Chapter 6 Note to the instructor: Many of the problems in this chapter are carried over from the previous edition. The solutions have changed slightly due to some minor changes. First, the calculation of the endurance limit of a rotating-beam specimen S e is given by S e = 0.5 S ut instead of S e = 0.504 S ut . Second, when the fatigue stress calculation is made for deterministic problems, only one approach is given, which uses the notch sensitivity factor, q, together with Eq. (6-32). Neuber’s equation, Eq. (6-33), is simply another form of this. These changes were made to hope- fully make the calculations less confusing, and diminish the idea that stress life calculations are precise. 6-1 H B = 490 Eq. (2-17): S ut = 0.495(490) = 242.6 kpsi > 212 kpsi Eq. (6-8): S e = 100 kpsi Table 6-2: a = 1.34, b =−0.085 Eq. (6-19): k a = 1.34(242.6) −0.085 = 0.840 Eq. (6-20): k b = 1/4 0.3 −0.107 = 1.02 Eq. (6-18): S e = k a k b S e = 0.840(1.02)(100) = 85.7 kpsi Ans . 6-2 (a) S ut = 68 kpsi, S e = 0.5(68) = 34 kpsi Ans . (b) S ut = 112 kpsi, S e = 0.5(112) = 56 kpsi Ans . (c) 2024T3 has no endurance limit Ans. (d) Eq. (6-8): S e = 100 kpsi Ans . 6-3 Eq. (2-11): σ F = σ 0 ε m = 115(0.90) 0.22 = 112.4 kpsi Eq. (6-8): S e = 0.5(66.2) = 33.1 kpsi Eq. (6-12): b =− log(112.4/33.1) log(2 · 10 6 ) =−0.084 26 Eq. (6-10): f = 112.4 66.2 (2 · 10 3 ) −0.084 26 = 0.8949 Eq. (6-14): a = [0.8949(66.2)] 2 33.1 = 106.0 kpsi Eq. (6-13): S f = aN b = 106.0(12 500) −0.084 26 = 47.9 kpsi Ans. Eq. (6-16): N = σ a a 1/b = 36 106.0 −1/0.084 26 = 368 250 cycles Ans. budynas_SM_ch06.qxd 01/29/2007 18:21 Page 147
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Chapter 6
Note to the instructor: Many of the problems in this chapter are carried over from the previousedition. The solutions have changed slightly due to some minor changes. First, the calculationof the endurance limit of a rotating-beam specimen S′
e is given by S′e = 0.5Sut instead of
S′e = 0.504Sut . Second, when the fatigue stress calculation is made for deterministic problems,
only one approach is given, which uses the notch sensitivity factor, q, together with Eq. (6-32).Neuber’s equation, Eq. (6-33), is simply another form of this. These changes were made to hope-fully make the calculations less confusing, and diminish the idea that stress life calculations areprecise.
Material and condition: 1095 HR and from Table A-20 Sut = 120, Sy = 66 kpsi.
Design factor: n f = 1.6 per problem statement.
Life: (1150)(3) = 3450 cycles
Function: carry 10 000 lbf load
Preliminaries to iterative solution:
S′e = 0.5(120) = 60 kpsi
ka = 2.70(120)−0.265 = 0.759
I
c= πd3
32= 0.098 17d3
M(crit.) =(
6
24
)(10 000)(12) = 30 000 lbf · in
The critical location is in the middle of the shaft at the shoulder. From Fig. A-15-9: D/d =1.5, r/d = 0.10, and Kt = 1.68. With no direct information concerning f, use f = 0.9.
For an initial trial, set d = 2.00 in
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Chapter 6 153
kb =(
2.00
0.30
)−0.107
= 0.816
Se = 0.759(0.816)(60) = 37.2 kpsi
a = [0.9(120)]2
37.2= 313.5
b = −1
3log
0.9(120)
37.2= −0.15429
Sf = 313.5(3450)−0.15429 = 89.2 kpsi
σ0 = M
I/c= 30
0.098 17d3= 305.6
d3
= 305.6
23= 38.2 kpsi
r = d
10= 2
10= 0.2
Fig. 6-20: q =̇ 0.87
Eq. (6-32): K f =̇ 1 + 0.87(1.68 − 1) = 1.59
σa = K f σ0 = 1.59(38.2) = 60.7 kpsi
n f = Sf
σa= 89.2
60.7= 1.47
Design is adequate unless more uncertainty prevails.
Thus the design is controlled by the threat of fatigue equally at the fillet and the hole; theminimum factor of safety is n f = 1.61. Ans.
6-24 (a) Curved beam in pure bending where M = −T
throughout. The maximum stress will occur at theinner fiber where rc = 20 mm, but will be com-pressive. The maximum tensile stress will occur atthe outer fiber where rc = 60 mm. Why?
Inner fiber where rc = 20 mm
rn = h
ln(ro/ri )= 5
ln (22.5/17.5)= 19.8954 mm
e = 20 − 19.8954 = 0.1046 mm
ci = 19.8954 − 17.5 = 2.395 mm
A = 25 mm2
σi = Mci
Aeri= −T (2.395)10−3
25(10−6)0.1046(10−3)17.5(10−3)(10−6) = −52.34 T (1)
where T is in N . m, and σi is in MPa.
σm = 1
2(−52.34T ) = −26.17T , σa = 26.17T
For the endurance limit, S′e = 0.5(770) = 385 MPa
ka = 4.51(770)−0.265 = 0.775
de = 0.808[5(5)]1/2 = 4.04 mm
kb = (4.04/7.62)−0.107 = 1.07
Se = 0.775(1.07)385 = 319.3 MPa
For a compressive midrange component, σa = Se/n f . Thus,
26.17T = 319.3/3 ⇒ T = 4.07 N · m
T T
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Chapter 6 163
Outer fiber where rc = 60 mm
rn = 5
ln(62.5/57.5)= 59.96526 mm
e = 60 − 59.96526 = 0.03474 mm
co = 62.5 − 59.96526 = 2.535 mm
σo = − Mci
Aeri= − −T (2.535)10−3
25(10−6)0.03474(10−3)62.5(10−3)(10−6) = 46.7 T
Comparing this with Eq. (1), we see that it is less in magnitude, but the midrange compo-nent is tension.
σa = σm = 1
2(46.7T ) = 23.35T
Using Eq. (6-46), for modified Goodman, we have
23.35T
319.3+ 23.35T
770= 1
3⇒ T = 3.22 N · m Ans.
(b) Gerber, Eq. (6-47), at the outer fiber,
3(23.35T )
319.3+
[3(23.35T )
770
]2
= 1
reduces to T 2 + 26.51T − 120.83 = 0
T = 1
2
(−26.51 +
√26.512 + 4(120.83)
)= 3.96 N · m Ans.
(c) To guard against yield, use T of part (b) and the inner stress.
ny = 420
52.34(3.96)= 2.03 Ans.
6-25 From Prob. 6-24, Se = 319.3 MPa, Sy = 420 MPa, and Sut = 770 MPa
Decision: Depending on availability, (1) select t = 10 mm, recalculate n f and R, anddetermine whether the reduced reliability is acceptable, or, (2) select t = 11 mm orlarger, and determine whether the increase in cost and weight is acceptable. Ans.
6-34
Rotation is presumed. M and Sut are given as deterministic, but notice that σ is not; there-fore, a reliability estimation can be made.
From Eq. (6-70):
S′e = 0.506(110)LN(1, 0.138)
= 55.7LN(1, 0.138) kpsiTable 6-10:
ka = 2.67(110)−0.265LN(1, 0.058)
= 0.768LN(1, 0.058)
Based on d = 1 in, Eq. (6-20) gives
kb =(
1
0.30
)−0.107
= 0.879
Conservatism is not necessary
Se = 0.768[LN(1, 0.058)](0.879)(55.7)[LN(1, 0.138)]
S̄e = 37.6 kpsi
CSe = (0.0582 + 0.1382)1/2 = 0.150
Se = 37.6LN(1, 0.150)
Fig. A-15-14: D/d = 1.25, r/d = 0.125. Thus Kt = 1.70 and Eqs. (6-78), (6-79) andTable 6-15 give
Note: The correlation method uses only the mean of Sut ; its variability is already includedin the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, en-gineers state, “For a Design Load of M, the reliability is 0.991.” They are in fact referringto a Deterministic Design Load.
6-35 For completely reversed torsion, ka and kb of Prob. 6-34 apply, but kc must also be con-sidered.
Fig. A-15-15: D/d = 1.25, r/d = 0.125, then Kts = 1.40. From Eqs. (6-78), (6-79) andTable 6-15
Kts = 1.40LN(1, 0.15)
1 + (2/
√0.125
)[(1.4 − 1)/1.4](3/110)
= 1.34LN(1, 0.15)
τ = Kts16T
πd3
τ = 1.34[LN(1, 0.15)]
[16(1.4)
π(1)3
]
= 9.55LN(1, 0.15) kpsi
From Eq. (5-43), p. 242:
z = −ln
[(22.2/9.55)
√(1 + 0.152)/(1 + 0.1952)
]√
ln [(1 + 0.1952)(1 + 0.152)]= −3.43
From Table A-10, pf = 0.0003
R = 1 − pf = 1 − 0.0003 = 0.9997 Ans.
For a design with completely-reversed torsion of 1400 lbf · in, the reliability is 0.9997. Theimprovement comes from a smaller stress-concentration factor in torsion. See the note atthe end of the solution of Prob. 6-34 for the reason for the phraseology.
6-38 This is a very important task for the student to attempt before starting Part 3. It illustratesthe drawback of the deterministic factor of safety method. It also identifies the a priori de-cisions and their consequences.
The range of force fluctuation in Prob. 6-23 is −16 to +4 kip, or 20 kip. Repeatedly-applied Fa is 10 kip. The stochastic properties of this heat of AISI 1018 CD are given.
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Chapter 6 175
Function Consequences
Axial Fa = 10 kip
Fatigue load CFa = 0
Ckc = 0.125
Overall reliability R ≥ 0.998; z = −3.09with twin fillets CK f = 0.11
R ≥√
0.998 ≥ 0.999
Cold rolled or machined Cka = 0.058surfaces
Ambient temperature Ckd = 0
Use correlation method Cφ = 0.138
Stress amplitude CK f = 0.11
Cσa = 0.11
Significant strength Se CSe = (0.0582 + 0.1252 + 0.1382)1/2
= 0.195
Choose the mean design factor which will meet the reliability goal
Cn =√
0.1952 + 0.112
1 + 0.112= 0.223
n̄ = exp[−(−3.09)
√ln(1 + 0.2232) + ln
√1 + 0.2232
]n̄ = 2.02
Review the number and quantitative consequences of the designer’s a priori decisions toaccomplish this. The operative equation is the definition of the design factor
This thickness separates S̄e and σ̄a so as to realize the reliability goal of 0.999 at eachshoulder. The design decision is to make t the next available thickness of 1018 CD steelstrap from the same heat. This eliminates machining to the desired thickness and the extracost of thicker work stock will be less than machining the fares. Ask your steel supplierwhat is available in this heat.
6-39
Fa = 1200 lbf
Sut = 80 kpsi
(a) Strength
ka = 2.67(80)−0.265LN(1, 0.058)
= 0.836LN(1, 0.058)
kb = 1
kc = 1.23(80)−0.078LN(1, 0.125)
= 0.874LN(1, 0.125)
S′a = 0.506(80)LN(1, 0.138)
= 40.5LN(1, 0.138) kpsi
Se = 0.836[LN(1, 0.058)](1)[0.874LN(1, 0.125)][40.5LN(1, 0.138)]
6-40 Each computer program will differ in detail. When the programs are working, the experi-ence should reinforce that the decision regarding n̄ f is independent of mean values ofstrength, stress or associated geometry. The reliability goal can be realized by noting theimpact of all those a priori decisions.
6-41 Such subprograms allow a simple call when the information is needed. The calling pro-gram is often named an executive routine (executives tend to delegate chores to others andonly want the answers).
6-42 This task is similar to Prob. 6-41.
6-43 Again, a similar task.
6-44 The results of Probs. 6-41 to 6-44 will be the basis of a class computer aid for fatigue prob-lems. The codes should be made available to the class through the library of the computernetwork or main frame available to your students.
6-45 Peterson’s notch sensitivity q has very little statistical basis. This subroutine can be used toshow the variation in q , which is not apparent to those who embrace a deterministic q .