WAQAR ALI MS MECHANICAL ENGINEERING UNIVERSITY OF LAHORE SUBJECT:-TO DETERMINE THE VALUE OF STIFFNESS MATRIX WITH THE HELP OF AN EXAMPLE (ASSIGNMENT NO 3) Determine the elements in the stiffness matrix for an angle-ply lamina containing 60 vol % of T-300 carbon fibers in an epoxy matrix. Consider fiber orientation angles of both 45° and - 45° for the fiber, E f = 220 GPa and V f = 0.2, and for the matrix, E m = 3.6 GPa and V m = 0.35 SOLUTION Step 1: Calculate E 11 , E 22 , V 12 , V 21 , and G 12 using Equations below. E 11 = (220)(0.6) + (3.6)(1- 0.6) = 133.44 GPa, E 22 =(220)(3.6)/ (220)(1-0.6) + (3.6)(0.6) = 8.78 GPa, V 12 = (0.2)(0.6) + (0.35)(1-0.6) = 0.26, V 21 = 8.78/ 133.44 (0.26) = 0.017 To calculate G 12 , we need to know the values of Gf and Gm. Assuming isotropic relationships, we estimate G f = E f /2(1 + V f ) = 220/ 2(1 + 0.2) = 91.7 GPa, G m = E m /2(1+ V m ) = 3.6/ 2(1+ 0.35) = 1.33 GPa Therefore. G 12 = (91.7) (1.33)/ (91.7) (1-0.6) + (1.33) (0.6) = 3.254 GPa Note that the T-300 carbon fiber is not isotropic, and therefore, the calculation of G f based on the isotropic assumption will certainly introduce error. Since the actual value of Gf is not always available, the isotropic assumption is often made to calculate G f . Step 2: Calculate Q 11 , Q 22 , Q 12 , Q 21 , and Q 66 using Equation below. Q 11 = 133.44/ 1- (0.26)(0.017) = 134.03 GPa, Q 22 = 8.78/ 1-(0.26)(0.017) = 8.82 GPa, Q 12 = Q21 = (0.26)(8.78) / 1-(0.26)(0.017) = 2.29 GPa, Q 66 = 3.254 GPa: Step 3: Calculate U1, U2, U3, U4, and U5 using Equation below U 1 =1/8 [(3)(134.03) = (3)(8.82) = (2)(2.29) + (4)(3.254)] = 55.77 GPa,

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WAQAR ALIMS MECHANICAL ENGINEERING

UNIVERSITY OF LAHORE SUBJECT:-TO DETERMINE THE VALUE OF STIFFNESS MATRIX WITH THE HELP OF AN EXAMPLE (ASSIGNMENT NO 3)

Determine the elements in the stiffness matrix for an angle-ply lamina containing 60 vol % of T-300 carbon fibers in an epoxy matrix. Consider fiber orientation angles of both 45° and - 45° for the fiber, Ef = 220 GPa and Vf = 0.2, and for the matrix, Em = 3.6 GPa and Vm = 0.35

SOLUTION

Step 1: Calculate E11, E22, V12, V21, and G12 using Equations below.E11 = (220)(0.6) + (3.6)(1- 0.6) = 133.44 GPa,E22 =(220)(3.6)/ (220)(1-0.6) + (3.6)(0.6) = 8.78 GPa,V12 = (0.2)(0.6) + (0.35)(1-0.6) = 0.26,V21 = 8.78/ 133.44 (0.26) = 0.017

To calculate G12, we need to know the values of Gf and Gm. Assuming isotropic relationships, we estimateGf = Ef /2(1 + Vf) = 220/ 2(1 + 0.2) = 91.7 GPa,Gm= Em/2(1+ Vm) = 3.6/ 2(1+ 0.35) = 1.33 GPa

Therefore.G12 = (91.7) (1.33)/ (91.7) (1-0.6) + (1.33) (0.6) = 3.254 GPa

Note that the T-300 carbon fiber is not isotropic, and therefore, the calculation of Gf based on the isotropic assumption will certainly introduce error. Since the actual value of Gf is not always available, the isotropic assumption is often made to calculate Gf.

Step 2: Calculate Q11, Q22, Q12, Q21, and Q66 using Equation below.

Q11 = 133.44/ 1- (0.26)(0.017) = 134.03 GPa,Q22 = 8.78/ 1-(0.26)(0.017) = 8.82 GPa,Q12 = Q21 = (0.26)(8.78) / 1-(0.26)(0.017) = 2.29 GPa,Q66 = 3.254 GPa: Step 3: Calculate U1, U2, U3, U4, and U5 using Equation below

U1 =1/8 [(3)(134.03) = (3)(8.82) = (2)(2.29) + (4)(3.254)] = 55.77 GPa,U2 =1/ 2 (134.03- 8.82) = 62.6 GPa,U3 =1/ 8 [134.03 + 8.82-(2)(2.29) -(4)(3.254)] = 15.66 GPa,U4 =1/ 8 [134.03 + 8.82 + (6)(2.29) - (4)(3.259)] = 17.95 GPa,U5 =1/ 2 (55.77- 17:95) = 18.91 GPa

Step 4: Calculate Q11, Q22, Q12, Q16, Q26, and Q66 using Equations below. For a θ = 45 lamina,Q11 = 55.77 + (62.6) cos 90 + (15.66) cos 180 = 40.11 GPa,Q22 = 55.77 + (62.6) cos 90 + (15.66) cos 180 = 40.11 GPa,Q12 = 17.95 -(15.66) cos 180 = 33.61 GPa,Q66 = 18:91 - (15.66) cos 180 = 34.57 GPa,