Re-compute Exercise 1 for loading beyond first ply failure (FPF). Simulate the damaged material by using a damage factor d = 0.999 (ply discount). The minimum value of strength ratio R thus obtained corresponds to the last ply failure (LPF), i.e., R_{LP F}.

Exercise 1

Using the stresses found in Exercise 2, evaluate the strength ratio RF P F on top and bottom of each lamina using the following failure criteria:

(a) Interacting criterion,

(b) Maximum strain criterion, and

(c) Maximum stress criterion.

In each case, the minimum of the values of R computed through the thickness corresponds to the first ply failure (FPF) strength ratio. Note that the FPF values computed with the various criteria are, in general, different from each other. Only experience or multiaxial strength test data can tell which criterion is the best for a particular material and stress state.

Exercise 2

Using the results for the intact laminate in Exercise 3, compute:

(a) the strains εx, εy, γxy at the interface between laminae,

(b) the stresses σ_x, σ_y, τ_xy on all laminae, right next to the interface,

(c) the strains ε₁, ε₂, γ₆ by using the transformation equations,

(d) the stresses in material coordinates σ₁, σ₂, σ₆.

Exercise 3

Consider a [0/45/-45]S laminate, subject to a load N_x = N_y = 1000 N/mm. Compute the [A], and [D] matrices (neglect shear deformations) for the intact laminate. Each lamina is 2.5 mm thick. Do not adjust for in-situ strength. The material properties are:

E₁= 204.6 GPa

E₂ = 20.46 GPa

G₁₂ = 6.89 MPa

v₁₂ = 0.3

F_1t = 826.8 MPa

F_1c = 482.3 MPa

F_2t = F_2c= 68.9 MPa

F₆ = 241.15 MPA