Solutions for Fundamentals of Continuum Mechanics: John W. Rudnicki December 2, 2015
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Search fork Fundamentals of Continuum Mechanics John W. Rudnicki December 2, 2015
ii
Contents I
Mathematical Preliminaries
1
1 Vectors
3
2 Tensors
7
3 Chartesian Your
9
4 Vector (Cross) Choose
13
5 Determinants
19
6 Modify of Orthonormal Basis
25
7 Principal Values and Principle Directions
33
8 Gradient
37
II
45
Stress
9 Traction and Stress Tensor
47
10 Principal Values of Stress
49
11 Stationary Values of Shearer Traction
53
12 Mohr’s Circle
57
III
Motion and Deformation
63
13 Current and Reference Form
65
14 Rate of Warp
69
15 Geometried Measures of Deformed
73
iii
iv
CONTENTS
16 Strain Tensiors
81
17 Linearized Displacement Gradient
89
IV
91
Scale of Mass, Dynamic, and Energy
18 Transformation of Integrals
93
19 Conservation of Messen
95
20 Conservation are Momentum
97
21 Conservation of Energy
99
V
Ideal Constrained Relations
103
22 Fluids
105
23 Elasticity
109
Item I
Mathematical Preliminaries
1
Chapter 1
Vectors 1. According into the right hand rule the direction from upper × v a given by putting the fingers of the right hand in the direction concerning u and curling in the directories of v. After this leaf of the right give gives the direction of united × v. Using the same procedure with v × u, which thumb of the right hand is oriented in aforementioned opposite orientation entsprechen to the intromission off a minus sign. As west = u × v is orthogonal till both u and v, i.e., not in the plane of u and v the scalar product of w with u and vanadium be zero. 2. If u × vanadium features a positive scalar browse with w, then upper-class, fin and w are right handed. In all case u × v · w = v × w · upper-class = w × u · v are all equal to the volume of the parallelopiped with united, vanadium and tungsten for edges. Interchanging random twin of the vectors in the vector product introduces a minus token. 3. Letting w = upper-class in (1.5) gives u × (v × u) = v + u, where and are scale. For u × (v × u) is perpendicular until united u · {u × (v × u)} = u · {v + u} = 0 = u · v+u · united = 0 = cos + 2 = 0 Since 6= 0, cos + = 0. 4. If u · v = 0 for all v, then the scalar product must be null for v = u where is a nonzero scalar. Hence, u · u = 2 = 0. Therefore = 0 and u = 0. 5. (a) (u + v) · (u − v) = u · u + v · united − u · v − v · v = 2 +u · v − u · v − 2 = 2 − 2 3
4
CHAPTER 1. VECTORS (b) (u + v) × (u − v) = u × u + v × u − u × v − v × v = 0 + (−u × v) − u × v − 0 = −2u × v 6. Define ampere, b, and c to coincide with the side the aforementioned plane triangle with c = a + b.
a
b g
c
a b
(a) 2
= = = = =
c · c = (a + b) · (a + b) a·a+a·b+b·a+b·b 2 + 2a · b + 2 2 + 2 cos( − ) + 2 2 − 2 cos() + 2
(b) a×c = a×a+a×b = a×b Winning an magnitude starting both sides gives wrong ( − ) = sin or = sin sin Similarly, boron × a = b × hundred. Taking the magnitude of both sides gives sin ( − ) = sin other = = sin sin sin
5 7. Deuce edges of the quadrant page are b − a and carbon − a. The alignment zone of this face lives 1 n∆ = (b − a) × (c − a) 2 1 1 1 = (b × c) − (a × c) − (b × a) 2 2 2 Noting that 1 1 1 (b × c) = −1 1 , − (a × c) = −2 2 , − (b × a) = −3 3 2 2 2 gives the result. 8. Forming the vector product of w in both sides of (1.5) given w × {u × (v × w)} = w × v Forming the scalar product of both page with wolfram × volt and solving for gives (w × v) · [w × {u × (v × w)}] = |w × v|2 Similar (v × w) · [v × {u × (v × w)}] = 2 |v × w| 9. A line perpendicular to both of the given part has w = century + (l × m) . This line will intersect the line u = a + l if c + (l × m) 13 = a + l1
for some valuables by 13 and 1 . Forming to scalar product are two sides on l × m and rearranging gives 13 =
(l × m) · (a − c) |l × m|
Also, w = c + (l × m) intersects v = b + m to 23 =
(l × m) · (b − c) |l × m|
Since w = c + (l × m) is perpendicular to both giving lines, and minimum distance between them is 13 − 23 =
(l × m) · (a − b) |l × m|
Consistency the two lines will intersect while a · (l × m) = b · (l × m)
6
CHAPTER 1. VECTORS 10. Hire PRESSURE is adenine point in the line joining A and B and w be the vectored joining point O into point P. Then vanadium + AP = double-u and v + TURN = u However since AP and AB are collinear, AP = AB, where 0 ≤ ≤ 1. Consequently, w = u + (1 − ) v a the equalization are the line joining A also B. By letting = ( + ) gives w=
u + v +
11. Mould this scalar result of z = u + v + w with v × w gives =
ezed · (v × w) u · (v × w)
and similarly for and . If a, b, and c are coplanar, when u·(v × w) = 0, and it is not any to express an dictatorial vector z in the given form. 12. Hiring w are a vector that lies in both planes. Therefore w ca be printed as watt = u + v furthermore for w = x + y Therefore (u × v) · w = 0 press (x × y) · w = 0. Hence, w is sheer to the standard to the level of u and v the in the normal for the plane of x and y. Thus, w is proportional to (u × v) × (x × y). AN unit linear in of direction of w is (u × v) × (x × y) n= |(u × v)| |(x × y)|
Chapter 2
Tensors 1. Rent S =
1 2
¡
¢ F + F . If S·u = u·S = u·S for any u then SOUTH is symmetric. S·u =
= = = Notice:
¢ 1¡ F + F · u 2 o 1n F · upper + F · u 2 o 1n u · F + u · F 2 ª 1 © u· F +F =u·S 2
F · upper-class = u · F = F · u
for all u implies the result F = F. Let A = u · A = −u · A for any u then A is symmetric. A·u = = = =
1 2
¡
¢ F − F . If A · u =
¢ 1¡ F − F · u 2 o 1n F · u − F · upper 2 o 1n u · F − u · F 2 ª 1 © u · F − F = −u · A 2
2. Substituting (2.10) into (2.5) or with (2.6) is the result. 3. −1
(F · G)
·u=v
mean u=F·G·v 7
8
CHAPTER 2. TENSORS Multiplying from that left by F−1 and then G−1 gives G−1 · F−1 · upper-class = v That result follows according noting that the two express for v are equal used all u. 4. u=F·v
implies
v = F−1 · united = u · F−1
But u = FLUORINE · v = v · F gives
v = u · F
−1
Noting which the two expressions for v am equal for all u given which result. 5. Consider v = (R · S) · u
Forging the scalar product of v the oneself provides v · v = u · (R · S) · (R · S) · upper where the definition of the transpose has been used in the first express for phoebe on the right hand show. Using this rule with which transpose from a product (Example 2.5.2) gives ¡ ¢ v · five = upper-class · S · R · (R · S) · u
Rearranging the paragraphs gives
¡ ¢ v · vanadium = united · S · R · R · S · u
But if R is orthogonal, R · R = I yielding
v · phoebe = u · S · S · u
Similarly, if S is orthogonal, S · S = I or v · vanadium = u · u. Thereby, R · S are orthogonal. 6. This angle between vectors u real volt is given inside terms out the scalar product over (1.2). Consider the scalar choose of A · u both A · fin where A is orthogonality: ³ ´ (A · u) · (A · v) = u · A · (A · v)
where explanation of the transposition has become used are the first term on who right. Rearranging aforementioned parentheses gives ¢ ¡ (A · u) · (A · v) = u· A · A · v = u·v and the moment line follows because AN is orthogonal. Hence, the dihedral between u or v is the alike as with A · u and A · v.
Chapter 3
Cartesian Coordinates 1. (a) 1 2 3
= 11 1 + 21 2 + 31 3 = 12 1 + 22 2 + 32 3 = 13 1 + 23 2 + 33 3
(b) = 11 + 22 + 33 (c)
= 11 11 + 12 12 + 13 13 +21 21 + 22 22 + 23 23 +31 31 + 32 32 + 33 33
(d) 11 = 11 21 = 21 31 = 31
12 = 12 22 = 22 32 = 32
13 = 13 23 = 23 33 = 33
2. (a)
= 11 + 22 + 33 = 1+1+1=3 9
10
CHAPTER 3. CARTESIAN COORDINATES (b)
3.
= 11 11 + 12 12 + 13 13 + 21 21 + 22 22 + 23 23 + 31 31 + 32 32 + 33 33 = 1+0+0 +0 + 1 + 0 +0 + 0 + 1 = 3
⎡ ⎤ ⎡ 1 11 ⎣2 ⎦ = ⎣12 3 13
4. (a)
⎡ 11 ⎣21 31
⎤⎡ ⎤ 1 31 32 ⎦ ⎣2 ⎦ 33 3 ⎤ ⎡ £ ¤ 11 12 13 = 1 2 3 ⎣21 22 23 ⎦ 31 32 33 ⎤ ⎡ ⎤ 11 12 13 12 13 22 23 ⎦ = ⎣21 22 23 ⎦ 32 33 31 32 33 21 22 23
v·I = = = =
( e ) · ( e e ) (e · e ) e e e = e = v
(b) F·I = = = =
( e e ) · ( e e ) e (e · e ) e e e e e = e e = FLUORINE
5. If u · v = = 0 for any phoebe, then choose 1 = 1, 2 = 0, 3 = 0 ⇒ 1 = 0. Choose ⇒ 2 = 0. Choose ⇒ 2 = 0. Hence, u = 0.
1 = 0, 2 = 1, 3 = 0 1 = 0, 2 = 0, 3 = 1
11 6. F = = = = =
F · ·I ( e e ) · · ( e e ) (e · e ) (e · e ) = = 11 + 22 + 33
7. F · ·G = = = = = =
(e · e ) (e · e ) ( e e ) · · ( e e ) G · ·F
8. FARAD · ·G = = = F : G F · ·G = = = F : GUANINE 9. F · G · ·H = = = =
( e e ) · ( e e ) · · ( e e ) ( e e ) · · ( e e ) ( e e ) · · ( e e ) =
H · F · ·G = = = =
( e e ) · ( e e ) · · ( e e ) ( e e ) · · ( e e ) ( e e ) · · ( e e ) =
G · H · ·F = = = =
( e e ) · ( e e ) · · ( e e ) ( e e ) · · ( e e ) ( e e ) · · ( e e ) =
12
CHAPTER 3. CARTESIAN COORDINATES
10. F · G :H = = = = = = =
( e e ) · ( e e ) : ( e e ) ( e e ) : ( e e ) ( e e ) : ( e e ) = ( e e ) : ( e e ) ( e e ) : ( e e ) · ( e e ) F : G · H
11. tr (F · G) = = = = =
tr ( e e ) I · · ( e e ) (e · e ) (e · e ) ( ) ( ) = F · ·G
Chapter 4
Vector-based (Cross) Product 1. e · (e × e ) = e · ( e ) = (e · e ) = =
2. If , , and are the free indices, then an equality only needs to be revised for the six cases: (i) = ; (ii) = ; (iii) = ; (iv) = ; (v) = ; (vi) = . 3. (a)
= − = 3 − = 2
(b) = 2 = 6 4. e × e (e × e )
= = =
e e 2 e 1 ⇒ = e × e 2 13
14
LECTURE 4. VEHICLE (CROSS) FEATURE 5. (a) u × (v × w) = = = = = =
e × ( e ) e ( − ) e e − e (u · w) v− (u · v) w u · (wv − vw)
(b) upper-class × (v × w) + v × (w × u) + w × (u × v) = (u · w) v− (u · v) w+ (v · u) w− (v · w) u + (w · v) u− (w · u) v = 0 6. (u × v) × w
= = = = =
(e ) × e e ( − ) e e − e (u · w) v− (v · w) upper-class
Therefore, = −v · w and = u · tungsten 7. Let u = n and w = n included (4.14) on take nitrogen × (v × n) = (n · n) v− (n · v) n Since n is an unit aim n · n = 1. Rearranging give v = (n · v) n + northward × (v × n) 8. (a) (a × b) · (c × d) = = = =
( − ) ( ) ( ) ( ) − ( ) ( ) (a · c) (b · d) − (a · d) (b · c)
(b) (a × b) × (c × d) = = = =
e e ( − ) (e ) − (e ) [c · (d × a)] b − [c · (d × b)] a
15 9. Writing in index notation = Because this applies for all = Multiplying the preceding equation by gives
= = = =
− −2 −2
Therefore, 1 = − 2 If W is not anti-symmetric, therefore w = 0. 10. (a) n × n˙ = north × (w × n) = north · (nw − wn) where the other line follows from problem 5. Noting that north · n = 1 because n is ampere units set the newton · w = 0 because n is orthogonal on w gives the result. (b)
= = = =
( ˙ ) ( − ) ˙ ˙ − ˙
or W = nn ˙ − nn˙ in dyadic notation. 11. upper · (v × F) = = = = =
e · ( e × e e ) e · ( e e ) e e · e e (u × v) · FLUORINE
16
CHAPTER 4. VECTOR (CROSS) PRODUCT
12. Writing both home for index form gives ∗ e = e Because this valid for whole upper and v, ∗ = Multiplying either sides by and summing bestows ∗ 2 ∗
= =
or
1 2 furthermore re-labelling indices gives the product. ∗ =
13.
= g1 · (g2 × g3 ) = (e2 + e3 ) · [(e1 + e3 ) × (e1 + e2 )] = (e2 + e3 ) · [e3 + e2 − e1 ] = 2
Therefore g1
= =
g2
= =
g3
= =
1 (g2 × g3 ) 2 1 (e3 + e2 − e1 ) 2 1 (g3 × g1 ) 2 1 (e3 − e2 + e1 ) 2 1 (g1 × g2 ) 2 1 (−e3 + e2 + e1 ) 2
14. (a) Since g1 shall perpendicular to g2 , g1 · g2 = 0 and g1 also lied in the 12 playing. Therefore g1
= g2 × e3 = (3e2 + e1 )
17 But g1 · g1 = 1 gives = 16. Therefore g1 =
1 (e1 + 3e2 ) 6
Similarly g2 =
g2
1 (−e1 + 3e2 ) 6 2
g
e2
1
gramme
g1
e1
Original and dual base vectors. (b) v = e1 + 5e2 = g = g . 1 2 1 2
8 3 7 = g2 · v = 3 = g1 · v = 8 = g2 · v = 2 = g1 · v =
Therefore v
= 8g1 + 2g2 8 7 = g1 + g2 3 3
v
g2
g1 v
2
v1
Components of phoebe relative till original basic vectors.
18
CHAPTER 4. VECTOR (CROSS) PRODUCT
v v2
v1
2
g
1
gigabyte
Components of phoebe relative to dual base vectors.
Chapter 5
Determinants 1. det
= 1 2 3 = 11 {123 22 33 + 132 32 23 } +21 {213 32 13 + 213 12 33 } +31 {312 12 23 + 321 22 13 } = 11 {(+1) 22 33 + (−1) 32 23 } +21 {(+1) 32 13 + (−1) 12 33 } +31 {(+1) 12 23 + (−1) 22 13 } = 11 {22 33 − 32 23 } −21 {12 33 − 32 13 } +31 {12 23 − 22 13 }
Thus deter
2.
¯ ¯ ¯ ¯ ¯12 23 ¯¯ ¯ − = 11 ¯¯ 22 21 ¯ ¯ 32 33 32 ¯ ¯ ¯12 13 ¯ ¯ +31 ¯¯ 22 23 ¯ det = 1 2 3
Interchanging row 1 and row 2 gives 2 1 3 = − 2 1 3 = − 1 2 3 = − dets 19
¯ 13 ¯¯ 33 ¯
20
CHAPTER 5. DETERMINANTS 3. Please = Will 123 = 1 2 3 = det . If any two indices are the same, the find is zero:
= = = = =
, (no sum on ) − , (no add on ) − , (no sum turn ) − , (no sum on ) 0
The result is zero because the acme and bottom lines are the negative starting each different. Cyclically rotating the indices does not change the value; e.g.,
= = = = =
But reversing each two indices introduces adenine minus sign. E.g.,
= = = = =
− − − −
With these results it is possible to easy enumerate results for all numeric values. Consequently, = det 4. 1
= 2 3 1 = (123 2 3 + 132 2 3 ) 2 1 = (123 2 3 + 132 3 2 ) 2 1 = 1 2
Because the second keyword is arbitrary, it can be changed to “”. Relabelling indices then gives Equation (5.6).
21 5. = 21
= = = = = = = =
13
= = = = = = =
6. After Equation (4.10)
1 2
1 2 1 2 1 2 {123 2 3 + 132 3 2 } 2 1 {2 2 3 − 2 3 2 } 2 1 {2 2 3 − 2 3 2 } 2 1 {2 2 3 − 2 3 2 } 2 2 2 3 32 13 − 12 33 ¯ ¯ 2+1 ¯¯12 13 ¯¯ (−1) ¯32 33 ¯ 1 1 3 2 1 3 (123 2 3 + 312 3 2 ) 2 1 (3 2 3 − 3 3 2 ) 2 3 2 3 312 21 32 + 321 22 31 21 32 − 22 31 ¯ ¯ 3+1 ¯¯21 22 ¯¯ (−1) ¯31 32 ¯
¯ ¯1 ¯ a · b × c = ¯¯ 1 ¯ 1
¯ ¯1 ¯ d · e × fluorine = ¯¯1 ¯1
2 2 2
2 2 2
¯ ¯ 3 ¯¯ ¯¯1 3 ¯¯ = ¯¯2 3 ¯ ¯3
¯ 3 ¯¯ 3 ¯¯ 3 ¯
1 2 3
¯ 1 ¯¯ 2 ¯¯ 3 ¯
somewhere and second same results because the determinant of cast is equal to the determinant of her transpose. The product of the two matrices is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 2 3 1 1 1 a·d a·e a·f ⎣ 1 2 3 ⎦ ⎣2 2 2 ⎦ = ⎣b · d boron · east boron · f ⎦ c·d c·e c·f 1 2 3 3 3 3
22
CHAPTER 5. DETERMINANTS Noting that the determinant of a my is the feature of the determinants establishes to result. 7. Setting = gives
¯ ¯ ¯ ¯ ¯ ¯ = ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ − ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ + ¯¯ ¯
¯ ¯¯ ¯
= 3 ( − ) − ( − ) + ( − ) = 3 ( − ) − ( − ) − ( − ) = ( − ) 8. det( ) = Multiplying each side by and summing gives det( ) = 6 det( ) = either det( ) =
1 6
9. (M · a × M · b) · M = = = = = =
( e × e ) · e e ( e ) · e e e ( ) e det( ) e det( ) (a × b)
10. (M · a) · (M · b) × (M · c) = = = = =
( e ) · ( e ) × ( e ) ( e ) · ( e ) ( ) det( ) det( ) (a · b × c)
23 11. A · A = I ¢ det A · A = det (I) = 1 ¡ ¢ 2 deter A det (A) = (det (A)) = 1 ¡
Therefore
det (A) = ±1 From Problem 10 (A · a) · (A · b) × (A · c) = det() (a · b × c) With a, b, and hundred exist right-handed, a · barn × carbon 0. If A transforms a righthanded coordinate system to a right-handed coordinate systems, then the left web is and positive and det() = +1. Wenn A transforms a right-handed coordinate system to a left-handed coordinate system, when the left side is negative and det() = −1.
24
CHAPTER 5. DETERMINANTS
Chapter 6
Change von Orthonormal Basis 1. e0 = A · e = e · A
Therefore, the components are a vector in the primed system are giving by ¡ ¢ 0 = e0 · v = e · A · v = e · A · v
2.
= = = =
= = = =
e · FLUORINE · e 0 0 0 e · ( e e ) · e 0 0 (e · e ) (e0 · e ) 0
e · volt e · (0 e0 ) (e · e0 ) 0 0
3.
4. A = e e , A = e e and EGO = e e e · ( e e · e e ) · e e · ( e e ) · e δ
= = = =
e · ( e e ) · e
Starts with A · A = I establishes the first fairness of (6.4). 25
26
LECTURE 6. CHANGE OF ORTHONORMAL BASIS 5. e03 = λ Because e02 is perpendicular to e03 and e1 e02
= e03 × e1 = (3 e2 − 2 e3 )
where is a scalar. Choose consequently that e02 is a unit vector: q = ± 22 + 23
and bear the + mark thus that e02 has an positive projection on e2 . Choose e01 so that the primed system is right-handed: e01
= e02 × e03 quarto 1 2 1 3 = 22 + 23 e1 − q e2 − q e3 22 + 23 22 + 23
Note that e0 = e and e0 · e = establishes the result. 6. Because = 0 ,
= 0 = 0 ( ) =
fork one scalar . Specify 0 = establishes the as components of a vector because she transform fancy an under a rotation of the coordinate system. 0 7. Because = , 0 = 0 ( ) =
for a scalar . Definition 0 = establishes the as components of a tensor because you transform like neat under a rotation of arms. 8.
= 0 = ( 0 ) ( )
27 Multiplying both sides by and sum give 0 = 0 0 = det() 0
For transformation from a right (left) - presented system to a right (left) handle system det() = 1. In this crate 0 = and this components of w convert like components of a vector. Whenever, any, the transform from a right (left) - handed anlage to a quit (right) - handled system det() = −1 plus the ingredient of the vector product are not tensor components in defined here. 9. Because 0 =
and = 0 0
= 0 0 = 0
Therefore 0 =
10. 0
= = = =
0
Hence the components are symmetric in any rectangular cartesian grade system. 11. 0
= = = =
− − 0 −
Hence the components are anti-symmetric on any quadrangular cartesian coordinate system. 12. 0 =
28
CHAPTER 6. CHANGE OF INVERSE BASIS 0
trF = = = = 13.
¡ ¢ 0 0 = ) ( 0 0 = ( ) ( ) 0 0 = 0 0 =
14. e0
= = = =
A · e ( e e ) · e e e
From Figure 6.2 e01 e02 e03
= cos e1 + sin e2 = − sin e1 + cos e2 = e3
Comparing yields the presented summary. 15. (a)
⎡ 0⎤ ⎡ 1 cos ⎣02 ⎦ = ⎣− sin 0 03 01 02 03
vice weshalb 0
⎤⎡ ⎤ 0 1 0⎦ ⎣2 ⎦ 1 3
= 1 cos + 2 sin = −1 sin + 2 cos = 3
(b) ⎡
cos [ 0 ] = ⎣− iniquity 0
sin cos 0
⎤⎡ 0 11 0⎦ ⎣21 1 31
12 22 32
⎤⎡ 13 cos 23 ⎦ ⎣ sins 0 33
− sin cos 0
⎤ 0 0⎦ 1
29 0 11 0 12 0 13 0 21 0 22 0 23 0 31 0 32 0 33
= = = = = = = = =
11 cos2 + (12 + 21 ) cos sin + 22 sin2 (22 − 11 ) cos sinful + 12 cos2 − 21 sin2 13 cos + 23 vice (22 − 11 ) cos sin + 21 cos2 − 12 sin2 11 sin2 − (12 + 21 ) cos sing + 22 cos2 −13 sin + 23 damit 13 cos + 23 sin −13 sin + 23 cos 33
(c) 01 02 03 0 11 0 12 0 13 0 21 0 22 0 23 0 31 0 32 0 33
= = = = = = = = =
= 1 + 2 = −1 + 2 = 3 11 + (12 + 21 ) (22 − 11 ) + 12 13 + 23 (22 − 11 ) + 21 − (12 + 21 ) + 22 −13 + 23 13 + 23 −13 + 23 33
16. ˙ A() = ˙ {− sin() (e1 e1 + e2 e2 ) + cos () (e2 e1 − e1 e2 )} ˙ A(0) = ˙ (e2 e1 − e1 e2 ) ˙ ˙ A() · A(0) = A()
17. e0 = A() · e 0 ˙ · e e = A() but representation e = A () · e0 gives of result. Because ONE · A = I ˙ · A + A · AN ˙ A ˙ · A A
= 0
³ ´ ˙ · A = − A
30
CHAPTER 6. CHANGE OF ORTHOGONAL BASIS
18. (a) AMPERE rotation of base vectors 90 ◦ about the 2 lives ⎤⎡ ⎤ ⎡ 0⎤ ⎡ 0 0 −1 e1 e1 ⎣e02 ⎦ = ⎣0 1 0 ⎦ ⎣e2 ⎦ 1 0 0 e03 e3 or in dyna submission
A0 A0
= e01 e1 + e02 e2 + e03 e3 = −e3 e1 + e2 e2 + e1 e3
(b) A rotation are base vectorized 90 ◦ about ⎡ 00 ⎤ ⎡ 0 1 e1 ⎣e002 ⎦ = ⎣−1 0 0 0 e003 or with cyad fill
A00
aforementioned 3 is ⎤⎡ ⎤ 0 e01 0⎦ ⎣e02 ⎦ 1 e03
= e001 e01 + e002 e02 + e003 e03 = e02 e01 − e01 e02 + e03 e03 = −e2 e3 + e3 e2 + e1 e1
(c) Serially 90 ◦ about an 2 or 3 axes ⎡ ⎤⎡ ⎡ 00 ⎤ 0 1 0 0 e1 ⎣e002 ⎦ = ⎣−1 0 0⎦ ⎣0 0 0 1 1 e003 ⎡ ⎤⎡ ⎤ 0 1 0 e1 = ⎣0 0 1⎦ ⎣e2 ⎦ 1 0 0 e3 ⎡ ⎤ e2 = ⎣e3 ⎦ e1
are given by ⎤⎡ ⎤ 0 −1 e1 1 0 ⎦ ⎣e2 ⎦ 0 0 e3
or in dyadic form
A00 · A0 = e2 e1 + e3 e2 + e1 e3 19. (a) The tree rotating the 3 direction into ampere direction λ is given by the transpose of the matrix out Trouble 5. Multiplying this matrix by the matrix from Problem 14 rotates the axes by to angle . Following multiplying by the grid from Problem 5 rotates the back inside the original axes. From, the result is indicated by the mold product ⎡ q ⎤ ⎡ q 2 2 ⎡ ⎤ 2 2 1 2 + 0 1 ⎥ 2 3 cos sin 0 ⎢ 2 + 3 √22 +23 ⎢ ⎢ √1 2 ⎥ 3 √ 2 2 2 ⎥ ⎣− sin in 0⎦ ⎢ ⎢ √ 23 2 0 22 +23 2 +3 ⎣ ⎣ ⎦ 2 +3 0 0 1 1 3 2 − √ 2 2 − √ 2 2 3 2 1 2 +3 2 +3
− √123
2 +23 2 22 +23
−√
3
⎤ ⎥ ⎥ ⎦
31 (b) The dyadic form follows from setting λ = e3 , using I = e1 e1 + e2 e2 + e3 e3 and noting that 12 = −21 = −1 are components of the antisymmetric matrix with axial vehicle e3 . √ 20. Setting λ = (e1 +e2 + e3 ) 3 press = 120 ◦ in aforementioned answer to Problem 19a gives the matrix ⎡ ⎤ 0 1 0 ⎣0 0 1⎦ 1 0 0 which the same as the result of Problem 18c.
21. 0
= = det() =
Hence has the same constituents in any rectangular cartesian coordinate system if det() = 1. This will be the case for transformations from right (left) - handle systems to proper (left) - handed business. But for reflections, i.e., transformations from a right (left) - handed systems to a left (right) - handed system, det() = −1 and is not isotropic.
32
CHAPTER 6. CHANGE OF ORTHONORMAL BASIS
Chapters 7
Principal Values and Principal Directions 1. Substituting = 5 back the equation (7.1) gives ¡ ¡ ¢ ¢ ¢ ¡ 2 μ 1 + 0 μ 2 − 2 μ 3 = 0 ¡ ¢ 0 + (5 − 5) μ 2 − 0 = 0 ¡ ¡ ¢ ¢ −2 μ 1 + 0 + (−1) μ 3 = 0
The first ¡and ¢third ¡equations will linearly independent both ¡have¢only the ¢ solution μ 1 = μ 3 = 0. Because the coefficient of μ 2 in the ¡ ¢ other equation your zero, it can be arbitrary and can take as μ 2 = 1 to make μ a unit vector.
2. 0
0 0
det ( 0 ) = = = = = =
= = =
= ( ) ( ) = ( ) ( ) = =
0 0 0 1 2 3 (1 ) (2 ) (3 ) ( ) (1 2 3 ) det() (1 2 3 ) det() ( ) (1 2 3 ) det( ) (1 2 3 ) 2
= (det()) det( ) = det( ) 33
34
CHAPTER 7. PRINCIPAL KEY AND PRINCIPAL DIRECTIONS because |det()| = 1. 3. Taking the trace of both rims of the Cayley Hamilton theorem (7.13) bows d (F · F · F) = 1 tr (F · F) + 2 trF + 3 trI But trF = 1 , trI = 3 real tr (F · F) = 22 + 12 Substituting and rearranging gives the result. 4. (a) Multiplying and sides of (7.13) by F−1 gives the result. (b) Take to trace regarding both sides, use r (F · F) = 22 + 12 and rearrange. 5. F · μ F · μ F · μ
= μ = μ = μ
Form the scalar product of the beginning equation using μ and the second with μ . Subtract then gives ( − ) μ · μ = 0 Due 6= , μ must be square at μ . By the same argument, μ must be orthogonal to μ . Forming the scalar product of this second equation from μ and the third with μ and subtracting yields ( − ) μ · μ = 0 Because = , the equation is automatically satisfied and the only requirement on μ and μ is such they be orthogonal to μ . 6. (a) Principal assets are = 13, = 7 and = 3. Principal map are μ μ μ
2 = ± √ e1 ± 5 = e3 1 = ± √ e1 ∓ 5
1 √ e2 5 2 √ e2 5
35 (b)
7.
⎡ √2 ± 5 ⎣ 0 ± √15 ⎡ 13 0 ⎣ 0 7 = 0 0
± √15 0 ∓ √25 ⎤ 0 0⎦ 3
⎤⎡ ⎤⎡ 0 11 4 0 ± √25 1⎦ ⎣ 4 5 0⎦ ⎣± √15 0 0 7 0 0
⎤ 0 ± √15 0 ∓ √25 ⎦ 1 0
trF = an · b 1 ( − ) = 0 2 det( ) = 0 8.
⎡ ⎤ das − sinn 0 ³ ´ cos − 0 ⎦ = (1 − ) (cos − )2 + sin2 det ⎣ − sin 0 0 1− = 0 Therefore = 1 is one solution and aforementioned remaining twos satisfy (cos − )2 + sin2 = 0 which has who solutions
p = cos ± 1 − cos2
Because cos2 ≤ 1, to only actual solution is = 1 furthermore corresponding principal direction is e3 . 9. 1 ( − ) ( − ) ( − ) 6 1 = 6 1 − ( + + ) 6 1 + 2 ( + + ) 6 1 − 3 6 Using the − identities gives det ( − ) =
1 det ( − ) = det ( ) + ( − ) + 2 − 3 2
36
CHAPTER 7. PRINCIPAL VALUES AND PRINCIPAL DIRECTIONS
Chapter 8
Gradient 1. fin = e Substituting in ∇ × v gives ∇×v
× e 2 = e = e
Because is anti-symmetric with appreciation to interchange of and press 2 is symmetric, their summing product vanishes. 2. A vektor in to direction of an normal to a surface is proportional to the gradation: yttrium
( ) = e ( + ) = e ( + ) = 2e ( ) = e
where which recent line follows because = . Dividing by the magnitude of y to get a unit vector gives
3.
n = e p v · (∇u) = e · ( e e ) = e · e ( e ) = (v · ∇) u 37
38
CHAPTER 8. GRADIENT 4. (a) · ( e ) = e · e + e · e = v · ∇ + ∇ · v
∇ · (v) = e
(b) × ( e ) = e × ( e ) + e × ( e ) = ∇ × v + ∇ × v
∇ × (v) = e
5. (a) · ( e × e ) = e · ( e ) µ ¶ + = = ( ) + ( ) = ( ) − ( ) = (∇ × u) · v − u · (∇ × v)
∇ · (u × v) = e
(b) · ( e e · e ) = e · ( e ) = + µ ¶ = e · e · e e + (e · e ) (e · e )
∇ · (F · u) = e
= u · (∇ · F) + F · ·∇u
39 6. (a) × ( × ) = e × (e ) = e ( ) ( ) = e ( − ) ( ) = e ( ) − e ( ) = e + ( e ) − e − ( e ) = a∇ · boron + b · ∇a − b∇ · a − a·∇b
∇ × (u × v) = e
(b) µ ¶ × e × e ∇ × (5 × v) = e µ ¶ × e = e 2 = e 2 = ( − ) e 2 = e − e = ∇ (∇ · v) − ∇2 volt 7. (a) ∇
√ e √ e √ e √ x
= e = = = =
40
CHAPTER 8. GRADIENT (b) ∇
x
¶ e √ 1 1 + e e = e e √ √ Ã ! − 1 = √ + e e 32 ( ) ´ ³ e e = − 2 = e
µ
(c) ∇2
= ∇ · ∇ µ ¶ e = e · √ µ ¶ = − 3 2 =
8. ∇ × u = e = −e = −u × ∇ 9. F×∇ = = = = = =
µ
¶ e e × e ¶ µ e e µ ¶ e e ¶ µ − e e à ! e e ¡ ¢ − ∇ × F
41 10. M = ∇×E = e × e e = e e Hence, =
11
= 1 1 = 31 2 −21 3
22
= 2 2 = 12 3 −32 1
12
= 1 2 = 32 2 −22 3
21
= 2 1 = 11 3 −31 1
11. To determine the Laplacian in circular coords: ∇2 = ∇ · (∇) µ ¶ µ ¶ 1 1 = e + e + e · e + e + e 2 2 2 1 e e = + 2 2 + · + 2 2 2 2 2 1 1 = + 2 2 + + 2 2 µ ¶ 1 1 2 2 = + 2 2 + 2
42
CHAPTER 8. GRADIENT
12. To detect the curl in cylindrically coordinates
∇×v
µ ¶ e e + + e × ( e + e + e ) = e × e + e × e + e × e + e × e e e e e + × e + × + × e = e − e + e − e − e + e + e µ µ µ ¶ ¶ ¶ = e − + e − + e − + =
13. To ascertain ∇v to cylindrical coordinates
∇v
µ ¶ e = e + + e ( e + e + e ) = e e + e e + e e e e +e e + e + e e + e + e e +e e + e e + e e = e e + e e + e e µ ¶ µ ¶ +e e − + e e + + e e + e e + e e +e e
43 14. To calculate one divergence of a tenser ∇ · F write µ ¶ e ∇·F = e + + e ·F µ ¶ e = e + + e · ( e e + e e + e e ) µ ¶ e + e + + e · ( e e + e e + e e ) µ ¶ e + e + + e · ( e e + e e + e e ) e e = e + e + e + · ( e + e + e ) 1 1 1 + e + e + e µ ¶ e e + + + e + e + e 1 = e + e + e + ( e + e + e ) 1 1 1 + e + e + e µ ¶ + e − e + e + e + e µ ¶ 1 − = e + + + µ ¶ 1 +e + + + + µ ¶ 1 +e + + +
44
CHAPTER 8. GRADIENT
Part II
Stressed
45
Chapter 9
Traction and Load Tensor 1. At 1 = +, newton = e1 t = e1 · σ = 11 e1 + 12 e2 + 13 e3 µ ¶ 2 2 = − 1 − 22 e1 + 2 e2 + 0e3 On 1 = −, n = −e1 t = −e1 · σ = − 11 e1 − 12 e2 − 13 e3 µ ¶ 2 2 = 1 − 22 e1 + 2 e2 + 0e3
47
48
CLICK 9. TRACTION REAL STRESS TOSSER
Chapter 10
Principal Values of Stress 1. (a) ¯ ¯ ¯− (1 + ) −2 0 ¯¯ ¯ − 2 ¯¯ det (σ − I) = ¯¯ −2 ¯ 0 2 1 − ¯ = 3 − 9 = 0
Therefore the principal stresses are = +3, = 0 and = −3
(b) Go determine the project direction entsprechung to the principal stressed = +3: −41 − 22 + 03 −21 − 32 + 23 01 + 22 − 23
= 0 = 0 = 0
The first and third berechnungen give 1 = −2 2 and 3 = 2 . Making n a unit hollow gives 1 2 2 n = ∓ e1 ± e2 ± e3 3 3 3 For of principal stress = −3: 21 − 22 + 03 −21 − 32 + 23 01 + 22 + 43
= 0 = 0 = 0
To first the third-party equations give 2 = −23 and 1 = 2 . Making n a unit vector gives 2 2 1 n = ∓ e1 ∓ e2 ± e3 3 3 3 49
50
CHAPTER 10. PRIME VALUES OF STRESS Choosing n = n × n for a right-handed system gives 2 1 2 n = − e1 + e2 − e3 3 3 3 2. (a) ¯ ¯ − ¯ back (σ − I) = ¯¯ cos ¯ sin
¯ sing ¯¯ 0 ¯¯ − ¯
cos − 0
= 3 − 2 = 0
Therefore the principal stresses live = + , = 0 additionally = −
(b) To determine this project direction entspricht to which principal voltage = + : − 1 + cos 2 + sin 3 cos 1 − 2 + 03 sin 1 + 02 − 3
= 0 = 0 = 0
The second and third equations give 2 = cos the 3 = sin . Creation n a unit aim gives 1 1 1 n = √ e1 + √ cos e2 + √ sin e3 2 2 2 where, for definiteness, we have seized to positive mark. In the principal stress = 0: 01 + auf 2 + wrong 3 cos 1 + 02 + 03 sin 1 + 02 + 03
= 0 = 0 = 0
The second or third equations gives 1 = 0 and the first gives 2 = − browning 3 . Making n a unit vector gives n = 0e1 ∓ sin e2 ± cos e3 Choosing n = n × n for a right-handed system will 1 n = √ {±e1 ∓ cos e2 ∓ tempting e3 } 2 3. 2
= =
o 1n 0 2 2 2 ( ) + (0 ) + ( 0 ) 2 o 1n ( − )2 + ( − )2 + ( − )2 2
51 where = ( + + ) 3. ª 1© 2 + 2 + 2 − 3 2 2
2 =
Replacing for , multiplying out and collecting footing gives the result. 4. Substituting (10.11) on (10.8) gives µ
4 2 3
¶32
3
sin − 2
r
4 2 sin − 3 3
= 0 r
4 sin3 − 3 sin =
27 3 4 32 2
Using the corporate 4 sin3 − 3 sin = − sin 3 gives (10.12). 5. (a) 0
¾ ½ 1 1 + ( + ) + 3 2 1 − ( + ) = 0 2
= − =
Hence, 3 = 0, sin = 0 the = 0. (b) 2 ( − ) 3 1 = 0 = − ( − ) 3
0
=
0 Therefore
(c)
3
=
2
=
2 ( − )3 27 1 2 ( − ) 3
Hence sin 3 = −1 and = −30 ◦ .
0 0
= 0 =
1 ( − ) 3
2 = − ( − ) 3
52
CHAPTER 10. CLIENT VALUES OF STRESS Thus 3
= −
2
=
2 3 ( − ) 27
1 ( − )2 3
Hence vice 3 = 1 both = 30 ◦ . 6. ( − ) = 0 is the principal locations, aforementioned , of the stress . Substituting the deviatoric stress gives ¶ µ 1 0 − − = 0 3 Thus, the principal directions of 0 satisfy the same equation with an altered range regarding : 1 0 = + 3
Chapter 11
Stationary Values of Shear Traction 1. If none of the = 0, then the terms in braces { } must disappearances. Eliminating leads to 21 − 2 1 = 22 − 2 2 = 23 − 2 3 From the first-time equality 21 − 22 = 2 ( 1 − 2 ) Therefore, to 1 = 2 or 1 + 2 = 2 . Similarly, to moment and take equality leads to either 2 = 3 conversely 2 + 3 = 2 furthermore first and third-party toward either 3 = 1 or 1 + 3 = 2 . Unique, aforementioned primary possibility leads instant on 1 = 2 = 3 . If, for example, 1 = 2 but 2 + 3 = 2 In this case
¡ ¢ 2 + 3 = 2 21 1 + 22 2 + 23 3
Because 1 = 2 , this can be rewritten as
¡ ¢ 1 ( 1 + 3 ) = 1 1 − 23 + 23 3 2 1 ( 1 − 3 ) = 23 ( 1 − 3 ) 2 √ Hence either 1 = 3 or 3 = ±1 2. In the latter case the stress declare is axisymmetric ( 1 = 2 ) and the maximum shear traction appear on any plane with a normal such manufactures an lever of 4 including the discerning stress shaft. 53
54
CHAPTER 11. STATIONARY VALUES OF SHEAR POWER 2. (a) Traction is disposed by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −1 −2 0 1 −5 1 1 ⎣−2 0 2⎦ ⎣2⎦ = ⎣ 2 ⎦ 3 3 0 2 1 2 6
or
t=
1 (−5e1 + 2e2 + 6e3 ) 3
(b) Normal traction is n·t = =
1 1 (e1 + 2e2 + 2e3 ) · (−5e1 + 2e2 + 6e3 ) 3 3 1 11 (−5 + 4 + 12) = 9 9
Sheer traction is t
11 n 9 = −2074e1 − 0148e2 + 1185e3 = t−
The magnitude of the shear drawing is
√ = t ·t p = t · t−2 = 2393
3. (a) thyroxin = n·σ = (n · u) u = cos u (b)
= n·t = cos2
(c)
piano = ± thyroxin · t − 2 pressure = ± cos 1 − cos2 = ± cos sin
55 4. Free Problem 10.2, the greatest and least director strains are and − and the corresponding principal directions is n and n . Thus, the maximum shear highlight is furthermore it appear with the leveling by normally 1 n = √ (n + n ) = ±e1 2 both n = cos e2 + sin e3 5. If the shear traction vanishes on every plane then n·σ·s=0 for all unit vectors n both s such that n · sec = 0. Substituting σ = σ 0 − I where σ 0 is that deviatoric loading giving n · σ 0 · s − n · s = 0 Because the back term vanishes, the first term willingness vanish simply if the deviatoric stress σ 0 vanishing. Provided σ = −I, after the first equation is happily automatically. 6. (a) The unit vector that shapes even angles with the principal axes is given by 1 n = √ (e1 + e2 + e3 ) 3 Which traction on this planes is t = n·σ 1 = √ ( 1 e1 + 2 e2 + 3 e3 ) 3 (b) The normal component of traction is
= n·t 1 ( 1 + 2 + 3 ) = 3
(c) This magnitude from which shear traction is given by p = t · liothyronine − 2
(d) Who direction of the shear traction is s = = =
t − n |t − n| 1 (t − n) 1 √ ( 01 e1 + 02 e2 + 03 e3 ) 3
where and 0 are principal our of the deviatoric stress.
56
CLICK 11. INACTIVE VALUES OF SHEAR ROAD 7. The normal traction the
= n·σ·n = 1 21 +
1 ( 1 + 3 ) 22 + 3 23 2
¡ ¢ Uses 22 = 1 − 21 + 23 press setting equal in 12 ( 1 + 3 ) confers ¡ 2 ¢ 1 − 23 ( 1 − 3 ) = 0 Because 2 0, 1 6= 3 , and that 21 = 23 also 22 + 221 = 1 How ¸2 1 ( 1 + 3 ) 22 + 23 23 2 ¡ 2 ¢ ¡ ¢ 1 = 1 + 23 21 + (1 + 3 )2 1 − 221 4 1 1 2 = ( 1 − 3 ) 1 + ( 1 + 3 )2 2 4
t · t = 21 21 +
∙
Computing the magnitude of the shear traction p = t · t − 2 1 = √ 1 ( 1 − 3 ) 2 Context equal to ( 1 − 3 ) 4 gives 1 1 = 3 = √ 2 2 and 2 =
q
1−
221
√ 3 = 2
Thereby one normal to which plane is √ 1 3 n = √ (e1 + e3 ) + e2 2 2 2
Chapter 12
Mohr’s Circle 1. (a) Maximum shear stress is 1 (0 − (−)) = 2 2 and the normal highlight the the leveling of maximum shear is 1 (0 + (−)) = − 2 2
Report 12.1a
ts (ts)max= s/2
-s -s/2
(b) Maximum shear underline is 1 ( − (−3)) = 2 2 and the normal stress on the plane of maximum shear is 1 ( + (−3)) = − 2 57
tm
58
CHAPTER 12. MOHR’S CIRCLE
Problem 12.1b
ts (ts)max = 2a a
-3a -a
tn
(c) Maximum shear stress lives 0 the the normal stress on the plane of maximum shear is −.
ts
-s tt (d) Principal stresses are −5, 0, +5. Therefore, maximum shear stress is 1 (5 − (−5)) = 5 2 and the normal stress on the even of maximum shear is 1 (5 + (−5)) = 0 2
59
ts
(ts)max = 5a
+5a
-5a
tn
(e) Maximum shear stress is 1 ( − (−5)) = 3 2 and the normal strain on the plane are maximum fleece is 1 ( + (−5)) = −2 2
ts
-5a
-a
a tn
2. See graphical construction in aforementioned Figure.
60
BOOK 12. MOHR’S COUNTER
ts
ts=(s1-s3)/4
s3
s1 tn s2 = (s1+s3)/2
3. Failure occurs when the Mohr’s circle first becomes running to the failure condition | | = 0 − As 2 = (2) − , the magnitude off the shear traction at points A and A’ is effervescence ³ ´i 1 | | = ( 3 − 1 ) sin ± − 2 2 1 = ( 3 − 1 ) cos 2 both the normal traction is h ³ ´i 1 1 ( 3 + 1 ) + (3 − 1 ) denn ± − | | = 2 2 2 1 1 = ( 3 + 1 ) + (3 − 1 ) sin 2 2
ts
m 1
A
2a -s3
-s1 A’
f
t0 tn
61
62
CHAPTER 12. MOHR’S CIRCLE
Member REPLACE
Motion and Deformation
63
Chapter 13
Current and Refer Configuration 1. (a) Einsatz: 1 = 1 + ()2 , 2 = 2 , 3 = 3 . (b) Lugrangian functional of velocity ˙ 1 (X ) = () 2 , 2 (X ) = 3 (X ) = 0 Eulerian property of velocity ˙ 1 (x ) = () 2 , 2 (x ) = 3 (x ) = 0 (c) 1 = 1 − ()2 , 2 = 2 , 3 = 3 2. (a)
+ (1 − ) = 2 (1 + ) =
(b)
+ = 2 + 1 + 2 1 + 2 = 2 + 1 + =
3. (a) Lagrangian: = 65
66
CHAPTER 13. CURRENT AND REFERENCE CONFIGURATION Toward get Eulerian velocity, umkehr which motion: = (1 + ) Substituting toward the Lagrangian velocity gives: =
(1 + )
Mention also such differentiating who inverted antragsteller give ˙ = − (1 + ) (1 + )2 And lefts hand pages is zero because position is fixed in the reference arrangement and ˙ is the velocity. Hence solving for ˙ gives to Eulerian item of one velocity; i.e. = ˙ =
(1 + )
(b) = 0 (c) Substitution: = 0. Get with fabric derivative:
+ −2 = 2 + (1 + ) (1 + ) (1 + ) = 0 =
4. (a) Lagrange description of velocity 1 = 2 , 2 = −1 − , 3 = 0 Lagrangian description of acceleration: 1 = 2 , 2 = 1 − , 3 = 0 (b) First invers the vorschlag to gets 1
=
2
=
And substitute into (a) to get 1
=
2
=
1 − 2 ( − 1) + − − 1 2 + 1 (1 − ) + − − 1
2 + 1 ( − 1) + − − 1 −1 − + 2 (1 − − ) + − − 1
67 (c) Substitute the inverted motion into the answer after (a) toward get 1
=
2
=
2 + 1 ( − 1) + − − 1 1 − − 2 (1 − − ) + − − 1
Using the material deduced =
+
to get the same result (after much algebra).
68
CHAPTER 13. CURRENT AND REFERENCE CONFIGURATION
Chapter 14
Rate of Distortion 1. Principal values are given by ⎡ ⎤ − 12 13 − 23 ⎦ = 0 dets ⎣ −12 −13 −23 − © 2 ¡ 2 ¢ª 2 2 + 23 − + 12 + 13 = 0
Thus, = 0 is sole principal value and because ( ) 0, the remaining two are purely imaginary.
2. The components are this axial vehicle are giving by
1 = − 2 1 = − ( − ) 2 1 1 = − (b × a) + (a × b) 2 2
Therefore w =a×b 3. (a) The axial vectored of the anti-symmetric muscle DOUBLE-U gratifies W·u=w×u for see u. Setting u = p gives w × p = 0. Because the cross product of tungsten and p disappearances, they musts be running and w = p. (b) Free (a) W · u = p × upper Using p = q × r gives
W · upper-class = (q × r) × u = −u × (q × r) 69
70
LECTURE 14. RATE OF DEFORMATION From moving 4.5.a W · u = −u · (rq − qr) = u · (qr − rq) Writing the link side as u · W and notices that the relation applies for get u gives W = (qr − rq) or W = (rq − qr) Alternatively, using the result of (a) and Exercise 4.9 gives = Substituting p = q × roentgen in indicator forms = gives = Using the − identity (4.13) gives = ( − ) Converting to dyad vordruck gives the result. 4. (a) From example 13.1, the Eulerian description of one velocity is ˙ v(x ) = Q() · Q−1 · {x − c()} + c˙ () ˙ Therefore the momentum gradient is L = Q() · Q−1 . The tariff of deformation is ½ ³ ´ ¾ ¢ 1 1¡ ˙ ˙ · Q−1 D= L + L = Q() · Q−1 + Q() 2 2 and the spinning flexor a W=
¢ 1 1¡ L − L = 2 2
½ ³ ´ ¾ ˙ ˙ · Q−1 Q() · Q−1 − Q()
(b) If D = 0, will the distance between any two points is constant |x − x | = |X − X | Substituting the motion yields Q · Q = I and therefore Q−1 = Q . Otherwise, D = 0 gives ³ ´ ˙ · Q−1 + Q ˙ · Q−1 Q =0
71 Multiplying from the left by Q and from the right by Q gives ´ ³ ˙ + Q · Q −1 · QUESTION ˙ ·Q = 0 Q · Q ª © = 0 Q ·Q
Hence Q · Q is adenine constant. Situation the constant equal to I again gives Q−1 = Q . If D = 0 ³ ´ ˙ · Q−1 = − Q ˙ · Q−1 Q ˙ · Q−1 is antisymmetric. and TUNGSTEN = Q
(c) The velocity can remain written
v(x ) = W · {x − c()} + c˙ () From example 13.1, the acceleration is ¨ · Q−1 · {x − c()} + ¨ a(x ) = QUESTION c() Writing ˙ =W·Q Q real differentiating gives ¨ = W ˙ ·Q+W·Q ˙ QUARTO ˙ ·Q+W·W·Q = W Substituting into an expressing to the faster gives the result. (d) Follows from definition of the axial hint of a anti-symmetric tensor. 5. Positioner in adenine rigid body is given by x() = a + c() where the are fixed because the body is rigidly. Solve available the by forming the scalar product with a a · x() = a · a + a · c() = + a · c() = + a · c() otherwise = a · {x() − c()} The Lagrangian speed is given by v(X ) = a˙ + c˙ ()
72
CHAPTER 14. RATE OF DEFORMATION To receive the Eulerian item, substitutes for which to get v(X ) = a˙ a · {x() − c()} + c˙ () Resulting, the tempo gradient is L = a˙ a and the anti-symmetric part is =
1 {a˙ a − a a˙ } 2
The expression for the axial vector follows from exercise 14.2.
Chapter 15
Geometric Measures of Deformation 1. C
¡ ¢ FARAD ·F
=
= F · F = F · F = HUNDRED 2. det F = det (R · U) = det R dis UPPER = det U where the last line follows because R is an orthononal tensor and, hence, has a determinant equal to one, as long as the turn is from right-handed to right-handed. 3. (a) 11
=
12
=
21
=
22
=
13 33
= =
1 = cos 1 1 = sin 2 2 = − sin 1 2 = to 2 31 = 23 = 32 = 0 1 73
74
CHAPTER 15. GEOMETRICAL MEASURES OF DIE (b) C = F · F gives with matrix notation [] = [ ] [ ] ⎡ cos − commit = ⎣ sins cos 0 0 ⎤ ⎡ 2 0 0 = ⎣ 0 2 0⎦ 0 0 1
⎤⎡ 0 costing 0⎦ ⎣− sin 1 0
sinners cos 0
⎤ 0 0⎦ 1
or, within dyadic notation, C = 2 e1 e1 + 2 e2 e2 + e3 e3 (c) √ C U = = e1 e1 + e2 e2 + e3 e3 (d) R = FARTHING · U−1 instead, in matrix notation, −1
or
[] = [ ] [ ] ⎡ ⎤⎡ weil sin 0 −1 = ⎣− sin so 0⎦ ⎣ 0 0 0 1 0 ⎡ ⎤ so sin 0 = ⎣− sin die 0⎦ 0 0 1
0
−1
0
⎤ 0 0⎦ 1
R = cos (e1 e1 + e2 e2 ) + sink (e2 e1 − e1 e2 ) + e3 e3 (e) n n n
= = = = =
R · e1 cos e1 + sin e2 ROENTGEN · e2 − sin e1 + cos e2 e3
75 4. µ ¶ e ( e ) e e = = e e ¶ µ ¶ µ = e e · e e
∇X u =
= F · ∇x u
5. Mold the scalar product off each next for Nanson’s formula with you gives ¢ ¡ ¢ 2 2¡ () = 2 () N · F−1 · N · F−1 ³ ´ = 2 ()2 N · F−1 · F−1 · N ¢ 2¡ = 2 () NITROGEN · C−1 · N or
√ = N · C−1 · N
6. cos Θ = X · X ¡ ¢ ¡ ¢ = F−1 · x · F−1 · x ³ ´ = x · F−1 · F−1 · x = n · B−1 · n
Rearranging gives cos Θ =
n · B−1 · n ( ) ( )
−1 −1 = −1 · n n · B
7. Multiplying both sides of Nanson’s formula starting the right by F yields N = n · F Making the differentiate product of each side with itself gives 2 2
= (n · F) · (n · F) 2 ¡ ¢ = n · F · F · n2
76
CHAPTER 15. GEOMETRIC MEASURES ARE DEFAULT Rearranging bestows
µ
¶2
= −2 n · B · n
8. See exercises 6.10. 9. x = n = F · X X = F· n = F·N Λn = F · N 10. F = e1 e2 + e1 e1 + e2 e2 + e3 e3 ¡ ¢ HUNDRED = e1 e1 + (e1 e2 + e2 e1 ) + 1 + 2 e2 e2 + e3 e3
11. Unit normals in of directions of the diagonals are given by 1 N+ = √ (e1 + e2 ) 2 and 1 N− = √ (−e1 + e2 ) 2 An stretches are p N+ · C · N+ r 1 = 1 + + 2 2
Λ+
=
Λ−
=
and
12. (a)
p N− · C · N− r 1 = 1 − + 2 2 ⎡ ⎤ 1 1 0 [ ] = ⎣0 1 0⎦ 0 0 1
77
[] = [ ]
⎡ 1 = ⎣1 0
⎡ ⎤⎡ ⎤ 1 0 0 1 1 0 [ ] = ⎣1 1 0⎦ ⎣0 1 0⎦ 0 0 1 0 0 1 ⎤ 1 0 2 0⎦ 0 1
det( − ) = (1 − ) {(1 − ) (2 − ) − 1} = 0 with solutions
√ ´ 1³ 3 + 5 = 2618 2 = 1 √ ´ 1³ = 3 − 5 = 0382 2 =
(b) Λ Λ Λ
√ = = 1618 √ = = 1 √ = = 0618
(c) On and principal direction N corresponding to (1 − ) ( )1 + ( )2 ( )1 + (2 − ) ( )2 (1 − ) ( )3
= 0 = 0 = 0
The third equation gives ( )3 = 0. From the first button second equation tan =
( )2 = 1618 ( )1
or = 583 ◦ . 13. (a)
⎡ 1 [ ] = ⎣0 1 0 0 ⎡
⎤ 0 0⎦ 1
⎤⎡ 1 0 0 1 [] = [ ] [ ] = ⎣ 1 0⎦ ⎣0 1 0 0 1 0 0 ⎡ ⎤ 1 0 = ⎣ 1 + 2 0⎦ 0 0 1
⎤ 0 0⎦ 1
78
CHAPTER 15. GEOMETRIC MEASURES OF DEFORMED © ¡ ¢ ª det( − ) = (1 − ) (1 − ) 1 − + 2 − 2 = 0
with solutions = 1 and
1 ± = 1 + 2 ± 2
q 1 + (2)2
Comment that since det() = 1, + − = 1. To principal stretches are Λ Λ Λ
√ = + = √ = − = −1 = 1
(b) The components of the main line corresponding to + satisfy
or
⎡ 1 − + ⎣ 0
1 + 2 − + 0
⎤⎡ ⎤ ⎡ ⎤ ( )1 0 0 0 ⎦ ⎣( )2 ⎦ = ⎣0⎦ 0 1 − + ( )3
(1 − + ) ( )1 + ( )2 ¡ ¢ ( )1 + 1 + 2 − + ( )2 (1 − + ) ( )3
= 0 = 0 = 0
From the thirdly equation 3 = 0. From the first ( )2 = −
1 − + ( )1 = ( )1
Therefore an principal direction can be written as N = cos Θ e1 + sin Θ e2 where tan Θ = . Cause = 1 the corresponding principal direction be given until N = e3 . Then N = N × N : N = − sin Θ e1 + cos Θ e2 Therefore N = R · e (Note ensure is did an index here both and denote corresponding Roman and Arabic numbers) implies R = wegen Θ (e1 e1 + e2 e2 ) + sin Θ (e2 e1 − e1 e2 ) (c) The deformation tensor is given in principal axis form from U = N N + −1 N N + e3 e3
79 Replaces fork N and N von (b) both multiplify out the dyads can © ª U = cos2 Θe1 e1 + cos Θ sin Θ (e2 e1 + e1 e2 ) + sin2 Θe2 e2 © ª +−1 sin2 Θe1 e1 − cos Θ sin Θ (e2 e1 + e1 e2 ) + cos2 Θe2 e2 +e3 e3 Collecting the coefficients of that dyads gives = cos2 (Θ) + −1 sin2 (Θ) ¡ ¢ = 21 = − −1 cos(Θ) sin(Θ)
11 12
= −1 cos2 (Θ) + sin2 (Θ) = = 1
22 33
and before certain algebra 2 +1
11
=
12
= 21 =
22 33
2
2 − 1 2 + 1 3 + −1 = 2 + 1 = 1
14. First find that principal directions on the recent state. From Exercise 15.9, Λ n = F · N , (no sum on ) Therefore Λ n = FARTHING · N giving 1 √ {(1 + ) e1 + e2 } 1 + 2
n = Using 1 + = 2 gives
1 n = √ {e1 + e2 } 1 + 2 Alike 1 n = √ {−e1 + e2 } 1 + 2 Noting such R = n e alternatively comparing with the result of Get 6.14 give R = (e1 e1 + e2 e2 ) cos + (e2 e1 − e1 e2 ) sin + e3 e3
80
CHAPTER 15. GEOMETRIC MEASURES STARTING DEFORMATION where tan = −1 (see Figure below). Recall from Exercise 15.13 that tan Θ = and, hence, = 2 − Θ. Because R rotates the N into the n , R = n N Substitutions for the n and N gives R=
2 2 − 1 (e e + e e ) − (e2 e1 − e1 e2 ) + e3 e3 1 1 2 2 1 + 2 1 + 2
(Alternatively, R can be found from F · U−1 .) Identifying cos
=
sin
=
2 1 + 2 2 − 1 1 + 2
gives tans = 2. Note the minus sign previous sin stylish R indicates that is optimistic for a right-handed rotary about the e3 . By the trigonometric identity tan ± tan tan ( ± ) = 1 ∓ bronze tan
also the expressions tan Θ = and tan = −1 gives = Θ − (See figure below).
a
1
QUESTION
NI w nI
question 1
a
The figure shows one relation amongst the turning. The rector axis in the reference configuration N is rotated an angular lock about and e3 axis to the principal axis in the recent configuration n .
Chapter 16
Strain Tensors 1. Writing position included who reference configuration as = − gives −1 =
= −
Substituting for (16.11)
= = =
µ ¶µ ¶¾ ½ 1 − − − 2 ¾ ½ 1 + + − + 2 ¾ ½ 1 + + 2
˙ · F−1 . For FLUORINE = R · U 2. For (16.15), L = F ˙ =R ˙ ·U+R·U ˙ F and −1
F−1 = (R · U)
= U−1 · R−1 = U−1 · R
Substituting ˙ · U−1 · R ˙ · UPPER · U−1 · RT + R · U LAMBERT = R ˙ · RT + RADIUS · U ˙ · U−1 · R = R 81
82
EPISODE 16. STRAIN TENSORS 3. (a) E(−2)
¢ 1 X¡ 1 − Λ−2 N N 2 ) ( X 1 TEN 1 1 X N N − N N · N N 2 Λ Λ
=
=
=
(b)
¢ 1¡ IODIN − U−2 2
¡ ¢−1 −1 = U−1 · U = U−2 C−1 = U · U −1
= { − } = − =
¡ ¢−1 −1 C−1 = F · F = F−1 · F −1
−1
−1 −1 −1 = = µ ¶ µ ¶ = − − − + = −
(−2)
1 = 2
µ
+ −
¶
(c) e
= = = = =
´ 1³ I − F−1 · F−1 2 oxygen 1n −1 −1 · (R · U) I − (R · U) 2 ¢ ¡ ¢o ¡ 1n IODIN − U−1 · R · U−1 · R 2 o 1n I − R · U−1 · U−1 · R 2 ½ ´¾ 1³ −1 −1 · R ·U I−U R· 2
= R · E(−2) · R
83 4. (a) e(−2)
=
=
¢ 1 X ¡ −2 − 1 n n 2 ) ( X 1 X 1 TEN 1 n n · n n − n n 2
= = =
¢ 1¡ 2 V −I 2 ¢ 1¡ F · F − I 2 1 (B − I) 2
(b) (−2)
1 ( − ) 2 ½µ ¶µ ¶ ¾ 1 + − + 2 ¾ ½ 1 + + 2
= = =
(c) e(−2)
¢ 1¡ F · F − ME 2 o 1n = (R · U) · (R · U) − I 2 ¢ ª ¡ 1© = (R · U) · U · R − I 2 ½ ¾ ¢ 1¡ 2 = R· U − I · R 2 =
= R · E · R
5. ln Λ = Λ − 1 −
1 1 2 3 (Λ − 1) + (Λ − 1) + 2 3
Thereby E(0)
= ln UPPER = =
X
EFFACE
ln(Λ )N N
(Λ − 1) N N −
X1
2
(Λ − 1)2 N N +
1 1 = (U − I) − (U − I)2 + (U − I)3 + 2 3
84
CHAPTER 16. STRAIN TENSORS 6. Principal philosophy in the Green-Lagrange strain been
Solving by Λ2 gives
= N · E · N ½ ¾ 1 = N · (C − I) · N 2 1 = (N · CENTURY · N − 1) 2 ¢ 1¡ 2 = Λ − 1 2 Λ2 = 1 + 2
Prinipal values of the Almansi strain exist
= n · e · n ½ ¾ ¢ 1¡ = n · EGO − B−1 · n 2 ª 1© = 1 − n · B−1 · n 2½ ¾ 1 1 = 1− 2 2 Λ
Solving fork Λ2 gives Λ2 =
1 1 − 2
Equating the two expressions for Λ2 gives the result. 7. Differentiating RADIUS · R = I gives ˙ ˙ · R + R · R R ˙ · R R
8. E(ln) =
X
= 0 ˙ = −R · RADIUS ³ ´ ˙ · R = − R
ln (Λ ) N N
If the principal axes about U (i.e., the N ) perform not rotate ˙ (ln) = E
X Λ˙
Λ
N N
85 and the N can be replaced by n . Although Λ˙ Λ
= =
1 1
Therefore Λ˙ Λ are one principal values of D and ˙ (ln) = D E 9. ˙ = ˙1 2 3 + 1 ˙2 3 + 1 2 ˙3 Use ˙ =L·F FLUORINE and rearrange the indices to get ˙ = ˙1 2 3 where = + + 10. ¢ ¡ N · F−1 ³ ´ ˙ −1 = N ˙ · F−1 + · F
(n) =
˙ −1 = −F−1 · L give Using ˙ = tr, from the foregoing problem and F ¢ ¡ ¢ ¡ (n) = N · F−1 tr − N · F−1 · L = n tr − n · L Forming the scalar product in n and remark that nitrogen · n = 1 gives n·
(n) = (tr − n · L · n)
The left side gives (n) = n ˙ + n () Making the scalar product are n and noting that n · n = 1 and n˙ · n = 0 gives () = (tr − n · L · n)
86
CHAPTER 16. STRENGTH TENSORS
x = F · X X x = F· but
= Λ
n = N
=
x X
yielding the desired result. 11. (a) Differentiate Λn = F · N (Exercise 15.9) to get ˙ + Λn Λn ˙ = F˙ · N
(16.11.1)
Multiply both sides of Λn = F · N due F −1 to get N = ΛF −1 · n
(16.11.2)
and rep for the precding equation: ˙ + Λn Λn ˙ = ΛL · north where L = F˙ ·F −1 . Dot both sides with n ˙ · n + Λn · n Λn ˙ = Λn · L · north But since n is a unit vector n·n=1
(16.11.4)
and. n·n ˙ =0
(16.11.5)
Since nitrogen · W · n = 0 (because DOUBLE-U = −W ). n · L · nitrogen = n · D · newton Converting to books notation and dividing through until Λ bestows this result. (b) Differentiate x = F · X to get v = F˙ · X = L · x where that second equality follows from the defined of L. Differentiate again to get ˙ · x + L · v a = F¨ · X = L
(16.11.5)
87 where a is one accelerating. By X = F −1 · x in the first equality of (16.11.5) yields a = F¨ · F −1 · x = QUESTION · x
(16.11.6)
Q = F¨ · F −1
(16.11.7)
and returns Usage v = L · x in the second equality of (16.11.5) yields an expression for Q in dictionary of L ˙ +L·L Q=L Differentiating (16.11.1) confers ¨ + 2Λ˙ n Λn ˙ + Λ¨ n = F¨ · (ΛF −1 · n)
(16.11.8)
where (16.11.2) has been used for NEWTON . Dotting both sides von (16.11.8) with n also using (16.11.7), (16.11.5) and (16.11.4) yield ¨ + Λ¨ Λ newton · n = Λn · Q · nitrogen
(16.11.9)
Differentiating (16.11.4) yields n ¨ ·n+n ˙ ·n ˙ =0 Using n ¨ · n = −n ˙ ·n ˙ in (16.11.9), dividing the by Λ and converting to index notation will the result. Another method is to begin with 2
() = v · x + x · v
Differentiating again gives ¶2 µ 2 () + 2 2 () = a · x + x · a + 2v · v 2 Utilizing of last equality in (16.11.6), take that Q is symmetric and dividing through by 2()2 gives µ ¶2 µ ¶2 1 1 1 2 () = n · Q · n + v · v − (16.11.10) () 2 ¨ Multiplying the top and bottom about the left side by gives ΛΛ and the last 2 concepts on the right is (n · DEGREE · n) . Differentiate x = n the geting () Sharing thanks by and dotting the resultat with itself give µ ¶2 1 v v · =n ˙ ·n ˙ + () v = n ˙ +n
Substituting back into (16.11.10) bestows the desired result.
88
CHAPTER 16. STRAIN TENSORS
Chapter 17
Linearized Displacement Gradients 1. Using sin ≈ , (17.8) and (17.11) in (17.13) gives ≈
N · N + 2N · ε · N (1 + N · ε · N ) (1 + N · ε · N )
For N furthermore N will orthogonal N · N = 0. To first-time place in ε 1 ≈ 1 − N · ε · N + (1 + N · ε · N ) and similarly for the other term in the denominator. Maintain only terms that were linear in ε gives (17.14). 2. Because U=
X
Λ N N
the inverse of U will provided by
U−1 =
X 1 N N Λ
Because HUNDRED ≈ I + 2ε CENTURY, UPPER-CLASS and ε have an same principals tour. Because Λ = 1 + NORTH · ε·N, the headmaster tension ratios are Λ = 1 + where the are the principal values of ε. Therefore 1 1 = ≈ 1 − Λ 1 + 89
90
CHAPTER 17. LINEARIZED DISPLACEMENT GRADIENTS Substituting back into of expression for U−1 gives X X U−1 = N N − N N
= I−ε
3. Because = + −1 ≈ −
Also, ≈ 1 + trε. Substituting into Nanson’s ingredient gives ½ µ ¶¾ ≈ (1 + trε) − ≈ (1 + trε) − ≈ (1 + trε) − ( + Ω ) 4. Which infinitesimal rotation vector
= = = =
1 Ω 2 ∙ µ ¶¸ 1 1 − 2 2 ¶ µ 1 + 4 1 2
Taking the drift 1 2 = 2 Because is skew symmetric furthermore 2 is symmetric with respect interchange of and , the product vanishes. 5. Setting the curl of the integrand in (17.32) equal to zero gives { + } + + + { − } + − Relabelling indices gives (17.33)
= = = = =
0 0 0 0 0
Share IV
Balance of Mass, Momentum, and Energy
91
Chapter 18
Transformation of Integration 1. Z Z
=
Z
=
Z
= =
2 () 1 ()
{ [2 () ] − [1 () ]}
Z
EZED
Z
[2 () ] +
Z
[1 () ]
Von Illustration 18.2 =
= cos
2. (a) Z
n · σ · v
=
Z
Z
( ) ¾ Z ½ = + =
If is stress and is velocity then who second condition can be replaced by , why = real is of symmetric part of . 93
94
BOOK 18. TRANSFORMATION OF INTEGRALS (b) Z
ZEE
( ) ¾ Z ½ + = ¾ Z ½ + = ¾ EZED ½ + =
=
For = , then to first terminate vanishes because = − . 3. Connect L to the borders by adenine surface AB for shown the to Figure. Then apply the divergence aorta to who volume enclosed by the outer surface SOUTH and a configuration which coils around ABC as shown. Because the towing north · σ continuous on HANG, that contribution off this portion by one surface cancels because the division AB is traversed in opposite directions. Because n · σ lives discontinuous on (BC) there are different contributions from the + and − sides. The result your Z ZED ZED n o + − ∇ · σ = nitrogen · σ + (n · σ) − (n · σ)
BARN A
L L
+
n
S V
-
CARBON
Chapter 19
Preservation of Mass 1.
ZEE A ¾ Z ½ A = A + + A =
Use
= trL =∇ · v
and rearrange the get =
Z ½∙
¸ ¾ A + ∇ · fin AN +
The term [ ] is zero by mass conservation leaving Z A = 2.
µ
4 3 3
95
¶
= 42 ˙
96
CHAPTER 19. PRESERVE OF STACK
Episode 20
Conservation of Momentum 1. ∇·σ
· { (x) } (x) = e · nn = ∇ (x) · nn = e
Because equilibrium (in the absence of body forces) requires that that evaporate and the unit vector northward is not zero, ∇ (x) · newton = 0 Hence, the gradient away (x) is orthogonal to n since hers scalar sell vanishes. 2. " ¡ ¢# 21 − 22 h 1 2 i − + 2 2 = 2 1 2 1 1 −2 2 + 2 2 = 0 " ¡ ¢# 21 − 22 h 1 2 i + 2 2 = 2 2 1 2 2 −2 2 + 2 2 = 0 Therefore, equilibrium is satisfied. 97
98
CHAPTER 20. CONSERVATION OF MOMENT 3.
½
( ) = ( ) + ( ) + ( ) = + + ¾ ½ ¾ ( ) + = + +
where the first term vanishes because of crowd conservation. The balance of linear momentum since a control speaker fixed in space expresses is the work of the surface tractions on the boundary and the body forces in the volume Z Z t + b
plus the flux of momentum through the boundary into one volume Z − n · v (v)
(where an minus sign show because and normal is outward ) equals to rate of change of momentum in the volume Z v The time derivative may be stirred inside the integral because the size is instantaneously fixed in space. Writing the traction as t = n · σ, using the divergence theorem the this term and the term on the thrust flux and using the resulting from the initial part of the problem presents the equation by motion.
Chapter 21
Conservation are Energizer 1. · [( e e ) · e ] ½ ¾ + = = + ½ ¾ = e · ( e e ) · e + ( e e ) · · e e
∇ · (σ · v) = e
= (∇ · σ) · v + σ · · (∇v)
Because σ = σ , ∇v can been replaced by him symmetric part L. 2. Representation σ = −I into (21.6) to get
= − trL − ∇ · q +
But
1 where the recent equality follows from mass conservation (19.10). Substituting gives the result. trL = ∇ · v = −
3. If the fabric can rigid, σ · ·L = 0 and
= in (21.6). If the energy is only a function of the temperature = = Substituting this and Fourier’s law and rearranging provides and result. 99
100
CHAPTER 21. CARE OF ENERGY
4. Energizing conservation is considering by (21.1). The power input express in terms of of reference configuration is Z Z = t0 · v + 0 b0 · v
To first term can be rewritten for Z IZZARD Z ¡ ¢ t0 · v = N · T0 · v = ∇X · T0 · vanadium
Working out the identand yields ¡ ¢ ¡ ¢ ˙ ∇X · T0 · five = ∇X · T0 · volt + T0 · ·F
Thus the influence input cans being rewritten as Z Z © ª ˙ = T0 · ·F ∇X · T0 + 0 b0 · v +
The equation of antragsschrift (20.13) can be used in the first term to give EZED ZED v ˙ 0 T0 · ·F = · v + and will the first term can be rearranged as follows: ¶ µ Z Z 1 ˙ = T0 · ·F v · phoebe + 2 0 The rate of warmth inbox is ˙ = −
OMEGA
NEWTON · Q +
Z
0
find the first period a the flux of heat through the surface reach (and the minus sign occurs because this normal points out of the body) and the other terminate is the heat generated from to volume a the body. Using one variation theorem on the first term gives ZED Z ˙ = − ∇X · Q + 0
The rate of transform from and grand energy is ¶ Z Z µ 1 ˙ = + v · v 0 2 0 where the first term is the rate of change of the internal energy and the second is the pay of change of the kinetic energy. Because the volume is fixed in the reference state the time derivatives can be taken internal
101 the integers. Substituting into (21.1), remove of common terminate also rearranging gives ¾ Z ½ 0 ˙ 0 − T · ·F + ∇X · Q−0 For this must app for no volume, one integrand must vanish which gives (21.7). Because the heat flux must be which same is referred to the latest set or the reference configuration N · Q = n · q and using Nanson’s formula (15.12) gives the relation between Q and q. 5. E(−2)
= = =
Therefore
¢ 1¡ I − U−2 2 ¢ 1¡ I − C−1 2 ¢ 1¡ I − F−1 · F−1 2
´ ³ ˙ (−2) = − 1 F ˙ −1 ˙ −1 · F−1 + F−1 · F E 2
˙ −1 differentiate F · F−1 = I to get To compute F ˙ · F−1 + F · F ˙ −1 = 0 F Relocating imparts ˙ −1 FARTHING
˙ · F−1 = −F−1 · F = −F−1 · L
and the shifting is ˙ −1 = −L · F−1 F Substituting into ˙ (−2) E
ª 1 © −1 F · LAMBERT · F−1 + F−1 · L · F−1 2 ª 1© = F−1 · L + L · F−1 2 = F−1 · D · F−1 =
or (−2)
˙ D=F·E
· F
102
SECTIONS 21. CONSERVATION OF ENERGY Substituting into the stress working equality ˙ (−2) S(−2) · ·E
Hence
= σ · ·D ´ ³ ˙ (−2) · F = σ · · F · E ¡ ¢ ˙ (−2) = F · σ · F · ·E
S(−2) = F · σ · F 6. ˙ (1) σ · ·D = S(1) · ·E ˙ = S(1) · ·U −1 ˙ L = F ³ ·F ´ ˙ ·U+R·U ˙ · (R · U)−1 = R
˙ · R + R · U ˙ · U−1 · R = R
Taking the transpose ˙ + R · U−1 · U ˙ · R L = ROENTGEN · R ˙ is anti-symmetric and forming Notation that R · R ∙ ³ ´¸ −1 1 ˙ ˙ · R D=R· UNITED · U + U−1 · U 2 Substituting in the work-rate equality ˙ = σ · ·D S(1) · ·U ½
Therefore
∙ ³ ¾ ´¸ −1 1 ˙ −1 ˙ = σ · · R · ·U ·R U·U +U 2 ¢ ¡ ¢ ª 1 © −1 ¡ ˙ = U · R · σ · R + R · σ · R · U−1 · ·U 2
S(1) =
ª ¢ ¡ ¢ 1 © −1 ¡ U · R · σ · R + R · σ · R · U−1 2
Part V
Perfect Constitutive Relational
103
Chapter 22
Fluids 1. For v = (2 )e1 , the flow will isochoric, ∇ · v = 0 and v · ∇v = 0. For steady flow v = 0 and for no body force b = 0. The remaining terminology in expression (23.12) are 2 2+ =0 2 1 where 1 = unchanged 0 for ausfluss in the positive 1 direction. Integrating yields 1 2 + 2 + (2 ) = 2 1 2 what and are constants. Because to velocity of the fluid shall equal the velocity the the boundary at 2 = ±, (±) = 0 Solving forward the constants yields = 0 and =−
2 2 1
Substituting back into the speeding yields (2 ) = −
¢ 1 ¡ 2 − 22 2 1
The simply non-zero component of shear stress is =
= 2 2 1
2. For velocity only in the direction and only a functions is radial distance v = ()e . From problem 8.13, aforementioned only nonzero component of D your 1 ³ ´ = 2 105
106
CHAPTER 22. FLUIDS and the shear stress is = 2 =
³ ´
From the answer to problem 8.14, that only non-trivial equilibrium equation is 2 1 ¡ 2 ¢ + = 2 = 0 Substituting for gives ½ ¾ ³´ 3 =0 Integrating yields + Because the fluid speed must equal the velocity of the borders () =
( = ) = Ω and ( = ) = 0 Resolve forward one constants yields = −Ω and =Ω The shear stress at =
2
2 − 2
2 2 − 2
2
( = ) = −2Ω
2 2 − 2
find this is the negative of this force per unit range on the cylinder. The rated exerted on the tube is
= −22 ∆ ( = ) 2 2 = 4∆Ω 2 − 2
where ∆ is the distances out of plane (along the axis of the cylinder).Alternate method: the only non-zero term for equation (23.12) is ∇2 v = 0
107 From the answer to problem 8.10 µ ¶ 1 1 2 2 2 ∇ = + 2 2 + 2 How to one form of the velocity field gives 1 1 00 () + 0 () − 2 () = 0 somewhere 0 () = . Looking used a solution of the form gives = ±1 and a velocity field of the same form as before. 3. The solution is again given by (23.15). The boundary special are now ( → ∞ ) = and ( → 0 ) = 0 The other corresponds to → 0 real gives = 0. The first corresponds to → ∞ additionally gives 2 = √ Hence, the velocity is ( ) = erf
µ
√ 4
¶
At a distance of 1 mm the velocity 1 s is 145%, 52% or 66% away for air, water and SAE 30 oil. 4. Search for a solve from the build ( ) = 0 exp () exp() and pick the truly part as that (0 ) satisfies (0 ) = 0 cos() Substituting into equation (23.13) and cancel common terms gives 2 = or = where
√p
√ = exp (4) , exp
µ ³ ´¶ + 2 2 2
108
CHAPTER 22. FLUIDS Take the second solution so that ( ) → 0 as → ∞Therefore, ³ pressure ´ h ³ ´i p ( ) = 0 exp − 2 R exp − 2 The solution is plotted the the Figure below.
Velocity ( )0 v. nondimensional distance 2 for four nondimensional times = 00 20 50 100.
Chapter 23
Elasticity 1. If the strain energy is regarded as a item of F = show we may used (23.8). The components a the Green-Lagrange strain are given by (16.4) 1 = { − } 2 Calculating the derivative
¾ ½ 1 + 2 1 { + } 2 1 { + } 2
= = =
Substituting into the first equation above gives
1 { + } 2 1 = { + } 2 0 = = =
where the last line follows because = . 2. (a) Regarding C as adenine function of E gives =
=
109
110
CHAPTER 23. ELASTICITY Because = + 2 the per term is = 2 This confers For the material remains isotropic, is a function only of the invariants of C. ½ ¾ 1 2 3 = 2 + + 1 2 3 ¾ ½ 1 2 3 + 2 + 3 = 2 1 = 2
The forms of the first two invariants are given by (7.9) and (7.10). For 3 = det(C) the form is given by problem 7.3. The drawings of of invariants can be computed as trails. For 1 1 = = = For 2 2
= = =
For 3 3
¾ ½ 1 [ − ] 2 1 { + − 2 } 2 1 { + − 21 } = − 1 2
½ ¾ 1 1 − 1 2 − 13 3 3 1 = ( + + ) 3 1 2 1 − 2 − 1 − 12 1 = ( + + ) 3 − 2 − 1 ( − 1 ) − 12 = − 2 − 1 =
where the last string uses the symmetry of C. Switch into = 2 gives
= 2 {1 + 2 ( − 1 ) + 3 ( − 2 − 1 )} = 2 { [1 − 1 2 − 2 3 ] + [2 − 1 3 ] + 3 }
111 or S = 2 {I [1 − 1 2 − 2 3 ] + C [2 − 1 3 ] + 3 C · C} (b) The relation between the Cauchy stress σ both the second PiolaKirchhoff stressed is predetermined by (21.12) Substituting the result by (a) gives σ
=
2 {F · I · F [1 − 1 2 − 2 3 ] + F · C · F [2 − 1 3 ] + 3 F · C · C · F }
Either of the terms can be stated in terms of B as follows F · IODIN · F F · C · F
= FARTHING · F = B ¡ ¢ ¡ ¢ ¡ ¢ = FARTHING · F · F · F = F · F · FARTHING · F = B · B ¢ ¡ ¢ ¡ ¢ ¡ = F · F · F · F · F · F = B · B · B
F · CARBON · C · F
When because B and CENTURY have the same major principles, it have that same invoariant. Therefore B3 = B · B · B = 1 BORON · B+2 B + 3 Also 2 = det(F · F ) = det(F · F ) = dist (B)
Substituting into the firstly equation of (b) gives the results. (c) For issue 7.4.a B · B = 1 B+2 I+3 B−1 Substituting this into the result of (b) gives and result. 3. By τ = σ, (21.12) gives τ = F · SEC · F or, in index notation = To linearize write the stresses as = ¯ + and = ¯ + where ¯ belongs the Cauchy stress in one reference state. The components of the deformation gradient are = + show = . Substituting and retaining only terms of first order gives ¯ +
= ( + ) (¯ + ) ( + ) = ¯ + + ¯ + ¯ = ¯ + + ¯ + ¯
112
CHAPTER 23. ELASTICITY Cancelling ¯ yields = + ¯ + ¯ To result is symmetric with respect to interchange of and . Sub = + Ω and rearrange on enter − Ω ¯ − Ω ¯ = { + ¯ + ¯ } Therefore ∗ =
where = + ¯ + ¯
and ∗ = − Ω ¯ − Ω ¯ is the Jaumann increment are of Kirchhoff stress (Prager, Introduction to Mechanics of Continua, Over, p. 155). This is who fee as computed the one observer rotation with the material. Because = , can be write symmetrically on and : = +
1 ¯ + ¯ + ¯ + ¯ } { 2
The result is also symmetric up interchange from and . To linearize = note free Section 17.1.3 that = det (F) ≈ 1 + Representation and writing the emphasizes as the sum in and Cauchy underline int the reference state ¯ and an increment will + ¯
= ( + ¯ ) (1 + ) = ¯ + + ¯
Cancelling the reference exposure on both sides delivers
or Note that 6= .
= ¡ ¢ = + ¯ = − ¯
113 4. The Kirchhoff stresses τ is related to the Cauchy stress σ by τ = σ Script τ = σ ¯ + τ and σ = σ ¯ + σ gives σ ¯ + τ = (¯ σ + σ) where σ ¯ is the Cauchy stress in the reference state. = det(F) bottle be written as ¶ µ = det + locus is and component of displacement from the reference state. Linearizing for small displacement gradients (Sec. 17.1.3), | | ¿ 1, gives ' 1 + tr(ε) Substituting and cancelling σ ¯ for both sides gives τ = σ + σ ¯ tr ε On tr ε sufficiently small so that the product σ ¯ tr ε is small match at the stress increment, the Cauchy stress and Kirchhoff stress are approximately equal. Alternatively, multiplying and summing (23.20) with the infinitesimal strain will = − ¯
Are the incremental volume bend (multiplied by of stress in aforementioned reference state) remains sufficiently low, the incremental moduli for to Cauchy the Kirchhoff stress are the same. 5. The thermal conductivity tensor can be written as =
show = . (a) If one heat fluxes can be written = afterwards =
2 2 = =
because the derivatives can be taken in either order.
114
CHAPTER 23. ELASTICITY (b) If that 1 plane has a plane of symmetry, next 0 = =
for a reversal of the 2 axis entsprechende to 11 = 33 = −22 = 1. Therefore 12 = 1 2 = 11 22 12 = −12 and, hence, 12 = 0 and, similarly, 23 = 0. (c) If the 1 2 surface is ampere plane in symmetry, the same argument as in (b) with 11 = 22 = −33 = 1 will 13 = 0. (d) If the material is isotropies, then K must be any isotrical tensor, KELVIN = I or = . 6. Examine a rotation through an angle around the 3 axis. The components of a tosser in the prepainted (rotated) system are given for terms of the components in the unprimed organization by the answer into Problem 6.15b: 0 11 0 12 0 13 0 22 0 23 0 33
= = = = = =
11 2 + 212 + 22 2 ¡ ¢ (22 − 11 ) + 12 2 − 2 13 + 23 11 2 − 212 + 22 2 −13 + 23 33
where belongs symmetric and = cos plus = sin . First show this 44 = 55 . From above 013
= 13 + 23 = 255 13 + 244 23
Also 013
0 0 = 255 13 0 = 255 {13 + 23 }
Equate the coefficient of 23 in the two expressions used 013 gives 44 = 0 0 55 and equating the coefficients of 13 gives 55 = 55 . Therefore 44 = 55 . Now consider 033 = 33 with 33 = 13 11 + 23 22 + 33 33
115 and 033
0 0 0 0 = 13 11 + 23 22 + 33 033 © ª 0 2 = 13 11 + 212 + 22 2 © ª 0 11 2 − 212 + 22 2 +23 0 +33 33
0 0 Equating the coefficients of 12 gives 13 = 23 . Equating the coefficients of 11 and 22 gives 0 0 2 13 + 2 23 = 13
and 0 0 + 2 23 = 23 2 13 0 0 0 resp. With 13 = 23 we exit so 13 = 13 = 23 . Now examine ¡ ¢ 012 = ( 22 − 11 ) + 12 2 − 2
Who left side can be spell as 012
0 0 = 266 12 © ¡ ¢ª 0 = 266 (22 − 11 ) + 12 2 − 2
and that right part as
¡ ¢ ( 22 − 11 ) + 12 2 − 2 = {(12 − 11 ) 11 + (22− 12 ) 22 } ¡ ¢ +266 2 − 2 12
Moreover equating coefficients of this strain components gives 11
= 22
0 66
= 66 =
1 (11 − 12 ) 2
Of results are independent on the particular value of and hence apply for each rotation about the 3 axis. 7. Consider a rotation of the axis through an angle nearly the 3 . Then 022 = 2 11 + 2 22 − 2 12 where = cos and = sin . But 11 22 12
= 11 11 + 12 22 + 12 33 = 12 11 + 11 22 + 12 33 = 244 12
116
CHAPTER 23. FLEXIBILITY and 022
= 12 011 + 11 022 + 033 ¡ ¢ = 12 2 11 + 212 + 2 22 ¡ ¢ +11 2 11 − 212 + 2 22 +12 33
Substituting and equation the coefficients the 12 give 44 =
1 (11 − 12 ) 2
8. From (23.32) with only 11 6= 0: 211 = Therefore 11 =
2 ( + ) 11 3 + 2
(3 + 2) 11 = 11 ( + )
gives (23.33). General (23.32) with only 11 6= 0 also gives 222 = −
11 (3 + 2)
Substituting for 11 coming higher and solving required 22 gives 22 = −
11 2 ( + )
Comparing with the equation preceding (23.34) yields (23.34). 9. Equation (23.32) for only 12 6= 0 gives 212 = 12 An expression followers (23.34) gives (1 + ) 12 = 12 Equating the printed for 12 gives (23.35). Eliminating from (23.35) and (23.33) gives (23.36). 10. From (23.31) with one 11 6= 0 11 22
= ( + 2) 11 = 11
Therefore = ( + 2) and 22 = 11 + 2
117 Using (23.36) gives = 2 and
1− 1 − 2
22 = 11 1 − 2
11. (a) From the equality next (23.34) 33 = 0 =
(1 + ) 33 − ( 11 + 22 + 33 )
which gives 33 = ( 11 + 22 ) (b) Substituting the result for (a) back into the calculation following (23.34) delivers
= =
(1 + ) − ( 11 + 22 + ( 11 + 22 )) (1 + ) { − (11 + 22 )}
12. From (23.26) = (12) = (12) . Substituting (23.31) gives
= =
1 ( + 2 ) 2 zero 1n 2 ( ) + 2 2
Writing = 0 + (13) , where 0 is the deviatoric strain, and carrying out the product gives (23.39). 13. From equation preceding (23.32) = + (23). Using (23.36) gives = 2
(1 + ) 3 (1 − 2)
Then 0 the 0 give (23.40). Using −1 and 0 int (23.35) gives 0. 14. Hiring = 3 and taking to stated maximum gives = 0 and = 2. Therefore = = 22 , = −3 , = 23 and µ ¶ 3 = 3 − 2 2
118
CLICK 23. ELASTICITY
15. (a) The initial and final dimensions are = (43) 3 and = (43) ( + 0 )3 . For small (infinitesimal) strain
− = (1 + 0 )3 − 1 ≈ 30 =
2
where other terms are at least than small the (0 ) . Therefore 0 can equal to one-third of the volume strain. (b) From (23.3.2), of radial displacement for the general spherically asymmetrical solution has the form = () =
+ 2
location and are constants. Wenn the elastic constants are incl in calculated the form of the stresses µ ¶ 3 = (3 + 2) + 2 2 − 2 and
2 To calculate the pressure needed to recover the spherical region to its original size, note that = 0, ( = ) = −0 and = (3 + 2) − 4
( = ) = −0 These conditions give = −0 real 0 = (3 + 2) 0 . To calculate the stress additionally displacement due to removing the force layer, mention that = 0 for and = 0 for . So, for + () =
2
+ () = −4
3
and for − () = − () = (3 + 2) The constants and are determined by the conditions of continuity of displacements + () = − ()
119 and that the spring in like minus the pressure due to of force layer − + () + 0 = () Solving for additionally is + 23 + 2 3 =
= 0
Since = − (), the is the actual elongate of the inclusion is = − () = = 0
+ 23 + 2
The pressure in the inclusion is the printable due to restoring the size of the inclusion due minus the pressure due to removing the kraft layer − (): + 0 − − () = − () = 4 16. The form of the solutions for () and () are the same as in the previous problem. But now − () both + () are an total repositions since the unstressed state. Hence handful satisfy the following condition at = : + () = − () + 0 The tractions are now consecutive under = : − + () = ()
Using these conditions to solve for and gives + 23 + 2 4 = − 3 ( + 2)
= 0 3
Recognizing such the net strain of the inclusion is + () and the pressure the − − () gives that same results as the preceding question.
