Page 208 - Mechanics Analysis Composite Materials

Chapter 4. Mechanics of a composite layer 193

Using the last equation of Eqs. (4.134) we can find the shear strain as
sinylz= cos$. After some rearrangements with the aid of Eqs. (4.141) and
(4.142) we arrive at

(4.143)
For 4 = 0, axes 1 and 2 coincide, respectively, with axes x and y (see Fig. 4.60), and
Eq. (4.143) yields

(4.144)

Using this result to express E,,, we can write Eqs. (4.141H4.143) in the following
final form:

(1 +E\)? = (1 +E,)~COS~~+(I sin^++(^ +&,)(I +q.)sinyr,.sin2+.
(I +E~Y= (1 +~,)~sin’4+(1+E,,)~COS~+- (I +~,.)sin?/,,.sin26:
1
sin-y12=
(1 + El)(1 + ez)
x ([(~+c,.)~-(l+E.r)2~sin4cos#+ (1 +€,)(I +E,.)sing,,cos24} .
(4.145)

As follows from Fig. 4.60 and the last equation of Eqs. (4.134), dsl =
ds; sin $ = ds; cosyl2.So in accordance with Eqs. (4.134) and (4.135)
I +e; = (I +Ez)COSy,2 .

Using Eqs. (4.145) to transform this equation we get

(4.146)

To express Q,’ in terms of Q, and strains referred to the global coordinate frame x, y,
consider Eq. (4.136). After rather cumbersome transformation with the aid of
Eqs. (4.141) and (4.142) we obtain