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Shaft Design Dr Hengan Ou Coates B104 [email protected]

Adapted from Dr Mike Johnson

Overview of Lecture • Introduction to shaft function, types and applications; • Methods to evaluate shaft loading and to determine shaft diameter using ASME code;

• Considerations for proper shaft connections & design features to prevent fatigue; • Calculation of shaft deflection & critical speed.  Effective learning should be achieved in conjunction with Bearings & Mechanics of Solids lectures

Learning Objectives • To understand shaft function, types, connections and applications; • To select appropriate methods for shaft connections;

• To analyse shaft loading; • To be able to determine shaft diameter using ASME method; • To design features for preventing fatigue; • To calculate shaft deflection & critical speed • To be able to use the general principles in real shaft design problems

• Function:

Function & Types

 a slender component of circular cross-section that rotates and transmits power from a driving device;  a means to carry gears, pulleys & usually connected by couplings;  a means to provide necessary shaft-hub connections.

• Types:

 Plain shaft

 Stepped shaft  Crankshaft  Spline shaft  Camshaft Camshaft

Spline shaft

Applications YamahaR6 engine crankshaft (60Nm & 91KW)

Main shaft

High speed shaft Rolls-Royce Trent XWB Engine

http://www.rolls-royce.com/civil/ products/largeaircraft/trent_xwb/

Modular drive train http://www.nrel.gov/wind Arm driven by turbine shaft

Crankshaft of the world most powerful diesel engine for large container ships (7x106Nm&80MW)

http://www.amusingplanet.com/2013/03 Vestas V90-3MW wind turbine /the-largest-and-most-powerfuldiesel.html http://www.vestas.com/

Ring gear, Sun gear, 3 4

Planet gear, 1

Planet gear, 2

Gear, 6 to generato r

Gear, 5

A 3MW compound planetary helical gearbox,

What are the important questions to be answered in designing a shaft?

Using Yamaha R6 engine camshaft & crankshaft as two examples

Yamaha R6 Camshaft

Yamaha R6 Engine (60Nm & 91KW)

http://youtu.be/OGj8OneMjek

Yamaha R6 crankshaft

Shaft design considerations, cont’d • Rolls-Royce Trent 900 three shaft jet engine

Shaft design considerations • • • •

Function & loading conditions Size & connection of components Material selection & treatments Deflection & rigidity Fan

A pair of angular contact ball bearings

Square key

• • • •

Stress & strength Critical speed Manufacturing constraints Other design considerations Cylindrical roller bearing

End plate

shaft

Pulley of a Belt Drive

A fan supported by two bearings, SKF

Shaft-hub connections

Pinned

Woodruff Key

Square Keys

Spline Shaft integral with gear

Shaft-hub connections Set Screw

Dowel pin

Key

Woodruff key

Circlip

Lock nut

Location of Bearing on Shaft Other means of locating outer/inner rings of bearings

Lock nut & lock washer

Cover plate, circlip & lock nut End plates

Location of Bearing on Shaft Rotating ring:

interference fit

Stationary ring:

interference fit or push fit

Axial location: - Axial location of both rings against abutment faces Shoulder fillet radius  corner radius of bearing Shoulder height  2~2.5 corner radius of bearing

Partial view of a SKF bevel gearbox

Axial location of roller bearings

Shaft Design Design procedure 1. Determine shaft speed

2. Determine torque to be transmitted 3. Determine shaft loadings 4. Selection & position of bearings

Iterative Process!

5. Determine shaft diameter – ASME Design Code 6. Design necessary design features 7. Understand Macaulay’s method and Castigliano’s theorem for shaft deflections

8. Understand Rayleigh-Ritz Equation for critical speed

Shaft Loading • Axial stresses:

– due to self-weight in vertical shafts; – due to axial restraint at bearings and associated axial loads

• Bending stresses: – Due to self-weight, tensile forces in belt drives, gear forces, mounted component weights (e.g. gear, flywheel)

– dynamic forces which can lead to fatigue and resonance

• Shear stresses:

– due to torsional load

Example One: Shaft Loading Resultant force, R1

P1 Pulleys

Belts

P2

Total belt force, P

Torque, T

Axial force, F

Fans with two bearing housings, SKF

L

F

T, ω

Axial resultant force, FR

L1

P

R1 FR

T,ω

R2

Torque, T

R1

Resultant force, R2 P M=PL1

R2

T=P/ω

Exercise one • Determining input shaft loading (bending moment & torque diagrams) FR

d

Output Torque, Tout

R1 L Meshing force, F

R1

FT

Input Torque, Tin

R2

SKF spur gearbox

R2

T

Exercise one with solution • Determining input shaft loading (bending moment & torque diagrams) FR

Output Torque, Tout

R1 L M=½FL

Meshing force, F

Input Torque, Tin

T=½dFT R1

T

d

FT

R2

SKF spur gearbox

R2

F=(FR2+FT2)1/2

Shaft diameter Use the ASME design code for transmission shafting Reserve factor (often use 2)  32ns  d    

2

M 3  T    4   y e 

1/3

         2

Max bending moment on shaft Endurance limit stress

Max torque on shaft Yield strength of shaft material

Shaft diameter Endurance limit stress is related to the ultimate tensile strength σUTS decrease for each cycle of loading

After more than 106 cycles, reduction in σUTS stops.

σUTS

σe

100

Material will last for “infinite” cycles so long as the endurance limit stress, σe, isn’t exceeded

101

102

103

104

105

106

107

Number of cycles

Most steels have this behaviour -> Often used for shafts

108

Shaft diameter Endurance limit stress, σe, is affected by factors such as loading, reliability and stress concentrations, etc

 e  kakbkc kd kekf kg

' e

' Where,  e - Endurance limit of test specimen ka = surface factor kb = size factor (=1) Check handouts for specific kc = reliability factor values of all the factors kd = temperature factor (=1) ke = duty cycle factor (=1) kf = fatigue stress concentration factor kg = miscellaneous effects factor (=1)

Shaft diameter

 e  kakbkc kd kekf kg

' e

 e'  0.504  UTS for  UTS  1400 MPa  e'  700 MPa for  UTS  1400 MPa

Select factors from graphs, tables, empirical formulas as given in handouts

Where ka = surface factor kb =1 kc = reliability factor 1 kd = 1 kf  1  qK t  1 ke = 1 kf = stress concentration factor kg = 1

Shaft fatigue Features on shaft cause stress concentrations -> fatigue failure Observe “best practice” to minimise stress: Poor Fatigue Strength

Shoulders Sharp Corner

Improved Large fillet radius

Undercut fillet with collar Undercut radiused fillets

Shaft fatigue Features on shaft cause stress concentrations -> fatigue failure Observe “best practice” to minimise stress: Poor Fatigue Strength

Holes

Improved Enlarged section a hole

Stress relieving grooves

Shaft fatigue Features on shaft cause stress concentrations -> fatigue failure Observe “best practice” to minimise stress: Poor Fatigue Strength

Splines

Improved

Increase shaft diameter

Radius fillets

Shaft fatigue Features on shaft cause stress concentrations -> fatigue failure Observe “best practice” to minimise stress: Poor Fatigue Strength

Fitted Assemblies

Improved Radius

Increase dia

Add grooves

Shaft fatigue Features on shaft cause stress concentrations -> fatigue failure Observe “best practice” to minimise stress: Poor Fatigue Strength

Keyways

Improved Increase diameter

Add Radi

Exercise two: air motor shaft •Determining air motor shaft diameter Shaft material: mild steel σuts = 600 MPa σy = 350 MPa.

Assuming a case of cantilever beam

F=pA=80N

R2

Reserve factor, ns=2 Assume:

ka kb kc kd ke k f k g  0.3

R1

Calculate the camshaft diameter:  32ns d    

2

M 3  T    4   y e 

Δ=6 mm 2

   

1/3

   

l=10 mm

output

Exercise two: air motor shaft

•Determining air motor shaft diameter Shaft material: mild steel

Assuming a case of cantilever beam

F=pA=80 N

σuts = 600 Mpa, σy = 350 MPa. Reserve factor, ns=2

R2

ka kb kc kd ke k f k g  0.3  e'  0.504  UTS  302.4MPa  e  k a kb kc k d ke k f k g e'  0.3  302.4  90.7 MPa  32ns d    

M  3 T         e  4  y  2

2

   

R1

Δ=6 mm

 32  2   79.2 109   3.1416 

13

l=10 mm M=Fl=80x0.01=0.8 Nm

1/ 3

 32  2  0.8  2 3  0.48  2       6  6   3.1416  90.7 10  4  350 10    

output

1

3

 5.6 103 m  5.6mm

T=FΔ=80x6=0.48Nm

Shaft Critical Speed Centre of mass

Centre of rotation

Centre of Mass should equal Centre of Rotation (but in practice it doesn’t) Imbalance causes a deflection (centrifugal force, mrω2 normally balanced by flexural rigidity, EI)

At the critical speed (natural frequency) shaft is unstable (deflection increases significantly to break shaft, damage bearing and cause destructive vibration, “shaft whirl”)

Shaft Critical Speed First Critical Speed

C  Shaft with a single mass

g

 st

rad s 

where, g – acceleration of gravity (m/s2), δst- static deflection of shaft (m).

Rayleigh-Ritz equation Shaft with multiple masses

Rule of Thumb Operational speed of shaft should be ½ the critical speed

C 

w  g w

i i 2 i i

rad s 

where, wi – weight of node i (N), δi - static deflection of node i (m).

Shaft deflection Shaft deflections required to determine critical speed

• Macaulay’s method for beam bending (MM2MS2): d2y M  2 dx EI

y

x

0

M  0  EI dx dx  C1 x  C2 x

Useful for plain shaft

• Castigliano’s theorem: The deflection of an elastically deformed body is equal to the partial derivative of strain energy wrt the force applied at that point.

U i  Fi

δi

Fi

Shaft deflection F

Castigliano’s theorem: •

Axial loading of a bar: F 2L U 2 EA

•

U

  F 2 L  FL    A  F  2 EA  EA

δ δA

FA

Bending of a plain shaft:

U 

L

0

2 3 L F 2 x2 M2 F L 2 dx  2 dx  0 2 EI 8EI 96 EI

  F 2 L3  FL3     F  96 EI  48EI

1 F 2

FA

F R1

δ

L

R2

Shaft deflection Types of beams

Max deflection

PL3  max  3EI

PL3  max  48EI

Pb 2 L  max  3EI

Deflection at any point x

Px 3 3L  x   6 EI

Px  12 EI

 3L2 2   x   4 

For 0  x  a : Pbx 2  x  a2 6aEI





For 0  z  b : Pbx 3  z  b2 L  b z  2b 2 L 6 EI





Summary • To be familiar with shaft function, types, connections and applications; • To select appropriate methods for shaft connections; • To analyse shaft loading;

• To be able to determine shaft diameter using ASME method; • To design features for preventing fatigue; • To determine shaft deflection & critical speed.

A worked example • Determining Shaft diameter of a transmission shaft with belt & spur gear drives Example, Childs pp98-101

Calculate the minimum shaft diameter Material:

817M40 hot rolled alloy steel σUTS = 1000 MPa σy = 770 MPa. Brinell hardness is approximately 220 BHN. Reserve factor, ns=2

Features: Radii on fillets is 3 mm. Reliability required is 90% Output: 8 kW at 900 RPM with a maximum Torque of ??

PW   T Nm1 / s  2 1 / s   nrpm 60 T Nm 

60 103 PkW  2 nrpm

• Determining Shaft loading • Resolve loads in vertical and horizontal planes Loads in Vertical Plane

A

R1V

Ft

Fr

13.3 Nm

Loads in Horizontal Plane

A

C

B

9.81 Nm

R2V

T=84.9 Nm

A

T

R1H

Ft

C

R1H

158.5 Nm

C B

R2H

B 52.6 Nm

R2H

T

• Calculate Bending Moments The vertical bending moments are calculated as: MBV  R1V L1  111.1  0.12  13.3 Nm MCV  R1V (L1  L2 )  ((Ff  mg g)L2 )  111.1 (0.12  0.08)  (321.9  8  9.81)0.08  9.81 Nm

The horizontal bending moments are calculated as: MBH  R1H L1  438.5  0.12  52.62 Nm MCH  R1H (L1  L2 )  (Ft L2 )  438.5  0.2  884.4  0.08  158.5 Nm

The resultant bending moments are calculated as: Maximum Bending

MB  (13.3)2  (52.62)2  54.27 Nm MC  (9.81)2  (158.5)2  158.8 Nm

Moment on Shaft

• Determining shaft diameter • Calculate the endurance stress limit

Check handouts for the specific values of factors

 e  ka kb kc kd ke k f k g e'  0.405  0.856  0.897 11 0.629 1 504  98.6 MPa

• Calculate minimum diameter from the ASME equation:  32ns d    

2

M 3  T    4   y e 

2

   

   

1/3

 32  2 3  84.9   158.8        6 6 98 . 6  10 4 770  10         0.032 m 2

2

   

1/3

Choose standard diameter of 35 mm