Method for designing discrete layering angles of laminated board

[0032] Example 1: (a) Divide a square planar structure with a length and width of 1m into a square grid of 16×16; every 4×4 units constitute a design area, and all units have the same laying angle; the left side of the structure Laterally fixed support, the load is the concentrated load F=1000N acting on the lower right corner of the structure; orthotropic material property: E x =146.86GPa,E y =10.62GPa,G xy =5.45GPa, v xy =0.33; the number of ply angles m=4, the ply angles are 90° / 45° / 0° / -45° respectively; given the initial value of the design variable x ik =0.

[0033] For m ply angle design problems, the number of design variables required for each unit Establish the finite element model of laminate design space and the initial value of design variable x ik =0, where i represents the unit number, and k represents the kth design variable corresponding to the unit; the stiffness matrix of unit i under the given j-th ply angle is K i (j) ; Use the following formula

[0050] Example 2: (a) Divide the rectangular planar structure with a length and width of 6m and 1m into a square grid of 240×40; every 10×10 units constitute a design area, and all units have the same ply angle ;The structure is simply supported, and the load is a concentrated load acting on the symmetry axis of the structure F=1000N; Orthotropic material properties Orthotropic material properties: E x =146.86GPa,E y =10.62GPa,G xy =5.45GPa, v xy =0.33; the number of ply angles m=36, the ply angles are

[0051] 90° / 85° / 80° / 75° / 70° / 65° / 60° / 55° / 50° / 45° / 40° / 35° / 30° / 25° / 20° / 15° / 10° / 5° / 0° / -5° / -10° / -15° / -20° / -25° / -30° / -35° / -40° / -45° / -50° / -55° / -60° / -65° / -70° / -75° / -80° / -85°; given initial value of design variable x ik =0; use the following formula to set the coefficient s jk value of

[0052] s jk = 1