10.3 Belts, Roller Chains & Flexible Drives

10.3 Belts, Roller Chains & Flexible Drives

Flexible drives carry power between shafts that are too far apart for gears, absorb shock, and tolerate some misalignment. The machine designer sizes them from a few relationships: the capstan (belt-friction) equation, centrifugal tension, wrap angle, belt or chain length, and, for chains, the polygonal (chordal) speed variation.

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Flexible Drives: Flat Belts, V-Belts & Synchronous Belts

Flexible drive media transmit mechanical power between separated shafts through friction grip or positive mechanical toothed meshing:

1. Flat Belts:
   • Mechanics: Pure friction drive on smooth, crowned cylindrical pulleys (pulley crown ensures lateral tracking).
   • Characteristics: Operates at very high speeds ($> 40\,\text{m/s}$), quiet, absorbs shock via elastic stretch. Requires high initial tension $F_i$, resulting in large overhung shaft and bearing radial loads.
2. V-Belts:
   • Mechanics: Operates in trapezoidal sheave grooves with included groove angle $\beta \approx 34^\circ - 40^\circ$ (half-angle $\beta/2 \approx 17^\circ - 20^\circ$).
   • The Wedging Effect: Lateral compression against the tapered groove sidewalls multiplies normal contact force without increasing belt tension:

$$N = \frac{F_n}{\sin(\beta/2)}$$

This multiplies effective friction by a factor of $1/\sin(\beta/2) \approx 1/\sin(18^\circ) \approx 3.23$, delivering over three times the torque capacity of a flat belt under identical tension. 3. Synchronous (Timing) Belts:

• Mechanics: Positive mechanical engagement between molded elastomeric belt teeth (trapezoidal or curvilinear HTD profile) and matching axially grooved pulleys.
• Characteristics: Zero kinematic slip; maintains precise angular phase synchronization. Requires minimal initial tension, substantially reducing bearing radial loads.

Flexible Drive Comparison Matrix

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The Belt Tension Formulation

Consider a belt traveling at pitch velocity $v$ (in $\text{m/s}$) around a sheave with wrap angle $\phi$ (in radians). As belt speed increases, centrifugal force throws the belt outward, creating a centrifugal tension $F_c$ that reduces effective normal pressure against the sheave:

$$F_c = m v^2$$

where $m$ is the linear mass density of the belt (in $\text{kg/m}$ or $\text{lbm/ft}$). The governing Euler-Eytelwein capstan relationship governing limiting tight-side tension $F_1$ and slack-side tension $F_2$ is:

$$\frac{F_1 - m v^2}{F_2 - m v^2} = \exp\left( \frac{f \phi}{\sin(\beta/2)} \right)$$

For flat belts, set $\sin(\beta/2) = \sin(90^\circ) = 1.0$, reducing the exponent to $f\phi$. For V-belts, $\beta/2$ is the groove half-angle ($17^\circ - 20^\circ$).

Wrap Angle Calculation

For an open belt drive with large sheave diameter $D$, small sheave diameter $d$, and center-to-center distance $C$, the wrap angle on the smaller (limiting) sheave is:

$$\phi = \pi - 2 \arcsin\left( \frac{D - d}{2 C} \right) \quad [\text{radians}]$$

Transmitted Power & Initial Tension

• Net Transmitted Force: $F_{\text{net}} = F_1 - F_2$
• Transmitted Mechanical Power ($P$):

$$P = (F_1 - F_2) v$$

• Initial Static Installation Tension ($F_i$) (assuming linear belt elasticity):

$$F_i = \frac{F_1 + F_2}{2} - F_c$$

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Roller Chain Drives & Chordal Action

Roller chains transmit power through positive engagement between cylindrical rollers and sprocket teeth, governed by ANSI standard B29.1:

• Chain Pitch ($p$): Distance between adjacent pin centers. ANSI numbering designates pitch in eighths of an inch: ANSI #40 has $p = 4/8\,\text{in} = 0.500\,\text{in}$; ANSI #60 has $p = 6/8\,\text{in} = 0.750\,\text{in}$; ANSI #80 has $p = 8/8\,\text{in} = 1.000\,\text{in}$.
• Sprocket Pitch Diameter ($d$): Because chain links form a regular polygon around the sprocket:

$$d = \frac{p}{\sin(180^\circ / N)}$$

where $N$ is the number of sprocket teeth.

Chordal Action (Polygonal Effect)

As the sprocket rotates, the chain link chord swings through an arc, causing the effective pitch radius to oscillate cyclically between pitch radius $R = d/2$ and inscribed radius $r = R \cos(180^\circ / N)$:

                    Sprocket Tooth Centerline
                             / \
                            /   \
               Link Pin -> o-----o <- Link Pin
                           |  p  |
                           |     |
                           +--+--+
                              |
                     Chordal Pitch Circle

This geometry induces a continuous cyclical variation in linear chain velocity:

$$\frac{v_{\max} - v_{\min}}{v_{\max}} = 1 - \cos\left( \frac{180^\circ}{N} \right)$$

• With $N = 10$ teeth, linear velocity oscillates by 4.9%, causing severe chain whip, chordal impact shock, high noise, and rapid roller fatigue.
• With $N = 17$ teeth, speed fluctuation drops to 1.7%.
• With $N = 25$ teeth, speed fluctuation drops below 0.8%.

Belt and Chain Length

For an open belt drive, the exact pitch length is

$$L = \sqrt{4C^2 - (D - d)^2} + \frac{1}{2}\left(D\,\theta_D + d\,\theta_d\right), \qquad \theta_d = \pi - 2\sin^{-1}\frac{D - d}{2C}, \quad \theta_D = \pi + 2\sin^{-1}\frac{D - d}{2C}$$

and a close approximation is $L \approx 2C + \tfrac{\pi}{2}(D + d) + \tfrac{(D - d)^2}{4C}$.

For a roller chain, the length in pitches (links) is

$$\frac{L}{p} = \frac{2C}{p} + \frac{N_1 + N_2}{2} + \frac{(N_2 - N_1)^2}{4\pi^2 (C/p)}$$

Round up to an even number of links so that no offset (half) link is needed.

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Worked Example: V-Belt Drive and Chain Selection Checks

Part A: V-Belt Tensions

A single V-belt transmits $P = 7.5\,\text{kW}$ from a $d = 160\,\text{mm}$ driver sheave turning at $1{,}750\,\text{rpm}$ to a $D = 400\,\text{mm}$ driven sheave. The center distance is $C = 600\,\text{mm}$, the belt mass is $m = 0.20\,\text{kg/m}$, the groove angle is $\beta = 36^\circ$, and $f = 0.15$.

1. Belt speed: $v = \pi d n / 60 = \pi (0.160)(1{,}750)/60 = 14.66\,\text{m/s}$.
2. Wrap on the small sheave: $\phi = \pi - 2\sin^{-1}\!\left[(400 - 160)/(2 \times 600)\right] = \pi - 2(0.2014) = 2.739\,\text{rad}$ ($156.9^\circ$).
3. Friction exponent: $f\phi/\sin(\beta/2) = (0.15)(2.739)/\sin 18^\circ = 1.329$, so the tension ratio is $e^{1.329} = 3.78$.
4. Centrifugal tension: $F_c = m v^2 = 0.20(14.66)^2 = 43.0\,\text{N}$.
5. Net (effective) pull: $F_1 - F_2 = P/v = 7{,}500/14.66 = 511.6\,\text{N}$.
6. Solve $(F_1 - F_c) = 3.78(F_2 - F_c)$ together with step 5: $F_2 - F_c = 511.6/2.78 = 184.1\,\text{N}$, giving $F_2 = 227.1\,\text{N}$ and $F_1 = 738.7\,\text{N}$.
7. Required installation tension: $F_i = (F_1 + F_2)/2 - F_c = 482.9 - 43.0 = 439.9\,\text{N}$.
8. Pitch length: $\sqrt{4(0.6)^2 - (0.24)^2} + \tfrac{1}{2}[0.400(3.544) + 0.160(2.739)] = 1.176 + 0.928 = 2.104\,\text{m}$; the approximation gives $1.200 + 0.880 + 0.024 = 2.104\,\text{m}$.

The driven sheave turns at about $1{,}750 \times 160/400 = 700\,\text{rpm}$ (slightly less because of belt creep), and the shafts carry a radial load of roughly $F_1 + F_2 \approx 966\,\text{N}$.

Part B: Roller Chain Length and Chordal Variation

An ANSI No. 60 chain ($p = 0.750\,\text{in}$) connects a 17-tooth driver to a 51-tooth sprocket at a center distance of $C = 30\,\text{in}$ ($C/p = 40$).

• Length: $L/p = 2(40) + (17 + 51)/2 + (51 - 17)^2/(4\pi^2 \times 40) = 80 + 34 + 0.73 = 114.7$, so specify 116 links (the next even number).
• Driver pitch diameter: $d = p/\sin(180^\circ/17) = 0.750/0.1838 = 4.08\,\text{in}$.
• Chordal speed variation: $1 - \cos(180^\circ/17) = 1.7\%$, acceptable for an industrial drive.