Please answer the ff in complete solution and draw an fbd use moment distribution method only. Determine the reactions of the given beamusing moment distribution method. w == 25kN/m, P₁ = 25 kN, and P₂ = 25 kNRARBMBRc||=W=8 m21Numerical answer in 4 decimal places andpositive values only:B 2mP₁I2mP22m CkN (Downward)kN-mkN (Upward)kN (Upward) Sign convention-consider clockwise rotation is positive, counterclockwise is negative.MARMBAFixed End Moments (FEM) - moment at the wall or fixed jointMBAMABStiffness Factor - required moment M to rotate the end A of the beam 0₁ = 1 rad.4EI(absolute) K=far end fixed3EIK=far end hingedM1M' = M, far end fixed(relative) K =, E is constantL'3EIM(relative) K =Carry-Over Moment - the moment induced at the fixed end of a beam by the action of the momentapplied at the other end.4L-, far end hingedBM' = 0, far end hingedDistribution Factor, DF - the distribution factor at a joint of a member is the ratio of the bendingstiffness of the member to the sum of bending stiffness of all the members connected to the joint.кDF=;ΣΚNote: DF = 1, for hinge or roller supportDF = 0, for fixed support

Answered: Image /qna-images/answer/f0e1f6ad-d7ae-4448-8694-6d061d6d58cf.jpg

Determine the reactions of the given beam using moment distribution method. w = = 25 kN/m, P₁ = 25 kN, and P₂ = 25 kN RA RB _ || MB Rc _|| = W Numerical answer in 4 decimal places and positive values only: = 8 m 21 B 2m P₁ I 2m P2 2m C kN (Downward) kN-m kN (Upward) kN (Upward)

Determine the reactions of the given beam using moment distribution method. w = = 25 kN/m, P₁ = 25 kN, and P₂ = 25 kN RA RB _ || MB Rc _|| = W Numerical answer in 4 decimal places and positive values only: = 8 m 21 B 2m P₁ I 2m P2 2m C kN (Downward) kN-m kN (Upward) kN (Upward)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

System of Forces

A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum changes. If the body is at rest, under the action of force, its state of rest changes to motion. In other words, the magnitude of the momentum of the body changes from zero to a measurable value. Similarly, if the body possesses some initial velocity, it has some associated momentum, the effect of force can cause the body to halt or come to rest, that is, the momentum reduces to zero.

Question

Please answer the ff in complete solution and draw an fbd use moment distribution method only.

Determine the reactions of the given beam
using moment distribution method. w =
= 25
kN/m, P₁ = 25 kN, and P₂ = 25 kN
RA
RB
MB
Rc
||
=
W
=
8 m
21
Numerical answer in 4 decimal places and
positive values only:
B 2m
P₁
I
2m
P2
2m C
kN (Downward)
kN-m
kN (Upward)
kN (Upward)

Sign convention-consider clockwise rotation is positive, counterclockwise is negative.
MAR
MBA
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

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