixed end moments are considered to be positive.)

4.

Find the balancing moments at each joint required to provide equilibrium. (a) At each joint, evaluate the unbalanced moments. (b) Distribute the balancing moments to the members connected to the joint according to the distribution factors.

5.

Carry over one-half of each distributed moment to the opposite end of the member. (a) A carry-over moment has the same sign as the distributed moment. (b) There is no carry-over from fixed support.

6.

Repeat steps 4 & 5 until the unbalanced moments are acceptably small (say 1% of the final moment) or until all the free joints are balanced.

7.

Determine the final member end moments by summing up the fixed end moment and all the distributed and carry-over moments at each member end.

8.

Self-check:

The final moments must satisfy equilibrium at all the joints.

Calculate the Support Reactions After the member end moments have been found, support reactions may be found as follows: (1) Member end shears may be found by considering the equilibrium of each member of the structure. (2) Support reactions may be found by considering the equilibrium of each joint of the structure.

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Draw the Shear Force Diagram Having calculated the support reactions, draw the shear force diagram from the left hand side to right hand side following the loading direction as in the determinate beams. Locate the positions of the zero shear force as these points will be the positions for the maximum or minimum bending moments. Draw the Bending Moment Diagram The member end moments will be the support moments. As for the span moments, determine the positions of the zero shear force and hence calculate the span moments by the methods of “free body diagram” or “the area of shear force diagram”.

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Example 1

Analyze the 2-span beam shown below by moment distribution method. EI is constant for both spans.

40 kN 4.5 kN/m A

D 2m

B

C

2m

8m

Solution

1. Calculate the distribution factors at all joints that are free to rotate, i.e. joint B. K 4EI / 4 2 4 EI / 8 1 = , DFBC = = DFBA = BA = ΣK B (4 EI / 4 + 4 EI / 8 ) 3 (3 * 4 EI / 8) 3 2. Assuming all joints to be locked, the fixed end moments are calculated. PL 40 * 4 =− = −20 kNm, FEM BA = 20 kNm FEM AB = − 8 8 2 2 wL 4.5 * 8 FEM BC = − =− = −24 kNm, FEM CB = 24 kNm 12 12

3. Begin the actual moment distribution process.

1

2

Joint Member DF FEM

A AB ---20

BA 2/3 20

BC 1/3 -24

C CB --24

Dist.

---

2.67

1.33

---

CO

1.34

---

---

0.67

Dist.

---

---

---

---

Final Mt.

-18.66

22.67

-22.67

24.67

B

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4.

The final moment at the end of each member is obtained by adding the moments developed during each of the preceding steps.

40 kN 4.5 kN/m 22.67 22.67

18.66 19

D

A

24.67

21

B

17.75

B 21 22.67

C 18.25

17.75 22.67

B 38.75

5. Shear forces can be determined by considering the equilibrium of the members. Consider member AB, Take moment about B, VAB * 4 + 22.67 – 18.66 – 40*2 = 0 ∴ VAB = 19 kN ΣY = 0, VAB + VBA – 40 = 0 ∴ VBA = 21 kN Consider member BC, Take moment about C, VBC * 8 + 24.67 – 22.67 – 4.5*8*4 = 0 ∴ VBC = 17.75 kN ΣY = 0, VBC + VCB – 4.5*8 = 0 ∴ VCB = 18.25 kN Consider joint B, ΣY = 0, VBA + VBC – VB = 0 V B = 21 + 17.75 = 38.75 kN

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Shear Force and Bending Moment Diagrams

19

3.944 m

19 17.75

C A

D

-21

B -21

-18.25

Shear Force (kN)

-24.67

-22.67

-18.66

D B

A

C 12.33

19.34

Bending Moment (kNm)

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Example 2 Analyze the 2-span beam shown below by moment distribution method. constant for both spans.

EI is

5 kN/m B

EI

A

C

EI

6m

9m

Solution Distribution Factors:

DFBA =

Stiffness of member BA = 4EI/6 = 0.667EI Stiffness of member BC = 4EI/9 = 0.444EI

0.667 EI = 0.60, (0.667 + 0.444)EI

DFBC = 1 − 0.6 = 0.4

Fixed End Moments:

For member AB: FEM AB FEM BA

For member BC: FEM BC

FEM CB Moment Distribution: Joint Member DF 1 FEM Dist. 2 CO Dist. Final Mt.

5*62 wl 2 =− =− = −15 kNm 12 12 2 5*62 wl = = = 15 kNm 12 12 5* 92 wl 2 =− =− = − 33.75 kNm 12 12 wl 2 5 * 9 2 = = = 33.75 kNm 12 12 A AB ---15 --5.625 ---9.375

B BA 0.6 15 11.25 ----26.25

BC 0.4 -33.75 7.5 -----26.25

C CB --33.75 --3.75 --37.5

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A

5kN/m

9.375

5kN/m

26.25

A 12.1875 12.1875

37.5 C

26.25

B

B

17.8125 17.8125

B 21.25

C 23.75 23.75

21.25

39.0625 21.25 12.1875

A

Shear Force (kN)

C

B 2.4375m -17.8125 4.25m

-23.75 -37.5 -26.25

-9.375 A

C

B 5.4785

Bending Moment (kNm) 18.9

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Example 3A

Determine the internal moments at each support of the beam shown.

20 kN A 3m

E

20 kN 8 kN/m

B 3m

C 6m

F 3m

D 3m

EI constant

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Solution

20 kN

20 kN A

E

3m

8 kN/m

B

C

3m

F 3m

6m

D 3m

EI constant

Lock all joints

Unlock joint B

K K / ΣK --FEM -15

1

1 0.5 0.5

0.5 0.5

15 -24

24 -15

DM

To eliminate this artificial moment, apply +9 to joint B. After the process of distribution, joint B ...