10 Velocity Gradient, Rate of Deformation, Vorticity

FIGURE 4.7

Differential velocity field at point p.



Consider the velocity components at two neighboring points p and q. Let

the particle currently at p have a velocity vi, and the particle at q a velocity

vi + dvi as shown in Figure 4.7. Thus the particle at q has a velocity relative

to the particle at p of

dvi =



∂vi

dxj or dv = L ⋅ dx

∂x j



(4.10-5)



Note that



∂vi

∂ vi ∂ X A d  ∂ xi  ∂ X A

=

= 

∂ xj ∂ XA ∂ xj

dt  ∂ X A  ∂ x j





(4.10-6a)



or in symbolic notation

˙

L = F ⋅ F –1



(4.10-6b)



where we have used the fact that material time derivatives and material

gradients commute. Therefore,

˙

F =L⋅F



(4.10-7)



Consider next the stretch ratio Λ = dx/dX where Λ is as defined in Eq 4.8ˆ

4, that is, the stretch of the line element dX initially along N and currently

ˆ

along n . By the definition of the deformation gradient, dxi = xi,A dXA, along

with the unit vectors ni = dxi /dx and NA = dXA /dX we may write

dx ni = xi,A dX NA

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which becomes (after dividing both sides by scalar dX)

ni Λ = xi,A NA



ˆ

ˆ

nΛ = F ⋅ N



or



(4.10-8)



If we take the material derivative of this equation (using the symbolic notation for convenience),

˙

ˆ

ˆ

ˆ ˙ ˙ ˆ

ˆ

nΛ + nΛ = F ⋅ N = L ⋅ F ⋅ N = L ⋅ nΛ

so that

˙

ˆ ˆ˙

ˆ

n + nΛ / Λ = L ⋅ n



(4.10-9)



ˆ

By forming the inner product of this equation with n we obtain

˙

ˆ ˆ ˆ ˆ˙

ˆ

ˆ

n ⋅ n + n ⋅ nΛ / Λ = n ⋅ L ⋅ n

˙

ˆ ˆ

ˆ ˆ

But n ⋅ n = 1 and so n ⋅ n = 0, resulting in

˙

˙

ˆ

ˆ

Λ / Λ = n ⋅ L ⋅ n or Λ / Λ = vi, j ni n j



(4.10-10)



which represents the rate of stretching per unit stretch of the element that

ˆ

ˆ

originated in the direction of N , and is in the direction of n of the current

configuration. Note further that Eq 4.10-10 may be simplified since W is

skew-symmetric, which means that

Lijninj = (Dij + Wij)ninj = Dijninj

and so

˙

ˆ

ˆ

Λ/Λ = n⋅ D⋅ n



or



˙

Λ / Λ = Dij ni n j



(4.10-11)



ˆ ˆ

For example, for the element in the x1 direction, n = e1 and

 D11

˙ / Λ = [1, 0, 0]  D

Λ

 12

 D13





D12

D22

D23



D13  1

 

D23  0 = D11

D33  0

 



˙

˙

ˆ ˆ

ˆ ˆ

Likewise, for n = e2 , Λ / Λ = D22 and for n = e 3 , Λ / Λ = D33 . Thus the

diagonal elements of the rate of deformation tensor represent rates of extension, or rates of stretching in the coordinate (spatial) directions.

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In order to interpret the off-diagonal elements of the rate of deformation

tensor, we consider two arbitrary differential vectors dxi(1) and dxi( 2 ) at p. The

material derivative of the inner product of these two vectors is (using the

superpositioned dot to indicate differentiation with respect to time of the

quantity in the brackets to the left of the dot),



[dx



( 1)

i



] [



]



•



[



•



dxi( 2 ) = dxi(1) dxi( 2 ) + dxi(1) dxi( 2 )



]



•



= dvi(1) dxi( 2 ) + dxi(1) dvi( 2 ) = vi, j dx (j1) dxi( 2 ) + dxi(1) vi, j dx (j2 )



(



)



= vi, j + v j,i dxi(1) dx (j2 ) = 2 Dij dxi(1) dx (j2 )



(4.10-12)



But dxi(1) dxi( 2 ) = dx (1) dx ( 2 ) cos θ, and



[dx



( 1)



] [



]



•



[



•



]



•



˙

dx ( 2 ) cos θ = dx (1) dx ( 2 ) cos θ + dx (1) dx ( 2 ) cos θ − dx (1) dx ( 2 )θ sin θ



[



] [



]



•

  dx (1) •



dx ( 2 ) 





( 1)

(2)

˙

= 

+

 cos θ − θ sin θ  dx dx

dx ( 2 ) 

  dx (1)











(4.10-13)



Equating Eqs 4.10-12 and 4.10-13, this gives



( 1)

i



2Dij dx dx



(2)

j



[



] [



]



•



  dx (1) •

dx ( 2 ) 





( 1)

(2)

˙

= 

+

 cos θ − θ sin θ  dx dx (4.10-14)

dx ( 2 ) 



  dx (1)









If dxi(1) = dxi( 2 ) = dxi , then θ = 0, and cos θ = 1, sin θ = 0 and dx (1) = dx ( 2 ) = dx

so that Eq 4.10-14 reduces to

Dij



dxi dx j

( dx )•

= Dij ni n j =

dx dx

dx



(4.10-15)



which is seen to be the rate of extension per unit length of the element currently

in the direction of ni (compare with Eq 4.10-11). If, however, dxi(1) is perpenπ

dicular to dxi( 2 ) so that θ =

, cos θ = 0, sin θ = 1, then Eq 4.10-14 becomes

2

˙

ˆ

ˆ

2 Dij ni(1) n(j2 ) = n1 ⋅ 2 D ⋅ n 2 = −θ



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(4.10-16)



This rate of decrease in the angle θ is a measure of the shear rate between

ˆ

ˆ

the elements in the directions of n1 and n 2 . In the engineering literature it

is customary to define the rate of shear as half the change (increase or decrease)

between two material line elements instantaneously at right angles to one

ˆ

ˆ

ˆ

ˆ

another. Thus for n1 = e1 and n 2 = e 2 ,

1 ˙

ˆ

ˆ

− θ12 = e1 ⋅ D ⋅ e 2 = D12

2

and, in general, the off-diagonal elements of the rate of deformation tensor

are seen to represent shear rates for the three pairs of coordinate axes.

Because D is a symmetric, second-order tensor, the derivation of principal

values, principal directions, a Mohr's circles representation, a rate of deformation deviator tensor, etc., may be carried out as with all such tensors.

Also, it is useful to develop the relationship between D and the material

derivative of the strain tensor E. Recall that

2E = C – I = F T ⋅ F – I

so that, using Eq 4.10-7,

T

˙

˙

˙

2 E = F T ⋅ F + F T ⋅ F = ( L ⋅ F ) ⋅ F + F T ⋅ (L ⋅ F )



(



)



= F T ⋅ LT ⋅ F + F T ⋅ L ⋅ F = F T ⋅ LT + L ⋅ F = F T ⋅ (2 D) ⋅ F

or

˙

E = FT ⋅ D ⋅ F



(4.10-17)



Note also that from ui + Xi = xi we have ui,A + δi,A = xi,A and if the displacement

gradients ui,A are very small, ui,A << 1 and may be neglected, then δi,A ≈ xi,A

(I ≈ F), and of course, F T = I T = I. At the same time for ui,A very small in

magnitude, by Eq 4.7-4, E ≈ ε and Eq 4.10-17 reduces to

˙

ε = I⋅D⋅I = D



(4.10-18)



for the infinitesimal theory. Finally, taking the material derivative of the

difference (dx)2 – (dX)2 = dX ⋅ 2E ⋅ dX, and noting that [(dx)2 – (dX)2]• = [(dx)2]•

since [(dX)2]• = 0, we obtain

˙

[(dx) ]2 • dX ⋅ 2E ⋅ dX = dX ⋅ F T 2D ⋅ F ⋅ dX = dx ⋅ 2D ⋅ dx

=

⋅



(4.10-19)



which shows that the local motion at some point x is a rigid body motion if

and only if D = 0 at x.

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˙

Solving Eq 4.10-9 for ni and using Eq 4.10-11, we may write



(



)



˙

˙

ni = vi, j n j − ni Λ / Λ = Dij + Wij n j − Dqk nq nk ni

If now ni is chosen along a principal direction of D so that Dij n(p) = D(p)ni(p)

j

(p = 1, 2, 3) where D(p) represents a principal value of D, then

(

(

˙

ni = D( p ) ni( p ) + Wij n(j p ) − D( p ) nq p ) nq p ) ni( p ) = Wij n(j p )



(4.10-20)



(p) (p)

since nq nq = 1. Because a unit vector can change only in direction, Eq 4.1020 indicates that Wij gives the rate of change in direction of the principal axes

of D. Hence the names vorticity or spin given to W. Additionally, we associate

with W the vector



wi =



1

εijkvk,j

2



or w =



1

curl v

2



(4.10-21)



called the vorticity vector, by the following calculation,

1

ε pqi wi = 2 ε pqiε ijk vk,j =



=



1

2



(v



q,p



1

2



(δ



)



δ − δ pkδ qj vk,j



pj qk



)



− v p,q = Wqp



(4.10-22)



Thus if D ≡ 0 so that Lij = Wij, it follows that dvi = Lijdxj = Wijdxj = εjikwkdxj

and since εjik = –εijk = εikj,

dvi = εijkwj dxk



or dv = w × dx



(4.10-23)



according to which the relative velocity in the vicinity of p corresponds to a

rigid body rotation about an axis through p. The vector w indicates the

angular velocity, the direction, and the sense of this rotation.

To summarize the physical interpretation of the velocity gradient L, we

note that it effects a separation of the local instantaneous motion into two parts:

1. The so-called logarithmic rates of stretching, D(p), (p = 1,2,3), that is, the

eigenvalues of D along the mutually orthogonal principal axes of D, and

d (lnΛ )

˙

Λ/Λ =

= Dij ni( p ) n(jp ) = ni( p ) D( p ) ni( p ) = D( p )

dt

2. A rigid body rotation of the principal axes of D with angular velocity w.

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FIGURE 4.8

Area dS° between vectors dX(1) and dX(2) in the reference configuration becomes dS between

dx(1) and dx(2) in the deformation configuration.



4.11 Material Derivative of Line Elements, Areas, Volumes

Consider first the material derivative of the differential line element dx = F ⋅

˙

dX. Clearly, (dx)• = F ⋅ dX and by Eq 4.10-7,

˙

(dx)• = F ⋅ dX = L ⋅ F ⋅ dX = L ⋅ dx



or



(dxi)• = vi,jdxj



(4.11-1)



Note further that from Eq 4.11-1 the material derivative of the dot product

dx ⋅ dx is

(dx ⋅ dx)• = 2dx ⋅ (dx)• = dx ⋅ 2L ⋅ dx = dx ⋅ 2(D + W) ⋅ dx = dx ⋅ 2D ⋅ dx

in agreement with Eq 4.10-19.

It remains to develop expressions for the material derivatives of area and

volume elements. Consider the plane area defined in the reference configu(

(

ration by the differential line elements dXA1) and dX A2 ) as shown in

Figure 4.8. The parallelogram area dS° may be represented by the vector

o

(

(

dS A = ε ABC dX B1) dX C2 )



(4.11-2)



As a result of the motion x = x(X,t) this area is carried into the current area

dSi shown in Figure 4.8, and given by

(

(

(

dSi = εijk dx (j1) dxk2 ) = ε ijk x j,B dX B1) xk,C dX C2 )



which upon multiplication by xi,A results in

(

(

(

(

xi,AdSi = ε ijk xi,A x j,B xk,C dX B1) dX C2 ) = ε ijk FiA FjB FkC dX B1) dX C2 )

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(4.11-3)



Recall that det F = J (the Jacobian) and from Eq 2.4-12



ε ijk FiA FjB FkC = ε ABCdet F = ε ABC J

Therefore, by inserting this result into the above equation for xi,AdSi and

multiplying both sides by XA,q, we obtain

(

(

xi,A X A,q dSi = ε ABC JdX B1) dX C2 ) X A,q



But xi,AXA,q = δiq so that

o

δiqdSi = dSq = XA,q JdS A



(4.11-4)



which expresses the current area in terms of the original area.

To determine the material derivative of dSi, we need the following identity:

˙

(det A)• = tr( A ⋅ A −1 )det A



(4.11-5)



where A is an arbitrary tensor. Substituting F for A, we obtain



(



)



˙

˙

(det F)• = J = (det F)tr F ⋅ F −1 = J tr (L)

or

˙

J = Jvi,i = J div v



(4.11-6)



Noting that Eq 4.11-4 may be written

o

dS q = JX A,q dS A



and using symbolic notation to take advantage of Eq 4.11-5, we obtain

dS = J(F –1)T ⋅ dS° = JdS° ⋅ F –1

and so

dS ⋅ F = JdS°

which upon differentiating becomes

˙

˙ ˙

dS ⋅ F + dS ⋅ F = JdSo = J (tr L)dSo

˙

˙

dS + dS ⋅ F ⋅ F –1 = J (tr L)dSo ⋅ F –1 = (tr L)dS

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FIGURE 4.9

Volume of parallelepiped defined by vectors dX(1), dX(2), and dX(3) in the reference configuration

deforms into the volume defined by paralellepiped defined by vectors dx(1), dx(2), and dx(3) in

the deformed configuration.



and finally

˙

dS = (tr L) dS – dS ⋅ L



or



˙

dSi = vk,k dSi − dS j v j,i



(4.11-7)



which gives the rate of change of the current element of area in terms of the

current area, the trace of the velocity gradient, and of the components of L.

Consider next the volume element defined in the referential configuration

by the box product

(

(

(

dV° = dX(1) ⋅ dX(2) × dX(3) = εABC dX A1) dX B2 ) dX C3 ) = [dX(1), dX(2), dX(3)]



as pictured in Figure 4.9, and let the deformed volume element shown in

Figure 4.9 be given by

(

dV = dx(1) ⋅ dx(2) × dx(3) = εijk dxi(1) dx (j2 ) dxk3 ) = [dx(1), dx(2), dx(3)]



For the motion x = x(X,t), dx = F ⋅ dX so the current volume is the box product

(

(

(

dV = [F ⋅ dX(1), F ⋅ dX(2), F ⋅ dX(3)] = εijkxi,A xj,Bxk,C dX A1) dX B2 ) dX C3 )



= det F [dX(1), dX(2), dX (3)] = JdV°



(4.11-8)



which gives the current volume element in terms of its original size. Since

J ≠ 0 (F is invertible), we have either J < 0 or J > 0. Mathematically, J < 0 is

possible, but physically it corresponds to a negative volume, so we reject it.

Henceforth, we assume J > 0. If J = 1, then dV = dV° and the volume magnitude is preserved. If J is equal to unity for all X, we say the motion is

isochoric.

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FIGURE P4.1

Unit square OABC in the reference configuration.



To determine the time rate of change of dV, we take the material derivative

as follows:

˙

(dV)• = JdV o = Jtr(L) dV ° = Jvi,i dV ° = vi,i dV



(4.11-9)



Thus, a necessary and sufficient condition for a motion to be isochoric is that

vi,i = div v = 0



(4.11-10)



In summary, we observe that the deformation gradient F governs the

stretch of a line element, the change of an area element, and the change of

a volume element. But it is the velocity gradient L that determines the rate

at which these changes occur.



Problems

4.1 The motion of a continuous medium is specified by the component

equations

x1 =



1

2



(X1 + X2)et +



x2 =



1

2



(X 1 + X 2)e t –



x3 = X 3

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1

2



1

2



(X1 – X2)e –t

(X 1 – X 2)e –t



(a) Show that the Jacobian determinant J does not vanish, and solve

for the inverse equations X = X(x, t).

(b) Calculate the velocity and acceleration components in terms of

the material coordinates.

(c) Using the inverse equations developed in part (a), express the velocity and acceleration components in terms of spatial coordinates.

Answer: (a) J = cosh2 t – sinh2 t = 1

X1 =



1

2

1

2



X2=



1

2



(x1 + x 2)e –t +



(x1 – x 2)et

1

2



(x1 + x 2) e –t –



(x1 – x 2) et



X3= x3

(b) v1 =

v2=



1

2

1

2



(X1 + X2)et –



1

2



(X1 – X2)e –t

1

2



(X1 + X2)et +



(X1 – X2)e –t



v3= 0

a1 =

a2=



1

2

1

2



(X1 + X2)et +



1

2



(X1 – X2)e –t



(X1 + X2)et –



1

2



(X1 – X2)e –t



a3= 0

(c) v1 = x2, v2 = x1, v3 = 0

a1 = x1, a2 = x2, a3 = 0

4.2 Let the motion of a continuum be given in component form by the

equations

x1 = X1 + X2t + X3t2

x 2 = X2 + X3t + X1t2

x 3 = X3 + X1t + X2t2

(a) Show that J ≠ 0, and solve for the inverse equations.

(b) Determine the velocity and acceleration

(1) at time t = 1 s for the particle which was at point (2.75, 3.75,

4.00) when t = 0.5 s.

(2) at time t = 2 s for the particle which was at point (1, 2, –1)

when t = 0.



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Answer: (a) J = (1 – t3)2

X1 = (x1 – x2 t)/(1 – t3)

X2 = (x2 – x3 t)/(1 – t 3)

X3 = (x 3 – x1 t)/(1 – t 3)

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

(b) (1) v = 8e1 + 5e 2 + 5e 3 , a = 6e1 + 2e 2 + 4e 3

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

(2) v = −2e1 + 3e 2 + 9e 3 , a = −2e1 + 2e 2 + 4e 3

4.3 A continuum body has a motion defined by the equations

x1 = X1 + 2X2t2

x2 = X2 + 2X1t2

x3 = X3

(a) Determine the velocity components at t = 1.5 s of the particle

which occupied the point (2, 3, 4) when t = 1.0 s.

(b) Determine the equation of the path along which the particle designated in part (a) moves.

(c) Calculate the acceleration components of the same particle at time

t = 2 s.

Answer: (a) v1 = 2, v2 = 8, v3 = 0

(b) 4x1 – x2 = 5 in the plane x3 = 4

(c) a1 = 4/3, a2 = 16/3, a3 = 0.

4.4 If the motion x = x(X,t) is given in component form by the equations

x1 = X1(1+ t),



x2 = X2(1+ t)2,



x3 = X3(1 + t2)



determine expressions for the velocity and acceleration components

in terms of both Lagrangian and Eulerian coordinates.

Answer: v1 = X1 = x1/(1+ t)

v2 = 2X2(1+ t) = 2x2/(1+ t)

v3 = 2X3t = 2x3t/(1+ t2)

a1 = 0

a2 = 2X2 = 2x2/(1+ t)2

a3 = 2X3 = 2x3/(1+ t2)



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