9.2 Power Transmission Shafts, Keys & Rigid/Flexible Couplings

9.2 Power Transmission Shafts, Keys & Rigid/Flexible Couplings

Power transmission shafts are rotating machine elements used to transmit mechanical energy and torque between drivers (motors, turbines) and driven components (gears, pulleys, sprockets). Shaft design requires evaluating static shear and bending stresses, dynamic shock factors, lateral deflection limits, critical whirling speeds, and keyway attachment stress concentrations.

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1. Shaft Design under Combined Torsion and Bending (ASME Shaft Code)

Shafts in mechanical power systems are almost never subjected to pure torsion alone; belt tension, gear mesh forces, and overhung weight create transverse bending moments ($M_b$) simultaneous with torsional torque ($T_t$).

Maximum Shear Stress Theory (ASME Code Formulation)

Under the ASME Code for Design of Transmission Shafting, the allowable shear stress $\tau_{allow}$ is governed by material yield ($S_y$) or ultimate tensile strength ($S_{ut}$):

$\tau_{allow} = \min(0.30 S_y, 0.18 S_{ut})$

If keyways are present, $\tau_{allow}$ is reduced by 25% (i.e., multiplied by 0.75).

To account for dynamic fatigue and impact shock loading, combined numerical factors are introduced:

• $K_m$: Combined numerical bending shock and fatigue factor.
• $K_t$: Combined numerical torsional shock and fatigue factor.

Equivalent Torque ($T_e$) and Equivalent Bending Moment ($M_e$)

1. Equivalent Torque ($T_e$): $T_e = \sqrt{(M_b K_m)^2 + (T_t K_t)^2}$
2. Equivalent Bending Moment ($M_e$): $M_e = \frac{1}{2} \left[ M_b K_m + \sqrt{(M_b K_m)^2 + (T_t K_t)^2} \right]$

Solid Shaft Sizing Equation

Equating maximum torsional shear stress to allowable shear stress $\tau_{allow} = \frac{16 T_e}{\pi d^3}$:

$d = \left[ \frac{16}{\pi \tau_{allow}} \sqrt{(M_b K_m)^2 + (T_t K_t)^2} \right]^{1/3}$

Hollow Shaft Sizing Equation

For a hollow shaft with inside diameter $d_i$ and outside diameter $d_o$, defining diameter ratio $k = \frac{d_i}{d_o}$ ($0 < k < 1$):

$d_o = \left[ \frac{16}{\pi \tau_{allow} (1 - k^4)} \sqrt{(M_b K_m)^2 + (T_t K_t)^2} \right]^{1/3}$

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2. Critical Speed of Shafts (Whirling Speed)

When a shaft rotates, unavoidable mass eccentricities generate centrifugal forces that bend the shaft dynamically. At specific rotational speeds (critical speeds or whirling speeds), the rotational frequency matches the shaft's natural frequency of lateral vibration, causing severe resonance, large deflections, and potential failure.

Single Concentrated Load

For a shaft supporting a single load with static deflection $\delta$ (measured in meters or cm):

$\omega_c = \sqrt{\frac{g}{\delta}} \quad (\text{rad/s})$

$N_c = \frac{30}{\pi} \sqrt{\frac{g}{\delta}} \approx \frac{300}{\sqrt{\delta_{cm}}} \quad (\text{rpm})$

Multiple Loads: Dunkerley's Empirical Formula

For a shaft carrying multiple concentrated loads (or pulleys/gears) with individual critical speeds $N_1, N_2, \dots, N_n$, and a self-weight critical speed $N_s$:

$\frac{1}{N_c^2} = \frac{1}{N_s^2} + \frac{1}{N_1^2} + \frac{1}{N_2^2} + \dots + \frac{1}{N_n^2}$

Note: Dunkerley's equation always gives a conservative (slightly lower) estimate of the actual fundamental critical speed.

Rayleigh-Ritz Method

The Rayleigh energy method equates maximum kinetic energy to maximum strain energy:

$\omega_c^2 = g \frac{\sum_{i=1}^n W_i y_i}{\sum_{i=1}^n W_i y_i^2}$

Where $W_i$ is the weight of load $i$, and $y_i$ is the static deflection under load $i$.

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3. Keys and Keyways Stress Analysis

Keys are demountable machinery components inserted between a shaft and a hub (gear, pulley) to prevent relative rotational motion and transmit torque.

  Shaft & Key Cross-Section:
        +-------+
        | Key   |  <- Height h (Width w)
  +-----+-------+-----+
  |  Keyway in Hub    |
  |-------------------|  <- Shaft Diameter d
  |  Shaft Body       |
  +-------------------+

Key Types

• Square Key: Width $w = h = d/4$. Equal shear and crushing resistance when yield strength in shear is half compressive yield strength.
• Flat Key: Width $w = d/4$, Height $h = 2w/3 = d/6$. Used for larger shafts where keyway depth must be minimized.
• Woodruff Key: Semi-circular disk fitting into a matching semicircular keyway milled into the shaft. Self-aligning; ideal for tapered shaft ends.
• Feather Key: Fastened to either shaft or hub, allowing axial sliding while preventing relative rotation.

Stress Calculations for Keys

Given torque $T$, shaft diameter $d$, key width $w$, key height $h$, and key active length $L$:

1. Tangential Force on Key ($F$): $F = \frac{2 T}{d}$
2. Shear Stress Failure Mode (Key Shear Area $A_s = w \cdot L$): $\tau = \frac{F}{A_s} = \frac{2 T}{d \cdot w \cdot L} \le \tau_{allow}$
3. Crushing (Compressive) Stress Failure Mode (Key Contact Area $A_c = \frac{h}{2} \cdot L$): $\sigma_c = \frac{F}{A_c} = \frac{2 T}{d \left(\frac{h}{2}\right) L} = \frac{4 T}{d \cdot h \cdot L} \le \sigma_{c,allow}$

Equal Strength Condition

For a square key ($w = h$), setting allowable compressive stress to twice allowable shear stress ($\sigma_{c,allow} = 2 \tau_{allow}$) makes the key equally resistant to shear failure and crushing failure.

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4. Rigid and Flexible Shaft Couplings

Couplings connect two coaxial shafts to transmit torque.

Rigid Couplings

Used when shafts are perfectly aligned in a rigid structure. They transmit no axial compliance or angular flexibility.

• Flange Coupling: Flanged hubs keyed to shaft ends and bolted together around a bolt circle diameter $D_b$.
  • Bolt Shear Stress: $\tau_b = \frac{8 T}{\pi n d_b^2 D_b} \le \tau_{allow}$ (for $n$ bolts of diameter $d_b$).
  • Flange Hub Shear Stress: $\tau_f = \frac{2 T}{\pi d^2 t_f}$ (where $t_f$ is flange thickness).
• Sleeve / Muff Coupling: Hollow cylinder fitted over shaft ends with a single long key.

Flexible Couplings

Designed to accommodate shaft misalignment (angular, parallel radial offset, and axial movement) while dampening torsional vibration and shock loads.

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5. Worked Shaft & Key Design Calculation

Problem Statement

A solid transmission shaft delivers $50 \text{ kW}$ of mechanical power at $500 \text{ rpm}$. The shaft is supported by bearings and experiences a maximum combined bending moment $M_b = 1200 \text{ N}\cdot\text{m}$. According to design codes, the fatigue shock factors are $K_m = 1.5$ and $K_t = 1.0$. The allowable shear stress for the shaft material is $\tau_{allow} = 50 \text{ MPa}$.

The shaft is keyed to a driving pulley using a standard square key with width $w = 14 \text{ mm}$ and height $h = 14 \text{ mm}$. The key material has an allowable shear stress $\tau_{key} = 60 \text{ MPa}$ and allowable crushing stress $\sigma_{c,key} = 120 \text{ MPa}$.

Calculate:

1. The transmitted torque $T_t$.
2. The required solid shaft diameter $d$.
3. The minimum required key length $L$ to prevent key shear failure.
4. The minimum required key length $L$ to prevent key crushing failure.

Step-by-Step Solution

Step 1: Compute Transmitted Torque $T_t$ $\omega = \frac{2 \pi N}{60} = \frac{2 \pi (500)}{60} = 52.360 \text{ rad/s}$

$T_t = \frac{P}{\omega} = \frac{50,000 \text{ W}}{52.360 \text{ rad/s}} = 954.93 \text{ N}\cdot\text{m} = 954,930 \text{ N}\cdot\text{mm}$

Step 2: Compute Equivalent Torque $T_e$ and Shaft Diameter $d$ Bending moment $M_b = 1200 \text{ N}\cdot\text{m} = 1,200,000 \text{ N}\cdot\text{mm}$.

$T_e = \sqrt{(M_b K_m)^2 + (T_t K_t)^2} = \sqrt{(1,200,000 \times 1.5)^2 + (954,930 \times 1.0)^2}$

$T_e = \sqrt{(1,800,000)^2 + (954,930)^2} = \sqrt{3.24 \times 10^{12} + 0.911891 \times 10^{12}} = \sqrt{4.151891 \times 10^{12}}$

$T_e = 2,037,619 \text{ N}\cdot\text{mm}$

Using ASME shaft sizing equation: $d = \left[ \frac{16 T_e}{\pi \tau_{allow}} \right]^{1/3} = \left[ \frac{16 \times 2,037,619}{\pi \times 50} \right]^{1/3} = \left[ \frac{32,601,904}{157.0796} \right]^{1/3} = [207,550.2]^{1/3}$

$d = 59.208 \text{ mm}$

Select standard shaft diameter: $d = 60 \text{ mm}$.

Step 3: Calculate Key Length $L$ for Shear Failure Using standard shaft diameter $d = 60 \text{ mm}$ and torque $T_t = 954,930 \text{ N}\cdot\text{mm}$:

$\tau = \frac{2 T_t}{d \cdot w \cdot L} \implies L_{shear} = \frac{2 T_t}{d \cdot w \cdot \tau_{key}}$

$L_{shear} = \frac{2 \times 954,930}{60 \times 14 \times 60} = \frac{1,909,860}{50,400} = 37.89 \text{ mm}$

Step 4: Calculate Key Length $L$ for Crushing Failure $\sigma_c = \frac{4 T_t}{d \cdot h \cdot L} \implies L_{crush} = \frac{4 T_t}{d \cdot h \cdot \sigma_{c,key}}$

$L_{crush} = \frac{4 \times 954,930}{60 \times 14 \times 120} = \frac{3,819,720}{100,800} = 37.89 \text{ mm}$

Conclusion: The required minimum key length is $37.89 \text{ mm}$ (standard commercial length choice: $40 \text{ mm}$). Notice that because $\sigma_{c,key} = 2 \tau_{key}$ and $w = h$, the shear and crushing lengths match perfectly.