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一、建立振动轮的有关力学模型设振动轮上分配重量为M,振动轮质量为M_1(M_1不包括偏心轴质量),偏心轴质量为m,偏心轴的偏心距为e,偏心轴的静偏心矩为u(u=m·e),偏心轴的转速为ω,振动轮的直径为D,振动轮的宽度为B,偏心轴的转动惯量为I,压实过程中土壤的下沉量为Se。当偏心轴以转速ω高速旋转时,产生离心力为: F=mew~2 (2) 只有当F/Mg>1时,振动轮才有可能跳离地面,离心力的垂直分力为:
First, establish the relevant mechanical model of the vibrating wheel Set the weight on the vibratory wheel to be M, the mass of the vibrating wheel to be M_1 (M_1 does not include the mass of the eccentric shaft), the mass of the eccentric shaft is m, the eccentricity of the eccentric shaft is e, the static of the eccentric shaft The eccentric moment is u (u=m·e), the eccentric shaft rotation speed is ω, the vibration wheel diameter is D, the vibration wheel width is B, the eccentric shaft’s moment of inertia is I, and the soil subsidence during the compaction process For Se. When the eccentric shaft rotates at high speed ω, the centrifugal force is: F=mew~2 (2) Only when F/Mg>1, the vibrating wheel may jump off the ground. The vertical component of the centrifugal force is: