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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
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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
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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
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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
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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
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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
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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
:
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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.
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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.
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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
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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
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Shaft Layout: Axial Layout and Axial Support
Image Source: Shigley’s Mechanical Engineering Design
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)
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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
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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.
Image Source: Shigley’s Mechanical Engineering Design
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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:
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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).
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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
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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( )
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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
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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.
Image Source: Shigley’s Mechanical Engineering Design
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
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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
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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