Politecnico di Torino

Facolta` di IngegneriaCorso di Laurea in Ingegneria Matematica

Continuum Mechanics03BOWNG

Professor:Prof. Luigi Preziosi

Student:Nathan Quadrio

A.Y. 2013/2014

Chapter 1 : Finite and Infinitesimal Deformations

Exercise 1.12

Given the deformation gradient

F =(xi

X l

)=

2 12

2

21

2

2

(1)apply the polar decomposition theorem.

First of all one may compute detF to check that is strictly greater thanzero according to the finite deformation hypothesis. Since detF = 1 > 0 weknow from the polar decomposition theorem that there exists two tensorsU and V, symmetrical and positive definite and one proper rotation R suchthat

F = RU = VR (2)

where U,V and R are univocally defined by the relations

U =FTF V =

FFT

R = FU1 = V1F.

Knowing this facts, one could easily compute

U =

(2 00 1/2

)V =

12

2

(17

1515

17

)

R = FU1 =

22

(1 11 1

)As expected the proper rotation matrix R is orthogonal since R RT = I.The importance of this results lies in the fact that every deformation can bedecompose in a dilation along the eigenvectors at first, in rigid translationthen and in a rigid rotation in the end.

1

Chapter 2 : Streamlines and Pathlines

Exercise 2.4

Given the velocity field

v(X, t) =

X2Y (1 + t)3Z(1 + t)2

v(x, t) = 11 + t

x2y3z

deduce the streamlines and the pathlines associated to the system.

Pathlines are the trajectories that individual fluid particles follow, i.e.the trajectories of every particle of C. Using the lagrangian velocity one candetermine them as

x(X, t) = X + t

0v(X, )d.

So, in this case, one has:

x = X + t0 Xds = X(1 + t) y = Y + t0 2Y (1 + s)ds = Y (1 + t)2 z = Z + t0 3Z(1 + s)2ds = Z(1 + t)3

In this way one obtains the equations of motion that describe particle tra-jectories. In figure 1 streamlines from time t = 0 to time t = 2 are reported.

11.5

22.5

3

2

4

6

8

5

10

15

20

25

x axisy axis

z a

xis

(a) Trajectory of the material point(X,Y, Z) = (1, 1, 1).

11.5

22.5

3

105

05

100

5

10

15

20

25

30

x axisy axis

z a

xis

(b) Trajectory of the material points(X,Y, Z) = (1, Y, 1), Y (1, 1).

Figure 1: Pathlines from time t = 0 to time t = 2.

Streamlines are a family of curves that are instantaneously tangent tothe velocity vector of the flow, i.e are the integral curves of the velocity fieldin eulerian coordinates in a fixed time t = t0. Let = (s) be a genericstreamline, one has:

s= v((s), t0)

2

Streamlines are tangent to the velocity field everywhere. Fixed t = t0, let = (x, y, z)T the curve passing through (x0, y0, z0)T for s = 0, one has:

dxds =

x1+t0

dyds =

2y1+t0

dzds =

3z1+t0

(s = 0) = (x0, y0, z0)

This it will be: dxx

= 11+t0dsdyy

= 21+t0dsdzz

= 31+t0ds

(s = 0) = (x0, y0, z0)log x = 11+t0 s+ Cxlog y = 21+t0 s+ Cylog z = 31+t0 s+ Cz(s = 0) = (x0, y0, z0)x = Kxe

11+t0

s

y = Kye2

1+t0s

z = Kze3

1+t0s

(s = 0) = (x0, y0, z0)

And given the initial condition it will become:x = x0e

11+t0

s

y = y0e2

1+t0s

z = z0e3

1+t0s

In a non-parametric form it will be for every t:{xx0

= ( yy0 )12

zz0

= ( yy0 )32

in figure 2 will be represented the velocity field and the streamlines in the yxand yz plane respectively. One can see how streamline are always tangentto the velocity field.

3

1.5 1 0.5 0 0.5 1 1.51.5

1

0.5

0

0.5

1

1.5xy plane

y axis

x a

xis

(a) Velocity field in the yx-plane at t = 1.

1.5 1 0.5 0 0.5 1 1.51.5

1

0.5

0

0.5

1

1.5yz plane

y axis

z a

xis

(b) Velocity field in the yz-plane at t = 1.

Figure 2: Velocity field and streamlines at t = 1.

Chapter 6 : Elastic Energy

Exercise 6.5-3

Compute how much energy requires an neo-Hookean elastic solid for a simpleshear deformation like:

x = X + Yy = Yz = Z

Assuming that the solid in incompressible, one knows that the strainenergy density function for a neo-Hookean solid is given by

W =

2(trC 3),

where is the shear modulus (proper of the material) and and C is the leftCauchy-Green deformation tensor.As one has a simple shear deformation it is known that the deformationgradient F is:

F =(xi

X l

)=

1 00 1 00 0 1

So, one can easily compute the left Cauchy-Green deformation tensor as:

C = FTF =

1 0 1 + 2 00 0 1

and obtain

trC = 1 + 1 + 2 + 1 = 3 + 2

In the end, for a simple shear deformation one has:

W =

2(trC 3) =

22

4

Chapter 7 : Non-Newtonian Poiseuille motion

Exercise 7.13

Study the motion of a fluid in a cylinder for a non-newtonian power-lawfluid.

Figure 3: Stress and shear rate dependency.

In this exercise it will be considered a Poiseuille stationary motion in acylinder of non-newtonian power law fluid. That is a non-Newtonian fluid,in which there is a power law between the stress-tensor T and the shear-rate, such as

Tzr = K()n,

also known as Ostwald de Waele relationship, whereK is the flow consistencyindex and n is the flow behavior index. As one can see in Figure 3 1 differentvalues of n correspond to different behavior of the fluids. In fact, for n = 1the fluid is newtonian, for n > 1 is dilatant and for n < 1 it is said to bepseudoplastic or viscoplastic.

Figure 4: Cylindrical pipe.

A cylindrical pipe, where the radius of the pipe is R and the length is L,will be considered (cfr. Figure 4). Some further hypothesis have been donein order to provide a solution:

stationary motion, so vt = 0;1Image taken from Appunti di Meccanica dei Continui

5

p = p0 Gz so G = p0pLL = pL ; v = vz(r)ez (vz it will be notice by only v).

The axial momentum equation gives

dpdz

+1r

r(rTzr) = 0

r(rTzr) = p

Lr

After the integration, one gets to

rTzr = pL

r2

2+A

If r = 0,0 = 0 +A = A = 0

So one obtainsTzr = p2L r

Then the power law will be applied, so

K(v

r)n = p

2Lr

After the integration, one gets to

v = ( p

2LK

) 1n r

1n

+1

1n + 1

+B

Applying the boundary condition of no-slip, one obtains

v =( p

2LK

) 1n R

1n

+1 r 1n+11n + 1

Now, one can compute the flow

= R

02pirv dr =

= R

02pir

( p2LK

) 1n R

1n

+1 r 1n+11n + 1

dr =

= 2pi( p

2LK

) 1n

R0rR

1n

+1 r 1n+11n + 1

dr =

6

2

1.5

1

0.5

0

0.5

1

1.5

20 0.2 0.4 0.6 0.8 1 1.2 1.4

Poiseuille velocity profiles

Figure 5: Velocity profiles for n < 1 (green), n = 1 (blue), and n > 1 (red).

= 2pi( p

2LK

) 1n[ R 1n+1r2

2( 1n + 1) r

1n

+3

( 1n + 3)(1n + 1)

]r=Rr=0

=

= 2pi( p

2LK

) 1n( R 1n+3

2( 1n + 1) R

1n

+3

( 1n + 3)(1n + 1)

)Finally, after some simplifications, one has

= A( p

2LK

) 1n( n

1 + 3n

)R

1n

+1

where A represents the section area of the pipe.

1 Maxwell Fluids

Exercise 8.2

Determine the constitutive equation of two parallel Maxwell elements.

Problem : Non-Material Surfaces

Exercise 2.14

Demonstrate that the cylindric surface x2 +y2 = R2(1+t)2 is not materialfor the deformation

x = X1+ty = Y (1 + t)z = Z

and compute the progress and the propagation velocities.

7

For simplicity this analysis will be done in two dimensions (x, y) since thethird will not influence the results. In Lagrangian coordinates the surfacecan be written as:

S(X,Y, t) = X2

(1 + t)4+ Y 2 R2 = 0

Since there is time dependence one can deduce that the surface is not built-in with the material points so is not a material surface for the considereddeformation. One can see that in Figure 6, but in the latter an analytical

1

0.5

0

0.5

1

1

0.5

0

0.5

10

0.2

0.4

0.6

0.8

1

(a) t = 0.

2.5 21.5 1

0.5 00.5 1

1.5 22.5

2

1

0

1

20

0.2

0.4

0.6

0.8

1

(b) t = 0.5.

21

01

2

21.5

10.5

00.5

11.5

20

0.2

0.4

0.6

0.8

1

(c) t = 1.

Figure 6: Time evolution of the deformed and the non-deformed surface.

proof will be given.If a surface is material then the propagation speed is zero, i.e.

v N = 0.

To compute this propagation speed one can parametrize the surface S(x, y, t)as:

(u, t) =

{x = R(1 + t) cosuy = R(1 + t) sinu

8

In lagrangian coordinates that will be

(u, t) = 1((u, t)) =

{x = R(1 + t)2 cosuy = R sinu

Nowv =

t

= (2R(1 + t) cosu, 0) = (2X

(1 + t), 0)

while

N =Grad F|Grad F | =

(2X

(1+t)4

2Y

)1

4X2

(1+t)8+ 4Y 2

=

(X

(1+t)4

Y

)1

X2

(1+t)8+ Y 2

Then

v N = 2X2

(1 + t)51

X2

(1+t)8+ Y 2

6= 0

In conclusion, since the propagation speed is not zero one can deduce thatthe surface is not material for the deformation.

In a similar fashion one can compute the progress speed defined as

vn = v n.

Nowv =

t

= (R cosu,R sinu) = (x

(1 + t),

y

(1 + t))

while

n =f|f | =

(xy

)1

x2 + y2

Then

vn = v n = 1x2 + y2

(x2

(1 + t)+

y2

(1 + t)) =

(1 + t)

x2 + y2

Other Exercises

Exercise 7.8 - Cylindric Couette Motion

Determine the stationary velocity profile of a fluid which flows between twocoaxial cylinders rotating about their own axis, as represented in Figure 7 2.

2Image taken from Appunti di Meccanica dei Continui

9

Figure 7: Cylindric Couette flux.

As represented in Figure 7, let the radius of the cylinders be Ri < Rerespectively. Inside the two cylinders there is an incompressible fluid ofviscosity and density . Both cylinders are rotating with velocity iand e respectively. Assuming the motion stationary and in cylindricalcoordinates the velocity profile and the pressure will be computed. Remindthat in cylindrical coordinates velocity and pressure are expressed as

v(r, z) = u er + v e, P (r, z).

Considering the previous assumptions, one can write the continuity equationin cylindrical coordinates:

1r

(ru)= 0

which, integrated with respect to r, gives:

u =A

r

Since a viscous fluid is assumed, one has u(Re) = 0 which implies u = 0.The momentum balance equations give

v2r = Pr(r

(1rr (rv)

)+

2vz2

)= 0

Pz = 0

From the first equation one can conclude that v is a function of only r, sothe middle equation can be integrated and it gives:

v(r) =A

2r +

B

r

Applying no-slip conditions on the two cylinders v(Ri) = iRi e v(Re) =eRe the values of the two constant are obtained:

A = 2eR2e iR2iR2e R2i

10

B = (i e) R2eR

2i

R2e R2iKnowing the velocity profile one can easily compute the pressure using thefirst equation of the system above:

P (r) = P0 + (A2

8r2 +AB log r B

2

2r2)

In Figure 8 two example of velocity profiles are reported.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

(a) Ri = 1, Re = 2, i = 1, e = 0.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

(b) Ri = 1, Re = 2, i = 1, e = 1.

Figure 8: Velocity profiles.

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