Determine the reactions of the given beam using slope deflection method. w = 25 kN/m and P = 45 kN RA= MA= Ra Numerical answer in 4 decimal places and positive values only: - ↓↓ Mec kN (Upward) kN-m (ccw) kN (Upward) W kN-m (ccw) ↓↓ 6m B 2m P
Determine the reactions of the given beam using slope deflection method. w = 25 kN/m and P = 45 kN RA= MA= Ra Numerical answer in 4 decimal places and positive values only: - ↓↓ Mec kN (Upward) kN-m (ccw) kN (Upward) W kN-m (ccw) ↓↓ 6m B 2m P
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Please answer in complete solution and draw an fbd if needed.
Note:Please use the second picture as a guide for the formulas. Thank you!°
![Determine the reactions of the given beam using slope deflection method. w = 25 kN/m and P = 45 kN
RA =
Numerical answer in 4 decimal places and positive values only:
MA =
RB =
MBC =
kN (Upward)
kN-m (ccw)
kN (Upward)
W
kN-m (ccw)
6m
B
2m
P](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff60c02a2-b6aa-4cbf-943d-25d00ff95caf%2F80e9dfab-5707-4eda-98e1-fedb7184c931%2Fxnjrv6r_processed.png&w=3840&q=75)
Transcribed Image Text:Determine the reactions of the given beam using slope deflection method. w = 25 kN/m and P = 45 kN
RA =
Numerical answer in 4 decimal places and positive values only:
MA =
RB =
MBC =
kN (Upward)
kN-m (ccw)
kN (Upward)
W
kN-m (ccw)
6m
B
2m
P
![wa 23a 2. SaL+6L3
2
12L²
14
PL
8
Pab²
L²
Mb(b-2a)
L²
wy
L
M
L
L
P
L
L
L
L
b
Fixed End Moment Formulas
12
W
W[2
8
wa (a²+ 41².4aL)
8L2
wa ³ (4L-3a)
121²
Pa²b
Ma(2b-a)
L
12:
10 3
3 PL
16
Pab(L+b)
22²
3a 1.5a
HO
120
2
-a-
L
MK
L
L
L
L
L
b
18
W2
HA
W
MAB
deflection -0₂-²
curve -L-
P
hum
cord
El is constant
positive sign convention
Fig. 11-2, Structural Analysis,R.C.Hibbeler 8th ed
MAB = 2E20A + 0B - 3+ FEM AB
MBA= -2E20B +0A-33
A
(FEM) AB
-3+ FEM BA
Sign convention: All clockwise internal moments and end rotation are positive.
MBA
For end span with far end pinned or roller support
P
mini
2EI
L
W
B
Fig.11-Sb, Structural Analysis, R.C.Hibbeler, 8th ed
3EI
MN=[N+ FEMN
General Form
MNEF= [20N+0F-34+ FEMNF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff60c02a2-b6aa-4cbf-943d-25d00ff95caf%2F80e9dfab-5707-4eda-98e1-fedb7184c931%2Fjzcawk_processed.png&w=3840&q=75)
Transcribed Image Text:wa 23a 2. SaL+6L3
2
12L²
14
PL
8
Pab²
L²
Mb(b-2a)
L²
wy
L
M
L
L
P
L
L
L
L
b
Fixed End Moment Formulas
12
W
W[2
8
wa (a²+ 41².4aL)
8L2
wa ³ (4L-3a)
121²
Pa²b
Ma(2b-a)
L
12:
10 3
3 PL
16
Pab(L+b)
22²
3a 1.5a
HO
120
2
-a-
L
MK
L
L
L
L
L
b
18
W2
HA
W
MAB
deflection -0₂-²
curve -L-
P
hum
cord
El is constant
positive sign convention
Fig. 11-2, Structural Analysis,R.C.Hibbeler 8th ed
MAB = 2E20A + 0B - 3+ FEM AB
MBA= -2E20B +0A-33
A
(FEM) AB
-3+ FEM BA
Sign convention: All clockwise internal moments and end rotation are positive.
MBA
For end span with far end pinned or roller support
P
mini
2EI
L
W
B
Fig.11-Sb, Structural Analysis, R.C.Hibbeler, 8th ed
3EI
MN=[N+ FEMN
General Form
MNEF= [20N+0F-34+ FEMNF
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