Materials Science · Cracks, fracture and toughness
Cracks and stress concentration
Learn why the stress near a hole or crack differs from force divided by area. Compare nominal and local stress, calculate the effect of tip shape, and recognise when an elastic peak stops being a useful failure measure.
Useful background: Elasticity, plasticity and tensile strength. Each worked case explains the quantities it uses.
Back to the chapterAfter reading, you can
- Explain how a free boundary redistributes stress while the total applied force still balances.
- Distinguish gross-section average, net-section average and local peak stress, using a concentration factor with the matching reference area.
- Use the ideal elliptical-hole model to calculate a shape effect and explain its assumptions.
- Distinguish a finite-radius stress concentration from an ideal sharp-crack field, including the limits imposed by yielding, measurement and numerical resolution.
A crack redirects force
A hole or crack removes load-carrying material and bends the surrounding force paths. Stress can become much larger near a sharp crack tip than the remote average. Nominal force divided by total area therefore misses a consequential part of the problem.
Imagine pulling a thin strip with a small hole in its middle. Far from the hole, an approximately uniform axial stress is a useful description. At the hole, there is no material to carry traction across the empty opening. The surrounding material must carry the resultant force. A force-path sketch can help visualise that redistribution, but its lines are not physical fibres and are not a measured stress map.
The solution must satisfy both force balance and compatible deformation: connected regions cannot deform independently. The material law relates their strains and stresses. Those requirements create a varying field, not simply a uniform stress over a smaller area. This is why accounting for missing area is necessary but does not, by itself, calculate the peak.
Three stresses answer three different questions
Let F be the axial tensile force, W the strip width and t its thickness. On a cross-section well away from a central circular hole of diameter d, the gross area is A_g = Wt and the gross-section average is σ_g = F/(Wt). Across the centre of the hole, the remaining net area is A_n = (W − d)t and its axial average is σ_n = F/[(W − d)t]. Both expressions are averages of the axial stress over their respective load-carrying sections.
The local stress is the field at a specified position. Its largest relevant tensile value, σ_max, generally exceeds the net average because the load is not distributed uniformly beside the hole. A stress concentration factor is a ratio: K_t = σ_max/σ_nom. It is dimensionless. The symbol σ_nom must be defined; a factor based on σ_g cannot be combined with σ_n.
The two conventions describe one physical peak: K_t,g σ_g = K_t,n σ_n. Dividing by σ_g gives K_t,g = K_t,n W/(W − d). Thus a disagreement between two quoted factors may be a reference-area disagreement rather than a disagreement about the stress field. The chart or calculation must also match the hole geometry, loading and boundary conditions.
| Quantity | Definition | What it describes |
|---|---|---|
| Gross average σ_g | F/(Wt) | Average well away from the hole |
| Net average σ_n | F/[(W − d)t] | Average over the remaining section through the hole |
| Local peak σ_max | K_t,g σ_g = K_t,n σ_n | Largest relevant tensile stress from the selected elastic geometry model |
A round hole provides a checkable model
For an isolated circular hole in an effectively infinite, thin, homogeneous and isotropic plate under uniform remote uniaxial tension, the linear-elastic solution gives a maximum tensile stress of three times the remote stress: K_t,g = 3. The hole boundary carries no applied traction. The tensile maxima lie at the two sides of the hole perpendicular to the pulling direction. This result is a solution of that boundary-value problem, not a rule for every drilled or loaded hole.
The peak is local. Put R for the hole radius and x for distance from its centre along the line perpendicular to the tension. On that line, outside the hole, the load-direction stress satisfies σ_load/σ_∞ = 1 + (R/x)²/2 + 3(R/x)⁴/2. Here σ_∞ is the uniform remote stress. At x = R the ratio is 3; far away the added terms vanish. This checks that the formula recovers both the edge peak and the remote state.
A small hole and a larger geometrically similar hole have the same dimensionless elastic concentration in this infinite-plate model. They do not necessarily give the same component strength: finite ligaments, flaws, plasticity and crack growth introduce other questions. A finite-width plate needs a factor for its actual geometry. A pin pressing on the hole wall also changes the traction boundary and is not the unloaded-hole problem.
| x/R | σ_load/σ_∞ | Interpretation |
|---|---|---|
| 1 | 3.000000 | At the hole edge |
| 2 | 1.218750 | One radius beyond the edge |
| 3 | 1.074074 | Two radii beyond the edge |
| 4 | 1.037109 | Approaching the remote value of 1 |
An ideal sharp crack needs a different description
Linear-elastic crack models describe a stress field that grows strongly near an ideal sharp tip. Real tips have finite structure and a process zone where plasticity or other damage occurs. Use the model outside that zone under suitable conditions rather than interpreting an infinite ideal stress as a literal measurement.
In the elliptical-hole expression, letting ρ approach zero makes the predicted peak unbounded. A finite peak-to-average factor is therefore not a useful description of a perfectly sharp crack. Opening mode, also called mode I, means the two crack faces separate normal to the crack plane. In its elastic crack field, the leading stress varies as K_I/√r multiplied by factors for direction; r is distance from the tip. K_I, called the stress intensity factor, measures the amplitude of that field and has units of stress times square root of length. It is not the dimensionless K_t.
The elastic near-tip description needs a region outside the process zone and small compared with the relevant crack and component dimensions. Inside the zone, local plastic flow, bond separation or other damage changes the response. It is also wrong to assume that every stress component in a general three-dimensional plastic zone is simply capped at the uniaxial yield stress: the full stress state and constitutive response matter.
Whether a crack grows requires a driving-force and fracture-resistance assessment under the relevant loading, constraint and environment. This lesson establishes why the nominal stress alone is insufficient. Strength, toughness and fracture resistance develops the resistance distinction, and Specimen geometry and size effects develops how component dimensions enter the assessment.
Check what a calculated or measured peak means
A finite-radius elastic solution is useful only while its material and deformation assumptions are acceptable. If it predicts local yielding, continuing to multiply the force by the same elastic factor does not calculate the subsequent plastic stress distribution. Local yielding is also not identical to immediate collapse of the entire strip. Brittle fracture, fatigue initiation, stable damage and plastic collapse require different assessments.
Spatial resolution matters. A strain sensor averages over a finite footprint, so a reading taken near a sharp gradient can miss the actual peak. In a numerical model with a finite, resolved radius, mesh refinement should approach a finite elastic solution. In an ideal sharp-crack model, the reported nearest-point stress can rise as the mesh samples smaller r even while a correctly evaluated fracture parameter approaches a stable value. Treating the largest mesh number as a mesh-independent material strength confuses the model with the material.
Before accepting a comparison, write down the load case, geometry and radius; name the nominal area; identify the constitutive model and predicted yielding; and check the relevant failure mechanism. A stronger material may change resistance without changing the elastic geometric factor for this simple isotropic plane problem. A better shape may reduce the concentration without establishing an acceptable lifetime. The next lessons connect these separate pieces.
Worked examples
One strip, two nominal conventions and one peak
A long, thin plate has W = 100 mm, t = 2 mm and a central hole d = 10 mm. It carries F = 20,000 N in uniform axial tension at its distant ends. For this finite-width elastic model, a matching chart supplies the rounded net-section factor K_t,n = 2.72. Calculate the gross average, net average and predicted peak. What changes if the material begins to yield locally at about 250 MPa in the applicable stress state?
- Find the two areas: A_g = 100 × 2 = 200 mm² and A_n = (100 − 10) × 2 = 180 mm². Keep this supplied finite-width factor separate from the infinite-plate factor of 3.
- Calculate σ_g = 20,000/200 = 100 N/mm² = 100 MPa. Calculate σ_n = 20,000/180 ≈ 111.11 MPa. The smaller remaining area must carry the same axial force, so its average is larger.
- Use the convention attached to the factor: σ_max = 2.72 × 111.11 ≈ 302.22 MPa. The equivalent gross-section factor is 302.22/100 ≈ 3.0222. Either convention gives the same peak when used consistently.
- Multiplying 2.72 by the gross average would instead give 272 MPa, which is a bookkeeping error, not another physical prediction. The net-section factor was paired with the wrong area.
- The elastic prediction is above the stipulated local-yield level. It flags that the assumed wholly elastic state cannot be used unchanged at the peak. It does not calculate the ensuing plastic redistribution, prove immediate rupture or establish a safe allowable load.
100 MPa gross average, about 111.11 MPa net average and about 302 MPa elastic peak. The stipulated yielding calls for a suitable further assessment; elastic K_t alone does not establish component failure.
Change radius without changing the loading
In the ideal elliptical-hole family, keep a = 4 mm and σ_∞ = 10 MPa. Increase the tip radius from ρ = 0.25 mm to 1 mm. Predict the change before calculating it.
- A larger root radius means a less sharp tip at fixed a, so expect a smaller elastic peak. This is an elliptical-hole comparison, not a claim about every notch.
- At ρ = 0.25 mm, b = √(aρ) = √(4 × 0.25) mm = 1 mm. Thus K_t,g = 1 + 2 × 4/1 = 9 and σ_max = 90 MPa.
- At ρ = 1 mm, b = √(4 × 1) mm = 2 mm. Thus K_t,g = 5 and σ_max = 50 MPa.
- Quadrupling ρ halves the square-root contribution K_t,g − 1, from 8 to 4; it does not halve the entire factor. The unitless ratio a/ρ makes the expression dimensionally consistent.
- The removed elliptical area πab also changes, from 4π to 8π mm². The comparison isolates an elastic shape effect; a real component still needs its altered geometry and failure mode assessed.
The model peak falls from 90 to 50 MPa. The factor becomes 5 rather than 4.5 because it is K_t,g − 1 that halves.
Closer to an ideal tip is not an unlimited prediction
For one angular direction in a specified opening-mode crack model, the leading elastic stress is 80 MPa at r = 0.20 mm. Suppose that field is applicable outside a process zone extending to r = 0.02 mm. Estimate the leading stress at 0.05 mm. Can the same calculation establish the actual stress at 0.0125 mm?
- In this specified leading field, stress is proportional to 1/√r. Taking a ratio cancels its common amplitude and direction factor.
- At 0.05 mm, the distance is one quarter as large. The stress ratio is √(0.20/0.05) = 2, giving 160 MPa. Both specified sample distances are outside the supplied process zone.
- Blind extrapolation to 0.0125 mm would give 80√(0.20/0.0125) = 320 MPa, but that point lies inside the stipulated 0.02 mm zone.
- The value 320 MPa is an extrapolation of a model outside its stated domain. It is not a predicted measurement there. The process-zone constitutive response and crack-growth behaviour require additional information.
160 MPa is the leading-field estimate at the admitted distance of 0.05 mm. The actual stress at 0.0125 mm has not been determined.
Try the reasoning
A different thin plate has W = 60 mm, t = 2 mm, d = 12 mm and F = 6000 N. A geometry-matched elastic calculation supplies K_t,g = 3.14. Find σ_g, σ_n, σ_max and the equivalent K_t,n. Identify the error in multiplying 3.14 by σ_n.
Show the answer and reasoning
σ_g = 50 MPa; σ_n = 62.5 MPa; σ_max = 157 MPa; K_t,n = 2.512. Using 3.14 with the net average would incorrectly give 196.25 MPa.
The gross and net areas are 120 and 96 mm². The supplied factor uses 6000/120, so its peak is 3.14 × 50. Divide that same peak by 62.5 to convert conventions. Both correct ratios describe the same field; changing the reference area does not create a second peak.
Keep a = 4 mm in the ideal transverse elliptical-hole model. A designer claims that increasing ρ from 0.25 to 1 mm halves K_t. Calculate the two factors and explain the mistaken step.
Show the answer and reasoning
The factors are 9 and 5. The square-root contribution halves, but the entire factor does not.
K_t,g = 1 + 2√(a/ρ). Its variable part changes from 8 to 4, with the constant 1 retained. This also distinguishes a tip-curvature change from simply changing a whole part’s scale.
For the ideal circular-hole model, the remote stress is 40 MPa. Use the supplied dimensionless table to estimate the load-direction stress at x = R and x = 2R. Explain why a sensor covering a region around 2R cannot directly report the edge peak.
Show the answer and reasoning
120 MPa at R and 48.75 MPa at 2R. The positions sample different parts of the field, and a finite sensor averages over its footprint.
Multiply the same remote stress by 3 and 1.21875. The sharp spatial decrease is the relationship the diagram represents. Neither the reading at a nearby position nor a footprint average is automatically the maximum at the boundary.
An ideal sharp-crack simulation reports a growing nearest-point stress each time its mesh is refined, while its independently evaluated K_I settles to a stable value. Give a model-based explanation and name the extra evidence needed to assess crack growth.
Show the answer and reasoning
Refinement samples the singular elastic field closer to the ideal tip; the nearest-point stress is not a converged finite material strength. Crack-growth assessment needs the relevant resistance and a model whose process-zone, geometry, loading and constraint assumptions are acceptable.
K_I describes field amplitude rather than a finite tip peak. Its numerical convergence is useful evidence about the selected elastic model, not proof that the model applies or that the component can tolerate the crack.
Two otherwise identical isotropic plates have different yield and fracture resistance. Under the same small-strain, wholly elastic tensile load, their circular-hole models both give K_t,g = 3. A learner concludes their failure loads and service lives must be equal. Explain what the common factor establishes and what remains unknown.
Show the answer and reasoning
The common factor establishes the same normalised elastic concentration for this geometry and loading. It does not establish equal yield onset, crack resistance, fatigue response or service life.
A geometric elastic amplification and a material’s resistance are separate inputs. A valid failure or lifetime comparison must select the actual mechanism and conditions rather than substitute K_t for strength or toughness.
What to carry forward
- Force balance fixes a section resultant; it does not make stress uniform near a hole or crack.
- Always pair a concentration factor with its stated nominal area and loading model.
- Relative tip sharpness can matter strongly: in the ideal transverse ellipse, K_t,g = 1 + 2√(a/ρ).
- A sharp-crack field, a local yielding calculation and a crack-growth assessment answer different questions.
- Check model domain and spatial resolution before interpreting a very large calculated or measured stress.
References
Teaching resources and research supporting this lesson’s explanations and examples.