MECH 401 Mechanical Design Applications Dr. M. O’Malley– Master Notes Spring 2007 Dr. D. M. McStravick Rice University

Reading Homework Test Fundaments Test 2/22/07 Chapter 6 HW 4 available, due 2-9-07 Test Fundaments Test 2/22/07

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

More Fatigue Failure Examples (ASM)

Fatigue Failure Examples Miscellaneous Fatigue Failures [Text fig 6-3 to 6-8] Jacob Neu chair failure See types of fatigue failures: figure 6-2

Fatigue Failure

Jim Neu Fatigue Failure

Various Fatigue Failure Modes: Fig. 6-2 bud21932_0602.jpg

Fatigue Thus far we’ve studied static failure of machine elements The second major class of component failure is due to dynamic loading 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

Approach to fatigue failure in analysis and design Fatigue-life methods (6-3 to 6-6) Stress-life method (rest of chapter 6)

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 Not accurate for low-cycle Most traditional We will come back to this method later

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

The 3 major methods Stress-life Strain-life Based on stress levels only Not used 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

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

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

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

Stress Life Approach Fatigue strength and endurance limit Estimating FS and EL Modifying factors

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)

Experimental Determination of S-N Curve

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

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

S-N Plot with Endurance Limit

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 < 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

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

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

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]

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

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 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

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

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

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 Pages 1006-1014 q and qshear Notch sensitivity factor Find using figures 6-20 and 6-21 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

Example AISI 1020 as-rolled steel Machined finish Find alternating 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? Considerations? Infinite life, steel Modification factors Stress concentration (hole) Find s’nom (without SC)

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

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

Example, cont. Now, estimate Se Steel: Se’ = 0.5 Sut for Sut < 1400 MPa (eqn. 6-8) 740 MPa for Sut > 1400 MPa AISI 1020 As-rolled Sut = 448 MPa Se’ = 0.5(448) = 227 MPa [Eqn. 6-8]

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-Budynas (see examples)

Example, cont. Modification factors Surface: ka = aSutb (Eq. 6-19) a and b from Table 6-2 Surface was Machined ka = (4.45)(448)-0.265 = 0.88

Example, cont. Size: kb Load: kc Axial loading kb = 1 (Eq. 6-21) kc = 0.85 (Eq. 6-26)

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:

Fluctuating Fatigue Failures 2 primary classifications of fatigue Alternating – no DC component Fluctuating – non-zero DC component

Alternating vs. fluctuating

Alternating Stresses sa characterizes alternating stress

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

Alternating vs. Fluctuating

Modified Goodman Diagram For a given midrange stress

Fluctuating Fatigue 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

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

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.

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

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

Example Given: Find: 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

Stress Concentration Factor

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 6-20) kf = 2.14 nominal

Example, cont. Find strength Eqn. 6-8: S’e = .5 Sut Modification factors

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

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

Example, cont. However, we used Fa = constant sa = constant Even 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, 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 6-14

Spares