Moment Distribution1

8/11/2019 Moment Distribution1

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Moment-Distribution

The method of moment distribution relies on a series of calculations that are repeated and that with everycycle come closer to the final situation. In this way we are able to avoid solving simultaneous equations.Inspection of the slope-deflection equations shows us that the final end-moments depend on 4 effects

namely, A, B, ABand the fied end moments, !"#. By using moment-distribution we are able to investigateeach effect separately. The following beam will be used to illustrate moment-distribution.

$otation is possible at both B and %

$otation at B and % are prevented and the load is applied.!"# will result. These are called the initial moments.

Allow B to rotate until moment equilibrium is reached.$otation at B will induce a moment at %.

Allow % to rotate until moment equilibrium is reached. Therotation of % will induce a moment at B.

$epeat this process until moment equilibrium is reached atthe nodes.

Assume that the sum of the initial moments at the node B is equal to #&.

$otation will ta'e place until moment equilibrium is attained, i.e., sum moments #B( &.

Therefore)*here are the moments as a result of the

rotation at B, B, and are called thedistribution moments. $emember that all the other rotations and sway are prevented.

. In a similar fashionBut)

+olve for

B.

+olve the distribution moments.

'BAis the stiffness of themember BA at the node B. Itis also the moment thatwould be induced if a unit rotation were applied at B in the member BA and the rotation at A was ero.

If B rotates a bending moment will be induced at A and %. Assume a rotation Band calculate the moment atA.

Moment-distribution Page 1 of 13

&&D D

BA BC M M M+ + =

D D

BA BC M and M

( ) 4

D AB AB BBA B

AB AB

EI EI M

L L

= =( )

4D BC BC B

BC B

BC BC

EI EI M

L L

= =

&&D D

BA BC M M M+ + =

&

44BBCAB

AB BC

M

EIEI

L L

=

+

&

& &

4

44

AB

D BA BAAB

BA

BCAB BA BC B

AB BC

EIM

k M k M LM

EIEI k k k

L L

= = =

++

&

& &

4

44

BC

D BC BC BC

BC

BCAB BA BC B

AB BC

EIM

L k M k M M

EIEI k k k

L L

= = =

++


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, but bear in mind that

The distributed bending moment is halfthe value of the distributed bendingmoment at B. This is called the carry-over factor, %BA( .

The same solution may be obtained if one remembers that the stiffness of a member is the moment that isinduced if a unit rotation is applied at the node.


( )D AB

AB B

AB

ELM

L

= &

44BBCAB

AB BC

M

EIEI

L L

=

+ &

44

AB

D ABAB

BCAB

AB BC

EIM

LM

EIEI

L L

= +

( ) ( )- - 4

- - /,&AB AB ABAB AB A

AB AB AB

EI EI EI M k

L L L

= = = =

( ) ( )- - -

/,&AB AB ABBA AAB AB AB

EI EI EI

ML L L

= = =

/

BA

AB

AB

MC

M= =


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"ample)

0se the method of moment-distribution to determine the bending moment diagramme of the following beam.

1istribution at A and B

+tiffness of membersat A)

1istribution!actors

+tiffness of members at B)

Initial #oments)


4 4/,&

4

AB

AB

AB

EI EIk

L

= = =

/,&/,&

/,&

AB

AB

kD

k= = =

/,&k=

4 4/,&

4

BA

BA

AB

EI EIk

L

= = =

/,&&,2&

/,22223

AB

AB

kD

k= = =

4 4 &,222232

BC

BC

BC

EI EIk

L

= = =

&,22223&,4&

/,22223

BC

BC

kD

k= = =

/,22223k=

& /& 4,& .

5 5AB AB

W LM FEM kN m

= = = = +

& /& 4,& .

5 5BA BA

W LM FEM kN m

= = = = - -

& 2/,& .

/- /-BC BC

w LM FEM kN m

= = + = + =- -

& 2/,& .

/- /-CB CB

w LM FEM kN m

= = = =


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1istribution of the moments)

%arry over factors

Action #AB #BA #B% #%B

1istribution factors /,&& &,2& &,4&

Initial moments 6 ,&&& - ,&&& 6 /,&&& - /,&&&

Allow rotation of A - ,&&& - ,&&

Allow rotation of B - ,& - 4,&& -7,&&& -/,&&

Allow rotation of A 6 ,& 6 /,/

Allow rotation of B - &,773 - &,23 - &,4& - &,

Allow rotation of A 6 &,773 6 &,/255

Allow rotation of B - &,&/ - &,/&/7 - &,&23 - &,&775

Allow rotation of A 6 &,&/ 6 &,&

- &,&/7 - &,&/&

&,&&& - //,43 6 //,43 - /2,38

Members with a hinge on one side:

Assume a member with a hinge at B.

+tiffness ( moment required toinduce a unit rotation at A)

$edo eample / using the stiffness of amember with a hinge.

#BA #B% #%B

1ist !actors &,84/ &,43&8

Init moments - 3,&& 6 /,&&& - /,&&&

$otate B - 7,83&2 - 7,84 - /,3243

-//,43&2 6 //,43&2 -/2,3243


( )7 7

AB AB

AB AB A

AB AB

EI EI k M

L L

= = =

& /

AB AB BA

M FEM FEM =

7 7&,3&

4

BA

BA

AB

EI EIk

L

= = =

&,3&&,84/

/,4/223

AB

AB

kD

k= = =

4 4 &,222232

BC

BC

BC

EI EIk

L

= = =

&,22223&,43&8

/,4/223

BC

BC

kD

k= = =

/,4/223k=

& / / 3,& .

5 5BA BA AB

W L W LM FEM FEM kN m

= = + =

- -

& 2/,& .

/- /-BC BC

w LM FEM kN m

= = + = + =- -

& 2/,& .

/- /-CB CB

w LM FEM kN m

= = = =


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"ample 7)

1etermine the bending momentdiagramme of the following structure.

$otation willoccur at B and%.

+tiffness at B

Initial moments)

#AB #B" #B% #%B #%1 #1%

&,33 &,/5/5 &,4 &,2&& &,4&&

6/&,&&&& -/&,&&&

-,33 -/.5/5 -.4 -,35

67,5/5 63,273 6,&8/ 6,4

-/,&4/ -&,284 -,&57 -/,&4/

6&,7/ 6&,2 6&,4/2 6&,&5

-&,&5 -&,&3 -&,/3& -&,&5

6&,&/ 6&,&74

-7,57 -,28 62,4 -,4/ 6,4/ 6,37

Structural Frames with Sway.

!rames with a sway mechanism may be tac'led by preventing the sway and calculating the force required toprevent the sway, call this 9. Arbitrary sway is then applied to the structure and the force that leads to thearbitrary sway is calculated, call this :. Apply the super-position equation as neither of the forces are reallythere.

9 6 : ( &

!inal bending moment ( Bending moment with sway prevented 6 times bending moment with arbitrarysway.


7 7 /,

4

BA

BA

AB

EI EIk

L

= = =

/,&,33

,

BA

BA

kD

k= = =

7 7 /,&7

BE

BE

BE

EI EIk

L

= = =

/,&&,/5/5

,

BE

BE

kD

k= = =

4 4 7 7,&&4

BC

BCBC

EI EIk

L

= = =

7,&&,4

,

BC

BC

kD

k= = = ,k=4 4 7

7,&4

CB

CB

CB

EI EIk

L

= = =

7,&&,2&&

,&

CB

CB

kD

k= = =

4 4 ,&4

CD

CD

CD

EI EIk

L

= = =

,&&,4&&

,&

CD

CD

kD

k= = =

,&k=

& -& 4/&,& .

5 5BC BC

W LM FEM kN m

= = + = + = +

& /& 4/&,& .

5 5CB CB

W LM FEM kN m

= = = =


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"ample 4)

1etermine the bending moment diagramme of the following sway structure. The support A is a hinge, " isfied and 1 is a roller. There is a hinge in B% at %.

!orce 9 prevents thesway and force :induces the arbitrarysway.Apply force 9 to preventthe sway. $otation willoccur at B and at %.



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#BA #B% #%1#%" #"%

&,73 &,2 &,2 &,4444

/,&&& -&,&&& 6/&,&&& 64,&&& 64,&&

67,&&& 6,&&& -/5,58& -/,//& -3,

6/,&&& -/,&&& -5,58& 65,58& 6/2,44

- 4 : 6 7 /& @/2,44 6 ,&23 /&@ /,45/3 8 ( &

: ( /,373 '

+uperpositionequation)

9 6 : ( & ( &,227/7

!inal bending moments)

#!( # sway prevented6 # arbitrary sway

#BA #B% #%1 #%" #"%

-/2,53 6/2,53 & & & # sway prevented

68,843 -8,843 -,58 6,58 6/&,8& # arbitrary sway

-2,85 62,85 -,58 6,58 6/&,8& #!

!inal Bending #oment1iagramme.

Structure withDis"lacemento# a Su""ort

"ample )

1etermine the bending moment diagramme of the structure if " ( && C9a, I ( /& /& -2m4and the support" moves & mm to the right.


= &/OM


10/13


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#AB #BA #B% #%B #%1 #%"

&,452 &,3/4 &,442 &,74&8 &,&4

6/3&,&&& 6//&,&&& -3,&&& -8&,&&&

-72,47/ -3,52 -83,/75 -45,28

67,4/ 643,&5 67,7&3 6/,/5&

-,&4 -/&,&8& -/7,4/ -2,32

6/,8 67,&5 6,87 6/,73

-&,75 -&,2 -&,534 -&,473

6&,&88 6&,/88 6&,/48 6&,&58

-&,&4 -&,&3

-4/,5&4 -57,248 657,248 6/&4,2&3 -73,/ -23,72

Final Bending MomentDiagramme



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Moment Distribution1

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