Hertz's Solution for the Elastic Half-Space Under an Axisymmetric Pressure

Contents

What is quoted here
Superposition and the contact conditions
The elliptic pressure and its displacement
Matching, and the coefficient $\tfrac{4}{3}$
Checks and consequences

This appendix completes the derivation of Phenomenon 34.70 of Continuum Mechanics and Elasticity. It is deliberately not a second derivation of that phenomenon. The inline derivation in the chapter already establishes the exponent, and establishes it exactly: from the geometry of two nearly touching spheres, the single strain scale of the contact region and the linearity of the elastic law it obtains $F\propto E^{*}R^{1/2}\delta^{3/2}$, with $a=\sqrt{R\delta}$, and nothing about that argument is a guess. What it cannot reach is the pure number in front and the shape of the pressure distribution, because both require actually solving the elastic problem. Those two things — the coefficient $\tfrac{4}{3}$ of Equation (34.60) and the elliptic profile $p(r)=p_{0}\sqrt{1-r^{2}/a^{2}}$ — are what this section supplies, together with a proof of the two-body combination rule for $E^{*}$, which the chapter's derivation asserts in a clause and proves nowhere.

What is quoted here

One result is imported, and everything else is derived from it.

Theorem A58.1 (Boussinesq's point-force solution, quoted).

Let an isotropic elastic half-space $z\ge0$ with Young's modulus $E$ and Poisson's ratio $\nu$ (Definition 34.31) be loaded by a normal force $P$ concentrated at the origin of its free surface, the rest of the surface being free of traction and the displacement vanishing at infinity. Then the normal displacement of the surface at distance $r$ from the load is

\begin{equation}\tag{A58.1} u_{z}(r)=\frac{1-\nu^{2}}{\pi E}\,\frac{P}{r} =\frac{P}{\pi E^{*}r}\ec\qquad \frac{1}{E^{*}}\equiv\frac{1-\nu^{2}}{E}\ec \end{equation}

directed into the half-space [Landau:1986]. Here $P$ carries $\mathrm{N}$, $E$ and $E^{*}$ carry $\mathrm{Pa}$ and $u_{z}$ carries $\mathrm{m}$. Rests on Equation (34.21), Definition 34.31 and Theorem 34.18.

Remark A58.2 (What is quoted here).

Theorem A58.1 is Boussinesq's, published in 1885, and it is the one step of this section that is not carried out from the material of this treatise. Obtaining it needs a general representation of the solutions of the Navier–Cauchy equations of elastostatics in terms of harmonic functions — the Papkovich–Neuber representation — and the potential theory of Section 14.4.3 carries the harmonic functions but not that representation. It is therefore a debt owed by Partial Differential Equations, and it is named here rather than passed over; the statement above is proved in Landau and Lifshitz's treatment of the elastic half-space [Landau:1986], from which the form Equation (A58.1) is taken.

Two features of Equation (A58.1) can be checked here without the representation, and are worth checking, because between them they fix everything except the pure number $1/\pi$. Dimensions: $P/(E^{*}r)$ is $\mathrm{N}/(\mathrm{Pa}\cdot\mathrm{m}) =\mathrm{m}$, as a displacement must be. Scale invariance: the problem has no length in it at all — a half-space is scale-invariant, a point load introduces no length, and linear elasticity has no intrinsic scale — so $u_{z}$ must be a homogeneous function of $r$, and linearity in $P$ together with the dimensions forces the exponent $-1$. The modulus combination $E/(1-\nu^{2})$ appears rather than $E$ because the material beneath a surface load is laterally confined, which is the same statement as the one made for a plate in Remark A57.4.

Superposition and the contact conditions

Lemma A58.3 (Surface displacement under a distributed pressure).

Let a pressure $p(\rho)\ge0$ act on the disc $\rho\le a$ of the free surface of the half-space, and vanish outside it. Then the surface normal displacement is

\begin{equation}\tag{A58.2} u_{z}(r)=\frac{1}{\pi E^{*}}\,\Phi(r)\ec\qquad \Phi(r)=\int_{\rho\le a}\frac{p\left(\rho\right)}{s}\,\dd A'\ec \end{equation}

where $s$ is the distance in the surface plane from the field point to the element $\dd A'$. The integral converges although the kernel is singular, because $\dd A'=s\,\dd s\,\dd\varphi$ in polar coordinates centred on the field point. Rests on Theorem A58.1.

Proof.

Derives Lemma A58.3. The elastic problem is linear (Equation (34.21)) and the boundary conditions are linear in the applied traction, so displacements superpose. Treating the load on the element $\dd A'$ as a point force $P=p\,\dd A'$ and summing Equation (A58.1) gives Equation (A58.2). In polar coordinates $(s,\varphi)$ about the field point the element is $s\,\dd s\,\dd\varphi$, so the integrand is $p\,\dd s\,\dd\varphi$ with no singularity at all; the integral exists for any bounded $p$.

∎

Definition A58.4 (The Hertz contact problem).

Two elastic bodies of radii $R_{1}$ and $R_{2}$, with moduli $(E_{1},\nu_{1})$ and $(E_{2},\nu_{2})$, are pressed together along their line of centres by a normal force $F$, their centres approaching by $\delta$. Let $a$ be the radius of the circle of contact and $p(r)$ the contact pressure. The problem is to find $p$, $a$ and the relation between $F$ and $\delta$, subject to

contact inside the circle: for $r\le a$ the sum of the two surface displacements closes the initial gap,

\begin{equation}\tag{A58.3} u_{z}^{(1)}(r)+u_{z}^{(2)}(r)=\delta-\frac{r^{2}}{2R}\ec \qquad\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\ec \end{equation}
separation outside: $p(r)=0$ for $r>a$, and the surfaces there do not overlap;
no adhesion and no edge singularity: $p(r)\ge0$ throughout, and $p$ is bounded as $r\to a$.

The contact is taken to be frictionless, so that the traction transmitted across it is purely normal; and $a\ll R$, so that near the contact each body is indistinguishable from a half-space and Theorem A58.1 applies to it. Rests on Theorem A58.1 and Phenomenon 34.70.

Remark A58.5 (The third condition is not decoration).

Conditions (1) and (2) alone do not determine the answer. The same geometry admits the flat-punch solution $p(r)\propto\left(1-r^{2}/a^{2}\right)^{-1/2}$, which is a perfectly good solution of the elastic equations, produces a constant surface displacement inside the circle, and diverges at the rim; it is the right answer for a rigid flat cylindrical punch pressed into a half-space, where the contact radius is fixed by the punch and the material at the rim is genuinely singular. What excludes it here is condition (3): the contact circle of two smooth bodies is not fixed externally — it is whatever radius the load produces — so the surfaces must meet the rim tangentially and the pressure must go to zero there, not to infinity. Condition (3) also excludes the tensile solutions that would hold the surfaces together outside the circle, which is the assumption of no adhesion; it fails for very small or very compliant bodies, where surface energy matters, and Hertz's relation is then replaced by an adhesive contact theory. Stating the third condition is what turns the profile below from a lucky guess into the unique answer. Rests on Definition A58.4.

The elliptic pressure and its displacement

The whole solution rests on one integral, and it is worth doing in full.

Lemma A58.6 (The displacement produced by an elliptic pressure).

Let $p(\rho)=p_{0}\sqrt{1-\rho^{2}/a^{2}}$ on $\rho\le a$ and zero outside. Then for every field point inside the circle, $r\le a$,

\begin{equation}\tag{A58.4} \Phi(r)=\frac{\pi^{2}p_{0}}{4a}\left(2a^{2}-r^{2}\right)\ec \qquad\text{hence}\qquad u_{z}(r)=\frac{\pi p_{0}}{4E^{*}a}\left(2a^{2}-r^{2}\right)\ep \end{equation}

The displacement is therefore quadratic in $r$ inside the contact circle — which is exactly the shape Equation (A58.3) demands. Rests on Lemma A58.3 and Definition A58.4.

Proof.

Derives Lemma A58.6. Place the field point at $(r,0)$ with $0\le r\le a$ and use polar coordinates $(s,\varphi)$ centred on it, so that the element at $(s,\varphi)$ sits at $\left(r+s\cos\varphi,\ s\sin\varphi\right)$ and is at distance

\begin{equation}\tag{A58.5} \rho^{2}=r^{2}+2rs\cos\varphi+s^{2}\ec \end{equation}

measured from the centre of the disc. By Lemma A58.3 the kernel cancels the Jacobian, so

\begin{equation}\tag{A58.6} \Phi(r)=\int_{0}^{2\pi}\!\!\dd\varphi\int_{0}^{\ell(\varphi)} p\left(\rho\right)\dd s =\frac{p_{0}}{a}\int_{0}^{2\pi}\!\!\dd\varphi \int_{0}^{\ell(\varphi)} \sqrt{a^{2}-r^{2}-2rs\cos\varphi-s^{2}}\ \dd s\ec \end{equation}

where $\ell(\varphi)$ is the distance from the field point to the rim along the direction $\varphi$, that is the value of $s$ at which $\rho=a$.

Completing the square. Put

\begin{equation}\tag{A58.7} u=s+r\cos\varphi\ec\qquad A^{2}=a^{2}-r^{2}\sin^{2}\varphi\ec\qquad a^{2}-r^{2}-2rs\cos\varphi-s^{2}=A^{2}-u^{2}\ec \end{equation}

which is checked by expanding $A^{2}-u^{2}$. The rim, $\rho=a$, is $u=A$; the field point, $s=0$, is $u=r\cos\varphi$. Since $u$ and $s$ differ by a constant at fixed $\varphi$, $\dd u=\dd s$, and

\begin{equation}\tag{A58.8} \int_{0}^{\ell(\varphi)}\sqrt{A^{2}-u^{2}}\,\dd s =\int_{r\cos\varphi}^{A}\sqrt{A^{2}-u^{2}}\,\dd u =\left[\frac{u}{2}\sqrt{A^{2}-u^{2}} +\frac{A^{2}}{2}\arcsin\frac{u}{A}\right]_{u=r\cos\varphi}^{u=A}\ec \end{equation}

the antiderivative being elementary. At the upper limit the first term vanishes and $\arcsin1=\pi/2$, giving $\pi A^{2}/4$. At the lower limit,

\begin{equation}\tag{A58.9} A^{2}-r^{2}\cos^{2}\varphi =a^{2}-r^{2}\sin^{2}\varphi-r^{2}\cos^{2}\varphi=a^{2}-r^{2}\ec \end{equation}

a constant independent of $\varphi$ — the geometric fact that makes the calculation work. Writing $B=\sqrt{a^{2}-r^{2}}$, the inner integral is

\begin{equation}\tag{A58.10} \frac{\pi A^{2}}{4}-\frac{rB\cos\varphi}{2} -\frac{A^{2}}{2}\arcsin\frac{r\cos\varphi}{A}\ep \end{equation}

The angular integral. Integrate Equation (A58.10) over $\varphi\in[0,2\pi)$ term by term. The second term integrates to zero, $\int_{0}^{2\pi}\cos\varphi \,\dd\varphi=0$. The third also integrates to zero: under $\varphi\mapsto\pi-\varphi$ the quantity $A$ is unchanged, since it depends on $\sin^{2}\varphi$, while $\cos\varphi$ changes sign and $\arcsin$ is odd, so the integrand is odd about that involution. The first term gives

\begin{equation}\tag{A58.11} \frac{\pi}{4}\int_{0}^{2\pi} \left(a^{2}-r^{2}\sin^{2}\varphi\right)\dd\varphi =\frac{\pi}{4}\left(2\pi a^{2}-\pi r^{2}\right) =\frac{\pi^{2}}{4}\left(2a^{2}-r^{2}\right)\ec \end{equation}

using $\int_{0}^{2\pi}\sin^{2}\varphi\,\dd\varphi=\pi$. Multiplying by the prefactor $p_{0}/a$ of Equation (A58.6) gives the first of Equation (A58.4), and Equation (A58.2) then gives the second.

∎

Matching, and the coefficient $\tfrac{4}{3}$

Proposition A58.7 (The two-body compliance rule).

Under Definition A58.4 the two bodies carry the same contact pressure $p(r)$, so their surface displacements add, and

\begin{equation}\tag{A58.12} u_{z}^{(1)}(r)+u_{z}^{(2)}(r)=\frac{\Phi(r)}{\pi E^{*}}\ec \qquad \frac{1}{E^{*}}=\frac{1-\nu_{1}^{2}}{E_{1}} +\frac{1-\nu_{2}^{2}}{E_{2}}\ec \end{equation}

which is the combination appearing in Equation (34.60). The geometry likewise combines: the gap between the two undeformed surfaces is $r^{2}/2R$ with $1/R=1/R_{1}+1/R_{2}$, so the pair is equivalent to a single sphere of radius $R$ pressed against a rigid plane by the same force. Rests on Lemma A58.3 and Definition A58.4.

Proof.

Derives Proposition A58.7. The pressures are equal. By Newton's third law the traction each body exerts on the other is equal and opposite at every point of the common contact area, so the same scalar $p(r)$ appears in Equation (A58.2) for both. Applying that formula to each body with its own modulus and adding,

\begin{equation}\tag{A58.13} u_{z}^{(1)}+u_{z}^{(2)} =\left(\frac{1-\nu_{1}^{2}}{\pi E_{1}} +\frac{1-\nu_{2}^{2}}{\pi E_{2}}\right)\Phi(r)\ec \end{equation}

which is Equation (A58.12). It is the compliances $1/E^{*}_{i}$ that add and not the moduli, because the two bodies are loaded in series by a common force and their deflections accumulate; that is the clause the chapter's inline derivation asserts without proof, and Equation (A58.13) is the proof. A rigid body has $E_{i}\to\infty$ and contributes nothing, so a single elastic sphere against a rigid plane has $E^{*}=E/(1-\nu^{2})$.

The geometry. Measure heights from the plane of first contact. To leading order in $r/R_{i}$ the surface of a sphere of radius $R_{i}$ lies at height $r^{2}/2R_{i}$ below its own pole — the expansion $R_{i}-\sqrt{R_{i}^{2}-r^{2}}=r^{2}/2R_{i}+O(r^{4}/R_{i}^{3})$ — so the gap between the two undeformed surfaces at distance $r$ from the axis is $r^{2}/2R_{1}+r^{2}/2R_{2}=r^{2}/2R$. Both the elastic response and the geometry therefore depend on the pair only through $E^{*}$ and $R$.

∎

Theorem A58.8 (Hertz's solution).

The unique solution of Definition A58.4 is the elliptic pressure distribution

\begin{equation}\tag{A58.14} p(r)=p_{0}\sqrt{1-\frac{r^{2}}{a^{2}}}\ec\qquad p_{0}=\frac{2E^{*}a}{\pi R}=\frac{3F}{2\pi a^{2}}\ec \end{equation}

with contact radius, approach and force related by

\begin{equation}\tag{A58.15} a=\sqrt{R\delta}\ec\qquad \delta=\frac{a^{2}}{R}\ec\qquad F=\frac{4}{3}\,E^{*}R^{1/2}\delta^{3/2}\ec \end{equation}

which is Equation (34.60) complete with its coefficient. Rests on Lemma A58.6, Proposition A58.7 and Definition A58.4.

Proof.

Derives Theorem A58.8. Matching. Insert Equation (A58.4) into Equation (A58.12) and set the result equal to the required displacement Equation (A58.3):

\begin{equation}\tag{A58.16} \frac{\pi p_{0}}{4E^{*}a}\left(2a^{2}-r^{2}\right) =\delta-\frac{r^{2}}{2R} \qquad\text{for all }r\le a\ep \end{equation}

Both sides are polynomials in $r^{2}$ of degree one, so Equation (A58.16) holds identically if and only if the constant terms and the coefficients of $r^{2}$ agree separately:

\begin{equation}\tag{A58.17} \frac{\pi p_{0}a}{2E^{*}}=\delta\ec \qquad \frac{\pi p_{0}}{4E^{*}a}=\frac{1}{2R}\ep \end{equation}

That an elliptic pressure produces a displacement of exactly the required quadratic form, with two free constants $p_{0}$ and $a$ to match two coefficients, is what makes the problem solvable in closed form; it is the content of Lemma A58.6 and it is why Hertz's guess was the right one.

Solving. The second of Equation (A58.17) gives $p_{0}=2E^{*}a/(\pi R)$, the first of Equation (A58.14). Substituting into the first of Equation (A58.17),

\begin{equation}\tag{A58.18} \delta=\frac{\pi a}{2E^{*}}\cdot\frac{2E^{*}a}{\pi R} =\frac{a^{2}}{R}\ec \end{equation}

which is $a=\sqrt{R\delta}$, the contact radius quoted in Phenomenon 34.70.

The force. Integrating the pressure over the contact circle, with the substitution $t=r^{2}/a^{2}$,

\begin{equation}\tag{A58.19} F=\int_{0}^{a}p_{0}\sqrt{1-\frac{r^{2}}{a^{2}}}\,2\pi r\,\dd r =\pi a^{2}p_{0}\int_{0}^{1}\sqrt{1-t}\,\dd t =\frac{2}{3}\pi a^{2}p_{0}\ec \end{equation}

so $p_{0}=3F/(2\pi a^{2})$: the peak pressure is $\tfrac{3}{2}$ times the mean pressure $F/\pi a^{2}$, which is the second equality of Equation (A58.14). Eliminating $p_{0}$ between Equation (A58.19) and Equation (A58.14), and then $a$ by Equation (A58.18),

\begin{equation}\tag{A58.20} F=\frac{2}{3}\pi a^{2}\cdot\frac{2E^{*}a}{\pi R} =\frac{4}{3}\,\frac{E^{*}a^{3}}{R} =\frac{4}{3}\,\frac{E^{*}\left(R\delta\right)^{3/2}}{R} =\frac{4}{3}\,E^{*}R^{1/2}\delta^{3/2}\ep \end{equation}

Uniqueness and admissibility. The profile Equation (A58.14) is non-negative and vanishes at $r=a$, so it satisfies condition (3) of Definition A58.4; outside the circle the deformed surfaces separate, since the elastic displacement there falls faster than the parabolic gap grows. That it is the only admissible solution follows from linearity together with the positive definiteness of the elastic energy. Suppose two admissible pressure distributions produced the same $\delta$ over the same contact circle. Their difference drives an elastic field whose surface traction is purely normal (the contact is frictionless), whose normal displacement vanishes inside the circle, and whose traction vanishes outside it. Its stored energy is $\tfrac{1}{2}\oint\sigma_{ik}n_{k}u_{i}\,\dd S$ by Equation (34.18) with $w_{i}=u_{i}$ and no body force, and every term of that integral has one vanishing factor, so the energy is zero. Since the energy density is positive definite (Proposition 34.30), the difference field has zero strain and the two pressures coincide.

The coefficient $\tfrac{4}{3}$ of Equation (A58.20) is exactly and only what the scaling argument of the chapter left open, and it is assembled from three pure numbers: the $\pi$ of the Boussinesq kernel Equation (A58.1), the $\pi^{2}/4$ of the integral Equation (A58.4), and the $\tfrac{2}{3}$ of Equation (A58.19).

∎

Checks and consequences

Example A58.9 (The numbers of the chapter, recomputed).

Example 34.71 presses two steel balls of radius $10\,\mathrm{mm}$ together with $1\,\mathrm{N}$, with $E=210\,\mathrm{GPa}$ and $\nu=0.29$. Then Equation (A58.12) gives

\begin{equation}\tag{A58.21} \frac{1}{E^{*}}=\frac{2\left(1-\nu^{2}\right)}{E} =\frac{2\left(1-0.0841\right)}{2.10\times 10^{11}\,\mathrm{Pa}} =8.72\times 10^{-12}\,/\mathrm{Pa}\ec\qquad E^{*}=1.15\times 10^{11}\,\mathrm{Pa}\ec \end{equation}

and $R=5.0\,\mathrm{mm}$ by Equation (A58.3). Inverting Equation (A58.20) with $\delta=a^{2}/R$ gives $a=\left[3FR/(4E^{*})\right]^{1/3}$, so

\begin{equation}\tag{A58.22} a=\left[\frac{3\left(1\,\mathrm{N}\right) \left(5.0\times 10^{-3}\,\mathrm{m}\right)} {4\left(1.15\times 10^{11}\,\mathrm{Pa}\right)}\right]^{1/3} =3.2\times 10^{-5}\,\mathrm{m}=32\,\mu\mathrm{m}\ec \end{equation}

whence $\delta=a^{2}/R=2.0\times 10^{-7}\,\mathrm{m}=0.20\,\mu\mathrm{m}$ and, by Equation (A58.14), $p_{0}=3F/(2\pi a^{2})=4.7\times 10^{8}\,\mathrm{Pa}=470\,\mathrm{MPa}$. All three reproduce Example 34.71, which is the arithmetic check a reader can run in a minute. Note also the ratio $a/R=6.4\times 10^{-3}$, which is what licenses the half-space approximation of Definition A58.4. Rests on Theorem A58.8 and Phenomenon 34.70.

Remark A58.10 ($E^{*}$ in two roles, and one trap).

The symbol $E/(1-\nu^{2})$ occurs twice in Continuum Mechanics and Elasticity and means two different things, which is worth stating plainly because the two are easily read as one quantity. In Definition 34.63 it is a single-body stiffness: the plane-strain modulus of the plate material, appearing because a bent plate cannot contract transversely. In Equation (34.60) the same expression is one term of a two-body sum, Equation (A58.12), and the contact modulus $E^{*}$ is the harmonic-type combination of both bodies' plane-strain moduli. The underlying modulus is the same and its physical reason is the same — lateral confinement — but a contact between two steel bodies has $E^{*}$ equal to half the plate value for steel, and substituting one for the other misstates the contact stiffness by a factor of two. Rests on Theorem A58.8 and Definition 34.63.

Remark A58.11 (The dangerous stress is below the surface).

The peak pressure is at the centre of the contact, on the surface. The peak shear stress is not: evaluating the subsurface stress field from the same Boussinesq kernel — an elementary but long computation, not carried out here — places the maximum of $\tfrac{1}{2}(\sigma_{1}-\sigma_{3})$ on the axis at a depth of about $0.48\,a$ for $\nu\approx0.3$, where it reaches about $0.31\,p_{0}$. Since ductile yielding is governed by the maximum shear stress and not by the pressure (Proposition 34.74), first yield under a Hertzian contact begins at that depth, inside apparently undamaged material. For the balls of Example A58.9 the peak shear is $0.31\times470\,\mathrm{MPa}=146\,\mathrm{MPa}$ at a depth of $15\,\mu\mathrm{m}$. That is the mechanism behind rolling-contact fatigue: a bearing race accumulates plastic damage at a fixed small depth under every pass of a ball, and eventually a subsurface crack reaches the surface and detaches a flake. It is why such a failure is a spall and not wear, and why it is invisible until the moment it is not — the point made at the end of Example 34.71 and taken up in Section 34.8. Rests on Theorem A58.8 and Proposition 34.74.

Remark A58.12.

Hertz's Solution for the Elastic Half-Space Under an Axisymmetric Pressure discharges the derivation owed at Phenomenon 34.70 of Continuum Mechanics and Elasticity, and only that part of it which the inline derivation cannot reach: the elliptic pressure profile Equation (A58.14), the coefficient $\tfrac{4}{3}$ of Equation (34.60), and the two-body compliance rule of Proposition A58.7. The exponent $\tfrac{3}{2}$ and the relation $a=\sqrt{R\delta}$ are proved in the chapter, by an argument that needs no elasticity solution at all, and are not reproved here. One result is imported and named as such (Remark A58.2): Boussinesq's point-force solution Equation (A58.1), a debt owed by Partial Differential Equations, whose Papkovich–Neuber representation this treatise does not carry. Everything else — the potential integral of Lemma A58.6 in particular — is carried out in full from the isotropic Hooke law Equation (34.21) proved in Isotropic and Cubic Elastic Tensors.