08 dmcslecturenotes-chapter6abweb
DESCRIPTION
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MECH 401 Mechanical Design Applications Dr. M. O’Malley– Master
Notes
Spring 2008
Dr. D. M. McStravick
Rice University
Reading Chapter 6
Homework 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
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
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:
As the number of cycles approaches infinity (N ∞), Sf(N) = Se (for iron or Steel)
a
f NS
)(
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
a
f NS
)(
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 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
a
f NS
)(
a
eS
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 SutNote: ASTM # for gray cast iron is the min Sut
S-N Plot with Endurance Limit
a
eS
a
f NS
)(
a
eS
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 < 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
For aluminum and copper
bSa
S
Sb
aNNS
ut
e
ut
bf
39.0loglog
9.0log
3
1
'
bSa
S
Sb
aNNS
ut
e
ut
bf
39.0loglog
9.0log
3
1
'
For 103 < N < 106
For N < 108
Where Se’ is the value of S’f at N = 108
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
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 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
Ground, machined, cold-drawn, hot-rolled, as-forged Size (kb)
Different factors depending on loading Bending and torsion (see pg. 280) 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 (Not significant factor for T<250 C [482 F])
Reliability (ke) 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: s’ > Sf(N) Factor of safety or Life of the part: h = Sf(N) / s’ Where b = - 1/3 log (0.9 Sut / Se) log (a) = log (0.9 Sut) - 3b
b
aN
1
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
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
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:
h = 1.8 Infinite life
Design Equation: h = Se / s’
Se because infinite life
Example, cont.
h = Se / s’ What do we need?
Se
s’ Considerations?
Infinite life, steel Modification factors Stress concentration
(hole) Find s’nom (without SC)
FF
hdb
P
A
Pnom 2083
101260
Example, cont.
Now add SC factor:
From Fig. 6-20, r = 6 mm Sut = 448 MPa = 65.0 ksi q ~ 0.8
nomtnomf kqk 11
Example, cont.
From Fig. A-15-1, Unloaded hole d/b = 12/60 = 0.2 kt ~ 2.5
q = 0.8 kt = 2.5
s’nom = 2083 F
FF
kq nomt
4583
208315.28.01
11
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
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 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 = aSut
b (Eq. 6-19) a and b from Table 6-2 Machined
ka = (4.45)(448)-0.265 = 0.88
Example, cont.
Size: kb Axial loading kb = 1 (Eq. 6-21)
Load: kc
Axial loading kc = 0.85 (Eq. 6-26)
Example, cont.
Temperature: 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:
kN 4.218.14583
10x177
8.14583
177
6
F
F
MPaSe
Fluctuating Fatigue Failures
Alternating vs. fluctuatingAlternating Fluctuating
I
MrA
P
a
m
Alternating Stresses
sa characterizes alternating stress
Fluctuating stresses
Mean Stress
Stress amplitude
Together, sm and sa characterize fluctuating stress
2' minmax
m
2minmax'
a
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 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 < 1 Failure
)S(or max ucycam S
1
ut
m
f
a
SS
Modified Goodman Langer Equations
Fluctuating stresses, cont.
Relationship is easily seen by plotting:
Goodman Line
(safe stress line)
Safe design region(for arbitrary fluctuationsin sm and sa )
1ut
m
f
a
SS
1
ut
m
f
a
SS
Note: sm + sa = smax
sm + sa > Syt (static failure by yielding)
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.
Special cases of fluctuating stresses Case 1: sm fixed
Case 2: sa fixed
a
aS
m
mS
Special cases of fluctuating stresses Case 3: sa / sm fixed
Case 4: both vary arbitrarily
m
m
a
a SS
ut
m
f
a
SS
1
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: h for infinite life, assuming
Fmean is constant
Example, cont.
Find sm and sa
MPa28
MPa228
MPa200
Nm 8003.0100.67x14
1
4
1
22
Nm 7023.010x36.94
1
4
1
22
m 009.02
mx1016.318107512
1
12
1
12
1
max
maxmaxmax
max
3max
maxmax
3
max
48333
ma
mm
mm
m
I
yMI
yM
LFLF
M
LFLF
M
hy
hdwbhI
I
My
Stress Concentration Factor
Example, cont.
Since this is uniaxial loading, m = 200 MPa a = 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
11 tf kqk
nominal
MPa 42820014.2
MPa 602814.2
nom
nom
mfmm
afaa
k
k
Example, cont.
Find strength Eqn. 7-8: S’e = .504Sut
Modification factors
86.0
24.1
808.0
mm 51d2.8
:19)-(7Equation
:Size
107.0
2
1
eq
b
eqb
eq
k
dk
hbd
MPa1400S since MPa 700~ ut eS
201.0
995.0
271
:Surface
a
buta
k
b
a
aSk25)-7 (Eq. 1
Bending
:Load
ck
MPa12170086.0201.0 eS
Example, cont.
Design criteria Goodman line:
For arbitrary variation in sa and sm,
nSS ut
m
e
a /1
121
1400
11400121
ma
25.1
1400
428
121
601
1
1400121
ma
Example, cont.
However, we know that Fmean = constant from problem statement sm = constant
4.160
84
MPa84
11400
428
121
1
a
a
a
a
ut
m
e
a
S
S
S
SS
S
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 sa’
6. Compare sa’ with Sa to find the factor of safetyAdditional details are in Section 6-14
More Fatigue Failure Examples