Use the Moment Distribution Method to analyse the frame when subjected to the lo
ID: 1867102 • Letter: U
Question
Use the Moment Distribution Method to analyse the frame when subjected to the loading shown. Dimensions and load values given in the table below, which depend on your student ID, are identical to those of e Task 4.1. Determine the member forces and support reactions required in the table below. Enter your answers in the space provided. Dimensions: a= 4. 2m El= 260000 kNm² KN Applied actions: = W= 255 KN 27 kN/m a- b - - H Provide Answer for the followings to an accuracy of 2.5% Result required Absolute Value Unit Direction Member CF - Moment at F: Mec= Member CF - Moment at C: Mee Member BC - Moment at C: Mce= Member BC - Moment at B: Mgc= Member AB - Moment at B: MBA= Member AB - Moment at Point Force F: MF= Member BE - Moment at B: MBE Horizontal Reaction at F: Fx= Vertictal Reaction at A: Ay=Explanation / Answer
STEP - 1,Fixed End Moment Calculation
In Beam AB
MFAB = -Pab2 / L2 = -(255x4.2x3.62) / (4.2+3.6)2 = -228.14 KN-m
MFBA = Pba2 / L2 = (255x3.6x4.22) / (3.6+4.2)2 = 266.17 KN-m
Now in Beam CF
MFCF = -wL2 / 12= -(27x32) / 12 = -20.25 KN-m
MFFC = wL2 / 12= (27x32) / 12 = 20.25 KN-m
In the members BE, BC and CD there is no Load/Moment/Deflection, so
MFBE = MFEB = MFBC = MFCB = MFCD = MFDC = 0
STEP - 2 , Calculation of Distribution Factor
Joints
Members
Relative Stiffness(K)
Summation of ’K’
D.F =K/(summation of ’K’)
B
BA
(3/4)X(I/L)=0.096I
0.496I
0.193
BE
(3/4)X(I/L)=0.25I
0.504
BC
I/L=0.15I
0.303
C
CB
I/L=0.15I
0.59I
0.25
CD
(3/4)X(I/L)=0.11I
0.19
CF
I/L=0.33I
0.56
STEP - 3 ,Moment Distribution
JOINTS
A
B
C
F
D
E
MEMBERS
AB
BA
BE
BC
CB
CD
CF
FC
DC
EB
D.F
1
0.193
0.504
0.303
0.25
0.19
0.56
1
1
1
F.E.M
-228.14
266.17
0
0
0
0
-20.25
20.25
0
0
BAL
228.14
C.O
114.07
BAL
-73.4
-191.64
-115.2
5.06
3.85
11.34
C.O
2.53
-57.6
5.67
BAL
-0.49
-1.275
-0.76
14.4
10.94
32.26
C.O
7.2
-0.38
16.13
BAL
-1.39
-3.63
-2.18
0.095
0.07
0.213
C.O
0.047
-1.09
0.1
FINAL
0
304.96
-196.55
-108.36
-39.52
14.86
23.35
42.15
0
0
* all moments are in KN-m
now, in memeber AB,for moment at point force
due to simply supported action , moment at point force = Pab/L = (255x3.6x4.2) / 7.8 = 494.31 KN-m (sagging)
and end moments at A and B, from the table
MFAB = 0 , MFBA = 304.96 (Clockwise)
End moments vary linerly in the span, so at Point Force = (304.96/7.8)x4.2 =164.21 KN-m
and the direction of this moment is clockwise, that means hogging
so moment at pont force, MF = 494.31 - 164.21 = 330.1 KN-m
due to simply supported action, vertical reaction at A, V1 = Pb/L = (255x3.6)/7.8 = 117.7 KN (upward)
also due to end moment, vertical reaction at A, V2 = (0-304.96) / 7.8 = -39 KN
so total vertical reaction at A, V = V1 + V2 = 117.7-39 = 78.7 KN (upward)
Now in Beam FC,
due to simply supported action Horizontal Reaction at F, H1 = wL/2 = (24X3)/2 = 36 KN (Leftward)
also End momets, MFCF = 23.35 and MFFC = 42.15
so due to end momets Horizontal Reaction will be, H2 = (42.15+23.35) / 3 = 21.83 KN (Leftward)
so total Horizontal Reaction at F , H = H1 + H2 = 34 + 21.83 =55.83 KN (Leftward)
Joints
Members
Relative Stiffness(K)
Summation of ’K’
D.F =K/(summation of ’K’)
B
BA
(3/4)X(I/L)=0.096I
0.496I
0.193
BE
(3/4)X(I/L)=0.25I
0.504
BC
I/L=0.15I
0.303
C
CB
I/L=0.15I
0.59I
0.25
CD
(3/4)X(I/L)=0.11I
0.19
CF
I/L=0.33I
0.56
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