chapter 6.6: moment distribution method · pdf file“moment distribution method” 2....
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Chapter 6.6:
Moment Distribution Method
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“Moment Distribution Method”
1. Stiffness, K – the amount of force/moment necessary to
produce a unit displacement/rotation.
Absolute Stiffness = 4EI / L *Prismatic
Relative Stiffness = I / L
* K for overhang is always zero
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“Moment Distribution Method”
2. Distribution Factor, DF – a value which determines the
appropriate amount of moment to be distributed to a
member.
Where: K – stiffness of the member
∑jtK – sum of stiffness of all members meeting
at a joint
* DF for hinge/roller = 1, for fixed support =0
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“Moment Distribution Method”3. Carry-Over Factor, COF
For prismatic members: COF = 1 / 2
4. Fixed-End Moments, FEM – moments developed at the ends of amember due to applied loads considering the beam is fixed at bothends.
*For overhangs: No need to assume fixed End
5. Distributed Moment, DM
DM = (∑jtM)(-DF)
CO = ½ (DM)
FM = ∑FEM + ∑CO+ ∑DM
*Counterclockwise positive
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“Fixed-End Moments (FEM)”
ω
L
-ωL2/12+ωL2/12
*Sign Convention: Counterclockwise Positive
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“Fixed-End Moments (FEM)”
*Sign Convention: Counterclockwise Positive
P
a b
+Pab2/L2 -Pa2b/L2
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“Fixed-End Moments (FEM)”
*Sign Convention: Counterclockwise Positive
ω
L
+ωL2/30 -ωL2/20
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“Fixed-End Moments (FEM)”
*Sign Convention: Counterclockwise Positive
ω
P
L
Use Double-Integration Method to get FEM
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Example 1: Determine the reactions at the supports.
4m
3 kN-m
A B
5 kN/m 20 kN
6m 6m 6m
C D
2m
E
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Example 1: Determine the reactions at the supports.
10m
300 kN-m
A
B
144 kN/m 1200 kN
4m 5m 10m
C D E
5m
200 kN
I = 20 I = 20 I = 20I = 30