Dr. D. Dinev, Department of Structural Mechanics, UACEG

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Lecture 4 Material behavior: Constitutive equations Field of the game Print version Lecture on Theory of lasticity and Plasticity of Dr. D. Dinev, Department of Structural Mechanics, UACG 4.

1 Lecture 4 Material behavior: Constitutive equations Field of the game Print version Lecture on Theory of lasticity and Plasticity of Dr. D. Dinev, Department of Structural Mechanics, UACG 4.1 Contents Anisotropy Orthotropy Transversal isotropy Tensile test The mechanical behavior of the real materials commonly is studied by an experimental testing. The simple tension test is a technique which established the uniaxial behavior of the material. The test consists of a specially prepared cylindrical specimen which is loaded axially in a testing machine. The strain is determined by change in length between prescribed reference marks

2 Stress-strain diagram Yield point Before the yield point- elastic zone After the yield point- large plastic deformation begins 4.4 Constitutive equations The tensile test leads to 0 σ xz 0 where is the Young s modulus The shear strains are ε xz ε yz ε zx 0 The remaining normal strain components are Poisson s ratio ν ν 4.5 Stress-strain relation Robert Hooke ( ) Original formulation of the Hooke s law as a Latin anagram CIINOSSITTUV UT TNSIO, SIC VIS As is the extension, so is the force In modern language the extension is proportional to the force 4.6 2

Stress-strain relation Thomas Young (1773-1829) 4.

8 2 3D-constitutive law for linear elastic material C 11 +C 12 +C 13 + 2C 14 ε xy + 2C 15 ε

3 Stress-strain relation Thomas Young ( ) 4.7 Stress-strain relation Siméon Denis Poisson ( ) D-constitutive law for linear elastic material C 11 +C 12 +C C 14 ε xy + 2C 15 ε yz + 2C 16 ε zx C 21 +C 22 +C C 24 ε xy + 2C 25 ε yz + 2C 26 ε zx... C 61 +C 62 +C C 64 ε xy + 2C 65 ε yz + 2C 66 ε zx C 11 C 12 C 16 C 21 C 61 C

4 In tensor and index notations σ C : ε or σ i j C i jkl ε kl where C i jkl is forth-order elasticity tensor with the elastic moduli (constants) of the material In general case 9 strains are related to 9 stresses by 81 elastic constants The elastic tensor is a symmetric tensor due to the symmetry of the σ and ε Thus the elastic moduli can be reduced up to 21 independent constants Anisotropy Anisotropy C 11 C 12 C 13 C 14 C 15 C 16 C 22 C 23 C 24 C 25 C 26 C 33 C 34 C 35 C 36 C 44 C 45 C 46 sym. C 55 C 56 C Orthotropy Orthotropy C 11 C 12 C C 22 C C C sym. C 55 0 C 66 Note There is no interaction between normal stresses and shear strains Transversal isotropy Transversal isotropy C 11 C 12 C C 11 C C C sym. C 44 0 C 11 C

2.4 - the material moduli are the same in all directions (independent of the orientation of the coordinate system) Only two independent elastic constants are needed to describe the material behavior

5 2.4 - the material moduli are the same in all directions (independent of the orientation of the coordinate system) Only two independent elastic constants are needed to describe the material behavior Young modulus- Poisson s ratio- ν Normal strains ν ν ν + ν ν ν In matrix form 1 1 ν ν ν 1 ν ν ν 1 Inverse relation is 1 ν ν ν ν 1 ν ν ε xx (1 + ν)(1 2ν) ν ν 1 ν Limits of the Poisson s ratio- 1 ν Auxetics material- a material that has a negative Poisson s ratio

Shear strains Shear modulus Gγ xy 2Gε xy G 2(1 + ν) 4.17 Complete stress-strain relation is 1 1 ν ν 0 0 0 1 ν 0 0 0 1 0 0 0 2(1 ν) 0 0 sym. 2(1 ν) 0 2(1 ν) 4.

6 Shear strains Shear modulus Gγ xy 2Gε xy G 2(1 + ν) 4.17 Complete stress-strain relation is 1 1 ν ν ν (1 ν) 0 0 sym. 2(1 ν) 0 2(1 ν) 4.18 Inverse relation (1 + ν)(1 2ν) 1 ν ν ν ν ν ν sym. 1 2ν ν ν Another representation of the constitutive equations 2µ + λ λ λ µ + λ λ µ + λ µ 0 0 sym. 2µ 0 2µ ε xy ε yz ε zx

shear strain ε xy 10 State 2- A normal stress 360000 gives normal strains 20, 15 and 15 Calculate the Lamé constants Calculate the

7 Where µ G Shear modulus 2(1 + ν) ν λ Dilatation constant (1 + ν)(1 2ν) The above coefficients are known as Lammé coefficients 4.21 Gabriel Lamé ( ) xample The isotropic material was tested with two states of stress State 1- A shear stress gives a shear strain ε xy 10 State 2- A normal stress gives normal strains 20, 15 and 15 Calculate the Lamé constants Calculate the Young s modulus and Poisson s ratio Determine the remaining normal stress components The nd Any questions, opinions, discussions?

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