If all the forces determined in exercise 1 are increased


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 Nx, Ny, Nxy, Mx, My, Mxy required to produce a curvature κx = 0.00545 mm-1, κy = -0.00486 mm-1. Neglect small numbers and shear deformation.

Exercise 2

Consider a [0/45/-45]S laminate, subject to a load Nx = N= 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

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Physics: If all the forces determined in exercise 1 are increased
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