Moment distribution method

39
BY DR. MAHDI DAMGHANI Moment distribution method

Transcript of Moment distribution method

Page 1: Moment distribution method

BYDR. MAHDI DAMGHANI

Moment distribution method

Page 2: Moment distribution method

Introduction

It was developed by Prof. Hardy Cross (one of America's most brilliant engineers) in 1932

It is also known as Hardy cross method

This method was originally though of to analyse reinforced concrete structures

A very powerful method of analysing indeterminate continuous beams and frames

Page 3: Moment distribution method

Indeterminate structure

How many unknowns does the beam have?How many equations do we have (knowns) to

solve the unknowns?

15 kN/m 10 kN/m150 kN

8 m 6 m 8 m

AB C

DI I I

3 m

Page 4: Moment distribution method

Example

Page 5: Moment distribution method

Back to our problem

Fix all the joints (assume all the joints in the structure have no ability to rotate)

Calculate the Fixed End Moments (FEM)Allow the joints that were fixed artificially to

rotate freelyThe unbalanced moments created must be

balanced out based on the relative stiffness of members and carry over factor

Sum up the moments

Page 6: Moment distribution method

See it in practice

15 kN/m 10 kN/m150 kN

8 m 6 m 8 m

AB C

DI I I

3 m

8 m

-80 kN.m +80 kN.m15 kN/m

A B6 m

-112.5kN.m 112.5 kN.m

B C 8 m

-53.33 kN.m10 kN/m

C D

150 kN53.33 kN.m

3 m

We are taking anti-clockwise moments as negative

Page 7: Moment distribution method

How fixed end moment are calculated

In beam ABFixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.mFixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m

In beam BCFixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62

= -112.5 kN.mFixed end moment at C = + (Pab2)/l2 = + (150)(3)(3)2/62

= + 112.5 kN.mIn beam ABFixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.mFixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m

You can get FEMs from next slide for various types of loadings

Page 8: Moment distribution method

Fixed End Moment Table-

-

-

-

-

-

-

Page 9: Moment distribution method

Fixed End Moment Table

Page 10: Moment distribution method

Release the fixed joints

Now we allow those joints that were artificially fixed to rotate freely

Due to the joint release, the fixed end moments on either side of joints B, C and D act in the opposite direction now, and cause a net unbalanced moment to occur at the joint

15 kN/m 10 kN/m

8 m 6 m 8 m

A B C DI I I

3 m

150 kN

Released moments -80.0

-112.5 +53.33 -53.33+112.5

unbalanced moment +32.5 -59.17 -53.33

Page 11: Moment distribution method

Unbalanced moment

The joint moments are distributed to either side of the joint B, C or D, according to their relative stiffnesses

These distributed moments also modify the moments at the opposite side of the beam span, i.e. at joint A in span AB, at joints B and C in span BC and at joints C and D in span CD.

Modification is dependent on the carry-over factor (which is equal to 0.5 in this case)15 kN/m 10 kN/m

8 m 6 m 8 m

A B C DI I I

3 m

150 kN

Released moments -80.0

-112.5 +53.33 -53.33+112.5

unbalanced moment +32.5 -59.17 -53.33

Page 12: Moment distribution method

Essential concepts

Before continuing we need to know about Stiffness Distribution factor Carry-over factor

Page 13: Moment distribution method

Stiffness

Stiffness = Resistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions

Note: The above stiffness is obtained assuming that the opposite support is fixed, if it is not the case you may use

LEIK 4

Flexural stiffness of a member

LEI

LEIK 34

43

Page 14: Moment distribution method

Distribution factor

For each member at each node is

Note: unbalanced moment is distributed between members based on DF of each member, i.e.

KKDF member

member

Summation of flexural stiffness of all connected members at a particular joint

memberunbalancedmember DFMM

Page 15: Moment distribution method

Carry-over factor

If MA is applied to the beam below it causes;

Then, half of this moment goes to end B;

A

MAMB

A BA

RA RB

L

E, I – Member properties

EILM A

A 4

AB MM )21(

Carry over factor

Page 16: Moment distribution method

Back to our problem

Calculate distribution factor for all members

00.1

4284.0500.0667.0

500.0

5716.0500.0667.0

667.0

5716.0667.05.0

667.0

4284.0667.05.0

5.0

0.0)(5.0

5.0

DC

DCDC

CDCB

CDCD

CDCB

CBCB

BCBA

BCBC

BCBA

BABA

wallBA

BAAB

K

KDF

EIEIEI

KK

KDF

EIEIEI

KK

KDF

EIEIEI

KK

KDF

EIEIEI

KK

KDF

stiffnesswallEI

KK

KDF

Page 17: Moment distribution method

FEM

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

Page 18: Moment distribution method

Distribution of FEM

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

Page 19: Moment distribution method

Carry-over factor

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

Page 20: Moment distribution method

Re-calculate unbalancing moment and redistribution

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

Page 21: Moment distribution method

Carry-over factor effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725

Page 22: Moment distribution method

Calculate unbalancing moment and distribute it

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725

Page 23: Moment distribution method

Carry-over effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395

Page 24: Moment distribution method

Re-calculate unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5

Page 25: Moment distribution method

Distribute the unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5+1.365

+1.825

+1.871

+1.406

-2.395

Page 26: Moment distribution method

Carry-over effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5+1.365

+1.825

+1.871

+1.406

-2.395+0.68

2+0.93

5+0.91

2-1.197 0.703

Page 27: Moment distribution method

Unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5+1.365

+1.825

+1.871

+1.406

-2.395+0.68

2+0.93

5+0.91

2-1.197 0.703

+0.935

-0.285

Page 28: Moment distribution method

Distribute the unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5+1.365

+1.825

+1.871

+1.406

-2.395+0.68

2+0.93

5+0.91

2-1.197 0.703

+0.935

-0.285-0.4 -0.535 +0.16 +0.12

5-0.703

Page 29: Moment distribution method

Sum up the moments

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

95

DistributionCarry over (M*0.5) -

16.89+9.29

5-26.665 -12.695

-32.5+13.91

+18.59

+59.17-

33.78-25.39 -53.33

Unbalancing moment

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

+6.955

+3.61 +4.95 +4.8305

+6.347 +3.725+4.95 +11.1

77-2.1186

-2.8314

-6.382 -4.79 -3.725-

1.0593-3.191 -

1.4157-

1.8625-2.395-3.191 -3.278 +2.39

5+1.365

+1.825

+1.871

+1.406

-2.395+0.68

2+0.93

5+0.91

2-1.197 0.703

+0.935

-0.285-0.4 -0.535 +0.16 +0.12

5-0.703

81.69MM A

81.10197.99

BR

BL

MM

90.9790.97

BR

BL

MM 0DM

Page 30: Moment distribution method

Example 2

Draw the moment diagram for the beam below

Page 31: Moment distribution method

Answer 2

Page 32: Moment distribution method

Example 3

The beam below is simply supported at A, cantilevered at D. IAB=8500cm4, IBC=6500cm4 and ICE=5500cm4

Page 33: Moment distribution method

Solution 3

Page 34: Moment distribution method

Distribution factors

Page 35: Moment distribution method

FEM calculation

Page 36: Moment distribution method

Table

Page 37: Moment distribution method

Bending moment diagram

Page 38: Moment distribution method

Example 4

Page 39: Moment distribution method

Example 5