Materials Science · Polymers, elastomers and soft materials

Additives, degradation and material history

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

Assess formulation and history when interpreting polymer changes.
Explain representative additive functions and distinguish chemical chain changes from physical ageing.
Close a component mass balance and choose the correct denominator for an aged formulation.
Use complementary, matched observations to discriminate mechanisms without inferring chemistry from stiffness alone.
Recognise why a surface measurement, changed exposure or accelerated test may misrepresent a complete product.

A polymer product is often a formulation

Fillers, plasticisers, stabilisers and pigments can alter response. Their amount, distribution and compatibility matter. A resin name alone leaves much of the product’s structure and behaviour unspecified.

A stored flexible sheet becomes harder to bend. Three explanations are plausible: it has lost a mobile additive, its chemical connections have changed, or its chain arrangement has evolved without new bonds. The observation tells us that response changed. To choose between explanations, we must identify the starting formulation and the evidence collected after storage.

A plasticiser is a component used to change molecular mobility and response. A compatible small molecule can change interactions and the temperature range in which segmental rearrangement is accessible. In a suitable formulation this increases flexibility; it does not create a universal rule that every added molecule lowers every modulus. Concentration, chemistry, temperature and separation into different phases matter. Polymer chains and molecular architecture supplies the distinction between chain connections and conformation.

A filler is a dispersed second material. Particles or fibres can transfer load through their interfaces, obstruct a transport path or alter optical response. Their shape, size, distribution and adhesion matter; the same mass fraction of poorly bonded aggregates need not reinforce like well-dispersed particles. A pigment changes optical absorption or scattering. A colour function does not by itself certify reinforcement, ultraviolet protection or chemical stability.

An additive acts on a particular mechanism

A stabiliser slows a specified change under specified conditions. In a representative radical oxidation pathway, an unpaired-electron species reacts with a chain or with oxygen. Further reactions can generate hydroperoxides and new radicals, allowing the chemical process to continue. Chain scission cuts backbone connections; new interchain bonds can instead form during other steps. These processes may occur together, so a net response does not count either process separately.

Some antioxidants intercept chain-carrying radicals; other additives help remove hydroperoxides through pathways that generate fewer new radicals. They address different steps. Protection depends on the actual formulation and exposure, and an additive can be transformed, redistributed or lost. “Contains stabiliser” is therefore not a measured remaining service life. This is a mechanism explanation, not a chemical preparation or handling procedure.

Moisture also has different possible roles. Absorbed molecules can change interactions and mobility without cutting chains. In susceptible chemical linkages, hydrolysis is a reaction that breaks bonds through interaction with water. Mere water uptake does not prove hydrolysis, and drying does not automatically repair irreversible chemical or mechanical damage. The dedicated Moisture, light and polymer degradation lesson develops those environmental pathways further.

Ageing can change the network

Light, heat, oxygen or moisture can promote chain scission, crosslinking or additive migration, depending on chemistry and conditions. A changed colour is an observation, not a complete degradation diagnosis. Connect chemical evidence, mechanical response and exposure history.

Chemical ageing changes molecular structure through reactions. Cutting a load-bearing strand can remove one route for carrying load; forming a new connection can constrain rearrangement. Neither effect alone predicts every property, because cracks, morphology, additives and the measurement timescale also matter. A specimen can become stiffer while becoming less tolerant of damage.

Physical ageing of a glassy polymer is a different change: a nonequilibrium arrangement evolves without requiring new covalent connections. Changes in packing and available rearrangements can change volume, enthalpy and the measured response. This does not mean that every physical change is reversible or that a heating treatment restores a real damaged product. Crystallization, loss of solvent and physical ageing of a glass are also distinct processes.

Use the phrase material history concretely: manufacturing and cooling state, exposure time, temperature, humidity, atmosphere, mechanical constraint and contact with other substances. Retest at a defined state. Viscoelasticity, temperature and time explains why changing the loading duration alone can change the measured response, even if no ageing reaction has occurred.

Track components before calculating percentages

Consider a wholly invented dry specimen initially containing 70.0 g polymer, 20.0 g unbound plasticiser and 10.0 g inert filler. Its total mass is 100.0 g. The symbols m_P, m_A and m_F denote these three masses; total m = m_P + m_A + m_F. Here a mass fraction w_A = m_A/m uses the complete specimen as denominator.

After storage, an independent component-specific measurement establishes that 18.0 g of the original plasticiser has left. For this teaching balance only, stipulate no polymer or filler loss, no uptake, no chemical reaction and no retained surface liquid. The remaining masses are 70.0, 2.00 and 10.0 g, so m = 82.0 g. These stipulations make the ledger solvable; total mass alone would not establish them.

The plasticiser fraction is now 2.00/82.0 = 0.0243902, or 2.439%. It is not 2.00%: that would divide by the original mass. The filler fraction has risen from 10.0% to 10.0/82.0 = 12.195%, even though no filler was added. A changing denominator can mimic enrichment. The first figure shows absolute component masses before it shows fractions.

The amount of original additive lost is 18.0/20.0 = 90.0%. This is a different question from the specimen mass loss of 18.0/100.0 = 18.0%. State the reference mass whenever reporting a loss. No modulus is calculated from these percentages; the relation between composition and response would need separate evidence.

Original mass-conservation exercise, with all exchanges specified. These are invented components and measurements, not a real formulation or storage prescription.
State	Polymer (g)	Plasticiser (g)	Filler (g)	Total (g)
Initially	70.0	20.0	10.0	100.0
Remaining	70.0	2.00	10.0	82.0
Left specimen	0	18.0	0	18.0

Original invented dry mass balance, drawn on one horizontal scale of 4.2 pixels per gram. Initially the specimen contains 70 grams polymer, 20 grams unbound plasticiser and 10 grams inert filler, total 100 grams. Independently established loss of 18 grams plasticiser leaves 70, 2 and 10 grams, total 82 grams. Polymer is blue, plasticiser orange, filler diagonally hatched violet; every amount is also written. Filler remains 10 grams while its current specimen fraction rises from 10 to 12.195 percent; plasticiser falls to 2.439 percent. The loss is 90 percent of the original additive, or 18 percent of the original specimen. No other exchange or reaction is allowed in this fictional ledger; no response is derived from the proportions.

A measured change in response needs complementary evidence

For a separate invented small-strain tensile comparison, test fresh and aged specimens at the same temperature and elapsed loading time. Each has actual gauge length L = 50.0 mm and area A = 10.0 mm² at that test. A 10.0 N axial force gives nominal stress σ = F/A = 1.00 MPa. Fresh extension is 0.100 mm and aged extension is 0.0500 mm; strain ε = ΔL/L is respectively 0.00200 and 0.00100.

The ratios σ/ε are 500 and 1000 MPa. These are apparent tensile stiffness measures for that stated observation and assumed small-strain range. The calculation establishes a doubled response ratio; it does not identify plasticiser loss, new links or crystallization. It also says nothing about fracture energy. Deformation and relative strain, together with Applied force and stress over an area, supplies the loading quantities.

If a component-specific composition measurement also confirms the earlier 18.0 g additive loss, that supports additive depletion as a plausible contributor. To claim it is the sole cause, one would need to rule out consequential chemical, structural and geometry changes. Conversely, unchanged total mass is not evidence that nothing changed: a separate specimen could gain 5.00 g from its environment while losing 5.00 g of volatile matter. The net balance is zero despite two exchanges.

Select evidence for the alternatives: a calibrated component analysis for additive amount; a chemical method sensitive to the proposed new groups or connections; structural evidence for ordered regions; and response measurements with matched state, time and geometry. Each has a sampling depth, resolution and possible interference. “No detected change” means no resolved change for that method, rather than proof that every molecular feature is identical.

Which question does an observation answer? These are interpretation limits, not a list of interchangeable tests.
Observation	Can establish, with suitable calibration	Cannot establish alone
Whole-specimen mass	Net mass exchange	Identity of every leaving or entering species
Component-specific assay	Amount of the identified additive in its sampled material	Every chemical or structural cause of stiffening
Chemical spectrum	Resolved chemical features within its sensitivity and sampled region	Unique mechanical consequence or complete network topology
Matched load/extension	Response at the stated state and time	A unique chemical degradation mechanism

A surface value need not represent the entire specimen

An unbound additive can move through a polymer toward its surface and then transfer into the surroundings. Internal transport and surface removal are separate steps. Their relative rates, the external environment and changing concentration can create gradients. A slower loss in a sealed volume need not mean that the polymer chemistry became intrinsically more durable.

Use a separate invented three-region accounting problem. Define the retained base as polymer plus inert filler, excluding the mobile additive. The two surface regions each contain 25.0 g of this base; the central region contains 50.0 g. These are supplied base masses, not volumes inferred from a picture. The measured additive-to-base ratio r is 0.0500 g/g in each surface region and 0.250 g/g in the core.

The additive masses are 1.25, 12.5 and 1.25 g. Their sum is 15.0 g per 100.0 g retained base, so the whole-specimen additive mass fraction is 15.0/(100.0 + 15.0) = 0.130435 = 13.0435%. A surface-only measurement would instead give 0.0500/(1 + 0.0500) = 4.7619% of that local specimen mass. The difference is a sampling issue, not an arithmetic contradiction.

The second figure uses cumulative retained-base mass along the ordered regions as its horizontal coordinate. It does not claim equal thicknesses, uniform density, an actual diffusion solution or a measured time history. Convert each local ratio to an actual additive mass and sum before dividing by total mass. Averaging percentages with unrelated denominators can give the wrong bulk answer.

Likewise, oxygen consumption near an exposed surface can exceed the rate at which oxygen reaches the interior, leaving different chemical states across the specimen. A bulk mechanical result and a surface chemical observation then sample different distributions. Geometry and exposure must be matched before transporting an accelerated result to a different product.

Original invented three-region additive account, not a diffusion solution. The horizontal axis orders regions by cumulative retained base mass, excluding additive: 25 grams surface, 50 grams core and 25 grams surface. The blue stepped profile with circles is 5, 25 and 5 grams additive per 100 grams local base. The orange dashed line is the base-mass-weighted mean ratio of 15 grams per 100 grams base. Actual additive masses are 1.25, 12.5 and 1.25 grams. The complete specimen contains 15 grams additive and 100 grams base, so its additive fraction is 13.0435 percent; the local surface whole-region fraction is 4.7619 percent. The horizontal coordinate is not physical depth; there is no time, diffusivity or thickness prediction.

Keep the diagnosis narrower than the evidence

A useful report identifies the formulation and starting state, records relevant exposure and loading history, closes the mass account where possible, and states which proposed mechanism each observation supports or leaves unresolved. Correlation between stiffening and storage time is a prompt for this comparison; it is not a reaction equation.

Repeat comparisons should distinguish fresh specimens from pieces that were already heated, stretched, extracted or otherwise changed during a previous test. Sampling preparation can remove a mobile component or alter the structure being investigated. This lesson gives no extraction recipe or rejuvenation procedure.

A higher test temperature can change transport, chemical pathways, morphology and the additive’s distribution at once. A faster change is not automatically the same change with a faster clock. Laboratory acceleration and real-service evidence develops the separate requirements for an extrapolation. Neither an intact appearance nor a nominal additive name is a lifetime or component-safety qualification.

Worked examples

Work through the case

A flexible polymer becomes stiffer during storage. Must crystallinity have increased?

Consider loss of plasticiser.
Consider chemical crosslinking or structural ageing.
Use measurements that distinguish these mechanisms.
Under the supplied matched tensile comparison, calculate stress 10.0/10.0 = 1.00 MPa and strains 0.100/50.0 = 0.00200 and 0.0500/50.0 = 0.00100. The apparent ratios double from 500 to 1000 MPa.
The independently established component loss supports depletion as a contributor in the mass-ledger case. It does not mathematically exclude simultaneous structural or chemical changes. A stiffness value contains no direct count of crystals or crosslinks.

No. Several formulation and structural changes can produce stiffening. The matched response and component evidence narrow the explanation, while sole causation remains unproved.

Separate loss of additive from enrichment of filler

Use the 100.0 g initial and 82.0 g remaining mass ledger. A report says “18% of the plasticiser was lost, leaving 2% plasticiser and adding filler to 12.195%”. Correct each statement.

The lost 18.0 g is 18.0% of initial specimen mass but 90.0% of the original 20.0 g plasticiser.
The remaining plasticiser is 2.00 g out of 82.0 g: 2.439% by current specimen mass.
Filler remains exactly 10.0 g. Its fraction rises to 12.195% solely because the total mass decreased.
The three remaining fractions are 85.3659% polymer, 2.4390% plasticiser and 12.1951% filler, summing to 100% within rounding.

90.0% of the additive was lost; the current specimen contains 2.439% plasticiser and the same 10.0 g filler. No filler addition is implied.

Recover the complete additive fraction from three regions

Use the supplied surface/core base masses 25.0/50.0/25.0 g and ratios 0.0500/0.250/0.0500. What whole-specimen fraction follows, and why is the mean of local percentages a different calculation?

Multiply each ratio by its own base mass: 1.25 + 12.5 + 1.25 = 15.0 g additive.
The total specimen mass is 100.0 g base + 15.0 g additive = 115.0 g, so its additive fraction is 13.0435%.
Local whole-region fractions are 4.7619%, 20.0% and 4.7619%. Weighting these by base fractions 0.25/0.50/0.25 gives 12.3810%, which uses the wrong weights.
Correct whole-region masses are 26.25, 62.50 and 26.25 g. Weighting local fractions by those masses divided by 115.0 g recovers 13.0435%.

13.0435% of the whole specimen mass. Average component amounts first, or use weights matching the denominators of local fractions.

No net mass change still permits exchange

A separate dry specimen starts at 100.0 g, loses 5.00 g of material and gains 5.00 g from its surroundings. Its stiffness changes. Does the final mass exclude migration or chemical ageing?

The mass balance gives 100.0 − 5.00 + 5.00 = 100.0 g.
A balance reading measures the net result of both exchanges. It does not identify the species, their locations or whether reactions accompanied uptake.
Independent collection/composition and chemical evidence would be needed to assign mechanisms. Do not transfer the no-uptake stipulation of the earlier teaching ledger to this different specimen.

No. Equal mass in and out hides exchange; final mass alone cannot identify the ageing mechanism.

Try the reasoning

Does an additive have the same effect at every concentration?
Show the answer and reasoning
Not necessarily.
Interactions and distribution can change. Compatibility, phase separation, dispersion and the test state matter. A concentration change needs a measured response, rather than a universal monotonic rule.
Why retain storage history?
Show the answer and reasoning
The tested state evolves with exposure.
As-made data may not describe an aged formulation. Include formulation, cooling, temperature, atmosphere, humidity, contacts, loading and prior tests so a later comparison identifies what actually changed.
A new specimen starts with the same 70.0/20.0/10.0 g components but loses only 5.00 g plasticiser with no other exchange. Find its current plasticiser fraction and the fraction of original additive lost.
Show the answer and reasoning
15.0/95.0 = 15.7895% plasticiser by current specimen mass; 5.00/20.0 = 25.0% of the original additive lost.
Its final mass is 95.0 g. The two percentages have different denominators. The earlier 82.0 g balance belongs to the 18.0 g loss case.
A supplied aged specimen has actual L = 40.0 mm and A = 8.00 mm². At the matched observation time, F = 10.0 N produces extension 0.0500 mm. Is its apparent σ/ε lower than the earlier aged value?
Show the answer and reasoning
No: σ = 1.25 MPa, ε = 0.00125 and σ/ε = 1000 MPa, the same supplied aged ratio.
Equal extension or force does not mean equal strain or stress when geometry changes. The ratio remains a time- and state-specific response, not a diagnosis of chemistry.
Keep the three base masses 25.0/50.0/25.0 g and surface ratio 0.0500, but change the supplied core ratio to 0.150 g/g. Find the whole-specimen additive fraction.
Show the answer and reasoning
Additive = 1.25 + 7.50 + 1.25 = 10.0 g; fraction = 10.0/110.0 = 9.0909%.
The new core composition changes the complete mass account. The surface fraction is unchanged, so a surface-only observation would miss this difference.
Two aged specimens show the same small hardness increase. May one infer the same number of chain cuts and new crosslinks?
Show the answer and reasoning
No. The observation does not uniquely separate the competing processes.
Different amounts of scission, new connections, physical ageing, additives or local morphology can yield similar net responses. A mechanical measurement is useful evidence but needs mechanism-sensitive complements and matched sampling.
A pigmented product shows no visible colour change. Is degradation excluded?
Show the answer and reasoning
No. Colour is only one optical observation.
Chemical or mechanical changes may occur below its sensitivity, in an interior region or without changing the visible pigment response. Conversely, a colour change alone does not identify chain scission.
A thin ventilated specimen loses additive faster during a hotter test than a thick specimen in a confined atmosphere. May the ratio of loss times be called a universal temperature acceleration factor?
Show the answer and reasoning
No. Geometry, boundary conditions and possibly the controlling mechanism changed too.
Internal transport, surface removal and evolving composition must be separated before a temperature-only relation is justified. No component lifetime follows from that unmatched comparison.

What to carry forward

Identify the formulation, not just the base polymer.
Diagnose ageing through mechanisms and evidence.
Close component masses and state the denominator before comparing aged percentages.
Chemical scission, added connections and glass physical ageing have different meanings and may coexist.
Match time, state, geometry and sampling depth; surface and bulk observations need not represent the same material.
Neither a net stiffness change, unchanged mass nor unchanged colour uniquely identifies an ageing mechanism.

References

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