Large deformation plasticity and MPM (working paper)

8/14/2019 Large deformation plasticity and MPM (working paper)

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A Material Point Method Formulation for Plasticity

Biswajit Banerjee

November 14, 2006

1 Governing Equations

The equations that govern the motion of an elastic-plalstic solid include the balance laws for mass, momentum, and

energy. Kinematic equations and constitutive relations are needed to complete the system of equations. Physical

restrictions on the form of the constitutive relations are imposed by an entropy inequality that expresses the secondlaw of thermodynamics in mathematical form.

The balance laws express the idea that the rate of change of a quantity (mass, momentum, energy) in a volume must

arise from three causes:

1. the physical quantity itself flows through the surface that bounds the volume,

2. there is a source of the physical quantity on the surface of the volume, or/and,

3. there is a source of the physical quantity inside the volume.

Let be the body (an open subset of Euclidean space) and let be its surface (the boundary of).

Let the motion of material points in the body be described by the map

x = (X) = x(X) (1)

where X is the position of a point in the initial configuration and x is the location of the same point in the deformed

configuration. The deformation gradient (F) is given by

F =x

X=0 x . (2)

1.1 Balance Laws

Let f(x, t) be a physical quantity that is flowing through the body. Let g(x, t) be sources on the surface of the bodyand let h(x, t) be sources inside the body. Let n(x, t) be the outward unit normal to the surface . Let v(x, t) bethe velocity of the physical particles that carry the physical quantity that is flowing. Also, let the speed at which the

bounding surface is moving be un (in the direction n).

Then, balance laws can be expressed in the general form ([1])

d

dt

f(x, t) dV

=

f(x, t)[un(x, t) v(x, t) n(x, t)] dA +

g(x, t) dA +

h(x, t) dV . (3)

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Note that the functions f(x, t), g(x, t), and h(x, t) can be scalar valued, vector valued, or tensor valued -depending on the physical quantity that the balance equation deals with.

It can be shown that the balance laws of mass, momentum, and energy can be written as (see Appendix):

+

v = 0 Balance of Mass

v b = 0 Balance of Linear Momentum = T Balance of Angular Momentum

e : (v) + q s = 0 Balance of Energy.

(4)

In the above equations (x, t) is the mass density (current), is the material time derivative of , v(x, t) is theparticle velocity, v is the material time derivative ofv, (x, t) is the Cauchy stress tensor, b(x, t) is the body forcedensity, e(x, t) is the internal energy per unit mass, e is the material time derivative ofe, q(x, t) is the heat fluxvector, and s(x, t) is an energy source per unit mass.

With respect to the reference configuration, the balance laws can be written as

det(F) 0 = 0 Balance of Mass0 x0 PT 0 b = 0 Balance of Linear Momentum

F P = PT FT Balance of Angular Momentum0 e PT : F+0 q 0 s = 0 Balance of Energy.

(5)

In the above, P is the first Piola-Kirchhoff stress tensor, and 0 is the mass density in the reference configuration.The first Piola-Kirchhoff stress tensor is related to the Cauchy stress tensor by

P = det(F) F1 . (6)We can alternatively define the nominal stress tensor Nwhich is the transpose of the first Piola-Kirchhoff stress tensor such that

N := PT = det(F) (F1 )T = det(F) FT . (7)

Then the balance laws become

det(F) 0 = 0 Balance of Mass

0 x 0 N 0 b = 0 Balance of Linear Momentum

F NT = N FT Balance of Angular Momentum

0 e N : F+0 q 0 s = 0 Balance of Energy.

(8)

The gradient and divergence operators are defined such that

v =

3i,j=1

vixjei ej = vi,jei ej ; v =

3i=1

vixi= vi,i ; S=

3i,j=1

Sijxjei = ij,j ei . (9)

where v is a vector field, BS is a second-order tensor field, and ei are the components of an orthonormal basis inthe current configuration. Also,

0 v =3

i,j=1

viXj

EiEj = vi,jEiEj ; 0 v =3

i=1

viXi

= vi,i ; 0 S=3

i,j=1

SijXj

Ei = Sij,j Ei (10)

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where v is a vector field, BS is a second-order tensor field, and Ei are the components of an orthonormal basis inthe reference configuration.

The contraction operation is defined as

A : B =3

i,j=1 Aij Bij = Aij Bij . (11)1.2 The Clausius-Duhem Inequality

The Clausius-Duhem inequality can be used to express the second law of thermodynamics for elastic-plastic

materials. This inequality is a statement concerning the irreversibility of natural processes, especially when energy

dissipation is involved.

Just like in the balance laws in the previous section, we assume that there is a flux of a quantity, a source of the

quantity, and an internal density of the quantity per unit mass. The quantity of interest in this case is the entropy.

Thus, we assume that there is an entropy flux, an entropy source, and an internal entropy density per unit mass ()

in the region of interest.

Let be such a region and let be its boundary. Then the second law of thermodynamics states that the rate ofincrease of in this region is greater than or equal to the sum of that supplied to (as a flux or from internalsources) and the change of the internal entropy density due to material flowing in and out of the region.

Let move with a velocity un and let particles inside have velocities v. Let n be the unit outward normal tothe surface . Let be the density of matter in the region, q be the entropy flux at the surface, and r be theentropy source per unit mass. Then (see [1, 2]), the entropy inequality may be written as

d

dt

dV

(un v n) dA +

q dA +

r dV . (12)

The scalar entropy flux can be related to the vector flux at the surface by the relation q =

(x)

n. Under the

assumption of incrementally isothermal conditions (see [3] for a detailed discussion of the assumptions involved),

we have

(x) =q(x)

T; r =

s

Twhere q is the heat flux vector, s is a energy source per unit mass, and T is the absolute temperature of a materialpoint at x at time t.

We then have the Clausius-Duhem inequality in integral form:

d

dt

dV

(un v n) dA

q nT

dA +

s

TdV . (13)

We can show that (see Appendix) the entropy inequality may be written in differential form as

q

T

+

s

T. (14)

In terms of the Cauchy stress and the internal energy, the Clausius-Duhem inequality may be written as (see

Appendix)

(e T ) :v q TT

. (15)

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1.3 Constitutive Relations

A set of constitutive equations is required to close to system of balance laws. For large deformation plasticity, we

have to define appropriate kinematic quantities and stress measures so that constitutive relations between them may

have a physical meaning.

1.3.1 Thermoelastic Relations

Let the fundamental kinematic quantity be the deformation gradient (F) which is given by

F =x

X=0 x ; detF > 0 .

A thermoelastic material is one in which the internal energy (e) is a function only ofF and the specific entropy (),that is

e = e(F, ) .

For a thermoelastic material, we can show that the entropy inequality can be written as (see Appendix)

e

T

+

e

F FT

: F+

q TT

0 . (16)

At this stage, we make the following constitutive assumptions:

1. Like the internal energy, we assume that and T are also functions only ofF and , i.e.,

= (F, ) ; T = T(F, ) .

2. The heat flux q satisfies the thermal conductivity inequality and ifq is independent of and F, we have

q T 0 = ( T) T 0 = 0

i.e., the thermal conductivity is positive semidefinite.

Therefore, the entropy inequality may be written as

e

T

+

e

F FT

: F 0 .

Since and F are arbitrary, the entropy inequality will be satisfied if and only if

e

T = 0 = T =e

and e

F FT = 0 = = e

F FT .Therefore,

T =e

and =

e

F FT . (17)

Given the above relations, the energy equation may expressed in terms of the specific entropy as (see Appendix)

T = q + s . (18)4


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Effect of a rigid body rotation of the internal energy:If a thermoelastic body is subjected to a rigid body rotation Q, then its internal energy should not change. After a rotation, the new deformationgradient (F) is given by

F= Q F .

Since the internal energy does not change, we must have

e = e(F, ) = e(F, ) .

Now, from the polar decomposition theorem, F= RUwhereRis the orthogonal rotation tensor (i.e., RRT

= RT

R= 1 ) andU is thesymmetric right stretch tensor. Therefore,

e(Q RU, ) = e(F, ) .

Now, we can choose any rotation Q. In particular, if we choose Q= RT, we have

e(RT RU, ) = e(1 U, ) = e(U, ) .

Therefore,

e(U, ) = e(F, ) .

This means that the internal energy depends only on the stretch Uand not on the orientation of the body.

1.3.2 Alternative strain and stress measures

The internal energy depends on F only through the stretch U. A strain measure that reflects this fact and alsovanishes in the reference configuration is the Green strain

E=1

2(FT F 1 ) = 1

2(U2 1 ) . (19)

Recall that the Cauchy stress is given by

= e

F FT .

We can show that the Cauchy stress can be expressed in terms of the Green strain as (see Appendix)

= F e

E FT

. (20)

Recall that the nominal stress tensor is defined as

N= detF ( FT) .

From the conservation of mass, we have 0 = detF. Hence,

N=0

FT . (21)

The nominal stress is unsymmetric. We can define a symmetric stress measure with respect to the reference

configuration call the second Piola-Kirchhoff stress tensor (S):

S := F1 N= P FT = 0F1 FT . (22)

In terms of the derivatives of the internal energy, we have

S=0

F1

F e

E FT

FT = 0 e

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and

N=0

F e

E FT

FT = 0 F e

E.

That is,

S= 0e

Eand N= 0 F

e

E. (23)

1.3.3 Stress Power

The stress power per unit volume is given by :v. In terms of the stress measures in the referenceconfiguration, we have

:v =

F e

E FT

: (F F1) .

Using the identity A : (B C) = (A CT) : B, we have

:v = F eE FT FT : F = F eE : F =

0N : F .

We can alternatively express the stress power in terms ofSand E. Taking the material time derivative ofEwe

have

E=1

2(FT F+ FT F) .

Therefore,

S : E=1

2[S : (FT F) + S : (FT F)] .

Using the identities A : (B C) = (A CT) : B = (BT A) : Cand A : B = AT : BT and using the symmetryofS, we have

S : E= 12

[(S FT) : FT + (F S) : F] = 12

[(F ST) : F+ (F S) : F] = (F S) : F .

Now, S= F1 N. Therefore, S : E= N : F. Hence, the stress power can be expressed as

:v = N : F = S : E . (24)

If we split the velocity gradient into symmetric and skew parts using

v = l = d+w

where d is the rate of deformation tensor and w is the spin tensor, we have

:v = : d+ : w = tr(T d) + tr(T w) = tr( d) + tr( w) .

Since is symmetric and w is skew, we have tr( w) = 0. Therefore, :v = tr( d). Hence, we may alsoexpress the stress power as

tr( d) = tr(NT F) = tr(S E) . (25)

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1.3.4 Helmholtz and Gibbs free energy

Recall that

S= 0e

E.

Therefore,

e

E=

1

0S .

Also recall thate

= T .

Now, the internal energy e = e(E, ) is a function only of the Green strain and the specific entropy. Let us assume,that the above relations can be uniquely inverted locally at a material point so that we have

E= E(S, T) and = (S, T) .

Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions ofSand T,

or Eand T, i.e.,

e = e(E, ) = e(S, T) = e(E, T) ; = (S, T) = (E, T) ; and S= S(E, T)

We can show that (see Appendix)

d

dt(e T ) = T + 1

0S : E or

d

dt= T + 1

0S : E . (26)

andd

dt(e T 1

0S : E) = T 1

0S : E or

dg

dt= T +

1

0S : E . (27)

We define the Helmholtz free energy as

= (E, T) := e T . (28)

We define the Gibbs free energy as

g = g(S, T) := e + T + 10S : E . (29)

The functions (E, T) and g(S, T) are unique. Using these definitions it can be showed that (see Appendix)

E

= 10S(E, T) ;

T= (E, T) ; g

S= 1

0E(S, T) ; g

T= (S, T) (30)

andS

T= 0

Eand

E

T= 0

S. (31)

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1.3.5 Specific Heats

The specific heat at constant strain (or constant volume) is defined as

Cv :=e(E, T)

T. (32)

The specific heat at constant stress (or constant pressure) is defined as

Cp :=e(S, T)

T. (33)

We can show that (see Appendix)

Cv = T

T= T

2

T2(34)

and

Cp = T

T+

1

0S :

E

T= T

2g

T2+ S :

2g

ST. (35)

Also the equation for the balance of energy can be expressed in terms of the specific heats as (see Appendix)

Cv T = ( T) + s +

0TS : E

Cp

1

0S : E

T = ( T) + s

0TE : S

(36)

where

S :=S

Tand E :=

E

T. (37)

The quantity S is called the coefficient of thermal stress and the quantity E is called the coefficient of thermal

expansion.

The difference between Cp and Cv can be expressed as

Cp Cv =1

0

S T S

T

:

E

T. (38)

However, it is more common to express the above relation in terms of the elastic modulus tensor as (see Appendix

for proof)

Cp Cv =1

0S : E +

T

0E : C : E (39)

where the fourth-order tensor of elastic moduli is defined as

C :=S

E= 0

2

EE. (40)

For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff

material model, we can show that (see Appendix)

Cp Cv =1

0

tr(S) + 9 2 K T

.

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1.3.6 Kinematic Relations

Additive split of the rate of deformation tensor: The rate of deformation tensor (d) is given by

d =1

2 v + (v)T

(41)where v is the velocity. We assume that the rate of deformation can be additively decomposed into an elastic part, a

plastic part, and a thermal part:

d = de + dp + dth . (42)

The thermal part of the rate of deformation is computed separately and subtracted from d to give

d := d dth = ee + dp .For simplicity, we use the symbol d instead of d in the following development.We split the rate of deformation tensors into volumetric and deviatoric parts:

d = 13

1 + ; de = 13

tr(de) 1 + e ; dp = 13

tr(dp) 1 + p (43)

where = tr(d), the deviatoric part of the rate of deformation tensor is = d 13 tr(d) 1 , and 1 is thesecond-order identity tensor.

1.3.7 Stress

To deal with nearly incompressible behavior, we introduce an isomorphic split of the Cauchy stress tensor into a

volumetric and a deviatoric part of the form ([4]):

= m1 + ss = p 1 + s (44)where

m =1

3tr() =

3p ; s = s ; p = 1

3tr() ; s = 1

3tr() 1 ; 1 = 11 ; s = ss ; A = A : A .

(45)

Note that the following identities hold:

m = : 1 ; s = : s ; 1 : 1 = 1 ; s : s = 1 ; 1 : s = 0 ; 1 : s = 0 ; s : s = 0 . (46)The rate form of equation (44) is

= m 1 + s s+ s s = p 1 + s . (47)In the following development, we enforce frame indifference while evaluating the constitutive relation by assuming

that the stresses and rate of deformation have been rotated to the reference configuration using

s = RT s R; = RT R

where the rotation tensor Ris obtained from a polar decomposition of the deformation gradient.

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1.3.8 Elastic Relations:

The constitutive model is assumed to consist of two parts. The first is a hypoelastic model for the deviatoric part of

the stress and the second is a equation of state for the mean stress.

The deviatoric stress is given by the rate equation

s = C(m, T) : e s = C(p,T) : e (48)

where s is the deviatoric stress rate, C is a pressure and temperature dependent fourth order elastic modulus tensor,

and e is the deviatoric part ofde (the elastic part of the rate of deformation tensor).

Let us assume that the deviatoric elastic response of the material is isotropic, i.e.,

s = 2 (m, T) e s = 2 (p,T) e . (49)

Remark: Note that since the shear modulus depends on m, the rate ofs should have a contribution that containsm. We ignore this contribution in the subsequent development.

The mean stress (m) is calculated using a Mie-Gruneisen equation of state of the form ([5, 6])

m =

3 p =

3 0 C20 (1 Je)[1 0(1 Je)][1 S(1 Je)]2 2

3 0 e ; J

e := detFe (50)

where C0 is the bulk speed of sound, 0 is the initial mass density, 2 0 is the Gruneisens gamma at the referencestate, S = dUs/dUp is a linear Hugoniot slope coefficient, Us is the shock wave velocity, Up is the particlevelocity, and e is the internal energy density (per unit reference volume), Fe is the elastic part of the deformationgradient. For isochoric plasticity,

Je = J = det(F) =0

.

The change in internal energy is computed using

e = 0

CvdT 0 [Cv(T) T Cv(T0) T0] (51)

where T0 is the reference temperature and Cv is the specific heat at constant volume. We assume that Cp and Cv areequal.

The rate equation (47) has three rate terms. We need constitutive relations for each of these three terms.

The rate form of the pressure equation is obtained by taking the material time derivative of (50) to get

m = 3 p =

3 0 C20 J

e [1 + (S

2 0)(1

Je)]

[1 S(1 Je)]3 Je

Je 23 0 e (52)Now,

Je

Je= tr(de) = e = de : 1 =

3 de : 1 ; e = 0 Cv(T) T .

For isochoric plasticity, the above relations become

J

J= tr(de) = tr(d) = = d : 1 =

3 d : 1 .

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Hence,

m =

3 p = 3 K(Je) de : 1 23 0 Cv 0 T (53)where

K(Je) =0 C

20 J

e [1 + (S 2 0)(1 Je)][1

S(1

Je)]3

. (54)

The stress rate s is given by

s =

t(s : s) = s : s = 2 (m, T) e : s .

Therefore,

s = 2 (m, T) e : s . (55)

The rate ofs is obtained by noting that

s =

s

s s : ss

s .

Therefore,s = 2 (m, T)

s[e (e : s)s] . (56)

A slightly different, but equally valid, decomposition of the stress rate can be obtained by setting

s = 2 (m, T) de : s . (57)

and

s = 2 (m, T)s

de (de : 1 )1 (de : s)s . (58)

We use the above decomposition in the following development.

1.3.9 Plastic Relations:

We consider yield functions of the form

f(m, s, , p, T) 0 (59)where is the total strain rate, p is the plastic strain, and T is the temperature. Here the strain rates and the plasticstrain are defined as

:=

2

3d : d ; p :=

2

3dp : dp ; p :=

t0

p dt

. (60)

As a particular case, we consider a class of isotropic yield functions of the von Mises form

f(m, s, , p, T) =3

22s 2y (, p, T)

2(m, T)

20(61)

where y is the yield stress, is the shear modulus, and 0 is a reference shear modulus.

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Remark: This form of the yield function can be used for deformation induced anisotropic plasticity by converting

it to the form used by Maudlin and Schiferl ([7]). In addition, the effect of porosity can be incorporated using

Brannons approach ([4]).

At yield, we have

3

2 2s

2y (, p, T)

2(m, T)

20 = 0 .

Taking a time derivative of the above equation gives us

3 s s 2 y2

20

yt

2 2y

20

mm = 0 . (62)

We assume that the plastic part of the rate of deformation can be obtained from a plastic potential in the form

(associated flow rule)

dp = f

= f (63)

The unit outward normal to the yield surface is given by

n = 1

f

=

1

f ; = f

= f ; n : n = 1 = dp = n . (64)

Therefore,

p :=

2

3 ; p :=

2

3

t0

dt

.

Note thatm

= 1 ; s

= s . (65)Assuming that , p, and T remain unchanged during the elastic part of the deformation, we have

f =f

=

f

m1 + f

ss = fm 1 + fs s ; = (fm)2 + (fs)2 . (66)

For the yield function (61), we have

fm = 2 2y (, p, T)(m, T)

20

m; fs = 3 s . (67)

Therefore, equation (62) can be written as

fs s 2 y2

2

0

yt

+ fm m = 0 . (68)

Also,

dp =

2 2y (, p, T)

(m, T)

20

m

1 + 3 s s . (69)This implies that

dp = (dp : 1 )1 + (dp : s)s ; dp : 1 = 2 2y (, p, T) (m, T)20 m ; dp : s = 3 s ;12


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Recall,

= m 1 + s s+ s sm = 3 K(J

e) de :

1 2

3 0 Cv 0 T

s = 2 (m, T) de : ss = 2 (m, T)

s

de (de : 1 )1 (de : s)s

Using the decomposition de = d dp, we get

m = 3 K(Je) (d dp) : 1 23 0 Cv 0 T

s = 2 (m, T) (d dp) : ss = 2 (m, T)

s

d dp (d : 1 )1 + (dp : 1 )1 (d : s)s+ (dp : s)s

or,

= m 1 + s s+ s sm = 3 K(J

e)

d : 1 + 2 2y (, p, T) (m, T)20 m

2

3 0 Cv 0 T

s = 2 (m, T) (d : s 3 s) = 2 (m, T) ( : s 3 s)s = 2 (m, T)

s[ (d : s)s] = 2 (m, T)

s[ ( : s)s]

(70)

Now, the consistency condition requires that

t[f(m, s, , p, T)] = 0

or,f

mm +

f

ss +

f

t+

f

p +

f

TT = 0

or,

fm m + fs s + fd + fp + ft T = 0 (71)

where

fd :=f

= 2 y

2

20

y

; =

2

3

d : d

d ; d = RT 1

2[a + (a)T] R

fp :=

f

p = 2 y

2

20

y

p ; ft :=

f

T = 2 y

2

20

y

T .

Note that the above equation is the same as equation (68). Also note that,

m = 3 K [d : 1 fm] G(T) T ; s = 2 [ : s fs]where

G(T) := 2

3 0 Cv(T) 0 .

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Plugging into the consistency condition, we get

3 K fm [d : 1 fm] fm G(T) T + 2 fs [ : s fs] + fd + fp + ft T = 0or,

(3 K f 2m + 2 f2

s ) = 3 K fm d : 1 + 2 fs : s+ fd + fp + [ft fm G(T)] T = 0or, =

3 K fm d : 1 + 2 fs : s+ fd + fp + [ft fm G(T)] T3 K f 2m + 2 f

2s

. (72)

1.4 Mass Balance:

The mass balance equation is

+ v = 0 (73)where is the current mass density, and is the material time derivative of .

1.5 Momentum Balance:

The balance of linear momentum can be written as

+ b = v (74)where is the Cauchy stress, b is the body force per unit volume, and v is the acceleration.

The momentum equation can be written as

(p 1 + s) + b = v

or, (p 1 ) + s+ b = v . (75)

1.6 Energy Balance:

The balance of energy can be written as

: d q + h = e (76)where q is the heat flux per unit area, h is the heat source per unit mass, and e is the specific internal energy. A dotover a quantity represents the material time derivative of that quantity.

We convert the energy equation into an approximate equation for heat conduction that includes heat source termdue to plastic dissipation ([8]):

Cv T ( T) : dp = h (77)where Cv is the specific heat at constant volume, is a second order tensor of thermal conductivity, and is theTaylor-Quinney coefficient. Expressing the stress and the plastic part of rate of deformation into volumetric and

deviatoric parts, we get

Cv T ( T) (p 1 + s) :

1

3tr(dp) 1 + p

= h

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or,

Cv T ( T)

1

3p tr(dp)1 : 1 +

1

3tr(dp) s : 1 +p 1 : p + s : p

= h

or,

Cv T ( T) (p tr(dp) + s : p) = h (78)

2 Weak Forms

2.1 Rate of deformation:

The weak form of the kinematic relation is given byW :

d 1

2

v + (v)T

d = 0 (79)

where W is an arbitrary second-order tensor valued weighting function. We write the rate of deformation and the

weighting function in terms of their volumetric and deviatoric components to get

dev(W) +

1

3tr(W) 1

:

+

1

3 1 1

2

v + (v)T

d = 0

or,

dev(W) : +

1

3 dev(W) : 1 1

2dev(W) :

v + (v)T

+

1

3tr(W) 1 : +

1

9 tr(W)1 : 1 1

6tr(W)1 :

v + (v)T

d = 0

or,

dev(W) : 1

2dev(W) :

v + (v)T

+

1

3 tr(W) 1

6tr(W)

tr(v) + tr((v)T)

d = 0

or,

dev(W) : ( d) + 1

3 tr(W) 1

3tr(W) v

d = 0

or,

1

3 dev(W) : 1 +

1

3 tr(W) 1

3tr(W) v

d = 0

or,

13 tr(W) 13 tr(W) v d = 0Define w := 1/3 tr(W). Then the weak form becomes

w

v

d = 0 (80)

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2.2 Constitutive Relations:

The weak forms of the constitutive relations are given by

W :

s C(p,T) : e d = 0 (81)

and

wp 0C

20 (1 J)[1 0(1 J)]

[1 S(1 J)]2 2 0e

d = 0 (82)

where W is a tensor valued weighting function and w is a scalar weighting function.


To derive the weak form of the momentum equation, we multiply the momentum equation with a vector-valued

weighting function (w) and integrate over the domain (). The weighting function (w) satisfies velocity boundaryconditions on the parts of the boundary where velocities are prescribed. Then we get,

w [ (p 1 ) + s+ b] d =

w v d . (83)

Using the identity v ( S) = (ST v) S :v, where S is a second-order tensor valued field and v is avector valued field, we get (using the symmetry of the stress tensor)

{ [(p 1 ) w] p 1 :w + (s w) s :w + w b} d =

w v dor,

{ (p w) p w + (s w) s :w + w b} d =

w v d . (84)Applying the divergence theorem, we have

{p w s :w + w b} d +

n (p w) d +

n (s w) d =

w v d

or,

{ w v +p w + s :w} d =

w b d +

p n w d +

n (s w) d

or,

{ w v +p w + s :w} d =

w b d +

p n w d +

(sT n) w d

or,

{ w v +p w + s :w} d =

w b d +

p n w d +

(T np n) w d (85)

or,

{ w v +p w + s :w} d =

w b d +

(T n) w d . (86)In the above, n is the outward normal to the surface .

If the applied surface traction is t = T n, we get (since w is zero on the part of the boundary where velocitiesare specified)

{ w v +p w + s :w} d =

w b d +

t

t w d . (87)

16


17/42

2.4 Energy Balance:

The weak form of the heat conduction equation is given by

w Cv T ( T) (p tr(dp) + s : p) d = w h d .

Using the identity ( v) = ( v) () v we getw Cv T d

{ [w( T)] (w) T} d

w [p tr(dp) + s : p] d =

w h d .

Using the divergence theorem, we have

w Cv T + (w) T d = w h + [p tr(dp) + s : p] d +

n [w( T)] d .If a heat flux (q = n ( T)) is specified on part of the boundary (q), the weak form becomes

w Cv T + (w) T d = w h + [p tr(dp) + s : p] d +

q

w q d . (88)3 MPM Discretization

3.1 Rate of Deformation:

The weak form of the rate of deformation equation is

w v d = 0 . (89)We assume a discontinuous approximation for over each cell and a weighting function w of the form

(x) ncp

q=1

qq(x) ; w(x) =

ncpp=1

wpp(x) (90)

where ncp is the number of particles in a grid cell, w

p are the weighting functions at the particle locations, and

p(x) are the basis functions. The velocities are approximated in the standard MPM way using

v(x)

ng

g=1 vgNg(x) (91)where ng is the number of grid points, vg are the velocities at the grid vertices, and Ng(x) are the grid basisfunctions.

17


18/42

Remark: The approximations forw and need not be a sum over the particles. We could alternatively assume aconstant value over each cell or a sum over the grid nodes with the same basis functions as the grid basis functions.

We have chosen a sum over particles in a cell to keep our formulation consistent with the standard MPM procedure.

Plugging the approximations into the weak form, we get the following equations that are valid over each cell

c

ncpp=1

wpp(x)ncp

q=1

qq(x) ncgg=1

vgNg(x) d = 0or,

ncpp=1

wp

c

p(x)

ncp

q=1

qq(x) ncg

g=1

vg Ng

d = 0 .

The abitrariness ofwp gives us a system ofnc

p equations

cp(x)

ncp

q=1qq(x)

ncg

g=1vg Ng

d = 0 ; p = 1 . . . ncp

or,ncp

q=1

c

p(x)q(x) d

q

ncgg=1

c

p(x)Ng d

vg = 0 ; p = 1 . . . ncp

or,ncp

q=1

c

p(x)q(x) d

q =

ncgg=1

c

p(x)Ng d

vg ; p = 1 . . . ncp .

If we identify the basis functions p with the particle characteristic functions and use the property ([9], p. 485)

p(x) = 1 ifx p0 otherwise

(92)

we get pc

p(x)p(x) d

p =

ncgg=1

pc

p(x)Ng d

vg ; p = 1 . . . ncp .

Define the volume of the particle inside the cell as

Vpc :=

pc

p(x)p(x) d . (93)

We then have

Vpc p =

ncgg=1

pc

p(x)Ng d

vg ; p = 1 . . . ncp .

Define the gradient weighting function as

Npg :=

pc

p(x)Ng d . (94)

18


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Then,

Vpc p =

ncgg=1

Npg vg ; p = 1 . . . ncp .

or,

p =1

Vpc

ncgg=1

Npg vg ; p = 1 . . . ncp (95)

3.2 Constitutive Relations:

The weak form of the constitutive relation for pressure is

w(x)

p(x) 0C

20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x)

d = 0 . (96)

We use a discontinuous approximation for the pressure field (p) and a weighting function w of the form

p(x) ncp

q=1

pqq(x) ; w(x) =

ncpp=1

wp p(x) . (97)

After plugging these approximations into the weak form, we get the following relations over each cell

c

ncpp=1

wp p(x)

ncp

q=1

pqq(x) 0C20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x) d = 0

or,

ncpp=1

wp

c

p(x)

ncpq=1

pqq(x) 0C20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x) d = 0 .

Invoking the arbitrariness ofwp we get

c

p(x)

ncp

q=1

pqq(x) 0C20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x) d = 0 ; p = 1 . . . ncp

or,

ncpq=1

c

p(x) q(x) d pq = c

p(x)0C20 (1 J(x))[1 0(1 J(x))][1 S(1 J(x))]2 2 0e(x)

d ; p = 1 . . . ncpOnce again, if we identify the basis functions p with the particle characteristic functions and use the property

p(x) =

1 ifx p0 otherwise

(98)

19


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and we getpc

p(x) p(x) d

pp =

pc

p(x)

0C

20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x)

d ; p = 1 . . . ncp

or,

Vpc pp =

pc

p(x)0C20 (1 J(x))[1 0(1 J(x))]

[1 S(1 J(x))]2 2 0e(x)

d ; p = 1 . . . ncp .

Identifying the volume averaged J(x) and e(x) as p and ep, respectively, we get

Vpc pp = Vpc0C

20 (1 p)[1 0(1 p)]

[1 S(1 p)]2 2 0ep ; p = 1 . . . nc

p

or,

pp =0C

20 (1 p)[1 0(1 p)]

[1 S(1 p)]2 2 0ep ; p = 1 . . . nc

p . (99)

The weak form of the constitutive relation for the deviatoric stress rate is given byW(x) :

s(x) C(p,T,x) : e(x)

d = 0 . (100)

We assume that the weighting function is

W(x) =

npq=1

Wq q(x) (101)

and plug these into the weak form to get the following relations

npq=1

Wq q(x) : s(x) C(p,T,x) : e(x) d = 0or,

npq=1

Wq :

q(x)


d

= 0 .

The arbitrariness ofWq givesq

q(x)


d = 0 ; q = 1 . . . np

Using arguments similar to those used for the pressure constitutive relation, we get

sq = Cq(p,T) :

eq ; q = 1 . . . np (102)

20


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The weak form of the momentum balance relation is

[(x) w(x) v(x) +p(x) w + s(x) :w] d =

(x) w(x) b(x) d +

tt(x) w(x) d . (103)

The first step in the MPM discretization is to convert the integral over into a sum of integrals over particles. Toachieve this we assume that the particle characteristic functions have the form

p(x) =

1 ifx p0 otherwise

(104)

Then the weak form of the momentum balance equation can be written as

npp=1

p

p(x) [(x) w(x) v(x) +p(x) w + s(x) :w] d =

npp=1

p

p(x) (x) w(x) b(x) d + t

t(x) w(x) d .

The weighting function is

w(x) =

ngg=1

wgNg(x) . (105)

The velocity is approximated as

v(x) ng

h=1

vhNh(x) . (106)

The material time derivative ofv is also approximated in a similar manner (see [10]). Thus,

v(x) ng

h=1

vhNh(x) . (107)

Plugging these into the left hand side of the momentum equation we get

LHS =

npp=1

p

p(x)

(x)

ngg=1

wgNg(x)

ng

h=1

vhNh(x)

+p(x)

ng

g=1

wgNg(x)

+s(x) :

ng

g=1wgNg(x)

d

or,

LHS =

ngg=1

wg np

p=1

ngh=1

p

p(x) (x) Ng(x) Nh(x) d

vh

+ng

g=1

wg np

p=1

p

p(x) p(x)Ng d

+ ngg=1

wg np

p=1

p

p(x) s(x) Ng d

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or,

LHS =

ngg=1

wg np

p=1

ngh=1

p


vh

+

p

p(x) p(x)Ng d + p

p(x) s(x) Ng dSimilarly, the right hand side of the weak form of momentum balance can be written as

RHS =

npp=1

p

p(x) (x)

ngg=1

wgNg(x)

b(x) d + t

t(x) ng

g=1

wgNg(x)

dor,

RHS =

ng

g=1wg

np

p=1

pp(x) (x) Ng(x) b(x) d

+

ng

g=1wg

t

t(x) Ng(x) d

or,

RHS =

ngg=1

wg

npp=1

p

p(x) (x) Ng(x) b(x) d

+

t

t(x) Ng(x) d

Combining the left and right hand sides and invoking the arbitrariness ofwg, we get

npp=1

ngh=1

p


vh

+

pp(x) p(x)Ng d + p

p(x) s(x) Ng d=

np

p=1

p

p(x) (x) Ng(x) b(x) d

+

t

t(x) Ng(x) d ; g = 1 . . . ng

If(x), p(x), s(x), and b(x) are assumed to be constant over each particle, we can replace these quantities withthe values at the particles to get

npp=1

ngh=1

p

p

p(x) Ng(x) Nh(x) d

vh

+

pp p p(x)Ng d + sp p p(x)Ng d =

npp=1

p bp

p

p(x) Ng(x) d

+

t

t(x) Ng(x) d ; g = 1 . . . ng .

Then, using the definition of the gradient weighting function (94), and defining

Npg :=

p

p(x) Ng(x) d (108)

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we get

ngh=1

npp=1

p

p

p(x) Ng(x) Nh(x) d

vh +

npp=1

ppNpg

+

npp=1

sp Npg

=

np

p=1

p bp Npg + t t(x) Ng(x) d ; g = 1 . . . ngDefine the mass matrix as

mgh :=

npp=1

p

p

p(x) Ng(x) Nh(x) d . (109)

Define the internal force vector as

fintg :=

npp=1

[ppNpg + sp Npg] . (110)

Define the body force vector as

fbodg :=

np

p=1

p bp Npg . (111)

Also define the external force vector as

fextg :=

t

t(x) Ng(x) d (112)

Then we get the semidiscrete system of equations

ngh=1

mghvh + fintg = f

bodg + f

extg ; g = 1 . . . ng (113)

3.4 Energy Balance:

The weak form of the energy balance equation is

w(x) (x) Cv(x) T(x) +w (x) T d = (114)w(x) (x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)] d +

q

w(x) q(x) d .As before, we express this equation as a sum over particles to get

npp=1

p

p(x)w(x) (x) Cv(x) T(x) +w (x) T d =np

p=1

p

p(x) w(x) (x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)] d + q

w(x) q(x) dWe approximate T using

T(x) ng

g=1

Tg Ng(x) . (115)

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The material time derivative ofT is then given by

T(x) ng

g=1

Tg Ng(x) . (116)

The weighting functions are also chosen to be of the same form:

w(x) = ngg=1

wg Ng(x) . (117)Plugging these into the weak form, we get

npp=1

p

p(x)

ngg=1

wg Ng(x) (x) Cv(x)

ngh=1

Th Nh(x)

d

+

np

p=1

pp(x)

ng

g=1wg Ng(x)

(x) ngh=1

Th Nh(x)

d

=

npp=1

p

p(x)

ngg=1

wg Ng(x) (x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)] d+

q

ngg=1

wg Ng(x) q(x) d

For simplicity, let us assume that (x) is isotropic, i.e., (x) = (x). Then,

ng

g=1wg

np

p=1 pp(x) Ng(x) (x) Cv(x)

ngh=1

Th Nh(x)

d

+

ngg=1

wg

npp=1

p

p(x) (x)Ng ng

h=1

Th Nh

d

=

ngg=1

wg np

p=1

p

p(x) Ng(x)

(x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)]

d

+

ngg=1

wg q

Ng(x) q(x) d

Invoking the arbitrariness of wg, we getnpp=1

p

p(x) Ng(x) (x) Cv(x)

ngh=1

Th Nh(x)

d +

npp=1

p

p(x) (x)Ng ng

h=1

Th Nh

d

=

npp=1

p

p(x) Ng(x)

(x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)]

d

+

q

Ng(x) q(x) d ; g = 1 . . . ng

24


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or,

ngh=1

npp=1

p

p(x) Ng(x) Nh(x) (x) Cv(x) d

Th + ngh=1

npp=1

p

p(x) (x)Ng Nh dTh

=

npp=1

p

p(x) Ng(x) (x) h(x) + [p(x) tr(dp(x)) + s(x) : p(x)] d+

q

Ng(x) q(x) d ; g = 1 . . . ng

If we assume that (x), Cv(x), (x), h(x), p(x), s(x), and dp(x) are constant over each particle, we get

ngh=1

npp=1

p Cvp

p

p(x) Ng(x) Nh(x) d

Th + ngh=1

npp=1

p

p

p(x)Ng Nh dTh

=

np

p=1

p hp p p(x) Ng(x) d +np

p=1

pp tr(dp

p)p p(x) Ng(x) d+

npp=1

sp : pp

p

p(x) Ng(x) d +

q

Ng(x) q(x) d ; g = 1 . . . ng

Using the definition of the particle weighting functions (108), we have

ngh=1

npp=1

p Cvp

p

p(x) Ng(x) Nh(x) d

Th + ngh=1

npp=1

p

p

p(x)Ng Nh dTh

=

np

p=1 p hp Npg +np

p=1 pp tr(dpp) Npg +np

p=1 sp : pp Npg + q Ng(x) q(x) d ; g = 1 . . . ngor,

ngh=1

npp=1

p Cvp

p

p(x) Ng(x) Nh(x) d

Th + ngh=1

npp=1

p

p

p(x)Ng Nh dTh

=

npp=1

p hp + pp tr(d

pp) + sp :

pp

Npg +

q

Ng(x) q(x) d ; g = 1 . . . ng .

Define the thermal inertia matrix as

Mgh :=

npp=1

p Cvp

p

p(x) Ng(x) Nh(x) d , (118)

the thermal stiffness matrix as

Kgh :=

npp=1

p

p

p(x)Ng Nh d , (119)

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26/42

the thermal source vector as

Sg :=

npp=1

p hp + pp tr(d

pp) + sp :

pp

Npg , (120)

and the external heat flux vector as

qextg := q

Ng(x) q(x) d . (121)

Then we get the following semidiscrete system of equations:

ngh=1

Mgh Th +

ngh=1

Kgh Th = Sg + qextg ; g = 1 . . . ng . (122)

4 Appendix

1. The integral

F(t) =Zb(t)a(t) f(x, t) dx

is a function of the parameter t. Show that the derivative ofF is given by

dF

dt=

d

dt

Zb(t)a(t)

f(x, t) dx

!=

Zb(t)a(t)

f(x, t)

tdx + f[b(t), t]

b(t)

t f[a(t), t]

a(t)

t.

This relation is also known as the Leibnitz rule.

The following proof is taken from [11].

We have,

dF

dt= lim

t0

F(t + t) F(t)

t.

Now,

F(t + t) F(t)t

= 1t

"Zb(t+t)

a(t+t)f(x, t + t) dx

Zb(t)

a(t)f(x, t) dx

#

1

t

Zb+ba+a

f(x, t + t) dx

Zba

f(x, t) dx

=1

t

Za+aa

f(x, t + t) dx +

Zb+ba

f(x, t + t) dx

Zba

f(x, t) dx

=1

t

Za+aa

f(x, t + t) dx +

Zba

f(x, t + t) dx +

Zb+bb

f(x, t + t) dx

Zba

f(x, t) dx

=

Zba

f(x, t + t) f(x, t)

tdx +

1

t

Zb+bb

f(x, t + t) dx 1

t

Za+aa

f(x, t + t) dx .

Since f(x, t) is essentially constant over the infinitesimal intervals a < x < a + a and b < x < b + b, we may write

F(t + t) F(t)t

Zba

f(x, t + t) f(x, t)t

dx + f(b, t + t) bt

f(a, t + t) at

.

Taking the limit as t 0, we get

limt0

"F(t + t) F(t)

t

#= lim

t0

"Zba

f(x, t + t) f(x, t)

tdx

#+ limt0

"f(b, t + t)

b

t

# limt0

"f(a, t + t)

a

t

#

or,

dF(t)

dt=

Zb(t)a(t)

f(x, t)

tdx + f[b(t), t]

b(t)

t f[a(t), t]

a(t)

t.

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2. Let (t) be a region in Euclidean space with boundary (t). Let x(t) be the positions of points in the region and let v(x, t) be the velocity field inthe region. Let n(x, t) be the outward unit normal to the boundary. Let f(x, t) be a vector field in the region (it may also be a scalar field). Show that

d

dt

Z(t)

f dV

!=

Z(t)

f

tdV +

Z(t)

(v n)f dA .

This relation is also known as the Reynolds Transport Theorem and is a generalization of the Leibnitz rule.

This proof is taken from [12] (also see [13]).

Let 0 be reference configuration of the region (t). Let the motion and the deformation gradient be given by

x = (X, t) ; F(X, t) =0 .

Let J(X, t) = det[F(x, t)]. Then, integrals in the current and the reference configurations are related by

Z(t)

f(x, t) dV =

Z0

f[(X, t), t] J(X, t) dV =

Z0

f(X, t) J(X, t) dV .

The time derivative of an integral over a volume is defined as

d

dt

Z(t) f(x, t) dV

!= lim

t0

1

t

Z(t+t) f(x, t + t) dV Z(t) f(x, t) dV

!.

Converting into integrals over the reference configuration, we get

d

dt

Z(t)

f(x, t) dV

!= lim

t0

1

t

Z0

f(X, t + t) J(X, t + t) dV

Z0

f(X, t) J(X, t) dV

.

Since 0 is independent of time, we have

d

dt

Z(t)

f(x, t) dV

!=

Z0

"limt0

f(X, t + t) J(X, t + t) f(X, t) J(X, t)

t

#dV

=

Z0

t[f(X, t) J(X, t)] dV

= Z0

t

[f(X, t)] J(X, t) + f(X, t)

t

[J(X, t)] dVNow, the time derivative ofdetF is given by (see [13], p. 77)

J(X, t)

t=

t(detF) = (detF)( v) = J(X, t) v((X, t), t) = J(X, t) v(x, t) .

Therefore,

d

dt

Z(t)

f(x, t) dV

!=

Z0

t[f(X, t)] J(X, t) + f(X, t) J(X, t) v(x, t)

dV

=

Z0

t[f(X, t)] + f(X, t) v(x, t)

J(X, t) dV

=

Z(t)

f(x, t) + f(x, t) v(x, t)

dV

where f is the material time derivative off. Now, the material derivative is given by

f(x, t) =f(x, t)

t+ [f(x, t)] v(x, t) .

Therefore,

d

dt

Z(t)

f(x, t) dV

!=

Z(t)

f(x, t)

t+ [f(x, t)] v(x, t) + f(x, t) v(x, t)

dV

or,

d

dt

Z(t)

f dV

!=

Z(t)

f

t+f v + f v

dV .

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28/42

Using the identity

(vw) = v( w) +v w

we then have

d

dt

Z(t)

f dV

!=

Z(t)

f

t+ (fv)

dV .

Using the divergence theorem and the identity (ab) n = (b n)a we have

d

dt

Z(t)

f dV

!=

Z(t)

f

tdV +

Z(t)

(fv) n dV =

Z(t)

f

tdV +

Z(t)

(v n)f dV .

3. Show that the balance of mass can be expressed as:

+ v = 0

where (x, t) is the current mass density, is the material time derivative of, and v(x, t) is the velocity of physical particles in the body bounded by the surface .

Recall that the general equation for the balance of a physical quantity f(x, t) is given by

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density (x, t). Since mass is neithercreated or destroyed, the surface and interior sources are zero, i.e., g(x, t) = h(x, t) = 0 . Therefore, we have

d

dt

Z

(x, t) dV

=

Z

(x, t)[un(x, t) v(x, t) n(x, t)] dA .

Let us assume that the volume is a control volume (i.e., it does not change with time). Then the surface has a zero velocity (un = 0) and we get

Z

tdV =

Z

(v n) dA .

Using the divergence theorem Z v dV =

Z

v n dA

we getZ

tdV =

Z ( v) dV.

or, Z

t+ ( v)

dV = 0 .

Since is arbitrary, we must have

t+ ( v) = 0 .

Using the identity

( v) = v + v

we have

t+ v + v = 0 .

Now, the material time derivative of is defined as =

t+ v .

Therefore,

+ v = 0 .

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4. Show that the balance of linear momentum can be expressed as:

v b = 0

where (x, t) is the mass density, v(x, t) is the velocity, (x, t) is the Cauchy stress, and b is the body force density.

Recall the general equation for the balance of a physical quantity

d

dt

Z

f(x, t) dV

=Z

f(x, t)[un(x, t) v(x, t) n(x, t)] dA +Z

g(x, t) dA +Z

h(x, t) dV .

In this case the physical quantity of interest is the momentum density, i.e., f(x, t) = (x, t) v(x, t). The source of momentum flux at the surface isthe surface traction, i.e., g(x, t) = t. The source of momentum inside the body is the body force, i.e., h(x, t) = (x, t) b(x, t). Therefore, we have

d

dt

Z

v dV

=

Z

v[un v n] dA +

Z

t dA +

Z

b dV .

The surface tractions are related to the Cauchy stress by

t = n .

Therefore,d

dt

Z

v dV

=

Z

v[un v n] dA +

Z n dA +

Z

b dV .

Let us assume that is an arbitrary fixed control volume. Then,Z

t( v) dV =

Z

v (v n) dA +

Z n dA +

Z

b dV .

Now, from the definition of the tensor product we have (for all vectors a)

(uv) a = (a v) u .

Therefore, Z

t( v) dV =

Z

(vv) n dA +

Z n dA +

Z

b dV .

Using the divergence theorem Z v dV =

Z

v n dA

we have Z

t( v) dV =

Z [ (vv)] dV +

Z dV +

Z

b dV

or, Z

t( v) + [( v)v] b

dV = 0 .

Since is arbitrary, we have

t( v) + [( v)v] b = 0 .

Using the identity

(uv) = ( v)u + (u) v

we get

tv +

v

t+ ( v)(v) +( v) v b = 0

or,

t+ v

v +

v

t+( v) v b = 0

Using the identity

( v) = v + v ()

we get

t+ v

v +

v

t+ [v + v ()] v b = 0

From the definition

(uv) a = (a v) u

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we have

[v ()] v = [v ()] v .

Hence,

t+ v

v +

v

t+ v v + [v ()] v b = 0

or,

t + v + v

v +

v

t + v v b = 0 .

The material time derivative of is defined as

=

t+ v .

Therefore,

[ + v]v + v

t+ v v b = 0 .

From the balance of mass, we have

+ v = 0 .

Therefore,

v

t+ v v b = 0 .

The material time derivative ofv is defined as

v =

v

t +v v .

Hence,

v b = 0 .

5. Show that the balance of angular momentum can be expressed as:

= T

We assume that there are no surface couples on or body couples in . Recall the general balance equation

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

In this case, the physical quantity to be conserved the angular momentum density, i.e., f = x ( v). The angular momentum source at the surfaceis then g = x t and the angular momentum source inside the body is h = x ( b). The angular momentum and moments are calculated with

respect to a fixed origin. Hence we have

d

dt

Zx ( v) dV

=

Z

[x ( v)][un v n] dA +

Z

x t dA +

Zx ( b) dV .

Assuming that is a control volume, we have

Zx

t( v)

dV =

Z

[x ( v)][v n] dA +

Z

x t dA +

Zx ( b) dV .

Using the definition of a tensor product we can write

[x ( v)][v n] = [[x ( v)]v] n .

Also, t = n. Therefore we have

Zx

t( v)

dV =

Z

[[x ( v)]v] n dA +Z

x ( n) dA +Zx ( b) dV .

Using the divergence theorem, we get

Zx

t( v)

dV =

Z [[x ( v)]v] dV +

Z

x ( n) dA +

Zx ( b) dV .

To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

Z

x ( n) dA

i

=

Z

eijk xj kl nl dA =

Z

Ail nl dA =

ZA n dA

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where [ ]i represents the i-th component of the vector. Using the divergence theorem

ZA n dA =

Z A dV =

Z

Ail

xldV =

Z

xl(eijk xj kl) dV .

Differentiating,

ZA n dA =

Z

eijk jl kl + eijk xj kl

xl

dV =

Z

eijk kj + eijk xj kl

xl

dV =

Z

eijk kj + eijk xj [ ]l

dV .

Expressed in direct tensor notation, ZA n dA =

Z

h[E : T]i + [x ( )]i

idV

where E is the third-order permutation tensor. Therefore,

Z

x ( n) dA

i

==

Z

h[E : T]i + [x ( )]i

idV

or, Z

x ( n) dA ==

Z

hE : T + x ( )

idV .

The balance of angular momentum can then be written as

Zx

t( v)

dV =

Z [[x ( v)]v] dV +

Z

hE : T + x ( )

idV +

Zx ( b) dV .

Since is an arbitrary volume, we have

x

t( v)

= [[x ( v)]v] + E : T + x ( ) + x ( b)

or,

x

t( v) b

= [[x ( v)]v] + E : T .

Using the identity,

(uv) = ( v)u + (u) v

we get

[[x ( v)]v] = ( v)[x ( v)] + ([x ( v)]) v .

The second term on the right can be further simplified using index notation as follows.

[([x ( v)]) v]i = [([ (xv)]) v]i =

xl( eijk xj vk) vl

= eijk

xlxj vk vl +

xj

xlvk vl + xj

vk

xlvl

= (eijk xj vk)

xlvl

+ (eijk jl vk vl) + eijk xj

vk

xlvl

= [(xv)( v) + vv + x (v v)]i

= [(xv)( v) + x (v v)]i .

Therefore we can write

[[x ( v)]v] = ( v)(x v) + ( v)(xv) + x (v v) .

The balance of angular momentum then takes the form

x

t( v) b

= ( v)(x v) ( v)(xv) x (v v) + E : T

or,

x

t( v) + v v b

= ( v)(x v) ( v)(xv) + E : T

or,

x

v

t+

tv + v v b

= ( v)(x v) ( v)(xv) + E : T

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The material time derivative ofv is defined as

v =v

t+v v .

Therefore,

x [ v b] = x

tv + ( v)(x v) ( v)(xv) + E : T .

Also, from the conservation of linear momentum

v b = 0 .

Hence,

0 = x

tv + ( v)(x v) + ( v)(xv) E : T

=

t+ v + v

(xv) E : T .

The material time derivative of is defined as

=

t+ v .

Hence,

( + v)(xv) E : T = 0 .

From the balance of mass

+ v = 0 .

Therefore,E : T = 0 .

In index notation,

eijk kj = 0 .

Expanding out, we get

12 21 = 0 ; 23 32 = 0 ; 31 13 = 0 .

Hence,

= T

6. Show that the balance of energy can be expressed as:

e : (v) + q s = 0

where (x, t) is the mass density, e(x, t) is the internal energy per unit mass, (x, t) is the Cauchy stress, v(x, t) is the particle velocity, q is theheat flux vector, and s is the rate at which energy is generated by sources inside the volume (per unit mass).

Recall the general balance equation

d

dt

Z

f(x, t) dV

=

Z


Z

g(x, t) dA +

Z

h(x, t) dV .

In this case, the physical quantity to be conserved the total energy density which is the sum of the internal energy density and the kinetic energy

density, i.e., f = e + 1/2 |v v|. The energy source at the surface is a sum of the rate of work done by the applied tractions and the rate of heatleaving the volume (per unit area), i.e, g = v t q n where n is the outward unit normal to the surface. The energy source inside the body is thesum of the rate of work done by the body forces and the rate of energy generated by internal sources, i.e., h = v (b) + s.

Hence we have

d

dt Z e +1

2v v dV = Z e +

1

2v v (un v n) dA + Z(v t q n) dA + Z (v b + s) dV .

Let be a control volume that does not change with time. Then we get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

(v n) dA +

Z

(v t q n) dA +

Z

(v b + s) dV .

Using the relation t = n, the identity v ( n) = (T v) n, and invoking the symmetry of the stress tensor, we get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

(v n) dA +

Z

( v q) n dA +

Z

(v b + s) dV .

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We now apply the divergence theorem to the surface integrals to get

Z

t

e +

1

2v v

dV =

Z

e +

1

2v v

v

dA +

Z ( v) dA

Z q dA +

Z

(v b + s) dV .

Since is arbitrary, we have

t

e +1

2 v v

=

e +1

2 v vv

+ ( v) q + (v b + s) .

Expanding out the left hand side, we have

t

e +

1

2v v

=

t

e +

1

2v v

+

e

t+

1

2

t(v v)

=

t

e +

1

2v v

+

e

t+

v

t v .

For the first term on the right hand side, we use the identity ( v) = v + v to get

e +

1

2v v

v

=

e +

1

2v v

v +

e +

1

2v v

v

= e +1

2

v v v + e +1

2

v v v + e +1

2

v v v=

e +

1

2v v

v +

e +

1

2v v

v + e v +

1

2(v v) v

=

e +

1

2v v

v +

e +

1

2v v

v + e v + (vT v) v

=

e +

1

2v v

v +

e +

1

2v v

v + e v + (v v) v .

For the second term on the right we use the identity (ST v) = S :v + ( S) v and the symmetry of the Cauchy stress tensor to get

( v) = :v + ( ) v .

After collecting terms and rearranging, we get

t

+ v + ve + 12v v+ v

t+ v v b v+ e

t+e v+ :v+ q s = 0 .

Applying the balance of mass to the first term and the balance of linear momentum to the second term, and using the material time derivative of the

internal energy

e =e

t+e v

we get the final form of the balance of energy:

e :v + q s = 0 .

7. Show that the Clausius-Duhem inequality in integral form:

d

dt

Z

dV

Z

(un v n) dA

Z

q n

TdA +

Z

s

TdV .

can be written in differential form as

qT

!+ s

T.

Assume that is an arbitrary fixed control volume. Then un = 0 and the derivative can be take inside the integral to give

Z

t( ) dV

Z

(v n) dA

Z

q n

TdA +

Z

s

TdV .

Using the divergence theorem, we get

Z

t( ) dV

Z ( v) dV

Z

q

T

!dV +

Z

s

TdV .

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Since is arbitrary, we must have

t( ) ( v)

q

T

!+

s

T.

Expanding out

t +

t () v ( v)

q

T

!+

s

T

or,

t +

t v v ( v)

q

T

!+

s

T

or,

t+ v + v

+

t+ v

q

T

!+

s

T.

Now, the material time derivatives of and are given by

=

t+ v ; =

t+ v .

Therefore,

( + v) + q

T! +

s

T

.

From the conservation of mass + v = 0 . Hence,

q

T

!+

s

T.

8. Show that the Clausius-Duhem inequality

q

T

!+

s

T

can be expressed in terms of the internal energy as

(e T ) :v q T

T

.

Using the identity ( v) = v + v in the Clausius-Duhem inequality, we get

q

T

!+

s

Tor

1

T q q

1

T

!+

s

T.

Now, using index notation with respect to a Cartesian basis ej ,

1

T

!=

xj

`T1

ej =

`T2

Txj

ej = 1

T2T .

Hence,

1

T q +

1

T2q T +

s

Tor

1

T( q s) +

1

T2q T .

Recall the balance of energy

e :v + q s = 0 = e :v = ( q s) .

Therefore,

1

T( e :v) +

1

T2q T = T e :v +

q T

T.

Rearranging,

(e T ) :v q T

T.

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9. For thermoelastic materials, the internal energy is a function only of the deformation gradient and the temperature, i.e., e = e(F, T). Show that, forthermoelastic materials, the Clausius-Duhem inequality

(e T ) :v q T

T

can be expressed as

e

T

+

eF

FT

: F q TT

.

Since e = e(F, T), we have

e =e

F: F+

e

.

Therefore,

e

F: F+

e

T

:v

q T

Tor

e

T

+

e

F: F :v

q T

T.

Now,v = l = F F1. Therefore, using the identity A : (B C) = (A CT) : B, we have

:v = : (F F1) = ( FT) : F .

Hence,

e

T

+

e

F: F ( FT) : F

q T

T

or,

e

T

+

e

F FT

: F

q T

T.

10. Show that, for thermoelastic materials, the balance of energy

e :v + q s = 0 .

can be expressed as

T = q + s .

Since e = e(F, T), we have

e = eF

: F+ e

.

Plug into energy equation to get

e

F: F+

e

:v + q s = 0 .

Recall,e

= T and

e

F= FT .

Hence,

( FT) : F+ T :v + q s = 0 .

Now,v = l = F F1. Therefore, using the identity A : (B C) = (A CT) : B, we have

:v = : (F F1) = ( FT) : F .

Plugging into the energy equation, we have

:v + T :v + q s = 0

or,

T = q + s .

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11. Show that, for thermoelastic materials, the Cauchy stress can be expressed in terms of the Green strain as

= F e

E FT .

Recall that the Cauchy stress is given by

= e

F FT = ij =

e

Fik FTkj =

e

Fik Fjk .

The Green strain E = E(F) = E(U) and e = e(F, ) = e(U, ). Hence, using the chain rule,

e

F=

e

E:

E

F=

e

Fik=

e

Elm

Elm

Fik.

Now,

E =1

2(FT F 1) = Elm =

1

2(FTlp Fpm lm) =

1

2(Fpl Fpm lm) .

Taking the derivative with respect to F, we get

E

F=

1

2

FT

F F+ FT

F

F

=

Elm

Fik=

1

2

Fpl

FikFpm + Fpl

Fpm

Fik

.

Therefore,

=1

2

e

E:

FT

F F+FT

F

F

FT = ij =

1

2

e

Elm

Fpl

FikFpm + Fpl

Fpm

Fik

Fjk .

Recall,A

A

Aij

Akl= ik jl and

AT

A

Aji

Akl= jk il .

Therefore,

ij =1

2

e

Elm

`pi lk Fpm + Fpl pi mk

Fjk =

1

2

e

Elm(lk Fim + Fil mk)

Fjk

or,

ij =1

2

e

EkmFim +

e

ElkFil

Fjk = =

1

2

"F

e

E

T+F

e

E

# FT

or,

=1

2 F

"e

E

T+

e

E

# FT .

From the symmetry of the Cauchy stress, we have

= (F A) FT and T = F (F A)T = F AT FT and = T = A= AT .

Therefore,

e

E=

e

E

Tand we get

= F e

E FT .

12. For thermoelastic materials, the specific internal energy is given by

e = e(E, )

whereE is the Green strain and is the specific entropy. Show that

d

dt(e T ) = T +

1

0S : E and

d

dt(e T

1

0S : E) = T

1

0S : E

where 0 is the initial density, T is the absolute temperature, Sis the 2nd Piola-Kirchhoff stress, and a dot over a quantity indicates the material timederivative.

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Taking the material time derivative of the specific internal energy, we get

e =e

E: E+

e

.

Now, for thermoelastic materials,

T =e

and S= 0e

E

.

Therefore,

e =1

0S : E+ T . = e T =

1

0S : E .

Now,d

dt(T ) = T + T .

Therefore,

e d

dt(T ) + T =

1

0S : E =

d

dt(e T ) = T +

1

0S : E .

Also,

d

dt

1

0S : E

!=

1

0S : E+

1

0S : E .

Hence,

e d

dt(T ) + T =

d

dt

1

0S : E

!

1

0S : E =

d

dt

e T

1

0S : E

!= T

1

0S : E .

13. For thermoelastic materials, show that the following relations hold:

E=

1

0S(E, T) ;

T= (E, T) ;

g

S=

1

0E(S, T) ;

g

T= (S, T)

where (E, T) is the Helmholtz free energy and g(S, T) is the Gibbs free energy.

Also show thatS

T = 0

E andE

T = 0

S .

Recall that

(E, T) = e T = e(E, ) T .

and

g(S, T) = e + T +1

0S : E .

(Note that we can choose any functional dependence that we like, because the quantities e, , E are the actual quantities and not any particularfunctional relations).

The derivatives are

E=

e

E=

1

0S;

T= .

andg

S=

1

0

S

S: E =

1

0E ;

g

T= .

Hence,

E=

1

0S(E, T) ;

T= (E, T) ;

g

S=

1

0E(S, T) ;

g

T= (S, T)

From the above, we have

2

T E=

2

ET=

E=

1

0

S

T.

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and2g

T S=

2g

ST=

S=

1

0

E

T.

Hence,

S

T= 0

Eand

E

T= 0

S.

14. For thermoelastic materials, show that the following relations hold:

e(E, T)

T= T

T= T

2

T2

ande(S, T)

T= T

T+

1

0S :

E

T= T

2g

T2+S :

2g

ST.

Recall,

(E, T) = = e T = e(E, T) T (E, T)

and

g(S, T) = g = e + T +1

0S : E = e(S, T) + T (S, T) +

1

0S : E(S, T) .

Therefore,e(E, T)

T=

T+ (E, T) + T

T

ande(S, T)

T=

g

T+ (S, T) + T

T+

1

0S :

E

T.

Also, recall that

(E, T) =

T=

T=

2

T2,

(S, T) =g

T=

T=

2g

T2,

and

E(S, T) = 0g

S=

E

T= 0

2g

ST.

Hence,

e(E, T)

T= T

T= T

2

T2

and

e(S, T)

T= T

T+

1

0S :

E

T= T

2g

T2+S :

2g

ST.

15. For thermoelastic materials, show that the balance of energy equation

T = q + s

can be expressed as either

Cv T = ( T) + s +

0T

S

T: E

or

Cp

1

0S :

E

T

!T = ( T) + s

0T

E

T: S

where

Cv =e(E, T)

Tand Cp =

e(S, T)

T.

If the independent variables are E and T, then

= (E, T) = =

E: E+

TT .

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On the other hand, if we consider Sand T to be the independent variables

= (S, T) = =

S: S+

TT .

Since

E

= 1

0

S

T

;

T

=Cv

T

;

S

=1

0

E

T

; and

T

=1

TCp

1

0

S :E

T!

we have, either

= 1

0

S

T: E+

Cv

TT

or

=1

0

E

T: S+

1

T

Cp

1

0S :

E

T

!T .

The equation for balance of energy in terms of the specific entropy is

T = q + s .

Using the two forms of , we get two forms of the energy equation:

0T

S

T : E+ Cv T = q + s

and

0T

E

T: S+ Cp T

0S :

E

TT = q + s .

From Fouriers law of heat conduction

q = T .

Therefore,

0T

S

T: E+ Cv T = ( T) + s

and

0T

E

T: S+ Cp T

0S :

E

TT = ( T) + s .

Rearranging,

Cv T = ( T) + s +

0T

S

T: E

or,

Cp

1

0S :

E

T

!T = ( T) + s

0T

E

T: S.

16. For thermoelastic materials, show that the specific heats are related by the relation

Cp Cv =1

0

S T

S

T

!:

E

T.

Recall that

Cv :=e(E, T)

T= T

T

and

Cp :=e(S, T)

T= T

T+

1

0S :

E

T.

Therefore,

Cp Cv = T

T+

1

0S :

E

T T

T.

Also recall that

= (E, T) = (S, T) .

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Therefore, keeping Sconstant while differentiating, we have

T=

E:

E

T+

T.

Noting thatE = E(S, T), and plugging back into the equation for the difference between the two specific heats, we have

Cp Cv = T E

: ET

+ 10S : E

T.

Recalling that

E=

1

0

S

T

we get

Cp Cv =1

0

S T

S

T

!:

E

T.

17. For thermoelastic materials, show that the specific heats can also be related by the equations

Cp Cv =1

0

S :E

T

+E

T

: 2

EE

:E

T =

1

0

S :E

T

+T

0

E

T

: S

E

:E

T .

Recall that

S= 0

E= 0 f(E(S, T), T) .

Recall the chain rule which states that if

g(u, t) = f(x(u, t), y(u, t))

then, if we keep u fixed, the partial derivative ofg with respect to t is given by

g

t=

f

x

x

t+

f

y

y

t.

In our case,

u = S, t = T, g(S, T) = S, x(S, T) = E(S, T), y(S, T) = T, and f = 0 f.

Hence, we have

S= g(S, T) = f(E(S, T), T) = 0 f(E(S, T), T) .

Taking the derivative with respect to T keepingSconstant, we have

g

T=7

0

S

T= 0

2664 fE : ET + fT

71

T

T

3775

or,

0 =f

E:

E

T+

f

T.

Now,

f=

E=

f

E=

2

EEand

f

T=

2

T E.

Therefore,

0 =2

EE:

E

T+

2

T E=

E

E

:

E

T+

T

E

.

Again recall that,

E=

1

0S.

Plugging into the above, we get

0 =2

EE:

E

T+

1

0

S

T=

1

0

S

E:

E

T+

1

0

S

T.

40


41/42

Therefore, we get the following relation for S/T:

S

T= 0

2

EE:

E

T=

S

E:

E

T.

Recall that

Cp Cv =1

0S T S

T : E

T

.

Plugging in the expressions for S/T we get:

Cp Cv =1

0

S+ T 0

2

EE:

E

T

:

E

T=

1

0

S+ T

S

E:

E

T

:

E

T.

Therefore,

Cp Cv =1

0S :

E

T+ T

2

EE:

E

T

:

E

T=

1

0S :

E

T+

T

0

S

E:

E

T

:

E

T.

Using the identity (A : B) : C= C : (A : B), we have

Cp Cv =1

0S :

E

T+ T

E

T:

2

EE:

E

T

=

1

0S :

E

T+

T

0

E

T:

S

E:

E

T

.

18. Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the St-Venant Kirchhoff model,

i.e,

E = 1 and C = 1 1 + 2 I

where is the coefficient of thermal expansion and 3 = 3 K 2 where K, are the bulk and shear moduli, respectively.

Show that the specific heats related by the equation

Cp Cv =1

0

tr(S) + 9 2 K T .

Recall that,

Cp Cv =1

0S : E +

T

0E : C : E .

Plugging the expressions ofE and C into the above equation, we have

Cp Cv =1

0S : ( 1) +

T

0( 1) : ( 1 1 + 2 I) : ( 1)

=

0tr(S) +

2 T

01 : ( 1 1 + 2 I) : 1

=

0tr(S) +

2 T

01 : ( tr(1 ) 1 + 2 1)

=

0tr(S) +

2 T

0(3 tr(1) + 2 tr(1 ))

=

0tr(S) +

3 2 T

0(3 + 2)

= tr(S)

0+

9 2 K T

0.

Therefore,

Cp Cv =1

0

tr(S) + 9 2 K T .

41


42/42

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Cambridge, UK, 2002.

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Large deformation plasticity and MPM (working paper)

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