120 7 Laminate Analysis-Part I Strains(eps_ro,eps-yo,gam_ryo,kap_ro,kap-yo,kap_ryo,z)-This function calcu- lates the three strains er,Ey,and yry at any point P on the normal line given the three strainsand the three curvaturesat point po,and the distance z between P and po.There are seven input arguments to this function. The function returns the 3 x 1 strain vector. The following is a listing of the MATLAB source code for this function: function y Strains(eps_xo,eps_yo,gam_xyo,kap_xo,kap_yo,kap_xyo,z) %Strains This function returns the strain vector at any point P along the normal line at distance z from point Po which % lies on the reference surface.There are seven input arguments for this function-namely the three strains and three curvatures at point Po and the distance z. 名 The size of the strain vector is 3 x 1. epsilonx eps_xo +z kap_xo; epsilony eps_yo z kap_yo; gammaxy gam_xyo +z kap_xyo; y [epsilonx epsilony gammaxy]; MATLAB Example 7.1 Consider a graphite-reinforced polymer composite laminate with the elastic con- stants as given in Example 2.2.The laminate has total thickness of 0.500 mm and is stacked as a [0/90s laminate.The four layers are of equal thickness.It is deformed so that at a point (r,y)on the reference surface,we have the following strains and curvatures: €2=400×10-6 9=2g=k2=K9=k2y=0 0 Use MATLAB to determine the following: (a)the three components of strain at the interface locations. (b)the three components of stress in each layer.Plot the stress distribution along the depth of the laminate for each component. (c)the force and moment resultants in the laminate. (d)the three components of strain at the interface locations with respect to the principal material system. (e)the three components of stress in each layer with respect to the principal material system. Solution This example is solved using MATLAB.First the strains are calculated at the five interface locations using the MATLAB function Strains as follows:
120 7 Laminate Analysis – Part I Strains(eps xo, eps yo, gam xyo, kap xo, kap yo, kap xyo, z) – This function calculates the three strains εx, εy, and γxy at any point P on the normal line given the three strains ε0 x, ε0 y, γ0 xy and the three curvatures κ0 x, κ0 y, κ0 xy at point P0, and the distance z between P and P0. There are seven input arguments to this function. The function returns the 3 × 1 strain vector. The following is a listing of the MATLAB source code for this function: function y = Strains(eps_xo,eps_yo,gam_xyo,kap_xo,kap_yo,kap_xyo,z) %Strains This function returns the strain vector at any point P % along the normal line at distance z from point Po which % lies on the reference surface. There are seven input % arguments for this function - namely the three strains % and three curvatures at point Po and the distance z. % The size of the strain vector is 3 x 1. epsilonx = eps_xo+z* kap_xo; epsilony = eps_yo+z* kap_yo; gammaxy = gam_xyo+z* kap_xyo; y = [epsilonx ; epsilony ; gammaxy]; MATLAB Example 7.1 Consider a graphite-reinforced polymer composite laminate with the elastic constants as given in Example 2.2. The laminate has total thickness of 0.500 mm and is stacked as a [0/90]S laminate. The four layers are of equal thickness. It is deformed so that at a point (x, y) on the reference surface, we have the following strains and curvatures: ε 0 x = 400 × 10−6 ε 0 y = γ0 xy = κ0 x = κ0 y = κ0 xy = 0 Use MATLAB to determine the following: (a) the three components of strain at the interface locations. (b) the three components of stress in each layer. Plot the stress distribution along the depth of the laminate for each component. (c) the force and moment resultants in the laminate. (d) the three components of strain at the interface locations with respect to the principal material system. (e) the three components of stress in each layer with respect to the principal material system. Solution This example is solved using MATLAB. First the strains are calculated at the five interface locations using the MATLAB function Strains as follows:
7.2 MATLAB Functions Used 121 >>eps11on1=Strains(400e-6,0,0,0,0,0,-0.250e-3) epsilon1 1.0e-003* 0.4000 0 0 >>epsi1on2=Strains(400e-6,0,0,0,0,0,-0.125e-3) epsilon2 1.0e-003* 0.4000 0 0 >>epsi1on3=Strains(400e-6,0,0,0,0,0,0) epsilon3 1.0e-003* 0.4000 0 0 >>epsi1on4=Strains(400e-6,0,0,0,0,0,0.125e-3) epsilon4 1.0e-003* 0.4000 0 0 >>epsi1on5=Strains(400e-6,0,0,0,0,0,0.250e-3) epsilon5 1.0e-003* 0.4000 0 0
7.2 MATLAB Functions Used 121 >> epsilon1 = Strains(400e-6,0,0,0,0,0,-0.250e-3) epsilon1 = 1.0e-003 * 0.4000 0 0 >> epsilon2 = Strains(400e-6,0,0,0,0,0,-0.125e-3) epsilon2 = 1.0e-003 * 0.4000 0 0 >> epsilon3 = Strains(400e-6,0,0,0,0,0,0) epsilon3 = 1.0e-003 * 0.4000 0 0 >> epsilon4 = Strains(400e-6,0,0,0,0,0,0.125e-3) epsilon4 = 1.0e-003 * 0.4000 0 0 >> epsilon5 = Strains(400e-6,0,0,0,0,0,0.250e-3) epsilon5 = 1.0e-003 * 0.4000 0 0
122 7 Laminate Analysis-Part I Next,the reduced stiffness [Q}in GPa is calculated for this material using the MATLAB function ReducedStiffness as follows: >>Q=ReducedStiffness(155.0,12.10,0.248,4.40) Q= 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 The transformed reduced stiffnesses [Q]in GPa for the four layers are now cal- culated using the MATLAB function Qbar as follows: >Qbar1 =Qbar(Q,0) Qbar1 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 >Qbar2 Qbar(Q,90) Qbar2 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 >Qbar3 Qbar(Q,90) Qbar3 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 >Qbar4 Qbar(Q,0) Qbar4 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next,the stresses in each layer are calculated in MPa.Note that the stress vector is calculated twice for each layer-once at the top of the layer and once at the bottom of the layer
122 7 Laminate Analysis – Part I Next, the reduced stiffness [Q} in GPa is calculated for this material using the MATLAB function ReducedStiffness as follows: >> Q = ReducedStiffness(155.0, 12.10, 0.248, 4.40) Q = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 The transformed reduced stiffnesses [Q¯] in GPa for the four layers are now calculated using the MATLAB function Qbar as follows: >> Qbar1 = Qbar(Q,0) Qbar1 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 >> Qbar2 = Qbar(Q,90) Qbar2 = 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 >> Qbar3 = Qbar(Q,90) Qbar3 = 12.1584 3.0153 -0.0000 3.0153 155.7478 0.0000 -0.0000 0.0000 4.4000 >> Qbar4 = Qbar(Q,0) Qbar4 = 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Next, the stresses in each layer are calculated in MPa. Note that the stress vector is calculated twice for each layer – once at the top of the layer and once at the bottom of the layer
7.2 MATLAB Functions Used 123 >sigmala Qbar1*epsilon1*1e3 sigmala 62.2991 1.2061 0 >sigmaib Qbar1*epsilon2*1e3 sigmalb 62.2991 1.2061 0 >sigma2a =Qbar2*epsilon2*1e3 sigma2a 4.8634 1.2061 -0.0000 >sigma2b Qbar2*epsilon3*1e3 sigma2b 4.8634 1.2061 -0.0000 >sigma3a Qbar3*epsilon3*1e3 sigma3a 4.8634 1.2061 -0.0000 >sigma3b =Qbar3*epsilon4*1e3 sigma3b 4.8634 1.2061 -0.0000
7.2 MATLAB Functions Used 123 >> sigma1a = Qbar1*epsilon1*1e3 sigma1a = 62.2991 1.2061 0 >> sigma1b = Qbar1*epsilon2*1e3 sigma1b = 62.2991 1.2061 0 >> sigma2a = Qbar2*epsilon2*1e3 sigma2a = 4.8634 1.2061 -0.0000 >> sigma2b = Qbar2*epsilon3*1e3 sigma2b = 4.8634 1.2061 -0.0000 >> sigma3a = Qbar3*epsilon3*1e3 sigma3a = 4.8634 1.2061 -0.0000 >> sigma3b = Qbar3*epsilon4*1e3 sigma3b = 4.8634 1.2061 -0.0000
124 7 Laminate Analysis-Part I >sigma4a =Qbar4*epsilon4*1e3 sigma4a 62.2991 1.2061 0 >sigma4b Qbar4*epsilon5*1e3 sigma4b 62.2991 1.2061 0 Next,we setup the y-axis for the three plots: >y=[0.2500.1250.12500-0.125-0.125-0.250] y= 0.2500 0.1250 0.125000-0.1250 -0.1250 -0.2500 The distribution of the stress o along the depth of the laminate is now plotted as follows (see Fig.7.4): >>x [sigma4b(1)sigma4a(1)sigma3b(1)sigma3a(1)sigma2b(1) sigma2a(1)sigma1b(1)sigmala(1)] X= 62.2991 62.29914.8634 4.8634 4.8634 4.8634 62.2991 62.2991 >plot(x,y) >xlabel('\sigma_x (MPa)') >ylabel(‘z(mm)') The distribution of the stress ou along the depth of the laminate is now plotted as follows (see Fig.7.5): >x [sigma4b(2)sigma4a(2)sigma3b(2)sigma3a(2)sigma2b(2) sigma2a(2)sigma1b(2)sigmala(2)] X= 1.2061 1.20611.20611.20611.20611.20611.2061 1.2061 >plot(x,y) >ylabel(‘z(mm)') >xlabel('\sigma_y (MPa)')
124 7 Laminate Analysis – Part I >> sigma4a = Qbar4*epsilon4*1e3 sigma4a = 62.2991 1.2061 0 >> sigma4b = Qbar4*epsilon5*1e3 sigma4b = 62.2991 1.2061 0 Next, we setup the y-axis for the three plots: >> y = [0.250 0.125 0.125 0 0 -0.125 -0.125 -0.250] y = 0.2500 0.1250 0.1250 0 0 -0.1250 -0.1250 -0.2500 The distribution of the stress σx along the depth of the laminate is now plotted as follows (see Fig. 7.4): >> x = [sigma4b(1) sigma4a(1) sigma3b(1) sigma3a(1) sigma2b(1) sigma2a(1) sigma1b(1) sigma1a(1)] x = 62.2991 62.2991 4.8634 4.8634 4.8634 4.8634 62.2991 62.2991 >> plot(x,y) >> xlabel(‘\sigma_x (MPa)’) >> ylabel(‘z (mm)’) The distribution of the stress σy along the depth of the laminate is now plotted as follows (see Fig. 7.5): >> x = [sigma4b(2) sigma4a(2) sigma3b(2) sigma3a(2) sigma2b(2) sigma2a(2) sigma1b(2) sigma1a(2)] x = 1.2061 1.2061 1.2061 1.2061 1.2061 1.2061 1.2061 1.2061 >> plot(x,y) >> ylabel(‘z (mm)’) >> xlabel(‘\sigma_y (MPa)’)