Specific properties and component performance

A fibre is still not the complete component

A cable, laminate or reinforced part adds terminations, load-sharing geometry and environmental exposure. Connections can govern usable force even when the fibre itself is strong. Stiffness, creep, fatigue and damage can impose other constraints.

A small tension link holds two instrument parts a prescribed distance apart. Its free fibre body joins two terminations, the end regions that transfer force to the supports. Its function requires carrying a force and keeping the support-to-support movement within a limit. A very light fibre that stretches too far, slips at an end or needs heavy fittings may not give the lightest satisfactory link.

Follow one straight axial load path. In quasistatic equilibrium without distributed axial loads, the same tensile force passes through the left termination, free body and right termination. Their force capacities do not add: they are consecutive elements, not three parallel cables. Their axial extensions do add when the two supports separate. Bending, knots, pulley contact, eccentricity, impact and cable self-weight need different or additional accounts.

Connections matter experimentally as well as in a finished component. In a 2013 study of assembled CNT sheets, changing the gripping arrangement changed the observed delamination and measured tensile response. Grip displacement also differed from direct specimen deformation; yarn strain measurements had additional limitations. This is evidence about those tests, not a numerical efficiency factor for a new cable. A test fixture can hide or introduce a governing path.

Define a body and connection family before scaling it

All numerical inputs below are invented for two candidate families, A and B. They describe uniform dry assembled fibre bodies, not individual nanotube walls. The compared load is P = 1.00 N, quasistatically applied along the axis at the same stipulated temperature and surrounding environment. Free-body reference length L ranges from 0.20 to 2.00 m. The body linear density λ, mass per unit length, is continuously selectable from 0.10 to 2.50 tex. One tex is 1 g/km = 10⁻⁶ kg/m, so λ in tex times L in metres is body mass in milligrams.

The body’s supplied static capacity is Rλ, where R is a matched force-at-endpoint divided by body linear density, in N/tex. Its meaning depends on the stated specimen length and protocol. Here R is stipulated constant within the chosen family and range; length effects, scatter and progressive failures are not inferred away. A real fibre must establish the appropriate length-dependent response rather than borrowing an ideal-tube number.

Let η be the dimensionless retained fraction of body capacity under the specified end-load transfer, with 0 < η ≤ 1. It represents the weaker end’s proportional transfer limit. A separate supplied force ceiling C represents another connection limit that does not increase when more body fibre is added. Both terminations together have fixed mass M and fixed total series compliance J in this illustrative family. Compliance is extension per applied force; J includes both end regions, while the free length L excludes them.

The table supplies a separate linear body stiffness input H, axial rigidity per unit linear density, in N/tex. Axial rigidity is the coefficient relating axial force to engineering strain: F = Hλ ε, with ε = δ_body/L. Thus the small-extension body stiffness is Hλ/L. H and R have the same units here but answer different questions: H gives the force per strain; R supplies the stipulated static endpoint. The linear relation is assumed only in the load/strain range used for the service calculations.

The model holds η, R, H, C, M and J fixed while λ and L vary over the stated range. End mass includes all end hardware and retained material, including any extra fibre inside it; that material is not counted again in λL. No separate sheath or distributed added mass is present. If a real termination must grow with body size, or if adding strands changes contact sharing or stiffness, this fixed-family model must be revised.

Derive the useful index

An ideal static tensile tie can benefit from strength/density; another geometry may require a different combination. Include porosity, matrix and connection mass consistently. Do not substitute one constituent’s specific strength for a finished assembly’s measured performance.

For the stated serial path, the model’s component capacity is F_cap = min(ηRλ, C). The min operation selects the smaller entry. Because η ≤ 1, the proportional connection limit is already no greater than the body’s Rλ capacity. In a more detailed model with unequal end limits, take the minimum over the body and both ends explicitly. Adding their force capacities would count the same load several times.

Total component mass is m_total = λL + M when λ is in kg/m and M in kg. If λ is in tex and M in mg, the same numerical relation gives m_total in mg for L in m. A capacity-to-mass comparison for one chosen component is therefore F_cap/m_total, not simply R. Keep its length, connections, force endpoint and mass boundary with that ratio.

For a required force P, the static screen F_cap ≥ P requires both C ≥ P and λ ≥ P/(ηR). If C < P, no increase of λ can pass the stated screen. With C ≥ P, continuously selectable λ and the stated lower λ bound inactive, the force-screen minimum total mass is m_min = PL/(ηR) + M, using the tex/mg convention above. The corresponding force-per-mass is P/m_min. Fixed connection mass makes this a function of P and L as well as ηR.

At P = 1.00 N, A needs λ = 1/(0.800 × 2.00) = 0.625 tex; B needs 1/(0.800 × 1.25) = 1.00 tex. The body-only mass at a fixed length favours A. But the connections add 0.600 mg for A versus 0.100 mg for B. At L = 0.20 m, complete masses are 0.725 and 0.300 mg. At L = 2.00 m they are 1.850 and 2.100 mg if only the static-force screen is applied.

The force-screen mass curves cross where 0.625L + 0.600 = 1.00L + 0.100, giving L = 4/3 m. That is a model crossover between two component families, not a material constant or an optimum length for a real cable. Specific strength and the scope of a reported result prepares the consistent body denominator; here the connections change the performance boundary.

A link may carry the force and still move too far

Now impose a second, independent condition: total support-to-support extension at P must be no more than δ_lim = 20.0 mm. The free body is linear under this comparison, and both connections act as reversible linear compliances. There is no initial slack, slip, creep, heating or residual length change in this model. The total displacement is δ_total = PL/(Hλ) + JP. The first term is the free body’s extension; the second is the two terminations’ combined extension.

Only the remaining displacement budget δ_lim − JP is available to the body. If it is positive, the extension constraint requires λ ≥ PL/[H(δ_lim − JP)]. Use the same length units in L and δ. If δ_lim ≤ JP, a finite compliant body at positive P and L cannot satisfy the limit: even infinitely stiff body material cannot remove the connection movement. This is a displacement boundary, not a new strength value.

Combine both screens by choosing λ_min = max[0.10 tex, P/(ηR), PL/{H(δ_lim − JP)}], then check C ≥ P and λ_min ≤ 2.50 tex. The max operation chooses the largest lower bound; adding the two requirements would overcount the same body. Equality is permitted in this exact illustrative screen. No reliability margin or uncertainty allowance has been supplied.

At L = 1.00 m, A’s connections use 4.00 mm of the 20.0 mm budget, leaving 16.0 mm. Its extension requirement is λ ≥ 1.00/(100 × 0.0160) = 0.625 tex, equal to its force requirement. B’s connections use 1.00 mm, leaving 19.0 mm; it needs λ ≥ 1.00/(50.0 × 0.0190) = 1.05263 tex. Its force-only body of 1.00 tex would give 20.0 mm body extension plus 1.00 mm end extension and miss the condition.

At L = 2.00 m, extension governs both families: A needs 1.25 tex, for 2.500 mg body plus 0.600 mg ends; B needs 2.10526 tex, for 4.21053 mg body plus 0.100 mg ends. Both lie within the stated selectable range and satisfy their force ceilings at P = 1.00 N. Their total masses are now 3.100 and 4.31053 mg. A longer body requires more mass per length to hold the same absolute movement, so in this regime its body mass grows as L², rather than the force-only L dependence.

Improve the quantity that is actually governing

At L = 2.00 m, doubling A’s R alone would reduce its force requirement from 0.625 to 0.3125 tex. The extension requirement remains 1.25 tex, so selected λ and the 3.100 mg total mass are unchanged. Its static capacity at that selected λ remains capped by C = 1.50 N. This is a precise case in which an improved body endpoint changes neither the selected mass nor the governing component ceiling.

Instead, doubling A’s H while holding all other supplied inputs fixed reduces the two-metre extension requirement to 0.625 tex. It then coincides with the original force requirement; total mass falls to 0.625 × 2.00 + 0.600 = 1.850 mg. This hypothetical change is not proof that a real process can double stiffness without changing strength, density, contacts or durability.

Under both screens the mass crossover also moves. Above L = 1.00 m, the two candidate mass expressions are 0.625L² + 0.600 and (20/19)L² + 0.100 in mg. They are equal at L ≈ 1.08131 m, within that regime. B is lighter at one metre; A is lighter at two metres. A strength-only comparison would place their equality at 1.33333 m and would select an under-stiff body for some lengths.

For a positioning link, this displacement criterion can be consequential even at a force well below the supplied static ceiling. A hoisting line, a pulley-routed tendon or an electrically functional yarn introduces other load directions, bend radii, contact, thermal or functional constraints. The same R or even the same axial component ratio cannot be transferred to those applications without an appropriate model and evidence.

Test the component and the source of its movement

A useful component comparison records the force history, free body length and mass, both end constructions and their mass, the measured support separation, and any separately observed body extension or end movement. Match the environment and loading direction. Locate the failure or slip rather than labelling every force drop a body-strength limit. A measured effective η is specific to that connection/specimen family and protocol; it is not an intrinsic constant of carbon.

Total extension at one load alone cannot identify H and J. For example, two hypothetical one-metre bodies at λ = 1.00 tex can both extend 20.0 mm at 1.00 N: H = 100 N/tex with J = 10.0 mm/N, or H = 62.5 N/tex with J = 4.00 mm/N. Their body extensions are 10.0 and 16.0 mm. An independent body-gauge measurement can discriminate them while the full assembly measurement still controls a support-to-support requirement. The two parameter pairs are separate examples, not fitted values for A or B.

Static breaking capacity does not determine cyclic life, creep under a sustained load, damage after exposure or residual movement after unloading. A historical aramid-rope fatigue/weathering investigation reported that inadequate end fittings prevented thorough evaluation of the ropes. This is a limited analogy showing why an intended durability question can remain unanswered when a different assembly path fails first; it supplies no CNT lifetime or present qualification rule.

Appropriate evidence follows the actual task: repeated connected specimens for scatter, body and end movement for stiffness attribution, defined loading/unloading for residual extension, and the relevant duration, environment and cycling for durability. Record specimens that fail at connections as part of the assembly evidence rather than silently retaining only body fractures. These calculations pass only two invented exact screens; they establish no safe service load, approved cable design or transferable component lifetime.

Worked examples

Try the reasoning

1. Does highest specific fibre strength guarantee lowest complete cable mass?

   Show the answer and reasoning

   No.

   Terminations and other constraints contribute.

2. Why assess repeated loading?

   Show the answer and reasoning

   Service may be fatigue-limited.

   Static capacity does not establish cyclic life.

3. For family B, can adding fibre meet a required 1.30 N while the same C = 1.20 N connection ceiling applies?

   Show the answer and reasoning

   No. The fixed connection ceiling is below the requirement for every λ.

   The component capacity is the smaller of ηRλ and C. More body mass cannot raise the stipulated C; a changed connection family would need its own mass, compliance and evidence.

4. Keep family A at P = 1.00 N but require total extension no more than 3.00 mm at any positive L. Can arbitrarily high body stiffness solve the stated model?

   Show the answer and reasoning

   No. The two ends alone contribute JP = 4.00 mm.

   The body term is positive for finite λ and H. A smaller movement limit requires a changed connection response, not only a stronger or stiffer free body.

5. Find the minimum mass of each family at L = 1.00 m with both original screens. Which is lighter?

   Show the answer and reasoning

   A: λ = 0.625 tex and 1.225 mg total. B: λ = 1.052632 tex and 1.152632 mg total. B is lighter.

   A’s force and extension requirements coincide. B’s 1.00 tex force-only body would give 21.0 mm total movement; use the extension requirement, then add its 0.100 mg ends.

6. For the two-metre link, relax δ_lim to 25.0 mm while retaining P = 1.00 N and all family inputs. Recalculate the selected λ and mass.

   Show the answer and reasoning

   A: λ = 0.952381 tex and 2.504762 mg. B: λ = 1.666667 tex and 3.433333 mg.

   The body budgets are now 21.0 and 24.0 mm. Required λ values are 2/[100(0.0210)] and 2/[50(0.0240)], both larger than the unchanged force requirements and within the declared range. A still has the lower total mass in this changed case.

7. For the 0.20 m A link, reduce η from 0.800 to 0.500 while R, H, C, M and J stay as supplied. What minimum λ and total mass pass both original screens?

   Show the answer and reasoning

   λ = 1.00 tex and total mass 0.800 mg; total extension is 6.00 mm at 1.00 N.

   The force requirement becomes 1/(0.500 × 2.00) = 1.00 tex, exceeding the 0.125 tex extension requirement. Body mass is 0.200 mg, plus 0.600 mg ends. Connection transfer can dominate despite unchanged bare-body R.

8. A calculation uses M = 0.600 mg for A but adds it twice after computing λL. What is wrong, and what must be measured if an end design changes with body size?

   Show the answer and reasoning

   M already includes both terminations, so it is added once. A size-dependent end design needs its own total retained mass, capacity/transfer and compliance account.

   The free body excludes end-region material. Omitting connection mass and double-counting it are different boundary errors; neither is repaired by dividing by fibre density.

9. In the L ≥ 1.00 m regime, equate the two both-screen mass expressions. Use A: 0.625L² + 0.600 mg; B: (20/19)L² + 0.100 mg. Where do they cross, and is that regime consistent?

   Show the answer and reasoning

   L = √(76/65) ≈ 1.08131 m, with mass about 1.33077 mg for each. The result is above 1.00 m, so the assumed regime is consistent.

   Their L² coefficient difference is 20/19 − 5/8 = 65/152. Equality requires L² = (1/2)/(65/152) = 76/65. A root must be checked against the piece of the model used to derive it.

10. At L = 2.00 m, family B’s body/connection data are known only for λ ≤ 2.00 tex instead of the originally stipulated 2.50 tex. May the 2.105263 tex solution still be accepted without more evidence?

    Show the answer and reasoning

    No. It lies outside the newly stated family domain.

    With a 2.00 tex body, δ_total = 2/(50 × 2) + 0.001 = 21.0 mm, above the limit. An algebraic extrapolation past a validated range is not evidence that the new larger assembly retains the same parameters.

What to carry forward

• Follow load into and through the assembly.
• Use a design-specific mass and performance boundary.
• Serial capacities take a minimum; component extensions and retained masses add under their stated boundaries.
• With fixed end mass, even a strength-limited mass comparison depends on load and free length.
• Choose the larger body requirement when force and extension constraints both apply, then check connection ceilings and the family domain.
• Improving an inactive property can leave selected component mass unchanged.
• Static endpoint, body-gauge motion and support motion are different evidence; none alone supplies cyclic or environmental life.