INVESTIGATION OF THE DEGREE OF NONUNIFORMITY OF LOAD DISTRIBUTION ALONG THE TOOTH LENGTH ON A CLOSED TEST RIG

V. R. VORONOV

(Presented by the scientific seminar of the Department of Applied Mechanics)

For testing gear transmissions, the most widespread use has been made of rigs with a closed loop [1—4].

Loading of gear wheels during operation according to a closed scheme can be carried out by means of the forces of elastic deformation of a pre-twisted torsion shaft, but changing the load in test stands of this type is associated with a considerable expenditure of time and requires shutting down the rig for this period. Therefore, in our case a method of loading was used, proposed by V. N. Kudryavtsev [2], which makes it possible to start the rig in an unloaded state and to change the load while running.

Fig. 1. Diagram of the rig

The rig described consists of an electric motor (A 42/4, with power \(N = 2.8\) kW at \(n = 1420\) rpm) and four single-stage reducers, I, II, III, IV, connected to one another by couplings 3, 6, 9 and by cardan shaft 13 in a closed scheme (Fig. 1).

The housing of the loading reducer is suspended on ball bearings in the stand in such a way that it can rotate relative to the suspension axis 1, 2.

When the load \(Q\) is applied to lever 16 of loading reducer 1, the reducer housing turns in the direction of loading, and gear \(Z_{\text{k}}\) tends to turn relative to pinion \(Z_{\text{sh}}\). Rotation will occur only within the limits of the elastic deformations of the shafts and the taking-up of clearances.

The teeth of gears \(Z_{\text{sh}}\) and \(Z_{\text{k}}\), and of all other pairs of the closed loop, are pressed against one another and loaded by the corresponding forces, while the shafts are loaded by moments.

When the installation is set in rotation by the electric motor, a power flow will circulate through the closed loop in the direction indicated by the arrow (Fig. 1). The electric motor is suspended on spherical ball bearings in the stands and can freely rotate relative to the suspension axis 1, 2. The power of the electric motor is expended only to cover the losses in the installation. By measuring the reactive moment on the motor stator, the expended power can be determined. To prevent damage to the electric motor, protection is provided. If any component in the closed loop breaks, its closure is disrupted, the reducer housing will turn and, through a rod, press the “stop” button of the magnetic starter (Fig. 2), and the electric motor will be disconnected. If jamming occurs in the installation,

Fig. 2. Electrical diagram

the load on the electric motor increases sharply, and its stator, which until then had been in equilibrium under the action of the weight \(P\) located on the pan 19 of the motor-weighing device, will turn and, through a system of levers and a rod, press the “stop” button and disconnect the motor. For protection of the electric motor—

...of the motor in the event of a phase failure, an IT-816/1 maximum-current relay is installed on the panel (Fig. 2).

In order to make it possible to test gears while maintaining specified artificial misalignments of the shaft axes, reducers II and III provide for mounting the shafts on spherical ball bearings installed in special cups with eccentric bores. Reducers II and III are connected to each other by a double universal joint, which permits operation of the installation with a misalignment of the axes of shafts 5 and 7. To place the loading reducer in a horizontal position, a special disk coupling 9 is provided. The reducer housings are welded and split, in order to simplify assembly and disassembly of the gear transmissions.

Comparative tests of gears were carried out on this experimental installation to determine the influence of the design of the gear rim on the distribution of load along the length of the contact lines and on the load-carrying capacity of the transmission.

The investigations were carried out on gears of different designs (Fig. 3).

The experimental pinions and gears were made of the same material, steel 3, without heat treatment, and had a hardness of HB = 118–120.

All pinions and gears were cut with a hob in two passes and meet accuracy grade 8 according to GOST 3675-56, while the finish of the working tooth surface corresponds to the sixth class. The width of the pinion is 50 mm, and that of the gear is 46 mm.

Fig. 3. Experimental gears

The tests were carried out under a constant load on the lever of the loading reducer, \(Q = 20\) kg, which corresponds to contact stresses \(\sigma_k = 3000\) kg/cm².

The installation operated for 8–10 hours per day, after which it was stopped for inspection and photographing of the working tooth surfaces and, if necessary, for adding oil to the specified level.

During the tests, measurements were taken of the torque on the motor-weights, the temperature of the oil bath in each reducer, and the number of loading cycles of the teeth of the experimental gears. The gear wheels of reducers I and IV were installed without misalignment, while in reducers II and III an artificial shaft misalignment of 0.005 radian was introduced.

The tests showed that wheels whose teeth are cut on a solid disk are very sensitive to misalignments. Owing to the low compliance of the body of a solid wheel, the load along the tooth length is distributed nonuniformly. Therefore, nonuniform wear is observed, as well as the formation of pitting cavities on the side of greater load intensity. The pitting cavities gradually spread over a large part of the tooth, but the nonuniformity of load distribution along the tooth length does not disappear even after a large number of loading cycles. Gear wheels with a disk located in the middle of the wheel width, having a more compliant rim, reduce the nonuniformity of load distribution along the tooth length, which, in turn, affects the increase in the load-carrying capacity of the gear transmission.

When wheels with a disk shifted toward one edge of the wheel are installed, the nonuniformity of load distribution along the tooth length depends on the side from which the torque is applied. A gear wheel with the disk shifted to one side should be installed so that the disk is located on the side opposite to the application of the torque. In this case, there is a considerable leveling of the nonuniformity of load distribution along the tooth length caused by shaft misalignment.

Determination of Efficiency and Torques

Let us take the direction of rotation of shaft 1 to be clockwise (Fig. 4). In this case, the gear wheels of the closed rig will rotate in the direction indicated in Fig. 4.

Fig. 4. Diagram of moment distribution in the loading reducer

Accordingly, pinions \(Z_{\mathrm{sh}2}\), \(Z_{\mathrm{sh}4}\) and wheels \(Z_{\mathrm{k}1}\) and \(Z_{\mathrm{k}3}\) will be under the operating conditions of driving wheels, since the direction of their rotation coincides with the direction of the forces with which they act on the meshing wheels, while pinions \(Z_{\mathrm{sh}1}\), \(Z_{\mathrm{sh}3}\) and wheels \(Z_{\mathrm{k}2}\) and \(Z_{\mathrm{k}4}\) will be under the operating conditions of driven wheels.

Consider the equilibrium condition of the loading gearbox (Fig. 4)

\[ M_{\mathrm{н}} + M_1 - M_{15} - M_2 = 0, \tag{1} \]

where \(M_{\mathrm{н}} = Q \cdot L\) is the moment applied to the housing of loading gearbox I;
\(Q\) is the weight of the load on the lever of the loading gearbox;
\(L\) is the distance from the point of application of the force to the axis of rotation of gearbox I;
\(M_1\) is the moment supplied from the electric motor to shaft 1;
\(M_{15}\) is the moment with which shaft 15 acts on wheel \(Z_{\mathrm{k}_1}\);
\(M_2\) is the moment with which shaft 2 acts on gear \(Z_{\mathrm{sh}_1}\).

Denote by \(M_{\mathrm{sh}_1}\) the moment with which gear \(Z_{\mathrm{sh}_1}\) acts on shaft 2. Then

\[ M_2 = M_{\mathrm{sh}_1} + M_1. \tag{2} \]

From (1) and (2) we have

\[ M_{\mathrm{н}} - M_{15} - M_{\mathrm{sh}_1} = 0. \tag{3} \]

But

\[ M_{\mathrm{sh}_1} = M_{15} \cdot \frac{Z_{\mathrm{sh}_1}}{Z_{\mathrm{k}_1}} \cdot \eta_{\mathrm{I}}, \tag{4} \]

since gear \(Z_{\mathrm{sh}_1}\) is the driven gear; \(\eta_{\mathrm{I}}\) is the efficiency of the loading gearbox,

\[ M_{\mathrm{k}_1} = M_{15} = \frac{M_{\mathrm{sh}_1}}{\eta_{\mathrm{I}} \cdot \dfrac{Z_{\mathrm{sh}_1}}{Z_{\mathrm{k}_1}}}; \tag{5} \]

From (3) and (5) we have

\[ M_{\mathrm{н}} = \frac{M_{\mathrm{sh}_1}}{\dfrac{Z_{\mathrm{sh}_1}}{Z_{\mathrm{k}_1}} \cdot \eta_{\mathrm{I}}} + M_{\mathrm{sh}_1}. \]

Let us now express the moments on all shafts of the closed loop through the known moment

\[ M_{\mathrm{sh}_1} = \frac{M_{\mathrm{н}}}{1 + \dfrac{Z_{\mathrm{k}_1}}{Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}}}}. \tag{6} \]

From equalities (5) and (6) we obtain

\[ M_{\mathrm{k}_1} = \frac{M_{\mathrm{н}}}{\left(1 + \dfrac{Z_{\mathrm{k}_1}}{Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}}}\right)} \cdot \frac{Z_{\mathrm{k}_1}}{Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}}}. \tag{7} \]

The remaining moments are determined from the following relations:

\[ M_{\mathrm{k}_4} = \frac{M_{\mathrm{k}_1}}{\eta_{14} \cdot \eta_{12}}; \qquad M_{\mathrm{k}_4} = M_{\mathrm{sh}_1} \cdot \frac{Z_{\mathrm{k}_1}}{Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}} \cdot \eta_{12} \cdot \eta_{14}}, \tag{8} \]

\[ M_{\mathrm{sh}_4} = \frac{M_{\mathrm{k}_4}}{\dfrac{Z_{\mathrm{k}_4}}{Z_{\mathrm{sh}_4}} \cdot \eta_{\mathrm{IV}}}; \qquad M_{\mathrm{sh}_4} = M_{\mathrm{sh}_1} \cdot \frac{Z_{\mathrm{k}_1} \cdot Z_{\mathrm{sh}_4}} {Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}} \cdot \eta_{\mathrm{IV}} \cdot \eta_{12} \cdot \eta_{14} \cdot Z_{\mathrm{k}_4}}, \tag{9} \]

\[ M_{\mathrm{sh}_3} = \frac{M_{\mathrm{sh}_4}}{\eta_{9}}; \qquad M_{\mathrm{sh}_3} = M_{\mathrm{sh}_1} \cdot \frac{Z_{\mathrm{k}_1} \cdot Z_{\mathrm{sh}_4}} {Z_{\mathrm{sh}_1} \cdot \eta_{\mathrm{I}} \cdot \eta_{\mathrm{IV}} \cdot \eta_{9} \cdot \eta_{12} \cdot \eta_{14} \cdot Z_{\mathrm{k}_4}}. \tag{10} \]

Since wheel \(Z_{\text{k}3}\) is the driving one, then

\[ M_{\text{k}3} = \frac{M_{\text{sh}3}} {\eta_{\mathrm{III}}\cdot \dfrac{Z_{\text{sh}3}}{Z_{\text{k}3}}}; \qquad M_{\text{k}3} = \frac{M_{\text{sh}1}\cdot Z_{\text{k}1}\cdot Z_{\text{sh}4}\cdot Z_{\text{k}3}} {Z_{\text{sh}1}\cdot Z_{\text{k}4}\cdot Z_{\text{sh}3}\cdot \eta_{\mathrm{I}}\cdot \eta_{\mathrm{III}}\cdot \eta_{\mathrm{IV}}\cdot \eta_9\cdot \eta_{12}\cdot \eta_{14}}, \tag{11} \]

\[ M_{\text{k}2}=\frac{M_{\text{k}3}}{\eta_6}; \qquad M_{\text{k}2}= M_{\text{sh}1}\cdot \frac{Z_{\text{k}1}\cdot Z_{\text{sh}4}\cdot Z_{\text{k}3}} {Z_{\text{sh}1}\cdot Z_{\text{k}4}\cdot Z_{\text{sh}3}\cdot \eta_{\mathrm{I}}\cdot \eta_{\mathrm{III}}\cdot \eta_{\mathrm{IV}}\cdot \eta_6\cdot \eta_9\cdot \eta_{12}\cdot \eta_{14}}. \tag{12} \]

Then

\[ M_{\text{sh}2}= \frac{M_{\text{k}2}} {\dfrac{Z_{\text{k}2}}{Z_{\text{sh}2}}\cdot \eta_{\mathrm{II}}}; \tag{13} \]

\[ M_{\text{sh}2} = M_{\text{sh}1}\cdot \frac{Z_{\text{k}1}\cdot Z_{\text{sh}4}\cdot Z_{\text{k}3}\cdot Z_{\text{sh}2}} {Z_{\text{sh}1}\cdot Z_{\text{k}4}\cdot Z_{\text{sh}3}\cdot Z_{\text{k}2}\cdot \eta_{\mathrm{I}}\cdot \eta_{\mathrm{II}}\cdot \eta_{\mathrm{III}}\cdot \eta_{\mathrm{IV}}\cdot \eta_6\cdot \eta_9\cdot \eta_{12}\cdot \eta_{14}} . \]

Moment on shaft 2

\[ M_2=\frac{M_{\text{sh}2}}{\eta_3}; \tag{14} \]

\[ M_2= \frac{M_{\text{sh}1}\cdot Z_{\text{k}1}\cdot Z_{\text{sh}4}\cdot Z_{\text{k}3}\cdot Z_{\text{sh}2}} {Z_{\text{sh}1}\cdot Z_{\text{k}4}\cdot Z_{\text{sh}3}\cdot Z_{\text{k}2}\cdot \eta_{\mathrm{I}}\cdot \eta_{\mathrm{II}}\cdot \eta_{\mathrm{III}}\cdot \eta_{\mathrm{IV}}\cdot \eta_3\cdot \eta_6\cdot \eta_9\cdot \eta_{12}\cdot \eta_{14}} . \]

Taking into account that

\[ \eta_{\mathrm{I}}\cdot \eta_{\mathrm{II}}\cdot \eta_{\mathrm{III}}\cdot \eta_{\mathrm{IV}}\cdot \eta_3\cdot \eta_6\cdot \eta_9\cdot \eta_{12}\cdot \eta_{14} = \eta_{\text{c.c.}}, \tag{15} \]

and

\[ Z_{\text{sh}1}=Z_{\text{sh}2}=Z_{\text{sh}3}=Z_{\text{sh}4} \quad \text{and} \quad Z_{\text{k}1}=Z_{\text{k}2}=Z_{\text{k}3}=Z_{\text{k}4}, \]

we have

\[ M_2= \frac{M_{\text{n}}} {\left(1+\dfrac{Z_{\text{k}1}}{Z_{\text{sh}1}\cdot \eta_{\mathrm{I}}}\right)\cdot \eta_{\text{c.c.}}}, \tag{16} \]

or, from (6) and (16),

\[ M_2=\frac{M_{\text{sh}1}}{\eta_{\text{c.c.}}}, \tag{17} \]

therefore, from (2) and (17), we obtain

\[ M_1=M_2\cdot (1-\eta_{\text{c.c.}}). \tag{18} \]

Whence

\[ M_1= \frac{M_{\text{n}}\cdot (1-\eta_{\text{c.c.}})} {\left(1+\dfrac{Z_{\text{k}1}}{Z_{\text{sh}1}\cdot \eta_{\mathrm{I}}}\right)\cdot \eta_{\text{c.c.}}}. \tag{19} \]

But from (18)

\[ \eta_{\text{c.c.}}= \frac{M_2-M_1}{M_2} \tag{20} \]

or

\[ \eta_{\text{c.c.}}= \frac{M_{\text{sh}1}}{M_{\text{sh}1}+M_1}. \tag{21} \]

From equations (6) and (21) we obtain an expression for determining the efficiency of the closed loop:

\[ \eta_{\mathrm{z.k.}}= \frac{M_{\mathrm{n}}} {M_{\mathrm{n}}+M_{1}\cdot\left(1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}\right)} . \tag{22} \]

If the coaxiality of shafts 2 and 4, 5 and 7, 8 and 10, 11 and 15 is maintained, the efficiencies of the connecting couplings may be taken as approximately equal to unity; therefore

\[ M_{\mathrm{sh}1}= \frac{M_{\mathrm{n}}} {1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}}; \]

\[ M_{2}=M_{\mathrm{sh}2};\qquad M_{\mathrm{sh}2}= \frac{M_{\mathrm{n}}} {\left(1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}\right)\cdot\eta_{\mathrm{z.k.}}}. \]

\[ M_{\mathrm{k}2}=M_{\mathrm{k}3};\qquad M_{\mathrm{k}2}= \frac{M_{\mathrm{n}}\cdot\eta_{\mathrm{II}}\cdot Z_{\mathrm{k}2}} {\left(1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}\right)\cdot\eta_{\mathrm{z.k.}}\cdot Z_{\mathrm{sh}2}}; \tag{23} \]

\[ M_{\mathrm{sh}3}=M_{\mathrm{sh}4};\qquad M_{\mathrm{sh}3}= \frac{M_{\mathrm{n}}\cdot\eta_{\mathrm{II}}\cdot\eta_{\mathrm{III}}} {\left(1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}\right)\cdot\eta_{\mathrm{z.k.}}}; \tag{24} \]

\[ M_{\mathrm{k}4}=M_{\mathrm{k}1};\qquad M_{\mathrm{k}4}= \frac{M_{\mathrm{n}}\cdot\eta_{\mathrm{II}}\cdot\eta_{\mathrm{III}}\cdot\eta_{\mathrm{IV}}\cdot Z_{\mathrm{k}4}} {\left(1+\frac{Z_{\mathrm{k}1}}{Z_{\mathrm{sh}1}\cdot\eta_{\mathrm{I}}}\right)\cdot\eta_{\mathrm{z.k.}}\cdot Z_{\mathrm{sh}4}}, \tag{25} \]

where

\[ \eta_{\mathrm{z.k.}}=\eta_{\mathrm{I}}\cdot\eta_{\mathrm{II}}\cdot\eta_{\mathrm{III}}\cdot\eta_{\mathrm{IV}}. \]

With the intermediate test gearboxes II and III disconnected, when 3 and 9 are connected directly by an intermediate shaft, and if the coaxiality of shafts 2 and 10, 11 and 15 is maintained, we have

\[ \eta_{\mathrm{z.k.}}=\eta_{\mathrm{I}}\cdot\eta_{\mathrm{IV}}, \]

from which \(\eta_{\mathrm{I}}\) and \(\eta_{\mathrm{IV}}\) can be determined; after this, by connecting gearboxes II and III, their efficiencies can be determined and, using the formulas given above, the moments on the shafts of the installation can be calculated. Knowing the torques, one can find the forces on the teeth of the gear wheels of the closed loop.

References

1. R. M. Brumberg. Testing gear transmissions by the closed-loop method. Bulletin of Engineers and Technicians, no. 4, 1936.
2. V. N. Kudryavtsev. Closed installations for testing gear transmissions. Bulletin of Mechanical Engineering, no. 10, 1951.
3. Yu. S. Semenov. A closed installation for testing gear transmissions under variable load. Proceedings of TPI, vol. 96, part I, 1960.
4. G. K. Trubin. Contact Fatigue of Materials for Gear Wheels. Mashgiz, 1962.