Steel Reinforcement
2.4 Reinforcing Bars under Cyclic Loading
2.4.2 Low-Cycle Fatigue
depends on the chemical content of the steel, and is not observed for some steels. This behavior is not considered in routine analysis or design.
FIGURE 2.11 Low-cycle fatigue tests of A615 reinforcing bars under (a) constant strain amplitude and (b) variable strain amplitude. (After Brown and Kunnath, 2000.)
Figure 2.12 shows fatigue-life data for constant-amplitude cycling of the type shown in Figure 2.11a. Note that the trends are slightly different for the different bar sizes, indicating that fatigue life varied slightly with bar size.
FIGURE 2.12 Fatigue-life data for A615 Grade 60 (420) reinforcing bars in constant-amplitude tests. Strain amplitude is equal to half the strain range [that is, εa = (εmax – εmin)/2]. (After Brown and Kunnath, 2000.)
We can use the Coffin–Manson relation (Manson, 1953; Coffin, 1954) to represent the data trend, as
In Eq. (2.3), εa is the strain amplitude [defined as half the strain range, that is, εa = 0.5(εmax – εmin)] during a cycle, Nf is the number of full cycles (therefore, 2Nf = number of half cycles = number of reversals) to failure, and M and m are material constants to be determined from experiments.
Brown and Kunnath (2004) reported different values for M and m for different bar sizes. Here we adopt M = 0.11 and m = – 0.44, which was reported as the best fit for No. 7 (25) bars.
The fatigue-life data indicate that the maximum strain capacity under cyclic loading is less than it is under monotonic loading. We can use Eq.
(2.3) to calculate the strain amplitude εa at fracture as a function of the number of full cycles Nf. Figure 2.13 plots the result.
FIGURE 2.13 Number of full cycles to failure as function of strain amplitude.
Earthquake loading involves cyclic histories that are more complicated than the constant-amplitude example of Figure 2.11a. For more complicated histories, we can adopt Miner’s rule (Miner, 1945), which is based on the assumption that damage accumulates linearly during cyclic loading.
Accordingly, if a reinforcing bar can sustain 2Nf half cycles under fully reversed strain cycles to amplitude εa, then the fraction of total damage in half cycle i at amplitude εai is given by
in which 2Nfi = number of half cycles to failure at strain εai and Di = damage accumulated during a half cycle at strain εai. Total damage is the linear sum of damage in the individual half cycles, as
Failure is predicted when the damage index D = 1. To use Eq. (2.4), we must first calculate the quantity 2Nfi. We do this by solving Eq. (2.3) for 2Nfi as
For each half cycle to strain amplitude εai, the value of 2Nfi is calculated and inserted into Eq. (2.4) to determine the damage Di for that half cycle, then values of Di for different half cycles are summed according to Eq. (2.5) to determine total damage. According to the model, failure occurs for the half cycle in which the damage index D reaches or exceeds 1.0.
Example 2.1. A reinforcing bar has strain history as shown in Figure 2.14.
Is the bar likely to fracture?
FIGURE 2.14 Strain history.
Solution
For the first two cycles, εa = 0.02. Therefore, from Eq. (2.6), the number of half cycles to fracture is Therefore, the damage at the end of four half cycles is For the second two cycles, εa = 0.04, and the number of half cycles to fracture at that amplitude is Therefore, the accumulated damage at the end of these
cycles is Because D < 1.0, fracture is
unlikely.
We can use Eq. (2.5) to investigate more complicated loadings. For example, a common laboratory testing protocol is to cycle a test specimen through three deformation cycles at constant amplitude ∆, followed successively by three cycles at each of 2∆, 3∆, 4∆, etc. until failure. Strain amplitude in such tests generally does not scale linearly, but let us assume that it does as a simplifying assumption. Suppose we start with an initial strain amplitude εa = 0.002, with three full cycles. For each half cycle, we use Miner’s rule to calculate damage accumulation. We then increment the strain amplitude to 0.004, 0.006, 0.008, etc. until failure (D≥1). We can repeat the process for different values of the strain amplitude increment εa. Figure 2.15 presents the results for strain amplitude increments εa = 0.002, 0.005, 0.01, and 0.02. Interestingly, the analytical model tells us that bar fracture could occur at strains ranging from approximately 0.025 to 0.045.
From such results we can see that the maximum deformation capacity of a structural component subjected to cyclic loading will depend on the loading history if its response is sensitive to reinforcing bar fracture.
FIGURE 2.15 Failure strain predictions from Eq. (2.5).
An important observation in fatigue problems is that fatigue life depends on two aspects of the strain history. One is the number of cycles. The other is the strain range, that is, the difference between the maximum and minimum strain in a cycle. In Eq. (2.6), we defined εai as half the strain range, that is, εai = 0.5(εmax – εmin). Thus, according to Eq. (2.6), a bar cycled between strains 0.0 and +0.02 (εa = 0.01) will have a longer fatigue life than one cycled between strains -0.02 and +0.02 (εa = 0.02).
To extend the procedures to more complicated loadings, another algorithm will be required. Consider the reinforcing bar strain history of Figure 2.16a and corresponding stress–strain history of Figure 2.16b. As the bar is strained from a to e, we assume that it accumulates an amount of damage associated with a half cycle having a strain range along path ae. We also need to account for the damage resulting from the relatively smaller- amplitude cycle bcd that occurs along the path ae; we will assume that the damage is equal to the sum of damage for a half cycle having strain range bc and for another half cycle having strain range cd. Strain reversal from e to f causes damage associated with a half cycle having strain range ef. Finally, another small half cycle of strain range fg finishes the history. For each of these half cycles, we can calculate the strain amplitude εai = half the strain range for that half cycle, use Eqs. (2.6) and (2.4) to calculate damage in that
half cycle, and then sum over all half cycles using Eq. (2.5) to calculate the damage index.
FIGURE 2.16 Illustration of the rainflow-counting method: (a) strain history; (b) stress–strain history; (c) rainflow-counting diagram.
The approach just described is known as the rainflow-counting method (ASTM E1049-85, 2005), because it is analogous to rain flowing over a series of rooftops. To use the analogy, first rotate the strain history so it appears as a series of inclined surfaces (or pagoda rooftops) and imagine rain flowing down each one (Figure 2.16c). Rain flow begins at each peak and ends when one of the following conditions is met:
1. The end of the strain history is reached.
2. Rain flows opposite a peak having amplitude greater than the one from which it started.
3. Rain flow is interrupted by a flow that started at an earlier peak.
For example, a raindrop starting at peak a (Figure 2.16c) would flow along rooftop ab, drop down to rooftop ce, flow along de, drop off e, and stop at point 1 because it is opposite a peak (f) having amplitude in the opposite direction greater than the one from which it started. Another flow starts at b and must stop opposite peak e for the same reason. A third flow starts at c and must stop at d because it is interrupted by flow coming from above, and so on. The strain amplitude of each of these paths (half cycles) is determined, and then Miner’s rule is used as described above to sum the damage. Computer algorithms for automatic computation using the rainflow- counting method are available.
We can apply the rainflow-counting algorithm to any strain history. In
reference to Figure 2.17, the algorithm predicts failure at strains of 0.035 for Case A and 0.06 for Case B, whereas Case C is essentially the same as monotonic loading.6
FIGURE 2.17 Strain histories with amplitude incremented by 0.005 after three cycles: (a) equal positive and negative strain amplitudes; (b) zero minimum strain; (c) minimum strain equal to maximum strain minus 0.005.
References
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1See Section 2.2.2 for discussion of reinforcement grades and ASTM designations.
2Engineering stress is defined as the ratio of the applied force to the initial area As. True stress is the ratio of the applied force to the instantaneous cross-sectional area As(1–0.5ε1)2.
3Personal communication from Robert Smith, ERICO, 2008.
4In many beams, the area of top reinforcement exceeds the area of bottom reinforcement at the face of the column. Under earthquake loading, the top reinforcement yields in tension for loading in one direction. Upon moment reversal, the bottom reinforcement yields in tension, but its force capacity is insufficient to yield the larger area of top reinforcement. Therefore, the top reinforcement primarily experiences tensile strain with only moderate compressive stress reversals.
5Strain-aging was noted by Bauschinger (Martens, 1899) and Richart et al. (1929). Restrepo- Posada et al. (1994) provide more recent discussion.
6Note that these methods are not intended to be used for monotonic loading, and do not necessarily produce sensible results. According to the rainflow-counting algorithm, under monotonic loading the number of half cycles is 2Nf = 1 and the strain amplitude, defined as half the strain range, is εa = εu/2. Using M = 0.11 as recommended here, Eq. (2.3) gives εa = 0.11, or εu = 0.22. Usually the elongation capacity under monotonic loading is considerably less than this value.