How Chain Alignment Changes Polymer Fibre Tensile Strength

A polymer filament is stretched until it becomes longer and thinner. You might expect it to snap more easily because there’s less material across its width. Yet the stretched filament can become stronger when pulled along its length.

The reason is molecular chain alignment. Drawing a polymer fibre makes more of its long molecular chains, or sections of those chains, point along the direction of the pull. This helps the fibre transfer and resist tensile forces. Its tensile strength can increase even though its diameter decreases.

But here’s the catch: a fibre with higher tensile strength doesn’t necessarily require a greater total force to break. To understand why, we need to separate what happens to the molecules from what a testing machine actually measures.

01Why aligned polymer chains resist pulling more effectively

Imagine a handful of long shoelaces tangled together. Pull the bundle from opposite ends. Some sections tighten immediately, but others must straighten or shift before they carry much of the load.

Now imagine arranging those shoelaces so that most run along the direction of the pull. More sections can resist stretching directly.

That gives us a useful first model for polymer fibres.

A polymer consists of long molecules containing repeating units joined by covalent bonds. In an initially poorly oriented fibre, many chain segments curve, fold, or point away from the fibre’s length.

When the fibre is drawn, more segments become oriented along its length.

What changes at the molecular level?

Three connected effects help explain the increase in longitudinal tensile strength.

1. More direct load transfer along polymer backbones. Aligned chain segments can resist an axial pull through their strong covalent backbones, rather than first needing to straighten or rearrange.
2. More effective transfer between neighbouring chains. Intermolecular attractions, entanglements, and chain segments connecting different regions help transfer the load through the fibre.
3. Changes in ordered regions. Drawing can also orient existing crystalline regions or promote crystallisation, depending on the polymer and processing conditions.

The shoelace analogy has limits. Polymer chains aren’t separate cords held together by ordinary friction. They are extremely long molecules, their neighbouring segments interact through intermolecular forces, and some regions contain closely packed, ordered structures.

Also, drawing does not normally make each individual covalent bond stronger. The main change is how the existing molecular structure is arranged and how the whole material responds to a force.

Nylon shows why intermolecular forces still matter

Nylon contains amide groups, including the linkage \(\ce{-C(=O)-NH-}\). These groups allow hydrogen bonding between suitable sites on neighbouring polymer chains.

When nylon chains are more favourably oriented, their covalent backbones can carry a greater share of an axial load, while hydrogen bonding and other interactions help transfer forces between chains.

Don’t conclude that alignment simply creates more hydrogen bonds. The number and arrangement of hydrogen bonds can change during processing, but better axial strength does not require every chain to form additional bonds.

This is why molecular structure and molecular arrangement must be treated as different ideas.

02Molecular structure, processing, and bulk properties are not the same thing

A useful HSC Chemistry explanation keeps these three levels separate.

The direction of the explanation is:

Chemical structure determines what interactions are possible. Processing influences how the chains are arranged. Together, they affect the measured properties.

An important extra term is crystallinity.

Crystalline regions contain chain segments packed in an ordered arrangement. Amorphous regions lack that regular long-range packing.

A polymer fibre can contain both.

Chain alignment and crystallinity are related, but they aren’t interchangeable. An amorphous region can have chains preferentially oriented in one direction without becoming a perfectly ordered crystal.

This becomes important in experiments. If drawing changes both orientation and crystallinity, an increase in strength cannot automatically be attributed to alignment alone.

03How an experiment can test whether drawing strengthens a fibre

Suppose we want to investigate one question:

Does drawing a nylon filament increase its tensile strength along the drawing direction?

We could use matched filaments from the same batch of nylon, leave some as-spun, and draw the others under controlled conditions.

Stage 1: Draw one set of filaments

Picture the drawing apparatus from left to right.

A spool supplies the initially spun filament to a first roller. The filament then passes through a drawing zone to a second roller rotating at a higher surface speed.

The second roller takes up material faster than the first supplies it. As a result, the filament stretches between them.

The elongated filament is collected on a take-up spool.

Temperature and drawing speed matter because polymer chains need enough mobility to rearrange without simply causing uncontrolled damage.

For a controlled comparison, we’d keep the polymer grade, sample preparation, and relevant testing conditions consistent.

Stage 2: Pull each specimen until it fails

A tensile-testing machine can measure how a filament responds to a pulling force.

The specimen is held vertically between two grips. The upper grip is connected to a force sensor, called a load cell. The lower grip moves away at a controlled rate, increasing the extension and tensile force.

The machine records force as the specimen extends.

We also measure the specimen’s cross-sectional area before the tensile test. For a circular filament, this can be calculated from its measured diameter.

What result should we expect?

If drawing improves axial load transfer, the drawn material should generally withstand a higher tensile stress under suitable conditions.

But comparing breaking forces alone is not a reliable test of that prediction.

Stage 3: Correct the first experimental mistake

Suppose our machine gives these results.

Illustrative data only – not measurements from a published experiment.

The drawn filament breaks under a smaller force.

Does that mean drawing weakened the polymer?

No. The drawn filament is also considerably thinner. We’re comparing different amounts of material.

We need a quantity that accounts for cross-sectional area.

Tensile stress measures force per unit area

Engineering tensile stress is calculated using:

\[ \sigma=\frac{F}{A_0} \]

Here:

• \(\sigma\) is tensile stress, measured in pascals (\(\mathrm{Pa}\)) or megapascals (\(\mathrm{MPa}\)).
• \(F\) is the applied tensile force, measured in newtons (\(\mathrm{N}\)).
• \(A_0\) is the cross-sectional area measured before the tensile test, in \(\mathrm{m^2}\) or \(\mathrm{mm^2}\).

We take pulling forces and tensile stresses as positive magnitudes.

A useful unit relationship is:

\[ 1\ \mathrm{N\,mm^{-2}}=1\ \mathrm{MPa} \]

The maximum engineering stress is called the ultimate tensile strength. It is calculated from the maximum force and the original area of that test specimen.

The force at rupture can sometimes differ from the maximum force reached earlier in the test. In the numerical examples here, assume the maximum force occurs at rupture.

04Why drawing ratio cannot tell you strength by itself

Drawing ratio describes how much a material segment is elongated during processing.

\[ \lambda=\frac{L_{\mathrm{after}}} {L_{\mathrm{before}}} \]

Here, \(\lambda\) is the dimensionless drawing ratio, and \(L_{\mathrm{before}}\) and \(L_{\mathrm{after}}\) are the lengths of the same material segment before and after drawing.

If a segment doubles in length, its drawing ratio is \(2\).

For a simple calculation, suppose the polymer’s volume remains constant during drawing and the cross-section remains uniform.

Because volume equals area multiplied by length:

\[ A_{\mathrm{before}}L_{\mathrm{before}} = A_{\mathrm{after}}L_{\mathrm{after}} \]

Rearranging gives:

\[ A_{\mathrm{after}} =\frac{A_{\mathrm{before}}}{\lambda} \]

Notice what this equation doesn’t tell us: the new tensile strength.

Drawing ratio predicts a change in geometry under our volume assumption. The resulting strength must be determined experimentally or supplied separately.

05The contrasting case: What happens if you pull across the aligned chains?

So far, we’ve pulled in the same direction that the polymer chains were oriented.

Now imagine drawing a sheet of polyester, then cutting two narrow test strips from it.

One strip runs parallel to the drawing direction. The other runs perpendicular to it.

Both strips come from the same sheet, so they have the same chemical composition and, assuming a uniform sheet, essentially the same overall crystalline fraction.

Would they necessarily have the same tensile strength?

No.

The parallel strip can take greater advantage of polymer backbones oriented along the tensile force.

The perpendicular strip is loaded across that preferred orientation. Its response depends more heavily on interactions between chains and regions, their connectivity, and other aspects of the microstructure.

This behaviour is called anisotropy: a material has different properties when measured in different directions.

A polymer fibre can therefore have excellent longitudinal tensile strength without being equally strong under transverse loading.

This also exposes a common misconception. A higher degree of chain alignment does not necessarily make a polymer stronger in every direction.

Stronger is not the same as stiffer or tougher

Three material properties are worth keeping separate.

A highly drawn fibre may become stronger and stiffer along its length while also having a smaller extension at break.

That doesn’t automatically tell us whether its toughness has increased. We’d need more information about how stress changes throughout the deformation.

06What would make the experimental conclusion convincing?

The original breaking-force comparison couldn’t separate strength from filament thickness.

Correcting for cross-sectional area fixes that specific problem. It still doesn’t establish which molecular change caused the strength increase.

For a stronger investigation, we’d use several independently tested specimens at each processing condition and collect structural evidence alongside the mechanical measurements.

For example, polarised-light measurements can provide information about molecular orientation through birefringence, the difference in how light travels through a material along different directions. X-ray diffraction can provide information about crystalline ordering and orientation.

Neither measurement should be treated as a magic alignment detector without interpreting what contributes to its signal.

The following distinctions help when assessing polymer experiments:

A good explanation doesn’t claim more than the evidence supports.

If the polymer chemistry is unchanged, you can discuss changes in molecular arrangement. If the processing history or environmental conditions differ, identify those differences before assigning a cause.

07Questions and solutions

The first question checks the basic calculation. The later questions require you to combine molecular structure, processing, experimental evidence, and the behaviour of an entire specimen.

Unless stated otherwise, all numerical data are illustrative, areas refer to cross-sections before tensile testing, forces are positive tensile magnitudes, and maximum force occurs at rupture.

Question 1

Two nylon fibres have the same cross-sectional area of \(0.10\ \mathrm{mm^2}\).

An as-spun fibre breaks at \(4.0\ \mathrm{N}\), while a drawn fibre breaks at \(6.0\ \mathrm{N}\).

a) Calculate the tensile strength of each fibre.

b) Explain one molecular reason why drawing may have increased the tensile strength, assuming both fibres have the same chemical structure.

Solution 1

The drawn fibre has a tensile strength of \(60\ \mathrm{MPa}\), compared with \(40\ \mathrm{MPa}\) for the as-spun fibre. Drawing has increased the tensile strength by \(50\%\).

Part a

Using \(\sigma_{\max}=F_{\max}/A_0\):

For the as-spun fibre:

\[ \begin{aligned} \sigma_{\mathrm{U}} &=\frac{4.0\ \mathrm{N}} {0.10\ \mathrm{mm^2}}\\ &=40\ \mathrm{MPa} \end{aligned} \]

For the drawn fibre:

\[ \begin{aligned} \sigma_{\mathrm{D}} &=\frac{6.0\ \mathrm{N}} {0.10\ \mathrm{mm^2}}\\ &=60\ \mathrm{MPa} \end{aligned} \]

The percentage increase is:

\[ \frac{60-40}{40}\times100=50\% \]

Part b

Drawing can orient more polymer chain segments along the fibre axis.

This allows the applied tensile load to be carried more effectively along the covalent backbones and transferred through the surrounding molecular structure.

The important point is that the arrangement has changed. We don’t need to suggest that drawing created stronger covalent bonds or changed the nylon repeating units.

Because both samples have equal cross-sectional areas, comparing their breaking forces also gives the correct strength ranking in this case.

Question 2

A researcher compares two nylon-6,6 filaments from the same production batch.

The as-spun specimen is tested dry. The drawn specimen is tested after exposure to moisture.

The test temperature and pulling rate are otherwise identical.

A separate optical measurement indicates greater molecular orientation in the drawn specimen.

Student A argues that drawing weakened the nylon because the maximum force decreased.

Student B argues that the experiment proves molecular alignment caused an increase in tensile strength.

a) Calculate the two tensile strengths and evaluate Student A’s argument.

b) Evaluate Student B’s conclusion.

c) Describe how you would improve the experiment to investigate the effect of chain alignment more reliably.

Solution 2

The drawn, damp filament has the higher measured tensile strength, \(100\ \mathrm{MPa}\) compared with \(60\ \mathrm{MPa}\). Student A has confused breaking force with tensile strength, while Student B has made a causal claim that the experiment cannot fully support.

Part a

For the as-spun specimen:

\[ \begin{aligned} \sigma_{\mathrm{U}} &=\frac{9.0\ \mathrm{N}} {0.15\ \mathrm{mm^2}}\\ &=60\ \mathrm{MPa} \end{aligned} \]

For the drawn specimen:

\[ \begin{aligned} \sigma_{\mathrm{D}} &=\frac{8.0\ \mathrm{N}} {0.080\ \mathrm{mm^2}}\\ &=100\ \mathrm{MPa} \end{aligned} \]

Student A’s argument is incorrect because the drawn sample has a smaller cross-sectional area.

Although its maximum force is lower, its measured force per unit area is higher.

Part b

Student B is partly correct about the observation but goes too far in explaining its cause.

The experiment establishes that the drawn, damp specimen has greater axial tensile strength under its test conditions. The optical result also indicates greater molecular orientation.

However, moisture exposure differs between the specimens.

Water can interact with amide groups in nylon and influence chain mobility and mechanical behaviour. The size and even the details of that effect depend on the nylon and its conditions.

Drawing might also have changed crystallinity or introduced defects.

Because more than one relevant factor differs, the strength difference cannot be assigned entirely to alignment.

Part c

A better investigation would:

• Prepare matched specimens from the same polymer batch and control their moisture content.
• Use consistent test temperature, pulling rate, gauge length, and sample preparation.
• Measure each specimen’s cross-sectional area before tensile testing.
• Test multiple specimens at each drawing condition and compare the results and variation.
• Measure orientation and, where possible, crystallinity independently.

The essential principle is to distinguish correlation from causation.

Greater alignment and greater tensile strength in the same specimen are consistent with an alignment-based explanation. They are not, on their own, proof that alignment is the only mechanism involved.

Question 3

A polyester film is drawn horizontally. Narrow strips are then cut from the same, approximately uniform sheet.

A parallel strip is pulled along the drawing direction. A perpendicular strip is pulled across it.

Both strips have an initial cross-sectional area of \(0.15\ \mathrm{mm^2}\).

A third strip is taken from an undrawn film of the same polymer. It has the same area.

The following maximum forces are measured:

Assume the undrawn film has approximately the same tensile strength in all in-plane directions, and all specimens are tested identically.

a) Calculate the tensile strength of each strip.

b) Evaluate the claim that drawing has made the polymer stronger in every direction.

c) A designer needs a strip of the specified area to withstand a static tensile force of \(20\ \mathrm{N}\) without breaking. Which orientation meets this requirement under the simplified test conditions?

d) Explain why these results challenge a model in which only the total crystalline fraction determines tensile strength.

Solution 3

Drawing has increased tensile strength parallel to the draw axis but reduced it in the perpendicular direction. Only the parallel-cut strip meets the \(20\ \mathrm{N}\) requirement under the stated simplified conditions.

Part a

Use \(\sigma_{\max}=F_{\max}/A_0\).

For the parallel strip:

\[ \begin{aligned} \sigma_{\parallel} &=\frac{24\ \mathrm{N}} {0.15\ \mathrm{mm^2}}\\ &=160\ \mathrm{MPa} \end{aligned} \]

For the perpendicular strip:

\[ \begin{aligned} \sigma_{\perp} &=\frac{9.0\ \mathrm{N}} {0.15\ \mathrm{mm^2}}\\ &=60\ \mathrm{MPa} \end{aligned} \]

For the undrawn strip:

\[ \begin{aligned} \sigma_{\mathrm{U}} &=\frac{12\ \mathrm{N}} {0.15\ \mathrm{mm^2}}\\ &=80\ \mathrm{MPa} \end{aligned} \]

Part b

The claim is incorrect.

The parallel tensile strength has doubled, from \(80\) to \(160\ \mathrm{MPa}\).

The perpendicular tensile strength has fallen from \(80\) to \(60\ \mathrm{MPa}\), a reduction of \(25\%\).

The drawn sheet is anisotropic. Its tensile strength depends on the direction of loading.

A student might expect that if drawing improves chain packing, the material must improve in every direction. But the increased alignment particularly favours force transmission along the preferred chain direction.

Part c

The parallel strip reaches a maximum force of \(24\ \mathrm{N}\), so it is the only option that exceeds the required \(20\ \mathrm{N}\).

The perpendicular strip reaches \(9.0\ \mathrm{N}\), and the undrawn strip reaches \(12\ \mathrm{N}\).

This establishes only the theoretical ranking under the supplied static test conditions. A real design would also require an appropriate safety margin and consideration of long-term loading, defects, and environmental conditions.

Part d

The parallel and perpendicular specimens were cut from the same approximately uniform drawn film.

Their overall crystalline fractions should therefore be approximately equal.

Yet their tensile strengths differ substantially.

A model based only on the amount of crystalline material cannot explain this directional difference.

The orientation of chains and crystalline regions relative to the pulling direction provides an important additional explanation. Other directional structural features may also contribute.

The key evidence is that chemical composition and approximate crystalline fraction remain constant while loading direction changes.

Question 4

A materials scientist prepares four idealised polymer specimens with controlled differences in molecular orientation and crystallinity.

Assume their chemical composition, average chain length, cross-sectional area, defect population, and tensile-testing conditions are identical. Orientation and crystalline fraction have been characterised independently.

The values below are illustrative.

a) Identify two comparisons that investigate the effect of molecular orientation while holding crystalline fraction constant. Calculate the strength increase in each.

b) Suppose orientation and crystallinity contribute independently, so their strength increases simply add together. Use A, B, and C to predict the tensile strength of D under this model.

c) Compare your prediction with the observed value for D. What does the difference suggest?

d) Explain why the results do not justify the statement that increasing crystallinity always increases tensile strength by the same amount.

Solution 4

Both chain orientation and crystalline fraction influence tensile strength in these illustrative specimens, but their effects are not simply additive. The data indicate an interaction between the two structural features.

Part a

Compare A with B because they have the same crystalline fraction of \(25\%\).

The increase associated with greater orientation is:

\[ 85-50=35\ \mathrm{MPa} \]

Next, compare C with D because both have a crystalline fraction of \(45\%\).

The increase associated with greater orientation is:

\[ 110-65=45\ \mathrm{MPa} \]

In both comparisons, the more highly oriented specimen is stronger.

Notice that the increase differs: \(35\ \mathrm{MPa}\) in one comparison and \(45\ \mathrm{MPa}\) in the other.

Part b

Under the proposed independent-effects model, start with the tensile strength of A, \(50\ \mathrm{MPa}\).

Increasing orientation from low to high at \(25\%\) crystallinity adds \(35\ \mathrm{MPa}\).

Increasing crystallinity from \(25\%\) to \(45\%\) at low orientation adds:

\[ 65-50=15\ \mathrm{MPa} \]

The additive prediction for D is therefore:

\[ 50+35+15=100\ \mathrm{MPa} \]

Part c

The observed value for D is \(110\ \mathrm{MPa}\), which is \(10\ \mathrm{MPa}\) higher than the additive prediction.

With the question’s assumptions, this shows that the two factors do not act as independent, fixed additions to tensile strength.

A plausible molecular explanation involves the interaction between oriented chain segments, ordered crystalline regions, and chains linking different regions.

For example, orienting chains may improve load transfer more effectively when the surrounding crystalline structure provides favourable connections and constraints.

The data demonstrate the interaction, but do not establish that particular microscopic explanation uniquely. Additional structural measurements would be needed.

Part d

At low orientation, increasing crystalline fraction from \(25\%\) to \(45\%\) increases tensile strength by:

\[ 65-50=15\ \mathrm{MPa} \]

At high orientation, the same crystalline-fraction change increases strength by:

\[ 110-85=25\ \mathrm{MPa} \]

The effects differ.

Therefore, even within these controlled illustrative samples, the strength change associated with crystallinity depends on the accompanying chain orientation.

The original tempting model treated orientation and crystalline fraction as two independent numbers that could be added to predict strength.

The results show why that simplification can fail: the way different parts of a polymer’s microstructure work together can be as important as the amount of each type of region.

Question 5

A nylon monofilament contains two clearly marked regions, P and Q, connected end-to-end.

The filament has a uniform pulling force along its length, but the two regions have different cross-sectional areas and molecular orientations.

Initially:

• Region P has cross-sectional area \(0.12\ \mathrm{mm^2}\).
• Region Q has cross-sectional area \(0.20\ \mathrm{mm^2}\).
• Region Q has tensile strength \(60\ \mathrm{MPa}\).
• The complete filament breaks in region P at \(9.0\ \mathrm{N}\).

A technician investigates two alternative drawing treatments on separate fresh copies of this original filament.

In both treatments, only P is drawn. Q is unchanged. Assume P conserves volume, remains uniform after drawing, and undergoes no chemical change.

Treatment M – moderate drawing

Region P is drawn to \(1.5\) times its original length.

The complete filament is then tested. It breaks in region Q at \(12\ \mathrm{N}\), with no prior failure in P.

Treatment H – heavier drawing

Region P is drawn to \(2.0\) times its original length.

An independent test establishes that material processed under Treatment H has an axial tensile strength of \(160\ \mathrm{MPa}\).

Assume all regions behave as specified, the grips do not slip, and there are no additional failures at the boundary between P and Q.

a) Determine the original tensile strength of P.

b) Determine what the Treatment M result tells you about the tensile strength of the moderately drawn P. Explain why you cannot calculate its exact value from the breaking force of the complete filament.

c) Predict the breaking force and failure region for Treatment H.

d) Explain how Treatment H can produce a filament with a lower breaking force than Treatment M, even though Treatment H raises P’s tensile strength above its original value.

e) Can the available evidence establish that Treatment H produces a higher tensile strength in P than Treatment M? Can modifying P alone make the complete filament support more than \(12\ \mathrm{N}\)?

Solution 5

The original tensile strength of P is \(75\ \mathrm{MPa}\). Treatment M establishes that P’s tensile strength exceeds \(150\ \mathrm{MPa}\), but not its exact value. Treatment H produces a filament that breaks in P at \(9.6\ \mathrm{N}\). No treatment of P alone can raise the complete filament’s maximum force above the unchanged Q limit of \(12\ \mathrm{N}\).

The difficult part is recognising that a specimen’s breaking force is controlled by whichever region reaches its own limit first. A region that survives the test has not revealed its maximum capacity.

Part a: Original tensile strength of P

The original filament breaks in P at \(9.0\ \mathrm{N}\).

Because P is the region that fails, its local tensile strength is:

\[ \begin{aligned} \sigma_{\mathrm{P,original}} &=\frac{9.0\ \mathrm{N}} {0.12\ \mathrm{mm^2}}\\ &=75\ \mathrm{MPa} \end{aligned} \]

Check that this is consistent with Q surviving.

The maximum force Q can withstand is:

\[ \begin{aligned} F_{\mathrm{Q,max}} &=\sigma_{\mathrm{Q}}A_{\mathrm{Q}}\\ &=60\ \mathrm{N\,mm^{-2}}\\ &\quad\times0.20\ \mathrm{mm^2}\\ &=12\ \mathrm{N} \end{aligned} \]

So Q can withstand more than the \(9.0\ \mathrm{N}\) that causes P to fail.

The original result is consistent.

Part b: What Treatment M reveals

With volume conserved, drawing P by a factor of \(1.5\) changes its area to:

\[ \begin{aligned} A_{\mathrm{P,M}} &=\frac{0.12\ \mathrm{mm^2}}{1.5}\\ &=0.080\ \mathrm{mm^2} \end{aligned} \]

The complete filament now breaks in Q at \(12\ \mathrm{N}\).

At the instant Q fails, P is also experiencing a tensile force of \(12\ \mathrm{N}\), because the two regions are connected in series.

But P has not failed.

Therefore, P’s tensile strength must exceed the stress it experiences at that force:

\[ \begin{aligned} \sigma_{\mathrm{P,M}} &>\frac{12\ \mathrm{N}} {0.080\ \mathrm{mm^2}}\\ &>150\ \mathrm{MPa} \end{aligned} \]

The strict inequality follows from the stated observation that Q fails first while P remains intact, assuming the idealised measurements and failure conditions.

This is a lower bound, not an exact measurement of P’s tensile strength.

A tempting error is to divide \(12\ \mathrm{N}\) by P’s area and report \(150\ \mathrm{MPa}\) as its measured ultimate strength.

But P didn’t fail. It might have withstood \(160\), \(180\), or another value greater than \(150\ \mathrm{MPa}\) before reaching its own limit.

To measure P’s exact ultimate strength, we’d need a suitable specimen that actually fails in material processed like P, rather than in Q.

Part c: Predict Treatment H’s failure

Treatment H doubles P’s original length.

Its new area is:

\[ \begin{aligned} A_{\mathrm{P,H}} &=\frac{0.12\ \mathrm{mm^2}}{2.0}\\ &=0.060\ \mathrm{mm^2} \end{aligned} \]

Its measured tensile strength is \(160\ \mathrm{MPa}\).

Therefore, P’s maximum force under Treatment H is:

\[ \begin{aligned} F_{\mathrm{P,H,max}} &=160\ \mathrm{N\,mm^{-2}}\\ &\quad\times0.060\ \mathrm{mm^2}\\ &=9.6\ \mathrm{N} \end{aligned} \]

Q still withstands up to \(12\ \mathrm{N}\).

Since \(9.6\ \mathrm{N}\) is smaller than \(12\ \mathrm{N}\), region P reaches its failure limit first.

The Treatment H filament therefore breaks in P at \(9.6\ \mathrm{N}\).

Part d: Why stronger material can produce a weaker complete filament

Treatment H raises P’s tensile strength from \(75\) to \(160\ \mathrm{MPa}\).

However, it also reduces P’s area from \(0.12\) to \(0.060\ \mathrm{mm^2}\).

Its maximum force is the product of both quantities:

\[ F_{\max}=\sigma_{\max}A_0 \]

Treatment M produces a complete filament that carries \(12\ \mathrm{N}\) before Q fails.

Treatment H produces a filament that carries only \(9.6\ \mathrm{N}\) before P fails.

There is no contradiction.

The heavy-drawn P has greater intrinsic tensile strength than the original P, but its smaller cross-section limits the total force it can carry.

This is also why a whole-fibre failure result cannot always be interpreted as the intrinsic strength of every part of that fibre.

Part e: What remains unknown, and the limit imposed by Q

Treatment M tells us:

\[ \sigma_{\mathrm{P,M}}>150\ \mathrm{MPa} \]

Treatment H tells us:

\[ \sigma_{\mathrm{P,H}}=160\ \mathrm{MPa} \]

We cannot establish that H has the higher tensile strength.

For example, M might have a strength of \(155\ \mathrm{MPa}\), in which case H is stronger. But M might instead have a strength of \(180\ \mathrm{MPa}\), in which case H is weaker.

Both possibilities fit the available results.

Finally, regardless of how strong P becomes, Q still reaches its tensile limit at \(12\ \mathrm{N}\).

Because P and Q are in series, they carry the same pulling force. Any attempt to load the complete filament beyond \(12\ \mathrm{N}\) would exceed Q’s capacity under the stated assumptions.

Therefore, strengthening P alone cannot make the complete filament withstand more than \(12\ \mathrm{N}\).

The surprising lesson is that the best treatment for one section of a fibre need not give the strongest complete fibre. Geometry, local molecular arrangement, and the location of the limiting region all matter.

08The next connection: Choosing a polymer and choosing how to process it

When you compare nylon with polyester, or two fibres made from the same polymer, make two separate decisions.

First, examine the molecular structure: the repeating units, covalent backbones, and intermolecular attractions. These determine what kinds of interactions the material can form.

Then examine the processing history: chain orientation, crystalline arrangement, fibre diameter, and possible defects. These help determine how the finished material uses those molecular interactions when loaded.

A strong HSC explanation connects these levels without mixing them up. If two fibres have the same chemistry but different tensile strengths, processing may explain the difference. If both the chemistry and processing change, you need further evidence before deciding which was responsible.

That distinction is the starting point for understanding why the same family of polymers can be manufactured into materials with very different mechanical properties.