Investigation of the dynamics of a brake electric motor with a mechanical amplifier
Abstract Generated abstract
This paper analyzes the dynamics of a brake electric motor equipped with an electromagnetic brake using a mechanical amplifier formed by beveled washers and balls. It derives expressions for the brake amplification coefficient, the critical bevel angle at which jamming occurs, and the time intervals of the braking process, which is divided into stages covering electromagnet current decay, axial disk motion, free disk rotation, and amplified braking. The authors also formulate equations for the initial release stage, when deformation of the friction linings is removed and the motor shaft is freed. The resulting calculation method indicates that mechanical amplification can increase braking torque, allow adjustment through washer bevel angle, and potentially reduce active and structural material requirements.
Full Text
STUDY OF THE DYNAMICS OF A BRAKE ELECTRIC MOTOR WITH A MECHANICAL AMPLIFIER
V. B. TARMAEV, E. M. GUSELNIKOV
Presented by the scientific seminar of the Departments of Electrical Machines and General Electrical Engineering
At present, to ensure rapid stopping of various drives, brake electric motors with built-in electromagnetic brakes are widely used [1]. These designs of electromagnetic brakes have the following shortcomings: in order to obtain small values of the drive braking time, it is necessary to increase the tractive force of the electromagnet or the diameter of the brake disk, i.e., to increase the overall dimensions of the brake; as the tractive force of the electromagnet is increased, the braking process is accompanied by jolts, impacts, and increased wear of the friction surfaces. To eliminate these shortcomings, it is proposed to use brakes with mechanical amplifiers. An example of an electromagnetic brake with a mechanical amplifier, made in the form of two washers 3 and 4 and balls 5, is shown in Fig. 1. The braking process proceeds as follows. After the supply is removed from the electric motor and electromagnet 2, springs 7 press brake disk 8 against friction linings 6, which are fastened to the brake wheel-fan 9. The angular displacement of the brake disk during braking provides additional pressing of the disk against the wheel, owing to the bevels on the end faces of washers 3, 4 and ball 5. Release is effected by means of the electromagnet. When voltage is applied to coil 2, under the action of the magnetic flux, brake disk 8 overcomes the force of power springs 7 and is attracted to bearing shield 12, freeing the electric-motor shaft for rotation.
To determine the effectiveness of amplification, the concept of the brake amplification coefficient [2] is adopted
\[ s = \frac{M_{TS}}{M_T}, \tag{1} \]
where \(S\) is the amplification coefficient,
\(M_T\) is the braking torque without amplification,
\(M_{TS}\) is the braking torque with amplification.
For the electromagnetic-brake design under consideration, the amplification coefficient can be calculated by the following formula:
\[ s = \frac{1}{ 1 - \mu_1 \cdot \frac{R}{R_{\mathrm{sh}}} \cdot \frac{\cos \alpha - \mu_2 \sin \alpha}{\sin \alpha + \mu_2 \cos \alpha} }, \tag{2} \]
where \(R\) is the friction radius,
Fig. 1
\(R_{\mathrm{sh}}\) — radius of the position of the balls relative to the electric-motor shaft,
\(\mu_1\) — coefficient of friction between the friction lining and the brake disk,
\(\mu_2\) — coefficient of rolling friction on the bevels of the washers,
\(\alpha\) — angle of inclination of the washer bevel.
When the denominator of expression (2) is equal to zero, the amplification coefficient tends to infinity. From this condition the critical angle of the washer bevel, at which jamming occurs, is determined:
\[ \operatorname{tg}\alpha_{\mathrm{cr}} = \frac{\mu_1 \dfrac{R}{R_{\mathrm{sh}}} - \mu_2} {1 + \mu_1 \mu_2 \dfrac{R}{R_{\mathrm{sh}}}} . \tag{3} \]
Expressions (2) and (3) can be considerably simplified if specific values of the coefficient of rolling friction are substituted into them:
\[ s = \frac{1} {1 - \dfrac{\mu_1 R}{\operatorname{tg}\alpha \, R_{\mathrm{sh}}}}, \tag{4} \]
\[ \operatorname{tg}\alpha_{\mathrm{cr}} = \mu_1 \frac{R}{R_{\mathrm{sh}}}. \tag{5} \]
The braking process of an electromagnetic brake with amplification is divided in time into four stages:
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Power is removed from the electromagnet winding; the magnitude of the attraction force of the electromagnet is greater than the magnitude of the force of the power springs. The brake disk is stationary.
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The attraction force of the electromagnet becomes equal to the magnitude of the force of the power springs; with a slight excess of the force of the power springs, the brake disk begins to move in the axial direction until it comes into contact with the friction linings (the gap between the brake disk and the brake wheel–fan is eliminated).
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Under the action of friction forces, the brake disk turns until the balls come into contact with the bevels of the washers (elimination of the gap between the balls and the bevels of the washers). The magnitude of the electromagnetic force is equal to zero.
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Motion of the brake disk after the balls have come into contact with the bevels of the washers.
The first stage of the braking process, neglecting the influence of eddy currents, is described by the following equation:
\[ U = \rho \cdot i + r \cdot i + L_1 \cdot \frac{di}{dt}, \tag{6} \]
where \(U\) is the supply voltage of the electromagnet,
\(L_1\) is the inductance of the coil at the final gap \(\delta_1\),
\(\rho = f(t)\) is the resistance of the electric arc when the contacts of the control circuit of the electromagnet winding open,
\(r\) is the active resistance of the winding.
Assume that \(\rho = \rho_{\mathrm{cp}} = \mathrm{const}\). Then the solution of equation (6) will be as follows:
\[ i = \left( \frac{U}{r} - \frac{U}{\rho_{\mathrm{cp}} + r} \right) \cdot e^{-\frac{t}{T_1}} + \frac{U}{\rho_{\mathrm{cp}} + r}, \tag{7} \]
where
\[ T_1 = \frac{L_1}{\rho_{\mathrm{cp}} + r}. \]
Since \(\rho_{\mathrm{cp}} \gg r\), equation (7) takes the form
\[ i \simeq \frac{U}{r} \cdot e^{-\frac{t}{T_1}}. \tag{8} \]
From equation (8) we find the time interval from the moment the electromagnet winding is switched off to the moment the brake disk begins to move, i.e., the magnetic flux decreases to the value at which the opposing force becomes equal to the electromagnetic force:
\[ t_1 = \frac{L_1}{\rho_{\mathrm{cp}} + r} \cdot \ln \frac{U}{i_{\mathrm{tr}} \cdot r}. \tag{9} \]
In this case the value of the starting current will be equal to
\[ i_{\mathrm{tr}} = \sqrt{ \frac{2P_{\mathrm{pr}} \cdot \delta_1}{L_1} }, \tag{10} \]
where \(P_{\mathrm{pr}}\) is the force of the power spring,
\(\delta_1\) is the magnitude of the air gap with the brake disk attracted.
The second stage of the braking process is characterized by the axial displacement of the brake disk and is described by the following equation:
\[ m \frac{d^2 x}{dt^2} + c_1 \frac{dx}{dt} + c_2 x = P_{\mathrm{pr}} - F_{\mathrm{e}}(t), \tag{11} \]
where \(m\) is the mass of the brake disk,
\(x\) is the displacement of the disk in the axial direction;
\(F_{\mathrm{e}}(t)\) is the law of variation of the attractive force of the electromagnet upon switching off;
\(c_1\) is the resistance (proportional to the first power of the velocity) exerted during displacement of the brake disk;
\(c_2\) is the stiffness of the springs.
Neglecting the forces of rolling friction of the balls on the brake disk \(\left(c_1 \frac{dx}{dt}\right)\), the stiffness of the force springs \(c_2x\), and also taking the magnitude of the electromagnetic force to be constant and equal to the force at the moment of switching off, the time of motion of the brake disk can be determined from the formula [4]
\[ t_2 = \sqrt{\frac{2m\delta}{P_{\mathrm{pr}} - F_{\mathrm{e}}}}, \tag{12} \]
where \(\delta\) is the magnitude of the clearance between the brake disk and the brake wheel.
The velocity of the disk at the end of the second stage will be equal to
\[ v_2 = \frac{P_{\mathrm{pr}} - F_{\mathrm{e}}}{m}\, t_2 . \tag{13} \]
In the third stage of braking, the disk comes into contact with the friction linings of the wheel, and their deformation occurs. In this case a friction force \(\mu_1 N_1\) arises, which turns the brake disk in the direction of rotation of the brake wheel. The calculation scheme of the third stage is shown in Fig. 2a. The equations of motion of the brake disk for the third stage have the form
\[ I \frac{d^2 \varphi}{dt^2} = \mu_1 N_1 R; \tag{14} \]
\[ m \frac{d^2 x}{dt^2} = -N_1 + P_{\mathrm{pr}}, \tag{15} \]
where \(I\) is the moment of inertia of the brake disk,
\(N_1\) is the normal force between the brake disk and the friction linings,
\[ N_1 = kx, \tag{16} \]
where \(k\) is the stiffness of the friction lining.
The angle of free rotation of the brake disk corresponds to the magnitude of the initial clearance between the brake disk and the wheel:
\[ \varphi_0 = \frac{\delta}{R_{\mathrm{sh}}\tan \alpha}. \tag{17} \]
The time for eliminating the clearance between the balls and the bevels of the washers is determined from the solution of equations (14–17):
\[ t_3 = \sqrt{\frac{2I\delta}{P_{\mathrm{pr}} R R_{\mathrm{sh}} \mu \tan \alpha}} . \tag{18} \]
Fig. 2
The speed of the brake disk at the end of the third stage is
\[ \omega_3=\sqrt{\frac{2P_{\mathrm{pr}}R\mu_1\delta}{IR_{\mathrm{sh}}\tan\alpha}} . \tag{19} \]
The calculation scheme for the fourth stage is shown in Fig. 2b. The motion of the brake disk is characterized by the following system of equations:
\[ \left. \begin{aligned} I\frac{d^2\varphi}{dt^2} &= \mu_1N_1R - N_2R_{\mathrm{sh}}\sin\alpha;\\ m\frac{d^2x}{dt^2} &= -N_1 + N_2\cos\alpha + P_{\mathrm{pr}};\\ N_1 &= kx;\\ x &= \varphi R_{\mathrm{sh}}\tan\alpha . \end{aligned} \right\} \tag{20} \]
Solving the system of equations with respect to \(\varphi\), we obtain
\[ \varphi_4=A\sin(pt+\Psi)+E, \tag{21} \]
where
\[ A=\sqrt{\left(\frac{P_{\mathrm{pr}}}{kR_{\mathrm{sh}}\tan\alpha}\right)^2+\left(\frac{\omega_3}{k}\right)^2}; \tag{22} \]
\[ p=\sqrt{\frac{k\tan\alpha\,R_{\mathrm{sh}}\left(R_{\mathrm{sh}}\sin\alpha-\mu_1R\cos\alpha\right)} {I\cos\alpha+mR_{\mathrm{sh}}^{2}\tan\alpha\sin\alpha}}; \tag{23} \]
\[ \Psi=\operatorname{arctg}\frac{kP_{\mathrm{pr}}}{R_{\mathrm{sh}}\tan\alpha\cdot\omega_3}; \tag{24} \]
\[ E=\frac{P_{\mathrm{pr}}\sin\alpha} {k\tan\alpha\left(R_{\mathrm{sh}}\sin\alpha-\mu_1R\cos\alpha\right)} . \tag{25} \]
From equation (21) it follows that during the fourth stage of the braking process the magnitude of the force on the friction lining oscillates about the mean value
\[ N_1=kx=kR_{\mathrm{sh}}\tan\alpha\,\varphi =\frac{P_{\mathrm{pr}}}{1-\mu_1\frac{R}{R_{\mathrm{sh}}}\cdot\frac{1}{\tan\alpha}} =P_{\mathrm{pr}}\cdot s, \tag{26} \]
where \(s\) is the amplification factor.
Assuming that the reduced torque on the shaft of the electric motor is equal to \(M_c\) and that the angular velocity of the motor shaft at the start of braking is equal to \(\omega_0\), the time of the fourth braking stage is determined by the formula
\[ t_4=\frac{I_M\cdot\omega_0}{P_{\mathrm{pr}}\cdot s\,R\mu_1-M_c}, \tag{27} \]
where \(I_M\) is the reduced moment of inertia of the mechanism.
The release process, like the braking process, is divided into four stages, proceeding in the reverse sequence.
The most important is the first stage, when the friction linings restore their deformations and the electric-motor shaft is released. Therefore we shall confine ourselves to considering only this stage of release. The calculation scheme is shown in Fig. 26. The initial position of this process is characterized by the maximum deformations, and the final position by zero deformations of the friction lining. Let us write the equations of motion of the brake disk for this stage of release:
\[ \left. \begin{aligned} I\frac{d^2\varphi}{dt^2} &= \mu_1N_1R-N_2R_{\mathrm{sh}}\sin\alpha;\\[4pt] m\frac{d^2x}{dt^2} &= -N_1+N_2\cos\alpha+P_{\mathrm{pr}}-F_{\mathrm{e}0};\\[4pt] F_{\mathrm{e}0} &= \frac{I_yL_0}{2\delta}\left(1-e^{-\frac{t}{\tau_0}}\right)^2 =F_{\mathrm{em}}\left(1-e^{-\frac{t}{\tau_0}}\right)^2;\\[4pt] N_1&=kx;\\[4pt] x&=\varphi\cdot R_{\mathrm{sh}}\tan\alpha, \end{aligned} \right\} \tag{28} \]
where \(L_0\) is the inductance of the electromagnet coil at a gap equal to \(\delta\),
\[ \tau_0=\frac{L_0}{r} \]
is the time constant,
\[ I_y=U/r \]
is the steady-state value of the current of the electromagnet winding.
At this stage, the axial displacement \(x\) corresponds to deformations of the friction lining. The solution of the system will be as follows:
\[ \begin{aligned} x_{01}={}& \frac{M_{c0}}{k\mu_1R}\cdot\cos pt +\frac{2F_{\mathrm{em}}\cdot R_{\mathrm{sh}}\cdot\sin\alpha\cdot e^{-\frac{t}{\tau_0}}} {k\left(R_{\mathrm{sh}}\sin\alpha-\mu_1R\cos\alpha\right) +\dfrac{I\cos\alpha+mR_{\mathrm{sh}}^2\sin\alpha\cdot\tan\alpha} {R_{\mathrm{sh}}\tan\alpha\cdot\tau_0^2}} \\[6pt] &-\frac{\left(F_{\mathrm{em}}-P_{\mathrm{pr}}\right)\cdot R_{\mathrm{sh}}\sin\alpha} {k\left(R_{\mathrm{sh}}\sin\alpha-\mu_1R\cdot\cos\alpha\right)} \\[6pt] &-\frac{F_{\mathrm{em}}\cdot R_{\mathrm{sh}}\cdot\sin\alpha\cdot e^{-\frac{2t}{\tau_0}}} {k\left(R_{\mathrm{sh}}\sin\alpha-\mu_1R\cdot\cos\alpha\right) +\dfrac{4\left(I\cos\alpha+mR_{\mathrm{sh}}^2\sin\alpha\cdot\tan\alpha\right)} {R_{\mathrm{sh}}\tan\alpha\cdot\tau_0^2}} . \end{aligned} \tag{29} \]
where \(M_{\mathrm{co}}\) is the torque on the motor shaft at the moment of release. The duration of this stage of release is determined from equation (29) under the condition \(x_{01}=0\).
Thus, the use of brakes with mechanical amplifiers in brake electric motors is promising and makes it possible to save active and structural materials. The braking torque can be varied by changing the angle of inclination of the bevel of the washers. The developed dynamic calculation makes it possible to describe in detail the processes of braking and release and to determine the response time of the brake.
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