Design Procedure 7.3: Estimation of Endurance Limit
7.10 Influence of Nonzero Mean Stress
accumulation at a particular stress level is independent of the stress history. This is not strictly true, as it is well-known that more severe loadings cause disproportionate damage, espe- cially if they occur early in the part’s life. Despite these short- comings, the linear damage rule remains popular, largely be- cause it is so simple.
IfNt0is the total number of cycles to failure when there are different cyclic patterns (all of which are completely re- versed), the ratio of the number of cycles at a specific stress level to the total number of cycles to failure is
αi= n0i
Nt0 or n0i=αiNt0. (7.25) Substituting Eq. (7.25) into Eq. (7.24), it is predicted that fail- ure will occur if
Xαi
Ni0 ≥ 1
Nt0. (7.26)
Example 7.7: Cumulative Damage
Given: Consider a steel with an ultimate strength of 440 MPa, so thatS0l = 330MPa andSe0 = 200MPa. A compli- cated loading cycle is applied, so that the stress is 175 MPa for 20% of the time, 220 MPa for 30%, 250 MPa for 40%, and 275 MPa for 10%.
Find:The number of cycles until cumulative failure.
Solution:First of all, note from Eqs. (7.11) and (7.12) that bs=−1
3log S0l
Se0
=−1 3log
330 200
=−0.0725,
C¯= log
"
(Sl0)2
Se0
#
= log 3302
200
= 2.736.
Note thatS01 = 175 MPa is less thanSe0, so thatN10 = ∞, implying that at this stress level failure will not occur. From Eq. (7.14) for the other three fatigue strength levels,
N20 = S
0 f10−
C¯1/bs
= (220)(10)−2.736
−1/0.0725
= 2.683×105cycles, N30 =(250)(10)−2.736
−1/0.0725
= 4.601×104cycles, N40 =(275)(10)−2.736−1/0.0725
= 1.236×104cycles.
Making use of Eq. (7.26) gives α1
N10 +α2
N20 + α3
N30 + α4
N40 = 1 Nt0; 0.2
∞ + 0.3
2.683×105 + 0.4
4.601×104 + 0.1
1.236×104 = 1 Nt0. This is solved asNt0= 55,860cycles.
7.10 Influence of Nonzero Mean Stress
Other than in classifying cyclic behavior, completely reversed (σm= 0) stress cycles have been assumed. Many machine ele- ments involve fluctuating stresses about a nonzero mean. The experimental apparatus used to generate the results shown in Fig. 7.7 cannot apply mean and alternating stresses, so other test approaches are needed (such as tension-compression or combined stresses). If a material has been extensively char- acterized, the effect of a nonzero mean stress can be incor- porated through aHaigh diagram, also called aconstant life diagramas shown in Fig. 7.15. Because such data are not gen- erally available, the influence of nonzero mean stress must be estimated by using one of several empirical relationships that determine failure at a given life when alternating and mean stresses are both nonzero.
7.10.1 Ductile Materials
Figure 7.16 shows how four empirical relationships estimate the influence of nonzero mean stress on fatigue life for duc- tile materials loaded in tension. The ordinate has both the yield strength, Syt, and endurance limit,Se, indicated, and the abscissa shows both the yield and ultimate strengths. The yield line shown is for reference purposes, as it indicates fail- ure during the first loading cycle. Commonly used criteria for fatigue failure with a non-zero mean stress are shown in the figure.
The effects of stress concentrations on ductile materials (see Section 7.7) require further explanation. If there is a stress concentration and the yield stress is exceeded, then the de- formed geometry is difficult to obtain. Recall from Eq. (3.21) that stress is proportional to strain for linearly elastic mate- rials. However, during plastic deformation, stresses are re- lated to strain increment. If the yield stress is exceeded, it is possible that plastic strains are negligible, but it is also possi- ble that they are so large as to effectively eliminate the stress concentration effect. The amount of plastic strain that occurs depends on the nature of a part’s supports, its ductility and strain hardening capability, and applied loads.
However, consider the situation where the mean stress is small; in Fig. 7.16 this would suggest that the endurance limit should not be exceeded, consistent with the approach in Section 7.8. Since the endurance limit is much smaller than the yield strength, it is reasonable to apply stress con- centration effects to the alternating stress. Applying stress concentration effects to the mean stress has little impact be- cause it was assumed these stresses are small. On the other hand, if the mean stress is large, then plastic deformation will relieve stress concentrations, and applying the fatigue stress concentration factor based on the original geometry to the al- ternating stress results in a conservative approach. Therefore, a common modeling simplification is to apply stress concen- trations to alternating stresses but not mean stresses, and this is the approach followed in this text.
Gerber Line
TheGerber lineis sometimes called theGerber parabolic rela- tionshipbecause the equation is
Kfnsσa
Se + nsσm
Sut
2
= 1. (7.27)
where
Influence of Nonzero Mean Stress 177
R = -0.5 104cycles
105 cycles 106 cycles σmax/Sut
1.0 0.8 0.6 0.4 0.2
0.0-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0
R = 0.0
R = 0.5
R = 1.0
σmin/Sut
R = -1.0 σa
Sut σm
Sut
0.8
0.6
0.4
0.2
0.8
0.6
0.4 0.2
Figure 7.15: A typical Haigh diagram showing constant life curves for different combinations of mean and alternating stresses.
Se = modified endurance limit, Pa Sut= ultimate strength in tension, Pa ns = safety factor
σa = alternating stress, Pa σm= mean stress, Pa
Kf= fatigue stress concentration factor
This line passes through the central portion of the experimen- tal failure points and hence should be the best predictor of failure, but the parabolic nature of the equation complicates mathematics.
Goodman Line
TheGoodman line proposes connecting the modified en- durance limit on the alternating stress axis with the ultimate strength in tension on the mean stress axis in Fig. 7.16 by a straight line, or
Kfσa
Se
+ σm
Sut
= 1 ns
. (7.28)
Note the linearization of Eq. (7.28) relative to Eq. (7.27). Equa- tion (7.28) fits experimental data reasonably well and is sim- pler to use than Eq. (7.27). The starting and ending points for the Goodman and Gerber lines are the same in Fig. 7.16, but between these points the Goodman line is linear and the Gerber line is parabolic.
Example 7.8: Effect of Nonzero Mean Stress
Given: A straight, circular rotating beam with a 30-mm di- ameter and a 1-m length has an axial load of 30,000 N applied at the end and a stationary radial load of 400 N. The beam is cold-drawn and the material is AISI 1040 steel. Assume that ks=kr=kt=km= 1.
Find:The safety factor for infinite life by using the Goodman line.
Solution:From Table A.1 for AISI 1040 steel,Su= 520 MPa.
From Fig. 7.11a for the cold-drawn process and Su = 520 MPa, the surface finish factor is 0.78. From Eq. (7.18), the modified endurance limit while making use of Eq. (7.6) is
Goodman line Gerber line Yield line
Alternating stress, σa
Mean stress, σm
0 Syt
Sut Se
Soderberg line
Syt
Figure 7.16: Influence of nonzero mean stress on fatigue life for tensile loading as estimated by four empirical relation- ships.
Se = kfkskrktkmSe0
= (0.78)(1)(1)(1)(1)(0.45)(520)
= 182.5MPa.
The bending stress from Eq. (4.45) gives the alternating stress that the beam experiences as
σa=M c
I =(64)(400)(1)(0.03/2) π(0.03)4
= 150.9MPa.
The mean stress due to the axial load is σm= Pa
A =(30,000)(4)
π(0.03)2 = 42.44MPa.
For an unnotched beamKf = 1. From Eq. (7.28), Kfσa
Se + σm
520= 1 ns. Therefore,
(1)(150.9)
182.5 +42.44 520 = 1
ns
; ns= 1
0.9084= 1.101.
Using the Goodman line, the safety factor for infinite life is 1.101.
Soderberg Line
TheSoderberg lineis conservative and is given as Kfσa
Se +σm
Syt = 1
ns. (7.29)
Note from Fig. 7.16 and Eqs. (7.28) and (7.29) that the ultimate strength in the Goodman relationship has been replaced with the yield strength in the Soderberg relationship.
Yield Line
To complete the possibilities, theyield lineis given. It is used to define yielding on the first cycle, or
σa
Syt
+σm
Syt
= 1 ns
. (7.30)
This completes the description of the theories presented in Fig. 7.16.
Modified Goodman Diagram
The Goodman relationship given in Eq. (7.28) is modified by combining fatigue failure with failure by yield. The complete modified Goodman diagramis shown in Fig. 7.17.1 Thus, all points inside a modified Goodman diagram ABCDEFGH correspond to fluctuating stresses that should cause neither fatigue failure nor yielding. The word “complete” is used to indicate that the diagram is valid for both tension and com- pression. The word “modified” designates that the Good- man line shown in Fig. 7.16 has been modified in Fig. 7.17;
that is, in Fig. 7.16 the Goodman line extends from the en- durance limit on the alternating stress ordinate to the ulti- mate strength on the mean stress abscissa. In the modified Goodman diagram in Fig. 7.17, the Goodman line is modi- fied such that for stresses larger than the yield strength the yield line BC is used. Thus, the modified Goodman diagram combines fatigue criteria as represented by the Goodman line and yield criteria as represented by the yield line. Note in Fig. 7.17 that lines AB, DE, EF, and HA are Goodman lines and that lines BC, CD, FG, and GH are yield lines. The static load is represented by line CG. Table 7.5 gives the equations and range of applicability for the construction of the Good- man and yield lines of the complete modified Goodman di- agram. Note that when the ultimate and yield strengths are known for a specific material, as well as the corresponding endurance limit for a particular part made of that material, the modified Goodman diagram can be constructed.
1Equation (7.28) is referred to in the technical literature as either theGoodmanor themodified Goodmanrelationship, since it was the combined contribution of multiple researchers. Figure 7.17 is often referred to as the modified Goodman diagram, as is done in this text, but is called anenhanced modified Goodman diagramin some sources.
Table 7.5: Equations and range of applicability for construc- tion of complete modified Goodman diagram.
e g n a R n
o it a u q E e
n i L
AB σmax = Se
Kf + σm 1− Se
SuKf 0≤ σm ≤ Sy− Se/Kf 1− Se
KfSu
BC σmax = Sy
Sy − Se
Kf
1− Se
KfSu
≤ σm ≤ Sy
CD σmin = 2σm − Sy
Sy − Se
Kf
1− Se
KfSu
≤ σm ≤ Sy
DE σmin = 1 + Se
KfSu σm − Se
Kf 0≤ σm ≤
Sy − Se
Kf 1− Se
KfSu
EF σmin = σm − Se
Kf
Se
Kf − Sy ≤ σm ≤ 0
FG σmin = −Sy −Sy ≤ σm ≤ Se
Kf − Sy
GH σmax = 2σm + Sy −Sy ≤ σm ≤ Se
Kf − Sy
HA σmax = σm + Se
Kf
Se
Kf − Sy ≤ σm ≤ 0
As an example of the way the modified Goodman dia- gram aids in visualizing the various combinations of fluctu- ating stress, consider the mean stress indicated by point L in Fig. 7.17. The Goodman criterion indicates that this stress can fluctuate between points M and N. This fluctuation is sketched on the right of the figure.
Also shown in Fig. 7.17 are the four regions of mean stress on the abscissa. Table 7.6 gives the failure equation for each of these regions as well as the validity limits for each equation. Table 7.6 is an extremely valuable guide when ap- plying the modified Goodman diagram.
Example 7.9: Safety Factor Using the Modified Goodman Criterion
Given:For the beam given in Example 7.5 the bending mo- ment varies between 50 and 200 Nm.
Find:Using the modified Goodman relationship, determine the safety factor guarding against fatigue failure.
Solution: From Example 7.5,Sut = 395MPa andSy= 295 MPa. Also, it was determined in Example 7.5 thatSe = 119 MPa andKf = 3.0. The nominal stresses at the groove (d= 0.046m) are given by
σmin= 32Mmin
πd3 = 32(50)
π(0.046)3 = 5.232MPa, σmax= 32Mmin
πd3 = 32(200)
π(0.046)3 = 20.93MPa.
Therefore, from Eqs. (7.1) and (7.3),
Influence of Nonzero Mean Stress 179
a b
E
G F
H
A Su
Sy
D L
N
M 0
B C
σmax
σmax
σmin
σmin
σm +σ
–σm σm
–σ
c d
45°
45°
Su
Sy
Su
–Se/Kf
Sy
–Sy
Se/Kf
Figure 7.17: Complete modified Goodman diagram, plotting stress as ordinate and mean stress as abscissa.
Table 7.6: Failure equations and validity limits of equations for four regions of complete modified Goodman relationship.
Region in Failure Validity limits
Fig. 7.16 equation of equation
a σmax − 2σm = Sy/ns −Sy ≤ σm ≤ Se
Kf −Sy
b σmax − σm = Se
nsKf
Se
Kf −Sy ≤ σm ≤ 0
c σmax + σm KSe
fSu −1 = nsSKef 0≤ σm ≤
Sy − Se
Kf
1− Se
Kf Su
d σmax = Sy
ns
Sy − Se
Kf 1− Se
Kf Su
≤ σm ≤ Sy
–4.00 0.5 1.0 1.5
–3.0 –2.0 –1.0 0 1.0 Mean stress ratio, σm/Su
Alternating stress ratio, σa/Su
= (0.4)(0.9) = 0.36 Se
—Su
Figure 7.18: Alternating stress ratio as a function of mean stress ratio for axially loaded cast iron.
σa=σmax−σmin
2 = 20.93−5.232
2 = 7.849MPa, σm=σmax+σmin
2 = 20.93 + 5.232
2 = 13.08MPa.
In applying the modified Goodman approach, Table 7.6 is very valuable. Note the requirement for Region a to be valid, then note that
Se
Kf −Sy=119
3.0 −295 =−255MPa.
However, sinceσm = 20.93MPa, this cannot be true and therefore Region a is not valid. Similarly, Region b cannot be valid sinceσm>0. Note that
Sy− Se
Kf
1− Se
KfSu
=
295−119 3.0 1− 119
(3.0)(395)
= 284MPa.
Sinceσm<284MPa, Region c is valid and the failure equa- tion from Table 7.6 is
σmax+σm
Se
KfSut−1
=Se
nsKf; 20.93 + 13.08
119 (3.0)(395)−1
= 119 (3.0)ns
. This is solved asns= 4.22.
7.10.2 Brittle Materials
Until recently, the use of brittle materials in a fatigue envi- ronment has been limited to gray cast iron in compression.
Now, however, carbon fibers and ceramics have had signifi- cant use in fatigue environments. Figure 7.18 shows the al- ternating stress ratio as a function of the mean stress ratio for axially loaded cast iron. The figure is skewed since the com- pressive strength is typically several times greater than the tensile strength.
In Fig. 7.18, the dimensionalization of the alternating and mean stresses is with respect to the ultimate strength rather than to the yield strength as done for ductile materials. Also, the compressive mean stress permits large increases in alter- nating stress.
For brittle materials a stress raiser increases the likeli- hood of failure under either steady or alternating stresses, and it is customary to apply a stress concentration factor to both. Thus, designers apply the fatigue stress concentration factorKf to the alternating component of stress for ductile materials but apply the stress concentration factorKcto both
the alternating and mean components of stress for brittle ma- terials.
For a single normal stress in brittle materials, the equa- tion for the safety factor with a steady stressσmand with the ultimate tensile strengthSutas the basis of failure is
ns= Sut
Kcσm. (7.31)
With an alternating stress, σa, and a modified endurance limit,Se:
ns= Se
Kcσa. (7.32)
For a single shear stress on a brittle component and with a steadyshear stress,τm, the safety factor is
ns= Sut
Kcsτm
1 + Sut
Suc
. (7.33)
With an alternating shear stress, τa, and a modified en- durance limit,Se:
ns= Se
Kcsτa
1 + Sut
Suc
. (7.34)