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DISPLACEMENT MEDTHOD DISPLACEMENT MEDTHOD OF ANALYSIS: OF ANALYSIS: MOMENT DISTRIBUTION MOMENT DISTRIBUTION Member Stiffness Factor ( Member Stiffness Factor ( K K ) ) Distribution Factor (DF) Distribution Factor (DF) Carry-Over Factor Carry-Over Factor Distribution of Couple at Node Distribution of Couple at Node Moment Distribution for Beams Moment Distribution for Beams General Beams General Beams Symmetric Beams Symmetric Beams Moment Distribution for Frames: No Moment Distribution for Frames: No Sidesway Sidesway Moment Distribution for Frames: Moment Distribution for Frames: Sidesway Sidesway
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moment distribution method

Oct 30, 2014

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ppt on moment distribution teaching aid
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Page 1: moment distribution method

DISPLACEMENT DISPLACEMENT MEDTHOD OF ANALYSIS: MEDTHOD OF ANALYSIS:

MOMENT DISTRIBUTIONMOMENT DISTRIBUTION

Member Stiffness Factor (Member Stiffness Factor (KK))Distribution Factor (DF)Distribution Factor (DF)Carry-Over FactorCarry-Over FactorDistribution of Couple at NodeDistribution of Couple at NodeMoment Distribution for BeamsMoment Distribution for BeamsGeneral BeamsGeneral BeamsSymmetric BeamsSymmetric BeamsMoment Distribution for Frames: No SideswayMoment Distribution for Frames: No SideswayMoment Distribution for Frames: SideswayMoment Distribution for Frames: Sidesway

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General Principles and DefinitionsGeneral Principles and DefinitionsThus the Moment Distribution Method (also known as the Cross Method) became Thus the Moment Distribution Method (also known as the Cross Method) became

the preferred calculation technique for reinforced concrete structures. the preferred calculation technique for reinforced concrete structures. The description of the moment distribution method by Hardy Cross is a little The description of the moment distribution method by Hardy Cross is a little

masterpiece. He wrote: "Moment Distribution. The method of moment masterpiece. He wrote: "Moment Distribution. The method of moment distribution is this:distribution is this:

Imagine all joints in the structure held so that they cannot rotate and compute Imagine all joints in the structure held so that they cannot rotate and compute the moments at the ends of the members for this condition;the moments at the ends of the members for this condition;

at each joint distribute the unbalanced fixed-end moment among the connecting at each joint distribute the unbalanced fixed-end moment among the connecting members in proportion to the constant for each member defined as "stiffness";members in proportion to the constant for each member defined as "stiffness";

multiply the moment distributed to each member at a joint by the carry-over multiply the moment distributed to each member at a joint by the carry-over factor at the end of the member and set this product at the other end of the factor at the end of the member and set this product at the other end of the member; member;

distribute these moments just "carried over"; distribute these moments just "carried over"; repeat the process until the moments to be carried over are small enough to be repeat the process until the moments to be carried over are small enough to be

neglected; andneglected; and add all moments - fixed-end moments, distributed moments, moments carried add all moments - fixed-end moments, distributed moments, moments carried

over - at each end of each member to obtain the true moment at the end." over - at each end of each member to obtain the true moment at the end." [Cross 1949:2] [Cross 1949:2]

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1. Restrain all possible displacements.

2. Calculate Distribution Factors:

The distribution factor DFi of a member connected to any joint J is

                                              

where S is the rotational stiffness , and is given by

                                      

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3. Determine carry-over factors

The carry-over factor to a fixed end is always 0.5, otherwise it is 0.0.

4. Calculate Fixed End Moments. (Table 3.1).

These could be due to in-span loads, temperature variation and/or• relative displacement between the ends of a member.

5. Do distribution cycles for all joints simultaneously

Each cycle consists of two steps:1. Distribution of out of balance moments Mo,2.Calculation of the carry over moment at the far end of each member.

The procedure is stopped when, at all joints, the out of balance moment is a negligible value. In this case, the joints should be balanced and no carry-over moments are calculated.

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6. Calculate the final moment at either end of each member.

This is the sum of all moments (including FEM) computed during the distribution cycles.

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ExampleExample

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Stiffness-Factor ModificationStiffness-Factor Modification

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Example

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Symmetric Beam and LoadingSymmetric Beam and Loading

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Symmetric Beam with Symmetric Beam with Antisymmetric LoadingAntisymmetric Loading

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Moment Distribution for frames:Moment Distribution for frames: No sidesway No sidesway

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Moment Distribution for frames: Moment Distribution for frames: sideswaysidesway

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