If all the forces determined in Exercise 1 are increased proportionally, use the interacting criterion to determine

(a) the first ply failure load, and

(b) the last ply failure load assuming d = 0.999.

Exercise 1

Consider a [±45] laminate with the same material properties of Exercise 2. Determine the value of the forces N_x, N_y, N_xy, M_x, M_y, M_xy required to produce a curvature κ_x = 0.00545 mm^-1, κ_y = -0.00486 mm^-1. Neglect small numbers and shear deformation.

Exercise 2

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 = F2c =  68.9 MPa
F₆ = 241.15 MPa