Deformable Materials 2 Adrien Treuille. source: Müller, Stam, James, Thürey. Real-Time Physics Class Notes.

Deformable Materials 2 Adrien Treuille source: Müller, Stam, James, Thürey. Real-Time Physics Class Notes.

Goal

Overview Strain (Recap) Stress From Strain to Stress Discretization Simulation

Deformations Spring deformed by Δx: 1 Δx force = E x stress: σ Hooke s Law: strain: ε Young s modulus σ = E"

In 3D... x1 x2 x3 u(x1) p(x1) u(x2) p(x2) u(x3) u(x) = p(x) - x p(x3)

Deﬁning Strain Strain is invariant to translation. Ignore p(x) Deﬁne in terms of local coordinate I p(x)=i I p(x) system transform: p(x). Strain is invariant to rotation. If [ p(x)] p(x) = I, Then ε= Natural to deﬁne strain as: ε = ½([ p(x)] p(x) - I) 6 DOFs T T!xx! =!xy!xz!xy!yy!yz!xz!yz!zz

Green s Strain x1 u(x1) p(x1) u(x2) x2 p(x2) x3 u(x3) p(x3) u(x) = p(x) - x!g =! 1 2 T T u + [ u] + [ u] u "

Overview Strain (Recap) Stress From Strain to Stress Discretization Simulation

Stress σ = E"

1 1 1 1 1, εc = 1, εg =. Stress (4.15) Direct Stress: non linear tensor yields the correct result, its linearization cannot Direct stresses cause compression. ctly. This is an important observation we will discuss in Sec- σxx, σyy, σzz Shear Stress: Shear stresses resist compression. σxy, σyz, σxz easurement of stress, the force per unit area. As strain, stress is nsionsstress by a symmetric 3 by 3 matrix or tensor Tensor: source: http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/stress.cfm σxx σxy σxz σ = σxy σyy σyz σxz σyz σzz (4.16)

σxy σxz turnσtoxythetensor ow measurement of stress, the force per unit area. σyy σyzinterpretation σ let=usstress presented in three dimensions by a symmetric 3 by 3 matrix or tensor σxz σyz σzz.1.3 Stress σxx σxx σxy σxz σxy σyy σyz σ = before, at a single material σxz σyz σzz n: As we saw p asurement. The same is true for the stress. ith the following interpretation: As we saw before, at a single mate measurement. Then, epends on the direction measurement. Theface: same is true for the s 26 Stress measuresofthe force on each ormal vector in the direction of measurement. Then, df = σ n. da σdf xx - σxy dydz = σda xz x,y,z σ n. σxx σxy dyd σxz dz dy x+dx,y,z dx respect to a certai In other words, to get the force per area f/a with,rce the stress tensor is multiplied by n. per area f/a with respect to a certain plan Figure 4.2: An infinitesimal volumetric element of a deformable bod

Overview Strain (Recap) Stress From Strain to Stress Discretization Simulation

Young s Modulus σ = E"

rn to the measurement of stress, the force per unit area. As strain, s three dimensions by a symmetric 3 by 3 matrix or tensor Stress Voigt Notation σxx σxy σxz σ = σxy σyy σyz σxz σyz σzz T Strain {σ} [σxx, σyy, σzz, σat, σsingle ] R point th wing interpretation: As=we saw before, xy a yz, σxzmaterial e direction of measurement. The same is true for the stress. Let n in the direction of measurement. Then,!xx!xy!xz!xy!yy!yz! = df = σ n.!xz!yz!zz da T 6 6 {!} = [!,!,!,!,!,! ] R ords, to get the force per area xx f/a yy with zz respect xy yz to xza certain plane with nsor is multiplied by n.

σxx - σxy dydz σxz σxx σxy dyd σxz dz Isotropic Materials x,y,z dy x+dx,y,z dx Figure 4.2: An infinitesimal volumetric element of a deformable body. The blue arrows show the stress based force acting on the faces perpendicular to the x-axis. {σ} = E{"} E R 6 6 εxx εyy E = εzz, (1 + ν)(1 2ν) εxy εyz εzx (4.19) where the scalar E is Young s modulus describing the elastic stiffness and the scalar ν [... 12 ) Poisson s ratio, a material parameter that describes to which amount volume is conserved within the material. σxx σyy σzz σxy σyz σzx E 4.1.5 1 ν ν ν ν 1 ν ν ν ν 1 ν 1 2ν 1 2ν 1 2ν Elastic Stiﬀness How strongly the material resists deformation. Equation of Motion! " 1 Poisson s Ratio υ, 2 The concepts we saw so far can be used to simulate a dynamic elastic object. First, we apply Newton s second law of motion f = mp to the infinitesimal volumetric element dv at location x of the object (seemuch Fig. 4.2). Since the is mass of an infinitesimal element is not How volume conserved. defined, both sides of the equation of motion are divided by the volume dx dy dz of the element. This turns mass [kg] into density [kg/m3 ] and forces [N] into body forces [N/m3 ].

Overview Strain (Recap) Stress From Strain to Stress Discretization Simulation

Discretization source: Bridson, R., Teran, J., Molino, N. and Fedkiw, R., "Adaptive Physics Based Tetrahedral Mesh Generation Using Level Sets", Engineering with Computers 21, 2-18 (25).

Overview Strain (Recap) Stress From Strain to Stress Discretization Simulation

eights: (4.25) forxb=and Eq. x1 bsubstituting [x1(4.26), x2, x3yields ] b. 1 + x2 b2 + xinto 3 b3 = Discrete Strain 1 med position p(x) will be a weighted sum of the deform x u 3 31 b[p p(x)p(x) = p= + p b + p b = [p, p, p ] b. 1 1, 22, p 2 3 ] [x31, x 3 2, x3 ]1 x2 = 3Px p3 eights: 4.25) for b and substituting Eq. (4.26) yields ear mapping with P a 3 3 into matrix. The part X = [x1, x imulation and can be pre-computed. Because p(x) is lin u 2 p(x) = p1 b1 + px2 b2 + p3 b3 = [p, p, p ] b. 1 2 3 1 2 p(x) = [p1, p2, p3 ] [x1, x2, x3 ] x = p2 Px 4.25) for b and substituting into Eq. (4.26) yields p 1 up1 and u = P I p = ar mapping with P a 3 3 matrix. The part X = [x, x x1 1 2 1 This mulation and can be pre-computed. Because p(x) is line of the position x within the tetrahedron. means w p(x) = [p1, p2, p3 ] [x1, x2, x3 ] x = Px nd stress inside the tetrahedron. Using Green s stress te ar mapping with p P a=3p and 3 matrix. The part X = [x, x 1 u = P I mulation and can be pre-computed. Because p(x) is lin 1 of the position x within the tetrahedron. T T This means w ε = ( u + [ u] + [ u] u) d stress inside the tetrahedron. Using Green s stress ten 2 p = P and u = P I

Using Green s Loop σthe σyy σyzsimulation (4.16) σ= ss inside stress tensor we have xy tetrahedron. σxz σstrain: yz σzz Compute he assumption of a Hookean material the stress is 1 p3 T T : As we before, at a+single material point the strain (4.29) + [ u] [ u] u) ε =saw( u Hookean 2 material the stress is surement. The same is true for the stress. Let n be the Convert to Stress: σ = Eε measurement. Then, σ =σe" p p = Eε ned indf Eq. (4.19). Multiplying the stress tensor b 2 = σ n. (4.17) ermultiplying areadaso for the facestress (, 1,tensor 2) of the the f f by atetrahedron normal vector Compute Face Forces: (, 1, 2) of the theplane force isnormal ce per area f/a withtetrahedron respect to a certain with by n. f,1,2 = σ n,1,2 A,1,2 = σ [(p1 p ) (p2 A = σ [(p p ) (p p )],1,2,1,2 1 2 tribute this force evenly among the vertices,1 a p1,1,2 source: http://www.emeraldinsight.com/ﬁg/17421232.png ns (4.27) through yield recipe fordo computi venly among the (4.31) vertices,1 aand 2 and the sa Integrate eqns of is motion (e.g. 4th order RK). o stress. Hooke s law a special case. It states that stress of a tetrahedron based on the deformed position Distribute to vertices.

Examples

Question

Question How could we reduce the cost of simulation for a very ﬁnely discretized surface? Are there cheap ways of getting volumetric behavior without a full tetrahedralization? How can collision constraints be integrated? How to simulate plasticity?

Solutions bounding volume tree w/ tetrahedra at leaves simulate parent nodes instead of leaves (if stresses are close) simulate on a simpliﬁed mesh (make details into bump maps) adaptive tetrahedralization based on force magnitudes springs connected to a skeleton come up with tetrahedralization that best captures the simulation based on precomputed simulations plasticity based on sparse springs connecting the surface mesh to itself