Rainflow counting algorithm

Algorithm

1. Reduce the time history to a sequence of (tensile) peaks and (compressive) valleys.
2. Imagine that the time history is a template for a rigid sheet (pagoda roof).
3. Turn the sheet clockwise 90° (earliest time to the top).
4. Each tensile peak is imagined as a source of water that "drips" down the pagoda.
5. Count the number of half-cycles by looking for terminations in the flow occurring when either:
6. It reaches the end of the time history;
7. It merges with a flow that started at an earlier tensile peak; or
8. It flows when an opposite tensile peak has greater magnitude.
9. Repeat step 5 for compressive valleys.
10. Assign a magnitude to each half-cycle equal to the stress difference between its start and termination.
11. Pair up half-cycles of identical magnitude (but opposite sense) to count the number of complete cycles. Typically, there are some residual half-cycles.

Example

The stress history in Figure 2 is reduced to peaks and valleys in Figure 3.

Half-cycle (A) starts at tensile peak (1) and terminates opposite a greater tensile stress, peak (2); its magnitude is 16 MPa.

Half-cycle (B) starts at tensile peak (4) and terminates where it is interrupted by a flow from an earlier peak, (3); its magnitude is 17 MPa.

Half-cycle (C) starts at tensile peak (5) and terminates at the end of the time history.

Similar half-cycles are calculated for compressive stresses (Figure 4) and the half-cycles are then matched.

Block Loading example

There are many cases in which a structure will undergo periodic loading. Assume that a specimen is loaded periodically until failure. The number of blocks endured before failure can be determined easily by using the Palmgren-Miner rule of block loading. The actual load history is shown in Figure 5.

If all of the similar loads are grouped together, it forms a series of block loads as shown in Figure 6.

The Palmgren-Miner rule can be expressed as

B ( N 1 N f 1 + N 2 N f 2 + . . . + N k N f k ) = 1

where,

B = number of blocks N_k = number of cycles of loading condition, k N_fk =number of cycles to failure for loading condition, k

In this example, each N_k=1 because there is one instance of each load for every period of loading. To find N_f (number of loads to failure) for each load the Goodman-Basquin relation can be used

N f = 1 2 ( σ a σ f ′ 1 1 − σ m σ u ) 1 b

where,

σ_a = stress amplitude σ'_f = fatigue strength coefficient (material property) σ_m = mean stress σ_ult = ultimate stress (material property) b = fatigue strength exponent (material property)

Assumptions

There are two key assumptions made in order to rearrange the loads into blocks. These assumptions may affect the validity of the procedure depending on the situation.

The loads are independent.

The order of loading does not matter.