Elasticity Theory Solution of the Problem on Bending of a Narrow Multilayer Cantilever with a Circular Axis by Loads at its End

An exact solution of the problem on plane bending of a narrow multilayer cantilever bar with a curved circular axis by tangential and normal loads distributed on its free end face is presented. The natural (for a bar structure) cylindrical circular coordinate system is used to describe the structure and geometry of the bar. The solution is obtained by directly integrating the equations of a plane elasticity theory problem using an analytical description of the mechanical characteristics of the discrete-inhomogeneous multilayer bar. Physical relations that take into account the cylindrical orthotropy of the material of bar layers and the conditions of absolutely rigid contact of layers are used in constructing the solution. The theoretical relations are realized for the test problem on determining the strain-stress state of a four-layer cantilever with a semiring axis. The solution obtained allows one to predict the strength and rigidity, to develop optimum design techniques, and to construct analytical solutions of different problems on bending of multilayer curved bars.

Introduction

In the modern designs of various purpose, layered composite elements are increasingly being used. However, the development of deformation mechanics and calculation and optimization methods for composite elements noticably lag behind the modern level of creation technologies of composites. This complicates a wide introduction of composite materials into the designing practice.

Bars working in bending are one of the most widespread elements in building and mechanical engineering. However, their analytical modeling is developed rather poorly in comparison with that of composite plates and shells, to which a number of known fundamental works [1,2,3,4,5,6,7,8] have been devoted. To a smaller degree, models and exact solutions to the problems of bending of rectilinear composite bars have been developed [5, 9,10,11,12]. But for curvilinear composite bars, known are only some works [10, 13,14,15], which are devoted to exact solutions for anisotropic inhomogeneous bars with a circular axis.

The problem on bending of a continuous inhomogeneous bar with a circular axis by forces and moments at its ends is solved in [10]. The author considered the case where the bar material is cylindrically orthotropic and showed that an exact solution of the problem is possible if its elastic constants vary according to a power law along the height of bar cross section. Later, in [13], the problem on pure bending of an isotropic circular bar of a continuously inhomogeneous material whose elastic constants vary along the bar linearly or sinusoidally as functions of the angular coordinate was considered. In the solution procedure, the distributions of components of the stress-strain state (SSS) were assumed constant across the width of bar cross section, which corresponds to the plane problem of elasticity theory.

In [14], on the basis of the known Golovin solution for pure bending of a homogeneous circular bar [16], a discrete model of pure bending for a multilayer composite bar with isotropic layers is constructed. Similarly, in [15], using the solutions given in [10], a discrete model of bending for a multilayer circular bar with anisotropic layers is constructed for the case where moments and transverse forces are applied to its ends. In these works, for constructing models, the discrete approach where the layers of a composite bar are consider separately was used, and their joint work was modeled by coordinating boundary conditions on interfaces. Such an approach allows one found exact solutions of problems for multilayer bars; however, it leads to some mathematical complexities in their realization and use for solving problems with more complex combinations of loads.

In [12], with the example of solution of the problem on bending of a multilayer rectilinear bar, it is shown that an exact solution can be found for the problem by the methods of elasticity theory on the basis of a continual approach, which allows one to obtain relations for SSS components at once for all the package of bar layers. The same approach can also be employed in solving problems on bending of multilayer bars with a circular axis.

The purpose of the given work is to found an analytical solution of the problem on elastic bending of a multilayered circular bar by normal and tangential loads applied to its ends. To do this, a continual approach considering the absolutely rigid contact between layers and anisotropy of their materials will be used. This will allow us to close, to some degree, the gap in the corresponding division of the mechanics of elements and to construct a basis for solving a wide range of applied problems.

Statement of the problem

Let us consider a multilayer bar with a circular axis of radius r_C and a rectangular cross section of width b and height h ( b ≪ h) (Fig. 1). The bar consists of m concentric cylindrical layers ${P}_k\left(k=\overline{1,m}\right)$ with inside and outside radii r_bd,k−1 and r_{bd, k}, respectively. The layers are rigidly connected together at all points of their boundary surfaces.

Fig. 1.

Schematic of loads and structure of cross section of the layered composite bar considered. Explanations in the text.

The bar is related to a rectangular coordinates system xyz , as shown in Fig. 1. The coordinate plane xOz is the symmetry plane of the bar. To describe the geometry of the bar and to obtain relations for components of its SSS, we use a cylindrical coordinates system rθy (see Fig. 1a).

The bar is subjected to a system of normal ${p}_{\theta}^{T_1}(r)$ and tangential ${p}_{\theta}^{T_1}(r)$ loads applied to its end Τ₁ with the coordinate θ₁ = 0 (see Fig. 1a). Its longitudinal cylindrical surfaces Π_ϛ (ϛ = 1, 2 ) and lateral sides are free from loads.

At the end Τ₂ with the coordinate θ = θ₂, some kinematic and static conditions are given which agree with the type of fastening of beam end sufficient to ensure the static equilibrium of the bar.

The layers P_k of the bar are made of homogeneous materials with the cylindrical orthotropy [10], where, for any point of a layer, one planes of its elastic symmetry coincides with the cross section of the bar, the second one is parallel to the xOz plane, and the anisotropy axis coincides with the Oy axis. The elastic characteristics of each kth layer of the bar make up the set of elastic constants

$$ \left\Vert {s}_a^{\left[k\right]}\right\Vert =\left\Vert {E}_r^{\left[k\right]},{E}_{\theta}^{\left[k\right]},{E}_y^{\left[k\right]},{G}_{r\theta}^{\left[k\right]},{G}_{\theta y}^{\left[k\right]},{G}_{y r}^{\left[k\right]},{v}_{r\theta}^{\left[k\right]},{v}_{\theta r}^{\left[k\right]},{v}_{y\theta}^{\left[k\right]},{v}_{\theta y}^{\left[k\right]},{v}_{r y}^{\left[k\right]},{v}_{y r}^{\left[k\right]}\right\Vert . $$

(1)

The elastic characteristics of the bar material as a whole are piecewise constant functions ${\mu}_a^S$ of the coordinate r, which, as in [12], can be written with the help of Heaviside functions H (r):

$$ {\mu}_a^S=\sum \limits_{k=1}^m{S}_a^{\left[k\right]}\left[H\left(r-{r}_{bd,k-1}\right)-H\left(r-{r}_{bd,k}\right)\right], $$

(2)

where r_{bd 0} = r₁ and r_bd,m = r₂ are the radii of curvature of the longitudinal surfaces Π_ϛ.

Let us consider the basic solution stages of the problem on determination of the SSS of the circular multilayer curvilinear bar with orthotropic layers.

General solution of the problem

The initial data of the problem ( b ≪ h) allow us to neglect the change in SSS components across the width b of the bar and to determine them using the system of static, physical, and geometrical equations of plane problem of the linear elasticity theory in the polar coordinate system:

$$ {\displaystyle \begin{array}{l}\frac{\partial {\sigma}_r}{\partial_r}+\frac{1}{r}\frac{\partial {\tau}_{r\theta}}{\partial \theta }+\frac{\sigma_r-{\sigma}_{\theta }}{r}=0\\ {}\frac{\partial {\tau}_{r\theta}}{\partial r}+\frac{1}{r}\frac{\partial {\sigma}_{\theta }}{\partial \theta }+\frac{2{\tau}_{\theta r}}{r}=0,\end{array}}\left\}{\displaystyle \begin{array}{l}{\varepsilon}_r=\frac{\sigma_r}{\mu_r^E}-\frac{\mu_{\theta r}^v{\sigma}_{\theta }}{\mu_{\theta}^E},\\ {}{\varepsilon}_{\theta }=\frac{\sigma_{\theta }}{\mu_{\theta}^E}-\frac{\mu_{r\theta}^v{\sigma}_r}{\mu_r^E},\\ {}{\gamma}_{r\theta}=\frac{1}{\mu_{r\theta}^G}{\tau}_{r\theta},\end{array}}\right\}{\displaystyle \begin{array}{l}{\varepsilon}_r=\frac{\partial {u}_r}{\partial r},\\ {}{\varepsilon}_{\theta }=\frac{1}{r}\left(\frac{\partial {\mu}_{\theta }}{\partial \theta }+{u}_r\right),\\ {}{\gamma}_{r\theta}=\frac{1}{r}\left(\frac{\partial {\mu}_r}{\partial \theta }-{u}_{\theta}\right)+\frac{\partial {\mu}_{\theta }}{\partial r}.\end{array}}\Big\} $$

(3)

According to the accepted system of external loads on the longitudinal surfaces Π_ϛ (ϛ = 1,2) and the end Τ₁ of the bar, the static boundary conditions are

$$ {\displaystyle \begin{array}{l}{\left.{\sigma}_{\theta}\right|}_{\theta =0}=-{p}_{\theta}^{T_1},\\ {}{\left.{\sigma}_r\right|}_{r={r}_{\varsigma }}=0,\end{array}}\kern0.5em {\displaystyle \begin{array}{l}{\left.{\tau}_{\theta r}\right|}_{\theta =0}=-{p}_r^{T_1},\\ {}{\left.{\tau}_{r\theta}\right|}_{r={r}_{\varsigma }}=0.\end{array}} $$

(4)

Integrating the first and second relations of (4) over the cross section, with account of relations between the internal forces N_θ, Q_r, and M_y and loads at the beam end, we arrive at the static conditions at the end Τ₁

$$ {\displaystyle \begin{array}{c}b\underset{r_1}{\overset{r_2}{\int }}{\sigma}_{\theta }{\left|{}_{\theta =0} dr=b\underset{r_1}{\overset{r_2}{\int }}\left(-{p}_{\theta}^{T_1}\right) dr={N}_{\theta}\right|}_{\theta =0},\\ {}b\underset{r_1}{\overset{r_2}{\int }}{\tau}_{\theta r}{\left|{}_{\theta =0} dr=b\underset{r_1}{\overset{r_2}{\int }}\left(-{p}_r^{T_1}\right) dr={Q}_r\right|}_{\theta =0},\\ {}{\left.b\underset{r_1}{\overset{r_2}{\int }}\Big[{\sigma}_{\theta}\right|}_{\theta =0}\left(r-{r}_C\right)\left] dr=b\underset{r_1}{\overset{r_2}{\int }}\right[-{p}_r^{T_1}\Big(r-{\left|{r}_C\left)\right] dr={M}_y\right|}_{\theta =0}.\end{array}} $$

(5)

Solving static equations (3) for the normal stresses, with account of boundary conditions (4) on the longitudinal sides and paired shear stresses, we have

$$ {\displaystyle \begin{array}{c}{\sigma}_r=-\frac{1}{r}\underset{r_1}{\overset{r}{\int }}\left({\left.\frac{\partial {\tau}_{r\theta}}{\partial r} d\theta +2\underset{0}{\overset{\theta }{\int }}{\tau}_{r\theta} d\theta -{\sigma}_{\theta}\right|}_{\theta =0}\right) d r,\\ {}{\left.{\sigma}_{\theta }=-\underset{0}{\overset{\theta }{\int }}\left(r\frac{\partial {\tau}_{r\theta}}{\partial r}+2{\tau}_{\theta r}\right) d\theta +{\sigma}_{\theta}\right|}_{\theta =0}.\end{array}} $$

(6)

An approximate form of solutions for the shear stresses τ_rθ in (6) can be found from the distribution of the transverse force Q_r, which, for the bar considered, can be written as follows:

$$ {Q}_r=-b\left(\sin \theta \underset{r_1}{\overset{r_2}{\int }}{p}_{\theta}^{{\mathrm{T}}_1} dr+\cos \theta \underset{r_1}{\overset{r_2}{\int }}{p}_r^{{\mathrm{T}}_1} dr\right)=b\underset{r_1}{\overset{r_2}{\int }}{\tau}_{\theta r} dr. $$

(7)

According to relation (7), the possible form of solution for the shear stresses τ_rθ is

$$ {\tau}_{r\theta}={R}_{r\theta}^{\tau 1}\sin \theta +{R}_{r\theta}^{\tau 2}\cos \theta, $$

(8)

where ${R}_{r\theta}^{\tau 1}={R}_{r\theta}^{\tau 1}(r)$ and ${R}_{r\theta}^{\tau 2}={R}_{r\theta}^{\tau 2}(r)$ are unknown distribution functions of shear stresses along the height of cross section.

Inserting Eq. (8) into the fourth boundary condition of (4) and considering that it has to be satisfied for all points of the longitudinal surfaces Π_ϛ, we arrive at the following boundary conditions for the functions ${R}_{r\theta}^{\tau 1}$ and ${R}_{r\theta}^{\tau 2}$:

$$ {R}_{r\theta}^{\tau 1}{\left|{}_{r={r}_{\varsigma }}=0,\kern1em {R}_{r\theta}^{\tau 2}\right|}_{r={r}_{\varsigma }}=0. $$

(9)

Inserting Eq. (8) into solutions (6), after some transformations with account of Eqs. (9), we have

$$ {\displaystyle \begin{array}{c}{\sigma}_r={R}_{r\theta}^{\tau 1}\cos \theta -{R}_{r\theta}^{\tau 2}\sin \theta +{R}_r^{\sigma },\\ {}{\sigma}_{\theta }=\frac{1}{r}\frac{d}{dr}\left({r}^2{R}_{r\theta}^{\tau 1}\cos \theta -{r}^2{R}_{r\theta}^{\tau 2}\sin \theta \right)+\frac{d}{dr}\left(r{R}_r^{\sigma}\right),\end{array}} $$

(10)

where

$$ {\left.{R}_r^{\sigma }=-{R}_{r\theta}^{\tau 1}-\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{R}_{r\theta}^{\tau 1} dr+\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{\sigma}_{\theta}\right|}_{\theta =0} dr. $$

(11)

Inserting the second relation of (10) into the third boundary condition of (4), after transformations with account of Eqs. (9), we find boundary conditions for the function ${R}_r^{\sigma }$:

$$ {\left.{R}_r^{\sigma}\right|}_{r={r}_{\varsigma }}=0,\kern1em \zeta =1,2. $$

(12)

Using integral conditions (5), together with Eqs. (8) and (11), the following conditions for the unknown functions are obtained:

$$ {\displaystyle \begin{array}{c}\underset{r_1}{\overset{r_2}{\int }}{R}_{r\theta}^{\tau 1} dr=\frac{1}{b}{N}_{\theta }{\left|{}_{\theta =0},\kern1em \underset{r_1}{\overset{r_2}{\int }}{R}_{r\theta}^{\tau 2} dr=\frac{1}{b}{Q}_r\right|}_{\theta =0},\\ {}\underset{r_1}{\overset{r_2}{\int }}\left(r{R}_r^{\sigma}\right) dr=-\frac{1}{b}\left({r}_C{N}_{\theta }{\left|{}_{\theta =0}+{M}_y\right|}_{\theta =0}\right).\end{array}} $$

(13)

Thus, the further solution of the problem consists in determination of functions ${R}_{r\theta}^{\tau 1}$, ${R}_{r\theta}^{\tau 2}$, and ${R}_r^{\sigma }$, agreeing with boundary and integral conditions (9), (12), and (13). To determine these functions, we use the remaining equations of (3).

Solving the first and second geometrical relations of (3) for the displacements u_r and u_θ, we have

$$ {\left.{u}_r=\underset{r_1}{\overset{r}{\int }}{\varepsilon}_r dr+{u}_r\right|}_{r={r}_1},\kern1em {u}_{\theta }=r\underset{0}{\overset{\theta }{\int }}{\varepsilon}_{\theta } d\theta -\underset{0}{\overset{\theta }{\int }}\underset{r_1}{\overset{r}{\int }}{\varepsilon}_r dr d\theta -\underset{0}{\overset{\theta }{\int }}{u}_r{\left|{}_{r={r}_1} d\theta +{u}_{\theta}\right|}_{\theta =0}. $$

(14)

Inserting Eqs. (14) into the last geometrical relation of (3), leads to the compatibility conditions for strains and displacements

$$ {\displaystyle \begin{array}{c}{r}^2\underset{0}{\overset{\theta }{\int }}\frac{\partial {\varepsilon}_{\theta }}{\partial r} d\theta +\underset{r_1}{\overset{r}{\int }}\frac{\partial {\varepsilon}_r}{\partial \theta } d r+\underset{0}{\overset{\theta }{\int }}\underset{r_1}{\overset{r}{\int }}{\varepsilon}_r d r d\theta -r\underset{0}{\overset{\theta }{\int }}{\varepsilon}_r d\theta \\ {}+\frac{{\left.d{u}_r\right|}_{r={r}_1}}{d\theta}+\underset{0}{\overset{\theta }{\int }}{u}_r{\left|{}_{r={r}_1} d\theta +r\frac{{\left.d{u}_{\theta}\right|}_{\theta =0}}{d r}-{u}_{\theta}\right|}_{\theta =0}=r{\gamma}_{r\theta}.\end{array}} $$

(15)

Expression (15) is the governing equation for solving the problem considered. To put it into a final form, physical relations (3) and the preliminary solutions (8) and (10) for stresses are employed.

Inserting Eqs. (8) and (10) into physical relations (3) leads to relations for the linear and shear strains in the form

$$ {\displaystyle \begin{array}{c}{\varepsilon}_r={R}_r^{\varepsilon 1}\cos \theta +{R}_r^{\varepsilon 2}\sin \theta +{R}_r^{\varepsilon 3},\kern1.1em {\varepsilon}_{\theta }={R}_{\theta}^{\varepsilon 1}\cos \theta +{R}_{\theta}^{\varepsilon 2}\sin \theta +{R}_{\theta}^{\varepsilon 3},\\ {}{\gamma}_{r\theta}=\frac{R_{r\theta}^{\tau 1}\sin \theta }{\mu_{r\theta}^G}+\frac{R_{r\theta}^{\tau 2}\cos \theta }{\mu_{r\theta}^G},\end{array}} $$

(16)

where the following designations for the components of distribution of the linear strains along the height of bar cross section have been introduced:

$$ {\displaystyle \begin{array}{c}{R}_r^{\varepsilon 1}=-\frac{\mu_{\theta r}^v}{\mu_{\theta}^E}\frac{1}{r}\frac{d\left({r}^2{R}_{r\theta}^{\tau 1}\right)}{dr}+\frac{R_{r\theta}^{\tau 1}}{\mu_{\theta}^E},\kern1em {R}_r^{\varepsilon 2}=\frac{\mu_{\theta r}^v}{\mu_{\theta}^E}\frac{1}{r}\frac{d\left({r}^2{R}_{r\theta}^{\tau 2}\right)}{dr}-\frac{R_{r\theta}^{\tau 2}}{\mu_r^E},\\ {}{R}_r^{\varepsilon 3}=-\frac{\mu_{\theta r}^v}{\mu_{\theta}^E}\frac{1}{r}\frac{d\left(r{R}_r^{\sigma}\right)}{dr}+\frac{1}{\mu_r^E}{R}_r^{\sigma },\kern1em {R}_{\theta}^{\varepsilon 1}=\frac{1}{\mu_{\theta}^E}\frac{1}{r}\frac{d\left({r}^2{R}_{r\theta}^{\tau 1}\right)}{dr}-\frac{\mu_{r\theta}^v{R}_{r\theta}^{\tau 1}}{\mu_r^E},\\ {}{R}_{\theta}^{\varepsilon 2}=-\frac{1}{\mu_{\theta}^E}\frac{1}{r}\frac{d\left({r}^2{R}_{r\theta}^{\tau 2}\right)}{dr}+\frac{\mu_{r\theta}^v{R}_{r\theta}^{\tau 2}}{\mu_r^E},\kern1em {R}_{\theta}^{\varepsilon 2}=\frac{1}{\mu_{\theta}^E}\frac{d\left({r}^2{R}_r^{\sigma}\right)}{dr}-\frac{\mu_{r\theta}^v}{\mu_r^E}{R}_r^{\sigma }.\end{array}} $$

Inserting relations (16) into Eq. (15) and performing necessary transformations, we obtain the integrodifferential equation

$$ {\displaystyle \begin{array}{c}\left[{r}^2\frac{d{R}_{\theta}^{\varepsilon 1}}{d r}-r{R}_r^{\varepsilon 1}-\frac{r{R}_{r\theta}^{\tau 1}}{\mu_{r\theta}^G}\right]\sin \theta -\left[{r}^2\frac{d{R}_{\theta}^{\varepsilon 2}}{d r}-r{R}_r^{\varepsilon 2}-\frac{r{R}_{r\theta}^{\tau 2}}{\mu_{r\theta}^G}\right]\cos \theta \\ {}+\left[{r}^2\frac{d}{d r}\left({R}_{\theta}^{\varepsilon 3}-\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 3} d r\right)\right]\theta -\left[{r}^2\frac{d}{d r}\left(-\frac{{\left.{u}_{\theta}\right|}_{\theta =0}}{r}-{R}_{\theta}^{\varepsilon 2}+\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 3} d r\right)\right]\\ {}{\left.+\frac{{\left.d{u}_r\right|}_{r={r}_1}}{d\theta}+\underset{0}{\overset{\theta }{\int }}{u}_r\right|}_{r={r}_1} d\theta =0.\end{array}} $$

(18)

A nontrivial solution of Eq. (18) for the unknown functions is possible only if the expressions in square brackets in it are equal to some constants:

$$ {\displaystyle \begin{array}{c}{r}^2\frac{d{R}_{\theta}^{\varepsilon 1}}{dr}-r{R}_r^{\varepsilon 1}-\frac{r{R}_{r\theta}^{\varepsilon 1}}{\mu_{r\theta}^G}={K}_1,\kern1em {r}^2\frac{d{R}_{\theta}^{\varepsilon 2}}{dr}-r{R}_r^{\varepsilon 2}+\frac{r{R}_{r\theta}^{\varepsilon 2}}{\mu_{r\theta}^G}={K}_2,\\ {}{r}^2\frac{d}{dr}\left({R}_{\theta}^{\varepsilon 3}-\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 3} dr\right)={K}_3,\kern1em {r}^2\frac{d}{dr}\left(-\frac{{\left.{u}_{\theta}\right|}_{\theta =0}}{r}-{R}_{\theta}^{\varepsilon 2}+\frac{1}{r}\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 3} dr\right)={K}_4.\end{array}} $$

(19)

With account of Eqs. (19), Eq. (18) takes the form

$$ {\left.\frac{{\left.d{u}_r\right|}_{r={r}_1}}{d\theta}+\underset{0}{\overset{\theta }{\int }}{u}_r\right|}_{r={r}_1}\; d\theta =-{K}_1\sin \theta +{K}_2\cos \theta -{K}_3\theta +{K}_4. $$

(20)

Thus, the further solution of the problem is reduced to the search for solutions of the ordinary differential equations (19) and (20). Let us consider the basic approaches to finding the general and particular solutions of the given equations.

Inserting relations (17) into the first equation of (19), we arrive at the governing equation for the function ${R}_{r\theta}^{\tau 1}$ in the form

$$ {\displaystyle \begin{array}{c}r\frac{d}{dr}\left[\frac{r}{\mu_{\theta}^E}\frac{d{R}_{r\theta}^{\tau 1}}{dr}+\left(\frac{2}{\mu_{\theta}^E}-\frac{\mu_{r\theta}^v}{\mu_r^E}\right){R}_{r\theta}^{\tau 1}\right]\\ {}+\frac{\mu_{\theta r}^v}{\mu_{\theta}^E}r\frac{d{R}_{r\theta}^{\tau 1}}{dr}+\left(\frac{2{\mu}_{\theta r}^v}{\mu_{\theta}^E}-\frac{1}{\mu_r^E}-\frac{1}{\mu_{r\theta}^G}\right){R}_{r\theta}^{\tau 1}=\frac{K_1}{r}.\end{array}} $$

(21)

Equation (21) is a homogeneous differential equation with coefficients varying stepwise on the borders of layers of the bar. Within the limits of a kth homogeneous layer at r ∈ (r_{db, k − 1}, r_{db, k}) , the given equation is transformed into a special form of the Cauchy–Euler, namely,

$$ {r}^2\frac{d^2{R_{r\theta}^{\tau 1}}^{\left[k\right]}}{d{r}^2}+{\alpha}_1^{\left[k\right]}r\frac{d{R_{r\theta}^{\tau 1}}^{\left[k\right]}}{dr}+{\alpha}_0^{\left[k\right]}{R_{r\theta}^{\tau 1}}^{\left[k\right]}=\frac{K_1{E}_{\theta}^{\left[k\right]}}{r},k=\overline{1,m}, $$

(22)

where

$$ {\displaystyle \begin{array}{cc}{\alpha}_0^{\left[k\right]}=2{\nu}_{\theta r}^{\left[k\right]}-\frac{E_{\theta}^{\left[k\right]}}{E_r^{\left[k\right]}}-\frac{E_{\theta}^{\left[k\right]}}{G_{r\theta}^{\left[k\right]}},& {\alpha}_1^{\left[k\right]}=3+{\nu}_{\theta r}^{\left[k\right]}-\frac{E_{\theta}^{\left[k\right]}{\nu}_{r\theta}^{\left[k\right]}}{E_r^{\left[k\right]}}.\end{array}} $$

(23)

The general solution of Eq. (22) is

$$ {R}_{r\theta}^{\tau 1\left[k\right]}={C}_{11}^{\left[k\right]}{r}^{\kappa_1^{\left[k\right]}}+{C}_{21}^{\left[k\right]}{r}^{\kappa_2^{\left[k\right]}}+\frac{K_1{E}_{\theta}^{\left[k\right]}}{\alpha_0^{\left[k\right]}-{\alpha}_1^{\left[k\right]}+2}\frac{1}{r},k=\overline{1,m}, $$

(24)

where ${C}_{11}^{\left[k\right]}$ and ${C}_{21}^{\left[k\right]}$[ ] are unknown constants of integration; ${\kappa}_1^{\left[k\right]}$ and ${\kappa}_2^{\left[k\right]}$ are roots of the characteristic equation

$$ {\kappa}_{1,2}^{\left[k\right]}=\frac{1}{2}\left(1-{\alpha}_1^{\left[k\right]}\pm \sqrt{{\left({\alpha}_1^{\left[k\right]}-1\right)}^2-4{\alpha}_0^{\left[k\right]}}\right). $$

(25)

We should note that, for the case of an isotropic layer, where ${E}_r^{\left[k\right]}={E}_{\theta}^{\left[k\right]}={E}^{\left[k\right]}$, ${\nu}_{r\theta}^{\left[k\right]}={\nu}_{\theta r}^{\left[k\right]}={\nu}^{\left[k\right]}$, and ${G}_{r\theta}^{\left[k\right]}={G}^{\left[k\right]}={E}^{\left[k\right]}/\left(2+2{\nu}^{\left[k\right]}\right)$, solution (24) is transformed to

$$ {R}_{r\theta}^{\tau 1\left[k\right]}={C}_{11}^{\left[k\right]}r+{C}_{21}^{\left[k\right]}\frac{1}{r^3}-\frac{K_1{E}^{\left[k\right]}}{4}\frac{1}{r}. $$

(26)

The function ${R}_{r\theta}^{\tau 1}$ for all the bar can be collected from solutions (24) in the same way as the functions of elastic constants (2):

$$ {R}_{r\theta}^{\tau 1}=\sum \limits_{k=1}^m{R}_{r\theta}^{\tau 1\left[k\right]}\left[H\left(r-{r}_{bd,k-1}\right)-H\left(r-{r}_{bd,k}\right)\right]. $$

(27)

Similarly to Eq. (21), the second equation of (19), with account of Eqs. (17), takes the following form within the limits of a kth layer:

$$ {r}^2\frac{d^2{R}_{r\theta}^{\tau 2\left[k\right]}}{d{r}^2}+{\alpha}_1^{\left[k\right]}r\frac{d{R}_{r\theta}^{\tau 2\left[k\right]}}{dr}={\alpha}_0^{\left[k\right]}{R}_{r\theta}^{\tau 2\left[k\right]}=-\frac{K_2{E}_{\theta}^{\left[k\right]}}{r},k=\overline{1,m}. $$

(28)

The general solution of Eq. (28) is similar to solutions (24) and (26) for orthotropic and isotropic layers, respectively:

$$ {R}_{r\theta}^{\tau 2\left[k\right]}={C}_{12}^{\left[k\right]}{r}^{\kappa_1^{\left[k\right]}}+{C}_{22}^{\left[k\right]}{r}^{\kappa_2^{\left[k\right]}}-\frac{K_2{E}_{\theta}^{\left[k\right]}}{\alpha_0^{\left[k\right]}-{\alpha}_1^{\left[k\right]}+2}\frac{1}{r},\kern0.36em k=\overline{1,m}, $$

(29)

$$ {R}_{r\theta}^{\tau 2\left[k\right]}={C}_{12}^{\left[k\right]}r+{C}_{22}^{\left[k\right]}\frac{1}{r^3}+\frac{K_2{E}^{\left[k\right]}}{4}\frac{1}{r},\kern0.36em k=\overline{1,m}, $$

(30)

For all the package of layers, the solution is written in the form

$$ {R}_{r\theta}^{\tau 2}=\sum \limits_{k=1}^m{R}_{r\theta}^{\tau 2\left[k\right]}\left[H\left(r-{r}_{bd,k-1}\right)-H\left(r-{r}_{bd,k}\right)\right]. $$

(31)

We should note that the function ${R}_{r\theta}^{\tau 2}$ determines the stress components caused by the tangential load. As can be seen, in solution (30) for an isotropic layer, the fundamental system is similar to the solution of the problem on bending of an isotropic circular bar by a transverse force [16]. To some part, this confirms the validity of the solution construc ted.

Integrating the third equation of (19), we have

$$ r{R}_{\theta}^{\varepsilon 3}-\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 3} dr=-{K}_3+r{K}_5. $$

(32)

We transform Eq. (32) to the form

$$ {\left.r\left({R}_{\theta}^{\varepsilon 3}-{K}_5\right)-\int {R}_r^{\varepsilon 3} dr=-\Big({K}_3+\left(\int {R}_r^{\varepsilon 3} dr\right)\right|}_{r={r}_1}\Big)=\mathrm{const}. $$

(33)

Integrodifferential equation (33) with account of Eqs. (17) can have a solution only if ${\left.{K}_3+\left(\int {R}_r^{\varepsilon 3} dr\right)\right|}_{r={r}_1}$ whence

$$ {\left.{K}_3=-\frac{1}{E_r^{\left[1\right]}}\left(\int {R}_r^{\sigma } dr\right)\right|}_{r={r}_1}. $$

(34)

Inserting Eqs. (17) into Eq. (32) with account of Eq. (34), it becomes

$$ r\left(\frac{1}{\mu_{\theta}^E}\frac{d}{dr}\left(r{R}_r^{\sigma}\right)-\frac{\mu_{r\theta}^{\nu }}{\mu_r^E}{R}_r^{\sigma}\right)-\int \left(\frac{1}{\mu_r^E}{R}_r^{\sigma }-\frac{\mu_{\theta r}^{\nu }}{\mu_{\theta}^E}\frac{d}{dr}\left(r{R}_r^{\sigma}\right)\right)\; dr={K}_5r. $$

(35)

Differentiating Eq. (35), we obtain a linear differential equation that, within the limits of any kth layer, take the form

$$ {r}^2\frac{d^2}{d{r}^2}\left(r{R}_r^{\sigma \left[k\right]}\right)+{\beta}_1^{\left[k\right]}r\frac{d}{dr}\left(r{R}_r^{\sigma \left[k\right]}\right)+{\beta}_0^{\left[k\right]}\left(r{R}_r^{\sigma \left[k\right]}\right)={K}_5{E}_{\theta}^{\left[k\right]}r,k=\overline{1,m}, $$

(36)

where

$$ {\displaystyle \begin{array}{cc}{\beta}_0^{\left[k\right]}=-\frac{E_{\theta}^{\left[k\right]}}{E_r^{\left[k\right]}},& {\beta}_1^{\left[k\right]}=1+{\nu}_{\theta r}^{\left[k\right]}\frac{E_{\theta}^{\left[k\right]}{\nu}_{r\theta}^{\left[k\right]}}{E_r^{\left[k\right]}}.\end{array}} $$

(37)

The general solution of Eq. (36) is found to be

$$ {R}_r^{\sigma \left[k\right]}={C}_{13}^{\left[k\right]}{r}^{\kappa_3^{\left[k\right]}-1}+{C}_{23}^{\left[k\right]}{r}^{\kappa_4^{\left[k\right]}-1}+\frac{K_5{E}_{\theta}^{\left[k\right]}}{\beta_0^{\left[k\right]}+{\beta}_1^{\left[k\right]}},\kern0.36em k=\overline{1,m}, $$

(38)

where ${\kappa}_3^{\left[k\right]}$ and ${\kappa}_4^{\left[k\right]}$ are roots of the characteristic equation

$$ {\kappa}_{3,4}^{\left[k\right]}=\frac{1}{2}\left[1-{\beta}_3^{\left[k\right]}\pm \sqrt{{\left({\beta}_1^{\left[k\right]}-1\right)}^2-4{\beta}_0^{\left[k\right]}}\right]. $$

(39)

We should note that, for an isotropic layer, the form of the solution varies, as ${\beta}_{0,1}^{\left[k\right]}=\mp 1$ in this case, to which there corresponds the solution

$$ {R}_r^{\sigma \left[k\right]}={C}_{13}^{\left[k\right]}+{C}_{23}^{\left[k\right]}\frac{1}{r^2}+\frac{K_5{E}_{\theta}^{\left[k\right]}}{2}\ln r. $$

(40)

According to Eqs. (10) and (13), the function ${R}_r^{\sigma }$ is a component of distribution of the normal stresses σ_θ and σ_r caused by the resultant of moment of the normal load at the beam end Τ₁. By the corresponding transformations (10) and (40) the known Golovin solution for the pure bending of a circular bar [16] can be obtained, which once again indirectly confirms the validity of the solution found.

For all the bar, the solution for the function ${R}_r^{\sigma }$ is similar to (27):

$$ {R}_r^{\sigma }=\sum \limits_{k=1}^m{R}_r^{\sigma \left[k\right]}\left[H\left(r-{r}_{bd,k-1}\right)-H\left(r-{r}_{bd,k}\right)\right]. $$

(41)

We should note that Eqs. (19) also remain unchanged in the case where layer materials are continuously inhomogeneous in the radial direction: ${S}_a^{\left[k\right]}={S}_a^{\left[k\right]}(r)$. However, in this case, their solutions in a finite form can be obtained only in special cases of ${S}_a^{\left[k\right]}(r)$, which, with reference to a homogeneous circular bar are considered in [10].

The solution of the last equation of (19) can be found for all the package of bar layers by a direct integration. With the use of Eqs. (17), the solution of the given equation takes the form

$$ {\left.{u}_{\theta}\right|}_{\theta =0}=-r{R}_{\theta}^{\varepsilon 2}+\underset{r_1}{\overset{r}{\int }}{R}_r^{\varepsilon 2} dr+r{K}_6+{K}_4. $$

(42)

To solve Eq. (20), we introduce the replacement

$$ {\left.\underset{0}{\overset{\theta }{\int }}{u}_r\right|}_{r={r}_1} d\theta ={R}^u, $$

(43)

and Eq. (20) becomes

$$ \frac{d{R}^u}{d\theta}+{R}^u=-{K}_1\sin \theta +{K}_2\cos \theta -{K}_3\theta +{K}_4. $$

(44)

We should note that replacement (43) suggested the natural condition R^u|_θ = 0 = 0 for the function R^u, in view of which the general solution of Eq. (44) is

$$ {R}^u=\frac{1}{2}{K}_1\theta \cos \theta +\frac{1}{2}{K}_2\theta \sin \theta -{K}_3\theta +{K}_4\left(1-\cos \theta \right)+{K}_7\sin \theta . $$

(45)

Differentiating Eq. (45) with account of Eqs. (43) and (34) leads to the relation

$$ {\displaystyle \begin{array}{c}{\left.{u}_r\right|}_{r={r}_1}=\frac{1}{2}{K}_1\left(\cos \theta -\theta \sin \theta \right)+\frac{1}{2}{K}_2\left(\cos \theta +\theta \sin \theta \right)\\ {}{\left.+\frac{1}{E_r^{\left[k\right]}}\left(\int {R}_r^{\sigma \left[1\right]} dr\right)\right|}_{r={r}_1}+{K}_4\sin \theta +{K}_7\cos \theta .\end{array}} $$

(46)

Thus, the general solutions for all the unknown functions ${\left.{R}_{r\theta}^{\tau 1},{R}_{r\theta}^{\tau 2},{R}_r^{\sigma },{u}_{\theta}\right|}_{\theta =0}$, and ${\left.{u}_r\right|}_{r={r}_1}$ of the problem considered have been found. These solutions, together with Eqs. (8), (10), (14), (16), and (17), allow one to determine relations for all SSS components of the multilayer bar considered. However, they will contain 6m+ 6 unknown constants of integration. A part of them can be found from boundary conditions (9) and (12) and integral conditions (13). The others are determined from the condition of absolutely rigid contact between layers and kinematic conditions at the fixed bar end.

$$ {\displaystyle \begin{array}{c}\begin{array}{cc}{R}_{r\theta}^{\tau 1\left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_{r\theta}^{\tau 1\left[k+1\right]}\right|}_{r={r}_{db,k}},& {R}_{r\theta}^{\tau 2\left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_{r\theta}^{\tau 2\left[k+1\right]}\right|}_{r={r}_{db,k}},\end{array}\\ {}{R}_r^{\sigma \left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_r^{\sigma \left[k+1\right]}\right|}_{r={r}_{db,k}},k=\overline{1,m-1}.\end{array}} $$

(47)

We require that the displacements u_r and u_θ also be continuous on the interfaces of layers P_k. According to the first ratio of (14), the solution for u_r is continuous at any values of unknown constants in the solutions for ${R}_{r\theta}^{\tau 1},{R}_{r\theta}^{\tau 2}$, and ${R}_r^{\sigma }$, and the solution for u_θ, in view of the last equation of (19), is continuous if continuous are ε_θ and ${\left.{u}_{\theta}\right|}_{u_{\theta }=0}$ :

$$ {\varepsilon}_{\theta}^{\left[k\right]}{\left|{}_{r={r}_{db,k}}={\varepsilon}_{\theta}^{\left[k+1\right]}\right|}_{r={r}_{db,k}},k=\overline{1,m-1} $$

(48)

Inserting Eqs.(16) into Eqs. (48) and comparing the coefficient of identical functions of θ, it was established that

$$ {\displaystyle \begin{array}{c}\begin{array}{cc}{R}_{\theta}^{\varepsilon 1\left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_{\theta}^{\varepsilon 1\left[k+1\right]}\right|}_{r={r}_{db,k}},& {R}_{\theta}^{\varepsilon 2\left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_{\theta}^{\varepsilon 2\left[k+1\right]}\right|}_{r={r}_{db,k}},\end{array}\\ {}{R}_{\theta}^{\varepsilon 3\left[k\right]}{\left|{}_{r={r}_{db,k}}={R}_{\theta}^{\varepsilon 3\left[k+1\right]}\right|}_{r={r}_{db,k}},k=\overline{1,m-1}.\end{array}} $$

(49)

According to Eq. (42), continuity of the function ${\left.{u}_{\theta}\right|}_{u_{\theta }=0}$ is fulfilled automatically if condition (49) holds.

Solutions (24) and (29), together with conditions (9), (47), and (49) and relations (17), show that the functions ${R}_{r\theta}^{\tau 1}$ and ${R}_{r\theta}^{\tau 2}$ are linearly dependent, i.e.,

$$ {R}_{r\theta}^{\tau 1}=\psi {R}_{r\theta}^{\tau 2}, $$

(50)

where ψ is a constant.

Applying Eq. (50) to the first two integral conditions (13), we have

$$ \psi ={Q}_r{\left|{}_{\theta =0}/{N}_{\theta}\right|}_{\theta =0}. $$

(51)

Relations (50) and (51) reduce the number of unknown constants by one third, to 4m+ 5 . As a result, only one of the functions ${R}_{r\theta}^{\tau 1}$ and ${R}_{r\theta}^{\tau 2}$ has to be determined. However, it is necessary to take into account some loading cases at which Eq. (50) loses its meaning:

• if N_θ|_θ = 0 = 0, then $\psi =\infty \Rightarrow {R}_{r\theta}^{\tau 1}=0\wedge {R}_{r\theta}^{\tau 2}\ne 0$;

• if Q_r|_θ = 0 = N_θ|_θ = 0 = 0, then $\psi =0/0\Rightarrow {R}_{r\theta}^{\tau 1}={R}_{r\theta}^{\tau 2}\ne 0$.

Conditions (9), (12), (13), (47), and (49), with account of Eqs. (50) and (51), allow one to find all unknown constants for the functions ${R}_{r\theta}^{\tau 1},{R}_{r\theta}^{\tau 2}$, and ${R}_r^{\sigma }$. The remaining constants K₄,K₆, and K₇ are determined from the kinematic conditions at the beam end Τ₂.

For the case considered, where the work of the circular cantilever is modeled, the fastening can be ensured forbidding circumferential displacements u_θ for two points of the end Τ₂ and the radial displacement u_r of one of its points (Fig. 2), for example,

Fig. 2.

Modeling the fastening of the cantilever bar.

$$ {\displaystyle \begin{array}{ccc}{\left.{u}_r\right|}_{\theta ={\theta}_2,r={r}_1}=\mid 0,& {\left.{u}_{\theta}\right|}_{\theta ={\theta}_2,r={r}_1}=\mid 0,& {\left.{u}_{\theta}\right|}_{\theta ={\theta}_2,r={r}_2}=\mid 0,\end{array}} $$

(52)

From the viewpoint of static conditions, possible are also other combination of fixed points of the end or the longitudinal sides, but their analysis is already the subject of investigation of applied applications of the solution found.

Thus, we have obtained a system of analytical relations (8), (10), (14), (16), (17), (24), (27), (29), (31), (38), (41), (42), and (46) and conditions (9), (12), (13), (47), (49), (50), (51), and (52) that allows one to determine all SSS components for the problem on bending of a circular multilayer bar by normal and tangential loads at its ends. This solution is exact for the corresponding plane problem of elasticity theory, provided that the loads on the end Τ₁ are distributed according to the relations found, namely,

$$ {\displaystyle \begin{array}{cc}{p}_{\theta}^{{\mathrm{T}}_1}=-\left[\frac{1}{r}\frac{d}{dr}\left({r}^2{R}_{r\theta}^{\tau 1}\right)+\frac{d}{dr}\left(r{R}_r^{\sigma}\right)\right],& {p}_r^{{\mathrm{T}}_1}={R}_{r\theta}^{\tau 2}\end{array}} $$

(53)

For other types of load distribution, according to the Saint Venant principle, the solution of the problem will differ from the exact decision only the on a small distance from the loaded bar end.

The relations obtained for SSS components become considerably simpler if layer materials are not compliant to the strains of transverse shear and compression, ${E}_r^{\left[k\right]},{G}_{r\theta}^{\left[k\right]}\to \infty$, ${\nu}_{r\theta}^{\left[k\right]},{\nu}_{\theta r}^{\left[k\right]}\to 0$, $k=\overline{1,m}$, which corresponds to the hypothesis of plane cross sections of the engineering theory of bending. In that case, solutions for the governing functions ${R}_{r\theta}^{\tau 1},{R}_{r\theta}^{\tau 2}$, and Rr σ with account of boundary conditions (9) and (12) and integral conditions (13), take the form

$$ {\displaystyle \begin{array}{c}{R}_{r\theta}^{\tau 1}=-\frac{F_{\theta }}{b{D}_{-1}}\frac{1}{r^2}\left(\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr-\frac{B_0}{B_1}\underset{r_1}{\overset{r}{\int }}\left({\mu}_{\theta}^Er\right) dr\right),\\ {}{R}_{r\theta}^{\tau 2}=-\frac{F_r}{b{D}_{-1}}\frac{1}{r^2}\left(\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr-\frac{B_0}{B_1}\underset{r_1}{\overset{r}{\int }}\left({\mu}_{\theta}^Er\right) dr\right),\\ {}{R}_r^{\sigma }=-\frac{r_C{F}_{\theta }+M}{b{D}_0}\frac{1}{r}\left(\underset{r_1}{\overset{r}{\int }}\frac{\mu_{\theta}^E}{r} dr-\frac{B_{-1}}{B_0}\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr\right),\\ {}{\left.{u}_{\theta}\right|}_{\theta =0}=-\frac{F_r}{b{D}_{-1}}\left(1-\frac{B_0}{B_r}r\right)+{K}_4-{C}_7r,\\ {}{\left.{u}_r\right|}_{r={r}_1}=\frac{F_{\theta}\left(\cos \theta -\sin \theta \right)}{2b{D}_{-1}}-\frac{F_{\theta}\left(\theta \cos \theta +\sin \theta \right)}{2b{D}_{-1}}-\frac{r_C{F}_{\theta }+M}{b{D}_0}+{K}_4\sin \theta +{K}_7\cos \theta, \end{array}} $$

(54)

where F_r, F_θ, and M are resultant of the loads ${p}_r^{T1}$ and ${p}_{\theta}^{T1}$ , reduced to the bar axis (see Fig. 1). In this case, stress relations (8) and (10) remain unchanged, but displacement relations (14) become

$$ {\displaystyle \begin{array}{c}{\left.{u}_r\right|}_{r={r}_1}=\frac{F_{\theta}\left(\cos \theta -\sin \theta \right)}{2b{D}_{-1}}-\frac{F_r\left(\theta \cos \theta +\sin \theta \right)}{2b{D}_{-1}}-\frac{r_C{F}_{\theta }+M}{b{D}_0}+{K}_4\sin \theta +{K}_7\cos \theta, \\ {}{u}_{\theta }=r\left[\frac{-{F}_{\theta}\sin \theta +{F}_r\left(1-\cos \theta \right)}{b{D}_{-1}}\left(\frac{1}{r}-\frac{B_0}{B_1}\right)+\frac{r_C{F}_{\theta }+M}{b{D}_0}\left(\frac{1}{r}-\frac{B_{-1}}{B_0}\right)\theta \right]-\underset{0}{\overset{\theta }{\int }}{u}_r{\left|{}_{r={r}_1} d\theta +{u}_{\theta}\right|}_{\theta =0}.\end{array}} $$

(55)

In relations (54) and (55), the following designations for the definite integrals have been used:

$$ {\displaystyle \begin{array}{c}\begin{array}{ccc}{B}_0=\underset{r_1}{\overset{r_2}{\int }}{\mu}_{\theta}^E dr,& {B}_1=\underset{r_1}{\overset{r_2}{\int }}\left({\mu}_{\theta}^Er\right) dr,& {B}_{-1}=\underset{r_1}{\overset{r_2}{\int }}\frac{\mu_{\theta}^E}{r} dr,\end{array}\\ {}{D}_{-1}=\underset{r_1}{\overset{r_2}{\int }}\left(\frac{1}{r^2}\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr-\frac{B_0}{B_1}\frac{1}{r^2}\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr\right) dr,\\ {}{D}_0=\underset{r_1}{\overset{r_2}{\int }}\left(\frac{1}{r_1}\underset{r_1}{\overset{r}{\int }}\frac{\mu_{\theta}^E}{r} dr-\frac{B_{-1}}{B_0}\underset{r_1}{\overset{r}{\int }}{\mu}_{\theta}^E dr\right) dr.\end{array}} $$

(56)

As is seen, if the stresses of transverse shear and compression are neglected, unknown remain only three constants of integration, which simplifies the use of the solution obtained. But this can lead to significant discrepancies in determining SSS of the bar if values of the ratios ${E}_{\theta}^{\left[k\right]}/{E}_r^{\left[k\right]}$ and ${E}_{\theta}^{\left[k\right]}/{G}_{r\theta}^{\left[k\right]}$ of its layer materials are great.

Example of realization of the relations found

Let us consider an application of the received solution found with an example of a composite multilayer cantilever in the form of semiring (Fig. 3). The radius of curvature of fibers on the internal longitudinal surface of the bar is r₁ = 50 mm. The resultant the normal and tangential loads at the bar end (Fig. 3), reduced to the geometrical center C (r_C = 80 mm), are F_r = 750 N and F_θ = 2750 N; M = 450 N · m.

Fig. 3.

Schematic of the circular bar (a) and structure of its cross section (b) (dimensions in millimeters).

The mechanical characteristics of layer materials of the bar (Fig. 3b) are as follows:

• glass-fiber plastic (P₁, P₃) ${E}_{\theta}^{\left[k\right]}={E}_{\theta}^{\left[3\right]}=36.8\ \mathrm{GPa},{E}_r^{\left[k\right]}={E}_r^{\left[3\right]}=11.0\ \mathrm{GPa},{G}_{r\theta}^{\left[k\right]}={G}_{r\theta}^{\left[3\right]}=4.5\ \mathrm{GPa},{\nu}_{\theta r}^{\left[k\right]}={\nu}_{\theta r}^{\left[3\right]}=0.351$, and ${\nu}_{\theta r}^{\left[k\right]}={\nu}_{\theta r}^{\left[3\right]}=0.105$;

• FBS plywood (P₂): ${E}_{\theta}^{\left[2\right]}=14.8\ \mathrm{GPa},{E}_r^{\left[2\right]}=1.13\ \mathrm{GPa},{G}_{\theta r}^{\left[2\right]}=1.35\ \mathrm{GPa},{\nu}_{\theta r}^{\left[2\right]}=0.445,$ and ${\nu}_{r\theta}^{\left[2\right]}=0.034$;

• aluminum alloy (P4): ${E}_{\theta}^{\left[4\right]}={E}_r^{\left[4\right]}=70\ \mathrm{GPa},{G}_{r\theta}^{\left[4\right]}=26.9\ \mathrm{GPa}$, and ${\nu}_{\theta r}^{\left[4\right]}={\nu}_{r\theta}^{\left[4\right]}=0.034$

We should note that the solutions found do not directly contain conditions for determining the radius r_C of bar axis. Therefore, this quantity can be considered arbitrary, but the static equivalence of resultants has to be obeyed when the point of reduction, C, is changed.

Realization of the solution obtained for determining SSS of the circular cantilever required the following basic steps. First, for the section shown on Fig. 3, according to Eq. (1), functions for the mechanical characteristics ${\mu}_a^S$ were generated:

$$ {\displaystyle \begin{array}{c}{\mu}_a^S={\mu}_a^{\left[1\right]}\left[H\left(r-0.050\right)-H\left(r-0.054\right)\right]+{S}_a^{\left[2\right]}\left[H\left(r-0.054\right)-H\left(r-0.103\right)\right]\\ {}+{S}_a^{\left[3\right]}\left[H\left(r-0.103\right)-H\left(r-0.107\right)\right]+{S}_a^{\left[4\right]}\left[H\left(r-0.107\right)-H\left(r-0.110\right)\right],\end{array}} $$

where ${S}_a^{\left[k\right]}={E}_r^{\left[k\right]},{E}_{\theta}^{\left[k\right]},{G}_{r\theta}^{\left[k\right]},{\nu}_{r\theta}^{\left[k\right]}$, and ${\nu}_{\theta r}^{\left[k\right]},k=\overline{1,4}.$

Further, coefficients (23) and (37) and the roots (25) and (39) of characteristic equations for all bar layers were determined, with use of which the general solutions for the governing functions ${R}_{r\theta}^{\tau 1\left[k\right]}$ and ${R}_{r\theta}^{\tau 2\left[k\right]}$, and ${R}_r^{\sigma \left[k\right]}$ were found. For the orthotropic layers, relations (24), (29), and (38) were used, and for the isotropic ones — (26), (30), and (40). Inserting the solutions found into conditions (47), (49), (9), (12), and (13), the 4m+ 5 unknown constants for the functions ${R}_{r\theta}^{\tau 1\left[k\right]}$ and ${R}_r^{\sigma \left[k\right]}$ were detemined.

The solutions ${R}_{r\theta}^{\tau 1\left[k\right]}$ and ${R}_r^{\sigma \left[k\right]}$, with the use of Eqs. (27), (29), and (41), were collected in the functions ${R}_{r\theta}^{\tau 1}$ and ${R}_r^{\sigma }$ for all the package of bar layers, and, according to Eqs. (50) and (51), the function ${R}_{r\theta}^{\tau 2}$ was found. Graphs of the functions ${R}_{r\theta}^{\tau 1}$ and ${R}_{r\theta}^{\tau 2}$, and ${R}_r^{\sigma }$ are given on Fig. 4.

Fig. 4.

Graphs of governing functions.

Inserting the functions ${R}_{r\theta}^{\tau 1},{R}_{r\theta}^{\tau 2}$, and ${R}_r^{\sigma }$ into solutions (8) and (10), the distribution functions of shear τ_rθ and normal σ_r and σ_θ stresses were determined, whose graphs in some sections are shown on Fig. 5.

Fig. 5.

Distribution graphs of elastic modulus and SSS components. Explanations in the text.

For clearnessness, the graphs of the longitudinal elastic modulus ${\mu}_{\theta}^E$ is shown to the left of stress distribution graphs. The dashed lines designate the curves constructed with use of solutions (54) for the case where layer materials are not compliant to the strains of transverse shear and compression.

The distribution graphs of stresses along the bar height (see Fig. 5) indicate the continuity conditions for the required functions σ_r and τ_rθ on layer interfaces are obeyed. The stresses σ_θ change stepwise on them, but within the limits of a layer, they, as expected, exhibit a variation of hyperbolic type. We should note that neglecting the strains of transverse shear and compression affects the qualitative and quantitative character of distribution of the stresses σ_r and τ_rθ only weakly. But it affects the normal circumferential stresses σ_θ rather significantly. In particular, for the internal (r₁ = 0.05 m) and external (r₂ = 0.11 m) fibers in the initial section (θ = 0), the refinement reaches 15 and 27%, respectively, and in the fixed section — 198 and 17%, respectively.

When SSS of the bar had been determined, its components were found with account of relations (16) and (17), which allowed us to determine the general solution for displacements (42), (46), and (14). The unknown constants K₄, K₆, and K₇ were found with the help of kinematic conditions (52).

The graphs of displacements for some sections are depicted on Fig. 6, where the dashed lines show the distributions of displacements according to Eqs. (54) and (55) in the case of neglecting the stresses transverse shear and compression. In contrast to the stress graphs, the graphs on Fig. 6 point to a significant dependence, up to the change of sign, of radial and circumferential displacements on the shear and compression strains. In addition, the curving (deplanation) of cross sections of the bar is observed, which grows with distance from the loaded end Τ₁, when the fastening is approached.

Fig. 6.

Graphs of displacements in the loaded and fixed sections. Explanations in the text.

The effect of transverse shear and compression strains on the overall strain state of the bar can be estimated by the data of Fig. 7, where the continuous lines show the full displacements, magnified 50×, for points of the external surface and layers interfaces, and the dashed ones — their positions before deformation. These data testify to the significant effect of the shear and compression strains on both the magnitude and character of distribution of full displacements. The distinction for the maximum full displacement of external longitudinal full displacements fibers in the initial (θ = 0) section is 103%.

Fig. 7.

Graphs of full displacements of bar surfaces (50× magnification): solution using the hypothesis of plane cross sections (a) and the exact solution (b).

Conclusion

Thus, the solution of the problem on planar bending of a multilayer cantilever bar with a circular axis under the action normal and a tangent loads at its free end has been obtained in the case of elastic deformation of layer materials. The relations (8), (10), (14), (16), (17), (24), (26), (29), (30), (38), (40), (42), and (46) found for SSS components make an exact solution of the problem of elasticity theory if the distribution of external loads obey relations (53).

The solution was realized for the case of bending of a four-layer cantilever bar with isotropic and orthotropic layers. The results obtained show that the distributions of radial normal and shear stresses differ insignificantly from those found using the hypothesis of plane cross sections. But the maximum magnitude and distribution character of the circumferential normal stresses, as well of strains and displacements, greatly depend on the compliance of layer materials and the strains of transverse shear and compression.

The solution constructed allows one to solve problems on predicting the strength and rigidity of multilayer curvilinear bars with a circular axis, their optimum design, and also open the possibility of solving more complex problems of deformation and development of applied calculation procedures for composite elements.