How to deal with loading which is not constant amplitude Simple Tools … Faster Solutions
How to deal with loading which is not constant amplitude
Simple Tools … Faster Solutions
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)