Fatique Lecture Notes

1

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Fatique FailureLecture NotesPrepared by

H. Orhan YILDIRAN

MECE 304 Mechanical Machine Elements-Fatique Failure

LECTURE NOTES- MECE 304 Mechanical Machine Elements

Chapter 3- Fatique failure

(Notes from: Chapter 6 in Budynas R.G., Nisbett J.K., Shigleys Mechanical Engineering Design,

Mc Graw Hill, 8th Edition)

Spring Semester 2007/2008

Halil Orhan YILDIRAN, MS2

3


6-1 Fatigue in MetalsIn stress-strain testing diagram, the load is applied gradually, to give

sufficient time for the strain to fully develop and the specimen is tested to

destruction, and so the stresses are applied only once. Testing of this kind is

applicable, to what are known as static conditions.

Fatique loading conditions produces stresses that vary with time or they fluctuate between different levels. These stresses are called variable,

repeated, alternating, or fluctuating stresses.

In a fatigue failure;

*Maximum stresses of failure are well below the ultimate strength of the

material, and quite frequently below the yield strength.

*Repeated a very large number of times.

4


Figure 6-2 Schematics of fatigue fracture surfaces produced in

smooth and notched components with round and rectangular cross sections under various loading conditions

5


Figure 6-2 Schematics of fatigue fracture surfaces(Contn’d)

6


Figure 6-2 Schematics of fatigue fracture surfaces(Contn’d)

7


Figure 6-5 Fatique fracture surface of a connecting rod

8


Figure 6-6 Fatique fracture surface of a piston rod.

9


6-4 The Stress Life Method in Fatique Failure analysisTo determine the strength of materials under the action of fatigue loads,

specimens are subjected to repeated or varying forces of specified

magnitudes while the cycles or stress reversals are counted to destruction.

The most widely used fatigue-testing device is the R. R. Moore high-speed

rotating-beam machine. This machine subjects the specimen to pure

bending by means of weights. The specimen, shown in Fig. 6–9, is very carefully machined and polished.

Figure 6-9 Test specimen

geometry for R. R. Moore

high-speed rotating-beam machine

10


The ordinate of the S-N diagram is called the fatigue strength Sf ; stated with the number of cycles N to which this strength corresponds.

In the case of the steels, a knee occurs in the graph, and beyond this knee

failure will not occur, no matter how great the number of cycles. The strength

corresponding to the knee is called the endurance limit Se, or the fatigue

limit. The graph of Fig. 6–10 never does become horizontal for nonferrous

metals and alloys.

Figure 6–10 An S-N diagram

plotted from the results of completely reversed axial

fatigue tests.

Material: UNS G41300

steel, normalized;Sut = 810 Mpa; maximumSut = 1050 Mpa

11


6-7 The Endurance LimitThe determination of endurance limits by fatigue testing is now routine,

though a lengthy procedure. Generally, stress testing is preferred to strain

testing for endurance limits.

For preliminary and prototype design and for some failure analysis as well, a

quick method of estimating endurance limits is needed.

Figure 6-17 Graph of Se

vs Sut

12


For steels, observation of Fig. 6–17, obtained from simple tensile test and rotating beam test, we will estimate the endurance limit as

0.5Sut for Sut ≤ 1400 MPa

S’e =700 MPa for Sut > 1400 MPa

Notes:*Heat treatment effects Se of steels. Ductile microstructures have higher

S’e/Sut ratios

*Endurance limit for cast iron (polished or machined) are given in the Table A-22.

*Aluminum alloys do not have endurance limits. See Table A-22 for fatique

strength of some aliminum alloys

13


6-8 Fatique strength (Sf): From Figure 6-10 for low cycles (N=10^3 cycles)

At 10^3 cycles

For steels with HB ≤ 500

And for b

Where σ’F is fatique strength coefficient (p.269), f is the fraction ofSut represented by (S’f)10^3 cycles

14


For an actual mechanical component, S’e is reduced to Se (see Sec. 6–9)

which is less than 0.5 Sut.

For the actual mechanical component, following equation can be written

Figure 6-18 f vs Sut at

10^3 cycles

for Se=S’e=0.5

Sut

15


Where a and b are constants

For a completely reversed stress σa given, with Sf=σa Eq.6-13

becomes

For low cycle fatique (1≤N≤10^3 cycles)

16


SAMPLE PROBLEM 6-2 Given a SAE 1050 HR steel, estimate:

(a) the rotating-beam endurance limit at 10^6 cycles.(b) the endurance strength of a polished rotating-beam specimen

corresponding to 10^4 cycles to failure

(c) the expected life of a polished rotating-beam specimen under a completely reversed stress of 385 Mpa.

17


6-9 Endurance Limit Modifying Factors

Rotating-beam specimen used in the laboratory to determine endurance

limits is prepared very carefully and tested under closely controlled

conditions.In general endurance limit of a mechanical or structural member in use, can

not match the values obtained in the laboratory. Some differences include;

• Material: composition, basis of failure, variability• Manufacturing: method, heat treatment, fretting corrosion, surface

condition, stress concentration

• Environment: corrosion, temperature, stress state, relaxation times

• Design: size, shape, life, stress state, stress concentration, speed, fretting, galling

Marin equation for modifying endurance limit:

18


*Surface factor ka

Constants a and b are to be found from Table 6-2

Table 6-2 Constants a

and b for

surface modification

factor

SAMPLE PROBLEM 6-3 A steel has a minimum ultimate strength of 520

MPa and a machined surface. Estimate ka.

19


*Size factor kb,

For bending and torsion

For axial loading

What are the effects of rotating/non rotating and circular/ and

noncircular cross sections?

For rotating round sections (Define 95 % stress area A0.95σ of a ring

with do=d, di=0.95d)

20


For nonrotating solid or hollow rounds

Effective size of a round, corresponding to a nonrotating solid or hollow round.

For a rectangular section of dimensions h × b

SAMPLE PROBLEM 6-4 A steel shaft loaded in bending is 32 mm in

diameter, abutting a filleted shoulder 38 mm in diameter. The shaft

material has a mean ultimate tensile strength of 690 Mpa. Estimate the Marin size factor kb if the shaft is used in

(a) A rotating mode.

(b) A nonrotating mode

21


Table 6-3 Effective

diameter (de) for common nonrotating

shapes

22


*Loading factor kc

*Temperature factor kd (kd=ST/SRT)

Table 6-4 Effect of

Operating

Temperature on the

Tensile Strength ofSteel.* (ST = tensile

strength at operating

temperature;SRT = tensile strength

at room temperature

SAMPLE PROBLEM 6-5 A SAE 1035 steel has a tensile strength of 490 MPa and is to be used for a part that sees 230°C in service.

Estimate the Marin temperature modification factor and (Se)230

23


Reliability factor ke

Table 6-5 Reliability Factors ke Corresponding to 8 Percent Standard

Deviation of the Endurance Limit

24


Miscellaneous effect factor kfResidual stresses ; Generally, if the residual stress in the surface of the part

is compression, the endurance limit is improved. Fatigue failures appear to be

tensile failures, or at least to be caused by tensile stress, Operations such as shot peening, hammering, and cold rolling build

compressive stresses into the surface of the part and improve the endurance

limit significantly. Of course, the material must not be worked to exhaustion.

Directional characteristics of the operations. Rolled or drawn parts, for

example, have an endurance limit in the transverse direction that may be 10 to 20 percent less than the endurance limit in the longitudinal direction.

Case-hardened parts may fail at the surface or at the maximum core radius, depending upon the stress gradient. Figure 6–19 shows the typical triangular

stress distribution of a bar under bending or torsion. Also plotted as a heavy

line in this figure are the endurance limits Se for the case and core. For this

example the endurance limit of the core rules the design because the figure shows that the stress σ or τ, whichever applies, at the outer core radius, is appreciably larger than the core endurance limit.

25


Other factors for miscellaneous effects factor kf

*Corrosion:Parts operation in corrosive atmospheres have a lowered fatique

resistance.

There is no fatique limit.

Minimize factors effecting fatique life

*Electrolytic plating: Cr, Ni and Cd reduce strength up to 50%. Zinc plating

has no effect on fatique strength.

*Metal spraying : Reduces up to 14 %

*Cyclic frequency: Shorter the life

*Frettage corrosion: Microscopic motions of tigthly fitted parts. Depends on

material of mating parts, ranges in strength reduction are 24 to 90 %

26


6-10 Stress Concentration and Notch Sensitivity

Some materials are not fully sensitive to the presence of irregularities

or discontinuities, such as holes, grooves, or notches, as with the case of stress concentration factor Kt (or Kts). So we introduce an another

factor Kf, fatique stress concentration factor and

Notch sensitivity q is defined by the equation

Or if we know q ,

27


Figure 6-20 Notch-sensitivity charts for steels and UNS A92024-T wrought

aluminum alloys subjected to reversed bending or reversed axial loads

SAMPLE PROBLEM 6-6 A steel shaft in bending has an ultimate

strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate Kf using:Figure 6–20.

28


Figure 6-21 Notch-sensitivity curves for materials in reversed torsion.

29


Neuber equation, which is given by the formula below is the bases for Figure 6-21

where √a is defined as the Neuber constant and is a material constant.

Equating Eqs. (6–31) and (6–33) yields the notch sensitivity equation

Notes:

*for cast iron take q=0.20

*For steel see Table 6-15*for simple loading reduce endurance limit or multiply stresses by Kf. For

complex loading stresses are multiplied by Kf.

30


SAMPLE PROBLEM 6-9 Figure 6–22a shows a rotating shaft simply

supported in ball bearings at A and D and loaded by a nonrotating force F

of 6.8 kN. Using ASTM “minimum” strengths, estimate the life of the part.

Figure 6-22 (a) Shaft drawing showing alldimensions in millimeters; all fillets 3-mm radius. The shaft rotates and the load isstationary; material is machined from AISI 1050 cold-drawn steel. (b) Bending momentdiagram

31


6-11 Fluctuating Stresses

Figure 6-23 Stress time

relation for sinusoidal fluctuating stress

R=Stress Ratio

A=Amplitude Ratio

32


6-12 Fatique failure Criteria for Fluctuating Stress

Figure 6-24

Modified Goodman

diagram showing all the

strengths and the limiting values of all the stress

components for a

particular midrange stress

33


Figure 6-7 Fatigue diagram showing various criteria of failure.

34


Sodeberg line criterion Eq.

Mod.Goodman criterion

Eq.

Gerber criterion Eq.

ASME elliptic criterion Eq.

Langer (yielding) criterion

Eq.

35


For design purposes

36


Tables 6-6 to 6-8. First row fatique criterion,second row static Langer criterion, third row intersection of static and fatique criteria

37


38


39


SAMPLE PROBLEM 6-10 A 140 mm diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 170kN. Because of the ends, and the fillet

radius, a fatigue stress-concentration factor Kf is 1.85 for 10^6 or larger

life. Find Sa and Sm and the factor of safety guarding against fatigue

and first-cycle yielding, using (a) the Gerber fatigue line and (b) the

ASME-elliptic fatigue line.

Figure 6-28 Principal points A, B, C, and D on

the designer’s diagram

drawn for Gerber,

Langer, and load line

40


6-13 Torsional fatique strength in Fluctuation stresses

In construction of Goodman diagram use

6-14 Combination of loading Models with alternating and midrange components

41


6-15 Varying Fluctuating stresses:

On a part suppose fully reversed σ1 act for n1 cycles, σ2 act for n2 cycles

etc. Than we have

Where

ni is the number of cycles of stress level σi

Ni is the number of cycles to failure at stress level σi

Minor’s Rule