1 Question 3:2 The beam is subjected to the uniform distributed load shown. Draw the shear andmoment diagrams for the beam.345Take:C kN/m1m670.751AB mA =B =C =8kN/m910111213141516 Solution:17 Equation of Equilibrium:18190.5(A+1) m20212223242628293031323334353637383940414243444516vl0וזןA meter(a)M = 0;Fac (3/5)A-0.5C (A+1)² = 0F₁ = 0; A, +Fc (3/5)-C (A+1)=0V(K)M(KAM)v3MIx2x2v2C(A+1) KNFBC(d)AyFBCA,4 mC kN/mAM(b)KNKNV, (kN) =x₂ (m) =V₂(KN)=-ON) V₂(KN)=M₁(kNm) =M, (kNm) =1mIm

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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

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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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and after loading.

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