Materials Science · Graphene and carbon nanotubes: theory and measurement

Theoretical tensile limits of ideal defect-free structures

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

State what an ideal tensile prediction assumes.
Connect a declared energy–strain branch with nominal sheet force, tangent stiffness and a conditional axial instability.
Distinguish initial elastic stiffness from a tensile maximum and identify missing atomic deformation modes.
Convert the same sheet force using different explicit thickness conventions without inventing a change in physical load.
State which model, boundary and specimen differences prevent a direct comparison or transfer of a tensile prediction.

An ideal structure is a defined model

An atomistic tensile calculation can start with a defect-free tube or sheet, a chosen interatomic description and specified boundary conditions. Its limiting response belongs to that model, temperature and loading protocol. It is not a measured property of a manufactured yarn.

A small-strain elastic modulus describes how steeply force or stress initially rises. It does not say how far a structure can be stretched before its response changes or a new deformation becomes unstable. Extrapolating the initial slope to a large strain assumes that the same straight-line law remains valid. Strong bonds can still have a nonlinear force response as their lengths and angles change.

An atomistic calculation assigns positions and an energy description to atoms. It can follow a chosen stretch while allowing specified internal positions and transverse dimensions to relax. The derivative of the resulting energy along that path gives a force-related response. Electronic-structure methods and classical interatomic potentials estimate that energy differently; a model that reproduces an equilibrium spacing or initial stiffness has not thereby been tested for stretched bonds and failure.

Initially defect-free describes the starting structure, not a promise that it can never form a defect. An ideal calculation can lose stability and nucleate a different structure. Temperature, loading direction, imposed lateral strain versus lateral relaxation, allowed cell size and deformation patterns all belong to the prediction. Graphene is a sheet network; a curved tube with specified chirality and end constraints is a different model. The calculation must establish any transfer, rather than inheriting a universal number from the word carbon.

A simplified energy branch makes the quantities explicit

Use a completely original scalar teaching model, not a fitted graphene or nanotube potential. A flat sheet strip has reference length L₀ = 1.00 µm and reference width w₀ = 1.00 µm. Its width is held fixed, it stretches uniformly, and only the axial strain ε = (L − L₀)/L₀ is allowed to vary. No atoms, thermal motion, bending, lateral relaxation or nonuniform modes are resolved. This one-coordinate restriction will matter to the stability conclusion.

Let W be stored energy per reference sheet area, in J/m². Supply W(ε) = K a²[1 − cos(πε/a)]/π² for 0 ≤ ε ≤ a, with K = 300 N/m and dimensionless a = 0.400. Both are invented coefficients. The initial area is A₀ = L₀w₀ and total stored energy is U = A₀W. W is an energy density for this branch; it is not the energy needed to create a unit area of crack surface.

Define T = F/w₀, axial force per reference width, in N/m. Work in a small additional stretch is F dL = F L₀ dε. The energy change is dU = L₀w₀(dW/dε)dε. Equating the two gives T = dW/dε = (Ka/π)sin(πε/a). If differentiation is new, read dW/dε as the local slope of the W-versus-ε graph: energy change divided by a sufficiently small strain change.

The slope of T is K_t = dT/dε = K cos(πε/a), called the tangent stiffness per reference width here. Near zero strain, sin(πε/a) is approximately πε/a, giving T ≈ Kε and initial slope K. Thus K has units N/m, while an ordinary three-dimensional elastic modulus has units Pa. A sheet force per width does not need an assumed atomic wall thickness.

The first T maximum occurs when πε/a = π/2, so ε_peak = a/2 = 0.200 and T_peak = Ka/π ≈ 38.20 N/m. At that point K_t = 0. Beyond it this mathematical branch decreases. The descending continuation in the figure is useful for diagnosing the model, not evidence that an actual sheet can remain uniformly intact through that interval.

Original scalar branch with K = 300 N/m and a = 0.400. Rounded values describe the stipulated model, not measured or computed graphene properties.
ε	W / J m⁻²	T / N m⁻¹	K_t / N m⁻¹
0	0	0	300
0.100	1.4245	27.0095	212.132
0.200	4.8634	38.1972	0
0.300	8.3024	27.0095	−212.132

A force maximum is a conditional stability result

Now impose a constant axial force F on this same uniform one-coordinate strip. The external force can do work as the strip extends. Its total potential energy, apart from an arbitrary constant, is Π(ε) = A₀W(ε) − F L₀ε. Equilibrium requires the local slope to vanish: dΠ/dε = 0, or T = F/w₀. A small allowed perturbation is stable if it raises Π, making equilibrium a local energy minimum.

The curvature of Π is A₀K_t. A point before the peak has positive curvature and is locally stable against this axial perturbation. A point on the descending branch has negative curvature and is unstable under that force control: a small strain change can lower the potential energy. The zero tangent at the peak marks the loss of positive axial stiffness. At the exact peak the strict positive-curvature test is marginal, so it does not establish a stable reserve of load. This is a local equilibrium test on the declared branch; it does not assert a global energy minimum over unrestricted strains or additional structures.

With F/w₀ = 30.0 N/m, the branch equation gives sin(πε/a) = 30π/(300 × 0.400) = π/4. There are two solutions in 0 ≤ ε ≤ 0.400: approximately 0.1150 and 0.2850. They carry the same nominal sheet force, but the first has positive tangent stiffness and the second negative tangent stiffness. An equilibrium equation alone does not tell which is stable.

At F/w₀ = 40.0 N/m there is no equilibrium on this branch because 40 exceeds 38.1972 N/m. Calling it a slightly larger elastic stretch would ignore the nonlinear law. If displacement were instead rigidly prescribed, the axial strain degree of freedom would be fixed; the force-control argument would no longer test that allowed perturbation. Other nonuniform or atomic modes could still be unstable. Boundary conditions select which stability question is being asked.

A smooth branch can conceal an earlier atomic instability

An atomistic structure has many possible displacement patterns. A uniform stretch curve tests only one path through them. A lattice vibration pattern, often called a phonon mode in a periodic crystal, can soften: the energy cost of a small displacement in that pattern decreases. If an allowed pattern lowers the energy, the original structure is not stable against it even when the uniform tensile curve still rises. This is the physical reason to examine modes beyond the single scalar derivative.

First-principles work on strained graphene has explicitly checked such patterns along specified loading paths. A later study showed that allowing an enlarged periodic cell admitted a distortion under near-biaxial stretching that a smaller primitive cell excluded. The larger cell lost stability before the smaller cell’s smooth tensile maximum. This is a model-space difference, not an experimental defect introduced into the sheet.

The scalar model has no lattice coordinates and cannot discover that distortion. Its 38.20 N/m maximum therefore is only its own axial force-control limit. It cannot establish the ideal strength of graphene, identify a bond-breaking path or exclude an earlier bending, shear, edge or lattice instability. A numerical calculation needs adequate allowed modes as well as well-converged energy values.

The definition of stress also matters at large strain. T here uses reference width; dividing force by a current width would define another force-per-width quantity if the width changed. Likewise nominal stress and true stress use different area conventions. A peak of one curve cannot silently be treated as a peak of another after a strain-dependent area change. This toy model fixes width to keep that issue separate.

Change a reporting convention without changing the force

For the original strip, F_peak = T_peak w₀ = 38.1972 N/m × 10⁻⁶ m = 38.1972 µN. Suppose a report defines an equivalent three-dimensional nominal stress using an assigned constant thickness h: σ_eq = F/(w₀h) = T/h. That thickness is an explicit reporting convention in this sheet account, not an additional measured force or a claim that every atom has a sharply bounded wall.

With h = 0.300 nm, σ_eq,peak ≈ 127.324 GPa and the equivalent initial modulus K/h = 1000 GPa. With h = 0.400 nm, those become 95.493 GPa and 750 GPa. The actual force, strain, energy and sheet initial stiffness K remain unchanged. A larger reported stress produced only by a smaller chosen denominator is not stronger bonding.

The ratio of this reported peak stress to its initial modulus is T_peak/K = a/π ≈ 0.127324 for either constant h. It is not the peak strain 0.200: multiplying the initial modulus by that large strain would use a linear extrapolation past its validity. Equal initial stiffness also does not imply equal peak force. In a separate model with the same K but a = 0.300, the peak is at ε = 0.150 with T_peak ≈ 28.65 N/m.

Compare an ideal prediction with an experiment only after matching structure, loading mode, quantity and conventions. A particular experiment on multiwalled nanotubes observed outer-wall fracture and subsequent inner-section pullout under its mounting conditions. Treating the whole apparent cylinder as if every wall shared the tensile load would change both the mechanical question and its stress denominator. That result is a measured specimen account, not this sheet model or a universal rule for every multiwall tube.

A zero-temperature quasistatic branch also has no finite-time activation law. A simulation at a stated rapid strain rate allows a different time for thermally activated rearrangements than a slow experiment. Changing temperature or rate therefore needs its own model or observations; this scalar energy curve cannot supply a lifetime, a probability of failure or a production allowable.

Same original strip and same T_peak = 38.1972 N/m. Only the assigned constant thickness changes; these are equivalent nominal quantities.
Assigned h	Equivalent initial modulus K/h	Equivalent peak stress T_peak/h	Physical peak force
0.300 nm	1000 GPa	127.324 GPa	38.1972 µN
0.400 nm	750 GPa	95.493 GPa	38.1972 µN

Three original calculated plots for the fictional uniform strip with K = 300 N/m, reference length and width 1 micrometre, width held fixed and only uniform axial strain allowed. Horizontal axes are dimensionless strain from 0 to 0.400. Panel A plots stored energy per reference area W in J/m², rising from 0 to 9.7268; its local slope gives the nominal reference-width force T. W is not fracture-surface energy. Panel B plots T in N/m for original a = 0.400 and separate a = 0.300 with the same initial slope. Peaks are respectively 38.1972 at strain 0.200 and 28.6479 at strain 0.150. A circle marks the original a = 0.400 peak and its legend; a triangle marks the a = 0.300 peak and its legend, distinguishing the model series beyond colour. Rising parts are solid; mathematical post-peak continuations are dashed and do not represent surviving sheets. A dotted 30 N/m load line intersects the original branch at strain 0.1150167 and 0.2849833, marked by diamonds. Panel C plots force-controlled potential per reference area Φ = W − 30ε in J/m² for the original branch. The first equilibrium is a local minimum, marked by a circle; the second is a local maximum, marked by a square. These are local stability tests for the one allowed axial perturbation under fixed force, not global energy minima or tests of missing lattice modes. Fixed displacement changes the perturbations permitted by the model.

The calculation contains conventions

Check strain rate, size, chirality, wall definition and the quantity used to identify instability. Different energy models can produce different predictions. Numerical agreement between methods helps but does not replace relevant experimental validation.

Numerical convergence checks test whether a selected calculation is adequately resolved. Comparing different energy descriptions tests sensitivity to model choice. Neither automatically verifies missing deformation modes or validates a model against a corresponding experiment. A prediction can be internally consistent and still concern different temperature, geometry, constraints or response from the observation.

For a manufactured yarn, defects, alignment, contacts, load transfer, voids and specimen-scale failure introduce further questions. Keep the ideal single-structure result as a conditional reference and build the missing evidence at the actual scale. It is useful for explaining mechanisms and testing models; a high predicted number alone is not a cable, a component test or a guaranteed upper bound on every differently constrained experiment.

Worked examples

Work through the case

A defect-free model predicts a high failure stress. What is needed before using it as a yarn allowable?

Establish real tube state and defects.
Model and measure tube-to-tube load transfer.
Assess manufacturing variation and yarn-scale tests.

A justified bridge from the ideal structure to the real architecture; the prediction alone is insufficient.

Follow the work from stored energy to force

For the original scalar strip, find the peak strain, sheet force per width, actual force and stored energy. How does a straight-line extrapolation differ?

The first sine maximum is at ε = a/2 = 0.200. It gives T_peak = 300 × 0.400/π = 38.1972 N/m.
The reference width is 10⁻⁶ m, so F_peak = 38.1972 µN. The reference area is 10⁻¹² m².
At the peak cos(π/2) = 0, giving W = 300 × 0.400²/π² = 4.8634 J/m² and U = 4.8634 × 10⁻¹² J. This is stored energy on the branch, not fracture-surface energy.
The initial straight-line estimate would be Kε = 60.0 N/m at ε = 0.200, overestimating the nonlinear value by about 57.1% relative to 38.1972. An initial slope is not a large-strain failure law.

Peak strain 0.200, T_peak≈ 38.20 N/m, F_peak≈ 38.20 µN and stored U≈ 4.863×10⁻¹² J. The initial linear extrapolation gives 60.0 N/m and does not describe the supplied branch.

Solve equilibrium, then test the allowed perturbation

For F = 30.0 µN on the 1.00 µm-wide strip, find both scalar equilibrium strains and their force-control stability.

F/w₀ = 30.0 N/m, so sin(πε/a) = π/4. Take θ = arcsin(π/4)≈ 0.90334 radians.
The two solutions are ε₁ = aθ/π≈ 0.11502 and ε₂ = a(π−θ)/π≈ 0.28498.
Their tangent stiffnesses are ±K√[1−(π/4)²]≈ ±185.70 N/m. Positive A₀K_t makes the first a local potential-energy minimum; negative curvature makes the second unstable to the allowed axial perturbation.
This test is for fixed force and one uniform degree of freedom. It does not prove stability against absent lattice, bending or nonuniform modes.

Approximately 0.1150 and 0.2850 strain. The lower branch is locally stable to the scalar axial perturbation; the descending branch is unstable under the specified force control.

Keep the small-strain slope and change the nonlinear branch

A second invented branch retains K = 300 N/m but has a = 0.300. Compare its peak strain and force with the original. Does the common initial slope establish the same limiting response?

Both have T≈ Kε near zero, so their initial sheet slope is 300 N/m.
The second peak strain is a/2 = 0.150, compared with 0.200. Its peak T is 300 ×0.300/π≈ 28.6479 N/m.
For equal 1.00 µm reference width, its peak force is 28.6479 µN. It is 25% lower than 38.1972 µN, while the initial slopes remain identical.
The coefficients are separate stipulated models, not two measured carbon materials. Fitting a common elastic slope cannot choose between their nonlinear predictions.

The changed peak is 0.150 strain and 28.65 µN for this width. Equal initial stiffness does not establish equal ideal-model tensile limits.

Try the reasoning

Does computed mean experimentally made?
Show the answer and reasoning
No.
Computation establishes a conditional prediction.
Why state strain rate in simulations?
Show the answer and reasoning
Fast simulated loading can differ from experiments.
Thermally activated failure may depend on time.
Keep a = 0.400 but halve K to 150 N/m. What happens to initial sheet stiffness, peak strain and peak force for the original strip?
Show the answer and reasoning
Initial stiffness becomes 150 N/m; peak strain stays 0.200; peak force halves to about 19.10 µN.
K multiplies the whole energy and force branches. The sine maximum is set by a, so changing only the energy scale does not change its strain position.
Double only the strip’s reference length to 2.00 µm while keeping width and the same W(ε). Compare its small-extension axial spring stiffness, peak force and stored energy at the peak.
Show the answer and reasoning
Small-extension stiffness halves; peak force remains 38.1972 µN; stored energy doubles.
Near zero, F≈ K w₀ΔL/L₀, so the force-versus-extension stiffness is K w₀/L₀. Peak force depends on width and T_peak. U = L₀w₀W doubles with length. A weakest-link or finite-temperature size effect is absent from this homogeneous model.
Use the same original T_peak but choose a constant equivalent thickness 0.600 nm. Find the reported peak stress and initial equivalent modulus. Did physical load capacity double or halve?
Show the answer and reasoning
About 63.662 GPa and 500 GPa. Physical peak force is unchanged at 38.1972 µN.
Divide T and K by 0.600×10⁻⁹ m. Compared with the 0.300 nm convention, both reported 3 D quantities halve because the denominator doubled; the same sheet force is being described.
Two periodic atomistic calculations use the same energy description. A small cell gives a smooth rising curve to a high peak; a larger cell admits a distortion that lowers energy at an earlier strain. Must the larger-cell result be a defective sample or a numerical error?
Show the answer and reasoning
No. The small cell may exclude a physically relevant deformation pattern.
Convergence within a constrained model is not completeness of its allowed modes. Check the actual distortion, boundary conditions, mode stability and numerical convergence. Starting defect-free does not prevent a structure from losing stability.
Under the original scalar model and fixed 1.00 µm reference width, can F = 40.0 µN be carried by another strain in 0≤ε≤0.400?
Show the answer and reasoning
No equilibrium on this supplied branch carries it.
It requires T = 40.0 N/m, exceeding the branch maximum 38.1972. The model cannot determine the later fracture or alternative-structure outcome, only the failure of that equilibrium claim.
A colleague uses the negative tangent on the descending branch to declare that a perfectly displacement-fixed test has the identical one-coordinate instability as a force-controlled test. What is missing?
Show the answer and reasoning
The set of allowed perturbations and the external constraint energy differ.
If axial displacement is rigidly fixed, that uniform strain cannot vary as it did in the force-control potential. Other structural modes or finite machine compliance may still destabilise the state; the scalar force-control test does not decide them.
The static W(ε) curve contains no temperature or time variable. Can it predict a 20-year lifetime at a load below the peak?
Show the answer and reasoning
No.
A static local branch does not supply thermally activated transitions, defect evolution, environmental reactions, load-history dependence or probabilities. A sub-peak scalar equilibrium is not a service-life calculation.

What to carry forward

Label ideal predictions and conventions.
Do not transfer them directly to bulk products.

References

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