4 Plane Stress 4.1 Basic Equations In the analysis of fiber-reinforced composite materials,the assumption of plane stress is usually used for each layer (lamina).This is mainly because fiber- reinforced materials are utilized in beams,plates,cylinders,and other struc- tural shapes which have at least one characteristic geometric dimension in an order of magnitude less than the other two dimensions.In this case,the stress components 03,T23,and 713 are set to zero with the assumption that the 1-2 plane of the principal material coordinate system is in the plane of the layer (lamina)-see [1].Therefore,the stresses 01,02,and 7i2 lie in a plane,while the stresses o3,T23,and T13 are perpendicular to this plane and are zero(see Fig.4.1) Using the assumption of plane stress,it is seen that the stress-strain re- lations of Chap.2 are greatly simplified.Setting o3 =723=T13 =0 in(2.1) leads to the following: E1 S11 S12 S13 0 0 0 01 E2 S12 S22 S23 0 0 0 02 3 S13 S23 S33 0 0 0 0 (4.1) Y23 0 0 0 S44 0 0 0 13 0 0 0 0 S 0 Y12 0 0 0 0 S66」 T12 As a result of the plane stress assumption and using (4.1),we conclude that: 723=0 (4.2) Y13=0 (4.3) e3=S1301+S2302卡0 (4.4)
4 Plane Stress 4.1 Basic Equations In the analysis of fiber-reinforced composite materials, the assumption of plane stress is usually used for each layer (lamina). This is mainly because fiberreinforced materials are utilized in beams, plates, cylinders, and other structural shapes which have at least one characteristic geometric dimension in an order of magnitude less than the other two dimensions. In this case, the stress components σ3, τ23, and τ13 are set to zero with the assumption that the 1-2 plane of the principal material coordinate system is in the plane of the layer (lamina) – see [1]. Therefore, the stresses σ1, σ2, and τ12 lie in a plane, while the stresses σ3, τ23, and τ13 are perpendicular to this plane and are zero (see Fig. 4.1). Using the assumption of plane stress, it is seen that the stress-strain relations of Chap. 2 are greatly simplified. Setting σ3 = τ23 = τ13 = 0 in (2.1) leads to the following: ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ ε1 ε2 ε3 γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ S11 S12 S13 000 S12 S22 S23 000 S13 S23 S33 000 000 S44 0 0 0000 S55 0 00000 S66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 0 0 0 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ (4.1) As a result of the plane stress assumption and using (4.1), we conclude that: γ23 = 0 (4.2) γ13 = 0 (4.3) ε3 = S13σ1 + S23σ2 = 0 (4.4)
48 4 Plane Stress Fig.4.1.An infinitesimal fiber-reinforced composite element in a state of plane stress Therefore (4.1)reduces to the following equation: S11 S12 0 01 S12 S22 (4.5) 0 0 S66J】 T12 The 3 x 3 matrix in (4.5)is called the reduced compliance matrir.The inverse of the reduced compliance matrix is the reduced stiffness matrix given as follows: {- Q11 (4.6) where the elements Qij are given as follows: Q11= S22 (4.7a) 5S11S22-S2 Q12=- S12 (4.7b) S11S22-S3 Q22- Su S11S22-Si2 (4.7c 1 Q66= S66 (4.7d)
48 4 Plane Stress Fig. 4.1. An infinitesimal fiber-reinforced composite element in a state of plane stress Therefore (4.1) reduces to the following equation: ⎧ ⎨ ⎩ ε1 ε2 γ12 ⎫ ⎬ ⎭ = ⎡ ⎣ S11 S12 0 S12 S22 0 0 0 S66 ⎤ ⎦ ⎧ ⎨ ⎩ σ1 σ2 τ12 ⎫ ⎬ ⎭ (4.5) The 3 × 3 matrix in (4.5) is called the reduced compliance matrix. The inverse of the reduced compliance matrix is the reduced stiffness matrix given as follows: ⎧ ⎨ ⎩ σ1 σ2 τ12 ⎫ ⎬ ⎭ = ⎡ ⎣ Q11 Q12 0 Q12 Q22 0 0 0 Q66 ⎤ ⎦ ⎧ ⎨ ⎩ ε1 ε2 γ12 ⎫ ⎬ ⎭ (4.6) where the elements Qij are given as follows: Q11 = S22 S11S22 − S2 12 (4.7a) Q12 = − S12 S11S22 − S2 12 (4.7b) Q22 = S11 S11S22 − S2 12 (4.7c) Q66 = 1 S66 (4.7d)
4.2 MATLAB Functions Used 49 4.2 MATLAB Functions Used The four MATLAB functions used in this chapter to calculate the reduced compliance and stiffness matrices are: ReducedCompliance(E1,E2,NU12,G12)-This function calculates the re- duced compliance matrix for the lamina.Its input consists of four arguments representing the four elastic constants E,E2,v12,and G12.See Problem 4.1. ReducedStiffness(E1,E2,NU12,G12)-This function calculates the reduced stiffness matrix for the lamina.Its input consists of four arguments represent- ing the four elastic constants E1,E2,v12,and G12.See Problem 4.2. ReducedIsotropicCompliance(E,NU)-This function calculates the reduced isotropic compliance matrix for the lamina.Its input consists of two arguments representing the two elastic constants E and v.See Problem 4.3. ReducedIsotropicStiffness(E,NU)-This function calculates the reduced isotropic stiffness matrix for the lamina.Its input consists of two arguments representing the two elastic constants E and v.See Problem 4.4. The following is a listing of the MATLAB source code for each function: function y ReducedCompliance(E1,E2,NU12,G12) %ReducedCompliance This function returns the reduced compliance % matrix for fiber-reinforced materials. % There are four arguments representing four % material constants.The size of the reduced % compliance matrix is 3 x 3. y=[1/E1-NU12/E10;-NU12/E11/E20;001/G12]: function y ReducedStiffness(E1,E2,NU12,G12) %ReducedStiffness This function returns the reduced stiffness % matrix for fiber-reinforced materials. % There are four arguments representing four % material constants.The size of the reduced % stiffness matrix is 3 x 3. NU21=NU12*E2/E1; y=[E1/(1-NU12*NU21)NU12*E2/(1-NU12*NU21)0;NU12*E2/(1-NW12*NU21) E2/(1-NU12*NU21)0;00G12]; function y ReducedIsotropicCompliance(E,NU) %ReducedIsotropicCompliance This function returns the % reduced isotropic compliance % matrix for fiber-reinforced materials. There are two arguments representing % two material constants.The size of the reduced compliance matrix is 3 x 3. y=[1/E-NW/E0;-NU/E1/E0;002*(1+U)/E;
4.2 MATLAB Functions Used 49 4.2 MATLAB Functions Used The four MATLAB functions used in this chapter to calculate the reduced compliance and stiffness matrices are: ReducedCompliance(E1, E2, NU12, G12) – This function calculates the reduced compliance matrix for the lamina. Its input consists of four arguments representing the four elastic constants E1, E2, ν12, and G12. See Problem 4.1. ReducedStiffness(E1, E2, NU12, G12) – This function calculates the reduced stiffness matrix for the lamina. Its input consists of four arguments representing the four elastic constants E1, E2, ν12, and G12. See Problem 4.2. ReducedIsotropicCompliance(E, NU) – This function calculates the reduced isotropic compliance matrix for the lamina. Its input consists of two arguments representing the two elastic constants E and ν. See Problem 4.3. ReducedIsotropicStiffness(E, NU) – This function calculates the reduced isotropic stiffness matrix for the lamina. Its input consists of two arguments representing the two elastic constants E and ν. See Problem 4.4. The following is a listing of the MATLAB source code for each function: function y = ReducedCompliance(E1,E2,NU12,G12) %ReducedCompliance This function returns the reduced compliance % matrix for fiber-reinforced materials. % There are four arguments representing four % material constants. The size of the reduced % compliance matrix is 3 x 3. y = [1/E1 -NU12/E1 0 ; -NU12/E1 1/E20;00 1/G12]; function y = ReducedStiffness(E1,E2,NU12,G12) %ReducedStiffness This function returns the reduced stiffness % matrix for fiber-reinforced materials. % There are four arguments representing four % material constants. The size of the reduced % stiffness matrix is 3 x 3. NU21 = NU12*E2/E1; y = [E1/(1-NU12*NU21) NU12*E2/(1-NU12*NU21) 0 ;NU12*E2/(1-NU12*NU21) E2/(1-NU12*NU21) 0 ; 0 0 G12]; function y = ReducedIsotropicCompliance(E,NU) %ReducedIsotropicCompliance This function returns the % reduced isotropic compliance % matrix for fiber-reinforced materials. % There are two arguments representing % two material constants. The size of % the reduced compliance matrix is 3 x 3. y = [1/E -NU/E 0 ; -NU/E 1/E0;00 2*(1+NU)/E];
50 4 Plane Stress function y ReducedIsotropicStiffness(E,NU) %ReducedIsotropicStiffness This function returns the % reduced isotropic stiffness % matrix for fiber-reinforced materials. % There are two arguments representing % two material constants.The size of % the reduced stiffness matrix is 3 x 3. y [E/(1-NU+NU)NU*E/(1-NU*NU)0 NU*E/(1-NU*NU)E/(1-NU*NU)00 0E/2/(1+NU)]; Example 4.1 Derive the following expressions for the elements Qii of the 3 x 3 reduced stiffness matrix where Cij are the elements of the 6 x 6 stiffness matrix of (2.3) Q11=C11- C (4.8a C33 Q12=C12- C13C23 (4.8b) C33 C Q22=C22-C3s (4.8c Q66=C66 (4.8d) Solution For the case of plane stress,set o3 =T23=T13 =0 in(2.3)to obtain (while using the symmetric form of the [C]matrix). 01 「C11C12 C13 0 0 0 E1 02 C12 C22 C23 0 0 0 2 0 C13 C23 C33 0 0 0 63 0 0 0 0 C44 0 0 (4.9) 23 0 0 0 0 0 C 0 Y13 T12) 0 0 0 0 0 C66」 Y12 We can therefore write the following three equations based on the first,second, and sixth rows of (4.9): 01=C11s1+C122+C13e3 (4.10a) 02=C251+C22E2+C233 (4.10b) T12=C6612 (4.10c) In addition,we can write the following relation based on the third row of (4.9)
50 4 Plane Stress function y = ReducedIsotropicStiffness(E,NU) %ReducedIsotropicStiffness This function returns the % reduced isotropic stiffness % matrix for fiber-reinforced materials. % There are two arguments representing % two material constants. The size of % the reduced stiffness matrix is 3 x 3. y = [E/(1-NU*NU) NU*E/(1-NU*NU) 0 ; NU*E/(1-NU*NU) E/(1-NU*NU) 0 ; 0 0 E/2/(1+NU)]; Example 4.1 Derive the following expressions for the elements Qij of the 3 × 3 reduced stiffness matrix where Cij are the elements of the 6 × 6 stiffness matrix of (2.3). Q11 = C11 − C2 13 C33 (4.8a) Q12 = C12 − C13C23 C33 (4.8b) Q22 = C22 − C2 23 C33 (4.8c) Q66 = C66 (4.8d) Solution For the case of plane stress, set σ3 = τ23 = τ13 = 0 in (2.3) to obtain (while using the symmetric form of the [C] matrix). ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ σ1 σ2 0 0 0 τ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ C11 C12 C13 000 C12 C22 C23 000 C13 C23 C33 000 000 C44 0 0 0000 C55 0 00000 C66 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ ε1 ε2 ε3 γ23 γ13 γ12 ⎫ ⎪⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎪⎭ (4.9) We can therefore write the following three equations based on the first, second, and sixth rows of (4.9): σ1 = C11ε1 + C12ε2 + C13ε3 (4.10a) σ2 = C12ε1 + C22ε2 + C23ε3 (4.10b) τ12 = C66γ12 (4.10c) In addition, we can write the following relation based on the third row of (4.9):
4.2 MATLAB Functions Used 51 0=C13e1+C23e2+C333 (4.11) Solving (4.11)for 3 to obtain: 3=- C33 C22 C1361-Ca3 (4.12) Substitute (4.12)into(4.10a,b)and simplify to obtain the following relations: 01= cu- C13 1+ C92 C13C23 (4.13a) 02= Cua-Co 3C23 2 -C3 (4.13b) T万2=C66Y12 (4.13c) Rewriting (4.13a,b,c)in matrix form we obtain (see (4.6)): }-俗年 (4.14) where the elements Qii are given by (see (4.8a,b,c,d).): Q11=C1- (4.15a) C33 Q12=C12- C13C23 (4.15b) C33 Q22=C22- C33 (4.15c Q66=C66 (4.15d) MATLAB Example 4.2 Consider a layer of graphite-reinforced composite material 200mm long, 100mm wide,and 0.200mm thick.The layer is subjected to an inplane ten- sile force of 4kN in the fiber direction which is perpendicular to the 100-mm width.Assume the layer to be in a state of plane stress and use the elastic constants given in Example 2.2.Use MATLAB to determine the transverse strain E3. Solution This example is solved using MATLAB.First,the full 6 x 6 compliance matrix is obtained as follows using the MATLAB function OrthotropicCompliance of Chap.2
4.2 MATLAB Functions Used 51 0 = C13ε1 + C23ε2 + C33ε3 (4.11) Solving (4.11) for ε3 to obtain: ε3 = −C13 C33 ε1 − C23 C33 ε2 (4.12) Substitute (4.12) into (4.10a,b) and simplify to obtain the following relations: σ1 = C11 − C2 13 C33 ε1 + C12 − C13C23 C33 ε2 (4.13a) σ2 = C12 − C13C23 C33 ε1 + C22 − C2 23 C33 ε2 (4.13b) τ12 = C66γ12 (4.13c) Rewriting (4.13a,b,c) in matrix form we obtain (see (4.6)): ⎧ ⎨ ⎩ σ1 σ2 τ12 ⎫ ⎬ ⎭ = ⎡ ⎣ Q11 Q12 0 Q12 Q22 0 0 0 Q66 ⎤ ⎦ ⎧ ⎨ ⎩ ε1 ε2 γ12 ⎫ ⎬ ⎭ (4.14) where the elements Qij are given by (see (4.8a,b,c,d).): Q11 = C11 − C2 13 C33 (4.15a) Q12 = C12 − C13C23 C33 (4.15b) Q22 = C22 − C2 23 C33 (4.15c) Q66 = C66 (4.15d) MATLAB Example 4.2 Consider a layer of graphite-reinforced composite material 200 mm long, 100 mm wide, and 0.200 mm thick. The layer is subjected to an inplane tensile force of 4 kN in the fiber direction which is perpendicular to the 100-mm width. Assume the layer to be in a state of plane stress and use the elastic constants given in Example 2.2. Use MATLAB to determine the transverse strain ε3. Solution This example is solved using MATLAB. First, the full 6×6 compliance matrix is obtained as follows using the MATLAB function OrthotropicCompliance of Chap. 2
52 4 Plane Stress >>S=0 rthotropicComp1 iance(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 00.3125 0 0 0 0 0 0 0.2273 0 0 0 0 0 0 0.2273 Using the third row of(4.1),we obtain the following expression for the trans- verse strain E3 (see (4.4)): E3=S1301+S2302 (4.16) Next,the stresses o1 and o2 are calculated in GPa as follows: >s1gma1=4/(100*0.200) sigma1 0.2000 >sigma2 =0 sigma2 0 Finally,the transverse strain s3 is calculated using (4.16)as follows: >epsilon3 S(1,3)*sigma1 S(2,3)*sigma2 epsilon3 -3.2000e-004 Thus,we obtain the transverse strain E3=-3.2 x 10-4. MATLAB Example 4.3 Consider the graphite-reinforced composite material of Example 2.2. (a)Use MATLAB to determine the reduced compliance and stiffness matrices. (b)Use MATLAB to check that the two matrices obtained in (a)are indeed inverses of each other by multiplying them together to get the identity matrix
52 4 Plane Stress >> 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 Using the third row of (4.1), we obtain the following expression for the transverse strain ε3 (see (4.4)): ε3 = S13σ1 + S23σ2 (4.16) Next, the stresses σ1 and σ2 are calculated in GPa as follows: >> sigma1 = 4/(100*0.200) sigma1 = 0.2000 >> sigma2 = 0 sigma2 = 0 Finally, the transverse strain ε3 is calculated using (4.16) as follows: >> epsilon3 = S(1,3)*sigma1 + S(2,3)*sigma2 epsilon3 = -3.2000e-004 Thus, we obtain the transverse strain ε3 = −3.2 × 10−4. MATLAB Example 4.3 Consider the graphite-reinforced composite material of Example 2.2. (a) Use MATLAB to determine the reduced compliance and stiffness matrices. (b) Use MATLAB to check that the two matrices obtained in (a) are indeed inverses of each other by multiplying them together to get the identity matrix
Problems 53 Solution This example is solved using MATLAB.First,the reduced compliance matrix is obtained as follows using the MATLAB function ReducedCompliance. >S =ReducedCompliance(155.0,12.10,0.248,4.40) S= 0.0065 -0.0016 0 -0.0016 0.0826 0 0 00.2273 Next,the reduced stiffness matrix is obtained as follows using the MATLAB function ReducedStiffness: >Q=ReducedStiffness(155.0,12.10,0.248,4.40) 0= 155.7478 3.0153 0 3.0153 12.1584 0 0 0 4.4000 Finally,the two matrices are multiplied with each other to get the identity matrix in order to show that they are indeed inverses of each other. >>S*Q ans 1.0000 0 0 -0.0000 1.0000 0 0 0 1.0000 Problems Problem 4.1 Write the reduced compliance matrix for a fiber-reinforced composite material in terms of the four elastic constants E1,E2,v12,and G12. Problem 4.2 Write the reduced stiffness matrix for a fiber-reinforced composite material in terms of the four elastic constants E1,E2,v12,and G12
Problems 53 Solution This example is solved using MATLAB. First, the reduced compliance matrix is obtained as follows using the MATLAB function ReducedCompliance. >> 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 reduced stiffness matrix is obtained as follows using the MATLAB function ReducedStiffness: >> 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 Finally, the two matrices are multiplied with each other to get the identity matrix in order to show that they are indeed inverses of each other. >> S*Q ans = 1.0000 0 0 -0.0000 1.0000 0 0 0 1.0000 Problems Problem 4.1 Write the reduced compliance matrix for a fiber-reinforced composite material in terms of the four elastic constants E1, E2, ν12, and G12. Problem 4.2 Write the reduced stiffness matrix for a fiber-reinforced composite material in terms of the four elastic constants E1, E2, ν12, and G12
54 Plane Stress Problem 4.3 Write the reduced compliance matrix for an isotropic fiber-reinforced compos- ite material in terms of the two elastic constants E and v. Problem 4.4 Write the reduced stiffness matrix for an isotropic fiber-reinforced composite material in terms of the two elastic constants E and v. MATLAB Problem 4.5 Consider the glass-reinforced polymer composite material of Problem 2.7. (a)Use MATLAB to determine the reduced compliance and stiffness matrices. (b)Use MATLAB to check that the two matrices obtained in (a)are indeed inverses of each other by multiplying them together to get the identity matrix. MATLAB Problem 4.6 Consider the layer of composite material of Example 4.2.Suppose that the layer is subjected to an inplane compressive force of 2.5kN in the 2-direction instead of the 4kN force in the 1-direction.Use MATLAB to calculate the transverse strain E3 in this case. MATLAB Problem 4.7 Consider the isotropic material aluminum with E=72.4 GPa and v=0.3. (a)Use MATLAB to determine the reduced compliance and stiffness matrices. (b)Use MATLAB to check that the two matrices obtained in (a)are indeed inverses of each other by multiplying them together to get the identity matrix. MATLAB Problem 4.8 Suppose in Example 4.2 that the fibers are perpendicular to the 200-mm direction.Use MATLAB to calculate the transverse strain sg in this case. MATLAB Problem 4.9 Write two MATLAB functions called ReducedStiffness2 and ReducedIsotrop- icStifness?where the reduced stiffness matrix in each case is determined by taking the inverse of the reduced compliance matrix
54 Plane Stress Problem 4.3 Write the reduced compliance matrix for an isotropic fiber-reinforced composite material in terms of the two elastic constants E and ν. Problem 4.4 Write the reduced stiffness matrix for an isotropic fiber-reinforced composite material in terms of the two elastic constants E and ν. MATLAB Problem 4.5 Consider the glass-reinforced polymer composite material of Problem 2.7. (a) Use MATLAB to determine the reduced compliance and stiffness matrices. (b) Use MATLAB to check that the two matrices obtained in (a) are indeed inverses of each other by multiplying them together to get the identity matrix. MATLAB Problem 4.6 Consider the layer of composite material of Example 4.2. Suppose that the layer is subjected to an inplane compressive force of 2.5 kN in the 2-direction instead of the 4 kN force in the 1-direction. Use MATLAB to calculate the transverse strain ε3 in this case. MATLAB Problem 4.7 Consider the isotropic material aluminum with E = 72.4 GPa and ν = 0.3. (a) Use MATLAB to determine the reduced compliance and stiffness matrices. (b) Use MATLAB to check that the two matrices obtained in (a) are indeed inverses of each other by multiplying them together to get the identity matrix. MATLAB Problem 4.8 Suppose in Example 4.2 that the fibers are perpendicular to the 200-mm direction. Use MATLAB to calculate the transverse strain ε3 in this case. MATLAB Problem 4.9 Write two MATLAB functions called ReducedStiffness2 and ReducedIsotropicStifness2 where the reduced stiffness matrix in each case is determined by taking the inverse of the reduced compliance matrix
Problems 55 Problem 4.10 Consider a layer of fiber-reinforced composite material that is subjected to both temperature and moisture variations.Write the 3 x 3 reduced stress- strain equations that correspond to (4.5)and (4.6).See Problems 2.9 and 2.10 of Chap.2
Problems 55 Problem 4.10 Consider a layer of fiber-reinforced composite material that is subjected to both temperature and moisture variations. Write the 3 × 3 reduced stressstrain equations that correspond to (4.5) and (4.6). See Problems 2.9 and 2.10 of Chap. 2
