3.1 Shaft Sizing, Torsional Deflection & Critical Speeds

Q: A 75 hp motor drives a process blower shaft at 1160 rpm. What is the steady torque delivered to the shaft in inch-pounds?

4,075 in-lbf. Using the US Customary torque equation: T = (63,025 * Power) / Speed = (63,025 * 75) / 1160 = 4,726,875 / 1160 = 4,074.89 in-lbf (approx 4,075 in-lbf). The value 3,400 in-lbf uses 5252 with an incorrect inch conversion; 4,890 in-lbf is an arithmetic error; and 8,150 in-lbf is double the correct torque.

Q: A rotating shaft carries two pulleys. Acting alone on the bare shaft, the first pulley produces a critical speed of 1500 rpm, while the second pulley produces a critical speed of 2000 rpm. Neglecting the mass of the shaft, what is the combined fundamental critical speed using Dunkerley's equation?

1,200 rpm. Applying Dunkerley's approximation: 1 / (N_cr)^2 = 1 / (N_1)^2 + 1 / (N_2)^2 = 1 / (1500)^2 + 1 / (2000)^2 = 1 / 2,250,000 + 1 / 4,000,000 = (4.444e-7) + (2.500e-7) = 6.944e-7 (1/rpm^2). Taking the square root and reciprocal: N_cr = 1 / sqrt(6.944e-7) = 1200 rpm. The value 1,750 rpm is an invalid arithmetic average, and 2,500 rpm adds critical speeds in quadrature.

Q: What is the maximum permissible shaft slope at a standard spur or helical gear mesh to avoid edge loading and premature tooth pitting?

0.001 rad. Standard machine design practice specifies a maximum permissible shaft slope at spur and helical gear meshes of 0.001 rad (approx 0.057 degrees or 3.4 arcminutes) to ensure uniform line contact across the tooth face width. A tighter limit of 0.0005 rad applies to bevel gears, while 0.004 rad is acceptable for deep groove ball bearings.

Key Takeaways

Power, rotational speed, and torque are fundamental kinematic quantities linked by $T = \frac{63025 \times P_{\text{hp}}}{N}$ (in-lbf) and $T = \frac{5252 \times P_{\text{hp}}}{N}$ (ft-lbf) in US Customary, and $T = \frac{9550 \times P_{\text{kW}}}{N}$ (N-m) in SI metric units.
Rotating power transmission shafts experience fully reversed alternating bending stresses ( $\sigma_a = \frac{32 K_f M_a}{\pi d^3}$ , $\sigma_m = 0$ ) due to stationary transverse loads and steady mean torsional shear stresses ( $\tau_m = \frac{16 K_{fs} T_m}{\pi d^3}$ , $\tau_a = 0$ ), requiring combined fatigue sizing via the ASME DE-Goodman criterion.
Shaft rigidity constraints frequently govern minimum shaft diameter over fatigue strength: maximum permissible slope at spur/helical gear meshes is $0.001\text{ rad}$ ( $0.057^\circ$ ), at bevel gears is $0.0005\text{ rad}$ , and at deep groove ball bearings is $0.004\text{ rad}$ ( $14'$ ), while torsional deflection should not exceed $1.0^\circ/\text{m}$ ( $0.08^\circ/\text{ft}$ ).
Keyway stress concentrations ( $K_t \approx 2.1$ to $3.0$ ) significantly reduce shaft fatigue strength, and parallel square keys must be sized against both pitch-plane shear ( $\tau = \frac{2T}{d w L}$ ) and compressive side-bearing failure ( $\sigma_b = \frac{4T}{d h L}$ ), where bearing stress always governs for square keys of equal material strength.
Dynamic whirling resonance occurs at the critical speed $\omega_{cr} = \sqrt{k/m} = \sqrt{g/\delta_{st}}$ ; multi-mass shaft critical speeds are bounded below by Dunkerley's approximation ( $\frac{1}{\omega_{cr}^2} = \sum \frac{1}{\omega_i^2}$ ) and bounded above by the Rayleigh-Ritz energy method.

Last updated: August 2026

Power Transmission Shaft Design & Critical Speeds

Power transmission shafts are the core rotating mechanical elements used to transmit rotary power and torque from prime movers (electric motors, internal combustion engines, turbines) to driven industrial machinery such as gears, pulleys, sprockets, and pumps. Designing a safe, durable, and reliable shaft requires the mechanical engineer to satisfy multiple concurrent design criteria: kinematic power-torque-speed conversions, combined multi-axial fatigue sizing under fluctuating loads, geometric stress concentrations (keyways, shoulders, retaining ring grooves), torsional and flexural rigidity limits, and dynamic critical whirling speeds.

+---------------------------------------------------------------------------------------------------+
|                         POWER TRANSMISSION SHAFT DESIGN FRAMEWORK                                 |
|                                                                                                   |
|   [KINEMATIC INPUTS]      [COMBINED SHAFT STRESSES]          [FATIGUE & STRENGTH SIZING]          |
|   - Power P (hp or kW)    - Reversed Bending (Sigma_a)       - ASME DE-Goodman Criterion          |
|   - Speed N (rpm)      -> - Steady Torsion (Tau_m)        -> - DE-Gerber / ASME Elliptic          |
|   - Torque T (in-lb/N-m)  - Axial Thrust Forces (Sigma_ax)   - Static Yield (Langer Line) Check   |
|                                     |                                     |                       |
|                                     v                                     v                       |
|                       [DEFLECTION & RIGIDITY LIMITS]         [CRITICAL WHIRLING DYNAMICS]         |
|                       - Torsional Windup (Theta <= 1 deg/m)  - Jeffcott Single-Disk Rotor         |
|                       - Gear Mesh Slope (Theta < 0.001 rad)  - Dunkerley Lower-Bound Sum          |
|                       - Bearing Slope (Theta < 0.004 rad)    - Rayleigh-Ritz Upper-Bound Method   |
+---------------------------------------------------------------------------------------------------+

1. Power, Torque, and Rotational Speed Kinematics

Rotary mechanical power $P$ is the product of transmitted torque $T$ and angular velocity $\omega$ :

P = T \omega = T \left( \frac{2 \pi N}{60} \right)

Where:

$P = \text{Power (Watts in SI; } \text{ft}\cdot\text{lbf/s or horsepower in US Customary)}$
$T = \text{Transmitted torque (}\text{N}\cdot\text{m in SI; } \text{in}\cdot\text{lbf or } \text{ft}\cdot\text{lbf in US Customary)}$
$N = \text{Rotational shaft speed in revolutions per minute (rpm)}$
$\omega = \text{Angular velocity in radians per second (rad/s)} = \frac{2\pi N}{60} \approx 0.10472 N$

Rapid Conversion Matrix for PE Exam Calculations

Unit System	Power Input Unit	Formula for Transmitted Torque ( $T$ )	Torque Output Unit
SI Metric	Kilowatts ( $\text{kW}$ )	$T = \frac{9549.3 \times P_{\text{kW}}}{N} \approx \frac{9550 \times P_{\text{kW}}}{N}$	$\text{N}\cdot\text{m}$
SI Metric	Watts ( $\text{W}$ )	$T = \frac{P_{\text{W}}}{\omega} = \frac{60 P_{\text{W}}}{2\pi N}$	$\text{N}\cdot\text{m}$
US Customary	Horsepower ( $\text{hp}$ )	$T = \frac{63025 \times P_{\text{hp}}}{N}$	$\text{in}\cdot\text{lbf}$
US Customary	Horsepower ( $\text{hp}$ )	$T = \frac{5252.1 \times P_{\text{hp}}}{N}$	$\text{ft}\cdot\text{lbf}$

Note

Derivation of Constants: In US Customary units, $1\text{ hp} = 550\text{ ft}\cdot\text{lbf/s} = 33000\text{ ft}\cdot\text{lbf/min} = 396000\text{ in}\cdot\text{lbf/min}$ . Dividing by $2\pi\text{ rad/rev}$ yields $\frac{33000}{2\pi} = 5252.1\text{ ft}\cdot\text{lbf}\cdot\text{rpm/hp}$ and $\frac{396000}{2\pi} = 63025.4\text{ in}\cdot\text{lbf}\cdot\text{rpm/hp}$ . In SI, $\frac{1000\text{ W/kW} \times 60\text{ s/min}}{2\pi} = 9549.3\text{ N}\cdot\text{m}\cdot\text{rpm/kW}$ .

2. Multi-Axial Combined Fatigue Sizing (ASME DE-Goodman)

In typical industrial shafts, transverse forces from gears, belt pulleys, and chain sprockets induce bending moments that remain stationary in space. As the shaft rotates through $360^\circ$ , outer fibers alternate between maximum tension and maximum compression every half-revolution. Consequently, the normal bending stress is fully reversed alternating fatigue stress ( $\sigma_a = \sigma_b$ , $\sigma_m = 0$ ). Concurrently, the transmitted torque delivers a steady mean shear stress ( $\tau_m = \tau_t$ , $\tau_a = 0$ ).

+---------------------------------------------------------------------------------------------------+
|                     STRESS STATE IN A ROTATING POWER TRANSMISSION SHAFT                           |
|                                                                                                   |
|   REVERSED BENDING STRESS (Alternating):            STEADY TORSIONAL STRESS (Mean):               |
|            +Sigma_max (Tension)                              +Tau_mean                            |
|               /\      /\                                         -----------------------          |
|              /  \    /  \                                                                         |
|   ----------+----+--+----+-----------               ------------------------------------          |
|              \  /    \  /                                                                         |
|               \/      \/                                     [Alternating Torsion Tau_a = 0]      |
|            -Sigma_max (Compression)                                                               |
|   [Mean Bending Sigma_m = 0, Alternating Sigma_a = M*c/I]   [Mean Torsion Tau_m = T*r/J]          |
+---------------------------------------------------------------------------------------------------+

Stress Components for Solid Circular Shafts

For a solid round shaft of diameter $d$ , the alternating and mean stress components are:

\sigma_a = K_f \frac{32 M_a}{\pi d^3}, \quad \sigma_m = K_f \frac{32 M_m}{\pi d^3}

\tau_a = K_{fs} \frac{16 T_a}{\pi d^3}, \quad \tau_m = K_{fs} \frac{16 T_m}{\pi d^3}

Where $K_f$ and $K_{fs}$ are the fatigue stress concentration factors for bending and torsion, defined by $K_f = 1 + q(K_t - 1)$ where $q$ is the material notch sensitivity.

Distortion Energy (Von Mises) Equivalent Stresses

Combining normal and shear components into equivalent von Mises alternating stress $\sigma_a'$ and mean stress $\sigma_m'$ :

\sigma_a' = \sqrt{\sigma_a^2 + 3 \tau_a^2} = \frac{16}{\pi d^3} \sqrt{4(K_f M_a)^2 + 3(K_{fs} T_a)^2}

\sigma_m' = \sqrt{\sigma_m^2 + 3 \tau_m^2} = \frac{16}{\pi d^3} \sqrt{4(K_f M_m)^2 + 3(K_{fs} T_m)^2}

The ASME / DE-Goodman Sizing Criterion

The Modified Goodman fatigue failure criterion states:

\frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_{ut}} = \frac{1}{n_f}

Substituting the equivalent stress definitions yields the closed-form ASME / DE-Goodman shaft diameter sizing equation:

d = \left[ \frac{16 n_f}{\pi} \left( \frac{\sqrt{4(K_f M_a)^2 + 3(K_{fs} T_a)^2}}{S_e} + \frac{\sqrt{4(K_f M_m)^2 + 3(K_{fs} T_m)^2}}{S_{ut}} \right) \right]^{1/3}

The Standard Rotating Shaft Case ( $M_a = M, M_m = 0$ and $T_m = T, T_a = 0$ )

Under steady torque and rotating bending, the sizing equation reduces to:

d = \left[ \frac{16 n_f}{\pi} \left( \frac{2 K_f M}{S_e} + \frac{\sqrt{3} K_{fs} T}{S_{ut}} \right) \right]^{1/3}

Static Yield Check (Langer Line)

To prevent localized plastic yielding during startup spikes or severe torque transients, the first-cycle yield check must be satisfied:

\sigma_{\max}' = \sigma_a' + \sigma_m' \le \frac{S_y}{n_y} \implies n_y = \frac{S_y}{\sigma_a' + \sigma_m'} \ge 1.0

3. Marin Endurance Limit Modifying Factors

The laboratory rotating-beam endurance limit $S_e' \approx 0.5 S_{ut}$ (for steels with $S_{ut} \le 200\text{ kpsi}$ or $1400\text{ MPa}$ ) must be adjusted for actual component conditions via the Marin equation:

S_e = k_a k_b k_c k_d k_e k_f S_e'

+---------------------------------------------------------------------------------------------------+
|                                 MARIN FACTORS SUMMARY FOR SHAFTING                                |
|                                                                                                   |
|   k_a (Surface Finish):   k_a = a * (S_ut)^b                                                      |
|                           Machined / Cold-Drawn: a = 4.51 MPa (2.70 kpsi), b = -0.265             |
|                           Ground: a = 1.58 MPa (1.34 kpsi), b = -0.085                            |
|                                                                                                   |
|   k_b (Size Factor):      For 2.79 mm <= d <= 51 mm:  k_b = 1.24 * d^(-0.107)                     |
|                           For 51 mm < d <= 254 mm:    k_b = 1.51 * d^(-0.157)                     |
|                           For 0.11 in <= d <= 2.0 in: k_b = 0.879 * d^(-0.107)                    |
|                                                                                                   |
|   k_c (Load Factor):      Bending = 1.0; Axial = 0.85; Pure Torsion = 0.59                        |
|                           (Note: Use k_c = 1.0 in DE-Goodman because Von Mises handles shear)     |
|                                                                                                   |
|   k_d (Temperature):      k_d = 1.0 for T <= 450°C (750°F)                                        |
|                                                                                                   |
|   k_e (Reliability):      90% = 0.897; 95% = 0.868; 99% = 0.814; 99.9% = 0.753                    |
|                                                                                                   |
|   k_f (Misc Effects):     Corrosion, plating, fretting, residual stresses                         |
+---------------------------------------------------------------------------------------------------+

4. Keys, Keyways & Geometric Stress Concentrations

Keys transmit torque from the shaft to the hub of a mounted component (gear, pulley, coupling).

+---------------------------------------------------------------------------------------------------+
|                          PARALLEL SQUARE KEY SHEAR AND BEARING FORCES                             |
|                                                                                                   |
|                      +------------------------+                                                   |
|                      |       HUB (GEAR)       |                                                   |
|                      +------------------------+                                                   |
|                             | F_bearing (Top Half: h/2)                                           |
|                        +----+----+                                                                |
|                        |  KEY    | <--- Shear Plane along pitch line (Width w * Length L)         |
|                        +----+----+                                                                |
|                             | F_bearing (Bottom Half: h/2)                                        |
|                      +------------------------+                                                   |
|                      |     SHAFT (Radius r)   |  Torque T = F * (d/2)                             |
|                      +------------------------+                                                   |
+---------------------------------------------------------------------------------------------------+

1. Key Shear Stress Failure Mode

The transmitted torque $T$ produces a tangential force $F = \frac{2T}{d}$ along the shear plane of width $w$ and length $L$ :

\tau_{\text{key}} = \frac{F}{A_{\text{shear}}} = \frac{2 T}{d \cdot w \cdot L} \le \frac{S_{sy}}{n_s} = \frac{0.577 S_y}{n_s}

\implies L_{\text{shear}} \ge \frac{2 T n_s}{d \cdot w \cdot (0.577 S_y)}

2. Key Compressive Bearing Stress Failure Mode

The tangential force crushes the side contact face of the key over depth $h/2$ and length $L$ :

\sigma_{b,\text{key}} = \frac{F}{A_{\text{bearing}}} = \frac{2 T}{d \cdot (h/2) \cdot L} = \frac{4 T}{d \cdot h \cdot L} \le \frac{S_y}{n_b}

\implies L_{\text{bearing}} \ge \frac{4 T n_b}{d \cdot h \cdot S_y}

Important

Governing Failure Mode for Square Keys: For a standard square key where width $w = h$ , bearing stress is exactly twice the shear stress ( $\sigma_b = 2\tau$ ). Under Distortion Energy theory, compressive yield strength $S_y$ is $1.732$ times shear yield strength ( $S_{sy} = 0.577 S_y$ ). Because stress increases by $2.0\times$ while allowable strength increases by only $1.732\times$ , compressive bearing stress always governs the required key length when the key and shaft are of equal strength.

Keyway Stress Concentration Factors on the Shaft

Profile (End-Milled) Keyway: $K_t \approx 2.14$ (bending), $K_t \approx 3.00$ (torsion).
Sled-Runner Keyway: $K_t \approx 1.70$ (bending), $K_t \approx 1.30$ (torsion) — cut with a disc milling cutter providing gradual runout radius.

5. Torsional Deflection & Shaft Rigidity Limits

Shaft sizing is frequently governed by stiffness constraints rather than fatigue strength to avoid premature gear tooth wear or bearing binding.

+---------------------------------------------------------------------------------------------------+
|                             SHAFT RIGIDITY & SLOPE LIMIT GUIDELINES                               |
|                                                                                                   |
|   LOCATION / COMPONENT               MAXIMUM PERMISSIBLE SLOPE / DEFLECTION                       |
|   --------------------------------   ----------------------------------------------------------   |
|   Spur & Helical Gear Meshes         Theta < 0.001 rad (0.057 deg) - Prevents tooth edge loading  |
|   Bevel Gear Meshes                  Theta < 0.0005 rad (0.029 deg) - Prevents apex dislocation   |
|   Deep Groove Ball Bearings          Theta < 0.004 rad (14 arcmin) - Prevents race pinching       |
|   Cylindrical Roller Bearings        Theta < 0.001 rad (3.5 arcmin) - Prevents edge stress        |
|   Spherical Roller Bearings          Theta < 0.026 to 0.052 rad (1.5 deg to 3.0 deg)              |
|   Torsional Deflection (Windup)      Theta_t <= 1.0 deg/m (0.08 deg/ft) for line shafting         |
|   Mid-Span Lateral Deflection        delta_max <= 0.0005 in/in of bearing span                    |
+---------------------------------------------------------------------------------------------------+

Torsional Deflection Formula

The angle of torsional twist $\theta_t$ in a circular shaft of length $L$ , shear modulus $G$ , and polar moment of inertia $J = \frac{\pi d^4}{32}$ is:

\theta_t = \frac{T L}{G J} = \frac{32 T L}{\pi G d^4} \quad (\text{radians})

\theta_t (\text{degrees}) = \frac{584 T L}{G d^4}

6. Critical Speeds & Dynamic Whirling

Rotating shafts with attached masses exhibit lateral dynamic instability at specific critical whirling speeds where centrifugal forces equal internal elastic restoring forces.

+---------------------------------------------------------------------------------------------------+
|                                JEFFCOTT ROTOR WHIRLING MODEL                                      |
|                                                                                                   |
|                       Geometric Center (O) ------- r -------> Center of Mass (G)                  |
|                                |                                   |                              |
|                       Elastic Restoring Force: F_s = k*r     Centrifugal Force: F_c = m*(r+e)*w^2 |
|                                                                                                   |
|   EQUILIBRIUM: k*r = m*(r+e)*w^2  ===>  Whirl Radius: r = e * (w / w_n)^2 / [1 - (w / w_n)^2]     |
+---------------------------------------------------------------------------------------------------+

Single-Mass (Jeffcott Rotor) Formulation

For a concentrated mass $m$ causing static gravity deflection $\delta_{st} = \frac{mg}{k}$ :

\omega_{cr} = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{\delta_{st}}} \quad (\text{rad/s})

N_{cr} = \frac{60}{2\pi} \sqrt{\frac{g}{\delta_{st}}} = \frac{30}{\pi} \sqrt{\frac{g}{\delta_{st}}} \quad (\text{rpm})

In SI Units ( $g = 9.81\text{ m/s}^2$ , $\delta_{st}$ in mm): $N_{cr} \approx \frac{948.7}{\sqrt{\delta_{st}\text{ (mm)}}}$ .
In US Customary Units ( $g = 386.4\text{ in/s}^2$ , $\delta_{st}$ in inches): $N_{cr} \approx \frac{187.7}{\sqrt{\delta_{st}\text{ (in)}}}$ .

Multi-Mass Shafts: Dunkerley and Rayleigh-Ritz Methods

Dunkerley's Approximation (Lower-Bound Estimate): Sums reciprocal squares of the individual critical speeds: $\frac{1}{\omega_{cr}^2} = \frac{1}{\omega_0^2} + \frac{1}{\omega_1^2} + \frac{1}{\omega_2^2} + \dots + \frac{1}{\omega_n^2}$ Where $\omega_0$ is the bare shaft critical speed, and $\omega_i = \sqrt{g/\delta_{ii}}$ is the critical speed of mass $i$ alone on the shaft.
Rayleigh-Ritz Energy Method (Upper-Bound Estimate): Equates maximum kinetic energy to maximum elastic potential energy: $\omega_{cr} = \sqrt{\frac{g \sum_{i=1}^n w_i y_i}{\sum_{i=1}^n w_i y_i^2}}$ Where $w_i = m_i g$ is the weight of mass $i$ , and $y_i$ is the static lateral deflection at mass $i$ under all combined loads.

7. Step-by-Step Worked Problem: Complete Shaft & Keyway Sizing

Problem: A solid AISI 1045 cold-drawn steel shaft ( $S_{ut} = 630\text{ MPa}$ , $S_y = 530\text{ MPa}$ ) transmits $37.5\text{ kW}$ at $1800\text{ rpm}$ from an electric motor to a spur gear. At the gear location, stationary transverse tooth loads create a bending moment $M = 280\text{ N}\cdot\text{m}$ . The gear is mounted with a profile keyway ( $K_f = 2.0$ , $K_{fs} = 1.6$ ). Corrected endurance limit at the keyway is $S_e = 190\text{ MPa}$ . Using a design factor of safety $n_f = 2.0$ :

Calculate transmitted torque $T$ .
Determine the minimum required shaft diameter $d$ using the ASME DE-Goodman equation.
Size a square steel key ( $w = h = 10\text{ mm}$ , $S_y = 400\text{ MPa}$ ) with factor of safety $n = 2.5$ .

+---------------------------------------------------------------------------------------------------+
|                                 STEP-BY-STEP SOLUTION WORKFLOW                                    |
|                                                                                                   |
|   STEP 1: Calculate Transmitted Torque                                                            |
|           omega = 2 * pi * 1800 / 60 = 188.50 rad/s                                               |
|           T = P / omega = 37,500 W / 188.50 rad/s = 198.94 N-m                                    |
|           (Shortcut: T = 9549.3 * 37.5 / 1800 = 198.94 N-m)                                       |
|                                                                                                   |
|   STEP 2: ASME DE-Goodman Sizing (M_a = 280 N-m, M_m = 0; T_m = 198.94 N-m, T_a = 0)              |
|           Fatigue Bending Term = 2 * K_f * M_a / S_e = 2 * 2.0 * 280 / (190 * 10^6)               |
|                                = 1120 / 1.90 * 10^8 = 5.8947 * 10^(-6) m^3                        |
|           Mean Torsion Term    = sqrt(3) * K_fs * T_m / S_ut                                      |
|                                = 1.7321 * 1.6 * 198.94 / (630 * 10^6)                             |
|                                = 551.33 / 6.30 * 10^8 = 0.8751 * 10^(-6) m^3                      |
|           Sum of Terms         = (5.8947 + 0.8751) * 10^(-6) = 6.7698 * 10^(-6) m^3               |
|           Multiply by (16 * n_f / pi) = 16 * 2.0 / pi = 10.1859                                   |
|           d^3 = 10.1859 * 6.7698 * 10^(-6) = 6.8957 * 10^(-5) m^3                                |
|           d = (6.8957 * 10^(-5))^(1/3) = 0.0410 m = 41.0 mm  ===> Standard: 45 mm (1.75 in)       |
|                                                                                                   |
|   STEP 3: Keyway Length Sizing (w = h = 10 mm = 0.010 m, d = 0.045 m, S_y = 400 MPa, n = 2.5)     |
|           Shear Strength S_sy = 0.577 * 400 = 230.8 MPa                                           |
|           Shear Mode:   L_s = 2 * T * n / (d * w * S_sy)                                          |
|                             = 2 * 198.94 * 2.5 / (0.045 * 0.010 * 230.8 * 10^6) = 9.58 mm        |
|           Bearing Mode: L_b = 4 * T * n / (d * h * S_y)                                           |
|                             = 4 * 198.94 * 2.5 / (0.045 * 0.010 * 400 * 10^6)  = 11.05 mm       |
|           CONCLUSION: Bearing governs; specify standard key length L = 12 mm or 15 mm.            |
+---------------------------------------------------------------------------------------------------+

8. Exam Tips & Common Traps

Tip

Quick Reference Rules:

In DE-Goodman equations for rotating shafts, the alternating bending moment has factor $2$ (because $\sigma_a = \frac{32 M}{\pi d^3} = \frac{16}{\pi d^3}(2M)$ ), while the steady torque has factor $\sqrt{3}$ (from Von Mises $\tau \sqrt{3}$ ). Remember: $2 M$ for bending, $\sqrt{3} T$ for torsion.
When calculating keyway bearing stress, contact height is $h/2$ , not $h$ , because half the key sits in the shaft and half in the hub.

Warning

Common Pitfalls:

Mixing Radians and Degrees: Torsional windup formula $\theta = TL/GJ$ yields radians. Multiply by $180/\pi \approx 57.3$ to convert to degrees.
Dunkerley vs Direct Sum: Never add critical speeds directly ( $N_1 + N_2$ ). You must sum the reciprocals of squares ( $1/N_1^2 + 1/N_2^2$ ).

Test Your Knowledge

A 75 hp motor drives a process blower shaft at 1160 rpm. What is the steady torque delivered to the shaft in inch-pounds?

3,400 in-lbf

4,075 in-lbf

4,890 in-lbf

8,150 in-lbf

Test Your Knowledge

Why does compressive bearing stress govern the required length of a square key (w = h) over shear stress when the key and shaft are fabricated from the same ductile steel?

Because key shear strength is twice the compressive yield strength.

Because bearing contact area is only half the shear area, doubling bearing stress, while compressive yield strength is only 1.732 times shear yield strength.

Because the shaft keyseat experiences torsional stress concentrations that do not affect the key.

Because centrifugal force reduces the effective shear area along the pitch plane.

Test Your Knowledge

A rotating shaft carries two pulleys. Acting alone on the bare shaft, the first pulley produces a critical speed of 1500 rpm, while the second pulley produces a critical speed of 2000 rpm. Neglecting the mass of the shaft, what is the combined fundamental critical speed using Dunkerley's equation?

1,200 rpm

1,750 rpm

2,500 rpm

857 rpm

Test Your Knowledge

What is the maximum permissible shaft slope at a standard spur or helical gear mesh to avoid edge loading and premature tooth pitting?

0.0001 rad

0.001 rad

0.004 rad

0.026 rad

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PE Mechanical

3.1 Shaft Sizing, Torsional Deflection & Critical Speeds

Key Takeaways

Power Transmission Shaft Design & Critical Speeds

1. Power, Torque, and Rotational Speed Kinematics

Rapid Conversion Matrix for PE Exam Calculations

2. Multi-Axial Combined Fatigue Sizing (ASME DE-Goodman)

Stress Components for Solid Circular Shafts

Distortion Energy (Von Mises) Equivalent Stresses

The ASME / DE-Goodman Sizing Criterion

The Standard Rotating Shaft Case (Ma=M,Mm=0M_a = M, M_m = 0Ma​=M,Mm​=0 and Tm=T,Ta=0T_m = T, T_a = 0Tm​=T,Ta​=0)

Static Yield Check (Langer Line)

3. Marin Endurance Limit Modifying Factors

4. Keys, Keyways & Geometric Stress Concentrations

1. Key Shear Stress Failure Mode

2. Key Compressive Bearing Stress Failure Mode

Keyway Stress Concentration Factors on the Shaft

5. Torsional Deflection & Shaft Rigidity Limits

Torsional Deflection Formula

6. Critical Speeds & Dynamic Whirling

Single-Mass (Jeffcott Rotor) Formulation

Multi-Mass Shafts: Dunkerley and Rayleigh-Ritz Methods

7. Step-by-Step Worked Problem: Complete Shaft & Keyway Sizing

8. Exam Tips & Common Traps

The Standard Rotating Shaft Case ( $M_a = M, M_m = 0$ and $T_m = T, T_a = 0$ )