Question 3: The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take: A = C kN/m 2 B = 1.5 m C = kN /m B В т A m 1m Solution: Equation of Equilibrium: C(A+1) kN C kNlm 0.5(A+1) m Ax 13 4 Fec AM Im A meter (a) (b) 2MA = 0; Fec (3 /5)A – 0.5C(A +1)° = 0 F, = 0; A, + Fec (3/5)- C(A +1) = 0 FRC kN A, = kN kN V, = M(KN.M) m X2 v3 M1 V2 kN vl V, = kN x2 x2 м, M3 M, = v2 (C) (d)

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question 3:
The beam is subjected to the uniform distributed load shown. Draw the shear and
moment diagrams for the beam.
Take:
A =
C kN/m
2
B =
1.5
m
C =
kN /m
B
В m
A m
1m
Solution:
Equation of Equilibrium:
C(A+1) kN
C kNlm
0.5(A+1) m
Ax
4
AM
Im
Fec
A meter
(a)
(b)
2MA = 0; Fec (3 /5)A – 0.5C(A +1)° = 0
F, = 0; A, + Fec (3/5)- C(A +1) = 0
kN
A, =
kN
kN
V, =
M(KN.M)
m
X2
v3
M1
V2
kN
vl
V, =
kN
x2
x2
м,
M3
M, =
v2
(C)
(d)
Transcribed Image Text:Question 3: The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take: A = C kN/m 2 B = 1.5 m C = kN /m B В m A m 1m Solution: Equation of Equilibrium: C(A+1) kN C kNlm 0.5(A+1) m Ax 4 AM Im Fec A meter (a) (b) 2MA = 0; Fec (3 /5)A – 0.5C(A +1)° = 0 F, = 0; A, + Fec (3/5)- C(A +1) = 0 kN A, = kN kN V, = M(KN.M) m X2 v3 M1 V2 kN vl V, = kN x2 x2 м, M3 M, = v2 (C) (d)
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