Measured strength of individual nanotubes

A small experiment can answer a precise question

Yu and colleagues reported tensile tests on individual multiwalled carbon nanotubes in Science in 2000. The study examined breaking behaviour under a specific gripping and observation arrangement. It is empirical evidence for those specimens, distinct from an ideal-lattice calculation.

A tensile test pulls a specimen along its length. The test does not directly display “strength”: it records observables such as sensor motion and images. Force is inferred through a sensor model, elongation through identified gauge endpoints, and stress through an explicitly chosen loaded area. Each inference has conditions. A smooth-looking graph cannot supply a missing endpoint or identify which wall broke.

In this study a tube bridged two atomic-force-microscope tips inside a scanning electron microscope. One support was relatively rigid; the other deflected as a force sensor. Video supplied length changes, and transmission-electron microscopy examined fragments after failure. These complementary observations linked an individual specimen to its failure evidence. The worked record below is invented; it does not reproduce their data or calibration.

Separate sensor deflection from specimen extension

Consider a straight, aligned specimen whose gauge endpoints remain fixed to the two grips while it is intact. All tracked features lie in the image plane. Use a single calibrated x axis pointing from the compliant grip towards the driven rigid grip. Motions are measured from the same initially unloaded frame. Let Δx_R be the motion of the driven gauge end, Δx_S that of the compliant gauge end, and Δx_B that of a reference on the compliant sensor base. R, S and B label features, not three independent force sensors.

The sensor bends relative to its own base. Its incremental deflection is q = Δx_S − Δx_B. For this invented, quasistatic, linear sensor, an independently calibrated coefficient k = 0.200 N/m applies to this loading point and direction. With no initial force and no other axial forces, the tensile-force magnitude is F = kq for positive q. Because 1 nm = 10⁻⁹ m, multiplying N/m by nm gives nN. Inertia, sensor nonlinearity and a changing loading point are excluded from this model.

The intact gauge length changes by ΔL = Δx_R − Δx_S. Its original length is L₀ = 2000 nm, so engineering strain is ε = ΔL/L₀. The two differences answer different questions: q measures bending of the force sensor; ΔL measures separation of the material gauge endpoints. Drive motion alone includes motion of the compliant end and is not specimen elongation.

This treatment assumes no slipping at either gauge endpoint and no unaccounted change of viewing geometry. A measured image distance is a projection: a tube rotating out of plane can change its projected length without the same material extension. Calibration must apply to the imaging conditions and direction used. A valid sensor coefficient also belongs to a specified geometry and mode of loading; a nominal catalogue stiffness is not automatically that coefficient.

Read a record before fitting a material property

The table gives exact stipulated values for a teaching calculation, with no uncertainty model implied. The motions are image-derived changes from the unloaded frame, not absolute positions. The specimen is visibly continuous through 3.0 s. At 3.1 s the images show a gap inside the gauge, both attachments remain in place, and the sensor has relaxed to q = 0 in this ideal record.

At 3.0 s, q = 105 − 15 = 90 nm and F = (0.200 N/m)(90 × 10⁻⁹ m) = 18 × 10⁻⁹ N = 18 nN. The gauge extension is 165 − 105 = 60 nm; ε = 60/2000 = 0.030 = 3.0%. Ignoring base motion would give 21 nN. Calling all 165 nm of drive motion “extension” would give 8.25% instead of 3.0%. These are errors in the measurement model, not new nanotube properties.

The intact sampled points with nonzero extension have F/ΔL = 0.300 N/m. This is a force–extension ratio for the measured gauge under the stated model. It is not a Young modulus: a modulus requires stress versus strain, a justified loaded area and an appropriate material-response interpretation. The data also give an average engineering-strain rate of 0.010 s⁻¹ over each of the three intact one-second intervals; they do not determine the instantaneous rate during the final unobserved interval.

After the gauge breaks, R and S no longer bound one continuous material segment. Their separation can still be measured, but assigning (169 − 17)/2000 = 7.6% to the broken specimen would mistake an empty-span change for material strain. Do not extend an intact-specimen formula beyond its domain.

Ask which correction changes which result

A common frame translation adds the same offset to Δx_R, Δx_S and Δx_B at a given time. It cancels in both differences. This is why tracking a reference can distinguish sensor bending from common image motion. Cancellation does not correct rotation, unequal axis scales, distortion, endpoint misidentification or motion out of the image plane.

There is another useful, conditional cancellation. Suppose the 3.0 s record originally used 10 nm per pixel: R, S and B moved 16.5, 10.5 and 1.5 pixels, and L₀ spanned 200 pixels. Revising the single common conversion to 11 nm per pixel changes q from 90 to 99 nm and ΔL from 60 to 66 nm, while L₀ becomes 2200 nm. Thus ε remains 3.0%. With the separately calibrated k explicitly held fixed, F rises from 18 to 19.8 nN. A dimensionless ratio can cancel a shared scale factor while a dimensional force estimate still changes.

That cancellation is not a universal uncertainty statement. If k was itself obtained using image-derived cantilever dimensions, the same calibration change could also alter k. Correlations then belong in the complete measurement model. Similarly, a 5 nm displacement bound is not automatically the uncertainty of an entire test. If an analysis explicitly supplies |q| ≤ 5 nm with k fixed at 0.200 N/m, it implies |F| ≤ 1 nN for that residual signal; it does not prove exactly zero force.

A load drop needs a cause and a time window

The record establishes that the specimen was intact at 3.0 s and broken by 3.1 s. With the stated synchronized observations and no reconnection, rupture occurred within (3.0 s, 3.1 s]. The largest recorded intact load was 18 nN. It need not be the instantaneous rupture force or the highest force reached between frames. For example, an unseen rise followed by rupture could fit the same sampled endpoints. The height of the plotted axis supplies no physical upper bound.

A force drop alone would be ambiguous. Loss of an attachment, slipping of a gripped section, specimen rupture and an instrumental change can all reduce sensor deflection. Images showing an interior gauge gap with both attachments retained support a different event classification from images showing an intact tube released at one end. Even then, what happens to individual walls requires adequate structural evidence.

Keep unsuccessful and interrupted tests in the outcome record. Suppose three attempts yield: T1, gauge rupture with an 18 nN last intact sample; T2, attachment loss after a 14 nN maximum observed load with no gauge rupture identified; T3, an intact test stopped at 16 nN. Their arithmetic mean, 16 nN, is a mean of three reported load maxima or last samples. It is not a mean nanotube breaking force. Neither discarding T2 silently nor relabelling it as specimen fracture repairs the claim.

A defensible specimen-level report identifies the gauge and attachments, force and image calibrations, time resolution, area convention, loading history and observed failure category. Temperature, surrounding medium or pressure, alignment and specimen structure also define the tested conditions. A stopped test gives survival under its recorded history, not a material allowable or an unlimited lifetime. Broader population claims require an appropriate sampling and outcome-analysis design.

The force path matters

In a multiwalled tube, the walls need not share force equally. A failure involving an outer wall differs from simultaneous rupture of every wall. Interpret stress with the area convention and observed failure mechanism used in the study; do not turn one result into a universal carbon-nanotube strength.

The study reported longer combined fragments after some breaks and interpreted these with outer-wall rupture and inner-wall pullout. This supports a wall-specific failure account; it does not directly measure force in every wall. Longer fragments do not mean carbon was created or that their sum is an elastic gauge extension. Nested material and material initially within attachments must be distinguished from a continuous original gauge.

An area convention converts a force to a reported stress, not to an observed wall identity. The companion lesson Cross-sectional area, gauge length and test conditions develops that conversion. Here the main conclusion is about evidence: preserve the observed load path and state which further assumptions connect it to a wall stress. A stronger material claim cannot be obtained merely by dividing by a smaller area.

Worked examples

Try the reasoning

1. Does a measured individual tube establish yarn performance?

   Show the answer and reasoning

   No.

   A yarn adds alignment, porosity and interfaces.

2. Why examine the failure images?

   Show the answer and reasoning

   They help identify the loaded and broken structure.

   Force alone does not reveal load sharing.

3. At 3.0 s, an image-registration offset adds 10 nm to all three recorded motions R, S and B. What force and strain does the model infer?

   Show the answer and reasoning

   18 nN and 3.0%, unchanged.

   The new values are 175, 115 and 25 nm. Then q = 115 − 25 = 90 nm and ΔL = 175 − 115 = 60 nm. Only the common translation cancels; distortion or a changed viewing angle would need a different correction.

4. A different intact specimen has L₀ = 2500 nm, k = 0.150 N/m, and motions R = 190 nm, S = 130 nm, B = 70 nm. Infer its force and strain. Does a greater drive motion imply a greater strain than the original 3.0 s frame?

   Show the answer and reasoning

   9 nN and 2.4%; compare gauge extension divided by the correct original length.

   q = 60 nm gives F = 0.150 × 60 = 9 nN. ΔL = 60 nm, so ε = 60/2500 = 0.024. The original 3.0 s strain was 3.0% for the same extension and a shorter gauge. Drive motion alone is insufficient even when compared between tests.

5. For the same original pixel record, use 8 nm per pixel instead of 10, with independently calibrated k still 0.200 N/m. Find q, ΔL, L₀, F and ε.

   Show the answer and reasoning

   72 nm, 48 nm, 1600 nm, 14.4 nN and 3.0%.

   Multiply every original image-derived length by 0.8: the pixel differences are 9 and 6, and the original gauge is 200 pixels. Force changes with q, but 48/1600 = 0.030. This assumes one common scale and unchanged k, not independent arbitrary changes in each length.

6. A post-event analysis supplies a bound |q| ≤ 5 nm, with k = 0.200 N/m treated as fixed. What can be said about force? Does it identify the original rupture force?

   Show the answer and reasoning

   The residual magnitude is at most 1 nN under that bound; the rupture force remains unresolved.

   |F| ≤ k|q| = 0.200 × 5 nN = 1 nN. A bounded signal is not exactly zero. A post-event force bound also says nothing quantitative about an unobserved pre-rupture peak.

7. The force drops, but images show a continuous tube and a detached right-hand attachment. Can this be entered as a nanotube tensile-fracture result?

   Show the answer and reasoning

   No; it is an attachment-loss outcome with the recorded load history.

   The image contradicts gauge rupture as the identified event. Retain the attempt and what was actually observed. Neither the force drop nor the experimenter’s intended endpoint overrides the observed failure category.

8. Could an unrecorded 25 nN load followed by rupture between 3.0 and 3.1 s produce the same sampled table? Does the graph ceiling rule it out?

   Show the answer and reasoning

   The samples alone do not rule it out; the plotting limit is not a physical bound.

   The table states endpoint observations, not the continuous intervening force path. A new bound would require further validated dynamic, rate or measurement information. Do not manufacture an upper bound from display choices.

9. At 3.1 s the endpoint-motion difference is 152 nm. Why is 152/2000 not the strain of the original intact tube?

   Show the answer and reasoning

   The observed gap interrupts the material gauge; endpoint separation includes empty space.

   The arithmetic 7.6% is a ratio of positions to original length, but the physical identification of those positions as endpoints of one continuous deforming segment has failed. A formula needs its material domain as well as correct units.

10. From 2.0 to 3.0 s, intact strain rises from 2.0% to 3.0%. What average strain rate follows? Does it resolve the final event rate or establish behaviour at ten times that rate?

    Show the answer and reasoning

    0.010 s⁻¹ over that interval; neither further inference follows.

    The strain increment is 0.030 − 0.020 = 0.010 over 1.0 s. No intermediate final-event samples or changed-rate experiment were supplied. An interval average is not an instantaneous rupture history or a rate-transfer law.

What to carry forward

• Use individual-specimen measurements as individual-specimen evidence.
• Retain force path, area and failure observations.
• Force-sensor motion, drive motion and material-gauge extension are distinct observables.
• Calibration cancellations are conditional on the whole measurement model.
• Identify event time, failure category and unresolved quantities before reporting a strength statistic.