Shaft Design

Shaft vs Axle

The main difference between a shaft and an axle, is that a shaft is a rotating member whereas an axle is not

Shafts:

• Transmit power
• Are subject to bending and torque
• Usually circular

Main focus of this course:

• Material selection
• Geometry
• Stress and strength (static and fatigue)

Extra factors:

• Deflection and rigidity
• Vibration

The General Process for Analysis

• Identify critical sections
• Analyze stress in each critical section for static and dynamic/fatigue failure
• Consider stress concentration factors in addition to loading

A note on deflection

Analysis with deflection is usually done in software and occurs after stress analysis

Materials

Steels

• Stiffness is constant
• Strength varies with steel choice which impacts static and dynamic failure
• Fatigue failure will reduce as strength increases but only to a certain point where endurance limits and notch sensitivity counteract improvements
• Common steel choices include
  • AISI 1020-1050 CD or HD
    • Use CD if your diameter is less than 3 inches
  • AISI 1340-50, 3140-50, 4140, 4340, 4340, 5140, 8650
    • Higher strength
  • Carburizing grades of AISI 1020, 4320, 4820, 8620
    • Surface hardened
    • Only used under specific conditions like: Journal of a bearing surface?

Layout

Overall shape is a stepped cylinder

Axial Layouts

• Keep load bearing components close to bearings
• Minimize shaft length
• Keep cantilever as small as possible
• Axial loads are transferred into the shaft then intot he ground
  • Shoulders, retaining rings, and pins are ways you can do this See slides

Torque Transmission

This is the main purpose of most shafts (to transmit torque from some input to some output)

These components are meant to fail if torque gets too high protecting more expensive equipment.

Common Torque Transfer Elements

• Keys
• Splines
• Pins
• Set Screws
• Press Fits
• Shrink Fits
• Tapered Fits

Shaft Design for Stresses

Shafts usually support Axial Loads, Bending Moments, and Torques
Critical locations are usually on the outer surface and there could be more than one and axial loads are typically negligible.

Finding Critical Locations

Find points where bending moments are large:

• Use shear force and bending moment diagrams to assess
• If forces are in two planes you need 2 diagrams then calculate the resultant Torques:
• Use FBDs to assess torques

Shaft Stresses

As a shaft rotates, stresses will fluctuate this results in completely reversed stress which means you need to determine mean stress and the stress amplitude

All three loading factors could be present which means you need to combine loading.

See The Stress-Life Method and S-N for constant amplitude loading definitions.

*Completely reversed stress means that average stress is 0

Shaft Specific Stress Equations

$$\text{Assuming negligible axial stress}$$ $${\sigma }_a=K_f\frac{M_{a}c}{I}{\sigma }_m=K_f\frac{M_{m}c}{I}$$ $${\tau }_a=K_{fs}\frac{T_{a}r}{J}{\tau }_m=K_{fs}\frac{T_{m}r}{J}$$ $$\begin{array}{rl}r& =\text{shaft radius}\\ d& =\text{shaft diameter}\\ T_m& =\text{mean torque}\\ I& =\text{second moment of area}\\ M_m& =\text{mean bending moment (not constant because it is fluctuating)}\\ c& =\text{distance from centroidal axis}\\ J& =\text{polar second moment of area}\\ T_a& =\text{alternating/amplitude torque}\\ M_a& =\text{alternating/amplitude bending moment}\\ K_f\text{ and }{K}_{fs}& =\text{fatigue stress-concentration factor}\end{array}$$

Where a and m refer to amplitude and midrange stresses respectively.

Assuming a solid shaft with a round cross section

$${\sigma }_a=K_f\frac{32M_a}{\pi d^3}{\sigma }_m=K_f\frac{32M_m}{\pi d^3}$$ $${\tau }_a=K_{fs}\frac{16T_a}{\pi d^3}{\tau }_m=K_{fs}\frac{16T_m}{\pi d^3}$$

Von Mises (Distortion Energy) Stresses

$${\sigma }_a^{\prime }={\left({\sigma }_a^2+3{\tau }_a^2\right)}^{1/2}={\left[{\left(\frac{32K_f{M}_a}{\pi d^3}\right)}^2+3{\left(\frac{16K_{fs}{T}_a}{\pi d^3}\right)}^2\right]}^{1/2}$$ $${\sigma }_m^{\prime }={\left({\sigma }_m^2+3{\tau }_m^2\right)}^{1/2}={\left[{\left(\frac{32K_f{M}_m}{\pi d^3}\right)}^2+3{\left(\frac{16K_{fs}{T}_m}{\pi d^3}\right)}^2\right]}^{1/2}$$

Designing Shafts

There are two scenarios:

1. Where the shaft diameter is known which is similar to what we did before where we follow the steps in Endurance Limit Modifying Factors
2. Where the shaft diameter is unknown where we have different criteria equations

Shaft Specific Failure Criterion

$$A=\sqrt{4{\left(K_f{M}_a\right)}^2+3{\left(K_{fs}{T}_a\right)}^2}$$ $$B=\sqrt{4{\left(K_f{M}_m\right)}^2+3{\left(K_{fs}{T}_m\right)}^2}$$

DE-Goodman

$$n=\frac{\pi d^3}{16}{\left(\frac{A}{S_e}+\frac{B}{S_{ut}}\right)}^{-1}d={\left[\frac{16n}{\pi }\left(\frac{A}{S_e}+\frac{B}{S_{ut}}\right)\right]}^{1/3}$$

DE-Morrow

$$n=\frac{\pi d^3}{16}{\left(\frac{A}{S_e}+\frac{B}{{\tilde{\sigma }}_f}\right)}^{-1}d={\left[\frac{16n}{\pi }\left(\frac{A}{S_e}+\frac{B}{{\tilde{\sigma }}_f}\right)\right]}^{1/3}$$

Where:

$${\tilde{\sigma }}_f\text{ is true fracture strength}$$ $${\sigma }_f^{\prime }\text{ is fatigue strength coefficient (for steels these two are almost the same)}$$

For steel:

$${\sigma }_f^{\prime }=S_{ut}+50\text{kpsi}$$ $${\sigma }_f^{\prime }=S_{ut}+345\text{MPa}$$

DE-Gerber

$$\frac{1}{n}=\frac{8A}{\pi d^3{S}_e}\left(1+{\left[1+{\left(\frac{2B{S}_e}{A{S}_{ut}}\right)}^2\right]}^{1/2}\right)$$ $$d={\left[\frac{8nA}{\pi S_e}\left(1+{\left[1+{\left(\frac{2B{S}_e}{A{S}_{ut}}\right)}^2\right]}^{1/2}\right)\right]}^{1/3}$$

DE-SWT

$$n=\frac{\pi d^3}{16}\frac{S_e}{{\left(A^2+AB\right)}^{1/2}}$$ $$d={\left[\frac{16n}{\pi S_e}{\left(A^2+AB\right)}^{1/2}\right]}^{1/3}$$

Shaft Specific Yield Check

Von Mises maximum stress calculation:

$${\sigma }_{\text{max}}^{\prime }={\left[{\left({\sigma }_m+{\sigma }_a\right)}^2+3{\left({\tau }_m+{\tau }_a\right)}^2\right]}^{1/2}$$ $${\sigma }_{\text{max}}^{\prime }={\left[{\left(\frac{32K_f\left(M_m+M_a\right)}{\pi d^3}\right)}^2+3{\left(\frac{16K_{fs}\left(T_m+T_a\right)}{\pi d^3}\right)}^2\right]}^{1/2}$$

Yield check:

$$n_y=\frac{S_y}{{\sigma }_{\text{max}}^{\prime }}$$

Quick, conservative yield check:

$$n_y=\frac{S_y}{\left({\sigma }_a^{\prime }+{\sigma }_m^{\prime }\right)}$$

Designing shafts with keys

You need to consider the different ways the shaft can fail.

• Based off of the size of the shaft diameter, you choose the material and machining type
• You choose the key size also based off of shaft diameter from a table
• Considering the different ways the key can fail:
  • Shearing
  • Crushing
  • Here we solve for key length using F, n, t, $S_{sy}$ where $S_{sy}=\frac{S_y}{\sqrt{3}}$ by using these equations for each one respectively $n={\frac{Sy}{\sqrt{3}{\tau }_{xy}}}_{}$ and $n={\frac{Sy}{\sigma }}_{}$
• Solving in this case, means choosing the machining type, material, key-size, and length