Materials Science · Graphene and carbon nanotubes: theory and measurement

Specific strength and the scope of a reported result

Compare maximum tensile force divided by linear density, correct a mismatch between pore-inclusive and solid-only density conventions, and keep a density-normalised result attached to its actual specimen, endpoint and uncertainty.

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

Compare density-normalised nanotube results at matching structural levels.
Convert force and linear density into specific tensile strength with consistent SI and tex units.
Reconcile pore-inclusive and solid-only stress/density pairs without mistaking a changed convention for improved performance.
Interpret a supplied uncertainty budget and a denominator error without turning specimen measurements into component allowables.

A ratio retains the scope of its inputs

Two assembled nanotube fibres can reach different maximum tensile forces while the fibre with less mass in the same gauge length has the larger specific tensile strength. Before ranking them, identify the tensile endpoint and check what mass the denominator represents. “Strength and strength relative to density” introduces the ratio; here the question is whether the reported inputs describe the same tested object.

Specific strength divides a defined tensile strength by a defined density. For a hollow or porous structure, area and volume conventions must be mutually consistent. Changing an effective wall thickness can change both stress and density estimates. For the same uniform gauge section, use σ = F/A and ρ = m/(AL): then σ/ρ = FL/m = F/μ, where μ = m/L is mass per length. The common area cancels only when the definitions describe the same specimen volume.

For the calculations here, F_max is the maximum axial tensile force in the recorded test. A₀ and L₀ are the reference area and gauge length before deformation; m is the mass assigned to that same reference gauge. Thus σ_max=F_max/A₀ and ρ₀=m/(A₀L₀) give R=σ_max/ρ₀=F_maxL₀/m=F_max/μ₀, with μ₀=m/L₀. A specific tensile-strength result requires that defined strength endpoint. A breaking force, yield force and maximum force need not be interchangeable.

The cancellation uses a uniform reference gauge or a justified representative linear density. If mass per length varies along a fibre, a long-piece average need not describe the tested segment; its weakest region and the failure location still matter. Mass includes the constituents actually assigned to that fibre, including retained coating or other non-load-bearing material if present. Quietly removing such mass while retaining the whole-fibre force changes the claimed property. “Cross-sectional area, gauge length and test conditions” explains the accompanying geometric and test record.

Calculate the ratio without choosing a circular area

Linear density is mass per length, not volume density. The textile unit tex means grams per kilometre: 1 tex = 1 g/km = 10⁻⁶ kg/m. Therefore 1 N/tex = 10⁶ N·m/kg. In SI units, Pa/(kg/m³) also reduces to N·m/kg. Matching dimensions do not make a strength ratio a measured energy absorption: rupture work requires a force–extension history, not just its maximum.

The table supplies central values for two invented, uniform assembled fibres. Both are tested axially with a 20.0 mm reference gauge under the same stipulated rate, temperature, surrounding gas and gripping assessment; their reported endpoint is F_max. These are teaching numbers, not measurements from a published nanotube study or design allowables.

Fibre A gives 0.480/0.200 = 2.40 N/tex. Fibre B gives 0.600/0.300 = 2.00 N/tex. B reaches the greater absolute force, but A has the greater nominal specific tensile strength. The comparison neither ranks every specimen made by either process nor establishes which fibre meets a particular component requirement.

Invented same-condition fibre results. Each gauge mass is consistent with its supplied linear density and 20.0 mm reference length.
Fibre	F_max (N)	μ₀ (tex)	Gauge mass (μg)	R (N/tex)
A	0.480	0.200	4.00	2.40
B	0.600	0.300	6.00	2.00

Change both geometric bases together

Use fibre A again. Assign a gross reference cross-section A_g = 200 μm² that includes its empty pores. For this invented two-component geometry, the solid occupies A_s = 100 μm² along the whole gauge and the pores have negligible mass. The solid fraction is f=A_s/A_g=0.500. “Solid-only” here means this explicitly assigned material volume; it is not a universal density of nanotubes or proof that every tube carries equal load.

With the same μ₀=2.00×10⁻⁷ kg/m, the gross density is ρ_g=μ₀/A_g=1000 kg/m³. The solid-only density is ρ_s=μ₀/A_s=2000 kg/m³. The force normalisations are σ_g=F_max/A_g=2.40 GPa and σ_s=F_max/A_s=4.80 GPa. Both compatible pairs give R=2.40×10⁶ N·m/kg=2.40 N/tex. The solid-only average is a bookkeeping stress, not a resolved local stress in every tube.

The relation is general within this stated geometry: ρ_g=fρ_s and σ_g=fσ_s, so σ_g/ρ_g=σ_s/ρ_s. Mixing σ_g with ρ_s instead gives fR, here 1.20 N/tex. It understates the consistently defined fibre ratio by 50%. Mixing σ_s with ρ_g would give R/f=4.80 N/tex, twice the consistent value. Neither mixed quotient is the same-volume σ/ρ result.

The diagram shows these same central values. The two compatible points lie on one constant-ratio line; the mixed point lies below it. Changing an area convention has changed neither the measured force nor the fibre mass. Physically densifying or otherwise processing a fibre is different: it can change contacts, alignment, defects and F_max as well as geometry, so a genuine property change requires a new measurement.

Original invented fibre A has maximum axial tensile force 0.480 newton and reference linear density 0.200 tex. Same reference gauge and mass apply to every point. Horizontal axis is the density assigned by a stated volume convention, in kilograms per cubic metre, from zero to 2400; vertical axis is the corresponding force-per-area normalisation, in gigapascals, from zero to six. A hollow circle marks the gross pore-inclusive pair: density 1000, stress 2.40. A square marks the solid-only pair: density 2000, stress 4.80. Both lie on the dashed constant-ratio guide for 2.40 newtons per tex. An orange cross at density 2000 and stress 2.40 combines the gross stress with solid-only density and instead gives 1.20 newtons per tex. The two matching points are different calculations for the same fibre, not different specimens or a measured densification path. The guide is a mathematical ratio relation; no measurement error bars or local nanotube stresses are represented.

Cancellation of area does not cancel measurement uncertainty

For R=F_max/μ₀, let u_F and u_μ denote supplied standard uncertainties: uncertainties expressed on an estimated-standard-deviation scale. A first-order propagation for small, uncorrelated input uncertainties gives u_R/R≈√[(u_F/F_max)²+(u_μ/μ₀)²]. The force and linear-density inputs must already include the relevant calibration, segment representativeness and model contributions for this budget to be complete. A plausible-looking number is not a substitute for that evaluation.

For invented fibre A, stipulate complete u_F=0.012 N and u_μ=0.010 tex for these two input estimates, with zero covariance. Their relative contributions are 0.025 and 0.050. Thus u_R=2.40√(0.025²+0.050²)=0.134 N/tex, about 0.13 N/tex after rounding. This is a standard uncertainty for the specified result and model, not a maximum error bound, a population spread or a guaranteed confidence interval. “Calibration, uncertainty and repeatability” develops those distinctions.

If σ and ρ were both calculated using the same uncertain A₀, they share that geometric input. Its common multiplicative effect cancels in this exact algebraic ratio. Treating those derived quantities as independent and adding the same area uncertainty twice would lose that cancellation. This does not erase force uncertainty, mass/length uncertainty, a nonrepresentative gauge, or genuinely different volume conventions.

Measurement error in the denominator is also not repaired by the cancellation. A mass-per-length estimate that is too small makes F_max/μ₀ too large. For example, a vibration-based estimate relies on a mechanical model of the vibrating specimen; a reduction based on a perfectly flexible-string model can require a stiffness correction for a real fibre. Direct mass/length estimates also require adequate sensitivity, reference length and representative sampling. Compare the actual method and uncertainty rather than assuming one instrument name guarantees an accurate denominator.

Match the compared objects

A tube-wall estimate, a bundle, a fibre and a composite have different architecture and void content. Use measured mass per length and force where helpful, and explain conversions. A spectacular microscopic ratio does not supply a macroscopic component allowable.

A fibre transfers force through contacts among finite, imperfectly aligned tubes and bundles. Some segments can straighten or slide while other segments carry more axial load. Density normalisation retains these architectural consequences in F_max; it does not repair them. In individual multiwalled-tube tests, outer-wall fracture followed by inner-wall pull-out likewise shows why a whole-tube or whole-fibre mass cannot silently be paired with a selected wall-strength value. “Measured strength of individual nanotubes” keeps that evidence at its actual scale.

A report should distinguish the nominal result for an individual specimen from an average and scatter across specimens, and distinguish a measured maximum from a service allowable. Keep gauge length, rate, environment, failure location, mass basis and uncertainty with the ratio. “Individual tubes, bundles, fibres and composites” develops the onward assembly distinction.

Worked examples

Work through the case

A fibre’s quoted density excludes its pores while its area includes them. Is its specific-strength comparison automatically consistent?

Identify the volume used for density.
Identify the area used for stress.
Reconcile the structural and geometric bases.

No. The numerator and denominator need compatible definitions.

Compare two fibres using the measured gauge mass

Use the invented table. Convert A’s 4.00 μg gauge mass and 20.0 mm reference length to linear density and specific tensile strength. Then compare A and B. What does the larger result establish?

Convert the reference quantities: m_A=4.00×10⁻⁹ kg and L₀=0.0200 m. Hence μ_A=m_A/L₀=2.00×10⁻⁷ kg/m=0.200 tex. Including material from a different segment would require a representativeness justification.
Use the defined maximum force: R_A=F_maxL₀/m_A=(0.480)(0.0200)/(4.00×10⁻⁹)=2.40×10⁶ N·m/kg=2.40 N/tex. No diameter was needed.
For B, μ_B=6.00×10⁻⁹/0.0200=3.00×10⁻⁷ kg/m=0.300 tex. R_B=0.600/0.300=2.00 N/tex. The nominal ratio R_A/R_B=1.20: A is 20% higher relative to B under these supplied conditions.
B nevertheless reaches 0.600 N rather than 0.480 N. The nominal per-mass comparison is not a statement that A carries the greater absolute force, a statistical conclusion about whole production populations, or an approved service allowable.

A gives 2.40 N/tex and B 2.00 N/tex. A’s nominal specific tensile strength is 20% higher; B’s measured maximum force is larger.

Correct the numerical consequence of a mixed density

A report pairs fibre A’s gross-area stress 2.40 GPa with its solid-only density 2000 kg/m³ and quotes 1.20 N/tex. Correct the comparison using either compatible basis, and interpret the change.

The written arithmetic is (2.40×10⁹)/(2000)=1.20×10⁶ N·m/kg. The error is its mismatched volume basis, not that division.
For the gross basis, use ρ_g=1000 kg/m³ with σ_g=2.40 GPa. Then R=2.40×10⁶ N·m/kg. For the solid-only basis, use ρ_s=2000 kg/m³ with σ_s=4.80 GPa and obtain the same R.
Both agree with the direct F_max/μ₀ value 2.40 N/tex. The correction doubles the reported mixed value; equivalently the mixed value was 50% below the consistent one. The percentages use different denominators.
No fibre was made stronger during this correction. The same force and mass have been reported on a mutually consistent basis. The solid-only stress is not evidence of uniform force sharing within the assembly.

The compatible result is 2.40 N/tex. Repairing the mismatched denominator changes the report, not the specimen.

Improve the input that currently contributes more

For A’s stipulated complete uncorrelated budget, compare the original u_R with the result of halving u_F alone and with halving u_μ alone. Hold all central values and other contributions fixed.

The original relative variance is (0.012/0.480)²+(0.010/0.200)²=0.003125. Its square root times 2.40 gives u_R=0.134164 N/tex.
Halving u_F to 0.006 N gives u_R=2.40√(0.0125²+0.050²)=0.123693 N/tex. The denominator contribution is still larger.
Halving u_μ to 0.005 tex gives u_R=2.40√(0.025²+0.025²)=0.084853 N/tex. This intervention reduces the combined standard uncertainty more in the supplied model.
These are conditional information improvements, not claims that a particular instrument or more repeats will achieve them. Shared corrections, correlations or an omitted representativeness effect would require a different budget.

Original u_R≈0.134 N/tex; halving force uncertainty gives ≈0.124 N/tex, while halving linear-density uncertainty gives ≈0.0849 N/tex.

Try the reasoning

Does normalising by density remove interface losses?
Show the answer and reasoning
No.
Load-transfer efficiency still affects strength.
What should accompany a headline ratio?
Show the answer and reasoning
Specimen level, definitions, conditions and uncertainty.
The ratio does not erase the source evidence.
For the same F_max=0.480 N, a linear-density estimate is 0.160 tex instead of the stipulated correct 0.200 tex. Calculate the reported and corrected ratios. Does a 20% denominator underestimate imply a 20% ratio overestimate?
Show the answer and reasoning
The reported ratio is 3.00 N/tex; the corrected ratio is 2.40 N/tex. The denominator is 20% low, but the ratio is 25% high relative to the corrected value.
The multiplier is 0.200/0.160=1.25, not 1.20. A known denominator error needs correction and a revised uncertainty evaluation; adding a symmetric uncertainty label to the uncorrected central value is not the same operation.
A different invented uniform fibre has maximum tensile force 0.360 N. The mass of its 30.0 mm reference gauge is 9.00 μg. Find μ₀ and R in SI and tex units. Can A’s 4.00 μg gauge mass be substituted because both fibres contain nanotubes?
Show the answer and reasoning
μ₀=(9.00×10⁻⁹)/(0.0300)=3.00×10⁻⁷ kg/m=0.300 tex. R=0.360/0.300=1.20 N/tex=1.20×10⁶ N·m/kg. A’s mass cannot be substituted.
The same tested object and reference gauge must define the mass per length. Shared constituent names do not establish identical linear density, architecture or failure force.
A second consistent two-component geometry has solid fraction f=0.400. Its solid-only density is 2000 kg/m³ and its gross-area maximum stress is 1.60 GPa. Calculate the gross density, solid-only average stress and compatible R. Identify the quotient obtained by pairing the gross stress with solid-only density.
Show the answer and reasoning
ρ_g=800 kg/m³; σ_s=4.00 GPa. Both compatible ratios give 2.00 N/tex. The mixed quotient gives 0.800 N/tex, only 40% of the compatible result.
Use ρ_g=fρ_s and σ_g=fσ_s. The mixed ratio is fR and is therefore 60% too low relative to R. This is a stated geometry/mass model, not a claim of uniform local stress or a universal nanotube density.
Suppose B has complete uncorrelated standard uncertainties u_F=0.018 N and u_μ=0.009 tex. Assume the A and B estimates also have zero cross-covariance. Find u_R for B and u_D for the nominal difference D=R_A−R_B. Does this alone identify a production-population advantage?
Show the answer and reasoning
u_R,B=2.00√(0.030²+0.030²)=0.084853 N/tex. D=0.400 N/tex and u_D=√(0.134164²+0.084853²)=0.158745 N/tex. No population advantage follows from these two specimen estimates alone.
The stipulated zero covariance permits adding the variances of the two result estimates. No specimen sampling plan, population distribution or statistical decision criterion is supplied. A shared force calibration or density model could correlate the results and change u_D.

What to carry forward

Use compatible geometric bases.
Keep specific properties attached to their specimen level.
For compatible reference-gauge definitions, R=F_max/μ₀; 1 N/tex equals 10⁶ N·m/kg.
Pore-inclusive stress requires pore-inclusive density. A change of convention alone does not improve the specimen.
State the tensile endpoint, mass basis and uncertainty; specimen measurements are not automatically population rankings or component allowables.

References

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