1. MECH 401 Mechanical Design Applications Dr. M. O’Malley– Master Notes
Spring 2008
Dr. D. M. McStravick
Rice University

2. Reading Homework Tests Chapter 6 HW 4 available, due 2-7
Fundamentals Exam will be in class on 2-21

4. 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

6. 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)

7. 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

8. More Fatigue Failure Examples (ASM)

9. More Fatigue Failure Examples (ASM)

10. 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)

11. Fatigue-life methods Three major methods
Stress-life
Strain-life
Linear-elastic fracture mechanics
Each predict life in number of cycles to failure, N, for a specified level of loading
Low-cycle fatigue
1N103 cycles
High-cycle fatigue
N>103 cycles

12. The 3 major methods Stress-life Strain-life
Based on stress levels only
Least accurate for low-cycle fatigue
Most traditional
Easiest to implement
Ample supporting data
Represents high-cycle applications adequately
Strain-life
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

13. Strain-life method
Fatigue failure almost always begins at local discontinuity
Notch, crack or other SC
When stress at discontinuity > elastic limit, plastic strain occurs
Fatigue fracture occurs for cyclic plastic strains
Can find fatigue life given strain and other cyclic characteristics
Often the designer does not have these a priori

14. Linear-elastic fracture mechanics method
Stage 1 – crystal slip through several contiguous grains
Stage 2 – crack extension
Stage 3 – fracture
Method involves
Determining stress intensity as function of crack length
From here, determine life
In reality, computer programs are used to calculate fatigue crack growth and therefore onset of failure

15. Fatigue analysis
Always good engineering practice to conduct a testing program on the materials to be employed in design and manufacture
Is actually a requirement in guarding against possibility of fatigue failure
Because of this necessity, it would really be unnecessary for us to proceed in study of fatigue failure except for one important reason:
The desire to know why fatigue failures occur so that the most effective method or methods can be used to improve fatigue strength
Stress-life method
Least accurate for low-cycle
Most traditional
We will come back to this method later

16. Fatigue analysis 2 primary classifications of fatigue
Alternating – no DC component
Fluctuating – non-zero DC component

17. 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:
As the number of cycles approaches infinity (N  ∞), Sf(N) = Se (for iron or Steel)

18. 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 s-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

19. Terminology and notation
Infinite life versus finite life
Infinite life
Implies N ∞
Use endurance limit (Se) of material
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)
Prime (‘) versus no prime
Variables 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

20. 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’ = Sut for Sut < 212 ksi (1460 MPa)
= 107 ksi (740 MPa) for all other values of Sut
Iron
Se’ = 0.4 Sut for Sut < 60 ksi (400 MPa)
= 24 ksi (160 MPa) for all other values of Sut

21. S-N Plot with Endurance Limit

22. 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) = 0.4 Sut for Sut < 48 ksi (330 MPa)
= 19 ksi (130 MPa) for all other values of Sut
Copper alloys
Se’ (Sf at 108 cycles) = 0.4 Sut for Sut < 40 ksi (280 MPa)
= 14 ksi (100 MPa) for all other values of Sut

23. 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 (Sf) is never less than the endurance limit (Se’)
For aluminum and copper, note that the fatigue strength (Sf) eventually goes to zero (failure!), but we will use the value of Sf at 108 cycles as our endurance limit (Se’) for these materials

24. 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 (Sf)
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 Sf 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]

25. Estimating Sf (N) For steel and iron For f=0.9 For aluminum and copper
For 103 < N < 106
For N < 108
Where Se’ is the value of
Sf at N = 108

26. 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
Strength reduction factors
Marin modification factors
These will account for differences between an ideal lab specimen and real life
Se = ka kb kc kd ke kf Se’
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 7-9 of Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide

27. Endurance limit modifying factors
Surface (ka)
Accounts for different surface finishes
Ground, machined, cold-drawn, hot-rolled, as-forged
Size (kb)
Different factors depending on loading
Bending and torsion (see pg. 329)
Axial (kb = 1)
Loading (kc)
Endurance limits differ with Sut based on fatigue loading (bending, axial, torsion)
Temperature (kd)
Accounts for effects of operating temperature
Reliability (ke)
Accounts for scatter of data from actual test results
We will probably not address ke
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

28. 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:
s’ > Sf(N)
Factor of safety:
h = Sf(N) / s’
Life of part
b = - 1/3 log (0.9 Sut / Se) log(a) = log (0.9 Sut) - 3b

29. Stress concentrations 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

30. SC factor – fatigue s = kfsnom+ = kfso t = kfstnom = kfsto
kf is a reduced value of kT and so 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

31. Fatigue SC factor kf = [1 + q(kt – 1)] kfs = [1 + qshear(kts – 1)]
kt or kts and nominal stresses
Pages
q and qshear
Notch sensitivity factor
Find using figures 7-20 and 7-21 in book (SMB) 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

32. Example AISI 1020 as-rolled steel Machined finish Find Fmax for:
Infinite life
Design Equation:
h = Se / s’
Se because infinite life

33. Example, cont. h = Se / s’ What do we need? Considerations?
Infinite life, steel
Modification factors
Stress concentration (hole)
Find s’nom (without SC)

34. Example, cont. Now add SC factor: From Fig. 7-20, r = 6 mm
Sut = 448 MPa = 65.0 ksi
q ~ 0.8

35. Example, cont. From Fig. A-15-1, q = 0.8 kt = 2.5 s’nom = 2083 F
Unloaded hole
d/b = 12/60 = 0.2
kt ~ 2.5
q = 0.8
kt = 2.5
s’nom = 2083 F

36. Example, cont. Now, estimate Se Steel:
Se’ = Sut for Sut < 1400 MPa (eqn. 7-8)
740 MPa else
AISI 1020 As-rolled
Sut = 448 MPa
Se’ = 0.506(448) = 227 MPa

37. 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 (Sf) is never less than the endurance limit (Se’)
For aluminum and copper, note that the fatigue strength (Sf) eventually goes to zero (failure!), but we will use the value of Sf at 108 cycles as our endurance limit (Se’) for these materials

38. 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
Strength reduction factors
Marin modification factors
These will account for differences between an ideal lab specimen and real life
Se = ka kb kc kd ke kf Se’
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 7-9 of Shigley-Mischke-Budynas (see examples)
If calculating fatigue strength for finite life, (Sf), use equations on previous slide

39. Example, cont. Modification factors Surface: ka = aSutb (Eq. 7-18)
a and b from Table 7-4
Machined
ka = (4.45)(448) = 0.88

40. Example, cont. Size: kb Load: kc Axial loading kb = 1 (Eq. 7-20)
kc = 0.85 (Eq. 7-25)

41. Example, cont. Temperature: Reliability: Miscellaneous:
kd = 1 (no info given)
Reliability:
ke = 1 (no info given)
Miscellaneous:
kf = 1
Endurance limit:
Se = kakbkckdkekfSe’ = (0.88)(0.85)(227) = 177 MPa
Design Equation:

42. Fluctuating Fatigue Failures

43. More Fatigue Failure Examples

44. Alternating vs. fluctuating

45. Alternating Stresses
sa characterizes alternating stress

46. Fluctuating stresses Mean Stress Stress amplitude
Together, sm and sa characterize fluctuating stress

47. Alternating vs. Fluctuating

48. Modified Goodman Diagram

49. 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

50. Fluctuating stresses, cont.
As with alternating stresses, fluctuating stresses have been investigated in an empirical manner
For sm < 0 (compressive mean stress)
sa > Sf Failure
Same as with alternating stresses
Or,
Static Failure
For sm > 0 (tensile mean stress)
Modified Goodman criteria
h < Failure

51. Fluctuating stresses, cont.
Note: sm + sa = smax
sm + sa > Syt (static failure by yielding)
Relationship is easily seen by plotting:
Goodman Line
Safe design region
(for arbitrary fluctuations
in sm and sa )
(safe stress line)
Important point: Part can fail because of fluctuations in either sa, sm, or both.
Design for prescribed variations in sa and sm to get a more exact solution.

52. Special cases of fluctuating stresses
Case 1: sm fixed
Case 2: sa fixed

53. Special cases of fluctuating stresses
Case 3: sa / sm fixed
Case 4: both vary arbitrarily

54. Example Given: Find: Sut = 1400 MPa Syt = 950 MPa
Heat-treated (as-forged)
Fmean = 9.36 kN
Fmax = kN
d/w = 0.133; d/h = 0.55
Find:
h for infinite life, assuming Fmean is constant

55. Example, cont.
Find sm and sa

56. Stress Concentration Factor

57. Example, cont. Since this is uniaxial loading,
sm = 200 MPa
sa = 28 MPa
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
nominal

58. Example, cont. Find strength Eqn. 7-8: S’e = .504Sut
Modification factors

59. Example, cont. Design criteria For arbitrary variation in sa and sm,
Goodman line:
For arbitrary variation in sa and sm,
121
1400

60. Example, cont.
However, we know that Fmean = constant from problem statement
sm = constant
Less conservative!

61. 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,
For the strength, use the fully corrected endurance limit for bending, Se
Apply the appropriate fatigue SC factors to the torsional stress, the bending stress, and the axial stress components
Multiply any alternating axial stress components by the factor 1/kc,ax
Enter the resultant stresses into a Mohr’s circle analysis to find the principal stresses
Using the results of step 4, find the von Mises alternating stress sa’
Compare sa’ with Sa to find the factor of safety
Additional details are in Section 7-14

62. Another Fatigue Failure Representation