Fatigue II

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Fatigue IILecture 11 Engineering 473 Machine Design
Finite Life EstimatesHow can the life of a part be estimated if the mean stressalternating stress pair lie above the Goodman line?
Alternating S yt Stress, a
SeInfinite Life Stress State
Finite Life (Cycles to failure?)
Mean Stress, m
S yt
Sut
Goodman Diagram
SN CurveThe SN curve gives the cycles to failure for a completely reversed (R=1) uniaxial stress state. What do you do if the stress state is not completely reversed?Completely reversed cyclic stress, UNS G41200 steelShigley, Fig. 76
Definitions r = max min max min a = 2Alternating Stress Stress Range
Stress Ratio
Amplitude Ratio
max + min m = 2Note that R=1 for a completely reversed stress state with zero mean stress.
Mean Stress
min R= max
a A= m
FluctuatingStress Failure Interaction CurvesThe interaction curves provide relationships between alternating stress and mean stress. When the mean stress is zero, the alternating component is equal to the endurance limit. The interaction curves are for infinite life or a large number of cycles.Shigley, Fig. 716
Goodman Interaction Linek f Sa S m + =1 Se SutAny combination of mean and alternating stress that lies on or below Goodman line will have infinite life. Factor of Safety Format
k f Sa Sm 1 + = Se Sut N fNote that the fatigue stress concentration factor is applied only to the alternating component.
Master Fatigue Plot
Constant cycles till failure interaction curves.
Shigley, Fig. 715
Equivalent Alternating StressaAlternating S yt Stress, a m =0
Alternating stress at zero mean stress that fails the part in the same number of cycles as the original stress state.
Se
105 cycles 106 cycles
Mean Stress, m
S yt
Sut
The red and blue lines are estimated fatigue interaction curves associated with a specific number of cycles to failure.
Number of Cycles to Failure
Once the equivalent alternating stress is found, the SN curve may be used to find the number of cycles to failure.
Equivalent Alternating Stress Formulak f a m 1 + = Se Sut N f k f a m 1 + = a =0 Sut N fm
Goodman Line Equivalent completely reversed (R = 1) stress that causes fatigue failure in the same number of cycles as the original a and m pair.
a
m =0
a
m =0
k f a = 1 m N f Sut
Example5 in 1 5 in 2
Pmax = 3000 lb Pmin = 2000 lbMaterial UNS G41200 Steel Notch sensitivity q=0.3
1.5 in. dia. 0.875 in. dia. 0.125 in. rad.
4 4 D 1 = (1.5) = 0.249 in 4 I1 = 64 64 4 4 I2 = D 2 = (0.875) = 0.088 in 4 64 64
I1 0.249 in 4 S1 = = = 0.332 in 3 c1 0.75 in I 2 0.088 in 4 S2 = = = 0.201 in 3 c 2 0.438 in
Example(Continued) 5 in 1 5 in 2
Pmax = 3000 lb Pmin = 2000 lbMaterial UNS G41200 Steel Notch sensitivity q=0.3
1.5 in. dia. 0.875 in. dia. 0.125 in. rad.
kf 1 q= k t 1
k f = 1 + q(k t 1)
D 1.5 in = = 1.71 d 0.875 in r 0.125 = = 0.143 d 0.875 Ref. Peterson
k t = 1.61
k f = 1 + q(k t 1) = 1 + 0.3(1.61 1) = 1.18
Example(Continued) 5 in 1 5 in 2 Material UNS G41200 Steel Notch sensitivity q=0.3
Pmax = 3000 lb Pmin = 2000 lb
1.5 in. dia. 0.875 in. dia. 0.125 in. rad. Section 1 (Base) max M1 (3000 lb )(10 in ) = = = 90.4 ksi 3 S1 0.332 in M1 (2000 lb )(10 in ) = = 60.2 ksi 3 S1 0.332 in
min =
max min a = = 15.1 ksi 2 max + min m = = 75.3 ksi 2
Example(Continued) 5 in 1 5 in 2
Pmax = 3000 lb Pmin = 2000 lbMaterial UNS G41200 Steel Notch sensitivity q=0.3
1.5 in. dia. 0.875 in. dia. 0.125 in. rad. Section 2 (Fillet) max M1 (3000 lb )(5 in ) = = = 74.6 ksi 3 S1 0.201 in M1 (2000 lb )(5 in ) = = 49.8 ksi 3 S1 0.201 in
min =
max min = 12.4 ksi 2 max + min m = = 62.2 ksi 2 a =
Example(Continued)
Sut = 116 ksi S = 30 ksi = Se e
k f a m 1 + = Se Sult N fCompletely reversed cyclic stress, UNS G41200 steelShigley, Fig. 76
Example(Continued) Section 1 (Base) max
(3000 lb)(10 in ) = 90.4 ksi M = 1= S1 0.332 in 3M1 (2000 lb )(10 in ) = = 60.2 ksi 3 S1 0.332 in
min =
max min a = = 15.1 ksi 2 max + min m = = 75.3 ksi 2Nf = 1 1.0(15.1 ksi ) 75.3 ksi + = 1.15 30 ksi 116 ksi
Sut = 116 ksi S = 30 ksi = Se e
k f a m 1 + = Se Sult N f
Part has finite life at base.
Example(Continued) Section 2 (Fillet) max M1 (3000 lb )(5 in ) = = = 74.6 ksi 3 S1 0.201 in M1 (2000 lb )(5 in ) = = 49.8 ksi 3 S1 0.201 in
min =
max min = 12.4 ksi 2 max + min m = = 62.2 ksi 2 a =Nf = 1 1.18(12.4 ksi ) 62.2 ksi + = 1.02 30 ksi 116 ksi
Sut = 116 ksi S = 30 ksi = Se e
k f a m 1 + = Se Sult N f
Part has finite life.
Calculation of Equivalent Alternating StressaBase a = 15.1 ksi m = 75.3 ksi m =0
k f a = 1 m N f SutFillet a = 12.4 ksi m = 62.2 ksi
(1.0)15.1 a =0 = m 1 75.3 1.0 116 = 43.0 ksi
(1.18)12.4 a =0 = m 1 62.2 1.0 116 = 31.5 ksi
Cycles to Failure Estimate
90 70 50 30 20 10
Base Fillet
Multiaxis Fluctuating Stress StatesEverything presented on fatigue has been based on experiments involving a single stress component. What do you do for problems in which there are more than one stress component?
Marin Load Factor, kcSe = k a k b k c k d S eThe endurance limit is a function of the load/stress component used in the test.
Axial loading Sut 220 ksi (1520 MPa) 0.923 1 Axial loading Sut > 220 ksi (1520 MPa) kc = Bending 1 0.577 Torsion and shear
Alternating and Mean Von Mises Stresses1. Increase the stress caused by an axial force by 1/kc. 2. Multiply each stress component by the appropriate fatigue stress concentration factor. 3. Compute the maximum and minimum von Mises stresses. 4. Compute the alternating and mean stresses based on the maximum and minimum values of the von Mises stress. 5. Use the Goodman alternating and mean stress interaction curve and SN curve to estimate the number of cycles to failure. Use the reversed bending endurance limit.
Complex LoadsA part is subjected to completely reversed
, F
2 13t, time
stresses as follows 1 for n1 cycles, 2 for n 2 cycles, 3 for n 3 cycles, M m for n m cycles,
What is the cumulative effect of these different load cycles?
Minors RuleCumulative Damage Law
n1 n 2 n 3 nm + + +K + =C N1 N 2 N 3 Nmn i number of cycles for stress level i N i cycles to failure at stress level i C Constant ranging from 0.7 to 2.2.C is usually taken as 1.0 Minors Rule is the simplest and most widely used Cumulative Damage Law
ExampleStress Cycles State (n) 1 2 3 Life (N)
n
N
1,000 2,000 5,000 10,000
0.5 0.5 1.1
10,000 100,000 0.1 Part will fail
Assignment(Problem No. 1)
A rotating shaft is made of 42 x 4 mm AISI 1020 colddrawn steel tubing and has a 6mm diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue failure when the shaft is subjected to a completely reversed torque of 120 Nm in phase with a completely reversed bending moment of 150 Nm. Use the stress concentration factor tables found in the appendices, and estimate the Marin factors using information in the body of the text.
Assignment(Problem No. 2)
A solid circular bar with a 5/8 inch diameter is subjected to a reversed bending moment of 1200 inlb for 2000 cycles, 1000 inlb for 100,000 cycles and 900 inlb for 10,000 cycles. Use the SN curve used in this lecture. Determine whether the bar will fail due to fatigue. Assume all Marin factors are equal to 1.0.
Assignment(Problem No. 3)
Same as Problem No. 2 except there is a constant axial force of 5,000 lb acting on the bar in addition to the completely reversed bending moment.