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 system. This transformation will be done for the state of plane stress using the standard transformation relations for stresses and strains given in introductory courses in mechanics of materials [1]. Consider an isolated infinitesimal element in the principal material coordinate 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 fiberreinforced 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 matrix [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);