§ 7 － 1 Introduction § 7 － 2 Motion Equation of a Mechanical System § 7 － 5 Introduction to Aperiodic speed Fluctuation and Its Regulation § 7 － 4 Periodic Speed Fluctuation and Its Regulation § 7 － 3 Solution of the Motion Equation of a Mechanical System Chapter 7 Chapter 7 Motion of Mechanical Systems and Its Regulation

一、 Introduction In the kinematic analysis of a mechanism, the motion of the input link should be given. In most cases, the input link is supposed to run at a constant speed. This is only an approximation to reality. The actual motion of the input link depends on the mass distribution of the mechanism and the external forces acting. on the. mechanism. In this chapter, we will study the actual motion of the mechanism according to the mass distribution of the mechanism and the external forces acting on it. § 7 － 1 § 7 － 1 Introduction

Start 二、 Three operating phases of a machine ω t Stop Running T T ωmωm 1. Starting phase 2. Steady working phase ω → ω m W d = W c + E 2)Angular velocity remains constant ω m ＝ constant W d = W c 1)Cyclical fluctuation ω(t)=ω(t+T p ) ω m ＝ constant W d = W c 3. Stopping phase E = -W c

三、 Driving and resistance forces act on machines 1.Driving force A different driving motor will provide a different driving force which is a function of different kinematic parameters. Function relations between the driving force and the kinematic parameters are called mechanical characteristic of driving motors. 2. Resistance force On studying the movement of a machine under external forces with analytical method, the driving force provided by the driving motor must be given as the analytical formula.

B A y O x φ1φ1 As slider crank mechanism is shown in Fig.. Let crank act as the input member.The angular velocities, masses, mass center locations, velocities of the mass center, and moments of inertia of the all links are given. The driving moment is M 1 and working resistance force is F 3. 一、 General Expression of the Equation of Motion According to the principle of work and energy, we have dE=dW=Ndt S2S2 S3S3 S1S1 J 2 m 2 m3m3 J1J1 ω1ω1 M1M1 F3F3 Equation of motion ： § 7 － 2 § 7 － 2 Motion Equation of a Mechanical System

Choose the crank as an equivalent link, the above Eq. Can be rewritten as. Ｏ A 1 φ1φ1 JeJe ω1ω1 MeMe J e —equivalent moment of inertia ， M e —equivalent moment of force Equivalent model S2S2 S3S3 S1S1 J 2 m 2 m3m3 J1J1 B A y O x φ1φ1 F3F3 ω1ω1 MeMe

For a planar mechanism consisting of n moving links, the general equation of motion is ： where α i is the angle between F i and v i. If the directions of M i and ω i are identical, then “ ＋ ”is used,otherwise, “ － ” is used. 二、 Dynamically Equivalent Model of a Mechanical System A mechanical system with one degree of freedom can be assumed as an imaginary link. Such an imaginary link is called the equivalent link, or the dynamically equivalent model, of a mechanical system.

Choose link 3 as an equivalent link, then S2S2 S3S3 S1S1 J 2 m 2 m3m3 J1J1 B A y O x φ1φ1 F3F3 ω1ω1 MeMe 3 v３v３ s3s3 meme FeFe m e —Equivalent mass ，Ｆ e —Equivalent force Equivalent model

The rotate link is the equivalent one. The sliding link is the equivalent one. Equivalent mass Equivalent force Equivalent moment of force Equivalent moment of inertia

三、 Other Forms of Motion Equation 1. The equations of motion in differential form 2. The equations of motion in moment of force form 3. The equations of motion in energy integral form

一、 The equivalent moment of inertia is a function of position and the equivalent moment of force is a function of position. 二、 The equivalent moment of inertia is a constant and the equivalent moment of force is a function of velocity. 三、 The equivalent moment of inertia is a function of position and the equivalent moment of force is a function of position and velocity. § 7 － 3 § 7 － 3 Solution of the Motion Equation of a Mechanical System

一、 Reasons for Periodic Speed Fluctuation The driving moment M d and the resistant moment M r acting on the machinery are a periodic function of rotating angle φ of driving motor. Thus, the equivalent moment of force is a periodic function of equivalent rotating angle φ. MdMd MrMr a b c d e a'a' φ MeMe The kinetic energy increases: The work done by the equivalent link: § 7 － 4 § 7 － 4 Periodic Speed Fluctuation and Its Regulation

MdMd MrMr a b c d e a'a' φ φ MeMe E a b c d e a'a' ab: M d < M r ， △ E < 0 deficiency of work“ － ”, ω↓ bc: M d > M r ， △ E > 0 excess work“+”, ω↑ cd: M d < M r ， △ E < 0 deficiency of work“ － ”, ω ↓ de: M d > M r ， △ E > 0 excess work“+”, ω↑ within a period ， W d =W r ， △ E=0 ； then This means that the work done by driving forces is equal to the work done by resistant forces within a period. This is the condition for a periodic steady working state.

二、 Regulation of Periodic Speed Fluctuation 1. Coefficient of Speed Fluctuation TT ω φ Ｏ ω min ω max Average angular speed The speed fluctuation of a machine may cause an extra dynamic load and vibration of the system and therefore should be controlled within some limits to ensure good working quality. The coefficient of speed fluctuation  is then δ≤[δ] The regulation of periodic speed fluctuation means limitation of the coefficient of speed fluctuation so that

2.Design method for flywheel （ 1 ） Principle A flywheel serves as a mechanical reservoir for storing mechanical energy. Its function is to store the extra energy when the available energy is in excess of the load requirements and to give away the same when the available energy is less than the required load. Point b ：  E min ，    min ，  W min ； φ E a b c d e a'a' E max E min Point c ：  E max ，    max ，  W max ； The maximum increment of work  W max

If J e =constant, then We can add a flywheel with enough moment of inertia to reduce the value of δ The coefficient of speed fluctuation is （ 2 ） Calculation of moment of inertia of a Flywheel If J e << J F, J e can be neglected. The equivalent moment of inertia of the flywheel J F can be calculated by

Rim type flywheel ： Rim1 、 Hub2 、 Web(spokes)3. (3) Determination of flywheel size The moment of inertia: G A D 2 ＝ 4gJ F where G A D 2 is flywheel moment, D is the mean diameter of the rim Thickness of rim is b and density isγ(N/m 3 ), then G A = π DHb γ HB ＝ G A / ( π D γ)

The aperiodic speed fluctuation is needed to regulate with a speed regulator. One kind of Centrifugal speed regulator is sketched in Fig. § 7 － 5 § 7 － 5 Introduction to Aperiodic speed Fluctuation and Its Regulation