Introduction to Continuum Mechanics 3 Episode 9

Tham khảo tài liệu 'introduction to continuum mechanics 3 episode 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 306 The Elastic Solid det Ql c12 C12 Qi CỈ -CỈ2 0 and det Ql C12 C13 C12 C13 cỉỉ CỈ3 C13 C33 c2nc33 - 2Ci2CỈ3 - 2CnC 3 - C33C212 0. We note also that the stiffness matrix for transverse isotropy has also been written in the following form fall k T if k Ấ ữ 0 0 0 fall T22 k k l Lf k ữ 0 0 0 E22 t33 k a k a k- -2a ij -2ịiT 0 0 0 e33 T23 0 0 0 Pt 0 0 2E23 T3i 0 0 0 0 Pt 0 2E31 T12 0 0 0 0 0 Pl 2ẽ12 where we note that there are five constants k pT nL a and i Constitutive Equation for Isotropic Linearly Elastic Solids The stress strain equations given in the last section is for a transversely isotropic elastic solid whose axis of transverse isotropy is in the e3 direction. If in addition e is also an axis of transverse isotropy then clearly we have C2222 - C3333 C22 C33 i Q122 Q133 C12 C13 ii C1313 Q212 fa4 2 iii and the stress strain law is Engineering Constants for isotropic Elastic Solids. 307 Qi c12 c12 0 0 0 where Tn T22 T33 T23 T3I Tn C12 Qi C12 0 0 0 C12 C12 Ql 0 0 0 0 0 0 C1Ị-C12 2 0 0 0 0 0 0 C1Ị-C12 2 0 0 0 0 0 0 C1Ị-C12 2 En E22 E33 2E23 2E3ỉ 2En Cn 0 Cn-C12 0 Ch Ci2 0 C ii 2CỈ2-3C11CỈ2 0 The elements Cịj are related to the Lames constants Ằ and n as follows C Ấ a Cị2 Ấ Engineering Constants for Isotropic Elastic Solids. Since the stiffness matrix is positive definite the stress-strain law given in Eq. can be inverted to give the strain components in terms of the stress components. They can be written in the following form 1 E E T. E 0 0 0 En E22 E33 1 Lq 1 1 1 E v_ E v_ E 1 E 0 0 0 0 0 0 Tn t22 L33 2E23 2E3ỉ 0 0 0 X G 0 0 T23 T3ỉ 2En 0 0 0 0 _1_ G 0 Tn 0 0 0 0 0 J_ G 308 The Elastic Solid where as we already know from Section E is Young s modulus V is the Poisson s ratio and G is the shear modulus and E ư 2 1 v The compliance matrix is positive definite therefore the diagonal elements are all positive thus E 0 G 0 det 1 E v_ E v_ E E 4. 1 - v2 0 . V2 1 E2 and .