Answer using moment distribution method. Use second picture as guide W= 25 P1 = 25 P2 = 25
Answer using moment distribution method. Use second picture as guide W= 25 P1 = 25 P2 = 25
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Answer using moment distribution method. Use second picture as guide
W= 25
P1 = 25
P2 = 25

Transcribed Image Text:Sign convention - consider clockwise rotation is positive, counterclockwise is negative.
MBA
MAR
Fixed End Moments (FEM) - moment at the wall or fixed joint
MBA
MAB
Stiffness Factor - required moment M to rotate the end A of the beam 0₁ = 1 rad.
4EI
(absolute) K =
far end fixed
3EI
K=far end hinged
M
1
M' = M, far end fixed
(relative) K ==, E is constant
L'
3EI
M
(relative) K =
Carry-Over Moment - the moment induced at the fixed end of a beam by the action of the moment
applied at the other end.
4L
-, far end hinged
B
M' = 0, far end hinged
Distribution Factor, DF - the distribution factor at a joint of a member is the ratio of the bending
stiffness of the member to the sum of bending stiffness of all the members connected to the joint.
к
DF=;
ΣΚ
Note: DF = 1, for hinge or roller support
DF = 0, for fixed support

Transcribed Image Text:يلبليستي
RA =
RB =
Numerical answer in 4 decimal places and positive values only:
Me =
Rc=
W
8 m
21
kN-m
kN (Downward)
kN (Upward)
| kN (Upward)
P1
B 2m
Pz
2m
2m C
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