Materials Science · Fatigue, creep and time-dependent damage

Fatigue crack growth and variable loading

Follow an existing crack through a bounded growth calculation, explain why the same applied cycle can act on a changed crack state, and compare that history with a simple fractional-life sum.

Useful background: Elasticity, plasticity and tensile strength; Cracks, fracture and toughness. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Calculate an average crack-growth rate and a Mode I stress-intensity range with a specified crack convention.
Use a conditioned Paris-type relation only within its stated range, with consistent coefficient units.
Explain how crack size, closure and an overload history can change later growth.
Calculate a linear damage sum and identify the sequence information and uncertainty it leaves out.

An initiated crack can grow cycle by cycle

Fracture-mechanics descriptions often relate growth per cycle to the range of stress intensity, with other variables retained. Here growth per cycle means the extra crack length divided by the number of load cycles, and the stress-intensity range is the maximum K minus the minimum K in a cycle. A power-law fit may represent an intermediate regime; it does not automatically cover thresholds, very short cracks or final rapid failure.

Imagine a tensile component with a crack already detected. Its next interval concerns propagation from that crack state; the cycles spent creating the crack are a separate part of the history. A growth calculation needs an initial crack size, shape and location. It does not recover an unknown initiation interval.

In many ductile metals, repeated local plastic deformation at the tip can blunt, resharpen and advance a crack even when the surrounding body remains mainly elastic. Crack faces and the deformed material left behind the advancing tip also change the local loading. This is a representative metallic mechanism, not a claim that every material grows cracks by identical atomic steps.

Let a be a stated crack dimension and N the number of complete load cycles. Over a measured interval, the average rate is Δa/ΔN. The notation da/dN describes a local slope of the a-versus-N curve. Both have length per cycle units. A finite measured average does not prove the rate was constant within the interval.

Connect the cycle to the existing crack

For an opening-mode crack in a predominantly elastic body, write K_I = Yσ√(πa). The dimensionless factor Y encodes geometry and the definitions of a and remote nominal stress σ. A central through crack in the ideal infinitely wide plate has total length 2a and Y = 1. An edge crack or a finite-width specimen needs its own geometry factor; changing the word crack does not leave Y unchanged.

For a fixed crack size within a tension–tension cycle, K_max = Yσ_max√(πa), K_min = Yσ_min√(πa), and ΔK = K_max − K_min = YΔσ√(πa). Here Δσ is the stress range, not its half-range amplitude. If σ is in MPa and a in metres, K is in MPa√m. The examples assume the crack advances negligibly during one cycle and use positive tensile stresses.

The intensity ratio R_K = K_min/K_max equals σ_min/σ_max in this simple proportional loading, with nonzero K_max. Changing R_K can change growth even at the same ΔK. More complicated residual or mixed-mode loading needs a more complete description. The formal applied range also need not equal the range for which crack faces are open.

At fixed Δσ and constant Y, increasing a by a factor of four doubles ΔK. A growing crack therefore sees changing driving force even under an unchanged nominal cycle. K_max must also be compared with appropriate fracture-resistance evidence when assessing rapid failure; a growth-range fit is not that final-failure criterion.

A power fit describes a selected growth regime

A common intermediate-regime fit is da/dN = C(ΔK)^m, often called a Paris-type relation. C and the dimensionless exponent m are fitted under specified material state, stress ratio, temperature, environment and loading conditions. They are not supplied by tensile strength. The fit organises evidence; it is not a derivation of one universal fatigue mechanism.

The units belong to the coefficient. For da/dN in m/cycle and ΔK in MPa√m, C has units (m/cycle)/(MPa√m)^m. Changing the numerical length or stress unit changes the numerical C. In a rate ratio using the same fit, C cancels: rate₂/rate₁ = (ΔK₂/ΔK₁)^m. Doubling the range with m = 3 gives eight times the rate, not twice the rate.

A measured long-crack threshold is an operational stress-intensity range associated with very low growth under stated test conditions and a stated rate criterion; it is not a universal boundary below which every crack stops. Near it, closure, environment, microstructure and test procedure can matter strongly. A crack can be short relative to grain or other microstructural dimensions, or to its local plasticity scale; there is no single universal millimetre cutoff for all those limits. Such cracks can behave differently from a long-crack fit or threshold. Near final fracture, K_max and extensive yielding may become consequential. Do not extend a power fit through all three situations or call an unresolved rate exactly zero.

For the original illustration here, choose m = 3 and C = 1.0 × 10^−10 (m/cycle)/(MPa√m)^3. Stipulate that this fit applies only to the current long-crack state and fixed conditions for 5 ≤ ΔK ≤ 12 MPa√m. The table and graph show those chosen values, not measured data for a named alloy.

Original illustrative intermediate-regime fit under fixed conditions. The displayed range is a stipulated model domain, not a measured threshold or a design allowable.
ΔK (MPa√m)	Rate (m/cycle)	Rate (μm/cycle)
5	1.25 × 10^−8	0.0125
6	2.16 × 10^−8	0.0216
8	5.12 × 10^−8	0.0512
10	1.00 × 10^−7	0.1000
12	1.728 × 10^−7	0.1728

Original chosen fit: rate=C(ΔK)³ with C=10^−10 in metre/MPa units, shown only from 5 to 12 MPa√m. Rates at 5, 6, 8, 10, 12 are 0.0125, 0.0216, 0.0512, 0.1000, 0.1728 μm/cycle. The range is stipulated, not a measured threshold or named-alloy dataset.

Update the state instead of freezing the first rate

For a short block containing ΔN cycles, a simple estimate is Δa ≈ [da/dN at the block start]ΔN. Add that increment to the existing a, then update Y, ΔK and the rate for the next block. This is a stepwise approximation: it treats the rate as constant within each block. Smaller blocks are needed when the geometry or rate changes appreciably.

Keep the crack convention throughout the calculation. If a denotes half of a central crack, an increment in a is the advance of each symmetric tip; the total crack length grows by 2Δa. A measured edge-crack length would use another definition. Confusing the two doubles the reported length change without changing the underlying calculation.

A future inspection comparison also has a measurement resolution and an uncertain initial crack state. Numerical stepping alone does not choose a safe interval or a population guarantee. It estimates propagation under the supplied model and history; missing short-crack, sequence or environmental physics remains missing even with very small steps.

Load order can matter

Overloads, residual stresses and crack closure can change later growth. A simple cumulative-damage rule adds fractions of nominal lives but does not capture every sequence effect. Evaluate variable-amplitude histories using evidence appropriate to the structure and mechanism.

Crack closure means that portions of the crack faces contact during part of the load cycle, including sometimes while the applied load remains tensile. A plastically deformed wake, rough faces or oxide debris can alter that contact. The load at which a particular measurement indicates opening is therefore a state-dependent quantity, not automatically the minimum applied load.

One illustrative opening-range model defines ΔK_eff = max[0, K_max − max(K_min, K_open)], where K_open is a supplied opening intensity. It counts the portion above the larger of the applied minimum and the opening level, with no negative open range. This parameter can help correlate certain growth records. It is not proof that all possible tip damage vanishes below one opening estimate.

A tensile overload can leave a larger plastically deformed region and a different residual/contact state. Some subsequent growth histories show retardation associated with that changed state; immediate growth during the overload and other acceleration effects can also occur. The amount and duration depend on material, constraint and history. An overload is not automatically beneficial.

If a separate effective-range fit is supplied, use coefficients calibrated against ΔK_eff. Do not silently reuse the C from a fit against applied ΔK. In the worked opening-state comparison, only the ratio of a separately stipulated cubic effective-range fit is used. A changed rate is a conditional model result, not a complete prediction of the overload history.

At identical current applied K_min=2 and K_max=12 MPa√m, a supplied K_open of 4 gives open range 8; K_open 6 gives open range 6. A separate cubic effective-range fit gives rate ratio 0.421875. The opening values are chosen history inputs, not predictions of overload magnitude or lifetime.

A fractional-life sum deliberately discards order

A different approximation starts from constant-amplitude lives. Let N_j be a nominal life to one specified endpoint at load level j under matched conditions, and n_j the cycles applied at that level. The linear sum is D = Σ n_j/N_j. Each term and D are dimensionless. A common model convention associates D = 1 with its predicted endpoint. This is an assumed accumulation rule, not a universal physical failure event.

For the illustrative lives N_L = 200,000 and N_H = 50,000 cycles, applying 20,000 low cycles and 10,000 high cycles gives D = 0.10 + 0.20 = 0.30. Reversing the two blocks gives the same sum. Addition cannot represent which block came first, what crack size it met, or the opening/residual state it left for the next block.

A value D = 0.30 is neither a 30% probability of failure nor a direct measurement of 30% of the crack length or microscopic damage. Its remaining nominal capacity belongs to the chosen linear model. The cycle endpoints, source-life scatter and material state also matter. A model that matches one sequence does not establish all reversed or irregular histories.

For an existing crack, combine a suitable crack-growth description with measurements of the evolving state and its actual loading history. For a specimen-life estimate, identify the life endpoint and test whether an accumulation approximation is supported. The two approaches answer related questions with different state variables; their numerical outputs cannot be exchanged without an explicit bridge.

Worked examples

Follow a short growth block from a known crack

In an ideal infinitely wide plate, a central through crack has half-length a = 1.00 mm. It cycles between remote tensile stresses 25 and 125 MPa with Y = 1. Use the illustrative applied-range fit above for a short block of 1000 cycles. Find ΔK, the initial rate and the approximate new half-length. Does the original rate remain unchanged as the crack grows?

Convert a to 0.00100 m. Then √(πa) ≈ 0.056050 √m. K_min ≈ 1.401 and K_max ≈ 7.006 MPa√m, giving ΔK ≈ 5.605 MPa√m. R_K = 25/125 = 0.20. The calculated range lies inside the stipulated fit interval.
Apply the fit: da/dN = 1.0 × 10^−10 × 5.605³ ≈ 1.761 × 10^−8 m/cycle. This is about 0.01761 μm/cycle. The stress range used was 100 MPa, rather than its 50 MPa amplitude.
Freezing that initial rate over the short block gives Δa ≈ 1.761 × 10^−5 m = 0.01761 mm. The updated half-length is about 1.01761 mm. Because a is a half-length, the total central-crack length increases by about 0.03522 mm.
Recompute the next block start: ΔK ≈ 5.654 MPa√m and rate ≈ 1.808 × 10^−8 m/cycle. The unchanged nominal load does not preserve the first rate. The block estimate is approximate because the rate already rose during those 1000 cycles.
As a separate state comparison, at a = 4.00 mm the range would be twice its initial value and the same fit would give eight times the initial rate. This remains within the chosen range interval, but no supplied fracture resistance or yielding check establishes component survival.

The initial ΔK is about 5.605 MPa√m and rate 1.761 × 10^−8 m/cycle. A short-block estimate gives a ≈ 1.01761 mm; the next initial rate is higher. This is propagation from a known crack, not a complete life prediction.

The same applied range can meet a different crack state

At the same current crack size, two illustrative histories now share K_min = 2 and K_max = 12 MPa√m. A supplied opening estimate is K_open = 4 after a baseline history and 6 after a history containing an overload. A separate effective-range fit is stipulated as rate ∝ (ΔK_eff)³ over both values. Compare the current open ranges and rates. What has not been predicted?

Both applied ranges are 12 − 2 = 10 MPa√m. Their current applied cycles therefore match, including R_K = 1/6.
The opening-range model gives baseline ΔK_eff = 12 − 4 = 8 and post-overload ΔK_eff = 12 − 6 = 6 MPa√m. The figure shows the different portions above the supplied opening levels.
Using the separately stipulated effective-range fit, rate_post/rate_base = (6/8)³ = 0.421875, about 0.422. That is approximately 58% lower at this comparison state. No applied-range coefficient was reused.
The opening levels are chosen inputs, not computed consequences of an overload magnitude. The comparison does not count crack advance during the overload, determine how long the changed opening persists, or establish total life. It illustrates how a history variable can matter after the applied cycle has returned to its earlier values.

Applied ΔK is 10 in both cases; the supplied open ranges are 8 and 6 MPa√m. The separate cubic model gives a current rate ratio about 0.422, with the full overload history still unpredicted.

Calculate what a linear sum can and cannot remember

Use the illustrative constant-amplitude lives N_L = 200,000 and N_H = 50,000 cycles. Compare D for 20,000 low cycles followed by 10,000 high cycles and the reverse order. A supplied illustrative crack-inspection record reports extensions 0.20 and 0.28 mm for those respective histories. What does that record test?

For either order, D = 20,000/200,000 + 10,000/50,000 = 0.10 + 0.20 = 0.30. The same terms occur in a different order.
The supplied extensions differ by 0.08 mm; 0.28/0.20 = 1.40. These are invented inspection values used to test the inference, not measured evidence for a named alloy.
The sum contains no order variable, so it cannot explain a difference that remains after accounting for measurement uncertainty and matched starting states. Crack size, residual/contact state or other evolving processes would need suitable evidence.
The inspection example does not prove closure was the cause or that all reversed histories differ by 40%. Nor does D = 0.30 certify either history safe. It exposes which information the assumed accumulation rule leaves out.

Both sums equal 0.30. The supplied differing extensions challenge treating that sum as a complete description of crack evolution; they do not uniquely identify the missing mechanism.

Try the reasoning

Under the same illustrative applied-range fit, ΔK changes from 6 to 9 MPa√m. Find the rate ratio and the new rate. Why is this not automatically the ratio of total lives?
Show the answer and reasoning
The rate ratio is (9/6)³ = 3.375. The new rate is 7.29 × 10^−8 m/cycle.
The two local rates are evaluated at stated states within the fit interval. Total propagation also changes crack geometry and driving force, and a total life includes its defined start/end and possibly initiation.
A measured crack dimension increases by 0.040 mm over 20,000 cycles. Find the average growth rate in m/cycle. A worksheet inserts 0.040 as metres. Identify the error.
Show the answer and reasoning
The mean rate is (4.0 × 10^−5 m)/20,000 = 2.0 × 10^−9 m/cycle. The worksheet makes the length 1000 times too large.
The numerator is a length increment, not the final crack size. This finite average can hide a changing rate within the block.
For the same geometry and crack size, a cycle has tensile stresses 40 and 140 MPa. A learner inserts its 50 MPa amplitude where Δσ is required by the applied-range model. With m = 3, how does the resulting local rate compare with the correctly calculated value?
Show the answer and reasoning
The calculated ΔK is halved, so the fitted rate is (1/2)³ = 1/8 of the correct value.
The range is 100 MPa. The calculation has also changed the model input while claiming to describe the same cycle; a coefficient fitted to range is not interchangeable with one fitted to amplitude.
Use the opening-range definition with K_min = 7, K_max = 12 and K_open = 6 MPa√m. What is ΔK_eff? Why would subtracting 6 be wrong here?
Show the answer and reasoning
ΔK_eff = 12 − max(7,6) = 5 MPa√m.
The applied cycle never reaches 6. Counting a range from 6 to 12 would include loading that this current cycle did not traverse. The outer maximum also prevents a negative open range in a state where opening is not reached.
A linear sum currently has D = 0.60. Under that model alone, how many further cycles at a level with N_j = 80,000 would bring D to 1? What does the answer leave uncertain?
Show the answer and reasoning
The model gives (1 − 0.60) × 80,000 = 32,000 additional cycles.
That is an accumulation-model estimate, not a measured remaining life, survival probability or recommended inspection interval. Sequence effects, source-life scatter and evolving crack state were not resolved by this arithmetic.
Two sequences have equal block counts and equal D. The final crack dimensions also agree, but no intermediate inspections were made. Does that prove the crack-opening states and growth rates matched throughout?
Show the answer and reasoning
No. A matching final value does not establish identical paths or internal states.
Different intermediate rates can lead to the same final dimension. Time-resolved crack-size and relevant opening/residual-state evidence would distinguish histories; endpoint agreement alone cannot validate every part of a sequence model.

What to carry forward

An existing-crack calculation needs a defined starting size and geometry; it does not supply an initiation history.
Applied ΔK uses the stress range, a consistent crack convention and a conditioned growth fit.
Growing geometry and a changed opening/residual state can alter later rates under repeated nominal loads.
A linear fractional-life sum is order independent by construction; its numerical value is not a probability or direct crack measurement.

References

Teaching resources and research supporting this lesson’s explanations and examples.