Page 311 - Mechanics Analysis Composite Materials

296 Mechanics and analysis of composite materials

For +45" and -45" layers, we get, respectively, Tu = 21.7 kNm and T, = 17.6 kNm.
Thus, Tu = 17.6 kNm.
As the second example, consider the cylindrical shell described in Section 5.1 1 (see
Fig. 5.19) and loaded with internal pressure p. Axial, p,,and circumferential, N,.,
stress resultants can be found as

N, =$pR, N,=pR ,

where R = 100 mm is the shell radius. Applying constitutive equations, Eqs. (5.74),
and neglecting the change of the cylinder curvature (ti,. = 0) we arrive at the
following equations for strains:

Using Eqs. (5.71) and (5.72) to determine strains and stresses in the principal
material coordinates of the plies, we get

ay)= [(Bz~- 2B12)(cos' 4; + VI' sin' 4;)
2B

(6.48)

Here, B =BllB'2 -B:', and membrane stiffnessesB,, for the shell under study are
presented in Section 5.1 1. Subscript "i" in Eqs. (6.48) indicates the helical plies for
which i = 1, 4, = 4 = 36" and circumferential plies for which i = 2 and 42= 90".
The problem that we consider is to find the ultimate pressurep,. For this purpose,
we use strength criteria in Eqs. (6.3), (6.4) and (6.16), and the following material
properties tf; = 1300 MPa, at = 27 MPa, 212 = 45 MPa.
Calculation with the aid of Eqs. (6.48) yields

Applying Eqs. (6.3) to evaluate the plies' strength along the fibers, we get

(ril) = G;, pu = 14.9 MPa,
- -+
I - IJ~, p,, = 11.2 MPa .