《纺织复合材料》课程参考文献（Mechanics of Composite Materials with MATLAB）05 Global Coordinate System

5 Global Coordinate System 5.1 Basic Equations In this chapter,we will refer the response of each layer (lamina)of material to the same global system.We accomplish this by transforming the stress-strain relations for the lamina 1-2-3 coordinate system into the global coordinate sys- tem.This transformation will be done for the state of plane stress using the standard transformation relations for stresses and strains given in introduc- tory courses in mechanics of materials [1]. Consider an isolated infinitesimal element in the principal material coor- dinate system (1-2-3 system)that will be transformed into the z-y-z global coordinate system as shown in Fig.5.1.The fibers are oriented at angle with respect to the +x axis of the global system.The fibers are parallel to the z-y plane and the 3 and z axes coincide.The orientation angle 6 will be considered positive when the fibers rotate counterclockwise from the +x axis toward the +y axis. The stresses on the small volume of element are now identified with respect to the r-y-z system.The six components of stress are now oz,oy,o=,Tyz, T,and Ty,while the six components of strain are ,y,,,Y,and Yry (see Fig.5.2). Note that in a plane stress state,it follows that the out-of-plane stress components in the z-y-z global coordinate system are zero,i.e.o:=Tyz= Tz2=0(see Problem 5.1). The stress transformation relation is given as follows for the case of plane stress: 2 n2 2mn n2 23 -2m (5.1) T12 -mn mn m2-n2 where m cos and n sin 0.The above relation is written in compact form as follows:

5 Global Coordinate System 5.1 Basic Equations In this chapter, we will refer the response of each layer (lamina) of material to the same global system. We accomplish this by transforming the stress-strain relations for the lamina 1-2-3 coordinate system into the global coordinate sys￾tem. This transformation will be done for the state of plane stress using the standard transformation relations for stresses and strains given in introduc￾tory courses in mechanics of materials [1]. Consider an isolated infinitesimal element in the principal material coor￾dinate system (1-2-3 system) that will be transformed into the x-y-z global coordinate system as shown in Fig. 5.1. The fibers are oriented at angle θ with respect to the +x axis of the global system. The fibers are parallel to the x-y plane and the 3 and z axes coincide. The orientation angle θ will be considered positive when the fibers rotate counterclockwise from the +x axis toward the +y axis. The stresses on the small volume of element are now identified with respect to the x-y-z system. The six components of stress are now σx, σy, σz, τyz, τxz, and τxy, while the six components of strain are εx, εy, εz, γyz, γxz, and γxy (see Fig. 5.2). Note that in a plane stress state, it follows that the out-of-plane stress components in the x-y-z global coordinate system are zero, i.e. σz = τyz = τxz = 0 (see Problem 5.1). The stress transformation relation is given as follows for the case of plane stress: ⎧ ⎨ ⎩ σ1 σ2 τ12 ⎫ ⎬ ⎭ = ⎡ ⎣ m2 n2 2mn n2 m2 −2mn −mn mn m2 − n2 ⎤ ⎦ ⎧ ⎨ ⎩ σx σy τxy ⎫ ⎬ ⎭ (5.1) where m = cos θ and n = sin θ. The above relation is written in compact form as follows:

58 5 Global Coordinate System Z.3 Fig.5.1.A infinitesimal fiber-reinforced composite element showing the local and global coordinate systems 2 y Fig.5.2.An infinitesimal fiber-reinforced composite element showing the stress components in the global coordinate system

58 5 Global Coordinate System Fig. 5.1. A infinitesimal fiber-reinforced composite element showing the local and global coordinate systems Fig. 5.2. An infinitesimal fiber-reinforced composite element showing the stress components in the global coordinate system

5.1 Basic Equations 59 }- (5.2) where [T is the transformation matrix given as follows: 3 2 2mn [T= n2 m2 -2mn (5.3) -mn mn m2-n2 The inverse of the matrix [T]is [T]-given as follows(see Problem 5.3): 「m2 n2 -2mn T]-1= n2 m2 2mn (5.4) m -mn m2-n2 where [T]is used in the following equation: (5.5) Similar transformation relations hold for the strains as follows: E1 E2 T (5.6) 312 =四1 (5.7) 12 Note that the strain transformation (5.6)and (5.7)include a factor of 1/2 with the engineering shear strain.Therefore (4.5)and (4.6)of Chap.4 are modified now to include this factor as follows: S12 (5.8) 0 11 Q12 Q22 (5.9) 0 2Q66

5.1 Basic Equations 59 ⎧ ⎨ ⎩ σ1 σ2 τ12 ⎫ ⎬ ⎭ = [T] ⎧ ⎨ ⎩ σx σy τxy ⎫ ⎬ ⎭ (5.2) where [T] is the transformation matrix given as follows: [T] = ⎡ ⎣ m2 n2 2mn n2 m2 −2mn −mn mn m2 − n2 ⎤ ⎦ (5.3) The inverse of the matrix [T] is [T] −1 given as follows (see Problem 5.3): [T] −1 = ⎡ ⎣ m2 n2 −2mn n2 m2 2mn mn −mn m2 − n2 ⎤ ⎦ (5.4) where [T] −1 is used in the following equation: ⎧ ⎪⎨ ⎪⎩ σx σy τxy ⎫ ⎪⎬ ⎪⎭ = [T] −1 ⎧ ⎪⎨ ⎪⎩ σ1 σ2 τ12 ⎫ ⎪⎬ ⎪⎭ (5.5) Similar transformation relations hold for the strains as follows: ⎧ ⎪⎨ ⎪⎩ ε1 ε2 1 2 γ12 ⎫ ⎪⎬ ⎪⎭ = [T] ⎧ ⎪⎨ ⎪⎩ εx εy 1 2 γxy ⎫ ⎪⎬ ⎪⎭ (5.6) ⎧ ⎪⎨ ⎪⎩ εx εy 1 2 γxy ⎫ ⎪⎬ ⎪⎭ = [T] −1 ⎧ ⎪⎨ ⎪⎩ ε1 ε2 1 2 γ12 ⎫ ⎪⎬ ⎪⎭ (5.7) Note that the strain transformation (5.6) and (5.7) include a factor of 1/2 with the engineering shear strain. Therefore (4.5) and (4.6) of Chap. 4 are modified now to include this factor as follows: ⎧ ⎪⎨ ⎪⎩ ε1 ε2 1 2 γ12 ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ S11 S12 0 S12 S22 0 0 0 1 2S66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ σ1 σ2 τ12 ⎫ ⎪⎬ ⎪⎭ (5.8) ⎧ ⎪⎨ ⎪⎩ σ1 σ2 τ12 ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ Q11 Q12 0 Q12 Q22 0 0 02Q66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ ε1 ε2 1 2 γ12 ⎫ ⎪⎬ ⎪⎭ (5.9)

60 5 Global Coordinate System Substitute(5.6)and (5.2)into (5.8)and rearrange the terms to obtain (also multiply the third row through by a factor of 2): 511 S12 516 512 522 526 (5.10) 516 526566 where the transformed reduced compliance matrix [S]is given by: 512 5161 S11 S12 0 [= 512 522 526 =[T] S12 S22 0 (5.11) 516 526 566 0 0S66 Equation(5.11)represents the complex relations that describe the response of an element of fiber-reinforced composite material in a state of plane stress that is subjected to stresses not aligned with the fibers,nor perpendicular to the fibers.In this case,normal stresses cause shear strains and shear stresses cause extensional strains.This coupling found in fiber-reinforced composite materials is called shear-ertension coupling. Similarly,we can derive the transformed reduced stiffness matrir [Q]by substituting(5.2)and(5.6)into(5.9)and rearranging the terms.We therefore obtain: Q12 Q16 Q12 02 (5.12) 16 26 where [Q]is given by: [Qu Q12 Q16 Q11 Q12 0 012 022 026 =T]-1 Q12 Q22 0 [TI (5.13) 16 Q26 Q66 0 0 Q66」 Equation (5.13) further supports the shear-extension coupling of fiber- reinforced composite materials.Note that the following relations hold between [可and[②]： [②=-1 (5.14a) [=[②1-1 (5.14b) 5.2 MATLAB Functions Used The four MATLAB functions used in this chapter to calculate the four major matrices are:

60 5 Global Coordinate System Substitute (5.6) and (5.2) into (5.8) and rearrange the terms to obtain (also multiply the third row through by a factor of 2): ⎧ ⎪⎨ ⎪⎩ εx εy γxy ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ S¯11 S¯12 S¯16 S¯12 S¯22 S¯26 S¯16 S¯26 S¯66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ σx σy τxy ⎫ ⎪⎬ ⎪⎭ (5.10) where the transformed reduced compliance matrix [S¯] is given by: [S¯] = ⎡ ⎢ ⎣ S¯11 S¯12 S¯16 S¯12 S¯22 S¯26 S¯16 S¯26 S¯66 ⎤ ⎥ ⎦ = [T] −1 ⎡ ⎢ ⎣ S11 S12 0 S12 S22 0 0 0 S66 ⎤ ⎥ ⎦ [T] (5.11) Equation (5.11) represents the complex relations that describe the response of an element of fiber-reinforced composite material in a state of plane stress that is subjected to stresses not aligned with the fibers, nor perpendicular to the fibers. In this case, normal stresses cause shear strains and shear stresses cause extensional strains. This coupling found in fiber-reinforced composite materials is called shear-extension coupling. Similarly, we can derive the transformed reduced stiffness matrix [Q¯] by substituting (5.2) and (5.6) into (5.9) and rearranging the terms. We therefore obtain: ⎧ ⎪⎨ ⎪⎩ σx σy τxy ⎫ ⎪⎬ ⎪⎭ = ⎡ ⎢ ⎣ Q¯11 Q¯12 Q¯16 Q¯12 Q¯22 Q¯26 Q¯16 Q¯26 Q¯66 ⎤ ⎥ ⎦ ⎧ ⎪⎨ ⎪⎩ εx εy γxy ⎫ ⎪⎬ ⎪⎭ (5.12) where [Q¯] is given by: [Q¯] = ⎡ ⎢ ⎣ Q¯11 Q¯12 Q¯16 Q¯12 Q¯22 Q¯26 Q¯16 Q¯26 Q¯66 ⎤ ⎥ ⎦ = [T] −1 ⎡ ⎢ ⎣ Q11 Q12 0 Q12 Q22 0 0 0 Q66 ⎤ ⎥ ⎦[T] (5.13) Equation (5.13) further supports the shear-extension coupling of fiber￾reinforced composite materials. Note that the following relations hold between [S¯] and [Q¯]: [Q¯]=[S¯] −1 (5.14a) [S¯]=[Q¯] −1 (5.14b) 5.2 MATLAB Functions Used The four MATLAB functions used in this chapter to calculate the four major matrices are:

5.2 MATLAB Functions Used 0 T(theta)-This function calculates the transformation matrix T]given the angle“theta'”.The orientation angle“theta”must be given in degrees.The returned matrix has size 3 x 3. Tinv(theta)-This function calculates the inverse of the transformation ma- trix[T]given the angle“theta”.The orientation angle“theta”must be given in degrees.The returned matrix has size 3 x 3. Sbar(S,theta)-This function calculates the transformed reduced compliance matrix [for the lamina.Its input consists of two arguments representing the reduced compliance matrix [S and the orientation angle "theta".The returned matrix has size 3 x 3. Qbar(Q,theta)-This function calculates the transformed reduced stiffness matrix [Q]for the lamina.Its input consists of two arguments representing the reduced stiffness matrix [Q]and the orientation angle "theta".The returned matrix has size3×3. The following is a listing of the MATLAB source code for each function: function y =T(theta) %T This function returns the transformation matrix T 名 given the orientation angle "theta". 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 [mtm ntn 2+m*nn*n m*m -2*m*n -m*n m*n m*m-n*n] function y Tinv(theta) XTinv This function returns the inverse of the % transformation matrix T % given the orientation angle "theta". 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*nn*n m*m 2*m*n m*n -m*n m*m-n+n]; function y Sbar(S,theta) %Sbar This function returns the transformed reduced % compliance matrix "Sbar"given the reduced % compliance matrix S and the orientation % angle "theta". % There are two arguments representing S and "theta" 名 The size of the matrix is 3 x 3. The angle "theta"must be given in degrees. m cos(theta*pi/180);

5.2 MATLAB Functions Used 61 T(theta) – This function calculates the transformation matrix [T] given the angle “theta”. The orientation angle “theta” must be given in degrees. The returned matrix has size 3 × 3. Tinv(theta) – This function calculates the inverse of the transformation ma￾trix [T] given the angle “theta”. The orientation angle “theta” must be given in degrees. The returned matrix has size 3 × 3. Sbar (S,theta) – This function calculates the transformed reduced compliance matrix [S¯] for the lamina. Its input consists of two arguments representing the reduced compliance matrix [S] and the orientation angle “theta”. The returned matrix has size 3 × 3. Qbar (Q,theta) – This function calculates the transformed reduced stiffness matrix [Q¯] for the lamina. Its input consists of two arguments representing the reduced stiffness matrix [Q] and the orientation angle “theta”. The returned matrix has size 3 × 3. The following is a listing of the MATLAB source code for each function: function y = T(theta) %T This function returns the transformation matrix T % given the orientation angle "theta". % 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]; function y = Tinv(theta) %Tinv This function returns the inverse of the % transformation matrix T % given the orientation angle "theta". % 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]; function y = Sbar(S,theta) %Sbar This function returns the transformed reduced % compliance matrix "Sbar" given the reduced % compliance matrix S and the orientation % angle "theta". % There are two arguments representing S and "theta" % The size of the matrix is 3 x 3. % The angle "theta" must be given in degrees. m = cos(theta*pi/180);

62 5 Global Coordinate System n sin(theta*pi/180); T=[m*mn知2*m知；n*nm*m-2*m知；-m*nm*nm*m-n*n]; Tinv [m*m n*n -2*m*nn*n m*m 2*m*n m*n -m*n m*m-n*n] y=Tinv*S+T; function y Qbar(Q,theta) %Qbar This function returns the transformed reduced stiffness matrix "Qbar"given the reduced 名 stiffness matrix Q and the orientation angle "theta". There are two arguments 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); T=[m*mnn2*m知；n*nm*m-2*m*知；-m*nm*nm*m-n*n]; Tinv [m*m n*n -2*m*nn*n m*m 2*m*n m*n -m*n m*m-n+n] y Tinv*Q*T; Example 5.1 Using(5.11),derive explicit expressions for the elements Sij in terms of S and8(use m and n forθ). Solution Multiply the three matrices in (5.11)as follows: 511 512 5161 「m2 n2 -2mn S11 S12 7 0 512 522 526 n2 m2 2mn S12 S22 0 516 526 566 mn -mn m2-n2 0 0 S66 23 (5.15) n2 2mn n2 m2 -2mn -mn mn m2-n2 The above multiplication can be performed either manually or using a com- puter algebra system like MAPLE or MATHEMATICA or the MATLAB Sym- bolic Math Toolbox.Therefore,we obtain the following expression: 511=S11m4+(2S12+S66)n2m2+S22n4 (5.16a) 512=(S11+S22-S66)n2m2+S12(n4+m4) (5.16b) 516=(2S11-2S12-S66)nm3-(2S22-2S12-S66)n3m (5.16c) 522=S11n4+(2512+S66)n2m2+S22m4 (5.16d) 526=(2S11-2S12-S66)n3m-(2S22-2512-S66)nm3 (5.16e 566=2(2S11+252-4S12-S66)n2m2+S66(n4+m4) (5.16f)

62 5 Global Coordinate System n = sin(theta*pi/180); T = [m*m n*n 2*m*n ; n*n m*m -2*m*n ; -m*n m*n m*m-n*n]; Tinv = [m*m n*n -2*m*n ; n*n m*m 2*m*n ; m*n -m*n m*m-n*n]; y = Tinv*S*T; function 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 arguments 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); T = [m*m n*n 2*m*n ; n*n m*m -2*m*n ; -m*n m*n m*m-n*n]; Tinv = [m*m n*n -2*m*n ; n*n m*m 2*m*n ; m*n -m*n m*m-n*n]; y = Tinv*Q*T; Example 5.1 Using (5.11), derive explicit expressions for the elements S¯ij in terms of Sij and θ (use m and n for θ). Solution Multiply the three matrices in (5.11) as follows: ⎡ ⎢ ⎣ S¯11 S¯12 S¯16 S¯12 S¯22 S¯26 S¯16 S¯26 S¯66 ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ m2 n2 −2mn n2 m2 2mn mn −mn m2 − n2 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ S11 S12 0 S12 S22 0 0 0 S66 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ m2 n2 2mn n2 m2 −2mn −mn mn m2 − n2 ⎤ ⎥ ⎦ (5.15) The above multiplication can be performed either manually or using a com￾puter algebra system like MAPLE or MATHEMATICA or the MATLAB Sym￾bolic Math Toolbox. Therefore, we obtain the following expression: S¯11 = S11m4 + (2S12 + S66)n2m2 + S22n4 (5.16a) S¯12 = (S11 + S22 − S66)n2m2 + S12(n4 + m4) (5.16b) S¯16 = (2S11 − 2S12 − S66)nm3 − (2S22 − 2S12 − S66)n3m (5.16c) S¯22 = S11n4 + (2S12 + S66)n2m2 + S22m4 (5.16d) S¯26 = (2S11 − 2S12 − S66)n3m − (2S22 − 2S12 − S66)nm3 (5.16e) S¯66 = 2(2S11 + 2S22 − 4S12 − S66)n2m2 + S66(n4 + m4) (5.16f)

5.2 MATLAB Functions Used 63 MATLAB Example 5.2 Consider a graphite-reinforced polymer composite lamina with the elastic con- stants as given in Example 2.2.Use MATLAB to plot the values of the six elements Sij of the transformed reduced compliance matrix [S]as a function of the orientation angle0 in the range-π/2≤0≤π/2. Solution This example is solved using MATLAB.First,the reduced 3 x 3 compliance matrix is obtained as follows using the MATLAB function ReducedCompliance of Chap.4. >>S=ReducedComp1 iance(155.0,12.10,0.248,4.40) S= 0.0065 -0.0016 0 -0.0016 0.0826 0 0 00.2273 Next,the transformed reduced compliance matrix [S]is calculated at each value of9 between-g0°andg0°in increments of 10°using the MATLAB function Sbar. >S1=Sbar(s,-90) S1= 0.0826 -0.0016 -0.0000 -0.0016 0.0065 0.0000 -0.0000 0.0000 0.2273 >S2=Sbar(s,-80) S2= 0.0909 -0.0122 -0.0452 -0.0122 0.0193 0.0712 -0.0226 0.0356 0.2061 >S3=Sbar(s,-70) S3= 0.1111 -0.0390 -0.0647 -0.0390 0.0528 0.1137 -0.0323 0.0568 0.1524

5.2 MATLAB Functions Used 63 MATLAB Example 5.2 Consider a graphite-reinforced polymer composite lamina with the elastic con￾stants as given in Example 2.2. Use MATLAB to plot the values of the six elements S¯ij of the transformed reduced compliance matrix [S¯] as a function of the orientation angle θ in the range −π/2 ≤ θ ≤ π/2. Solution This example is solved using MATLAB. First, the reduced 3 × 3 compliance matrix is obtained as follows using the MATLAB function ReducedCompliance of Chap. 4. >> S = ReducedCompliance(155.0, 12.10, 0.248, 4.40) S = 0.0065 -0.0016 0 -0.0016 0.0826 0 0 0 0.2273 Next, the transformed reduced compliance matrix [S¯] is calculated at each value of θ between −90◦ and 90◦ in increments of 10◦ using the MATLAB function Sbar . >> S1 = Sbar(S, -90) S1 = 0.0826 -0.0016 -0.0000 -0.0016 0.0065 0.0000 -0.0000 0.0000 0.2273 >> S2 = Sbar(S, -80) S2 = 0.0909 -0.0122 -0.0452 -0.0122 0.0193 0.0712 -0.0226 0.0356 0.2061 >> S3 = Sbar(S, -70) S3 = 0.1111 -0.0390 -0.0647 -0.0390 0.0528 0.1137 -0.0323 0.0568 0.1524

64 5 Global Coordinate System >S4=Sbar(S,-60) S4= 0.1315 -0.0695 -0.0454 -0.0695 0.0934 0.1114 -0.0227 0.0557 0.0914 >S5=Sbar(S,-50) S5= 0.1390 -0.0894 0.0065 -0.0894 0.1258 0.0685 0.0033 0.0342 0.0516 >S6=Sbar(S,-40) S6= 0.1258 -0.0894 0.0685 -0.0894 0.1390 0.0065 0.0342 0.0033 0.0516 >s7=sbar(s,-30) S7= 0.0934 -0.0695 0.1114 -0.0695 0.1315 -0.0454 0.0557 -0.0227 0.0914 >s8=Sbar(s,-20) S8=- 0.0528 -0.0390 0.1137 -0.0390 0.1111 -0.0647 0.0568 -0.0323 0.1524 >>S9=Sbar(S,-10) S9= 0.0193 -0.0122 0.0712 -0.0122 0.0909 -0.0452 0.0356 -0.0226 0.2061

64 5 Global Coordinate System >> S4 = Sbar(S, -60) S4 = 0.1315 -0.0695 -0.0454 -0.0695 0.0934 0.1114 -0.0227 0.0557 0.0914 >> S5 = Sbar(S, -50) S5 = 0.1390 -0.0894 0.0065 -0.0894 0.1258 0.0685 0.0033 0.0342 0.0516 >> S6 = Sbar(S, -40) S6 = 0.1258 -0.0894 0.0685 -0.0894 0.1390 0.0065 0.0342 0.0033 0.0516 >> S7 = Sbar(S, -30) S7 = 0.0934 -0.0695 0.1114 -0.0695 0.1315 -0.0454 0.0557 -0.0227 0.0914 >> S8 = Sbar(S, -20) S8 = 0.0528 -0.0390 0.1137 -0.0390 0.1111 -0.0647 0.0568 -0.0323 0.1524 >> S9 = Sbar(S, -10) S9 = 0.0193 -0.0122 0.0712 -0.0122 0.0909 -0.0452 0.0356 -0.0226 0.2061

5.2 MATLAB Functions Used 65 >S10 Sbar(S,0) S10= 0.0065 -0.0016 0 -0.0016 0.0826 0 0 0 0.2273 >S11=Sbar(s,10) S11= 0.0193 -0.0122 -0.0712 -0.0122 0.0909 0.0452 -0.0356 0.0226 0.2061 >S12=Sbar(s,20) S12= 0.0528 -0.0390 -0.1137 -0.0390 0.1111 0.0647 -0.0568 0.0323 0.1524 >S13=Sbar(s,30) S13= 0.0934 -0.0695 -0.1114 -0.0695 0.1315 0.0454 -0.0557 0.0227 0.0914 >S14=Sbar(s,40) S14= 0.1258 -0.0894 -0.0685 -0.0894 0.1390 -0.0065 -0.0342 -0.0033 0.0516 >S15=sbar(s,50) S15= 0.1390 -0.0894 -0.0065 -0.0894 0.1258 -0.0685 -0.0033 -0.0342 0.0516 >S16=Sbar(S,60)

5.2 MATLAB Functions Used 65 >> S10 = Sbar(S, 0) S10 = 0.0065 -0.0016 0 -0.0016 0.0826 0 0 0 0.2273 >> S11 = Sbar(S, 10) S11 = 0.0193 -0.0122 -0.0712 -0.0122 0.0909 0.0452 -0.0356 0.0226 0.2061 >> S12 = Sbar(S, 20) S12 = 0.0528 -0.0390 -0.1137 -0.0390 0.1111 0.0647 -0.0568 0.0323 0.1524 >> S13 = Sbar(S, 30) S13 = 0.0934 -0.0695 -0.1114 -0.0695 0.1315 0.0454 -0.0557 0.0227 0.0914 >> S14 = Sbar(S, 40) S14 = 0.1258 -0.0894 -0.0685 -0.0894 0.1390 -0.0065 -0.0342 -0.0033 0.0516 >> S15 = Sbar(S, 50) S15 = 0.1390 -0.0894 -0.0065 -0.0894 0.1258 -0.0685 -0.0033 -0.0342 0.0516 >> S16 = Sbar(S, 60)

66 5 Global Coordinate System S16= 0.1315 -0.0695 0.0454 -0.0695 0.0934 -0.1114 0.0227 -0.0557 0.0914 >S17=Sbar(s,70) S17= 0.1111 -0.0390 0.0647 -0.0390 0.0528 -0.1137 0.0323 -0.0568 0.1524 >S18=Sbar(s,80) S18= 0.0909 -0.0122 0.0452 -0.0122 0.0193 -0.0712 0.0226 -0.0356 0.2061 >>s19=sbar(s,90) S19= 0.0826 -0.0016 0.0000 -0.0016 0.0065 -0.0000 0.0000 -0.0000 0.2273 The z-axis is now setup for the plots as follows: >x=[-90-80-70-60-50-40-30-20-10010203040 5060708090] X -90-80-70-60-50-40-30-20-1001020304050 60708090 The values of 5u1 are now calculated for each value of 0 between-90 and 90°in increments of10°. >y1=[S1(1,1)S2(1,1)S3(1,1)S4(1,1)S5(1,1)S6(1,1)S7(1,1) S8(1,1)S9(1,1)S10(1,1)S11(1,1)S12(1,1)S13(1,1)S14(1,1) S15(1,1)16(1,1)S17(1,1)S18(1,1)S19(1,1)] y1= Columns 1 through 14

66 5 Global Coordinate System S16 = 0.1315 -0.0695 0.0454 -0.0695 0.0934 -0.1114 0.0227 -0.0557 0.0914 >> S17 = Sbar(S, 70) S17 = 0.1111 -0.0390 0.0647 -0.0390 0.0528 -0.1137 0.0323 -0.0568 0.1524 >> S18 = Sbar(S, 80) S18 = 0.0909 -0.0122 0.0452 -0.0122 0.0193 -0.0712 0.0226 -0.0356 0.2061 >> S19 = Sbar(S, 90) S19 = 0.0826 -0.0016 0.0000 -0.0016 0.0065 -0.0000 0.0000 -0.0000 0.2273 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 S¯11 are now calculated for each value of θ between −90◦ and 90◦ in increments of 10◦. >> y1 = [S1(1,1) S2(1,1) S3(1,1) S4(1,1) S5(1,1) S6(1,1) S7(1,1) S8(1,1) S9(1,1) S10(1,1) S11(1,1) S12(1,1) S13(1,1) S14(1,1) S15(1,1) 16(1,1) S17(1,1) S18(1,1) S19(1,1)] y1 = Columns 1 through 14