Design of belts

A belt transmits power by friction between the belt and the pulley. The tight side tension $T_1$ is greater than the slack side tension $T_2$, and their difference is the effective pull.

Derivation

Consider a small element of belt subtending angle $\delta\theta$ at the centre of the pulley (angle of contact $\theta$). Let the belt mass per metre be $m$ and speed $v$. The forces on the element are:

• tensions $T$ and $T + \delta T$ at its ends,
• normal reaction $R$ from the pulley,
• friction $\mu R$,
• centrifugal force $m v^2\,\delta\theta$ (equal to $T_c\,\delta\theta$, where $T_c = mv^2$).

Resolving radially:

$$R + T_c\,\delta\theta = (T + \delta T)\sin\frac{\delta\theta}{2} + T\sin\frac{\delta\theta}{2} \approx T\,\delta\theta$$

so $R = (T - T_c)\,\delta\theta$.

Resolving tangentially:

$$\mu R = (T + \delta T)\cos\frac{\delta\theta}{2} - T\cos\frac{\delta\theta}{2} = \delta T$$

Substituting $R$:

$$\mu (T - T_c)\,\delta\theta = \delta T \quad\Rightarrow\quad \frac{\delta T}{T - T_c} = \mu\,\delta\theta$$

Integrating from $T_2$ to $T_1$ over the angle $0$ to $\theta$:

$$\ln\frac{T_1 - T_c}{T_2 - T_c} = \mu\theta \quad\Rightarrow\quad \frac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$$

If the centrifugal tension is neglected (low speed, $T_c = 0$), this becomes $\dfrac{T_1}{T_2} = e^{\mu\theta}$, where $\theta$ is in radians and is taken for the smaller pulley (or the smaller $\mu\theta$ product).

Condition for maximum power

Power $P = (T_1 - T_2)\,v$. Put $T_1 = T_{max}$ (the maximum permissible tension). From the result above, $T_2 - T_c = (T_1 - T_c)\,e^{-\mu\theta}$, so $T_1 - T_2 = (T_1 - T_c)\left(1 - e^{-\mu\theta}\right)$ and

$$P = \left(T_{max} - m v^2\right)\left(1 - e^{-\mu\theta}\right) v$$

For maximum power, $dP/dv = 0$:

$$\frac{d}{dv}\left(T_{max}v - m v^3\right) = T_{max} - 3 m v^2 = 0$$ $$v = \sqrt{\frac{T_{max}}{3m}}$$

At this speed the centrifugal tension is $T_c = mv^2 = T_{max}/3$, i.e. one third of the maximum permissible tension.

Differences

Ratio of tensions in a V-belt

The belt sits in a groove whose included angle is $2\beta$ (usually 34 to 40 degrees). If $R$ is the total normal reaction from the two flanks, each flank carries a normal reaction $R_n$ with $R = 2R_n \sin\beta$. The friction on both flanks is $2\mu R_n = \dfrac{\mu R}{\sin\beta}$.

So the V-belt behaves like a flat belt with an effective coefficient of friction $\mu/\sin\beta$. Using the flat-belt derivation:

$$\frac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta/\sin\beta}$$

Why a V-belt transmits more power

For $\beta = 17^\circ$ to $20^\circ$, $\sin\beta \approx 0.3$ so the exponent is about three times larger. With $\mu = 0.3$ and $\theta = 160^\circ$ (2.79 rad), a flat belt gives $e^{0.84} = 2.3$ while a V-belt with $\beta = 20^\circ$ gives $e^{0.84/0.342} = e^{2.45} = 11.6$. A bigger ratio $T_1/T_2$ means a bigger difference of tension for the same tight side tension, and thus more power. The belt need not be tightly stretched, so bearing loads are smaller.

Data: $d = 0.4$ m, $D = 0.8$ m, $C = 2.5$ m, $n = 600$ rpm, $b = 0.15$ m, $t = 0.008$ m, $\rho = 1000$ kg/m$^3$, $\sigma = 2$ MPa, $\mu = 0.3$.

(a) Open belt

Length

$$L = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 5 + 1.885 + 0.016 = 6.901\ \text{m}$$

Angle of contact on the small pulley

$$\sin\alpha = \frac{D - d}{2C} = \frac{0.4}{5} = 0.08, \qquad \alpha = 4.589^\circ$$ $$\theta = 180^\circ - 2\alpha = 170.82^\circ = 2.981\ \text{rad}$$

Speed and masses

$$v = \frac{\pi d n}{60} = \frac{\pi \times 0.4 \times 600}{60} = 12.57\ \text{m/s}$$ $$m = \rho b t = 1000 \times 0.15 \times 0.008 = 1.2\ \text{kg/m}$$

Tensions

$$\begin{aligned} T_{max} &= \sigma b t = 2 \times 150 \times 8 = 2400\ \text{N} \\ T_c &= m v^2 = 1.2 \times 12.57^2 = 189.5\ \text{N} \\ T_1 - T_c &= 2400 - 189.5 = 2210.5\ \text{N} \end{aligned}$$

The tight side tension is the maximum permissible value, $T_1 = T_{max} = 2400$ N. Using $\dfrac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$:

$$e^{\mu\theta} = e^{0.3 \times 2.981} = e^{0.894} = 2.446$$ $$T_2 - T_c = \frac{2210.5}{2.446} = 903.7\ \text{N} \quad\Rightarrow\quad T_2 = 903.7 + 189.5 = 1093.2\ \text{N}$$

Power

$$P = (T_1 - T_2)\,v = (2400 - 1093.2) \times 12.57 = 16\,420\ \text{W} = 16.4\ \text{kW}$$

(b) Cross belt

$$\sin\alpha = \frac{D + d}{2C} = \frac{1.2}{5} = 0.24, \qquad \alpha = 13.89^\circ$$ $$\theta = 180^\circ + 2\alpha = 207.8^\circ$$ $$L = 2C + \frac{\pi}{2}(D + d) + \frac{(D + d)^2}{4C} = 5 + 1.885 + 0.144 = 7.029\ \text{m}$$

The angle of contact is the same on both pulleys, and larger than for the open belt, so a cross belt can transmit more power, but the belt rubs where it crosses and wears faster.

Answer: (a) L = 6.90 m; theta = 170.8 degrees; Tc = 189.5 N; T1 = 2400 N, T2 = 1093 N; P = 16.4 kW. (b) L = 7.03 m; theta = 207.8 degrees.

Pulley sizes and speed

$$D = d\,\frac{n_1}{n_2} = 140 \times \frac{1440}{480} = 420\ \text{mm}$$ $$v = \frac{\pi d n_1}{60} = \frac{\pi \times 0.14 \times 1440}{60} = 10.56\ \text{m/s}$$

(The speed is within the usual limit of 25 m/s for V-belts.)

Belt length

Trial centre distance $C = 800$ mm:

$$L = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 1600 + 879.6 + 24.5 = 2504\ \text{mm}$$

Nearest standard pitch length: 2500 mm.

Actual centre distance

$$C = B + \sqrt{B^2 - \frac{(D - d)^2}{8}}, \qquad B = \frac{L}{4} - \frac{\pi (D + d)}{8}$$ $$B = 625 - 219.9 = 405.1\ \text{mm}$$ $$C = 405.1 + \sqrt{405.1^2 - \frac{280^2}{8}} = 405.1 + \sqrt{164\,106 - 9800} = 405.1 + 392.8 = 797.9\ \text{mm}$$

Contact angle on the small pulley:

$$\theta = 180^\circ - 2\sin^{-1}\frac{D - d}{2C} = 180^\circ - 2\sin^{-1}\frac{280}{1596} = 159.8^\circ$$

This agrees with the 160 degrees assumed for the factor 0.95.

Number of belts

$$\text{Design power} = 1.2 \times 15 = 18\ \text{kW}$$ $$\text{Corrected rating per belt} = 3.4 \times 0.95 \times 1.04 = 3.36\ \text{kW}$$ $$\text{Number of belts} = \frac{18}{3.36} = 5.36 \;\Rightarrow\; 6\ \text{belts}$$

Answer: D = 420 mm; v = 10.56 m/s; L = 2500 mm; C = 798 mm; 6 B-section belts.

Data: $p = 19.05$ mm, $z_1 = 19$, $z_2 = 48$, $n_1 = 1000$ rpm, $P = 15$ kW, $C \approx 40p$, breaking load $= 31.1$ kN.

Driven speed

$$n_2 = n_1\,\frac{z_1}{z_2} = 1000 \times \frac{19}{48} = 395.8\ \text{rpm}$$

Pitch diameters

$$d_1 = \frac{p}{\sin(180^\circ/z_1)} = \frac{19.05}{\sin 9.474^\circ} = 115.7\ \text{mm}, \qquad d_2 = \frac{19.05}{\sin 3.75^\circ} = 291.3\ \text{mm}$$

Chain length in pitches

$$L_p = \frac{2C}{p} + \frac{z_1 + z_2}{2} + \left(\frac{z_2 - z_1}{2\pi}\right)^2 \frac{p}{C}$$ $$L_p = 80 + 33.5 + \left(\frac{29}{2\pi}\right)^2\frac{1}{40} = 80 + 33.5 + 0.53 = 114.03$$

Take 114 pitches (even number, so that no offset link is needed).

Exact centre distance

$$\frac{C}{p} = \frac{1}{4}\left[\left(L_p - \frac{z_1 + z_2}{2}\right) + \sqrt{\left(L_p - \frac{z_1 + z_2}{2}\right)^2 - 8\left(\frac{z_2 - z_1}{2\pi}\right)^2}\right]$$ $$\frac{C}{p} = \frac{1}{4}\left[80.5 + \sqrt{80.5^2 - 8(21.30)}\right] = \frac{1}{4}[80.5 + 79.43] = 39.98$$ $$C = 39.98 \times 19.05 = 761.7\ \text{mm}$$

Chain velocity and pull

$$v = \frac{z_1 p n_1}{60\times 1000} = \frac{19 \times 19.05 \times 1000}{60\,000} = 6.03\ \text{m/s}$$ $$F = \frac{P}{v} = \frac{15\,000}{6.03} = 2487\ \text{N}$$

Factor of safety

$$FS = \frac{31\,100}{2487} = 12.5$$

This is above the usual minimum (about 8 to 11 for ordinary speeds), so the chain is adequate. Chain drives give a constant average velocity ratio with no slip, but the velocity varies slightly with the chordal action, and they need lubrication.

Answer: n2 = 395.8 rpm; d1 = 115.7 mm, d2 = 291.3 mm; 114 pitches; C = 762 mm; v = 6.03 m/s; F = 2.49 kN; FS = 12.5.