Shaft Design appendix

1

Dr Andrei Lozzi Machine Design & CAD

Mech Design II

School of Aerospace Mechanical & Mechatronic Engineering

Shafts and Axles Shaft lecture AL 2016 1

Refs: R L Norton, Machine Design …, Prentice Hall.

N Reshetov, Machine Design, MIR Pub, Moscow.

J E Shigley et al, Mechanical Machine Design, ed 4 to 7, McGraw Hill.

1 Introduction Shafts and axles have very wide range of applications, consequently although there are

specific similarities, they can also vary widely in appearances. Figure 1 gives an idea of what they can

look like. There is a categorical difference between shafts and axles, it is that shafts rotate about their

long axis, whereas axle remain stationary and provide a pivot for other components to rotate on.

At times axles have been considered as just structural members and the methods of analysis that has

been developed for shafts have not been applied to them. Yet the effective difference between typical

examples of each, is that axles may be just subjected to fewer load cycles than shafts. The analysis then

can be the same for both, and in these notes the name ‘shaft’ will stand for both items, unless for some

particular reason the need arises to differentiate between them.

Fig 1. a) Plain constant diameter transmission shaft, with or without keyways. b) & c) Machined

stepped shafts for machine tools or gear boxes, with screw threads, keyways and ground lands to

possibly carry bearings or gears. d) A large steam turbine shaft, e) crankshaft, f) rotating railways shaft,

often referred to as an ‘axle’. g) Fixed axle as used on heavy trucks, trailers and aircraft landing gear.

Shaft Design appendix

2

F1

F2

F3 R1

R2

R3

a b c d

2 An Overview of the analysis. There are two objectives to be kept in mind in the analysis of most

shafts, The stress levels have to kept sufficiently low to provide the required service life, and the

various deflections have to be controlled to ensure that the shaft can function as expected. The loads

that usually produce the greatest stresses and strains in shafts are moments, caused by transverse forces

and reactions on the shaft, and the torque transmitted by the shaft. Secondary loads may be longitudinal

trust forces down the along axis of a shaft and contact stresses caused possibly by interference fits. Both

the maximum allowable stresses and maximum allowable deflections are used separately and

independently to determine shaft diameters. It is common that deflections requirements dictated larger

shaft diameters than the diameters required to give adequate fatigue life.

Fig 2 Above is almost a caricature of a loaded shaft. At station a there is a sprocket carrying the

tension F1 from a chain. Bearings at b and d locate the shaft but they fix the shaft at only those stations.

The helical gear at c transmits a transverse force F2 and an axial force F3 to the shaft. The shaft is of

greater diameter between bearings to reduce deflections there. The bending moments nearly completely

determine the lateral deflections at the gear and angular deflection θ at the bearing d. Note that single bearings should only be subjected to combinations of transverse and axial forces, no bending moment.

3 Stresses and strains. Stress analysis determines shaft diameters at one location at the time. It uses

moments, torques, surface finish, stress concentration and others factors, all of which must apply to the

location of interest, to determine the diameter of the shaft only at that location. In contrast deflections

analysis requires the overall shape and size of the shaft, as well as all the loads on it, be known

before the deflections of the shaft can be calculated. Hence the normal procedure is to use stress analysis

to determine the minimum diameter of the shaft at critical locations, followed by some projections as to a

reasonable shape everywhere else. The deflections of the proposed shaft are then calculated, again at one

location at the time. Finally and obviously, the size of the shaft everywhere along its length has to be at

least equal to the larger of the two diameters dictated by the stress and deflection analysis.

4 An example. In Fig 2, deflection analysis determined the minimum shaft diameter between b and

d to control the lateral and angular θc (shown but not labeled) deflection at the gear wheel c. Gear teeth have to make precise contact with each other or much of their capacity is wasted. Stress analysis

determined the minimum shaft diameters at the bearings, since the extra cost of larger bearings can

be significant, the shaft is shown machined down at the ends. Angular deflection at the bearings had to be checked, because many bearings will tolerate very little misalignment before failing.

Shaft Design appendix

3

x

xy(T)

X

Y

Z

O

F

T

x=0 x=x1

xy(F)

F

Shown in Fig 3 (and 4) is a short section of a shaft where the stresses at station x=0 are considered from

a force F and a torque T applied at x=x1. The force causes in part a moment about the Z axis, with its

associated normal stress x. We well know that x varies linearly between maxima at y=D/2 and 0 at

the Z axis. The torque T causes a shear stress xy(T) which is a maximum on the surface and points tangentially to the circumference. We know that:

I

DFx

x21 Eq 1

J

DTTxy

2)( Eq 2

Fig 3. At left, the stresses at x=0

are shown from loads applied to a

shaft at x=x1. One stress is

omitted from Fig3 for clarity, but

is shown by itself in Fig 4.

Fig 4. At right is the same piece of

shaft as shown above, but here the

transverse shear stress associated

with force F, xy(F) is shown. This stress reaches a maximum

on the neutral axis (Z axis) and

zero at y=D/2, in opposition to

the variation in xy.

5 Transverse Shear stress. This last stress xy(F), is nearly always omitted from any consideration, but it is worthwhile to see why this is so, and perhaps when it may not be so. We can gauge the relative

significance of this stress by comparing its magnitude to the magnitude of x. The transverse shear stress is a maximum on the neutral axis. For a circular section this is given as:

A

Fxy

3

4 , (where

4

2DA ) Eq 3

Since the bending moment, and its associated normal stress increases with x, and the transverse shear

stress remains constant, we can ask for what value of x1 will the two stresses, xy(F) and x be of equal magnitude, ie:

Shaft Design appendix

4

x

O

xy

xy

1 2

)(Fxyxy Eq 4

43

4

64

224

1

D

F

D

DFx

Eq5

that is where: 6

1

Dx , ie x1 is 0.17 of the shaft diameter! Eq 6

x is 10 times larger than xy(F) where x1 = 1.7 D, and 100 time where x1 = 17 D. Not only do transverse shear stresses go from their maximum to 0 exactly where normal stresses due to bending go

from 0 to their maximum, but also these shear stresses are very small except for shafts with exceedingly

short separation between loads and bearings. Proportionally transverse shear stress contribute more to

total deflection than they do to total stress, but usually by less than a few % points.

6 Stresses in Shafts. Given that the highest coexisting biaxial stresses in Fig 3 are: x and xy(T). These stresses are shown on a Mohr’s circle in Fig 5.

Fig 5. Mohr’s circle of the significant stresses

in a shaft caused by the shear force and torque

shown on Fig 3.

Giving the biaxial principal stresses:

21

2

2

2,122

xy

xx

Eq 7

These are substituted in the von Mises biaxial expression:

2

221

2

1

2' Eq 8

giving: 222' 3 xyx Eq 9

Since the force and torque may in general be made up of steady and alternating components, (Fm, Fa) and

(Tm, Ta), we can calculate the mean and alternating distortion energy stresses starting with Eq 1 & 2 and

getting:

222'

3 xymxmm Eq 10

222'

3 xyaxaa Eq 11

If there are mean and alternating axial forces on the shaft, the stresses resulting from those forces can be

added to the normal stresses in Eq 10 & 11.

7 The Goodman option. Substituting Eqs 1 & 2, with the appropriate expansions for I & J and using

the variable M= Fx1 for the moment, instead of F & x1 separately, into Eq 10 & 11. Then, substituting

these expanded von Mises stresses into the Goodman equation:

Shaft Design appendix

5

u

m

f

a

SSFS

''1

Eq 12

Rearranging the resulting expanded Goodman relation, to give diameter D for a required factor of safety

FS, after some time gives:

3

1

21

222

1

22

4

3

4

3

32

ut

mfsmmfm

f

afsaf

S

TkMk

S

TkMkFS

D

Eq 13

Where: Ma/m - moment, alternating or mean

Ta/m - torque, alternating or mean

kf/,fm - stress concentration factor for alternating or mean moment

kfs/,sfm - stress concentration factor for alternating or mean torque

Sf, Su - fatigue and ultimate tensile strength

If the average stress, or nominal stress, is below yield condition, then a stress concentration factor applies

to the mean stress (see Norton p385)

if: av kf < Sy, then: kfm = kf Eq 14

This is done on the grounds that in the stress concentration zone the stress cannot exceed the yield stress.

Thus a stress concentration factor can only apply if yielding does not occur. If yielding occurs in the

whole load cycle:

kfm = 1 Eq 15

Then the stress concentration factor can have no effect as at no time in the cycle will the stress be below

yield level.

The advantage of the Goodman condition is that substituting equations 1, 2 , 8, 9, 10, 11 into the linear

equation 12, gives an expression that can be rearranged to give shaft diameters directly, ie Eq 13.

8 The Gerber option. The principal drawback to the using of the Goodman line is that it does not

follow the experimental results at all well, for fatigue failures under combined mean and alternating

stresses. Some experimental data is shown in Fig 6, note that this straight line is not a good fit to the

distortion energy stresses at failure, ie the curved line above it. The Goodman line understates the

available strength by a variable amount, from about 20% to 50%. This may be too much to give

away to competitors whom are prepared to go the extra flops (floating point operations).

Fig 6. Experimental data for steel

specimen showing the Gerber

parabola and the Goodman line,

for tests between 106 to 108 cycles.

The Gerber curve and the ASME

ellipse both represent the average

values reasonably well,

Shaft Design appendix

6

An impediment to the use of the Gerber equation is that substituting equations 1, 2 , 8, 9, 10, 11 into the

Gerber parabola Eq 16, raises the expressions of the right hand sides of Eq 10 & 11 to the 2nd and 4th powers before the square root sign is lost. Extracting the shaft diameter as the dependent variable

becomes a little messy but it can look like this:

1

2''

u

m

e

a

SFS

S

FS Eq 16

letting: 222 )(4

3)( eafsaf STkMka and 222 43 umm STMb Eq 17a, b

we get:

21

2

22

6

42

322

aabab

bFSD

Eq 18

We can also use the numerical solver to determine the diameter that for the applied mean and alternating,

moments and torques, gives a point that meets the condition defined by Eq 16. What is required are the

equations mentioned above beginning with Eqs 1 & 2, showing their relationship to functions 8, 9, 10, 11

and finally 16.

9 Shaft material. In the proportions that we usually make shafts, the requirements that they should run

adequately true (ie straight) under load, results in thicker and stiffer shafts over much of their

lengths, than the need for an adequate fatigue life. As a consequence shafts are made from material of

relatively high modulus of elasticity, that is steel. Since this modulus does not change much with alloying

elements, high quality low to medium carbon steel is preferred. Stainless steel is employed where

chemical action is significant. Cast iron is used where the shafts may be small or complex (crankshafts)

and production costs may be significant. Copper and aluminum alloys are employed in rare circumstances

where those materials have particular advantages that outweigh their low modulus.

Small mass-produced shafts are made from cold rolled stock, that comes with a smooth bright finish.

These shafts are prone to warping or distorting when their surface is machined, as residual stresses are

partly relieved, but this stock makes for cheap and rapid production. Hot rolled or hot forged billets are

more commonly used for larger industrial machines. For precision, but usually limited production work,

ground shaft stock is available. This may be obtained in a range standard diameters and in considerable

lengths. It can be in a normalized state, ready for extensive machining, or in a high temper state, if no

more than limited grinding is to be done.

10 Deflection calculations. Beam deflection formulae due to bending, which may be found in

numerous texts on properties of solids, may be used to calculate the deflections of circular shafts. These

formulae deal with point and distributed loads, and a range of constraints at their reactions. Attached to

these notes is an example of such formulae. If a shaft has multiple loads at any location then the relevant

formulae may be used to arrive at the deflections at that location, one load at the time. The final deflection

is then the sum of the individual ones. If the loads are not coplanar it will be necessary to find the vector

sum of the deflections all points of interest along the shaft. Of course all this may be done provided that

the elastic limit is not exceeded.

These formulae typically use Macaulay functions or brackets to describe the discontinuous loads and

reactions very effectively. But there is a significant limitation in their use as they are they apply to shafts

of constant cross sections. That is, if a shaft has significant steps the application of this method requires

some careful considerations.

Shaft Design appendix

7

F

Energy = area

(F)/2

F

l

M

ds

d

11 Castigliano’s Energy method. There is an energy method that more readily deals with stepped

shafts and statically indeterminate designs such as the use of more than 2 bearings. The application of this

method is demonstrated with a tensile force in Figs 7a & 7b. A beam of uniform section shown on Fig 7a,

exhibits elongation when subjected to tension F. Assuming linear force/deflection response the energy U absorbed in that deformation is contained under the response curve (straight line) shown on Fig 7b.

The absorbed energy is:

Extension 2

FU Eq 19

where: AE

Fl Eq 20

We can use Eq 20 to replace either F or

from the expression for U in Eq 19

giving: ,2

2

EA

lFU or

l

AEU

2

2 Eq 21a, b

Fig 7a above, b below.

Note that:

EA

FL

F

U Eq 22

and Fl

AEU

Eq 23

The differential of energy with respect to (wrt) force gives

deformation, and the differential of energy wrt deformation

gives the force. In principle this can be done for any linear

means of absorbing energy. A more general form of Eq 21a,

applicable where the cross sectional area A and or the force

F are not constant:

l

dxAE

FU

0

2

2 Eq 24

Angular deflection

Fig 8. at left shows a bar being bent by moment M

through an angle d, in an arc of radius , over an arc distance ds. Over the infinitesimal distance ds

the energy absorbed is:

2

MddU Eq 25

ds

d , and M

EI Eq 26a, b

dsEI

MdU

2

2

Eq 27

dsEI

MU 2

2

Eq 28

Shaft Design appendix

8

F

x

For the simple case that is where M is constant, ds becomes l and: EI

lMU

2

2

Eq 29

EI

ML

M

U and M

U

Eq 30a, b

Lateral deflection Fig 9. At left is a beam deflected by a force F generating

a moment Fx. The stored energy is:

2

FxU Eq 31

EI

Fx Eq 32

Giving a stored energy in terms of the moment or

deflection:

22

22 EIor

EI

FxU Eq 33a, b

And differentials wrt to force or deflection give:

F

U , and F

U

Eq 34a, b

Over a distance x, if forces and diameters vary:

dxEI

FxU 2

2

Eq 35

12 Overview of Castigliano’s methods: For each driving influence (F, M & Fx) there is a deformation

(, & ) and energy is absorbed. Taking the angular deflection as an example: the rate change of energy wrt to the moment gives the angular deformation, at the point where the moment is applied. Thus for each

of these driving influences we can arrive at the deformation at the point were the force is applied.

The methods works if there are several forces and or moments coexisting, the differential wrt to a

particular moment or force will give the deformation due to all forces and moments, at the point of

application of that particular moment or force. This leads to the use of a dummy force or moment. If we

need to know the deformation at a particular location but there is no applied force there, we define one to

be there, carry out all the analysis, and finally we set that force or moment to 0. The final expression will

give the total deformation due to all other existing forces and moments at the required location. We will

do 2 examples to demonstrate the capacity of this method:

Example of Castigliano’s method

13 Stepped conveyor drive shaft. On Fig 10 we see a simplified drawing of a drive shaft for a

conveyor drive pulley. In the development of these pulleys 2 methods have been adopted, one uses

relatively thin flexible components the other thick stiff ones. The shaft on Fig 10 represents the stiff

approach, for which the angular deflection of the shaft at the keyways is to be limited to a gradient of

1/1200 (or the same amount in radians).

Here the shaft is assumed to be symmetric about the vertical centerline. The bearings provide the

reactions R and the cheeks in the pulley transmit the load F to the shaft at the center of the keyways. In

modern drive pulleys special torque couplings are used in place of keys.

Shaft Design appendix

9

R

F

R

F

m m

a O b l/2

D1 D2

Fig 10. A conveyor drive pulley shaft. The loads F from the pulley are transmitted to the shaft near the

centers of the keyways. At these points a hypothetical moment m is imposed on the shaft. By

differentiating the total deflection energy due to all bending moments, wrt m we can arrive at the angular

deflection at b, due to all the other moments that would coexist with m, were m not =0.

We need : m

U tm

, given : 12 nDD Eq 36a, b

Total deflection energy: 2/,,,02 lbxbaxaxt UUUU Eq 37

2

4

2

4

2

0

2

2222

l

b

b

a

a

t dxEIn

mbxFRxdx

Ein

Rxdx

EI

RxU Eq 38

Note that the integral for deflection energy changes form where there is a change in cross-section or

where a load in encounter. Now there are great simplifications that are applied to the integral Eq 37: We

differentiate Ut wrt m, and then set m=0. That means that the first 2 integrals between the limits: 0a,

and ab are =0 because they are not functions of m. In the last integral we can differentiate under the integral sign because x is not a function of m, and finally after we differentiate, only term which are not

function of m will remain, because those which include m will go to 0 after we set m=0.

2

42

2002

l

b

t dxEIn

mbxFRx

m

U, (m=0) Eq 39

2

42

22

l

b

m dxEIn

bxFRx Eq 40

blEIn

Fbm 24 Eq 41

If we have an indication of the ratio n of the 2 diameters (Eq 36b), we can calculate the diameter of the

smaller shaft D1 from I in Eq 41, and then the larger diameter D2. Thus as mentioned in section 3 on

page 2, we need a reasonable distribution of diameters for the whole shaft (calculated to meet stress

requirements) before we a can estimate deflections.

Shaft Design appendix

10

14 Three bearing shaft. The tables of deflection formulae relate to beams or shaft supported at 2

locations only, more than 2 reactions render the problem statically indeterminable. We can begin to

understand the behavior of multiple bearing shafts using Castigliano’s method. By differentiating the

deflection energy wrt to R2 at the center bearing of Fig 11, we can arrive at the deflection at the center

bearing. For any particular deflection we can calculate the loads at the other 2 bearings.

Fig 11. A 3 bearing shaft with a load between the first 2 bearings. If there is a lateral deflection of at the center bearing, then the load at the other 2 bearings can be arrived at.

Total deflection energy: cbxbaxaxt UUUU ,,,02 Eq 42

a b

b

c

b

t dxbxRaxFxRaxFxRxREI

U0

2

21

2

1

2

12

1 Eq 43

The deflection at b: dxbxbxRaxFxREIR

Uc

b

21

2

22

1 Eq 44

After more than a few hours, rearranging this equation to give the force R2 for a given deflection 2

gives:

3

3

2

2

6

6

bcc

b

bcc

aFEI

R

Eq 45

You may agree with me that the above equation indicates that if there is no deflection at the center

bearing then the reaction at the third bearing R3 will be zero. To arrive at a more balanced view we have

to take into account the stiffness of the 3 bearing mounts, because deflections will always take place.

R1 R2 R3

F f 2

O a b c

Shaft Design appendix

11

Shaft Design appendix

12

Shaft Design appendix

13

Shaft Design appendix

14

Shaft Design appendix

15

Shaft Design appendix

16