Fracture, fatigue, plasticity, and failure

Ask which change the model is following

A body can change shape permanently while remaining connected. An existing crack can extend even when most of the body is elastic. Repeated loading can accumulate damage through many cycles before rapid fracture occurs. Plasticity, fracture and fatigue describe related but different processes; failure additionally requires a stated criterion for what function or condition has been lost.

Start with a uniform bar of initial length L₀ = 0.100 m and area A₀ = 100 mm² = 1.00 × 10⁻⁴ m². Assign Young’s modulus E = 100 GPa and uniaxial yield stress σ_Y = 200 MPa. Extend it slowly with a prescribed strain history up to ε_max = 0.006, or 0.60%, then remove the tensile force. We will calculate how much it recovers using an elastic–perfectly plastic model.

A separate model is an ideal infinite plate containing a central through-crack. Its tensile stress will cycle between 5 and 50 MPa. The crack model shares the assigned E and yield stress for comparison, but has its own supplied crack resistance and growth coefficients. Neither example identifies a real material specimen or supplies a design limit.

Stress, strain, and Hooke’s law supplies the uniaxial elastic relation. Work, kinetic energy, potential energy, and power is a return path for mechanical work; the necessary graph areas are calculated here. Stress and strain tensors is optional preparation for why a crack-tip field differs from a uniform tensile stress. Integrals and differential equations are used only after the load path and crack geometry have been introduced; follow the diagrams and tables first.

Split total strain into elastic and plastic parts

Use engineering strain ε = (L − L₀)/L₀ and nominal tensile stress σ = F/A₀. Strain is dimensionless, while stress has unit Pa. Changes in cross-sectional geometry are neglected in this small-strain model. Write ε = ε_e + ε_p, where ε_e is the recoverable elastic part and ε_p is the plastic part. The elastic relation is σ = Eε_e, not σ = Eε for every loading history.

On the first monotonic tensile loading from zero, the model gives σ = Eε until σ reaches σ_Y. The yield strain is ε_Y = σ_Y/E = (200 × 10⁶)/(100 × 10⁹) = 0.002 = 0.20%. Beyond that point, continued prescribed extension increases ε_p while σ remains 200 MPa. Thus σ = min(Eε,σ_Y) on this particular initial loading branch.

At ε_max = 0.006, the elastic part is still σ_Y/E = 0.002. The plastic part is 0.006 − 0.002 = 0.004. The maximum extension is L₀ε_max = 0.600 mm, and the force is σ_YA₀ = 20,000 N. A plateau stress alone would not determine how far a perfectly plastic bar extends: the specified strain history determines the endpoint.

Perfect plasticity means that hardening is omitted. A real material can require increasing stress as plastic strain accumulates, or can soften and develop nonuniform deformation. Our assigned law has no necking, time dependence or crack-growth variable. Yield in this model marks the start of irreversible strain; it does not automatically break the bar.

Follow unloading to a nonzero residual strain

Unload from the maximum state while remaining on an elastic unloading branch. The plastic strain stays at 0.004, so σ = E(ε − 0.004). The stress–strain line has the original elastic slope E. At ε = 0.005, the stress is 100 MPa; at ε = 0.004, the stress is zero. Unloading changes ε_e, rather than undoing the accumulated ε_p.

The recovered extension is L₀(0.002) = 0.200 mm. The residual extension is L₀(0.004) = 0.400 mm, leaving length 100.400 mm after the force is removed. At the peak, total extension equals recovered plus residual extension: 0.600 = 0.200 + 0.400 mm. Residual strain is measured relative to the original length, not relative to the loaded length.

Reloading in tension initially retraces the same elastic segment from (ε,σ) = (0.004,0) to (0.006,200 MPa). Further prescribed extension produces more plastic strain at the same plateau stress in this perfectly plastic model. A graph therefore needs loading directions and history labels: the same total strain can correspond to different stress values on different branches.

Elasticity concerns recovery, while a linear relation is an additional constitutive assumption. Here the recoverable response is linear, but the complete loading–unloading path is history-dependent. Plastic deformation changes the reference state for the next elastic response. In this model the remaining strain does not disappear merely because the force has become zero.

Read work from the loading and recovery paths

For a uniform bar in this reference-area description, dW = F dL = A₀σL₀ dε. Divide by the initial volume V₀ = A₀L₀ to obtain work per reference volume: dw = σ dε. A stress–strain graph area therefore has unit Pa = J/m³. If its strain axis is plotted in percent, the numerical percent value must first be divided by 100 before calculating physical work.

The initial loading work density is the elastic triangle plus the plastic plateau area: w_in = ½σ_Yε_Y + σ_Y(ε_max − ε_Y). This is 200,000 + 800,000 = 1,000,000 J/m³. The maximum stored elastic energy density is w_e = σ_Y²/(2E) = 200,000 J/m³. It equals the magnitude of the work returned during elastic unloading to zero stress.

The difference w_in − w_e = σ_Yε_p = 800,000 J/m³ is plastic work not recovered as elastic mechanical work. Since V₀ = 1.00 × 10⁻⁵ m³, the whole-bar values are 10.0 J input, 2.0 J recovered and 8.0 J plastic work. This model’s energy accounting does not require the bar to fracture.

The same area can also be understood from F versus extension: ½(20,000 N)(0.000200 m) = 2.0 J for elastic loading, followed by (20,000 N)(0.000400 m) = 8.0 J on the plateau. Real plastic work can become heat and stored microstructural energy; the present calculation tracks mechanical recovery and does not assign a heat fraction.

Give an existing crack a geometry and an opening load

Now use the separate crack model. In an ideal infinite homogeneous isotropic plate, place a straight central through-crack along x, with endpoints at x = −a and +a. Its full length is 2a. Apply remote tension σ_∞ along y, normal to the crack plane. This is Mode I opening. A finite rectangle in the drawing is a window into the ideal infinite domain, not the boundary of a finite-width specimen.

In linear elastic fracture mechanics, the opening stress intensity is K_I = Y_gσ_∞√(πa), where Y_g is a dimensionless geometry/loading factor. For this central crack in an infinite plate, Y_g = 1. Another crack geometry generally needs another factor; the letter Y_g does not denote the bar’s yield stress σ_Y.

K_I has units of stress times square root of length, commonly MPa√m. It is not a stress in MPa and cannot be compared directly with σ_Y. Near an ideal sharp crack tip, the leading elastic stress field scales as K_I/√r, with angular factors, where r is distance from the tip. K_I sets the intensity of that field. The mathematical divergence as r approaches zero does not imply physically infinite stress.

At remote stress 50 MPa, a = 1.00 mm = 0.00100 m gives K_I = 50√(π × 0.00100) MPa√m ≈ 2.80250 MPa√m. At a = 4.00 mm, K_I ≈ 5.60499 MPa√m. Quadrupling the half-length doubles the stress intensity, although the remote stress is unchanged. The nominal remote stress is only one quarter of the assigned 200 MPa yield stress; that comparison does not describe the local crack-tip field.

Compare crack driving force with a conditioned resistance

Assume a plane-stress elastic field away from the small crack-tip process zone: stress normal to the plate faces is neglected. The Mode I energy release rate is G = K_I²/E. G measures available mechanical energy per increment of projected crack area, with unit J/m². The resistance accounts for the processes needed to create the two new crack faces and any associated irreversible deformation; it need not equal bond-breaking surface energy alone.

Convert K_I to Pa√m before using E = 1.00 × 10¹¹ Pa. For a = 1 mm, G = (2.802496 × 10⁶)²/(1.00 × 10¹¹) ≈ 78.5398 J/m². At a = 4 mm it is about 314.159 J/m². Thus G grows in direct proportion to a at fixed remote stress, while K_I grows with √a. Energy release rate has different dimensions from the work density in the bar.

Supply a configuration-specific Mode I initiation criterion K_c = 10.0 MPa√m. Within this plane-stress model it corresponds to G_c = K_c²/E = 1000 J/m². Both listed peak-load states lie below that supplied monotonic initiation criterion. Formally solving 50√(πa_c) = 10 gives a_c = 0.04/π m ≈ 12.7324 mm, or full length 25.4648 mm. This is a conditional model threshold, not a permissible real crack size.

The notation K_Ic is normally reserved for a qualified plane-strain fracture toughness. We do not infer that property from the assigned plane-stress K_c. Linear elastic fracture mechanics requires an elastic region controlled by K outside a sufficiently small nonlinear process zone. Close to the tip, yielding, damage and finite tip geometry can invalidate the singular elastic field; far away, the complete geometry controls the field. Large-scale plasticity, changing resistance or mixed-mode loading requires a different analysis.

Describe a cycle before counting damage

For the cracked plate, prescribe repeated tensile cycles with σ_max = 50 MPa and σ_min = 5 MPa. The stress range is Δσ = σ_max − σ_min = 45 MPa. The alternating amplitude is half that range, 22.5 MPa, and the mean is 27.5 MPa. The stress ratio is R = σ_min/σ_max = 0.10. A sinusoidal illustration could be σ(θ) = 27.5 + 22.5cosθ MPa as θ runs through one cycle; no cycle duration is assigned.

For a crack whose length changes little during one cycle, linearity gives K_max = 50√(πa), K_min = 5√(πa), and ΔK = K_max − K_min = 45√(πa), with stress in MPa and a in m. The factor relating minimum and maximum is also K_min/K_max = R. At a = 1 mm, ΔK ≈ 2.52225 MPa√m; at a = 4 mm, it is about 5.04449 MPa√m.

Fatigue concerns changes accumulated under repeated loading. It can involve crack initiation and later propagation. The following calculation begins with an already existing crack of half-length a₀ = 1 mm; it does not calculate how that crack first formed. A macroscopically elastic remote cycle can coexist with local irreversible processes near a defect or crack tip.

A monotonic initiation criterion and a cyclic growth law answer different questions. K_max below the supplied K_c does not force the fatigue growth rate to be zero. Conversely, a fatigue law fitted to an intermediate growth regime cannot describe every threshold, short crack or approach to rapid fracture. The loading ratio, environment and history are part of the conditions under which growth data can be used.

State the fatigue growth law and its units

The Paris relation is da/dN = C(ΔK)^m, where N counts cycles and da/dN is crack half-length growth per cycle. It describes a fitted power-law portion of fatigue crack-growth behaviour. C and m depend on the material and test conditions; they are not determined by E or σ_Y alone. Near-threshold and near-final-fracture behaviour can depart from this relation.

For an illustrative active growth interval only, 1.00 mm ≤ a ≤ 4.00 mm, assign m = 2 and C = 1.00 × 10⁻⁸ m/cycle/(MPa√m)². Use the given positive tensile cycle with fixed R = 0.10 and Y_g = 1. The law is stipulated to apply throughout this model interval. No measured threshold, material identity, environment, frequency dependence, initiation stage or crack-face contact model is supplied.

The coefficient’s units matter. Since (MPa√m)² has unit MPa²m, C can equivalently be written 1.00 × 10⁻⁸ MPa⁻² cycle⁻¹ for m = 2. If ΔK is instead evaluated in Pa√m, C must be 1.00 × 10⁻²⁰ m/cycle/(Pa√m)². Retaining the MPa-based number while switching to Pa would multiply the predicted rate by 10¹².

At a₀ = 0.001 m, the assigned law gives da/dN = C(2.522246 MPa√m)² ≈ 6.36173 × 10⁻⁸ m/cycle = 0.0636173 μm/cycle. At a = 0.004 m the rate is four times as large, about 0.254469 μm/cycle. Each increment remains small relative to a in this interval, allowing an averaged continuous-cycle differential equation.

Integrate growth over the stated interval

Substitute ΔK = Δσ√(πa) into the supplied m = 2 law. With C expressed for MPa√m and Δσ = 45 MPa, da/dN = Cπ(Δσ)²a = βa, where β = 2.025 × 10⁻⁵π cycle⁻¹ ≈ 6.361725 × 10⁻⁵ cycle⁻¹. The applied stress range stays constant, but the growing crack raises ΔK and therefore the growth rate.

Separate variables: da/a = β dN. Starting at a = a₀ when N = 0 gives ln(a/a₀) = βN and a(N) = a₀exp(βN). The cycle count to reach a_f = 4 mm is N₁→₄ = ln(4)/β ≈ 21,791.17 cycles in this averaged model. Each doubling from 1 to 2 mm and from 2 to 4 mm takes ln(2)/β ≈ 10,895.59 cycles, even though the second doubling adds twice as much crack length.

The graph also shows the tangent based on the initial rate: a_tangent(N) = a₀ + βa₀N. At N₁→₄ this line reaches only [1 + ln(4)] mm ≈ 2.38629 mm, while the integrated model reaches 4 mm. Extending that initial rate unchanged would predict about 47,157.02 cycles for the same 3 mm increase, missing the acceleration of growth. The tangent is a local comparison, not an alternative calibrated fatigue law.

Stop the growth calculation at the declared 4 mm endpoint. It is a chosen end of the illustrative growth interval, not fracture: K_max there is about 5.605 MPa√m, below the supplied K_c = 10 MPa√m. Extending the same Paris relation to a_c ≈ 12.7324 mm would leave the stated interval and cannot produce a justified physical failure life. The cycle count includes no crack-initiation stage.

Keep yielding, crack growth and failure criteria distinct

The bar model follows irreversible strain and recovery; it has no crack or fracture variable. The monotonic crack model follows an existing flaw’s driving force and conditioned initiation criterion; it has no cycle-count law. The fatigue model follows crack growth per cycle over a supplied interval; it has no initiation stage or final rapid-fracture regime. Combining the words does not combine these distinct models automatically.

Yield stress, stiffness and crack-growth resistance have different roles and dimensions. A large E does not by itself imply a large yield stress or toughness. Remote stress below σ_Y does not rule out an intense local crack-tip field or cyclic growth. Unloading can remove the elastic stress while leaving a plastic strain or an existing longer crack. An elastic–perfectly plastic bar, a crack-tip process zone and a fitted fatigue curve retain different information.

A failure criterion might concern permanent dimensional change, crack extension, instability or loss of load-carrying function. The present examples teach the mechanics of selected changes. A real prediction needs the actual geometry, loading history, constitutive response and appropriately conditioned measurements. Other mechanisms, such as buckling or time-dependent deformation, are outside these calculations.

Worked examples

Try it yourself

1. Load a fresh ideal bar only to ε_max = 0.0015, then unload to zero stress. What residual strain does the stated model give?

   Show the answer and reasoning

   Zero residual strain.

   The peak is below ε_Y = 0.002, so σ = Eε = 150 MPa and ε_p remains zero. The entire strain recovers on the elastic branch.

2. At total strain ε = 0.004, compare stress on the initial tensile loading path with stress after unloading from ε = 0.006.

   Show the answer and reasoning

   Initial loading: 200 MPa. Completed unloading: 0 MPa.

   On initial loading, ε_e = 0.002 and ε_p = 0.002. After the longer loading history and unloading, ε_e = 0 and ε_p = 0.004. Equal total strain does not specify the history-dependent stress.

3. At fixed remote stress and fixed central-crack geometry, increase half-length from a to 4a. How do K_I and the plane-stress G change?

   Show the answer and reasoning

   K_I doubles and G quadruples.

   K_I is proportional to √a; G = K_I²/E is proportional to a. The full crack length is 2a both before and after; that factor does not change the scaling.

4. The assigned m = 2 coefficient is 1.00 × 10⁻⁸ m/cycle/(MPa√m)². What coefficient must accompany a numerical ΔK in Pa√m?

   Show the answer and reasoning

   1.00 × 10⁻²⁰ m/cycle/(Pa√m)².

   The numerical stress intensity increases by 10⁶ when MPa becomes Pa. Its square increases by 10¹², so C must decrease by 10¹² to preserve the growth rate.

5. Does the computed 21,791.17-cycle interval include crack initiation and final fracture? Explain what its two endpoints mean.

   Show the answer and reasoning

   It includes neither. It follows an existing half-length 1 mm crack to the chosen 4 mm end of the assumed growth-law interval.

   The initial flaw is supplied, and the final K_max ≈ 5.605 MPa√m remains below the assigned K_c = 10 MPa√m. Extending the same law beyond 4 mm or converting the interval into a physical fatigue life would require additional justified information.

What to carry forward

• Total strain splits into elastic and plastic parts; elastic unloading can leave nonzero permanent strain without fracture.
• Stress–strain work density is an area in J/m³, while crack energy release rate is measured in J/m².
• Crack stress intensity depends on geometry, loading and crack size, and has unit MPa√m rather than MPa.
• A conditioned monotonic crack-initiation criterion does not replace a fatigue-growth law.
• A Paris-law calculation follows propagation only within its justified interval and loading conditions; it does not automatically predict initiation or final failure.