Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Whereas elongations and contractions in an isotropic material are independent of the
direction of the applied load, deformations of an orthotropic material depend very much on
the directions of the natural axes, as well as on the direction of the applied load.
An applied stress that is not in the direction of one of the natural axes of an orthotropic
material behaves as shown in Figure 13.17. A normal stress σ x applied in the x direction
produces an elongation ε x in the x direction, a contraction ε y in the y direction, and a shearing
deformation γ xy . Note that there will be a Poisson’s ratio ν xy relating the normal strain
in the y direction to the normal strain in the x direction that occurs in response to loading in
the x direction. The constant η xs is termed a shear coupling coefficient. This material constant,
the ratio of the shear strain γ xy to the normal strain ε x in the x direction, relates normal
stress in the x direction to shear strain in the x–y plane.
A shear stress τ xy produces an elongation ε x , a contraction ε y , and a shearing deformation
γ xy . The shear coupling coefficients η sx and η sy are the ratios of the normal strains
ε x and ε y to the shear strain γ xy , respectively, for the case of pure shear stress loading.
Thus, depending on the direction of the stress, there may exist a coupling among elongation,
contraction, and shearing deformation. By contrast, no coupling exists for an isotropic
material, regardless of the direction of stress.
Consider a thin piece of material (such as epoxy) reinforced by unidirectional fibers
(such as graphite) that is subjected to plane stress. The material that holds the fibers in
position is called the matrix. The combination of fibers embedded in a matrix creates a
material that is orthotropic (Figure 13.18). Axes x, y, and z are in the principal material
directions (i.e., the natural directions). Planes x–y, y–z, and x–z are planes of material symmetry.
For plane stress, σ z = τ xz = τ yz = 0.
To determine the stress–strain relations, assume that the behavior of the material is
linear elastic and use the principal of superposition. First, suppose that the orthotropic specimen
is subjected to a uniaxial loading of σ x in the x direction, as shown in Figure 13.19a.
Then the strains are
and
σ x
ε x =
E
ε =− v ε =−v
y xy x xy
where E x is the elastic modulus for loading in the x direction and ν xy is Poisson’s ratio for
loading in the x direction and strain in the y direction. For fibers directed as shown, ν xy is
often referred to as the major Poisson’s ratio. Strains do occur in the z direction, but since
the material is thin, ε z is of no particular interest.
Next, consider a uniaxial loading of σ y in the y direction as shown in Figure 13.19b.
The strains are now
and
ε
y
x
σ
=
E
ε =− v ε =−v
x yx y yx
y
y
σ x
E
σ
E
x
y
y
σ
ε = 1
12 σ
1 ε 1
E 2 = −
1
E1
y
y
x
x
2
2
τ 12
τ 12
σ x = σ 1
FIGURE 13.16 Orthotropic
material loaded along principal
material directions.
ε
σ
= x x
Ex
γ xy
y
y
τ xy
=
G xy
x
x
y
2
2
1
1
τ
γ 12 12
ν=
G 12
ν xy σ
ε x
y = –
Ex
εx=
η sx
FIGURE 13.17 Orthotropic
material loaded along nonprincipal
material directions.
Matrix
x
τ xy
Fibers
η xs
σ x
τ xy
τ
η xy sy
Gxy
FIGURE 13.18 Matrix and
fibers for a unidirectional layer.
1
1
Gxy
γ xy =
τ xy
ε y =
σ x
E x
where E y is the elastic modulus for loading in the y direction and ν yx is Poisson’s ratio for
loading in the y direction and strain in the x direction. For fibers directed as shown, ν yx is
often referred to as the minor Poisson’s ratio.
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