Fatigue Strength(6.4, 6.7-6.8, 6.11)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

Fatigue Strength1

Column Design2

Column Design3

Column Design4

Fatigue Strength (6.8)

Fatigue Strength5

Up to now, we have designed structures for static loads.

PP wd

t

ySmax

P

t

(σmax is also constant)

Fatigue Strength (6.8)

Fatigue Strength6

What if loading is not constant?

Even if σmax ≤ Sy, failure could occur if enough cycles are applied.

P

t

Fluctuating Stresses (6.11)

If σmin = - σmax, this is known as “fully-reversed” loading.

Fatigue Strength7

σ

t

σmax

σmin

max

min

minmax

minmax

)(2

1

)(2

1

R

galternatina

meanm

S-N Diagram (6.4)

Fatigue Strength8

Sf (fatigue strength) - stress level at which a corresponding number of cycles (N) will lead to failure (crack initiation)

Se (endurance limit) - stress level below which failure will never occur

Endurance Limit (6.7)

The simplest design rule to prevent fatigue failure is

This is a valid concept, but not quite so simple in reality.

Se is determined experimentally.

Simple approximate Se formulas exist for steel, but must be used carefully – better to have actual data.

where Sut = ultimate strength and Se’ = unmodified, laboratory determined value

Fatigue Strength9

eapplied S max

' 0.5 200 kpsi (1400 MPa)

' 100 kpsi > 200 kpsi

' 700 MPa > 1400 MPa

e ut ut

e ut

e ut

S S S

S S

S S

Endurance Limit (6.7)

For real design we will modify Se’ to account for the surface finish, stress concentration, temperature, etc.

These effects decrease the effective endurance limit.

Fatigue Strength10

High-cycle fatigue life (N > 1000 cycles) Typical S-N diagram for steel

Predicting Fatigue Life (6.8)

Fatigue Strength11

cNaS f

)(loglog

:c)ax(y line a ofEquation

cacaSl 3)10(log'log 3

cacaSe 6)10(log'log 6

'

)'(log

'

'log

3

1

2

e

l

e

l

S

Sc

S

Sa

(log Sf)

(log N)

Se’

Sl’

a-cf

acf

)(SN

NNS

1

63

10

or

cycles 1010for 10

Fatigue strength fraction

Fatigue strength12

Example Find Sf of 1020 hot-rolled steel if the required

life is 250,000 cycles, bending loads. Given: Sut = 57 ksi for 1020 steel

Note: For steel, Sl’ = 0.9Su (bending), 0.75Su (axial), and 0.72Su (torsion).

What is the life if Sf = 40 ksi?

Fatigue Strength13

High Cycle Fatigue(6.9-6.10)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

High Cycle Fatigue14

Modified Endurance Limit (6.9)

Modified endurance limit is defined as

ka = surface finish factor = aSutb

High Cycle Fatigue15

'e a b c d e f eS k k k k k k S

a b

Surface finish MPa (kpsi)

Ground 1.58 (1.34) -0.085

Machine or cold drawn

4.51 (2.7) -0.265

Hot rolled 57.7 (14.4) -0.718

As-Forged 272.0 (39.9) -0.995

Table 6-2 Surface finish factors ka

Modified Endurance Limit (6.9)

kb = size factor Axial loading

kb = 1

Bending and torsion kb = 0.879d-.107 (0.11 in ≤ d ≤ 2 in)

kb = 0.91d-.157 (2 < d < 10 in)

kb = 1.241d-.107 (2.79 ≤ d ≤ 51 mm)

kb = 1.51d-.157 (51 < d < 254 mm)

d is the diameter of the round bar or the equivalent diameter (de) of a non-rotating or non-circular bar (Table 6-3).

High Cycle Fatigue16

Modified Endurance Limit (6.9)

kc = loading factor 1 (bending) 0.85 (axial) 0.59 (torsion)

kd = temperature factor If endurance limit (Se’) is known, or use

equation

If Se’ is not known, use kd = 1 and temperature-corrected tensile strength (Sut) (see Example 6-5 in textbook)

High Cycle Fatigue17

3 5 2 8 3 12 40.975 0.432 10 0.115 10 0.104 10 0.595 10dk T T T T

(Table 6-4)Td

RT

Sk

S

Modified Endurance Limit (6.9)

ke = reliability factor

High Cycle Fatigue18

Table 6-5 Reliability factors ke

Survival Rate (%)

ke

50 1.00

90 0.89

95 0.87

98 0.84

99 0.81

99.9 0.75

99.99 0.70

Modified Endurance Limit (6.9)

kf = miscellaneous-effects factor Corrosion Electrolytic plating Metal Spraying Cyclic frequency Frettage corrosion

If none of the above conditions apply, kf = 1

High Cycle Fatigue19

Fatigue Stress Concentration Factor (6.10)

Kf = fatigue stress concentration factor

Kf = 1 + q(Kt – 1) q = notch sensitivity Kt = stress concentration factor

Kf can be used to reduce Se (multiply Se by 1/Kf) or to modify the nominal stress (σmax = Kfσnom).

High Cycle Fatigue20

Fatigue Stress Concentration Factor (6.10)

High Cycle Fatigue21

Figure 6-20 Notch sensitivity for bending and axial

Figure 6-21 Notch sensitivity for torsion

Example For the plate shown below, find the maximum

allowable load F for the plate to have infinite life. Given: 1018 cold-drawn steel, Sy = 373 MPa, Sut = 442

MPa

High Cycle Fatigue22

FF w =60 mm

d = 12 mm

t = 10 mm

F

t

Column Design23