Shafts and Shafts Components

Shafts and Shafts Components
Hussein Basma
Lebanese 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

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 (σ 𝑚 =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 :

 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
𝑆 𝑢𝑡 =735 MPa and a yield strength of 𝑆 𝑦 = 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 𝑆 𝑢𝑡 =735 MPa and
a yield strength of 𝑆 𝑦 = 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), 𝑞 𝑠ℎ𝑒𝑎𝑟 = 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) : 𝑆 𝑒’ = 0.5(735) = 367.5 MPa
Eq. (6–19): 𝑘 𝑎 =4.51(735)−0.265
Eq. (6–20): 𝑘 𝑏 =(
28
7.62
)−0.107
𝑘 𝑐= 𝑘 𝑑 = 𝑘 𝑓 =1
 Solution:

 Solution:
Table 6–6: 𝑘 𝑒 = 0.814
𝑆 𝑒 =0.787(0.870)0.814(367.5) = 205MPa
For a rotating shaft, the constant bending moment will create a
completely reversed
bending stress.
𝑀 𝑎= 142.4 N. m 𝑇 𝑚 = 124.3 N.m 𝑀 𝑚= 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]
1
2
205× 106 +
[3 1.39× 124.3 2]
1
2
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
𝑛 𝑦 =
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 𝑦 𝑎𝑙𝑙 is the allowable deflection and 𝑛 𝑑 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 ω2 𝑦𝑖. 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 as
1
ω11
2 = 𝑚1 𝛿11
 Similarly,
1
ω22
2 = 𝑚2 𝛿22 and
1
ω33
2 = 𝑚3 𝛿33
 Then the equation can be expressed as:

1
ω1
2 ≫
1
ω2
2 +
1
ω3
2 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
ω11
2 = 𝑚1 𝛿11, and 𝑚1=
𝑤1
𝑔
, then:
1
ω11
2 =
𝑤1
𝑔
𝛿11=
175×0.00119×10−3
9.81
1
ω22
2 = 𝑚2 𝛿22, and 𝑚2=
𝑤2
𝑔
, then:
1
ω22
2 =
𝑤2
𝑔
𝛿22=
275×0.00204×10−3
9.81
ω11= 217 rad/s
ω22= 132 rad/s

d) Using Dunkerley’s equation:
1
ω1
2 =
1
2172 +
1
1322 = 7.863× 10−5
Implies, ω1=113 rad/s
e) Using superposition:
δ 𝑐𝑐=
b𝑐x 𝑐(𝑙2−b𝑐𝑐
2
−x 𝑐𝑐
2
)
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
δ 𝑐𝑐 𝑤 𝑖𝑐
=
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

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