A MIXED PROBLEM FOR AN ELASTIC LAYER
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Submitted 1958-01-01 | RussiaRxiv: ru-195801.96757 | Translated from Russian

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THEORY OF ELASTICITY

Ya. S. Ufliand

A MIXED PROBLEM FOR AN ELASTIC LAYER

(Presented by Academician N. I. Muskhelishvili, 4 VIII 1958)

In the present paper an exact solution is given of a spatial problem of the theory of elasticity for an unbounded layer \((-\infty < x, y < \infty,\ 0 \leq z \leq h)\), on one boundary plane \((z = 0)\) of which elastic displacements \((u, v, w)\) are prescribed, and on the other, stresses \((\sigma_z, \tau_{zx}, \tau_{yz})\).

To solve the problem posed, we represent the displacements \(u, v, w\) in terms of the four harmonic functions \((\Phi_0, \Phi_1, \Phi_2, \Phi_3)\) of Papkovich–Neuber \((^{1,2})\).

\[ \begin{gathered} 2\mu u = -\frac{\partial F}{\partial x} + 4(1-\nu)\Phi_1,\qquad 2\mu v = -\frac{\partial F}{\partial y} + 4(1-\nu)\Phi_2,\\ 2\mu w = -\frac{\partial F}{\partial z} + 4(1-\nu)\Phi_3,\qquad F = \Phi_0 + x\Phi_1 + y\Phi_2 + z\Phi_3 \end{gathered} \tag{1} \]

\((\mu\) is the shear modulus, \(\nu\) is Poisson’s ratio).

We shall also give the expressions for the stresses entering the boundary conditions:

\[ \sigma_z = 2(1-\nu)\frac{\partial \Phi_3}{\partial z} - \frac{\partial^2 \Phi_0}{\partial z^2} + 2\nu\left( \frac{\partial \Phi_1}{\partial x} + \frac{\partial \Phi_2}{\partial y} \right) - \left( x\frac{\partial^2 \Phi_1}{\partial z^2} + y\frac{\partial^2 \Phi_2}{\partial z^2} + z\frac{\partial^2 \Phi_3}{\partial z^2} \right); \]

\[ \tau_{zx} = \frac{\partial \Phi}{\partial x} + 2(1-\nu)\frac{\partial \Phi_1}{\partial z}, \qquad \tau_{yz} = \frac{\partial \Phi}{\partial y} + 2(1-\nu)\frac{\partial \Phi_2}{\partial z}, \tag{2} \]

\[ \Phi = (1-2\nu)\Phi_3 - \frac{\partial \Phi_0}{\partial z} - \left( x\frac{\partial \Phi_1}{\partial z} + y\frac{\partial \Phi_2}{\partial z} + z\frac{\partial \Phi_3}{\partial z} \right). \]

Using the arbitrariness of one of the harmonic functions entering the solution, we supplement the boundary conditions of the problem

\[ \begin{gathered} u\big|_{z=0}=u_0(r,\varphi),\qquad v\big|_{z=0}=v_0(r,\varphi),\qquad w\big|_{z=0}=w_0(r,\varphi),\\ \sigma_z\big|_{z=h}=\sigma(r,\varphi),\qquad \tau_{zx}\big|_{z=h}=\tau_x(r,\varphi),\qquad \tau_{yz}\big|_{z=h}=\tau_y(r,\varphi) \end{gathered} \tag{3} \]

\((r,\varphi,z\) are cylindrical coordinates) by the following two additional conditions:

\[ F\big|_{z=0}=0,\qquad \Phi\big|_{z=h}=0. \tag{4} \]

In this case, for the functions \(\Phi_1\) and \(\Phi_2\) one immediately obtains separate boundary conditions:

\[ \begin{gathered} 2(1-\nu)\Phi_1\big|_{z=0}=\mu u_0(r,\varphi),\qquad 2(1-\nu)\frac{\partial \Phi_1}{\partial z}\bigg|_{z=h}=\tau_x(r,\varphi),\\ 2(1-\nu)\Phi_2\big|_{z=0}=\mu v_0(r,\varphi),\qquad 2(1-\nu)\frac{\partial \Phi_2}{\partial z}\bigg|_{z=h}=\tau_y(r,\varphi). \end{gathered} \tag{5} \]

If the functions \(\Phi_1\) and \(\Phi_2\) are regarded as found, then we arrive at a boundary-value problem for the harmonic functions \(\Phi_0\) and \(\Phi_3\) with mixed boundary conditions of the form\(^*\)

\[ \begin{gathered} \left.\Phi_0\right|_{z=0} = -\frac{\mu}{2(1-\nu)}(x u_0+y v_0) = F_1(r,\varphi), \\ \left[(3-4\nu)\Phi_3-\frac{\partial \Phi_0}{\partial z}\right]_{z=0} = 2\mu w_0+ \left(x\frac{\partial \Phi_1}{\partial z} + y\frac{\partial \Phi_2}{\partial z}\right)_{z=0} = F_2(r,\varphi), \\ \left[ 2(1-\nu)\frac{\partial \Phi_3}{\partial z} - z\frac{\partial^2\Phi_3}{\partial z^2} - \frac{\partial^2\Phi_0}{\partial z^2} \right]_{z=h} = \sigma+ \left(x\frac{\partial^2\Phi_1}{\partial z^2} + y\frac{\partial^2\Phi_2}{\partial z^2}\right)_{z=h} \\ \qquad - 2\nu\left( \frac{\partial \Phi_1}{\partial x} + \frac{\partial \Phi_2}{\partial y} \right)_{z=h} = F_3(r,\varphi), \\ \left[ (1-2\nu)\Phi_3 - z\frac{\partial \Phi_3}{\partial z} - \frac{\partial \Phi_0}{\partial z} \right]_{z=h} = \frac{x\tau_x+y\tau_y}{2(1-\nu)} = F_4(r,\varphi). \end{gathered} \tag{6} \]

The exact solution of the mixed problem of potential theory thus obtained can be found by means of the Hankel integral transform.

We shall seek the harmonic functions \(\Phi_0\) and \(\Phi_3\) in the form of the following expansions in Bessel functions:

\[ \begin{gathered} \Phi_0 = \sum_{n=-\infty}^{\infty} e^{in\varphi} \int_0^\infty \left(A_0^n \operatorname{ch}\lambda z+B_0^n \operatorname{sh}\lambda z\right) J_n(\lambda r)\,d\lambda, \\ \Phi_3 = \sum_{n=-\infty}^{\infty} e^{in\varphi} \int_0^\infty \left(A_3^n \operatorname{ch}\lambda z+B_3^n \operatorname{sh}\lambda z\right) J_n(\lambda r)\,d\lambda. \end{gathered} \tag{7} \]

Substituting (7) into (6) and representing the functions \(F_k(r,\varphi)\) \((k=1,2,3,4)\) in the form of the corresponding expansions \((^{3,4})\)

\[ \begin{gathered} F_k(r,\varphi) = \sum_{n=-\infty}^{\infty} e^{in\varphi} \int_0^\infty f_k^n(\lambda)J_n(\lambda r)\lambda\,d\lambda, \\ f_n^k(\lambda) = \frac{1}{2\pi} \int_0^\infty J_n(\lambda r)r\,dr \int_0^{2\pi} F_k(r,\varphi)e^{-in\varphi}\,d\varphi, \end{gathered} \tag{8} \]

we immediately obtain, for the unknown quantities \(A_0^n,\ B_0^n,\ A_3^n,\ B_3^n\), a system of linear algebraic equations.

Since from the first equation (6) the quantity \(B_0^n=\lambda f_1^n(\lambda)\) is found at once, and from the second equation (6) we have the relation \((3-4\nu)A_3^n-A_0^n=f_2^n(\lambda)\), the matter is in fact reduced to the solution of a system of two equations

\[ -\left[(1-2\nu)\operatorname{sh}\gamma+\gamma\operatorname{ch}\gamma\right]A_3^n + \left[2(1-\nu)\operatorname{ch}\gamma-\gamma\operatorname{sh}\gamma\right]B_3^n = \psi_1^n(\gamma), \]

\[ -\left[2(1-\nu)\operatorname{ch}\gamma+\gamma\operatorname{sh}\gamma\right]A_3^n + \left[(1-2\nu)\operatorname{sh}\gamma-\gamma\operatorname{ch}\gamma\right]B_3^n = \psi_2^n(\gamma) \tag{9} \]

with determinant

\[ D(\gamma)=(3-4\nu)\operatorname{sh}^2\gamma+\gamma^2+4(1-\nu)^2 \qquad (\gamma=\lambda h), \tag{10} \]

where the notation has been introduced

\[ \psi_1^n = \lambda f_1^n\operatorname{ch}\gamma - f_2^n\operatorname{sh}\gamma + \frac{1}{\lambda}f_3^n, \qquad \psi_2^n = \lambda f_1^n\operatorname{sh}\gamma - f_2^n\operatorname{ch}\gamma + f_4^n. \tag{11} \]

\[ \rule{0.18\textwidth}{0.4pt} \]

\(^*\) It is also required that, as \(r\to\infty\), the Papkovich–Neuber functions be of order \(1/r\), and their derivatives of order \(1/r^2\), which ensures the proper behavior of displacements and stresses at infinity. The functions prescribed in the right-hand sides of (3) must also satisfy the corresponding conditions at infinity.

As an example, let us consider a layer with a fixed base \((z=0)\), deformed by a tangential force \(T\) applied at the point \((0,0,h)\) in the direction of the \(Ox\) axis.

Since in the case under consideration \(u_0=v_0=w_0=\sigma=\tau_y=0\), we have \(\Phi_2=0\), while for \(\Phi_1\) we have the boundary conditions

\[ \left.\Phi_1\right|_{z=0}=0,\qquad \left.2(1-\nu)\frac{\partial \Phi_1}{\partial z}\right|_{z=h} =\tau_x(r,\varphi). \tag{12} \]

Distributing the force \(T\) over a circle of radius \(\varepsilon\), applying the Hankel transform and passing in the obtained solution to the limit as \(\varepsilon\to 0\), we find

\[ \Phi_1=\frac{T}{4\pi(1-\nu)} \int_0^\infty \frac{\operatorname{sh}\lambda z}{\operatorname{ch}\lambda h} J_0(\lambda r)\,d\lambda . \tag{13} \]

Substituting the expression found for \(\Phi_1\) into the right-hand sides of the basic system (6) and expanding the resulting functions in Hankel integrals, we find the following expressions for the functions \(\Phi_0\) and \(\Phi_3\):

\[ \Phi_0=\cos\varphi\int_0^\infty A_0^1\operatorname{ch}\lambda z\,J_1'(\lambda r)\,d\lambda, \]

\[ \Phi_3=\cos\varphi\int_0^\infty \left(A_3^1\operatorname{ch}\lambda z+B_3^1\operatorname{sh}\lambda z\right) J_1(\lambda r)\lambda\,d\lambda, \tag{14} \]

where

\[ A_0^1=(3-4\nu)A_3^1-\frac{Th\,\operatorname{sh}\gamma}{4\pi(1-\nu)\operatorname{ch}^2\gamma}, \]

and the quantities \(A_3^1\) and \(B_3^1\) must be found from system (9) with the following values of its right-hand sides:

\[ \psi_1^1=-\frac{Th}{4\pi(1-\nu)} \left[1+(1-2\nu)\frac{\operatorname{th}\gamma}{\gamma}\right], \qquad \psi_2^1=-\frac{Th}{4\pi(1-\nu)}\operatorname{th}\gamma . \tag{15} \]

Let us give the expression for the tangential stress \(\tau_0\equiv \tau_{zx}|_{z=0}\) in the fixed base:

\[ \tau_0=\frac{Th}{2\pi} \left\{ \frac{1}{r}\int_0^\infty \frac{\operatorname{sh}\gamma}{\operatorname{ch}^2\gamma} J_1(\lambda r)\lambda\,d\lambda + \int_0^\infty \frac{D_1(\gamma)}{D(\gamma)} \left[ \frac{J_1(\lambda r)}{r}\cos 2\varphi -\lambda J_0(\lambda r)\cos^2\varphi \right]\lambda\,d\lambda \right\}, \tag{16} \]

where the notation has been introduced

\[ D_1(\gamma)=2(1-\nu)\operatorname{sh}\gamma +\frac{\gamma}{\operatorname{ch}\gamma} -\frac{(1-2\nu)^2}{\gamma}\operatorname{th}\gamma\,\operatorname{sh}\gamma . \tag{17} \]

Physico-Technical Institute
Academy of Sciences of the USSR

Received
8 VII 1958

CITED LITERATURE

  1. P. F. Papkovich, Theory of Elasticity, 1939.
  2. G. Neuber, Stress Concentration, 1947.
  3. G. N. Watson, Bessel Functions, IL, 1949.
  4. I. Sneddon, Fourier Transforms, IL, 1955.

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A MIXED PROBLEM FOR AN ELASTIC LAYER