Fracture, fatigue, creep and failure evidence

Separate a failed function from a failure mechanism

Failure means loss of a required function: a leak, excessive deformation or unreliable motion can count even without complete separation. Fracture is the formation and growth of cracks that separate material. The final fracture may be sudden although damage developed slowly beforehand. Ductile tearing and brittle fracture describe different deformation and crack-growth behaviour; neither can be diagnosed from “it broke” alone.

The distinction has a physical basis. Brittle crack growth creates new surfaces with relatively little surrounding plastic deformation. In many ductile metals, local plastic flow absorbs energy; microscopic cavities can form, enlarge and join into a tear. These are different ways of separating material, rather than names for high and low strength. The same material can behave differently as temperature, loading rate, thickness and constraint change.

A tensile-strength result answers a particular uncracked-specimen question. It does not state how strongly an existing crack is driven or how much resistance that crack meets. A final rough fracture region can also be the last event after a long period of damage. First establish which function was lost, then reconstruct the sequence that produced the observed condition.

Why a modest average stress can drive a crack

Geometry matters alongside material. A crack or sharp notch concentrates stress locally. Dividing force by a gross area gives a nominal stress, useful for simple comparisons, but not the peak stress at a crack tip. Fracture resistance, flaw size, loading, thickness and environment can matter even when the nominal stress looks modest.

Imagine a very wide elastic plate with a central crack pulled open by tension perpendicular to the crack. The two crack faces carry no tensile traction, so load must transfer through the surrounding material. The elastic stress field rises sharply near each tip. An ideal sharp-tip solution even becomes singular; a real tip instead has a finite region of deformation or other microscopic processes. Treating the mathematical singularity as an actual infinite measured stress would be a mistake.

For a suitable opening-mode, predominantly elastic model, write K=Yσ√(πa). Here σ is the remote tensile stress, a is the defined crack-size parameter, and Y is a dimensionless factor for the crack and body geometry. For this ideal central crack, a is half the total crack length and Y=1. K is stress intensity, with units MPa√m when σ is in MPa and a in metres. It characterises the near-tip field; it is neither a stress in MPa nor a material strength.

Compare the calculated K with a fracture-resistance value K_c applicable to the same material, temperature, loading and constraint. A valid plane-strain test may establish the more specific property K_Ic; not every apparent toughness is that property. Large-scale plastic deformation, changing geometry or another loading mode can require another model. Even a successful comparison addresses only this fracture model, leaving yielding, fatigue and other requirements to check separately.

Fatigue: repeated local change can precede a sudden break

Fatigue involves damage under repeated or fluctuating loading, often with crack initiation followed by growth. It can occur at stress levels below a material’s one-pull tensile strength. Maximum stress, stress range, mean stress, surface condition and environment all influence behaviour. Counting cycles without describing their severity leaves the problem incomplete.

In a metal, repeated local plastic slip can accumulate irreversible changes near a surface feature, inclusion or geometric concentration even when most of the component responds elastically. A small crack may initiate there; an existing crack can instead begin the growth stage immediately. Each subsequent loading excursion can advance it. The final remaining section may then fracture rapidly by a ductile or brittle mechanism. Fatigue describes that damage history, not a requirement that the final separation be slow.

Describe one repeated tensile cycle by its maximum σ_max and minimum σ_min. The range is Δσ=σ_max−σ_min; the mean is (σ_max+σ_min)/2. A cycle from 20 to 100 MPa and one from 60 to 140 MPa both have an 80 MPa range, but their means are 60 and 100 MPa. Equal cycle counts and equal ranges therefore do not make the histories equivalent. Crack closure, environment and load sequence can change growth as well.

For a fixed geometry factor, a corresponding crack-driving range is ΔK=YΔσ√(πa). At fixed load range it grows as the crack lengthens. Over a restricted intermediate growth regime, experimental data may support a relation da/dN=C(ΔK)^m, where da/dN is crack extension per cycle and C and m are fitted for the stated conditions. This explains a common accelerating trend: a larger crack can produce a larger driving range and a larger growth increment. The relation does not describe initiation, every small crack, threshold behaviour or final instability; its constants are not supplied here, so it gives no numerical lifetime.

The original loading-history diagram distinguishes cycling from a sustained load. Its normalised traces are schematics for reading the history, not measured fatigue-life or creep-property curves. A cycle-count record becomes useful when it is paired with load severity, geometry, environmental exposure and observations of damage.

Creep: time changes strain while the load remains

Creep is time-dependent deformation under sustained stress. Temperature, stress and material condition affect it. It is especially important for metals at elevated temperatures relative to their melting temperatures, and can matter for polymers or some low-melting alloys around room temperature. Some materials also show delayed recovery; do not assume that every time-dependent extension is immediately permanent.

For crystalline metals, thermally activated atomic movement can redistribute material through grains or their boundaries. Diffusion can also assist dislocation motion past obstacles. Stress supplies a directional driving condition while these processes take time; this is why a sustained load can produce further shape change without adding loading cycles. Which process dominates depends on temperature, stress and microstructure. Polymer time dependence involves different molecular mobility and must not be assigned a metal creep mechanism just because both specimens lengthen.

On a strain-versus-time plot, an immediate elastic extension is distinct from the additional time-dependent strain. A primary creep region has a decreasing strain rate; a secondary region is approximately steady; a tertiary region accelerates as geometry or damage evolves. These descriptions identify curve behaviour rather than three universal timed stages. Under constant tensile force, thinning raises true stress, which can contribute to acceleration; holding true stress constant is a different experiment.

Estimate a mean creep rate from the change in creep strain divided by the elapsed time over a stated interval. Do not divide the total extension, including the initial elastic jump, by the whole test duration and call that a steady creep rate. Nor should a short secondary interval be projected indefinitely through later damage. A component can lose alignment or clearance through creep before it breaks.

Use observations to separate competing histories

Preserve the component, orientation, photographs and service records before altering a fracture surface. Note where a crack appears to start, earlier impacts, load cycles, temperature exposure and changes in maintenance or manufacture. A mark that resembles a fatigue feature is a clue requiring appropriate examination, not a complete diagnosis.

State what each observation distinguishes. An increasing crack-length record during cycling supports progressive crack growth; a final fracture surface alone does not supply a cycle-by-cycle growth law. Increasing strain under a documented sustained load supports a time-dependent deformation question; temperature drift, fixture movement and thermal expansion are alternative contributors to examine. Compare specimens and measurements under genuinely comparable conditions.

Several mechanisms can interact: corrosion can assist cracking, a joint can introduce bending, and creep or wear can change the load path. Compare hypotheses against the evidence and ask what further observation would distinguish them. A material certificate alone cannot explain a product’s failure history.

A useful investigation links mechanism, sequence and cause. A crack may start at a manufacturing flaw, grow during cycling and finish in overload; the existence of the flaw alone does not establish why the service demands exceeded what the component could tolerate. The observations and the inferred explanation belong in separate columns of the record.

Compare crack driving, loading history and deformation without confusing them

Start with a supplied ideal central-crack model, Y=1 and remote tensile stress σ=80 MPa. A first crack has half-length a=2 mm=0.002 m. Thus K=80√(π×0.002)=6.34 MPa√m. For a=8 mm=0.008 m, K=12.68 MPa√m. Multiplying a by four doubles K while leaving the nominal remote stress unchanged. The graph shows this square-root relationship, not a material test result.

Now supply a comparison resistance K_c=20 MPa√m for this invented model. At 8 mm, increasing σ from 80 to 120 MPa raises K to 19.02 MPa√m, a factor 1.5. Setting K=K_c and rearranging gives a_c=(K_c/(Yσ))²/π: 19.89 mm at 80 MPa and 8.84 mm at 120 MPa. These are model crossing sizes, not acceptable flaw limits or an inspection schedule. Notice the nonlinearity: a 50% stress increase reduces the crossing size to 1/2.25 of its earlier value.

An imaginary handle is pulled twice per minute during eight hours per day, five days per week. That is 2 × 60 × 8 × 5 = 4800 pulls per week. This is a count of assumed cycles, not a predicted fatigue life. The peak and minimum forces are still missing.

Suppose later measurements in a separate cycling study report crack half-lengths 1.00, 1.10 and 1.40 mm at 0, 10000 and 20000 cycles. Mean growth rates are 0.10/10000=0.000010 mm/cycle and 0.30/10000=0.000030 mm/cycle. The observed interval rate triples. This is consistent with acceleration, but three points do not establish a Paris exponent or identify its cause; load history and uncertainty must accompany the record.

A 500 N pull spread over 20 mm² gives nominal stress 25 MPa. If a simplified remaining net section were 10 mm², the corresponding average would be 50 MPa. A sharp crack needs more than this net-area calculation; neither number proves safety.

For a separate sustained-load coupon, retain an original unloaded gauge length of 100 mm. The supplied initial elastic extension is 0.20 mm. Total extensions are then 0.50 mm at 100 h and 0.90 mm at 300 h, with unchanged temperature and stress and no other strain contributions in this teaching account. Creep strains are (0.50−0.20)/100=0.003 and (0.90−0.20)/100=0.007. Their difference over 200 h gives mean rate 0.000020 h⁻¹, or 0.002 percentage points per hour. Assume the stated elastic jump is the complete extension immediately after loading, so initial creep strain is zero. The mean over 0–100 h is then 0.003/100=0.000030 h⁻¹. More time points would be needed to resolve how the rate changes within either interval; neither mean is an instantaneous rate or proof of a steady stage. A later rupture time is not determined.

For a different sustained-load coupon, length changes from 100 to 101 mm over a recorded interval: extension is 1 mm and strain is 1/100 = 1%. Record load, temperature and elapsed time before calling this a creep comparison. The cycle count and the strain measurement answer different questions.

Keep the conclusions separate: the elastic crack model compares driving force and resistance; the cycling record reveals growth versus cycles; the sustained-load record reveals deformation versus time. Together they direct different evidence questions. None turns the handle story into an approved product or a service-life prediction.

Try the reasoning

1. A reviewer reports K=80√(π×2)=200.5 MPa√m for the 2 mm half-crack. Identify the mistake and correct the calculation.

   Show the answer and reasoning

   The substituted length is in millimetres while the stated K unit requires metres. Use a=0.002 m: K=80√(π×0.002)=6.34 MPa√m. The wrong value is larger by √1000. The unit is part of the calculation, rather than a label added afterwards.

2. At fixed stress and fixed Y, a central crack grows from total length 4 mm to total length 16 mm. How does K change, and which a values belong in the formula?

   Show the answer and reasoning

   Half-lengths are 2 and 8 mm. Their ratio is four, so the K ratio is √4=2. Inserting total crack lengths into a formula defined for half-length would give both absolute values a factor √2 too large, although the ratio would accidentally remain correct.

3. Two specimens both experience 100000 cycles. One cycles from 0 to 80 MPa; the other from 40 to 120 MPa. Calculate range and mean, then assess the claim that their fatigue histories are identical.

   Show the answer and reasoning

   Both ranges are 80 MPa. Means are 40 and 80 MPa. Equal counts and ranges do not match the maximum or mean stress, and crack closure, geometry and environment may matter. The records do not establish equal fatigue damage or which specimen fails first.

4. A component grows a small crack over months, then breaks suddenly during one ordinary load cycle. Explain a compatible sequence and name one observation that would help test it.

   Show the answer and reasoning

   A crack could initiate or already exist, grow through fatigue, then leave a section that separates rapidly by ductile or brittle fracture. Dated crack measurements and corresponding load histories would support or challenge progressive growth. A sudden final event alone does not rule fatigue out.

5. A 50 mm coupon jumps elastically by 0.10 mm when loaded. Its total extensions are 0.20 mm after 10 h and 0.35 mm after 40 h at the same stress and temperature. Calculate the interval mean creep rate under the stated decomposition.

   Show the answer and reasoning

   Creep strains are (0.20−0.10)/50=0.002 and (0.35−0.10)/50=0.005. The interval rate is (0.005−0.002)/(40−10)=0.0001 h⁻¹. This is 0.01 percentage points per hour, an interval average rather than a complete creep law or a rupture prediction.

6. A held-load extension curve steepens while temperature also rises. A report calls this proof of tertiary creep damage. What competing explanation must be tested?

   Show the answer and reasoning

   Rising temperature can increase time-dependent deformation and add thermal expansion. Load, geometry, temperature and measurement-system changes must be separated before attributing the acceleration to damage. A slope change by itself identifies behaviour, not its unique mechanism.

7. An uncracked tensile coupon has higher yield strength after a processing change. What evidence is still needed before claiming that cracked components or cycling service improved?

   Show the answer and reasoning

   Yield strength does not specify crack-growth resistance, fatigue initiation or propagation, or creep behaviour. Relevant toughness and cyclic or sustained-load evidence must match the processed state, environment, temperature and component geometry. Improving one property can leave another unchanged or worsen it.

What to carry forward

A failure explanation must connect a physical mechanism to a loading history and discriminating evidence. Strength, crack-driving force, cycle-dependent growth and time-dependent strain answer related but different questions.