System Design

(a) Progression ratio and speeds

For $z$ speeds in GP: $n_{max}/n_{min} = \varphi^{\,z-1}$.

$$\varphi = \left(\frac{400}{125}\right)^{1/5} = 3.2^{0.2} = 1.262 \approx 1.26 \ (\text{R10 series})$$

Spindle speeds ($\times 1.26$ each step): 125, 160, 200, 250, 315, 400 rpm.

(b) Structure and ray diagram

Six speeds $= 3 \times 2$. Structural formula: 3(1) x 2(3), where the number in brackets is the characteristic $x$ (the step is $\varphi^{x}$).

• Group 1 (3 speeds, $x=1$): ratios $1,\ 1/\varphi,\ 1/\varphi^{2} = 1,\ 0.794,\ 0.630$.
• Group 2 (2 speeds, $x=3$): ratios $1,\ 1/\varphi^{3} = 1,\ 0.5$.

Check of range: group 1 $\varphi^{2}=1.6$, group 2 $\varphi^{3}=2$, both below the limit of 8, and no gear ratio is smaller than 1/4.

Belt drive: motor 1440 rpm to shaft I at 400 rpm, ratio $1440/400 = 3.6$.

 Shaft I     Shaft II      Shaft III (rpm)
  400 ----> 400 ---------> 400  (x1)
      \               \---> 200  (x0.5)
       \--> 315 --------> 315
        \            \---> 160
         \-> 250 -------> 250
                      \---> 125

Shaft II runs at 400, 315, 250 rpm. Each speed of shaft II gives two speeds on shaft III: the same speed (ratio 1) or three steps lower (ratio 0.5), giving 400, 315, 250 and 200, 160, 125 rpm.

(c) Gear tooth numbers

Stage 1 (sum 70): ratio $= z_1/z_2$.

Stage 2 (sum 60):

All pinions have more than 17 teeth (no undercutting for 20 degree teeth).

Check of actual spindle speeds: 400, 317.9, 251.2, 200.0, 159.0, 125.6 rpm. Errors against the standard speeds are below 1%, within the permitted $\pm 10(\varphi-1)\% = 2.6\%$.

Answer: $\varphi = 1.26$; speeds 125, 160, 200, 250, 315, 400 rpm; structure $3(1)\times2(3)$; teeth 35/35, 31/39, 27/43 and 30/30, 20/40.

Take $g = 9.81\ \text{m/s}^2$, $W = mg = 14715$ N.

(a) First-gear ratio

Slope angle $\theta = \tan^{-1}(0.3) = 16.70^\circ$; $\sin\theta = 0.2873$, $\cos\theta = 0.9578$.

Tractive force needed (gradient plus rolling resistance):

$$\begin{aligned} F &= W(\sin\theta + f\cos\theta) \\ &= 14715\,(0.2873 + 0.015\times0.9578) = 4440\ \text{N} \end{aligned}$$

Torque at the wheels: $T_w = F r = 4440 \times 0.32 = 1421\ \text{N m}$.

Overall ratio from engine to wheels:

$$G_o = \frac{T_w}{T_e\,\eta} = \frac{1421}{120\times0.88} = 13.45$$

First-gear ratio of the gearbox:

$$G_1 = \frac{G_o}{G_f} = \frac{13.45}{4.1} = 3.28$$

(b) Intermediate ratios (GP)

For four speeds, $G_1, G_2, G_3, G_4 = G_1,\ G_1/k,\ G_1/k^2,\ G_1/k^3$ with $G_4 = 1$:

$$k = (3.28)^{1/3} = 1.486$$

(c) Maximum speed in top gear

Engine speed $\omega_e = 5500\times2\pi/60 = 575.96$ rad/s.

$$v = \frac{\omega_e\, r}{G_4 G_f} = \frac{575.96\times0.32}{1\times4.1} = 44.95\ \text{m/s} = 161.8\ \text{km/h}$$

Traction check

Maximum tractive force available $= \mu\,(0.55\,W\cos\theta) = 0.7\times0.55\times14715\times0.9578 = 5426$ N.

Since $5426 > 4440$ N, the wheels do not slip on the 30% slope.

Answer: $G_1 = 3.28$; $G_2 = 2.21$, $G_3 = 1.49$; top speed about 162 km/h; traction is sufficient (5426 N available against 4440 N required).

Data

$P = 18$ kW, $N = 450$ rpm, $z = 28$, $m = 6$ mm, $\phi = 20^\circ$, $L = 600$ mm.

Torque and gear loads

$$T = \frac{P}{\omega} = \frac{18000}{2\pi\times450/60} = \frac{18000}{47.12} = 382.0\ \text{N m}$$

Pitch circle diameter $d_p = mz = 6\times28 = 168$ mm, so the pitch radius is 0.084 m.

$$F_t = \frac{T}{r} = \frac{382.0}{0.084} = 4547\ \text{N}, \qquad F_r = F_t\tan20^\circ = 1655\ \text{N}$$

Bending moments (load at mid-span, $M = FL/4$)

$$\begin{aligned} M_H &= \frac{4547\times0.6}{4} = 682.1\ \text{N m} \\ M_V &= \frac{1655\times0.6}{4} = 248.3\ \text{N m} \\ M &= \sqrt{682.1^2 + 248.3^2} = 725.9\ \text{N m} \end{aligned}$$

   Ft, Fr (down/side)
        |
   A ---+--- B        L = 600 mm
   R_A  300  R_B      M = FL/4 at mid-span

Allowable shear stress

$$\tau_{all} = 55\times0.75 = 41.25\ \text{MPa}$$

ASME code equation

$$d^3 = \frac{16}{\pi\,\tau_{all}}\sqrt{(K_bM)^2 + (K_tT)^2}$$ $$\sqrt{(1.5\times725.9)^2 + (1.2\times382.0)^2} = \sqrt{1088.9^2 + 458.4^2} = 1181.4\ \text{N m}$$ $$d^3 = \frac{16\times1181.4}{\pi\times41.25\times10^{6}} = 1.4587\times10^{-4}\ \text{m}^3$$ $$d = 0.0526\ \text{m} = 52.6\ \text{mm}$$

Standard size: 55 mm.

Answer: $d = 52.6$ mm, so use a 55 mm shaft.

A machine tool drive transmits power from the motor to the spindle or slide and gives the range of speeds and feeds needed for different materials, tool materials and diameters.

Requirements

• Cover the full range of cutting speeds, $v = \pi D N/1000$, for the smallest and largest diameters. The speed ratio $R_n = N_{max}/N_{min}$ is typically 20 to 100 or more.
• Enough power and torque at the spindle, constant power over most of the range.
• High rigidity and low vibration, accurate running of shafts and gears, low noise.
• Quick, easy change of speed, preferably during running or with levers; safe and simple to operate.
• Compact, economical, easy lubrication and maintenance; minimum number of gears and shafts.
• Stable speed under changing load.

Stepped vs stepless

Why geometric progression

If speeds are chosen as $N_1,\ N_1\varphi,\ N_1\varphi^2, \dots$ the percentage loss in cutting speed between steps, $(N_{j+1}-N_j)/N_{j+1} = (\varphi-1)/\varphi$, is the same at every step. The ideal speed for a job is therefore never more than a constant fraction away from an available speed. An arithmetic progression gives a large loss at low speeds and a wasteful closeness at high speeds. The GP also needs fewer gears and fits the standard preferred-number (Renard) series R5, R10, R20, R40 with $\varphi = 1.58, 1.26, 1.12, 1.06$.

An automobile gearbox changes the torque-speed ratio between engine and wheels and gives reverse and neutral.

Comparison

Working of the synchromesh

 Gear (with cone)  Blocker ring  Hub/sleeve
     |\               |\            |
     | \_cone_______  | friction    | shift
     |  engages      |  cone       | fork

1. The driver moves the gear lever; the fork pushes the sleeve toward the selected gear.
2. A blocker ring (with a cone surface) is pressed on the cone of the gear. Friction brings the speeds of the shaft hub and the gear to the same value.
3. The blocker ring prevents the sleeve from sliding on until the speeds match, so there is no clash.
4. After speeds are equal, the sleeve slides over the dog teeth of the gear and locks it to the shaft.

Requirements

• Minimum weight with high power: use of alloy steels, case-hardened ground gears, thin webs and light casings.
• High reliability and safety: redundancy, high factor on fatigue life, vibration and condition monitoring; the gearbox is flight-critical.
• Operation at high speed and temperature with forced oil lubrication and cooling.
• Compact size and the correct shaft geometry, with low noise and vibration.
• Ability to carry the propeller thrust and gyroscopic loads.

Why a reduction gearbox

A gas turbine runs efficiently at 10000 to 30000 rpm, but a large propeller must turn slowly (about 1000 to 2000 rpm) so that blade tip speed stays below sonic speed for efficiency and low noise. A reduction gearbox gives this speed ratio.

Why epicyclic

Power is shared among several planet gears, so each tooth carries less load; coaxial input and output; compact and light; balanced forces on the sun and carrier give lower bearing loads.

Numerical

Sun input, ring fixed, carrier output: speed ratio $= 1 + \dfrac{z_r}{z_s}$.

$$\frac{N_{in}}{N_{out}} = \frac{12000}{1500} = 8 = 1 + \frac{z_r}{z_s} \;\Rightarrow\; \frac{z_r}{z_s} = 7$$

For example $z_s = 18$, $z_r = 126$ (planet teeth $= (126-18)/2 = 54$).

Input torque $T_{in} = \dfrac{P}{\omega} = \dfrac{1\,500\,000}{12000\times2\pi/60} = 1193.7\ \text{N m}$.

Output torque (no loss) $= T_{in}\times 8 = 9549\ \text{N m}$.

Answer: $z_r/z_s = 7$; propeller torque about 9550 N m.

Need

When a vehicle turns, the outer wheel travels a longer path than the inner wheel, so it must rotate faster. If both rear wheels were rigidly fixed to one shaft, one would slip and wear the tyres. A differential lets the wheels turn at different speeds while still sharing the driving torque equally.

Construction

   Prop shaft ---> Pinion
                     |
                 Crown wheel (carries the cage)
                     |
   Left axle - Side gear \ / Side gear - Right axle
                    Spider (planet) pinions

The drive pinion meshes with the crown wheel. The crown wheel is bolted to the differential cage, which carries the spider shaft and two or four planet pinions. The planets mesh with two side (sun) gears splined to the left and right axle shafts.

Working

• Straight road: both wheels meet equal resistance, the planets do not spin on their own axes and the cage and both side gears rotate together.
• Turning: the inner wheel resists more, so the planets rotate about the spider axis, speeding up one side gear and slowing the other by the same amount.
• Torque is equal on both axles in a bevel-gear differential (a drawback on slippery surfaces, which limited-slip or locking differentials overcome).

Speed relation

Cage speed $N_c$, side gears $N_L$ and $N_R$. Relative to the cage the two side gears turn in opposite directions with equal ratio (planets are idlers): $N_L - N_c = -(N_R - N_c)$, therefore

$$N_L + N_R = 2N_c$$

Numerical

$N_c = 1800\times\dfrac{10}{40} = 450$ rpm.

$$N_R = 2\times450 - 400 = 500\ \text{rpm}$$

Answer: right wheel = 500 rpm (the outer wheel in the turn).