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

F1t = 826.8 MPa

F1c = 482.3 MPa

F2t = F2c= 68.9 MPa

F6 = 241.15 MPA