L7 - Shaft Design

8/16/2019 L7 - Shaft Design

1/22

Pre-Lecture Video: Shaft Manufacture

https://www.youtube.com/watch?v=pfOOHCTsEng

https://www.youtube.com/watch?v=iKizLfzz7GM

http://www.youtube.com/watch?v=hsMyfAjcoZs

https://www.youtube.com/watch?v=pfOOHCTsEnghttps://www.youtube.com/watch?v=iKizLfzz7GMhttp://www.youtube.com/watch?v=hsMyfAjcoZshttp://www.youtube.com/watch?v=hsMyfAjcoZshttp://www.youtube.com/watch?v=hsMyfAjcoZshttps://www.youtube.com/watch?v=iKizLfzz7GMhttps://www.youtube.com/watch?v=iKizLfzz7GMhttps://www.youtube.com/watch?v=iKizLfzz7GMhttps://www.youtube.com/watch?v=iKizLfzz7GMhttps://www.youtube.com/watch?v=pfOOHCTsEnghttps://www.youtube.com/watch?v=pfOOHCTsEng


2/22

Dr. Ferdinando [email protected]

Image Source: http://www.prlog.org/11380026-mechanical-design-services-mechanical-engineering-drawing-services.html

ENS3116/ENS5114 Advanced Mechanical Design

Lecture 7 – Shaft Design

mailto:[email protected]:[email protected]


3/22

Recap: Fatigue

Endurance StrengthEquations to estimate a steel’s endurance

limit based on the ultimate tensilestrength:

MPaS MPaS

MPaS S S

ut e

ut ut e

1400700

14005.0'

'

e

ut

S S f

a2)(

e

ut

S S f

b

log31

b f N aS

Calculates: Mean Fatigue Strength S f forgiven cycles, N [For high-cycle: 10 3 N 10 6]

Estimating Fatigue Strength/Cycles

Calculates: Cycles N, for given Mean FatigueStrength S f (or Reversed stress a)[For high-cycle: 10 3 N 10 6]

b

aS

N f /1

S

ut

S e

Low Cycle High Cycle

10 6

Number of stress cycles, NLog(N) F

a t i g u e

S t r e n g

t h ,

S f

L o g

( S f )

10 3


4/22

Recap Fatigue: Factors Affecting Fatigue Life

'e f ed cbae S k k k k k k S

S-N curves are generated using ideal test samples. There are a number of factors thataffect the endurance strength of the actual part.

Image Source: http://www.accutektesting.com/testing-services/mechanical-testing/rotating-beam/

Endurancelimit of part

Endurance

limit of testspecimen

Surface factor

Size factor

Load modification

Temperature factor

Reliability factor

Miscellaneous effects factor


5/22

Recap Fatigue: Stress-Concentration and Notch Sensitivity

It is useful to introduce a fatigue stress-concentration factor.

specimenfreenotchinstressspecimennotchedinstressmaximum

f K

Different materials show different sensitivities to the stresses raised by notches. Often thefatigue stress concentration factor is displayed in terms of a Notch Sensitivity q, thatincorporates the stress concentration factors K f and K t (not ALL materials are notch sensitive) .

1

1

t

f K

K q Rearranged: )1(1 t f K q K

Note: K t is a theoretical stress concentration factor that estimates the maximum stress at anotch based on a component’s nominal stress (without a notch).

Where:

60 MPa20 MPa


6/22

An example from Shigley textbook that showsthe geometric stress concentration factor (K t)

for a shouldered shaft under torsion.

An example from Shigley textbook thatshows the notch sensitivity (q) for materialsunder reversed torsion.

Recap Fatigue: Stress-Concentration and Notch Sensitivity


7/22

Recap Fatigue: Fluctuating Stress

minmax

r

RangeStress

Note: R=-1 for a completely reversedstress state with zero mean stress

max

min:

R RatioStress

2minmax

a

Stress g Alternatin

2

minmax

m

Stress Mean

m

a A Ratio Amplitude

:


8/22

Recap Fatigue: Modified Goodman Model

S e

S y

S ytS yc Mean Stress

Alternating Stress

TensionCompressio n

Modified GoodmanFailure Line

Yield (Langer)Failure Line

The relationship between mean andalternating stresses and failure is:

nS S ut m

e

a 1

Factor Safetyn

Stress Mean

Stress g Alternatin

Strength EnduranceS

StrengthTensileUltimateS

m

a

e

ut

:

:

:

:

:

n

S yma

Langer Static Yield relationship:

nS e

n

S y

n

S ycn

S yt n

S ut

ModifiedGoodman

Design LineSafe Design Area

'e f ed cbae S k k k k k k S Note:

σm

σa

Note: Any operating point below the Goodman or Langer lines should have infinite life.


9/22

Shaft Design: Shafts and Shaft Components

Image Source: http://bode.en.hisupplier.com/product-1001214-steel-shafts.html and http://www.traderscity.com/board/products-1/offers-to-sell-and-export-1/aisi-4140-annealed-steel-shaft-124714/

The following considerations should be made whenexamining shaft design:

o Material selectiono Geometric layouto Stress and strength

• Static• Fatigue

o Deflection and rigidity• Bending deflection• Torsional deflection• Slope at bearings and shaft-supports• Shear deflection due to transverse loading

o

Vibration due to natural frequency.

A complete shaft design has interdependence on thedesign of individual machine components/elements.


10/22

Shaft deflection is a function of material stiffness (Modulus ofElasticity). This is constant for steels, thus shaft rigidity is governedby shaft geometric dimensions.

Material and treatment choice affects shaft strength, which resistsapplied stresses. Many shafts are manufactured from low carbon,

cold-drawn or hot-rolled steel (such as 1020 to 1050 steels).

Shaft production quantities may also affect material selection:

Low quantities: Low carbon steel alloys are desirable for easier and faster

machining processesHigh quantities: Hot or cold forming, or casting are typical applied for a

range of material alloys, where machine elements can beintegral

Shaft Design: Shaft Materials


11/22

Shaft Design: Shaft Layout

Image Source: Shigley’s Mechanical Engineering Design

Shaft layout must accommodate the machineelements by providing radial, axial and/or torsionallocation.

The layout simply comprises of a stepped cylinder.These steps are known as shoulders and provideexcellent axial location of machine elements toresist thrust loads.

General shaft layout considerations whendesigning shafts:

Axial Layout of Components

Supporting Axial Loads

Providing Torque Transmission

Assembly and Disassembly


12/22

Shaft Layout: Axial Layout and Axial Support


General shaft layout considerations when designing shafts:

Axial Layout of Components

Should aim to support load-carrying components between bearings and not have themcantilevered

Pulleys and sprockets should be mounted outboard (cantilevered) for easy belt and chaininstallation

Typically 2 bearings are sufficient for most shaft support

Design the shaft as short as possible (within reason)

Locate load-carrying components near bearings

Supporting Axial Loads

Different shoulder types can be used based on the axial loado Near-zero axial load: Press fits, setscrews, pins, etc

o Low axial load: Retaining rings in grooves, sleeves, clamp-on collars, etc.

It is best to have one bearing resist the axial load (in both directions or a single direction)


13/22

Image Source: Shigley’s Mechanical Engineering Design and http://www.popularhotrodding.com and http://www.globalindustrial.com

Providing Torque Transmission

Shafts must provide torque transmission between the

shaft and machine elements. Methods include:o Keys

o Splines

o Setscrews

o Pins

o Press fits or shrink fits

o Tapered fits

Some torque transmission components are designedto fail for unexpected loads, thus protecting expensivemachine elements.

Splines are suitable for high torque applications andalso allow for axial motion whilst transmitting torque.

Shaft Layout: Torque Transmission and Assembly

Key

Pin

PinRound key

Roll-pinTapered Pin


14/22

Shaft Layout: Assembly and Disassembly

Assembly and Disassembly

Generally have the largest shaft diameter in the centre and then progressively smaller

diameter towards the ends.Use a retaining ring, sleeve or locknut in order to have a shoulder on both sides of a shaftelement.

Design the shaft such that machine elements onlypress along a short length of the shaft.

Allow access to at least one end of the shaft throughthe housing to access retaining rings, bearing pullersetc. for assembly and disassembly.



15/22

Shaft Design: Misc. Shaft Components

SetscrewsSetscrews rely on compression to generate clamping forces,dissimilar to bolts that rely on tension. A setscrew in a collar providesa holding power which can resist axial and torsional loads ( yourtextbook provides a table of holding power for various setscrewssizes ).

Keys and PinsKeys and pins are used to secure rotating elements to the shaft. Keysprovide only torque transmission between the shaft and shaft element,whilst pins provide both torque and thrust transfer ( your textbook

provides tables with common key and pin sizes ).

Retaining ringsRetaining rings readily replace shaft shoulders or sleeves for axiallocation of elements along a shaft. A radial groove cut into the shaftlocates the spring retainer.

Image Source: http://www.globalindustrial.com and training.bsc.com.au

Shafts comprise of various miscellaneous components to completetheir design, where such components include:


16/22

Worked Example: Shaft Key Design

Example source: Shigley’s Mechanical Engineering Design

A heat treated steel shaft has a minimum yield strength of 525MPa and adiameter of 36mm. The shaft rotates at 600RPM and transmits 30kWthrough a gear. Select a key for the gear.

Solution:Firstly calculate the load, F:

Angular velocity: = = 600 = 62.8 / Torque: = ⇒ = (30000)/62.8 = 478Nm Force: T = ⇒ = 478/0.018 = 26.6kN

The key shear strength can be determined from the normal yield strength, where the selected key steelhas a yield strength of 450MPa (notably less than the shaft):

= 0.577 = 0.577 450 = 259.7

Shear failure occurs along a-b, hence the shear stress is (for an initial guess of 10mm key):

.= = 259.7 3 = 86.6 = =26600

10 ⇒ = 30.7

Similarly, but checking the crushing resistance of the key by using one-half contact face:

.= = 450 3 = 150 =2 = 2(26600) 10 ⇒ = 35.5

Therefore, a 10mm square key must be approx. 36mm long, which is less than the upper limit of 1.5 shaftdiameters (54mm).


17/22

Shaft Design: Shaft Design and Stress

Shaft design considerations:Critical locations

Shaft stresses

Estimating stress concentrations

Critical locations

Should be identified to base the shaft design calculations i.e.stress assessment. Typically these locations are at the

o Outer surface (highest stress)

o Axial location where the bending moment is highest

o Section of shaft transmitting torque

o Stress concentration along the shaft

Torsional, bending and axial stresses should be considered forthe stress assessment at the critical location(s).

Image source: www.sportfishermen.com and www.yachtforums.com


18/22

Shaft Design: Shaft StressesShaft stresses

Bending, torsion and axial stresses can be calculated for both mean and alternating components. For a

solid round shaft, the bending and torsion stresses are:

σa =32

σ =32

τa =16

τ =16

Where, M m and M a are the mean and alternating bending momentsTm and T a are the mean and alternating torquesKf and K fs are the fatigue stress concentration factors for bending and torsion

These stresses can be combined to evaluate the alternating and mean Von Mises stresses:

σ′a = 3 =32

316

σ′ = 3 =32

316

Finally, these alternating and mean stresses can be combined with the modified Goodmanfailure model to solve for either the safety factor or diameter:

1=

16 14( ) 3( )

14( ) 3( )


19/22

Shaft Design: Shaft Static StressesShaft static stresses

The Modified Goodman fatigue failure model does not guard against yield failure. As a

result it is necessary to separately check for first load cycle yield failure. The Von Misesmaximum stress can be evaluated as follows:

σ′ ax = 3

= 32 ( + ) 3 16 ( + )

And then this maximum stress can be compared to the material yield strength:

= σ′ ax

h f h f d


20/22

Shaft Design: Shaft Design and StressEstimating stress concentrations

Fatigue stress analysis is highly dependent on stress concentrations. In the early design

stages, the stress concentrations are not known because the shaft dimensions are yet tobe determined. As a result, first iteration estimates are used for stress concentrations andthen replaced with actual values once the first shaft design iteration is complete.


Stress Concentration, K t: First estimate Bending Torsion AxialShoulder fillet-sharp (r/d=0.02) 2.7 2.2 3.0

Shoulder fillet-rounded (r/d=0.1) 1.7 1.5 1.9End-mill key (r/d = 0.02) 2.2 3.0

Sled runner key 1.7

Retaining ring groove 5.0 3.0 5.0

k d l h f


21/22

Worked Example: Shaft Design

Example source: Shigley’s Mechanical Engineering Design

A machined shoulder on a shaft has a small diameter of 28mm and a large diameter of 42mmand a 2.8mm fillet radius. The shaft undergoes a 142.4Nm bending moment and a 124.3Nmsteady torsion load. The steel shaft has an ultimate strength of 735MPa and a yield strength of

574MPa. For a reliability of 99%, determine the safety factor against fatigue failure and yield.Solution:D/d = 42/28 = 1.5 and r/d = 2.8/28 = 0.10, now using these values the stress concentration factors andnotch sensitivity can be found:

, = 1.68, , ℎ = 1.42 = 0.85, ℎ = 0.92

Therefore, stress concentration can be calculated to be:, = 1 0.85 1.68 1 = 1.58

, ℎ = 1 0.92 1.42 1 = 1.39 Specimen endurance limit is: ′ = 0.5 735 = 367.5

The correction factors are found to be:= 4.51(735)− . = 0.787

= 1.24(28)− .

= 0.870 = 0.814 = = = 1

Therefore, the corrected endurance limit for the part is: = 0.787 0.870 0.814 367.5 = 205

W k d E l Sh f D i


22/22

Worked Example: Shaft DesignSolution:For a rotating shaft, the constant bending moment becomes fully reversed:

= 142.4 , = 0 = 0 , = 124.3

Applying the Distortion Energy Theory Goodman failure model, the safety against fatigue is:

1=

16 14( ) 3( )

14( ) 3( )

1=

16(0.028)

1205×10 4(1.58(142.4))

1735×10 3(1.39(124.3)) = 0.615

∴ = 1.62

The yield safety factor is determined using the Von Mises maximum stress:

σ′ ax = 32 ( ) 3 16 ( ) = 32(1.58) (142.4)(0.028) 3 16(1.39) (124.3)(0.028) = 125.4

Thus the safety factor against yield is:

= σ′ ax=

574125.4 = 4.58

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L7 - Shaft Design