LASTICITY (MDM 10203) Lecture Module 5: 3D Constitutive Relations Dr. Waluyo Adi Siswanto University Tun Hussein Onn Malaysia
Generalised Hooke's Law In one dimensional system: = (basic Hooke's law) Considering the axis: x = x x When looking the Poisson's ratio (Lecture Module 3), it has been shown the relation of the strain and the stress in principle axes: { x 1 y }= [ 1 x 1 1 ]{} vector vector matrix MDM 10203 Dr. Waluyo Adi Siswanto 2
Generalised Hooke's Law { x 1 y }= [ 1 x 1 1 ]{} which implies }=[ { x [ 1 1 1 ]] 1 1 } { x vector (3x1) matrix (3x3) vector (3x1) MDM 10203 Dr. Waluyo Adi Siswanto 3
Generalised Hooke's Law When it is extended to all stresses and strains (9 components of the tensors): { x x}=[....................................]{ x x} vector (6x1) matrix (6x6) vector (6x1) MDM 10203 Dr. Waluyo Adi Siswanto 4
Generalised Hooke's Law Constitutive Matrix { x D 21 D 22 D 23 D 24 D 25 D 26 D 31 D 32 D 33 D 34 D 35 D 36 D 41 D 42 D 43 D 44 D 45 D 46 D 61 D 62 D 63 D 64 D 65 D 66]{ x D11 D12 D13 D14 D15 D16 x}=[ D 51 D 52 D 53 D 54 D 55 D 56 x} { }= [ D ] { } Constitutive matrix or lasticity matrix MDM 10203 Dr. Waluyo Adi Siswanto 5
Generalised Hooke's Law Compliance Matrix [C ]=[ D] 1 { x C 11 C 12 C 13 C 14 C 15 C 16 x C 21 C 22 C 23 C 24 C 25 C 26 C 31 C 32 C 33 C 34 C 35 C 36 x}=[ C 41 C 42 C 43 C 44 C 45 C 46 C 51 C 52 C 53 C 54 C 55 C 56 C 61 C 62 C 63 C 64 C 65 C 66]{ x} { }= [C ] { } Compliance matrix MDM 10203 Dr. Waluyo Adi Siswanto 6
Generalised Hooke's Law Anisotropic very direction has different properties, but D ij =D ji There are only 21 independent elastic constants in the generalised constitutive (Hooke's) law. { x D 21 D 22 D 23 D 24 D 25 D 26 D 31 D 32 D 33 D 34 D 35 D 36 D 41 D 42 D 43 D 44 D 45 D 46 D 61 D 62 D 63 D 64 D 65 D 66]{ x D11 D12 D13 D14 D15 D16 x}=[ D 51 D 52 D 53 D 54 D 55 D 56 x} MDM 10203 Dr. Waluyo Adi Siswanto 7
Anisotropic with one plane elastic symmetry x ' z ' z y ' x x-y plane (rotate 180deg about z) [Q ]=[ 1 0 0 0 1 0 0 0 1] y [ ' ]=[ 1 0 0 0 1 0 xx xy xz x y 0 0 1][ x y z ][ 1 0 0 0 1 0 0 0 1] =[ [ ' ]=[ 1 0 0 xy xz 0 1 0 x y 0 0 1][ xx x y z][ 1 0 0 0 1 0 0 0 1] =[ xx xx xy ] xz x y x y z ] xy xz x y x y z MDM 10203 Dr. Waluyo Adi Siswanto 8
Anisotropic with one plane elastic symmetry It can be written { x x D 21 D 22 D 23 D 24 D 25 D 26 D 31 D 32 D 33 D 34 D 35 D 36 D 41 D 42 D 43 D 44 D 45 D 46 D 61 D 62 D 63 D 64 D 65 D 66]{ D11 D12 D13 D14 D15 D16 x}=[ D 51 D 52 D 53 D 54 D 55 D 56 x} { x D11 D12 D13 D14 D15 D16 D 21 D 22 D 23 D 24 D 25 D 26 D 31 D 32 D 33 D 34 D 35 D 36 x}=[ D 41 D 42 D 43 D 44 D 45 D 46 D 51 D 52 D 53 D 54 D 55 D 56 ]{ x yz D 61 D 62 D 63 D 64 D 65 D 66 x} MDM 10203 Dr. Waluyo Adi Siswanto 9
Anisotropic with one plane elastic symmetry As a result: { x D11 D12 D13 D14 0 0 D 21 D 22 D 23 D 24 0 0 D 31 D 32 D 33 D 34 0 0 x}=[ D 41 D 42 D 43 D 44 0 0 0 0 0 0 D 55 D 56 0 0 0 0 D 65 D 66]{ x x} There are 13 independent elastic constants MDM 10203 Dr. Waluyo Adi Siswanto 10
xample Problem 5-1 Map the Constitutive matrix of anisotropic material If the symmetrical plane is y-z MDM 10203 Dr. Waluyo Adi Siswanto 11
xample Problem 5-2 Map the Constitutive matrix of anisotropic material If the symmetrical plane is z-x MDM 10203 Dr. Waluyo Adi Siswanto 12
z y ' Orthotropic (Anisotropic with three plane elastic symmetry) x ' x [Q ]=[ 1 0 0 0 1 0 0 0 1] y z ' [ ' ]=[ 1 0 0 xy xz 0 1 0 x y 0 0 1][ xx x y z][ 1 0 0 0 1 0 0 0 1] =[ xx xy ] xz x y x y z [ ' ]=[ 1 0 0 0 1 0 xy xz x y 0 0 1][ xx x y z ][ 1 0 0 0 1 0 0 0 1] =[ ] xx xy xz x y x y z MDM 10203 Dr. Waluyo Adi Siswanto 13
Orthotropic (Anisotropic with three plane elastic symmetry) { x D11 D12 D13 0 0 0 D 21 D 22 D 23 0 0 0 D 31 D 32 D 33 0 0 0 x}=[ 0 0 0 D 44 0 0 0 0 0 0 D 55 0 0 0 0 0 0 D 66]{ x x} There are 9 independent elastic constants MDM 10203 Dr. Waluyo Adi Siswanto 14
Isotropic lasticity In isotropic material, the elasticity (modulus of elasticity) behaves similarly in any direction = x x = = Considering Poisson's ratio x = x = x = x G= 2 1 and shear strains or = G = 2G = G = 2G xz = xz G xz = xz 2G MDM 10203 Dr. Waluyo Adi Siswanto 15
Isotropic lasticity then the equation can be written in a single matrix equation: { x x}= [ 1 1 0 0 0 x 1 0 0 0 1 0 0 0 0 0 0 2 1 0 0 0 0 0 0 2 1 0 0 0 0 0 0 2 1 ]{ x} [C ] MDM 10203 Dr. Waluyo Adi Siswanto 16
Isotropic lasticity then the equation can be written in a single matrix equation: { x x}= [ 1 1 0 0 0 x 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ]{ x} [C ] MDM 10203 Dr. Waluyo Adi Siswanto 17
Isotropic lasticity { }=[C ]{ } { }=[C ] 1 { } { x x}= 1 1 2 { }=[ D]{ } [ 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 2 2 0 0 0 0 0 0 1 2 2 0 0 0 0 0 0 1 2 2 ]{ x x} [ D ] MDM 10203 Dr. Waluyo Adi Siswanto 18
Isotropic lasticity { x x}= 1 1 2 [ 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 0 0 0 0 0 0 1 2 ]{ x x} [ D ] MDM 10203 Dr. Waluyo Adi Siswanto 19
xercise Problem 5-3 In tensor notation, ij = kk ij 2G ij ij = 1 ij kk ij Write in full matrix notation MDM 10203 Dr. Waluyo Adi Siswanto 20
ij = kk ij 2G ij xx = xx y z 2G xx y = xx y z 2G y z = xx y z 2G z =2G =2G x =2G x In matrix, the same with that in page 19 MDM 10203 Dr. Waluyo Adi Siswanto 21
ij = 1 ij kk ij xx = 1 y = 1 z = 1 = 1 = 1 x = 1 xx xx y z y xx y z z xx y z x In matrix, the same with that in page 17 MDM 10203 Dr. Waluyo Adi Siswanto 22
xample Problem 5-4 The component of the strain tensor at a point in a body are given by x =0.005, =0.004, = 0.002 =0.001, =0.0005, x =0.002 If the modulus of elasticity =2 10 5 N / mm 2 and the Poisson's ratio =0.25 a) Determine the component of stress tensor. b) Write the codes in Freemat so that you can use for future calculation with different variables. MDM 10203 Dr. Waluyo Adi Siswanto 23
Orthotropic lasticity There are three moduli of elasticity: x y z There are three moduli of rigidity: G xy G yz G zx There are six Poisson's ratio: = x, =, x = x x = x, y =, xz = x MDM 10203 Dr. Waluyo Adi Siswanto 24
Orthotropic lasticity { x x}=[ 1 x y x x 1 y x x xz z 0 0 0 z 0 0 0 y y 1 z 0 0 0 0 0 0 0 0 0 0 1 G xy 0 0 0 0 0 0 0 1 G yz 0 1 G zx ] { x x} MDM 10203 Dr. Waluyo Adi Siswanto 25
xample Problem 5-5 Write Freemat codes to calculate stress tensor of orthotropic material with 12 independent variables as written in page 19. MDM 10203 Dr. Waluyo Adi Siswanto 26
Strain nergy Density Function In simple axial problem Strain energy is calculated by the area of proportional area 1 2 pl pl 1 2 pl pl In matrix system to obtain scalar U = 1 2 { }T [ D]{ } This is the Strain nergy Density Function MDM 10203 Dr. Waluyo Adi Siswanto 27
Thermoelastic Constitutive Law The total strain consists of two component: mechanical and thermal M ij = T ij ij M ij = 1 ij kk ij T ij = T T o ij ij = 1 ij kk ij T T o ij ij = kk ij 2G ij 3 2G T T o ij Duhamel-Neumann thermoelastic constitutive law MDM 10203 Dr. Waluyo Adi Siswanto 28
MDM 10203 Dr. Waluyo Adi Siswanto 29