How to deal with loading which is not constant amplitude

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Constant amplitude fully reversed cycles in to S-N diagram

Log S

Log N

1

3 Mean stresses = 1

2

2

1

4 3 2 Palmgren Miner

1

5 Notches, etc, etc, etc..

Rainflow Cycle Counting

Block loading

Irregular sequences

Palmgren Miner

Eg, Goodman

Block Loading – Palmgren Miner Hypothesis

Palmgren Miner states failure when,

Or when,

nN∑ = 10.

0.1...3

3

2

2

1

1 =+++Nn

Nn

Nn

time

1aS 2aS3aS

1n 2n

3n cycles

Log S

Log N 2 N

1 S

1 N 3 N

2 S

3 S

Linear Non-interactive

S

N

100 MPa

60000

Material Life Curve!

Life!Accumulated damage! %!5!.!0!60000!300! =!=!∴!

Range!

300 Cycles!

∑!=!i! f!

i!N!N!Damage!

DAMAGE COUNTING WITH MINER

1.45 x 10-3

300

500 1.21 x 10-3

250

2500 0.98 x 10-3

203

15000 0.76 x 10-3

157

120300 0.68 x 10-3

140

400000 0.60 x 10-3

124

1000000 0.54 x 10-3

112

3000000 0.45 x 10-3

93

5000000

Strain range recorded

Derived stress range (MPa)

Number of occurrences

Part of a steel structure, located permanently in the sea, is known to be susceptible to fatigue damage. Strain gauges attached to this part were monitored continuously during the first year of service producing the following information.

Specimen tests on the same material showed that the fatigue limit in air was 156 MPa and that in seawater (no corrosion protection) it was 110 N/mm2. In addition it was found that stresses above either of these levels produced failure according to the following relationship.

3333.825.10196.2 −= SxNAfter 7 years the corrosion protection fails. Determine the expected fatigue life of this structure after failure of the corrosion protection.

Would the original total life of 20 years be achieved?

Example: Stress-Life Fatigue Life Calculation

During first seven years, damage occurs only at a strain range of 0.76 x 10-3 (157MPa) and above.

Total damage during this period = 0.0437 x 7 = 0.3059.

Therefore, remaining damage to be accumulated = 1 – 0.3059 = 0.6941.

n

N

n/N with CP

n/N without CP

1.45 x 10-3

300

500

49,700 0.01

0.01

1.21

250

2,500

225,000 0.0111

0.0111

0.98

203

15,000

103 x 106 0.0115

0.0115

0.76

157

120,300

10.8 x 106 0.0111

0.0111

0.68

140

400,000

27.4 x 106 0.0

0.0146

0.60

124

1,000,000

77.7 x 106 0.0

0.0129

0.54

112

3,000,000

187 x 106 0.0

0.0160

0.45

93

5,000,000

854 x 106 0.0

0.0

SΔεΔ

Example: Stress-Life Fatigue Life Calculation (cont)

0.0437 0.0872

   0 ⋅69410 ⋅ 0872

= 7⋅ 96 years

If corrosion fatigue occurs, amount of damage/annum = 0.0872. Therefore, remaining life after breakdown of protection is

NB. Life = remaining damage factor / damage per year

With corrosion protection the life would be 22.9 years.

Without corrosion protection the life would drop to 14.96 years.

The above example demonstrates the reduction due to subsequent corrosion.

Example: Stress-Life Fatigue Life Calculation (cont)