On the given beam, find the following unknowns use the SLOPE DEFLECTION METHOD. Assume that E is constant. Provide a thorough and well-organized solution. a. Fixed-end moment (FEM) on joint B of member AB in kN-m. Indicate negative sign if counterclockwise. FEMBA b. Moment at joint D in kN-m. Indicate negative sign if counter-clockwise. MDB c. Moment at joint B of member BC in kN-m. Indicate negative sign if counter-clockwise. MBC d. Vertical reaction at roller A in kN. RA e. Vertical reaction at roller C in kN. RC

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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On the given beam, find the following unknowns use the SLOPE DEFLECTION METHOD. Assume that E is constant. Provide a thorough and well-organized solution.

a. Fixed-end moment (FEM) on joint B of member AB in kN-m. Indicate negative sign if counterclockwise. FEMBA
b. Moment at joint D in kN-m. Indicate negative sign if counter-clockwise. MDB
c. Moment at joint B of member BC in kN-m. Indicate negative sign if counter-clockwise. MBC
d. Vertical reaction at roller A in kN. RA
e. Vertical reaction at roller C in kN. RC

Please DO NOT ANSWER WITH OTHER METHODS. 

5 m
21
27 kN/m
5 m
12 kN
3 m
21
5 m
I
5 m
11 kN
D
Transcribed Image Text:5 m 21 27 kN/m 5 m 12 kN 3 m 21 5 m I 5 m 11 kN D
(FEM) AB
(FEM)AB=
=
(FEM)AB=
|-} + {
2PL
9
(FEM) AB=
(FEM)AB=
(FEM)
(FEM) AB=
(FEM)AB
PL
8
=
Pb²a
[²
=
| - 4 + 4 + 4 + ¼ -
SPL
16
wL²
12
P
11w[²
192
W
20
P
W
SwL²
D
W
P
-L
P
(FEM)BA =
B
(FEM) RA= 1²
1/3
(FEM)BA =
B
11/2014
B
(FEM) BA=
11-
(FEM)BA =
B
(FEM)
(FEM)BA=
B
(FEM)BA=
B
PL
8
=
Pa²b
2PL
9
5PL
16
wL2
12
5wL²
192
wL²
30
SwL²
DC
3PL
(FEM) AB= 16
(FEM) AB= (b²a + b)
2
(FEM) AB=
(FEM) AB =
A
(FEM) AB=
A
(FEM) AB
W
(FEM) AB
(FEM) AB
13
12
PL
45PL
96
=
P
wL²
8
9w1,²
128
w[.²
WE
SwL²
P
W
2+
P
ļ
22
P
22
Transcribed Image Text:(FEM) AB (FEM)AB= = (FEM)AB= |-} + { 2PL 9 (FEM) AB= (FEM)AB= (FEM) (FEM) AB= (FEM)AB PL 8 = Pb²a [² = | - 4 + 4 + 4 + ¼ - SPL 16 wL² 12 P 11w[² 192 W 20 P W SwL² D W P -L P (FEM)BA = B (FEM) RA= 1² 1/3 (FEM)BA = B 11/2014 B (FEM) BA= 11- (FEM)BA = B (FEM) (FEM)BA= B (FEM)BA= B PL 8 = Pa²b 2PL 9 5PL 16 wL2 12 5wL² 192 wL² 30 SwL² DC 3PL (FEM) AB= 16 (FEM) AB= (b²a + b) 2 (FEM) AB= (FEM) AB = A (FEM) AB= A (FEM) AB W (FEM) AB (FEM) AB 13 12 PL 45PL 96 = P wL² 8 9w1,² 128 w[.² WE SwL² P W 2+ P ļ 22 P 22
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