Report

Design Stress & Fatigue MET 210W E. Evans P Parts Fail When? Crack initiation site P This crack in the part is very small. If the level of stress in the part is SMALL, the crack will remain stable and not expand. If the level of stress in the part is HIGH enough, the crack will get bigger (propagate) and the part will eventually fail. Design Factor • Analysis Failure Strength Factor of Saf ety Applied Stress Sy Example : N • Design Failure Strength Allowable Stress Design Factor Sy Example : ALLOW N Factors Effecting Design Factor • • • • • • Application Environment Loads Types of Stresses Material Confidence Factors Effecting Design Factor • Application • How many will be produced? • • • • • • What manufacturing methods will be used? Environment Loads Types of Stresses Material Confidence • What are the consequences of failure? •Danger to people •Cost • Size and weight important? • What is the life of the component? • Justify design expense? Factors Effecting Design Factor • Application • Environment • • • • Loads Types of Stresses Material Confidence • Temperature range. • Exposure to electrical voltage or current. • Susceptible to corrosion • Is noise control important? • Is vibration control important? • Will the component be protected? •Guard •Housing Factors Effecting Design Factor • Application • Environment • Loads • Types of Stresses • Material • Confidence • Nature of the load considering all modes of operation: • Startup, shutdown, normal operation, any foreseeable overloads • Load characteristic • Static, repeated & reversed, fluctuating, shock or impact • Variations of loads over time. • Magnitudes • Maximum, minimum, mean Factors Effecting Design Factor • Application • Environment • Loads • Types of • Material • Confidence • What kind of stress? • Direct tension or compression • Direct shear • Bending Stresses • Torsional shear • Application • Uniaxial • Biaxial • Triaxial Factors Effecting Design Factor • • • • Application Environment Loads Types of Stresses • Material • Confidence • Material properties • Ultimate strength, yield strength, endurance strength, • Ductility • Ductile: • Brittle: %E 5% %E < 5% • Ductile materials are preferred for fatigue, shock or impact loads. Factors Effecting Design Factor • • • • • Application Environment Loads Types of Stresses Material • Confidence • Reliability of data for • Loads • Material properties • Stress calculations • How good is manufacturing quality control • Will subsequent handling, use and environmental conditions affect the safety or life of the component? Design Factor Adapted from R. B. Englund Design Factor Predictions of Failure Static Loads • Brittle Materials: – Maximum Normal Stress – Modified Mohr - Uniaxial - Biaxial • Ductile Materials: – Yield Strength – Maximum Shear Strength – Distortion Energy - Uniaxial - Biaxial - Biaxial or Triaxial Predictions of Failure Fluctuating Loads • Brittle Materials: – Not recommended • Ductile Materials: – Goodman – Gerber – Soderberg Maximum Normal Stress •Uniaxial Static Loads on Brittle Material: –In tension: Kt d = Sut / N –In compression: Kt d = Suc / N Modified Mohr • Biaxial Static Stress on Brittle Materials 2 45° Shear Diagonal Sut Suc Sut 2 1 1, 2 Suc 1 Stress concentrations applied to stresses before making the circle Often brittle materials have much larger compressive strength than tensile strength Yield Strength Method • Uniaxial Static Stress on Ductile Materials – In tension: d = Syt / N – In compression: d = Syc / N For most ductile materials, Syt = Syc Maximum Shear Stress • Biaxial Static Stress on Ductile Materials tmax td = Sys / N = 0.5(Sy )/ N Ductile materials begin to yield when the maximum shear stress in a load-carrying component exceeds that in a tensiletest specimen when yielding begins. Somewhat conservative – use Distortion Energy for more precise failure estimate Distortion Energy • Static Shear Diagonal Biaxial or Triaxial Stress on Ductile Materials 2 Best predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses. Sy Sy Sy 1 ' 12 2 1 2 2 ’ = von Mises stress Sy Distortion Energy Failure: ’ > Sy Design: ’ d = Sy/N von Mises Stress • Alternate Form ' x y 3t 2 x 2 y For uniaxial stress when y = 0, • Triaxial Distortion Energy 2 xy ' 3t 2 x 2 xy (1 > 2 > 3) 2 ( 2 1 )2 (3 1 )2 (3 2 )2 ' 2 Fluctuating Stress • Varying stress with a nonzero mean. alternating = a Stress max max min mean 2 max min a 2 Stress Ratio, mean min min R max Time -1 R 1 Fluctuating Stress Example • Bending of Rocker Arm Valve Spring Force Valve Open Valve Closed • Tension in Valve Stem Valve Closed RBE 2/1/91 Valve Spring Force Valve Open Adapted from R. B. Englund Fatigue Testing • Bending tests – Spinning bending elements – most common – Constant stress cantilever beams Top View Front View Fixed Support Applied Deformation – Fully Reversed, R = -1 Test Data Stress, (ksi) Fatigue Testing Number of Cycles to Failure, N Data from R. B. Englund, 2/5/93 Endurance Strength • The stress level that a material can survive for a given number of load cycles. • For infinite number of cycles, the stress level is called the endurance limit. • Estimate for Wrought Steel: Endurance Strength = 0.50(Su) • Most nonferrous metals (aluminum) do not have an endurance limit. Typical S-N Curve Estimated Sn of Various Materials Actual Endurance Strength Sn’ = Sn(Cm)(Cst)(CR)(CS) Sn’ Sn Cm Cst CR CS = actual endurance strength (ESTIMATE) = endurance strength from Fig. 5-8 = material factor (pg. 174) = stress type: 1.0 for bending 0.8 for axial tension 0.577 for shear = reliability factor = size factor Actual Sn Example • Find the endurance strength for the valve stem. It is made of AISI 4340 OQT 900°F. From Fig. A4-5. Su = 190 ksi From Fig. 5-8. Sn = 62 ksi (machined) 62 ksi Actual Sn Example Continued Sn’ = Sn(Cm)(Cst)(CR)(CS) = 62 ksi(1.0)(.8)(.81)(.94) = 37.8 ksi Sn,Table 5-8 Wrought Steel Axial Tension Actual Sn’ Estimate Reliability, Table 5-1 Size Factor, Fig. 5-9 99% Probability Sn’ is at or above the Guessing: diameter .5” calculated value Goodman Diagram a Sy Yield Line Sn’ NO FATIGUE FAILURE REGION -Sy 0 FATIGUE FAILURE REGION Goodman Line a m 1 Sn S u Sy Su m Goodman Diagram Safe Stress Line K t a Sn a m 1 Su N Sy Yield Line Sn’ FATIGUE FAILURE REGION Goodman Line a m 1 Sn S u Sn’/N SAFE ZONE -Sy 0 Su/N Sy Su Safe Stress Line m Example: Problem 5-53. Find a suitable titanium alloy. N = 3 1.5 mm Radius 42 mm DIA 30 mm DIA F varies from 20 to 30.3 kN + FORCE MAX = 30.3 - 30.3 20 alt 5.15 kN 2 30.3 20 mean 25.15 kN 2 MIN = 20 TIME Example: Problem 5-53 continued. • Find the mean stress: 25,150 N m 3 5 .6M P a (30 mm)2 4 • Find the alternating stress: a 5,150 N (30 mm)2 4 7 .3 M P a • Stress concentration from App. A15-1: D 42 mm 1.4; d 30 mm r 1.5 mm .05 d 30 mm K t 2 .3 Example: Problem 5-53 continued. • Sn data not available for titanium so we will guess! Assume Sn = Su/4 for extra safety factor. • TRY T2-65A, Su = 448 MPa, Sy = 379 MPa K t a Sn m 1 Su N (Eqn 5-20) 2.3(7.3 MPa ) 35.6 MPa 1 .297 .8(.86)(448 MPa / 4) 448 MPa N Size Tension Reliability 50% 1 N 3 .3 6 .297 3.36 is good, need further information on Sn for titanium. Example: Find a suitable steel for N = 3 & 90% reliable. 3 mm Radius 50 mm DIA 30 mm DIA T varies from 848 N-m to 1272 N-m TORQUE + - MAX = 1272 N-m MIN = 848 N-m TIME 1272 848 alt 212 N m 2 1272 848 mean 1060 N m 2 T = 1060 ± 212 N-m Example: continued. • Stress concentration from App. A15-1: D 50 mm 1.667; d 30 mm r 3 mm .1 K t 1 .3 8 d 30 mm • Find the mean shear stress: ) Tm 1060 N m(1000 mm m tm 2 0 0M P a Zp (30 mm)3 16 • Find the alternating shear stress: Ta 212000 N mm ta 40M Pa 3 Zp 5301 mm Example: continued. • So, t = 200 ± 40 MPa. Guess a material. TRY: AISI 1040 OQT 400°F Su = 779 MPa, Sy = 600 MPa, %E = 19% • Verify that tmax Sys: Ductile tmax = 200 + 40 = 240 MPa Sys 600/2 = 300MPa • Find the ultimate shear stress: Sus = .75Su = .75(779 MPa) = 584 MPa Example: continued. • Assume machined surface, Sn 295 MPa (Fig. 5-8) • Find actual endurance strength: S’sn = Sn(Cm)(Cst)(CR)(CS) = 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa Sn Wrought steel Shear Stress Size – 30 mm 90% Reliability Example: continued. • Goodman: K t ta Ssn tm 1 S su N (Eqn. 5-28) 1.38( 40 MPa ) 200 MPa 1 .7606 132 MPa 584 MPa N 1 N 1 .3 1 .7606 No Good!!! We wanted N 3 Need a material with Su about 3 times bigger than this guess or/and a better surface finish on the part. Example: continued. • Guess another material. TRY: AISI 1340 OQT 700°F Su = 1520 MPa, Sy = 1360 MPa, %E = 10% Ductile • Find the ultimate shear stress: Sus = .75Su = .75(779 MPa) = 584 MPa • Find actual endurance strength: S’sn = Sn(Cm)(Cst)(CR)(CS) = 610 MPa(1.0)(.577)(.9)(.86) = 272 MPa Sn shear wrought size reliable Example: continued. • Goodman: K t ta Ssn tm 1 S su N (Eqn. 5-28) 1.38( 40 MPa ) 200 MPa 1 .378 272 MPa 1140 MPa N 1 N 2 .6 4 .378 No Good!!! We wanted N 3 Decision Point: • Accept 2.64 as close enough to 3.0? • Go to polished surface? • Change dimensions? Material? (Can’t do much better in steel since Sn does not improve much for Su > 1500 MPa Example: Combined Stress Fatigue RBE 2/11/97 Example: Combined Stress Fatigue Cont’d PIPE: TS4 x .237 WALL MATERIAL: ASTM A242 Equivalent Reversed, Repeated DEAD WEIGHT: SIGN + ARM + POST = 1000# (Compression) 45° Bending RBE 2/11/97 Repeated one direction Example: Combined Stress Fatigue Cont’d Stress Analysis: Dead Weight: P 1000 # 315.5 psi 2 A 3.17 in (Static) Vertical from Wind: P 200 # 63.09 psi 2 A 3.17 in (Cyclic) Bending: M 500# (60 in) 9345.8 psi 3 Z 3.21 in (Static) Example: Combined Stress Fatigue Cont’d Stress Analysis: Torsion: T 200# (100 in) t 3115.3 psi 3 ZP 2(3.21 in ) Stress Elements: STATIC: 315.5 psi (Cyclic) (Viewed from +y) CYCLIC: 9345.8 psi z x z x 63.09 psi – Repeated One Direction t = 3115.3 psi Fully Reversed Example: Combined Stress Fatigue Cont’d Mean Stress: Static Repeated / 2 - 8998.8 psi t (CW) TIME Stress 9345.8 -315.5 -31.5 + Alternating Stress: m MIN = -63.09 psi t (CW) a tmax tmax (0,-3115.3) 1 (-31.5,-3115.3) tmax 8998.8 psi 4499.4 psi 2 tmax 3115.34 psi Example: Combined Stress Fatigue Cont’d Determine Strength: Try for N = 3 some uncertainty Size Factor? OD = 4.50 in, Wall thickness = .237 in ID = 4.50” – 2(.237”) = 4.026 in Max. stress at OD. The stress declines to 95% at 95% of the OD = .95(4.50”) = 4.275 in. Therefore, amount of steel at or above 95% stress is the same as in 4.50” solid. ASTM A242: Su = 70 ksi, Sy = 50 ksi, %E = 21% t 3/4” Ductile Example: Combined Stress Fatigue Cont’d We must use Ssu and S’sn since this is a combined stress situation. (Case I1, page 197) Sus = .75Su = .75(70 ksi) = 52.5 ksi S’sn = Sn(Cm)(Cst)(CR)(CS) = 23 ksi(1.0)(.577)(.9)(.745) = 8.9 ksi Hot Rolled Surface Wrought steel Combined or Shear Stress Size – 4.50” dia 90% Reliability Example: Combined Stress Fatigue Cont’d “Safe” Line for Goodman Diagram: ta = S’sn / N = 8.9 ksi / 3 = 2.97 ksi tm = Ssu / N = 52.5 ksi / 3 = 17.5 ksi Alternating Stress, ta K t ta Ssn 10 S’sn tm 1 S su N 1.0(3115.3 psi) 4499.4 psi 1 .426 8900 psi 52500 psi N N 5 S’sn/N Su Kttalt 3115.3 0 0 5 tmean = 4499.4 10 15 Mean Stress, tm 1 2 .2 9 .426 Su/N 20