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@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

@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
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 but I don't know where I have gone wrong, please correct my mistakes using the SAME method 

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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