Materials Science · Graphene and carbon nanotubes: theory and measurement

Defects, temperature, strain rate and environment

Useful background: Size, surfaces and nanoscale behaviour; Elasticity, plasticity and tensile strength; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Identify variables separating ideal tubes from real tensile specimens.
Explain why losing an atom, rotating a bond and breaking a tube are different structural events.
Use a bounded competing-first-event model to compare loading time and temperature.
Choose matched observations that distinguish a changed event pathway from a changed total event frequency.

Defects can create a preferred failure site

Vacancies, chemical modification, bends and other imperfections alter the network and its loading. Their effects depend on position, density and tube structure. A broad label such as nanotube does not establish that a specimen is defect-free.

Consider one missing carbon atom in the interior of an otherwise three-neighbour network. Before local relaxation, removing that atom also removes its three incident bonds. The remaining atoms must transmit the applied force through a different connected network. Forces and bond extensions near the opening need not equal their values far away; additional bond breaking can extend the opening into a crack. This is a possible initiation pathway, not a rule that one missing atom subtracts a fixed percentage of strength.

The drawing holds all remaining positions fixed to isolate this change of connectivity. A real vacancy can reconstruct: neighbouring atoms move and can form new bonds. Its shape, location relative to the tensile direction, other nearby defects and the tube wrapping then matter. Counting missing atoms alone does not specify the relaxed structure or its failure load.

Atomistic fracture calculations have explicitly compared an intact tube with a tube missing one atom and its three bonds; in that chosen model the defective tube broke earlier, while much of its preceding force–extension response remained similar. The calculation did not explain the entire difference from the compared experiments. An apparently unchanged initial stiffness therefore need not establish an unchanged fracture endpoint.

Original schematic of the same finite hexagonal carbon network in two panels, not an atomistic calculation or a relaxed vacancy image. The left panel retains every drawn site and nearest-neighbour link. The right panel removes one interior site and its three incident links. The vacant site is an empty orange dashed circle with an X; the three deleted links are dashed orange. Remaining sites and links retain exactly the same positions and are drawn as light nodes and solid blue lines. The three neighbouring sites now each have one fewer connection. The patch remains connected through other routes. Outer edge sites have fewer neighbours because the drawing is finite, not because they are the highlighted vacancy. Link lengths are equal in this construction; no bond force, tensile direction, stress concentration factor or failure load is calculated. Real atoms can relax and reconstruct, so this frozen drawing is only a connectivity comparison.

A rearrangement is not the same event as rupture

Carbon atoms can also change neighbours without first removing an atom. A Stone–Wales rearrangement rotates a bond and changes four adjacent six-membered rings into two five-membered and two seven-membered rings. Ring size counts atoms around a closed bonded path. This changes the local network and can initiate plastic relaxation or later damage; it is not automatically a repair or the final separation of a tube.

To move from one bonded arrangement to another, the atoms may have to pass through a higher-energy intermediate configuration. The activation barrier is that energy rise from the current loaded state along the proposed path. Thermal motion gives repeated opportunities to cross a barrier. Mechanical loading can also change the barrier and the available paths. A favourable final arrangement alone does not say how quickly it will appear.

Calculations of nanotube yield have compared activated bond rearrangement with a competing brittle bond-breaking route near a mechanical threshold. Their relative importance depends on the stated structure and conditions. The following deliberately simpler model teaches event competition below an assumed immediate-breaking threshold. Its numerical rates and barriers are invented; neither route is calibrated to a real nanotube.

Failure has a timescale and environment

Temperature, loading rate and surroundings can affect accessible failure pathways. Grips and irradiation can alter the specimen too. Compare tests only after examining these variables; do not assign every lower value to one assumed defect.

In a 2000 individual multiwall-tube experiment, outer-wall breaking followed by inner-shell pullout was observed with the particular end attachments used. That identifies an engaged load path; it does not measure a universal vacancy density. Attachment failures also limited which specimens supplied the reported fracture records. Retain the mounting and tested structural level when comparing a calculation with a measured endpoint.

A separate 2014 study of suspended CVD graphene reported crack initiation and growth when relative humidity was increased from 20% to 60% at a maintained pressure difference in its controlled sequence. The authors proposed a role for moisture and polymer-transfer residues. The observation does not directly identify a unique reacting species or elementary reaction, and it is not a nanotube tensile-strength measurement. The overall fracture study retained seven selected datasets after substantial preparation and recording losses, so those records do not establish population-wide failure statistics.

Follow departures from one specified loaded state

Imagine an ensemble of identical initial loaded networks at a fixed temperature, force and environment. While a network remains in that state, let r be its conditional rate of rupture and s its conditional rate of a first irreversible rearrangement, both in s⁻¹. For a short interval Δt, the conditional chances are approximately rΔt and sΔt. Assume constant rates, no third exit, no return, and a negligible chance of two events in that short interval. These are whole-specimen rates; do not multiply them by an atom count again.

Let P₀(t) be the probability that neither event has yet occurred, starting with P₀(0) = 1. Each short step retains approximately the fraction 1 − (r + s)Δt of the networks still in the initial state. Repeating increasingly short steps gives P₀(t) = exp[−(r + s)t]. Here exp(z) means the exponential function e raised to z. The exponent is dimensionless because a rate in s⁻¹ multiplies time in seconds.

Among the first departures, the fraction taking rupture is r/(r + s), and the fraction taking rearrangement is s/(r + s). Thus Fᵣ(t) = [r/(r + s)][1 − P₀(t)] and Fₛ(t) = [s/(r + s)][1 − P₀(t)]. They count which event occurred first by time t, so P₀ + Fᵣ + Fₛ = 1. They do not double-count a specimen after a rearrangement.

The accounting stops at the first rearrangement. Its later rupture, renewed loading or further relaxation requires a new state model. Consequently P₀ is an unchanged-state probability, not the probability that a component remains usable; Fₛ is not guaranteed survival. A deterministic immediate-breaking limit, changing load, evolving defects or time-dependent chemistry also requires a different or extended model.

Temperature can change the mix of first events

Use the fictional reference rates r₀ = 0.0100 s⁻¹ and s₀ = 0.00200 s⁻¹ at T₀ = 300 K. Over 300–350 K only, stipulate effective barriers Eᵣ = 0.400 eV and Eₛ = 0.800 eV, with temperature-independent prefactors and the same defined initial loaded state. An electron volt, eV, is an energy unit; these are model barrier inputs, not measured carbon bond energies.

For either route i, the supplied activated law is λᵢ(T) = λᵢ(T₀) exp[(Eᵢ/kᴮ)(1/T₀ − 1/T)], where λᵢ means r or s and kᴮ = 8.617333262 × 10⁻⁵ eV/K is the rounded Boltzmann constant. It follows by dividing a constant-prefactor exp[−Eᵢ/(kᴮT)] law by its value at T₀. Use kelvin, not Celsius. Eᵢ/kᴮ has units K, so the exponent again has no units.

Both rates increase in this model, but the larger barrier gives the rearrangement rate a larger temperature multiplier. At 350 K it overtakes rupture as the more frequent first route. This does not establish that the warmer specimen is stronger, safer or more ductile: later outcomes after rearrangement have not been supplied. Real barriers and prefactors can also change with strain, local structure and surroundings.

The two plots hold different variables fixed. The upper plot changes time at 300 K; the lower plot changes temperature for a fixed ten-second hold. A point on a curve is a calculated probability under the stipulated model, not an extra observed specimen.

Invented rates and calculated first-event probabilities after 10 s. The three probability columns close to one before rounding; the rows are not experimental data.
T / K	r / s⁻¹	s / s⁻¹	P₀	Fᵣ	Fₛ
300	0.010000	0.002000	0.886920	0.094233	0.018847
330	0.040821	0.033327	0.476410	0.288254	0.235336
350	0.091192	0.166320	0.076144	0.327162	0.596693

Original fictional competing-first-event model. Unchanged initial state P zero uses a solid blue line and circles; rupture first F r uses a dashed orange line and squares; rearrangement first F s uses a dotted green line and triangles. The upper plot fixes 300 kelvin and varies time from zero to 100 seconds. Its probability axis runs from zero to one. Rates are r equals 0.0100 per second and s equals 0.00200 per second. At ten seconds P zero, F r and F s are 0.886920, 0.094233 and 0.018847; at 100 seconds they are 0.301194, 0.582338 and 0.116468. The lower plot fixes ten seconds and varies temperature from 300 to 350 kelvin with the supplied constant-prefactor barriers 0.400 and 0.800 electron volt. At 330 kelvin the three probabilities are 0.476410, 0.288254 and 0.235336; at 350 kelvin they are 0.076144, 0.327162 and 0.596693. A vertical dotted violet guide at 334.828 kelvin marks equal route rates and therefore equal first-route probabilities for this hold. Every plotted set closes to one. The rearranged-first curve stops classifying later outcomes; it is not a survival curve. No real nanotube rates, strength or lifetime are supplied.

A strain rate sets time only after the loading path is specified

Engineering strain ε is extension divided by initial gauge length; a strain increment Δε = 0.0200 is a two-percentage-point increment. A constant strain rate ε̇ is strain change per second. The time to traverse a specified interval is Δt = Δε/ε̇. Increasing ε̇ tenfold gives one tenth as much time in that same interval.

For a simple comparison, condition on reaching the start of a narrow interval with the same initial network unchanged. Stipulate the 300 K reference rates as effective constants throughout that interval, even though its strain is increasing. This is an additional approximation, not a result of the fixed-force hold model. With Δε = 0.0200, rates 0.00100 s⁻¹ and 0.0100 s⁻¹ take 20.0 s and 2.00 s. The slower traverse permits more first events in this conditional model.

In a real tensile ramp, the barriers generally vary with strain and the state can change before that interval. A faster test can also change temperature or the apparatus response. First-event probabilities in one interval do not give an entire distribution of fracture strengths. Compare the actual force/extension history, gauge length, temperature and preparation rather than treating strain rate as a universal strength multiplier.

Choose evidence that separates competing explanations

A record of unchanged specimens alone measures only the combined departure rate r + s within the assumed model. For example, (r,s) = (0.0100,0.00200) and (0.00400,0.00800) s⁻¹ give exactly the same P₀ at every time, yet different first-event types. Resolve a bond rearrangement separately from rupture, or the two explanations remain indistinguishable by that observable. Real finite counts introduce sampling uncertainty; the exact probabilities here are teaching inputs.

A useful matched comparison holds the tube structure, engaged walls, area convention, initial defect description, loading history and temperature as close as the inference requires. To investigate an environmental change, include specimens kept in the original environment for the same elapsed time and observation exposure. A within-specimen change can reveal timing, but an irreversible crack means the sequence cannot simply be reversed to recover the original control.

For a defect explanation, independently observing a local missing-atom structure and subsequent crack initiation near that site is stronger evidence than inferring its existence only from a low failure load. It still does not make every nearby chemical or mounting effect disappear. An observation method that changes the network needs a matched exposure control; a clean image is not proof of an unaltered specimen.

Graphene sheets and nanotube structure supplies the network and wall distinction; Cross-sectional area, gauge length and test conditions explains the reporting basis. Atomistic models and computational predictions explains what an energy calculation establishes. Those lessons support this comparison, while this lesson adds the time and competing-pathway question.

Discriminating observations for a model–test discrepancy. Each comparison addresses a particular alternative; no single row establishes an entire failure mechanism.
Proposed explanation	Useful matched observation	What remains unresolved
Changed event competition	Classify the first event as rupture or rearrangement at matched loading and time	Later fracture after rearrangement is outside the first-event account
Changed surroundings	Track crack evolution with a same-duration unchanged-environment and observation-exposure control	A humidity association alone does not identify an elementary reaction
A local defect starts the crack	Resolve the defect state and initiation location without assuming observation is passive	One location does not determine population defect density or bulk strength
Attachment or engaged-wall difference	Separate specimen fracture from grip detachment and shell pullout	The report still needs force, geometry and loading-history information

Worked examples

Work through the case

A measured tube fails below an ideal prediction. Does the difference uniquely measure defect density?

Compare the model and area conventions.
Check loading, temperature and specimen structure.
Several differences can influence failure.

No. The discrepancy does not identify one unique cause.

Calculate what a ten-second hold does and does not establish

Use r = 0.0100 s⁻¹ and s = 0.00200 s⁻¹ at 300 K. What are the probabilities of remaining unchanged, rupture first and rearrangement first after 10.0 s?

The combined departure rate is r + s = 0.0120 s⁻¹. Its dimensionless product with 10.0 s is 0.120; this is not itself the finite-time event probability.
P₀ = exp(−0.120) = 0.886920. The first-event probability is 1 − P₀ = 0.113080; replacing it by 0.120 would use a short-time approximation outside the exact calculation.
Partition the departures: r/(r + s) = 5/6 and s/(r + s) = 1/6. Therefore Fᵣ = 0.0942330 and Fₛ = 0.0188466.
The three terms add to one. The 1.88466% rearranged-first group is no longer in the initial state; its later rupture is not counted by either a new independent repeat of this calculation or an assumed survival label.

About 88.6920% unchanged, 9.42330% rupture first and 1.88466% rearrangement first. These are model ensemble probabilities, not a component lifetime guarantee.

A warmer model changes which first route is more frequent

For the supplied activated laws, raise T from 300 K to 350 K and hold for 10.0 s. Calculate the two rates and first-event probabilities. Does a larger rearrangement fraction prove improved tensile performance?

Compute 1/300 − 1/350 = 1/2100 K⁻¹. The rupture exponent is 0.400/(8.617333262 × 10⁻⁵ × 2100) = 2.210384; the rearrangement exponent is twice that.
The rate multipliers are 9.11922 and 83.1602. Thus r = 0.0911922 s⁻¹ and s = 0.166320 s⁻¹; neither value is itself a probability.
For 10.0 s, P₀ = exp[−(0.0911922 + 0.166320)10] ≈ 0.0761442. Partitioning departures gives Fᵣ ≈ 0.327162 and Fₛ ≈ 0.596693, using unrounded rates.
The rearrangement route is now more frequent as the first event. Both it and rupture remove networks from the initial state. No law for the rearranged state, failure force or usable strain was provided.

At 350 K the model gives about 7.61442% unchanged, 32.7162% rupture first and 59.6693% rearrangement first. The changed pathway mix does not establish better performance.

Change loading time without inventing a strength law

Conditional on an unchanged network entering the stated Δε = 0.0200 window, compare ε̇ = 0.00100 s⁻¹ and 0.0100 s⁻¹. Use the stipulated constant 300 K reference rates over the window.

The times are 0.0200/0.00100 = 20.0 s and 0.0200/0.0100 = 2.00 s. The strain increments are dimensionless.
For 20.0 s, the model gives P₀ = exp(−0.240) = 0.786628, Fᵣ = 0.177810 and Fₛ = 0.0355620.
For 2.00 s, P₀ = exp(−0.0240) = 0.976286, Fᵣ = 0.0197619 and Fₛ = 0.00395238.
The faster traverse leaves less opportunity for either initial-state exit. It does not state the loading behaviour outside the interval, nor the final rupture load of either group.

The first-event probability is about 21.3372% for the slow traverse and 2.37143% for the fast traverse, under this narrow-window approximation. This is not a universal rate–strength relation.

Use an unfamiliar probability record without claiming microscopic identity

A separate hypothetical exact probability record follows the same constant-rate first-event model: P₀(20 s) = 0.500, and two thirds of the first departures are rupture. Infer r and s. These are supplied exact model probabilities, not counts from a finite experiment.

Invert P₀ = exp[−(r + s)t]: r + s = −ln(P₀)/t. The natural logarithm ln is the inverse of exp; ln 2 = 0.693147.
The combined rate is ln 2/20 = 0.0346574 s⁻¹. A probability of 0.500 after 20 s is not a rate of 0.0250 s⁻¹.
Two thirds of departures gives r = (2/3)0.0346574 = 0.0231049 s⁻¹; s is the remaining 0.0115525 s⁻¹.
The classified first events separate two effective rates within the model. They do not reveal a vacancy count, an activation barrier or a reacting chemical species. A real finite record would also require its sampling and classification uncertainty.

Effective whole-specimen rates r ≈ 0.0231049 s⁻¹ and s ≈ 0.0115525 s⁻¹, conditional on the supplied model and exact probabilities.

Try the reasoning

Can the measurement method damage the sample?
Show the answer and reasoning
Yes.
Imaging or preparation may alter it.
Does faster always mean the same strength change?
Show the answer and reasoning
No.
The mechanism and structural state determine the trend.
Keep the 300 K reference rates and extend the unchanged-state hold to 100 s. Calculate P₀, Fᵣ and Fₛ. Is Fᵣ + Fₛ a rupture probability?
Show the answer and reasoning
P₀ ≈ 0.301194, Fᵣ ≈ 0.582338 and Fₛ ≈ 0.116468. Their sum is a first-departure probability of 0.698806, not the probability of rupture.
The exponent is −0.0120 × 100 = −1.20. Partition 1 − exp(−1.20) in the ratio 5:1. Rearrangement first and rupture first remain different endpoints.
Replace (r,s) = (0.0100,0.00200) by (0.00400,0.00800) s⁻¹. At 10 s, can P₀ distinguish the models? What happens to Fᵣ and Fₛ?
Show the answer and reasoning
P₀ stays 0.886920 in both models. The changed pair gives Fᵣ ≈ 0.0376932 and Fₛ ≈ 0.0753864.
The total rate remains 0.0120 s⁻¹, but rupture now accounts for one third rather than five sixths of departures. Classifying the first event supplies information absent from unchanged-state observations alone.
In a separate invented environmental change at 300 K, triple r to 0.0300 s⁻¹ while s remains 0.00200 s⁻¹. Calculate the 10 s probabilities. Must Fₛ increase because the environment is more damaging?
Show the answer and reasoning
P₀ ≈ 0.726149, Fᵣ ≈ 0.256735 and Fₛ ≈ 0.0171157. Fₛ is slightly lower than its original 0.0188466, even though its own conditional rate is unchanged.
The combined exponent is −0.320; rupture takes 15/16 of departures. More networks leave through rupture before they can rearrange. This invented rate change neither identifies environmental chemistry nor predicts later outcomes.
Under the 300 K reference model, when does P₀ fall to one half? Use ln 2 = 0.693147. Is that a median time to component failure?
Show the answer and reasoning
t = ln 2/0.0120 ≈ 57.7623 s. It is the model time to 50% first departures, not a median usable-component failure time.
Solve exp(−0.0120t) = 1/2. The departed group includes rearrangements with unspecified later behaviour.
Within the stipulated 300–350 K laws, find the temperature at which s = r. Use ln 5 = 1.609438 and the local constants.
Show the answer and reasoning
About 334.828 K.
The ratio is s/r = 0.200 exp[(0.800 − 0.400)(1/300 − 1/T)/kᴮ]. Set it to one: 1/T = 1/300 − kᴮ ln 5/0.400. Equality of first-route rates does not establish equal later fracture outcomes.
The strain increment becomes 0.0400 and the strain rate becomes 0.00200 s⁻¹. If the same effective reference rates are again explicitly stipulated over this new interval, how do its conditional probabilities compare with the original slow traverse?
Show the answer and reasoning
The time is again 20.0 s, so this stipulated model gives the same P₀ ≈ 0.786628, Fᵣ ≈ 0.177810 and Fₛ ≈ 0.0355620.
The time ratio Δε/ε̇ is unchanged. For a real wider strain interval, barrier constancy must be reassessed; the calculation does not infer it from equal elapsed time.
An image after a short hold shows no resolved crack. Does that establish P₀ = 1 or prove no bond rearrangement occurred?
Show the answer and reasoning
No.
One observed specimen is not an ensemble probability, and the image may not resolve a local rearrangement or a small crack. State the detection limit and observation exposure rather than changing the event definitions.
A humidity step coincides with crack growth in one previously loaded graphene membrane. What comparison would test elapsed time as an alternative, and what would the result still not identify?
Show the answer and reasoning
Use comparably prepared membranes held for the same loading duration, temperature and observation exposure without the humidity step. Even a reproducible difference would not by itself identify an elementary chemical reaction or a nanotube strength.
A crack changes its own state irreversibly, so merely switching the original specimen back is not a fresh initial-state control. Matching and independent chemical or structural evidence address distinct questions.

What to carry forward

Describe defects and protocol.
Investigate discrepancies through competing mechanisms.
A vacancy changes local connectivity; an irreversible bond rearrangement is distinct from rupture.
Rates have units of inverse time. First-event probabilities depend on both their sum and their competition.
Temperature or a faster ramp can change accessible paths without giving a universal strength multiplier.
An unchanged-state curve alone cannot identify the mix of exits, microscopic defect density or environmental chemistry.

References

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