Fatigue Failure
2103320 Des Mach Elem Mech. Eng. Department Chulalongkorn University
• Introduction
• Fatigue testing
• The Endurance Limit
• Endurance Limit modifying factors
• Fluctuating stresses
• Stress concentation
• Fatigue Failure Criteria
• Combinations of Loading Modes
Introduction
• Fatigue failure occurs by machine parts are subjected to time varying loading.
• Failure probably occurs although the actual maximum stresses below the yield strength.
• Fatigue failure occur suddenly without warning, and hence dangerous.
• The initiation of micro-cracks that are not normally
discernible to the naked eye.
• frequently initiate at the
discontinuity part that has high stress concentration.
Crack is enlarged by the repetitive load.
• Large crack
• The remaining material cannot support the loads, resulting in a sudden, fast fracture.
Fatigue testing
R.R. Moore rotating-beam machine
Test-specimen
Bending moment
ω
The Endurance Limit (1)
Tensile strength, S
ut: Maximum stress obtained from the tension test
Fatigue strength, S
f: The highest stress that a material can withstand for a given number of cycles without breaking
Endurance limit, S
e: the limiting value of stress at which failure occurs as the number of cycles become very large. For the stress below the endurance limit, failure never occurs.
When the material is subjected to the repetitive loading, the strength of material is
reduce.
The Endurance Limit (2)
′ =
Se
0.5(
Sut)
Sut≤ 200 kpsi (1400 MPa) 100 kpsi (700 MPa)
Sut> 200 kpsi
700 Mpa
Sut> 1400 MPa
is the endurance limit of the controlled laboratory specimen.
Modifying factors will be added to account for difference between the specimen and the actual machine part.
Se′
Graph of endurance limits versus tensile strengths from actual test results.
Endurance Limit Modifying Factors
e f e d c b a
e k k k k k k S
S = ′
Endurance limit obtained from laboratory test is modified by multiplying many factors
ka : surface condition modification factor kb : size modification factor
kc : load modification factor
kd : temperature modification factor ke : reliability factor
kf : miscellaneous-effects modification factor : endurance limit at the critical
location of a machine part : rotary-beam test specimen
endurance limit
Se′ Se
Surface factor, ka ka = aSutb
The surface of specimen is highly polished. For the other surface conditions, surface factor must be added.
Endurance Limit Modifying Factors
Size factor, kb
kb is obtained from Bending and torsion experiments and can be calculated from
b = k
mm 254 51
51 . 1
mm 51 2.79
)
62 . 7 / (
in 10 2
91
. 0
in 2 11
. 0 )
3 . 0 / (
157 . 0
107 . 0 157 . 0
107 . 0
≤
<
≤
≤
≤
<
≤
≤
−
−
−
−
d d
d d
d d
d d
for axial loading, there is no size effect, kb = 1 For nonrotating round bar and rectangular bar, use equivalent diameter de instead.
For nonrotating bar
d de =0.370
hb de =0.808
Modifying Factors, k b , k c , k d
Temperature factor, kd
The tensile strength is changed with temperature.
This effect is corrected by kd
4 12 3
8 -
2 5 3
) 10 ( 595 . 0 )
0.104(10
) 10 ( 115 . 0 )
10 ( 432 . 0 975 . 0
F F
F F
d
T T
T T
k
−
−
−
− +
− +
=
70 ≤ TF ≤ 1000°F Loading factor, kc
Fatigue test was done with rotating bending load. For the other loading conditions the correction should be added.
c =
k 1 : bending
0.85 : axial
0.59 : pure torsion torsion + bending, kc = 1
RT d ST
k = S
Effect of operating temperature on the tensile strength of steel.
ST = Tensile strength at operating temp.
SRT = Tensile strength at room temp.
Modifying Factors, k e , k f
Miscellaneous factor, kf
This factor is intended to account for the reduction in endurance limit due to all other effects such as the effect of manufacturing process, corrosion, frequency of loading. If there are no other effects use kf = 1.
Reliability factor, ke
Reliability factor ke corresponding to 8% standard deviation of the endurance limit.
Because of the scatter of the experimental data, reliability factor is added to compensate the uncertainty of endurance limit.
Fluctuating Stresses
• Fluctuating stresses in machinery often take the form of a sinusoidal pattern because of the nature of rotating
machinery.
• Other patterns are probable, but the shape of the wave is not important.
• The peaks values (maximum and minimum) are important.
2 min
max σ
σm = σ +
2 min
max σ
σa = σ −
Alternating component Midrange component
Amplitude ratio Stress ratio
max
σmin
= σ R
m
A a
σ
= σ
Stress concentrations
• In the development of the basic stress equations, it was assumed that no geometric irregularities occurred in the member.
• The existence of irregularities (holes, grooves, notches) increases stresses significantly
• This phenomena is called stress concentration.
0
σmax
= σ Kf
0
τmax
=τ Kfs
Stress concentration factor (Normal stress) Stress concentration factor
(Shear stress)
σ σ0 ( ) (w d)
w t
d w
F
= −
= −
t w F =σ t = thickness
σmax, τmax : actual maximum stress
σ0, τ0 : stress calculated from basic equation
Notch-sensitivity, q
Notch-sensitivity charts for steel and UNS A92024-T wrought aluminum alloys
subjected to reversed bending or reversed axial loads.
For larger notch radii, use the values of q corresponding to the r = 0.16 in (4 mm) ordinate.
• K
t(theoretical stress concentration factor) in the figures depends only on the geometry of the part, but the stress concentration also be affected with the notch-sensitivity of the materials.
• K
f(fatigue stress concentration factor is used instead.
) 1 (
1+ −
= t
f q K
K
σ
max = Kfσ
0Notch-sensitivity, q shear
) 1 (
1+ shear −
= ts
fs q K
K
For shear stress K
tsbecomes K
fs.
Notch-sensitivity charts for materials in reverse torsion.
For larger notch radii, use the values of qshear corresponding to the r = 0.16 in (4 mm).
0
max
τ
τ
= KfsStress Concentrations (1)
) 1 (
1+ −
= t
f q K
K
Kt Kts
) 1 (
1+ shear −
= ts
fs q K
K
Stress concentration factor Kt (bending) and Kts (torsion) for round shaft with shoulder fillet
Stress Concentrations (2)
) 1 (
1+ −
= t
f q K
K
Kt Kts
) 1 (
1+ shear −
= ts
fs q K
K
Stress concentration factor Kt (bending) and Kts (torsion) for grooved round bar
Stress Concentrations (3)
Kt Kts
Stress concentration factor Kt (bending) and Kts (torsion) of round shaft with flat-bottom groove
) 1 (
1+ −
= t
f q K
K Kfs =1+qshear(Kts −1)
Fatigue Failure Criteria (1)
Plot of fatigue failures for midrange stresses in both tensile and compressive regions.
Fatigue diagram showing various criteria of failure. (tensile side)
Fatigue Failure Criteria (2)
Fatigue failure criteria Equation Soderberg
Modified-Goodman Gerber
ASME-elliptic Langer static yield
n S
Sea my
= 1 +σ σ
n S
Sea utm
= 1 +σ σ
1
2
=
+
ut m e
a
S n S
nσ σ
1
2 2
=
+
y m e
a S
n S
nσ σ
: Alternating stress : Midrange stress Se : endurance limit Sy : yield strength Sut : tensile strength n : factor of safety
σa
σm
Fatigue failure occurs when
• The left hand side > the right hand side (for n = 1)
• n from the calculation less than 1
• The maximum stress is larger than the static yield criteria
n Sy
m
a +σ =
σ
Combinations of Loading Modes
1. Calculate σm, τm และ σa, τa
2. Consider Stress concentration factor 3. Calculate von Mises stress σ′m and σ′a
4. Substitute von Mises stress in fatigue criteria
Ex. Shaft subjected to bending + torsional shear stress
σm σa τm τa
m
Kfσ Kfσa Kfsτm Kfsτa
[( )2 ( )2 ( )2 6( 2 2 2)]1/2
2
1 σx σy σy σz σz σx τxy τyz τzx
σ′= − + − + − + + + σ′=(σx2 +3τxy2 )1/2
2 / 1 2
2 3( ) )
)
(( f m fs m
m K σ K τ
σ′ = +
2 / 1 2
2 3( ) )
)
(( f a fs a
a K σ K τ
σ′ = +
n S
Sea my
= 1 + ′
′ σ
σ Ex. Soderberg
σx
τxy
5. Check for static yield σa′ +σm′ = Sy n
Example
A rotating shaft is made of 42 × 4 mm AISI 1018 cold-drawn steel tubing and has a 6 mm diameter hold drilled transversely through it. Estimate the factor of safety guarding against fatigue and static failure using Gerber and Langer failure criteria for the following loading conditions:
(a) The shaft is subjected to a completely reversed torque of 120 Nm in phase with a completely reversed bending moment of 150 Nm.
(b) The shaft is subjected to pulsating torque fluctuating from 60 to 160 Nm and a steady bending moment of 150 Nm.
(Shigley’s Mechanical Engineering Design, Ninth Edition, Example 6-14) Given: S
ut= 440 MPa, S
y= 370 MPa
Equations for calculate stress and K
t, K
tsare given in Table A-16
Example
Example
