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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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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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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.
Moment and deflection diagrams.
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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.
Moment distribution method : no sideway frames
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 and deflection diagrams.
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