Design Stress & Fatigue

MET 210W

E. Evans

Parts Fail When?

Crack initiation site

P

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

• Design

ySN:Example

StressApplied

StrengthFailureSafetyofFactor

N

S:Example

FactorDesign

StrengthFailureStressAllowable

yALLOW

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 d = Sut / N

–In compression:

Kt d = Suc / N

45° Shear Diagonal

Modified Mohr

• Biaxial Static Stress on Brittle Materials

12

Stress concentrations applied to stresses before

making the circle

1

2

Sut

Sut

Suc

Suc

1, 2

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

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.

max d = Sys / N = 0.5(Sy )/ N

Somewhat conservative – use Distortion Energy for more precise failure estimate

Distortion Energy

Best predictor of failure for ductile materials under static loads or completely reversed normal, shear or combined stresses.

Shear Diagonal

1

2

2122

21 '

’ = von Mises stress

Failure: ’ > Sy

Design: ’ d = Sy/N

• Static Biaxial or Triaxial Stress on Ductile Materials

Sy

Sy

Sy

Sy

Distortion Energy

von Mises Stress• Alternate Form

For uniaxial stress when y = 0,

• Triaxial Distortion Energy(1 > 2 > 3)

222 3 xyyxyx'

22 3 xyx'

223

213

2122

2)()()('

Fluctuating StressS

tres

s

Timemin

mean

max

• Varying stress with a nonzero mean.

alternating = a2

minmaxmean

2minmax

a

Stress Ratio,

max

minR

-1 R 1

Fluctuating Stress Example

Valve Spring Force

Valve Spring ForceValve Open

Valve Closed

Valve Closed

Valve Open

• Bending of Rocker Arm

• Tension in Valve Stem

Adapted from R. B. Englund

RBE 2/1/91

Fatigue Testing

• Bending tests– Spinning bending elements – most common– Constant stress cantilever beams

Fixed SupportApplied Deformation – Fully Reversed, R = -1

Top View

Front View

Fatigue Testing

Number of Cycles to Failure, N

Str

ess,

(

ksi)

Data from R. B. Englund, 2/5/93

Test Data

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’ = 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.

62 ksi

From Fig. A4-5.Su = 190 ksi

From Fig. 5-8.Sn = 62 ksi (machined)

Actual Sn Example Continued

Sn’ = Sn(Cm)(Cst)(CR)(CS)

= 62 ksi(1.0)(.8)(.81)(.94) = 37.8 ksi

Size Factor, Fig. 5-9

Wrought Steel

Axial Tension

Reliability, Table 5-1

99% Probability Sn’ is at or above the calculated value

Sn,Table 5-8

Actual Sn’ Estimate

Guessing: diameter .5”

Sn’

Goodman Diagram

m

a

-Sy

Sy

Sy Su

FATIGUE FAILURE REGION

NO FATIGUE FAILURE REGION

Goodman Line

0

Yield Line

1

u

m

n

a

SS

Sn’/N

Sn’

Goodman Diagram

m

a

-Sy

Sy

Sy Su

FATIGUE FAILURE REGION

Goodman Line

0

Yield Line

1

u

m

n

a

SS

Su/N

NSS

K

u

m

n

at 1

Safe Stress Line

Safe Stress Line

SAFE ZONE

MAX = 30.3

Example: Problem 5-53. Find a suitable titanium alloy. N = 3

42 mm DIA 30 mm DIA

1.5 mm Radius

F varies from 20 to 30.3 kN+

-

FO

RC

E

MIN = 20

TIME

kN..

mean 15252

20330

kN..

alt 1552

20330

Example: Problem 5-53 continued.

• Find the mean stress:

• Find the alternating stress:

• Stress concentration from App. A15-1:

MPa35.6

2

m

)mm30(4

N150,25

MPa7.3

2

a

)mm30(4

N150,5

2.3Kt 05.mm30

mm5.1

d

r;4.1

mm30

mm42

d

D

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

3.36

297.

1N

297.N

1

MPa448

MPa6.35

)4/MPa448)(86(.8.

)MPa3.7(3.2

N

1

SS

K

u

m

n

at(Eqn 5-20)

Tension

Size

Reliability 50%

3.36 is good, need further information on Sn for titanium.

MAX = 1272 N-m

Example: Find a suitable steel for N = 3 & 90% reliable.

T varies from 848 N-m to 1272 N-m+

-

TO

RQ

UE

MIN = 848 N-m

TIME

mN10602

8481272mean

mN2122

8481272alt

50 mm DIA30 mm

DIA

3 mm Radius

T = 1060 ± 212 N-m

Example: continued. • Stress concentration from App. A15-1:

• Find the mean shear stress:

• Find the alternating shear stress:

MPa200

3

mmm

p

mm

)mm30(16

)1000(mN1060

Z

T

MPa40

3

p

aa

mm5301

mmN212000

Z

T

1.38Kt 1.mm30

mm3

d

r;667.1

mm30

mm50

d

D

Example: continued. • So, = 200 ± 40 MPa. Guess a material.

TRY: AISI 1040 OQT 400°F Su = 779 MPa, Sy = 600 MPa, %E = 19%

• Verify that max Sys:

max = 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:S’sn = Sn(Cm)(Cst)(CR)(CS)

= 295 MPa(1.0)(.577)(.9)(.86) = 132 MPa

Sn

Wrought steelShear Stress

90% ReliabilitySize – 30 mm

(Fig. 5-8)

Example: continued.

• Goodman:

1.31

7606.

1N

7606.N

1

MPa584

MPa200

MPa132

)MPa40(38.1

N

1

SS

K

su

m

sn

at(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

wrought

shear

reliable

size

Example: continued.

• Goodman:

2.64

378.

1N

378.N

1

MPa1140

MPa200

MPa272

)MPa40(38.1

N

1

SS

K

su

m

sn

at(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

RBE 2/11/97 Repeated one direction

PIPE: TS4 x .237 WALL

MATERIAL: ASTM A242 Equivalent

DEAD WEIGHT:SIGN + ARM + POST = 1000#

(Compression)

Reversed, Repeated

45°

Bending

Example: Combined Stress Fatigue Cont’dStress Analysis:

psi5.315in17.3

#1000

A

P2

Dead Weight:

psi09.63in17.3

#200

A

P2

Vertical from Wind:

(Static)

(Cyclic)

psi8.9345in21.3

)in60(#500

Z

M3

Bending:

(Static)

Example: Combined Stress Fatigue Cont’dStress Analysis:

Torsion:

(Cyclic)psi3.3115)in21.3(2

)in100(#200

Z

T3

P

Stress Elements: (Viewed from +y)

CYCLIC:STATIC: 315.5 psi

9345.8 psi

63.09 psi – Repeated One Direction

= 3115.3 psi Fully Reversedx

z

x

z

Example: Combined Stress Fatigue Cont’dMean Stress:

9345.8-315.5

-31.5

8998.8 psi

Static

Repeated / 2

Alternating Stress:

(CW)

1

max

psi4.44992

psi8.8998max

+TIME

-

Str

ess

MIN = -63.09 psi

m

a (CW)

(-31.5,-3115.3)

(0,-3115.3)max

psi34.3115max

Example: Combined Stress Fatigue Cont’dDetermine Strength:

Try for N = 3 some uncertainty

Size Factor? OD = 4.50 in, Wall thickness = .237 inID = 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 steelCombined or Shear Stress

90% ReliabilitySize – 4.50” dia

Example: Combined Stress Fatigue Cont’d“Safe” Line for Goodman Diagram:

a = S’sn / N = 8.9 ksi / 3 = 2.97 ksi

m = Ssu / N = 52.5 ksi / 3 = 17.5 ksi

Mean Stress, m

0 15105 200

5

10

Alte

rnat

ing

Str

ess,

a

2.29

426.

1N

426.N

1

psi52500

psi4.4499

psi8900

)psi3.3115(0.1

N

1

SS

K

su

m

sn

at

Su/N

S’sn/N

N = 1 FailNot Fail

Su

S’sn

mean = 4499.4

3115.3 Ktalt

N = 3Safe