Materials Science · Composites, fibres and interfaces

Damage, delamination and interface failure

Useful background: Units, loads, stress and strain; Atoms, bonding, structure and defects; Processing, structure and useful properties. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish constituent failure from interlaminar separation.
Trace how a matrix crack, fibre break or lost interface changes a remaining load path.
Explain why interlayer sliding can change bending and compression stability without the same change in axial stiffness.
Interpret a damage map and a matched residual-response record without inferring a universal strength loss from appearance or area.

Several damage paths can compete

A laminate can experience matrix cracking, fibre fracture, interfacial debonding or separation between layers called delamination. Each changes the remaining load paths. An intact outer surface does not prove an undamaged interior.

A ply, or lamina, is one layer of a laminate; its fibres can have a specified direction. The lay-up describes the sequence, thicknesses and orientations of the plies. A 0-degree ply has fibres parallel to a chosen reference direction; a 90-degree ply is across it. The same constituents and total fractions can therefore have different internal load paths.

A matrix crack interrupts a path through the surrounding material. Intact fibres crossing that crack may bridge it and transfer force through nearby interfaces. A broken fibre removes its own continuous axial path; neighbouring material must receive the force it formerly carried. Fibre–matrix debonding removes part of the tangential connection that loaded that fibre. Delamination separates neighbouring plies; their fibres need not all have broken.

At the edge of a separated region, force that previously passed through a distributed connection must be redirected through the remaining connected material. Stress and deformation can become nonuniform there. This is why naming the damage and locating it relative to fibres, surfaces and loading is more useful than calling the whole specimen simply “cracked”. Constituents and composite architecture, together with Load transfer through fibres and interfaces, supplies those distinct paths.

The order of damage matters

Some microdamage redistributes load before final failure; other damage rapidly destabilises the structure. Evaluate loading mode, lay-up, defects and residual stress. A single tensile strength cannot capture every impact, compression or interlaminar failure mode.

One possible impact sequence is local matrix cracking, a crack reaching an interface, and separation spreading along that interface. Another event may break fibres directly or leave several separated interfaces. The first visible mark does not determine that order, and this is not a universal sequence for every laminate. Fibre direction, matrix resistance, interface state, thickness, supports and impact history can change the outcome.

An interior separation can leave a thin group of plies, called a sublaminate, less supported by the material underneath. Under compression that group may deflect sideways locally: buckling. Opening and sliding at the separation edge can then change the force and energy available for further growth. The first buckling event, subsequent crack growth and final loss of load capacity are different events.

The retained schematic identifies a possible hidden separation; it is not a measured section, damage-size calibration or buckling solution. An intact outer surface is compatible with different interior states. Conversely, a large visible mark need not by itself quantify the remaining compressive response.

Stacked layers include a schematic internal separation. Delamination can change support between layers without requiring every fibre to break.

Axial stiffness can miss a lost bending connection

Use an original ideal comparison of two identical, straight, rectangular elastic strips. Each has width b = 20.0 mm, thickness h = 1.00 mm and length L = 100.0 mm. Its fictional axial modulus is E = 40.0 GPa = 40,000 N/mm². Compare perfect bonding over the complete interface with complete separation into two strips that can slide axially without friction. All dimensions are initial values; there are no cracks within either strip.

For a small uniform axial compression before any instability, stipulate symmetric end loading so both strips have the same axial strain and remain straight. Complete area is A = 2bh = 40.0 mm². The axial stiffness K = EA/L is 16,000 N/mm for either ideal state. A 100 N total force gives shortening 0.00625 mm and average stress 2.50 MPa. Losing the interlayer bond does not have to change this particular axial response when the two strips still receive the same end strain.

Bending is different. In the elementary beam picture, axial strain varies across thickness: one side lengthens while the other shortens. The neutral axis is the line through the section where that bending strain is zero. With complete bonding, the two strips act as one thickness 2h, with one neutral axis. With free interlayer sliding, each strip can bend about its own neutral axis. Material far from a shared neutral axis contributes strongly to the bending moment; removing that connection changes its contribution.

For a uniform rectangular section the second moment of area about its central bending axis is I = bH³/12, where H is the thickness in the bending direction. I has units of length to the fourth power; it is a geometrical quantity, not a mass moment or a material modulus. The bending rigidity EI relates bending moment to curvature in this linear slender-beam model. A larger EI needs more moment to obtain the same small curvature.

The bonded section has I_bond = b(2h)³/12 = 13.3333 mm⁴. The sum for freely sliding strips is I_sep = 2bh³/12 = 3.33333 mm⁴. Thus (EI)_sep/(EI)_bond = 1/4 even though total material, area and axial stiffness are unchanged. Equivalently, each bonded half contributes its own central I plus its area times (h/2)² because it lies away from the combined neutral axis; the offset contributions are absent when the strips slide freely. For this bending comparison, impose the same small curvature on both freely sliding strips through coordinated loading and end motion, then add their bending moments. Each strip bends about its own central neutral axis; complete separation alone does not impose a common curvature under arbitrary independent loading.

This compares two fully specified whole-strip ideal states under small bending, with equal constituent E and no transverse shear deformation, friction, bridging or partial connection. It does not assign a quarter of the bending rigidity to any real local delamination. Real plies may be anisotropic, coupled and unequally thick; contact, restraint and remaining bonded regions also matter.

Original whole-strip comparison, not measured laminate data. Both states contain the same two intact 20 mm × 1 mm strips with E = 40 GPa and L = 100 mm.
Response quantity	Perfectly bonded	Freely sliding strips
Complete area A	40.0 mm²	40.0 mm²
Uniform axial K = EA/L	16,000 N/mm	16,000 N/mm
I about operative neutral axis/axes	13.3333 mm⁴	3.33333 mm⁴ summed
Bending rigidity EI	533,333 N·mm²	133,333 N·mm² summed
Ratio to bonded bending rigidity	1	0.250

Original ideal cross-sections of two identical intact strips, with equal modulus E, width b and thickness h per strip. Perfect bonding makes one thickness 2h and one neutral axis. Complete frictionless axial sliding permits two separate central neutral axes; the visible gap is only a schematic separation. Orange dashed lines identify zero bending strain, not cracks. Bonded I is b multiplied by the cube of 2h, divided by 12; the separate central I values sum to 2b multiplied by the cube of h, divided by 12. Bending rigidity ratio is 1/4 while complete axial EA is unchanged under the stated uniform equal-strain straight loading. The normalised bending bars have widths 400 and 100 pixels on one scale. This whole-strip model is not a quarter-stiffness rule for actual local delamination and predicts no final failure load. The two freely sliding strips are stipulated to share the same small curvature through coordinated loading and end motion, so their moments can be added; arbitrary independent loads need not impose that curvature.

A stability model supplies an onset, not a breaking strength

For a sufficiently slender ideal elastic member, a small-deflection buckling model gives an onset force proportional to EI/L². Write P_cr = C EI/L², where C is a dimensionless number set by the specified end and mode constraints. This tells us that a longer unsupported path can become unstable at a smaller force, even when the material’s local stress is below its breaking threshold. It does not supply that breaking threshold.

If two compared whole-strip models have the same effective length and corresponding end/mode constraints, so the same C applies, their onset ratio is the bending-rigidity ratio: P_cr,sep/P_cr,bond = 1/4. If an independent comparison changes the effective length as well, use the complete ratio (EI)_2/(EI)_1 × (L₁/L₂)². Merely saying that both contain two layers does not fix either factor.

A local delamination is not generally a pair of independent full-length columns. Its ends remain connected to surrounding plies; contact, anisotropy, initial curvature, residual stress and load sharing can matter. Buckling can occur without immediate growth, and postbuckling can change the driving force for growth. A resistance criterion and a suitable nonlinear structural model would be needed to predict propagation or final failure.

This is why neither an unchanged tensile stiffness nor a first local bulge settles a real laminate’s residual compressive strength. Test state, supports, direction and the endpoint being reported must be identified. The ideal strip calculation explains a mechanism; it supplies no allowable load or handling decision for an impacted component.

A projected damage footprint is a different quantity from separated interface area

Consider two original invented rectangular separated regions on different interfaces. In a common plan view, each is 15.0 mm long by 10.0 mm wide, so each interface has 150 mm² separated area. Their projections overlap over 5.00 mm by 10.0 mm, or 50.0 mm². The area covered by either projection is therefore 150 + 150 − 50.0 = 250 mm².

Adding the two interface areas instead gives 300 mm². That count distinguishes the two depths, including their common projected region twice because two different interfaces are separated there. It does not double each interface to count both exposed faces. The union footprint, the sum over interfaces, crack-front length and number of damaged layers answer different questions. State which is being reported.

The map is a stipulated geometry, not a claim that one inspection reconstructs every plane. An indication in an ultrasound or radiographic record must be interpreted with its calibration, sampling direction, resolution and other possible causes of contrast. A reported footprint without resolved depth is not proof of one separated interface, and no detected indication is bounded by the method’s sensitivity.

Even if all geometric quantities were known, no universal residual-strength ratio follows from 250 or 300 mm². A near-surface separation may leave a thin sublaminate; another at a different depth may change support differently. Multiple matrix cracks and fibre breaks are also not represented by one clean embedded separation. An appropriate damage model and matched residual-response evidence are still needed.

Original stipulated two-depth separated-interface map, not an actual inspection. At 10 pixels per millimetre, rectangle A is 15 by 10 mm and rectangle B is also 15 by 10 mm, offset 10 mm horizontally. Their overlap is 5 by 10 mm, 50 square millimetres. Blue outline identifies A; orange hatching identifies B at a different depth. The projected union is 250 square millimetres, while the sum of separated areas over the two interfaces is 300 square millimetres. The depth schematic has arbitrary exaggerated vertical spacing. Each interface is counted once; the sum does not count both crack faces. These stipulated areas neither establish the resolution of an imaging method nor predict crack growth, constituent damage or remaining strength.

Match the residual-response comparison to the question

Use a separate wholly invented matched coupon comparison. Original and post-impact specimens have the same lay-up, complete nominal section 40.0 mm², temperature, conditioning and compression supports. Their small-load axial stiffnesses are both supplied as 16,000 N/mm. The undamaged specimen reaches the reported compressive failure endpoint at 20.0 kN and the impacted specimen at 14.0 kN. These are supplied observations, not predictions from the beam or damage-area models.

The retained endpoint force fraction is 14.0/20.0 = 0.700, or 70.0%; the reduction is 30.0% relative to the original endpoint. Dividing by 40.0 mm² gives nominal stresses 500 and 350 MPa. Equal small-load stiffness and a lower failure endpoint are compatible: they describe different parts of the response and do not uniquely identify the damaged mechanism.

A useful investigation links an internal-state observation, a measured response and a proposed mechanism. A resolved separation plus a local compressive bulge would support investigating loss of interlayer support. It would not exclude matrix cracks, fibre damage or another limiting region. State which evidence distinguishes competing explanations and which uncertainties remain.

Keep a calibrated specimen endpoint separate from a service allowable, population guarantee or diagnosis of a particular component. Repeated tests, variability, environmental history and relevant loading modes may change the conclusion. Controlled embedded flaws can help isolate a mechanism, while an impact can create several interacting damage types; matching only their projected areas does not make the damage states equivalent.

Worked examples

Work through the case

After impact, a laminate looks nearly intact but has reduced compressive capacity. What mechanism deserves investigation?

Look for internal delamination and matrix damage.
Damaged layers can lose support under compression.
Use appropriate inspection and residual-strength tests.
Distinguish the proposed sequence from a diagnosis: matrix cracks and interlayer separation can leave a less supported sublaminate, but appearance alone does not establish them.
Compare the actual loading direction, supports, geometry and state. The separate ideal strip case shows why the same small axial stiffness need not imply the same bending stability.

Internal damage can alter stability even when surface damage appears small. A measured internal state and a matched residual compression response are needed to test that mechanism.

Keep area, axial response and bending response separate

For the two fictional 20 mm-wide,1 mm-thick strips at E = 40 GPa and L = 100 mm, calculate axial K and the bonded/free-sliding bending-rigidity ratio. What onset ratio follows only if length and corresponding constraints also match?

Complete area 2 × 20 × 1 = 40 mm², so K = 40,000 × 40/100 = 16,000 N/mm in either stipulated straight axial state.
Bonded I = 20 × 2³/12 = 13.3333 mm⁴. Separate central moments sum to 2 × 20 × 1³/12 = 3.33333 mm⁴.
Multiply by the same E: EI values 533,333 and 133,333 N·mm². Their ratio is 0.250.
Only when the applicable C and effective L also match does P_cr have that same ratio. No actual compressive failure load has been calculated.

Same axial K = 16,000 N/mm; separated/bonded bending rigidity 0.250. The specified identical-constraint onset comparison also gives 0.250, with no general residual-strength prediction.

Count two interfaces and one projected union correctly

Two separated interface rectangles each measure 15.0 × 10.0 mm. Their common projection is 5.00 × 10.0 mm. A report calls the damage “300 mm²”. What convention would make that right, and what is the projected union?

Each rectangle has area 150 mm². Summing separated areas over the two different interfaces gives 300 mm².
The overlap has area 50.0 mm². Count it once in a plan-view union: 150 + 150 − 50.0 = 250 mm².
The common projected region contains two distinct separated interfaces. It contributes twice to the interface sum and once to the union.
Neither count measures all fibre/matrix damage or supplies a strength reduction. Resolve depth and state the convention.

300 mm² is the sum over the two stipulated interfaces; 250 mm² is their union footprint. Both are correct for different defined quantities.

Unchanged initial stiffness can accompany a lower failure endpoint

Use the separate matched 40 mm² coupon observations: K remains 16,000 N/mm, but the compressive endpoint falls from 20.0 to 14.0 kN. Find the retained endpoint and nominal stresses. Does this uniquely establish delamination?

The retained force ratio is 14.0/20.0 = 0.700. A 30.0% reduction uses the original 20.0 kN as denominator.
Nominal endpoint stresses are 20,000/40.0 = 500 MPa and 14,000/40.0 = 350 MPa.
At a small 100 N force, either supplied initial stiffness gives 0.00625 mm shortening. That common initial response does not contradict the different endpoints.
The observed endpoint change is real within this supplied record, while its cause remains to be discriminated by internal-state and loading evidence.

70.0% retained endpoint, with 500 versus 350 MPa nominal stresses. The observations do not uniquely identify delamination or establish a service allowable.

Try the reasoning

Is delamination identical to fibre breakage?
Show the answer and reasoning
No.
It separates layers rather than necessarily breaking fibres. Matrix cracks, fibre fracture and local fibre–matrix debonding may accompany it, but a diagram or measurement must identify those separately.
Why test residual properties after impact?
Show the answer and reasoning
Damage can change later performance.
The initial undamaged state is no longer sufficient. Match geometry, lay-up, supports, state, direction and the response endpoint. An unaltered initial slope does not guarantee an unaltered failure endpoint.
In a new ideal whole-strip comparison, split a uniform thickness 3h into three identical fully free-sliding strips of thickness h, with the same E and width. Find the separated/bonded bending-rigidity ratio and compare axial stiffness.
Show the answer and reasoning
3bh³/[b(3h)³] = 1/9; complete axial EA and K remain the same under the stipulated equal-strain straight state.
The common factor E/12 cancels in the bending ratio. Axial area remains 3bh. This is a complete-separation idealisation, not a formula for an arbitrary three-ply delamination.
For two specified ideal member models with the same applicable end/mode factor C, state 2 has one quarter of EI but half the effective length of state 1. What P_cr,2/P_cr,1 follows?
Show the answer and reasoning
(1/4) × (L₁/(L₁/2))² = 1.
The length change exactly offsets the rigidity change in this conditional onset comparison. A real local delamination need not preserve the same C or the assumed load sharing.
Keep two 150 mm² interface regions, but make their projected overlap 100 mm². Find the interface sum and union footprint.
Show the answer and reasoning
The interface sum remains 300 mm²; the union is 150 + 150 − 100 = 200 mm².
Changing overlap changes a plan projection without changing either supplied separated area. An area report needs its depth and counting convention.
A different specimen reaches 14.0 kN over nominal area 28.0 mm². Does its lower force than the original 20.0 kN,40.0 mm² coupon establish a 30% lower nominal stress endpoint?
Show the answer and reasoning
No. Both nominal endpoint stresses are 500 MPa.
14,000/28.0 = 500 N/mm² and 20,000/40.0 = 500 N/mm². Geometry differs; even equal nominal stress does not establish equal internal damage, constraint or component capacity.
Two specimens have the same 250 mm² projected indication but different residual compression responses. Must one response record be wrong?
Show the answer and reasoning
No. A common projected area does not fix depth, lay-up, matrix cracks, fibre damage, supports or history.
A clean embedded interface separation and distributed impact damage can share a footprint while changing different load paths. Identify the actual state and compare matched endpoints before assigning a mechanism.
An observed local bulge appears before a final compressive force drop. May its onset force be called the fibre tensile strength or the final laminate strength?
Show the answer and reasoning
Neither follows. The bulge is evidence of an instability under that compression state.
The fibre endpoint concerns a different local loading mode. Later delamination growth, contact, redistribution and other damage can govern the final laminate endpoint. A model for onset alone does not supply that later path.

What to carry forward

Name damage mechanisms separately.
Assess their interaction and residual performance.
A loss of connection changes force paths and support; equal material quantity or initial axial stiffness does not guarantee equal bending stability.
Separate buckling onset, delamination growth and final failure. State geometry, direction and constraints before using a ratio.
Distinguish projected union from summed areas on different interfaces. Neither area alone is a universal strength law.
Combine internal-state evidence with matched residual-response records and preserve the distinction between a specimen endpoint and a service allowable.

References

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