🎓 mecademyAI 정역학 Internal Forces and Moments Problem 7_50
Engineering Mechanics: Statics 9th Edition · Internal Forces and Moments · Problem 7_50
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Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_50

⚡ Mecademy AIENG정역학 · ch7  Problem Statement The double-axle front suspension is used on small trucks. In a test of the designed action, the frame must be jacked up so that in order to relieve the compression in the coil springs. Determine the value of when the jack is removed. Each spring has a stiffness of . The load is , and the central frame has a mass of . Each wheel and attached link have a mass of with a center of mass from the vertical centerline. Problem 7/50 (a) Analysis of Equilibrium Position ● Calculation Process 1. Identify Given Values and System Parameters: Uncompressed height: Spring stiffness: (per spring) Applied load on frame: Mass of frame : Mass of each wheel/link: Horizontal positions from centerline (CL): Inner pivot of lower link: Spring center: Wheel center: Center of mass of wheel/link: 2. Kinematic Analysis and Leverage Factor: Assume the wheels are on the ground and the frame moves vertically by an amount . The outer end of the lower link (at the wheel) remains at a constant height. The link rotates about this outer pivot. Fh=350 mm h 120 kN/mL12 kNF40 kg 35 kg680 mm h h = 0 350 mm=0.35 m k=120 kN/m=120,000 N/m L=12 kN=12,000 N Fm = F 40 kg→W = F 40×9.81=392.4 N m = w 35 kg→W = w 35×9.81= 343.35 N x = 1 400/2=200 mm=0.2 m x = s 500/2=250 mm=0.25 m x = w 750 mm=0.75 m x = G 680 mm=0.68 m δ=h − 0 h Downward displacement of the frame pivot: Distance between outer pivot and inner pivot: Downward displacement of the link at the spring seat (): Since t

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주어진 조건: . In, 350 mm, 120 kN, 40 kg, 680 mm, 0.35 m

구하는 것: (a) Analysis of Equilibrium Position ● Calculation Process 1

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📘 교재 정보

Engineering Mechanics: Statics · 9th Edition
저자: Meriam, Kraige & Bolton
출판사: Wiley
단원: Internal Forces and Moments