Kossa ContMech Problems 2012 Fall

7/27/2019 Kossa ContMech Problems 2012 Fall

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BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012

EXERCISE 1

The following two vectors (a and b) are given in the Cartesian coordinate system {ei}:

a = 7e1 + 2e2 5e3, b = 3e1 + 6e2 + 4e3. (1)

Calculate the following quantities:(a) a b,(b) a b,(c) a ,(d) ea (unit vector in the direction ofa),(e) (the angle between a and b),(f) a b.

SOLUTION

(a) The scalar (inner) product ofa and b is

a b = aibi = 7 (3) + 2 6 + (5) 4 = 29. (2)

(b) The cross product a b gives a new vector:

a b =e1 e2 e37 2 53 6 4

= aibjijkek= e1 (2

4

(

5)

6)

e2 (7

4

(

5)

(

3)) + e3 (7

6

2

(

3))

= 38e1 13e2 + 48e3. (3)Thus the new vector in matrix notation has the form

[a b] = 3813

48

. (4)

(c) The norm (lenght) of the given vectors:

a

=a

a =

aiai = 72 + 22 + (5)

2 =

78= 8.832, (5)

b =b b =

bjbj =

(3)2 + 62 + 42 =

61 = 7.810. (6)

(d) The unit vector in the direction ofa is computed by dividing a with its lenght a, i.e.

ea =a

a =778e1 +

278e2 5

78e3, [ea] =

0.7930.2260.566

. (7)

(e) The angle defined between a and b is calculated by the relation

= arccos

a b

a b

= arccos

2978 61

= 2.00471 rad = 114.861. (8)

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(f) Dyadic product ofa and b yields a 2nd-order tensor:

a b = aibjei ej (9)= a1b1e1 e1 + a1b2e1 e2 + a1b3e1 e3

+a2b1e2e1 + a2b2e2

e2 + a2b3e2

e3

+a3b1e3 e1 + a3b2e3 e2 + a3b3e3 e3, (10)where

[e1 e1] = 1 0 00 0 0

0 0 0

, [e1 e2] =

0 1 00 0 0

0 0 0

, [e1 e3] =

0 0 10 0 0

0 0 0

, (11)

[e2 e1] = 0 0 01 0 0

0 0 0

, [e2 e2] =

0 0 00 1 0

0 0 0

, [e2 e3] =

0 0 00 0 1

0 0 0

, (12)

[e3 e1] = 0 0 00 0 0

1 0 0

, [e3 e2] =

0 0 00 0 0

0 1 0

, [e3 e3] =

0 0 00 0 0

0 0 1

. (13)

The matrix representation of the dyad a b then

[a b] = 21 42 286 12 8

15 30 20

. (14)

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EXERCISE 2

Two 2nd-order tensors (A and B) are given in matrix form as

[A] = 7

9 8

3 2 55 1 6

, [B] =

2 6 5

6 1 25 2 9

. (15)

Determine the following quantities:(a) A : B,(b) det(AB),(c) B1,(d) A,(e) IB, IIB, I I I B,(f) symmetric and skewsymmetric parts ofA,(g) eigenvalues and eigenvectors ofB,(h) spectral representation ofB.

SOLUTION

(a) The double-contraction between A and B is calculated by

A : B = AijBij (16)

= A11 B11 + A12 B12 + A13 B13+A21 B21 + A22 B22 + A23 B23+A31 B31 + A32 B32 + A33 B33, (17)

= 7 2 + (9) (6) + 8 5+3 (6) + (2) (1) + 5 2+5 5 + 1 2 + 6 9, (18)

A : B = 183. (19)

The solution can be obtained using the identity:

A : B = trATB

= tr

ABT

= tr

108 17 8943 6 56

34 19 81

= 108 + (6) + 81 = 183. (20)

(b)

det(AB) =

108 17 8943 6 5634 19 81

(21)= 108 [(6) (81) (56)(19)] (17) [(43) (81) (56) (34)]

+(89)[(43)(19) (6) (34)] , (22)det(AB) = 34710. (23)

(c) The inverse ofB is computed by the fomrula

B1 =adjB

detB, (24)

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where the adjugate matrix is the transpose of the cofactor matrix:

[adjB] =

1 22 9

6 25 9

6 15 2

6 5

2 9 2 55 9

2

6

5 2 6 51 2

2 56 2

2 66 1

T

(25)

=

13 64 764 7 34

7 34 38

(26)

and

detB =

2 6 5

6

1 25 2 9

= 445. (27)

Therefore the invrese reads as

B1

=

1

(445)

13 64 764 7 34

7 34 38

0.0292135 0.14382 0.01573030.14382 0.0157303 0.0764045

0.0157303 0.0764045 0.0853933

. (28)

(d) The norm ofA is given by

A

=A : A = tr AAT = tr ATA 17.1464282. (29)

(e) The principal scalar invariants are:

IB = trB = 2 1 + 9 = 10, (30)

IIB =1

2

I2B tr

B2

=

1

2

102 tr

65 4 434 41 14

43 14 110

= 58, (31)

IIIB = detB =

2 6 5

6

1 2

5 2 9

=

445.

(f) The symmetric and skewsymmetric parts of tensor A are:

Asymm =1

2

A + AT

, [Asymm] =

7 3 6.53 2 3

6.5 3 6

, (32)

Askew =1

2

AAT , [Askew] =

0 6 1.56 0 2

1.5

2 0

. (33)

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(g) The Cardanos solution for the eigenvalues of real symmetric 3 3 matrix B can be deter-mined with the following algorithm:

P =1

3

(I2B 3IIB) =

274

3= 5.5176, (34)

D = 227

I3B 13 IBIIB + IIIB = 479527 = 177.593, (35)

i = 2Pcos

1

3arccos

D

2P3

+ (i 1) 2

3

+

1

3IB, i = 1...3. (36)

1 = 11.7075, 2 = 7.0778, 3 = 5.3703. (37)Eigenvector corresponding to 1 is calculated by

(B 1I)n1 = 0, n1 = n11e1 + n12e2 + n13e3, n1 = 1, (38)

9.7075 6 56 12.7075 25 2 2.7075

n11n12n13

= 000

, (39)

9.7075n11 6n12 + 5n13 = 0, (40)6n11 12.7075n12 + 2n13 = 0, (41)

5n11 + 2n12 2.7075n13 = 0. (42)Let n13 = 1, then

9.7075n11

6n12 =

5, (43)

6n11 12.7075n12 = 2, (44)5n11 + 2n12 = 2.7075. (45)

From the first equation we can write

n12 =5

6 9.7075

6n11. (46)

Substituting into the second equation we have

6n11

12.7075

5

6

9.7075

6

n11 = 2 (47)

from which

n11 = 0.5900, n12 = 0.1212. (48)Normalized eigenvector:

n1 =n1

n1 =1

(0.5900)2 + (0.1212)2 + 11(0.5900e1 0.1212e2 + 1e3) , (49)

[n1] = 0.50540.1038

0.8566

. (50)

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n2 can be calculated similarly to n1, whereas n3 have to be determined with n3 = n1 n2 inorder to have right-handed coordinate system ofn1,n2 and n3.Finally we have

[n2] =

0.6331

0.71920.2863

, [n3] = 0.5864

0.68700.4292

. (51)

(h) The spectral representation ofB reads as

B = 1n1 n1 + 2n2 n2 + 3n3 n3, (52)

where the following dyads are defined:

[n1

n1] =

0.255408 0.0524574 0.432924

0.0524574 0.010774

0.0889167

0.432924 0.0889167 0.733818

, (53)

[n2 n2] = 0.400774 0.455296 0.1812730.455296 0.517235 0.2059330.181273 0.205933 0.0819908

, (54)

[n3 n3] = 0.343818 0.402839 0.2516510.402839 0.471991 0.294850.251651 0.29485 0.184191

. (55)

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EXERCISE 3

Let C be a symmetric 3 3 matrix:

[C] = 33 17 1017 22 5

10 5 38

(56)

with eigenvalues and eigenvectors

1 = 48.6329, 2 = 37.9255, 2 = 6.44156, (57)

[n1] =

0.7559830.382872

0.530942

, [n2] =

0.2787090.545642

0.790314

, [n3] =

0.5922930.745442

0.305786

. (58)

Calculate the tensor U=C.

SOLUTION

First, it must to note that

C=

33

17 1017 22 510

5

38

. (59)

The matrix ofC in the Cartesian basis formed by its unit eigenvectors is given by

C = QT [C] [Q] =

1 0 00 2 0

0 0 3

, (60)

where the orthogonal rotation tensor Q is defined by

[Q] =

[n1] [n2] [n3]

=

0.755983 0.278709 0.5922930.382872 0.545642 0.745442

0.530942 0.790314 0.305786

. (61)

Q is an orthogonal tensor, i.e. QQT = I, Q1 = QT.The matrix of the tensor U in the basis of unit eigenvectors ofC (where C is represented bya diagonal matrix) is computed by

U

=

C

= 1 0 00 2 0

0 0

3

= 6.97373 0 00 6.15837 00 0 2.53802

. (62)U in the original basis is calculated with the formula

[U] = [Q]U QT

= 5.3543 1.83446 0.9829721.83446 4.26613 0.659494

0.982972 0.659494 6.0497

. (63)

Verification:

[U] [U] =

33 17 1017 22 510 5 38

[C] . (64)

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EXERCISE 4

Let be the symmetric 2nd-order tensor represented in a coordinate frame by the matrix

[] = 50 30 4030 100 2040 20 10

. (65)

Determine its spherical and deviatoric parts, respectively.

SOLUTION

Spherical part is computed by

p =1

3(tr) I, (66)

[p] = 13

(60)1 0 00 1 00 0 1

= 20 0 00 20 00 0 20

, (67)

whereas the deviatoric part is determined by the relation

s = p = 13

(tr) I, (68)

[s] =

50 30 4030 100 2040 20 10

20 0 00 20 0

0 0 20

=

70 30 4030 80 2040 20 10

. (69)

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EXERCISE 5

Let a new right-handed Cartesian coordinate system be given by the set of basis vectors

e1 = cose1 + sine2, (70)

e2 = sine1 + cose2. (71)(a) Find e3 in terms of the old set of basis vectors.(b) Find the othogonal matrix [Q] and express the new coordinates in terms of the old one forthe particular value = 30.(c) Express vector r = 5e1 e2 + 2e3 in the new coordinate system.

SOLUTION

(a) Since the coordinate system is right-handed, it follows that

e3 = e1 e2, (72)e3 =

e1 e2 e3

cos sin 0sin cos 0

= e3 (73)

(b) The orthogonal tensor Q is constructed by

[Q] =

[e1] [e2] [e3]

=

cos sin 0sin cos 0

0 0 1

=

32

12

012

32

00 0 1

. (74)

(c) The components of vector r in the new coordinate system is calculated according to

[r] =QT

[r] =

32

12

0

12

32

00 0 1

51

2

=

5312

5312

2

=

3.8303.366

2

, (75)

r = 3.83e1 3.366e2 + 2e3. (76)

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EXERCISE 6

Simplify the expression a (b c) utilizing the rules of the Einsteins summation notation.

SOLUTION

The expression above can be written as

(aiei) ((bjej) (ckek)) = aibjckei(ej ek) = aibjckjkmeiem = aibjckjkmimpep. (77)

Thus the pth component (a (b c))p reads as

(a (b c))p = aibjckjkmimp = aibjckjkmpim, (78)

wher the identity jkmimp = jkmpim was employed.Now we can use the identity jkmpim = jpki jikp, which yields

(a (b c))p = aibjck (jpki jikp) (79)= aibpci aibicp= (aici) bp (aibi) cp. (80)

Thus, it follows that

a (b c) = (a c) b (a b) c. (81)

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EXERCISE 7

Suppose that the scalar field

(x) =

x1 + 4x2x2

3(1)

describes a physical quantity. Compute the gradient of the above scalar field at pointP(4, 1,3).

SOLUTION

The gradient of the scalar field is given by

grad = = xa

ea =

x1e1 +

x2e2 +

x3e3. (2)

Using matrix notation we have

[grad] =

x1

x2

x3

=

,1,2

,2

=

1

2

x14x2

3

8x2x3

. (3)

Its numerical value at point P(4, 1,3) is

[gradP] =

1

244 (3)2

8(1)(3)

= 0.253624

. (4)

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EXERCISE 8

Determine the gradient of the scalar field

= x21

x2

7x3x1 (5)

at point P(1, 2, 1) along the direction defined by the unit vector

n =114

(2e1 + 3e2 e3) . (6)

SOLUTION

The gradient of the scalar field at point P is determined by

[grad] =

x1

x2

x3

=

2x1x2 7x3x2

1

7x1

, [gradP] =

111

7

. (7)

Thus, the directional derivative along n is given by

(gradP) n = (11e1 + 1e2 + 7e3) (2e1 + 3e2 e3) 114

= 2614

6.9488. (8)

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EXERCISE 9

Let the vector field u in the rectangular Cartesian coordinate system be the following:

u = (1 + x1) e1 + x1x2e2 + (x3 x1) e3, [u] = u1

u2u3

= 1 + x1

x1x2x3 x1

. (9)

Compute the gradient of the vector field u.

SOLUTION

The gradient ofu yields the 2nd-order tensor

gradu = u = (uaea)

xbeb

=

ua

xbea eb = ua,bea eb, (10)

gradu =u1

x1e1 e1 + u1

x2e1 e2 + u1

x3e1 e3 (11)

+u2

x1e2 e1 + u2

x2e2 e2 + u2

x3e2 e3 (12)

+u3

x1e3 e1 + u3

x2e3 e2 + u3

x3e3 e3, (13)

[gradu] =

u1

x1

u1

x2

u1

x3u2

x1

u2

x2

u2

x3u3

x1

u3

x2

u3

x3

= 1 0 0

x2 x1 01 0 1

. (14)

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EXERCISE 10

Determine the divergence of the vector field

v = x1x2x3 (x3e1 + x2e2 + x1e3) . (15)

at point P(1, 2, 3).

SOLUTION

The divergence of the vector field v is given by the relation

divv = v = vaxb

ea eb = vaxb

ab =va

xa= va,a (16)

=v1

x1+

v2

x2+

v3

x3, (17)

divv =

x2x2

3

+ (2x1x2x3) +

x21

x2

= x2 (x1 + x3)2 . (18)

At point P we have the numerical value

divvP = 2 (1 + 3)2 = 32. (19)

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EXERCISE 11

Calculate the divergence of the 2nd-order tensor field

[A] = x

2

1x2 2

7 4x2x3 5x32

1 x1x3

(20)

from the left at point P(2, 3, 4).

SOLUTION

The divergence ofA is computed as

divA = A = Axc

ec = Aabxc

ea (eb ec) = Aabxc

bcea = Aab,cbcea = Aab,bea. (21)

Using matrix notation we arrive at

[divA] =

A11

x1+

A12

x2+

A13

x3A21

x1+

A22

x2+

A23

x3A31

x1+

A32

x2+

A33

x3

=

2x1 1 + 00 + 4x3

0 + 0 + x1

=

2x1 14x3

x1

. (22)

At point P:

[divAP] = 316

2

. (23)

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EXERCISE 12

Determine the curl of the vector field given by its matrix representation in the {x , y , z } coor-dinate system

[v] =

xyzy2 + xzx y

. (24)

SOLUTION

The curl of the vector field v yield the new vector

curlv = v = ea vxa

=vb

xaea eb = vb

xaabcec = vb,aabcec. (25)

Using matrix notation:

[curlv] =

x 1xy 1

z xz

. (26)

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EXERCISE 13

Calculate the curl of the following 2nd-order tensor field:

[A] = xy y z

2

1 z 00 y y

. (27)

SOLUTION

The curl ofA is defined by

curlA = A =

xaea

(Abceb ec) = Abc

xaea (eb ec) (28)

=Abc

xaabded

ec = Abc,aabded

ec. (29)

In matrix representation:

[curlA] =

Ab1,aab1 Ab2,aab1 Ab3,aab1Ab1,aab2 Ab2,aab2 Ab3,aab2

Ab1,aab3 Ab2,aab3 Ab3,aab3

. (30)

Using the rule for the permutation symbol, we arrive at the matrix

[curlA] =

A31,2 A21,3 A32,2 A22,3 A33,2 A23,3A11,3

A31,1 A12,3

A32,1 A13,3

A33,1

A21,1 A11,2 A22,1 A12,2 A23,1 A13,2

. (31)

[curlA] =

0 0 1 1 1 00 0 0 0 2z 0

0 x 0 1 0 0

. (32)

Thus, the final result is

[curlA] =

0 0 10 0 2z

x

1 0

. (33)

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EXERCISE 14

Suppose that the motion of a continuous body is given by the mapping

(X, t) = (X1 + tX2)e1 + X2 t2X1 e2 + (X3) e3. (1)

The motion is illustrated in Figure 1.

Figure 1: Illustration of the motion

Determine the inverse mapping X= 1 (x, t).

SOLUTION

The motion is described by linear combinations. Thus it can be written as

x1x2

x3

=

X1 + tX2X2 t2X1

X3

=

1 t 0t2 1 0

0 0 1

X1X2

X3

. (2)

The inverse motion is derived by solving the following system of equations:

x1 = X1 + tX2, (3)

x2

= X2 t

2

X1

, (4)x3 = X3. (5)

Thus, the inverse mapping X= 1 (x, t) has the form

1 (x, t) =1

1 + t3(x1 tx2)E1 + 1

1 + t3

t2x1 + x2E2 + x3E3.

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EXERCISE 15

Suppose that the deformation of a continuous body at a fixed time is given by

(X) = (3

2X1

X2) e1 + 2 + 12

X1

1

2X2 e2 + (X3)e3. (6)

Determine the matrix representation of the deformation gradient and its inverse.

SOLUTION

The definition for the deformation gradient is

F = Grad (X) = X =

XaEa

. (7)

Thus, the matrix representation is given by

[F] =

x1X1

x1X2

x1X3

x2X1

x2X2

x2X3

x3X1

x3X2

x3X3

=

2 1 00.5 0.5 0

0 0 1

. (8)

The inverse deformation gradient can be obtained from the inverse motion. The inverse mappingfor this case is

1 (x) =13 13x1 + 23 x2E1 + 113 13x1 43x2E2 + (x3)E3. (9)

The inverse deformation gardient is computed by

F1 = grad1 (x) = 1 x = 1

xaea

. (10)

Thus

F1

=

X1x1

X1x2

X1x3

X2x1

X2x2

X2x3

X3x1

X3x2

X3x3

= 1/3 2/3 01/3 4/3 0

0 0 1

. (11)

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EXERCISE 16

Let the deformation described by the following mapping:

(X) = (1 + X1 + X2) e1 + (X2

2X1)e2 + (X3)e3. (12)

Calculate the stretch corresponding to the material line element (fibre), which is oriented inthe reference configuration along the unit vector

[N1] =1

5

21

0

. (13)

Calculate the stretch corresponding to the material line element (fibre), which is oriented inthe spatial configuration along the unit vector

[n2] =1

5

210

. (14)

SOLUTION

The deformation gradient and its inverse for this homogeneous deformation are

[F] =

1 1 02 1 00 0 1

, F1

=1

3

1 1 02 1 0

0 0 3

. (15)

The material fibre along N1 is transformed to the spatial configuration according to

FN1. (16)

Its length divided by the length ofN1 gives the stretch

N1 =|FN1||N1| =

N1 FT F N1 = 3

2

5 1.89737,

The material fibre along n2

is located in the reference configuration along the line element

F1n2. (17)

The stretch is defined by

n2 =|n2|F1n2 =

1n2 FT F1 n2

= 3

5

26 1.31559. (18)

The streches are illustrated in Figure 2.It must be noted, that this example is a homogeneous deformation. Consequently material

fibre with finite length deform (similarly to the infinitesimal material line element) with thedeformation gradient.

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Referenceconfiguration:

Spatialconfiguration:

Figure 2: Illustration of the stretches

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EXERCISE 17

Let a 2D motion given by the mapping

(X) = (X1X2) e1 + X1 + X2

2e2. (19)Determine the region, where the material points cannot correspond to a real physical deforma-tion.

SOLUTION

The deformation of a point can correspond to a real physical deformation if J > 0. Thedeformation gradient for this non-homogenous deformation is

[F] =

X2 X11 2X2

. (20)

Its determinant

J = detF = 2X22 X1. (21)

The non-admissible region is plotted in Figure 3, where the deformation of a square withdimension of 2x2 is also illustrated.

Figure 3: Illustration of the physically non-admissible region

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[E] =

1

2(C I)

=

1

18

11 6 06 17 0

0 0 0

0.611 0.333 00.333 0.944 0

0 0 0

. (33)

EulerAlmansi strain

e = 12

(i c) = eijei ej (34)

[e] =

1

2(i c)

=

1

968

223 27 027 331 0

0 0 0

0.231 0.028 00.028 0.342 0

0 0 0

. (35)

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EXERCISE 19

Let a non-homogeneous 2D deformation of a square with dimension 1 1 be given by thefollowing mapping:

x1 = 32

+ X1 + 32

15

X1 + 32

X22

, x2 = 32

+ X2 32

X21

. (36)

The deformation is illustrated in Figure 4.Determine the eigenvalues and eigenvectors of the left CauchyGreen deformation tensor atmaterial point P0(0.8, 0.2).

Figure 4: Illustration of the motion

SOLUTION

After the deformation, the material point corresponding to P0 has the following spatial coor-dinates:

x1 =3

2+ 0.8 +

3

2

1

50.8 +

3

20.2

2

= 2.6174, (37)

x2 =3

2

+ 0.2

3

2

0.82 = 0.74. (38)

P(2.6174, 0.74)

The deformation gradient is

[F] =

x1X1

x1X2

x2X1

x2X2

=

(1 + 0.12X1 + 0.9X2) (0.9X1 + 6.75X2)

3X1 1

. (39)

Its value corresponding to the material point P0 is

[F] =

1.276 2.072.4 1

. (40)

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The matrix representation of the left CauchyGreen deformation tensor is given by

[b] =FFT

=

5.91308 0.99240.9924 6.76

. (41)

Calculating its eigenvalues:

(5.91308 ) (6.76 ) (0.9924)2 = 0, (42)

1 = 7.4155, 2 = 5.2575. (43)

Eigenvectors are:

[b 1i] [n1] = [0] [n1] = 0.55115

0.8344

, (44)

[b 2i] [n2] = [0] [n2] = 0.83440.55115 . (45)The eigenvectors are illustrated in Figure 5.

Figure 5: Illustration of the eigenvectors ofb

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EXERCISE 20 Let a two-dimensional displacement field be given in the reference configu-ration as

U1 =

2X2, U2 = 2X1, [U] =

2X22X1

. (46)Determine the displacement field in the spatial configuration. Find the coordinates of the spatialdisplacement vector for the material point, which located at point P0 (1, 1) in the referenceconfiguration.

SOLUTION

The deformation mapping is

x1 = X1 + U1 = X1 2X2, x2 = X2 + U2 = X2 + 2X1. (47)

Thus, the inverse deformation mapping is given by

X1 =1

5(x1 + 2x2) , X2 =

1

5(x2 2x1) . (48)

The coordinates of the displacement field in the spatial configuration are computed as

u1 = x1 X1 (x) = 15

(4x1 2x2) , u2 = x2 X2 (x) = 15

(4x2 + 2x1) . (49)

Thus, the spatial displacement field in matrix form is

[u] = 15

4x1 2x24x2 + 2x1

. (50)

The displacement vector U corresponding to the material point P0 (1, 1) is

[UP0] =

22

. (51)

The matrial point P0 moves (according to the motion) to spatial location P(1, 3).Thus, the displacement vector u at point P(1, 3) is

[uP] =15

4 (1) 2(3)4 (3) + 2 (1)

= 1

5

1010

= 2

2

. (52)

The motion is illustrated in Figure 6.

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Figure 6: Illustration of the displacement vector

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EXERCISE 21

Consider a deformation (X) defined in components by

x1 = X2

1, x2 = X

2

3, x3 = X2X3. (1)

Determine the area change and volume change corresponding to material point P0 (2, 3, 1).Compute the ratio da/dA at this point for infinitesimal surface area with initial normal vectorE3.

SOLUTION

The deformation gradient is computed as

[F] =

x1X1

x1X2

x1X3

x2

X1

x2

X2

x2

X3x3X1

x3X2

x3X3

=

2X1 0 00 0 2X30 X3 X2

. (2)

Its determinant is

J = detF = 4X1X23 . (3)The inverse deformation gradient has the form

F

1=

1

2X10 0

0 X2

2X23

1

X3

01

2X30

. (4)

The area change at point P0 is determined according to the Nansons formula:

da = (detF)FTdA, (5)

[da] =

4X1X

2

3

1

2X10 0

0

X22X2

3

1

2X3

01

X30

[dA] , (6)

[da] =

2X

2

30 0

0 2X1X2 2X1X30 4X1X3 0

[dA] , (7)

[daP] =

2 0 00 12 4

0 8 0

[dA] . (8)

The ratio da/dA is given by

da

dA= J

NBN, (9)

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where N= E3 and the Piola deformation tensor is given by

B = F1FT, [B] =

1

4X21

0 0

0X2

2

4X43+

1

X23 X2

4X23

0 X24X2

3

1

4X23

. (10)

Thus it follows that

da

dA= J

NBN = 4X1X23

1

2X3

= 2X1X3 = 4. (11)

The volume change at P is

dvPdVP = detF

P = JP = 8. (12)

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EXERCISE 22

Consider the homogeneous deformation

x1 = X1 + tX2, x2 = X2

tX2, x3 = X3. (13)

Determine the angle of shear for the unit vector pair defined in the reference configuration by

NI =

10

0

, NII =

01

0

. (14)

SOLUTION


[F] =

1 t 00 1 t 0

0 0 1

. (15)

Its inverse transpose is

FT

=

1 0 0t

t 11

t 1 00 0 1

. (16)

The determinant of the deformation gradient is

J = detF = 1 t. (17)Thus, t < 1 has to be satisfied.The right CauchyGreen deformation tensor is given by

[C] =FTF

=

1 t 0t (1 t)2 + t2 0

0 0 1

. (18)

The GreenLagrange strain tensor is computed as

[E] =

1

2(C I)

=

0t

20

t

2t (t 1) 0

0 0 0

. (19)

The Cauchy deformation tensor is

[c] =b

1=F

T

F1

=

1t

t 1 0t

t 1t2 + 1

(t 1)2 00 0 1

. (20)

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The EulerAlmansi strain tensor is given by

[e] = 1

2

(i

c) =

0

t

2 2t 0t

2 2tt

(t 1)2

0

0 0 0

. (21)

Stretch ratios along NI and NII are

NI =NI C NI = 1, (22)

NII =NII C NII =

(1 t)2 + t2. (23)

Unit vectors in the spatial configuration directed along material line element dXI = dS0INIand dXII = dS0IINII, respecively are

nI =1

NIFNI =

10

0

, nII = 1

NIIFNII =

1t2 + (1 t)2

t1 t

0

. (24)

The angle of shear can be computed with the following formulae:

sin = sin

2

= cos =

NICNII

NINII=

2NIENIININII

= 2nIenII =t

t2 + (1 t)2. (25)

Note that in this example the initial angle is 0 =

2

.

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EXERCISE 23

Let a homogenous deformation be given by the following mapping:

x1 = X1 + tX2, x2 = X2

2tX1, x3 = X3. (26)

Determine the change of the angle defined by the material line elements dXI and dXII fort = 1. The direction of these material elements are given by the following unit vectors

NI =

10

0

, NII = 1

2

11

0

. (27)

Determine the values of parameter t > 0, for which the angle between dxI and dxII is themaximum.

SOLUTION


[F] =

1 t 02t 1 0

0 0 1

. (28)

The right CauchyGreen deformation tensor is

[C] = FTF =

1 + 4t2 t 0

t 1 + t2 0

0 0 1

. (29)

Stretch ratios:

NI =NI C NI =

1 + 4t2, (30)

NII =NII C NII =

1 t + 5t

2

2. (31)

The angle change is defined by 0 , where

cos0 = NINII

0 =

4

= 45 (32)

and

cos =NICNII

NINII=

1 + t (4t 1)1 + 4t2

2 t (2 5t) . (33)

For t = 1, its value is

cos|t=1 = 45

36.87. (34)

Consequently

0 = 4 4

5 8.13. (35)

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Figure 1: The evolution of cos and

The evolution of cos and are illustrated in Figure 1.The maximum angle is calculated according to

ddt

(t) = 0 (36)

2

4t2 + 1 3

5t2 2t + 2 = 0 (37)

t =1

2

6 2

0.2247. (38)

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EXERCISE 24

Consider a brick element with initial dimensions 2 3 1. Let the deformed body be given bythe following mapping:

x1 = X1 + X1X2, x2 = X2 X1, x3 = X3. (39)The deformed configuration in the {x1, x2} plane is illustrated in Figure 2.

Figure 2: Illustration of the deformed configuration in the {x1, x2} plane

Determine the change of the angle defined by the material line elements dXI and dXII atpoint P0 (2, 3, 1), where

NI =

10

0

, NII =

01

0

. (40)

SOLUTION


[F] =

1 + X2 X1 01 1 0

0 0 1

. (41)


[C] =FTF

= 1 + (1 + X2)

2 X1 (1 + X2)

1 0

X1 (1 + X2) 1 1 + X21 00 0 1

. (42)

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Stretch ratios:

NI =NI C NI =

1 + (1 + X2)

2, (43)

NII = NII C

NII = 1 + X

21

. (44)

Since 0 = /2, the angle of shear is defined by

sin = sin

2

= cos =

NICNII

NINII=

X1 (1 + X2) 11 + (1 + X2)

2

1 + X21

. (45)

At point P0 (2, 3, 1):

cos|P0 =785

40.6. (46)

Consequently the angle change is

0 = 90 40.6 = 49.4.

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EXERCISE 25

A non-homogeneous deformation is given by the mapping

x1 =

X2, x2 = X2X3, x3 = X3

X1. (47)

Determine the isochoric and volumetric parts of the deformation gradient at material pointP0 (2, 8, 4).

SOLUTION

The deformation gradient has the form

[F] =

0 1 00 X3 X21 0 1

. (48)

The volume change is

J = detF = X2. (49)

The isochoric and volumetric parts ofF are defined by

Fiso = J 1

3F, (50)

[Fiso] =1

3

X2

0 1 00 X3 X2

1 0 1

=

0 13X2

0

0 X33X2

3

X2

2

1

3X2

0 13X2

, (51)

Fvol = J1

3I, (52)

[Fvol] =3

X2

1 0 00 1 0

0 0 1

=

3

X2 0 00 3

X2 0

0 0 3

X2

. (53)

At P0 (2, 8, 4):

[Fiso] =

0 0.5 00 2 4

0.5 0 0.5

, [Fvol] =

2 0 00 2 00 0 2

. (54)

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BMEGEMMMW03 / Continuum Mechanics / Exercises 4 October 2012

EXERCISE 26

Let the deformation gradient be the following:

[F] =

1 1.5 0

0 1 00 0 1

. (1)

Determine the right stretch tensor U, the proper orthogonal tensor R and the left stretchtensor V using the polar decomposition theorem.

SOLUTION

The right CauchyGreen deformation tensor is computed by

[C] = FTF = 1 1.5 0

1.5 3.25 00 0 1

. (2)

Its eigenvalues are obtained as

det(C I) = 0 1 = 4, 2 = 1, 3 = 0.25. (3)

Consequently, the principal stretches are

1 =

1 = 2, 2 =

2 = 1 3 =

3 = 0.5. (4)

The unit eigenvectors ofC (and U) are calculated according to

(C iI) Ni = 0 (5)

N1

=

0.44720.8944

0

, N2

=

00

1

N3

=

0.89440.4472

0

.

The following basis tensors can be defined:

[M1] =N1 N1

=

0.2 0.4 00.4 0.8 0

0 0 0

, (6)

[M2] =N2 N2

=

0 0 00 0 0

0 0 1

, (7)

[M3] =N3 N3

=

0.8 0.4 00.4 0.2 0

0 0 0

. (8)

These basis tensors can be calculated in a different (but equivalent) way, using the followingformulae:

Ma = 1(a b) (a c)

(C bI) (C cI) , (9)

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M1 =1

(1 2) (1 3)(C 2I) (C 3I) (10)

=1

(4 1)(4 0.25) (C 1I) (C 0.25I) (11)

M2 = 1(2 3) (2 1)

(C 3I) (C 1I) (12)

=1

(1 0.25)(1 4) (C 0.25I) (C 4I) (13)

M3 =1

(3 1) (3 2)(C 1I) (C 2I) (14)

=1

(0.25 4)(0.25 1) (C 4I) (C 1I) (15)

Then, the right stretch tensor can be computed using the spectral representation

U = 1M1 + 2M2 + 3M3, (16)

[U] =

0.8 0.6 00.6 1.7 0

0 0 1

, U1 =

1.7 0.6 00.6 0.8 0

0 0 1

. (17)

The proper orthogonal tensor is given by

R = FU1, (18)

[R] =

0.8 0.6 00.6 0.8 0

0 0 1

, detR= 1, R1 = RT. (19)

Finally, the left stretch tensor can be obtained as

V = FR1 = FRT = RURT, (20)

[V] =

1.7 0.6 00.6 0.8 0

0 0 1

. (21)

Calculating V in a different way.The unit eigenvectors ofV are defined as

[n1] =

RN1

=

0.8944

0.44720

, [n2] =

RN2

=

0

01

, [n3] =

RN3

=

0.4472

0.89440

. (22)

Tha basis tensors in the spatial configuration:

[m1] = [n1 n1] =

0.8 0.4 00.4 0.2 0

0 0 0

, (23)

[m2] = [n2 n2] =

0 0 00 0 0

0 0 1

, (24)

[m3] = [n3 n3] = 0.2 0.4 00.4 0.8 0

0 0 0

. (25)

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V can be computed using the spectral representation

V = 1m1 + 2m2 + 3m3, (26)

[V] =

1.7 0.6 00.6 0.8 0

0 0 1

. (27)

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EXERCISE 27

Let the deformation be given with the following mapping:

x1 = X1 + X2, x2 =

X1 + X2, x3 = X3. (28)

Determine: U, R, V using the polar decomposition theorem.

SOLUTION

Deformation gradient:

[F] =

0 0

0 0

, J = 2. (29)

Right CauchyGreen deformation tensor:

[C] =F

TF

=

2

2 0 00 22 00 0 2

. (30)

Eigensystem ofC:

1 = 22, 1 = 2

2, 3 = 2, (31)

N1

=

10

0

, N2

=

01

0

, N3

=

00

1

. (32)

Principal stretches are

1 =

1 =

2, 2 =

2 =

2 3 =

3 = . (33)

The right stretch tensor:

U = 1N1 N1 + 2N2 N2 + 3N3 N3, (34)

[U] =

2 0 0

0

2 00 0

. (35)

The proper orthogonal tensor:

[R] = [F]U1 =

0 0

0 0

12

0 0

012

0

0 01

=1

2

1 1 01 1 0

0 0

2

. (36)

Left stretch tensor:

[V] = [R] [U] RT = 12 + + 0

+ + 0

0 0 2 . (37)

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EXERCISE 29

Compute the generalized material strain measures for the homogenous deformation

[F] = 1 1 02 1 0

0 0 1

. (48)

SOLUTION


[C] =F

TF

=

5 1 01 2 0

0 0 1

. (49)

Eigenvalues ofC

aredet(C I) = 0 (50)

1 =1

2

7 +

13 5.3028, 2 =

1

2

7

13 1.6972, 3 = 1.

Principal stretches are

1 =

1 2.3028, 2 =

2 1.3028 3 =

3 = 1. (51)

Unit eigenvectors ofC are

N1

0.95710.2898

0

, N2

0.28980.9571

0

N3 =

001

. (52)

Generalized material strain measures:

E(n) =

3=1

1

n(n

1) N N. (53)

E(2) =

3=1

1

2

1 1

2

N N,

E

(2)

=

0.3889 0.0556 00.0556 0.2222 0

0 0 0

, (54)

E(1) =

3=1

1 1

N N, E(1) =

0.5378

0.0925 0

0.0925 0.2604 00 0 0

, (55)

E(0) =

3=1

(ln) N N,E

(0)

=

0.7863 0.1580 00.1580 0.3123 0

0 0 0

, (56)

E(1) =

3=1

( 1) N N,E

(1)

=

1.2189 0.2774 00.2774 0.3868 0

0 0 0

, (57)

E(2)

=

3=1

1

2

2 1 N N, E(2) =

2 0.5 00.5 0.5 0

0 0 0

. (58)

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EXERCISE 30

Determine the Eulerian and Lagrangian velocity and acceleration fields, respectively, for the2D deformation

x1 = X1 + (sint) X2, x2 = X2 + tX2. (1)

At instant time t = 1, compute the velocity and acceleration of the material particle, whichinitially occupies the position P0 (1, 1).

SOLUTION

The material velocity field is computed by

V(X, t) = (X, t)

t X fixed =

(X1 + (sint) X2)

t(X2 + tX2)

t

=

X2costX2 . (2)

The material acceleration field is

A (X, t) =2 (X, t)

t2

X fixed

=V(X, t)

t

X fixed

=

(X2cost)

t(X2)

t

=

X2sint

0

. (3)

The inverse motion has the form

X1 = x1 x2sint

1 + t

, X2 =x2

1 + t

. (4)

The Eulerian velocity field is then given by

v (x, t) = V1 (x, t) , t

, [v (x, t)] =

x21 + tcostx2

1 + t

. (5)

The spatial acceleration field is determined by

a (x, t) = A 1 (x, t) , t , [a (x, t)] =

x21 + t

sint

0 . (6)

The velocity and acceleration of material point at t = 1 corresponding to P0 (1, 1) are

[V(XP0, 1)] =

1cos1

1

0.5403

1

, (7)

[A (XP0, 1)] =

1sin1

0

0.8415

0

. (8)

These vector components can be determined using the spatial description. For this reason firstwe need to obtain the spatial coordinates of material point P0 at t = 1:

[xP] =

1 + (sin1) 1

1 + 1 1

=

1 + sin1

2

1.8415

2

. (9)

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[v (xP, 1)] =

2

1 + 1cos1

2

1 + 1

0.5403

1

, (10)

[a (xP, 1)] =

2

1 + 1sin1

0

0.8415

0

. (11)

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EXERCISE 31

Compute the Eulerian and Lagrangian velocity and acceleration fields, respectively, for the 2Ddeformation

x1 = X1ln (e + t) , x2 = X2e2t

. (12)

Compute the velocity and acceleration of material point P0 (100, 2) at instant time t = 1.

SOLUTION

The material velocity field is computed by

V(X, t) = (X, t)

t

X fixed

=

(X1ln (e + t))

t(X2e

t)

t

=

X1

e + t2X2e

2t

. (13)

The material acceleration field is

A (X, t) =2 (X, t)

t2

X fixed

=V(X, t)

t

X fixed

(14)

=

X1

e + t

t

(2X2e2t)

t

=

X1(e + t)2

4X2e2t

. (15)


X1 =x1

ln (e + t), X2 = x2e

2t (16)

The Eulerian velocity field is then given by

v (x, t) = V1 (x, t) , t

, [v (x, t)] =

x1(e + t) ln (e + t)

2x2

. (17)

The spatial acceleration field is determined by

a (x, t) = A1 (x, t) , t

, [a (x, t)] =

x1(e + t)2 ln (e + t)

4x2

. (18)

The velocity and acceleration vector of material point P0 (100, 2) at t = 1 are

[V(XP0, 1)] =

100e + 1

2 2e21

26.89429.556

, (19)

[A (XP0, 1)] = 100

(e + 1)2

4 2e21 7.233

59.112

. (20)

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These vector components can be determined using the spatial description. For this reason firstwe need to compute the spatial coordinates of material point P0 at t = 1.

[xP] = 100ln (e + 1)

2e21 131.32614.778 . (21)


[v (xP, 1)] =

131.326(e + 1) ln (e + 1)

2 14.778

26.894

29.556

, (22)

[a (xP, 1)] =

131.326(e + 1)2 ln (e + 1)

4 14.778

7.233

59.112

. (23)

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EXERCISE 33

Consider the following 2D deformation

x1 = (1 + t) X1, x2 = X2 tX1. (33)

Let the distribution of a physical quantity in the reference configuration is given by

Q (X, t) = t2X1. (34)

Denote q(x, t) the distribution of this field written in the spatial configuration. Detrmine thematerial time derivative of the field q(x, t).

SOLUTION


X1 = x11 + t

, X2 = tx11 + t

+ x2. (35)

The velocity field in the material description is then

[V(X, t)] =

((1 + t) X1)

t(X2 tX1)

t

= X1

X1

. (36)

The field under consideration in the current configuration is computed as

q(x, t) = Q1

(x, t) , t

=

t2x11 + t . (37)

The velocity field in the spatial description is given by

v (x, t) = V1 (x, t) , t

, [v (x, t)] =

x11 + t

x11 + t

. (38)

The gradient of q with respect to the current coordinates is

[gradq(x, t)] = t2

1 + t0 . (39)

The material time derivative of q is then

q =q

t+ gradq v =

t2x1

(1 + t)2+

2tx11 + t

+t2

1 + t

x11 + t

, (40)

q =2tx11 + t

. (41)

The material time derivative of Q has the simple form

Q = Qt

= 2tX1. (42)

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EXERCISE 34

Let the deformation be given by

x1 = X1 + t, x2 = X2 tX1, x3 = X3. (43)

Compute the material time derivative of spatial vector field

[k] =

x2t2x1t

0

. (44)

SOLUTION

The inverse motion is

X1 = x1 t, X2 = x2 + x1t t2, X3 = x3. (45)

The material (Lagrangian) velocity field is

[V] =

1X1

0

. (46)

The spatial (Eulerian) velocity field is

[v] = 1t x1

0

. (47)The material time derivative ofk is computed by

k (x, t) =dk (x, t)

dt=

k (x, t)

t+ gradk v, (48)

where

[gradk] =

k1

x1

k1

x2

k1

x3k2x1

k2x2

k2x3

k3x1

k3x2

k3x3

= 0 t2 0t 0 0

0 0 0

. (49)

Therefore

k (x, t)

=

2x2tx1

0

+

0 t2 0t 0 0

0 0 0

1t x1

0

=

t3 x1t2 + 2x2tt + x1

0

. (50)

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EXERCISE 35

Let the deformation be given by

x1 = (1 + t) X1, x2 = X2 tX3, x3 = X3. (51)

Compute the material time derivative of spatial vector field

[f] =

x1 + x2tx2x3

x1

. (52)

SOLUTION


X1 = x1

(1 + t), X2 = x2 + tx3, X3 = x3. (53)

The material (Lagrangian) velocity field is

[V] =

X1X3

0

. (54)

The spatial (Eulerian) velocity field is

[v] =

x1

(1 + t)x3

0

. (55)The material time derivative off is computed by

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

dt=

f(x, t)

t+ gradf v, (56)

where

[gradf] =

f1x1

f1x2

f1x3

f2x1

f2x2

f2x3

f3x1

f3x2

f3x3

= 1 t 00 x3 x2

1 0 0

. (57)

Therefore

f(x, t) =

x20

0

+

1 t 00 x3 x2

1 0 0

x1(1 + t)x3

0

=

x1(1 + t)

+ x2 tx3

x23

x1(1 + t)

. (58)

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EXERCISE 36

The Lagrangian velocity field is given by

V(X, t) = X2E2 + 2tX1E2. (59)

Determine the motion, inverse motion, Eulerian velocity field and the Eulerian accelerationfield.

SOLUTION

The Lagrangian displacement field is obtained by

U(X, t) =

t0

VX, t

dt = tX2E2 + t

2X1E2. (60)

Thus, the motion is computed as

(X, t) = X+ U(X, t) , [ (X, t)] =

X1 tX2X2 + t2X1

X3

. (61)

The deformation gradient and the volume change are

[F] =

1 t 0t2 1 00 0 1

, J = 1 + t3. (62)


X1 =x1 + tx2

1 + t3, X2 =

x2 t2x1

1 + t3, X3 = x3. (63)

Therefore, the Eulerian velocity field is given by

v (x, t) = V1 (x, t) , t , [v (x, t)] =

t2x1 x2

1 + t3

2tx1 + tx2

1 + t3

0

. (64)

The Lagrangian acceleration field is

A (X, t) =V(X, t)

t= 2X1E2. (65)

The Eulerian acceleration field by definition is

a (x, t) = A

1 (x, t) , t

= 2

x1 + tx21 + t3

e2. (66)

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EXERCISE 37

The spatial displacement field for a 2D motion is given by

u1 = tx1, u2 = tx1 (1 + t) . (67)

Determine the Eulerian velocity field.

SOLUTION

The Eulerian velocity field is the material time derivative of the spatial displacement field.Thus we can write that

v = u =u

t+ gradu v. (68)

Using matrix notation:v1v2

=

u1/tu2/t

+

u1/x1 u1/x2u2/x1 u2/x2

v1v2

. (69)

Therefore, with regard to the unknown components v1 and v2, the above expression forms asystem of linear equations:

v1v2

=

x1

x1 (1 + 2t)

+

t 0

(t + t2) 0

v1v2

. (70)

v1

=

x1

tv1

, (71)v2 = x1 (1 + 2t)

t + t2

v1. (72)

The solutions are

v1 = x1

1 + t, v2 = (1 + t) x1. (73)

Remark: from (68), one can obtain the solution for v as

v = (I gradu)1u

t. (74)

Thus

[v] =

1 + t 0t + t2 1

1

x1x1 (1 + 2t)

, (75)

[v] =

11 + t

0

t 1

x1

x1 (1 + 2t)

=

x11 + t

(1 + t) x1

. (76)

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EXERCISE 38

Suppose that the motion of a continuous body is described by the mapping

x1 = (1 + t) X1, x2 = X2 + tX3, x3 = X3. (77)

Compute the quantities l, d and w.

SOLUTION


X1 =x1

1 + t, X2 = x2 tx3 X3 = x3. (78)

The Lagrangian velocity field is computed as

V(X, t) = (X, t)t

X fixed

[V(X, t)] = X1X3

0

. (79)The Eulerian velocity field is obtained as

v (x, t) = V1 (x, t) , t

[v (x, t)] =

x11 + t

x30

. (80)

Thus, the Eulerian velocity gradient is given by

l = gradv, [l] =

v1x1

v1x2

v1x3

v2x1

v2x2

v2x3

v3x1

v3x2

v3x3

=

1

1 + t0 0

0 0 10 0 0

. (81)

The rate of deformation and the spin tensors are calculated as the symmetric and skew-symmetric parts of l:

d =1

2

l + lT

, [d] =

1

1 + t 0 0

0 01

2

01

20

, (82)

w =1

2

l lT

, [w] =

0 0 0

0 01

2

0 1

20

. (83)

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EXERCISE 39

The Eulerian velocity field is given by its coordinates as

v1 = 3t2, v2 = x1 + tx2, x3 = tx3. (84)

Determine the Eulerian acceleration field.

SOLUTION

The Eulerian acceleration field is computed as the material time derivative ofv:

a = v =v

t+ (gradv)v =

v

t+ lv, (85)

where

[l] = 0 0 01 t 0

0 0 t

, vt

= 6tx2x3

. (86)Therefore, the Eulerian acceleration filed is given by

a =

6tx2x3

+

0 0 01 t 0

0 0 t

3t2x1 + tx2

tx3

=

6tx2 + 3t2 + t (x1 + tx2)

x3 + t2x3

. (87)

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BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012

EXERCISE 40

Let a homogenous deformation be given by

x1 = (1 + t) X1, x2 = X2 + tX1, x3 = X3. (1)

Compute the material time derivatives of stretch ratio corresponding to the material line ele-ment, which has the initial orientation

[N] =1

5

12

0

(2)

in the reference configuration.

SOLUTION


[F] =

1 + t 0 0t 1 0

0 0 1

. (3)

Right CauchyGreen deformation tensor is computed as

[C] =FTF

=

t2 + (1 + t)2 t 0

t 1 00 0 1

. (4)

The stretch ratio is calculated by the formula

n N =NCN =

5 + 2t (3 + t)

5. (5)

The spatial direction of material line element having oriantation N in the reference configura-tion is obtained by

n =1

nFN, [n] =

1

5 + 2t (3 + t)

1 + t2 + t

0

. (6)


X1 =x1

1 + t, X2 = x2

tx1

1 + t, X3 = x3. (7)

Lagrangian and Eulerian velocity fields:

[V(X, t)] =

X1X1

0

, [v (x, t)] =

x1

1 + tx1

1 + t0

. (8)

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Spatial (Eulerian) velocity gradient is computed as

[l] =

v1

x1

v1

x2

v1

x3v2

x1

v2

x2

v2

x3v3

x1

v3

x2

v3

x3

=

1

1 + t0 0

1

1 + t 0 00 0 0

. (9)

The symmetric part is the rate of deformation:

[d] =1

2[l] +

1

2

lT

=

1

1 + t

1

2 (1 + t)0

1

2 (1 + t)0 0

0 0 0

. (10)

Material time derivatives of stretch ratio:

n = n (ndn) =3 + 2t

5

5 + 2t (3 + t). (11)

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EXERCISE 41

A certain motion is described with the following mapping:

x1 = 2X1 + tX2, x2 = X2

X1, x3 = 3tX3. (12)

Determine J.

SOLUTION

The material time derivative of J is computed as

J = Jtrd = Jtrl. (13)

The deformation gardient and the volume change are:

[F] = 2 t 01 1 0

0 0 3t

, J = detF = 3t2 + 6t. (14)


X1 =x1 tx2

2 + t, X2 =

x1 + 2x22 + t

, X3 =x3

3t. (15)

The Lagrangian and the Eulerian velocity fields are

[V] = X20

3X3

, [v] = x1 + 2x2

2 +t

0x3

t

. (16)

The spatial velocity gradient:

[l] =

1

2 + t

2

2 + t0

0 0 0

0 01

t

, trl =

1

t+

1

2 + t. (17)

Therefore

J = Jtrl =

3t2 + 6t1

t+

1

2 + t

= 6 (1 + t) . (18)

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EXERCISE 42

Prove that the rate of deformation tensor corresponding to the motion

x1 = (t) X1, x2 = (t) X2, x3 = (t) X3 (19)

has the form

d =

i. (20)

SOLUTION


X1 =x1

(t)

, X2 =x2

(t)

, X3 =x3

(t)

. (21)

The velocity fields are

[V(X, t)] =

X1X2

X3

, [v (x, t)] =

x1

x2

x3

. (22)

Spatial velocity gradient is computed as

[l] =

v1

x1

v1

x2

v1

x3v2

x1

v2

x2

v2

x3v3

x1

v3

x2

v3

x3

=

0 0

0

0

0 0

. (23)

Its symmetric part is the rate of deformation:

d = 12l + lT

, [d] =

0 0

0 0

0 0

. (24)

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EXERCISE 43

A 2D motion of a continuum medium is given by

(X, t) = X1 t2 e1 + e

X2

t e2. (25)

Determine the material time derivative of the left CauchyGreen deformation tensor.

SOLUTION

Using the rule l = FF1 we can write

b =

FFT

= F FT + FF

T

= lF FT + FFTlT = lb + blT. (26)


X1 =x1

t2, X2 = ln

x2

t. (27)

The Lagrangian and Eulerian velocity fields, respectively:

[V] =

2tX1eX2

, [v] =

2x1tx2

t

. (28)


[l] =

2

t0

0 1t

. (29)

The deformation gradient and the left CauchyGreen deformation tensor:

[F] =

t2 00 eX2 t

, [b] =

t4 00 x2

2

. (30)

Substituting (29) and (30) into (26) we have

b =

4t3 0

02x2

2

t

. (31)

This result can be obtained by taking the material time derivative of b such as

b =b

t+ gradb v, (32)

where the third-order tensor gradb is filled out with 0 elements, except the (gradb)222

=b22

x2=

2x2. Therefore

b= 4t

3 0

0 0

+

4t3 0

0 2x2

2

t

=

4t3 0

0 2x2

2

t .

(33)

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EXERCISE 44

The Eulerian velocity field is defined by the equations,

v1 = tx1sinx3, v2 = 4tx2cosx3, v3 = 0. (34)

At spatial point P(2, 1, 0), at time t = 1, determine d, w, the stretch rate per unit length(stretching) along the direction [n] =

13

[1,1, 1] and the maximum stretching and the direc-tion in which it occurs.

SOLUTION

The spatial velocity gradient and its values at spatial point P at t = 1:

[l] =

tsinx3 0 tx1cosx30 4tcosx3

4tx2sinx3

0 0 0

, [l] =

0 0 20 4 00 0 0

. (35)

The rate of deformation tensor and the vorticity (spin) tensor are

[d] =

0 0 10 4 01 0 0

, [w] =

0 0 10 0 0

1 0 0

. (36)

The stretch rate per unit length in the direction n defines the stretching

n = n (ndn) =n

n = ndn =

2

3 . (37)

The maximum values ofn

nis the largest eigenvalue ofd. The direction in which it occurs is

the corresponding (unit) eigenvector. The eigenvalues and eigenvectors ofd are

1 = 4, 2 = 1, 3 = 1,

n1 = e2, n2 =1

2(e1 + e3) , n3 =

12

(e1 + e3) .

Consequently, the direction ofe2 designates the particular direction in which the stretching isthe maximum.

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EXERCISE 45

Prove that the spatial velocity field

[v] = 3x3 6x2

6x1 2x32x2 3x1

(38)

corresponds to a rigid body rotation. Determine the direction of the axis of spin.

SOLUTION


[l] =

0 6 36 0 23 2 0

. (39)

Since it is a skew-symmetric tensor, it follows that w = l and d = 0. The axis of spin is definedby the corresponding axial vector

[] =

w32w13

w21

=

23

6

. (40)

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EXERCISE 46

A steady spatial velocity field is given by

[v] = 2x2

2x2

2

x1x2x3

. (41)

Determine the stretching at P(2, 2, 2) along the direction n =1

2(e1 + e2).

SOLUTION

The velocity gradient and its values at point P:

[l] =

0 2 0

0 4x2 0x2x3 x1x3 x1x2

, [l] =

0 2 0

0 8 04 4 4

. (42)

The rate of deformation tensor:

d =1

2

l + lT

, [d] =

0 1 21 8 2

2 2 4

. (43)

Thus, the stretching is

n

n = ndn = 5. (44)

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EXERCISE 47

Let the components of the Cauchy stress tensor at a certain point be given in matrix form by

[] =

200 300 100300 0 0100 0 100

. (1)

Determine the components of the Cauchy traction vector tn and the length of the vector alongthe normal to the plane that passes through this point and that is parallel to the plane definedby (x1, x2, x3) x1 3x2 + 2x3 = 1. Calculate |tn| and the angle between tn and the normalof the plane.

SOLUTION

The unit normal vector of the plane is computed as

n = grad|grad| = || , (2)

where

[] =

x1

x2

x3

=

13

2

, || = 14. (3)

Therefore the unit normal of the plane is

[n] =

1

14 1

32

. (4)

The Cauchy traction vector by definition is computed according to

tn = n, [tn] =114

900300

100

240.53580.1784

26.7261

. (5)

The length of the traction vector and its normal component are the following:

|tn| =tntn =

65000 254.951,

n = tnn = 8007 114.2857. (6)

The length of the component parallel to the plane is

n =

|tn|2 2n =

50

7

1018 227.901. (7)

The angle between tn and the normal of the plane is calculated by

tnn = |tn| |n| cos = |tn| cos. (8)Thus

= arccostnn

|tn| = arccosn|tn| = arccos (0.448265) = 2.03562 = 116.632

.

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EXERCISE 48

Let a non-homogenous stress field be given by the components of the Cauchy stress tensor:

[] = 2x2x3 4x

2

20

4x2

2 0 2x10 2x1 0

. (9)

Determine the traction vector corresponding to the plane (x1, x2, x3) x21 + x22 + x3 = 1.5 atpoint P(0.5, 1.5,1).

SOLUTION

The components of the stress tensor at point P are

[] =3 9 09 0 10 1 0

. (10)

The unit normal vector of the plane at point P is computed as

n =grad

|grad| =|| , (11)

where

[

] =

x1

x2

x3

=

2x12x2

1 , []P =

131 . (12)

Therefore

[n] =111

13

1

. (13)

The Cauchy traction vector at this point is

tn = n, [tn] = 111

2483

7.232.410.90

. (14)

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EXERCISE 49

Let the components of the Cauchy stress tensor at a certain point be given in matrix form by

[] = 4 3 0

3 4 00 0 9

. (15)

Compute the spherical and deviatoric components of the stress tensor. Determine the principalstresses. Compute the components of the stress tensor in the Cartesian coordinate system givenby the (unit) basis vectors

e1 = e3, e2 =110

(3e1 + e2) , e3 =110

(e1 + 3e2) . (16)

SOLUTION

The spherical (or hydrostatic) component is computed by

p =1

3(tr) i = 3i, (17)

whereas the deviatoric part is determined as

s = p, [s] = 4 3 03 4 0

0 0 9

3 0 00 3 0

0 0 3

=

1 3 03 7 0

0 0 6

. (18)

One of the principal stresses is 9 because the traction vector on the plane with normal vectore3 is normal to the plane. The two remaining principal stresses are obtained by solving thefollowing quadratic equation

det

4 33 4

1 00 1

= 0, (19)

2 25 = 0. (20)Therefore the principal stresses are

1 = 9, 2 = 5, 3 = 5. (21)In order to obtain the components of in the new coordinate system, first we consruct theorthogonal transformation matrix as

[Q] = [e1, e2, e3] =110

0 3 10 1 3

10 0 0

. (22)

The components of the stress tensor in the new coordinate system is calculated by

[] =QT

[] [Q] = 1

10

0 0

103 1 01 3 0

4 3 03 4 00 0 9

0 3 10 1 310 0 0

, (23)

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[] =1

10

0 0 9

10

15 5 05 15 0

0 3 10 1 3

10 0 0

=

9 0 00 5 0

0 0 5

. (24)

Since [] is diagonal, it follows that e1, e2 and e3 are the unit eigenvectors of.

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EXERCISE 50

Let a homogenouos deformation be given by the mapping

x1 =

X2, x2 = 2X1, x3 = 2X3. (25)

Determine the first and the second Piola-Kirchhoff stress tensors, respectively, if the Cauchystress is given by

[] =

0 10 010 20 0

0 0 10

. (26)

Determine the first Piola-Kirchhoff traction vector for the plane which has the normal vector

[n] = 1

00

(27)

in the current configuration.

SOLUTION


[F] =

0 1 02 0 00 0 2

, J = detF = 4. (28)

Its inverse is

F1

=

0 0.5 01 0 0

0 0 0.5

. (29)

The first and the second Piola-Kirchhoff tensors are

P = JFT, S= JF1FT (30)

[P] =

20 0 040

40 0

0 0 20

, [S] =

20 20 0

20 0 0

0 0 10

. (31)

The normal vector of the plane in the reference configuration is obtained by

da = JFTdA (32)

dan = JdAFTN (33)

dAN= da1

JFTn N =

1

JFTn 1

JFTn

= FTnFTn , (34)

where

FTn = 01

0

N = 010

. (35)

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Therefore

TN = PN, [TN] =

0400

. (36)

The Cauchy traction vector is

tn = n, [tn] =

010

0

. (37)

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EXERCISE 51

Let the components of the first Piola-Kirchhoff tensor be given by

[P] = 20 0 0120 80 40

0 20 0

. (38)

The inverse motion is known as

X1 = x1 3x3, X2 = 1k

x2, X3 = x3. (39)

Determine the value of parameter k if the first scalar invariant of the Cauchy stress tensor is70. Compute the components of the Cauchy stress tensor.

SOLUTIONThe motion is

x1 = X1 + 3X3, x2 = kX2, x3 = X3. (40)

The deformation gradient has the form

[F] =

1 0 30 k 0

0 0 1

, J = detF = k. (41)

The Cauchy stress tensor by definition is

=1

JPFT, [] =

20/k 0 00 80 40/k

0 20 0

. (42)

Its first scalar invariant is

I = tr = 80 20k

k = 2080 I. (43)

If I = 70 then

k =20

80 70 = 2. (44)

The Cauchy stress tensor in this case is therefore

[] =

10 0 00 80 20

0 20 0

. (45)

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EXERCISE 52

Prove that the JaumannZaremba rate of the Cauchy stress is an objective Eulerian second-order tensor field.

SOLUTION

The Jaumann-Zaremba rate of the Cauchy stress tensor is defined as

JZ = w + w. (1)

The JZ-rate of is objective if the following condition is satisfied:

JZ = QJZ

QT. (2)

Expanding the left-hand side yields:

JZ

= w + w, (3)

where

=

QQT

= QQT +QQT +QQT

, (4)

w = QQT + QwQT. (5)

Thus

JZ = QQT +QQT +QQT

(6)

QQT +QwQTQQT + QQT QQT +QwQT (7)= QQT +QQT +QQ

T

(8)

QQTQQT QwQTQQT +QQTQQT +QQTQwQT. (9)

Using the identity QQT = I, we can simplify the result above as

JZ = QQT +QQT +QQT

(10)

QQT QwQT +QQTQQT +QwQT (11)

= QQT +QQT

QwQT +QQTQQT +QwQT. (12)

Since QQT = is a skew-symmetric tensor, it follows that T =

. Thus,QQT

T=

QQT

= QQT. Using this relation in the fourth term yields

JZ = QQT +QQT

QwQT QQTQQT

+QwQT (13)

= QQT +QQT

QwQT QQT

+QwQT (14)

= QQT QwQT +QwQT (15)

Therefore

JZ = Q ( w + w)QT. (16)

Thus, (2) is satisfied.

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EXERCISE 53

Prove that the GreenNaghdi rate of the Cauchy stress is an objective Eulerian second-ordertensor field.

SOLUTION

The GreenNaghdi rate of the Cauchy stress tensor is defined as

GN = RRT

+

RRT

. (17)

The GN-rate of is objective if the following condition is satisfied:

GN = QGN

QT. (18)

Expanding the left-hand side yields:

GN =

RR

T +

RR

T

, (19)

where =

QQT

= QQT +QQT +QQ

T

, (20)

R = QR, (21)R = QR+QR. (22)

Thus

GN = QQT +QQT +QQT

(23)

QR+QR

(QR)TQQT (24)

+QQT QR+QR

(QR)

T

(25)

= QQT +QQT +QQT

(26)

QRRTQT +QRRTQT

QQT (27)

+QQTQRRTQT +QRRTQT

(28)

= QQT +QQT +QQT

(29)

QQTQQT +QRRTQTQQT

(30)

+QQTQQT +QQTQRRTQT (31)= QQT +QQT +QQ

T

(32)

QQT +QRRTQT

+QQTQQT +QRRTQT

(33)

= QQT +QQT

QRRTQT +QQTQQT +QRRTQT. (34)

Since QQT = is a skew-symmetric tensor, it follows that T = . Thus,QQT

T=

QQT

= QQT. Using this relation in the fourth term yields

GN = QQT +QQT

QRRTQT QQTQQT

+ QRRTQT (35)

= QQT +QQTQRRTQT

QQT +QRRTQT (36)

= QQT QRRTQT +QRRTQT. (37)

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Therefore

GN = Q

RRT

+

RRT

QT. (38)

Thus, (18) is satisfied.

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EXERCISE 54

Let the deformation of a material body is described by the mapping

x1 = X1 + tX2, x2 = tX2, x3 = X3. (1)

At a particular material point the Cauchy stress tensor is given by its matrix as

[] =

10 0 100 30 010 0 0

. (2)

Prove the identity P : F =1

2S : C.

SOLUTION

We can rewrite the stress power (per unit reference volume) as

P : F = trPTF

= tr

PTFTFTF

= tr

F1P

TFTF

(3)

= trST

FTF

= tr

SFTF

. (4)

Since S is symmetric and C= FTF+ FTF, we can simplify the result above as

P : F = S :FTF

=

1

2S : C. (5)

Proof with numerical calculations. The deformation gradient, its inverse and its determinant:

[F] =

1 t 00 t 0

0 0 1

, F1 =

1 1 0

01

t0

0 0 1

, J = detF = t. (6)

The right CauchyGreen deformation tensor and its material time derivative:

[C] =FTF

=

1 t 0t 2t2 0

0 0 1

,

C

=

C

t

=

1 1 01 4t 0

0 0 0

. (7)

The inverse motion:

X1 = x1 x2, X2 =x2

t, X3 = x3. (8)

The Lagrangian and Eulerian velocity fields are

[V] =

X2X20

, [v] =

x2

tx2

t

0

. (9)

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The spatial velocity gradient and the material time derivative of the deformation gradient(which is the material velocity gradient):

[l] =

01

t0

0 1t

0

0 0 0

, F = [lF] =

0 1 0

0 1 00 0 0

. (10)

The first and the second PiolaKirchhoff stress tensors are

[P] =

JFT

=

10t 0 10t30t 30 010t 0 0

, (11)

[S] = F1P =

40t 30 10t

3030

t

0

10t 0 0

. (12)

Therefore we can prove that

P : F = 30, (13)1

2S : C= 30. (14)

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EXERCISE 55

Let the Cauchy stress distribution in a bricket domain with dimension

x1 = 0 . . . 2, x2 = 0 . . . 3, x3 = 0 . . . 4 (15)

be given as

[] =

5x

21 4 3x2

4 x1x2x3 x33x2 x3 0

. (16)

Prove the Gauss divergence theorem.

SOLUTION

The Gauss divergence theorem for this problem is written as(a)

nda =

(v)

divdv, (17)

(a)

tda =

(v)

divdv, (18)

where n is the outward unit normal field acting along the surface (a), while da and dv areinfinitesimal spatial surface and volume elements at x, respectively, whereas t denotes the

surface Cauchy traction vector atx

.The whole surface is composed of six surfaces with domains and unit normals

a1 : = {x2 = 0 . . . 3; x3 = 0 . . . 4} , (19)

a2 : = {x1 = 0 . . . 2; x3 = 0 . . . 4} , (20)

a3 : = {x1 = 0 . . . 2; x2 = 0 . . . 3} , (21)

a4 = a1, a5 = a2, a6 = a3, (22)

n1 = e1, n2 = e2, n3 = e3, n4 = e1, n5 = e2, n6 = e3. (23)

The surface traction vectors corresponding to the surfaces of the bricket domain are

t1 = (n1) |x1=2, [t1] =

2043x2

, t2 = (n2) |x2=3, [t2] =

43x1x3

x3

, (24)

t3 = (n3) |x3=4, [t3] =

3x24

0

, t4 = (n4) |x1=0, [t4] =

04

3x2

, (25)

t5 = (n5) |x2=0, [t5] =

40x3

, t6 = (n6) |x3=0, [t2] =

3x20

0

. (26)

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With this in hand, we can compute the left-hand side of (17):

(a)

nda =

40

30

t1dx2dx3 +

40

20

t2dx1dx3 +

30

20

t3dx1dx2 (27)

+

40

30

t4dx2dx3 +

40

20

t5dx1dx3 +

30

20

t6dx1dx2, (28)

(a)

nda

=

2404854

+

3248

16

+

2724

0

+

048

54

+

32016

+

270

0

, (29)

(a)

nda = 240e1 + 72e2 . (30)

In order to calculate the right-hand side in (17), first the divergence of the Cauchy stress needto be computed:

div = =

xc ec =

ab

xcea (eb ec) =

ab

xcbcea = ab,cbcea = ac,cea, (31)

[div] =

11

x1+

12

x2+

13

x321

x1+

22

x2+

23

x331

x1 +32

x2 +33

x3

=

10x11 + x1x3

0

. (32)

The right-hand side of (17) thus becomes:

(v)

divdv =

40

30

20

divdx1dx2dx3, (33)

(v)

divdv

=

4

0

3

0

20

2 + 2x3

0

dx2dx3 =

4

0

60

6 + 6x3

0

dx3 =

24072

0

, (34)

(v)

divdv = 240e1 + 72e2 . (35)

Results (30) and (35) are equal, thus we have proven relation (17).

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EXERCISE 56

Let a vector field be given as u = x1x2 (e1 + e2 + e3). Prove the Stokes theorem for therectangular plane surface defined as

a := {x1 = 0 . . . 2; x2 = 0 . . . 1} . (36)

SOLUTION

The Stokes theorem:c

uds =

(a)

(curlu)nda. (37)

The closed curve c is consist of four line elements:

c1 : = {x1 = 0 . . . 2} , c2 := {x2 = 0 . . . 1} , (38)

c3 : = {x1 = 0 . . . 2} , c4 := {x2 = 0 . . . 1} (39)

with infinitesimal tangent vectors:

ds1 = dx1e1 ds2 = dx2e2 ds3 = dx1e1 ds4 = dx2e2. (40)

The left-hand side of (37) is computed with the help of the following four integrals:

c

uds =

2

0

(uds1)|x2=0 +

1

0

(uds2)|x1=2 +

2

0

(uds3)|x2=1 +

1

0

(uds4)|x1=0 , (41)

c

uds =

20

(0) dx1 +

10

(2x2) dx2 +

20

(x1) dx1 +

10

(0) dx2, (42)

c

uds = 0 + 1 2 + 0, (43)

c

uds = 1 . (44)

In order to calculate the right-hand side in (37) first we need to compute the curl of the vectorfield u:

curlu = u = ea u

xa=

ub

xaea eb =

ub

xaabcec = ub,aabcec. (45)

[curlu] =

u3

x2

u2

x1u1

x3

u3

x1

u2x1

u1x2

=

x1x2

x2 x1

. (46)

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EXERCISE 57

Consider the motion

x1 = 1 +t

kX1, x2 = 21 +

t

kX2, x3 = 31 +

t

kX3, (50)

where k is a constant. From the conservation of mass and the initial condition (t = 0) = 0,determine as a function of 0, t, and k.

SOLUTION


[F] =

1 +

t

k

1 0 00 2 00 0 3

, J = detF = 6

(k + t)3

k3. (51)

Since0 = J it follows that

=0

J=

0k3

6 (k + t)3(52)

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EXERCISE 58

Consider the motion of a body described by the mapping

x1 = X1, x2 =X2

1 tX2, x3 = X3. (53)

Determine the material density as a function of spatial position vector x and time t withoutdirect use ofF.

SOLUTION

The inverse mapping:

X1 = x1, X2 =x2

1 + tx2, X3 = x3. (54)

The velocity field is

V(X, t) =X22

(1 tX2)2E2, v (x, t) = x

22e2. (55)

From the continuity equation we can write

+ divv = 0 d

dt= divv = 2x2, (56)

1

d = 2x2dt = 2

X2

1 tX2dt, (57)

0

1

d = 2

t0

X2

1 tX2dt, ln ln0 = 2ln (1 tX2) 2ln1 (58)

ln

0= ln(1 tX2)

2 (X, t) = 0 (1 tX2)2

. (59)

Using the inverse mapping we have

(x, t) =0

(1 + tx2)2 . (60)

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EXERCISE 59

The Eulerian velocity field of a plane motion is given by the components

v1 = (x1 x2) t, v2 = x1 x2, v3 = 0, (61)

where is a positive constant. Assume that the spatial mass density does not depend onthe current coordinates, so that grad = 0. Express so that the continuity mass equation issatisfied.

SOLUTION

The continuity mass equation in the spatial description can be written as

+ trl =

t+ (grad) v + trl = 0. (62)

In this example grad = 0, therefore equation (62) reduces to

t+ trl = 0. (63)

The spatial velocity gradient and its trace:

[l] =

t t 01 0

0 0 0

, trl = (1 t) . (64)

By integrationg (63) we arrive at0

1

d =

t0

(1 t) dt, ln

0=

t

t2

2

, (65)

(t) = 0 et t

2

2

. (66)

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EXERCISE 60

The velocity field of an incompressibleflow be given as

v =x1

r2e1 +

x2

r2e2, (67)

where r2 = x21 + x22. Does this velocity field satisfy the continuity equation?

SOLUTION

The continuity equation for incompressible material reduces to

+ divv = 0 = constant

divv = 0. (68)

[divv] =v1

x1

+v2

x2

= 1

x

2

1 + x

2

2

2x21

(x2

1 + x2

2)

2 + 1

x

2

1 + x

2

2

2x22

(x2

1 + x2

2)

2 = 0. (69)Therefore, it does.

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EXERCISE 61

A dynamical process is given by the deformation mapping

x1 = etX1 e

tX2, x2 = etX1 + e

tX2, x3 = X3, (70)

with the Cauchy stress distribution

[] =

x

21 x1x3 0

x1x3 x2 00 0 x33

. (71)

Determine the body force vector q so that Cauchys first equation of motion is satisfied.

SOLUTION

The Cauchys first equation of motion:

div + q = a q = a div. (72)

In order to calculate q, we need to compute the quantities , a and div.The inverse motion is

X1 =1

2et (x2 + x1) , X2 =

1

2et (x2 x1) , X3 = x3. (73)

The deformation gradient:

[F] =

et et 0et et 0

0 0 1

J = 2 =

1

20. (74)

The Lagrangian velocity and acceleration fields:

[V(X, t)] =

e

tX1 + etX2

etX1 etX2

0

[A (X, t)] =

e

tX1 etX2

etX1 + etX2

0

. (75)

The Eulerian acceleration field:

[a (x, t)] =

x1x2

0

. (76)

The divergence of the Cauchy stress:11

+12

+13

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Kossa ContMech Problems 2012 Fall

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