Materials Science · Ceramics, glasses and inorganic networks

Porosity, defects and brittle failure

A pore reduces the solid carrying load and can concentrate stress depending on shape and location. A sharp crack can be more severe than a rounded pore of similar overall size. Average porosity alone does not describe the most critical flaw.

Useful background: Atoms, bonding, structure and defects; Energy, stability and environmental context. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Connect pores and crack populations with ceramic strength evidence.
Separate total pore volume, access from an exterior surface and a through-connected path.
Use a mass–volume balance without equating bulk density with a complete defect description.
Compare defect shape and local loading without assigning a universal strength–porosity law.
Read scatter and fracture-origin evidence without treating a sample fraction or mean as a component guarantee.

Missing solid changes the load path

Two ceramic pieces have the same mass, outer volume and phase composition. A liquid enters pores in one but not the other. Their measured failure stresses also need not be equal. Which information did the density comparison leave out?

A pore is a region not occupied by the solid framework. Define total porosity φ = V_p/V_b, where V_p is all pore volume and V_b is the outer bulk volume, including the pores. Both volumes must refer to the same specimen state. A fraction of 0.04 is 4%, not 0.04%. The amount alone does not describe the shape, largest opening, spatial clustering or route to the exterior.

Open pores connect to an exterior surface; closed pores do not. An open dead-end pore can fill without providing a path between two opposite faces. Through connectivity is a further question, and a connected route alone does not determine its permeability, the ease of flow under an applied pressure difference. Narrow throats, tortuosity and the fluid’s properties matter. Controlled pores can be useful for a filter or catalyst support even when a dense structural support calls for a different pore state.

Count equal pore volumes, then test connectivity

The first figure is an original exact geometric model. Each specimen is a 10 mm cube divided into 5 × 5 × 5 cubic cells of side 2 mm. A cell is either completely solid or completely empty. All solid cells have the same stipulated pore-free density ρ_s = 3.20 g/cm³. The diagrams show every layer, not just one attractive section. They are not microscope data, an instrument resolution claim or an elastic stress model for cubic pores.

Model I has five separate interior empty cells. Model T has five face-connected empty cells forming a straight passage from the left face to the right face. In each, V_b = 1000 mm³ = 1.000 cm³, and V_p = 5 × 8 = 40 mm³ = 0.0400 cm³. Thus φ = 4.00% in both. I has no open pore volume; all of T’s pore volume is open and belongs to a through path. Connecting at a point or an edge is not treated as an open face in this model.

Neglect pore-gas mass. Each solid volume is V_s = V_b − V_p = 0.960 cm³, so each dry mass is ρ_s V_s = 3.072 g. Its dry bulk density is ρ_b = m/V_b = 3.072 g/cm³. Rearranging m = ρ_s(1 − φ)V_b gives φ = 1 − ρ_b/ρ_s. This inference needs the independent pore-free solid density: a changed phase mixture or an incorrect ρ_s can imitate a porosity change.

In an ideal filling comparison, a compatible liquid of density 1.00 g/cm³ completely fills all exterior-accessible empty cells, and the retained liquid amount is determined with no trapped gas, drainage, surface film, absorption into the solid or reaction. I gains no liquid mass; T gains 0.0400 g, or 40.0 mg. The liquid-filled mass of T is 3.112 g. These exact model balances are not a complete immersion-test procedure.

For a real specimen, incomplete penetration can undercount accessible pores, while retained surface liquid can overcount them. State the liquid, temperature, filling protocol and volume measurement. A measured open fraction also needs its operational method; it does not identify a through-flow path by itself. Combine suitable three-dimensional imaging or a flow test with the question actually being asked.

Two exact original geometric voxel models, not measured microstructures. Each bulk cube is 10 millimetres per side, divided into 125 cubic cells of side 2 millimetres. Every layer z from zero to four is shown. Within a grid, x increases rightward and y upward; coordinates are zero based. Hatched blue cells are void and plain cells are solid. In model I the empty coordinates are (1,1,1), (3,1,1), (1,3,1), (3,3,3) and (1,1,3): five isolated closed interior cells. In model T the empty coordinates are (x,2,2) for every x from zero to four: a face-connected channel between opposite x faces. Both have 40 cubic millimetres of void in 1000 cubic millimetres of bulk volume, so total porosity is 4.00 percent. With stipulated solid density 3.20 grams per cubic centimetre and negligible pore-gas mass, both dry masses are 3.072 grams and dry bulk densities 3.072 grams per cubic centimetre. Under the ideal filling assumptions, I takes up no liquid and T retains 40.0 milligrams of a liquid of density 1.00 gram per cubic centimetre. Equal total porosity does not establish equal access, flow or strength.

A small amount of void can still contain a severe shape

An empty pore surface cannot carry the solid’s traction across the opening. The surrounding framework redirects the load; nearby tensile stress is generally nonuniform. The nominal stress reported from a test geometry and the stress in a small solid ligament are different quantities. “Bonding and structures in ceramics” develops an intact-versus-hole section account. Here, equal total missing volume is the comparison, not another net-section calculation.

Compare two separate ideal planar profiles, each extruded through the same depth 0.50 mm. R is a circle of radius R = 0.20 mm. F is an ellipse with horizontal semiaxis a = 0.80 mm and vertical semiaxis b = 0.050 mm. A semiaxis is half the full span. Their void areas are πR² = πab = 0.0400π mm², and their void volumes are both 0.0200π ≈ 0.06283 mm³. This equal-volume comparison is independent of the coarse voxel models.

At the major-axis end of an ellipse, the geometric radius of curvature is ρ_tip = b²/a. It is the radius of the locally matching circle, not the ellipse’s half-width. Here it is 0.003125 mm = 3.125 μm, compared with 200 μm for the circular profile: 64 times smaller. The elliptical profile also extends farther along its major axis. Equal void volume has hidden different lengths and sharpness.

With remote tension directed vertically, opening across the long horizontal profile can produce a strong local tensile concentration at its ends in an appropriate elastic model. Rotating the flaw, moving it into a lower-stress region, changing neighbouring pores or introducing residual stress changes the comparison. A pore that touches another pore or breaks into a surface can leave a very small ligament or a sharp cusp; counting its volume alone misses that feature.

The curvature calculation supplies geometry, not a fracture stress, a toughness or a universal stress-concentration factor. A mathematically sharp crack requires its own driving-force and resistance description; a blunt pore is not automatically equivalent to that crack. The later “Cracks and stress concentration” lesson provides those models and limits. Ceramic fracture may occur before enough plastic rearrangement redistributes a local stress, but neither hardness nor porosity alone establishes the outcome.

Strength is a population question

Specimens can fail at different stresses because their flaw populations differ. Surface finishing, stressed volume and loading mode can change what is sampled. Use distributions and relevant specimen sizes rather than treating one exceptionally strong small specimen as the strength of every component.

Use the separate original failure record below. N and W are two small sets of six matched four-point-bend coupons. Their total porosity is supplied as 4.00%, with matched phase/grain state, outer geometry, nominal finishing specification, loading rate, direction, temperature and environment. The actual retained defects can still differ. Each listed σ_f is the test’s supplied nominal maximum tensile stress at fracture, in MPa; it is not a uniform stress everywhere in the beam or a tensile-strength conversion. These invented values are not predictions for I, T, R or F and are not a named ceramic datasheet.

N gives 140, 145, 150, 155, 160 and 165 MPa. W gives 100, 130, 150, 160, 180 and 195 MPa. Each sum is 915 MPa and each arithmetic mean is 152.5 MPa. Yet their ranges, maximum minus minimum, are 25 and 95 MPa. By a nominal applied stress of 135 MPa under this monotonic test convention, none of N and two of W have fractured. Equal means and porosity have not described equal lower tails.

The empirical fraction failed by a chosen stress is the number with σ_f at or below that stress divided by the number tested. At 135 MPa it is 0/6 for N and 2/6 ≈ 33.3% for W. Those are observed fractions of these records. Zero observed failures is not zero population failure probability, and six observations do not determine a reliable tail model. The graph shows individual values and an empirical staircase rather than fitting a universal distribution.

A larger similarly processed specimen can sample more opportunities for a critical flaw. In bending, only part of the body has the largest tensile stress; edges and surfaces can have their own flaw populations. Changing specimen volume, stressed surface, stress gradient or orientation changes which flaws compete. The relevant weakness is the combination of flaw severity, local loading and resistance, not necessarily the physically largest pore. A volume ratio alone cannot rescale this small dataset.

Original invented matched-coupon failure record. Sorted values are nominal maximum tensile stresses from the same four-point-bend method, not a universal strength–porosity fit. W-origin labels are separately supplied examination results, not inferred from the stresses.
Sorted rank	N: σ_f (MPa)	W: σ_f (MPa)	Supplied W origin assessment
1	140	100	Edge machining chip
2	145	130	Touching-pore cusp
3	150	150	Surface pore
4	155	160	Subsurface grinding crack
5	160	180	Unresolved
6	165	195	Fine microstructural region

The top panel shows a circular profile R of radius 0.20 millimetre and an elliptical profile F with semiaxes 0.80 and 0.050 millimetre, at 100 drawing pixels per millimetre. Both are extruded through 0.50 millimetre depth, giving void volume 0.0200 times pi, about 0.06283 cubic millimetre. The ellipse major-axis tip curvature radius is 3.125 micrometres, compared with 200 micrometres for the circle. This is geometry only, with no stress field or failure prediction. Separate lower panels show original invented matched-coupon fracture stresses in megapascals: N, 140, 145, 150, 155, 160 and 165; W, 100, 130, 150, 160, 180 and 195. Blue solid lines and circles denote N; orange dashed lines and squares denote W. Both arithmetic means are 152.5 megapascals; ranges are 25 and 95 megapascals. The empirical staircase counts values at or below the applied stress, with closed markers at the upper side of each jump. At 135 megapascals the observed failed fractions are zero of six and two of six; at 150 megapascals both are three of six. These are finite invented records, not fitted population laws, component qualification or results predicted from the drawn profiles or voxel models.

Finding a pore on a fracture is not finding the origin

Fractography examines the broken pieces and fracture surfaces to investigate where fracture began and how it spread. Preserve the overall pieces, their orientation and loading history before selecting a tiny feature. A fracture can pass through a pore that did not initiate it. Locating the origin and characterising that initiating defect are related but distinct tasks.

In the invented W record, independent examination has identified an edge machining chip at the origin of the 100 MPa specimen and a touching-pore cusp at the 130 MPa origin. These support different processing investigations, although both specimens have the same total porosity. The unresolved 180 MPa origin must remain unresolved. Do not force every low-strength result into a pore category or label a particle chemically from its dark appearance alone.

A real feldspathic-porcelain study found several competing origin types, including pores, machining damage and inclusions. A seemingly good fit of the combined strengths to one distribution concealed those different populations. Polishing also did not remove all earlier subsurface preparation damage in those specimens. This is a bounded experimental counterexample to a one-number interpretation, not a claim that polishing always weakens ceramics or that isolated internal pores are harmless in every material.

To test a proposed cause, compare origin location with the tensile stress field, distinguish a pre-existing flaw from damage created during final breakage, and retain actual preparation and environment. Surface finishing can remove some damage, introduce other defects or change residual stress. A changed strength record requires that evidence; a smoother-looking surface alone does not supply it.

Match the evidence to the function

For a filter, ask whether the relevant faces connect and what flow or particle-retention performance is measured. For a load-bearing part, ask which defect and local stress control failure under the actual service conditions. An open-pore fraction is not permeability, and neither total nor open porosity alone is strength.

For a real strength comparison retain specimen dimensions, loaded surface, orientation, finishing, residual stress, environment, rate and fracture origin alongside all valid results. Separate material scatter from test uncertainty and distinguish a sampled maximum from a lower-tail requirement. “Microscopy, diffraction and structure evidence” and “Sampling, scale and extrapolation limits” develop measurement and representativeness; “Lifetime evidence, scatter and extrapolation” develops population inference and finite records.

These geometric models and small supplied failure records teach how to find a missing comparison. They provide no minimum safe load, component failure probability, universal porosity coefficient or process recipe. The useful chain is pore amount and geometry → actual load path and flaw competition → measured failure evidence under matched conditions.

Worked examples

Work through the case

Two batches have the same pore fraction, but one contains a few large sharp defects. Should their strengths be assumed equal?

Match total porosity.
Compare defect shape, size distribution and location.
Assess the largest relevant flaws and loading.
In the separate N/W record both porosities are 4.00% and both means are 152.5 MPa, yet the observed minima are 140 and 100 MPa. Different shape, location and surface damage remain consequential, so density and a shared mean do not establish identical failure behaviour.

No. Equal pore fraction leaves consequential flaw differences unresolved. The separate small records demonstrate what those averages can hide, not a universal ranking for every future specimen.

Equal dry mass can conceal a different path

For I and T, calculate total porosity, dry mass and bulk density. Under the stated ideal filling conditions, calculate the liquid mass increase. Would zero liquid uptake prove zero total porosity?

The outer cube has volume 10³ = 1000 mm³ = 1.000 cm³. Five empty 2 mm cells have volume 5 × 2³ = 40 mm³ = 0.0400 cm³, so each total porosity is 40/1000 = 0.0400 = 4.00%.
The solid volume is 0.960 cm³, giving dry mass 3.20 × 0.960 = 3.072 g. Divide by bulk volume to get 3.072 g/cm³ in both. Recovering φ = 1 − 3.072/3.20 also gives 4.00%.
I’s five interior cells are closed; no liquid enters. T’s passage is accessible, so full filling adds 1.00 × 0.0400 = 0.0400 g = 40.0 mg. Its filled mass is 3.112 g.
I disproves the zero-uptake inference even in the perfect model. In a real test, inadequate penetration could also produce low uptake. Access, total void volume and a through route must be distinguished.

Both have 4.00% total porosity, 3.072 g dry mass and 3.072 g/cm³ dry bulk density. Ideal liquid gains are zero and 40.0 mg. Zero uptake does not establish zero total porosity.

The same missing volume does not give the same tip shape

For the circular and elliptical profiles, verify equal void volume through 0.50 mm depth. Compare their major-axis tip curvature radii. Can the radius ratio be used as a strength ratio?

The circle’s area is π(0.20)² = 0.0400π mm². The ellipse’s area is π(0.80)(0.050) = 0.0400π mm². Multiplying each by 0.50 mm gives 0.0200π ≈ 0.06283 mm³.
The circle’s boundary curvature radius is its 0.20 mm radius. For the supplied ellipse use b²/a: 0.050²/0.80 = 0.003125 mm = 3.125 μm.
The ratio is 0.20/0.003125 = 64. The smaller local radius and longer profile are hidden by the shared void volume. Their mechanical significance also depends on direction, nearby boundaries, local stress and the fracture mechanism.
No strength law is supplied. Neither dividing a failure stress by 64 nor declaring that the tip stress is 64 times larger follows from a curvature-radius ratio alone.

Both void volumes are about 0.06283 mm³. The ellipse’s major-axis tip radius is 64 times smaller; this geometric ratio is not a strength or stress ratio.

An equal mean need not meet the same screen

Calculate each mean and range in N/W. A limited arithmetic screen asks whether every listed specimen fractured at a nominal stress of at least 140 MPa. Which record passes, and what has not been qualified?

Each set sums to 915 MPa, so each mean is 915/6 = 152.5 MPa. N’s range is 165 − 140 = 25 MPa; W’s is 195 − 100 = 95 MPa.
All six N results are at least 140 MPa. W has four at or above 140 MPa and two below. The arithmetic screen therefore passes N and rejects W, despite equal means.
For the different endpoint 135 MPa, two W specimens and no N specimens have already fractured; these observed fractions are 2/6 and 0/6. Do not replace a minimum requirement with a high mean or maximum.
This is a screen of the listed records, not a guarantee for the untested population or an allowable service stress. Other specimen sizes, flaw distributions, uncertainty, environment and time under load are not resolved by these six endpoints.

The means are both 152.5 MPa, but ranges are 25 and 95 MPa. Only N passes the all-listed-results screen. Neither record alone qualifies a component or a population failure probability.

Try the reasoning

Why can polishing change measured strength?
Show the answer and reasoning
It changes surface flaws and possibly residual stress.
Surface condition can govern crack initiation. Removing roughness need not remove a deeper grinding crack; preparation and fracture-origin evidence are needed to distinguish removal, newly introduced damage and a residual-stress change.
Does a single failure stress describe scatter?
Show the answer and reasoning
No.
Multiple representative tests are needed. N and W share a mean but have different observed lower tails. Their small observed fractions do not establish a fitted distribution or population guarantee.
A new dry specimen has bulk volume 4.000 cm³, independently known pore-free solid density 3.20 g/cm³ and dry mass 12.032 g. Compatible liquid of density 1.00 g/cm³ fills all accessible pores with the ideal corrections specified above, adding 0.120 g. Find total, open and closed porosity. Does the open fraction prove a through-flow path?
Show the answer and reasoning
The solid volume is 3.760 cm³, so total pore volume is 0.240 cm³ and total porosity 6.00%. Accessible pore volume is 0.120 cm³, giving 3.00% open and 3.00% closed porosity. A through path is not established.
Use mass/solid density before subtracting from bulk volume. The mass increase measures only the stipulated filled accessible volume. Those pores could all be dead ends; their open fraction alone gives no permeability.
Starting from model I, remove one additional 2 mm cell at coordinates (0,1,1), adjacent to its existing (1,1,1) pore and the left exterior face. Use the zero-based coordinates defined in the figure. Find total, open and closed porosity under face connectivity. Is there now a left-to-right path?
Show the answer and reasoning
There are six empty cells: total porosity 6/125 = 4.80%. The new face cell and its one connected interior neighbour give 2/125 = 1.60% open porosity. Four isolated cells remain closed: 3.20%. There is no right-face connection.
Opening one access cell exposes an already existing void as well as adding new void volume. Total and open porosity therefore need not change by the same amount. The face-connected pair is a dead end; the remaining original voids stay separate.
A different equal-area ellipse has a = 0.40 mm and b = 0.10 mm. Compare its area and major-axis tip radius with R and F, at the same 0.50 mm depth. Must its failure stress lie halfway between theirs?
Show the answer and reasoning
Its area remains 0.0400π mm² and void volume 0.0200π mm³. Its tip radius is 0.0250 mm = 25.0 μm: eight times smaller than R’s radius and eight times larger than F’s. No halfway failure stress follows.
The product ab controls area while b²/a controls local curvature. Neither quantity is a supplied fracture criterion. A mechanical prediction still needs the actual loading, boundaries, defect interaction and resistance.
Read the N/W empirical staircases at 150 MPa, counting fracture at the threshold as failed. What fraction has failed in each? Does equality at that stress mean the distributions are identical?
Show the answer and reasoning
Three of six specimens have failed in each: 50%. The empirical records are still different.
At 135 MPa the fractions are 0 and 2/6, while their high ends also differ. One common point, like one common mean, does not establish identical distributions or a common population law.
A supplied examination record locates fracture initiation at an edge chip. A large smooth pore lies farther along the subsequently formed fracture surface. A report blames that pore because it is the largest photographed feature. What conclusion and follow-up are justified?
Show the answer and reasoning
The supplied origin evidence supports investigating the edge chip; the later encountered pore is not thereby the initiator. Check the overall fracture path, matching pieces, local loading and origin identification rather than selecting the largest image feature.
A crack can intersect defects after it has started. Feature size alone does not establish temporal order or a causal fracture origin. If origin evidence is uncertain, retain that uncertainty instead of claiming either mechanism proved.
A component has ten times a coupon’s bulk volume. A student divides the N mean of 152.5 MPa by ten to predict component strength. Another transfers the coupon’s zero observed failures at 135 MPa unchanged. Are either justified?
Show the answer and reasoning
Neither is justified. No volume-to-strength law or transferable population failure probability has been supplied.
The stressed surface/volume, stress gradient, actual flaw populations, size and surface interactions, environment and test uncertainty need a justified comparison. A larger body can expose more flaw opportunities without strength being inversely proportional to volume.

What to carry forward

Describe flaw populations beyond average porosity.
Keep specimen geometry and surface state with strength results.
Total pore fraction, exterior access and through connectivity are different descriptors.
An equal volume can hide different profile lengths, curvature, clustering and remaining ligaments.
A mean, maximum or observed sample fraction does not by itself describe a population lower tail.
Keep an identified fracture origin separate from a pore merely encountered by the final fracture surface.

References

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