The lamina of Example 3.6 is composed of glass fibers and an epoxy resin. The fibers have...

The lamina of Example 3.6 is composed of glass fibers and an epoxy resin. The fibers have a modulus of elasticity EF = 72.4 GPa, a shear modulus GF = 27.8 GPa, and a Poisson ratio νF = 0.30. The resin has a modulus of elasticity ER = 3.50 GPa, a shear modulus GR = 1.35 GPa, and a Poisson ratio νR = 0.30. The volume fraction of fibers is f = 0.70.

a. Determine the coefficients Cij of the lamina stress–strain relations [see Eqs. (m) and (n) of Example 3.6].

---

b. For a given load, the measured strain components were found to be

ϵxx = 500μ, ϵyy = 350μ, γxy = 1000μ

Determine the principal stresses and the orientation of the principal axes of stress.

Example 3.6

Stress–Strain Relations of a Fiber–Resin Lamina

A lamina (a thin plate, sheet, or layer of material) of a section of an airplane wing is composed of unidirectional fibers and a resin matrix that bonds the fibers. Let the volume fraction (the proportion of fiber volume to the total volume of the composite) be f. Determine the effective linear stress–strain relations of the lamina.

FIGURE E3.6 Lamina: fiber volume fraction = f, resin volume fraction= 1 − f.

Solution:

Let the modulus of elasticity and the Poisson ratio of the fibers be denoted EFand νF, respectively, and the modulus of elasticity and the Poisson ratio of the resin be ERand νR. Since the lamina is thin, the effective state of stress in the lamina is approximately one of plane stress in the x–y plane of the lamina (see Figure E3.6a). Hence, the stress–strain relations for the fibers and the resin are

where (σxxF, σyyF), (σxxR, σyyR), (ϵxxF,ϵyyF), and (ϵxxR,ϵyyR) denote stress and strain components in the fiber (F) and resin (R), respectively.

Since the fibers and resin are bonded, the effective lamina strain ϵxx (Figure E3.6a) is the same as that in the fibers and in the resin; that is, in the x direction,

ϵxx = ϵxxF = ϵxxR     (b)

In the y direction, the effective lamina strain ϵyy is proportional to the amount of fiber per unit length in the y direction and the amount of resin per unit length in the y direction. Hence,

ϵyy = fϵyyF + (1 − f) ϵyyR    (c)

Also, by equilibrium of the lamina in the x direction, the effective lamina stress σxx is

σxx = fσxxF + (1 − f) σxxR    (d)

In the y direction, the effective lamina stress σyy is the same as in the fibers and in the resin; that is,

σyy = σyyF = σyyR     (e)

Solving Eqs. (a) through (e) for ϵxx and ϵyy in terms of σxx and σyy, we obtain the effective stress–strain relations for the lamina as

where

To determine the shear stress–strain relation, we apply a shear stress σxy to a rectangular element of the lamina (Figure E3.6b), and we calculate the angle change γxy of the rectangle. By Figure E3.6b, the relative displacement b of the top of the element is

b = fγF + (1 − f) γR     (h)

where γF and γR are the angle changes attributed to the fibers and the resin, respectively; that is,

and GF and GR are the shear moduli of elasticity of the fiber and resin, respectively. Hence, the change γxy in angle of the element (the shear strain) is, with Eqs. (h) and (i),

By Eq. (j), the shear stress–strain relation is

σxy = Gγxy =2Gϵxy      (k)

where

Thus, by Eqs. (f), (g), (k), and (l), we obtain the stress–strain relations of the lamina, in the form of Eqs. 3.50, as

where

 (3.50)