84 6 Elastic Constants Based on Global Coordinate System function y =Etayxy(Sbar) %Etayxy This function returns the coefficient of % mutual influence of the first kind % ETAy,xy in the global coordinate system. % It has one argument -the reduced % transformed compliance matrix Sbar. Etayxy is returned as a scalar y Sbar(2,3)/Sbar(3,3); Example 6.1 Derive the expression for Ez given in (6.1). Solution From an elementary course on mechanics of materials,we have the following relation (assuming uniaxial tension with o0 and all other stresses zeros): =号 (6.14) However,from (5.10),we also have the following relation: Ex=S110 (6.15) Comparing (6.14)and (6.15),we conclude the following: 官=81 1 (6.16) Substituting for S1 from (5.16a)and taking the inverse of(6.16),we obtain the desired result as follows: 1 E B=m+(盒-2na)2m2+0m (6.17) In the above equation,we have substituted for the elements of the reduced compliance matrix with the appropriate elastic constants. MATLAB Example 6.2 Consider a graphite-reinforced polymer composite lamina with the elastic constants as given in Example 2.2.Use MATLAB to plot the values of the five elastic constants Er,vry,Ey,vur,and Gry as a function of the orientation angle 0 in the range -π/2≤0≤π/2
84 6 Elastic Constants Based on Global Coordinate System function y = Etayxy(Sbar) %Etayxy This function returns the coefficient of % mutual influence of the first kind % ETAy,xy in the global coordinate system. % It has one argument - the reduced % transformed compliance matrix Sbar. % Etayxy is returned as a scalar y = Sbar(2,3)/Sbar(3,3); Example 6.1 Derive the expression for Ex given in (6.1). Solution From an elementary course on mechanics of materials, we have the following relation (assuming uniaxial tension with σx = 0 and all other stresses zeros): εx = σx Ex (6.14) However, from (5.10), we also have the following relation: εx = S¯11σx (6.15) Comparing (6.14) and (6.15), we conclude the following: 1 Ex = S¯11 (6.16) Substituting for S¯11 from (5.16a) and taking the inverse of (6.16), we obtain the desired result as follows: Ex = 1 S¯11 = E1 m4 + E1 G12 − 2ν12 n2m2 + E1 E2 n4 (6.17) In the above equation, we have substituted for the elements of the reduced compliance matrix with the appropriate elastic constants. MATLAB Example 6.2 Consider a graphite-reinforced polymer composite lamina with the elastic constants as given in Example 2.2. Use MATLAB to plot the values of the five elastic constants Ex, νxy, Ey, νyx, and Gxy as a function of the orientation angle θ in the range −π/2 ≤ θ ≤ π/2.
6.2 MATLAB Functions Used 85 Solution This example is solved using MATLAB.The elastic modulus Er is calculated at each value of 0 between -90 and 900 in increments of 10 using the MATLAB function Ex. >Ex1=Ex(155.0,12.10,0.248,4.40,-90) Ex1= 12.1000 >>Ex2=Ex(155.0,12.10,0.248,4.40,-80) Ex2 11.8632 >Ex3=Ex(155.0,12.10,0.248,4.40,-70) Ex3 11.4059 >Ex4=Ex(155.0,12.10,0.248,4.40,-60) Ex4 11.2480 >>Ex5=Ex(155.0,12.10,0.248,4.40,-50) Ex5 11.9204 >Ex6=Ex(155.0,12.10,0.248,4.40,-40) Ex6= 14.1524 >Ex7=Ex(155.0,12.10,0.248,4.40,-30) Ex7 19.6820 >>Ex8=Ex(155.0,12.10,0.248,4.40,-20)
6.2 MATLAB Functions Used 85 Solution This example is solved using MATLAB. The elastic modulus Ex is calculated at each value of θ between −90◦ and 90◦ in increments of 10◦ using the MATLAB function Ex . >> Ex1 = Ex(155.0, 12.10, 0.248, 4.40, -90) Ex1 = 12.1000 >> Ex2 = Ex(155.0, 12.10, 0.248, 4.40, -80) Ex2 = 11.8632 >> Ex3 = Ex(155.0, 12.10, 0.248, 4.40, -70) Ex3 = 11.4059 >> Ex4 = Ex(155.0, 12.10, 0.248, 4.40, -60) Ex4 = 11.2480 >> Ex5 = Ex(155.0, 12.10, 0.248, 4.40, -50) Ex5 = 11.9204 >> Ex6 = Ex(155.0, 12.10, 0.248, 4.40, -40) Ex6 = 14.1524 >> Ex7 = Ex(155.0, 12.10, 0.248, 4.40, -30) Ex7 = 19.6820 >> Ex8 = Ex(155.0, 12.10, 0.248, 4.40, -20)
86 6 Elastic Constants Based on Global Coordinate System Ex8= 34.1218 >Ex9=Ex(155.0,12.10,0.248,4.40,-10) Ex9= 78.7623 >>Ex10=Ex(155.0,12.10,0.248,4.40,0) Ex10= 155 >Ex11=Ex(155.0,12.10,0.248,4.40,10) Ex11= 78.7623 >>Ex12=Ex(155.0,12.10,0.248,4.40,20) Ex12= 34.1218 >>Ex13=Ex(155.0,12.10,0.248,4.40,30) Ex13= 19.6820 >Ex14=Ex(155.0,12.10,0.248,4.40,40) Ex14= 14.1524 >Ex15=Ex(155.0,12.10,0.248,4.40,50) Ex15= 11.9204 >>Ex16=Ex(155.0,12.10,0.248,4.40,60)
86 6 Elastic Constants Based on Global Coordinate System Ex8 = 34.1218 >> Ex9 = Ex(155.0, 12.10, 0.248, 4.40, -10) Ex9 = 78.7623 >> Ex10 = Ex(155.0, 12.10, 0.248, 4.40, 0) Ex10 = 155 >> Ex11 = Ex(155.0, 12.10, 0.248, 4.40, 10) Ex11 = 78.7623 >> Ex12 = Ex(155.0, 12.10, 0.248, 4.40, 20) Ex12 = 34.1218 >> Ex13 = Ex(155.0, 12.10, 0.248, 4.40, 30) Ex13 = 19.6820 >> Ex14 = Ex(155.0, 12.10, 0.248, 4.40, 40) Ex14 = 14.1524 >> Ex15 = Ex(155.0, 12.10, 0.248, 4.40, 50) Ex15 = 11.9204 >> Ex16 = Ex(155.0, 12.10, 0.248, 4.40, 60)
6.2 MATLAB Functions Used 87 Ex16= 11.2480 >>Ex17=Ex(155.0,12.10,0.248,4.40,70) Ex17= 11.4059 >>Ex18=Ex(155.0,12.10,0.248,4.40,80) Ex18= 11.8632 >Ex19=Ex(155.0,12.10,0.248,4.40,90) Ex19= 12.1000 The x-axis is now setup for the plots as follows: >>x=[-90-80-70-60-50-40-30-20-100102030405060 708090] X= -90-80-70-60-50-40-30-20-100102030 405060708090 The values of Er are now calculated for each value of between-90 and 90 in increments of 10°. >y1 [Ex1 Ex2 Ex3 Ex4 Ex5 Ex6 Ex7 Ex8 Ex9 Ex10 Ex11 Ex12 Ex13 Ex14 Ex15 Ex16 Ex17 Ex18 Ex19] y1= Columns 1 through 14 12.100011.8632 11.4059 11.2480 11.920414.152419.6820 34.1218 78.7623155.0000 78.7623 34.1218 19.682014.1524 Columns 15 through 19 11.920411.248011.405911.8632 12.1000
6.2 MATLAB Functions Used 87 Ex16 = 11.2480 >> Ex17 = Ex(155.0, 12.10, 0.248, 4.40, 70) Ex17 = 11.4059 >> Ex18 = Ex(155.0, 12.10, 0.248, 4.40, 80) Ex18 = 11.8632 >> Ex19 = Ex(155.0, 12.10, 0.248, 4.40, 90) Ex19 = 12.1000 The x-axis is now setup for the plots as follows: >> x = [-90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90] x = -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 The values of Ex are now calculated for each value of θ between −90◦ and 90◦ in increments of 10◦. >> y1 = [Ex1 Ex2 Ex3 Ex4 Ex5 Ex6 Ex7 Ex8 Ex9 Ex10 Ex11 Ex12 Ex13 Ex14 Ex15 Ex16 Ex17 Ex18 Ex19] y1 = Columns 1 through 14 12.1000 11.8632 11.4059 11.2480 11.9204 14.1524 19.6820 34.1218 78.7623 155.0000 78.7623 34.1218 19.6820 14.1524 Columns 15 through 19 11.9204 11.2480 11.4059 11.8632 12.1000
88 6 Elastic Constants Based on Global Coordinate System 760 140 120 100 60 20 80-6040-20020406080100 (degrees) Fig.6.1.Variation of E versus 0 for Example 6.2 The plot of the values of Er versus 0 is now generated using the following commands and is shown in Fig.6.1.Notice that this modulus is an even function of 0.Notice also the rapid variation of the modulus as 0 increases or decreases from 0°. >plot(x,y1) >xlabel('\theta (degrees)'); >ylabel('E_x (GPa)'); Next,Poisson's ratio vry is calculated at each value of 0 between-90 and 90 in increments of 10 using the MATLAB function NUry. >NUxy1=NUxy(155.0,12.10,0.248,4.40,-90) NUxy1 0.0194 >NUxy2=NUxy(155.0,12.10,0.248,4.40,-80) NUxy2 0.0640 >NDxy3=NUxy(155.0,12.10,0.248,4.40,-70)
88 6 Elastic Constants Based on Global Coordinate System Fig. 6.1. Variation of Ex versus θ for Example 6.2 The plot of the values of Ex versus θ is now generated using the following commands and is shown in Fig. 6.1. Notice that this modulus is an even function of θ. Notice also the rapid variation of the modulus as θ increases or decreases from 0◦. >> plot(x,y1) >> xlabel(‘\theta (degrees)’); >> ylabel(‘E_x (GPa)’); Next, Poisson’s ratio νxy is calculated at each value of θ between −90◦ and 90◦ in increments of 10◦ using the MATLAB function NUxy. >> NUxy1 = NUxy(155.0, 12.10, 0.248, 4.40, -90) NUxy1 = 0.0194 >> NUxy2 = NUxy(155.0, 12.10, 0.248, 4.40, -80) NUxy2 = 0.0640 >> NUxy3 = NUxy(155.0, 12.10, 0.248, 4.40, -70)