2 Linear Elastic Stress-Strain Relations 2.1 Basic Equations Consider a single layer of fiber-reinforced composite material as shown in Fig.2.1.In this layer,the 1-2-3 orthogonal coordinate system is used where the directions are taken as follows: 1.The 1-axis is aligned with the fiber direction. 2.The 2-axis is in the plane of the layer and perpendicular to the fibers. 3.The 3-axis is perpendicular to the plane of the layer and thus also perpen- dicular to the fibers. Fig.2.1.A lamina illustrating the principle material coordinate system
2 Linear Elastic Stress-Strain Relations 2.1 Basic Equations Consider a single layer of fiber-reinforced composite material as shown in Fig. 2.1. In this layer, the 1-2-3 orthogonal coordinate system is used where the directions are taken as follows: 1. The 1-axis is aligned with the fiber direction. 2. The 2-axis is in the plane of the layer and perpendicular to the fibers. 3. The 3-axis is perpendicular to the plane of the layer and thus also perpendicular to the fibers. Fig. 2.1. A lamina illustrating the principle material coordinate system
10 2 Linear Elastic Stress-Strain Relations The 1-direction is also called the fiber direction,while the 2-and 3- directions are called the matrix directions or the transverse directions.This 1-2-3 coordinate system is called the principal material coordinate system.The stresses and strains in the layer (also called a lamina)will be referred to the principal material coordinate system. At this level of analysis,the strain or stress of an individual fiber or an element of matrix is not considered.The effect of the fiber reinforcement is smeared over the volume of the material.We assume that the two-material fiber-matrix system is replaced by a single homogeneous material.Obviously, this single material does not have the same properties in all directions.Such material with different properties in three mutually perpendicular directions is called an orthotropic material.Therefore,the layer (lamina)is considered to be orthotropic. The stresses on a small infinitesimal element taken from the layer are illustrated in Fig.2.2.There are three normal stresses o1,02,and 03,and three shear stresses T12,T23,and 713.These stresses are related to the strains 61,52,E3,12,723,and 13 as follows (see [1]): 2 23 Fig.2.2.An infinitesimal fiber-reinforced element showing the stresses
10 2 Linear Elastic Stress-Strain Relations The 1-direction is also called the fiber direction, while the 2- and 3- directions are called the matrix directions or the transverse directions. This 1-2-3 coordinate system is called the principal material coordinate system. The stresses and strains in the layer (also called a lamina) will be referred to the principal material coordinate system. At this level of analysis, the strain or stress of an individual fiber or an element of matrix is not considered. The effect of the fiber reinforcement is smeared over the volume of the material. We assume that the two-material fiber-matrix system is replaced by a single homogeneous material. Obviously, this single material does not have the same properties in all directions. Such material with different properties in three mutually perpendicular directions is called an orthotropic material. Therefore, the layer (lamina) is considered to be orthotropic. The stresses on a small infinitesimal element taken from the layer are illustrated in Fig. 2.2. There are three normal stresses σ1, σ2, and σ3, and three shear stresses τ12, τ23, and τ13. These stresses are related to the strains ε1, ε2, ε3, γ12, γ23, and γ13 as follows (see [1]): Fig. 2.2. An infinitesimal fiber-reinforced element showing the stresses
2.1 Basic Equations 11 E1 1/E1 -21/E2 -31/E3 0 0 01 -M2/E1 1/E2 -32/E3 0 0 0 02 Es -3/E1 -23/E2 1/E3 0 0 0 S 0 0 1/G23 0 (2.1) 23 0 0 1/G13 T13 12 0 0 0 0 1/G12」 T12 In(2.1),E1,E2,and E3 are the extensional moduli of elasticity along the 1,2,and 3 directions,respectively.Also,vij(i,j=1,2,3)are the different Poisson's ratios,while G2,G23,and G3 are the three shear moduli. Equation (2.1)can be written in a compact form as follows: {e}=[S{a} (2.2) where fe}and fo}represent the 6 x 1 strain and stress vectors,respectively, and [S]is called the compliance matrir.The elements of [S]are clearly ob- tained from(2.1),i.e.S11=1/E1,S12=-21/E2,..,S66=1/G12 The inverse of the compliance matrix [S]is called the stiffness matrix [C] given,in general,as follows: C11 C12 C13 0 0 0 E1 6 C21 C22 C23 0 0 0 03 C31 C32 C33 0 0 0 T23 0 0 0 CM 0 0 (2.3) Y23 T13 0 0 0 0 0 Y13 T12 0 0 0 0 0 C66 Y12 In compact form(2.3)is written as follows: {o}=[C{e} (2.4) The elements of [C]are not shown here explicitly but are calculated using the MATLAB function OrthotropicStiffness which is written specifically for this purpose. It is shown(see [1])that both the compliance matrix and the stiffness matrix are symmetric,i.e.C21=C12,C23 C32,C13 =C31,and similarly for S21,S23,and S13.Therefore,the following expressions can now be easily obtained: 1 C1=552s5a8-sa52a C12=5(51523-51253） C2=5(53351-513513) C13=5(512S23-S13.52）
2.1 Basic Equations 11 ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ ε1 ε2 ε3 γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1/E1 −ν21/E2 −ν31/E3 000 −ν12/E1 1/E2 −ν32/E3 000 −ν13/E1 −ν23/E2 1/E3 000 0 0 01/G23 0 0 0 0 0 01/G13 0 0 0 0 0 01/G12 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 σ3 τ23 τ13 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ (2.1) In (2.1), E1, E2, and E3 are the extensional moduli of elasticity along the 1, 2, and 3 directions, respectively. Also, νij (i, j = 1, 2, 3) are the different Poisson’s ratios, while G12, G23, and G13 are the three shear moduli. Equation (2.1) can be written in a compact form as follows: {ε} = [S] {σ} (2.2) where {ε} and {σ} represent the 6 × 1 strain and stress vectors, respectively, and [S] is called the compliance matrix. The elements of [S] are clearly obtained from (2.1), i.e. S11 = 1/E1, S12 = −ν21/E2,... , S66 = 1/G12. The inverse of the compliance matrix [S] is called the stiffness matrix [C] given, in general, as follows: ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 σ3 τ23 τ13 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ C11 C12 C13 000 C21 C22 C23 000 C31 C32 C33 000 000 C44 0 0 0000 C55 0 00000 C66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ ε1 ε2 ε3 γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ (2.3) In compact form (2.3) is written as follows: {σ} = [C] {ε} (2.4) The elements of [C] are not shown here explicitly but are calculated using the MATLAB function OrthotropicStiffness which is written specifically for this purpose. It is shown (see [1]) that both the compliance matrix and the stiffness matrix are symmetric, i.e. C21 = C12, C23 = C32, C13 = C31, and similarly for S21, S23, and S13. Therefore, the following expressions can now be easily obtained: C11 = 1 S (S22S33 − S23S23) C12 = 1 S (S13S23 − S12S33) C22 = 1 S (S33S11 − S13S13) C13 = 1 S (S12S23 − S13S22)
12 2 Linear Elastic Stress-Strain Relations 1 C33=s(S1s2-S1252) 1 C23=5(S12513-5235i) (2.5) C44三 S44 1 C55= 565 C66= 1 S S=S11S22S33-S11S23S23-S22S13S13-S33S12S12+2S12S23S13 It should be noted that the material constants appearing in the compliance matrix in (2.1)are not all independent.This is clear since the compliance matrix is symmetric.Therefore,we have the following equations relating the material constants: 12 21 E E2 43= 31 E·E3 (2.6) 岩詈 The above equations are called the reciprocity relations for the material constants.It should be noted that the reciprocity relations can be derived irrespective of the symmetry of the compliance matrix-in fact we conclude that the compliance matrix is symmetric from using these relations.Thus it is now clear that there are nine independent material constants for an orthotropic material. A material is called transversely isotropic if its behavior in the 2-direction is identical to its behavior in the 3-direction.For this case,E2=E3,v12 =vi3, and G12=G13.In addition,we have the following relation: G23= E2 2(1+23) (2.7) It is clear that there are only five independent material constants(E,E2, v12,v23,G12)for a transversely isotropic material. A material is called isotropic if its behavior is the same in all three 1-2-3 directions.In this case,E1=E2=E3=E,v12 v23 =v13 =v,and G12 G23 =G13 =G.In addition,we have the following relation: G E 2(1+w) (2.8) It is clear that there are only two independent material constants(E,v) for an isotropic material
12 2 Linear Elastic Stress-Strain Relations C33 = 1 S (S11S22 − S12S12) C23 = 1 S (S12S13 − S23S11) (2.5) C44 = 1 S44 C55 = 1 S55 C66 = 1 S66 S = S11S22S33 − S11S23S23 − S22S13S13 − S33S12S12 + 2S12S23S13 It should be noted that the material constants appearing in the compliance matrix in (2.1) are not all independent. This is clear since the compliance matrix is symmetric. Therefore, we have the following equations relating the material constants: ν12 E1 = ν21 E2 ν13 E1 = ν31 E3 (2.6) ν23 E2 = ν32 E3 The above equations are called the reciprocity relations for the material constants. It should be noted that the reciprocity relations can be derived irrespective of the symmetry of the compliance matrix – in fact we conclude that the compliance matrix is symmetric from using these relations. Thus it is now clear that there are nine independent material constants for an orthotropic material. A material is called transversely isotropic if its behavior in the 2-direction is identical to its behavior in the 3-direction. For this case, E2 = E3, ν12 = ν13, and G12 = G13. In addition, we have the following relation: G23 = E2 2(1 + ν23) (2.7) It is clear that there are only five independent material constants (E1, E2, ν12, ν23, G12) for a transversely isotropic material. A material is called isotropic if its behavior is the same in all three 1-2-3 directions. In this case, E1 = E2 = E3 = E, ν12 = ν23 = ν13 = ν, and G12 = G23 = G13 = G. In addition, we have the following relation: G = E 2(1 + ν) (2.8) It is clear that there are only two independent material constants (E, ν) for an isotropic material
2.2 MATLAB Functions Used 13 At the other end of the spectrum,we have anisotropic materials-these materials have nonzero entries at the upper right and lower left portions of their compliance and stiffness matrices. 2.2 MATLAB Functions Used The six MATLAB functions used in this chapter to calculate compliance and stiffness matrices are: OrthotropicCompliance(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13)- This function calculates the 6 x 6 compliance matrix for orthotropic materials. Its input are the nine independent material constants E,E2,E3,v12,v23, 13,G12,G23,andG13: OrthotropicStiffness(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13)-This function calculates the 6 x 6 stiffness matrix for orthotropic materials.Its input are the nine independent material constants E1,E2,E3,v12,v23,vi3, G12,G23,and G13. TransverselyIsotropicCompliance(E1,E2,NU12,NU23,G12)-This function calculates the 6 x 6 compliance matrix for transversely isotropic materials.Its input are the five independent material constants E1,E2,v12,v23,and Gi2. TransverselyIsotropicStiffness(E1,E2,NU12,NU23,G12)-This function cal- culates the 6 x 6 stiffness matrix for transversely isotropic materials.Its input are the five independent material constants E1,E2,v12,v23,and Gi2. IsotropicCompliance(E,NU)-This function calculates the 6 x 6 compliance matrix for isotropic materials.Its input are the two independent material constants E and v. IsotropicStiffness(E,NU)-This function calculates the 6 x 6 stiffness matrix for isotropic materials.Its input are the two independent material constants E and v. The following is a listing of the MATLAB source code for each function: function y OrthotropicCompliance(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13) %0rthotropicCompliance This function returns the compliance matrix % for orthotropic materials.There are nine % arguments representing the nine independent % material constants.The size of the compliance % matrix is 6 x 6. y=[1/E1-NU12/E1-NU13/E1000;-Nw12/E11/E2-NU23/E2000; -NU13/E1-NU23/E21/E3000;0001/G2300;00001/G130; 000001/G12];
2.2 MATLAB Functions Used 13 At the other end of the spectrum, we have anisotropic materials – these materials have nonzero entries at the upper right and lower left portions of their compliance and stiffness matrices. 2.2 MATLAB Functions Used The six MATLAB functions used in this chapter to calculate compliance and stiffness matrices are: OrthotropicCompliance(E1, E2, E3, NU12, NU23, NU13, G12, G23, G13) – This function calculates the 6×6 compliance matrix for orthotropic materials. Its input are the nine independent material constants E1, E2, E3, ν12, ν23, ν13, G12, G23, and G13. OrthotropicStiffness(E1, E2, E3, NU12, NU23, NU13, G12, G23, G13) – This function calculates the 6 × 6 stiffness matrix for orthotropic materials. Its input are the nine independent material constants E1, E2, E3, ν12, ν23, ν13, G12, G23, and G13. TransverselyIsotropicCompliance(E1, E2, NU12, NU23, G12) – This function calculates the 6×6 compliance matrix for transversely isotropic materials. Its input are the five independent material constants E1, E2, ν12, ν23, and G12. TransverselyIsotropicStiffness(E1, E2, NU12, NU23, G12) – This function calculates the 6×6 stiffness matrix for transversely isotropic materials. Its input are the five independent material constants E1, E2, ν12, ν23, and G12. IsotropicCompliance(E, NU) – This function calculates the 6 × 6 compliance matrix for isotropic materials. Its input are the two independent material constants E and ν. IsotropicStiffness(E, NU) – This function calculates the 6 × 6 stiffness matrix for isotropic materials. Its input are the two independent material constants E and ν. The following is a listing of the MATLAB source code for each function: function y = OrthotropicCompliance(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13) %OrthotropicCompliance This function returns the compliance matrix % for orthotropic materials. There are nine % arguments representing the nine independent % material constants. The size of the compliance % matrix is 6 x 6. y = [1/E1 -NU12/E1 -NU13/E1 0 0 0 ; -NU12/E1 1/E2 -NU23/E2 000; -NU13/E1 -NU23/E2 1/E3 000;000 1/G2300;0000 1/G13 0 ; 00000 1/G12];
14 2 Linear Elastic Stress-Strain Relations function y OrthotropicStiffness(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13) %OrthotropicStiffness This function returns the stiffness matrix for orthotropic materials.There are nine % arguments representing the nine independent % material constants.The size of the stiffness matrix is 6 x 6. x=[1/E1-NU12/E1-NU13/E1000;-o12/E11/E2-NU23/E2000; -U13/E1-NU23/E21/E3000;0001/G2300;00001/G130; 000001/G12]; y inv(x) function y TransverselyIsotropicCompliance(E1,E2,NU12,NU23,G12) %TransverselyIsotropicCompliance This function returns the % compliance matrix for % transversely isotropic % materials.There are five % arguments representing the % five independent material constants.The size of the 名 compliance matrix is 6 x 6. y=[1/E1-NU12/E1-NU12/E1000;-U12/E11/E2-NU23/E2000; -NU12/E1-NU23/E21/E2000;0002*(1+NU23)/E200; 00001/G120;000001/G12]; function y TransverselyIsotropicStiffness(E1,E2,NU12,NU23,G12) ransverselyIsotropicStiffness This function returns the % stiffness matrix for % transversely isotropic % materials.There are five % arguments representing the % five independent material constants.The size of the % stiffness matrix is 6 x 6. x=[1/E1-NU12/E1-NU12/E1000;-NU12/E11/E2-NU23/E2000: -N012/E1-NU23/E21/E2000;0002*(1+NU23)/E200; 00001/G120;000001/G12]; y inv(x); function y IsotropicCompliance(E,NU) %IsotropicCompliance This function returns the compliance matrix for isotropic 名 materials.There are two % arguments representing the two independent material % constants.The size of the compliance matrix is 6 x 6. y=[1/E-NW/E-W/E000;-NU/E1/E-NU/E000; -NU/E-NU/E1/E000;0002*(1+NU)/E00; 00002*(1+NU)/E0;000002*(1+NU)/E];
14 2 Linear Elastic Stress-Strain Relations function y = OrthotropicStiffness(E1,E2,E3,NU12,NU23,NU13,G12,G23,G13) %OrthotropicStiffness This function returns the stiffness matrix % for orthotropic materials. There are nine % arguments representing the nine independent % material constants. The size of the stiffness % matrix is 6 x 6. x = [1/E1 -NU12/E1 -NU13/E1 0 0 0 ; -NU12/E1 1/E2 -NU23/E2 000; -NU13/E1 -NU23/E2 1/E3 000;000 1/G2300;0000 1/G13 0 ; 00000 1/G12]; y = inv(x); function y = TransverselyIsotropicCompliance(E1,E2,NU12,NU23,G12) %TransverselyIsotropicCompliance This function returns the % compliance matrix for % transversely isotropic % materials. There are five % arguments representing the % five independent material % constants. The size of the % compliance matrix is 6 x 6. y = [1/E1 -NU12/E1 -NU12/E1 0 0 0 ; -NU12/E1 1/E2 -NU23/E2 000; -NU12/E1 -NU23/E2 1/E2 000;000 2*(1+NU23)/E2 0 0 ; 0000 1/G12 0 ; 0 0 0 0 0 1/G12]; function y = TransverselyIsotropicStiffness(E1,E2,NU12,NU23,G12) %TransverselyIsotropicStiffness This function returns the % stiffness matrix for % transversely isotropic % materials. There are five % arguments representing the % five independent material % constants. The size of the % stiffness matrix is 6 x 6. x = [1/E1 -NU12/E1 -NU12/E1 0 0 0 ; -NU12/E1 1/E2 -NU23/E2 000; -NU12/E1 -NU23/E2 1/E2 0 0 0 ; 0 0 0 2*(1+NU23)/E2 0 0 ; 0000 1/G12 0 ; 0 0 0 0 0 1/G12]; y = inv(x); function y = IsotropicCompliance(E,NU) %IsotropicCompliance This function returns the % compliance matrix for isotropic % materials. There are two % arguments representing the % two independent material % constants. The size of the % compliance matrix is 6 x 6. y = [1/E -NU/E -NU/E 0 0 0 ; -NU/E 1/E -NU/E000; -NU/E -NU/E 1/E 0 0 0;000 2*(1+NU)/E00; 0000 2*(1+NU)/E 0;00000 2*(1+NU)/E];
2.2 MATLAB Functions Used 15 function y IsotropicStiffness(E,NU) %IsotropicStiffness This function returns the stiffness matrix for isotropic % materials.There are two % arguments representing the two independent material constants.The size of the 名 stiffness matrix is 6 x 6. x=[1/E-NU/E-w/E000;-NU/E1/E-NU/E000; -NU/E-NU/E1/E000;0002*(1+N0U)/E00; 00002*(1+NU)/E0;000002*(1+NU)/E]; y=inv(x); Example 2.1 For an orthotropic material,derive expressions for the elements of the stiffness matrix Cii directly in terms of the nine independent material constants. Solution Substitute the elements of [S]from (2.1)into (2.5)along with using (2.6). This is illustrated in detail for Cl below.First evaluate the expression of S from (2.5)as follows: S=S11S22S33-S11S23S23-S22S13S13-S33S12S12+2S12S23S13 1111 -23 -32 E E2 Ex E1 E2 E3 1 -V3 -31 /1 E2 E 、E3 +2(器) 1-V23V32-V13V31-M221-2y12V2331 E1E2E3 1-0 EE2E3 (2.9a) where vo is given by V0=2332+y331+1221+212V2331 (2.9b) Next,C1l is calculated as follows
2.2 MATLAB Functions Used 15 function y = IsotropicStiffness(E,NU) %IsotropicStiffness This function returns the % stiffness matrix for isotropic % materials. There are two % arguments representing the % two independent material % constants. The size of the % stiffness matrix is 6 x 6. x = [1/E -NU/E -NU/E 0 0 0 ; -NU/E 1/E -NU/E000; -NU/E -NU/E 1/E 0 0 0;000 2*(1+NU)/E00; 0000 2*(1+NU)/E 0;00000 2*(1+NU)/E]; y = inv(x); Example 2.1 For an orthotropic material, derive expressions for the elements of the stiffness matrix Cij directly in terms of the nine independent material constants. Solution Substitute the elements of [S] from (2.1) into (2.5) along with using (2.6). This is illustrated in detail for C11 below. First evaluate the expression of S from (2.5) as follows: S = S11S22S33 − S11S23S23 − S22S13S13 − S33S12S12 + 2S12S23S13 = 1 E1 1 E2 1 E3 − 1 E1 −ν23 E2 −ν32 E3 − 1 E2 −ν13 E1 −ν31 E3 − 1 E3 −ν12 E1 −ν21 E2 +2 −ν12 E1 −ν23 E2 −ν31 E3 = 1 − ν23ν32 − ν13ν31 − ν12ν21 − 2ν12ν23ν31 E1E2E3 = 1 − ν0 E1E2E3 (2.9a) where ν0 is given by ν0 = ν23ν32 + ν13ν31 + ν12ν21 + 2ν12ν23ν31 (2.9b) Next, C11 is calculated as follows
16 2 Linear Elastic Stress-Strain Relations C11= 5(S22533-523523) E1E2E3「11 1-6 E2 E3 ()(】 (1-2332)E1 (2.9c) 1-0 Similarly,the following expressions for the other elements of [C can be derived: C2= (21+3123)E=(2+323)2 1-0 1-0 (2.9d) C13=1+212)B=M13+h223)E (2.9e) 1-0 1-0 C22= 1-h331)E2 (2.9f) 1-0 C23= g2+12的)色-23+2113)B (2.9g) 1-0 1-0 C33= 1-221)E (2.9h) 1-0 C44=G23 (2.9i) C55=G13 (2.9j) C66=G12 (2.9k) MATLAB Example 2.2 Consider a 60-mm cube made of graphite-reinforced polymer composite ma- terial that is subjected to a tensile force of 100kN perpendicular to the fiber direction,directed along the 2-direction.The cube is free to expand or con- tract.Use MATLAB to determine the changes in the 60-mm dimensions of the cube.The material constants for graphite-reinforced polymer composite material are given as follows [1]: E1=155.0GPa, E2=E3=12.10GPa 23=0.458, 2=13=0.248 G23=3.20GPa, G12=G13=4.40GPa Solution This example is solved using MATLAB.First,the normal stress in the 2- direction is calculated in GPa as follows:
16 2 Linear Elastic Stress-Strain Relations C11 = 1 S (S22S33 − S23S23) = E1E2E3 1 − ν0 1 E2 1 E3 − −ν23 E2 −ν32 E3 = (1 − ν23ν32) E1 1 − ν0 (2.9c) Similarly, the following expressions for the other elements of [C] can be derived: C12 = (ν21 + ν31ν23) E1 1 − ν0 = (ν12 + ν32ν13) E2 1 − ν0 (2.9d) C13 = (ν31 + ν21ν32) E1 1 − ν0 = (ν13 + ν12ν23) E3 1 − ν0 (2.9e) C22 = (1 − ν13ν31) E2 1 − ν0 (2.9f) C23 = (ν32 + ν12ν31) E2 1 − ν0 = (ν23 + ν21ν13) E3 1 − ν0 (2.9g) C33 = (1 − ν12ν21) E3 1 − ν0 (2.9h) C44 = G23 (2.9i) C55 = G13 (2.9j) C66 = G12 (2.9k) MATLAB Example 2.2 Consider a 60-mm cube made of graphite-reinforced polymer composite material that is subjected to a tensile force of 100 kN perpendicular to the fiber direction, directed along the 2-direction. The cube is free to expand or contract. Use MATLAB to determine the changes in the 60-mm dimensions of the cube. The material constants for graphite-reinforced polymer composite material are given as follows [1]: E1 = 155.0 GPa, E2 = E3 = 12.10 GPa ν23 = 0.458, ν12 = ν13 = 0.248 G23 = 3.20 GPa, G12 = G13 = 4.40 GPa Solution This example is solved using MATLAB. First, the normal stress in the 2- direction is calculated in GPa as follows:
2.2 MATLAB Functions Used 17 >s1gma2=100/(60*60) sigma2 0.0278 The stress vector is set up next as follows: >sigma [o sigma2 0000] sigma 00.02780000 The compliance matrix is then calculated using the MATLAB function Or- thotropicCompliance as follows: >>S=0 rthotropicCompliance(155.0,12.10,12.10,0.248,0.458,0.248, 4.40,3.20,4.40) S= 0.0065 -0.0016 -0.0016 0 0 0 -0.0016 0.0826 -0.0379 0 0 0 -0.0016 -0.0379 0.0826 0 0 0 0 0 0.3125 0 0 0 0 0 0 0.2273 0 0 0 0 0 0 0.2273 The stress vector is adjusted to be a 6 x 1 column vector as follows: >sigma sigma' sigma 0 0.0278 0 0 0 0 The strain vector is next obtained by applying (2.2)as follows: >epsilon S*sigma epsilon -0.0000 0.0023 -0.0011
2.2 MATLAB Functions Used 17 >> sigma2 = 100/(60*60) sigma2 = 0.0278 The stress vector is set up next as follows: >> sigma = [0 sigma2 0 0 0 0] sigma = 0 0.02780000 The compliance matrix is then calculated using the MATLAB function OrthotropicCompliance as follows: >> S = OrthotropicCompliance(155.0, 12.10, 12.10, 0.248, 0.458, 0.248, 4.40, 3.20, 4.40) S = 0.0065 -0.0016 -0.0016 0 0 0 -0.0016 0.0826 -0.0379 0 0 0 -0.0016 -0.0379 0.0826 0 0 0 0 0 0 0.3125 0 0 0 0 0 0 0.2273 0 0 0 0 0 0 0.2273 The stress vector is adjusted to be a 6 × 1 column vector as follows: >> sigma = sigma’ sigma = 0 0.0278 0 0 0 0 The strain vector is next obtained by applying (2.2) as follows: >> epsilon = S*sigma epsilon = -0.0000 0.0023 -0.0011
18 2 Linear Elastic Stress-Strain Relations 0 0 0 Note that the strain in dimensionless.Note also that s1l is very small but is not zero as it seems from the above result.To get the strain 1 exactly,we need to use the format command to get more digits as follows: >format short e >epsilon epsilon -4.4444e-005 2.2957e-003 -1.0514e-003 0 0 0 Finally,the change in length in each direction is calculated by multiplying the strain by the dimension in each direction as follows: >d1 epsilon(1)*60 d1= -2.6667e-003 >>d2=eps11on(2)*60 d2= 1.3774e-001 >>d3=eps11on(3)*60 d3= -6.3085e-002 Notice that the change in the fiber direction is -2.6667 x 10-3 mm which is very small due to the fibers reducing the deformation in this direction. The minus sign indicates that there is a reduction in this dimension along the fibers.The change in the 2-direction is 0.13774mm and is the largest change because the tensile force is along this direction.This change is positive indicating an extension in the dimension along this direction.Finally,the change in the 3-direction is -0.063085 mm.This change is minus since it indicates a reduction in the dimension along this direction
18 2 Linear Elastic Stress-Strain Relations 0 0 0 Note that the strain in dimensionless. Note also that ε11 is very small but is not zero as it seems from the above result. To get the strain ε11 exactly, we need to use the format command to get more digits as follows: >> format short e >> epsilon epsilon = -4.4444e-005 2.2957e-003 -1.0514e-003 0 0 0 Finally, the change in length in each direction is calculated by multiplying the strain by the dimension in each direction as follows: >> d1 = epsilon(1)*60 d1 = -2.6667e-003 >> d2 = epsilon(2)*60 d2 = 1.3774e-001 >> d3 = epsilon(3)*60 d3 = -6.3085e-002 Notice that the change in the fiber direction is −2.6667 × 10−3 mm which is very small due to the fibers reducing the deformation in this direction. The minus sign indicates that there is a reduction in this dimension along the fibers. The change in the 2-direction is 0.13774 mm and is the largest change because the tensile force is along this direction. This change is positive indicating an extension in the dimension along this direction. Finally, the change in the 3-direction is −0.063085 mm. This change is minus since it indicates a reduction in the dimension along this direction