Loading in 2 Seconds...

MECH 401 Mechanical Design Applications Dr. M. O’Malley – Master Notes

Loading in 2 Seconds...

0 Views

Download Presentation
##### MECH 401 Mechanical Design Applications Dr. M. O’Malley – Master Notes

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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**• 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**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)**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 • 1N103 cycles • High-cycle fatigue • N>103 cycles**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**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**• 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**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)**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**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’ = 0.504 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 7-9 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. 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**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 982-988 • 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**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)**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, • 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’ = 0.504 Sut for Sut< 1400 MPa (eqn. 7-8) 740 MPa else • AISI 1020 As-rolled • Sut = 448 MPa • Se’ = 0.506(448) = 227 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 (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**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**Example, cont.**• Modification factors • Surface: ka = aSutb (Eq. 7-18) • a and b from Table 7-4 • Machined • ka = (4.45)(448)-0.265 = 0.88**Example, cont.**• Size: kb • Axial loading • kb = 1 (Eq. 7-20) • Load: kc • Axial loading • kc = 0.85 (Eq. 7-25)**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:**Alternating vs. fluctuating**Alternating Fluctuating**Alternating Stresses**• sa characterizes alternating stress**Fluctuating stresses**• Mean Stress • Stress amplitude • Together, sm and sa characterize fluctuating stress**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