'08 DMcSLectureNotes-Chapter6abWeb

MECH 401 Mechanical Design Applications Dr. M. O’Malley– Master

Notes

Spring 2008

Dr. D. M. McStravick

Rice University

Reading Chapter 6

Homework HW 4 available, due 2-7

Tests Fundamentals Exam will be in class on 2-21

Nature of fatigue failure

Starts with a crack Usually at a stress concentration

Crack propagates until the material fractures suddenly

Fatigue failure is typically sudden and complete, and doesn’t give warning

Fatigue Failure Examples

Various Fatigue Crack Surfaces [Text fig. 6-2] Bolt Fatigue Failure [Text fig. 6-1] Drive Shaft [Text fig. 6-3] AISI 8640 Pin [Text fig. 6-4] Steam Hammer Piston Rod [Text fig. 6-6] Jacob Neu chair failure (in this classroom)

Fatigue Example 1

Fatigue Failure Example

Fatigue Failure Example

Fatigue Failure Example

Stamping Fatigue Failure Example

Schematic of Various Fatigue Failure

Jim Neu Chair Failure (Pedestal)

Fatigue Failure of Chair Shaft

Seat Fatigue Failure

Fatigue

Fatigue strength and endurance limit Estimating FS and EL Modifying factors

Thus far we’ve studied static failure of machine elements The second major class of component failure is due to dynamic loading

Variable stresses Repeated stresses Alternating stresses Fluctuating stresses

The ultimate strength of a material (Su) is the maximum stress a material can sustain before failure assuming the load is applied only once and held

A material can also fail by being loaded repeatedly to a stress level that is LESS than Su Fatigue failure

More Fatigue Failure Examples (ASM)

More Fatigue Failure Examples (ASM)

More Fatigue Failure Examples

Approach to fatigue failure in analysis and design Fatigue-life methods (6-3 to 6-6)

Stress Life Method (Used in this course) Strain Life Method Linear Elastic Fracture Mechanics Method

Stress-life method (rest of chapter 6) Addresses high cycle Fatigue (>103 ) Well Not Accurate for Low Cycle Fatigue (<103)

The 3 major methods

Stress-life Based on stress levels only Least accurate for low-cycle fatigue Most traditional

Easiest to implement Ample supporting data Represents high-cycle applications adequately

Strain-life More detailed analysis of plastic deformation at localized regions Good for low-cycle fatigue applications Some uncertainties exist in the results

Linear-elastic fracture mechanics Assumes crack is already present and detected Predicts crack growth with respect to stress intensity Practical when applied to large structures in conjunction with computer

codes and periodic inspection

Fatigue analysis

2 primary classifications of fatigue Alternating – no DC component

Fluctuating – non-zero DC component

Analysis of alternating stresses As the number of cycles

increases, the fatigue strength Sf (the point of failure due to fatigue loading) decreases

For steel and titanium, this fatigue strength is never less than the endurance limit, Se

Our design criteria is:

As the number of cycles approaches infinity (N ∞), Sf(N) = Se (for iron or Steel)

a

f NS

)(

Method of calculating fatigue strength Seems like we should be able to use graphs

like this to calculate our fatigue strength if we know the material and the number of cycles

We could use our factor of safety equation as our design equation

But there are a couple of problems with this approach S-N information is difficult to obtain and thus is

much more scarce than -s e information S-N diagram is created for a lab specimen

Smooth Circular Ideal conditions

Therefore, we need analytical methods for estimating Sf(N) and Se

a

f NS

)(

Terminology and notation

Infinite life versus finite life Infinite life

Implies N ∞ Use endurance limit (Se) of material Lowest value for strength

Finite life Implies we know a value of N (number of cycles) Use fatigue strength (Sf) of the material (higher than Se)

Prime (‘) versus no prime Strength variable with a ‘ (Se’)

Implies that the value of that strength (endurance limit) applies to a LAB SPECIMEN in controlled conditions

Variables without a ‘ (Se, Sf) Implies that the value of that strength applies to an actual case

First we find the prime value for our situation (Se’) Then we will modify this value to account for differences between a lab specimen and our

actual situation This will give us Se (depending on whether we are considering infinite life or finite life) Note that our design equation uses Sf, so we won’t be able to account for safety factors until

we have calculated Se’ and Se

a

f NS

)(

a

eS

Estimating Se’ – Steel and Iron

For steels and irons, we can estimate the endurance limit (Se’) based on the ultimate strength of the material (Sut)

Steel Se’ = 0.5 Sut for Sut < 200 ksi (1400 MPa)

= 100 ksi (700 MPa) for all other values of Sut

Iron Se’ = 0.4(min Sut)f/ gray cast Iron Sut<60 ksi(400MPa)

= 24 ksi (160 MPa) for all other values of SutNote: ASTM # for gray cast iron is the min Sut

S-N Plot with Endurance Limit

a

eS

a

f NS

)(

a

eS

Estimating Se’ – Aluminum and Copper Alloys For aluminum and copper alloys, there is no endurance limit Eventually, these materials will fail due to repeated loading To come up with an “equivalent” endurance limit, designers

typically use the value of the fatigue strength (Sf’) at 108 cycles

Aluminum alloys Se’ (Sf at 108 cycles) = 0.4 Sut for Sut < 48 ksi (330 MPa)

= 19 ksi (130 MPa) for all other values of Sut

Copper alloys Se’ (Sf at 108 cycles) = 0.4 Sut for Sut < 35 ksi (250 MPa)

= 14 ksi (100 MPa) for all other values of Sut

Constructing an estimated S-N diagram Note that Se’ is going to be our

material strength due to “infinite” loading

We can estimate an S-N diagram and see the difference in fatigue strength after repeated loading

For steel and iron, note that the fatigue strength (S’f) is never less than the endurance limit (Se’)

For aluminum and copper, note that the fatigue strength (S’f) eventually goes to zero (failure!), but we will use the value of S’f at 108 cycles as our endurance limit (Se’) for these materials

Estimating the value of Sf

When we are studying a case of fatigue with a known number of cycles (N), we need to calculate the fatigue strength (S’f)

We have two S-N diagrams One for steel and iron One for aluminum and copper

We will use these diagrams to come up with equations for calculating S’f for a known number of cycles

Note: Book indicates that 0.9 is not actually a constant, and uses the variable f to donate this multiplier. We will in general use 0.9 [so f=0.9]

Estimating Sf (N)

For steel and iron For f=0.9

For aluminum and copper

bSa

S

Sb

aNNS

ut

e

ut

bf

39.0loglog

9.0log

3

1

'

bSa

S

Sb

aNNS

ut

e

ut

bf

39.0loglog

9.0log

3

1

'

For 103 < N < 106

For N < 108

Where Se’ is the value of S’f at N = 108

5.7

Correction factors Now we have Se’ (infinite life) We need to account for differences between the lab specimen and a real

specimen (material, manufacturing, environment, design) We use correction factors

Strength reduction factors Marin modification factors

These will account for differences between an ideal lab specimen and real life Se = ka kb kc kd ke kf Se’

ka – surface factor kb – size factor kc – load factor kd – temperature factor ke – reliability factor Kf – miscellaneous-effects factor Modification factors have been found empirically and are described in section 6-9 of

Shigley-Mischke-Budynas (see examples) If calculating fatigue strength for finite life, (Sf), use equations on previous slide

Endurance limit modifying factors

Surface (ka) Accounts for different surface finishes

Ground, machined, cold-drawn, hot-rolled, as-forged Size (kb)

Different factors depending on loading Bending and torsion (see pg. 280) Axial (kb = 1)

Loading (kc) Endurance limits differ with Sut based on fatigue loading (bending, axial, torsion)

Temperature (kd) Accounts for effects of operating temperature (Not significant factor for T<250 C [482 F])

Reliability (ke) Accounts for scatter of data from actual test results (note ke=1 gives only a 50% reliability)

Miscellaneous-effects (kf) Accounts for reduction in endurance limit due to all other effects Reminder that these must be accounted for

Residual stresses Corrosion etc

Surface Finish Effect on Se

Temperature Effect on Se

Reliability Factor, ke

Steel Endurance Limit vs. Tensile Strength

Compressive Residual Stresses

Now what?

Now that we know the strength of our part under non-laboratory conditions…

… how do we use it? Choose a failure criterion Predict failure

Part will fail if: s’ > Sf(N) Factor of safety or Life of the part: h = Sf(N) / s’ Where b = - 1/3 log (0.9 Sut / Se) log (a) = log (0.9 Sut) - 3b

b

aN

1

Example Homework Problem 6-9 A solid rod cantilevered at one end. The rod is

0.8 m long and supports a completely reversing transverse load at the other end of +/- 1 kN. The material is AISI 1045 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end and assume f= 0.9.

Solution: -- See Board Work--

Stress concentration (SC) and fatigue failure Unlike with static loading, both ductile and

brittle materials are significantly affected by stress concentrations for repeated loading cases

We use stress concentration factors to modify the nominal stress

SC factor is different for ductile and brittle materials

SC factor – fatigue

s = kfsnom+ = kfso

t = kfstnom = kfsto

kf is a reduced value of kT and so is the nominal stress.

kf called fatigue stress concentration factor Why reduced? Some materials are not fully

sensitive to the presence of notches (SC’s) therefore, depending on the material, we reduce the effect of the SC

Fatigue SC factor

kf = [1 + q(kt – 1)] kfs = [1 + qshear(kts – 1)]

kt or kts and nominal stresses Table A-15 & 16 (pages 1006-1013 in Appendix)

q and qshear Notch sensitivity factor Find using figures 6-20 and 6-21 in book (Shigley) for steels

and aluminums Use q = 0.20 for cast iron

Brittle materials have low sensitivity to notches As kf approaches kt, q increasing (sensitivity to notches, SC’s) If kf ~ 1, insensitive (q = 0)

Property of the material

Example

AISI 1020 as-rolled steel Machined finish Find Fmax for:

h = 1.8 Infinite life

Design Equation: h = Se / s’

Se because infinite life

Example, cont.

h = Se / s’ What do we need?

Se

s’ Considerations?

Infinite life, steel Modification factors Stress concentration

(hole) Find s’nom (without SC)

FF

hdb

P

A

Pnom 2083

101260

Example, cont.

Now add SC factor:

From Fig. 6-20, r = 6 mm Sut = 448 MPa = 65.0 ksi q ~ 0.8

nomtnomf kqk 11

Example, cont.

From Fig. A-15-1, Unloaded hole d/b = 12/60 = 0.2 kt ~ 2.5

q = 0.8 kt = 2.5

s’nom = 2083 F

FF

kq nomt

4583

208315.28.01

11

Example, cont.

Now, estimate Se

Steel: Se’ = 0.5 Sut for Sut < 1400 MPa (eqn. 6-8)

700 MPa else AISI 1020 As-rolled

Sut = 448 MPa

Se’ = 0.50(448) = 224 MPa

Constructing an estimated S-N diagram Note that Se’ is going to be our

material strength due to “infinite” loading

We can estimate an S-N diagram and see the difference in fatigue strength after repeated loading

For steel and iron, note that the fatigue strength (S’f) is never less than the endurance limit (S’e)

For aluminum and copper, note that the fatigue strength (S’f) eventually goes to zero (failure!), but we will use the value of S’f at 108 cycles as our endurance limit (S’e) for these materials

Correction factors Now we have Se’ (infinite life) We need to account for differences between the lab specimen and a real

specimen (material, manufacturing, environment, design) We use correction factors

Strength reduction factors Marin modification factors

These will account for differences between an ideal lab specimen and real life Se = ka kb kc kd ke kf Se’

ka – surface factor kb – size factor kc – load factor kd – temperature factor ke – reliability factor Kf – miscellaneous-effects factor Modification factors have been found empirically and are described in section 6-9 of

Shigley-Mischke-Budynas (see examples) If calculating fatigue strength for finite life, (Sf), use equations on previous slide

Example, cont.

Modification factors Surface: ka = aSut

b (Eq. 6-19) a and b from Table 6-2 Machined

ka = (4.45)(448)-0.265 = 0.88

Example, cont.

Size: kb Axial loading kb = 1 (Eq. 6-21)

Load: kc

Axial loading kc = 0.85 (Eq. 6-26)

Example, cont.

Temperature: kd = 1 (no info given)

Reliability: ke = 1 (no info given)

Miscellaneous: kf = 1

Endurance limit: Se = kakbkckdkekfSe’ = (0.88)(0.85)(227) = 177 MPa

Design Equation:

kN 4.218.14583

10x177

8.14583

177

6

F

F

MPaSe

Fluctuating Fatigue Failures

Alternating vs. fluctuatingAlternating Fluctuating

I

MrA

P

a

m

Alternating Stresses

sa characterizes alternating stress

Fluctuating stresses

Mean Stress

Stress amplitude

Together, sm and sa characterize fluctuating stress

2' minmax

m

2minmax'

a

Alternating vs. Fluctuating

Modified Goodman Diagram

Fluctuating Stresses in Compression and Tension

Failure criterion for fluctuating loading Soderberg Modified Goodman Gerber ASME-elliptic Yielding

Points above the line: failure Book uses Goodman primarily

Straight line, therefore easy algebra Easily graphed, every time, for every problem Reveals subtleties of insight into fatigue problems Answers can be scaled from the diagrams as a check on the

algebra

Gerber Langer Plot for Fluctuating Stresses

Fluctuating stresses, cont.

As with alternating stresses, fluctuating stresses have been investigated in an empirical manner

For sm < 0 (compressive mean stress) sa > Sf Failure Same as with alternating stresses Or,

Static Failure

For sm > 0 (tensile mean stress) Modified Goodman criteria

h < 1 Failure

)S(or max ucycam S

1

ut

m

f

a

SS

Modified Goodman Langer Equations

Fluctuating stresses, cont.

Relationship is easily seen by plotting:

Goodman Line

(safe stress line)

Safe design region(for arbitrary fluctuationsin sm and sa )

1ut

m

f

a

SS

1

ut

m

f

a

SS

Note: sm + sa = smax

sm + sa > Syt (static failure by yielding)

Important point: Part can fail because of fluctuations in either sa, sm, or both. Design for prescribed variations in sa and sm to get a more exact solution.

Special cases of fluctuating stresses Case 1: sm fixed

Case 2: sa fixed

a

aS

m

mS

Special cases of fluctuating stresses Case 3: sa / sm fixed

Case 4: both vary arbitrarily

m

m

a

a SS

ut

m

f

a

SS

1

Example

Given: Sut = 1400 MPa Syt = 950 MPa Heat-treated (as-forged) Fmean = 9.36 kN Fmax = 10.67 kN d/w = 0.133; d/h = 0.55

Find: h for infinite life, assuming

Fmean is constant

Example, cont.

Find sm and sa

MPa28

MPa228

MPa200

Nm 8003.0100.67x14

1

4

1

22

Nm 7023.010x36.94

1

4

1

22

m 009.02

mx1016.318107512

1

12

1

12

1

max

maxmaxmax

max

3max

maxmax

3

max

48333

ma

mm

mm

m

I

yMI

yM

LFLF

M

LFLF

M

hy

hdwbhI

I

My

Stress Concentration Factor

Example, cont.

Since this is uniaxial loading, m = 200 MPa a = 28 MPa

We need to take care of the SC factors Su = 1400Mpa

kt ~ 2.2 (Figure A15-2) q ~ 0.95 (Figure 7-20) kf = 2.14

11 tf kqk

nominal

MPa 42820014.2

MPa 602814.2

nom

nom

mfmm

afaa

k

k

Example, cont.

Find strength Eqn. 7-8: S’e = .504Sut

Modification factors

86.0

24.1

808.0

mm 51d2.8

:19)-(7Equation

:Size

107.0

2

1

eq

b

eqb

eq

k

dk

hbd

MPa1400S since MPa 700~ ut eS

201.0

995.0

271

:Surface

a

buta

k

b

a

aSk25)-7 (Eq. 1

Bending

:Load

ck

MPa12170086.0201.0 eS

Example, cont.

Design criteria Goodman line:

For arbitrary variation in sa and sm,

nSS ut

m

e

a /1

121

1400

11400121

ma

25.1

1400

428

121

601

1

1400121

ma

Example, cont.

However, we know that Fmean = constant from problem statement sm = constant

4.160

84

MPa84

11400

428

121

1

a

a

a

a

ut

m

e

a

S

S

S

SS

S

Less conservative!

Combined loading and fatigue Size factor depends on loading SC factors also depend on loading Could be very complicated calculation to keep track of each load

case Assuming all stress components are completely reversing and

are always in time phase with each other,1. For the strength, use the fully corrected endurance limit for

bending, Se

2. Apply the appropriate fatigue SC factors to the torsional stress, the bending stress, and the axial stress components

3. Multiply any alternating axial stress components by the factor 1/kc,ax

4. Enter the resultant stresses into a Mohr’s circle analysis to find the principal stresses

5. Using the results of step 4, find the von Mises alternating stress sa’

6. Compare sa’ with Sa to find the factor of safetyAdditional details are in Section 6-14

More Fatigue Failure Examples