Upload
srinismech1830
View
222
Download
0
Embed Size (px)
Citation preview
7/27/2019 Kossa ContMech Problems 2012 Fall
1/84
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
2/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
(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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
3/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
4/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
4
7/27/2019 Kossa ContMech Problems 2012 Fall
5/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
(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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
6/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
7/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
8/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
8
7/27/2019 Kossa ContMech Problems 2012 Fall
9/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
9
7/27/2019 Kossa ContMech Problems 2012 Fall
10/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 6 September 2012
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)
10
7/27/2019 Kossa ContMech Problems 2012 Fall
11/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
12/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
13/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
14/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
4
7/27/2019 Kossa ContMech Problems 2012 Fall
15/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
16/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
17/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 13 September 2012
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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
18/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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.
1
7/27/2019 Kossa ContMech Problems 2012 Fall
19/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
20/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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.
3
7/27/2019 Kossa ContMech Problems 2012 Fall
21/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
Referenceconfiguration:
Spatialconfiguration:
Figure 2: Illustration of the stretches
4
7/27/2019 Kossa ContMech Problems 2012 Fall
22/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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
5
7/27/2019 Kossa ContMech Problems 2012 Fall
23/84
7/27/2019 Kossa ContMech Problems 2012 Fall
24/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
[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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
25/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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)
8
7/27/2019 Kossa ContMech Problems 2012 Fall
26/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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
9
7/27/2019 Kossa ContMech Problems 2012 Fall
27/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
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.
10
7/27/2019 Kossa ContMech Problems 2012 Fall
28/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 20 September 2012
Figure 6: Illustration of the displacement vector
11
7/27/2019 Kossa ContMech Problems 2012 Fall
29/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
30/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
31/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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
The deformation gradient is
[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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
32/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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
.
4
7/27/2019 Kossa ContMech Problems 2012 Fall
33/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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
The deformation gradient is
[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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
34/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
35/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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
The deformation gradient is
[F] =
1 + X2 X1 01 1 0
0 0 1
. (41)
The right CauchyGreen deformation tensor is
[C] =FTF
= 1 + (1 + X2)
2 X1 (1 + X2)
1 0
X1 (1 + X2) 1 1 + X21 00 0 1
. (42)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
36/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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.
8
7/27/2019 Kossa ContMech Problems 2012 Fall
37/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 27 September 2012
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)
9
7/27/2019 Kossa ContMech Problems 2012 Fall
38/84
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
39/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 4 October 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
40/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 4 October 2012
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
41/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 4 October 2012
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)
4
7/27/2019 Kossa ContMech Problems 2012 Fall
42/84
7/27/2019 Kossa ContMech Problems 2012 Fall
43/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 4 October 2012
EXERCISE 29
Compute the generalized material strain measures for the homogenous deformation
[F] = 1 1 02 1 0
0 0 1
. (48)
SOLUTION
The right CauchyGreen deformation tensor is
[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)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
44/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
45/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
Thus, it follows that
[v (xP, 1)] =
2
1 + 1cos1
2
1 + 1
0.5403
1
, (10)
[a (xP, 1)] =
2
1 + 1sin1
0
0.8415
0
. (11)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
46/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
The inverse motion has the form
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
47/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
Thus, it follows that
[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)
4
7/27/2019 Kossa ContMech Problems 2012 Fall
48/84
7/27/2019 Kossa ContMech Problems 2012 Fall
49/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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
The inverse motion has the form
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)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
50/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
51/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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
The inverse motion is
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)
8
7/27/2019 Kossa ContMech Problems 2012 Fall
52/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
The inverse motion has the form
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)
9
7/27/2019 Kossa ContMech Problems 2012 Fall
53/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
10
7/27/2019 Kossa ContMech Problems 2012 Fall
54/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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
The inverse motion has the form
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)
11
7/27/2019 Kossa ContMech Problems 2012 Fall
55/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 25 October 2012
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)
12
7/27/2019 Kossa ContMech Problems 2012 Fall
56/84
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
The deformation gradient is
[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)
The inverse motion has the form
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
57/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
58/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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)
The inverse motion is
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
59/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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
The inverse motion is
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)
4
7/27/2019 Kossa ContMech Problems 2012 Fall
60/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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)
The inverse motion has the form
X1 =x1
t2, X2 = ln
x2
t. (27)
The Lagrangian and Eulerian velocity fields, respectively:
[V] =
2tX1eX2
, [v] =
2x1tx2
t
. (28)
The spatial velocity gradient:
[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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
61/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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.
6
7/27/2019 Kossa ContMech Problems 2012 Fall
62/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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
The spatial velocity gradient:
[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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
63/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 8 November 2012
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)
8
7/27/2019 Kossa ContMech Problems 2012 Fall
64/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
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
.
1
7/27/2019 Kossa ContMech Problems 2012 Fall
65/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
66/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
67/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
[] =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.
4
7/27/2019 Kossa ContMech Problems 2012 Fall
68/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
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
The deformation gradient is
[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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
69/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
Therefore
TN = PN, [TN] =
0400
. (36)
The Cauchy traction vector is
tn = n, [tn] =
010
0
. (37)
6
7/27/2019 Kossa ContMech Problems 2012 Fall
70/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 15 November 2012
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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
71/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 22 November 2012
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.
1
7/27/2019 Kossa ContMech Problems 2012 Fall
72/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 22 November 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
73/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 22 November 2012
Therefore
GN = Q
RRT
+
RRT
QT. (38)
Thus, (18) is satisfied.
3
7/27/2019 Kossa ContMech Problems 2012 Fall
74/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
1
7/27/2019 Kossa ContMech Problems 2012 Fall
75/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
2
7/27/2019 Kossa ContMech Problems 2012 Fall
76/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
3
7/27/2019 Kossa ContMech Problems 2012 Fall
77/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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).
4
7/27/2019 Kossa ContMech Problems 2012 Fall
78/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
5
7/27/2019 Kossa ContMech Problems 2012 Fall
79/84
7/27/2019 Kossa ContMech Problems 2012 Fall
80/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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
The deformation gradient is
[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)
7
7/27/2019 Kossa ContMech Problems 2012 Fall
81/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
8
7/27/2019 Kossa ContMech Problems 2012 Fall
82/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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)
9
7/27/2019 Kossa ContMech Problems 2012 Fall
83/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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.
10
7/27/2019 Kossa ContMech Problems 2012 Fall
84/84
BMEGEMMMW03 / Continuum Mechanics / Exercises 29 November 2012
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