Chapter 6Fatigue Failure Resulting from

Variable Loading

Introduction

Nature of failure due to static load

Under static failure, the stress on the member is constant.

Prior to a static failure: a very large deformation will occur on the structure or machine members. Therefore the failure can be predicted by observation.

Nature of failure due to fatigue load

Alternating or fluctuating stresses on member will subject the member to fatigue failure. Fatigue failure happens at a sudden without any hint (i.e. without any deformation).

R. R Moore Rotating Beam

Testing specimen

Subject to pure bending (no traverse shear)

Perfectly machined and polished

No circumferential scratches

Fatigue Strength and Cycle Graph (S-N Graph)Material: Steel

Fat

igu

eS

tre

n gth

Sf

Endurance limit Se

Cycle

Life

High CycleLow Cycle

Infinite LifeFinite Life

NA

A

Cycle , N

103

Se

100 101 102 103 104 105 106 107 108

Terminology

N : Cycle : 1 rotation of the specimen = 1 cycle of alternating stress

Sf : fatigue strength is the limit of strength where failure occurs when the alternating

stress is above the fatigue strength. However, when knee occurs on graph, fatigue

strength becomes constant. The value of fatigue strength is called Endurance Limit

(Se). For steel, the knee occurs when cycle is larger than 106.

For non-ferrous metal and alloys: no endurance limit. However, the value of

endurance limit in Table A-22 is equivalent to the fatigue strength @ 5(108).

Cycle: Cycle is related to the cycle of the specimen: boundary is 103

Low cycle N < 103, High Cycle N P 103

Life: Life is related to the stress of the specimen: boundary is Se.

Finite Life : N < 106 Infinite Life: N P103

Endurance limit

Steel

Se’ 40% to 60% if Sut < 1425 MPa

Se’ = 724 MPa when Sut > 1425 MPa

0 560 1120 1680 21000

280

560

840

140

420

700

980

280 840 1400 1960

5.0

4.0

Tensile strength Sut, MPa

Endurance Limit

Se, MPa

6.0SS

u

'e =

Endurance limit- cont

Therefore, after simplifying the observation, the estimation of the endurance limit is

Eq 6.8 pp 274

Sut = 3.41 HB Mpa Eq 2.17 pp 37

Notes:

Various class of cast iron, polished and machined: Table A-22 pp 1023.

Aluminum alloy: does not have an endurance limit, fatigue strength from Table A-22

is set to 5(108) cycles

>≤

=MPa1400SMpa700

MPa1400SS5.0S

ut

utut'e

Fatigue Strength

To approximate relationship between S-N during high cycle, two stage approximation is

required which is

i) approximation of the fatigue strength fraction f @ 103

ii) approximation of the S –N during high cycle

i) Approximation of the fatigue strength f @10 3

Fatigue strength fraction:

Eq 6-10 pp 276

where Eq 6.12 pp 276

Eq 6.11 pp 276

ut

f

SS

f =

B

ut

F

Sf )102(

' 3⋅=σ

)N2log()S/'log(

Be

eF

⋅σ−=

Mpa345S' utF +=σ

Using the above formula, a graph is plotted generates Figure 6-18

ii) Approximation of S –N during high cycle

Calculate the fatigue strength when N is known

Eq 6.13 pp 277

Calculate the cycle when Sf is known

Eq 6.14 pp 277

where a: slope of the curve (log vs log)b: Sf @ 103

Eq 6.14 and 6.15 pp 277

f: fatigue strength fraction

bf aNS =

b

1

a

aN

σ=

e

ut

e

ut

S

Sfa

S

Sfb

2)(

)log(31

=

−=

ENDURANCE LIMIT (Se)

Se’ is the value from experiment of laboratory controlled environment (Refer R-R Moore test specimen). In reality, difference in material, manufacturing, environment and design exist. Therefore, modifying factor is introduced, using Marin equation.

where

S’e : endurance limit based on R.R. Moore experimentka: surface factor

kb: size factor

kc: loading factor

kd: temperature factorke: reliability factor

kf: miscellaneous-effects factor

efedcbae 'SkkkkkkS =

Surface Factor k a

Rotating specimen is highly polished without any circumferential scratches.

Therefore, surface factor for other finishes

Eq 6.19 pp 279

Factor a and b can be found in Table 6-2 pp 280

Size Factor k b

For rotating circular cross section subjected to bending or torsion

Eq 6.20 pp 280

kb = 1 (for axial load), please refer to kc

buta aSk =

≤<≤≤

=−

−

mm254d51d51.1

mm51d79.2d24.1k

157.0

107.0

b

For other condition such non-rotating circular shaft or non-circular cross section, effective

dimension de is used in Eq 6-20.

Effective dimension d e non-rotating solid and hollow circular shaft

Rotating solid or hollow circular shaft: 95% stress area is

Non-rotating solid or hollow circular shaft:

95% stress area is defined twice the area outside of two parallel chords having a spacing of 0.95D, where D is the diameter. Therefore the exact computation is

Therefore

[ ] 22295.0 d0766.0)d95.0(d

4A =−π=σ

295.0 D0105.0A =σ

Dd

Dd

DA

e

e

37.0

0105.00766.0

0105.022

295.0

==

=σ

Effective dimension de of rectangular cross section with h x b dimensions due to non-

rotating bending

95% stress area is

Therefore

Similar principle is applied to other common non-rotating structural shapes and effective dimension de is as in Table 6-3 pp 282

21

95.0 )hb(05.0A =σ

2

1

e )hb(808.0d =

Loading factor k c

Bending kc = 1

Axial kc = 0.85 Eq 6.26

Temperature Factor k d

ST: tensile strength at T temperature

SRT: tensile strength at room temperature

Refer to Table 6-4 pp 283: this table shows the endurance limit of the steel increases kd > 1 for 200C < T < 2500C. This data obtained from Figure 2-9 pp 39 (result from

145 tests of 21 carbon and alloy steel).

The mathematics representation of data using 4th order polynomial

RT

Td S

Sk =

C540T37T)10(246.6

T)10(5621.0T)10(03414T)10(6507.09877.0k0

C4c

12

3C

82C

5C

3d

≤≤−

+−+=−

−−−

Reliability Factor (k e)

This factor is intended to account the scattered data in Figure 6-17. Se’ = 0.5 Sut is

only mean of the data.

Refer to Table 6-5 pp 285

Miscellaneous-Effects Factor k f

This factor will take into account other factors that reduce the endurance limit such as

corrosion, electrolytic plating, metal spraying, cyclic frequency and/or frettagecorrosion.

Stress concentration and notch sensitivity

Stress increases at the vicinity of irregularities or discontinuities such as holes,

grooves or notches.

Theoretically, these discontinuities causes the stresses to increase significantly by

or Eq 6.30 pp 287

Kf: stress-concentration factor

Notch sensitivity q is defined

or Eq 6.31 pp 287

Kt: stress concentration factor

1K1K

qt

f

−−=

0fmax K σ=σ 0fsmax K τ=τ

1K1K

qts

fsshear −

−=

Therefore, fatigue stress concentration factor (Kf)

Eq 6.32 pp 287

q: notch sensitivity from Figure 6-20 pp 287

Eq 6.32 pp 287

qshear : notch sensitivity from Figure 6-21 pp 288

Note: If the value of notch sensitivity can not be retrieved from the respective figures, the most conservative assumption is that Kf = Kt or Kfs = Kts

Kt, Kts: can be obtained from Table A-13 to A-14

)1K(q1K tf −+=

)1K(q1K tsshearfs −+=

Stress concentration (K t or K ts)

0 0.05 0.10 0.15 0.20

r/d

0.25 0.301.0

1.4

1.8

2.2

2.6

3.0

D/d=1.5

1.54

Figure A-13-9

geometrical property to be used

in calculating the stress.

o

4

Roundshaft withshoulder fillet

Mcinbending, ,where

Id d

c ,andI2 64

σ =

π= = Kt

D= 30mm, d = 20mm and r = 2 mm and subjected to Moment M

Fluctuating Stresses

max

minRσσ=

Str

ess

Time

maxmaxmaxmax

min

m

a : amplitude stress

: midrange stress

Stress Ratio:

Amplitude Ratio:m

aAσσ=

Midrange Stress

Alternating Stress

Similarly for the fluctuating forces

Midrange Force

Alternating Force

These forces will generate respective midrange and alternating stresses.

2minmax

m

σ+σ=σ

2minmax σσσ −

=a

2FF

F minmaxm

+=

2FF

F minmaxa

−=

Assumption used in this book

• Avoid localized plastic strain at a notch:

and

• When plastic strain cannot be avoided at the notch, conservatively

and use Kfm = 1, that is

aofa K σ=σ 0mfm K σ=σ

aofa K σ=σ 0mm σ=σ

How to determine the midrange or alternating stress ?

Moment on rotating shaft.

When the shaft rotates in ccw direction, element A will rotate from position 1, 2, 3 and 4 for one full cycle. At the same time, the stresses of A will fluctuate from 0, +s0, 0 and -s0. Therefore, moment on rotating shaft will generate ssssa = Mc/I and ssssm = 0.

Stress

Torsion on rotating shaft

tttta = 0 and ttttm = Tr/J .

AI4

32 S

tres

s

Time

T

Fatigue Failure Criteria: graphical and empirical method

Load line

Sy

Se

Sy Sut

a

m

B: (S , S )a1 m1

A

B

C

D

C: (S , S )a2 m2

D: (S , S )a3 m3

A: ( , )a m

Langer Line

Gerber Line

Goodman Line

Graphical Method

Goodman line

Gerber Failure Theory

Langer Line

m

1m

a

1a SSn

σ=

σ=

m

2m

a

2a SSn

σ=

σ=

m

3m

a

3a SSn

σ=

σ=

Empirical Method

Goodman Line

Basic line equation

For a known (sa, sm) Eq 6.45 pp 298

Gerber Failure Locus (Parabolic Equation)

Basic parabolic equation

For a known (sa, sm) Eq 6.46 pp 298

1SS

SS

e

a

ut

m =+

n1

SS e

a

ut

m =σ+σ

1SS

SS

e

a

2

ut

m =+

1Sn

Sn

e

a

2

ut

m =σ

+

σ

Langer Line (failure due to yielding)

Basic line equation

For a known (sa, sm) Eq 6.48 pp 298

For other failure theories, the same principles can be applied. *In the following discussion, these three theories will be used.

1=+y

a

y

m

S

S

S

S

nS

n1

SS

ma

y

y

a

y

m

=σ+σ

=σ

+σ

Langer line vs. other theories(i.e. Modified Goodman and Gerber Line )

If the stresses on the member is plotted on the shaded area, the load line of the member will intersect with Langer line first. Therefore, failure due to yielding will occur

first. This is type of failure is called Langer-first-cycle failure . Empirically, it can be

determined when nlanger < nf(other).

Sy

Se

Sy Sut

a

m

Sy

Se

Sy Sut

a

m

Procedure in solving the fatigue problem

1. Determine sa, sm,ta and tm.

2. Combine the alternating stresses using DET

3. Combine the midrange stresses using DET

4. Use the theories to solve the problem

2a

2a)a(total 3τ+σ=σ

2m

2m)m(total 3τ+σ=σ