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 ma￾terial 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 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 = 155.0 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: