Design Stress & FatigueMET 210WE. Evans

Predictions of Failure Fluctuating LoadsBrittle Materials:Not recommended

Ductile Materials:GoodmanGerberSoderberg

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

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

SnGoodman Diagramsmsa-SySySySuFATIGUE FAILURE REGIONNO FATIGUE FAILURE REGIONGoodman Line0Yield Line

Sn/NSnGoodman Diagramsmsa-SySySySuFATIGUE FAILURE REGIONGoodman Line0Yield LineSu/NSafe Stress LineSafe Stress LineSAFE ZONE

MAX = 30.3Example: Problem 5-53. Find a suitable titanium alloy. N = 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.

MAX = 1272 N-mExample: Find a suitable steel for N = 3 & 90% reliable.T 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