FME461
Engineering Design II
Dr.Hussein Jama
[email protected]
Office 414
Lecture: Mon 8am -10am
Tutorial Tue 3pm - 5pm
10/1/2013
1
Semester outline
Date
Week
Topics
9th Sept
1
House keeping issues
Introduction to mechanical design
Assignment 1 is given out
16th
Sept
23rd
Sept
2
Ethics & safety
Various
3
Assignment 1 is due
Assignment 2 is given out
Static & Fatigue Failure
Various
Ch – 5 Shigley
Ch – 6 Shigley
30th
Sept
7th Oct
4
FMEA
Various
5
Continuous Assessment Test 1 (15%)
Assignment 2 is due
Assignment 3 is given out
14th Oct
6
Shafts and shaft components
Ch – 7 Shigley
21st Oct
7
Welding and permanent joints
Ch – 9 Shigley
28th Oct
8
Mechanical springs
Ch – 10 Shigley
4th Nov
11th Nov
9
10
Clutches & brakes
Belts and chains
Ch – 16 Shigley
Ch – 17 Shigley
18th Nov
11
Statistical consideration
Ch – 20 Shigley
25th Nov
12
Continuous Assessment Test 2 (15%)
13
Presentation of assignment 2
Assignment 2 is due
10/1/2013
2nd Dec
Reference
Reading
Ch 1 - Norton, Shigley
2
Discussion

Shigley Chapter 5 - Static failure criteria



Ductile materials
Brittle materials
Shigley – Chapter 6 Fatigue failure criteria
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3
Static & Fatigue Failure
Static load – a stationary load that is gradually
applied having an unchanging magnitude and
direction
Failure – A part is permanently distorted and will
not function properly
A part has been separated into two or more
pieces.
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4
Static failure theories



Maximum shear Stress theory
Distortion energy theory
Ductile Coulomb-Mohr theory
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5
Definitions
Material Strength
Sy = Yield strength in tension, Syt = Syc
Sys = Yield strength in shear
Su = Ultimate strength in tension, Sut
Suc = Ultimate strength in compression
Sus = Ultimate strength in shear = .67 Su
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6
Ductile and brittle materials
A ductile material deforms significantly before fracturing. Ductility is
measured by % elongation at the fracture point. Materials with 5% or
more elongation are considered ductile.
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Brittle material yields very little before fracturing, the
yield strength is approximately the same as the
ultimate strength in tension. The ultimate strength in
compression is much larger than the ultimate
strength in tension.
7
Failure theories – Ductile materials
• Maximum shear stress theory (Tresca 1886)
( maspecimen of the same material when that specimen x
)component > ( )obtained from a tension test at the yield point
Failure
= Sy
To avoid failure
=
Sy
2
(
)
max component
= Sy
max
=
Sy
2n
<
Sy
2
n = Safety
factor
Design equation
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=Sy
8
Max. Shear Theory

The maximum-shear-stress theory
predicts that yielding begins
whenever the maximum shear
stress in any element equals or
exceeds the maximum shear stress
in a tension test.
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9
Failure theories –Ductile materials
• Distortion energy theory (von Mises-Hencky)
Simple tension test → (Sy)t
Hydrostatic state of stress → (Sy)h
t
h
(Sy)h >> (Sy)t
Distortion contributes to
failure much more than
change in volume.
h
h
t
(total strain energy) – (strain energy due to hydrostatic stress) = strain energy
due to angular distortion > strain energy obtained from a tension test at the
yield point → failure
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10
Plane stress problems
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11
Stress components
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12
Failure theories – ductile materials
The area under the curve in the elastic region is called the Elastic Strain
Energy.
3D case
U=½ ε
UT = ½
ε +½
1 1
ε +½
2 2
ε
3 3
Stress-strain relationship
ε1 =
ε2 =
ε3 =
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UT =
1
2E
(
2
1
+
2
2
+
2
3
1
v
E
2
E
1
v
E
3
E
1
v
E
) - 2v (
2
1
E
2
+
v
v
3
E
3
E
2
v
1
E
3
+
2
3)
13
Failure theories – Ductile materials
Distortion strain energy = total strain
energy – hydrostatic strain energy
1
UT =
( 12 + 22 + 32) - 2v
2E
Substitute
Uh =
1
1
(
=
2
2
3=
2
h
+
Simplify and substitute
1
2E
3
Uh =
h
+
=
2
h
2E
(1 – 2v) =
(
1
2
+
1
3
) - 2v (
h
h
+
h
h+
+
=3
h
+
1
+
2
3
)
(1)
h
2
(
Ud = UT – Uh
2
+
2
3
+
h
h
h
)
into the above equation
2
(1 – 2v)
)
3
6E
Subtract the hydrostatic strain energy from the total energy to
obtain the distortion energy
10/1/2013 Ud = UT – Uh =
1+v
6E
(
1–
2
)
+ (
2
1–
2
)
+ (
3
2–
(2)
2
)
3
14
Failure theories – Ductile materials
Strain energy from a tension test at the yield point
1=
Sy and
2
=
U d = UT – Uh =
3
=0
1+v
6E
Substitute in equation (2)
(
2
)
+ (
2
1–
Utest = (Sy)
2
1–
2
)
+ (
3
2–
2
)
3
1+v
3E
To avoid failure, Ud < Utest
(
1–
2
)
+ (
2
2
)
+ (
3
1–
2
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2–
2 ½
)
3
<
Sy
15
Failure theories – ductile materials
(
2
)
+ (
2
1–
2
)
+ (
3
1–
2–
2 ½
)
3
2
2D case,
3
Sy
=0
(
2
1
–
1
2
+
2
2
½
) < Sy =
Where
′=
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<
Sy
n
is von Mises stress
Design equation
16
Failure theories -Ductile materials
Pure torsion,
=
1
(
3
2
=
=–
2
1
2
–
2
2
1
2
+
2
) = Sy2
Sys = Sy / √ 3 → Sys = .577 Sy
Sy
Relationship between yield strength in
tension and shear
If
y=
0, then
1,
2
[( x)/2]2 + (
x/2
=
xy)
2
the design equation can be written in terms of the dominant
component stresses (due to bending and torsion)
(
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x)
2
+ 3(
xy)
2
1/2
=
Sy
n
17
Summary – Ductile materials
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18
Design process
Maximum shear stress theory
Distortion energy theory
′=
Sy
max =
2n
Sy
n
•
Select material: consider environment, density, availability →
•
Choose a safety factor
n
Size
Weight
Sy , Su
Cost
The selection of an appropriate safety factor should be based
on the following:
 Degree of uncertainty about loading (type, magnitude and direction)
 Degree of uncertainty about material strength
 Uncertainties related to stress analysis
 Consequence of failure; human safety and economics
 Type of manufacturing process
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 Codes and standards
19
Flow process
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20
Design Process
 Use n = 1.2 to 1.5 for reliable materials subjected to
loads that can be determined with certainty.
 Use n = 1.5 to 2.5 for average materials subjected to
loads that can be determined. Also, human safety and
economics are not an issue.
 Use n = 3.0 to 4.0 for well known materials subjected to
uncertain loads.
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21
Design Process
Sy , Su
•
Select material, consider environment, density, availability →
•
Choose a safety factor
•
Formulate the von Mises or maximum shear stress in terms of size.
•
Use appropriate failure theory to calculate the size.
′=
•
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Sy
n
Sy
max =
2n
Optimize for weight, size, or cost.
22
Example – from Shigley
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23
Solution
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24
Failure theories- brittle materials
One of the characteristics of a brittle material is that the ultimate
strength in compression is much larger than ultimate strength in
tension.
Suc >> Sut
Mohr’s circles for compression and tension tests.
Suc
3
Stress
state
1
Sut
Tension test
Compression test
Failure envelope
The component is safe if the state of stress falls inside the failure envelope.
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25
1 > 3 and 2 = 0
Failure theories – brittle materials
Modified Coulomb-Mohr theory
3 or
2
3 or
Sut
Safe
Sut
Safe
I
Sut
Sut
1
Suc
1
Safe
-Sut
III
Suc
Cast iron data
II
-Sut
Safe
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2
Suc
Three design zones
26
Failure theories – brittle materials
Zone I
3
1> 0,
=
1
2 > 0 and
1
>
Sut
2
I
Sut
Design equation
n
1
II
-Sut
Zone II
III
1
>0,
1
=
2
Sut
n
< 0 and
2
< Sut
Suc
Design equation
Zone III
1
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>0,
2 < 0 and
2
> Sut
1 (
1
1
1
–
)– 2 =
n
Sut
Suc
Suc
Design equation
27
Summary – Brittle materials
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28
Example
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29
Solution
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30
Solution continued
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31
Static failure summary - Ductile
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32
Summary – Brittle materials
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33
Failure theories - Fatigue
It is recognised that a metal subjected to a
repetitive or fluctuating stress will fail at a
stress much lower than that required to cause
failure on a single application of load. Failures
occurring under conditions of dynamic loading
are called fatigue failures.
Fatigue failure is characterized by three stages
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
Crack Initiation

Crack Propagation

Final Fracture
34
Jack hammer component,
shows no yielding before
fracture.
Crack initiation site
Fracture zone
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Propagation zone, striation
35
Example
VW crank shaft – fatigue failure due to cyclic bending and torsional
stresses
Propagation
zone, striations
Crack initiation site
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Fracture area
36
928 Porsche timing pulley
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Crack started at the fillet
37
Fracture surface of a failed bolt. The
fracture surface exhibited beach marks,
which is characteristic of a fatigue failure.
25mm diameter steel pins from
agricultural equipment.
Material; AISI/SAE 4140 low
allow carbon steel
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38
bicycle crank spider arm
This long term fatigue crack in a high quality component took a
considerable time to nucleate from a machining mark between the spider
arms on this highly stressed surface. However once initiated propagation
was rapid and accelerating as shown in the increased spacing of the 'beach
marks' on the surface caused by the advancing fatigue crack.
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39
Crank shaft
Gear tooth failure
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40
Hawaii, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin
area rips off in mid-flight. Metal fatigue was the cause of the failure.
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41
Fracture surface characteristics
Mode of fracture
Typical surface characteristics
Ductile
Cup and Cone
Dimples
Dull Surface
Inclusion at the bottom of the dimple
Brittle Intergranular
Shiny
Grain Boundary cracking
Brittle Transgranular
Shiny
Cleavage fractures
Flat
Fatigue
Beachmarks
Striations (SEM)
Initiation sites
Propagation zone
Final fracture zone
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42
Fatigue failure – type of fluctuating stresses
Alternating stress
a=
min
a=
m=
2
=0
Mean stress
max /
2
max
m
=
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min
max
+
2
min
43
Fatigue Failure, S-N Curve
Test specimen geometry for R.R. Moore
rotating beam machine. The surface is
polished in the axial direction. A constant
bending load is applied.
Typical testing apparatus, pure bending
Motor
Load
Rotating beam machine – applies fully reverse bending stress
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44
Fatigue Failure, S-N Curve
Finite life
Infinite life
S′e
Se′ = endurance limit of the specimen
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45
Relationship Between Endurance Limit and
Ultimate Strength
Steel
Se′ =
0.5Sut
Sut ≤ 200 ksi (1400 MPa)
100 ksi
Sut > 200 ksi
700 MPa Sut > 1400 MPa
Cast iron
Se′ =
0.4Sut
Sut < 60 ksi (400 MPa)
24 ksi
Sut ≥ 60 ksi
Cast iron
160 MPa Sut < 400 MPa
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46
Relationship Between Endurance Limit and
Ultimate Strength
Aluminium
Se′ =
0.4Sut
Sut < 48 ksi (330 MPa)
19 ksi
Sut ≥ 48 ksi
130 MPa Sut ≥ 330 MPa
Copper alloys
Copper alloys
Se′ =
0.4Sut
Sut < 40 ksi (280 MPa)
14 ksi
Sut ≥ 40 ksi
100 MPa Sut ≥ 280 MPa
For N = 5x108 cycle
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47
For materials exhibiting a knee in the S-N curve at 106 cycles
S ′ = endurance limit of the specimen (infinite life > 106)
e
Se = endurance limit of the actual component (infinite life > 106)
S
103
Se
106
N
For materials that do not exhibit a knee in the S-N curve, the infinite
life taken at 5x108 cycles
Sf′ = fatigue strength of the specimen (infinite life > 5x108)
Sf = fatigue strength of the actual component (infinite life > 5x108)
S
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103
Sf
5x108
N
48
Correction factor’s for
specimen’s endurance limit
Se = Cload Csize Csurf Ctemp Crel (S′e)
Sf = Cload Csize Csurf Ctemp Crel (Sf′)
• Load factor, Cload
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(page 326, Norton’s 3rd ed.)
Pure bending
Cload = 1
Pure axial
Cload = 0.7
Pure torsion
Cload = 1 if von Mises stress is used, use
0.577 if von Mises stress is NOT used.
Combined loading
Cload = 1
49
Correction factor’s for
specimen’s endurance limit
• Size factor, Csize
(p. 327, Norton’s 3rd ed.)
Larger parts fail at lower stresses than smaller parts. This is
mainly due to the higher probability of flaws being present in
larger components.
For rotating solid round cross section
d ≤ 0.3 in. (8 mm)
Csize = 1
0.3 in. < d ≤ 10 in.
Csize = .869(d)-0.097
8 mm < d ≤ 250 mm
Csize = 1.189(d)-0.097
If the component is larger than 10 in., use Csize = .6
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50
Correction factor’s for specimen’s
endurance limit
For non rotating components, use the 95% area approach to calculate
the equivalent diameter. Then use this equivalent diameter in the
previous equations to calculate the size factor.
A95 = (π/4)[d2 – (.95d)2] = .0766 d2
d95 = .95d
d
Solid or hollow non-rotating parts
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dequiv = .37d
dequiv = (
A95
0.0766
)1/2
Rectangular parts
51
dequiv = .808 (bh)1/2
Correction factor’s for specimen’s
endurance limit
I beams and C channels
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52
Correction factor’s for specimen’s
endurance limit
• surface factor, Csurf
(p. 328-9, Norton’s 3rd ed.)
The rotating beam test specimen has a polished surface. Most
components do not have a polished surface. Scratches and
imperfections on the surface act like a stress raisers and reduce
the fatigue life of a part. Use either the graph or the equation with
the table shown below.
Csurf = A (Sut)b
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53
Correction factor’s for
specimen’s endurance limit
• Temperature factor, Ctemp
(p.331, Norton’s 3rd ed.)
High temperatures reduce the fatigue life of a component. For
accurate results, use an environmental chamber and obtain the
endurance limit experimentally at the desired temperature.
For operating temperature below 450 oC (840 oF) the
temperature factor should be taken as one.
Ctemp = 1
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for T ≤ 450 oC (840 oF)
54
Correction factor’s for
specimen’s endurance limit
• Reliability factor, Crel (p. 331, Norton’s 3rd ed.)
The reliability correction factor accounts for the scatter and
uncertainty of material properties (endurance limit).
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55
Fatigue Stress Concentration Factor, Kf
Experimental data shows that the actual stress concentration factor is not as
high as indicated by the theoretical value, Kt. The stress concentration factor
seems to be sensitive to the notch radius and the ultimate strength of the
material.
Notch sensitivity factor
Fatigue stress
Kf = 1 + (Kt – 1)q
concentration factor
rd
(p. 340, Norton’s 3 ed.)
Steel
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56
Fatigue Stress Concentration
Factor, Kf for Aluminum
(p. 341, Norton’s 3rd ed.)
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57
Design process
•
Determine the maximum alternating applied stress (
the size and cross sectional profile
•
Select material → Sy, Sut
•
Choose a safety factor → n
•
Determine all modifying factors and calculate the endurance
limit of the component → Se
•
Determine the fatigue stress concentration factor, Kf
•
Use the design equation to calculate the size
Kf
a)
in terms of
Se
a= n
•
Investigate different cross sections (profiles), optimize for size or weight
•
You may also assume a profile and size, calculate the alternating stress
and determine the safety factor. Iterate until you obtain the desired
58
safety factor
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Design for finite life
Sn = a (N)b equation of the fatigue line
A
S
B
Se
106
103
Point A
Point B
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N
Sn = .9Sut
Point A
Sn = .9Sut
N = 10
N = 103
Sn = Se
Sn = Sf
3
6
N = 10
Point B
N = 5x108
59
Design for finite life
Sn = a (N)b
log Sn = log a + b log N
Apply boundary conditions for point A and B to
find the two constants “a” and “b”
log .9Sut = log a + b log 10
a=
3
log Se = log a + b log 106
b=
N
Sn = Se ( 106 )
Calculate Sn
2
Se
1
3
log
.9Sut
Se
Se
log ( .9S )
ut
and replace Se in the design equation
Kf
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⅓
(.9Sut)
a=
Sn
n
Design equation
60
The effect of mean stress on
fatigue
life
Mean stress exist if the
loading is of a repeating or
fluctuating type.
a
Mean stress is not zero
Gerber curve
Alternating
stress
Se
Goodman line
Soderberg line
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Sy
Mean stress
Sut
m
61
The effect of mean stress on
fatigue life goodman diagram
a
Yield line
Alternating
stress
Goodman line
Safe zone
C
Sy
Sut
m
Mean stress
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62
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
a
Sy
Yield line
Goodman line
Safe zone
-
m
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- Syc
Safe zone
Sy
Sut
+
m
63
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
Fatigue,
m
≤0
Fatigue,
a
a
Se
a=
Se
nf
m
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Sut
m
Sut
m
=
>0
=
= 1
1
Infinite life
nf
Finite life
Yield
Syc
ny
Safe zone
-
Sn
+
m
Se
Yield
a+
a
+
m
Safe zone
C
a+
m
=
Sy
ny
- Syc
64
Applying Stress Concentration factor to Alternating and
Mean Components of Stress
•
Determine the fatigue stress concentration factor, Kf, apply directly to
the alternating stress → Kf
•
If Kf
max
a
< Sy then there is no yielding at the notch, use Kfm = Kf
and multiply the mean stress by Kfm → Kfm
•
m
> Sy then there is local yielding at the notch, material at the
notch is strain-hardened. The effect of stress concentration is reduced.
If Kf
max
Calculate the stress concentration factor for the mean stress using
the following equation,
Kfm =
Sy
Kf
a
m
Fatigue design equation
Kf a
Kfm m
=
+
Se
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Sut
1
nf
Infinite life
65
Combined loading
All four components of stress exist,
xa
alternating component of normal stress
xm
mean component of normal stress
xya
alternating component of shear stress
xym
mean component of shear stress
Calculate the alternating and mean principal stresses,
1a,
1m,
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2a
=(
xa /2)
(
2
/
2
)
+(
xa
2
)
xya
2m
=(
xm /2)
(
2
/
2
)
+(
xm
2
)
xym
66
Combined loading
Calculate the alternating and mean von Mises stresses,
a′ = (
2
1a
m′ = (
2
1m
+
+
2
2a -
1a
2
2m -
1m
1/2
)
2a
1/2
)
2m
Fatigue design equation
′a
Se
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+
′m
Sut
=
1
nf
Infinite life
67
Example 6-7
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68
Example 6-9
Shaft is
made from
SAE1050
steel and is
cold formed
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69
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70
Solution
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71
Solution
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72
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73