ON THE CONTACT PROBLEM OF THE THEORY OF ELASTICITY FOR BODIES OF FINITE DIMENSIONS
Unknown
Submitted 1963-01-01 | SovietRxiv: ru-196301.69558 | Translated from Russian

Abstract Generated abstract

The paper addresses a three-dimensional contact problem in linear elasticity for a finite rectangular parallelepiped placed in a rigid pit and pressed by a rigid punch over a prescribed contact region. Using a Papkovich-Neuber representation, the displacement field is expressed through two harmonic functions expanded in double Fourier series chosen to satisfy the rigid side and bottom constraints and the absence of tangential stresses. The remaining mixed boundary conditions on the upper face are reduced to an integral equation for an auxiliary function, identified as the contact stress distribution. Limiting cases are discussed, showing that the formulation passes to the known equations for an unbounded layer and, further, for an elastic half-space, with analogous extensions indicated for circular plates, cylinders, and plane rectangular problems.

Full Text

ELASTICITY THEORY

A. F. KHRUSTALEV

ON THE CONTACT PROBLEM OF THE THEORY OF ELASTICITY FOR BODIES OF FINITE DIMENSIONS

(Presented by Academician A. Yu. Ishlinskii, 13 III 1963)

A solution is considered for the contact spatial problem of the theory of elasticity for a rectangular parallelepiped \(|x| \leq a,\ |y| \leq b,\ 0 \leq z \leq c\) (the \(z\)-axis is directed downward), located in a rectangular pit of the same dimensions, whose walls and bottom are absolutely rigid. An absolutely rigid punch presses on the parallelepiped, and the contact area is a simply connected or multiply connected region \(D\). It is assumed that the normal stresses outside the contact region \(D\), as well as the tangential stresses on all faces of the parallelepiped, are absent.

In determining the stresses and displacements corresponding to the stated problem, we shall proceed from the general solution of the equilibrium equations in the Papkovich—Neuber form, from which a simpler solution is obtained, containing two harmonic functions \(\varphi(x; y; z)\) and \(\psi(x; y; z)\), through which the displacement vector is expressed as follows \((^{1-3})\).

\[ \mathbf{u}=\operatorname{grad}(\varphi+z\psi)-4(1-\nu)\psi\mathbf{k}. \tag{1} \]

Thus, the solution of the stated problem reduces to the determination of two harmonic functions \(\varphi\) and \(\psi\), which for \(z=0\) must satisfy the conditions

\[ \sigma_z=0 \qquad \text{outside the region } D; \tag{2} \]

\[ u_z=f(x,y) \qquad \text{inside the region } D \tag{3} \]

and the conditions of absence of tangential stresses on all faces of the parallelepiped. In addition, the normal components of the displacement vector on the faces \(x=\pm a,\ y=\pm b,\ z=c\) must be equal to zero.

To simplify the notation of the expressions obtained below, suppose that the displacements \(u_z=f(x;y)\), as well as the region \(D\), are symmetric with respect to \(x\) and \(y\).

It is not difficult to verify that the functions

\[ \varphi_{kn}= \left[ A_{kn}\operatorname{ch}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,z + B_{kn}\operatorname{sh}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,z \right] \cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b}; \tag{4} \]

\[ \psi_{kn}= \left[ C_{kn}\operatorname{ch}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,z + D_{kn}\operatorname{sh}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,z \right] \cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b}; \tag{5} \]

\[ k=0,1,2,\ldots;\quad n=0,1,2,\ldots, \]

are harmonic and even with respect to \(x\) and \(y\).

Assuming

\[ A_{kn}=-C_{kn}\frac{(1-2\nu)\operatorname{sh}2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c-2c\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}}{2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\operatorname{sh}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c}; \tag{6} \]

\[ B_{kn}=-\frac{(1-2\nu)C_{kn}}{\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}}; \qquad D_{kn}=C_{kn}\operatorname{cth}\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c, \tag{7} \]

we are readily convinced that the functions \(\varphi_{kn}\) and \(\psi_{kn}\) satisfy the conditions \(\tau_{xz}=\tau_{yz}=0\) for \(z=0,\ z=c\) and \(u_z=0\) for \(z=c\).

The functions (4) and (5) are such that they satisfy the conditions that the tangential stresses and normal displacements vanish on the faces \(x=\pm a,\ y=\pm b\), since in the corresponding expressions \(\sin k\pi\) occurs as a factor.

The functions \(\varphi\) and \(\psi\), which will satisfy the same conditions as \(\varphi_{kn},\ \psi_{kn}\), are constructed in the form

\[ \varphi=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\varphi_{kn}; \qquad \psi=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\psi_{kn}. \tag{8} \]

It remains to satisfy conditions (2) and (3), which, on the basis of (1) and (8), respectively take the form

\[ \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} Q(k;n)\cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b}=0 \quad \text{outside the region } D; \tag{9} \]

\[ \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} Q(k;n)\,q\left(\frac{k\pi}{a};\frac{n\pi}{b}\right) \cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b} = -\frac{\mu f(x;y)}{2(1-\nu)} \quad \text{inside the region } D. \tag{10} \]

Here

\[ Q(k;n)=C_{kn}\mu \frac{ \sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2} \operatorname{sh}2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c +2c\left[\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2\right] }{ \operatorname{sh}^2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2} }; \tag{11} \]

\[ q\left(\frac{k\pi}{a};\frac{n\pi}{b}\right)= \]

\[ = \frac{ \operatorname{sh}^2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c }{ \sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2} \operatorname{sh}2\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}\,c +2c\sqrt{\left(\frac{k\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2} }. \tag{12} \]

We shall seek the solution of equations (9) and (10) in the form

\[ Q(0;0)=\frac{1}{4ab}\iint_D \varphi_0(\xi;\eta)\,d\xi\,d\eta; \tag{13} \]

\[ Q(0;n)=\frac{1}{2ab}\iint_D \varphi_0(\xi;\eta)\cos\frac{n\pi\eta}{b}\,d\eta\,d\xi, \qquad n=1,2,\ldots; \tag{14} \]

\[ Q(k;0)=\frac{1}{2ab}\iint\limits_D \varphi_0(\xi;\eta)\cos\frac{k\pi \xi}{a}\,d\xi\,d\eta,\qquad k=1,2,\ldots; \tag{15} \]

\[ Q(k;n)=\frac{1}{ab}\iint\limits_D \varphi_0(\xi;\eta)\cos\frac{k\pi \xi}{a}\cos\frac{n\pi \eta}{b}\,d\xi\,d\eta; \quad k=1,2,\ldots;\ n=1,2,\ldots; \tag{16} \]

where \(\varphi_0(\xi;\eta)\) is such an unknown auxiliary function that

\[ \begin{aligned} &\sum_{k=0}^{\infty}\sum_{n=0}^{\infty} Q(k;n)\cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b} \\ &=\frac{1}{ab}\Bigg[ \frac{1}{4}\iint\limits_D \varphi_0(\xi;\eta)\,d\xi\,d\eta +\frac{1}{2}\sum_{n=1}^{\infty}\cos\frac{n\pi y}{b} \iint\limits_D \varphi_0(\xi;\eta)\cos\frac{n\pi \eta}{b}\,d\xi\,d\eta \\ &\quad+\frac{1}{2}\sum_{k=1}^{\infty}\cos\frac{k\pi x}{a} \iint\limits_D \varphi_0(\xi;\eta)\cos\frac{k\pi \xi}{a}\,d\xi\,d\eta \\ &\quad+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b} \iint\limits_D \varphi_0(\xi;\eta)\cos\frac{k\pi \xi}{a}\cos\frac{n\pi \eta}{b}\,d\xi\,d\eta \Bigg] \\ &= \begin{cases} 0, & \text{outside the region } D,\\ \varphi_0(x;y), & \text{inside the region } D, \end{cases} \tag{17} \end{aligned} \]

i.e., the function \(\varphi_0(x;y)\) admits an expansion in a double Fourier series.

From the boundary value of \(\sigma_z\) at \(z=0\) and equality (17) it follows that \(\sigma_z=\varphi_0(x;y)\) inside the region \(D\), i.e., \(\varphi_0(x;y)\) is the function of the distribution of contact stresses.

On the basis of (17) we conclude that condition (9) is satisfied identically.

Substituting expressions (13)—(16) for \(Q(k;n)\) into equation (10), we obtain

\[ \iint\limits_D M(x;y;\xi;\eta)\varphi_0(\xi;\eta)\,d\xi\,d\eta =-\frac{\mu f(x;y)}{2(1-\nu)}, \tag{18} \]

where

\[ \begin{aligned} M(x;y;\xi;\eta)=\frac{1}{ab}\Bigg[ &\frac{q(0;0)}{4} +\frac{1}{2}\sum_{n=1}^{\infty}q\left(0;\frac{n\pi}{b}\right) \cos\frac{n\pi y}{b}\cos\frac{n\pi \eta}{b} \\ &+\frac{1}{2}\sum_{k=1}^{\infty}q\left(\frac{k\pi}{a};\right) \cos\frac{k\pi x}{a}\cos\frac{k\pi \xi}{a} \\ &+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} q\left(\frac{k\pi}{a};\frac{n\pi}{b}\right) \cos\frac{k\pi x}{a}\cos\frac{n\pi y}{b} \cos\frac{k\pi \xi}{a}\cos\frac{n\pi \eta}{b} \Bigg]. \tag{19} \end{aligned} \]

For \(a\to\infty,\ b\to\infty\), i.e., in the case of an unbounded plane-parallel layer, kernel (19) takes the form

\[ M(x;y;\xi;\eta)= \int_{0}^{\infty}\cos\beta y\cos\beta\eta\,d\beta \int_{0}^{\infty}q(\alpha;\beta)\cos\alpha x\cos\alpha\xi\,d\alpha. \tag{20} \]

If in (20) one passes to the limit as \(c\to\infty\), taking into account that\({}^{(4)}\)

\[ \int_{0}^{\infty}\frac{\cos px}{\sqrt{q^2+x^2}}\,dx=K_0(pq); \qquad \int_{0}^{\infty}K_0(\beta x)\cos\alpha x\,dx =\frac{\pi}{2\sqrt{\alpha^2+\beta^2}}, \]

then we can easily verify that the integral equation (18) coincides with the known equation for the half-space \((^{1,2})\).

Axisymmetric problems for a circular plate, for bounded solid and hollow cylinders, and the plane problem for a rectangle under the corresponding mixed boundary conditions can be solved analogously.

Sevastopol Branch
of the Odessa Polytechnic Institute

Received
12 III 1963

REFERENCES

\(^{1}\) L. A. Galin, Contact Problems of the Theory of Elasticity, 1953.
\(^{2}\) A. I. Lur’e, Spatial Problems of the Theory of Elasticity, 1955.
\(^{3}\) P. F. Papkovich, Theory of Elasticity, 1939.
\(^{4}\) I. M. Ryzhik, I. S. Gradshteyn, Tables of Integrals, Sums, Series, and Products, 1951.

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ON THE CONTACT PROBLEM OF THE THEORY OF ELASTICITY FOR BODIES OF FINITE DIMENSIONS