Why Stronger Intermolecular Forces Don’t Guarantee Toughness

A polymer can withstand a much greater pulling force than another polymer, yet snap after stretching only a few millimetres. The weaker polymer might stretch considerably further before breaking, absorbing much more energy in the process.

This is why stronger intermolecular forces do not automatically mean greater toughness. Stronger attractions between polymer chains can increase cohesion and tensile strength, but they can also restrict chain movement. Toughness depends on both the stress a material can withstand and how much it can deform before fracture.

To understand the difference, we need to look beyond the strength of the attractions between chains and consider molecular structure, polymer processing, and the way the finished material behaves under an applied force.

01Strength and toughness are not the same property

Imagine testing two hypothetical polymers using a tensile testing machine.

• Polymer A reaches a maximum tensile stress of 60 MPa but breaks after stretching by just 5%.
• Polymer B reaches a maximum tensile stress of only 35 MPa but stretches by 30% before breaking.

Which polymer is tougher?

You might initially choose A because it withstands a greater stress. However, B could absorb considerably more energy before breaking.

The distinction comes down to what each property measures.

Toughness is particularly important because it considers the material’s behaviour throughout deformation, not just the maximum stress it reaches.

For a tensile test, toughness is represented by the area under the stress-strain curve:

\[ U_T=\int_0^{\epsilon_f}\sigma\,d\epsilon \]

Where:

• \(U_T\) is tensile toughness, measured in J/m³.
• \(\sigma\) is tensile stress, measured in pascals (Pa).
• \(\epsilon\) is strain, which is dimensionless.
• \(\epsilon_f\) is the strain at fracture.

A larger area means more energy was absorbed per unit volume before the specimen broke.

Think of the graph as having two important dimensions:

• How high does the curve rise? This indicates the stress the material can withstand.
• How far does the curve extend horizontally? This indicates how much it deforms before breaking.

A material that combines substantial strength with considerable deformation can have a large area under its curve.

This is the first important distinction: high tensile strength does not necessarily mean high toughness.

02How do scientists measure toughness?

To understand why stronger intermolecular forces do not guarantee greater toughness, consider how a tensile experiment actually works.

The tensile testing experiment

A tensile testing machine measures how a material responds when it is pulled apart.

The specimen is usually shaped like a narrow strip with wider ends, called a dog-bone specimen. This helps concentrate deformation in a defined central section.

One end is held by a stationary grip. The other is attached to a moving crosshead. A load cell measures the pulling force, while an extensometer measures the extension of the specimen’s central section.

The machine gradually pulls the specimen until it fractures.

Two measurements are essential:

1. The applied force, \(F\), measured in newtons (N).
2. The extension, \(\Delta L\), measured in metres (m).

These measurements are converted into stress and strain.

Engineering stress is calculated using:

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

Where \(A_0\) is the specimen’s original cross-sectional area, measured in m².

Engineering strain is calculated using:

\[ \epsilon=\frac{\Delta L}{L_0} \]

Where \(L_0\) is the original gauge length.

The machine records stress against strain as the specimen stretches.

What should we predict?

Suppose stronger intermolecular forces always produced greater toughness.

We might predict that the polymer with the greatest tensile strength would also have the largest area under its stress-strain curve.

But the experiment can show something different.

A polymer with strong attractions between its chains might withstand a large stress but fracture after relatively little deformation. Another polymer might withstand a smaller stress but stretch much further.

The second polymer can therefore absorb more energy before breaking.

Let’s calculate this using some illustrative experimental data.

03What happens between polymer chains?

Now we can connect the experimental observations to molecular structure.

A polymer consists of long molecules made from repeating units. Strong covalent bonds hold the atoms together along each polymer backbone.

Different polymer chains also attract one another through intermolecular forces, including:

• London dispersion forces.
• Dipole-dipole attractions.
• Hydrogen bonding.

These attractions help hold neighbouring chains together.

If you’re unsure about the difference between the bonds within a polymer chain and the attractions between chains, see Intermolecular vs Intramolecular Forces in HSC Chemistry.

Stronger attractions can increase cohesion

Imagine trying to slide two polymer chains past each other.

If the attractions between them are relatively weak, the chains can move past one another more easily.

If the attractions are stronger, more energy is needed to separate or slide the chains.

This can increase the material’s resistance to deformation. It can also increase the tensile stress required to cause failure.

However, here’s the catch.

A tough polymer needs to do more than resist movement. It must also be capable of deforming and redistributing stress before a crack causes it to fracture.

If the chains are held too rigidly in place, their ability to rearrange may be restricted.

The material might withstand a large stress initially but fail after relatively little deformation.

A useful analogy is a bundle of long cords.

If the cords have very little interaction, they slide apart easily and cannot support much load.

If they are held completely rigidly together, they have little opportunity to rearrange when the bundle is bent or stretched.

A bundle with enough interaction to remain cohesive, but enough freedom to rearrange, can absorb considerable energy.

The analogy has limits: real polymer chains are molecules, and entanglements are physical constraints on chain movement rather than literal knots.

The important point is that toughness requires a balance between cohesion and molecular mobility.

Why chain movement matters

When a polymer is stretched, several molecular processes can occur:

• Polymer chains can rotate and change their conformations.
• Flexible sections of chains can extend.
• Entangled chains can move relative to neighbouring chains.
• Some intermolecular attractions can be overcome and re-formed as the material deforms.

These processes can absorb energy.

If a material can undergo substantial deformation before fracture, it has more opportunity to absorb energy through these mechanisms.

By contrast, a polymer that fractures after very little deformation may absorb relatively little energy, even if its intermolecular attractions are strong.

This does not mean strong intermolecular forces always reduce toughness.

Stronger attractions can increase toughness if they improve cohesion without excessively restricting the movement needed for deformation.

The relationship is not a simple one-way rule.

04Molecular structure, processing and bulk properties

There is another important complication.

The forces between polymer chains are not the only factor determining toughness. We also need to consider how the polymer is manufactured and processed.

Three different levels of description are useful.

These levels are related, but they are not interchangeable.

The same polymer can have different properties

Imagine two samples made from exactly the same polymer.

Their repeat units and chemical composition are identical, so the types of intermolecular forces available between their chains are also the same.

However, the samples undergo different manufacturing processes.

One is moulded and cooled without being drawn.

The other is stretched during processing, aligning many of its polymer chains in a preferred direction.

What changes?

In the drawn sample, the chains become more aligned. When the material is pulled parallel to this direction, the aligned structure can transfer the applied load more effectively.

This can increase tensile strength along the direction of alignment.

However, the material may have less capacity to extend further before breaking. Its tensile toughness may therefore decrease.

This is an important distinction for HSC Chemistry:

Processing can change the arrangement of polymer chains without changing the chemical identity of their repeating units.

The finished material can therefore have different bulk properties even when its underlying polymer chemistry is unchanged.

For a closer look at this effect, read How Chain Alignment Changes Polymer Fibre Tensile Strength.

Other structural factors that influence toughness

Several additional features can affect how much energy a polymer absorbs before fracture.

1. Chain flexibility

Flexible backbones can allow greater molecular movement during deformation.

Rigid backbones may restrict this movement, although the overall effect also depends on intermolecular forces and polymer arrangement.

2. Molecular mass and entanglement

Longer polymer chains can become entangled with neighbouring chains.

These entanglements can help prevent chains from sliding past one another too easily, allowing the material to maintain cohesion during deformation.

3. Crystallinity

Crystalline regions contain more regularly packed polymer chains.

Greater crystallinity can increase stiffness and strength, but it can also restrict chain mobility.

Its effect on toughness depends on the polymer and its overall morphology. Neither high nor low crystallinity automatically guarantees greater toughness.

4. Cross-linking

Cross-links are covalent bonds that join different polymer chains.

They are not ordinary intermolecular attractions.

Cross-links restrict relative chain movement. A suitable cross-link density can provide useful elasticity and structural stability, but excessive cross-linking can limit deformation and reduce toughness.

5. Defects and voids

Small imperfections can concentrate stress in particular regions.

This can allow cracks to develop and propagate before the surrounding material has absorbed much energy.

The final toughness therefore depends on the combined effects of molecular structure, chain arrangement, processing and defects.

05Why a tensile test is not the whole story

So far, we’ve discussed toughness using the area under a stress-strain curve.

However, polymers do not always fail under slow, gradual pulling.

Think about a plastic component that is suddenly struck or dropped. The loading occurs much more quickly, and a crack may already be present.

Would the same polymer that performs best in a slow tensile test necessarily perform best during a sudden impact?

Not necessarily.

To investigate this, scientists can use a pendulum impact test, such as the Charpy test.

The Charpy impact experiment

The apparatus contains a heavy pendulum that swings down and strikes a horizontal specimen.

The specimen rests on two supports, with a notch in its centre. The notch faces away from the striker, creating a controlled location where a crack can begin.

The experiment follows a clear sequence:

1. The pendulum is raised to a known starting height.
2. The notched specimen is placed on the supports.
3. The pendulum is released and strikes the specimen.
4. The specimen fractures, absorbing some of the pendulum’s energy.
5. The pendulum continues swinging upwards, reaching a height lower than its original starting height.

The difference in the pendulum’s energy before and after impact provides an estimate of the energy absorbed by the specimen.

What does the experiment tell us?

The original tensile experiment measured how a specimen behaved under gradual, uniaxial pulling.

The impact experiment investigates how the specimen responds to sudden loading and fracture.

The notch is particularly important. It concentrates stress in a small region and provides a more controlled starting point for crack formation.

A material that stretches considerably under slow tension may still perform poorly when a crack is present and the loading occurs rapidly.

This does not mean the tensile experiment was incorrect. It simply answered a different question.

Impact behaviour can depend on:

• The rate at which the force is applied.
• The presence and shape of a notch.
• The temperature of the specimen.
• The orientation of the polymer chains.
• The ability of the material to deform around a developing crack.

Consequently, a polymer’s ranking in a slow tensile test may differ from its ranking in an impact test.

06How to decide what a polymer property tells you

When you encounter a question about polymer toughness, avoid jumping straight from intermolecular forces to the final property.

Instead, identify what is actually being measured.

Then connect the molecular explanation to the observed property.

For example:

• Stronger intermolecular attractions can increase cohesion between polymer chains.
• Increased cohesion can increase the stress required for chain separation or movement.
• However, restricted chain movement can reduce the material’s ability to deform.
• Reduced deformation can decrease the energy absorbed before fracture.
• The final toughness depends on the combined effects of cohesion, chain mobility and polymer structure.

The key is to avoid treating one molecular feature as the sole explanation for a bulk property.

A polymer with strong hydrogen bonding might be highly cohesive, but its toughness will also depend on its backbone flexibility, chain arrangement, crystallinity and processing history.

When comparing polymer applications, always identify the type of loading the material must withstand. A material designed to resist slow stretching may not be the best choice for a component exposed to sudden impact.

07Questions and solutions

Question 1

Two polymers are tested using a slow tensile test.

Their stress-strain curves are approximately triangular, starting at the origin and ending at fracture.

Calculate the approximate tensile toughness of each polymer and identify which is stronger and which is tougher.

Solution 1

Polymer Y is tougher, while Polymer X is stronger.

The approximate area under a triangular stress-strain curve is:

\[ U_T=\frac{1}{2}\sigma_{\max}\epsilon_f \]

For Polymer X:

\[ U_X=\frac{1}{2}\times48\times0.08 \]

\[ \boxed{U_X=1.92\text{ MJ/m}^3} \]

For Polymer Y:

\[ U_Y=\frac{1}{2}\times32\times0.24 \]

\[ \boxed{U_Y=3.84\text{ MJ/m}^3} \]

Polymer X has the greater maximum tensile stress, so it is stronger.

However, Polymer Y absorbs twice as much energy per unit volume before fracture, according to the triangular approximation.

The important distinction is that strength measures the maximum stress, whereas toughness accounts for the entire deformation process.

Question 2

Two samples are manufactured from the same polymer batch. They have the same repeating-unit chemistry and similar molecular mass distributions.

• Sample A is moulded and cooled without being drawn.
• Sample B is stretched along its length during processing and then annealed.

Both samples are tested in tension, parallel to the drawing direction of Sample B.

The following results are obtained:

Explain how Sample B can have greater tensile strength but lower toughness, despite having the same repeating-unit chemistry as Sample A.

Solution 2

Sample B is stronger but less tough because chain alignment improves its ability to withstand tensile stress along the drawing direction, while its capacity for further deformation is reduced.

Both samples have the same polymer chemistry, so their repeating units and the types of intermolecular forces available between their chains are unchanged.

However, their processing histories are different.

Drawing causes many polymer chains to become aligned along the direction of stretching. This improves the transfer of an applied tensile load along the aligned structure.

As a result, Sample B can withstand a higher maximum tensile stress.

However, its fracture strain is considerably lower:

\[ \epsilon_A=0.28 \]

\[ \epsilon_B=0.09 \]

Sample B fractures after much less deformation.

Its smaller area under the stress-strain curve confirms that it absorbs less energy per unit volume before breaking.

The annealing process may also change the packing and crystallinity of the chains, depending on the polymer and processing conditions.

Therefore, it would be incorrect to explain the difference simply by claiming that Sample B has stronger intermolecular forces.

The experiment demonstrates that processing can substantially change bulk properties without changing the repeating-unit chemistry.

One further limitation is that the results only describe loading parallel to the drawing direction. The polymer may behave differently when pulled perpendicular to that direction.

Question 3

A Charpy pendulum impact test is conducted on two notched polymer specimens, C and D.

Both specimens have identical dimensions and notch geometry. They are tested at the same temperature and impact speed, and both fracture completely.

The pendulum has a mass of 4.0 kg and an initial centre-of-mass height of 0.80 m.

The following measurements are recorded:

A blank swing establishes that 1.0 J is lost through friction and air resistance.

A separate tensile test shows that specimen D has a greater maximum tensile strength than specimen C.

(a) Calculate the energy absorbed by each specimen.

(b) Identify which specimen has greater impact resistance.

(c) Explain why this result does not contradict the tensile strength results.

Solution 3

Specimen C absorbs more energy during impact and therefore has greater impact resistance under the specified test conditions, despite specimen D having greater tensile strength.

(a) Calculate the energy absorbed.

The energy lost by the pendulum is:

\[ E=mg(h_1-h_2) \]

For specimen C:

\[ E_C=4.0\times9.8\times(0.80-0.35) \]

\[ E_C=17.64\text{ J} \]

Correcting for the 1.0 J apparatus loss:

\[ E_{C,\text{specimen}}=17.64-1.0 \]

\[ \boxed{E_C=16.64\text{ J}} \]

For specimen D:

\[ E_D=4.0\times9.8\times(0.80-0.48) \]

\[ E_D=12.544\text{ J} \]

Correcting for apparatus losses:

\[ E_{D,\text{specimen}}=12.544-1.0 \]

\[ \boxed{E_D=11.54\text{ J}} \]

(b) Compare impact resistance.

Specimen C absorbs 16.64 J, while specimen D absorbs approximately 11.54 J.

Therefore, specimen C has greater impact resistance in this test.

(c) Explain the difference.

There is no contradiction because tensile strength and impact resistance measure different aspects of material behaviour.

Tensile strength measures the maximum stress a material can withstand under tension.

The Charpy test measures the energy absorbed during sudden fracture of a notched specimen.

Specimen D can withstand a greater maximum tensile stress but absorb less energy during rapid fracture.

Specimen C may have a greater capacity for local deformation and energy absorption around the developing crack.

The important point is that a ranking from a slow tensile test cannot automatically be applied to a sudden impact test.

Question 4

A manufacturer produces a polymer sheet by drawing it in one direction.

Two strips are cut from the same sheet:

• Strip P is cut parallel to the drawing direction.
• Strip Q is cut perpendicular to the drawing direction.

Both strips are tested in tension along their respective lengths.

The following results are obtained:

A designer is considering two applications:

1. A strap that must carry a large tensile load along its length.
2. A thin panel that will be flexed and may eventually tear from a pre-existing notch.

Explain which strip is better supported by the data for the first application. Then evaluate whether the same data are sufficient to select a material for the second application.

Solution 4

Strip P is the better-supported choice for the tensile strap, but the supplied data are insufficient to establish which strip will resist tearing from a notch.

For the first application, the strap must carry a substantial tensile load along its length.

Strip P has a maximum tensile stress of 85 MPa, compared with only 28 MPa for Strip Q.

Therefore, P can withstand a greater tensile stress in the tested direction.

Its lower tensile toughness does not automatically make it unsuitable for this application because the main requirement is carrying a high tensile load.

For the second application, the situation is different.

Strip Q has a greater tensile toughness and fracture strain. It can absorb more energy during slow tensile deformation.

However, a notched panel may fail through crack initiation and propagation rather than uniform tensile deformation.

An unnotched tensile test does not directly measure resistance to tearing from an existing notch.

The designer should therefore conduct a suitable notched tear or impact test using the relevant loading direction and conditions.

The key issue is anisotropy: the properties of a drawn polymer can depend strongly on the direction in which it is loaded.

A material that performs well parallel to chain alignment may behave very differently perpendicular to it.

Question 5

Two candidate polymers are being considered for a snap-fit component.

Polymer H contains polar groups capable of forming hydrogen bonds between neighbouring chains.

Polymer D is non-polar, with London dispersion forces dominating its intermolecular attractions.

Their molecular mass distributions are similar, but their repeat-unit chemistry differs.

The following tests are conducted using identical specimen geometries within each test.

Test 1: Slow tensile test at 23°C

Fresh, unnotched specimens are used.

Test 2: Notched impact test at 5°C

Separate fresh specimens are used.

The manufacturer makes the following claim:

“Polymer H has stronger intermolecular forces, so it is tougher and should be selected.”

However, the component will experience two stages during normal use:

1. It will be bent slowly during installation at 23°C.
2. Once installed, it may experience a sudden impact at 5°C, concentrated near the same region that was bent during installation.

(a) Evaluate the manufacturer’s claim using both sets of experimental results.

(b) Explain why it is not possible to attribute Polymer H’s greater impact energy solely to its stronger intermolecular forces.

(c) Explain why testing fresh specimens separately may not provide enough information to select the best polymer for this application.

(d) Design a more appropriate experiment to compare the two polymers for their intended use.

Solution 5

The manufacturer’s claim is too broad: Polymer H performs better in the specified impact test, but Polymer D has greater tensile toughness. Neither polymer can be declared universally tougher from these results.

(a) Evaluate the claim.

Start by examining the slow tensile results.

Polymer H has a maximum tensile stress of 68 MPa, compared with 40 MPa for Polymer D.

Therefore, H is stronger under the specified tensile test.

However, Polymer D has a tensile toughness of 7.0 MJ/m³, compared with 2.8 MJ/m³ for H.

This means D absorbs more energy per unit volume before fracture during slow tensile deformation.

Now consider the impact results.

Polymer H absorbs 13 J during impact, whereas Polymer D absorbs only 7 J.

Therefore, H has greater impact resistance under the specified notched impact conditions.

The results show that the two polymers have different advantages:

• H has greater tensile strength and greater measured impact energy.
• D has greater tensile toughness and fracture strain in the slow tensile test.

The statement that H is simply tougher ignores the difference between the two testing conditions.

(b) Explain why intermolecular forces alone cannot explain the results.

Hydrogen bonding can increase the attraction between neighbouring polymer chains.

This may increase cohesion and make chain separation more difficult.

However, the two polymers have different repeat-unit chemistry, which can also affect:

• Backbone flexibility.
• Chain packing and crystallinity.
• Molecular mobility.
• The ability of chains to rearrange during deformation.
• The formation and propagation of cracks.

The information supplied does not establish that these other characteristics are identical.

Even if H absorbs more energy during impact, this does not prove that hydrogen bonding alone caused the difference.

To investigate the role of intermolecular forces, researchers would need to compare carefully controlled polymer samples while accounting for differences in molecular mass, chain architecture, crystallinity, orientation, cross-linking, defects and processing.

The test temperature and loading rate must also be controlled.

(c) Explain the limitation of testing fresh specimens separately.

The actual component experiences the two loading stages in sequence.

It is first bent slowly during installation. This deformation may leave residual strain or create microscopic damage near the region that is subsequently impacted.

The material’s response to the second force may therefore depend on what happened during the first stage.

However, the separate tensile and impact experiments use fresh specimens.

They do not reproduce the combined loading history experienced by the finished component.

This is an important limitation because the polymer that performs best when undamaged may not necessarily perform best after being bent during installation.

(d) Design a more appropriate experiment.

The most useful experiment would reproduce the component’s actual loading sequence.

First, manufacture identical snap-fit specimens from each polymer using controlled processing conditions.

Next:

1. Bend each specimen to the required installation displacement at 23°C.
2. Apply the same installation procedure to every specimen.
3. Condition the specimens to 5°C.
4. Apply a controlled impact at the region that experienced bending.
5. Record the energy absorbed, deformation and whether fracture occurs.

Include additional control groups that undergo impact without the initial bending stage.

This allows the manufacturer to determine whether the installation process changes subsequent impact resistance.

The final comparison should focus on the complete component and its actual loading conditions, rather than relying on a single measurement of tensile strength or intermolecular attraction.

If H performs better after the complete sequence, it may be the more suitable choice. If D performs better, the greater flexibility and deformation capacity may provide an advantage.

The correct selection depends on the combined requirements of the application.

This final example brings the central idea together: intermolecular forces influence polymer properties, but molecular structure, processing and loading conditions determine how those properties appear in the finished material.