ME2 Design & Manufacture

Shaft Design

Shafts

PLAIN TRANSMISSION

STEPPED SHAFT

MACHINE TOOL SPINDLE

RAILWAY ROTATING AXLE

NON-ROTATING TRUCK AXLE

CRANKSHAFT

CIRCLIPS

GEAR PULLEY

KEY

KEY

SHAFT HUBHUB

FRAME

WOODRUFF

PROFILED

P. Childs, 2014, Mechanical Design Engineering Handbook

Chapter 7

TRANSVERSELOAD

LOADTRANSVERSE

TRANSVERSELOAD

TORSIONALLOAD

AXIALLOAD

LOADTORSIONAL

AXIALLOAD

TWISTDUE TO

TORSIONALLOAD

DEFLECTION DUE TOBENDING MOMENT

Shaft Design Procedure Flow Chart for Shaft Strength & Rigidity (Beswarick 1994)

DETERMINEEXTERNAL LOADS

CHOOSE PRELIMINARYSHAFT DIMENSIONS

IDENTIFY CRITICALSHAFT SECTIONS

INTERNAL FORCESAND MOMENTS

COMBINEDSTRESSES

SET FACTOROF SAFETY

COMPARE FACTOREDSTRESSES WITH

MATERIAL STRENGTH

IS SHAFT SECTIONSATISFACTORY

SPECIFY SHAFT

TRANSVERSE FORCES,AXIAL FORCES ANDBENDING MOMENTS

DIRECT STRESS

CHOOSE MATERIAL

STRENGTH MODULUS

DEFLECTION

SHEAR FORCES ANDTWISTING MOMENTS

SHEAR STRESS

NO 1st OPTIONNO 2nd OPTION

YES

DETERMINE DETERMINEDETERMINE

DETERMINE

DETERMINE

DETERMINE

DETERMINE

DETERMINE

Determine External Loads

• Shaft rotational speed?

• Power or torque to be transmitted by the shaft?

• Belt / Chain tension?

• Gear & Pinion loading?

Choose (Preliminary) Shaft Dimensions

• Determine dimensions of components mounted on shaft

• Specify locations for each device

• Specify the locations of the bearings / support

• Propose a general form or scheme for geometry

• Size restrictions

• (Easily) available materials and/or components

Identify Critical Shaft Sections

Free Body Diagram:

• Determine magnitude of torques throughout shaft

• Determine forces exerted on shaft

Identify Critical Shaft Sections

• Where are the loads applied?

• Where are the dimensions smallest?

• Where are the stresses / deflections large?

• Stress-raisers?

– Slots, holes & keyholes

– Sharp corners

– Rough surfaces

Determine Internal Loads

Produce shearing force and bending

moment diagrams so that the

distribution of bending moments in

the shaft can be determined.

Shear and Moment Diagrams

𝑑2𝑦

𝑑𝑥2=𝑀

𝐸𝐼

𝑑𝑦

𝑑𝑥=

𝑀

𝐸𝐼𝑑𝑥 slope

𝑦 = 𝑀

𝐸𝐼𝑑𝑥 deflection

with: 𝑀 = 𝑉𝑑𝑥 and V = − 𝑞𝑑𝑥

M M+dM V V+dV

q

R

1V

R

R

1H

gm gR

2H

2V

Ft

Fr

1T

m gp

T2

Combining Normal Stresses

Vertically

Horizontally

1VR

A

L1

B L2

L

R2V

C

3

rF g+m g mpg

80

BEARING

120

GEAR

BEARING

DRIVEBELT

100

GEAR

1HR

A

21L B L

tF

3

R2H

L

C

T

Combining Normal Stresses

Vertical Bending Moments

Horizontal Bending Moments

1VR

A

L1

B L2

L

R2V

C

3

rF g+m g mpg

1HR

A

21L B L

tF

3

R2H

L

C

T

Vertically

Horizontally

A

5

B 3

C A

30

10

B C

Combining Normal Stresses

Vertical Bending Moments

Horizontal Bending Moments

21110522 .BM

13030322

.CM

Combined:

I

cM

A

5

B 3

C A

30

10

B C

A

30.1

11.2

B C

Normal stress & Shear stress

dx

dx

Normal Stress or Shear Stress?

Normal Stress or Shear Stress?

Shear stresses

• Shear stresses due to:

– Shear forces ( shear force diagram)

– Torque

• Power = Torque x Angular velocity

𝑃 = 𝑇 ∙𝑑𝜃

𝑑𝑡= 𝑇 ∙ 𝜔 = 𝑇 ∙ 2 ∙ 𝜋 ∙ 𝑓

• Shear stress: Torsion Formula 𝜏 =𝑇∙𝑟

𝐽

J: polar moment of inertia

r: radius

Mohr's Circle

Combining and visualising the normal and shear stress components

x

y

txy

txy

tx'y' x'

• Normal stresses σx & σy and

shear stress τ known.

• Average normal stress

𝜎𝑎𝑣𝑔 =𝜎𝑥 + 𝜎𝑦

2

• Actual combined stress

𝑅 =𝜎𝑥 + 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2

Mohr's Circle

Combine and visualise the normal and shear stress components

• Normal stresses σx & σy and

shear stress τ are known.

• Average normal stress

𝜎𝑎𝑣𝑔 =𝜎𝑥 + 𝜎𝑦

2

• Actual combined stress

𝑅 =𝜎𝑥 + 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2

• Principal stresses σ1 and σ2

t

avg

R

1 2

x y

txy

http://moodlepilot.imperial.ac.uk/pluginfile.php/12151/

mod_resource/content/1/out/index.html

Choose Material

• Maximum principal stresses

• Introduce safety factor

• Select a material to match design stress

– steel, low- or medium-carbon

– high quality alloy steel, usually heat treated (critical applications)

– brass, stainless steel (corrosive environments)

– aluminium (light weight)

– polyamide (Nylon®) or POM (Polyoxymethylene/Acetal, Delrin®)

small, light-duty shafts, electronics applications, food industry

eqyield n

Typical Safety Factors

1.25 to 1.5 reliable materials under controlled conditions subjected to

loads and stresses known with certainty

1.5 to 2

2 to 2.5

2.5 to 3

3 to 4 well-known materials

under uncertain conditions of load, stress and environment

untried materials

under mild conditions of load, stress and environment

Growing uncertainty

Fatigue - Correction Factors

with k < 1, and depending on:

• Surface

• Size

• Temperature

• Stress concentrations

• …

σe′ = k ∙ σe σe = 0.5 ∙ σuts

Shaft Design Procedure Flow Chart for Shaft Strength & Rigidity (Beswarick 1994)

• Analyse all the critical points on the shaft and

determine the minimum acceptable diameter

at each point to ensure safe design

• Determine the deflections of the shaft at critical

locations and estimate the critical frequencies

• Specify the final dimensions of the shaft

Critical Deflections for Efficiency & Performance

• Gears:

– deflection < 0.13 mm

– slope < 0.03°.

• Rolling element bearings:

– non self aligning - slope < 0.04°

– self aligning - slope < 2.5° - 3°

Shaft-Hub Connection

• Power transmitting components such as gears, pulleys and sprockets need to be mounted on shafts securely and located axially.

• In addition a method of transmitting torque between the shaft and the component is required.

• The hub of the component contacts with the shaft and can be attached to, or driven by the shaft by

– keys

– pins

– set screws

– press and shrink fits

– splines

– taper bushes

Shaft-Hub Connection after Hurst (1994)

Pin

Gru

b

scre

w

Cla

mp

Pre

ss fit

Shrink f

it

Splin

e

Key

Taper

Bush

High torque capacity x x x

Large axial loads x x x x

Axially compact x x x

Axial location provision x x

Easy hub replacement x x x

Fatigue x x x x

Accurate angular

positioning x x x x ()

Easy position

adjustment x x x x x

Example: What to do

when a shaft deflects too much

Choose the appropriate answer(s) from:

Use High Grade Steel, such as 30CrNiMo8

Increase the diameter of the shaft

Add bearings for extra support

Reduce the load bearing length of the shaft

Some general design considerations

IE

LF

3

3

Overhung layout

More robust layout

Ø=0.04 m

140 N 130 N 150 N

=0.15 m L 1

=0.08 m =0.14 m L 2

L 3

=0.07 m L 4

Example

Example

• As part of the preliminary design of a machine shaft, a check is to undertaken to determine the deflections

• The components on the shaft can be represented by three point masses.

• Assume the bearings are stiff and act as simple supports.

• The shaft diameter is 40 mm and the material is steel with a Young’s modulus of 200 GPa.

Example

=0.15 m

O

x

1 R

L 1

Ø=0.04 m

R 2

=0.08 m

140 N 130 N

=0.14 m L 2

1 W

L 3

2 W

150 N

=0.07 m L 4

3 W

Solution

Macaulay's Method

• Resolving vertical forces:

R1+R2=W1+W2+W3.

• Clockwise moments about O:

W1L1+W2(L1+L2)-R2(L1+L2+L3)+W3(L1+L2+L3+L4) =0

• Hence 321

43213212112

LLL

)L+L+L+(LW+)L+(LW+LWR

Solution cont.

• Calculating the moment at XX:

MXX = -R1x + W1[x-L1] + W2[x-(L1+L2)] - R2[x-(L1+L2+L3)]

• Relation between bending moment and deflection

• This equation can be integrated once to find

the slope θ = dy/dx

and twice to find the deflection y.

x

x

Mxx Vxx

Mdx

ydEI

2

2

Solution cont.

MXX = -R1x + W1[x-L1] + W2[x-(L1+L2)] - R2[x-(L1+L2+L3)]

x

x

Mxx Vxx

Mdx

ydEI

2

2

1

2

32122

2122

11

2

1 CLLLx2

RLLx

2

WLx

2

W

2

xR

dxMdx

dyEI

2

2

CxCLLLx6

RLLx

6

WLx

6

W

6

xR

xMdEIy

1

3

32123

2123

11

3

1

Note that in Macaulay's Method

terms within square brackets to be ignored

when the sign of the bracket goes negative.

Boundary conditions

Assuming: deflection at the bearings is zero

• y(x=0) = 0 → C2 = 0

• y(x=L1+L2+L3) = 0 →

321

3

36

3

326

3

3216

1

211

LLL

LLLLLLC

WWR

321

3

36

3

326

3

3216211

LLL

LLLLLLLLLx

2

R

LLx2

WLx

2

W

2

xR

dx

dyEI

WWR2

3212

2

2122

11

2

1

xLLL

LLLLLLLLLx

6

R

LLx6

WLx

6

W

6

xREIy

WWR3

3212

3

2123

11

3

1

321

3

36

3

326

3

3216211

Solving for deflections

Forces: W1=130 N, W2=140 N, W3=150 N,

Geometry: =4 mm,

Material E=200,000 MPa

Substitution of these values gives:

R1=79.2 N

R2=340.8 N

Deflections:

at x=0.15 m, y=5.110-3 mm

at x=0.29 m, y=2.810-3 mm

at x=0.44 m, y=-1.210-3 mm

=0.15 m

x

1 R

L 1

R 2

=0.08 m =0.14 m L 2

L 3

=0.07 m L 4

44744

mm57.12m102566.164

04.0

64

dI

Also check the slope of the shaft at the critical locations

Hollow v Solid

0

20

40

60

80

100

0 20 40 60 80 100

Re

lati

ve P

ola

r M

om

en

t o

f In

ert

ia [

%]

Wall Thickness / Shaft Radius [%]

Hollow v Solid

0

20

40

60

80

100

0 20 40 60 80 100

Re

lati

ve P

ola

r M

om

en

t o

f In

ert

ia [

%]

Relative Mass [%]

Hollow v Solid

0

20

40

60

80

100

0 20 40 60 80 100

Re

lati

ve P

ola

r M

om

en

t o

f In

ert

ia [

%]

Relative Mass [%]

Danger of buckling?

Some Concluding Remarks - I

Shaft Design Considerations

• size and spacing of components

• material selection, material treatments

• deflection and rigidity

• stress and strength

• frequency response

• assembly, manufacturing & servicing constraints

Some Concluding Remarks – II

1. Minimize deflections and stresses: short shaft, overhangs only if necessary

– Deflection of cantilever beam > deflection of simply supported beam

for the same dimensions and loading)

– But think about assembly and serviceability

2. Stress-raisers (i.e. keys, sharp corners) should not be placed in critical regions:

– minimize effects with a radius (standard values!) or a chamfer.

3. Low carbon steel is often as good as higher strength steels since deflection is

typical the design limiting issue.

4. Limiting deflections

– Gears: deflection < 0.13 mm and slope < 0.03°.

– Rolling element bearings

non self aligning: slope < 0.04°

self aligning: slope < 2.5° (depending on model / configuration)

5. Hollow shafts have better stiffness to mass (specific stiffness) and higher natural

frequencies than solid shafts, but are more expensive and typically have a larger

diameter.

6. Natural frequency of shaft should be >> highest excitation frequency in service.

Q&A 27 Oct 2014

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