fatique lecture notes
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Fatique FailureLecture NotesPrepared by
H. Orhan YILDIRAN
MECE 304 Mechanical Machine Elements-Fatique Failure
LECTURE NOTES- MECE 304 Mechanical Machine Elements
Chapter 3- Fatique failure
(Notes from: Chapter 6 in Budynas R.G., Nisbett J.K., Shigleys Mechanical Engineering Design,
Mc Graw Hill, 8th Edition)
Spring Semester 2007/2008
Halil Orhan YILDIRAN, MS2
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6-1 Fatigue in MetalsIn stress-strain testing diagram, the load is applied gradually, to give
sufficient time for the strain to fully develop and the specimen is tested to
destruction, and so the stresses are applied only once. Testing of this kind is
applicable, to what are known as static conditions.
Fatique loading conditions produces stresses that vary with time or they fluctuate between different levels. These stresses are called variable,
repeated, alternating, or fluctuating stresses.
In a fatigue failure;
*Maximum stresses of failure are well below the ultimate strength of the
material, and quite frequently below the yield strength.
*Repeated a very large number of times.
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MECE 304 Mechanical Machine Elements-Fatique Failure
Figure 6-2 Schematics of fatigue fracture surfaces produced in
smooth and notched components with round and rectangular cross sections under various loading conditions
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Figure 6-2 Schematics of fatigue fracture surfaces(Contn’d)
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Figure 6-2 Schematics of fatigue fracture surfaces(Contn’d)
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Figure 6-5 Fatique fracture surface of a connecting rod
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Figure 6-6 Fatique fracture surface of a piston rod.
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6-4 The Stress Life Method in Fatique Failure analysisTo determine the strength of materials under the action of fatigue loads,
specimens are subjected to repeated or varying forces of specified
magnitudes while the cycles or stress reversals are counted to destruction.
The most widely used fatigue-testing device is the R. R. Moore high-speed
rotating-beam machine. This machine subjects the specimen to pure
bending by means of weights. The specimen, shown in Fig. 6–9, is very carefully machined and polished.
Figure 6-9 Test specimen
geometry for R. R. Moore
high-speed rotating-beam machine
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The ordinate of the S-N diagram is called the fatigue strength Sf ; stated with the number of cycles N to which this strength corresponds.
In the case of the steels, a knee occurs in the graph, and beyond this knee
failure will not occur, no matter how great the number of cycles. The strength
corresponding to the knee is called the endurance limit Se, or the fatigue
limit. The graph of Fig. 6–10 never does become horizontal for nonferrous
metals and alloys.
Figure 6–10 An S-N diagram
plotted from the results of completely reversed axial
fatigue tests.
Material: UNS G41300
steel, normalized;Sut = 810 Mpa; maximumSut = 1050 Mpa
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6-7 The Endurance LimitThe determination of endurance limits by fatigue testing is now routine,
though a lengthy procedure. Generally, stress testing is preferred to strain
testing for endurance limits.
For preliminary and prototype design and for some failure analysis as well, a
quick method of estimating endurance limits is needed.
Figure 6-17 Graph of Se
vs Sut
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MECE 304 Mechanical Machine Elements-Fatique Failure
For steels, observation of Fig. 6–17, obtained from simple tensile test and rotating beam test, we will estimate the endurance limit as
0.5Sut for Sut ≤ 1400 MPa
S’e =700 MPa for Sut > 1400 MPa
Notes:*Heat treatment effects Se of steels. Ductile microstructures have higher
S’e/Sut ratios
*Endurance limit for cast iron (polished or machined) are given in the Table A-22.
*Aluminum alloys do not have endurance limits. See Table A-22 for fatique
strength of some aliminum alloys
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6-8 Fatique strength (Sf): From Figure 6-10 for low cycles (N=10^3 cycles)
At 10^3 cycles
For steels with HB ≤ 500
And for b
Where σ’F is fatique strength coefficient (p.269), f is the fraction ofSut represented by (S’f)10^3 cycles
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For an actual mechanical component, S’e is reduced to Se (see Sec. 6–9)
which is less than 0.5 Sut.
For the actual mechanical component, following equation can be written
Figure 6-18 f vs Sut at
10^3 cycles
for Se=S’e=0.5
Sut
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Where a and b are constants
For a completely reversed stress σa given, with Sf=σa Eq.6-13
becomes
For low cycle fatique (1≤N≤10^3 cycles)
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SAMPLE PROBLEM 6-2 Given a SAE 1050 HR steel, estimate:
(a) the rotating-beam endurance limit at 10^6 cycles.(b) the endurance strength of a polished rotating-beam specimen
corresponding to 10^4 cycles to failure
(c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 385 Mpa.
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6-9 Endurance Limit Modifying Factors
Rotating-beam specimen used in the laboratory to determine endurance
limits is prepared very carefully and tested under closely controlled
conditions.In general endurance limit of a mechanical or structural member in use, can
not match the values obtained in the laboratory. Some differences include;
• Material: composition, basis of failure, variability• Manufacturing: method, heat treatment, fretting corrosion, surface
condition, stress concentration
• Environment: corrosion, temperature, stress state, relaxation times
• Design: size, shape, life, stress state, stress concentration, speed, fretting, galling
Marin equation for modifying endurance limit:
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*Surface factor ka
Constants a and b are to be found from Table 6-2
Table 6-2 Constants a
and b for
surface modification
factor
SAMPLE PROBLEM 6-3 A steel has a minimum ultimate strength of 520
MPa and a machined surface. Estimate ka.
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*Size factor kb,
For bending and torsion
For axial loading
What are the effects of rotating/non rotating and circular/ and
noncircular cross sections?
For rotating round sections (Define 95 % stress area A0.95σ of a ring
with do=d, di=0.95d)
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For nonrotating solid or hollow rounds
Effective size of a round, corresponding to a nonrotating solid or hollow round.
For a rectangular section of dimensions h × b
SAMPLE PROBLEM 6-4 A steel shaft loaded in bending is 32 mm in
diameter, abutting a filleted shoulder 38 mm in diameter. The shaft
material has a mean ultimate tensile strength of 690 Mpa. Estimate the Marin size factor kb if the shaft is used in
(a) A rotating mode.
(b) A nonrotating mode
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Table 6-3 Effective
diameter (de) for common nonrotating
shapes
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*Loading factor kc
*Temperature factor kd (kd=ST/SRT)
Table 6-4 Effect of
Operating
Temperature on the
Tensile Strength ofSteel.* (ST = tensile
strength at operating
temperature;SRT = tensile strength
at room temperature
SAMPLE PROBLEM 6-5 A SAE 1035 steel has a tensile strength of 490 MPa and is to be used for a part that sees 230°C in service.
Estimate the Marin temperature modification factor and (Se)230
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Reliability factor ke
Table 6-5 Reliability Factors ke Corresponding to 8 Percent Standard
Deviation of the Endurance Limit
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Miscellaneous effect factor kfResidual stresses ; Generally, if the residual stress in the surface of the part
is compression, the endurance limit is improved. Fatigue failures appear to be
tensile failures, or at least to be caused by tensile stress, Operations such as shot peening, hammering, and cold rolling build
compressive stresses into the surface of the part and improve the endurance
limit significantly. Of course, the material must not be worked to exhaustion.
Directional characteristics of the operations. Rolled or drawn parts, for
example, have an endurance limit in the transverse direction that may be 10 to 20 percent less than the endurance limit in the longitudinal direction.
Case-hardened parts may fail at the surface or at the maximum core radius, depending upon the stress gradient. Figure 6–19 shows the typical triangular
stress distribution of a bar under bending or torsion. Also plotted as a heavy
line in this figure are the endurance limits Se for the case and core. For this
example the endurance limit of the core rules the design because the figure shows that the stress σ or τ, whichever applies, at the outer core radius, is appreciably larger than the core endurance limit.
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Other factors for miscellaneous effects factor kf
*Corrosion:Parts operation in corrosive atmospheres have a lowered fatique
resistance.
There is no fatique limit.
Minimize factors effecting fatique life
*Electrolytic plating: Cr, Ni and Cd reduce strength up to 50%. Zinc plating
has no effect on fatique strength.
*Metal spraying : Reduces up to 14 %
*Cyclic frequency: Shorter the life
*Frettage corrosion: Microscopic motions of tigthly fitted parts. Depends on
material of mating parts, ranges in strength reduction are 24 to 90 %
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6-10 Stress Concentration and Notch Sensitivity
Some materials are not fully sensitive to the presence of irregularities
or discontinuities, such as holes, grooves, or notches, as with the case of stress concentration factor Kt (or Kts). So we introduce an another
factor Kf, fatique stress concentration factor and
Notch sensitivity q is defined by the equation
Or if we know q ,
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Figure 6-20 Notch-sensitivity charts for steels and UNS A92024-T wrought
aluminum alloys subjected to reversed bending or reversed axial loads
SAMPLE PROBLEM 6-6 A steel shaft in bending has an ultimate
strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using:Figure 6–20.
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Figure 6-21 Notch-sensitivity curves for materials in reversed torsion.
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Neuber equation, which is given by the formula below is the bases for Figure 6-21
where √a is defined as the Neuber constant and is a material constant.
Equating Eqs. (6–31) and (6–33) yields the notch sensitivity equation
Notes:
*for cast iron take q=0.20
*For steel see Table 6-15*for simple loading reduce endurance limit or multiply stresses by Kf. For
complex loading stresses are multiplied by Kf.
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SAMPLE PROBLEM 6-9 Figure 6–22a shows a rotating shaft simply
supported in ball bearings at A and D and loaded by a nonrotating force F
of 6.8 kN. Using ASTM “minimum” strengths, estimate the life of the part.
Figure 6-22 (a) Shaft drawing showing alldimensions in millimeters; all fillets 3-mm radius. The shaft rotates and the load isstationary; material is machined from AISI 1050 cold-drawn steel. (b) Bending momentdiagram
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6-11 Fluctuating Stresses
Figure 6-23 Stress time
relation for sinusoidal fluctuating stress
R=Stress Ratio
A=Amplitude Ratio
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6-12 Fatique failure Criteria for Fluctuating Stress
Figure 6-24
Modified Goodman
diagram showing all the
strengths and the limiting values of all the stress
components for a
particular midrange stress
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Figure 6-7 Fatigue diagram showing various criteria of failure.
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Sodeberg line criterion Eq.
Mod.Goodman criterion
Eq.
Gerber criterion Eq.
ASME elliptic criterion Eq.
Langer (yielding) criterion
Eq.
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For design purposes
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Tables 6-6 to 6-8. First row fatique criterion,second row static Langer criterion, third row intersection of static and fatique criteria
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SAMPLE PROBLEM 6-10 A 140 mm diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 170kN. Because of the ends, and the fillet
radius, a fatigue stress-concentration factor Kf is 1.85 for 10^6 or larger
life. Find Sa and Sm and the factor of safety guarding against fatigue
and first-cycle yielding, using (a) the Gerber fatigue line and (b) the
ASME-elliptic fatigue line.
Figure 6-28 Principal points A, B, C, and D on
the designer’s diagram
drawn for Gerber,
Langer, and load line
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6-13 Torsional fatique strength in Fluctuation stresses
In construction of Goodman diagram use
6-14 Combination of loading Models with alternating and midrange components
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6-15 Varying Fluctuating stresses:
On a part suppose fully reversed σ1 act for n1 cycles, σ2 act for n2 cycles
etc. Than we have
Where
ni is the number of cycles of stress level σi
Ni is the number of cycles to failure at stress level σi
Minor’s Rule