Materials Science · Graphene and carbon nanotubes: theory and measurement

Cross-sectional area, gauge length and test conditions

Apply wall and solid reference areas to the same nanotube force record, separate gauge extension from fixture movement, and compare uncertainty and test conditions before interpreting reported GPa values.

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

Explain why nominal thickness and area affect reported nanostructure stress.
Use diameter and nominal thickness to distinguish an assigned annular wall area, an outer-diameter approximation and a solid reference area.
Calculate engineering gauge strain with a stated fixture-compliance correction and a conditional standard-uncertainty budget.
Recalculate unlike reports on one explicit convention and identify which physical comparisons remain unsupported.

A force measurement needs a geometric convention

A thin cylindrical wall can be assigned area approximately πdt using a stated diameter d and nominal wall thickness t when t is much smaller than d. Using the mid-wall diameter makes πdt the geometric annular area; using the outer diameter gives a thin-wall approximation. A solid-cylinder area πd²/4 gives a different number. Graphene can also be described using force per width instead of a nominal thickness.

A test first supplies observations such as force and a change of separation. A stress then assigns that force to a stated reference area: σ = F/A. “Applied force and stress over an area” develops this conversion. At the atomic scale, an assigned wall thickness is a continuum convention; it is not automatically a sharp physical boundary measured in the tensile test. Keep that choice alongside the force record.

For a circular annular-wall model, call the outer diameter d_o and the assigned wall thickness t. Its inner boundary is d_i = d_o − 2t, and its mid-wall diameter is d_m = d_o − t. Subtracting the two disk areas gives A_wall = π(d_o² − d_i²)/4 = πt(d_o − t) = πd_m t. That last expression is exact for this assigned annulus. The model requires 0 < 2t < d_o. Here d_i means the inner boundary of this one nominal wall; in a multiwalled tube it is not the innermost bore of the entire tube.

Using the outer diameter in πd_o t instead gives a slightly larger area than that annulus: the excess is πt². Its relative excess is t/(d_o − t). It is a thin-wall approximation when t is small relative to d_o. A solid reference area A_solid = πd_o²/4 includes the enclosed region and any hollow space; it is a different normalisation, not evidence that every enclosed region carries the measured force.

Consider invented values d_o = 20.0 nm, t = 0.34 nm and tensile force F = 400 nN. The nominal annulus has d_i = 19.32 nm and d_m = 19.66 nm. Its area is about 21.00 nm², whereas the solid reference area is 314.16 nm². Dividing the same force by them gives about 19.05 GPa and 1.273 GPa. The specimen did not become stronger when the denominator changed. The geometric worked case below also checks the outer-diameter approximation.

In a multiwalled tube, an outer-wall area needs a force-path interpretation. In tensile observations where the outer wall broke and inner sections pulled out, outer-wall normalisation was tied to that loading and failure evidence. Wall count alone does not establish equal load sharing. “Measured strength of individual nanotubes” develops that experimental distinction; none of the invented values here is a measurement from that study.

For a flat sheet loaded uniformly in its plane across width W, an alternative is force per width, N_2D = F/W, with units N/m. If a nominal thickness t is then chosen, the associated three-dimensional stress is σ = N_2D/t. A different t changes the quoted Pa value while F/W stays fixed. A direct sheet-tension test and a local indentation test also have different force paths and inference models; a total indentation force is not simply an axial force divided by an arbitrary sheet width.

Original invented circular reference-area comparison at outer diameter 20.0 nanometres, nominal one-wall thickness 0.34 nanometre and force 400 nanonewtons. Both circular drawings use the same diameter scale. The blue assigned annulus has outer radius 90 pixels and inner radius 86.94 pixels, corresponding to diameters 20.0 and 19.32 nanometres. Its thickness is a continuum convention, not an observed sharp atomic boundary; the inner circle is not the bore of a whole multiwalled tube. The orange filled disk is the solid reference area over the same outer diameter. The annular area is 20.9997 square nanometres; the solid reference area is 314.1593. Below, three bars share a linear zero-based stress axis from 0 to 20 gigapascals. Exact assigned annulus, outer-diameter approximation and solid reference stresses are 19.0479, 18.7241 and 1.27324 gigapascals. Labels and distinct bar patterns supplement colour. Changing normalisation did not change the force or establish which walls carried it.

Gauge and protocol add context

Report the free length, gripping, displacement measurement, rate, temperature and structural characterisation. An uncertain diameter or thickness can strongly affect the inferred stress. Compare forces and conventions before ranking reported GPa values.

Define the initial separation L₀ of the specimen points that delimit the gauge. If their separation increases by ΔL, the average engineering strain is ε = ΔL/L₀. The result is dimensionless: a 60 nm increase over 2000 nm gives 0.030, or 3.0%. It is an average over that gauge, not necessarily the highest local strain. “Deformation and relative strain” explains why the measurement interval matters.

The driven end of an apparatus can move farther than the specimen gauge extends. In a deliberately simple quasistatic series model, the same axial force F passes through the specimen and a stable linear fixture with compliance C_sys. Compliance is extension per force. If the measured change of end-to-end apparatus separation is Δu, then Δu = ΔL + C_sys F, so ΔL = Δu − C_sys F. All changes are measured from the same zero-force reference. The model excludes slip, evolving grip damage, slack take-up and unmeasured deformation outside the defined terms.

For an invented point before failure, take F = 200 nN, Δu = 80 nm, C_sys = 0.10 nm/nN and L₀ = 2000 nm. The fixture contributes 20 nm, leaving ΔL = 60 nm and ε = 3.0%. Assigning the whole 80 nm to the gauge would give 4.0%, one percentage point higher. The correction requires an independently justified fixture response; it cannot be chosen merely to obtain an expected modulus. A negative inferred extension would prompt investigation of the model, zero and records.

Direct imaging of gauge endpoints can instead measure their relative movement. It still needs spatial calibration, an identified reference state and adequate tracking. A force sensor may itself deflect; that deflection is not automatically specimen extension. In some devices specimen and support carry force in parallel rather than in series, so the appropriate background-force relation differs. Draw the actual force path before subtracting any apparatus signal.

Gauge length also sets engineering strain rate: with a fixed reference L₀, ε̇ = (rate of gauge extension)/L₀. The same gauge-extension speed of 20 nm/s gives 0.010 s⁻¹ over 2000 nm and 0.005 s⁻¹ over 4000 nm. Actuator speed need not equal gauge-extension speed. Changing the sampled length can also change which defects enter a failure test; neither a universal length correction nor a fixed rate-strength trend follows from geometry alone.

Separate uncertainty in the observations from a change of convention

A nominal thickness chosen to define stress is not automatically a random measurement error. For σ = F/(πd_m t) with t fixed by convention, small uncorrelated standard uncertainties in force and mid-wall diameter give (u_σ/σ)² ≈ (u_F/F)² + (u_d/d_m)². If d_m is obtained from d_o − t with that t fixed, its uncertainty follows the measured d_o. If thickness is instead an estimated physical input, or inputs share information, its uncertainty and covariance require the actual model.

For the 400 nN wall calculation, suppose u_F = 8 nN and u_d = 0.20 nm, stipulated uncorrelated, with nominal t = 0.34 nm fixed. The relative standard uncertainty is √[(8/400)² + (0.20/19.66)²] ≈ 0.02244. Thus u_σ ≈ 0.427 GPa around the nominal annular result 19.05 GPa. This is conditional on the area and measurement model. It does not quantify the separate uncertainty about whether one wall actually carries all the force. Choosing a different reference thickness is a stated recalculation, not a contribution to hide inside this budget.

The gauge-strain model has several sensitivities. For ε = (Δu − C_sys F)/L₀, a small change in Δu acts with coefficient 1/L₀; a change in C_sys acts with −F/L₀; a change in F acts with −C_sys/L₀; and a change in L₀ acts with −ε/L₀. Multiplying each input standard uncertainty by the magnitude of its coefficient expresses its contribution in strain units.

Use the invented complete input budget below for that four-input model. Inputs are stipulated uncorrelated and uncertainty is small enough for first-order propagation. Independent contributions add as variances, not as signed uncertainties: u_ε² ≈ (u_Δu/L₀)² + (F u_C/L₀)² + (C_sys u_F/L₀)² + (ε u_L/L₀)². Negative sensitivities do not make the squared contributions negative. Correlated inputs, drift, slip or a deficient fixture model need further evaluation rather than this root-sum-of-squares shortcut. “Calibration, uncertainty and repeatability” supplies the preceding uncertainty distinctions.

Here the variance is 0.00000210, so u_ε ≈ 0.001449. The central strain 0.03000 is 3.000%; its standard uncertainty is about 0.145 percentage points, not 0.145% of the strain. Relative to 0.03000, it is about 4.83%. No coverage probability or guaranteed error bound follows from this standard uncertainty.

Invented input budget for the stated pre-failure series model. Values and uncertainty contributions belong to this example, not to a named instrument or nanostructure.
Input estimate	Standard uncertainty	Magnitude of contribution to strain
Δu = 80 nm	2 nm	2/2000 = 0.001000
C_sys = 0.10 nm/nN	0.01 nm/nN	200 × 0.01/2000 = 0.001000
F = 200 nN	2 nN	0.10 × 2/2000 = 0.000100
L₀ = 2000 nm	20 nm	0.030 × 20/2000 = 0.000300

Original invented pre-failure series-model point. The upper force-path schematic has no geometric scale: the same axial force of 200 nanonewtons passes through a linear stable fixture and the specimen. A fixed support, fixture spring and specimen lie in series. Initial specimen gauge length is 2000 nanometres and fixture compliance is 0.10 nanometre per nanonewton. The middle stacked bar uses a linear displacement scale from 0 to 80 nanometres: fixture extension 20 plus gauge extension 60 equals apparatus movement 80. These segments encode displacement contributions, not the physical lengths of the drawn parts. The lower zero-based strain axis runs from 0 to 5 percent. The corrected gauge strain is a filled blue circle at 3.000 percent with a one-standard-uncertainty bar of 0.144914 percentage points on either side. A hollow orange diamond at 4.000 percent shows the erroneous assignment of all apparatus movement to the gauge, with no valid gauge-strain uncertainty attached. Inputs and their complete conditional budget are stipulated uncorrelated; the bar is not a confidence interval. Slip, changing grip response and slack are excluded from this teaching model.

Recalculate the convention, then examine the physical comparison

For two reports, first recover the force, diameter and thickness definitions and the gauge displacement actually used. Recalculate on a common explicit reference convention when the needed inputs exist. Preserve both the original report and that recalculation. Numerical agreement after normalisation only resolves that arithmetic question; it does not make different specimens or protocols identical.

Keep wall count and evidence of the loaded walls, gauge endpoints and length, strain definition, rate, temperature, surroundings, grip behaviour and uncertainty with the result. An individual tube, an assembled nanotube fibre and a sheet need different structural descriptions. A longer gauge or a different environment can introduce real changes as well as changes in how a signal is interpreted. “Defects, temperature, strain rate and environment” develops those physical variables.

In the comparison exercise below, two apparently different GPa values become equal on one chosen thin-wall reference area. Their gauge lengths and strain rates still differ. That does not show the differences have no effect; it shows that area conversion must precede any attempted explanation of a residual material difference. If wall loading is unverified, call the result a reference-area stress rather than an established intrinsic outer-wall strength.

Worked examples

Work through the case

An invented test gives failure force 10 nN. Compare stresses assigned using wall areas 10 nm² and 20 nm².

Convert 10 nN to 10⁻⁸ N and 10 nm² to 10⁻¹⁷ m².
The first stress is 10⁻⁸/10⁻¹⁷ = 10⁹ Pa = 1 GPa.
The second area is twice as large, so its stress is 0.5 GPa.

1 GPa versus 0.5 GPa, with exactly the same measured force. These invented values illustrate a convention, not nanotube test data.

Three reference areas for one force record

For the invented 400 nN force, d_o = 20.0 nm and fixed nominal t = 0.34 nm, calculate the assigned annulus, outer-diameter thin-wall and solid reference stresses. Which difference is a thin-wall approximation?

The inner boundary is d_i = 20.0 − 2 × 0.34 = 19.32 nm. The mid-wall diameter is d_m = 20.0 − 0.34 = 19.66 nm.
The assigned annulus is π × 19.66 × 0.34 = 20.9997 nm². The outer-diameter approximation is π × 20.0 × 0.34 = 21.3628 nm². The solid area is π × 20.0²/4 = 314.1593 nm².
One nN/nm² is 10⁻⁹ N/10⁻¹⁸ m² = 10⁹ Pa = 1 GPa. Thus the stresses are 400/20.9997 ≈ 19.05 GPa, 400/21.3628 ≈ 18.72 GPa and 400/314.1593 ≈ 1.273 GPa.
The outer-diameter area exceeds the annulus by πt² ≈ 0.3632 nm², about 1.73% of the annular area. Its stress is consequently slightly lower. The solid normalisation gives an area about 14.96 times the annulus, a change of convention rather than a thin-wall correction.
All three numbers use the same measured force. The nominal annulus makes a geometric assignment; whether the actual loaded structure corresponds to one wall needs separate evidence.

Approximately 19.05, 18.72 and 1.273 GPa. Only the first-to-second difference is the stated thin-wall approximation; the solid reference changes the normalisation.

A fixture correction and its uncertainty

Use the invented series-model point and complete input budget. Calculate gauge strain and its standard uncertainty, then assess a claim that reducing only displacement-readout uncertainty will halve the total strain uncertainty.

Fixture extension is C_sys F = 0.10 × 200 = 20 nm. The gauge extends 80 − 20 = 60 nm, so ε = 60/2000 = 0.03000 = 3.000%.
The four strain contributions are 0.001000, 0.001000, 0.000100 and 0.000300. Their squared sum is 1.00 × 10⁻⁶ + 1.00 × 10⁻⁶ + 1.00 × 10⁻⁸ + 9.00 × 10⁻⁸ = 2.10 × 10⁻⁶.
Taking the square root gives u_ε = 0.001449, equivalent to 0.145 percentage points. This budget is for a standard uncertainty; no multiplier or coverage claim has been added.
If u_Δu alone falls from 2 to 1 nm, its contribution falls to 0.000500. The other three remain unchanged. The new variance is 0.25 × 10⁻⁶ + 1.00 × 10⁻⁶ + 0.01 × 10⁻⁶ + 0.09 × 10⁻⁶ = 1.35 × 10⁻⁶, so u_ε ≈ 0.001162, or 0.116 percentage points.
Halving one readout contribution does not halve the combined uncertainty. The independently evaluated fixture response is now the largest supplied contribution. This conclusion belongs to the stated model and budget, not every nanoscale test.

Gauge strain is 3.000% with standard uncertainty about 0.145 percentage points. Improving only u_Δu to 1 nm reduces the standard uncertainty to about 0.116 percentage points, not half its original value.

Try the reasoning

Is nominal wall thickness necessarily directly measured in the test?
Show the answer and reasoning
Not always.
It can be a chosen convention that must be stated.
Why report gauge length?
Show the answer and reasoning
It defines strain and the sampled defect length.
Different lengths can change interpretation.
Invented report A uses F = 400 nN, d_o = 20 nm and πd_o t with nominal t = 0.34 nm, giving about 18.72 GPa. Report B uses F = 500 nN, d_o = 25 nm and a solid area, giving about 1.019 GPa. Recalculate B on A’s outer-diameter thin-wall convention. Both temperatures are 300 K, but A has a 2000 nm gauge at 0.010 s⁻¹ and B a 4000 nm gauge at 0.005 s⁻¹. What does the recalculation resolve?
Show the answer and reasoning
B becomes 500/(π × 25 × 0.34) ≈ 18.72 GPa, the same reference-area stress as A. The original numerical contrast arose from unlike area conventions; the specimens and test histories are not thereby equivalent.
B has 25% more force and 25% more thin-wall reference area, so their ratio is unchanged. Its solid-reference value was 500/(π × 25²/4) ≈ 1.019 GPa. Different gauge length and rate still need interpretation; temperature agreement does not remove those differences. Actual wall loading must also be established before calling the common number an intrinsic wall strength.
Keep the pre-failure series-model central values and other supplied uncertainties fixed, but increase u_C from 0.01 to 0.02 nm/nN. Find u_ε. Does choosing nominal wall thickness 0.68 rather than 0.34 nm change this gauge-strain result?
Show the answer and reasoning
u_ε ≈ 0.002258, or 0.226 percentage points. The gauge-strain model contains no wall thickness, so changing that stress convention does not change this result.
The compliance contribution doubles from 0.001 to 0.002. The strain variance becomes (0.001)² + (0.002)² + (0.0001)² + (0.0003)² = 5.10 × 10⁻⁶. Thickness can change a reported stress or modulus normalisation, but it is not an input to this displacement-based strain equation.
An invented flat sheet is in uniform axial tension with force 200 nN across width 100 nm. Find F/W and the stress assigned by nominal thicknesses 0.34 nm and 0.68 nm. Can the same calculation be applied directly to the total indentation force on a suspended membrane?
Show the answer and reasoning
F/W = 2 N/m; the assigned stresses are about 5.88 GPa and 2.94 GPa. A total indentation force requires its own membrane/contact model.
200 nN/100 nm = 2 N/m. Divide by thickness in metres: 2/(0.34 × 10⁻⁹) Pa, or half that when thickness doubles. Force per width remains 2 N/m. The calculation assumed a uniform axial force across a defined width; indentation has a different load direction and stress distribution.

What to carry forward

Keep force and area convention together.
Compare test geometry before numerical strength.
Use the assigned wall geometry consistently: mid-wall πd_m t is the annulus; outer-diameter πd_o t is a thin-wall approximation.
Gauge strain uses specimen endpoint separation. A fixture correction and its uncertainty need a justified force-path model.
A conditional measurement budget does not validate wall load sharing or make different gauges, rates and environments interchangeable.

References

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