The strength–density chart

Figure 6.6 shows the yield strength σ_y or elastic limit σ_el plotted against density ρ. The range of strength for engineering materials, like that of the modulus, spans about six decades: from less than 0.01 MPa for foams, used in packaging and energy-absorbing systems, to 10⁴ MPa for diamond, exploited in diamond tooling for machining and as the indenter of the Vickers hardness test. Members of each family again cluster together and can be enclosed in envelopes, each of which occupies a characteristic part of the chart.

Figure 6.6 The strength–density chart.

Comparison with the modulus–density chart (Figure 4.7) reveals some marked differences. The modulus of a solid is a well-defined quantity with a narrow range of values. The strength is not. The strength range for a given class of metals, such as stainless steels, can span a factor of 10 or more, while the spread in stiffness is at most 10%. Since density varies very little (Chapter 4), the strength bubbles for metals are long and thin. The wide ranges for metals reflect the underlying physics of yielding and present designers with an opportunity for manipulation of the strength by varying composition and process history. Both are discussed later in this chapter.

Polymers cluster together with strengths between 10 and 100 MPa. The composites CFRP and GFRP have strengths that lie between those of polymers and ceramics, as we might expect since they are mixtures of the two. The analysis of the strength of composites is not as straightforward as for modulus in Chapter 4, though the same bounds (with strength replacing modulus) generally give realistic estimates.

The modulus–strength chart

Figure 6.7 shows Young’s modulus, E, plotted against yield strength, σ_y or elastic limit σ_el. This chart allows us to examine a useful material characteristic, the yield strain, σ_y/E, meaning the strain at which the material ceases to be linearly elastic. On log axes, contours of constant yield strain appear as a family of straight parallel lines, as shown in Figure 6.7. Engineering polymers have large yield strains, between 0.01 and 0.1; the values for metals are at least a factor of 10 smaller. Composites and woods lie on the 0.01 contour, as good as the best metals. Elastomers, because of their exceptionally low moduli, have values of σ_y/E in the range 1 to 10, much larger than any other class of material.

Figure 6.7 The Young’s modulus–strength chart. The contours show the strain at the elastic limit, σ_y/E.

This chart has many other applications, notably in selecting materials for springs, elastic diaphragms, flexible couplings and snap-fit components. We explore these in Chapter 7.

Perfection: the ideal strength

The bonds between atoms, like any other spring, have a breaking point. Figure 6.8 shows a stress–strain curve for a single bond. Here an atom is assumed to occupy a cube of side a_o (as was assumed in Chapter 4) so that a force F corresponds to a stress . The force stretches the bond from its initial length a_o to a new length a, giving a strain (a − a_o)/a_o. When discussing the modulus in Chapter 4 we focused on the initial, linear part of this curve, with a slope equal to the modulus, E. Stretched farther, the curve passes through a maximum and sinks to zero as the atoms lose communication. The peak is the bond strength—if you pull harder than this it will break. The same is true if you shear it rather than pull it.

Figure 6.8 The stress–strain curve for a single atomic bond (it is assumed that each atom occupies a cube of side a_o).

The distance over which inter-atomic forces act is small—a bond is broken if it is stretched to more than about 10% of its original length. So the force needed to break a bond is roughly:

(6.5)

where S, as before, is the bond stiffness. On this basis the ideal strength of a solid should therefore be roughly

(remembering that E = S_o/a_o, equation (4.17))or

(6.6)

This doesn’t allow for the curvature of the force–distance curve; more refined calculations give a ratio of 1/15.

Figure 6.9 shows σ_y/E for metals, polymers and ceramics. None achieves the ideal value of 1/10; most don’t even come close. Why not? It’s a familiar story: like most things in life, materials are imperfect.

Figure 6.9 The ideal strength is predicted to be about E/15, where E is Young’s modulus. The figure shows σ_y/E with a shaded band at the ideal strength.

Crystalline imperfection: defects in metals and ceramics

Crystals contain imperfections of several kinds. Figure 6.10 introduces the broad families, distinguished by their dimensionality. At the top left are point defects. All crystals contain vacancies, shown in (a): sites at which an atom is missing. They play a key role in diffusion, creep and sintering (Chapter 13), but we don’t need them for the rest of this chapter because they do not influence strength. The others do.

Figure 6.10 Defects in crystals. (a) Vacancies–missing atoms. (b) Foreign (solute) atom on interstitial and substitutional sites. (c) A dislocation—an extra half-plane of atoms. (d) Grain boundaries.

No crystal is totally, 100%, pure and perfect. Some impurities are inherited from the process by which the material was made; more usually they are deliberately added, creating alloys: a material in which a second (or third or fourth) element is dissolved. ‘Dissolved’ sounds like salt in water, but these are solid solutions. Figure 6.10(b) shows both a substitutional solid solution (the dissolved atoms replace those of the host) and an interstitial solid solution (the dissolved atoms squeeze into the spaces or ‘interstices’ between the host atoms). The dissolved atoms or solute rarely has the same size as those of the host material, so it distorts the surrounding lattice. The red atoms here are substitutional solute, some bigger and some smaller than those of the host; the cages of host atoms immediately surrounding them, shown green, are distorted. If the solute atoms are particularly small, they don’t need to replace a host atom; instead, they dissolve interstitially like the black atoms in the figure, again distorting the surrounding lattice. So solute causes local distortion; this distortion is one of the reasons that alloys are stronger than pure materials, as we shall see in a moment.

Now to the key player, portrayed in Figure 6.10(c): the dislocation. ‘Dislocated’ means ‘out of joint’ and this is not a bad description of what is happening here. The upper part of the crystal has one more double-layer of atoms than the lower part (the double-layer is needed to get the top-to-bottom registry right). It is dislocations that make metals soft and ductile. Dislocations distort the lattice—here the green atoms are the most distorted—and because of this they have elastic energy associated with them. If they cost energy, why are they there? To grow a perfect crystal just one cubic centimeter in volume from a liquid or vapour, about 102³ atoms have to find their proper sites on the perfect lattice, and the chance of this happening is just too small. Even with the greatest care in assembling them, all crystals contain point defects, solute atoms and dislocations.

Most contain yet more drastic defects, among them grain boundaries. Figure 6.10(d) shows such boundaries. Here three perfect, but differently oriented, crystals meet; the individual crystals are called grains, the meeting surfaces are grain boundaries. In this sketch the atoms of the three crystals have been given different colours to distinguish them, but here they are the same atoms. In reality grain boundaries form in pure materials (when all the atoms are the same) and in alloys (when the mixture of atoms in one grain may differ in chemical composition from those of the next).

Now put all this together. The seeming perfection of the steel of a precision machine tool or of the polished case of a gold watch is an illusion: they are riddled with defects. Imagine all the frames of Figure 6.10 superimposed and you begin to get the picture. Between them they explain diffusion, strength, ductility, electrical resistance, thermal conductivity and much more.

So defects in crystals are influential. For the rest of this section we focus on getting to know just one of them: the dislocation.

Dislocations and plastic flow

Recall that the strength of a perfect crystal computed from inter-atomic forces gives an ‘ideal strength’ around E/15 (where E is the modulus). In reality the strengths of engineering materials are nothing like this big; often they are barely 1% of it. This was a mystery until halfway through the last century—a mere 60 years ago—when an Englishman, G. I. Taylor³, and a Hungarian, Egon Orowan⁴, realized that a ‘dislocated’ crystal could deform at stresses far below the ideal. So what is a dislocation, and how does it enable deformation?

Figure 6.11(a) shows how to make a dislocation. The crystal is cut along an atomic plane up to the line shown as ⊥—⊥, the top part is slid across the bottom by one full atom spacing, and the atoms are reattached across the cut plane to give the atom configuration shown in Figure 6.11(b). There is now an extra half-plane of atoms with its lower edge along the ⊥—⊥ line, the dislocation line—the line separating the part of the plane that has slipped from the part that has not. This particular configuration is called an edge dislocation because it is formed by the edge of the extra half-plane, represented by the symbol ⊥.

Figure 6.11 (a) Making a dislocation by cutting, slipping and rejoining bonds across a slip plane. (b) The atom configuration at an edge dislocation in a simple cubic crystal. The configurations in other crystal structures are more complex but the principle remains the same.

When a dislocation moves it makes the material above the slip plane slide relative to that below, producing a shear strain. Figure 6.12 shows how this happens. At the top is a perfect crystal. In the central row a dislocation enters from the left, sweeps through the crystal and exits on the right. By the end of the process the upper part has slipped by b, the slip vector (or Burger’s vector) relative to the part below. The result is the shear strain γ shown at the bottom.

Figure 6.12 An initially perfect crystal is shown in (a). The passage of the dislocation across the slip plan, shown in the sequence (b), (c) and (d), shears the upper part of the crystal over the lower part by the slip vector b. When it leaves the crystal has suffered a shear strain γ.

There is another way to make a dislocation in a crystal. After making the cut in Figure 6.11(a), the upper part of the crystal can be displaced parallel to the edge of the cut rather than normal to it, as in Figure 6.13. That too creates a dislocation, but one with a different configuration of atoms along its line—one more like a corkscrew than like a squashed worm—and for this reason it is called a screw dislocation. We don’t need the details of its structure; it is enough to know that its properties are like those of an edge dislocation except that when it sweeps through a crystal (moving normal to its line), the lattice is displaced parallel to the dislocation line, not normal to it. All dislocations are either edge or screw or mixed, meaning that they are made up of little steps of edge and screw. The line of a mixed dislocation can be curved but every part of it has the same slip vector b because the dislocation line is just the boundary of a plane on which a fixed displacement b has occurred.

Figure 6.13 A screw dislocation. The slip vector b is parallel to the dislocation line S—S.

It is far easier to move a dislocation through a crystal, breaking and remaking bonds only along its line as it moves, than it is to simultaneously break all the bonds in the plane before remaking them. It is like moving a heavy carpet by pushing a fold across it rather than sliding the whole thing at one go. In real crystals it is easier to make and move dislocations on some planes than on others. The preferred planes are called slip planes and the preferred directions of slip in these planes are called slip directions. A slip plane is shown in gray and a slip direction as an arrow on the FCC and BCC unit cells of Figure 4.11.

Slip displacements are tiny—one dislocation produces a displacement of about 10⁻¹⁰ m. But if large numbers of dislocations traverse a crystal, moving on many different planes, the shape of a material changes at the macroscopic length scale. Figure 6.14 shows just two dislocations traversing a sample loaded in tension. The slip steps (here very exaggerated) cause the sample to get a bit thinner and longer. Repeating this millions of times on many planes gives the large plastic extensions observed in practice. Since none of this changes the average atomic spacing, the volume remains unchanged.

Figure 6.14 Dislocation motion causes extension.

Why does a shear stress make a dislocation move?

Crystals resist the motion of dislocations with a friction-like resistance f per unit length—we will examine its origins in a moment. For yielding to take place the external stress must overcome the resistance f.

Imagine that one dislocation moves right across a slip plane, traveling the distance L₂, as in Figure 6.15. In doing so, it shifts the upper half of the crystal by a distance b relative to the lower half. The shear stress τ acts on an area L₁L₂, giving a shear force F_s = τL₁L₂ on the surface of the block. If the displacement parallel to the block is b, the force does work

Figure 6.15 The force on a dislocation. (a) Perspective view. (b) Plan view of slip plane.

(6.7)

This work is done against the resistance f per unit length, or f L₁ on the length L₁, and it does so over a displacement L₂ (because the dislocation line moves this far against f), giving a total work against f of fL₁L₂. Equating this to the work W done by the applied stress τ gives

(6.8)

This result holds for any dislocation—edge, screw or mixed. So, provided the shear stress τ exceeds the value f/b it will make dislocations move and cause the crystal to shear.

Line tension

The atoms near the core of a dislocation are displaced from their proper positions, as shown by green atoms back in Figure 6.10(c), and thus they have higher potential energy. To keep the potential energy of the crystal as low as possible, the dislocation tries to be as short as possible—it behaves as if it had a line tension, T, like an elastic band. The tension can be calculated but it needs advanced elasticity theory to do it (the books listed under ‘Further reading’ give the analysis). We just need the answer. It is that the line tension, an energy per unit length (just as a surface tension is an energy per unit area), is

(6.9)

where E, as always, is Young’s modulus. The line tension has an important bearing on the way in which dislocations interact with obstacles, as we shall see in a moment.

The lattice resistance

Where does the resistance to slip, f, come from? There are several contributions. Consider first the lattice resistance, f_i: the intrinsic resistance of the crystal structure to plastic shear. Plastic shear, as we have seen, involves the motion of dislocations. Pure metals are soft because the nonlocalised metallic bond does little to obstruct dislocation motion, whereas ceramics are hard because their more localised covalent and ionic bonds (which must be broken and reformed when the structure is sheared) lock the dislocations in place. When the lattice resistance is high, as in ceramics, further hardening is superfluous—the problem becomes that of suppressing fracture. On the other hand, when the lattice resistance f_i is low, as in metals, the material can be strengthened by introducing obstacles to slip. This is done by adding alloying elements to give solid solution hardening (f_ss), precipitates or dispersed particles giving precipitation hardening (f_ppt), other dislocations giving what is called work hardening (f_wh) or grain boundaries introducing grain-size hardening (f_gb). These techniques for manipulating strength are central to alloy design. We look at them more closely in the next section.

Plastic flow in polymers

At low temperatures, meaning below about 0.75 T_g, polymers are brittle. Above this temperature they become plastic. When pulled in tension, the chains slide over each other, unraveling, so that they become aligned with the direction of stretch, as in Figure 6.16(a), a process called drawing. It is harder to start drawing than to keep it going, so the zone where it starts draws down completely before propagating farther along the sample, leading to profiles like that shown in the figure. The drawn material is stronger and stiffer than before, by a factor of about 8, giving drawn polymers exceptional properties, but because you can only draw fibres or sheet (by pulling in two directions at once) the geometries are limited.

Figure 6.16 (a) Cold drawing—one of the mechanisms of deformation of thermoplastics.(b) Crazing—local drawing across a crack. (c) Shear banding.

Many polymers, among them PE, PP and nylon, draw at room temperature. Others with higher glass temperatures, such as PMMA, do not, although they draw well at higher temperatures. At room temperature they craze. Small crack-shaped regions within the polymer draw down. Because the crack has a larger volume than the polymer that was there to start with, the drawn material ends up as ligaments that link the craze surfaces, as in Figure 6.16(b). Crazes scatter light, so their presence causes whitening, easily visible when cheap plastic articles are bent. If stretching is continued, one or more crazes develop into proper cracks, and the sample fractures.

When crazing limits ductility in tension, large plastic strains may still be possible in compression by shear banding (Figure 6.16(c)). Within each band, shear takes place with much the same consequences for the shape of the sample as shear by dislocation motion. Continued compression causes the number of shear bands to increase, giving increased overall strain.

Strengthening metals

The way to make crystalline materials stronger is to make it harder for dislocations to move. As we have seen, dislocations move in a pure crystal when the force τb per unit length exceeds the lattice resistance f_i. There is little we can do to change this—it is an intrinsic property like the modulus E. Other strengthening mechanisms add to it, and here there is scope for manipulation. Figure 6.17 introduces them. It shows the view of a slip plane from the perspective of an advancing dislocation: each strengthening mechanism presents a new obstacle course. In the perfect lattice shown in (a) the only resistance is the intrinsic strength of the crystal; solution hardening, shown in (b), introduces atom-size obstacles to motion; precipitation hardening, shown in (c), presents larger obstacles; and in work hardening, shown in (d), the slip plane becomes stepped and threaded with ‘forest’ dislocations.

Figure 6.17 A ‘dislocation-eye’ view of the slip plane across which it must move.

Obstacles to dislocation motion increase the resistance f and thus the strength. To calculate their contribution to f, there are just two things we need to know about them: their spacing and their strength. Spacing means the distance L between them in the slip plane. The number of obstacles touching unit length of dislocation line is then

Each individual obstacle exerts a pinning force p on the dislocation line—a resisting force per unit length of dislocation—so the contribution of the obstacles to the resistance f is

Thus, the added contribution to the shear stress τ needed to make the dislocation move is (from equation (6.8))

(6.10)

The pinning is an elastic effect—it derives from the fact that both the dislocation and the obstacle distort the lattice elastically even though, when the dislocation moves, it produces plastic deformation. Because of this p, for any given obstacle in any given material, scales as Eb², which has the units of force. The shear stress τ needed to force the dislocation through the field of obstacles then has the form

(6.11)

where α is a dimensionless constant characterizing the obstacle strength.

Armed with this background we can explain strengthening mechanisms. We start with solid solutions.

Solution hardening

Solid solution hardening is strengthening by deliberate additions of impurities or, more properly said, by alloying (Figure 6.18(a)). The addition of zinc to copper makes the alloy brass—copper dissolves up to 30% zinc. The zinc atoms replace copper atoms to form a random substitutional solid solution. The zinc atoms are bigger that those of copper and, in squeezing into the copper lattice, they distort it. This roughens the slip plane, so to speak, making it harder for dislocations to move, thereby adding an additional resistance f_ss, opposing dislocation motion. The figure illustrates that the concentration of solute, expressed as an atom fraction, is on average:

Figure 6.18 (a) Solution hardening. (b) Precipitation or dispersion hardening. (c) Forest hardening (work hardening).

where L is the spacing of obstacles in the slip plane and b is the atom size. Thus,

Plugging this into equation (6.11) relates the contribution of solid solution to the shear stress required to move the dislocation:

(6.12)

τ_ss increases as the square root of the solute concentration. Brass, bronze and stainless steels, and many other metallic alloys, derive their strength in this way. They differ only in the extent to which the solute distorts the crystal, described by the constant α.

Dispersion and precipitate strengthening

A more effective way to impede dislocations is to disperse small, strong particles in their path. One way to make such a microstructure is to disperse small solid particles of a high melting point compound into a liquid metal, and to cast it to shape, trapping the particles in place—it is the way that metal–matrix composites such as Al–SiC are made. An alternative is to form the particles in situ by a precipitation process. If a solute (copper, say) is dissolved in a metal (aluminum, for instance) at high temperature when both are molten, and the alloy is solidified and cooled to room temperature, the solute precipitates as small particles, much as salt will crystallize from a saturated solution when it is cooled. An alloy of aluminum containing 4% copper, treated in this way, gives very small, closely spaced precipitates of the hard compound CuAl₂. Copper alloyed with a little beryllium, similarly treated, gives precipitates of the compound CuBe. Most steels are strengthened by precipitates of carbides, obtained in this way. The precipitates give a large contribution to f.

Figure 6.18(b) shows how particles obstruct dislocation motion. If the particles are too strong for the dislocation to slice through them, the force τb pushes the dislocation between them, bending it to a tighter and tighter radius against its line tension (equation (6.9)). The radius is at a minimum when it reaches half the particle spacing, L; after that it can expand under lower stress. It is a bit like blowing up a bicycle inner tube when the outer tire has a hole in it: once you reach the pressure that balloons the inner tube through the hole, the balloon needs a smaller pressure to get bigger still. The critical configuration is the semicircular one: here the total force τbL on one segment of length L is just balanced by the force 2T due to the line tension (equation (6.10)), acting on either side of the bulge, as in Figure 6.19. The dislocation escapes when

Figure 6.19 Successive positions of a dislocation as it bypasses particles that obstruct its motion. The critical configuration is that with the tightest curvature, shown in (b).

(6.13)

The obstacles thus exert a resistance of f_ppt = 2T/L. Precipitation hardening is an effective way to increasing strength: precipitate-hardened aluminum alloys can be 15 times stronger than pure aluminum.

Work hardening

The rising part of the stress–strain curve of Figure 6.1 is caused by work hardening: it is caused by the accumulation of dislocations generated by plastic deformation. The dislocation density, ρ_d, is defined as the length of dislocation line per unit volume (m/m³). Even in an annealed soft metal, the dislocation density is around 10¹⁰ m/m³, meaning that a 1 cm cube (the size of a cube of sugar) contains about 10 km of dislocation line. When metals are deformed, dislocations multiply, causing their density to grow to as much as 10¹⁷ m/m³ or more—100 million km per cubic centimeter. A moving dislocation now finds that its slip plane is penetrated by a forest of intersecting dislocations with an average spacing L = ρ_d^−1/2 (since ρ_d is a number per unit area). Figure 6.18(c) suggests the picture. If a moving dislocation advances, it shears the material above the slip plane relative to that below, and that creates a little step called a jog in each forest dislocation. The jogs have potential energy—they are tiny segments of dislocation of length b—with the result that each exerts a pinning force p = Eb²/2 on the moving dislocation. Assembling these results into equation (6.10) gives

(6.14)

The greater the density of dislocations, the smaller the spacing between them, and so the greater their contribution to τ_wh.

All metals work harden. It can be a nuisance: if you want to roll thin sheet, work hardening quickly raises the yield strength so much that you have to stop and anneal the metal (heat it up to remove the accumulated dislocations) before you can go on—a trick known to blacksmiths for centuries. But it is also useful: it is a potent strengthening method, particularly for alloys that cannot be heat-treated to give precipitation hardening.

Grain boundary hardening

Almost all metals are polycrystalline, made up of tiny, randomly oriented crystals, or grains, meeting at grain boundaries like those of Figure 6.10(d). The grain size, D, is typically 10–100 µm. These boundaries obstruct dislocation motion. A dislocation in one grain—call it grain 1—can’t just slide into the next (grain 2) because the slip planes don’t line up. Instead, new dislocations have to nucleate in grain 2 with slip vectors that, if superimposed, match that of the dislocation in grain 1 so that the displacements match at the boundary. This gives another contribution to strength, τ_gb, that is found to scale as D^−1/2, giving

(6.15)

where k_p is called the Petch constant, after the man who first measured it. For normal grain sizes τ_gb is small and not a significant source of strength, but for materials that are microcrystalline (D < 1 µm) or nanocrystalline (D approaching 1 nm) it becomes significant.

Relationship between dislocation strength and yield strength

To a first approximation the strengthening mechanisms add up, giving a shear yield strength, τ_y, of

(6.16)

Strong materials either have a high intrinsic strength, τ_i (like diamond), or they rely on the superposition of solid solution strengthening τ_ss, precipitates τ_ppt and work hardening τ_wh (like high-tensile steels). Nanocrystalline solids exploit, in addition, the contribution of τ_gb.

Before we can use this information, one problem remains: we have calculated the yield strength of one crystal, loaded in shear. We want the yield strength of a polycrystalline material in tension. To link them there are two simple steps. First, a uniform tensile stress σ creates a shear stress on planes that lie at an angle to the tensile axis; dislocations will first move on the slip plane on which this shear stress is greatest. Figure 6.20 shows how this is calculated. A tensile force F acting on a rod of cross-section A, if resolved parallel to a plane with a normal that lies at an angle θ to the axis of tension, gives a force F sin θ in the plane. The area of this plane is A/cos θ, so the shear stress is

Figure 6.20 The resolution of stress. A tensile stress σ gives a maximum shear stress τ = σ/2 on a plane at 45 degree to the tensile axis.

where σ = F/A is the tensile stress. The value of τ is plotted against θ in the figure. The maximum lies at an angle of 45 degrees, when τ = σ/2.

Second, when this shear stress acts on an aggregate of crystals, some crystals will have their slip planes oriented favorably with respect to the shear stress, others will not. This randomness of orientation jacks up the strength by a further factor of 1.5 (called the Taylor factor—see the footnote on p. 122). Combining these results, the tensile stress to cause yielding of a sample that has many grains is approximately three times the shear strength of a single crystal:

Thus, the superposition of strengthening mechanisms in equation (6.16) applies equally to the yield strength, σ_y.

Strength and ductility of alloys

Of all the properties that materials scientists and engineers have sought to manipulate, the strength of metals and alloys is probably the most explored. It is easy to see why—Table 6.1 gives a small selection of the applications of metals and their alloys. Their importance in engineering design is enormous. The hardening mechanisms often are used in combination. This is illustrated graphically for copper alloys in Figure 6.21. Good things, however, have to be paid for. Here the payment for increased strength is, almost always, loss of ductility so the elongation ε_f is reduced. The material is stronger but it cannot be deformed as much without fracture.

Table 6.1 Metal alloys with typical applications, indicating the strengthening mechanisms used

Symbols: ✓✓✓ = Routinely used. ✓ = Sometimes used.

Figure 6.21 Strengthening mechanisms and the consequent drop in ductility, here shown for copper alloys. The mechanisms are frequently combined. The greater the strength, the lower the ductility (the elongation to fracture, ε_f).

Strengthening polymers

In non-crystalline solids the dislocation is not a helpful concept. We think instead of some unit step of the flow process: the relative slippage of two segments of a polymer chain, or the shear of a small molecular cluster in a glass network. Their strength has the same origin as that underlying the lattice resistance: if the unit step involves breaking strong bonds (as in an inorganic glass), the materials will be strong and brittle, as ceramics are. If it only involves the rupture of weak bonds (the Van der Waals bonds in polymers, for example), it will be weak. Polymers too must therefore be strengthened by impeding the slippage of segments of their molecular chains. This is achieved by blending, by drawing, by cross-linking and by reinforcement with particles, fibres or fabrics.

A blend is a mixture of two polymers, stirred together in a sort of industrial food-mixer. The strength and modulus of a blend are just the average of those of the components, weighted by volume fraction (a rule of mixtures again). If one of these is a low molecular weight hydrocarbon, it acts as a plasticiser, reducing the modulus and giving the blend a leather-like flexibility.

Drawing is the deliberate use of the molecule-aligning effect of stretching, like that sketched in Figure 6.16(a), to greatly increase stiffness and strength in the direction of stretch. Fishing line is drawn nylon, Mylar film is a polyester with molecules aligned parallel to the film and geotextiles, used to restrain earth banks, are made from drawn polyethylene.

Cross-linking, sketched in Figures 4.18 and 4.19, creates strong bonds between molecules that were previously linked by weak Van der Waals forces. Vulcanized rubber is rubber that has been cross-linked, and the superior strength of epoxies derives from cross-linking.

Reinforcement is possible with particles of cheap fillers—sand, talc or wood dust. Far more effective is reinforcement with fibres—usually glass or carbon—either continuous or chopped, as explained in Chapter 4.