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The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take: 1 m C kN/m 0.75 1 B m A = B = C = kN/m Solution: Equation of Equilibrium: 0.5(A+1) m Ax Ayk vl וזן 2 C(A+1) KN A meter M=0; Fac (3/5) A-0.5C (A+1)² = 0 ΣF, = 0: A, +Fc (3/5)-C(A+1)=0 V(K) M(KNM) MI Fac x2 Ay 4 m Fac A₁ = C kN/m Am (b) KN KN V₁ (KN) = x₂ (m) = V₂ (KN)= -x) V₁ (KN) = 1m Im

The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take: 1 m C kN/m 0.75 1 B m A = B = C = kN/m Solution: Equation of Equilibrium: 0.5(A+1) m Ax Ayk vl וזן 2 C(A+1) KN A meter M=0; Fac (3/5) A-0.5C (A+1)² = 0 ΣF, = 0: A, +Fc (3/5)-C(A+1)=0 V(K) M(KNM) MI Fac x2 Ay 4 m Fac A₁ = C kN/m Am (b) KN KN V₁ (KN) = x₂ (m) = V₂ (KN)= -x) V₁ (KN) = 1m Im

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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1 Question 3: 2 The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. 3 4 5 Take: C kN/m 1 m 6 7 0.75 1 A B m A = B = C = 8 kN/m 9 10 11 12 13 14 15 16 Solution: 17 Equation of Equilibrium: 18 19 0.5(A+1) m 20 21 22 23 24 26 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 16 vl 0 וזן A meter (a) M = 0; Fac (3/5)A-0.5C (A+1)² = 0 F₁ = 0; A, +Fc (3/5)-C (A+1)=0 V(K) M(KAM) v3 MI x2 x2 v2 C(A+1) KN FBC (d) Ay FBC A, 4 m C kN/m AM (b) KN KN V, (kN) = x₂ (m) = V₂(KN)= -ON) V₂(KN)= M₁(kNm) = M, (kNm) = 1m Im
Transcribed Image Text:1 Question 3: 2 The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. 3 4 5 Take: C kN/m 1 m 6 7 0.75 1 A B m A = B = C = 8 kN/m 9 10 11 12 13 14 15 16 Solution: 17 Equation of Equilibrium: 18 19 0.5(A+1) m 20 21 22 23 24 26 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 16 vl 0 וזן A meter (a) M = 0; Fac (3/5)A-0.5C (A+1)² = 0 F₁ = 0; A, +Fc (3/5)-C (A+1)=0 V(K) M(KAM) v3 MI x2 x2 v2 C(A+1) KN FBC (d) Ay FBC A, 4 m C kN/m AM (b) KN KN V, (kN) = x₂ (m) = V₂(KN)= -ON) V₂(KN)= M₁(kNm) = M, (kNm) = 1m Im
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