复合材料结构力学例题-MATLAB.doc

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.mfunction 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 MATLAB2.2Example.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(45) woven fabric layer made of graphite fibers and epoxy resin.Solution:We approximate the woven fabric by a layer made of one 45and one -45 ply. The properties of these plies are taken to be those of the graphite epoxy unidirectional ply given in Example 2.2.For such a layer the elements of the stiffness matrix are calculated from:(1) Reduced stiffness matrix:% ReducedStiffness.mfunction 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-NU12*NU21) NU12*E2/(1-NU12*NU21) 0;NU12*E2/(1-NU12*NU21) E2/(1-NU12*NU21) 0;0 0 G12;(2) Transformed stiffness matrix:% Qbar.mfunction y=Qbar(Q,theta)%Qbar This function returns the transformed reduced stiffness matrix Qbar given the% reduced stiffness matrix Q and the orientation angle theta.% There are two argument representing Q and theta% The size of the matrix is 3 x 3.% The angle theta must be given in degrees.m=cos(theta*pi/180);n=sin(theta*pi/180);T1=m*m n*n -2*m*n;n*n m*m 2*m*n;m*n -m*n m*m-n*n;T2=m*m n*n m*n;n*n m*m -m*n;-2*m*n 2*m*n m*m-n*n;y=T1*Q*T2;(3) Transformed compliance matrix:The engineering constants of the woven fabric are:;2.4Example. The engineering constants of a graphite epoxy unidirectional ply are given as:,Determine whether or not this set of constant is valid.Solution: There are two ways to find the answer to this problem. Method 1. The compliance matrix is :% TransverselyIso1tropicCompliance.mfunction 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;The eigenvalues of this matrix are:Since every eigenvalue is positive, the specified set of engineering constants is valid.Method2. For a transversely isotropic material the engineering constants must satisfy the inequalities:In terms of the engineering constants these inequalities are: Since the inequalities are satisfied, the specified set of engineering constants is valid. The example illustrates that care should be taken to use proper values of the Poisson ratios. The and Poisson ratios (referred to as the major and minor Poisson ratios) must not be interchanged.3.1Example. Calculate the stiffnessand the compliance matrices of a laminate made of graphite epoxy unidirectional plies. The ply properties are given in Table 1.Table 1 Properties of the material used in the examplesLongitudinal Youngs modulus (GPa)14816.39Transverse Youngs modulus (GPa)9.6516.39Longitudinal shear modulus (GPa)4.5538.19Longitudinal Poissons modulus (GPa)0.30.801Thickness (mm)0.10.2Solution: The stiffness matrix of a unidirectional ply with the fibers in the 0-degree direction is . The stiffness matrix is given by:The stiffness matrix of a ply not in the 0-degree direction is Where and are given by:% Tstress.mfunction y=Tstress(theta)%T This function returns the transformation matrix T given the% orientation angle theta under plane-stress conditions.% There is only one argument representing theta% The size of the matrix is 3 x 3.% The angle theta must be given in degrees.m=cos(theta*pi/180);n=sin(theta*pi/180);y=m*m n*n 2*m*n;n*n m*m -2*m*n;-m*n m*n m*m-n*n;-% Tstrain.mfunction y=Tstrain(theta)%Tinv This function returns the inverse of transformation matrix T given the% orientation angle theta,under plane-strain conditions.% There is only one argument representing theta% The size of the matrix is 3 x 3.% The angle theta must be given in degrees.m=cos(theta*pi/180);n=sin(theta*pi/180);y=m*m n*n m*n;n*n m*m -m*n;-2*m*n 2*m*n m*m-n*n;We obtain the stiffness matrix of the 45-degree ply as follows:% UnidirectionalQbar.mfunction y=UnidirectionalQbar(Q,theta)%Qbar This function returns the transformed reduced stiffness matrix Qbar given the% reduced stiffness matrix Q and the orientation angle theta.% There are two argument representing Q and theta% The size of the matrix is 3 x 3.% The angle theta must be given in degrees.m=cos(theta*pi/180);n=sin(theta*pi/180);T1=m*m n*n -2*m*n;n*n m*m 2*m*n;m*n -m*n m*m-n*n;T2=m*m n*n m*n;n*n m*m -m*n;-2*m*n 2*m*n m*m-n*n;y=T1*Q*T2;-=The layup is shown in Fig 3.1. In calculating the matrices we treat the ten 0-degree plies as one layer and the ten 45-degree plies as another layer. The matrices are:(1)% Amatrix.mfunction y=Amatrix(A,Qbar,z1,z2)%Amatrix This function returns the A matrix after the layer k with% stiffness Qbar is assembled.% A-A matrix after layer k is assembled.% Qbar-Qbar matrix for layer k% z1-z(k-1) for layer k% z2-z(k) for layer k for i=1:3 for j=1:3 A(i,j)=A(i,j)+Qbar(i,j)*(z2-z1); endendy=A;(2)% Bmatrix.mfunction y=Bmatrix(B,Qbar,z1,z2)%Bmatrix This function returns the B matrix after the layer k with% stiffness Qbar is assembled.% B-B matrix after layer k is assembled.% Qbar-Qbar matrix for layer k% z1-z(k-1) for layer k% z2-z(k) for layer k for i=1:3 for j=1:3 B(i,j)=B(i,j)+Qbar(i,j)*(z22-z12); endendy=B;(3)% Dmatrix.mfunction y=Dmatrix(D,Qbar,z1,z2)%Dmatrix This function returns the D matrix after the layer k with% stiffness Qbar is assembled.% D-D matrix after layer k is assembled.% Qbar-Qbar matrix for layer k% z1-z(k-1) for layer k% z2-z(k) for layer k for i=1:3 for j=1:3 D(i,j)=D(i,j)+Qbar(i,j)*(z23-z13); endendy=D;-The compliance matrices are:Hence, we have:3.2 Example. Calculate the stiffnessand the compliance matrices of a laminate made of graphite epoxy unidirectional plies. The ply properties are given in Table 1.Solution: The unidirectional laminate is symmetrical, and the matrix is zero: By treating the 20 plies as a single layer, the and matrices are:Where is the thickness of the laminate.3.3 Example. Calculate the stiffness and the compliance matrices of (i) a laminated composite consisting of two layers of woven fabric, twelve layers of 0-degree unidirectional plies, and two layers of woven fabric ; and (ii) a laminate consisting five layers of woven fabric and ten layers of 0-degree unidirectional plies . The material properties are given in Table 3.1.Solution: First we consider the laminate with layup (Fig. 3.2). The laminate is symmetrical, and the matrix is zero:The compliance matrix of a unidirectional ply with the fibers in the 0-degree direction is . The stiffness matrix is :Figure 3.2 The layup of the laminate in Example 3.3. The superscript f denotes fabric.For a woven fabric the stiffness matrix is:In calculating the matrices we treat the twelve 0-degree plies as two layers and each adjacent woven fabric as one layer. The and matrices are:The distance (in meters) are , With these values yields:The compliance matrices and are:The compliance and stiffness matrices of the laminate are calculated similarly. The results are given as follows:4.1Example. A 0.7-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy. The material properties are given in Table 1. The layup is . The 0-degree plies are parallel to the short edge of the plate. The plate is simply supported along all four edges and is subjected to a uniformly distributed transverse load . Calculate the maximum deflection, the maximum bending moments, and the stresses and strains in each layer.Solution. The bending stiffnesses of the plates are and . We may treat this plate as long when the following condition is met:In the present problem the terms in this inequality are and .Thus, the preceding condition is satisfied and the long-plate expression may be used. The maximum deflection of a simply supported beam is The maximum deflection of the plate is obtained by replacing by . For the plate under consideration 、and , and we have The bending moments are For a long plate and are zero, and and are The maximum bending moment , which arises at , is From the above equations, we have The maximum bending moment (at is For the long plate the strains at are The strain distribution at is shown in Figure 4.6. The stresses are calculated by The stiffness matrices for the fabric and for the unidirectional layer are given by example 3.3. The stresses in the top layer (where ) at are The stresses in the other layers are calculated similarly. The results are shown in Figure 4.6.4.2 Example. A 0.7-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy. The material properties are given in Example 2.1. the layup is . The 0-degree plies are parallel to the short edge of the plate. The plate is simply supported along all four edges and is subjected to a uniformly distributed transverse load (Fig.4.8). Calculate the maximum deflection and the maximum bending moments.Solution: The deflection of the plate is 4.5 Example. A 0.7-m-long and 0.2-m-wide rectangular plate is made of graphite epoxy. The material properties are given in Example 2.1. the layup is . The 0-degree plies are parallel to the short edge of the plate. The plate is simply supported along all four edges. (Fig.4.15). the plate is subjected to uniform compressive loads along the long edges. Calculate the buckling load.Solution: From eq.(1), with and with the components of stiffness , we have (1) (2)The values of areThe lowest value is , which corresponds to . Thus, the lowest