🎓 mecademyAI 정역학 Distributed Forces: Centroids and Centers of Gravity Problem 5_103
Engineering Mechanics: Statics 9th Edition · Distributed Forces: Centroids and Centers of Gravity · Problem 5_103
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Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_103

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The quarter-circular cantilever beam is subjected to a uniform pressure on its upper surface as shown. The pressure is expressed in terms of the force per unit length of circumferential arc. Determine the reactions on the beam at its support in terms of the compression , shear , and bending moment . Problem 5/103 (a) Support Reactions at A 1. Formula: We represent the reactions at the fixed support as a compressive axial force , a shear force , and a bending moment . The beam is a quarter- circular arc of radius subjected to a uniform normal pressure (force per unit length). We define a coordinate system with the center of curvature at the origin , support at , and the free end at . The differential force acting on a segment at angle (measured from the vertical -axis) is: The equilibrium equations for the entire beam are: 2. Substitution: Substitute the expressions for the differential force and integrate from to : p AC A V A M A A C A V A M A rp O(0,0) A(0,r)B(r,0) ds=rdθθy dF=p(rdθ)(−sinθi−cosθj) F = ∑ x 0⇒R + Ax d F = ∫ x 0 F = ∑ y 0⇒R + Ay d F = ∫ y 0 M = ∑ A 0⇒M + A r × ∫ P/A dF=0 θ=0θ=π/2 F = x −p r sin θ d θ = ∫ 0 π/2 −p r [−cos θ ] = 0 π/2 −pr[0−(−1)]=−pr F = y −p r cos θ d θ = ∫ 0 π/2 −p r [sin θ ] = 0 π/2 −pr[1−0]=−pr The position vector from to a point on the arc is: The differential moment about is: 3. Calculation: Simplify the differential moment expression: Integrate to find the total moment of the load about : Apply

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주어진 조건: , a, 0 M

구하는 것: (a) Support Reactions at A 1

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Engineering Mechanics: Statics · 9th Edition
저자: Meriam, Kraige & Bolton
출판사: Wiley
단원: Distributed Forces: Centroids and Centers of Gravity