Shafts and Shafts Components

Shafts and Shafts Components

Hussein BasmaLebanese American University

Please note that this presentation is totally based on “Shigley’s Mechanical Engineering Design Book” as the only used reference including all the definitions, examples and details.

Introduction about shafts

Shaft materials.

Shaft layout

Shaft design for stress

Deflection consideration

Critical Speeds for shafts

Miscellaneous Shaft Components

Limits and fits

Outline

Introduction

Shafts are different than axels.

Shafts are rotating members, usually of circular cross section.

Axels are nonrotating members analyzed as static beams.

Shafts are used to transmit power or motion.

They are essential part in many machine designs.

Material selection.

Geometric layout.

Stress and strength

static strength

fatigue strength.

Deflection and rigidity

Vibration due to natural frequency

What to examine about shafts?

Shaft Materials

Deflection is affected by stiffness not strength.

Stiffness is related to the modulus of elasticity, thus the

material.

Modulus of elasticity is constant for all steels, thus material

decisions will not control rigidity.

Rigidity will be controlled only by geometric decisions.

Shaft Layout

The design of the shafts does not follow a certain rule.

It depends mainly on the application.

Certain conventions to follow in shaft design:

-Avoid long shafts to minimize deflection

-Support the loads between bearings

-Avoid using more than two bearings.

Shaft Layout

In many shaft applications the aim is to transmit torque from

one gear to another gear or pulley.

The shaft must be sized to support the torsional stress and

deflection.

To transmit this torque certain torque-transfer elements are

used such as:

Torque transmission

Keys

splines

setscrews

pins

Tapered fits

Shrink fits

Shaft Design for Stress

Locate the critical locations along the shaft.

Critical locations are on the outer surface where bending

moment is large, torque is present and stress concentrations

exist.

Bending moment is determined by shear and bending

moment diagrams. (1 or more planes could be needed)

A steady bending moment will produce a completely

reversed moment on rotating shaft (σ𝑚 =0)

Axial stresses can be neglected.

Critical Locations


Shaft stresses

We will deal with bending, torsion and axial stresses as

midrange and alternating components.

For analysis it is appropriate to combine the different stresses

into alternating and midrange von Mises stresses

(again we can neglect axial stresses):


P.S: For ductile materials the use of the stress-concentration

factors is sometimes optional.


Now we can take the acquired von Mises stresses and use

these values in any of the known fatigue failure criteria to

evaluate the factor of safety n or, for design purposes, the

diameter d.

Reminder of the fatigue failure criteria:

DE-Goodman

DE-Gerber

DE-ASME Elliptic

DE-Soderberg


DE-Goodman

Plugging in the values of the von Mises stresses will yield the

following equations, one solved for n and one for d :

DE-Gerber

DE-ASME Elliptic

DE-Soderberg

For rotating shafts with constant bending and torsion, the

bending stress is completely reversed and the torsion is

steady. ( )

We should always consider the possibility of static failure in

the first load cycle. ( check for yielding)

The ASME Elliptic criteria takes yield into account but it is

not entirely conservative.


For this purpose a von Mises maximum stress is calculated:


To check for yielding, we compare the maximum stress to

the yield strength:


Example:

At a machined shaft shoulder the small diameter d is 28 mm, the

large diameter D is 42 mm, and the fillet radius is 2.8 mm. The

bending moment is 142.4 N.m and the steady torsion moment is

124.3 N.m. The heat-treated steel shaft has an ultimate strength of

𝑆𝑢𝑡=735 MPa and a yield strength of 𝑆𝑦 = 574 MPa. Our

reliability goal is 0.99.

(a) Determine the fatigue factor of safety of the design using each

of the fatigue failure criteria described in this section.

(b) Determine the yielding factor of safety.


At a machined shaft shoulder the small diameter d is 28

mm, the large diameter D is 42 mm, and the fillet radius

is 2.8 mm. The bending moment is 142.4 N.m and the

steady torsion moment is 124.3 N.m. The heat-treated

steel shaft has an ultimate strength of 𝑆𝑢𝑡 =735 MPa and

a yield strength of 𝑆𝑦 = 574 MPa. Our reliability goal is

0.99


(a) D/d = 42/28 = 1.50, r/d 2.8/28= 0.10, 𝐾𝑡 = 1.68 (Fig. A–15–9).

𝐾𝑡𝑠 = 1.42 (Fig. A–15–8), q = 0.85 (Fig. 6–20), 𝑞𝑠ℎ𝑒𝑎𝑟 = 0.92

(Fig. 6–21).

From Eq. (6–32), 𝐾𝑓 = 1 + 0.85(1.68 − 1) = 1.58

𝐾𝑓𝑠 = 1 + 0.92(1.42 − 1) = 1.39

Eq. (6–8) : 𝑆𝑒’ = 0.5(735) = 367.5 MPa

Eq. (6–19): 𝑘𝑎 =4.51(735)−0.265

Eq. (6–20): 𝑘𝑏 =(28

7.62)−0.107

𝑘𝑐= 𝑘𝑑 = 𝑘𝑓 =1

Solution:


Solution:

Table 6–6: 𝑘𝑒 = 0.814

𝑆𝑒 =0.787(0.870)0.814(367.5) = 205MPa

For a rotating shaft, the constant bending moment will create a

completely reversed

bending stress.

𝑀𝑎= 142.4 N.m 𝑇𝑚 = 124.3 N.m 𝑀𝑚= 𝑇𝑎= 0

Applying Eq. (7–7) for the DE-Goodman criteria gives:

1

𝑛=

16

𝜋(0.028)3{[4 1.58× 142.4 2]

12

205× 106+[3 1.39× 124.3 2]

12

735× 106}= 0.615

n=1.62

Similarly, plugging the values of the mean and amplitude of

the torque and bending moment in the equations of the different

failure criteria, we get:

P.S: Note that the safety factor calculated using Soderberg

relation is the smallest one since this criteria is the most

conservative one.

(b) To calculate the yielding safety factor, we use:

σ′max= [32×1.58×142.4

𝜋 0.028 3

2+ 3

16×1.39×124.3

𝜋 0.028 3

2]1

2 = 125.4 MPa

𝑛𝑦 =574

125.4= 4.85


The stress concentration zones are mainly due to bearings,

gears fillets and change in shaft geometry or radius.

The stress concentration factor 𝐾𝑡depends on the type of

bearing or gear.

Figs. A-15-16 and A-15-17 give values of the concentration

factors of flat-bottomed grooves

Table 7-1 gives the typical stress-concentration factors for the

first iteration in the design of a shaft.

Estimating Stress Concentration:

Deflection Consideration

As we mentioned earlier, we need to know the complete shaft

geometry before we can perform deflection analysis.

Therefore we should design the dimensions at critical locations

to handle the stresses and estimate the other dimensions.

Once the complete geometry is known, we can perform the

deflection analysis.

Deflection should be checked at gears and bearing supports.

Chapter 4 deals with the methods used to calculate deflection,

including singularity functions and numerical integration.


For certain shafts of complex geometries we may need three

dimensional analysis and then use superposition to calculate the

overall deflection.

Deflection analysis is straight forward, but it incorporates long

procedures and tedious steps.

This can be simplified by using certain three dimensional

deflection analysis software.


Table 7-2 includes the maximum allowable deflection for

different types of bearings and gears.


Where 𝑦𝑎𝑙𝑙 is the allowable deflection and 𝑛𝑑 is the design

safety factor.

After calculating the deflection and slope in the shaft, we

compare it with the allowable values provided by the different

tables.

Sometimes, if the calculated deflection is greater than the

allowable values, changes in dimensions must be considered

and the diameter of the shaft must be changed according to this

equation:

The preceding equation can be applied on the slope as well:


Where (𝑠𝑙𝑜𝑝𝑒)𝑎𝑙𝑙 is the allowable slope.


Shearing deflections are usually 1% of the bending deflections

and are usually ignored.

In cases of short shafts, where 𝑙 𝑑 <10 , shearing deflections

become important.

The angular deflection is represented by:

For constant torque through a homogeneous material:

Critical Speeds of Shafts

A common problem encountered when dealing with rotating

shafts is the problem of critical speeds.

At certain speeds, when the frequency of rotation becomes close

to the shaft’s natural frequency, the shaft becomes unstable with

increasing deflections that may lead to failure.

For simple geometries, as a simply supported, constant-diameter

shaft, the first critical speed can be estimated using the following

equation:

where m the mass per unit length, γ is the specific weight.


For an ensemble of attachments, Rayleigh’s method for lumped

masses gives:

where 𝑤𝑖 is the weight of the ith location and 𝑦𝑖 is the

deflection at the ith body

At this stage, we will define something called the influence

coefficients.

An Influence coefficient is the transverse deflection at location i

on a shaft due to a unit load at location j on the shaft.

For a simply supported beams with a single unit load, the

influence coefficients are given by:




Assume three loads, the influence coefficients can be expressed:

From the influence coefficients we can calculate the deflections

𝑦1, 𝑦2 and 𝑦3 as follows:

The force 𝐹𝑖 arises from weights attached or centrifugal forces

𝑚𝑖 ω2𝑦𝑖. The equation can be expressed as:


Solving for the roots of this system will yield the following

relation:


If we had one mass 𝑚1, the critical speed denoted by ω11 will

be expressed as1

ω112 = 𝑚1𝛿11

Similarly,1

ω222 = 𝑚2𝛿22 and

1

ω332 = 𝑚3𝛿33

Then the equation can be expressed as:

1

ω12 ≫

1

ω22 +

1

ω32 since ω1 is way smaller than ω2 and ω3


This will simplify to:

The preceding equation can be extended to n-body shafts:

This equation is known as the Dunkerley’s equation.

Principle of superposition:


This principle includes calculating an equivalent load placed on

the center of the shaft. In other words, we transform each load

into an equivalent load placed at the center denoted by the

subscript c.

This equivalent load can be found from:


The critical speed ω 1 can be calculated after summing the

equivalent load of each load on the shaft:


Example:

Consider a simply supported steel shaft , with 25 mm diameter

and a 775 mm span between bearings, carrying two gears weighing

175 and 275 N.

(a) Find the influence coefficients.

(b) Find 𝑤𝑦 and 𝑤𝑦2 and the first critical speed using

Rayleigh’s equation.

(c) From the influence coefficients, find ω11 and ω22.

(d) Using Dunkerley’s equation, estimate the first critical speed.

(e) Use superposition to estimate the first critical speed.


Solution:

a) I = πd4

64=

π(25)4

64= 19175 𝑚𝑚4

6E Il = 6(207000)(19175)(775)= 18.5× 1012 N.𝑚𝑚3


δ11=600(175)(7752−6002−1752)

18.5×1012= 0.00119 𝑚𝑚

𝑁

δ12= δ21=275(175)(7752−2752−1752)

18.5×1012= 0.00129 𝑚𝑚

𝑁

δ22=275(500)(7752−2752−5002)

18.5×1012= 0.00204 𝑚𝑚

𝑁

b) 𝑦1= 𝑤1δ11+ 𝑤2δ12 = (175)(0.00119) + (275)(0.00129)= 0.56 mm

𝑦2= 𝑤1δ21+ 𝑤2δ22 = (175)(0.00129) + (275)(0.00204)= 0.79 mm


b) 𝑤𝑦 = 175(0.56)+275(0.79) = 315.3 N.mm

𝑤𝑦2= 175 (0.56)2+ 275(0.79)2= 226.5 N.𝑚𝑚2

Using Rayleigh’s equation,

ω = 9.81(315.3)×10−3

226.5×10−6= 117 rad/s

c)1

ω112 = 𝑚1𝛿11, and 𝑚1=

𝑤1

𝑔, then:

1

ω112 =

𝑤1

𝑔𝛿11=

175×0.00119×10−3

9.81

1

ω222 = 𝑚2𝛿22, and 𝑚2=

𝑤2

𝑔, then:

1

ω222 =

𝑤2

𝑔𝛿22=

275×0.00204×10−3

9.81

ω11= 217 rad/s

ω22= 132 rad/s

d) Using Dunkerley’s equation:


1

ω12 =

1

2172+

1

1322= 7.863× 10−5

Implies, ω1=113 rad/s

e) Using superposition:

δ𝑐𝑐=b𝑐x𝑐(𝑙2−b𝑐𝑐

2−x𝑐𝑐2)

18.5×1012

=387.5 387.5 7752−387.52−387.52

18.5×1012

= 0.00244 mm/N

𝑤1𝑐 =175(0.00119)

0.00244= 85.3 N

𝑤2𝑐 =275(0.00204)

0.00244= 229.9 N

Then, ω =𝑔

δ𝑐𝑐 𝑤𝑖𝑐=

9.81

0.00244(85.3+229.9)10−3= 112.9 𝑟𝑎𝑑/𝑠



Method Rayleigh Dunkerly Superposition

First Critical Speed 117 rad/sec 113 rad/sec 112.9 rad / sec

Since designers seek first critical speed at least twice the operating speed, so the difference has no effect on our design.

The critical speed calculated by Dunkerley and Superposition methods are always expected to be less than Rayleigh’s method because they neglected the effect of the other critical speeds