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Design Stress & Fatigue

Design Stress & Fatigue. MET 210W E. Evans. Parts Fail When?. P. Crack initiation site.

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Design Stress & Fatigue

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  1. Design Stress & Fatigue MET 210W E. Evans

  2. Parts Fail When? P Crack initiation site 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. P

  3. Design Factor • Analysis • Design

  4. Factors Effecting Design Factor • Application • Environment • Loads • Types of Stresses • Material • Confidence

  5. Factors Effecting Design Factor • Application • Environment • Loads • Types of Stresses • Material • Confidence • How many will be produced? • What manufacturing methods will be used? • What are the consequences of failure? • Danger to people • Cost • Size and weight important? • What is the life of the component? • Justify design expense?

  6. 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

  7. 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

  8. Factors Effecting Design Factor • Application • Environment • Loads • Types of Stresses • Material • Confidence • What kind of stress? • Direct tension or compression • Direct shear • Bending • Torsional shear • Application • Uniaxial • Biaxial • Triaxial

  9. Factors Effecting Design Factor • Application • Environment • Loads • Types of Stresses • Material • Confidence • Material properties • Ultimate strength, yield strength, endurance strength, • Ductility • Ductile: %E  5% • Brittle: %E < 5% • Ductile materials are preferred for fatigue, shock or impact loads.

  10. 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?

  11. Design Factor Adapted from R. B. Englund

  12. Design Factor

  13. Predictions of Failure Static Loads • Brittle Materials: • Maximum Normal Stress - Uniaxial • Modified Mohr - Biaxial • Ductile Materials: • Yield Strength - Uniaxial • Maximum Shear Strength - Biaxial • Distortion Energy - Biaxial or Triaxial

  14. Predictions of Failure Fluctuating Loads • Brittle Materials: • Not recommended • Ductile Materials: • Goodman • Gerber • Soderberg

  15. Maximum Normal Stress • Uniaxial Static Loads on Brittle Material: • In tension: Kts sd = Sut / N • In compression: Kts sd = Suc / N

  16. Modified Mohr • Biaxial Static Stress on Brittle Materials s2 Sut 45° Shear Diagonal s2 s1 Suc Sut s1 Stress concentrations applied to stresses before making the circle s1, s2 Often brittle materials have much larger compressive strength than tensile strength Suc

  17. Yield Strength Method • Uniaxial Static Stress on Ductile Materials • In tension: • s sd = Syt / N • In compression: •  sd = Syc / N For most ductile materials, Syt = Syc

  18. 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 tensile-test specimen when yielding begins. Somewhat conservative – use Distortion Energy for more precise failure estimate

  19. Distortion Energy • Static Biaxial or Triaxial Stress on Ductile Materials Shear Diagonal s2 Best predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses. Sy Sy s1 Sy s’ = von Mises stress Failure: s’ > Sy Design: s’  sd = Sy/N Sy Distortion Energy

  20. von Mises Stress • Alternate Form For uniaxial stress when sy = 0, • Triaxial Distortion Energy (s1 > s2 > s3)

  21. Fluctuating Stress • Varying stress with a nonzero mean. salternating = sa smax Stress Ratio, Stress smean Time -1  R  1 smin

  22. Fluctuating Stress Example • Bending of Rocker Arm Valve Spring Force Valve Open Valve Closed • Tension in Valve Stem Valve Closed Valve Spring Force Valve Open RBE 2/1/91 Adapted from R. B. Englund

  23. Fatigue Testing • Bending tests • Spinning bending elements – most common • Constant stress cantilever beams Top View Front View Applied Deformation – Fully Reversed, R = -1 Fixed Support

  24. Fatigue Testing Test Data Stress, s (ksi) Number of Cycles to Failure, N Data from R. B. Englund, 2/5/93

  25. 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.

  26. Typical S-N Curve

  27. Estimated Sn of Various Materials

  28. Actual Endurance Strength Sn’ = Sn(Cm)(Cst)(CR)(CS) Sn’ = actual endurance strength (ESTIMATE) Sn = endurance strength from Fig. 5-8 Cm = material factor (pg. 174) Cst = stress type: 1.0 for bending 0.8 for axial tension 0.577 for shear CR = reliability factor CS = size factor

  29. 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

  30. 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 Actual Sn’ Estimate Axial Tension Reliability, Table 5-1 Size Factor, Fig. 5-9 99% Probability Sn’ is at or above the calculated value Guessing: diameter  .5”

  31. Goodman Diagram sa Yield Line Sy FATIGUE FAILURE REGION Sn’ Goodman Line NO FATIGUE FAILURE REGION sm 0 Sy Su -Sy

  32. Goodman Diagram sa Safe Stress Line Yield Line Sy FATIGUE FAILURE REGION Sn’ Goodman Line Sn’/N SAFE ZONE sm 0 Su/N Sy Su -Sy Safe Stress Line

  33. F F Example: Problem 5-53. Find a suitable titanium alloy. N = 3 1.5 mm Radius 30 mm DIA 42 mm DIA F varies from 20 to 30.3 kN + MAX = 30.3 FORCE MIN = 20 TIME -

  34. Example: Problem 5-53 continued. • Find the mean stress: • Find the alternating stress: • Stress concentration from App. A15-1:

  35. Example: Problem 5-53 continued. • Sn data not available for titanium so we will guess! Assume Sn = Su/4 for extra safety factor. • TRYT2-65A, Su = 448 MPa, Sy = 379 MPa (Eqn 5-20) Size Reliability 50% Tension 3.36 is good, need further information on Sn for titanium.

  36. Example: Find a suitable steel for N = 3 & 90% reliable. 3 mm Radius 50 mm DIA 30 mm DIA T T T varies from 848 N-m to 1272 N-m + MAX = 1272 N-m TORQUE MIN = 848 N-m TIME - T = 1060 ± 212 N-m

  37. Example: continued. • Stress concentration from App. A15-1: • Find the mean shear stress: • Find the alternating shear stress:

  38. 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: tmax =200 + 40 = 240 MPa  Sys  600/2 = 300MPa • Find the ultimate shear stress: Sus = .75Su = .75(779 MPa) = 584 MPa Ductile

  39. Sn Size – 30 mm Wrought steel 90% Reliability Shear Stress Example: continued. • Assume machined surface, Sn 295 MPa • Find actual endurance strength: S’sn = Sn(Cm)(Cst)(CR)(CS) = 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa (Fig. 5-8)

  40. Example: continued. (Eqn. 5-28) • Goodman: 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.

  41. Example: continued. • Guess another material. TRY: AISI 1340 OQT 700°F Su = 1520 MPa, Sy = 1360 MPa, %E = 10% • 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 Ductile Sn shear size wrought reliable

  42. Example: continued. (Eqn. 5-28) • Goodman: • 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

  43. Example: Combined Stress Fatigue RBE 2/11/97

  44. Example: Combined Stress FatigueCont’d PIPE: TS4 x .237 WALL MATERIAL: ASTM A242 Equivalent DEAD WEIGHT: SIGN + ARM + POST = 1000# (Compression) Reversed, Repeated 45° Bending RBE 2/11/97 Repeated one direction

  45. Example: Combined Stress FatigueCont’d Stress Analysis: Dead Weight: (Static) Vertical from Wind: (Cyclic) Bending: (Static)

  46. z z x x Example: Combined Stress FatigueCont’d Stress Analysis: Torsion: (Cyclic) Stress Elements:(Viewed from +y) CYCLIC: STATIC: 315.5 psi 63.09 psi – Repeated One Direction 9345.8 psi t = 3115.3 psi Fully Reversed

  47. Example: Combined Stress FatigueCont’d + TIME 9345.8 -315.5 -31.5 8998.8 Static sm Stress Repeated / 2 - MIN = -63.09 psi psi t(CW) sa tmax t(CW) tmax (0,-3115.3) s s s1 (-31.5,-3115.3) Mean Stress: Alternating Stress:

  48. Example: Combined Stress FatigueCont’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

  49. Example: Combined Stress FatigueCont’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 Size – 4.50” dia Wrought steel 90% Reliability Combined or Shear Stress

  50. Example: Combined Stress FatigueCont’d S’sn N = 1 Fail Not Fail Su Kttalt 3115.3 N = 3 Safe tmean = 4499.4 “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 10 Alternating Stress, ta 5 S’sn/N 0 Su/N 0 5 10 15 20 Mean Stress, tm

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