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Fatigue II Fatigue II Lecture 11 Lecture 11 Engineering 473 Engineering 473 Machine Design Machine Design
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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

S-N CurveThe S-N 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. 7-6

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

Fluctuating-Stress 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. 7-16

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. 7-15

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 S-N 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. 7-6

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

Multi-axis 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 S-N 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 cold-drawn steel tubing and has a 6-mm 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 N-m in phase with a completely reversed bending moment of 150 N-m. 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 in-lb for 2000 cycles, 1000 in-lb for 100,000 cycles and 900 in-lb for 10,000 cycles. Use the S-N 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.