(Parte 1 de 5)


The behavior of composite materials whose micro- and macrostructures are much more complicated than those of traditional structural materials such as metals, concrete, and plastics is nevertheless governed by the same general laws and principles of mechanics whose brief description is given below.

2.1. Stresses

Consider a solid body referred by Cartesian coordinates as in Fig. 2.1. The body is fixed at the part Su of the surface and loaded with body forces qv having coordinate components qx, qy, and qz, and with surface tractions ps specified by coordinate components px, py, and pz. Surface tractions act on surface Sσ which is determined by its unit normal n with coordinate components lx, ly, and lz that can be referred to as directional cosines of the normal, i.e.,

Introduce some arbitrary cross section formally separating the upper part of the body from its lower part. Assume that the interaction of these parts in the vicinity of some point A can be simulated with some internal force per unit area or stress σ distributed over this cross section according to some as yet unknown law. Since the mechanics of solids is a phenomenological theory (see the closure of Section 1.1) we do not care about the physical nature of stress, which is only a parameter of our model of the real material (see Section 1.1) and, in contrast to forces F, has never been observed in physical experiments. Stress is referred to the plane on which it acts and is usually decomposed into three components – normal stress (σz in Fig. 2.1) and shear stresses (τzx and τzy in Fig. 2.1). The subscript of the normal stress and the first subscript of the shear stress indicate the plane on which the stresses act. For stresses shown in Fig. 2.1, this is the plane whose normal is parallel to the z-axis. The second subscript of the shear stress shows the axis along which the stress acts. If we single out a cubic element in the vicinity of point A (see Fig. 2.1), we should apply stresses to all its planes as in Fig. 2.2 which also shows notations and positive directions of all the stresses acting inside the body referred by Cartesian coordinates.

32 Advanced mechanics of composite materials Z pi sz tzx tzy s C v suz ux uz ux M uy

A B uy n(lx, ly, lz) ps (px, py, pz)

qv (qx, qy, qz)

Fig. 2.1. A solid loaded with body and surface forces and referred by Cartesian coordinates.

sy sx sy sz txy txz tzx sz tzy txy tyx tyz tzy tzx tyx tyz txz dx dy dz A

Fig. 2.2. Stress acting on the planes of the infinitely small cubic element.

Chapter 2. Fundamentals of mechanics of solids 3 2.2. Equilibrium equations

Now suppose that the body in Fig. 2.1 is in a state of equilibrium. Then, we can write equilibrium equations for any part of this body. In particular we can do this for an infinitely small tetrahedron singled out in the vicinity of point B (see Fig. 2.1) in such a way that one of its planes coincides with Sσ and the other three planes are coordinate planes of the Cartesian frame. Internal and external forces acting on this tetrahedron are shown in Fig. 2.3. The equilibrium equation corresponding, for example, to the x-axis can be written as

−σxdSx − τyxdSy − τzxdSz + pxdSσ + qxdV = 0

Here, dSσ and dV are the elements of the body surface and volume, whereas dSx = dSσlx, dSy = dSσly, and dSz = dSσlz. When the tetrahedron is infinitely diminished, the term including dV , which is of the order of the cube of the linear dimensions, can be neglected in comparison with terms containing dS, which is of the order of the square of the linear dimensions. The resulting equation is σxlx + τyxly + τzxlz = px (x,y ,z ) (2.2)

The symbol (x,y ,z ), which is widely used in this chapter, denotes permutation with the aid of which we can write two more equations corresponding to the other two axes changing x for y, y for z, and z for x. Consider now the equilibrium of an arbitrary finite part C of the body (see Fig. 2.1).

If we single this part out of the body, we should apply to it body forces qv and surface tractions pi whose coordinate components px, py, and pz can be expressed, obviously, by Eq. (2.2) in terms of stresses acting inside the volume C. Because the sum of the sy n(lx, ly, lz) ps (px, py, pz)

qv (qx, qy, qz) tyz tzx txz sx txy tyx tzy dy dx B

Fig. 2.3. Forces acting on an elementary tetrahedron.

34 Advanced mechanics of composite materials components corresponding, for example, to the x-axis must be equal to zero, we have ∫∫∫ v qxdv + s pxds = 0 where v and s are the volume and the surface area of the part of the body under consideration. Substituting px from Eq. (2.2) we get ∫∫ s (σxlx + τyxly + τzxlz)ds +

Thus, we have three integral equilibrium equations, Eq. (2.3), which are valid for any finite part of the body. To convert them into the corresponding differential equations, we use Green’s integral transformation

s (fxlx + fyly + fzlz)ds = ∫∫∫v which is valid for any three continuous, finite, and single-valued functions f (x, y, z) and allows us to transform a surface integral into a volume one. Taking fx = σx, fy = τyx, and fz = τzx in Eq. (2.4) and using Eq. (2.3), we arrive at

Since these equations hold true for whatever the part of the solid may be, provided only that it is within the solid, they yield

Thus, we have arrived at three differential equilibrium equations that could also be derived from the equilibrium conditions for the infinitesimal element shown in Fig. 2.2.

However, in order to keep part C of the body in Fig. 2.1 in equilibrium the sum of the moments of all the forces applied to this part about any axis must be zero. By taking moments about the z-axis we get the following integral equation v (qxy − qyx)dv + s (pxy − pyx)ds = 0

Using again Eqs. (2.2), (2.4), and taking into account Eq. (2.5) we finally arrive at the symmetry conditions for shear stresses, i.e.,

Chapter 2. Fundamentals of mechanics of solids 35

So, we have three equilibrium equations, Eq. (2.5) which include six unknown stresses σx,σ y,σ z and τxy,τ xz,τ yz. Eq. (2.2) can be treated as force boundary conditions for the stressed state of a solid.

2.3. Stress transformation

Consider the transformation of a stress system from one Cartesian coordinate frame to another. Suppose that the elementary tetrahedron shown in Fig. 2.3 is located inside the body and that point B coincides with the origin 0 of Cartesian coordinates x, y, and z in Fig. 2.1. Then, the oblique plane of the tetrahedron can be treated as a coordinate plane z′ = 0 of a new coordinate frame x′, y′, z′ shown in Fig. 2.4 and such that the normal element to the oblique plane coincides with the z′-axis, whereas axes x′ and y′ are located in this plane. Component px of the surface traction in Eq. (2.2) can be treated now as the projection on the x-axis of stress σ acting on plane z′ = 0.

Then, Eq. (2.2) can be presented in the following explicit form specifying projections of stress σ px = σxlz′x + τyxlz′y + τzxlz′z py = σylz′y + τzylz′z + τxylz′x pz = σzlz′z + τxzlz′x + τyzlz′y s (px, py, pz)

Fig. 2.4. Rotation of the coordinate frame.

36 Advanced mechanics of composite materials

Here, l are directional cosines of axis z′ with respect to axes x, y, and z (see Fig. 2.4 in which the corresponding cosines of axes x′ and y′ are also presented). The normal stress σz′ can be found now as σz′ = pxlz′x + pylz′y + pzlz′z

The final result was obtained with the aid of Eqs. (2.6) and (2.7). Changing x′ for y′, y′ for z′, and z′ for x′, i.e., performing the appropriate permutation in Eq. (2.8) we can write similar expressions for σx′ and σy′. The shear stress in the new coordinates is

2.4. Principal stresses

The foregoing equations, Eqs. (2.8) and (2.9), demonstrate stress transformations under rotation of a coordinate frame. There exists a special position of this frame in which the shear stresses acting on the coordinate planes vanish. Such coordinate axes are called the principal axes, and the normal stresses that act on the corresponding coordinate planes are referred to as the principal stresses.

To determine the principal stresses, assume that coordinates x′,y ′, and z′, in Fig. 2.4 are the principal coordinates. Then, according to the aforementioned property of the principal

directional cosines of the principal axis, i.e., taking lz′x = lpx,l z′y = lpy,l z′z = lpz we have from Eqs. (2.7)

(σx − σ)lpx + τxylpy + τxzlpz = 0 τxylpx + (σy − σ)lpy + τyzlpz = 0 τxzlpx + τyzlpy + (σz − σ)lpz = 0

These equations were transformed with the aid of symmetry conditions for shear stresses, Eq. (2.6). For some specified point of the body in the vicinity of which the principal stresses are determined in terms of stresses referred to some fixed coordinate frame x, y, z

Chapter 2. Fundamentals of mechanics of solids 37 and known, Eqs. (2.10) comprise a homogeneous system of linear algebraic equations.

Formally, this system always has the trivial solution, i.e., lpx = lpy = lpz = 0 which we can ignore because directional cosines should satisfy an evident condition following from

Eqs. (2.1), i.e.,

So, we need to find a nonzero solution of Eqs. (2.10) which can exist if the determinant of the set is zero. This condition yields the following cubic equation for σ

I3 = σxσyσz + 2τxyτxzτyz − σxτ2yz − σyτ2xz − σzτ2 xy are invariant characteristics (invariants) of the stressed state. This means that if we refer the body to any Cartesian coordinate frame with directional cosines specified by Eqs. (2.1), take the origin of this frame at some arbitrary point and change stresses in Eqs. (2.13) with the aid of Eqs. (2.8) and (2.9), the values of I1,I 2,I 3 at this point will be the same for all such coordinate frames. Eq. (2.12) has three real roots that specify three principal stresses σ1,σ 2, and σ3. There is a convention according to which σ1 ≥ σ2 ≥ σ3, i.e., σ1 is the maximum principal stress and σ3 is the minimum one. If, for example, the roots of Eq. (2.12) are 100MPa, −200MPa, and 0, then σ1 = 100MPa,σ 2 = 0, and σ3 =− 200MPa. To demonstrate the procedure, consider a particular state of stress relevant to several applications, namely, pure shear in the xy-plane. Let a thin square plate referred to coordinates x, y, z be loaded with shear stresses τ uniformly distributed over the plate thickness and along the edges (see Fig. 2.5). One principal plane is evident – it is plane z = 0, which is free of shear stresses. To find the other two planes, we should take in Eqs. (2.13) σx = σy = σz = 0,τ xz = τyz = 0, and τxy = τ. Then, Eq. (2.12) takes the form

The first root of this equation gives σ = 0 and corresponds to plane z = 0. The other two

To find the planes corresponding to σ1 and σ3 we should put lpz = 0, substitute σ =± τ into Eqs. (2.10), write them for the state of stress under study, and supplement this set with Eq. (2.1). The final equations allowing us to find lpx and lpy are

38 Advanced mechanics of composite materials

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(Parte 1 de 5)