@2) 40KN 75421M зий G 0.5m 0.25m 0.25m 0.75m [ ↓ ↓ ↓ ↓ ↓ D C y + M D RA RB EMA = 0. -3 +75 (0.75) (0.5 +0.5 + 0.75)-RB (1) - 40 (0.75) εFy = 0 2 Rg = 44.34375 N (1) 20 KA + 40+RB = 7500.75) - ·RA - 75 (0-15)-40-44.34375 RA 2 -28.09375 (4) RA= 28.09375) (↓) A+ x = 0 : M = O 2 x at x = 0.5 MO - RA CO·5) = 0 RACO ·5)=0 PAS -MX = 28.09375CO.5) = 14.04иNm at x = 0.5 Cincluding 3KN) ↑ RA G SUN & Moc 0.75 = A+ x = 0.75 : Mx 40 KN α+ x = 3 N t 0.25m RA 0.5m 0.25m J Mx MX-RA CO-75)+3=0 Mx=RACO·75)-3 Mx = 18.07 UNM 26 MX-RA (0.75)+3 -40 co·25) = 0 Mx - 28.07 uNm at x = 1 Mx = 0 -7 Mmax = 28.02am = X -24.07 kNm A beam ABC is pin-jointed at points A and B and is shown below. Calculate the maximum bending moment in kNm using anticlockwise moment as negative. Y X 3kN A 40kN 75 kN/m B C 0.5 m 0.25 m 0.25 m 0.75 m

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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I have attempted to answer it, but I do not know where I have gone wrong. Please answer and correct the below using the exact same method I have used( Section method of a pinned beam)

Correct answer = -24.07kNm

@2)
40KN
75421M
зий
G
0.5m
0.25m
0.25m
0.75m
[ ↓ ↓ ↓ ↓ ↓ D
C
y
+ M
D
RA
RB
EMA = 0.
-3 +75 (0.75) (0.5 +0.5 + 0.75)-RB (1) - 40 (0.75)
εFy = 0
2
Rg = 44.34375 N (1)
20
KA + 40+RB = 7500.75)
-
·RA - 75 (0-15)-40-44.34375
RA
2
-28.09375 (4)
RA= 28.09375) (↓)
A+ x = 0 :
M = O
2
x
at x = 0.5 MO - RA CO·5) = 0
RACO ·5)=0
PAS
-MX =
28.09375CO.5) = 14.04иNm
at x = 0.5 Cincluding 3KN)
↑
RA
G SUN
& Moc
0.75
=
A+ x = 0.75
: Mx
40 KN
α+ x =
3 N
t
0.25m
RA
0.5m 0.25m
J Mx
MX-RA CO-75)+3=0
Mx=RACO·75)-3
Mx = 18.07 UNM
26
MX-RA (0.75)+3 -40 co·25) = 0
Mx - 28.07 uNm
at x = 1 Mx = 0
-7
Mmax = 28.02am
=
X
-24.07 kNm
Transcribed Image Text:@2) 40KN 75421M зий G 0.5m 0.25m 0.25m 0.75m [ ↓ ↓ ↓ ↓ ↓ D C y + M D RA RB EMA = 0. -3 +75 (0.75) (0.5 +0.5 + 0.75)-RB (1) - 40 (0.75) εFy = 0 2 Rg = 44.34375 N (1) 20 KA + 40+RB = 7500.75) - ·RA - 75 (0-15)-40-44.34375 RA 2 -28.09375 (4) RA= 28.09375) (↓) A+ x = 0 : M = O 2 x at x = 0.5 MO - RA CO·5) = 0 RACO ·5)=0 PAS -MX = 28.09375CO.5) = 14.04иNm at x = 0.5 Cincluding 3KN) ↑ RA G SUN & Moc 0.75 = A+ x = 0.75 : Mx 40 KN α+ x = 3 N t 0.25m RA 0.5m 0.25m J Mx MX-RA CO-75)+3=0 Mx=RACO·75)-3 Mx = 18.07 UNM 26 MX-RA (0.75)+3 -40 co·25) = 0 Mx - 28.07 uNm at x = 1 Mx = 0 -7 Mmax = 28.02am = X -24.07 kNm
A beam ABC is pin-jointed at points A and B and is shown below. Calculate the maximum bending moment in kNm using
anticlockwise moment as negative.
Y
X
3kN
A
40kN
75 kN/m
B
C
0.5 m
0.25 m
0.25 m
0.75 m
Transcribed Image Text:A beam ABC is pin-jointed at points A and B and is shown below. Calculate the maximum bending moment in kNm using anticlockwise moment as negative. Y X 3kN A 40kN 75 kN/m B C 0.5 m 0.25 m 0.25 m 0.75 m
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