MECH 401
Mechanical Design Applications
Dr. M. O’Malley– Master Notes
Spring 2008
Dr. D. M. McStravick
Rice University

Reading


Homework


Chapter 6
HW 4 available, due 2-7
Tests

Fundamentals Exam will be in class on 2-21

Nature of fatigue failure

Starts with a crack



Usually at a stress concentration
Crack propagates until the material fractures
suddenly
Fatigue failure is typically sudden and
complete, and doesn’t give warning
Fatigue Failure Examples






Various Fatigue Crack Surfaces [Text fig. 6-2]
Bolt Fatigue Failure [Text fig. 6-1]
Drive Shaft [Text fig. 6-3]
AISI 8640 Pin [Text fig. 6-4]
Steam Hammer Piston Rod [Text fig. 6-6]
Jacob Neu chair failure (in this classroom)
Fatigue Example 1
Fatigue Failure Example
Fatigue Failure Example
Fatigue Failure Example
Stamping Fatigue Failure Example
Schematic of Various Fatigue Failure

Jim Neu Chair Failure (Pedestal)
Fatigue Failure of Chair Shaft
Seat Fatigue Failure
Fatigue





Fatigue strength and endurance limit
Estimating FS and EL
Modifying factors
Thus far we’ve studied static failure of machine elements
The second major class of component failure is due to dynamic loading






Variable stresses
Repeated stresses
Alternating stresses
Fluctuating stresses
The ultimate strength of a material (Su) is the maximum stress a
material can sustain before failure assuming the load is applied only
once and held
A material can also fail by being loaded repeatedly to a stress level that
is LESS than Su

Fatigue failure
More Fatigue Failure Examples (ASM)
More Fatigue Failure Examples (ASM)
More Fatigue Failure Examples
Approach to fatigue failure in analysis and
design

Fatigue-life methods (6-3 to 6-6)




Stress Life Method (Used in this course)
Strain Life Method
Linear Elastic Fracture Mechanics Method
Stress-life method (rest of chapter 6)


Addresses high cycle Fatigue (>103 ) Well
Not Accurate for Low Cycle Fatigue (<103)
The 3 major methods

Stress-life



Based on stress levels only
Least accurate for low-cycle fatigue
Most traditional




Strain-life




Easiest to implement
Ample supporting data
Represents high-cycle applications adequately
More detailed analysis of plastic deformation at localized regions
Good for low-cycle fatigue applications
Some uncertainties exist in the results
Linear-elastic fracture mechanics



Assumes crack is already present and detected
Predicts crack growth with respect to stress intensity
Practical when applied to large structures in conjunction with computer
codes and periodic inspection
Fatigue analysis

2 primary classifications of
fatigue


Alternating – no DC component
Fluctuating – non-zero DC
component
Analysis of alternating stresses




As the number of cycles
increases, the fatigue strength
Sf (the point of failure due to
fatigue loading) decreases
For steel and titanium, this
fatigue strength is never less
than the endurance limit, Se
Our design criteria is:
S f (N )

 a
As the number of cycles
approaches infinity (N  ∞),
Sf(N) = Se (for iron or Steel)
Method of calculating fatigue strength



Seems like we should be able to use graphs
like this to calculate our fatigue strength if
we know the material and the number of
cycles
We could use our factor of safety equation
as our design equation
But there are a couple of problems with this
approach


S-N information is difficult to obtain and thus is
much more scarce than -e information
S-N diagram is created for a lab specimen




Smooth
Circular
Ideal conditions
Therefore, we need analytical methods for
estimating Sf(N) and Se
S f (N )

 a
Terminology and notation

Infinite life versus finite life



Infinite life
Se

Implies N ∞



Use endurance limit (Se) of material
 a

Lowest value for strength
Finite life

Implies we know a value of N (number of cycles)

Use fatigue strength (Sf) of the material (higher than Se)

S f (N )
 a
Prime (‘) versus no prime






Strength variable with a ‘ (Se’)

Implies that the value of that strength (endurance limit) applies to a LAB SPECIMEN in
controlled conditions
Variables without a ‘ (Se, Sf)

Implies that the value of that strength applies to an actual case
First we find the prime value for our situation (Se’)
Then we will modify this value to account for differences between a lab specimen and our
actual situation
This will give us Se (depending on whether we are considering infinite life or finite life)
Note that our design equation uses Sf, so we won’t be able to account for safety factors until
we have calculated Se’ and Se
Estimating Se’ – Steel and Iron


For steels and irons, we can estimate the
endurance limit (Se’) based on the ultimate
strength of the material (Sut)
Steel


Se’
= 0.5 Sut
for
Sut < 200 ksi (1400 MPa)
= 100 ksi (700 MPa) for all other values of Sut
Iron

Se’
= 0.4(min Sut)f/ gray cast Iron Sut<60 ksi(400MPa)
= 24 ksi (160 MPa) for all other values of Sut
Note: ASTM # for gray cast iron is the min Sut
S-N Plot with Endurance Limit


S f (N )
 a
Se
 a
Se

 a
Estimating Se’ – Aluminum and Copper
Alloys





For aluminum and copper alloys, there is no endurance limit
Eventually, these materials will fail due to repeated loading
To come up with an “equivalent” endurance limit, designers
typically use the value of the fatigue strength (Sf’) at 108 cycles
Aluminum alloys
 Se’ (Sf at 108 cycles)
Copper alloys
 Se’ (Sf at 108 cycles)
= 0.4 Sut
for
Sut < 48 ksi (330 MPa)
= 19 ksi (130 MPa) for all other values of Sut
= 0.4 Sut
for
Sut < 35 ksi (250 MPa)
= 14 ksi (100 MPa) for all other values of Sut
Constructing an estimated S-N diagram

Note that Se’ is going to be our
material strength due to “infinite”
loading

We can estimate an S-N diagram
and see the difference in fatigue
strength after repeated loading

For steel and iron, note that the
fatigue strength (S’f) is never less
than the endurance limit (Se’)

For aluminum and copper, note
that the fatigue strength (S’f)
eventually goes to zero (failure!),
but we will use the value of S’f at
108 cycles as our endurance limit
(Se’) for these materials
Estimating the value of Sf


When we are studying a case of
fatigue with a known number of cycles
(N), we need to calculate the fatigue
strength (S’f)
We have two S-N diagrams




One for steel and iron
One for aluminum and copper
We will use these diagrams to come
up with equations for calculating S’f
for a known number of cycles
Note: Book indicates that 0.9 is not
actually a constant, and uses the
variable f to donate this multiplier. We
will in general use 0.9 [so f=0.9]
Estimating Sf (N)


For steel and iron
For f=0.9
S 'f N   aN b
For 103 < N < 106
1  0.9Sut 

b  - log 
3  Se 
log a   log 0.9Sut  - 3b

For aluminum and copper
S 'f N   aN b
For N < 108
1  0.9Sut 

b  - log 
3  Se 
Where Se’ is the value of
S’f at N = 108
log a   log 0.9Sut  - 3b
5.7
Correction factors



Now we have Se’ (infinite life)
We need to account for differences between the lab specimen and a real
specimen (material, manufacturing, environment, design)
We use correction factors




These will account for differences between an ideal lab specimen and real life
Se = ka kb kc kd ke kf Se’








Strength reduction factors
Marin modification factors
ka – surface factor
kb – size factor
kc – load factor
kd – temperature factor
ke – reliability factor
Kf – miscellaneous-effects factor
Modification factors have been found empirically and are described in section 6-9 of
Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide
Endurance limit modifying factors

Surface (ka)

Accounts for different surface finishes


Size (kb)

Different factors depending on loading



Accounts for effects of operating temperature (Not significant factor for T<250 C [482 F])
Reliability (ke)


Endurance limits differ with Sut based on fatigue loading (bending, axial, torsion)
Temperature (kd)


Bending and torsion (see pg. 280)
Axial (kb = 1)
Loading (kc)


Ground, machined, cold-drawn, hot-rolled, as-forged
Accounts for scatter of data from actual test results (note ke=1 gives only a 50% reliability)
Miscellaneous-effects (kf)


Accounts for reduction in endurance limit due to all other effects
Reminder that these must be accounted for



Residual stresses
Corrosion
etc
Surface Finish Effect on Se
Temperature Effect on Se
Reliability Factor, ke
Steel Endurance Limit vs. Tensile Strength
Compressive Residual Stresses
Now what?


Now that we know the strength of our part under
non-laboratory conditions…
… how do we use it?


Choose a failure criterion
Predict failure






Part will fail if:
’ > Sf(N)
Factor of safety
or Life of the part:
1
 = Sf(N) / ’
   b
N  
Where
a
b = - 1/3 log (0.9 Sut / Se)
log (a) = log (0.9 Sut) - 3b
Example Homework Problem 6-9

A solid rod cantilevered at one end. The rod is
0.8 m long and supports a completely
reversing transverse load at the other end of +/1 kN. The material is AISI 1045 hot-rolled
steel. If the rod must support this load for 104
cycles with a factor of safety of 1.5, what
dimension should the square cross section
have? Neglect any stress concentrations at the
support end and assume f= 0.9.

Solution: -- See Board Work--
Stress concentration (SC) and fatigue failure



Unlike with static loading, both ductile and
brittle materials are significantly affected by
stress concentrations for repeated loading
cases
We use stress concentration factors to modify
the nominal stress
SC factor is different for ductile and brittle
materials
SC factor – fatigue

 = kfnom+ = kfo

t = kfstnom = kfsto



kf is a reduced value of kT and o is the nominal
stress.
kf called fatigue stress concentration factor
Why reduced? Some materials are not fully sensitive
to the presence of notches (SC’s) therefore,
depending on the material, we reduce the effect of
the SC
Fatigue SC factor


kf = [1 + q(kt – 1)]
kfs = [1 + qshear(kts – 1)]

kt or kts and nominal stresses


Table A-15 & 16 (pages 1006-1013 in Appendix)
q and qshear



Notch sensitivity factor
Find using figures 6-20 and 6-21 in book (Shigley) for steels
and aluminums
Use q = 0.20 for cast iron



Brittle materials have low sensitivity to notches
As kf approaches kt, q increasing (sensitivity to notches, SC’s)
If kf ~ 1, insensitive (q = 0)

Property of the material
Example



AISI 1020 as-rolled steel
Machined finish
Find Fmax for:



 = 1.8
Infinite life
Design Equation:

 = Se / ’

Se because infinite life
Example, cont.


 = Se / ’
What do we need?



Considerations?




Se
’
Infinite life, steel
Modification factors
Stress concentration (hole)
Find ’nom (without SC)
 
 nom
P
P
F


 2083F
A b - d h 60 - 12 10
Example, cont.

Now add SC factor:
  1  qkt - 1 nom

   k f  nom

From Fig. 6-20,



r = 6 mm
Sut = 448 MPa = 65.0 ksi
q ~ 0.8
Example, cont.

From Fig. A-15-1,






Unloaded hole
d/b = 12/60 = 0.2
kt ~ 2.5
q = 0.8
kt = 2.5
’nom = 2083 F

   1  qkt - 1 nom
   1  0.82.5 - 12083F 
   4583F 
Example, cont.

Now, estimate Se



Steel:
Se’ = 0.5 Sut for Sut < 1400 MPa (eqn. 6-8)
700 MPa else
AISI 1020 As-rolled


Sut = 448 MPa
Se’ = 0.50(448) = 224 MPa
Constructing an estimated S-N diagram

Note that Se’ is going to be our
material strength due to “infinite”
loading

We can estimate an S-N diagram
and see the difference in fatigue
strength after repeated loading

For steel and iron, note that the
fatigue strength (S’f) is never less
than the endurance limit (S’e)

For aluminum and copper, note
that the fatigue strength (S’f)
eventually goes to zero (failure!),
but we will use the value of S’f at
108 cycles as our endurance limit
(S’e) for these materials
Correction factors



Now we have Se’ (infinite life)
We need to account for differences between the lab specimen and a real
specimen (material, manufacturing, environment, design)
We use correction factors




These will account for differences between an ideal lab specimen and real life
Se = ka kb kc kd ke kf Se’








Strength reduction factors
Marin modification factors
ka – surface factor
kb – size factor
kc – load factor
kd – temperature factor
ke – reliability factor
Kf – miscellaneous-effects factor
Modification factors have been found empirically and are described in section 6-9 of
Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide
Example, cont.


Modification factors
Surface:
ka = aSutb (Eq. 6-19)

a and b from Table 6-2
Machined

ka = (4.45)(448)-0.265 = 0.88

Example, cont.

Size:



Axial loading
kb = 1 (Eq. 6-21)
Load:


kb
kc
Axial loading
kc = 0.85 (Eq. 6-26)
Example, cont.

Temperature:


Reliability:


kf = 1
Endurance limit:


ke = 1 (no info given)
Miscellaneous:


kd = 1 (no info given)
Se = kakbkckdkekfSe’ = (0.88)(0.85)(227) = 177 MPa
Design Equation:

Se 177 MPa

 1.8




4583 F
177 x10 6
F
 21 .4 kN
4583 1.8
Fluctuating Fatigue Failures
Alternating vs. fluctuating
Alternating
Fluctuating
P
m 
A
Mr
a 
I
Alternating Stresses

a characterizes alternating stress
Fluctuating stresses

Mean Stress
 'm 

2
Stress amplitude
 
'
a

 max   min
 max -  min
2
Together, m and a
characterize fluctuating
stress
Alternating vs. Fluctuating
Modified Goodman Diagram

Fluctuating Stresses in Compression and
Tension
Failure criterion for fluctuating loading







Soderberg
Modified Goodman
Gerber
ASME-elliptic
Yielding
Points above the line: failure
Book uses Goodman primarily
 Straight line, therefore easy algebra
 Easily graphed, every time, for every problem
 Reveals subtleties of insight into fatigue problems
 Answers can be scaled from the diagrams as a check on the
algebra
Gerber Langer Plot for Fluctuating Stresses
Fluctuating stresses, cont.


As with alternating stresses, fluctuating stresses have been
investigated in an empirical manner
For m < 0 (compressive mean stress)





a > Sf
Failure
Same as with alternating stresses
Or,
 max   m -  a  S yc (or Suc )
For m > 0 (tensile mean stress)

Modified Goodman criteria

a
Sf


<1
m
Sut

1

Failure
Static Failure
Modified Goodman Langer Equations
Fluctuating stresses, cont.
Note: m + a = max

Relationship is easily
seen by plotting:
m + a > Syt (static failure by yielding)
Goodman Line
a
Safe design region
(for arbitrary fluctuations
in m and a )
a
Sf

m
Sut
Sf


m
Sut
1
1

(safe stress line)
Important point: Part can fail because of fluctuations in either a, m, or both.
Design for prescribed variations in a and m to get a more exact solution.
Special cases of fluctuating stresses

Case 1: m fixed


Sa
a
Case 2: a fixed

Sm
m
Special cases of fluctuating stresses

Case 3: a / m fixed


Sa
a

Sm
m
Case 4: both vary arbitrarily
1


a
Sf

m
Sut
Example

Given:







Sut = 1400 MPa
Syt = 950 MPa
Heat-treated (as-forged)
Fmean = 9.36 kN
Fmax = 10.67 kN
d/w = 0.133; d/h = 0.55
Find:

 for infinite life, assuming
Fmean is constant
Example, cont.

Find m and a
My
I
1
1
1
I  bh 3  w - d h 3  75 - 10183  3.16x10 -8 m 4
12
12
12
h
ymax   0.009 m
2
1
 F  L  1
M m   m    Fm L  9.36x103 0.3  702 Nm
4
 2  2  4

1
 F  L  1
M max   max    Fmax L  10.67x103 0.3  800 Nm
4
 2  2  4
M y
 m  m max  200 MPa
I
M y
 max  max max  228 MPa
I
 a   max -  m  28 MPa
Stress Concentration Factor
Example, cont.

Since this is uniaxial
loading,



m = 200 MPa
a = 28 MPa
nominal
We need to take care of
the SC factors

Su = 1400Mpa

kt ~ 2.2 (Figure A15-2)
q ~ 0.95 (Figure 7-20)
kf = 2.14

k f  1  qkt - 1
 a   a  k f  a
nom
 m   m  k f  m
 2.1428  60 MPa
nom
 2.14200  428 MPa
Example, cont.


Find strength
Eqn. 7-8: S’e = .504Sut
Se ~ 700 MPa since Sut  1400 MPa

Modification factors
Surface :
Size :
Load :
ka  aSut
Equation (7 - 19) :
Bending
kc  1 (Eq. 7 - 25)
b
a  271
2.8  d eq  51 mm
b  -0.995
d eq  0.808hb 
ka  0.201
kb  1.24d eq
kb  0.86
1
2
- 0.107
Se  0.2010.86700  121 MPa
Example, cont.

Design criteria

Goodman line:
a
Se


m
Sut
 1/ n
For arbitrary variation in
a and m,
a

m

121
a

m
121 1400
1
121 1400 
1 60 428


 121 1400
  1.25
1400
1
Example, cont.

However, we know that
Fmean = constant from
problem statement

m = constant
Sa  m

1
Se Sut
Sa
428

1
121 1400
Sa  84 MPa
S
84
 a 
 1 .4
 a 60
Less conservative!
Combined loading and fatigue
Size factor depends on loading
 SC factors also depend on loading
 Could be very complicated calculation to keep track of each load
case
 Assuming all stress components are completely reversing and
are always in time phase with each other,
1. For the strength, use the fully corrected endurance limit for
bending, Se
2. Apply the appropriate fatigue SC factors to the torsional stress,
the bending stress, and the axial stress components
3. Multiply any alternating axial stress components by the factor
1/kc,ax
4. Enter the resultant stresses into a Mohr’s circle analysis to find
the principal stresses
5. Using the results of step 4, find the von Mises alternating stress
a’
6. Compare a’ with Sa to find the factor of safety
Additional details are in Section 6-14

More Fatigue Failure Examples