Fatigue

Fatigue Lecture 10 Engineering 473 Machine Design

Load Histories and Design Objectives Failure σ, F

σ, F

Failure

t, time

t, time

Monotonic, Static, or Steady

Dynamic, Cyclic, or Unsteady

Design for Strength

Design for Life

Rotating Beam Fatigue Testing

Fatigue Dynamics, Inc. rotating beam test equipment.

Mott, Fig. 5-2 & 5-3

www.fdinc.com

S-N Curve

Completely reversed cyclic stress, UNS G41200 steel Shigley, Fig. 7-6

Fatigue Strength The Fatigue Strength, Sf(N), is the stress level that a material can endure for N cycles. The stress level at which the material can withstand an infinite number of cycles is call the Endurance Limit.

The Endurance Limit is observed as a horizontal line on the S-N curve. Shigley, Fig. 7-6

Representative S-N Curves

Note that non-ferrous materials often exhibit no endurance limit. Mott, Fig. 5-7

Endurance Limit Vs Tensile Strength Conservative Lower Bound for Ferrous Materials

S′e = 0.3Sut

S′e ≡ Endurance Limit of Test Specimen Sut ≡ Tensile Strength of Test Specimen Shigley, Fig. 7-7

Endurance Limit Multiplying Factors (Marin Factors)

Se = k a ⋅ k b ⋅ k c ⋅ k d ⋅ k e ⋅ S′e Se ≡ Endurance limit of part S′e ≡ Endurance limit of test specimen k a ≡ Surface factor k b ≡ Size factor

There are several factors that are known to result in differences between the endurance limits in test specimens and those found in machine elements.

k c ≡ Load factor k d ≡ Temperature factor k e ≡ Miscellaneous - effects factor

See sections 7-8 & 7-9 in Shigley for a discussion on each factor.

Mean Stress Effects • The S-N curve obtained from a rotating beam test has completely reversed stress states. • Many stress histories will not have completely reversed stress states.

Shigley, Fig. 7-12

Definitions Stress Range

σ r = σ max − σ min Alternating Stress

σ max − σ min σa = 2

Mean Stress

Stress Ratio

σ min R= σ max

Amplitude Ratio

σa A= σm

σ max + σ min σm = 2 Note that R=-1 for a completely reversed stress state with zero mean stress.

Mean Stress Fatigue Testing

Fatigue Dynamics, Inc., fluctuating fatigue stress testing equipment. www.fdinc.com

Fluctuating Stress Failure Data Note that a tensile mean stress results in a significantly lower fatigue strength for a given number of cycles to failure.

This plot shows the fatigue strength of several steels as a function of mean stress for a constant number of cycles to failure.

Note that a curved line passes through the mean of the data. Shigley, Fig. 7-14

Master Fatigue Plot

Shigley, Fig. 7-15

Fluctuating Stress Failure Interaction Curves

Shigley, Fig. 7-16

Soderberg Interaction Line k f Sa S m + =1 Se S yt

Any combination of mean and alternating stress that lies on or below the Solderberg line will have infinite life. Factor of Safety Format

k f Sa S m 1 + = Se S yt N f Note that the fatigue stress concentration factor is applied only to the alternating component.

Goodman Interaction Line k f Sa S m + =1 Se Sut

Any combination of mean and alternating stress that lies on or below the Goodman line will have infinite life. Factor of Safety Format

k f Sa Sm 1 + = Se Sut N f Note that the fatigue stress concentration factor is applied only to the alternating component.

Gerber Interaction Line 2

k f Sa æ S m ö + çç ÷÷ = 1 Se è Sut ø

Any combination of mean and alternating stress that lies on or below the Gerber line will have infinite life. Factor of Safety Format 2

k f N f Sa æ N f S m ö ÷÷ = 1 + çç Se è Sut ø Note that the fatigue stress concentration factor is applied only to the alternating component.

Modified-Goodman Interaction Line

The Modified-Goodman Interaction Line never exceeds the yield line.

Example No. 1 A 1.5-inch round bar has been machined from AISI 1050 colddrawn round bar. This part is to withstand a fluctuating tensile load varying from 0 to 16 kip. Because of the design of the ends and the fillet radius, a fatigue stress-concentration factor of 1.85 exists. The remaining Marin factors have been worked out, and are ka=0.797, kb=kd=1, and kc=0.923. Find the factor of safety using the Goodman interaction line.

Shigley, Example 7-5

Example No. 1 (Continued) Sut = 100. ksi S′e ≈ 0.50 ⋅ Sut = 50. ksi

σ max + σ min = 4.52 ksi σm = 2

π ⋅ d2 = 1.77 in 2 A= 4

Se = k a k b k c k dS′e

16 kip = 9.04 ksi σ max = 2 1.77 in σ min = 0 ksi σ max − σ min = 4.52 ksi σa = 2

= (0.797 )(1)(0.923)(1)(50 ksi )

Se = 36.8 ksi

Example No. 1 (Continued)

k f σa σm 1 + = Se Sut N f 1.85 ⋅ 4.52 ksi 4.52 ksi 1 + = 0.272 = 36.8 ksi 100. ksi Nf N f = 3.67

Example 5 in 1

Pmax = 1000 lb

5 in

Pmin = 350 lb

2

1.5 in. dia. 0.875 in. dia. 0.125 in. rad. Will the beam have infinite life? π 4 π 4 D 1 = (1.5) = 0.249 in 4 I1 = 64 64 π 4 π 4 I2 = D 2 = (0.875) = 0.088 in 4 64 64

Material UNS G41200 Steel Notch sensitivity q=0.3

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

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

Pmax = 1000 lb Pmin = 350 lb Material UNS G41200 Steel Notch sensitivity q=0.3

k t = 1.61

k f = 1 + q(k t − 1) = 1 + 0.3(1.61 − 1) = 1.18

Example (Continued) 5 in

Pmax = 1000 lb

5 in

1

Pmin = 350 lb

2

1.5 in. dia. 0.875 in. dia. 0.125 in. rad. Section 1 (Base) σ max

M1 (1000 lb )(10 in ) = = = 30.1 ksi 3 S1 0.332 in

σ min =

M1 (350 lb )(10 in ) = = 10.5 ksi 3 S1 0.332 in

Material UNS G41200 Steel Notch sensitivity q=0.3

σ max − σ min = 9.8 ksi σa = 2 σ max + σ min σm = = 20.3 ksi 2

Example (Continued) 5 in

Pmax = 1000 lb

5 in

1

Pmin = 350 lb

2

1.5 in. dia. 0.875 in. dia. 0.125 in. rad. Section 2 (Fillet) σ max

M1 (1000 lb )(5 in ) = = = 24.9 ksi 3 S1 0.201 in

σ min =

M1 (350 lb )(5 in ) = = 8.71 ksi 3 S1 0.201 in

Material UNS G41200 Steel Notch sensitivity q=0.3

σ max − σ min = 8.10 ksi σa = 2 σ max + σ min = 16.8 ksi σm = 2

Example (Continued) Section 1 (Base) σ max

(1000 lb)(10 in ) = 30.1 ksi M = 1= S1 0.332 in 3

σ min =

M1 (350 lb )(10 in ) = = 10.5 ksi 3 S1 0.332 in

Sut = 116 ksi S′e = 30 ksi = Se

k f σa σm 1 + = Se Sult N f

σ max − σ min σa = = 9.8 ksi 2 σ max + σ min σm = = 20.3 ksi 2 1.0(9.8 ksi ) 20.3 ksi + = 0.502 30 ksi 116 ksi 1 Nf = = 1.99 0.502 Part has infinite life.

Example (Continued) Section 2 (Fillet) σ max

(1000 lb)(5 in ) = 24.9 ksi M = 1= S1 0.201 in 3

σ min =

M1 (350 lb )(5 in ) = = 8.71 ksi 3 S1 0.201 in

Sut = 116 ksi S′e = 30 ksi = Se

k f σa σm 1 + = Se Sult N f

σ max − σ min σa = = 8.10 ksi 2 σ max + σ min σm = = 16.8 ksi 2 1.18(8.10 ksi ) 16.8 ksi + = 0.463 30 ksi 116 ksi 1 Nf = = 2.16 0.463

Part has infinite life.

Assignment Problem 1

Assignment (Continued) Problem 2