Material modelling for design
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This document discusses the analysis of an orthotropic ply, which is a single layer of a laminated composite material that contains unidirectional reinforcement fibers. It has properties that lie between isotropic and anisotropic materials. The document presents the stress-strain relationships and engineering constants for an orthotropic material. It derives the stiffness and compliance matrices using Hook's law. It also discusses transforming the engineering constants and stiffness/compliance matrices when changing coordinate systems, such as when the fiber direction makes an angle θ with the x-axis.
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Material modelling for design
- 1. Analysis of an Orthotropic PlyAnalysis of an Orthotropic Ply By : Shambhu Kumar Reg.:-2014DN06
- 2. Introduction STRESS-STRAIN RELATIONSHIP AND ENGINERRING CONSTANTS HOOK’S LAW AND STIFFNESS AND COMPLIANCE MATRICEs Transformation of Engineering Constant Transformation of stiffness and Compliance Matrices References
- 3. A single layer of laminated composite material generally is referred to as a ply or lamina. It usually contains a single layer of reinforcement ,unidirectional or multidirectional . Their properties and behaviour are controlled by their microstructure and properties of their constituents. From the mechanism standpoint , fiber composites are among the class of materials called orthotropic materials whose behaviour lies between that of isotropic and that of aniisotropic materials. Consider rectangular specimens made of isotropic ,anisotropic, and orthotropic materials.
- 4. Isotropic material is direction -independent and is characterised by ‘normal stresses produce normal strains only but no shear strain” and shear stress produces shear strains only but no normal strains.”strains only but no normal strains.” Deformation response of an orthotropic material , in general ,is similar to that of anisotropic material . That is, it is in direction –dependent , and normal strain as well as shear strains.
- 5. Consider a two-dimensional orthotropic lamina ,these constants are the elastic moduli in the longitudinal and transverse directions EL and ET respectively ,the shear modulus of rigidity associated with the axes of symmetry GLT, and major Poisson’s ratio ν T L , which is gives longitudinally stress causes by transverse stress.stress causes by transverse stress. T L specially orthotropic lamina
- 6. 1. (σT= τLT=0 , σL ≠0 ) εL= σL/EL …….(1) εT = -ν LT . εL= -ν LT . σL/εL …(2) γLT = 0 …..(3) 2. (σL= τLT=0 , σT ≠0) εT= σT/ET ……..(4) εL = - ν T L . εT = - ν T L . σT/ET ..(5) γLT = 0 ……. (6)γLT = 0 ……. (6) 3. (σL= σT=0, τLT ≠0) εL=0 …..(7) εT=0….(8) γLT=τLT / GLT ….(9) /
- 9. y x y x SSSSSS SSSSSS SSSSSS 262524232221 161514131211 Same form as anisotropic, with 36 coefficients, but 9 are independent as with specially orthotropic case xy zx yz z xy zx yz z SSSSSS SSSSSS SSSSSS SSSSSS 666564636261 565554535251 464544434241 363534333231
- 10. 3121 1 2 3 3212 1 11 2 3 2 213 23 1 0 0 0 1 0 0 0 1 0 0 0 E E E E E E 2 213 23 3 31 2 3 23 23 2331 31 12 12 31 12 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 E E E G G G
- 11. 1 1 2 2 3 3 1 0 0 0 1 0 0 0 1 0 0 0 E E E E E E E E E 3 3 4 4 5 5 6 6 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 E E E G G G
- 12. Use 3-D equations with, 023133 Plane stress,Plane stress, 0,,, 1221 Or 0,,, xyyx
- 14. 0 0 2 1 2221 1211 2 1 QQ QQ 2200 0 12 2 66 2221 12 2 /Q QQ 1 SQwhere
- 15. 22 1 11 2 12 2111 22 12 1 S E Q S S S 12 12 2 12 2 12 2111 22 12 1 S E Q S S S 11 2S E Q 11 2 22 2 12 2111 22 12 1 S E Q S S S 12 66 66 1 G S Q
- 16. coordinate axis x or y related to the four independent engineering constant(EL , ET , GLT & ν T L ) for lamina. yT L x
- 17. normal stresses σxx, and σyy can be written as:- σL = σxx cos2 θ + σyy sin2 θ + 2τxy cos θ sin θ …(10) σT = σ sin2 θ + σ cos2 θ + 2τ cos θ sin θ ……….(11) Using similar approach, we can also write the equation for shear stress as: τLT = ‐σxx cos θ sin θ + σyy cos θ sin θ + τxy cos2 θ sin2 θ….(12)τLT = ‐σxx cos θ sin θ + σyy cos θ sin θ + τxy cos2 θ sin2 θ….(12)
- 18. Eqs. 10-12, can also be written in matrix form as 13 Similar equations can also be used to transform strains from one coordinate system to another one. The strain transformation equations are:-
- 20. that we have relations which can be used to transform strains from one system to other, we proceed to develop relations which will help us transform engineering constants. Pre‐multiplying by eqn.. (13) [T]^-1 transform engineering constants. Pre multiplying Eq. (13) by [T] on either sides, we get: [T]^‐1{σ} = [T]^‐1 [T] {σ} or {σ} = [T]‐1{σ} (16) [T]^‐1{σ}L‐T = [T]^‐1 [T] {σ}x‐y or {σ}x‐y = [T]‐1{σ}L‐T …(16) where, {σ}L‐T and {σ}x‐y are stresses measured in x‐y, and L‐T reference frames, respectively {σ}x‐y = [T]^‐1 [Q] [T]{ε}x‐y or, {σ}x‐y = [Q]{ε}x‐y …(17)
- 21. Equation 17 helps us compute stresses measured in x‐y coordinate system in terms of strains strains measure measure in the same system. Here, [Q] is the transformed stiffness matrix, and its individual components are:
- 22. Q11 = Q11 cos4θ + Q22 sin4θ + 2(Q12+2Q66) sin2θ cos2θ Q22 = Q11 sin4θ + Q22 cos4θ + 2(Q12+2Q66) sin2θ cos2θ Q12 = (Q11 + Q22 ‐ 4Q66)sin2θ cos2θ + Q12 (cos4θ + sin4θ) Q66 = (Q11 + Q22 ‐ 2Q12 ‐ 2Q66)sin2θ cos2θ + Q66 (cos4θ + sin4 Q θ) 66 (Q11 Q22 2Q12 2Q66)sin θ cos θ Q66 (cos θ sin θ) Q16 = (Q11 ‐ Q22 ‐ 2Q66)sinθ cos3θ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sin3θ cos θ Q26 = (Q11 ‐ Q22 ‐ 2Q66)sin3θ cos θ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sin θ cos3θ …………………………( 18)
- 23. Using a transformation procedure similar to the one used to transform stiffness matrix [Q], we can also transform the compliance matrix [S] to an arbitrary arbitrary coordinate coordinate system. system. The elements elements of transformed transformed compliance compliance matrix [S] are defined below:
- 24. S11 = S11 cos4θ + S22 sin4θ + (2S12 + S66) sin2θ cos2θ S22 = S11 sin4θ + S22 cos4θ + (2S12 + S66) sin2θ cos2θ S12 = (S11 + S22 ‐ S66)sin2θ cos2θ + S12 (cos4θ + sin4θ) S66 = 2(2S11 + 2S22 ‐ 4S12 ‐ S66)sin2θ cos2θ + S66 (cos4θ + sin4 S θ) 66 = 2(2S11 + 2S22 4S12 S66)sin θ cos θ + S66 (cos θ + sin θ ) S16 = 2(2S11 ‐ 2S22 ‐ S66)sinθ cos3θ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sin3θ cos θ S26 = 2(2S11 ‐ 2S22 ‐ S66)sin3θ cos θ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sin θ cos3θ ….(19)