10.1 Belts, V-Belts, Chains, Wire Ropes & Flywheels

10.1 Belts, V-Belts, Chains, Wire Ropes & Flywheels

Quick Summary: Flexible power transmission components (belts, chains, wire ropes) transmit mechanical energy over long center distances via tension and friction, while flywheels absorb and release kinetic energy to stabilize cyclic shaft speed. Master the tension ratio formulas, centrifugal tension corrections, wedge action multipliers, chordal velocity variations, wire rope bending stresses, and flywheel rim stress equations.

Flexible Power Transmission Elements Overview

In mechanical design, flexible machine elements transfer power between non-coaxial shafts separated by moderate to long distances. Unlike rigid gear trains, flexible elements absorb shock loads, accommodate shaft misalignment, and offer economical power distribution.

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1. Flat Belt Drive Mechanics

Flat belt drives transfer torque across driving and driven pulleys via friction developed along the arc of contact $\theta$. Power capacity is governed by the ratio of tight-side tension $T_1$ to slack-side tension $T_2$.

Belt Tension Ratio Equation

Neglecting centrifugal force at low linear belt speeds ($V < 10\text{ m/s}$), the limiting tension ratio before slippage is derived from equilibrium on an infinitesimal belt element:

$$\frac{T_1}{T_2} = e^{\mu \theta}$$

Where:

• $T_1$ = Tight-side belt tension (N)
• $T_2$ = Slack-side belt tension (N)
• $\mu$ = Coefficient of friction between belt material and pulley surface
• $\theta$ = Contact (wrap) angle on the smaller pulley (radians)

Centrifugal Tension Correction

At high linear velocities ($V > 10\text{ m/s}$), mass inertia creates centrifugal tension $T_c$ along the belt arc. Centrifugal tension acts outward equally on both sides, pulling the belt away from the pulley rim without contributing to torque transmission:

$$T_c = m V^2$$

Where:

• $m$ = Mass per unit length of belt ($\text{kg/m} = \rho \cdot b \cdot t$)
• $V$ = Linear belt velocity ($\text{m/s} = \frac{\pi d N}{60}$)
• $\rho$ = Belt material mass density ($\text{kg/m}^3$)
• $b, t$ = Belt width and thickness (m)

Accounting for centrifugal tension, the modified tension ratio becomes:

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

Net power transmitted by the belt is:

$$P = (T_1 - T_2) V = (T_1 - T_c)\left(1 - e^{-\mu \theta}\right) V$$

Maximum Power Transmission Velocity

To maximize transmitted power for a given maximum allowable belt tension $T_{max}$, differentiate $P$ with respect to velocity $V$. Maximum power occurs when centrifugal tension equals one-third of the maximum allowable tension:

$$T_c = \frac{T_{max}}{3} \implies V_{opt} = \sqrt{\frac{T_{max}}{3 m}}$$

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2. V-Belt Drive Mechanics & Wedge Action

V-belts feature a trapezoidal cross-section that fits into matched V-grooved sheaves. The wedging action of the belt against the groove walls increases the normal contact force, providing significantly higher friction capacity without increasing initial tension.

Groove Geometry & Effective Friction

For a sheave groove angle $\beta$ (typically $34^\circ - 40^\circ$), the normal contact force on each side wall is $N_{wall} = \frac{N_{radial}}{2 \sin(\beta/2)}$. The virtual (effective) coefficient of friction $\mu'$ is:

$$\mu' = \frac{\mu}{\sin(\beta/2)}$$

Substituting $\mu'$ into the tension ratio formula yields:

$$\frac{T_1 - T_c}{T_2 - T_c} = e^{\mu' \theta} = e^{\frac{\mu \theta}{\sin(\beta/2)}}$$

Because $\sin(\beta/2) \approx \sin(19^\circ) = 0.3256$, the effective friction increases by a factor of roughly $3.0$, eliminating slip under high transient peak loads.

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3. Roller Chain Kinematics & Polygon Effect

Roller chains provide positive, non-slip power transmission through pin-and-bushing joints meshing with sprocket teeth. Pitch $p$ is the distance between adjacent pin centers.

Sprocket Pitch Diameter & Speed Ratio

From sprocket geometry with $Z$ teeth and pitch $p$:

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

The drive speed ratio is strictly inversely proportional to tooth counts:

$$\frac{N_1}{N_2} = \frac{Z_2}{Z_1} = \frac{d_2}{d_1}$$

Chordal Action (Polyhedral Effect)

As the chain feeds onto a sprocket, the pitch lines form a polygon rather than a true circle. This geometry causes cyclic variation in effective pitch radius between $R = \frac{d}{2}$ and $r = R \cos(180^\circ / Z)$.

• Maximum linear velocity: $V_{max} = \frac{\pi d N}{60}$
• Minimum linear velocity: $V_{min} = V_{max} \cos(180^\circ / Z)$
• Percentage speed fluctuation: $1 - \cos(180^\circ / Z)$

To keep velocity fluctuation below $1.5\%$, driving sprockets should have a minimum of $Z_1 \ge 17$ teeth (preferably $Z_1 \ge 21$ for high-speed operation).

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4. Wire Rope Mechanics & Fatigue Bending

Wire ropes consist of individual high-carbon steel wires twisted into strands, which are wrapped helically around a central core (fiber core FC or independent wire rope core IWRC).

Standard Construction Standardizations

• 6x19 Class: 6 strands of 19 wires each. Balances flexibility and abrasion resistance; used in hoists, elevators, and general engineering.
• 6x37 Class: 6 strands of 37 smaller wires. Extremely flexible; preferred for small-diameter sheaves and crane rigging.

Bending Stress Over Sheaves

When a wire rope of outer diameter $d$ passes around a sheave of pitch diameter $D$, individual outer wires of diameter $d_w$ suffer cyclic bending stresses:

$$S_b = \frac{E_w d_w}{D}$$

Where:

• $E_w$ = Modulus of elasticity of rope wire ($\approx 80,000 - 100,000\text{ MPa}$)
• $d_w$ = Diameter of individual wire ($\approx d/15$ for 6x19, $\approx d/22$ for 6x37)
• $D$ = Sheave pitch diameter (m)

Equivalent bending load on rope: $F_b = S_b A_r$, where $A_r \approx 0.40 d^2$ is the metallic wire area. The total effective tension is $F_{total} = F_{working} + F_b + F_{acceleration}$. Safety factor $FS = \frac{F_{ultimate} - F_b}{F_{working}} \ge 5.0$.

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5. Flywheels & Energy Storage Analysis

Flywheels act as mechanical energy reservoirs in reciprocating machinery (engines, punch presses, crushers) by absorbing energy during periods of excess power and delivering energy during high-load demands.

Coefficient of Speed Fluctuation ($C_s$)

$$C_s = \frac{\omega_1 - \omega_2}{\omega_0} = \frac{N_1 - N_2}{N_{avg}}$$

Where $\omega_0 = \frac{\omega_1 + \omega_2}{2}$ is mean angular velocity. Typical values range from $C_s = 0.005$ (precision AC generators) to $C_s = 0.05 - 0.10$ (punch presses).

Kinetic Energy Storage Equation

$$\Delta E_k = \frac{1}{2} I \left(\omega_1^2 - \omega_2^2\right) = I \omega_0^2 C_s = m k^2 \omega_0^2 C_s$$

Where $I = m k^2$ is mass moment of inertia, $m$ is flywheel mass (kg), and $k$ is radius of gyration (m).

Rim Tensile Stress (Thin Rim Approximation)

Centrifugal action induces hoop stress (tensile stress) in a thin flywheel rim:

$$\sigma = \rho V^2$$

Where $\rho$ is mass density (e.g., $7200\text{ kg/m}^3$ for gray cast iron) and $V = \omega R$ is peripheral speed. Cast iron flywheels are strictly limited to $V \le 30\text{ m/s}$ (limiting $\sigma \le 6.5\text{ MPa}$).

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6. Step-by-Step Worked Design Problem

Problem Statement

A flat belt drive connects a $15\text{ kW}$ electric motor turning at $1440\text{ rpm}$ to a machine shaft. Motor pulley diameter $d_1 = 250\text{ mm}$, driven pulley diameter $d_2 = 500\text{ mm}$, center distance $C = 1200\text{ mm}$. Belt thickness $t = 6\text{ mm}$, mass density $\rho = 1000\text{ kg/m}^3$, coefficient of friction $\mu = 0.30$, and maximum allowable belt stress $\sigma_{max} = 2.0\text{ MPa}$.

Calculate: (1) Belt velocity $V$, (2) Centrifugal tension $T_c$ per meter width, (3) Required belt width $b$ in millimeters.

Solution Steps

1. Linear Belt Velocity:

   $$V = \frac{\pi d_1 N_1}{60} = \frac{\pi \times 0.250 \times 1440}{60} = 18.85\text{ m/s}$$

2. Mass & Centrifugal Tension per Meter Width: Mass per meter length for $1.0\text{ m}$ width:

   $$m' = \rho \cdot t \cdot (1.0) = 1000 \times 0.006 \times 1.0 = 6.0\text{ kg/m}$$

   Centrifugal tension per meter width:

   $$T_c' = m' V^2 = 6.0 \times (18.85)^2 = 2131.9\text{ N/m}$$

3. Wrap Angle on Smaller Pulley:

   $$\theta = \pi - 2 \arcsin\left(\frac{d_2 - d_1}{2 C}\right) = 3.1416 - 2 \arcsin\left(\frac{0.500 - 0.250}{2.400}\right) = 3.1416 - 0.2090 = 2.9326\text{ rad}$$

4. Friction Tension Factor:

   $$e^{\mu \theta} = e^{0.30 \times 2.9326} = e^{0.8798} = 2.410$$

5. Net Required Transmitted Force:

   $$T_1 - T_2 = \frac{P}{V} = \frac{15,000}{18.85} = 795.8\text{ N}$$

6. Maximum Allowable Tight-Side Tension per Meter Width:

   $$T_{max}' = \sigma_{max} \cdot t = (2.0 \times 10^6\text{ N/m}^2) \times 0.006\text{ m} = 12,000\text{ N/m}$$

   Tension available for friction coupling:

   $$T_1' - T_c' = 12,000 - 2131.9 = 9868.1\text{ N/m}$$

   Net driving force per meter width:

   $$\Delta T' = (T_1' - T_c')\left(1 - e^{-\mu \theta}\right) = 9868.1 \times \left(1 - \frac{1}{2.410}\right) = 9868.1 \times 0.5851 = 5773.8\text{ N/m}$$

7. Required Belt Width:

   $$b = \frac{T_1 - T_2}{\Delta T'} = \frac{795.8\text{ N}}{5773.8\text{ N/m}} = 0.1378\text{ m} = 137.8\text{ mm}$$

   Selecting standard commercial size: $140\text{ mm}$ width.