The second planetary gear train considered is shown in Fig. 2.8a. The system consists of an input sun gear 1 and a planet gear 2 in mesh with 1 at B. Gear 2 is carried by the arm S fixed on the shaft of gear 3, as shown. Gear 3 meshes with the output gear 4 at F. The fixed ring gear 4 meshes with

Question

The second planetary gear train considered is shown in Fig. 2.8a. The system consists of an input sun gear 1 and a planet gear 2 in mesh with 1 at B. Gear 2 is carried by the arm S fixed on the shaft of gear 3, as shown. Gear 3 meshes with the output gear 4 at F. The fixed ring gear 4 meshes with the planet gear 2 at D.

There are four moving gears (1, 2, 3, and 4) connected by the following:

Four full joints ([latex]c_5 = 4[/latex]): one at A, between the frame 0 and the sun gear 1, one at C, between the arm S and the planet gear 2, one at E, between the frame 0 and the gear 3, and one at G, between the frame 0 and the gear 3.
three half joints ([latex]c_4 = 3[/latex]): one at B, between the sun gear 1 and the planet gear 2, one at D, between the planet gear 2 and the ring gear, and one at F, between the gear 3 and the output gear 4. The module of the gears m is 5 mm.

The system possesses one DOF,

[latex]m = 3n - 2c_5 - c_4[/latex] = 3 . 4 - 2 . 4 - 3 = 1. (2.15)

The sun gear has [latex]N_1 = [/latex] 19-tooth external gear, the planet gear has [latex]N_2 = [/latex] 28-tooth external gear, and the ring gear has [latex]N_5 = [/latex] 75-tooth internal gear. The gear 3 has [latex]N_3 = [/latex] 18-tooth external gear, and the output gear has [latex]N_4 = [/latex] 36-tooth external gear. The sun gear rotates with input angular speed [latex]n_1 = [/latex] 2970 rpm ([latex]\omega_1 = \omega_{10} = \pi n_1/30 = 311.018 ext{ rad/s}[/latex]). Find the absolute output angular velocity of the gear 4, the velocities of the pitch points B and F, and the velocity of joint C.

Accepted Answer

The velocity analysis is carried out using the contour equation method. The system shown in Fig. 2.8a has five elements (0, 1, 2, 3, 4) and seven joints. The number of independent loops is given by

[latex]n_c=l-p+1=7-5+1=3[/latex].

This gear system has three independent contours. The graph of the kinematic chain and the independent contours are represented in Fig. 2.8b.

The position vectors AB, AC, AD, AF, and AG are defined as follows:

AB [latex]= \left[x_B, y_B, 0 ight]=\left[x_B, r_1, 0 ight] = \left[x_B, m N_1 / 2,0 ight][/latex],

AC [latex]= \left[x_C, y_C, 0 ight]=\left[x_C, r_1 + r_2, 0 ight] = \left[x_C, m (N_1 + N_2) / 2,0 ight][/latex],

AD [latex]= \left[x_D, y_D, 0 ight]=\left[x_D, r_1 + 2r_2, 0 ight] = \left[x_D, m (N_1 + 2 N_2) / 2,0 ight][/latex],

AF [latex]= \left[x_F, y_F, 0 ight]=\left[x_F, r_3, 0 ight] = \left[x_F, m N_3 / 2,0 ight][/latex],

AG [latex]= \left[x_G, y_G, 0 ight]=\left[x_C, r_3 + r_4, 0 ight] = \left[x_G, m (N_1 + N_4) / 2,0 ight][/latex].

First Contour
The first closed contour contains the elements 0, 1, 2, and 0 (clockwise path). For the velocity analysis, the following vectorial equations can be written:

[latex]\begin{matrix}\omega _{10}+ \omega _{21}+ \omega _{02}= 0 \ A B imes \omega _{21}+ A D imes \omega _{02}= 0. \end{matrix}[/latex] (2.16)

Here the input angular velocity is

[latex]\omega _{10}=\left[\omega_{10}, 0,0 ight]=\left[\omega_1, 0,0 ight][/latex],

and the unknown angular velocities are

[latex]\omega _{21}=\left[\omega_{21}, 0,0 ight][/latex]

[latex]\omega _{02}=\left[\omega_{02}, 0,0 ight][/latex].

The sing of the relative angular velocities is selected positive, and then the numerical results will give the real orientation of the vectors.

Equation (2.16) becomes

[latex]\omega_1 ext{I}+\omega_{21} ext{I} +\omega_{02} ext{I} = 0[/latex]

[latex]\left|\begin{array}{ccc} ext{I} & J & k \ x_B & y_B & 0 \ \omega_{21} & 0 & 0 \end{array} ight| + \left|\begin{array}{ccc} ext{I} & J & k \ x_D & y_D & 0 \ \omega_{02} & 0 & 0 \end{array} ight| = 0. [/latex] (2.17)

Equation (2.17) projected on a "fixed" reference frame [latex]xOyz[/latex] is

[latex]\begin{matrix} \omega_1+\omega_{21}+\omega_{02}=0, \ y_B \omega_{21} + y_D \omega_{02} = 0. \end{matrix}[/latex] (2.18)

Equation (2.18) represents a system of two equations with two unknowns [latex]\omega_{21}[/latex] and [latex]\omega_{02}[/latex]. Solving the algebraic equations, we obtain the following value for the absolute angular velocity of planet gear 2:

[latex]\omega_{20}=-\omega_{02}=-\frac{N_1 \omega_1}{2 N_2}=-105.524 ext{ rad/s.}[/latex] (2.19)

Second Contour
The second closed contour contains the elements 0, 3, 2, and 0 (counterclockwise path). For the velocity analysis, the following vectorial equations can be written:

[latex]\begin{matrix} \omega_{30} + \omega_{23} + \omega_{02} = 0 \ AE imes \omega_{30} + AC imes \omega_{23} + AD imes \omega_{02} = 0. \end{matrix}[/latex] (2.20)

The unknown angular velocities are

[latex]\omega _{30}=\left[\omega_{21}, 0,0 ight][/latex]

[latex]\omega _{23}=\left[\omega_{23}, 0,0 ight][/latex].

When Eq. (2.20) is solved, the following value is obtained for the absolute angular velocity of the gear 3 and the arm S:

[latex]\omega_{30}=\frac{N_1 \omega_1}{2\left(N_1 + N_2 ight)} = 62.865 ext{ rad/s.}[/latex] (2.21)

Third Contour
The third closed contour contains the links 0, 4, 3, and 0 (counterclockwise path). The velocity vectorial equations are

[latex]\begin{matrix} \omega_{40} + \omega_{34} + \omega_{03} = 0 \ AG imes \omega_{40} + AF imes \omega_{34} + AE imes \omega_{03} = 0 \end{matrix}[/latex] (2.22)

or

[latex]\omega_{40} I +\omega_{34} I -\omega_{30} I = 0[/latex]

[latex]\left|\begin{array}{ccc} I & J & k \ x_G & y_G & 0 \ \omega_{40} & 0 & 0 \end{array} ight| + \left|\begin{array}{ccc} I & J & k \ x_F & y_F & 0 \ \omega_{34} & 0 & 0 \end{array} ight|= 0. [/latex] (2.23)

The unknown angular velocities are

[latex]\omega _{40}=\left[\omega_{40}, 0,0 ight][/latex]

[latex]\omega _{34}=\left[\omega_{34}, 0,0 ight][/latex].

The absolute angular velocity of the output gear 4 is

[latex]\omega_{40}=-\frac{N_1 N_3 \omega_1}{2 \left(N_1 + N_2 ight) N_4}=-31.432 ext{ rad/s.}[/latex] (2.24)

Linear Velocities of Pitch Points
The velocity of the pitch point B is

[latex]v_F=\omega_{40} r_4=2.828 ext{ m/s}[/latex].

The velocity of the joint C is

[latex]v_B=\omega_{10} r_1=14.773 ext{ m/s}[/latex],

and the velocity of the pitch point F is

[latex]v_C=\omega_{30} (r_1 + r_2) = 7.386 ext{ m/s}[/latex].

Gear Geometrical Dimensions
For standard external gear teeth the addendum is a = m.

Gear 1:

Pitch circle diameter [latex]d_1 = mN_1 = 95.0 ext{ mm}[/latex]

Addendum circle diameter [latex]d_{a1} = m(N_1 + 2) - 105.0 ext{ mm}[/latex]

Dedendum circle diameter [latex]d_{d1} = m(N_1-2.5) = 82.5 ext{ mm.}[/latex]

Gear 2:

Pitch circle diameter [latex]d_2 = mN_2 = 140.0 ext{ mm}[/latex]

Addendum circle diameter [latex]d_{a2} = m(N_2 + 2) = 150.0 ext{ mm}[/latex]

Dedendum circle diameter [latex]d_{d2} = m(N_2-2.5) = 127.5 ext{ mm.}[/latex]

Gear 3:

Pitch circle diameter [latex]d_3 = mN_3 = 90.0 ext{ mm}[/latex]

Addendum circle diameter [latex]d_{a3} = m(N_3 + 2) = 100.0 ext{ mm}[/latex]

Dedendum circle diameter [latex]d_{d3} = m(N_3-2.5) = 77.5 ext{ mm.}[/latex]

Gear 4:

Pitch circle diameter [latex]d_4 = mN_4 = 180.0 ext{ mm}[/latex]

Addendum circle diameter [latex]d_{a4} = m(N_4 + 2) = 190.0 ext{ mm}[/latex]

Dedendum circle diameter [latex]d_{d4} = m(N_4-2.5) = 167.5 ext{ mm.}[/latex]

Gear 5: (internal gear):

Pitch circle diameter [latex]d_5 = mN_5 = 375.0 ext{ mm}[/latex]

Addendum circle diameter [latex]d_{a5} = m(N_5-2) = 365.0 ext{ mm}[/latex]

Dedendum circle diameter [latex]d_{d5} = m(N_5 + 2.5) = 387.5 ext{ mm.}[/latex]

Number of Planet Gears
The number of necessary planet gears k is given by the assembly condition

[latex]\left(N_1+N_5 ight) / k= ext { integer, }[/latex]

and for the planetary gear train k = 2 planet gears. The vicinity condition between the sun gear and the planet gear

[latex]m\left(N_1+N_2 ight) \sin (\pi / k)>d_{a 2}[/latex]

is verified.

The group drawings for this mechanism with planetary gears are given in Fig. 2.9.

Question 5.2.5: The second planetary gear train considered is shown in Fig. ...

Related Answered Questions

Verified Answer:

A journal bearing has the diameter D = 2.5 in, the length L = 0.625 in, and the radial clearance c = 0.002 in, as shown in Fig. 5.17. The shaft rotates at n = 3600 rpm. The journal bearing supports a constant load W = 1500 lb. The lubricant film is SAE 40 oil at atmospheric pressure. The average ...

Verified Answer:

A shaft with a 120 mm diameter (Fig. 5.7) is supported by a bearing of 100 mm length with a diametral clearance of 0.2 mm and is lubricated by oil having a viscosity of 60 mPa s. The shaft rotates at 720 rpm. The radial load is 6000 N. Find the bearing coefficient of friction and the power loss. ...

Verified Answer:

The Saybolt kinematic viscosity of an oil corresponds to 60 s at 90°C (Fig. 5.2). What is the corresponding absolute viscosity in millipascal-seconds (or centipoises) and in microreyns? ...

Verified Answer:

Verified Answer:

Select a ball bearing for a machine for continuous 24-hour service. The machine rotates at the angular speed of 900 rpm. The radial force is Fr = 1 kN, and the thrust force is Fa = 1.25 kN, with light impact. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Two involute spur gears of module 5, with 19 and 28 teeth, operate at a pressure angle of 20°. Determine whether there will be interference when standard full-depth teeth are used. Find the contact ratio. ...

Verified Answer: