STUDY OF ASYNCHRONOUS MOTORS WITH A BUILT-IN BRAKE RELEASED BY THE LOAD

L. K. BURULKO, E. M. GUSELNIKOV, E. V. KONONENKO,
Yu. P. SHEVEL

(Submitted by the scientific seminar of the Departments of Electrical Machines and Electrical Engineering.)

In recent years asynchronous motors with a built-in brake have become increasingly widespread. Brakes not only ensure the safe operation of the actuating mechanism, but also have a substantial effect on its productivity. There are several designs of such motors. This paper considers a method for studying asynchronous motors with built-in brakes released by the load, whose design was developed at the Special Design Bureau of Electrical Machine Building (Tomsk) [1].

The operating principle of such a motor is based on the fact that release of the brake occurs only while the motor is operating, due to the load torque on the shaft. In an idle motor the brake is closed by the force of a spring. The design of the motor under consideration, with a built-in brake released by the load, is shown in Fig. 1. One of the distinctive features of the motor design is that the rotor core 1 can rotate relative to the shaft in both directions through an angle $\varphi_0$. For this purpose bronze-graphite bearings are used. The rotor core is rigidly connected with bushing 2, which makes it possible to move the braking device outside the motor housing. Bushing 2 is made with projections that enter the grooves of the hub of brake disk 3. The grooves of the hub are made wider than the projections of the bushing, which allows the driving part of the motor to rotate relative to the driven part through an angle $\pm \varphi_0$. Brake disk 3 is at the same time a fan and is mounted on a key, while having the possibility of axial displacement. On the hub of the brake disk, on axles 4, cams 5 are fastened, which enter the grooves of the cam washer (Fig. 1, view A). Spring 6 serves to create the braking torque; its force is regulated by nut 7.

When the motor is connected to the network, the rotor core, together with bushing 2 and the cam washer, turns relative to the hub of brake disk 3. The cams entering the grooves of the cam washer then rotate on their axes and, overcoming the resistance of the spring, begin to press the brake disk away from the braking surface of the bearing shield, and the brake is released. Axial displacement of the brake disk occurs until the projections of bushing 2 come into contact with the walls of the groove of the hub of brake disk 3. After this, the rotor core will rotate at the same speed as the motor shaft. Owing to the symmetry of the elements of the brake system, when the direction of rotation of the rotor is changed the processes proceed identically.

In order to study electromechanical transient pro-

Fig. 1. General view of the engine.

masses of such motors and the dynamic forces arising in this process, it is necessary to formulate equations describing the operation of the motors during starting and braking.

The voltage-balance equations of the stator and rotor windings of an asynchronous motor of the design under consideration do not differ in any way from the corresponding equations for a motor of normal construction. These equations, written in relative units in a synchronously rotating coordinate system, have the form [1, 2]:

\[ \left. \begin{aligned} \frac{d\Psi_{s\alpha}}{dt} &= U_{s\alpha}-r_s y_r \Psi_{s\alpha}+r_s y_m \Psi_{r\alpha}+\Psi_{s\beta},\\[4pt] \frac{d\Psi_{s\beta}}{dt} &= U_{s\beta}-r_s y_r \Psi_{s\beta}+r_s y_m \Psi_{r\beta}-\Psi_{s\alpha},\\[4pt] \frac{d\Psi_{r\alpha}}{dt} &= -r_r y_s \Psi_{r\alpha}+r_r y_m \Psi_{s\alpha}+\Psi_{r\beta}-\omega \Psi_{r\beta},\\[4pt] \frac{d\Psi_{r\beta}}{dt} &= -r_r y_s \Psi_{r\beta}+r_r y_m \Psi_{s\beta}-\Psi_{r\alpha}+\omega \Psi_{r\alpha} \end{aligned} \right\}, \tag{1} \]

where

\[ y_s=\frac{x_s}{x_s x_r-x_m^2}, \]

\[ y_r=\frac{x_r}{x_s x_r-x_m^2}, \]

\[ y_m=\frac{x_m}{x_s x_r-x_m^2}. \]

The electromagnetic torque is determined by the equality

\[ M=\frac{3}{2}y_m\left(\Psi_{r\alpha}\Psi_{s\beta}-\Psi_{s\alpha}\Psi_{r\beta}\right). \tag{2} \]

When operating from a network with constant voltage and frequency, the components of the voltage vector in the coordinate system under consideration are constant quantities and are equal to

\[ \begin{aligned} U_{s\alpha}&=U_m\cos\alpha_0,\\ U_{s\beta}&=U_m\sin\alpha_0, \end{aligned} \tag{3} \]

where
\(\alpha_0\)—is the initial phase.

In studying electromechanical transient processes, i.e., when the rotational speed changes (starting, braking), together with the voltage-balance equations of the electrical circuits it is necessary to consider the equation of motion of the rotor. The rotor core may occupy various positions relative to the shaft within the angle \(\pm\varphi_0\). When the angle \(\varphi\) changes, the torques acting on the motor shaft will also change. In this connection, the process of starting the motor with release of the brake may be subdivided into four stages.

First stage. Free rotation of the rotor core, during which the clearance is taken up. The angle of rotation of the core relative to the shaft \(\varphi\) varies within the limits from \(\varphi=0\) to \(\varphi=\pm\varphi_{01}\) (Fig. 2, a). The value of the angle \(\varphi_{01}\) is determined by the design of the elements of the braking system. At this stage the rotor core, freely rotating relative to the shaft, accelerates without load. The cams and the shaft are stationary,

Fig. 2. Position of the cam: a) at the beginning of the first stage of starting, b) at the end of the second stage of starting

and the brake is closed. Neglecting the friction torque of the core on the shaft, the equation of motion can be written in the form

\[ I_1 \frac{d^2 \varphi}{dt^2} = M, \tag{4} \]

where

\(I_1\) is the moment of inertia of the rotor core.

Second stage. Rotation of the rotor core and the cam rotation caused by it, corresponding to the selection of the clearance \(\Delta\) (Fig. 2). This stage is characterized by the fact that, under the influence of the motor torque (force \(F\)), the cam rotates about its axis \(B\) through an angle \(\beta_1\), while the angle \(\varphi\) changes within the limits from \(\varphi_{01}\) to \(\varphi_{02}\); at the same time the brake spring is compressed by an amount equal to \(y_1\).

The rotation of the core and the cam is opposed by the moment developed by the spring force. From (Fig. 3, a) it follows that

\[ M_{\mathrm{pr}} = N \cdot b \sin(\alpha - \beta), \tag{5} \]

where

\(b\) and \(\alpha\) are design parameters of the cam (Fig. 2, a).

The force developed by the spring can be determined as

\[ N = N_0 + cy, \tag{6} \]

where

\(N_0\) is the spring force at the initial position of the cam (Fig. 2, a);

\(c\) is the spring stiffness;

\(y\) is the compression of the spring during the rotation of the cam.

The angle of rotation of the cam \(\beta\) and the spring compression \(y\) occurring in this process depend on the angle of rotation of the rotor core \(\varphi\). The relationship between them, as well as the value \(\varphi_{02}\) (the angle corresponding to the completion of the second stage), depend on the design parameters of the braking system. These dependences can be determined using Figs. 4 and 5. \(\beta = f(\varphi)\) is determined from the expression

\[ R \operatorname{tg} \left[ \operatorname{arctg} \frac{a \sin \gamma}{R} - (\varphi - \varphi_{01}) \right] = a \sin(\gamma - \beta), \tag{7} \]

Here all quantities, except \(\beta\) and \(\varphi\), are constant and are determined by the design of the braking system. \(R = OO_1\) is the mean radius of the cam position relative to the motor shaft (Fig. 4).

Fig. 3. System of forces acting on the cam: a) during the second stage of starting, b) at the beginning of the third stage

Fig. 4. Diagram illustrating the step-by-step rotation of the rotor core relative to the shaft

The angle \(\beta_1\) depends on the magnitude of the gap \(\Delta\) and is found from the equality

\[ \Delta = a[\cos(\gamma-\beta_1)-\cos\gamma]. \tag{8} \]

Substituting the value of the angle \(\beta_1\) into expression (7), one can find \(\varphi=\varphi_{02}\).

The amount of spring compression can be determined from the expression

\[ y=b[\cos(\alpha-\beta)-\cos\alpha]. \tag{9} \]

From equations (7) and (9) one can find the dependence \(y=f(\varphi)\). In the second stage, a torque \(M\) is applied to the motor shaft through the axis of cam \(B\), under the action of which the shaft can begin to rotate. Rotation of the shaft in this case is hindered, first, by the load moment

Fig. 5. Position of the cam arms: a) at the beginning of the second stage (solid lines) and at the end (dashed); b) the same for the third stage

the spring moments \(M_c\) and, secondly, the braking friction moment \(M_t\), since the latter is closed at this stage.

The braking moment is determined by the magnitude of the axial force \(N_B\), equal in this case to the spring force,

\[ M_t = N_B R_t \mu \frac{1}{\sin \psi}, \tag{10} \]

where

\(R_t\) is the mean radius of the conical brake disk;
\(\mu\) is the coefficient of friction;
\(\psi\) is the cone angle of the rotor.

Then the equations of motion of the rotor in the second stage will have the form:

\[ I_1 \frac{d^2 \varphi}{dt^2} = M - M_{\mathrm{pr}}, \tag{11} \]

\[ (I_1 + I_2)\frac{d^2 \varphi_2}{dt^2} = M - M_t - M_c, \tag{12} \]

where

\(\varphi_2\) is the angle of rotation of the shaft;
\(I_2\) is the moment of inertia of the driven part (shaft, brake disk, and load).

Third stage. Rotation of the rotor core until the projections of the sleeve engage with the walls of the groove of the brake-disk hub. This stage is characterized by the appearance of the support reaction \(N_A\) (Fig. 3, b), and

this leads to a decrease in the pressing force in the brake and, under certain conditions, to release of the brake.

The magnitude of the force \(N_A\) can be determined as follows. At the end of the second stage, the equation of moment equilibrium for the cam will have the form (Fig. 3,a)

\[ F_1 a \cos(\gamma-\beta_1)=N_1 b\sin(\alpha-\beta_1), \tag{13} \]

where \(F_1, N_1\) are the forces acting on the cam at the end of the second stage. Since \(N_1=N_0+cy_1\), then

\[ F_1=\frac{M_1}{R}=(N_0+cy_1)\frac{b}{a}\frac{\sin(\alpha-\beta_1)}{\cos(\gamma-\beta_1)}. \tag{14} \]

Consequently, the magnitude of the force \(F_1\) and of the moment \(M_1\) is determined only by the design dimensions.

If the force \(F\) becomes greater than \(F_1\), then the support reaction \(N_A\) arises. Then, for the beginning of the third stage, the equation of moment equilibrium of the cam will have the form (Fig. 3,b)

\[ Fa\cos(\gamma-\beta_1)=N_A a\sin(\gamma-\beta_1)+N_1 b\sin(\alpha-\beta_1), \tag{15} \]

Subtracting equation (13) from (14), we obtain

\[ N_A=(F-F_i)\operatorname{ctg}(\gamma-\beta_1)=\frac{M-M_1}{R}\operatorname{ctg}(\gamma-\beta_1). \tag{16} \]

It follows from this expression that \(N_A\) is directly proportional to the difference of the moments \((M-M_1)\), and

\[ N_B=N-N_A. \tag{17} \]

If \(N>N_A\), then the engine operates with braking, \(\varphi=\varphi_{02}=\text{const}\). The magnitude of the braking moment depends on the value of \(N_B\). If \(N_A>N\), then the brake is released and the brake disk moves away. The distance \(x_1\) by which the brake disk moves away is determined by the design, including the angle \(\varphi_0\) (Fig. 4). From Fig. 5,b it can be determined that

\[ x=a[\cos(\gamma-\beta)-\cos(\gamma-\beta_1)]. \tag{18} \]

The dependence between the angles \(\beta\) and \(\varphi\) in the third stage is described by the same equation as in the second stage (7). The angle \(\varphi=\varphi_0\) corresponds to the angle \(\beta=\beta_2\). Substituting this value of \(\beta\) into (18), we find the magnitude \(x_1\).

The magnitude of the force \(N_A\) during the withdrawal of the brake disk is affected also by the acceleration imparted to the disk. Neglecting friction in the guides, the equation of axial displacement of the brake disk may be written in the form

\[ m\frac{d^2x}{dt^2}=N_A-N, \tag{19} \]

where

\(m\) is the mass of the brake disk.

Since the magnitude \(x_1\) is small \((0.5\text{–}1.0\ \text{mm})\), the dependence \(x=f(\varphi)\) can be approximated by an equation in the form of a straight line

\[ x=k(\varphi-\varphi_{02}). \]

Then

\[ \frac{d^2x}{dt^2}=k\frac{d^2\varphi}{dt^2}. \]

From expression (19) we find

\[ N_A=N+m\frac{d^2x}{dt^2}=N+m\kappa\frac{d^2\varphi}{dt^2}. \tag{20} \]

The moment produced by this force is equal to

\[ M_{N_A}=N_A a\sin(\gamma-\beta)= \left(N+m\kappa\frac{d^2\varphi}{dt^2}\right)a\sin(\gamma-\beta). \tag{21} \]

The rotation of the rotor core in the third stage is opposed by the moments produced by the forces \(\vec N\) and \(\vec N_A\) (Fig. 3, b). Then the equation of motion of the rotor core relative to the shaft has the form

\[ I_1\frac{d^2\varphi}{dt^2}=M-M_{\mathrm{pr}}-M_{\vec N_A} \tag{22} \]

or

\[ \left[I_1+m\cdot\kappa\cdot a\cdot\sin(\gamma-\beta)\right]\frac{d^2\varphi}{dt^2} = M-M_{\mathrm{pr}}-N a\cdot\sin(\gamma-\beta), \]

\[ (I_1+I_2')\frac{d^2\varphi}{dt^2} = M-M_{\mathrm{pr}}-N\cdot a\cdot\sin(\gamma-\beta), \tag{23} \]

where \(I_2'=m\cdot\kappa\cdot a\cdot\sin(\gamma-\beta)\) is the moment of inertia of the moving brake disk, reduced to the rotating rotor core.

The amount of compression of the spring in the third stage is determined by the expression

\[ y=x+b[\cos(\alpha-\beta)-\cos\alpha]. \tag{24} \]

The equation of motion of the rotor has the form (12). The moment \(M_T\) acts until \(N_B>0\).

Fourth stage. Acceleration of the motor. During this period the braking system operates as a coupling clutch. The rotor core and the motor shaft rotate at the same speed. The equation of motion in this case will have the form

\[ (I_1+I_2)\frac{d^2\varphi_2}{dt^2}=M-M_C. \tag{25} \]

In the steady-state process the asynchronous motor will operate in the fourth stage only in the case when the load moment \(M_C\) exceeds the minimum value at which the condition for release of the brake is satisfied,

\[ N_B\leq 0. \]

Taking into account (16) and (17), this condition can be written in the form

\[ \frac{M-M_1}{R}\operatorname{ctg}(\gamma-\beta_1)\geq N \tag{26} \]

or

\[ M_C\geq M_1+N\frac{R}{\operatorname{ctg}(\gamma-\beta_1)}. \tag{27} \]

If condition (27) is not satisfied, then in the steady-state process the motor will operate with braking, which corresponds to operation at the beginning of the third stage.

Braking of the motor, like starting, can be subdivided into the same stages. If the motor operated with the brake released, then at the beginning the brake disk approaches the stationary friction surface. In the following stages the motor is braked. The motions of the core-

of the rotor core and the motor shaft are described by equations corresponding to the different stages of starting, depending on the value of the angle \(\varphi\) and the condition \(M=0\).

The obtained system of equations describing the electromechanical transient processes is nonlinear. Therefore its solution can most simply be obtained by modeling on an analog computer. This makes it possible to determine the dynamic forces that arise, the starting and braking times, and other characteristics reflecting the dynamic properties of motors with built-in brakes.

The modeling of the voltage-balance equations (1), (2) is well known. The part of the model that makes it possible to solve the equations of motion using an MN-14 analog computer is shown in Fig. 6.

Amplifiers L37, L14, L15 solve the equations associated with the motion of the core relative to the shaft. The comparison devices assembled on amplifiers L47, L38, L48 and relays \(P_1, P_2, P_3, P_4\) make it possible to carry out the transition from one stage of core motion to another when

Fig. 6. Structural diagram of the model for solving the equations of motion

the angle changes from \(\varphi=0\) to \(\varphi=\varphi_0\). In this case, by switching on the relays as the preceding stages expire, the necessary initial conditions are established and the elements of the model circuit corresponding to the subsequent stage are switched on. The equation of motion of the driven part is solved by means of amplifiers L16, L56.

Let us consider the solution of the equation stage by stage. At the first stage, equation (4) is solved, and the end of the first stage is registered when \(\varphi_{01}-\varphi \le 0\) by switching on relays \(P_1\) and \(P_2\). This corresponds to the transition to the solution of equations (11) and (12). The end of the second stage is registered

by switching on relay \(P_3\), as a result of which the equation of motion of the armature takes the form (22). The end of the solution of equation (22) is recorded by switching on relay \(P_4\), which corresponds to the choice of angle \(\varphi_0\) and to the transition to the solution of the motor equation (25), in the event that the signal from amplifier \(L41\), proportional to the force \(N_{\mathrm{B}} = N - N_{\mathrm{A}}\), is less than or equal to zero, i.e., the condition for brake release (26) is satisfied and relays \(P_7, P_{10}\) operate. If the condition for brake release is not satisfied, then, by means of relays \(P_7\) and \(P_{10}\), a braking torque proportional to the force \(N_{\mathrm{B}}\) will act on the motor shaft. In this way the equations of motion during motor starting are solved.

Fig. 7. Oscillograms of motor starting and braking: a) at \(M_c = 0\); b) at \(M_c = M_n\)

For carrying out the braking process, it is necessary that the relay switching-off stages proceed in the reverse order. For this purpose, the winding of relay \(P_4\) is connected to the output of amplifier \(L49\) through the normally closed contact of relay \(P_9\). Relay \(P_9\) and amplifier \(L42\) constitute a comparison device, by means of which relay \(P_4\) is switched off and a signal is applied to amplifier \(L14\), the integration of which leads to the subsequent switching off of relays \(P_3 P_2 P_1\).

Oscillograms of motor starting and braking at different resisting torques are presented in Fig. 7.

REFERENCES

1. A. K. Abramov, V. S. Belyagin. Normally closed automatic brake. Inventor’s certificate No. 172163, Bulletin of Inventions No. 12, 1965.

2. M. M. Sokolov, L. P. Petrov, L. B. Masandilov, V. A. Ladenzon. Electromagnetic Transient Processes in an Asynchronous Electric Drive. “Energiya,” 1967.