Moment distribution method

BYDR. MAHDI DAMGHANI

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

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

DI I I

Example

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

See it in practice

15 kN/m 10 kN/m150 kN

8 m 6 m 8 m

DI I I

-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

150 kN53.33 kN.m

We are taking anti-clockwise moments as negative

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

Fixed End Moment Table-

Fixed End Moment Table

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

150 kN

Released moments -80.0

-112.5 +53.33 -53.33+112.5

unbalanced moment +32.5 -59.17 -53.33

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

150 kN

Released moments -80.0

-112.5 +53.33 -53.33+112.5

unbalanced moment +32.5 -59.17 -53.33

Essential concepts

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

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

LEIK 34

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

Carry-over factor

If MA is applied to the beam below it causes;

Then, half of this moment goes to end B;

E, I – Member properties

EILM A

AB MM )21(

Carry over factor

Back to our problem

Calculate distribution factor for all members

4284.0500.0667.0

5716.0500.0667.0

5716.0667.05.0

4284.0667.05.0

0.0)(5.0

wallBA

EIEIEI

stiffnesswallEI

0.428 0.5710 1

A B C D0.571 0.428

-80 +80 -112.5

+112.5

-53.33

+53.33

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

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

Re-calculate unbalancing moment and redistribution

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Carry-over factor effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Calculate unbalancing moment and distribute it

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Carry-over effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Re-calculate unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Distribute the unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Carry-over effect

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Distribute the unbalancing moment

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

Sum up the moments

-16.89

-17.37+7.2

2+9.66

1+9.91

8+7.45 +12.6

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

90.9790.97

MM 0DM

Example 2

Draw the moment diagram for the beam below

Answer 2

Example 3

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

Solution 3

Distribution factors

FEM calculation

Bending moment diagram

Example 4

Example 5

Moment distribution method

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