Design Stress & FatigueMET 210WE. Evans

Parts Fail When?Crack initiation sitePPThis 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 FactorAnalysis

Design

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidence

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceHow many will be produced?What manufacturing methods will be used?What are the consequences of failure?Danger to peopleCostSize and weight important?What is the life of the component?Justify design expense?

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceTemperature range.Exposure to electrical voltage or current.Susceptible to corrosionIs noise control important?Is vibration control important?Will the component be protected?GuardHousing

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceNature of the load considering all modes of operation:Startup, shutdown, normal operation, any foreseeable overloadsLoad characteristicStatic, repeated & reversed, fluctuating, shock or impactVariations of loads over time.MagnitudesMaximum, minimum, mean

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceWhat kind of stress?Direct tension or compressionDirect shearBendingTorsional shearApplicationUniaxialBiaxialTriaxial

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceMaterial propertiesUltimate strength, yield strength, endurance strength, DuctilityDuctile: %E 5%Brittle:%E < 5%Ductile materials are preferred for fatigue, shock or impact loads.

Factors Effecting Design FactorApplicationEnvironmentLoadsTypes of StressesMaterialConfidenceReliability of data forLoadsMaterial propertiesStress calculationsHow good is manufacturing quality controlWill subsequent handling, use and environmental conditions affect the safety or life of the component?

Design FactorAdapted from R. B. Englund

Design Factor

Predictions of Failure Static LoadsBrittle Materials:Maximum Normal Stress - UniaxialModified Mohr - Biaxial

Ductile Materials:Yield Strength - UniaxialMaximum Shear Strength - BiaxialDistortion Energy - Biaxial or Triaxial

Predictions of Failure Fluctuating LoadsBrittle Materials:Not recommended

Ductile Materials:GoodmanGerberSoderberg

Maximum Normal StressUniaxial Static Loads on Brittle Material:

In tension:Kt s sd = Sut / N

In compression:Kt s sd = Suc / N

Modified MohrBiaxial Static Stress on Brittle Materials45 Shear Diagonals1s2Stress concentrations applied to stresses before making the circles1s2SutSutSucSucs1, s2Often brittle materials have much larger compressive strength than tensile strength

Yield Strength MethodUniaxial Static Stress on Ductile Materials

In tension:s sd = Syt / N In compression: sd = Syc / NFor most ductile materials, Syt = Syc

Maximum Shear StressBiaxial Static Stress on Ductile MaterialsDuctile materials begin to yield when the maximum shear stress in a load-carrying component exceeds that in a tensile-test specimen when yielding begins.tmax td = Sys / N = 0.5(Sy )/ N

Somewhat conservative use Distortion Energy for more precise failure estimate

Distortion EnergyBest predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses.Shear Diagonals1s2s = von Mises stressFailure:s > SyDesign: s sd = Sy/N

Static Biaxial or Triaxial Stress on Ductile MaterialsSySySySyDistortion Energy

von Mises StressAlternate Form

For uniaxial stress when sy = 0,

Triaxial Distortion Energy(s1 > s2 > s3)

Fluctuating StressStressTimesminsmeansmaxVarying stress with a nonzero mean.salternating = saStress Ratio,-1 R 1

Fluctuating Stress ExampleValve Spring ForceValve Spring ForceValve OpenValve ClosedValve ClosedValve Open Bending of Rocker Arm Tension in Valve StemAdapted from R. B. EnglundRBE 2/1/91

Fatigue TestingBending testsSpinning bending elements most commonConstant stress cantilever beamsFixed SupportApplied Deformation Fully Reversed, R = -1Top ViewFront View

Fatigue TestingNumber of Cycles to Failure, NStress, s (ksi)Data from R. B. Englund, 2/5/93Test Data

Endurance StrengthThe 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 StrengthSn = Sn(Cm)(Cst)(CR)(CS)

Sn= actual endurance strength (ESTIMATE)Sn = endurance strength from Fig. 5-8Cm= material factor (pg. 174)Cst= stress type: 1.0 for bending0.8 for axial tension0.577 for shearCR = reliability factorCS = size factor

Actual Sn ExampleFind the endurance strength for the valve stem. It is made of AISI 4340 OQT 900F.

62 ksiFrom Fig. A4-5.Su = 190 ksiFrom Fig. 5-8.Sn = 62 ksi (machined)

Actual Sn Example ContinuedSn= Sn(Cm)(Cst)(CR)(CS)= 62 ksi(1.0)(.8)(.81)(.94) = 37.8 ksi

Size Factor, Fig. 5-9Wrought SteelAxial TensionReliability, Table 5-199% Probability Sn is at or above the calculated valueSn,Table 5-8Actual Sn EstimateGuessing: diameter .5

Goodman DiagramSnsmsa-SySySySuFATIGUE FAILURE REGIONNO FATIGUE FAILURE REGIONGoodman Line0Yield Line

Goodman DiagramSn/NSnsmsa-SySySySuFATIGUE FAILURE REGIONGoodman Line0Yield LineSu/NSafe Stress LineSafe Stress LineSAFE ZONE

Example: Problem 5-53. Find a suitable titanium alloy. N = 3MAX = 30.342 mm DIA30 mm DIA1.5 mm RadiusF varies from 20 to 30.3 kN+-FORCEMIN = 20TIME

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)TensionSizeReliability 50%3.36 is good, need further information on Sn for titanium.

Example: Find a suitable steel for N = 3 & 90% reliable.MAX = 1272 N-mT varies from 848 N-m to 1272 N-m+-TORQUEMIN = 848 N-mTIME50 mm DIA30 mm DIA3 mm RadiusT = 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 400F 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

Example: continued. Assume machined surface, Sn 295 MPa

Find actual endurance strength:Ssn = Sn(Cm)(Cst)(CR)(CS)= 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa

(Fig. 5-8)

Example: continued. Goodman:

(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 700F Su = 1520 MPa, Sy = 1360 MPa, %E = 10%Find the ultimate shear stress:Sus = .75Su = .75(779 MPa) = 584 MPaFind actual endurance strength:Ssn = Sn(Cm)(Cst)(CR)(CS)= 610 MPa(1.0)(.577)(.9)(.86) = 272 MPa

DuctileSnwroughtshearreliablesize

Example: continued. Goodman:

(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? (Cant 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 ContdRBE 2/11/97Repeated one directionPIPE: TS4 x .237 WALLMATERIAL: ASTM A242 EquivalentDEAD WEIGHT:SIGN + ARM + POST = 1000#(Compression)Reversed, Repeated45Bending

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

Example: Combined Stress Fatigue ContdStress Analysis:Torsion:(Cyclic)Stress Elements:(Viewed from +y)CYCLIC:STATIC:315.5 psi9345.8 psi63.09 psi Repeated One Directiont = 3115.3 psi Fully Reversed

Example: Combined Stress Fatigue ContdMean Stress:Alternating Stress:

Example: Combined Stress Fatigue ContdDetermine Strength:Try for N = 3 some uncertaintySize Factor? OD = 4.50 in, Wall thickness = .237 inID = 4.50 2(.237) = 4.026 inMax. 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 3/4Ductile

Example: Combined Stress Fatigue Contd

We must use Ssu and Ssn since this is a combined stress situation. (Case I1, page 197)

Sus = .75Su = .75(70 ksi) = 52.5 ksi

Ssn = Sn(Cm)(Cst)(CR)(CS)= 23 ksi(1.0)(.577)(.9)(.745) = 8.9 ksiHot Rolled SurfaceWrought steelCombined or Shear Stress90% ReliabilitySize 4.50 dia

Example: Combined Stress Fatigue Contd

Safe Line for Goodman Diagram:ta = Ssn / N = 8.9 ksi / 3 = 2.97 ksitm = Ssu / N = 52.5 ksi / 3 = 17.5 ksi

Mean Stress, tm015105200510Alternating Stress, taSu/NSsn/N