Moment Distribution Method1 [Compatibility Mode]

8/13/2019 Moment Distribution Method1 [Compatibility Mode]

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Structural Analysis IMOMENTS DISTRIBUTION METHOD 1

(NO SIDEWAY FRAMES)

Daniel Rumbi Teruna

School of Civil EngineeringUniversity of North Sumatera


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Moment Distribution Method : no sideway frames

Introduction

The moment distribution method is a unique method of structural analysis, in which

solution is obtained iteratively without ever formulating the equations for theunknowns. Consider a rigid jointed frame shown

.

Beam and column in afundamental configuration of

a moment applied at the end.

A Frame with rigid joint at b

bM


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Moment distribution method : no sideway frames

Suffices it to say that given the loading and support conditions shown below, the

rotation band member-end momentMbaat the near end, b, is proportional

The fundamental case and the reaction solutions.

We can write a similar equation forMbcof member bc.

(1)

(1)


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(4)

(5)

Sum of the moments in the member will be equal to external moment

bcbab MMM +=

( ) ( ) ( )

( ) bcbc

ba

babcbabcba M

EK

EKMEKEKMM =+=:

(2)

(3)

Subtituting eq.(3) into eq.(2) gives

( )

( )

( ) ( )

( ) bcbc

bcba

bc

bc

ba

bcb M

EK

EKEKM

EK

EKMM

+=+=

( )( ) ( ) bbcb

babc

bc

bc MDFM

EKEK

EKM =

+

=Or,

and ( )

( ) ( ) bbab

babc

ba

ba MDFM

EKEK

EKM =

+

= (5)


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Furthermore, the moment at the far end of member ab,Mabat a, is related to theamount of rotation at b by the following formula:

Similarly, for member bc,

As a result, the member-end moment at the far end is one half of the near end moment:

bccbbaab MMandMM

2

1,

2

1==

(6)

(7)

The ratio is known as carry-over factor,2/1/,/ =

bccbbaab

MMandMM

The is known as distribution factor of members baand bc, which adds

up to one or 100%bcba

DFandDF

(6)


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Example 1. Find all the member-end moments of the beam shown.EI is constant for all

members.

Solution.

(a) Unbalanced moment: At node b there is an externally applied moment

(EAM), which should be distributed as member-end moments (MEM) in the same sign.

(b) The distribution factors at node b:


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Tabulation: All the computing can be tabulated as shown below. The arrows indicate

the destination of the carryover moment. The dashed lines show how the distribution

factor (DF) is used to compute the distributed moment (DM).


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Post Moment-Distribution Operations. The moment and deflection diagrams are

shown below.

Moment and deflection diagrams.

Example 2. Find all the member-end moments of the beam shown.EI is constant for

all members.


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(a) Only nodes b is free to rotate. There is no externally applied moment at node b

to balance, but the transverse load between nodes create FEMs.

(b) FEM for member ab. The concentrated load of 4 kN creates FEMs at end aand end b. The formula for a single transverse load in the FEM table gives us:

(c) FEM for member bc. The distributed load of 3 kN/m creates FEMs at end b and end

c. The formula for a distributed transverse load in the FEM table gives us:


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(d) Compute DF at b:


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The shear forces at both ends of a member are computed from the FBDs of each

member. Knowing the member-end shear forces, the moment diagram can then be

drawn. The moment and deflection diagrams are shown below.

FBDs of the two members.



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Treatment of Hinged Ends. At a hinged end, the member-end moment (MEM) is

equal to zero or whatever an externally applied moment is at the end. During the

process of moment distribution, the hinged end may receive carried-over moment

from the neighboring node. That COM must then be balanced by distributing 100%of it at the hinged end. This is because the distribution factor of a hinged end is one

or 100%; the hinged end maybe considered to be connected to air which has zero

stiffness.


Example 3. Find all the member-end moments of the beam shown.EI is constant for

all members.


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Compute FEM of members aband bc

Member ab Member bc

Compute distribution factors at b

Assign DF at a and c: DF are one at a and zero c.


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+4


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The cycle of iteration is greatly simplified if we recognize at the very beginning of

moment distribution that the stiffness of a member with a hinged end is

fundamentally different from that of the standard model with the far end fixed

Member with a hinged end vs. the standard

model with the far end fixed.

Note that there is no carry-over-moment at the hinged end (Mba= 0) if we take the

member stiffness factor as 3EK instead of 4EK. We can thus compute the relative

distribution factors accordingly and when distribute the moment at one end of the

member, need not carryover the distributed moment to the hinged end.


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Compute DF at b:

+4


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Example 3. Find all the member-end moments of the frame shown.EI is constant for

all members.

A frame with two node

Only nodes b and c are free to rotate. There is

no side-sway because the support at c prevents

that. Only the transverse load between nodes a

and b will create FEMs at a and b.

Fixed end moments


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Compute DF at c:

Assign DF at a and d: DF is zero at a and d.

Compute DF at b:


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FBDs of the three members and node c.


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Moment Distribution Method1 [Compatibility Mode]

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