Stress cycles

Cumulative Damage. Pressure vessels may be subjected to a variety of stress cycles during service some of these cycles have ampHtudes below the fatigue (endurance) limit of the material and some have ampHtudes various amounts above it. The simplest and most commonly used method for evaluating the cumulative effect of these various cycles is a linear damage relationship in which it is assumed that, if cycles would produce failure at a... 

The linear damage rule takes no account of the orderin which the stress cycles are appHed. 

Fatigue properties in bending are most appropriate for copper aHoys as these are often used as spring contact components in beUows and electrical switches and coimectors. These articles are usuaHy designed for acceptable service Hves at a moderate to high number of stress cycles. 

More formally if a component or structure is subjected to repeated stress cycles, like the loading on the connecting rod of a petrol engine or on the wings of an aircraft - it may... 

Here is the number of cycles to fracture under the stress cycle in region i, and Nj/Nf is the fraction of the lifetime used up after N, cycles in that region. Failure occurs when the sum of the fractions is unity (eqn. (15.4)). This rule, too, is an empirical one. It is widely used in design against fatigue failure but if the component is a critical one. Miner s Rule should be checked by tests simulating service conditions. 

Fig. 28.5. Fatigue data for a typical structural steel in dry air. Note that, if the fatigue stress range is less than 440 MPa (the fatigue limit] the component should last indefinitely. The data relate to a fatigue stress cycle with a zero mean stress, which is what we have in the case of our tail drum.

Contain tensile residual stresses which are usually as large as the yield stress. Weld liable to fatigue even when applied stress cycle is wholly compressive. Reduce residual stresses by stress relieving, hammering or shot peening. 

Example 2.22 A certain grade of acrylic has a Kc value of 1.6 MN and the fatigue crack growth data as shown in Fig. 2.77. If a moulding in this material is subjected to a stress cycle which varies from 0 to 15 MN/m, estimate the maximum internal flaw size which can be tolerated if the fatigue endurance is to be at least 1(P cycles. 

In a small polypropylene component a tensile stress of 5.6 MN/m is applied for 1000 seconds and removed for 500 seconds. Estimate how many of these stress cycles could be permitted before the component reached a limiting strain of 1%. What is the equivalent modulus of the material at his number of cycles The creep curves in Fig. 2.5 may be used. 

A certain grade of PMMA has a K value of 1.6 MN m and it is known that under cyclic stresses, cracks grow at a rate given by (2 x 10 A/( ). If the intrinsic defects in the material are 50 mm long, how many hours will the material last if it is subjected to a stress cycle of 0 to 10 MN/m at a frequency of 1 Hz. 

As approaches the crack can also propagate by hydrogen embrittlement processes during the higher load parts of the stress cycle. This forms the basis of various models which have been developed to describe corrosion fatigue, probably the best-known of which are the superposition models due to Wei"". In its most recent version this model takes the form ... 

The fatigue strength of most TPs is about 20 to 30% of the ultimate tensile strength determined in the short-term test but higher for RPs. It decreases with increases in temperature and stress-cycle frequency and with the presence of stress concentration peaks, as in notched components. 

Endurance limit To develop S-N curves the fatigue specimen is loaded until, for example, the maximum stress in the sample is 275 MPa (40 ksi) (Fig. 2-43). At this stress level it may fail in only 10 cycles. These data are recorded and the stress level is then reduced to 206 MPa (30 ksi). Tliis specimen may not break until after 1,000 stress cycles at this rather low stress level. 

A different design approach is used in this case. Instead of assuming an apparent modulus of elasticity using a constant creep situation covering the life of the chair, it is better to determine the actual creep deflection over a typical stress cycle, the creep recovery over a non-use cycle, and so on until the creep is determined after a series of what might be considered typical hard usage cycles for the chair. The accumulated creep after a period of two weeks can be assumed to represent the base line for an apparent modulus of elasticity to determine the design life of the chair. 

During operation the shell, or components of the vessel, may be subjected to cyclic stresses. Stress cycling can arise from the following causes ... 

Applies to essentially noncorroded piping. Corrosion can sharply decrease cyclic life therefore, corrosion resistant materials should be considered where a large number of major stress cycles is anticipated. 

Fatigue tests are normally performed by applying one of the stress cycles described above until the test specimen fractures. The number of cycles to failure, Nf, at a... 

Figure 5.40 Variation of stress with time that accounts for fatigue failure by (a) a reversed stress cycle and (b) a repeated stress cycle. Reprinted, by permission, from W. Callister, Materials Science and Engineering An Introduction, 5th ed., p. 210. Copyright 2000 by John WUey Sons, Inc.

The overall response of the crystal to such a stress cycle is shown in Fig. 8.16. When the stress a0 is applied suddenly, the crystal instantaneously undergoes an ideally elastic strain following Eq. 8.62. As the stress is maintained, the crystal undergoes further time-dependent strain due to the re-population of the interstitials. When the stress is released, the ideally elastic strain is recovered instantaneously and the remaining anelastic strain will be recovered in a time-dependent fashion as the interstitials regain their random distribution. 

Figure 8.16 Strain vs. time for an anelastic solid during a stress cycle in which stress is...

The energy dissipated can be compared with the maximum elastic strain energy, W, which is stored in the material during the stress cycle. Because the elastic strain is proportional to the applied stress, W is equal to just half of the product of the maximum stress and strain (i.e., W = a0ei/2), and therefore... 

Solution. Using a torsion pendulum, find the anelastic relaxation time, r, by measuring the frequency of the Debye peak, cup, and applying the relation cupr = 1. Having r, the relationship between r and the C atom jump frequency F is found by using the procedure to find this relationship for the split-dumbbell interstitial point defects in Exercise 8.5. Assume the stress cycle shown in Fig. 8.16 and consider the anelastic relaxation that occurs just after the stress is removed. A C atom in a type 1 site can jump into two possible nearest-neighbor type 2 sites or two possible type 3 sites. Therefore,... 

Figure 9-1. Forms of strain and stress cycles, (a) Continuous constant amplitude (b) continuous decaying amplitude (c) successive half waves...

Figure 9-2. Sinusoidal stain and stress cycles. I strain, amplitude a II in-phase stress, amplitude b III out-of-phase stress, amplitude c IV total stress (resultant of II and III, amplitude d. a is the loss angle...

Based on our observation, a membrane degradation and failure mechanism under the RH cycling, a pure mechanical effect is theorized as the following sequence electrode-microcracking- - crazing initiation at the electrode/electrolyte interface - crack growth under stress cycling- -fast fracture/instability. 

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