Design of Shafts and Shaft Elements
Ratna Kumar Annabattula
208, Machine Design section
Department of Mechanical Engineering
IIT Madras
Office hours: Wednesday, 3:00 p.m. - 4:00 p.m.
email: ratna@iitm.ac.in
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Outline I
1 Shaft Basics
Shaft Materials
Shaft Layout
2 Shaft Loads
Shaft stresses
Stress concentration
Shaft Failure
3 Shaft Design
General guidelines
ASME shaft design
Equivalent stress in shafts
Design based on Goodman theory
Example Problem
Design data
Design of a shaft for deflection
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Outline II
4 Design of Keys
Types of Keys
Stresses in Keys
Key Materials
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Shaft Basics
Shaft Basics
What is a shaft?
A shaft usually is a rotating element which transmits torque from a
driving device
Shafts usually carry other machine elements such as gears,
pulleys, sheaves or sprockets
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Shaft Basics Shaft Materials
Shaft Materials
Materials used for shafts
Many shafts are made from low-carbon, cold-drawn or hot-rolled
steel such as AISI 1020-1050 steels
Significant strengthening and surface hardening is usually not
required
When used as journals, they need to be surface hardened
(Carburizing grades: AISI 1020, 4320, 4820, 8620)
Cold drawn steels for diameters less than 3 inches and need not
be machined all over
Hot-rolled steels should be machined all over
Cast-iron may be used in cases where the gears and shafts are
required to be cast together
In highly corrosive environment, stainless steel may be used
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Shaft Basics Shaft Layout
Shaft Layout
Figure 1 : Elements of a shaft assembly (source: Norton’s book)
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Shaft Basics Shaft Layout
Stress raisers
Various components on the
shaft assembly
Bearings
Clamp Collar
Key
Snap ring
Taper pin
Assembly of components on to shafts
Press and shrink fits
Set screws
Ease of assembly and disassembly for maintenance
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Shaft Loads
Shaft loads
Nature of loads on shafts
Fluctuating torque and fluctuating moment together with axial
loads
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Shaft Loads
Shaft loads
Nature of loads on shafts
Fluctuating torque and fluctuating moment together with axial
loads
A general case of complex multiaxial stress state
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Shaft Loads
Shaft loads
Nature of loads on shafts
Fluctuating torque and fluctuating moment together with axial
loads
A general case of complex multiaxial stress state
Simple multiaxial stress state: direction of principal alternating
stress remains constant
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Shaft Loads
Shaft loads
Nature of loads on shafts
Fluctuating torque and fluctuating moment together with axial
loads
A general case of complex multiaxial stress state
Simple multiaxial stress state: direction of principal alternating
stress remains constant
Most rotating shafts are loaded in bending and torsion
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Shaft Loads
Shaft loads
Nature of loads on shafts
Fluctuating torque and fluctuating moment together with axial
loads
A general case of complex multiaxial stress state
Simple multiaxial stress state: direction of principal alternating
stress remains constant
Most rotating shafts are loaded in bending and torsion
We will consider the case of fluctuating bending moment and
constant torque in this class, i.e., simple multiaxial case
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Shaft Loads Shaft stresses
Shaft stresses I
Normal stresses Shear stresses
Ma y Mm y Ta r Tm r
σa = kf , σm = kfm τa = kfs , τm = kfsm
I I J J
For a solid circular shaft, y = d/2 For a solid circular shaft, r = d/2
and I = πd 4 /64. and J = πd 4 /32.
32Ma 32Mm 16Ta 17Tm
σa = kf , σm = kfm τa = kfs , τm = kfsm
πd 3 πd 3 πd 3 πd 3
kf and kfm are bending fatigue kfs and kfsm are torsional fatigue
stress concentration factors for stress concentration factors for
the alternating and mean the alternating and mean
components. components.
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Shaft Loads Shaft stresses
Shaft stresses II
Axial stresses
A tensile axial stress: usually a mean component only
Fz 4Fz
σmaxial = kfm = kfm 2
A πd
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Shaft Loads Stress concentration
Stress-concentration effects
Stress-concentration with fluctuating stresses
The alternating component should be treated the same way as the
fully reversed case. The local alternating stress:
σa = kf σanom , kf = 1 + q(kt − 1)
The mean stress σm is treated based on the ductility (or
brittleness) of the material
If ductile, then amount of yielding possible at the notch should be
considered
If brittle, full value of geometric stress concentration factor (kt ) is
used as kf .
The mean fatigue stress concentration factor (kfm ) depends on the
level of mean stress σm at the notch compared to the yield
strength
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Shaft Loads Stress concentration
Mean fatigue stress concentration factor I
Relationships for mean fatigue stress concentration factor
If kf |σmaxnom | < Sy , then : kfm = kf
Sy − kf σanom
If kf |σmaxnom | > Sy , then : kfm =
|σmnom |
If kf |σmaxnom − σminnom | > 2Sy , then : kfm = 0
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Shaft Loads Stress concentration
Mean fatigue stress concentration factor II
How do you decide on kfm ?
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Shaft Loads Stress concentration
Mean fatigue stress concentration factor
Evolution of kfm as a function of σmaxnom
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Shaft Loads Shaft Failure
Shaft failure in combined loading
Combined
torsion and
bending of
ductile
materials in
fatigue is found
to follow elliptic
relationship.
Gough ellipse
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Shaft Design
Shaft design
Key points J
Shafts should be designed for both stress and deflection
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Shaft Design
Shaft design
Key points J
Shafts should be designed for both stress and deflection
Stresses can be calculated at various locations by knowing the
approximate cross section at the point
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Shaft Design
Shaft design
Key points J
Shafts should be designed for both stress and deflection
Stresses can be calculated at various locations by knowing the
approximate cross section at the point
Deflection calculation requires the knowledge of the entire
geometry
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Shaft Design
Shaft design
Key points J
Shafts should be designed for both stress and deflection
Stresses can be calculated at various locations by knowing the
approximate cross section at the point
Deflection calculation requires the knowledge of the entire
geometry
Hence, the shafts are first designed for stress and then for
deflection
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
Avoid overhangs
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
Avoid overhangs
Locate stress raisers away from regions of large bending moment
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
Avoid overhangs
Locate stress raisers away from regions of large bending moment
Deflections at gear mount positions should not exceed 0.005
inches
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
Avoid overhangs
Locate stress raisers away from regions of large bending moment
Deflections at gear mount positions should not exceed 0.005
inches
Operational/forcing frequency of the shaft should be far away (10
times smaller) from its natural frequency
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Shaft Design General guidelines
Shaft design
General guidelines
Minimize deflection and stress: the length should be kept as short
as possible
Avoid overhangs
Locate stress raisers away from regions of large bending moment
Deflections at gear mount positions should not exceed 0.005
inches
Operational/forcing frequency of the shaft should be far away (10
times smaller) from its natural frequency
Refer to Norton’s text book on several other guidelines.
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Shaft Design ASME shaft design
ASME shaft design I
ASME elliptic failure locus equation J
Assumption: Reversed bending and steady torsion
σa 2 τm 2
n + n =1
Se τy
√
Note that τy = Sy / 3 and hence,
σa 2 √ τm 2
n + n 3 =1
Se Sy
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Shaft Design ASME shaft design
ASME shaft design II
AMSE (elliptic) shaft diameter J
1/3
s
kf M a 2 kfsm Tm 2
16n
d= 4 + +3
π Se Sy
Note J
For a rotating shaft with constant bending moment and torsion, the
bending stress is completely reversed and the torsion is steady. In that
case, the above equation can be simplified by setting Mm = 0 and
Ta = 0.
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Shaft Design Equivalent stress in shafts
Equivalent or von-Mises stress due to combined
loading
" 2 2 #1/2
1/2 32kf Ma 16kfs Ta
σae = σa2 + 3τa2 = +3
πd 3 πd 3
" 2 2 #1/2
e
2
1/2 32kfm Mm 16kfsm Tm
σm = σm + 3τm2 = +3
πd 3 πd 3
Note:
e should have contribution from σ
Ideally, σm maxial also which is neglected
in the above equation.
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Shaft Design Design based on Goodman theory
Design based on Goodman diagram
J
Goodman line
1 σe σe
= a + m
n Se Sut
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Shaft Design Design based on Goodman theory
Design based on Goodman diagram
Goodman line J
1 σe σe
= a + m
n Se Sut
Substituting the expressions for σae and σm
e from the previous slide and
rearranging the terms, we get
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Shaft Design Design based on Goodman theory
Design based on Goodman diagram
Goodman line J
1 σe σe
= a + m
n Se Sut
Substituting the expressions for σae and σm
e from the previous slide and
rearranging the terms, we get
Diameter of the solid shaft J
q
16n 1
d= 4(kf Ma )2 + 3(kfs Ta )2 +
π Se
1/3
1
q
2
4(kfm Mm ) + 3(kfsm Tm )2
Sut
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Shaft Design Design based on Goodman theory
Points to be taken care I
Points of concern Fatigue Failure Surfaces
Possibility of yielding in
first loading cycle
should be checked
Soderberg and ASME
takes care of yielding to
some extent
Gerber and Modfied
Goodman do not guard
against yielding
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Shaft Design Design based on Goodman theory
Points to be taken care II
J
Von-mises maximum stress:
q
e
σmax = (σm + σa )2 + 3(τm + τa )2
Yield safety factor:
Sy
ny = e
σmax
e
A highly conservative estimate of σmax is
e
σmax ≤ σae + σm
e
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Shaft Design Example Problem
Example-1
At a machined shaft shoulder the small diameter d is 1.100 in, the
large diameter D is 1.65 in, and the fillet radius is 0.11 in. The bending
moment is 1260 lbf-in and the steady torsion moment is 1100 lbf-in.
The heat-treated steel shaft has an ultimate strength of Sut = 105 kpsi
and a yield strength of Sy = 82 kpsi. The 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.
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Shaft Design Example Problem
Kt for torsion and bending
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Shaft Design Example Problem
Notch sensitivity factor for reversed torsion and bending
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
Se0 = 0.5Sut =
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
Se0 = 0.5Sut = 52.5 kpsi
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
Se0 = 0.5Sut = 52.5 kpsi
Creliab
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1 Csize
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
Se0 = 0.5Sut = 52.5 kpsi
Creliab
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Shaft Design Example Problem
Solution
D/d = 1.5, r /d = 0.1 Csize
kt = 1.68, kts = 1.42
q = 0.85, qshear = 0.92
kf = 1 + q(kt − 1) = 1.58
kfs = 1 + qshear (kts − 1) =
1.39
Se0 = 0.5Sut = 52.5 kpsi b
Csurf = ASut
Creliab
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Shaft Design Example Problem
Solution Contd..
Se = Csurf Csize Creliab Se0 = (0.787)(0.87)(0.814)(52.5) = 29.3 kpsi
The shaft is rotating with a constant bending moment, i.e., Mm = 0
Steady torque is applied, implies Ta = 0
Ma = 1260 lbf-in and Tm = 1100 lbf-in
Applying Goodman criterion for multi-axial fatigue:
q
1 16 1 1
q
= 2
4(kf Ma ) + 3(kfsm Tm ) 2
n πd 3 Se Sut
Hence, n = 1.62.
Based on ASME elliptic, n = 1.88
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Shaft Design Example Problem
Yield factor of safety
Sy
ny = e
σmax
e
σmax = 18300psi
Hence, ny = 4.48
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Shaft Design Example Problem
Shaft Design: Fully Reversed Bending and Steady Torque I
Design a shaft to support the attachments shown in the Figure
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Shaft Design Example Problem
Shaft Design: Fully Reversed Bending and Steady Torque II
Example 9.1 in Norton’s book
Design a shaft to support the attachments shown in the Figure with a
preliminary design of shaft configuration. The shaft is expected to transmit 2
hp at 1725 rpm. The torque and the force on the gear are both constant with
time.
Assume that
No applied axial loads
Steel as a shaft material
stress-concentration factor of 3.5 for the step radii in bending, 2 in
torsion and 4 at the keyways.
Assume that the ratio of tensions on tight side to slack side on the v-belt
sheave is 5.
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Shaft Design Example Problem
Shaft Design: Fully Reversed Bending and Steady Torque III
Roller Bearings Standard Dimensions
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Shaft Design Design of a shaft for deflection
Design for deflection
Maximum Ranges for Slope and Transverse Deflection
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Shaft Design Design of a shaft for deflection
Design for deflection
Shaft as a beam for deflection
Due to change of cross section, M/EI changes along the length
Integrate M/EI twice to find deflection at any given point
Shaft as a torsion bar
T`
θ=
GJ
The torsional stiffness is
T GJ
kt = =
θ `
For a stepped shaft, the torsional stiffness is more complicated
Stepped shaft can be assumed as torsional springs in series
1 1 1 1
= + +
kteff kt1 kt2 kt3
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Design of Keys
Keys and Keyways I
Key as defined by ASME
A demountable machine element which, when assembled into
keyseats, provides a positive means for transmitting torque between
the shaft and hub
Types of keys
Parallel Key Tapered Key Woodruff Key
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Design of Keys
Keys and Keyways II
Square and Rectangular Key Sizes
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Design of Keys Types of Keys
Parallel Keys
Most commonly used
Made from cold-rolled bar stock (negatively toleranced)
Clearance fit between key and keyseat in general with some
exceptions
The clearance between key and keyseat leads to impact and high
stress called back lash when torque changes from positive to
negative
A set screw in the hub placed at 90◦ to the key prevents backlash
Length of the key ` ≤ 1.5d, where d is shaft diameter
For higher strength, more keys may be used at different
orientations
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Design of Keys Types of Keys
Tapered Keys
The tapered key is a locking key which holds the key axially
through friction
The gib head is optional
Tapered keys orient the radial clearance to one side thus creating
eccentricity between the shaft and hub
Woodruff Keys
Used on small shafts
Self-aligning and hence are preferred for tapered shafts
Semi-circular shape leads to a larger keyseat depth which resists
key rolling
Weakens the shaft due to larger keyseat depth
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Design of Keys Stresses in Keys
Stresses in Keys
Modes of failure in keys
Shear failure: shearing across its width at the interface between
the shaft and the hub
Average stress due to direct shear
F
τxy = , Ashear = `w
Ashear
Bearing failure: crushing either side in compression
Average bearing stress is defined as
F
σx = , Asquare
bearing = (h/2)`
Abear
Since compressive stresses do not cause fatigue failure, bearing
failure may be considered static
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Design of Keys Key Materials
Key Materials
Ductile materials due to shear loading of keys
Soft low-carbon steel is the most common material
Brass/Stainless steel under corrosive environment
Square and rectangular keys are merely cut from cold-rolled bar
stock
Tapered and Woodruff keys are also made from soft cold-rolled
steel
Lower strength materials than that of shaft
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Design of Keys Key Materials
Key Design I
Few design variables for key sizing
Shaft diameter at the key seat determines the key width
Key height is determined by width
The design variables are: the length and number of keys
Key is sized such that it fails before the keyseat or other locations
in the shaft
Key is relatively inexpensive and easy to replace if the keyseat is
undamaged
Focus on stress concentration factors due to sharp corners
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Design of Keys Key Materials
Key Design II
Stress concentration factors for an end milled key seat in bending
and torsion
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Design of Keys Key Materials
Example Problem - Key Design
Design the keys for the shaft in example problem in slide 30
Solve Norton’s example problem-9.4
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Design of Keys Key Materials
Other Design Consideration
Design of spline shafts
Design shafts for vibration
Critical speed of shafts
Lateral vibration
Shaft whirl
Torsional vibration
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Design of Keys Key Materials
References
1 Robert L Norton, Machine Design, An Integrated Approach
2 R. G. Budyanas and J. K. Nisbett, Shigley’s Mechanical
Engineering Design
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