Chapter 19Hooke's Law, Elastic Moduli, and Material Testing

19.1 Hooke's Law

The British physicist Robert Hooke first showed in 1678, from experiments on the relation between the extension and the load of a coil spring, that “within the elastic range, stress and strain are directly proportional.” This is called Hooke's law. You are probably familiar from high school physics with the proportional relation $F=kx$ ($k$ is the spring constant) between the extension $x$ of a spring and the load $F$. (In high school, the restoring force that a spring exerts on an object is written $F=-kx$, and the magnitude of the force applied to stretch the spring is written $F=kx$. These are the same relation, differing only in how the direction of the force is viewed.) What Hooke discovered is a more general fact: a proportional relation of the same form—“when a force is applied, the deformation grows in proportion to the force, and when the force is removed, the body returns to its original shape”—holds not only for a special component such as a spring but, as long as we stay within the elastic range, for the deformation of every solid.

Using the stress $\sigma$ (the internal force per unit area) defined in Section 18.3 of the previous chapter and the strain $\varepsilon$ (the dimensionless ratio of deformation) defined in Section 18.4, Hooke's law can be written as follows.

Depending on what we substitute for “stress” and “strain” in Eq. \eqref{eq:19-hooke-verbal}, there are several kinds of elastic modulus—the ratio of normal stress to longitudinal strain gives Young's modulus, the ratio of shear stress to shear strain gives the shear modulus, and so on (we define them in turn in Section 19.2). In every case, a “large elastic modulus” means a “small strain for the same stress”—that is, “hard to deform; stiff.” Conversely, a material with a small elastic modulus (such as rubber) deforms greatly even under a small stress.

19.2 Young's Modulus, Shear Modulus, and Bulk Modulus

The deformations treated in mechanics of materials fall broadly into three types—(i) “stretching and shrinking” caused by tension or compression, (ii) “sliding” caused by shear, and (iii) “change of volume” caused by hydrostatic pressure (Figure 19.2). Even for the same material, the ratio of stress to strain takes a different value when the manner of deformation differs, so a separate elastic modulus is defined for each—Young's modulus $E$, shear modulus $G$, and bulk modulus $K$. Let us look at them one by one.

19.2.1 Young's modulus $E$

Let us write this out concretely for a bar subjected to a tensile load $P$. When a load $P$ is applied to a bar of cross-sectional area $A_0$ and length $l_0$, the normal stress is $\sigma=P/A_0$ by the definition in Section 18.3; if the bar elongates by $\lambda$, the longitudinal strain is $\varepsilon=\lambda/l_0$ by the definition in Section 18.4.

Relation to the spring constant. Solving Eq. \eqref{eq:19-E-def} for $P$, we obtain

This has the same form as the spring formula $F=kx$ of high school physics. That is, a bar of cross-sectional area $A_0$, length $l_0$, and Young's modulus $E$ can be regarded as a spring with spring constant $k=EA_0/l_0$ (the unit is $[\mathrm{Pa\cdot m^2/m}]=[(\mathrm{N/m^2})\cdot\mathrm{m}]=\mathrm{N/m}$, which agrees with the unit of the spring constant). The thicker the bar (the larger $A_0$ is), the shorter it is (the smaller $l_0$ is), and the stiffer the material is (the larger $E$ is), the larger the spring constant becomes. Whereas the spring constant $k$ is a quantity that also depends on the shape of the bar ($A_0$ and $l_0$), Young's modulus $E$ is a quantity determined by the material alone—even for the same steel, $k$ changes if we change the thickness or the length, but $E$ does not.

19.2.2 Shear modulus $G$

The shear strain $\gamma$ defined in Section 18.5 is likewise proportional to the shear stress $\tau$.

19.2.3 Bulk modulus $K$

Finally, consider a stress that acts uniformly and perpendicularly over the entire surface of a body, regardless of direction—for example, the hydrostatic pressure on an object submerged in water. In this case the body does not stretch or shrink; rather, its volume as a whole changes.

19.3 Advanced Topic: Relations among the Elastic Constants — Why Are There Only Two Independent Elastic Constants?

So far, four elastic constants have appeared for the isotropic elastic body (Definition 18.2, “isotropic body,” in Section 18.1)—Young's modulus $E$, the shear modulus $G$, the bulk modulus $K$, and Poisson's ratio $\nu$. These may look like four unrelated constants, but in fact only two of them can be chosen independently, and the remaining two can be calculated from those two. In this section we derive the relations $G=E/\{2(1+\nu)\}$ and $K=E/\{3(1-2\nu)\}$ and confirm the meaning of the fact that “there are two independent elastic constants.” The content of this section is what is normally needed in order to understand the elastic moduli in a self-contained way.

19.3.1 The generalized Hooke's law — when stresses act in several directions at once

All of the definitions of $E,G,K$ so far considered “the simple case in which stress acts in only one direction.” However, to derive the two relations we are about to see, we need to know the strain when normal stresses act simultaneously in two or three directions. So let us stop here for a moment and extend Hooke's law to the case in which stresses in several directions act at the same time.

19.3.2 Relation between Young's modulus and the shear modulus $G=E/\{2(1+\nu)\}$

Let us show that the shear modulus (the elastic modulus for shear) $G$ is in fact determined by Young's modulus $E$ and Poisson's ratio $\nu$ alone. The key is the fact that the deformation called “pure shear” is, seen from another point of view, the same as “a combination of tension and compression in directions rotated by 45°.”

19.3.3 Relation between Young's modulus and the bulk modulus $K=E/\{3(1-2\nu)\}$

Next, consider the case in which a cubic element receives normal stresses of equal magnitude $\sigma$ from all three directions (a hydrostatic-type deformation). Setting $\sigma_x=\sigma_y=\sigma_z=\sigma$ in the generalized Hooke's law \eqref{eq:19-generalized-hooke}, we obtain:

19.3.4 Only two independent elastic constants / the possible range of Poisson's ratio

Looking at Formula 19.1 and Formula 19.2, we see that $G$ and $K$ are both written in terms of $E$ and $\nu$ alone. That is, for an isotropic elastic body, once any two of the four constants $E,G,K,\nu$ (in many cases $E$ and $\nu$, which are easy to measure) are fixed, the remaining two are determined automatically by Formula 19.1 and Formula 19.2—this is the meaning of “an isotropic elastic body has only two independent elastic constants.”

Let us also confirm that “any two may be chosen.” Suppose, for example, that we have measured $G$ and $K$. By Formula 19.1 and Formula 19.2, $E=2G(1+\nu)=3K(1-2\nu)$. Solving the right-hand equality for $\nu$, from $2G+2G\nu=3K-6K\nu$ we get $(2G+6K)\nu=3K-2G$, that is,

($E$ is found by substituting the $\nu$ just obtained into $E=2G(1+\nu)$). Thus $E$ and $\nu$ are determined from $G$ and $K$. In the same way, from any pair, the remaining two are determined.

19.4 Material Testing

In the sections so far, we have defined the elastic moduli $E,G,K$, constants that express “how much a material deforms within the elastic range.” However, to actually select materials and design a structure, it is necessary to know not only “what the elastic moduli are” but also the mechanical properties of the material in general, such as “up to what stress it can be used safely” and “how much force can be applied before it breaks.” Tests that investigate these properties experimentally are collectively called material testing.

Corresponding to the classification by loading rate (static loads and dynamic loads) seen in Section 18.2 of the previous chapter, material tests fall broadly into two groups.

Of these, the most fundamental test in the mechanics of materials, and the one used directly to measure elastic moduli and strength, is the tensile test. A round-bar or plate-shaped specimen is mounted in a universal testing machine (a device that clamps both ends of the specimen in grips and records the load and elongation while pulling at a constant speed), and while the load $P$ is gradually increased, the elongation $\lambda$ relative to the gauge length $l_0$ (the distance between two reference points marked on the specimen in advance) is measured. Here, rather than recording the load and elongation as they are, it is standard to use the quantities divided by the initial dimensions (the undeformed cross-sectional area $A_0$ and the gauge length $l_0$).

A graph in which the load $P$ applied to the specimen (or the nominal stress $\sigma$) is plotted against the nominal strain $\varepsilon$ on the horizontal axis is called a stress–strain diagram. It is one of the most important figures in the mechanics of materials, one from which the mechanical properties of a material can be read at a glance. Within the range of elastic deformation, as we saw in the preceding sections, $\sigma$ and $\varepsilon$ are linearly related (Hooke's law), but beyond a certain stress the slope of the graph gradually becomes gentler.

Beyond the proportional limit, the graph changes into a curve whose slope gradually becomes gentler. When mild steel (the most commonly used steel, with a low carbon content) is subjected to a tensile test, still more characteristic changes—the elastic limit, the yield point, the tensile strength, and fracture—appear in succession after the proportional limit is exceeded. The detailed definitions of these, and the difference between nominal stress and true stress, are treated as more advanced material in Section 19.5. Here we first show the overall picture of a typical stress–strain diagram for mild steel (Figure 19.5).

19.5 Advanced Topic: Details of the Stress–Strain Diagram and True Stress and True Strain

Beyond the proportional limit lies important information that bears directly on the design of structures: yielding, tensile strength, and fracture. In this section we define in turn each point (①–⑥) of the stress–strain diagram of mild steel in Figure 19.5, and then go on to look at the difference between the two kinds of stress and strain, “nominal” and “true,” and at the contrast with brittle materials such as ceramics.

19.5.1 Definitions of the points on the stress–strain diagram

For metals in which the upper and lower yield points do not appear clearly (many nonferrous metals and alloys), the proof stress, defined as follows, is used instead of the “yield point” as a guide to the onset of plastic deformation.

19.5.2 Nominal stress and nominal strain, and true stress and true strain

The nominal stress $\sigma=P/A_0$ that we saw in Definition 19.6 is a quantity that keeps being divided by the cross-sectional area $A_0$ before deformation, even as the deformation proceeds. The stress actually felt at a point inside the specimen, however, should be $P/A$, obtained by dividing by the true cross-sectional area at each moment, $A$. In particular, in the latter half of the tensile test (after necking has progressed), the area $A$ of the narrowest cross section has become much smaller than $A_0$, so $P/A_0$ and $P/A$ take very different values.

In the plastic deformation of metals, the atoms merely rearrange themselves and the volume hardly changes (the slight change in volume due to elastic deformation can be neglected). Therefore, while the specimen as a whole is stretching uniformly (until necking begins), we may set $AL=A_0l_0$ (constant volume), so that $A=A_0l_0/L=A_0/(1+\varepsilon)$, and the true stress and true strain can be rewritten in terms of the nominal stress and nominal strain.

19.5.3 Work hardening, Hollomon's equation, and necking

An empirical formula is known that expresses quantitatively, in terms of the true stress and true strain, the work hardening described in Section 19.5.1—the phenomenon in which the further plastic deformation proceeds, the harder it becomes for the material to deform.

Even though work hardening is present, why does the nominal stress decrease once the tensile strength is exceeded? The answer is “necking.” Once the tensile strength is exceeded, an extremely slight variation in the cross-sectional area of the specimen (a perfectly uniform bar cannot be made in reality) triggers the deformation to begin concentrating at the location with the smallest cross-sectional area. This is necking.

After necking begins, the stress distribution within the specimen is no longer uniform—the central region of the narrowest cross section of the neck receives tensile stress not only in the tensile direction but also in the transverse directions, because the thicker surrounding portions prevent lateral contraction. Such a state in which stress acts in all three directions is called a triaxial stress state. As a result, the mean stress at the minimum cross section,$\bar\sigma=P/A_{\min}$, is estimated to be higher than the true flow stress $\sigma_{\mathrm{eq}}$ (the value under the assumption of uniform tension). The correction that compensates for this effect is the Bridgman correction. Let $a$ be the radius of the minimum cross section of the neck and $R$ the radius of curvature of the neck profile; then the correction factor is given by $B=(1+2R/a)\ln(1+a/2R)\ (\ge1)$ and $\sigma_{\mathrm{eq}}=\bar\sigma/B$. As $R/a$ becomes larger (the gentler the neck), $B\to1$ and the correction becomes unnecessary; the sharper the neck (the smaller $R/a$), the larger the correction becomes. We do not go into the derivation of this correction in this book. An outline of what happens after necking that leads to fracture (the formation, growth, and coalescence of voids—tiny cavities that form inside the material) is given in Section 21.7.

19.5.4 Ductile fracture and brittle fracture, and the contrast with ceramics

The sequence “elastic deformation → yielding → work hardening → necking → fracture” seen so far is behavior characteristic of ductile materials such as steel and aluminum. They do not break immediately even when the tensile strength is exceeded, but fracture only after large plastic deformation (elongation and reduction of area); a visible warning in the form of “large deformation” therefore appears before a structure fails.

In contrast, brittle materials such as glass and alumina (aluminum oxide, a representative ceramic) show no yielding, and the straight line of elastic deformation continues unchanged until fracture. In this case the tensile strength is the fracture stress itself, and fracture occurs suddenly without warning. Note that ceramics are weak in tension but strong in compression (in compression, the force acts in the direction that closes cracks, so cracks have difficulty spreading). The reason metals and ceramics behave so differently lies in the difference in bonding at the atomic and molecular level. Metallic bonding is like metal ions immersed in a “sea of electrons” that is not strongly tied to any particular pair of atoms, so dislocations move relatively easily and plastic deformation (deformation in which atomic planes slip while the crystal structure is preserved) occurs readily. On the other hand, the situation is different for covalent bonds and ionic bonds, which account for most ceramics. A covalent bond is strongly directional along particular directions, so to move a dislocation, bonds must be broken and rearranged. An ionic bond has no directionality, but when atomic planes slip, ions of the same sign come to be adjacent and Coulomb repulsion arises, so slip is difficult. Consequently, in both cases dislocations hardly move at room temperature, and with almost no plastic deformation the interatomic bonds break immediately after elastic deformation (the mechanism by which bonds break is studied in detail in Chapter 21).

Application: The atomistic origin of the elastic modulus

Recall the potential energy curve $U(r)$ as a function of the interatomic distance $r$ (a Lennard-Jones-type potential, etc.), which we saw in Section 4.6, “Application: Interatomic potential and bond length.” The atoms stay near the equilibrium distance $r_0$ at which $U(r)$ takes its minimum value (a stable equilibrium point in the sense of Theorem 4.4 of Chapter 4). If an external force changes the interatomic distance slightly from $r_0$ by $x=r-r_0$, the interatomic force can be approximated as

$$ f(r)=-\frac{\dd U}{\dd r}\approx-\frac{\dd U}{\dd r}\bigg|_{r_0}-\left.\frac{\dd^2U}{\dd r^2}\right|_{r_0}x=-\left.\frac{\dd^2U}{\dd r^2}\right|_{r_0}x $$

where we used the Taylor expansion (University Mathematics, Section 4.2) and the fact that $r_0$ is a minimum point, so that $\dd U/\dd r|_{r_0}=0$. This is precisely the form of Hooke's law $f=-kx$, and the quantity corresponding to the spring constant is the curvature (second derivative) of the potential curve at its bottom, $k=\dd^2U/\dd r^2|_{r_0}$.

How, then, is the macroscopic Young's modulus $E$ determined from this microscopic spring constant $k$? Let us estimate it with a simple model in which the atoms are arranged in a cubic lattice with edge length $r_0$. The cross-sectional area carried by a single row of atoms (a chain of bonds) is $r_0^2$. When the interatomic distance is stretched by $x$, the force in one bond is $kx$, the strain is $\varepsilon=x/r_0$, and the stress is the force per bond divided by the area it carries, $\sigma=kx/r_0^2$. Therefore we obtain

\begin{equation} E=\frac{\sigma}{\varepsilon}=\frac{kx/r_0^{2}}{x/r_0}=\frac{k}{r_0}=\frac{1}{r_0}\left.\frac{\dd^2U}{\dd r^2}\right|_{r_0} \label{eq:19-E-atomic} \end{equation}

(Allowing for the crystal structure and the directions of the bonds produces deviations of a factor of several, but the order of magnitude is correct.) For example, for iron ($E=2\times10^{11}\ \mathrm{Pa}$, nearest-neighbor interatomic distance $r_0\approx2.5\times10^{-10}\ \mathrm{m}$), $k=Er_0\approx50\ \mathrm{N/m}$, which is much stiffer than a coil spring on a desk (a few $\mathrm{N/m}$). Eq. \eqref{eq:19-E-atomic} shows that the elastic modulus is nothing other than how sharply the bottom of the interatomic potential curves, measured on a macroscopic scale (Figure 19.8). The Young's modulus of a material is larger the stronger its bonds are and the more sharply and steeply the bottom of its potential rises (the larger its curvature). That the Young's modulus of diamond (whose covalent bonds are extremely strong and directional) exceeds $E\approx1000\ \mathrm{GPa}$, while that of polymer materials such as polyethylene, whose molecules are held together only by weak van der Waals forces, remains at about $E\sim1\ \mathrm{GPa}$, is due precisely to this difference in bond strength (the curvature of the potential).

19.6 Summary and Exercises

19.6.1 Summary

• Hooke's law (Law 19.1): Within the elastic range, stress and strain are directly proportional, and the constant of proportionality is called an elastic modulus. The ratio of normal stress to longitudinal strain is Young's modulus (modulus of longitudinal elasticity) $E$ (Definition 19.1), the ratio of shear stress to shear strain is the shear modulus (modulus of transverse elasticity) $G$ (Definition 19.2), and the ratio of hydrostatic normal stress to volumetric strain is the bulk modulus $K$ (Definition 19.3). For tension of a round bar, $E=l_0P/(\lambda A_0)$ (Eq. \eqref{eq:19-E-def}), and the bar can be regarded as a spring with spring constant $k=EA_0/l_0$ (Eq. \eqref{eq:19-bar-spring}).
• From the generalized Hooke's law (Eq. \eqref{eq:19-generalized-hooke}) we obtain $G=E/\{2(1+\nu)\}$ (Formula 19.1) and $K=E/\{3(1-2\nu)\}$ (Formula 19.2). For an isotropic elastic body only two elastic constants are independent ($E,\nu$ are also determined from $G,K$), and the range of possible values of $\nu$ is $-1\lt\nu\lt1/2$.
• Material testing is classified into static-load tests (tension, compression, torsion, flexure, hardness, creep) and dynamic-load tests (fatigue, impact), and is specified in detail by the Japanese Industrial Standards (JIS). The nominal stress is $\sigma=P/A_0$ (Definition 19.6), and the proportional limit is $\sigma_{\mathrm p}$ (Definition 19.7). The stress–strain diagram of mild steel follows a characteristic shape consisting of the proportional limit, elastic limit, upper and lower yield points, yield plateau, work hardening, tensile strength, necking, and fracture (Figure 19.5).
• For materials without a distinct yield point, the 0.2% proof stress (Definition 19.10) is used. In contrast to the nominal stress and nominal strain, which are obtained by dividing by the cross-sectional area before deformation, the true stress and true strain (Definition 19.13), which are obtained by dividing by the cross-sectional area at each instant, are related by $\sigma_{\mathrm t}=\sigma(1+\varepsilon)$ and $\varepsilon_{\mathrm t}=\ln(1+\varepsilon)$ (Eq. \eqref{eq:19-true-from-nominal}) as long as the specimen stretches uniformly (until necking begins). Work hardening is well approximated by Hollomon's equation $\sigma_{\mathrm t}=K_{\mathrm H}\varepsilon_{\mathrm t}^{\,n}$ (Law 19.2; the strength coefficient is written $K_{\mathrm H}$ to distinguish it from the bulk modulus $K$), and from Considère's condition $\dd\sigma_{\mathrm t}/\dd\varepsilon_{\mathrm t}=\sigma_{\mathrm t}$ (Eq. \eqref{eq:19-considere}) it follows that necking begins at the true strain $\varepsilon_{\mathrm t}=n$.
• Metals (ductile materials) fracture after large plastic deformation, whereas ceramics (brittle materials) remain in elastic deformation and fracture almost linearly (Figure 19.7)—this reflects the difference between metallic bonding on the one hand and, on the other, strongly directional covalent bonding or ionic bonding, in which slip brings ions of the same sign next to each other. Microscopically, Young's modulus corresponds to the curvature at the bottom of the interatomic potential curve (Chapter 4, Eq. \eqref{eq:19-E-atomic}).

Related simulator: Fracture Mechanics (Griffith and Weibull) and the corresponding derivation page (you can check the continuation of the nominal stress, true stress, Hollomon's equation, Considère's condition, and the stress inside the neck treated in this chapter by simulating an actual tensile test; the simulator writes the strength coefficient as $K$; it is used in earnest in Chapters 21 and 22).

19.6.2 Exercises

19.6.3 References