The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take : A = C kN/m m 3 B = 2.25 m C = 3 kN/ m B В m A m 1 m Solution : |Equation of Equilibrium: C kNlm C(A+1) kN 0.5(A+1) m Ax AM Im Ay Fec A meter (a) (b)

International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter9: Moments And Products Of Inertia Of Areas
Section: Chapter Questions
Problem 9.16P: Figure (a) shows the cross-sectional dimensions for the structural steel section known as C1020...
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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
m
3
B =
2.25
m
C =
3
kN/ m
B
В т
A m
Solution :
Equation of Equilibrium:
C kNIm
C(A+1) kN
0.5(A+1) m
Ax
AM
Im
Ay
Fec
A meter
(a)
(b)
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 m 3 B = 2.25 m C = 3 kN/ m B В т A m Solution : Equation of Equilibrium: C kNIm C(A+1) kN 0.5(A+1) m Ax AM Im Ay Fec A meter (a) (b)
Ум -0; Fsc(3 /5)4 - 0.5C (4 +1)' %3Dо
УF - 0; А, +Fsc (3/5)-с (4 +1)- о
FBC
kN
BC
Ar
kN
V, (kN)=
x, (m)=
M(KN.m)
v3
M1
vl
v,(kN)=
-Lim) v,(kN):
х2
x2
M,(kNm) =
M,(kNm) =
M3
v2
(C)
(d)
Transcribed Image Text:Ум -0; Fsc(3 /5)4 - 0.5C (4 +1)' %3Dо УF - 0; А, +Fsc (3/5)-с (4 +1)- о FBC kN BC Ar kN V, (kN)= x, (m)= M(KN.m) v3 M1 vl v,(kN)= -Lim) v,(kN): х2 x2 M,(kNm) = M,(kNm) = M3 v2 (C) (d)
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