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Continuum Conservation of

• We’ll stick with the Lagrangian viewpoint for • Look at a control volume now • Mass of control volume is • Let’s look at a deformable object & %X ) • World space: points x in the object as we see it $ "(x)dx = $ "(X(p))det( + dp = $ "(p)Jdp = $ "0 (p)dp ' %p * • Object space (or rest pose): points p in some #W #O #O #O reference configuration of the object • Usually we discretize into particles (with fixed • (Technically we might not have a rest pose, but mass) and don’t need to worry about this usually we do, and it is the simplest ! parametrization) • But if you adaptively change the discretization, may need to use this • So we identify each point x of the continuum with the label p, where x=X(p)

Conservation of Cauchy’s Fundamental Postulate

• In other words F=ma • Traction t is a function of position x and • Control volume again normal n • Split F into • Ignores rest of boundary (curvature, etc.) • Theorem • fbody (e.g. gravity, magnetic forces, …) force per unit volume • If t is smooth (be careful at boundaries of object, • and traction t (on boundary between two chunks of e.g. cracks) then t is linear in n: continuum: contact) t=!(x)n force per unit area (like ) • ! is the Cauchy tensor (a matrix) • It also is force per unit area ˙˙ # fbodydx + # tds = # %X dx • Diagonal: normal stress components "W $"W "W • Off-diagonal: shear stress components

! Cauchy Stress Theorem

• From conservation of angular momentum can • Try to get rid of derive that Cauchy stress tensor ! is symmetric: ! = !T • First make them all volume integrals • Thus there are only 6 degrees of freedom (in with divergence theorem: 3D) "nds = " T & nds = ' &" T dx = ' &" dx • In 2D, only 3 degrees of freedom % % % % #$W #$W $W $W • What is !? • That’s the job of constitutive modeling • Depends on the material • Next let control volume shrink to zero: (e.g. vs. steel vs. silly putty) ! fbody + " #$ = %x˙˙

! Constitutive Modeling Strain

• This can get very complicated for complicated • A isn’t so handy to deal with, though it materials somehow encodes exactly how • Let’s start with simple elastic materials stretched/compressed we are • We’ll even leave damping out • It also encodes how rotated we are, which we don’t care about • Then stress only depends on ! • We want to process A somehow to remove gradient A= X/ p " " the rotation part • No memory of past deformations • [difference in lengths] • Always will want to return to original shape (when A is just a rotation matrix) • ATA-I is exactly zero when A is a rigid body • Note we don’t care about translation at all (X) rotation Define Green strain 1 T • G = 2 (A A " I)

! Why the half?? Cauchy strain tensor

• [Look at 1D, small deformation] • Look at “small displacement” • A=1+# • Not only is the shape only slightly deformed, but it T 2 2 only slightly rotates • A A-I = A -1 = 2#+# ! 2# (e.g. if one end is fixed in place) • Therefore G ! #, which is what we expect • Then displacement x-p has gradient D=A-I • Note that for large deformation, Green strain • Then G = 1 DT D + D + DT grows quadratically 2 ( ) And for small displacement, first term - maybe not what you expect! • negligible • Whole cottage industry: defining strain T • Cauchy strain " = 1 D + D differently 2 ( ) !• Symmetric part of deformation gradient • Rotation is skew-symmetric part

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Linear Young’s modulus

• Very nice thing about Cauchy strain: it’s • Obvious first thing to do: if you pull on linear in deformation material, resists like a spring: !=E# • No quadratic dependence • E is the Young’s modulus • Easy and fast to deal with • Let’s check that in 1D (where we know what • Natural thing is to make a linear should happen with springs) relationship with Cauchy stress ! " #$ = %x˙˙ • Almost: • Then the full equation is linear! (close enough) & ( ( &X ++ * E* '1-- = %x˙˙ &x ) ) &p ,,

! Poisson Ratio What is Poisson’s ratio?

• Real materials are essentially incompressible • Has to be between -1 and 0.5 (for large deformation - neglecting foams and • 0.5 is exactly incompressible other weird composites…) • [derive] • For small deformation, materials are usually somewhat incompressible • Negative is weird, but possible [origami] ! • Imagine stretching block in one direction • Rubber: close to 0.5 • Measure the contraction in the perpendicular • Steel: more like 0.33 directions • Metals: usually 0.25-0.35 • Ratio is %, Poisson’s ratio $ • [draw ; " = # 22 ] • [cute: cork is almost 0] $11

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Putting it together Inverting…

E"11 = #11 $%# 22 $%# 33 & 1 # ) " = E( I + 1%1+, E"22 = $%#11 + # 22 $%# 33 ' 1+ # (1+ #)(1$ 2#) * E" = $%# $%# + # 33 11 22 33 • For convenience, relabel these expressions • Can invert this to get normal stress, but • $ and µ are called E# what about shear stress? ! the Lamé " = 1+ # 1$ 2# • [draw sheared block] coefficients ( )( ) ! • [incompressibility] E • When the dust settles, µ = 2(1+ #) E"ij = (1+ #)$ ij i % j % ij = "&kk'ij + 2µ&ij

! ! Rayleigh damping

• Putting it together and assuming • We’ll need to look at strain rate constant coefficients, simplifies to • How fast object is deforming

"v˙ = f body + #$%kk + 2µ$ &% • We want a damping force that resists change in deformation = fbody + #$ & $x + µ($ & $x + $$ & x) • Just the time derivative of strain = fbody + (# + µ)'x + µ$$ & x • For Rayleigh damping of linear elasticity • A PDE! " damp = #$˙ % + 2&$˙ ! • We’ll talk about solving it later ij kk ij ij

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Problems (Almost) Linear Elasticity

• Linear elasticity is very nice for small • Use the same constitutive model as before, deformation but with Green strain tensor • Linear form means lots of tricks allowed for speed- • This is the simplest general-purpose elasticity up, simpler to code, … model • But it’s useless for large deformation, or even • Note this is different from “Green elasticity” zero deformation but large rotation • Slightly better foundation: write down a potential • (without hacks) (a scalar function of strain) and take the • [draw tangent to circle] gradient to get stress • Thus we need to go back to Green strain 2D Elasticity Finite Volume Method

• Let’s simplify life before starting numerical • Simplest approach: finite volumes methods • We picked arbitrary control volumes before • The world isn’t 2D of course, but want to track • Now pick fractions of triangles from a triangle only deformation in the plane mesh • Split each triangle into 3 parts, one for each corner • Have to model why • E.g. Voronoi regions • Plane stress: very thin material, !3•=0 • Be consistent with mass! [explain, derive #3• and new law, note change in • Assume A is constant in each triangle (piecewise incompressibility singularity] linear deformation) • Plane strain: very thick material, #3•=0 • [ out] [explain, derive ! ] 3• • Note that exact choice of control volumes not critical - constant times normal integrates to zero