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# Axles and shafts

## 1. AXLES AND SHAFTS

## 2. AXLES AND SHAFTS

Links intended to carry rotating elements(pulleys, sprockets, pinions, gears, halfcouplings, etc.) are called as axles or shafts.

## 3. AXLES

Axles are intended to support rotating parts thatdo not transmit torques and are subjected to

bending only.

immovable axle

movable axle

## 4. SHAFTS

Shafts are designed to carry links whichtransmit torques and experience both bending

and torsion.

## 5. CLASSIFICATION OF SHAFTS

1. According to purposeShafts of various drives (gear drives, belt

drives, chain drives and so on);

Main shafts of mechanisms and machines

whose function is to carry not only drive

elements but other elements that do not

transmit torques such as rotors, flywheels, turbine disks, etc.

## 6. CLASSIFICATION OF SHAFTS

2. According to the shape• Straight shafts;

• Cranked shafts;

• Flexible shafts.

## 7. CLASSIFICATION OF SHAFTS

3. According to the construction• Shafts of constant cross section (without steps);

• Shafts of variable cross section (of stepped

configuration);

• Shafts made solid with gears or worms.

## 8. CLASSIFICATION OF SHAFTS

4. According to the shape of the cross section• Shafts with solid circular cross section;

• Shafts with hollow circular cross section;

• Shafts with keyways;

• Shafts with splines;

• Shafts with rectangular cross section.

## 9. SHAFTS

Portion of the shaft which is in contact with abearing is called journal. We will distinguish

between end journal, neck journal and thrust

journal.

## 10. CALCULATION OF SHAFTS

Shafts may be calculated for:• Strength;

• Rigidity;

• Oscillations.

## 11. CALCULATION OF SHAFTS FOR STRENGTH

Calculation of shafts for strength isdivided into 3 stages:

1. Determination of the minimum diameter

of the shaft;

2. Designing the shaft construction;

3. Strength analysis of the shaft.

## 12. DETERMINATION OF THE MINIMUM DIAMETER OF THE SHAFT

Minimum diameter of the shaft is determinedtaking into account torsion stresses only. In

order to compensate neglect of bending

stresses the allowable torsion stress is

assumed as down rated ([ =20…40 MPa).

T

=

;

Wp

p×d

Wp =

.

16

T

d min = 3

.

0.2 × [ ]

3

## 13. DESIGNING THE SHAFT CONSTRUCTION

Input shaftHalf coupling

Bearing

d3

t1

d4

d3

t2

d2

t1

d1

d4

Pinion

Seal

d 1 = d min ;

d 2 = d 1 + 2 × t1 ;

d3 = d2 + 2 × t2 ;

d 4 = d 3 + 2 × t1 .

Bearing

d, mm

20…50

55…120

t1, mm

2; 2.5

5

t2 , mm

1; 1.5

2.5

## 14. SEALS

Seals are divided into:• Commercial seals

(Lip-type seals);

• Labyrinth seals;

• Groove seals;

• Combined seals.

Rubbing element

Steel ring of L-shaped

cross-section

Coil spring

## 15. DESIGNING THE SHAFT CONSTRUCTION

Input bevel pinion shaftBevel pinion

d3

Input worm shaft

Worm

d4

d 2’

Bearing

Bearing

d3

d2

d1

Half coupling

Slotted nut

Seal

d3

d1

d2

d2 ’

Bearing

d4

Half coupling

Slotted nut

Seal

## 16. DESIGNING THE SHAFT CONSTRUCTION

Intermediate shaftBearing

Bearing

Gear

d1

d2

d3

d2

d1

Pinion

## 17. DESIGNING THE SHAFT CONSTRUCTION

Output shaftGear

Half coupling

Bearing

Bearing

d3

d5

d4

d3

d2

d1

Seal

## 18. SPUR GEAR

• Thickness of the rimδ = (3…4)·m;

• Thickness of the web

C = (0.2…0.3)·bg;

• Diameter of the hub

dhub=(1.5…1.7)·dshaft;

• Length of the hub

lhub=(1.2…1.5)·dshaft;

• Diameter of the hole

dhole=(D0-dhub)/4;

• Diameter of the hole centre line

Dc=(D0+dhub)/4; ;

• Fillet radii R ≥ 6 mm;

• Angle γ ≥ 7º.

## 19. WORM GEAR

Thickness of the bronze ring δ1= 2·m;

Thickness of the steel rim

δ2= 2·m;

Thickness of the web

C = 0.2…0.3)·bg;

Diameter of the hub dhub=(1.5…1.7)·dshaft;

Length of the hub

lhub=(1.2…1.5)·dshaft;

Diameter of the screw ds=(1.2…1.4)·m;

Length of the screw ls=(0.3…0.4)·bg;

Diameter of the hole dhole=(D0-dhub)/4;

Diameter of the hole centre line

Dc=(D0+dhub)/4;

• Width and height of the collar

h = 0.15·bg; t = 0.8·h ;

• Fillet radii R ≥ 6 mm

• Angle γ ≥ 7º.

## 20. SKETCH LAYOUT

Double stage spur gear speed reducer20

3…4

lhalf coupl

a

ssg

w

I

20

I

10

10

10

0.5B

10

1.25m

m

m

1.25m

0.5B

awhsg

3…4

10

20

B

10

10

20

lhub

## 21. SKETCH LAYOUT

Double stage coaxial spur gear speed reducer20

0.5B

10

10

aw

10

3…4

lhalf coupl

10

10

10

10

10

10

10

0.5B

3…4

10

20

20

20

lhub

## 22. SKETCH LAYOUT

Bevel gearsdeg

me

bg

1.2me

dep

C

lhub

10

dhub

## 23. SKETCH LAYOUT

Double stage bevel and spur gear speed reducerawhsg

lhub

3…4

20

10

0.5B

10

(1.5…2.3)f

10

f

20

10

10

10

10

10

## 24. STRENGTH ANALYSIS OF THE SHAFT

For single stagespeed reducers

Fc = 125 T ;

For double stage

speed reducers

Fc = 250 T .

Y

Z

X

A

RxA

RyA

Fa

d/2 Ft B Fr

RzA

RxC

RyC

C

D

Fc

T

## 25. STRENGTH ANALYSIS OF THE SHAFT

1. Draw the analytical model in the vertical plane and transfer all forcesto the shaft;

2. Determine vertical support reactions RyA and RyC. For this purpose we

set up equations of moments relative to points A and C. For checking

we will write equation of forces that are parallel to Y axis;

3. Plot the bending moment diagram in the vertical plane;

4. Draw the analytical model in the horizontal plane and transfer all

forces to the shaft;

5. Determine horizontal support reactions RxA and RxC. For that we set

up equations of moments relative to points A and C. For checking we

write equation of forces that are parallel to X axis;

6. Plot the bending moment diagram in the horizontal plane;

2

2

7. Plot the total bending moment diagram ( M å = M x + M y );

8. Plot the twisting moment diagram;

2

2

9. Plot the reduced moment diagram( M red = M t + 0.75 × T ).

## 26. STRENGTH ANALYSIS OF THE SHAFT

d1. M a = Fa × .

2

2. å M A = 0 : -Fr × a - M a + R yС × (a + b) = 0;

F × a + Ma

R yC = r

;

a+b

å M c = 0 : -R yA × (a + b)+ Fr × b - M a = 0;

F × b - Ma

R yA = r

;

a+b

å Fyi = 0 : R yA - Fr + R yC = 0.

Checking:

3. 0 £ x £ a;

M y (0) = 0;

M y = R yA × x;

M y (a) = R yA × a;

a £ x £ a + b;

M y = R yA × x + M a - Fr × (x - a);

4.

5.

d

T = Ft × .

2

å M A = 0 : Ft × a + RxС × (a + b) - Fc × (a + b + c) = 0;

å Mc = 0 :

6.

-RxA × (a + b) - Ft × b - Fc × c = 0;

M y (a) = R yA × a + M a ; M y (a + b) = 0.

- Ft × a + Fc × (a + b + c)

;

a+b

-F × b - Fc × c

RxA = t

;

a+b

Checking: å Fxi = 0 : RxA + Ft + RxC - Fc = 0.

0 £ x £ a; M x = RxA × x; M x (0) = 0; M x (a) = RxA × a;

0 £ x £ c; M x = Fc × x; M x (0) = 0;

M x (c) = Fc × c .

RxC =

## 27. STRENGTH ANALYSIS OF THE SHAFT

M å = M x2 + M y2 ;7.

8.

9.

T

M red = M t2 + 0.75 × T 2 ;

Calculation for static strength

σb =

M

£ [σ b ];

W

M = M red max ;

σb =

M red max

0.1 × d

3

π×d3

W=

;

32

£ [σ b ] , where

Mred max is the reduced

moment at the critical

section;

d is diameter of the shaft at

the critical section;

= …120 MPa.

## 28. STRENGTH ANALYSIS OF THE SHAFT

Calculation of the shaft for fatigue strengthChanging of bending stresses

Safety factor

S=

Sσ × S τ

Sσ2 + Sτ2

³ S = 1.5...2.5.

Safety factor for bending

Sσ =

Changing of torsion stresses

σ lim

Kσ

× σ peakσ+ ψmean

×σ

Kd × KF

;

Safety factor for torsion

Sτ =

τ lim

Kτ

× τ peakτ+ ψmean

×τ

Kd × KF

.

## 29. STRENGTH ANALYSIS OF THE SHAFT

Calculation of the shaft for fatigue strengthτ lim

σ lim

Sτ =

Sσ =

;

Kτ

Kσ

× τ peakτ+ ψmean

×τ

× σ peakσ+ ψmean

×σ

Kd × KF

K ×K

d

.

F

lim, lim – limit of endurance in bending and in torsion

σ lim = 0.43 × σ ult - for carbon steels;

σ lim = 0.35 × σ ult + 120 - for alloy steels;

τ lim = (0.2...0.3) × σ ult .

peak, peak – variable (peak) components of bending and torsion stresses

Må

Må

σ max - σ min

σ peak =

= σ max =

=

;

3

2

W

0.1 × d

τ peak

τ max + τ min τ max

T

T

=

=

= 0.5 ×

= 0.5 ×

.

3

2

2

Wp

0.2 × d

## 30. STRENGTH ANALYSIS OF THE SHAFT

Calculation of the shaft for fatigue strengthσ lim

τ lim

Sσ =

;

Sτ =

.

Kσ

Kτ

× σ peakσ+ ψmean

×σ

× τ peakτ+ ψmean

×τ

Kd × KF

Kd × KF

mean, mean– constant (mean) components of bending and torsion stresses

σ max + σ min

σ mean =

= 0;

2

τ max + τ min τ max

T

T

τ mean =

=

= 0.5 ×

= 0.5 ×

.

3

2

2

Wp

0.2 × d

– factors of constant components of bending and torsion stresses

= = - for carbon steels;

= = - for alloy steels.

K – effective stress concentration factors;

K d – scale factor;

K F - surface roughness factor.

## 31. STRENGTH ANALYSIS OF THE SHAFT

The most typical stress concentrations of the shaft• Filleted transition regions;

• Grooves;

• Radial holes;

• Keyed and splined portions;

• Threaded portions;

• Interference fits.

## 32. RIGIDITY ANALYSIS OF THE SHAFT

Flexural rigidityBasic criteria of flexural rigidity are:

• Maximum deflection (sag) y of the shaft;

•Angle of rotation of support sections.

Flexural rigidity conditions

y £ y ;

θ £ θ ,

where

[y] is the maximum safe sag; is the maximum safe angle of

rotation.

[y]= 0.01m – for shafts of spur gears and worm gear drives;

[y]= 0.005m – for shafts of bevel gear, hypoid gear and hourglass

worm gear drives;

[y]= (0.0002…0.0003)l – for general purpose shafts used in machine

tools;

[ ]= 0.001 rad – for shafts mounted in sliding contact bearings;

[ ]= 0.005 rad – for shafts mounted in radial ball bearings.

## 33. RIGIDITY ANALYSIS OF THE SHAFT

Flexural rigidityF × a × b × (l + b)

;

6× E ×J ×l

F × a × b × (l + a)

θB =

;

6× E ×J ×l

θC = θB ;

θA =

F × b × (l 2 - b 2 - 3d 2 )

θD =

;

6× E ×J ×l

F × a × (l 2 - a 2 - 3e 2 )

θE =

;

6×E ×J ×l

θH =

F × a × b × (b - a )

.

3× E ×J ×l

F × a × b × c × (l + a)

;

6× E ×J ×l

F × b × d × (l 2 - b 2 - d 2 )

yD =

;

6× E ×J ×l

F × a × e × (l 2 - a 2 - e 2 )

yE =

;

6× E ×J ×l

yC =

F × a 2 × b2

yH =

.

3× E ×J ×l

E is modulus of elasticity of the shaft material; J is centroidal moment of inertia.

## 34. RIGIDITY ANALYSIS OF THE SHAFT

Flexural rigidityθA =

F1 × c × l

;

6×E×J

θB =

F1 × c × l

;

3× E ×J

θC =

F1 × c × (2 × l + 3 × c)

;

6×E×J

F1 × c × (3 × d 2 + l 2 )

θD =

;

6× E ×J ×l

F1 × c 2 × (l + c)

yC =

;

3× E ×J

F1 × c × d × (l 2 + d 2 )

yD =

.

6× E ×J ×l

## 35. RIGIDITY ANALYSIS OF THE SHAFT

Torsional rigidityBasic criterion of torsional rigidity is the angle of twist.

Torsional rigidity condition

j £ j ,

where [j is the maximum safe angle of twist.

T ×l

j=

,

G × Jp

where T is torque; l is length of the shaft; G is shear modulus;

Jp= pd4/32 is polar moment of inertia.

## 36. CALCULATION OF THE SHAFT FOR OSCILLATIONS

G- static deflection;

lst =

c

Fc = mω

× 2 (y

× + e);

Fel = c × y;

Fc = Fel ;

m × w 2 × ( y + e ) = c × y;

m ×w 2 × e

y=

=

2

c - m ×w

e

c

m ×w 2

- dynamic

deflection

-1

m × w 2 = c - condition of resonance.

w cr =

c

m

- critical angular velocity.

## 37. CALCULATION OF THE SHAFT FOR OSCILLATIONS

p ×n30

30

c 30

c × g 30

g

w=

Þ ncr =

× w cr =

×

=

×

=

×

;

30

p

p

m p

m×g p

lst

30

g

- critical rotational speed,

ncr =

×

p

lst

where

g = 9.81 m / sec 2 - free fall acceleration;

G

lst =

- static deflection;

c

48 × E × J

c=

- rigidity of the shaft;

3

L

E - modulus of elasticity of the shaft material;

L - distance between shaft supports;

p × d 4 - shaft moment of inertia.

J=

64

## 38. CALCULATION OF THE SHAFT FOR OSCILLATIONS

if n £ 0.7 × ncr - rigid shafts;if n ³ 1.2 × ncr - flexible shafts.

y=

e

c

-1

2

m ×w

.

if w ® ¥ ,

y ® - e.

In this case we deal with shaft self-centering.