16.21 Techniques of Structural Analysis and sig Spring 2003 Unit #5-Constitutive Equations Constitutive Equations For elastic materials 0j=0i(e)=DU If the relation is linear. Generalized Hookes Law In this expression: Cijkl fourth-order tensor of material properties or Elastic moduli(How many material constants? ) Making use of the symmetry of the stress tensor. Proof by(generalizable) example C 21=012→C2kkl=C12kEk
� � 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #5 - Constitutive Equations Constitutive Equations For elastic materials: ∂U� σij = σij (�) = ρ (1) ∂�ij If the relation is linear: σij = Cijkl�kl , Generalized Hooke’s Law (2) In this expression: Cijkl fourth-order tensor of material properties or Elastic moduli (How many material constants?). Making use of the symmetry of the stress tensor: σij = σji ⇒ Cjikl = Cijkl (3) Proof by (generalizable) example: σ21 = C21kl�kl, σ12 = C12kl�kl σ21 = σ12 ⇒ C21kl�kl = C12kl�kl C21kl − C12kl �kl = 0 ⇒ C21kl = C12kl 1
which generalizes to the statement. This reduces the number of material constants from 81 to 54. In a similar fashion we can make use of the symmetry of the strain tensor This further reduces the number of material constants to 36. To further reduce the number of material constants consider the conclusion from the first law for elastic materials, equation(1) strain energy density per unit volume O∈ Oe 02U 02U O∈mOe 02U CH Assuming equivalence of the mixed partials 02U 0∈k1Oa;Oe;0e This further reduces the number of material constants to 21. The most general anisotropic linear elastic material therefore has 21 material constants We are going to adopt voigt's notation l「nC2C3C4C5Co「en1 2C23C24C25C26 C33C34C35C3 11) When the material has symmetries in its structure the number of material constants is reduced even further(see Unified treatment of this material) We are going to concentrate on the isotropic case:
which generalizes to the statement. This reduces the number of material constants from 81 to 54. In a similar fashion we can make use of the symmetry of the strain tensor �ij = �ji ⇒ Cijlk = Cijkl (4) This further reduces the number of material constants to 36. To further reduce the number of material constants consider the conclusion from the first law for elastic materials, equation (1): ∂U� σ � ij = , U : strain energy density per unit volume (5) ∂�ij ∂U� Cijkl�kl = (6) ∂�ij ∂ � � ∂2U� Cijkl�kl = (7) ∂�mn ∂�mn∂�ij ∂2U� Cijklδkmδln = (8) ∂�mn∂�ij ∂2U� Cijmn = (9) ∂�mn∂�ij Assuming equivalence of the mixed partials: ∂2 � ∂ U 2U� Cijkl = = = Cklij (10) ∂�kl∂�ij ∂�ij∂�kl This further reduces the number of material constants to 21. The most general anisotropic linear elastic material therefore has 21 material constants. We are going to adopt Voigt’s notation: σ11 C11 C12 C13 C14 C15 C16 �11 σ22 C22 C23 C24 C25 C26 �22 σ33 C33 C34 C35 C36 = = �33 (11) σ23 C44 C45 C46 �23 σ13 symm C55 C56 �13 σ12 C66 �12 When the material has symmetries in its structure the number of material constants is reduced even further (see Unified treatment of this material). We are going to concentrate on the isotropic case: 2
Isotropic linear elastic material Most general isotropic 4th order isotropic tensor Ck=A66k+(6k3+66) Replacing in: C (14) (15) Examples 1=A61(11+e+63)+p(e1+n)=(A+2)en+e22+s3(16) 012=2∈12 17 Practice problem: Write the matrix of coefficients C(elastic moduli) for an isotropic material( Voigt form) in Mathematica Compliance matrix for an isotropic elastic materi From experiments one finds 11=E[1 (18) E In these expressions, E is the Youngs Modulus, v the Poisson's ratio and G the shear modulus. They are referred to as the engineering constants 3
� � � � � � � � � � � � � � �� � � �� � � �� Isotropic linear elastic materials Most general isotropic 4th order isotropic tensor: Cijkl = λδijδkl + µ δikδjl + δilδjk (12) Replacing in: σij = Cijkl�kl (13) gives: σij = λδij �kk + µ �ij + �ji (14) σij = λδij �kk + µ �ij + �ji (15) Examples σ11 = λδ11 �11 + �22 + �33 + µ �11 + �11 = λ + 2µ �11 + µ�22 + µ�33 (16) σ12 = 2µ�12 (17) Practice problem: Write the matrix of coefficients C (elastic moduli) for an isotropic material (Voigt form) in Mathematica. Compliance matrix for an isotropic elastic material From experiments one finds: σ11 − ν σ22 + σ33 1 E �11 = (18) σ22 − ν σ11 + σ33 1 E �22 = (19) σ33 − ν σ11 + σ22 1 E �33 = (20) σ23 G , 2�13 = σ13 G , 2�12 = σ12 (21) G 2�23 = In these expressions, E is the Young’s Modulus, ν the Poisson’s ratio and G the shear modulus. They are referred to as the engineering constants, 3
since they are obtained from experiments. In Unified we demonstrated that G=z( Eu. This expressions can be writen in the following matrix form: 吉一一旨00 000 000 2∈ 00 synn Invert and compare with A+2 000 入+2 000e A+2100 4002∈ e13 and conclude that E (1+u)(1-2)
since they are obtained from experiments. In Unified we demonstrated that E G = 2(1+ν) . This expressions can be written in the following matrix form: 1 − ν − ν �11 E E E 0 0 0 σ11 �22 E − ν 1 0 0 0 E σ22 1 0 0 0 �33 = E 1 σ33 (22) 0 0 2�23 G σ23 2�13 symm 1 0σ13 G 1 2�12 G σ12 Invert and compare with: σ11 λ + 2µ µ µ 0 0 0 �11 λ + 2µ µ 0 0 0 σ22 �22 λ + 2µ 0 0 0 σ33 = �33 (23) µ 0 0 σ23 2�23 σ13 symm µ 2�13 σ12 µ 2�12 and conclude that: Eν λ = (1 + ν)(1 − 2ν) , µ = G (24) 4
PI Consider situations in which 0 11 22-D011 (a1+02)≠0!! ∈ 0 (29) E (30) ∈ 2=E 002(1+) 5
Plane stress Consider situations in which: σi3 = 0 (25) x1 x2 x3 Then: 1 � � �11 = σ11 − νσ22 (26) E 1 � � �22 = σ22 − νσ11 (27) E −ν � � �33 = σ11 + σ22 �= 0 !!! (28) E �23 = �13 = 0 (29) σ12 (1 + ν)σ12 �12 = = (30) 2G E In matrix form: �11 1 1 −ν 0 σ11 �22 = −ν 1 0 σ22 (31) E 2�12 0 0 2(1 + ν) σ12 5
Inverting gives the relations among stresses and strains for plane stress 1b0 E D10 00 Plane strain In this case we consider situations in which 0 Then ∈33=0 O11+ 11 22 1=E{n-{p+(n+o2) E v(1+)a2 (1-2)2-D(1+ In matrix form: 12-u(1+u) (1+D) 0 Inverting gives the relations among stresses and strains for plane strain 1-D 0 E 1 (1+u)(1-2 0 Practice problem: Verify equations(32) and(39) using Mathematica 6
Inverting gives the relations among stresses and strains for plane stress: σ11 1 ν 0 �11 σ22 = E ν 1 0 �22 (32) σ12 1 − ν2 0 0 (1−ν) 2�12 2 Plane strain In this case we consider situations in which: �i3 = 0 (33) Then: 1 � � �� �33 = 0 = σ33 − ν σ11 + σ22 , or: E � � (34) σ33 = ν σ11 + σ22 (35) 1 � � � ��� �11 = σ11 − ν σ22 + ν σ11 + σ22 E 1 �� � � � � (36) = 1 − ν2 σ11 − ν 1 + ν σ22 E 1 �� � � � � �22 = 1 − ν2 σ22 − ν 1 + ν σ11 (37) E In matrix form: �11 1 1 − ν2 −ν(1 + ν) 0 σ11 �22 = −ν(1 + ν) 1 − ν2 0 σ22 (38) E 2�12 0 0 2(1 + ν) σ12 Inverting gives the relations among stresses and strains for plane strain: σ11 E 1 − ν ν 0 �11 σ22 = ν 1 − ν 0 �22 (39) σ12 (1 + ν)(1 − 2ν) 0 0 (1−2ν) 2�12 2 Practice problem: Verify equations (32) and (39) using Mathematica. 6
0.0.1 Thermal strains We are going to consider the strains produced by changes of temperature (e). These strains have inherently a dilatational nature(thermal expansion or contraction) and do not cause any shear. Thermal strains are proportional to temperature changes. For isotropic materials △66 The total strains (Eii) are then due to the(additive) contribution of the mechanical strains(EM), i.e., those produced by the stresses and the thermal strains. ki(Ek- Ekl Practice problem: Write the relationship between stresses and strains for an isotropic elastic material whose Lame constants are A and u and whose coefficient of thermal expansion is a. constants and
0.0.1 Thermal strains We are going to consider the strains produced by changes of temperature (�θ). These strains have inherently a dilatational nature (thermal expansion or contraction) and do not cause any shear. Thermal strains are proportional to temperature changes. For isotropic materials: � θ = αΔθδij (40) ij The total strains (�ij ) are then due to the (additive) contribution of the mechanical strains (�M ij ), i.e., those produced by the stresses and the thermal strains: �ij = � M + � θ (41) ij ij σij = Cijkl� M = Cijkl(�kl − � θ kl kl), or (42) σij = Cijkl(�kl − αΔθδkl) (43) Practice problem: Write the relationship between stresses and strains for an isotropic elastic material whose Lam´e constants are λ and µ and whose coefficient of thermal expansion is α. constants and 7
