On One Case of Bending of Thin Plates

M. G. Pinsky

1. Let us consider the bending of a plate by a moment applied to a rigid central circular washer, with the outer edge of the plate clamped (Fig. 1). The problem in this formulation was considered by F. M. Dimentberg in [1],

Fig. 1

In that work the author arrived, as it seems to us, at an incorrect result, having made an error in setting up the boundary conditions on the contour of the washer.

As is known, the deflection of the plate \(W\) in this case satisfies the biharmonic equation

\[ \Delta \Delta W = 0 \tag{1} \]

or, in polar coordinates, has the form:

\[ \left( \frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \theta^2} \right) \left( \frac{\partial^2 w}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial w}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2 w}{\partial \theta^2} \right)=0. \tag{1'} \]

We carry out the solution in polar coordinates \(r\) and \(\theta\).
We shall seek the solution of the basic equation (1) in the form:

\[ W(r,\theta)=f(r)\cos \theta, \tag{2} \]

where

\[ f(r)=C_1 r+C_2 r^3+C_3 r\ln r+C_4 r^{-1}. \tag{3} \]

The boundary conditions have the form:

\[ \text{1) for } r=a,\qquad W=\frac{\partial W}{\partial r}=0; \tag{4} \]

2) when the washer is rotated through some angle \(\varphi_0\), the equation of the line of deflections for \(r=b\) has the form:

\[ W(b,\theta)=w_0\cos\theta, \tag{5} \]

where

\[ w_0=b\varphi_0. \]

We shall study the elastic curve along the radius \(\theta=0\).
In doing so we obtain a representation of the entire elastic surface of the plate as a whole.

The arbitrary constants entering into (3) will be determined by us from 4 boundary conditions on the inner and outer contours.

The boundary conditions take the following form:

1) For the outer contour

\[ C_1+3C_2a^2+C_3(1+\ln a)-C_4a^{-2}=0. \]

\[ C_1+C_2a^2+C_3\ln a+C_4a^{-2}=0. \tag{6} \]

2) For the inner contour

\[ W\big|_{r=b}=b\varphi_0, \tag{7} \]

\[ \frac{\partial W}{\partial n}\bigg|_{r=b}=-\varphi_0. \]

Attention should be paid to the last of the equalities (7).

The derivative of the deflection is taken along the inward normal \(\dfrac{\partial w}{\partial n}\), and, consequently, it is necessary to take into account the minus sign before the expression for the derivative \(\dfrac{\partial w}{\partial n}\).

It seems to us that the results obtained in paper (1) proved to be incorrect precisely because this circumstance was not taken into account.

Equations (7) give

\[ C_1+C_2b^2+C_3\ln b+C_4b^{-2}=\varphi_0, \]

\[ C_1+3C_2b^2+C_3(1+\ln b)-C_4b^{-2}=-\varphi_0. \tag{7′} \]

Solving the system of equations (6) and (7′), we obtain the values of the constants:

\[ C_1= \frac{ b^2\left[2(a^2\ln b-b^2\ln a)-a^2+b^2\right] }{ \left[(a^2+b^2)\ln\frac{a}{b}-a^2+b^2\right] }\,\varphi_0, \]

\[ C_2= \frac{ b^2\ln\dfrac{a}{b} }{ (a^2+b^2)\left[(a^2+b^2)\ln\dfrac{a}{b}-a^2+b^2\right] }\,\varphi_0, \]

\[ C_3= \frac{ -2b^2 }{ \left[(a^2+b^2)\ln\dfrac{a}{b}-a^2+b^2\right] }\,\varphi_0, \tag{8} \]

\[ C_4= \frac{ a^2b^2\left(a^2\ln\dfrac{a}{b}-a^2+b^2\right) }{ (a^2+b^2)\left[(a^2+b^2)\ln\dfrac{a}{b}-a^2+b^2\right] }\,\varphi_0. \]

Taking into account (2), (3), and (8), we write the expression for the deflection of the plate

\[ W= \frac{ \varphi_0\cos\theta }{ (a^2+b^2)\left[(a^2+b^2)\ln\dfrac{a}{b}-a^2+b^2\right] } \cdot \left\{ b^2\left[2(a^2\ln b-b^2\ln a)+(a^2+b^2)\right]r \right. \]

\[ \left. +\,b^2\ln\frac{a}{b}\,r^3 -2b^2(a^2-b^2)r\ln r +a^2b^2\left(a^2\ln\frac{a}{b}-a^2+b^2\right)r^{-1} \right\}. \tag{9} \]

We determine the angle \(\varphi_0\) from the equilibrium conditions, for which we shall compute the bending moment and the reaction at the place where the plate adjoins the washer

\[ M_\theta=-D\left[(6C_2r+C_3r^{-1}+2C_4r^{-3})- \frac{2\mu\varphi_0}{(a^2+b^2)\ln\dfrac{a}{b}-a^2+b^2}\right]\cos\theta . \tag{10} \]

\[ V_{(r)}=Q_\rho+\frac{\partial M_{\rho\theta}}{r\partial\theta} =-D\left[(8C_2-2C_3r^{-2}+2\varphi_0(1-\mu)r^{-2}\right]\cos\theta . \tag{11} \]

Let us write the sum of the moments with respect to the diameter \(\theta=\dfrac{\pi}{2}\)

\[ M=\int_0^{2\pi}\left[M_\theta(b)+bV_r(b)\right]b\cos\theta\,d\theta, \]

where \(M\) is the external moment applied to the washer.

Taking (10) and (11) into account, we finally obtain the expression for the angle of rotation of the washer

\[ \varphi_0= -\frac{M}{2\pi D}\cdot \frac{[1-\beta^4]\ln\beta+(1-\beta^2)^2} {\{(1-2\mu)[(1-\beta^4)\ln\beta+(1-\beta^2)^2]+(1+7\beta^4)\ln\beta-(1-\beta^2)^2\}} . \tag{12} \]

From (12) it is seen that the angle of rotation of the washer depends only on the ratio

\[ \beta=\frac{b}{a}. \]

Substituting \(\varphi_0\) from (12) into (9), we finally obtain for the deflection

\[ \begin{aligned} W(r,\theta)= -\frac{M\cos\theta}{2\pi D} \Bigg\{& \left[2\beta_i^2(\ln b-\beta_i^2\ln a)+\beta_i^2-\beta_i^4\right]r -\frac{\beta_i^2\ln\beta_i}{a^2}r^3 \\ &-2\beta_i^2(1-\beta_i^2)r\ln r -\beta_i^2(a^2\ln\beta_i+a^2-b^2)r^{-1} \Bigg\} \times \\ &\times \frac{1} {\{(1-2\mu)[(1-\beta_i^4)\ln\beta_i+(1-\beta_i^2)^2]+(1+7\beta_i^4)\ln\beta_i-(1-\beta_i^2)^2\}} . \end{aligned} \tag{13} \]

The expression for the bending moment will be:

\[ \begin{aligned} M_\theta=\frac{M\cos\theta}{2\pi} \Bigg\{& \frac{b^2}{a^4}\ln\frac{a}{b}\,r -2r^{-1}\left[ b^2(a^2-b^2)+\mu(a^2-b^2) \left((a^2+b^2)\ln\frac{a}{b}-a^2+b^2\right)\frac{1}{a^4} \right] \\ &+2a^2b^2r^{-3}\left(a^2\ln\frac{a}{b}-a^2+b^2\right) \Bigg\} \times \\ &\times \frac{1} {\{(1-2\mu)[(1-\beta^4)\ln\beta+(1-\beta^2)^2]+(1+7\beta^4)\ln\beta-(1-\beta^2)^2\}} . \end{aligned} \tag{14} \]

The greatest moment \(M_\theta\) will be at the place where the plate adjoins the washer, at \(r=b\).

Let us write the expression for the bending moment at \(\theta=0^\circ\)

\[ M_\theta(b)=\frac{M}{2\pi b} \left\{ 2(1+3\beta^4)+\beta^2(1-\beta^2)+2(1-\beta^2) +2\mu\left[(1-\beta^4)\ln\beta+(1-\beta^2)^2\right] \right\}\times \]

\[ \times \frac{1} {\{(1-2\mu)[(1-\beta_i^4)\ln\beta+(1-\beta^2)^2]+(1+7\beta^2)\ln\beta+(1-\beta^2)^2\}} . \tag{15} \]

From expression (15) it is clear that the bending moment at the contour of the washer does not depend on Poisson’s ratio \(\mu\), but depends on \(\beta=\dfrac{b}{a}\).

The value of the bending moment \(M_r\) for \(r=b\) will be:

\[ M_r(b)=-\frac{M\cos\theta}{2\pi b} \left\{ \mu\left[6\beta^4\ln\beta+2\beta^2(1-\beta^2)+2(1+\ln\beta-\beta^2)\right] -2(1-\beta^4)\ln\beta+(1-\beta^2)^2 \right\}\times \]

\[ \times \frac{1}{ (1-2\mu)\left[(1-\beta^4)\ln\beta+(1-\beta^2)^2\right] +(1+7\beta^2)\ln\beta-(1-\beta^2)^2 }. \tag{16} \]

The value of the bending moment \(M_\theta\) on the outer contour \(r=a\) we write in the form:

\[ M_\theta(a)=-\frac{M}{2\pi a} \left\{ 8\beta^2\ln\beta+4(1-\beta^2)\beta^2 +\mu\left[(1-\beta^4)\ln\beta+(1-\beta^2)^2\right] \right\}\times \]

\[ \times \frac{1}{ (1-2\mu)\left[(1-\beta^4)\ln\beta+(1-\beta^2)^2\right] +(1+7\beta^4)\ln\beta-(1-\beta^2)^2 }. \tag{17} \]

Having calculated the magnitude of the transverse shear force, we obtain

\[ Q_r=-\frac{M}{2\pi b^2} \left[ 4\beta^2(1-\beta^2) +2(1-\mu)(1-\beta^4)\ln\beta +(1-\beta^2)^2 -8\beta^4\ln\beta \right]\times \]

\[ \times \frac{1}{ (1-2\mu)\left[(1-\beta^4)\ln\beta+(1-\beta^2)^2\right] +(1+7\beta^4)\ln\beta-(1-\beta^2)^2 }. \tag{18} \]

Having expressions for \(\varphi_0\), \(w\), \(M_\theta\), \(M_r\), \(V_r\), \(Q_r\), we can obtain the solution for an infinite plate with a rigid central circular washer.

Let \(a\to\infty\) (i.e. \(\beta\to0\)).

Substituting the value \(\beta=0\) into expressions (12), (13), (15), (16), (11), and (18) and passing to the limit, we obtain the corresponding expressions for a plate with a small circular washer, i.e. an unbounded plate.

\[ \varphi_0=\frac{M}{4\pi D(1-\mu)}, \tag{19} \]

\[ W=\frac{Mb^2\cos\theta}{4\pi D(1-\mu)\,r}, \tag{20} \]

\[ M_\theta(b)=-\frac{M\cos\theta}{2\pi b}, \tag{21} \]

\[ M_r(b)=\frac{M\cos\theta}{2\pi b}, \tag{22} \]

\[ V_r=\frac{M\cos\theta}{2\pi b^2}, \tag{23} \]

\[ Q_r=-\frac{M\cos\theta}{2\pi b^2}. \tag{24} \]

To simplify the calculations, we shall put the expression for \(\varphi_0\) and \(M_\theta\) on the inner contour \(K\) into a form convenient for computation.

\[ \varphi_0=-\frac{M}{E h^3}\,\gamma, \tag{25} \]

where \(\gamma\) for steel \((\mu=0.28)\) is given in the table.

Fig. 2

For the bending moment \(M_\theta(b)\) we obtain an expression of the form

\[ M_\theta=-\frac{M\cos\theta}{2\pi b}\,m_\theta, \tag{26} \]

where \(m_\theta\) for steel is given in the table and in Fig. 2.

Table

II. We investigated the bending of an infinite plate with a rigid circular washer of radius \(b\), subjected to the action of a moment applied to the washer, by means of functions of a complex variable.

The elastic surface of the plate satisfies the biharmonic equation (1).

As is known, the solution of such an equation can be represented by means of two analytic functions \(\varphi(z)\) and \(\chi(z)\) of the complex variable by E. Goursat’s formula (2) in the following form:

\[ W=z\overline{\varphi(z)}+\bar z\,\varphi(z)+\chi(z)+\overline{\chi(z)}, \tag{II.1} \]

where \(\varphi(z)\) and \(\chi(z)\) are functions of the complex variable

\[ z=x+iy, \]

and \(\overline{\varphi(z)}\) and \(\overline{\chi(z)}\) are conjugate functions of the complex variable

\[ \bar z=x-iy. \]

The problem consists in determining a biharmonic function outside the contour of the washer for known conditions on the contour \((S)\) of the washer, i.e., for given values of the function itself and of its derivative along the normal to the contour. Since the functions \(\varphi(z)\) and \(\chi(z)\) are regular in the region of the plate, these functions can be represented inside this region by Laurent series, and uniquely so. Bearing this in mind, we represent the functions \(\varphi(z)\) and \(\chi(z)\) by series of the form:

\[ \varphi(z)=\sum_{0}^{\infty} a_k z^{-k} \]

\[ \chi(z)=\sum_{0}^{\infty} b_k z^{-k} \tag{II.2} \]

The coefficients of the series of the functions (II.2) must satisfy the condition that the displacement be equal to zero at an infinite distance from the washer,

\[ W\big|_{R\to\infty}=0, \tag{II.3} \]

where \(R\) is the distance to an arbitrary point of the plate. Since at any point of the plate \(z=Re^{i\nu}\), then, for

\[ R\to\infty \quad \text{and} \quad z\to\infty . \]

Let us consider what restrictions are imposed on the coefficients of the series in order that condition (II.3) be fulfilled.

Writing equality (II.1), taking (II.2) into account and passing to the limit as \(z\to\infty\), we obtain

\[ \lim\left[ z\bar a_0+\frac{z\bar a_1}{\bar z} +\frac{z\bar a_2}{\bar z^2}+\ldots +\bar z a_0+\frac{\bar z}{z}a_1 +\frac{\bar z}{z^2}a_2+\ldots +b_0+b_1 z^{-1}+\ldots+\bar b_0+\bar b_1\bar z^{-1} \right]=0. \tag{II.4} \]

Equality (II.4) will hold if the following equalities are satisfied:

\[ z\bar a_0+\bar z a_0=0, \]

\[ \frac{z}{\bar z}\bar a_1+\frac{\bar z}{z}a_1=0, \]

\[ b_0+\bar b_0=0. \]

It is easy to show that in this case

\[ a_0=\bar a_0=a_1=\bar a_1=0. \]

After these preliminary remarks we can represent the analytic functions in the form of Laurent series:

\[ \varphi(z)=\sum_{2}^{S} a_k z^{-k},\quad \bar{\varphi}(z)=\sum_{2}^{S}\bar{a}_k z^{-k}. \tag{II.5} \]

We represent the function \(\chi(z)\) in the form

\[ \chi(z)=b_0+\sum_{1}^{S} b_k z^{-k}, \]

\[ \bar{\chi}(z)=\bar{b}_0+\sum_{1}^{S}\bar{b}_k z^{-k}. \tag{II.6} \]

Let us write the boundary conditions on the contour of the washer.

1. The displacement on the contour of the washer can be written in the form

\[ W_s=b\varphi_0\cos\theta \tag{II.6'} \]

or

\[ \left[z\bar{\varphi}(\bar z)+z\varphi(z)+\chi(z)+\bar{\chi}(\bar z)\right]_s = b\varphi_0\,\frac{\sigma+\sigma^{-1}}{2}, \tag{II.6''} \]

where \(\sigma=e^{i\theta}\) is the unit normal vector.

1. To formulate the second condition on the contour of the washer, we take into account that the derivative along the external normal will be [3]

\[ \frac{\partial W}{\partial n} = n\,\frac{\partial W}{\partial z} + \bar n\,\frac{\partial W}{\partial \bar z}, \tag{II.7} \]

where

\[ n=e^{i\theta}=\sigma \text{ is the unit normal vector,} \]

\[ \bar n=e^{-i\theta}=\bar\sigma \text{ is the conjugate quantity.} \]

Then, from geometrical considerations, it may be written that on the contour

\[ \left|\frac{\partial w}{\partial n}\right| = \left|\frac{w_s}{b}\right| = \left|\varphi_0\cos\theta\right|. \]

Taking into account that the derivative is taken along the internal normal, we finally write the boundary condition on the contour in the form:

\[ \left[ z\bar{\varphi}(\bar z)+z\varphi(z)+\varphi'(z)+\bar{\varphi}'(\bar z)+z\chi'(z)+\bar z\,\bar{\chi}'(\bar z) \right]_s = -\varphi\,\frac{z+z^{-1}}{2}. \tag{II.8} \]

In equation (II.6′) we substitute the values (II.8) of the functions (II.5) and (II.6), taking into account that on the contour of the washer

\[ z=b\sigma=be^{i\theta}, \]

then finally we write condition (II.6′) as:

\[ \sum_{2}^{S} a_k b^{-(k-1)}\sigma^{-(k+1)} + \sum_{2}^{S}\bar a_k b^{-(k-1)}\sigma^{k+1} + \sum_{1}^{S} b_k b^{-k}\sigma^{-k} + \sum_{1}^{S}\bar b_k b^{-k}\sigma^k = \]

\[ = b\varphi_0\,\frac{\sigma+\sigma^{-1}}{2}. \tag{II.9} \]

To determine the coefficients \(b_k\), we equate the coefficients of like powers of \(\sigma\). This gives us the following equalities:

\[ b_1=\bar b_1=\frac{b^2}{2}\zeta_0, \]

\[ b_2=\bar b_2=0. \]

Equating the coefficients of \(\sigma^k\) and \(\sigma^{-k}\), we obtain

\[ \begin{aligned} b_k&=-b^2a_{k-1},\\ \bar b_k&=-b^2\bar a_{k-1}. \end{aligned} \tag{II.10} \]

Noting that the expansion for \(\varphi(z)\) is of the form (II.5), that is, begins with \(k=2\), we write the second boundary condition (II.8) in the form:

\[ \begin{aligned} &\sum_2^S \bar a_k b^{-k}\sigma^{k+1} -\sum_2^S ka_k b^{-k}\sigma^{-(k+1)} -\sum_1^S kb_k b^{-(k+1)}\sigma^k \\ &\quad -\sum_2^S k\bar a_k b^{-k}\sigma^{k+1} +\sum_2^S a_k b^{-k}\sigma^{-(k+1)} -\sum_1^S k\bar b_k b^{-(k+1)}\sigma^k = -\zeta_0\,\frac{\sigma+\sigma^{-1}}{2}. \end{aligned} \tag{II.11} \]

Equating the coefficients of \(\sigma^k\) and \(\sigma^{-k}\), we obtain

\[ \frac{a_{k-1}}{b^{k-1}} -\frac{(k-1)a_{k-1}}{b^{k-1}} -\frac{kb_k}{b^{k+1}}=0, \]

\[ \frac{\bar a_{k-1}}{b^{k-1}} -\frac{(k-1)\bar a_{k-1}}{b^{k-1}} -\frac{k\bar b_k}{b^{k+1}}=0. \tag{II.12} \]

From (II.10) and (II.12) we obtain

\[ \bar a_{k-1}=a_{k-1}=0 \qquad (k>1), \]

and, consequently, from (II.10) we obtain

\[ \bar b_k=b_k=0 \qquad (k>1). \]

Thus, the function \(\varphi(z)\) is identically equal to zero, and, consequently, the displacement of the plate is represented in the form

\[ W=b_0+\bar b_0+\frac{b_1}{z}+\frac{\bar b_1}{\bar z}. \tag{II.13} \]

Taking into account \(b_0+\bar b_0=0\), we obtain the deflection

\[ W=\frac{b^2\zeta_0}{z^2}\,x, \tag{II.14} \]

where

\[ x=R\cos\theta \]

or

\[ W=\frac{b^2\zeta_0\cos\theta}{R}. \tag{II.14'} \]

Let us determine the magnitudes of the moments \(M_\theta\), \(M_r\), and \(M_{r\theta}\); for this purpose we turn to the expressions due to S. G. Lekhnitskii [4]

\[ M_\theta - M_r + 2iM_{r\theta} = 4(1-\mu)D e^{i\theta}\left[z\omega''(z)+\chi''(z)\right] \]

\[ M_r+M_\theta=-8(1-\mu)D\operatorname{Re}[\omega'(z)]. \]

We must obtain the values of the moments on the contour of the washer for \(z=be^{i\theta}=b\zeta\).

After simple transformations, separating the real and imaginary parts, we obtain

\[ M_\theta-M_r=4(1-\mu)D\frac{\vartheta_0}{b}\cos\theta, \]

\[ M_\theta+M_r=0. \tag{II.15} \]

Solving (II.15), we obtain the values of the bending moments

\[ M_\theta=2(1-\mu)D\frac{\vartheta_0}{b}\cos\theta, \]

\[ M_r=-2(1-\mu)D\frac{\vartheta_0}{b}\cos\theta. \tag{II.16} \]

Equating the imaginary parts, we obtain

\[ M_{r\theta}=-2(1-\mu)D\frac{\vartheta_0}{b}\sin\theta. \tag{II.17} \]

We shall consider the equilibrium of a rigid circular washer under the action of bending and twisting moments, as well as transverse forces.

The reaction at the place where the washer adjoins the plate is expressed as

\[ V_r=Q_r+\frac{1}{r}\frac{\partial M_{r\theta}}{\partial\theta}\bigg|_{r=b} \]

or

\[ V_r=-2(1-\mu)D\frac{\vartheta_0}{b^2}\cos\theta. \]

Let us write the condition of equilibrium of the washer

\[ M=\int_0^{2\pi}\left[M_\theta\cos\theta+V_rb\cos\theta\right]b\,d\theta, \tag{II.18} \]

whence we obtain

\[ M=4(1-\mu)\pi D\vartheta_0, \]

where \(M\) is the external moment applied to the washer, or the angle of rotation of the washer is

\[ \vartheta_0=-\frac{M}{4\pi(1-\mu)D}. \tag{II.19} \]

For steel \((\mu=0.3)\) we obtain

\[ \vartheta_0=-1.24\,\frac{M}{Eh^3}. \]

Expression for the bending moments

\[ M_\theta=-\frac{M\cos\theta}{2\pi b}, \tag{II.20} \]

\[ M_r=\frac{M\cos\theta}{2\pi b}, \tag{II.21} \]

For the deflection of the plate we finally obtain

\[ W=\frac{Mb^2\cos\theta}{4\pi(1-\mu)D\cdot R}. \tag{II.22} \]

For a steel plate the deflection at the contour of the washer will be

\[ W=1.24\,\frac{Mb\cos\theta}{Eh^3}. \]

The shearing force at the place where the washer adjoins the plate is represented as

\[ V_r(b)=-\frac{M\cos\theta}{2\pi b^2}. \tag{II.23} \]

The expressions (II.19), (II.20), (II.21), (II.22), (II.23) obtained by us coincide, respectively, with expressions (19), (21), (22), (20), and (23), obtained in I from the solution for a finite circular plate in the limiting transition, when the outer radius \(a\) of the plate increases without bound while the radius of the washer remains unchanged.

We arrived at these results by entirely different methods; this agreement of the results confirms the correctness of the calculations.

The author expresses his gratitude to Prof. Dr. V. I. Blokh for discussion of the results.

REFERENCES

1. F. M. Dimentberg. On one case of bending of a circular plate. Vestnik inzhenerov i tekhnikov, No. 7, pp. 415–418, 1938.

2. V. I. Smirnov. A Course of Higher Mathematics, vol. III, p. 402.

3. A. I. Lur’e. Some problems on the bending of a circular plate. Prikladnaya matematika i mekhanika, issue I, vol. IV, pp. 93–102, 1940.

4. S. G. Lekhnitskii. On certain questions connected with the theory of bending of thin plates—PMM, issue 2, vol. II, pp. 181–210, 1938.