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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 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)

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
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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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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