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1 Notations 2
2 Completely Reversed or Cyclic Stresses 3
3 Fatigue and Endurance Limit 3
4 Effect of Loading on Endurance Limit and Load Factor 4
5 Effect of Surface Finish on Endurance Limit and Surface Finish Factor 4
6 Effect of Size on Endurance Limit and Size Factor 4
7 Effect of Miscellaneous Factors on Endurance Limit 5
8 Relation Between Endurance Limit and Ultimate Tensile Strength 5
9 Factor of Safety for Fatigue Loading 5
10 Stress Concentration 5
11 Theoretical or Form Stress Concentration Factor 6
12 Stress Concentration due to Holes and Notches 6
13 Methods of Reducing Stress Concentration 7
14 Factors to be Considered while Designing Machine Parts to Avoid Fatigue Failure 8
15 Stress Concentration Factor for Various Machine Members 8
16 Fatigue Stress Concentration Factor 12
17 Notch Sensitivity 12
18 Combined Steady and Variable Stress 12
19 Gerber Method for Combination of Stresses 12
20 Goodman Method for Combination of Stresses 13
21 Soderberg Method for Combination of Stresses 14
22 Combined Variable Normal Stress and Variable Shear Stress 14
23 Application of Soderbergs Equation 15
24 Examples 17
25 References 31
26 Contacts 31
1 Notations
• σm = Mean or average stress (tensile or compres-sive).
• σmax = Maximum applied stress.
• σmin = Minimum applied stress.
• σv = Reversed stress component or alternating orvariable stress.
• σ′e = Endurance limit for any stress range
• σe = Endurance limit for completely reversedstresses
• R = Stress ratio.
• Kb = Load correction factor for the reversed orrotating bending load. Its value is usually takenas unity.
• Ka = Load correction factor for the reversed axialload. Its value may be taken as 0.8.
• Ks = Load correction factor for the reversed tor-sional or shear load. Its value may be taken as 0.55for ductile materials and 0.8 for brittle materials
• σeb = Endurance limit for reversed bending load.
• σea = Endurance limit for reversed axial load.
• τe = Endurance limit for reversed torsional orshear load.
• Ksur = Surface finish factor.
• σel = Surface finish endurance limit.
• Ksz = Size factor.
• σe2 = Size endurance limit.
• Kr = Reliability factor.
• KT = Temperature factor.
• Ki = Impact factor.
• F.S. = Factor of safety.
• σy = Yield point stress.
• Kt = Theoretical stress concentration factor foraxial or bending loading.
• Kts = Theoretical stress concentration factor fortorsional or shear loading.
• σu = Ultimate point stress.
• Kf = Fatigue stress concentration factor.
• τy = Yield point shear.
• τu = Ultimate point shear.
• σneb = Equivalent normal stress due to reversedbending loading.
• σnea = Equivalent normal stress due to reversedaxial loading.
• σne = Total equivalent normal stress.
• τes = Equivalent shear stress due to reversed tor-sional or shear loading
• Wm = Mean or average load.
• Wv = Variable load.
• A = Cross-sectional area.
• Mm = Mean bending moment.
• Mv = Variable bending moment.
• Z = Section modulus.
• σb = Working or design bending stress.
• Tm = Mean or average torque.
• Tv = Variable torque.
• d = Diameter of the shaft.
• τm = Mean or average shear stress.
• τv = Variable shear stress.
• τ = Working or design shear stress.
• Kfs = Fatigue stress concentration factor for tor-sional or shear loading.
• k = Ratio of inner diameter to outer diameter.
2 Completely Reversed or Cyclic Stresses
For each revolution of the beam, the stresses are re-versed from compressive to tensile. The stresses whichvary from one value of compressive to the same value oftensile or vice versa, are known as completely re-versed or cyclic stresses.Notes:
1. The stresses which vary from a minimum value toa maximum value of the same nature, (i.e. tensileor compressive) are called fluctuating stresses.
Figure 1: Reversed or cyclic stresses.
2. The stresses which vary from zero to a certain maximum value are called repeated stresses.
3. The stresses which vary from a minimum value to a maximum value of the opposite nature (i.e. from acertain minimum compressive to a certain maximum tensile or from a minimum tensile to a maximumcompressive) are called alternating stresses.
3 Fatigue and Endurance Limit
It has been found experimentally that when a material is subjected to repeated stresses, it fails at stressesbelow the yield point stresses. Such type of failure of a material is known as fatigue. The failure is caused bymeans of a progressive crack formation which are usually fine and of microscopic size. The failure may occureven without any prior indication. The fatigue of material is effected by the size of the component, relativemagnitude of static and fluctuating loads and the number of load reversals.
Figure 2: Time-stress diagrams.
Mean or average stress,
σm =σmax + σmin
2Reversed stress component or alternating or variable stress,
σv =σmax − σmin
2
Stress ratio,
R =σmaxσmin
For completely reversed stresses, R = −1 and for repeated stresses, R = 0. It may be noted that R cannot begreater than unity. Endurance limit,
σ′
e =3σe
2 −R
4 Effect of Loading on Endurance Limit and Load Factor
Endurance limit for reversed bending load,
σeb = σeKb = σe
Endurance limit for reversed axial load,σea = σeKa
Endurance limit for reversed torsional or shear load,
τe = σeKs
5 Effect of Surface Finish on Endurance Limit and Surface Finish Factor
Figure 3: Surface finish factor for various surface conditions.
Endurance limit,
σel = σebKsur = σeKbKsur = σeKsur ...(For reversed bending load)= σeaKsur = σeKaKsur ...(For reversed axial load)= τeKsur = σeKsKsur ...(For reversed torsional or shear load)
Note: The surface finish factor for non-ferrous metals may be taken as unity.
6 Effect of Size on Endurance Limit and Size Factor
Endurance limit,
σe2 = σelKsz = σebKsurfKsz = σeKbKsurfKsz = σeKsurKsz ...(For reversed bending load)
= σeaKsurKsz = σeKaKsurKsz ...(For reversed axial load)= τeKsurKsz = σeKsKsurKsz ...(For reversed torsional or shear load)
Notes:
1. The value of size factor is taken as unity for the standard specimen having nominal diameter of 7.657 mm.
2. When the nominal diameter of the specimen is more than 7.657 mm but less than 50 mm, the value of sizefactor may be taken as 0.85.
3. When the nominal diameter of the specimen is more than 50 mm, then the value of size factor may betaken as 0.75.
7 Effect of Miscellaneous Factors on Endurance Limit
Endurance limit,
σ′
e = σebKsurKszKrKTKi ...(For reversed bending load)= σeaKsurKszKrKTKi ...(For reversed axial load)= τeKsurKszKrKTKi ...(For reversed torsional or shear load)
Note: In solving problems, if the value of any of the above factors is not known, it may be taken as unity.
8 Relation Between Endurance Limit and Ultimate Tensile Strength
Figure 4: Endurance limit of steel corresponding to ultimate tensile strength.
σe = 0.5σu ...(For steel)= 0.4σu ...(For cast steel)= 0.35σu ...(For cast iron)= 0.3σu ...(For non-ferrous metals and alloys)
9 Factor of Safety for Fatigue Loading
F.S. =Endurance limit stress
Design or working stress=σeσd
Note:σe = 0.8 to 0.9 σy ...(For steel)
10 Stress Concentration
Whenever a machine component changes the shapeof its cross-section, the simple stress distribution nolonger holds good and the neighborhood of the discon-tinuity is different. This irregularity in the stress dis-tribution caused by abrupt changes of form is calledstress concentration. It occurs for all kinds ofstresses in the presence of fillets, notches, holes, key-ways, splines, surface roughness or scratches etc. Figure 5: Stress concentration.
11 Theoretical or Form Stress Concentration Factor
The theoretical or form stress concentration factor is defined as the ratio of the maximum stress in a member (ata notch or a fillet) to the nominal stress at the same section based upon net area. Mathematically, theoreticalor form stress concentration factor,
Kt =Maximum stress
Nominal stress
The value of Kt depends upon the material and geometry of the part.Notes:
1. In static loading, stress concentration in ductile materials is not so serious as in brittle materials, becausein ductile materials local deformation or yielding takes place which reduces the concentration. In brittlematerials, cracks may appear at these local concentrations of stress which will increase the stress over therest of the section. It is, therefore, necessary that in designing parts of brittle materials such as castings,care should be taken. In order to avoid failure due to stress concentration, fillets at the changes of sectionmust be provided.
2. In cyclic loading, stress concentration in ductile materials is always serious because the ductility of thematerial is not effective in relieving the concentration of stress caused by cracks, flaws, surface roughness,or any sharp discontinuity in the geometrical form of the member. If the stress at any point in a memberis above the endurance limit of the material, a crack may develop under the action of repeated load andthe crack will lead to failure of the member.
12 Stress Concentration due to Holes and Notches
The maximum stress is given by
σmax = σ
(1 +
2a
b
)and the theoretical stress concentration factor,
Kt =σmaxσ
= 1 +2a
b
When a/b is large, the ellipse approaches a crack transverse to the load and the value of Kt becomes very large.When a/b is small, the ellipse approaches a longitudinal slit and the increase in stress is small. When the holeis circular, then a/b= 1 and the maximum stress is three times the nominal value.
Figure 6: Stress concentration due to holes.
The stress concentration in the notched tension mem-ber, is influenced by the depth a of the notch and ra-dius r at the bottom of the notch. The maximum stress,which applies to members having notches that are smallin comparison with the width of the plate, may be ob-tained by the following equation,
σmax = σ
(1 +
2a
r
)Figure 7: Stress concentration due to notches.
13 Methods of Reducing Stress Concentration
Figure 8: Methods of reducing stress concentration.
Figure 9: Methods of reducing stress concentration in cylindrical members with shoulders.
Figure 10: Methods of reducing stress concentration in cylindrical members with holes.
Figure 11: Methods of reducing stress concentration in cylindrical members with holes.
Figure 12: Methods of reducing stress concentration of a press fit.
14 Factors to be Considered while Designing Machine Parts to Avoid FatigueFailure
The following factors should be considered while designing machine parts to avoid fatigue failure:
1. The variation in the size of the component should be as gradual as possible.
2. The holes, notches and other stress raisers should be avoided.
3. The proper stress de-concentrators such as fillets and notches should be provided wherever necessary.
4. The parts should be protected from corrosive atmosphere.
5. A smooth finish of outer surface of the component increases the fatigue life.
6. The material with high fatigue strength should be selected.
7. The residual compressive stresses over the parts surface increases its fatigue strength.
15 Stress Concentration Factor for Various Machine Members
16 Fatigue Stress Concentration Factor
When a machine member is subjected to cyclic or fatigue loading, the value of fatigue stress concentrationfactor shall be applied instead of theoretical stress concentration factor. Since the determination of fatiguestress concentration factor is not an easy task, therefore from experimental tests it is defined as
Kf =Endurance limit without stress concentration
Endurance limit with stress concentration
17 Notch Sensitivity
In cyclic loading, the effect of the notch or the fillet isusually less than predicted by the use of the theoreti-cal factors as discussed before. The difference dependsupon the stress gradient in the region of the stress con-centration and on the hardness of the material. Theterm notch sensitivity is applied to this behavior.It may be defined as the degree to which the theoret-ical effect of stress concentration is actually reached.The stress gradient depends mainly on the radius ofthe notch, hole or fillet and on the grain size of thematerial.
q =Kf − 1
Kt − 1Kf = 1 + q (Kt − 1) ...(For tensile or bending stress)
Kfs = 1 + q (Kts − 1) ...(For shear stress)
18 Combined Steady and Variable Stress
The failure points from fatigue tests made with dif-ferent steels and combinations of mean and variablestresses are plotted in Fig.13 as functions of variablestress (σv) and mean stress (σm). The most significantobservation is that, in general, the failure point is littlerelated to the mean stress when it is compressive butis very much a function of the mean stress when it istensile. In practice, this means that fatigue failures arerare when the mean stress is compressive (or negative).Therefore, the greater emphasis must be given to thecombination of a variable stress and a steady (or mean)tensile stress.There are several ways in which problems involving thiscombination of stresses may be solved, but the follow-ing are important from the subject point of view :
1. Gerber method,
2. Goodman method, and
3. Soderberg method.
Figure 13: Combined mean and variable stress.
19 Gerber Method for Combination of Stresses
According to Gerber, variable stress,1
F.S.=
(σmσu
)2
F.S.+σvσe
Considering the fatigue stress concentration factor (Kf ),
1
F.S.=
(σmσu
)2
F.S.+σv Kf
σe
20 Goodman Method for Combination of Stresses
According to Goodman, variable stress,1
F.S.=σmσu
+σvσe
Considering the fatigue stress concentration factor (Kf ),
1
F.S.=σmσu
+σv Kf
σe
Considering the load factor, surface finish factor and size factor,
1
F.S.=
σmσu
+σv Kf
σeb Ksur Ksz
...(For ductile materials subjected to reversed bending loading)
1
F.S.=
σm Kt
σu+
σv Kf
σeb Ksur Ksz
...(For brittle materials subjected to reversed bending loading)
1
F.S.=
σmσu
+σv Kf
σea Ksur Ksz
...(For ductile materials subjected to reversed axial loading)
1
F.S.=
σm Kt
σu+
σv Kf
σea Ksur Ksz
...(For brittle materials subjected to reversed axial loading)
1
F.S.=
τmτu
+τv Kf
τe Ksur Ksz
...(For ductile materials subjected to reversed torsional or shear loading)
1
F.S.=
τm Kts
τu+
τv Kf
τe Ksur Ksz
...(For brittle materials subjected to reversed torsional or shear loading)
where suffix ’s’ denotes for shear.Note: For reversed torsional or shear loading, the values of ultimate shear strength (τu) and endurance shearstrength (τe) may be taken as follows:
τu = 0.8σu; and τe = 0.8σe
Figure 14: Goodman method.
21 Soderberg Method for Combination of Stresses
According to Soderberg, variable stress,1
F.S.=σmσy
+σvσe
Considering the fatigue stress concentration factor (Kf ),
1
F.S.=σmσy
+σv Kf
σe
Considering the load factor, surface finish factor and size factor,
1
F.S.=
σmσy
+σv Kf
σeb Ksur Ksz
...(For ductile materials subjected to reversed bending loading)
1
F.S.=
σmσy
+σv Kf
σea Ksur Ksz
...(For ductile materials subjected to reversed axial loading)
1
F.S.=
τmτy
+τv Kf
τe Ksur Ksz
...(For ductile materials subjected to reversed torsional or shear loading)
Note: The Soderberg method is particularly used for ductile materials.
Figure 15: Soderberg method.
22 Combined Variable Normal Stress and Variable Shear Stress
When a machine part is subjected to both variable normal stress and a variable shear stress; then it is designedby using the following two theories of combined stresses :
1. Maximum shear stress theory, and
2. Maximum normal stress theory.
Equivalent normal stress due to reversed bending,
σneb = σm +σv σy Kfb
σeb Ksur Ksz
Equivalent normal stress due to reversed axial loading,
σnea = σm +σv σy Kfa
σea Ksur Ksz
Total equivalent normal stress,
σne = σneb + σnea =σyF.S.
Equivalent shear stress due to reversed torsional or shear loading,
τes = τm +τv τy Kfs
τe Ksur Ksz
The maximum shear stress theory is used in designing machine parts of ductile materials. According to thistheory, maximum equivalent shear stress,
τes(max) =1
2
√(σne)
2 + 4 (τes)2 =
τyF.S.
The maximum normal stress theory is used in designing machine parts of brittle materials. According to thistheory, maximum equivalent normal stress,
σes(max) =1
2σne +
1
2
√(σne)
2 + 4 (τes)2 =
σyF.S.
23 Application of Soderbergs Equation
∵1
F.S.=
σmσy
+σv Kf
σe=σm σe + σv σy Kf
σy σe
∴ F.S. =σy σe
σm σe + σv σy Kf
=σy
σm σe + σvσeσy Kf
Working or design stress,
= σm σe +σvσe
σy Kf
Let us now consider the use of Soderberg’s equation to a ductile material under the following loading conditions.
1. Axial loadingIn case of axial loading, we know that the mean or average stress,
σm =Wm
A
and variable stress,
σv =Wv
AWorking or design stress,
=Wm
A+σyσe
Wv
AKf =
Wm + σyσeWv Kf
A
∴ F.S. =σy A
Wm + σyσeWv Kf
2. Simple bendingIn case of simple bending, we know that the bending stress,
σb =My
I=M
Z
Mean or average bending stress,
σm =Mm
ZVariable bending stress,
σv =Mv
ZWorking or design bending stress,
σb =Mm
Z+σyσe
Mv
ZKf =
Mm + σyσeMv Kf
Z
∴ F.S. =σy Z
Mm + σyσeMv Kf
For circular shafts,
Z =πd3
32
σb =32
πd3
[Mm +
σyσe
Mv Kf
]F.S. =
σy32πd3
[Mm + σy
σeMv Kf
]3. Simple torsion of circular shafts
In case of simple torsion, we know that the torque,
T =π
16τd3 or τ =
16T
πd3
Mean or average shear stress,
τm =16Tmπd3
Variable shear stress,
τm =16Tmπd3
Working or design shear stress,
τ =16Tmπd3
+τyτeKfs
16Tvπd3
=16
πde
[Tm +
τyτeKfs Tv
]Note: For shafts made of ductile material, τy = 0.5σy , and τe = 0.5σe may be taken.
4. Combined bending and torsion of circular shaftsIn case of combined bending and torsion of circular shafts, the maximum shear stress theory may be used.According to this theory, maximum shear stress,
τmax =τyF.S.
=1
2
√(σb)
2 + 4τ 2
=1
2
√[32
πd3
{Mm +
σyσe
Kf Mv
}]2+ 4
[16
πd3
{Tm +
τyτeKfs Tv
}]2=
16
πd3
√[Mm +
σyσe
Kf Mv
]2+
[Tm +
τyτeKfs Tv
]The majority of rotating shafts carry a steady torque and the loads remain fixed in space in both directionand magnitude. Thus during each revolution every fiber on the surface of the shaft under-goes a completereversal of stress due to bending moment. Therefore for the usual case when Mm = 0, Mv = M, Tm = Tand Tv = 0, the above equation may be written as
τyF.S.
=16
πd3
√[σyσe
Kf M
]2+ T 2
Note: The above relations apply to a solid shaft. For hollow shaft, the left hand side of the aboveequations must be multiplied by (1 − k4).
24 Examples
25 References
1. R.S. KHURMI, J.K. GUPTA, A Textbook Of Machine Design
26 Contacts
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