Continuum Mech Problems

BMEGEMMMW03 / Continuum Mechanics / Exercises

Continuum Mechanics

EXAMPLES

by Dr. Attila KOSSA

http://www.mm.bme.hu/˜kossa

Budapest University of Technology and EconomicsDepartment of Applied Mechanics

DRAFT version under continuous preparation

Any kind of comments are welcome at

[email protected]

Version (year-month-day):2014 - 09 - 10

11 : 40

1


EXERCISE 1

The following two vectors are given:

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

[a] =

72−5

, [b] =

−364

. (2)

Compute the following quantities: a · b, a × b, ‖a‖ , ea (unit vector in the direction ofa), ψ (the angle between a and b), a⊗ b.

SOLUTION

a · b = 7 · (−3) + 2 · 6 + (−5) · 4 = −29, (3)

a× b =

∣∣∣∣∣∣

e1 e2 e3

7 2 −5−3 6 4

∣∣∣∣∣∣

(4)

= +e1 (2 · 4− (−5) · 6)− e2 (7 · 4− (−5) · (−3)) + e3 (7 · 6− 2 · (−3)) (5)

= 38e1 − 13e2 + 48e3, (6)

‖a‖ =√a · a =

√

72 + 22 + (−5)2 =√78 ≈ 8.83176, (7)

‖b‖ =√b · b =

√

(−3)2 + 62 + 42 =√61 ≈ 7.81025, (8)

ea =a

‖a‖ =7√78

e1 +2√78

e2 −5√78

e3, [ea] ≈

0.7925940.226455−0.566139

, (9)

a · b = ‖a‖ ‖b‖ cosψ (10)

ψ = arccos

(a · b

‖a‖ ‖b‖

)

= arccos

( −29√78 ·

√61

)

= 2.00471rad = 114.861, (11)

a⊗ b = a1b1e1 ⊗ e1 + a1b2e1 ⊗ e2 + a1b3e1 ⊗ e3

+a2b1e2 ⊗ e1 + a2b2e2 ⊗ e2 + a2b3e2 ⊗ e3

+a3b1e3 ⊗ e1 + a3b2e3 ⊗ e2 + a3b3e3 ⊗ e3, (12)

where

[e1 ⊗ e1] =

1 0 00 0 00 0 0

, [e1 ⊗ e2] =

0 1 00 0 00 0 0

, [e1 ⊗ e3] =

0 0 10 0 00 0 0

, (13)

[e2 ⊗ e1] =

0 0 01 0 00 0 0

, [e2 ⊗ e2] =

0 0 00 1 00 0 0

, [e2 ⊗ e3] =

0 0 00 0 10 0 0

, (14)

[e3 ⊗ e1] =

0 0 00 0 01 0 0

, [e3 ⊗ e2] =

0 0 00 0 00 1 0

, [e3 ⊗ e3] =

0 0 00 0 00 0 1

, (15)

2


[a⊗ b] =

−21 42 28−6 12 815 −30 −20

. (16)

3


EXERCISE 2

The following two tensors are given:

[A] =

7 −9 83 −2 55 1 6

, [B] =

2 −6 5−6 −1 25 2 9

. (17)

Compute the following quantities:(a) A : B; (b) det (A ·B); (c) B−1; (d) ‖A‖; (e) IB, IIB, IIIB; (f) symmetric and skewsym-metric parts of A; (g) eigenvalues and eigenvectors of B; (h) spectral representation of B.

SOLUTION

(a)

A : B = A11 · B11 + A12 · B12 + A13 · B13 + A21 · B21 + A22 ·B22 + A23 · B23

+A31 · B31 + A32 · B32 + A33 · B33, (18)

A : B = 7 · 2 + (−9) · (−6) + 8 · 5 + 3 · (−6) + (−2) · (−1) + 5 · 2+5 · 5 + 1 · 2 + 6 · 9, (19)

A : B = 183 , (20)

or using the identity

A : B = tr[AT ·B

]= tr

[A ·BT

]= tr

108 −17 8943 −6 5634 −19 81

= 108 + (−6) + 81 = 183. (21)

(b)

det [A ·B] =

∣∣∣∣∣∣

108 −17 8943 −6 5634 −19 81

∣∣∣∣∣∣

(22)

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

+ (89) [(43) (−19)− (−6) (34)] , (23)

det [A ·B] = 34710 . (24)

(c)

B−1 =adj [B]

det [B], (25)

4


where

[adj [B]] =

∣∣∣∣

−1 22 9

∣∣∣∣

−∣∣∣∣

−6 25 9

∣∣∣∣

∣∣∣∣

−6 −15 2

∣∣∣∣

−∣∣∣∣

−6 52 9

∣∣∣∣

∣∣∣∣

2 55 9

∣∣∣∣

−∣∣∣∣

2 −65 2

∣∣∣∣

∣∣∣∣

−6 5−1 2

∣∣∣∣

−∣∣∣∣

2 5−6 2

∣∣∣∣

∣∣∣∣

2 −6−6 −1

∣∣∣∣

T

(26)

=

∣∣∣∣

−1 22 9

∣∣∣∣

−∣∣∣∣

−6 52 9

∣∣∣∣

∣∣∣∣

−6 5−1 2

∣∣∣∣

−∣∣∣∣

−6 25 9

∣∣∣∣

∣∣∣∣

2 55 9

∣∣∣∣

−∣∣∣∣

2 5−6 2

∣∣∣∣

∣∣∣∣

−6 −15 2

∣∣∣∣

−∣∣∣∣

2 −65 2

∣∣∣∣

∣∣∣∣

2 −6−6 −1

∣∣∣∣

=

−13 64 −764 −7 −34−7 −34 −38

(27)

and

det [B] =

∣∣∣∣∣∣

2 −6 5−6 −1 25 2 9

∣∣∣∣∣∣

= −445. (28)

Therefore

[B−1

]=

1

(−445)

−13 64 −764 −7 −34−7 −34 −38

≈

0.0292135 −0.14382 0.0157303−0.14382 0.0157303 0.07640450.0157303 0.0764045 0.0853933

.

(d)

‖A‖ =√A : A =

√

tr[A ·AT

]=

√

tr[AT ·A

]≈ 17.1464282. (29)

(e)

IB = trB = 2− 1 + 9 = 10, (30)

IIB =1

2

(I2B − tr

[B2

])=

1

2

102 − tr

65 4 434 41 −1443 −14 110

= −58, (31)

IIIB = det [B] =

∣∣∣∣∣∣

2 −6 5−6 −1 25 2 9

∣∣∣∣∣∣

= −445.

(f)

Asymm =1

2

(A+AT

), [Asymm] =

7 −3 6.5−3 −2 36.5 3 6

, (32)

Askew =1

2

(A−AT

), [Askew] =

0 −6 1.56 0 2

−1.5 −2 0

. (33)

5


(g) The Cardano’s 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 =2

27I3B − 1

3IBIIB + IIIB =

−4795

27≈ −177.593, (35)

λi = 2P cos

[1

3arccos

[D

2P 3

]

+ (i− 1)2π

3

]

+1

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

λ1 = 11.7075, λ2 = −7.0778, λ3 = 5.3703. (37)

Eigenvector corresponding to λ1 is calculated by

[B − λ1δ]n1 = 0, n1 = n11e1 + n12e2 + n13e3, ‖n1‖ = 1, (38)

−9.7075 −6 5−6 −12.7075 25 2 −2.7075

n11

n12

n13

=

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

6n11

)

= −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]T =

[0.5054 −0.1038 0.8566

]. (50)

6


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

[n2]T =

[−0.6331 −0.7192 0.2863

], [n3]

T =[−0.5864 0.6870 0.4292

]. (51)

(h)

B = λ1n1 ⊗ n1 + λ2n2 ⊗ n2 + λ3n3 ⊗ n3, (52)

where

[n1 ⊗ n1] =

0.255408 −0.0524574 0.432924−0.0524574 0.010774 −0.08891670.432924 −0.0889167 0.733818

, (53)

[n2 ⊗ n2] =

0.400774 0.455296 −0.1812730.455296 0.517235 −0.205933−0.181273 −0.205933 0.0819908

, (54)

[n3 ⊗ n3] =

0.343818 −0.402839 −0.251651−0.402839 0.471991 0.29485−0.251651 0.29485 0.184191

. (55)

7


EXERCISE 3

Let C be a symmetric 3x3 tensor of the second rank:

[C] =

33 −17 10−17 22 510 5 38

(56)

with eigenvalues and eigenvectors

λ1 = 48.6329λ2 = 37.9255λ2 = 6.44156

, [n1] =

0.755983−0.3828720.530942

, [n2] =

−0.2787090.5456420.790314

, [n3] =

−0.592293−0.7454420.305786

.

(57)

Calculate the tensor U =√C.

SOLUTION

First, it must to note that

√C 6=

√33

√−17

√10√

−17√22

√5√

10√5

√38

!!!!!! (58)

The tensor C in the basis of its unit eigenvectors is given by

[

C]

= [Q]T [C] [Q] =

λ1 0 00 λ2 00 0 λ3

, (59)

where Q is defined by

[Q] =[[n1] [n2] [n3]

]=

0.755983 −0.278709 −0.592293−0.382872 0.545642 −0.7454420.530942 0.790314 0.305786

. (60)

Q is an orthogonal tensor, i.e.

QQT = δ, Q−1 = QT . (61)

The tensor U in the basis of unit eigenvectors of C (where C is diagonal) is computed by

[

U]

=[√

C]

=

√λ1 0 00

√λ2 0

0 0√λ3

=

6.97373 0 00 6.15837 00 0 2.53802

. (62)

U in the original basis is calculated with the formula

[U ] = [Q][

U]

[Q]T =

5.3543 −1.83446 0.982972−1.83446 4.26613 0.6594940.982972 0.659494 6.0497

. (63)

Verification:

[U ] [U ] =

33 −17 10−17 22 510 5 38

≡ [C] . (64)

8


EXERCISE 4

Let σ be the symmetric tensor

[σ] =

50 −30 −40−30 −100 20−40 20 −10

. (65)

Determine its spherical and deviatoric parts, respectively.

SOLUTION

Spherical part is

p =1

3(trσ) δ, [p] =

1

3(−60)

1 0 00 1 00 0 1

=

−20 0 00 −20 00 0 −20

, (66)

while the deviatoric part is computed by

s = σ − p, (67)

[s] =

50 −30 −40−30 −100 20−40 20 −10

−

−20 0 00 −20 00 0 −20

=

70 −30 −40−30 −80 20−40 20 10

. (68)

9


EXERCISE 5

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

e1 = cosφe1 + sinφe2, (69)

e2 = − sinφe1 + cosφe2. (70)

(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 ifφ = 30.(c) Express vector r = 5e1 − e2 + 2e3 in the new CS.

SOLUTION

(a) Since the CS is right-handed, it follows

e3 = e1 × e2, e3 =

∣∣∣∣∣∣

e1 e2 e3

cosφ sinφ 0− sin φ cosφ 0

∣∣∣∣∣∣

= e3 (71)

(b)

[Q] =[[e1] [e2] [e3]

]=

cosφ − sinφ 0sin φ cos φ 00 0 1

=

√32

−12

012

√32

00 0 1

. (72)

(c)

[r] = [Q]T [r] =

√32

12

0

−12

√32

00 0 1

5−12

=

5√3−12

−5√3−12

2

≈

3.830−3.366

2

, (73)

r = 3.83e1 − 3.366e2 + 2e3. (74)

10


EXERCISE 6

Simplify the expression a× (b× c) utilizing the rules of the Einstein’s summation notation.

SOLUTION

The expression above can be written as

(aiei)×((bjej)× (ckek)) = aibjckei×(ej × ek) = aibjckǫjkmei×em = aibjckǫjkmǫimpep. (75)

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

(a× (b× c))p = aibjckǫjkmǫimp = aibjckǫjkmǫpim, (76)

where the identity ǫjkmǫimp = ǫjkmǫpim was employed.Now we can use the identity ǫjkmǫpim = δjpδki − δjiδkp, which yields

(a× (b× c))p = aibjck (δjpδki − δjiδkp) (77)

= aibpci − aibicp

= (aici) bp − (aibi) cp. (78)

Thus, it follows that

a× (b× c) = (a · c) b− (a · b) c. (79)

11


EXERCISE 7

Suppose that the scalar field

Φ =√x1 + 4x2x

23 (80)

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

SOLUTION

gradΦ = ∇Φ =∂Φ

∂xaea =

∂Φ

∂x1e1 +

∂Φ

∂x2e2 +

∂Φ

∂x3e3, (81)

[gradΦ] =

∂Φ

∂x1∂Φ

∂x2∂Φ

∂x3

=

1

2√x1

4x238x2x3

, (82)

[gradΦP ] =

1

2√4

4 (−3)2

8 (1) (−3)

=

1

436−24

(83)

12


EXERCISE 8

Determine the gradient of the scalar field

Φ = x21x2 − 7x3x1 (84)

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

n =1√14

(2e1 + 3e2 − e3) . (85)

SOLUTION

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

[gradΦ] =

∂Φ

∂x1∂Φ

∂x2∂Φ

∂x3

=

2x1x2 − 7x3x21

−7x1

, [gradΦP ] =

−1117

. (86)

Thus, the directional derivative along n is given by

(gradΦP ) · n = (−11e1 + 1e2 + 7e3) · (2e1 + 3e2 − e3)1√14

= − 26√14

≈ −6.9488. (87)

13


EXERCISE 9

Let the vector field u in the rectangular Cartesian CS be the following

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

u1u2u3

=

1 + x1x1x2x3 − x1

(88)

Compute the gradient of the above vector field.

SOLUTION

gradu = u⊗∇ = (uaea)⊗(

∂

∂xbeb

)

=∂ua∂xb

ea ⊗ eb = ua,bea ⊗ eb, (89)

gradu =∂u1∂x1

e1 ⊗ e1 +∂u1∂x2

e1 ⊗ e2 +∂u1∂x3

e1 ⊗ e3 (90)

+∂u2∂x1

e2 ⊗ e1 +∂u2∂x2

e2 ⊗ e2 +∂u2∂x3

e2 ⊗ e3 (91)

+∂u3∂x1

e3 ⊗ e1 +∂u3∂x2

e3 ⊗ e2 +∂u3∂x3

e3 ⊗ e3, (92)

[gradu] =

∂u1∂x1

∂u1∂x2

∂u1∂x3

∂u2∂x1

∂u2∂x2

∂u2∂x3

∂u3∂x1

∂u3∂x2

∂u3∂x3

=

1 0 0x2 x1 0−1 0 1

. (93)

14


EXERCISE 10

Determine the divergence of the following vector field at point P(1,2,3):

v = x1x2x3 (x3e1 + x2e2 + x1e3) . (94)

SOLUTION

divv = v · ∇ =∂va∂xb

ea · eb =∂va∂xb

δab =∂va∂xa

= va,a =∂v1∂x1

+∂v2∂x2

+∂v3∂x3

, (95)

divv =(x2x

23

)+ (2x1x2x3) +

(x21x2

)= x2 (x1 + x3)

2 , (96)

divvP = 2 (1 + 3)2 = 32. (97)

15


EXERCISE 11

Calculate the divergence of the tensor field

[A] =

x21 −x2 27 4x2x3 5x32 1 x1x3

(98)

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

SOLUTION

divA = A · ∇ =∂A

∂xc· ec =

∂Aab

∂xcea (eb · ec) =

∂Aab

∂xcδbcea = Aab,cδbcea = Aab,bea. (99)

[divA] =

∂A11

∂x1+∂A12

∂x2+∂A13

∂x3∂A21

∂x1+∂A22

∂x2+∂A23

∂x3∂A31

∂x1+∂A32

∂x2+∂A33

∂x3

=

2x1 − 1 + 00 + 4x3

0 + 0 + x1

=

2x1 − 14x3x1

(100)

[divAP ] =

3162

. (101)

16


EXERCISE 12

Determine the curl of the following vector

[v] =

xyzy2 + xzx− y

. (102)

SOLUTION

curlv = ∇× v = ea ×∂v

∂xa=∂vb∂xa

ea × eb =∂vb∂xa

ǫabcec = vb,aǫabcec. (103)

[v] =

−x− 1xy − 1z − xz

. (104)

17


EXERCISE 13

Calculate the curl of the following tensor

[A] =

xy y z2

1 z 00 y y

(105)

SOLUTION

curlA = ∇×A = ea ×∂A

∂xa=∂Abc

∂xaea × (eb ⊗ ec) =

∂Abc

∂xaǫabded ⊗ ec (106)

= Abc,aǫabded ⊗ ec (107)

[curlA] =

Ab1,aǫab1 Ab2,aǫab1 Ab3,aǫab1Ab1,aǫab2 Ab2,aǫab2 Ab3,aǫab2Ab1,aǫab3 Ab2,aǫab3 Ab3,aǫab3

(108)

[curlA] =

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

A11,3 −A31,1 A12,3 − A32,1 A13,3 − A33,1

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

(109)

[curlA] =

0− 0 1− 1 1− 00− 0 0− 0 2z − 00− x 0− 1 0− 0

(110)

[curlA] =

0 0 10 0 2z−x −1 0

(111)

18


EXERCISE 14

Suppose that the motion of a continuous body is given by

χ (X , t) =

X1 + tX2︸︷︷︸

x1

e1 +

X2 − t2X1︸︷︷︸

x2

e2 +

X3︸︷︷︸

x3

e3. (112)

The motion is illustrated in Figure 1:

Figure 1: Illustration of the motion in terms of the parameter t

Determine the inverse of the mapping.

SOLUTION

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

x1x2x3

=

X1 + tX2

X2 − t2X1

X3

=

1 t 0−t2 1 00 0 1

X1

X2

X3

= [M ]

X1

X2

X3

. (113)

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

x1 = X1 + tX2, (114)

x2 = X2 − t2X1, (115)

x3 = X3. (116)

The inverse motion has the form

χ−1 (x, t) =1

1 + t3(x1 − tx2) e1 +

1

1 + t3(t2x1 + x2

)e2 + x3e3.

19


EXERCISE 15

Suppose that the deformation of a continuous body is given by

χ (X) = (3− 2X1 −X2) e1 +

(

2 +1

2X1 −

1

2X2

)

e2 + (X3) e3. (117)

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

SOLUTION

The definition for the deformation gradient is

F = Gradχ (X) = χ⊗∇X = χ⊗(

∂

∂Xa

Ea

)

(118)

= (x1e1 + x2e2 + x3e3)⊗(

∂

∂Xa

Ea

)

, (119)

[F ] =

∂x1∂X1

∂x1∂X2

∂x1∂X3

∂x2∂X1

∂x2∂X2

∂x2∂X3

∂x3∂X1

∂x3∂X2

∂x3∂X3

=

−2 −1 00.5 −0.5 00 0 1

. (120)

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

χ−1 (x) =

(

−1

3− 1

3x1 +

2

3x2

)

E1 +

(11

3− 1

3x1 −

4

3x2

)

E2 + (x3)E3. (121)

The inverse deformation gardient is computed by

F−1 = gradχ−1 (x) = χ−1 ⊗∇ = χ−1 ⊗(

∂

∂xaea

)

(122)

= (X1E1 +X2E2 +X3E3)⊗(

∂

∂xaea

)

, (123)

[F−1

]=

∂X1

∂x1

∂X1

∂x2

∂X1

∂x3∂X2

∂x1

∂X2

∂x2

∂X2

∂x3∂X3

∂x1

∂X3

∂x2

∂X3

∂x3

=

−1/3 2/3 0−1/3 −4/3 00 0 1

. (124)

20


EXERCISE 16

Let the deformation described by the following mapping:

χ (X) = (1 +X1 +X2) e1 + (X2 − 2X1) e2 + (X3)e3. (125)

Calculate the stretch in the reference configuration in the direction

[N 1] =1√5

210

, (126)

and the stretch in the spatial configuration in the direction

[n2] =1√5

210

. (127)

SOLUTION

The deformation gradient and its inverse are the following

[F ] =

1 1 0−2 1 00 0 1

,[F−1

]=

1

3

1 −1 02 1 00 0 3

. (128)

The stretches are calculated by the formulas

λN1=

√

N 1 · F T · F ·N 1 = 3

√

2

5≈ 1.89737, (129)

λn2=

1√

n2 · F−T · F−1 · n2

= 3

√

5

26≈ 1.31559. (130)

The meaning of the streches is illustrated in Figure 2:

21


Figure 2: Meaning of the stretch measures22


EXERCISE 17

Let a 2D motion given by

χ (X) = (X1X2) e1 +(X1 +X2

2

)e2. (131)

Determine the region, where the material points cannot correspond to a real physical deforma-tion.

SOLUTION

The deformation of a point can be admissible if J > 0. The deformation gradient for thisnon-homogenous deformation is

[F ] =

[X2 X1

1 2X2

]

, J = detF = 2X22 −X1. (132)

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

Figure 3:

23


EXERCISE 18

Suppose thet the motion of a continuous body is given by

χ (X) =

(

X1 −1

3(X2 −X1)

)

e1 +

(

X2 +2

3(X1 +X2)

)

e2 + (X3)e3. (133)

Determine the following deformation and strain measures, respectively: left and right Caucy-Green deformation tensors; Piola deformation tensor; Cauchy deformation tensor; Green-Lagrange strain tensor; Almansi-Euler strain tensor.

SOLUTION

The deformation gardient and its inverse are

[F ] =1

3

4 −1 02 5 00 0 3

,[F−1

]=

1

11

7.5 1.5 0−3 6 00 0 11

. (134)

Right Cauchy-Green deformation tensor:

C = F TF , C = CIJEI ⊗EJ (135)

[C] =[F TF

]=

1

9

20 6 06 26 00 0 9

≈

2.222 0.667 00.667 2.889 00 0 1

. (136)

Piola deformation tensor:

B = C−1 = F−1F−T , B = BIJEI ⊗EJ (137)

[B] =[C−1

]=

1

242

117 −27 0−27 90 00 0 242

≈

0.483 −0.112 0−0.112 0.372 0

0 0 1

. (138)

Left Cauchy-Green deformation tensor:

b = FF T , b = bijei ⊗ ej (139)

[b] =[FF T

]=

1

9

17 3 03 29 00 0 9

≈

1.889 0.333 00.333 3.222 00 0 1

. (140)

Cauchy deformation tensor

c = b−1 = F−TF−1, c = cijei ⊗ ej (141)

[c] =[b−1

]=

1

484

261 −27 0−27 153 00 0 484

≈

0.539 −0.056 0−0.056 0.316 0

0 0 1

. (142)

Green-Lagrange strain

E =1

2(C − I) = EIJEI ⊗EJ (143)

24


[E] =

[1

2(C − I)

]

=1

18

11 6 06 17 00 0 0

≈

0.611 0.333 00.333 0.944 00 0 0

. (144)

Almansi-Euler strain

e =1

2(i− c) = eijei ⊗ ej (145)

[e] =

[1

2(i− c)

]

=1

968

223 27 027 331 00 0 0

≈

0.231 0.028 00.028 0.342 00 0 0

. (146)

25


EXERCISE 19

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

x1 =3

2+X1 +

3

2

(1

5X1 +

3

2X2

)2

, (147)

x2 =3

2+X2 −

3

2X2

1 . (148)

The deformation is illustrated in Figure 4.

Figure 4: Illustration of the deformation

Determine the eigenvalues and eigenvectors of the left Cauchy-Green deformation tensor atmaterial point P0(0.8, 0.2).

SOLUTION

After the deformation, the material point at P0 has the following spatial coordinates:

x1 =3

2+ 0.8 +

3

2

(1

50.8 +

3

20.2

)2

= 2.6174, (149)

x2 =3

2+ 0.2− 3

20.82 = 0.74. (150)

The deformation gradient is

[F ] =

∂x1∂X1

∂x1∂X2

∂x2∂X1

∂x2∂X2

=

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

−3X1 1

]

. (151)

Its value corresponding to the material point P0 is

[F ] =

[1.276 2.07−2.4 1

]

. (152)

26


The matrix representation of the left Cauchy-Green deformation tensor is

[b] =[FF T

]=

[5.91308 −0.9924−0.9924 6.76

]

. (153)

Calculating its eigenvalues:

(5.91308− λ) (6.76− λ)− (−0.9924)2 = 0, (154)

λ1 = 7.4155, λ2 = 5.2575. (155)

Eigenvectors:

[b− λ1i] [n1] = [0] ⇒ [n1] =

[−0.551150.8344

]

, (156)

[b− λ2i] [n2] = [0] ⇒ [n2] =

[−0.8344−0.55115

]

. (157)

The eigenvectors are illustrated in Figure 5.

Figure 5: Eigenvectors of the the left Cauchy-Green deformation tensor at material point P

27


EXERCISE 20

Consider a deformation ϕ (X) defined in components by

x1 = X21 , x2 = X2

3 , x3 = X2X3. (158)

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 ] =

∂x1∂X1

∂x1∂X2

∂x1∂X3

∂x2∂X1

∂x2∂X2

∂x2∂X3

∂x3∂X1

∂x3∂X2

∂x3∂X3

=

2X1 0 00 0 2X3

0 X3 X2

. (159)

Its determinant is

J = detF = −4X1X23 . (160)

The inverse deformation gradient has the form

[F−1

]=

1

2X10 0

0 − X2

2X23

1

X3

01

2X30

. (161)

The area change at point P0 is determined according to the Nanson’s formula:

da = (detF )F−TdA, (162)

[da] = −4X1X23

1

2X1

0 0

0 − X2

2X23

1

2X3

01

X30

[dA] , (163)

[da] =

−2X23 0 0

0 2X1X2 −2X1X3

0 −4X1X3 0

[dA] , (164)

[daP ] =

−2 0 00 −12 40 8 0

[dA] . (165)

The ratio da/dA is given by

da

dA= J

√NBN , (166)

28


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

B = F−1F−T , [B] =

1

4X21

0 0

0X2

2

4X43

+1

X23

− X2

4X23

0 − X2

4X23

1

4X23

. (167)

Thus it follows that

da

dA= J

√NBN = −4X1X

23

(1

2X3

)

= −2X1X3 = 4. (168)

The volume change at P is

dvPdVP

= detF P = JP = 8. (169)

29


EXERCISE 21

Consider the homogeneous deformation

x1 = X1 + tX2, x2 = X2 − tX2, x3 = X3. (170)

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

N I =

100

, N II =

010

. (171)

SOLUTION


[F ] =

1 t 00 1− t 00 0 1

. (172)

Its inverse transpose is

[F−T

]=

1 0 0t

t− 1

−1

t− 10

0 0 1

. (173)

The determinant of the deformation gradient is

J = detF = 1− t. (174)

Thus, t < 1 has to be satisfied.The right Cauchy–Green deformation tensor is given by

[C] =[F TF

]=

1 t 0

t (1− t)2 + t2 00 0 1

. (175)

The Green–Lagrange strain tensor is computed as

[E] =

[1

2(C − I)

]

=

0t

20

t

2t (t− 1) 0

0 0 0

. (176)

The Cauchy deformation tensor is

[c] =[b−1

]=

[F−TF−1

]=

1t

t− 10

t

t− 1

t2 + 1

(t− 1)20

0 0 1

. (177)

30


The Euler–Almansi strain tensor is given by

[e] =

[1

2(i− c)

]

=

0t

2− 2t0

t

2− 2t

−t(t− 1)2

0

0 0 0

. (178)

Stretch ratios along NI and NII are

λNI=

√

N I ·C ·N I = 1, (179)

λNII=

√

N II ·C ·N II =

√

(1− t)2 + t2. (180)

Unit vectors in the spatial configuration directed along material line element dXI = dS0IN I

and dXII = dS0IIN II , respecively are

nI =1

λNI

FN I =

100

, nII =1

λNII

FN II =1

√

t2 + (1− t)2

t1− t0

. (181)

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

sinγ = sin(π

2− α

)

= cosα =N ICN II

λNIλNII

=2N IEN II

λNIλNII

= 2nIenII =t

√

t2 + (1− t)2. (182)

Note that in this example the initial angle is α0 =π

2.

31


EXERCISE 22

Let a homogenous deformation be given by the following mapping:

x1 = X1 + tX2, x2 = X2 − 2tX1, x3 = X3. (183)

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

N I =

100

, N II =1√2

110

. (184)

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

SOLUTION


[F ] =

1 t 0−2t 1 00 0 1

. (185)

The right Cauchy–Green deformation tensor is

[C] =[F TF

]=

1 + 4t2 −t 0−t 1 + t2 00 0 1

. (186)

Stretch ratios:

λNI=

√

N I ·C ·N I =√1 + 4t2, (187)

λNII=

√

N II ·C ·N II =

√

1− t+5t2

2. (188)

The angle change is defined by α0 − α, where

cosα0 = N IN II ⇒ α0 =π

4= 45 (189)

and

cosα =N ICN II

λNIλNII

=1 + t (4t− 1)√

1 + 4t2√

2− t (2− 5t). (190)

For t = 1, its value is

cosα|t=1 =4

5⇒ α ≈ 36.87. (191)

Consequently

α0 − α =π

4− 4

5≈ 8.13. (192)

32


The evolution of cosα and α are illustrated in Figure 6.

Figure 6: The evolution of cosα and α

The maximum angle is calculated according to

d

dtα (t) = 0 (193)

2

4t2 + 1− 3

5t2 − 2t+ 2= 0 (194)

t =1

2

(√6− 2

)

≈ 0.2247. (195)

33


EXERCISE 23

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. (196)

The deformed configuration in the x1, x2 plane is illustrated in Figure 7.Determine the change of the angle defined by the material line elements dXI and dXII atpoint P0 (2, 3, 1), where

N I =

−100

, N II =

0−10

. (197)

Figure 7: Illustration of the deformed configuration in the x1, x2 plane

SOLUTION


[F ] =

1 +X2 X1 0−1 1 00 0 1

. (198)

The right Cauchy–Green deformation tensor is

[C] =[F TF

]=

1 + (1 +X2)2 X1 (1 +X2)− 1 0

X1 (1 +X2)− 1 1 +X21 0

0 0 1

. (199)

34


Stretch ratios:

λNI=

√

N I ·C ·N I =

√

1 + (1 +X2)2, (200)

λNII=

√

N II ·C ·N II =√

1 +X21 . (201)

Since α0 = π/2, the angle of shear is defined by

sinγ = sin(π

2− α

)

= cosα =N ICN II

λNIλNII

=X1 (1 +X2)− 1

√

1 + (1 +X2)2√

1 +X21

. (202)

At point P0 (2, 3, 1):

cosα|P0=

7√85

⇒ α ≈ 40.6. (203)

Consequently the angle change is

α0 − α = 90 − 40.6 = 49.4.

35


EXERCISE 24

A non-homogeneous deformation is given by the mapping

x1 = −X2, x2 = X2X3, x3 = X3 −X1. (204)

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 X2

−1 0 1

. (205)

The volume change is

J = detF = X2. (206)

The isochoric and volumetric parts of F are defined by

F iso = J− 1

3F , (207)

[F iso] =1

3√X2

0 −1 00 X3 X2

−1 0 1

=

0 − 13√X2

0

0 X3

3√X2

3

√

X22

− 13√X2

0 13√X2

, (208)

F vol = J1

3I, (209)

[F vol] = 3

√

X2

1 0 00 1 00 0 1

=

3√X2 0 00 3

√X2 0

0 0 3√X2

. (210)

At P0 (2, 8, 4):

[F iso] =

0 −0.5 00 2 4

−0.5 0 0.5

, [F vol] =

2 0 00 2 00 0 2

. (211)

36


EXERCISE 25

Let the deformation gradient be the following:

[F ] =

1 1.5 00 1 00 0 1

. (212)

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

SOLUTION

The right Cauchy-Green deformation tensor:

[C] =[F TF

]=

1 1.5 01.5 3.25 00 0 1

. (213)

Its eigenvalues are

det (C − µδ) = 0 ⇒ µ1 = 4, µ2 = 1, µ3 = 0.25. (214)

Principal stretches are

λ1 =√µ1 = 2, λ2 =

√µ2 = 1 λ3 =

√µ3 = 0.5. (215)

Unit eigenvectors of C (and U) are

(C − µiδ) ·N i = 0 (216)

[N 1] =

0.44720.8944

0

, [N 2] =

001

[N 3] =

0.8944−0.4472

0

. (217)

Basis tensors are

[M 1] = [N 1 ⊗N 1] =

0.2 0.4 00.4 0.8 00 0 0

, (218)

[M 2] = [N 2 ⊗N 2] =

0 0 00 0 00 0 1

, (219)

[M 3] = [N 3 ⊗N 3] =

0.8 −0.4 0−0.4 0.2 00 0 0

. (220)

These basis tensors can be calculated in different way, using the following formulas:

Ma =1

(µa − µb) (µa − µc)(C − µbδ) (C − µcδ) , (221)

M 1 =1

(µ1 − µ2) (µ1 − µ3)(C − µ2δ) (C − µ3δ) (222)

=1

(4− 1) (4− 0.25)(C − 1δ) (C − 0.25δ) (223)

37


M 2 =1

(µ2 − µ3) (µ2 − µ1)(C − µ3δ) (C − µ1δ) (224)

=1

(1− 0.25) (1− 4)(C − 0.25δ) (C − 4δ) (225)

M 3 =1

(µ3 − µ1) (µ3 − µ2)(C − µ1δ) (C − µ2δ) (226)

=1

(0.25− 4) (0.25− 1)(C − 4δ) (C − 1δ) (227)

The right Cauchy stretch tensor is

U = λ1M 1 + λ2M 2 + λ3M 3, (228)

[U ] =

0.8 0.6 00.6 1.7 00 0 1

, (229)

[U−1

]=

1.7 −0.6 0−0.6 0.8 00 0 1

. (230)

The rotation tensor is

R = FU−1, [R] =

0.8 0.6 0−0.6 0.8 00 0 1

, detR = 1, R−1 = RT . (231)

Computing the left stretch tensor:

F = V R ⇒ V = FR−1, [V ] =

1.7 0.6 00.6 0.8 00 0 1

. (232)

Calculating V in a different way:

[n1] = [RN 1] =

0.89440.4472

0

, [n2] = [RN 2] =

001

, [n3] = [RN 3] =

0.4472−0.8944

0

. (233)

[m1] = [n1 ⊗ n1] =

0.8 0.4 00.4 0.2 00 0 0

, (234)

[m2] = [n2 ⊗ n2] =

0 0 00 0 00 0 1

, (235)

[m3] = [n3 ⊗ n3] =

0.2 −0.4 0−0.4 0.8 00 0 0

. (236)

V = λ1m1 + λ2m2 + λ3m3, [V ] =

1.7 0.6 00.6 0.8 00 0 1

. (237)

38


Calculating V in a different way:

V = RURT , [R] [U ][RT

]=

1.7 0.6 00.6 0.8 00 0 1

. (238)

39


EXERCISE 26

Let the deformation be given with the following mapping:

x1 = αX1 + βX2, x2 = −αX1 + βX2, x3 = γX3. (239)

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

SOLUTION

Deformation gradient:

[F ] =

α β 0−α β 00 0 γ

, J = 2αβγ. (240)

Right Cauchy-Green deformation tensor:

[C] =[F TF

]=

2α2 0 00 2β2 00 0 γ2

. (241)

Eigensystem of C:

µ1 = 2α2, µ1 = 2β2, µ3 = γ2, [N 1] =

100

, [N 2] =

010

, [N 3] =

001

. (242)


λ1 =√µ1 =

√2α, λ2 =

√µ2 =

√2β λ3 =

√µ3 = γ. (243)

The right Cauchy-Green deformation tensor:

U = λ1N 1 ⊗N 1 + λ2N 2 ⊗N 2 + λ3N 3 ⊗N 3, [U ] =

√2α 0 0

0√2β 0

0 0 γ

. (244)

Rotation tensor:

[R] = [F ][U−1

]=

α β 0−α β 00 0 γ

1√2α

0 0

01√2β

0

0 01

γ

=1√2

1 1 0−1 1 0

0 0√2

. (245)

Left stretch tensor:

[V ] = [R] [U ][RT

]=

1√2

α + β −α + β 0−α + β α + β 0

0 0√2γ

. (246)

40


EXERCISE 27

Compute the Hencky strain tensor in the reference configuration for the deformation

[F ] =

1 1 0−2 1 00 0 1

. (247)

SOLUTION

The right Cauchy-Green deformation tensor is

[C] =[F TF

]=

5 −1 0−1 2 00 0 1

. (248)

Eigenvalues of C are

det (C − µδ) = 0 ⇒ µ1 =1

2

(

7 +√13)

, µ2 =1

2

(

7−√13)

, µ3 = 1. (249)


λ1 =√µ1 =

1

2

(

1 +√13)

, λ2 =√µ2 = −1

2

(

1−√13)

λ3 =√µ3 = 1. (250)

Unit eigenvectors of C (and U) are

[N 1] =

− 3 +√13

√

26 + 6√13√

2

13 + 3√13

0

, [N 2] =

√1

2− 3

2√13√

1

2+

3

2√13

0

[N 3] =

001

. (251)

The right stretch tensor is

U = λ1N 1 ⊗N 1 + λ2N 2 ⊗N 2 + λ3N 3 ⊗N 3, (252)

[U ] =

8√13

− 1√13

0

− 1√13

5√13

0

0 0 1

≈

2.2188 −0.2774 0−0.2774 1.3868 0

0 0 1

. (253)

The Hencky strain tensor in the reference configuration (after simplification) is

lnU = (lnλ1)N 1 ⊗N 1 + (lnλ2)N 2 ⊗N 2 + (lnλ3)N 3 ⊗N 3. (254)

[lnU ] =

ArcTanh(

52√13

)

√13

+ln 3

2−2ArcCoth

(√13)

√13

0

−2ArcCoth(√

13)

√13

−ArcTanh

(5

2√13

)

√13

+ln 3

20

0 0 0

(255)

≈

0.7863 −0.1580 0−0.1580 0.3123 0

0 0 1

. (256)

41


EXERCISE 28

Compute the generalized material strain measures for the homogenous deformation

[F ] =

1 1 0−2 1 00 0 1

. (257)

SOLUTION

The right Cauchy-Green deformation tensor is

[C] =[F TF

]=

5 −1 0−1 2 00 0 1

. (258)

Eigenvalues of C are

det (C − µδ) = 0, (259)

µ1 =1

2

(

7 +√13)

≈ 5.3028, µ2 =1

2

(

7−√13)

≈ 1.6972, µ3 = 1.


λ1 =√µ1 ≈ 2.3028, λ2 =

√µ2 ≈ 1.3028 λ3 =

√µ3 = 1. (260)

Eigenvectors of C are

[N 1] ≈

−0.95710.2898

0

, [N 2] ≈

0.28980.9571

0

[N 3] =

001

. (261)

Generalized material strain measures:

E(n) =3∑

α=1

1

n(λnα − 1)Nα ⊗Nα. (262)

E(−2) =

3∑

α=1

1

2

(

1− 1

λ2α

)

Nα ⊗Nα,[

E(−2)]

=

0.3889 −0.0556 0−0.0556 0.2222 0

0 0 0

, (263)

E(−1) =3∑

α=1

(

1− 1

λα

)

Nα ⊗Nα,[

E(−1)]

=

0.5378 −0.0925 0−0.0925 0.2604 0

0 0 0

, (264)

E(0) =

3∑

α=1

(lnλα)Nα ⊗Nα,[

E(0)]

=

0.7863 −0.1580 0−0.1580 0.3123 0

0 0 0

, (265)

E(1) =3∑

α=1

(λα − 1)Nα ⊗Nα,[

E(1)]

=

1.2189 −0.2774 0−0.2774 0.3868 0

0 0 0

, (266)

E(2) =

3∑

α=1

1

2

(λ2α − 1

)Nα ⊗Nα,

[

E(2)]

=

2 −0.5 0−0.5 0.5 00 0 0

. (267)

42


EXERCISE 29

Let a two-dimensional displacement be given in the reference configuration as

U1 = −2X2, U2 = 2X1, (268)

[U ] =

[−2X2

2X1

]

Determine the coordinates of the displacement of the material point P0 (1, 1) in the current(spatial or deformed) configuration.

SOLUTION

The deformation mapping is

x1 = X1 + U1 = X1 − 2X2, x2 = X2 + U2 = X2 + 2X1. (269)

The inverse deformation mapping is

X1 =1

5(x1 + 2x2) , X2 =

1

5(x2 − 2x1) . (270)

The coordinates of the displacement field in the spatial configuration are

u1 = x1 −X1 =1

5(4x1 − 2x2) , u2 = x2 −X2 =

1

5(4x2 + 2x1) , (271)

[u] =1

5

[4x1 − 2x24x2 + 2x1

]

.

The displacement vector U at point P0 (1, 1) is

[U ]P =

[−22

]

. (272)

The displacement vector u at point P (−1, 3) (which is the same material point as P0 (1, 1) inthe reference configuration) is

[u]P =1

5

[4 (−1)− 2 (3)4 (3) + 2 (−1)

]

=1

5

[−1010

]

=

[−22

]

. (273)

The motion is illustrated in Figure 8.

43


Figure 8:

44


EXERCISE 30

A certain motion of a continuum body in the reference configuration is given by the displacementfield

U1 = k (X1 +X3) , U2 = k (X2 −X1) , U3 = 0. (274)

Determine the coordinates of the displacement field in the spatial (deformed) configuration.

SOLUTION

The motion is computed as

x1 = X1 + k (X1 +X3) , x2 = X2 + k (X2 −X1) , x3 = X3. (275)

Its inverse is

X1 =1

1 + k(x1 − kx3) , (276)

X2 =1

(1 + k)2(kx1 + (1 + k) x2 − k2x3

), (277)

X3 = x3. (278)

The coordinates of the displacement field in the spatial configuration are

u1 = x1 −X1 =k

1 + k(x1 + x3) , (279)

u2 = x2 −X2 =k

(1 + k)2((1 + k) x2 − x1 + kx3) , (280)

u3 = 0. (281)

45


EXERCISE 31

Determine the Euler and Lagrange velocity and acceleration field, respectively, for the 2Ddeformation

x1 = X1 + (sin t)X2, x2 = X2 + tX2. (282)

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

SOLUTION

The material velocity field is computed by

V (X, t) =∂χ (X, t)

∂t

∣∣∣∣X fixed

(283)

[V (X, t)] =

∂ (X1 + (sin t)X2)

∂t∂ (X2 + tX2)

∂t

=

[X2 cos tX2

]

. (284)

The material acceleration field is

A (X, t) =∂χ2 (X, t)

∂t2

∣∣∣∣X fixed

=∂V (X, t)

∂t

∣∣∣∣X fixed

(285)

[A (X, t)] =

∂ (X2 cos t)

∂t∂ (X2)

∂t

=

[−X2 sin t

0

]

. (286)


X1 = x1 −x2 sin t

1 + t, X2 =

x21 + t

(287)

The Euler velocity field is

v (x, t) = V(χ−1 (x, t) , t

)[v (x, t)] =

x21 + t

cos tx2

1 + t

. (288)

The spatial acceleration field is determined by

a (x, t) = A(χ−1 (x, t) , t

)[a (x, t)] =

[

− x21 + t

sin t

0

]

. (289)

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

[V (XP0, 1)] =

[1 cos 1

1

]

≈[0.5403

1

]

, (290)

[A (XP0, 1)] =

[−1 sin 1

0

]

≈[−0.8415

0

]

. (291)

46


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

[xP ] =

[1 + (sin 1) 11 + 1 · 1

]

=

[1 + sin 1

2

]

≈[1.8415

2

]

, (292)

[v (xP , 1)] =

2

1 + 1cos 1

2

1 + 1

≈

[0.5403

1

]

, (293)

[a (xP , 1)] =

[

− 2

1 + 1sin 1

0

]

≈[−0.8415

0

]

. (294)

47


EXERCISE 32

Compute the Euler and Lagrange velocity and acceleration field, respectively, for the 2D de-formation

x1 = X1 ln (2 + t) , x2 = X2e2t. (295)

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) =∂χ (X, t)

∂t

∣∣∣∣X fixed

, (296)

[V (X, t)] =

∂ (X1 ln (2 + t))

∂t∂ (X2e

t)

∂t

=

X1

2 + t2X2e

2t

. (297)

The material acceleration field is

A (X, t) =∂χ2 (X, t)

∂t2

∣∣∣∣X fixed

=∂V (X, t)

∂t

∣∣∣∣X fixed

, (298)

[A (X, t)] =

∂

(X1

2 + t

)

∂t∂ (2X2e

2t)

∂t

=

−X1

(2 + t)2

4X2e2t

. (299)


X1 =x1

ln (2 + t), X2 = x2e

−2t (300)


v (x, t) = V(χ−1 (x, t) , t

), [v (x, t)] =

[ x1(2 + t) ln (2 + t)

2x2

]

. (301)

The spatial acceleration field is determined by

a (x, t) = A(χ−1 (x, t) , t

), [a (x, t)] =

−x1(2 + t)2 ln (2 + t)

4x2

. (302)

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

[V (XP0, 1)] =

[ 100

2 + 12 · 2e2·1

]

≈[33.33329.556

]

, (303)

[A (XP0, 1)] =

−100

(2 + 1)2

4 · 2e2·1

≈[−11.11159.112

]

. (304)

48


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

[xP ] =

[100 ln (2 + 1)

2e2·1

]

≈[109.86114.778

]

, (305)

[v (xP , 1)] =

109.861

(2 + 1) ln (2 + 1)2 · 14.778

≈[33.33329.556

]

, (306)

[a (xP , 1)] =

−109.861

(2 + 1)2 ln (2 + 1)4 · 14.778

≈[−11.11159.112

]

. (307)

49


EXERCISE 33

Suppose that the motion of a continuous medium is given by the 1D mapping

x = (1 + t)X (308)

and the temperature distribution along X in the reference description is described by

Φ (X, t) = Xt2 (309)

Compute the material time derivative of the temperature field φ (x, t) written in the spatialconfiguration.

SOLUTION

The inverse motion is

X =x

1 + t(310)

The temperature field φ (x, t) written in the spatial configuration is

φ (x, t) =xt2

1 + t(311)

The velocity field in the material description is

V (X, t) = X (312)

The Euler velocity filed is

v (x, t) = V(χ−1 (x, t) , t

)=

x

1 + t(313)

The material time derivative of φ (x, t) is calculated by

dφ (x, t)

dt=∂φ (x, t)

∂t+ gradφ (x, t) · v (x, t) = (2 + t) xt

(1 + t)2+

t2

1 + t· x

1 + t=

2xt

1 + t(314)

50


EXERCISE 34

Consider the following 2D mapping

x1 = tX1, x2 = X2 + (1 + t)X1. (315)

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

Q (X, t) = X2t. (316)

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

SOLUTION


X1 =x1t, X2 = x2 −

(1 + t)

tx1. (317)


[V (X, t)] =

∂ (tX1)

∂t∂ (X2 + (1 + t)X1)

∂t

=

[X1

X1

]

. (318)

The field under consideration in the current configuration is computed as

q (x, t) = Q(χ−1 (x, t) , t

)= x2t− (1 + t) x1. (319)

Therefore, the velocity field in the spatial description is

v (x, t) = V(χ−1 (x, t) , t

), [v (x, t)] =

x1tx1t

. (320)

The gradient of q with respect to the current coordinates is

[gradq (x, t)] =

[−1 − tt

]

. (321)

The material time derivative of q is

q =∂q

∂t+ gradq · v = (x2 − x1) +

(

−x1t(1 + t) + x1

)

, (322)

q = x2 −t+ 1

tx1. (323)

The material time derivative of Q has the simple form

Q =∂Q

∂t= X2. (324)

51


EXERCISE 35

Let the deformation be given by

x1 = X1 + t, x2 = X2 − tX1, x3 = X3. (325)

Compute the material time derivative of spatial vector field

[k] =

x2t2

x1t0

. (326)

SOLUTION


X1 = x1 − t, X2 = x2 + x1t− t2, X3 = x3. (327)

The material velocity field is

[V ] =

1−X1

0

. (328)


[v] =

1t− x1

0

. (329)

The material time derivative of f is computed by

dk (x, t)

dt=∂k (x, t)

∂t+ gradk · v, (330)

where

[gradk] =

∂k1∂x1

∂k1∂x2

∂k1∂x3

∂k2∂x1

∂k2∂x2

∂k2∂x3

∂k3∂x1

∂k3∂x2

∂k3∂x3

=

0 t2 0t 0 00 0 0

. (331)

Therefore

[dk (x, t)

dt

]

=

2x2tx10

+

0 t2 0t 0 00 0 0

·

1t− x1

0

=

t3 − x1t2 + 2x2t

t+ x10

. (332)

52


EXERCISE 36

Let the deformation be given by

x1 = (1 + t)X1, x2 = X2 − tX3, x3 = X3. (333)

Compute the material time derivative of spatial vector field

[f ] =

x1 + x2tx2x3x1

. (334)

SOLUTION


X1 =x1

(1 + t), X2 = x2 + tx3, X3 = x3. (335)

The material velocity field is

[V ] =

X1

−X3

0

. (336)


[v] =

x1(1 + t)−x30

. (337)

The material time derivative of f is computed by

df (x, t)

dt=∂f (x, t)

∂t+ gradf · v, (338)

where

[gradf ] =

∂f1∂x1

∂f1∂x2

∂f1∂x3

∂f2∂x1

∂f2∂x2

∂f2∂x3

∂f3∂x1

∂f3∂x2

∂f3∂x3

=

1 t 00 x3 x21 0 0

. (339)

Therefore

[df (x, t)

dt

]

=

x200

+

1 t 00 x3 x21 0 0

·

x1(1 + t)−x30

=

x1(1 + t)

+ x2 − tx3

−x23x1

(1 + t)

. (340)

53


EXERCISE 37

The Lagrangian velocity field is given by

V (X, t) = −X2E2 + 2tX1E2. (341)

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

SOLUTION

The Lagrangian displacement field is obtained by

U (X, t) =

t∫

0

V(X, t

)dt = −tX2E2 + t2X1E2. (342)

Thus, the motion is computed as

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

X1 − tX2

X2 + t2X1

X3

. (343)

The deformation gradient and the volume change are

[F ] =

1 −t 0t2 1 00 0 1

, J = 1 + t3. (344)


X1 =x1 + tx21 + t3

, X2 =x2 − t2x11 + t3

, X3 = x3. (345)

Therefore, the Eulerian velocity field is given by

v (x, t) = V(ϕ−1 (x, t) , t

), [v (x, t)] =

t2x1 − x21 + t3

2tx1 + tx21 + t30

. (346)

The Lagrangian acceleration field is

A (X , t) =∂V (X, t)

∂t= 2X1E2. (347)

The Eulerian acceleration field by definition is

a (x, t) = A(ϕ−1 (x, t) , t

)= 2

x1 + tx21 + t3

e2. (348)

54


EXERCISE 38

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

u1 = −tx1, u2 = −tx1 (1 + t) . (349)

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. (350)

Using matrix notation:

[v1v2

]

=

[∂u1/∂t∂u2/∂t

]

+

[∂u1/∂x1 ∂u1/∂x2∂u2/∂x1 ∂u2/∂x2

] [v1v2

]

. (351)

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

]

. (352)

v1 = −x1 − tv1, (353)

v2 = −x1 (1 + 2t)−(t+ t2

)v1. (354)

The solutions are

v1 = − x11 + t

, v2 = − (1 + t) x1. (355)

Remark: from (350), one can obtain the solution for v as

v = (I − gradu)−1 ∂u

∂t. (356)

Thus

[v] =

[1 + t 0t+ t2 1

]−1 [ −x1−x1 (1 + 2t)

]

, (357)

[v] =

[ 1

1 + t0

−t 1

] [−x1

−x1 (1 + 2t)

]

=

[

− x11 + t

− (1 + t) x1

]

. (358)

55


EXERCISE 39

Suppose that the motion of a continuous body is described by

x1 = (1 + t)X1, x2 = X2 + tX3, x3 = X3. (359)

Compute the quantities l, d and w respectively.

SOLUTION


X1 =x1

1 + t, X2 = x2 − tx3 X3 = x3 (360)


V (X, t) =∂χ (X, t)

∂t

∣∣∣∣X fixed

[V (X, t)] =

X1

X3

0

. (361)


v (x, t) = V(χ−1 (x, t) , t

)[v (x, t)] =

x11 + tx30

. (362)

The velocity gradient is

l = gradv, [l] =

∂v1∂x1

∂v1∂x2

∂v1∂x3

∂v2∂x1

∂v2∂x2

∂v2∂x3

∂v3∂x1

∂v3∂x2

∂v3∂x3

=

1

1 + t0 0

0 0 10 0 0

. (363)

The rate of deformation and the vorticity tensors are

d =1

2

(l + lT

), [d] =

1

1 + t0 0

0 01

2

01

20

, (364)

w =1

2

(l − lT

), [w] =

0 0 0

0 01

2

0 −1

20

. (365)

56


EXERCISE 40

The Euler velocity field is given by its coordinates as

v1 = 3t2, v2 = x1 + tx2, v3 = −tx3. (366)

Determine the Euler acceleration field.

SOLUTION

The Euler acceleration field is computed as

a = v =∂v

∂t+ gradv · v =

∂v

∂t+ l · v, (367)

where

[l] =

0 0 01 t 00 0 −t

,

[∂v

∂t

]

=

6tx2−x3

. (368)

Therefore:

a =

6tx2−x3

+

0 0 01 t 00 0 −t

·

3t2

x1 + tx2−tx3

=

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

−x3 + t2x3

. (369)

57


EXERCISE 41

Let a homogenous deformation be given by

x1 = (1 + t)X1, x2 = X2 + tX1, x3 = X3. (370)

Compute the material time derivatives of stretch ratio along material direction given by theunit vector

[N ] =1√5

120

. (371)

SOLUTION

• The deformation gradient is

[F ] =

1 + t 0 0t 1 00 0 1

. (372)

Right Cauchy-Green deformation tensor is

[C] =[F TF

]=

t2 + (1 + t)2 t 0t 1 00 0 1

. (373)

The stretch ratio is calculated by

λn ≡ λN =√NCN =

√

5 + 2t (3 + t)

5(374)

The direction of material line element having oriantation N is the reference configurationis

n =1

λnFN , [n] =

1√

5 + 2t (3 + t)

1 + t2 + t0

(375)


X1 =x1

1 + t, X2 = x2 −

tx11 + t

, X3 = x3 (376)

Velocity field:

[V (X, t)] =

X1

X1

0

, [v (x, t)] =

x11 + tx1

1 + t0

(377)

58


Velocity gradient:

[l] =

∂v1∂x1

∂v1∂x2

∂v1∂x3

∂v2∂x1

∂v2∂x2

∂v2∂x3

∂v3∂x1

∂v3∂x2

∂v3∂x3

=

1

1 + t0 0

1

1 + t0 0

0 0 0

(378)

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

(379)

Material time derivatives of stretch ratio:

λn = λn (ndn) =3 + 2t√

5√

5 + 2t (3 + t). (380)

59


EXERCISE 42

A certain motion is described with the following equations:

x1 = 2X1 + tX2, x2 = X2 −X1, x3 = 3tX3. (381)

Determine J .

SOLUTION

The material time derivative of J is computed as

J = Jtrd = Jtrl. (382)

The deformation gardient and the volume change:

[F ] =

2 t 0−1 1 00 0 3t

, J = detF = 3t2 + 6t. (383)

The inverse motion:

X1 =x1 − tx22 + t

, X2 =x1 + 2x22 + t

, X3 =x33t. (384)

The Lagrange and the Euler velocity fields, respectively:

[V ] =

X2

03X3

, [v] =

x1 + 2x22 + t0x3t

. (385)

The velocity gradient:

[l] =

1

2 + t

2

2 + t0

0 0 0

0 01

t

, trl =

1

t+

1

2 + t. (386)

Therefore:

J = Jtrl =(3t2 + 6t

)(1

t+

1

2 + t

)

= 6 (1 + t) . (387)

60


EXERCISE 43

Prove that the rate of deformation tensor of the motion

x1 = α (t)X1, x2 = α (t)X2, x3 = α (t)X3 (388)

has the form

d =α

αδ (389)

SOLUTION


X1 =x1α (t)

, X2 =x2α (t)

, X3 =x3α (t)

(390)

The velocity field is

[V (X, t)] =

αX1

αX2

αX3

, [v (x, t)] =

α

αx1

α

αx2

α

αx3

(391)

Velocity gradient:

[l] =

∂v1∂x1

∂v1∂x2

∂v1∂x3

∂v2∂x1

∂v2∂x2

∂v2∂x3

∂v3∂x1

∂v3∂x2

∂v3∂x3

=

α

α0 0

0α

α0

0 0α

α

(392)

Its symmetric part is the rate of deformation:

d =1

2

(l + lT

), [d] =

α

α0 0

0α

α0

0 0α

α

. (393)

61


EXERCISE 44

A 2D motion of a continuum medium is given by

χ (X , t) =(X1 · t2

)e1 +

(eX2 · t

)e2. (394)

Determine the material time derivative of the left Cauchy–Green deformation tensor.

SOLUTION

Using the rule l = F F−1 we can write

b =

·(FF T

)= F F T + F F

T= lFF T + FF T lT = lb+ blT . (395)


X1 =x1t2, X2 = ln

x2t. (396)

The Lagrangean and Eulerian velocity field, respectively:

[V ] =

[2tX1

eX2

]

, [v] =

2x1tx2t

. (397)

The velocity gradient:

[l] =

2

t0

01

t

. (398)

The deformation gradient and the left Cauchy–Green deformation tensor:

[F ] =

[t2 00 eX2 · t

]

, [b] =

[t4 00 x22

]

. (399)

Substituting (398) and (399) into (395) we have

[

b]

=

4t3 0

02x22t

. (400)

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

b =∂b

∂t+ gradb · v, (401)

where the third-order tensor gradb is filled out of 0 elements, except the 222 element which is

(gradb)222 =∂b22∂x2

= 2x2. Therefore

[

b]

=

[4t3 00 0

]

+

4t3 0

02x22t

=

4t3 0

02x22t

. (402)

62


EXERCISE 45 The Eulerian velocity field is defined by the equations,

v1 = tx1sinx3, v2 = 4tx2cosx3, v3 = 0. (403)

At point P (−2, 1, 0), at time t = 1, determine d, w, the stretch rate per unit length in the

direction [n] =1√3[1,−1, 1] andthe maximum stretch rate per unit length and the direction in

which it occurs, respectively.

SOLUTION

The velocity gradient and its values at point P :

[l] =

tsinx3 0 tx1cosx30 4tcosx3 −4tx2sinx30 0 0

, at point P : [l] =

0 0 −20 4 00 0 0

. (404)

The rate of deformation and the vorticity tensor, respectively:

[d] =

0 0 −10 4 0−1 0 0

, [w] =

0 0 −10 0 01 0 0

. (405)

The stretch rate per unit length in the direction n:

λn = λn (ndn) =⇒ λnλn

= ndn =2

3. (406)

The maximum values ofλnλn

is the largest eigenvalue of d. The direction in which it occurs is

the corresponding eigenvector. The eigenvalues and eigenvectors of d are

µ1 = 4, µ2 = 1, µ3 = −1,

n1 = e2, n2 =1√2(−e1 + e3) , n3 =

1√2(e1 + e3) .

63


EXERCISE 46

Prove that the velocity field

[v] =

3x3 − 6x26x1 − 2x32x2 − 3x1

(407)

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

SOLUTION

The spatial velocity gradient:

[l] =

0 −6 36 0 −2−3 2 0

. (408)

Since it is a skew-symmetric tensor, it follows that w = l and d = 0. The axis of spin:

[ω] =

w32

w13

w21

=

236

. (409)

64


EXERCISE 47

A steady spatial velocity field is given by

[v] =

2x22x22

x1x2x3

. (410)

Determine the rate of extension at P (2, 2, 2) in the direction n =1√2(e1 + e2).

SOLUTION

The velocity gradient and its values at point P :

[l] =

0 2 00 4x2 0

x2x3 x1x3 x1x2

, at point P : [l] =

0 2 00 8 04 4 4

. (411)

The rate of deformation:

d =1

2

(l + lT

), [d] =

0 1 21 8 22 2 4

. (412)

λnλn

= ndn = 5. (413)

65


EXERCISE 48

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

[σ] =

200 300 −100300 0 0−100 0 100

. (414)

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

SOLUTION

The eigen normal vector of the plane is computed as

n =gradφ

|gradφ| =φ|φ| , (415)

where

[φ] =

∂φ

∂x1

∂φ

∂x2

∂φ

∂x3

=

1−32

, |φ| =√14. (416)

Therefore

[n] =1√14

1−32

. (417)

The Cauchy traction vector by definition is computed by

tn = σn, [tn] =1√14

−900300100

≈

−240.53580.178426.7261

. (418)

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

|tn| =√tntn =

√65000 ≈ 254.951,

σn = tnn = −800

7≈ −114.2857. (419)

The length of the component parallel to the plane φ is

τn =

√

|tn|2 − σ2n =

50

7

√1018 ≈ 227.901. (420)

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

tnn = |tn| |n| cosϕ = |tn| cosϕ (421)

ϕ = arccostnn

|tn|= arccos

σn|tn|

= 2.03562 = 116.632. (422)

66


EXERCISE 49

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

[σ] =

2x2x3 4x22 04x22 0 −2x10 −2x1 0

. (423)

Determine the traction vector corresponding to the plane φ (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

[σ] =

−3 9 09 0 −10 −1 0

. (424)

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

n =gradφ

|gradφ| =φ|φ| , (425)

where

[φ] =

∂φ

∂x1

∂φ

∂x2

∂φ

∂x3

=

2x12x21

, [φ]P =

131

. (426)

Therefore

[n] =1√11

131

. (427)

The Cauchy traction vector at this point is

tn = σn, [tn] =1√11

248−3

≈

7.232.41−0.90

. (428)

67


EXERCISE 50

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

[σ] =

4 3 03 −4 00 0 9

. (429)

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 basis vectors

e1 = e3, e2 =1√10

(3e1 + e2) , e3 =1√10

(−e1 + 3e2) . (430)

SOLUTION

The spherical (or hydrostatic) component is computed by

p =1

3(trσ) δ = 3δ, (431)

while the deviatoric part is determined as

s = σ − p, [s] =

4 3 03 −4 00 0 9

−

3 0 00 3 00 0 3

=

1 3 03 −7 00 0 6

. (432)

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

]

− λ

[1 00 1

])

= 0, (433)

λ2 − 25 = 0. (434)

Therefore the principal stresses are

σ1 = 9, σ2 = 5, σ3 = −5. (435)

In order to obtain the components of σ in the new coordinate system, first we consruct theorthogonal transformation matrix as

[Q] = [e1, e2, e3] =1√10

0 3 −10 1 3√10 0 0

. (436)

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

[σ] =[QT

][σ] [Q] =

1

10

0 0√10

3 1 0−1 3 0

4 3 03 −4 00 0 9

0 3 −10 1 3√10 0 0

, (437)

[σ] =1

10

0 0 9√10

15 5 05 −15 0

0 3 −10 1 3√10 0 0

=

9 0 00 5 00 0 −5

. (438)

Therefore e1, e2 and e3 are the unit eigenvectors of σ.

68


EXERCISE 51

Let a homogenouos deformation be given by

x1 = −X2, x2 = 2X1, x3 = 2X3. (439)

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

[σ] =

0 10 010 20 00 0 10

. (440)

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

[n] =

100

(441)

in the current configuration.

SOLUTION


[F ] =

0 −1 02 0 00 0 2

, J = detF = 4. (442)

Its inverse is

[F−1

]=

0 0.5 0−1 0 00 0 0.5

. (443)

The first and the second Piola-Kirchhoff tensors are

P = JσF−T , S = JF−1σF−T (444)

[P ] =

20 0 040 −40 00 0 20

, [S] =

20 −20 0−20 0 00 0 10

. (445)

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

da = JF−TdA0 (446)

dan = JdA0F−TN (447)

dA0N = da1

JF Tn ⇒ N =

1JF Tn

∣∣ 1JF Tn

∣∣, (448)

where

1

JF Tn =

0−0.25

0

⇒ N =

0−10

. (449)

69


Therefore

TN = PN , [TN ] =

0400

. (450)

The Cauchy traction vector is

tn = σn, [tn] =

0100

. (451)

70


EXERCISE 52

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

[P ] =

−20 0 0−120 80 400 20 0

. (452)

The inverse motion is known as

X1 = x1 − 3x3, X2 =1

kx2, X3 = x3. (453)

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.

SOLUTION

The motion is

x1 = X1 + 3X3, x2 = kX2, x3 = X3. (454)

The deformation gradient has the form

[F ] =

1 0 30 k 00 0 1

, J = detF = k. (455)

The Cauchy stress tensor by definition is

σ =1

JPF T , [σ] =

−20/k 0 00 80 40/k0 20 0

. (456)

Its first scalar invariant is

Iσ = trσ = 80− 20

k⇒ k =

20

80− Iσ(457)

If Iσ = 70 then

k =20

80− 70= 2. (458)

The Cauchy stress tensor in this case is

[σ] =

−10 0 00 80 200 20 0

. (459)

71


EXERCISE 53

Prove that the Jaumann–Zaremba 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

σJZ = σ −wσ + σw. (460)

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

˚σJZ = Q(σJZ

)QT . (461)

Expanding the left-hand side yields:

˚σJZ = ˙σ − wσ + σw, (462)

where

˙σ =˙(

QσQT)= QσQT +QσQT +QσQ

T, (463)

w = QQT +QwQT . (464)

Thus

˚σJZ = QσQT +QσQT +QσQT

(465)

−(

QQT +QwQT)

QσQT +QσQT(

QQT +QwQT)

(466)

= QσQT +QσQT +QσQT

(467)

−QQTQσQT −QwQTQσQT +QσQT QQT +QσQTQwQT . (468)

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

˚σJZ = QσQT +QσQT +QσQT

(469)

−QσQT −QwσQT +QσQT QQT +QσwQT (470)

= QσQT +QσQT −QwσQT +QσQT QQT +QσwQT . (471)

Since QQT = Ω is a skew-symmetric tensor, it follows that ΩT = −Ω. Thus,(

QQT)T

=

QQT= −QQT . Using this relation in the fourth term yields

˚σJZ = QσQT +QσQT −QwσQT −QσQTQQ

T+QσwQT (472)

= QσQT +QσQT −QwσQT −QσQ

T+QσwQT (473)

= QσQT −QwσQT +QσwQT (474)

Therefore

˚σJZ = Q (σ −wσ + σw)QT . (475)

Thus, (461) is satisfied.

72


EXERCISE 54

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

SOLUTION

The Green–Naghdi rate of the Cauchy stress tensor is defined as

σGN = σ −(

RRT)

σ + σ(

RRT)

. (476)

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

˚σGN = Q(σGN

)QT . (477)

Expanding the left-hand side yields:

˚σGN = ˙σ −(˙RR

T)

σ + σ(˙RR

T)

, (478)

where

˙σ =˙(

QσQT)= QσQT +QσQT +QσQ

T, (479)

R = QR, (480)˙R = QR +QR. (481)

Thus

˚σGN = QσQT +QσQT +QσQT

(482)

−(

QR+QR)

(QR)T QσQT (483)

+QσQT(

QR +QR)

(QR)T (484)


(485)

−(

QRRTQT +QRRTQT)

QσQT (486)

+QσQT(

QRRTQT +QRRTQT)

(487)


(488)

−(

QQTQσQT +QRRTQTQσQT)

(489)

+(

QσQT QQT +QσQTQRRTQT)

(490)


(491)

−(

QσQT +QRRTσQT)

+(

QσQT QQT +QσRRTQT)

(492)

= QσQT +QσQT −QRRTσQT +QσQT QQT +QσRRTQT . (493)

Since QQT = Ω is a skew-symmetric tensor, it follows that ΩT = −Ω. Thus,(

QQT)T

=

QQT= −QQT . Using this relation in the fourth term yields

˚σGN = QσQT +QσQT −QRRTσQT −QσQTQQ

T+QσRRTQT (494)

= QσQT +QσQT −QRRTσQT −QσQ

T+QσRRTQT (495)

= QσQT −QRRTσQT +QσRRTQT . (496)

73


Therefore

˚σGN = Q(

σ −(

RRT)

σ + σ(

RRT))

QT . (497)

Thus, (477) is satisfied.

74


EXERCISE 55

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

x1 = X1 + tX2, x2 = tX2, x3 = X3. (498)

The Cauchy stress at this point is

[σ] =

10 0 −100 30 0

−10 0 0

. (499)

Prove the identity P : F =1

2S : C.

SOLUTION

We can write the stress power as

P : F = tr(

P T F)

= tr(

P TF−TF T F)

= tr((

F−1P)T

(

F T F))

, (500)

P : F = tr(

ST(

F T F))

= tr(

S(

F T F))

. (501)

Since S is symmetric and C = F TF + F T F , we can write

P : F = S :(

F T F)

=1

2S : C.

The deformation gradient, its inverse and its determinant:

[F ] =

1 t 00 t 00 0 1

,[F−1

]=

1 −1 0

01

t0

0 0 1

, J = detF = t. (502)

The right Caucy–Green deformation tensor and its material time derivative:

[C] =[F TF

]=

1 t 0t 2t2 00 0 1

,[

C]

=

[∂C

∂t

]

=

1 1 01 4t 00 0 0

. (503)

The inverse motion:

X1 = x1 − x2, X2 =x2t, X3 = x3. (504)

The Lagrangean and Eulerian velocity fields are

[V ] =

X2

X2

0

, [v] =

x2tx2t0

. (505)

75


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

[l] =

01

t0

01

t0

0 0 0

,

[

F]

= [lF ] =

0 1 00 1 00 0 0

. (506)

The first and the second Piola–Kirchoff stress tensors are

[P ] =[JσF−T

]=

10t 0 −10t−30t 30 0−10t 0 0

, (507)

[S] =[F−1P

]=

40t −30 −10t

−3030

t0

−10t 0 0

. (508)

Therefore we can compute

P : F = 30, (509)1

2S : C = 30. (510)

76


EXERCISE 56

Let the Caucy stress distribution in a bricket domain with dimension

x1 = 0 . . . 2, x2 = 0 . . . 3, x3 = 0 . . . 4 (511)

be given as

[σ] =

5x21 4 −3x24 x1x2x3 x3

−3x2 x3 0

. (512)

Prove the Gauss’ divergence theorem.

SOLUTION

The Gauss’ divergence theorem for this problem is written as

∫

(a)

σnda =

∫

(v)

divσdv (513)

∫

(a)

tda =

∫

(v)

divσdv

where n is the outward unit normal field acting along the surface (a), while da and dv areinfinitesimal surface and volume elements at x, respectively, whereas t is the surface Cauchytraction vector at x.The whole surface is composed of six surfaces with domains and unit normals

a1 : = x2 = 0 . . . 3; x3 = 0 . . . 4 , (514)

a1 : = x1 = 0 . . . 2; x3 = 0 . . . 4 , (515)

a3 : = x1 = 0 . . . 2; x2 = 0 . . . 3 (516)

a4 = a1 a5 = a2 a6 = a3 (517)

n1 = e1 n2 = e2 n3 = e3 n4 = −e1 n5 = −e2 n6 = −e3 (518)

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

t1 = (σn1) |x1=2 [t1] =

204

−3x2

; t2 = (σn2) |x2=3 [t2] =

43x1x3x3

t3 = (σn3) |x3=4 [t3] =

−3x240

; t4 = (σn4) |x1=0 [t4] =

0−43x2

t5 = (σn5) |x2=0 [t5] =

−40

−x3

; t6 = (σn6) |x3=0 [t2] =

3x200

77


With this in hand, we can compute the left-hand side in (513):

∫

(a)

σnda =

4∫

0

3∫

0

t1dx2dx3 +

4∫

0

2∫

0

t2dx1dx3 +

3∫

0

2∫

0

t3dx1dx2

+

4∫

0

3∫

0

t4dx2dx3 +

4∫

0

2∫

0

t5dx1dx3 +

3∫

0

2∫

0

t6dx1dx2 (519)

∫

(a)

σnda

=

24048−54

+

324816

+

−27240

+

0−4854

+

−320

−16

+

2700

∫

(a)

σnda = 240e1 + 72e2 . (520)

In order to calculate the right-hand side in (513), first the divergence of the Cauchy stress needto be computed:

divσ = σ · ∇ =∂σ

∂xc· ec =

∂σab∂xc

ea (eb · ec) =∂σab∂xc

δbcea = σab,cδbcea = σac,cea

[divσ] =

∂σ11∂x1

+∂σ12∂x2

+∂σ13∂x3

∂σ21∂x1

+∂σ22∂x2

+∂σ23∂x3

∂σ31∂x1

+∂σ32∂x2

+∂σ33∂x3

=

10x11 + x1x3

0

. (521)

The right-hand side in (513):

∫

(v)

divσdv =

4∫

0

3∫

0

2∫

0

divσdx1dx2dx3 (522)

∫

(v)

divσdv

=

4∫

0

3∫

0

202 + 2x3

0

dx2dx3 =

4∫

0

606 + 6x3

0

dx3 =

240720

∫

(v)

divσdv = 240e1 + 72e2 . (523)

78


EXERCISE 57

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 (524)

SOLUTION

The Stokes’ theorem:∮

c

uds =

∫

(a)

(curlu)nda. (525)

The closed curve c is consist of four line elements:

c1 : = x1 = 0 . . . 2 c2 := x2 = 0 . . . 1c3 : = x1 = 0 . . . 2 c4 := x2 = 0 . . . 1

with infinitesimal tangent vectors:

ds1 = dx1e1 ds2 = dx2e2 ds3 = −dx1e1 ds4 = −dx2e2.

The left-hand side of (525) is computed with the following four integral

∮

c

uds =

2∫

0

(uds1)|x2=0 +

1∫

0

(uds2)|x1=2 +

2∫

0

(uds3)|x2=1 +

1∫

0

(uds4)|x1=0

∮

c

uds =

2∫

0

(0) dx1 +

1∫

0

(2x2) dx2 +

2∫

0

(−x1) dx1 +1∫

0

(0) dx2

∮

c

uds = 0 + 1− 2 + 0∮

c

uds = −1 (526)

In order to calculate the right-hand side in (525) first we need to compute the curl of the vectorfield:

curlu = ∇× u = ea ×∂u

∂xa=∂ub∂xa

ea × eb =∂ub∂xa

ǫabcec = ub,aǫabcec. (527)

[curlu] =

∂u3∂x2

− ∂u2∂x1

∂u1∂x3

− ∂u3∂x1

∂u2∂x1

− ∂u1∂x2

=

x1−x2

x2 − x1

.

The right-hand side in (525):∫

(a)

(curlu)nda =

∫

(a)

(x2 − x1) da

79


since the normal of the surface is n = e3. Therefore

∫

(a)

(curlu)nda =

1∫

0

2∫

0

(x2 − x1) dx1dx2 =

1∫

0

(2x2 − 2) dx2 = −1, (528)

∫

(a)

(curlu)nda = −1 . (529)

80


EXERCISE 58

Consider the motion

x1 =

(

1 +t

k

)

X1, x2 = 2

(

1 +t

k

)

X2, x3 = 3

(

1 +t

k

)

X3 (530)

where k is a constant. From the conservation of mass and the initial condition ρ (t = 0) = ρ0,determine ρ as a function of ρ0, t, and k.

SOLUTION


[F ] =

(

1 +t

k

)

1 0 00 2 00 0 3

, J = detF = 6(k + t)3

k3. (531)

Sinceρ0 = Jρ it follows that

ρ =ρ0J

=ρ0k

3

6 (k + t)3(532)

81


EXERCISE 59 Consider the motion of a body described by the mapping

x1 = X1, x2 =X2

1− tX2

, x3 = X3. (533)

Determine the material density as a function of position vector x and time t without using F .

SOLUTION

The inverse mapping:

X1 = x1, X2 =x2

1 + tx2, X3 = x3.

The velocity field is

V (X, t) =X2

2

(1− tX2)2E2, v (x, t) = x22e2. (534)

From the continuity equation we can write

ρ+ ρdivv = 0 ⇒ dρ

dt= −ρdivv = −2ρx2 (535)

1

ρdρ = −2x2dt = −2

X2

1− tX2

dt (536)

ρ∫

ρ0

1

ρdρ = −2

t∫

0

X2

1− tX2dt, lnρ− lnρ0 = 2ln (1− tX2)− 2ln1 (537)

lnρ

ρ0= ln (1− tX2)

2 ⇒ ρ (X , t) = ρ0 (1− tX2)2 (538)

Using the inverse mapping we have

ρ (x, t) =ρ0

(1 + tx2)2 (539)

82


EXERCISE 60

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

v1 = (αx1 − x2) t, v2 = x1 − αx2, v3 = 0,

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

SOLUTION

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

ρ+ ρtrl =∂ρ

∂t+ gradρ · v + ρtrl = 0. (540)

In this example gradρ = 0, therefore equation (540) reduces to

∂ρ

∂t+ ρtrl = 0. (541)

The spatial velocity gradient and its trace:

[l] =

αt −t 01 −α 00 0 0

, trl = −α (1− t) . (542)

Integration (541) yields

ρ∫

ρ0

1

ρdρ = α

t∫

0

(1− t) dt, lnρ

ρ0= α

(

t− t2

2

)

(543)

ρ (t) = ρ0 · eα(

t− t2

2

)

. (544)

83


EXERCISE 61

The velocity field of an incompressible flow be given as

v =x1r2

e1 +x2r2

e2 (545)

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

SOLUTION

The continuity equation for incompressible material reduces to

ρ+ ρdivv = 0 ρ = constant−−−−−−−−−→ divv = 0. (546)

[divv] =∂v1∂x1

+∂v2∂x2

=

(1

x21 + x22− 2x21

(x21 + x22)2

)

+

(1

x21 + x22− 2x22

(x21 + x22)2

)

= 0. (547)

84


EXERCISE 62

A dynamical process is given by the deformation mapping and the Cauchy stress respectivelyas

x1 = etX1 − e−tX2, x2 = etX1 + e−tX2, x3 = X3, (548)

[σ] =

x21 x1x3 0x1x3 x2 00 0 x33

. (549)

Determine the body force vector q so that Cauchy’s first equation of motion is satisfied.

SOLUTION

The Cauchy’s first equation of motion:

divσ + q = ρa ⇒ q = ρa− divσ. (550)

In order to calculate q, we need to compute ρ, a and divσ.The inverse motion is

X1 =1

2e−t (x2 + x1) , X2 =

1

2et (x2 − x1) , X3 = x3.

The deformation gradient:

[F ] =

et −e−t 0et e−t 00 0 1

⇒ J = 2 ⇒ ρ =1

2ρ0. (551)

The Lagrangean velocity and acceleration fields:

[V (X, t)] =

etX1 + e−tX2

etX1 − e−tX2

0

[A (X, t)] =

etX1 − e−tX2

etX1 + e−tX2

0

. (552)

The Eulerian acceleration field:

[a (x, t)] =

x1x20

. (553)

The deivergence of the Cauchy stress:

[divσ] =

∂σ11∂x1

+∂σ12∂x2

+∂σ13∂x3

∂σ21∂x1

+∂σ22∂x2

+∂σ23∂x3

∂σ31∂x1

+∂σ32∂x2

+∂σ33∂x3

=

2x11 + x33x23

. (554)

Therefore the body force:

[q] =1

2ρ0

x1x20

−

2x11 + x33x23

(555)

85

Documents

Continuum Mech Problems