复合材料的matlab计算,复合材料结构力学例题-MATLAB.doc

复合材料结构力学例题-MATLAB

2.1Example.Calculate the elements of the stiffness and compliance matrices of a graphite expoxy unidirectional ply.The engineering constants are given as :

,,,,

Solution:

(1)compliance with MATLAB

% TransverselyIso1tropicCompliance.m

function y=TransverselyIso1tropicCompliance(E1,E2,NU12,NU23,G12)

%TransverselyIsotropicCompliance This function returns the compliance matrix for

% Transversely isotropic matrials.There are five arguments

% representing the five independent material

% constants.The size of the compliance matrix is 6x6.

y=[1/E1 -NU12/E1 -NU12/E1 0 0 0;-NU12/E1 1/E2 -NU23/E2 0 0 0;-NU12/E1 -NU23/E2 1/E2 0 0 0;0 0 0 2*(1+NU23)/E2 0 0;0 0 0 0 1/G12 0;0 0 0 0 0 1/G12];

(2)Stiffness with MATLAB

2.2Example.Calculate the elements of the stiffness and compliance matrices of a graphite expoxy unidirectional ply under plane-stress condition.The engineering constants are given as :

,,,,

Solution: the result is the same as example2.1.

2.3Example.Esitmate the plane-stress stiffness matrix and the engineering constants of a(?5) woven fabric layer made of graphite fibers and epoxy resin.

Solution:

We approximate the woven fabric by a layer made of one 45癮nd one -45?ply. The properties of these plies are taken to be those of the graphite epoxy unidirectional ply given in Example 2.2.

siFor such a layer the elements of the stiffness matrix are calculated from:

Reduced stiffness matrix:

% ReducedStiffness.m

function y=ReducedStiffness(E1,E2,NU12,G12)

%ReducedStiffness This function returns the reduced stiffness matrix

% for fiber-reinforced materials.

% There are four arguments representing four materials

% constants.The size of the reuduced stiffness matrix

% is 3 x 3.

NU21=NU12*E2/E1;

y=[E1/(1-NU12NU21) NU12E2/(1-NU12NU21) 0;NU12E2/(1-NU12NU21) E2/(1-NU12NU21) 0;0 0 G12];

Transformed st