Design Stress & Fatigue MET 210W E. Evans

Parts Fail When? P 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

Design Factor Analysis Design

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

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?

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 Stresses Material Confidence What kind of stress? Direct tension or compression Direct shear Bending 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: %E  5% Brittle: %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 - Uniaxial Modified Mohr - Biaxial Ductile Materials: Yield Strength - Uniaxial Maximum Shear Strength - Biaxial Distortion Energy - 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 s  sd = Sut / N In compression: Kt s  sd = Suc / N

Stress concentrations applied to stresses before making the circle 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

For most ductile materials, Syt = Syc 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

tmax  td = Sys / N = 0.5(Sy )/ N 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

Distortion Energy s2 s1 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

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

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

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

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

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

Endurance Strength = 0.50(Su) 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’ = 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

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 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”

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

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

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

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

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 (Eqn 5-20) Size Tension Reliability 50% 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 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

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

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

Assume machined surface, Sn  295 MPa Find actual endurance strength: 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) Sn Wrought steel Shear Stress 90% Reliability Size – 30 mm

Goodman: Example: continued. No Good!!! We wanted N  3 (Eqn. 5-28) 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% 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

Goodman: Example: continued. No Good!!! We wanted N  3 (Eqn. 5-28) 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 DEAD WEIGHT: SIGN + ARM + POST = 1000# (Compression) Reversed, Repeated 45° Bending RBE 2/11/97 Repeated one direction

Example: Combined Stress Fatigue Cont’d Stress Analysis: Dead Weight: (Static) Vertical from Wind: (Cyclic) Bending: (Static)

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

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

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 Size – 4.50” dia Wrought steel 90% Reliability Combined or Shear Stress

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 10 N = 1 Fail Not Fail Su S’sn Alternating Stress, ta 5 tmean = 4499.4 3115.3 Kttalt S’sn/N N = 3 Safe 5 10 15 Su/N 20 Mean Stress, tm