Question 3:The beam is subjected to the uniform distributed load shown. Draw the shear andmoment diagrams for the beam.Take:A =C kN/m2B =1.5mC =kN /mBВ mA m1mSolution:Equation of Equilibrium:C(A+1) kNC kNlm0.5(A+1) mAx4AMImFecA meter(a)(b)2MA = 0; Fec (3 /5)A – 0.5C(A +1)° = 0F, = 0; A, + Fec (3/5)- C(A +1) = 0kNA, =kNkNV, =M(KN.M)mX2v3M1V2kNvlV, =kNx2x2м,M3M, =v2(C)(d)

Answered: Image /qna-images/answer/68190f00-f388-4cda-9acb-4a763728d7b0.jpg

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

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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