'08 DMcSLectureNotes-Chapter6abWeb.PPT

8/10/2019 '08 DMcSLectureNotes-Chapter6abWeb.PPT

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MECH 401

Mechanical Design Applications

Dr. M. OMalleyMaster Notes

Spring 2008

Dr. D. M. McStravickRice University


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Reading

Chapter 6

Homework

HW 4 available, due 2-7

Tests

Fundamentals Exam will be in class on 2-21


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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 doesnt give warning


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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)


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Fatigue Example 1


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Fatigue Failure Example


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Stamping Fatigue Failure Example


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Schematic of Various Fatigue Failure


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Fatigue Failure of Chair Shaft


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Seat Fatigue Failure


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Fatigue

Fatigue strength and endurance limit

Estimating FS and EL

Modifying factors

Thus far weve 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 onlyonce and held

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


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More Fatigue Failure Examples (ASM)


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More Fatigue Failure Examples (ASM)


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More Fatigue Failure Examples


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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 computercodes and periodic inspection


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Fatigue analysis

2 primary classifications of

fatigue

Alternatingno DC component

Fluctuatingnon-zero DC

component


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Analysis of alternating stresses

As the number of cyclesincreases, the fatigue strengthSf(the point of failure due tofatigue loading) decreases

For steel and titanium, thisfatigue strength is never lessthan the endurance limit, Se

Our design criteria is:

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

a

f NS

)(


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Method of calculating fatigue strength

Seems like we should be able to use graphslike this to calculate our fatigue strength ifwe know the material and the number ofcycles

We could use our factor of safety equationas our design equation

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

much more scarce than -einformation

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

)(


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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< 200ksi (1400MPa)

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

Iron Se = 0.4(min Sut)f/ gray cast Iron Sut


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S-N Plot with Endurance Limit

a

eS

a

f NS

)(

a

eS


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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, designerstypically use the value of the fatigue strength (Sf) at 10

8cycles

Aluminum alloys

Se (Sfat 108cycles) = 0.4 Sut for Sut< 48 ksi (330 MPa)

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

Copper alloys

Se (Sfat 108cycles) = 0.4 Sut for Sut< 35 ksi (250 MPa)= 14 ksi (100 MPa) for all other values of Sut


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Constructing an estimated S-N diagram

Note that Se is going to be ourmaterial strength due to infiniteloading

We can estimate an S-N diagramand see the difference in fatiguestrength after repeated loading

For steel and iron, note that thefatigue strength (Sf) is never lessthan the endurance limit (Se)

For aluminum and copper, notethat the fatigue strength (S

f)

eventually goes to zero (failure!),but we will use the value of Sfat108cycles as our endurance limit(Se) for these materials


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Estimating the value of Sf

When we are studying a case offatigue with a known number of cycles(N), we need to calculate the fatiguestrength (Sf)

We have two S-N diagrams

One for steel and iron One for aluminum and copper

We will use these diagrams to comeup with equations for calculating Sffor a known number of cycles

Note: Book indicates that 0.9 is notactually a constant, and uses thevariable fto donate this multiplier.We will in general use 0.9 [so f=0.9]


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Estimating Sf(N)

For steel and iron

For f=0.9

For aluminum and copper

bSa

S

Sb

aNNS

ut

e

ut

b

f

39.0loglog

9.0log

3

1

'

-

-

bSa

SSb

aNNS

ut

e

ut

b

f

39.0loglog

9.0log31

'

-

-

For 103< N < 106

For N < 108

Where Se is the value ofSfat N = 10

8

5.7


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Correction factors

Now we have Se (infinite life)

We need to account for differences between the lab specimen and a realspecimen (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= kakbkckdkekfSe kasurface factor

kbsize factor

kcload factor

kdtemperature factor

kereliability factor

Kfmiscellaneous-effects factor

Modification factors have been found empirically and are described in section 6-9 ofShigley-Mischke-Budynas (see examples)

If calculating fatigue strength for finite life, (Sf), use equations on previous slide


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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 Sutbased on fatigue loading (bending, axial, torsion)

Temperature (kd) Accounts for effects of operating temperature (Not significant factor for T


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Surface Finish Effect on Se


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Temperature Effect on Se


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Reliability Factor, ke


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Compressive Residual Stresses


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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:

> Sf(N)

Factor of safety or Life of the part:

= Sf(N) /

Where

b = - 1/3 log (0.9 Sut/ Se) log (a) = log (0.9 Sut) - 3b

b

aN

1


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Example Homework Problem 6-9

A solid rod cantilevered at one end. The rod is

0.8 m long and supports a completelyreversing transverse load at the other end of +/-1 kN. The material is AISI 1045 hot-rolledsteel. If the rod must support this load for 104

cycles with a factor of safety of 1.5, whatdimension should the square cross sectionhave? Neglect any stress concentrations at thesupport end and assume f= 0.9.

Solution: -- See Board Work--


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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 brittlematerials


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SC factorfatigue

= kfnom+ =kfo

t= kfstnom = kfsto

kfis a reduced value of kTand ois the nominalstress.

kfcalled fatigue stress concentration factor

Why reduced? Some materials are not fully sensitiveto the presence of notches (SCs) therefore,depending on the material, we reduce the effect ofthe SC


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Fatigue SC factor

kf= [1 + q(kt1)]

kfs= [1 + qshear(kts1)] ktor kts and nominal stresses

Table A-15 & 16 (pages 1006-1013 in Appendix)

q and qshear Notch sensitivity factor

Find using figures 6-20and 6-21in book (Shigley) for steelsand aluminums

Use q = 0.20 for cast iron

Brittle materials have low sensitivity to notches

As kfapproaches kt, q increasing (sensitivity to notches, SCs)

If kf~ 1, insensitive (q = 0)

Property of the material


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Example, cont.

= Se/

What do we need?

Se

Considerations? Infinite life, steel

Modification factors

Stress concentration (hole)

Find nom(without SC)

F

F

hdb

P

A

Pnom 2083

101260

-

-


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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


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Example, cont.

From Fig. A-15-1, Unloaded hole

d/b = 12/60 = 0.2

kt~ 2.5

q = 0.8

kt= 2.5

nom= 2083 F

F

Fkq nomt

4583

208315.28.0111

--


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Example, cont.

Now, estimate Se Steel:

Se = 0.5 Sutfor Sut< 1400 MPa (eqn. 6-8)

700 MPa else AISI 1020 As-rolled

Sut= 448 MPa

Se = 0.50(448) = 224MPa


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Constructing an estimated S-N diagram

Note that Se is going to be ourmaterial strength due to infiniteloading

We can estimate an S-N diagramand see the difference in fatiguestrength after repeated loading

For steel and iron, note that thefatigue strength (Sf) is never lessthan the endurance limit (Se)

For aluminum and copper, notethat the fatigue strength (Sf)

eventually goes to zero (failure!),but we will use the value of Sfat108cycles as our endurance limit(Se) for these materials


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Correction factors

Now we have Se (infinite life)

We need to account for differences between the lab specimen and a realspecimen (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= kakbkckdkekfSe kasurface factor

kbsize factor

kcload factor

kdtemperature factor

kereliability factor

Kfmiscellaneous-effects factor

Modification factors have been found empirically and are described in section 6-9 ofShigley-Mischke-Budynas (see examples)

If calculating fatigue strength for finite life, (Sf), use equations on previous slide


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Example, cont.

Modification factors

Surface: ka= aSutb(Eq. 6-19)

a and b from Table 6-2

Machined

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


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Example, cont.

Size: kb

Axial loading

kb= 1 (Eq. 6-21)

Load: kc Axial loading

kc= 0.85 (Eq. 6-26)


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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:

kN4.21

8.14583

10x177

8.14583177

6

F

FMPaSe


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Fluctuating Fatigue Failures


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Alternating vs. fluctuating

Alternating Fluctuating

I

Mr

A

P

a

m

Al


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Alternating Stresses

acharacterizes alternating stress

Fl i


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Fluctuating stresses

Mean Stress

Stress amplitude

Together, m

and a

characterize fluctuating

stress

2' minmax

m

2

minmax'

-a

Al i Fl i


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Alternating vs. Fluctuating

M difi d G d Di


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Modified Goodman Diagram

Fl i S i C i d


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Fluctuating Stresses in Compression and

Tension

F il i i f fl i l di


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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 thealgebra

G b L Pl f Fl i S


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Gerber Langer Plot for Fluctuating Stresses

Fl i


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Fluctuating stresses, cont.

As with alternating stresses, fluctuating stresses have beeninvestigated in an empirical manner

For m< 0 (compressive mean stress) a> Sf Failure

Same as with alternating stresses

Or,

Static Failure

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

< 1 Failure

)S(ormax ucycam S-

1ut

m

f

a

SS

M difi d G d L E i


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Modified Goodman Langer Equations


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S i l f fl t ti t


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Special cases of fluctuating stresses

Case 1: mfixed

Case 2: afixed

a

aS

m

mS

S i l f fl t ti t


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Special cases of fluctuating stresses

Case 3: a/ mfixed

Case 4: both vary arbitrarily

m

m

a

a SS

ut

m

f

a

SS

1

E l


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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:

for infinite life, assuming

Fmeanis constant

E pl t


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Example, cont.

Find mand a

MPa28

MPa228

MPa200

Nm8003.0100.67x14

1

4

1

22

Nm7023.010x36.941

41

22

m009.02

mx1016.318107512

1

12

1

12

1

max

maxmaxmax

max

3

maxmax

max

3

max

48333

-

--

-

ma

mm

mm

m

I

yMI

yM

LFLF

M

LFLFM

hy

hdwbhI

I

My

Str C tr ti F t r


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Stress Concentration Factor

E mple ont


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Example, cont.

Since this is uniaxialloading,

m= 200 MPa

a= 28 MPa

We need to take care ofthe SC factors

Su = 1400Mpa

kt~ 2.2 (Figure A15-2) q ~ 0.95 (Figure7-20)

kf= 2.14

11 - tf kqk

nominal

MPa42820014.2

MPa602814.2

nom

nom

mfmm

afaa

k

k


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Example cont


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Example, cont.

Design criteria Goodman line:

For arbitrary variation in

aand m,

nSS ut

m

e

a /1

121

1400

11400121

ma

25.1

1400

428

121

601

1

1400121

ma

Example cont


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Example, cont.

However, we know thatFmean= constant from

problem statement

m= 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


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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 loadcase

Assuming all stress components are completely reversing andare 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 factor1/kc,ax

4. Enter the resultant stresses into a Mohrs circle analysis to findthe principal stresses

5. Using the results of step 4, find the von Mises alternating stressa

6. Compare a with Sato find the factor of safety

Additional details are in Section 6-14


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