Elasticity Theory Solution of the Problem on Plane Bending of a Narrow Layered Cantilever Beam by Loads at Its Free End

An exact elasticity theory solution for the problem on plane bending of a narrow layered composite cantilever beam by tangential and normal loads distributed on its free end is presented. Components of the stress-strain state are found for the whole layers package by directly integrating differential equations of the plane elasticity theory problem by using an analytic representation of piecewise constant functions of the mechanical characteristics of layer materials. The continuous solution obtained is realized for a four-layer beam with account of kinematic boundary conditions simulating the rigid fixation of its one end. The solution obtained allows one to predict the strength and stiffness of composite cantilever beams and to construct applied analytical solutions for various problems on the elastic bending of layered beams.

Introduction

Determination of the stresses and strains caused by external loads on elastic heterogeneous bodies of a complex structure by integrating elasticity theory equations is an insoluble problem in most cases. This has led to the development of numerical modeling, various refined analytical theories of strength of materials, and simplified elasticity theory solutions for calculating composite materials and structures. The development of such approximate theories of bending of beams, plates, and shells is described in many scientific publications including monographs and fundamental reviews [1,2,3,4,5,6,7,8,9,10,11]. According to review [9], the iterative analytical theory has achieved a special evolution in the mechanics of layered composite systems, in particular, for inhomogeneous composite (layered) beams [5]. For certain problems, this theory gives a practically exact description of the stress-strain state (SSS).

The number of exact solutions of elasticity theory equations of problems on bending composite and anisotropic beams is limited in the scientific literature. For example, some exact solutions of bending problems for composite beams are presented in [12,13,14,15,16], which have not found widespread practical applications due to the limited set of combinations of possible boundary conditions and structures of composites considered.

The numerical modeling is highly flexible and sufficiently accurate, but it does not allow one to establish a functional relationship between the components of SSS. This limits its employment for creating new composite systems, where an analytical modeling is more convenient, but less developed theoretically, and significantly narrows the range of solvable practical problems.

Problem Statement

Let us consider a straight layered beam with a structure and cross-sectional dimensions constant along its length. The composite beam, having a total height h and width b , is a package of layers (Fig. 1) made of dissimilar materials. The coordinate plane XOZ coincides with the symmetry plane of the beams and is its principal plane of stiffness.

Fig. 1.

Schematic of loads and structure of longitudinal (a) and transverse (b) sections of the layered composite bar.

Each kth layer of thickness ${t}_k={z}_{t_{k+1}}-{z}_{t_k}$ of the beam is made of a homogeneous transversely isotropic material whose isotropy plane coincides with its cross section and has constant mechanical properties over its volume. The elastic characteristics of the whole beam are piecewise constant functions of the coordinate z , which can be represented as the sum of finite functions

$$ {\mu}_a^S=\sum \limits_{k-1}^n\left({S}_a^{\left[k\right]}{p}_k\right), $$

(1)

where $\left\Vert {S}_a^{\left[k\right]}\right\Vert =\left\Vert {E}_x^{\left[k\right]},{E}_{yz}^{\left[k\right]},{G}_{zx}^{\left[k\right]},{G}_{zy}^{\left[k\right]},{v}_{zy}^{\left[k\right]},{v}_x^{\left[k\right]}\right\Vert$ are the elastic constants of a kth layer of the beam; p _k  = p _k (z) is the indicator function of a kth layer of the beam,

$$ {p}_k=\left\{\begin{array}{l}1,\kern1em z\in \left({z}_{t_k},{z}_{t_{k+1}}\right),\\ {}0,\kern1em z\notin \left({z}_{t_k},{z}_{t_{k+1}}\right).\end{array}\right. $$

(2)

We assume that the contact between beam layers is absolutely rigid, i.e., with no mutual displacement or separations.

All surfaces of the beam, except its ends, are free from external loads. To its end x = 0 , uniformly distributed tangential q_xz1 = q_xz1 (z) and normal q_x1 = q_x1 (z) loads (see Fig. 1a) are applied. We assume that the materials of all layers operate in the elastic stage under the action of the load applied. In such a loading, the following static boundary conditions have to be satisfied at the free end of the beam:

$$ {\displaystyle \begin{array}{ll}{\left.{\sigma}_x\right|}_{x={x}_1}=-{q}_{x1},& {\left.{\tau}_{xz}\right|}_{x={x}_1}=-{q}_{xz1},\\ {}{\left.{\sigma}_z\right|}_{\begin{array}{l}z={z}_1,\\ {}z={z}_2\end{array}}=0,& {\left.{\tau}_{zx}\right|}_{\begin{array}{l}z={z}_1,\\ {}z={z}_2\end{array}}=0.\end{array}} $$

(3)

Integrating the first and second relations of (3) over the beam cross section we obtain the integral form of static conditions at its end x = 0

$$ \underset{A}{\int}\left({\left.{\sigma}_x\right|}_{x=0}\right) dA={N}_{x1},\kern0.5em \underset{A}{\int}\left({\left.{\tau}_{xz}\right|}_{x=0}\right) dA=-{Q}_{z1},\kern0.5em \underset{A}{\int}\left({\left.{\sigma}_x\right|}_{x=0}z\right) dA=-{M}_{y1}, $$

(4)

where N_x1 , Q_z1 , and M_y1 are the internal forces and moment in the initial cross section.

At the beam end x = l , kinematic conditions that correspond to a certain type of end fixation (not necessarily absolutely rigid one) and is sufficient to ensure the static equilibrium of the beam are given.

The width b of beam cross section is so small compared with the total height h of the layer package that the uneven distribution of stresses, strains, and displacements across the width of beam cross sections can be neglected. In view of the absence of external loads on beam sides, the stress -strain state of the beam can be considered plane.

Thus, we have the problem on determination of the stress-strain state of a layered composite cantilever beam under the conditions of plane bending by a load applied to its free end and elastic behavior of layer materials.

General solution of the problem

The assumptions about the behavior of layer materials and the distribution of SSS components across the width of beam cross sections make it possible to use the known system of equations of the plane problem of linear elasticity theory for solving the problem considered.

Solving the system of equilibrium equations of the plane problem for the normal stresses, with account of boundary conditions (3), we obtain

$$ {\displaystyle \begin{array}{cc}{\sigma}_x=-\underset{0}{\overset{x}{\int }}\frac{\partial {\tau}_{zx}}{\partial z} dx+{\left.{\sigma}_x\right|}_{x=0},& {\sigma}_z=-\underset{z_1}{\overset{z}{\int }}\frac{\partial {\tau}_{xz}}{\partial x} dz.\end{array}} $$

(5)

In relations (5), the static boundary conditions at the loaded end of the beam are not considered, because, at this stage, it would be impossible to satisfy the compatibility conditions for displacements.

Since the transverse force Q _z in this problem is constant along the beam, it can be assumed that the required function of tangential stresses does not depend on the coordinate x , namely,

$$ {\tau}_{xz}={\tau}_{xz}(z). $$

(6)

With account of Eq. (6), system (5) can be transformed to the form

$$ {\displaystyle \begin{array}{cc}{\sigma}_x=-\frac{d{\tau}_{xz}}{dz}x+{\left.{\sigma}_x\right|}_{x=0},& {\sigma}_z=0.\end{array}} $$

(7)

Employing Hooke’s law and considering Eqs. (1), (6), and (7), the normal and shear strains are written as

$$ {\displaystyle \begin{array}{ccc}{\varepsilon}_x=-\frac{1}{\mu_x^E}\left(\frac{d{\tau}_{xz}}{dz}x-{\left.{\sigma}_x\right|}_{x=0}\right),& {\varepsilon}_z=\frac{\mu_x^{\nu }}{\mu_x^E}\left(\frac{d{\tau}_{xz}}{dz}x-{\left.{\sigma}_x\right|}_{x=0}\right),& {\gamma}_{xz}=\frac{\tau_{xz}}{\mu_{xz}^G}.\end{array}} $$

(8)

Using the first and second relations of (8) in the corresponding Cauchy relations, the relations for displacements are obtained:

$$ {\displaystyle \begin{array}{cc}u=\frac{1}{\mu_x^E}\left(-\frac{d{\tau}_{xz}}{dz}\frac{x}{2}+{\left.{\sigma}_x\right|}_{x=0}\right)x+{\left.u\right|}_{x=0},& w=\underset{z_1}{\overset{z}{\int }}\frac{\mu_x^{\nu }}{\mu_x^E}\left(\frac{d{\tau}_{xz}}{dz}x-{\left.{\sigma}_x\right|}_{x=0}\right) dz+{\left.w\right|}_{z={z}_1}.\end{array}} $$

(9)

Inserting Eqs. (9) and the third relation of (8) into the Cauchy relation for the shear strains γ _xz , we obtain the governing equation of the problem in the form

$$ \frac{{\left. dw\right|}_{z={z}_1}}{dx}-\left[\frac{d}{dz}\left(\frac{1}{\mu_x^E}\frac{d{\tau}_{xz}}{dz}\right)\right]\frac{x^2}{2}+\left[\frac{d}{dz}\left(\frac{{\left.{\sigma}_x\right|}_{x=0}}{\mu_x^E}\right)\right]x+\left[\underset{z_1}{\overset{z}{\int }}\left(\frac{\mu_x^{\nu }}{\mu_x^E}\frac{d{\tau}_{xz}}{dz}\right) dz-\frac{\tau_{xz}}{\mu_{xz}^G}+\frac{{\left. du\right|}_{x=0}}{dz}\right]=0. $$

(10)

An analysis of Eq. (10) leads to the conclusion that it has a nontrivial solution if the expressions in square brackets are unknown constants C₀ , C₁ , and C₂ :

$$ {\displaystyle \begin{array}{ccc}\frac{d}{dz}\left(\frac{1}{\mu_x^E}\frac{d{\tau}_{xz}}{dz}\right)={C}_0,& \frac{d}{dz}\left(\frac{{\left.{\sigma}_x\right|}_{x=0}}{\mu_x^E}\right)={C}_1,& \underset{z_1}{\overset{z}{\int }}\left(\frac{\mu_x^{\nu }}{\mu_x^E}\frac{d{\tau}_{xz}}{dz}\right) dz-\frac{\tau_{xz}}{\mu_{xz}^G}+\frac{{\left. du\right|}_{x=0}}{dz}={C}_2.\end{array}} $$

(11)

With account of Eqs. (11), Eq. (10) turns into the inhomogeneous linear ordinary differential equation

$$ \frac{{\left. dw\right|}_{z={z}_1}}{dx}-{C}_0\frac{x^2}{2}+{C}_1x+{C}_2=0. $$

(12)

The general solution of the first equation of (11), with account of boundary conditions (3), is

$$ {\tau}_{xz}={C}_0\left(\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz-\frac{B_1}{B_0}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^E\right) dz\right), $$

(13)

where the following designations have been introduced for the definite integrals:

$$ {\displaystyle \begin{array}{cc}\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^E\right) dz={B}_0,& \underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^Ez\right) dz={B}_1.\end{array}} $$

(14)

In Eqs. (14) and henceforth, the subscripts of constants indicate the order of the integrands in the definite integrals of functions of mechanical characteristics.

The constant B₁ , depending the coordinate system chosen, can take a positive or negative value and can also be zero. To simplify the further theoretical relations, the origin of coordinates of the beam is chosen in such a way that

$$ {B}_1=\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^Ez\right) dz=0. $$

(15)

The coordinate system in which condition (15) is obeyed will be called the principal one, and its axes in beam cross section — the principal stiffness axes of cross section.

Inserting Eq. (13) into the second integral of condition (4) and solving the resulting equation with account of Eq. (15), we obtain

$$ {C}_0=-\frac{Q_{z1}}{bB_2}, $$

(16)

where

$$ {B}_2=\underset{z_1}{\overset{z_2}{\int }}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dzdz. $$

(17)

With account of relations (16) and (15), the solution for tangential stresses (13) takes the final form

$$ {\tau}_{xz}=-\frac{Q_{z1}}{bB_2}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz. $$

(18)

Integration of the second equation of (11) gives

$$ {\left.{\sigma}_x\right|}_{x=0}={\mu}_x^E\left({C}_1z+{C}_3\right). $$

(19)

Inserting Eq. (19) into the first and third integral conditions of (4) and solving the resulting system of equations with account of Eq. (15), we obtain

$$ {\displaystyle \begin{array}{cc}{C}_1=-\frac{M_{y1}}{bB_2^{\ast }},& {C}_3=\frac{N_{x1}}{bB_0}.\end{array}} $$

(20)

where

$$ {B}_2^{\ast }=\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^E{z}^2\right) dz. $$

(21)

Between the constants B₂ and ${B}_2^{\ast }$ , the following relation is valid:

$$ {B}_2=\underset{z_1}{\overset{z_2}{\int }}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz dz=z\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right){\left. dz\right|}_{z={z}_1}^{z={z}_2}-\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^E{z}^2\right) dz=-\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^E{z}^2\right) dz=-{B}_2^{\ast }. $$

(22)

We should noted that the product bB₂ is an analog of the cross-sectional bending stiffness of a beam in the strength of materials.

Solution (19), with account of Eqs. (20) and (22), takes the form

$$ {\left.{\sigma}_x\right|}_{x=0}=\frac{\mu_x^E}{bB_2}\left({M}_{y1}z+\frac{B_2}{B_0}{N}_{x1}\right). $$

(23)

Inserting Eqs. (23) and (18) into Eq. (7), gives the solution for normal stresses

$$ {\sigma}_x=\frac{\mu_x^E}{bB_2}\left({Q}_{z1} zx+{M}_{y1}z+\frac{B_2}{B_0}{N}_{x1}\right). $$

(24)

According to (24), the normal stresses in cross sections of the layered beam vary linearly along the layer thickness and stepwise at layer interfaces in proportion to the elastic moduli ${\mu}_x^E$ of layer materials.

It follows from solutions (18) and (24) for stresses that the first and second boundary conditions of (3) are satisfied only in the case where

$$ {\displaystyle \begin{array}{cc}{q}_{xz1}=\frac{Q_{z1}}{bB_2}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz,& {q}_{x1}=-\frac{\mu_x^E}{bB_2}\left({M}_{y1}z+\frac{B_2}{B_0}{N}_{x1}\right).\end{array}} $$

(25)

Therefore, solutions (18) and (24) will be exact only if the external load on the beam varies by law (25). However, according to the Saint-Venant principle, the distribution of stresses at a sufficient distance from the loaded end of the beam will correspond to Eqs. (18) and (24) whatever the distribution of the external load, but only in the case of equality of the internal forces induced.

Next, let us determine the strain state of the beam.

With account of Eq. (18), the solution of the third equation of (11) is

$$ {\left.u\right|}_{x=0}=-\frac{Q_{z1}}{bB_2}\underset{z_1}{\overset{z}{\int }}\left(\frac{1}{\mu_{xz}^G}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz-\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz\right) dz+{C}_2\left(z-{z}_1\right)+{\left.u\right|}_{x=0,z={z}_1}. $$

(26)

Inserting Eqs. (18), (23), and (26) into the first relation of (9) leads to the general solution for longitudinal displacements

$$ {\displaystyle \begin{array}{c}u=\frac{Q_{z1}}{bB_2}\left[\frac{x^2z}{2}-\underset{z_1}{\overset{z}{\int }}\left(\frac{1}{\mu_{xz}^G}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz-\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz\right) dz\right]\\ {}+\frac{M_{y1}}{bB_2} xz+\frac{N_{x1}}{bB_0}x+{C}_2\left(z-{z}_1\right)+{\left.u\right|}_{x=0,z={z}_1}.\end{array}} $$

(27)

We should note that longitudinal displacements in cross sections of the beam, besides the constant and linear components, also contain a nonlinear cubic component, whose part in the general distribution of displacements is proportional to the transverse force.

Using Eq. (27) in the corresponding Cauchy relation gives the relation for longitudinal strains

$$ {\varepsilon}_x=\frac{1}{b}\left(\frac{Q_{z1}}{B_2} xz+\frac{M_{y1}}{B_2}z+\frac{N_{x1}}{B_0}\right). $$

(28)

Strains (28) of the layered composite beam in this problem remain continuous and, unlike the longitudinal displacements in its cross sections, vary linearly

The general solution of Eq. (12), with account of Eqs. (16) and (20), takes the form

$$ {\left.w\right|}_{z={z}_1}=-\frac{Q_{z1}{x}^3}{6{bB}_2}-\frac{M_{y1}{x}^2}{2{bB}_2}-{C}_2x+{\left.w\right|}_{z={z}_1,x=0}. $$

(29)

Inserting Eqs. (18), (23), and (29) into the second relation of (9), gives the relation for the transverse displacements of longitudinal fibers of the beam

$$ w=-\frac{1}{bB_2}\left[{Q}_{z1}\left(\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz x+\frac{x^3}{6}\right)+{M}_{y1}\left(\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz+\frac{x^2}{2}\right)\right]-\frac{N_{x1}}{bB_0}\underset{z_1}{\overset{z}{\int }}{\mu}_x^{\nu } dz-{C}_2x+{\left.w\right|}_{z={z}_1,x=0}. $$

(30)

With account of Eq. (30), the transverse strains of the layered beam are expressed as

$$ {\varepsilon}_z=-\frac{\mu_x^{\nu }}{b}\left(\frac{Q_{z1}}{B_2}{z}^2+\frac{M_{y1}}{B_2}z+\frac{N_{x1}}{B_0}\right). $$

(31)

According to Eq. (31), the transverse strains change stepwise at layer interfaces in proportion to the change in Poisson’s ratio ${\mu}_x^{\nu }$.

We should note that the solutions for displacements (27) and (30) and longitudinal strains (28) are continuous, and thus the conditions of an for absolutely rigid contact between layers are satisfied without using any additional conditions.

Partial solution of the problem for displacements. The unknown constants C₂ , ${\left.u\right|}_{x=0,z={z}_1}$, and ${\left.w\right|}_{z={z}_1,x=0}$ in relations for displacements (27) and (30) can be determined using kinematic boundary conditions corresponding to the chosen type of fixation of the beam end x = l . However, we should note that the conditions of absolutely rigid fixation cannot be satisfied with the use of this solution, because it is impossible to satisfy the kinematic conditions

$$ {\displaystyle \begin{array}{cc}{\left.u\right|}_{x=l}=0,& {\left.w\right|}_{x=l}=0\end{array}} $$

(32)

at any values of the constants C₂ , ${\left.u\right|}_{x=0,z={z}_1}$, and ${\left.w\right|}_{z={z}_1,x=0}$.

According to solution (27), the fixed beam end will deform; therefore the required constants can determine the location of only some points of the beam that will remain immov able or partially movable.

Many different variants of kinematic conditions can be prescribed for determining the constants C₂ , ${\left.u\right|}_{x=0,z={z}_1}$, and ${\left.w\right|}_{z={z}_1,x=0}$ that roughly simulate the rigid fixation of the end of a layered cantilever beam. Their analysis and limits of its application in solving applied problems is the subject of an independent study. Here, we will consider only one of the possible kinematic conditions, to which there corresponds the fixation scheme shown in Fig. 2. These kinematic conditions are

$$ {\displaystyle \begin{array}{ccc}{\left.u\right|}_{x=l,z={z}_1}=0,& {\left.u\right|}_{x=l,z={z}_2}=0,& {\left.w\right|}_{x=l,z={z}_1}=0.\end{array}} $$

(33)

Fig. 2.

Modeling the fixation of beam end.

Note that the reactive forces q_x2 and q_xz2 , which have to arise in the fixed cross section according to the static boundary conditions, are not shown in Fig. 2. These reactive loads, corresponding to solutions (24) and (18), have to balance the external force factors acting on the bar. Therefore, reactions at supports in the scheme do not arise.

Inserting relations (27) and (30) into conditions (33) and solving the equations obtained for the unknown constants, we obtain

$$ {\displaystyle \begin{array}{c}\begin{array}{cc}{\left.u\right|}_{x=0,z={z}_1}=-\frac{1}{b}\left(\frac{Q_{z1}{l}^2{z}_1}{2{B}_2}+\frac{M_{y1}{lz}_1}{B_2}+\frac{N_{x1}l}{B_0}\right),& {C}_2=-\frac{1}{b}\left[\frac{Q_{z1}}{B_2}\left(\frac{l^2}{2}-\frac{D_2}{h}\right)+\frac{M_{y1}l}{B_2}\right],\end{array}\\ {}{\left.w\right|}_{z={z}_1,x=0}=-\frac{1}{b}\left[\frac{Q_{z1}l}{B_2}\left(\frac{l^2}{3}-\frac{D_2}{h}\right)+\frac{M_{y1}{l}^2}{2{B}_2}\right],\end{array}} $$

(34)

where h = z₂ − z₁ is the height of cross section of the beam.

In relations (34),

$$ {D}_2=\underset{z_1}{\overset{z_2}{\int }}\left(\frac{1}{\mu_{xz}^G}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz-\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz\right) dz. $$

(35)

With account of Eq. (34), the relations for displacements (27) and (30) become

$$ {\displaystyle \begin{array}{c}u=\frac{Q_{z1}}{bB_2}\left[\frac{\left({x}^2-{l}^2\right)z}{2}-\underset{z_1}{\overset{z}{\int }}\left(\frac{1}{\mu_{xz}^G}\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^Ez\right) dz-\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz\right) dz+\frac{D_2\left(z-{z}_1\right)}{h}\right]\\ {}+\left(\frac{M_{y1}z}{bB_2}+\frac{N_{x1}}{bB_0}\right)\left(x-1\right),\\ {}w=-\frac{Q_{z1}}{bB_2}\left[\frac{x^3}{6}-\frac{xl^2}{2}+x\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz+\frac{D_2\left(x-l\right)}{h}+\frac{l^3}{3}\right]\\ {}-\frac{M_{y1}}{bB_2}\left[\underset{z_1}{\overset{z}{\int }}\left({\mu}_x^{\nu }z\right) dz+\frac{{\left(x-l\right)}^2}{2}\right]-\frac{N_{x1}}{bB_0}\underset{z_1}{\overset{z}{\int }}{\mu}_x^{\nu } dz.\end{array}} $$

(36)

According to the second relation of (36), the deflection — the vertical displacement of the initial cross section at the level of beam axis — is

$$ f={\left.w\right|}_{x=0,z=0}=-\frac{Q_{z1}{l}^3}{3{bB}_2}\left(1+\frac{3{D}_2}{hl^2}\right)-\frac{M_{y1}{l}^2}{2{bB}_2}\left(1+\frac{2}{l^2}\underset{z_1}{\overset{z_0}{\int }}\left({\mu}_x^{\nu }z\right) dz\right)-\frac{N_{x1}}{bB_0}\underset{z_1}{\overset{z_0}{\int }}{\mu}_x^{\nu } dz. $$

(37)

In the case of a homogeneous cantilever beam, relation (37) takes the forma

$$ f=\frac{Q_{z1}{l}^3}{3{E}_x{J}_y}\left(1+\frac{E_x-{\nu}_x{G}_{xz}}{4{G}_{xz}}\frac{h^2}{l^2}\right)+\frac{M_{y1}{l}^2}{2{E}_x{J}_y}\left(1+\frac{\nu_x}{4}\frac{h^2}{l^2}\right)-\frac{N_{x1}{\nu}_xh}{2{E}_xA}, $$

(38)

where J _y  = bh³/12.

It is seen that relation (38) contains both “classical” components and components considering the effect of shear and compression strains and the ratio between the height of cross section to the length of the beam on its deflection. At the same time, if the bar material is completely rigid in shear (G _xz → ∞) and compression (ν _x → 0), this relation takes the form known from the course in the strength of materials, indirectly confirming the correctness of the expressions obtained.

Example of realization of the relations obtained. Let us consider employment of the solution obtained with the example of a composite cantilever beam (Fig. 3). Its length is l = 0.3 m, and resultants of the tangential and normal loads distributed at on its free end are Q_z1 = −750N, ${N}_{x_1}$ = −2750 N, and M_y1 = −50 N · m. The mechanical characteristics of beam phase materials are as follows.

• Glass-fiber plastic ( Ρ₁, Ρ₃ ): ${E}_x^{\left[1\right]}={E}_x^{\left[3\right]}$ = 57 GPa, ${G}_{zx}^{\left[1\right]}={G}_{zx}^{\left[3\right]}$ = 5.2 GPa, and ${\nu}_x^{\left[1\right]}={\nu}_x^{\left[3\right]}$ = 0.40;

• wood ( Ρ₂ ): ${E}_x^{\left[2\right]}$ = 15.4 GPa, ${G}_{zx}^{\left[2\right]}$ = 0.98 GPa, and ${\nu}_x^{\left[2\right]}$ = 0.43;

• aluminum alloy ( Ρ₄ ): ${E}_x^{\left[4\right]}$ = 70 GPa, ${G}_{zx}^{\left[4\right]}$ = 26.9 GPa, and ${\nu}_x^{\left[4\right]}$ = 0.34.

Fig. 3.

Structure of beam cross section (dimensions in mm).

Solutions (18), (24), and (36) were obtained in the principal coordinate system, where B₁ = 0 . Therefore, at first, using the initial data on the structural configuration of the composite beam, we determine the location of origin of the principal coordinate system along the height of beam cross section.

Let us suppose that B₁ ≠ 0 in some auxiliary coordinate system Y′O′Z (see Fig. 3) and the required principal system is shifted relative to it by ${z}_{B_1}$ = z′ − z. Then, we have for the principal system that

$$ {B}_1=\underset{z_1}{\overset{z_2}{\int }}\left({\mu}_x^Ez\right) dz=\underset{z_1^{\prime }}{\overset{z_2^{\prime }}{\int }}\left({\mu}_x^{\prime E}{z}^{\prime}\right) dz-{z}_{B_1}\underset{z_1^{\prime }}{\overset{z_2^{\prime }}{\int }}{\mu}_x^{\prime E} dz={B}_1^{\prime }-{z}_{B_1}{B}_0^{\prime }=0, $$

whence it follows that

$$ {z}_{B_1}={B}_1^{\prime }/{B}_0^{\prime }. $$

(39)

In the coordinate system Y′O′Z , the point with coordinates y = 0 and z = ${z}_{B_1}$ can be called the center of stiffness of cross section of the discrete-inhomogeneous composite beam.

According to the data of Fig. 1b and expressions (1), we can write for the auxiliary coordinate system

$$ {\displaystyle \begin{array}{cc}{B}_0^{\prime }=\sum \limits_{k=1}^n\left({E}_x^{\left[k\right]}{t}_k\right),& {B}_1^{\prime }=\frac{1}{2}\sum \limits_{k=1}^n\left[{E}_x^{\left[k\right]}{t}_k\left({z}_{t_{k+1}}^{\prime }+{z}_{t_k}^{\prime}\right)\right].\end{array}} $$

(40)

Using Eq. (40), it was found that ${B}_0^{\prime }$ = 1.4206 · 10⁹ N/m and ${B}_1^{\prime }$ = 4.6787 · 10⁷ N for the cross section considered in the auxiliary coordinate system Y′O′Z . Using these characteristics and Eq. (39), the coordinate of the center of stiffness of cross section was found to be ${z}_{B_1}^{\prime }$ = 0.0329.

The next step in realization of the solution obtained is formation of the functions of mechanical characteristics (1). For their construction, an analytical representation of indicator functions (2) have to be known for each layer. These functions can be written using shifted Heaviside functions:

$$ {p}_k\left(z,{z}_{t_k},{z}_{t_{k+1}}\right)=\mathrm{H}\left(z-{z}_{t_k}\right)-\mathrm{H}\left(z-{z}_{t_{k+1}}\right), $$

(41)

where H = H(z) is the Heaviside function.

For the cross section shown on Fig. 3, the functions of mechanical characteristics ${\mu}_a^S$ (1) are found with the help of Eq. (41) as

$$ {\displaystyle \begin{array}{c}{\mu}_a^S={S}_a^{\left[1\right]}\left[\mathrm{H}\left(z+0.0329\right)-\mathrm{H}\left(z+0.0289\right)\right]+{S}_a^{\left[2\right]}\left[\mathrm{H}\left(z+0.0289\right)-\mathrm{H}\left(z-0.0201\right)\right]\\ {}+{S}_a^{\left[3\right]}\left[\mathrm{H}\left(z-0.0201\right)-\mathrm{H}\left(z-0.0241\right)\right]+{S}_a^{\left[4\right]}\left[\mathrm{H}\left(z-0.0241\right)-\mathrm{H}\left(z-0.02271\right)\right],\end{array}} $$

(42)

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

Using Eq. (42) in Eqs. (14), (17), and (35), the following stiffness characteristics of the inhomogeneous section required for calculation were determined:

$$ {\displaystyle \begin{array}{cccc}{B}_0=1.4206\cdot {10}^9\mathrm{N}/\mathrm{m},& {B}_1=0\;\mathrm{N},& {B}_2=-6.3304\cdot {10}^5\mathrm{N}\cdot \mathrm{m},& {D}_2=-5.8976\cdot {10}^{-4}{\mathrm{m}}^3.\end{array}} $$

Note that the value of B₀ does not depend on location of the coordinate system, i.e., on the coordinate z , and ${B}_0^{\prime }$ = B₀ = 1.4206 · 10⁹ N/m for the layered cross section accepted.

The distributions of strains and stresses in Fig. 4 (the top row) and displacements (the bottom row) are plotted using relations (18), (24) and (36). For clarity, the graph of function of the longitudinal elastic modulus ${\mu}_x^E$ is shown to the left of the graphs of strain and stress distributions; the dashed lines correspond to the case where ${G}_{xz}^{\left[k\right]}\to \infty$ and ${\nu}_x^{\left[k\right]}\to 0$.

Fig. 4.

Distributions of elastic moduli and components of SSS. Explanations in the text.

The graphs of stress and strain distributions presented have the form expected, which confirms that the condition of an absolutely rigid contact between phases is fulfilled in the solution obtained. The distributions of normal and tangential stresses within layers are “classical” — linear and quadratic, respectively, — with jump-like changes at layer interfaces. Note that the a compliance of layer materials in shear and transverse compression have no influence on the behavior of the stress distributions. At the same time, these material properties of materials significantly affect the distribution of displacements, which manifests itself in curved (warped) cross sections, decreased transverse displacements at the level of stiffness center of cross section, and significantly increased deflection of the beam.

Conclusions

In the present work, a solution of the problem on plane bending of a narrows layered composite cantilever beam subjected to normal and tangential loads at its free end is obtained. This solution includes relations (18), (24), (28), (31), and (36), which satisfy the system of static, geometric, and physical equations of the plane problem of linear elasticity theory. The solution is the exact if the load is distributed according to relations (25).

The results obtained indicate that the stress state of a layered cantilever beam with loads at its free end does not depend on the compliance of layer materials in shear and transverse compression. However, these materials properties have a significant effect on the distribution of longitudinal and transverse displacements and, especially, deflections, which requires attention in calculations of composite members.

The solution derived can be used both for a direct prediction of the strength and stiffness of composite cantilevers and for development of applied techniques to calculate composite structural members.