Lai et al, Introduction to Continuum Mechanics
CHAPTER 3 3.1 Consider the motion: x1 = (1 + kt) X1 / (1 + kto ), x2 = X2 , x3 = X3 . (a) Show that reference time is t = t o . (b) Find the velocity field in spatial coordinates. (c) Show that the velocity field is identical to that of the following motion: x1 = (1 + kt) X1 , x2 = X2 , x3 = X3 . ---------------------------------------------------------------------------------- Ans. (a) At t = t o , x1 = X1 , x2 = X2 , x3 = X3 . Thus, t = t o is the reference time. (b) In material description, v1 = kX1 / (1 + kto ), v2 = v3 = 0 . Now, from
1x =
(1 + k)t X 1 / (1 + kot) ,
X1 = kx1 / (1 + kt ), v2 = v3 = 0 . → X1 = (1 + kot) 1x/ (1 + k)t , therefore, → v1 = kkX (c) For x1 = (1 + kt) X1 , x2 = X2 , x3 = X3 , → v1 = kX1 , v2 = v3 = 0 → v1 = kx1 / (1 + kt), v2 = v3 = 0 , which are the same as the velocity components in (b).
___________________________________ ______________________________________________________ ______________________________ ___________ 3.2 Consider the motion: x1 = α t + X1 , x2 = X2 , x3 = X3 , where the material coordinates X i designate the position of a particle at t = 0 . (a) Determine the velocity and acceleration acceleration of a particle in both a material material and a spatial description. description. (b) If the temperature field in spatical spatical description is given by θ = Ax1 , what is its material description? Find the material material derivative derivative of
θ , using both descriptions descriptions of the temperature. (c) Do part (b) if the temperature field is is θ = Bx2 ---------------------------------------------------------------------------------- Ans. (a) Material description: v1 = Dx1 / Dt = ( ∂ x1 / ∂ t ) X −fixed = α , v2 = v3 = 0 , i
a1 = Dv1 / Dt = ( ∂ v1 / ∂ t ) X −fixed = 0, a2 = a3 = 0 . i
Spatial description: The same as above v1 = α , v2 = v3 = 0, a1 = a2 = a3 = 0 .. (b) The material description of θ is θ = A(α t + X1 ) . Using the material description: θ = A(α t + X1 ) → Dθ / Dt = (∂ / ∂ t) ⎡⎣ A(α t+ X1 ) ⎤⎦ = Aα . Using the spatical description: θ = Ax1 →
∂θ ∂θ ∂θ ∂θ + v1 + v2 + v3 = 0 + α ( A) + (0)(0) + (0)(0) = Aα . ∂ 1x ∂ 2x ∂ 3x Dt ∂ t (c) Using the material description: θ = BX2 → D θ / Dt= (∂ / ∂ )t( BX2 ) = 0 . Using the spatical description: θ = Bx2 → ∂θ ∂θ ∂θ Dθ ∂θ = + v1 + v2 + v3 = 0 + α (0) + ( 0 )( )( B) + ( 0 )( )( 0 ) = 0 . ∂ 1x ∂ 2x ∂ 3x Dt ∂ t Dθ
=
___________________________________ ______________________________________________________ ____________________________________ _________________ 3.3 Consider the motion 2 2
coordinates. (a) at t = 0 , the x1 = X1 , x2 = β X1 t + X2 , x3 = X3 , where X i are the material coordinates. corner cornerss of a unit unit square square are at
(0,0,0), B(0,1 (0,1,0), C(1,1,0) ,1,0) and D (1,0, 0) . Determine the position A
of ABCD at t = 1 and sketch the new shape of the square. (b) Find the velocity v and the acceleration in a material description and (c) Find the spatial velocity field. -----------------------------------------------------------------------------------------
Copyright 2010, Elsevier Inc 3-1
Lai et al, Introduction to Continuum Mechanics Ans. For the material line AB,
(
X1 , X2 , X3 ) = ( 0, X2 , 0 ) ; at t = 1 ,
( x1 , x2 , x3 ) = ( 0, X2 , 0)
( x1 , x2 , x3 ) = ( X1 , β X12 + 1, 0 )
For the material line BC,
( X1 ,
X2 , X3 ) = ( X1 , 1, 0) ; at t = 1 ,
For the material line AD,
( X1 ,
X2 , X3 ) = ( X1 , 0, 0) ; at t = 1 ,
For the material line CD,
( X1 , X 2 , X 3 ) = (1, X 2 , 0) ; at t = 1 , ( x1 ,
( x1 , x2 , x3 ) = ( X1 , β X12 , 0 ) x2 , x3 ) = (1, β + X2 , 0)
The shape of the material square at t = 1 is shown in the figure.
⎛ ∂ xi ⎞ ⎛ ∂v ⎞ → v1 = v3 = 0, v2 = 2 β X12 t; a1 = a3 = 0, a2 = 2β X12 , ai = ⎜ i ⎟ ⎟ ⎝ ∂t ⎠ X i −fixed ⎝ ∂t ⎠ X i −fixed
(b) vi = ⎜
2 2 (c) Since x1 = X 1 , in spatial descrip. v1 = v3 = 0, v2 = 2 β x1 t; a1 = a3 = 0, a2 = 2 β x1
___________________________________ ______________________________________________________ _______________________________ ____________ 3.4 Consider the motion: x1 = β X22 t2 + X1 , x2 = kX2 t + X2 , x3 = X3 (a) At t = 0 , the corner cornerss of a unit unit square square are at
(0,0,0), B(0,1 (0,1, 0), C(1,1, ,1, 0) and D(1,0, 0) . Sketch A
the deformed shape of the square at t = 2 . (b) Obtain the spatial description of the velocity field. (c) Obtain the spatial description of the acceleration field. -------------------------------------------------------------------------------------------- Ans. (a) For material line AB, ( X1 , X2 , X3 ) = ( 0, X2 , 0) ; at t = 2 , For material line BC, For material line AD, For mat. line CD,
( x1 , x2 , x3 ) = ( 4 β X22 , 2 kX2 +
( X1 , X2 , X3 ) = ( X1 , 1, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 4 β + X1 , 2 k + 1, 0) . ( X1 , X2 , X3 ) = ( X1 , 0, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( X1 , 0, 0) .
( X1 , X 2 , X 3 ) = (1, X 2 , 0) ; at t = 2 , (
4
B’ 2k
B
A
C’ C
D
x1
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(
)
2 x1 , x2 , x3 ) = 4 β X2 + 1, 2 kX2 + X2 , 0 .
The shape of the material square at t = 2 is shown in the figure. x2
)
X2 , 0 .
Lai et al, Introduction to Continuum Mechanics Ans. For the material line AB,
(
X1 , X2 , X3 ) = ( 0, X2 , 0 ) ; at t = 1 ,
( x1 , x2 , x3 ) = ( 0, X2 , 0)
( x1 , x2 , x3 ) = ( X1 , β X12 + 1, 0 )
For the material line BC,
( X1 ,
X2 , X3 ) = ( X1 , 1, 0) ; at t = 1 ,
For the material line AD,
( X1 ,
X2 , X3 ) = ( X1 , 0, 0) ; at t = 1 ,
For the material line CD,
( X1 , X 2 , X 3 ) = (1, X 2 , 0) ; at t = 1 , ( x1 ,
( x1 , x2 , x3 ) = ( X1 , β X12 , 0 ) x2 , x3 ) = (1, β + X2 , 0)
The shape of the material square at t = 1 is shown in the figure.
⎛ ∂ xi ⎞ ⎛ ∂v ⎞ → v1 = v3 = 0, v2 = 2 β X12 t; a1 = a3 = 0, a2 = 2β X12 , ai = ⎜ i ⎟ ⎟ ⎝ ∂t ⎠ X i −fixed ⎝ ∂t ⎠ X i −fixed
(b) vi = ⎜
2 2 (c) Since x1 = X 1 , in spatial descrip. v1 = v3 = 0, v2 = 2 β x1 t; a1 = a3 = 0, a2 = 2 β x1
___________________________________ ______________________________________________________ _______________________________ ____________ 3.4 Consider the motion: x1 = β X22 t2 + X1 , x2 = kX2 t + X2 , x3 = X3 (a) At t = 0 , the corner cornerss of a unit unit square square are at
(0,0,0), B(0,1 (0,1, 0), C(1,1, ,1, 0) and D(1,0, 0) . Sketch A
the deformed shape of the square at t = 2 . (b) Obtain the spatial description of the velocity field. (c) Obtain the spatial description of the acceleration field. -------------------------------------------------------------------------------------------- Ans. (a) For material line AB, ( X1 , X2 , X3 ) = ( 0, X2 , 0) ; at t = 2 , For material line BC, For material line AD, For mat. line CD,
( x1 , x2 , x3 ) = ( 4 β X22 , 2 kX2 +
( X1 , X2 , X3 ) = ( X1 , 1, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 4 β + X1 , 2 k + 1, 0) . ( X1 , X2 , X3 ) = ( X1 , 0, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( X1 , 0, 0) .
( X1 , X 2 , X 3 ) = (1, X 2 , 0) ; at t = 2 , (
4
B’ 2k
B
A
C’ C
D
x1
Copyright 2010, Elsevier Inc 3-2
(
)
2 x1 , x2 , x3 ) = 4 β X2 + 1, 2 kX2 + X2 , 0 .
The shape of the material square at t = 2 is shown in the figure. x2
)
X2 , 0 .
Lai et al, Introduction to Continuum Mechanics
⎛ ∂ xi ⎞ ⎛ ∂vi ⎞ →, , ai = ⎜ ⎟ ⎟ ⎝ ∂t ⎠ X − fixed ⎝ ∂t ⎠ X −fixed
b) vi = ⎜
i
2
2
v1 = 2 β X 2 t , v2 = kX 2 , v3 = 0; a1 = 2 β X 2 , a2 = a3 = 0 .
i
(c) x2 = ( kt+ 1) X2 → , v1 =
2 2x 2
2 β
(1 + kt )
t , v2 =
k2 x
(1 + kt )
, v3 = 0; a1 =
2β 22x
(1 + kt )
2
, a2 = a3 = 0 .
____________________________________ ______________________________________________________ _____________________________ ___________ 3.5 Consider the motion: x1 = k ( s + X1 ) t + X1 , x2 = X2 , x3 = X3 . (a) For this motion, repeat part (a) of the previous problem. (b) Find the velocity and acceleration as a function of time of a particle that is initially at the orgin. (c) Find the velocity and acceleration as a function of time of the particles that are passing through the origin. ---------------------------------------------------------------------------------- Ans. a) For material line AB, ( X1 , X2 , X3 ) = ( 0, X2 , 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2 ks, X2 , 0) . For material line BC, For material line AD, For material line CD,
( X1 , X2 , X3 ) = ( X1 , 1, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2 ks+ 2 kX1 + X1 , 1, 0) . ( X1 , X2 , X3 ) = ( X1 , 0, 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2 ks+ 2 kX1 + X1 , 0, 0) . ( X1 , X 2 , X 3 ) = (1, X 2 , 0) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2 ks+ 2 k+ 1, X2 , 0) .
The shape of the material square at t = 2 is shown in the figure.
x2 2k s B
2k(s+1)
B’
C’
C
1 A
A’
D’
D
x1
⎛ ∂ xi ⎞ ⎛ ∂vi ⎞ and ai = ⎜ → , v1 = k ( s + X1 ) , v2 = 0, v3 = 0; a1 = a2 = a3 = 0 . ⎟ ⎟ ⎝ ∂t ⎠ X − fixed ⎝ ∂t ⎠ X −fixed
(b) vi = ⎜
i
i
Thus, for the particle
( X1 , X2 , X3 ) = ( 0, 0, 0 ) ,
and a1 = 0, a2 = 0, a3 = 0 v1 = ks, v2 = 0, v3 = 0 and
(c) x1 = k ( s + X1 ) t + X1 → 1x = ks k st+ ( kt+ 1) X1 → X1 = ( x1 − kst ) / (1 + kt ) , thus, in spatial descriptions,
⎧⎪
kst⎫ ⎪
k ( s + x1 ) , v2 = 0, v3 = 0 and and a1 = 0, a2 = 0, a3 = 0 . ⎬= + + 1 1 k t k t ( ) ( ) ⎩⎪ ⎭⎪ (1 + kt ), v2 = 0, v3 = 0 and a1 = 0, 0, a2 = 0, a3 = 0 . At the position ( x1 , x2 , x3 ) = ( 0, 0, 0) , v1 = ks / (1 v1 = k ⎨ s +
x− 1
____________________________________ _______________________________________________________ _____________________________ __________ 3.6 The position at time t of a particle initially at
( X1 , X 2 , X3 ) is given by
x1 = X1 − 2 β X2 t , x2 = X2 − kX3 t, x3 = X3 , where β = 1 and k = 1 . 2 2
(a) Sketch the deformed shape, at time t = 1 of the material line OA which was a straight line at 0,0,0 ) and and the the poin pointt A at ( 0,1 0,1,0 ) . (b) Find Find the veloci velocity ty at t = 2 , of the t = 0 with the point O at ( 0,0,0 particle which was at (1, 3,1) 3,1) at t = 0 . (c) Find the velocity of the particle which which is at (1, 3,1) 3,1) at t = 2 .
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Lai et al, Introduction to Continuum Mechanics -------------------------------------------------------------------------------------------- Ans. With β = 1 and k = 1 ,
2 2
x1 = X1 − 2 X 2 t , x2 = X 2 − X 3 t, x3 = X 3
For the material line OA : ( X1 , X2 , X3 ) = (0, X2 , 0) : at t = 1 , x1 = −2 X22 , x2 = X2 , x3 = 0 . Thus, the deformed shape of the material line at t = 1 is a parabola given in the figure shown. 2
A
A’
x1
O
2 (b) v1 = Dx1 / Dt Dt = −4 X 2 t , v2 = Dx2 / Dt = − X 3 , v3 = Dx3 / Dt D t = 0
2 For the particle ( X1 , X 2 , X3 ) = (1, 3,1) , at t = 2 , v1 = −4(3) (2) = −72, v2 = −1, v3 = 0.
(c) The particle, which is at ( x1 , x2 , x3 ) = (1, 3,1) at t = 2 , has the material coordinates given by the following equations: 1 = X1 − 8 X22 , 3 = X2 − 2 X3 , 1 = X3 → 201 201, X 1 =
X2 = 5,
X3 = 1
→ v1 = −4 X 22 t = −4(5)2 (2) = −200, v2 = − X 3 = −1, v3 = 0. ____________________________________ _______________________________________________________ _____________________________ __________ 3.7 The position at time t of a particle initially at ( X1 , X 2 , X3 ) is given by: x1 = X1 + k ( X1 + X 2 ) t, x2 = X 2 + k ( X1 + X 2 ) t, x3 = X3 ,
(a) Find the velocity at t = 2 , of the particle which was at (1,1 (1,1,, 0) at the reference time t = 0 . (b) Find the velocity of the particle which is at (1,1 (1,1,, 0) at t = 2 . ----------------------------------------------------------------------------------------Dt = k ( X1 + X 2 ) , v2 = Dx2 / Dt = k ( X1 + X 2 ) , v3 = Dx3 / Dt D t = 0. Ans. (a) v1 = Dx1 / Dt For the particle ( X1 , X 2 , X3 ) = (1,1, 0) , at t = 2 , v1 = k (1 + 1) = 2k , v2 = k (1 (1 + 1) = 2 k , v3 = 0 (b) The particle, which is at
( x1 , x2 , x3 ) = (1,1, 0) at t = 2 , has the material coordinates given by
the following equations: 1 = X1 + 2 k ( X1 + X2 ) , 1 = X2 + 2 k( X1 + X2 ) , 0 = X3 . →
X=
1
1 1 + 4k
,
X=
2
1 1 + 4k
,
X= 0 ,
3
→ v1 = v2 = k ( X 1 + X 2 ) =
2 k 1 + 4 k
, v3 = 0
____________________________________ _______________________________________________________ _____________________________ __________ 3.8 The position at time t of a particle initially at
( X1 , X2 , X3 ) is given by
2 2 x1 = X1 + β X2 t , x2 = X2 + kX2 t, x3 = X3 , where β = 1 and k = 1 .
(a) for the particle which was initially at (1,1,0), what are its positions in the following instant of time: t = 0, t = 1, t = 2 . (b) Find the initial position for a particle which is at (1,3,2) at t = 2 . (c) Find the acceleration at t = 2 of the particle which was initially at (1,3,2) and (d) find the acceleration of a particle which is at (1,3,2) at t = 2 . -------------------------------------------------------------------------------------------2 2
Ans. With β = 1 and k = 1 , x1 = X1 + X 2 t , x2 = X2 + X2 t, x3 = X3
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Lai et al, Introduction to Continuum Mechanics (a) t = 0 → ( x1 , x2 , x3 ) = ( X1 , X 2 , X3 ) = (1,1, 0) , 2
t = 1 → ( x1 , x2 , x3 ) = ( X1 + X2 , X 2 + X2 , X3 ) = (2, 2,0) 2
t = 2 → ( x1 , x2 , x3 ) = ( X1 + 4 X 2 , X 2 + 2 X2 , X3 ) = (5,3, 0)
(b) x1 = X1 + X22 t2 x2 = X2 + X2 t, x3 = X3 , at t = 2 → 1 = X1 + 4 X 22 , 3 = 3 X 2 , 2 = X 3
→ X1 = −3, X2 = 1, X3 = 2 . (c) x1 = X1 + X 22 t2 x2 = X2 + X2 t, x3 = X3 → v1 = 2 X22 t, v2 = X2 , v3 = 0 . 2
→ a1 = 2 X 22 , a2 = 0, a3 = 0 . For ( X1 , X2 , X3 ) = (1,3, 2 ) , → a1 = 2 ( 3) = 18, a2 = a3 = 0 at any time. (d) The initial position of this particle was obtained in (b), i.e., → X1 = −3,
X2 = 1,
X3 = 2 .
Thus, → a1 = 2 X 22 = 2(1)2 = 2, a2 = 0, a3 = 0 . _________________________________________________________________ (a) Show that the velocity field vi = kxi / (1 + kt ) corresponds to the motion xi = Xi (1 + kt)
3.9
and (b) find the acceleration of this motion in material description. ---------------------------------------------------------------------------------------- Ans. (a) From xi = Xi (1 + kt) and Xi = xi / (1 + kt) → iv = kXi = kx ) . i / (1 + kt (b) vi = kX i → ai = 0 , or
⎛ ∂v ⎞ ai = ⎜ i ⎟ ⎝ ∂t ⎠ ix−
2 2 kx j k δ ij ∂vi k xi k xi kxi k + vj =− + =− + = 0. 2 ∂ + + + + + 1 1 1 1 1 x kt kt kt kt kt ( ) ( ) ( ) ( ) ( ) + 1 kt j ( ) fixed
_________________________________________________________________ 3.10 Given the two dimensional velocity field: v x = −2 y, vy = 2 x . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------∂v x ∂v ∂v + v x x+ v y x= 0 + ( − 2 y ) (0) + ( 2 x)( − 2 ) = − 4 x , Ans. (a) a x= ∂t ∂x ∂y a y=
(b)
∂v ∂t
dx dt dx
y
+v
∂v x
∂x
= −2 and y
y
+v dy dt dy
∂v y
∂y
y
= 0 + ( − 2 y ) (2) + ( 2 x)( 0 ) = − 4 y , i.e., a = −4 xe x − 4 ye y
= 2 →x
dy dx 2
=−
d x
x y
→ dy
+ xdx
= 0 , → x2 + y2 = constant= X2 + Y2 , ydy 2
d x
= −2 and =2 → = −2 = − 2 ( 2 )x→ + 4 =x 0 y x dt dt dt dt dt → x = Asin 2 t + Bcos 2 t and y = − Acos 2 t + Bsin 2 t , where A = − Y, B = X .
Or,
_________________________________________________________________ 3.11 Given the two dimensional velocity field: v x = kx, vy = − ky . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------∂v x ∂v ∂v + v x x+ v y x= 0 + ( kx ) ( k ) + ( − ky)( 0) = k 2 x Ans. (a) a x= ∂t ∂x ∂y
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Lai et al, Introduction to Continuum Mechanics
a y=
(b)
∂v
y
∂t
dx dt
+v
∂v x
∂x x
= kx →
∫
y
+v
dx x
X
∂v y
∂y
= 0 + ( kx ) (0) + ( − ky)( − k ) = k 2 y , That is, a = k 2 ( xe x + ye y )
y
t
= ∫ kdt → ln x − ln X = kt → ln 0
Similarly derivation gives → y= Ye−kt . Or,
x X
= kt → x= Xekt .
xy= XYwhere ( X , Y ) are material coordinates.
_________________________________________________________________ 3.12 Given the two dimensional velocity field: v x = k ( x2 − y2 ), vy = − 2 kxy . Obtain the acceleration field. ----------------------------------------------------------------------------------------∂v x ∂v ∂v + v x x+ v y x= 0 + k x2 − y2 (2 kx) + (− 2 kxy)(− 2 ky) = 2 xk2 ( x2 + y2 ) . Ans. a x= ∂t ∂x ∂y
(
a y=
∂v ∂t
y
+v
∂v x
∂x
(
y
+v
∂v y
∂y
That is, a = 2k 2 x 2 + y 2
(
y
)
)
= 0 + k x2 − y2 ( − 2 ky) − 2 kxy(− 2 kx) = 2 yk2 ( x2 + y2 ) .
) ( xe
x
+ ye y )
_________________________________________________________________
∂v + ( ∇v ) v is Dt ∂ t nonlinear. That is, if we consider two velocity fields v A and v B , then a A + a B ≠ aA+B , where 3.13 In a spatial description, the equation to evaluate the acceleration
A
Dv
=
B
a and a denote respectively the acceleration fields corresponding to the velocity fields
v A and v B each existing alone, aA+B denotes the acceleration field corresponding to the combined
velocity field v A +v B . Verify this inequality for the velocity fields: v
A
= −2 x2e1 + 2 1xe2 , v B = 2 2xe1 − 2 1xe2
------------------------------------------------------------------------------------------- Dv ∂v = + ( ∇v ) v Ans. From Dt ∂ t 0 0 −2 ⎤ ⎡ −2 2x⎤ ⎡ −4 1 x⎤ ⎡0 ⎤ ⎡ 0 2 ⎤ ⎡ 2 2x⎤ ⎡ −4 1 x⎤ ⎡aA ⎤ = ⎡⎢ ⎤⎥ + ⎡⎢ = , ⎡a B ⎤ = ⎢ ⎥ + ⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎣ ⎦ ⎣0 ⎦ ⎣ 2 0 ⎦ ⎣ 2 1 x⎦ ⎣ −4 2x⎦ ⎣ ⎦ ⎣0 ⎦ ⎣ −2 0 ⎥⎦ ⎢⎣ −2 1 ⎥⎦x = ⎢⎣−4 2x⎥⎦
→ a A = −4 1xe1 − 4 x2e 2 , aB = −4 1xe1 − 4 2xe2 → a A + aB = −8 x1e1 − 8 x2e2 . On the other hand, v A +v B =0, so that aA+B = 0 . Thus, aA + a B ≠ a A+B _________________________________________________________________ 3.14 Consider the motion: x1 = X1 , x2 = X2 + ( sin π t)( sin π X1 ) , x3 = X3 (a) At t = 0 , a material filament coincides with the straight line that extends from ( 0, 0, 0) to
(1, 0,0 ) . Sketch the deformed shape of this filament at t = 1 / 2, t = 1 and t = 3 / 2 . (b) Find the velocity and acceleration in a material and a spatial description.
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Lai et al, Introduction to Continuum Mechanics ---------------------------------------------------------------------------------------- Ans. (a) Since x1 = X1 and x3 = X3 , therefore there is no motion of the particles in the x1 and x3 directions. Every particle moves only up and down in the x2 direction.
When t = 1 / 2 → x2 = X 2 + sin π X1 , t = 1 → x2 = X 2 , t = 3 / 2 → x2 = X2 − sin π X1 The deformed shapes of the material at three different times are shown in the figure.
y
t=1/2 t=0, t=1 (1,0)
x
t=3/2
(b) v1 = 0, v2 = π ( cos π t )( sin π X1 ) , v3 = 0 ,
2 a1 = 0, a2 = −π ( sin π t )( sin π X1 ) , a3 = 0
Since x1 = X 1 , the spatial descriptions are of the same form as above except that X 1 is replaced with x1 . _________________________________________________________________ 3.15 Consider the following velocity and temperature fields: 2 2 v = α ( x1e1 + x2 e2 ) / ( x1 + x2 ),
Θ = k( x12 + x22 )
(a) Write the above fields in polar coordinates and discuss the general nature of the given velocity field and temperature field (e.g.,what do the flow and the isotherms look like?) (b) At the point A (1,1,0 ) , determine the acceleration and the material derivative of the temperature field. ----------------------------------------------------------------------------------------2 2 2 Ans. (a) In polar coordinates, x1e1 + x2 e2 = rer , where r = x1 + x2 and er is the unit vector in
the r direction, so that v =
α
2
er , Θ =kr . Thus, the given velocity field is that of a two
r dimensional source flow from the origin, the flow is purely radial with radial velocity inversely
proportional to the radial distance from the origin. With Θ=kr 2 , the isotherms are circles. α (b) From vr = and vθ = 0 , and Eq. (3.4.12) r
∂vr ∂vr vθ ∂vr vθ 2 α 2 ⎛ α ⎞⎛ α ⎞ + vr + − = 0 + ⎜ ⎟⎜ − 2 ⎟ + 0 + 0 = − 3 . ar = ∂t ∂r r ∂θ r r ⎝ r ⎠⎝ r ⎠ ∂v ∂v v ∂v v v aθ = θ + vr θ + θ θ + r θ = 0 . ∂t ∂r r ∂θ r 2 3 2 3 2 That is, a = −α / r er . At the point A(1,1,0), r = 2 , a = −α / ( 2) e r = −α 2 / 4e r . ∂Θ vθ ∂Θ DΘ ∂Θ ⎛ α ⎞ = + vr + = 0 + ⎜ ⎟ ( 2 kr ) = 2α k . ∂ r Dt ∂ t r ∂θ ⎝ r⎠ _________________________________________________________________ 3.16 Do the previous problem for the following velocity and temperature fields:
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Lai et al, Introduction to Continuum Mechanics
v=
α ( − x2e1 + x1e2 ) 2 x1
+
2 x2
(
, Θ =k x12 + x22
)
-----------------------------------------------------------------------------------------2
2
2
Ans. With x1 = rcos θ , x2 = rsin θ and x1 + x2 = r , we have v=
α ( - x2 e1 + 1xe2 ) 2 1x+
2 2x
=
α r( − sin θ e1 + cos θ e2 ) α
=
2
r
r
eθ and Θ =kr 2
Particles move in concentric circles with their speed inversely proportional to r . Isotherms are circles. α (b) With vr = 0, vθ = , we have, from Eq.(3.4.12). r 2
v ∂v v v ∂vr ∂vr vθ ∂vr vθ 2 ∂v ∂v α 2 ⎛α ⎞ 1 + vr + − = − ⎜ ⎟ = − 3 , aθ = θ + vr θ + θ θ + r θ = 0 ar = ∂t ∂r ∂t ∂r r ∂θ r r ∂θ r ⎝ r ⎠ r r 2 3 2 i.e., a = −α / r er . At the point A, r = 2 , therefore, a = − 2α / 4er DΘ ∂Θ ⎛ α ⎞ = + v ⋅ ∇Θ = 0 + ⎜ eθ ⎟ ⋅ 2kr er = 0 . Dt ∂ t ⎝ r⎠
_________________________________________________________________ 1 3.17 Consider : x = X + X1k e1 . let dX( ) = dS1 / 2
(
)
( e1 +e2 )
2 & dX( ) = dS2 / 2
(
) ( −e1 +e2 )
be
differential material elements in the undeformed configuration. (a) Find the deformed elements (1) (2 ) dx and d x . (b) Evaluate the stretches of these elements ds1 / dS1 and ds2 / dS2 and the change in the angle between them. (c) Do part (b) for k = 1 and k = 10 −2 and (d) compare the results of part (c) to that predicted by the small strain tensor E . ------------------------------------------------------------------------------------------⎡1 + k 0 0 ⎤
⎢
Ans. (a) x1 = X1 + kX1 , x2 = X2 , x3 = X3 → [ F ] = 0 ⎢
⎢⎣ 0
⎡1 + k dS ⎛ ⎞ 1 ⎢ ⎡ dx ⎢⎣ ⎥⎦ = ⎜⎝ 2 ⎟⎠ ⎢ 0 ⎢⎣ 0 ⎡1 + k dS ⎛ ⎞ 2 ( ) 2 ⎢ ⎡ dx ⎤ = ⎜ ⎥⎦ ⎝ 2 ⎟⎠ ⎢ 0 ⎣⎢ ⎢⎣ 0 (1) ⎤
(b)
ds1 dS1
=
1
⎥
0 , dx = Fd X → ⎥
0 1 ⎥⎦
0 0 ⎤ ⎡1 ⎤ 1 0 0 1 0
⎥ ⎢1 ⎥ → dx (1) = ⎛ dS1 ⎞ ⎡ 1 + k e + e ⎤ . ) 1 2⎦ ⎜ ⎟ ⎣( ⎥⎢ ⎥ 2⎠ ⎝ 1 ⎥⎦ ⎢⎣0 ⎥⎦ 0 ⎤ ⎡ − 1⎤ ⎥⎢ ⎥ ( 2 ) ⎛ dS ⎞ 0 1 → dx = ⎜ 2 ⎟ ⎡⎣ − (1 + k ) e1 + e2 ⎤⎦ . ⎥⎢ ⎥ ⎝ 2⎠ 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ 0
⎛ 1 ⎞ 2 =⎜ (1 + k ) + 1 . ⎟ dS2 ⎝ 2 ⎠ ds2
Let γ be the decrease in angle (from 90o ), then ( π / 2 ) − γ is the angle between the two deformed differential elements. Thus,
Copyright 2010, Elsevier Inc 3-8
Lai et al, Introduction to Continuum Mechanics 2 1 2 − (1 + k ) + 1 1 ⎛ dS1 ⎞⎛ dS2 ⎞ ⎡ ⎛ π ⎞ dx( ) ⋅ d x( ) 2 ⎤ = → cos ⎜ − γ ⎟ = ⎜ ⎟⎜ ⎟ − (1 + k ) + 1⎦⎥ = 2 ds1ds2 ds1ds2 ⎝ 2 ⎠⎝ 2 ⎠ ⎣⎢ ⎝2 ⎠ 1 1 + + ( k ) 2
sin γ =
− (1 + k ) + 1 2
(1 + k ) + 1
(c) For k = 1 , For k = 10−2 ,
ds1 dS1 ds1 dS1
. ds2
=
dS2
2 (1 + k ) + 1
2
, sin γ = −
3 5
.
⎛ 1 ⎞ ⎛ 1 ⎞ 2 =⎜ (1 + k ) + 1 ≈ ⎜ ⎟ 2 + 2 k = 1 + k = 1.01 = 1.005 . ⎟ dS2 ⎝ 2 ⎠ ⎝ 2⎠
2
sin γ =
5
ds2
=
− (1 + k ) + 1
=
≈
−2k − k −0.01 = = → γ = −0.0099 radian ( − sign indicates increase in 2 + 2 k 1 + k 1.01
angle).
⎡k ⎢ (d) u = x − X = kX1e1 → u1 = kX1 , u2 = u3 = 0 , → [∇u ] = 0 ⎢ ⎢⎣ 0 ⎡ k 0 0 ⎤ ⎡1 ⎤ 1 1 1 ' e1′ = ( e1 + e2 ) → E 11 = [1 1 0] ⎢⎢0 0 0 ⎥⎥ ⎢⎢ 1⎥⎥ = [1 1 2 2 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣0 ⎥⎦ ′ = E 11
ds − dS
dS Also with
=
k
2
→
ds dS
= 1+
0 0⎤
⎡ k 0 0 ⎤ ⎥ ⎢ ⎥ 0 0 → [E ] = 0 0 0 , ⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎥⎦ ⎡ k ⎤ ⎢ ⎥ k 0] 0 = , ⎢ ⎥ 2 ⎢⎣ 0 ⎥⎦
k
2
= 1.005 , same as the result of part (c).
⎡ − k ⎤ k ⎥⎢ ⎥ ⎢ ⎥ ′ = E− 0 0 1 = [1 1 0 ] 0 = − → 2 12 e′2 = ( -e1 + e2 ) ⎥⎢ ⎥ 2 ⎢ ⎥ 2 2 ⎢⎣ 0 ⎥⎦ 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ Thus, the decrease in angle = −k , or the increase in angle is 0.01 ≈ 0.0099 . 1
⎡k 1 ′ = E [1 1 0 ] ⎢0 → 12 ⎢ 2 ⎢⎣ 0
0 0 ⎤ ⎡ −1⎤
1
_________________________________________________________________ 3.18 Consider the motion: x = X + AX , where A is a small constant tensor (i.e., whose components are small in magnitude and independent of X i ). Show that the infinitesimal strain tensor is given by E = ( A + A T ) / 2 . ---------------------------------------------------------------------------------------- Ans. u = x − X = AX → ∇u = ∇ ( AX ) . Since A is a constant, therefore,
∇u = ∇ ( AX ) = A ( ∇X ) . Now, [∇X] = ⎡⎣∂ X i / ∂X j ⎤⎦ = ⎡⎣δ ij ⎤⎦ = [I ] → ∇u = A → E = ( A + A T ) / 2 _______________________________________________________________________ 3.19 At time t , the position of a particle, initially at x1 = X1 + kX3 , x2 = X2 + kX2 , x3 = X3 ,
( X1 , X2 , X3 ) is defined by:
5 k = 10− . (a) Find the components of the strain tensor
and (b) find the unit elongation of an element initially in the direction of e1 + e2 .
Copyright 2010, Elsevier Inc 3-9
k
Lai et al, Introduction to Continuum Mechanics ---------------------------------------------------------------------------------------- Ans. (a) u1 = x1 − X 1 = kX 3 , u2 = x2 − X2 = kX2 , u3 = x3 − X3 = 0
⎡0 0 k ⎤ ⎡ 0 0 k / 2 ⎤ T u u ∇ + ∇ [ ] [ ] ⎥ k 0 → [∇u ] = ⎢⎢0 k 0 ⎥⎥ → [E ] = = ⎢⎢ 0 ⎥ 2 ⎢⎣0 0 0 ⎥⎦ ⎢⎣ k / 2 0 0 ⎥⎦ ⎡ 0 0 k / 2 ⎤ ⎡1 ⎤ 5 k 10− 1 1 ⎢ ⎥ ⎢ ⎥ (b) Let e1′ = ( e1 + e2 ) → E 11′ = e1′ ⋅ Ee1′ → E11′ = [1 1 0] ⎢ 0 k 0 ⎥ ⎢ 1⎥ = = 2 2 2 2 ⎢⎣ k / 2 0 0 ⎥⎦ ⎢⎣0 ⎥⎦ _________________________________________________________________ 3.20 Consider the displacements: u1 = k (2 X12 + X1 X 2 ), u2 = kX 22 , u3 = 0, k = 10 −4 . (a) Find the unit elongations and the change of angles for two material elements (1) (2 ) dX = dX1e1 and dX = dX 2 e 2 that emanate from a particle designated by X = e1 + e2 . (b) Sketch deformed positions of these two elements. ----------------------------------------------------------------------------------------⎡ 4kX1 + kX 2 kX1 0 ⎤ Ans. (a) [∇u ] =
⎢ ⎢ ⎢⎣
0
2kX 2
0
0
⎥ ⎥ 0 ⎥⎦
0 ,
⎡5k k 0 ⎤ ⎡ 5k k / 2 0⎤ At ( X1 , X2 , X3 ) = (1,1, 0) , [∇u ] = ⎢ 0 2 k 0 ⎥ → [E ] = ⎢ k/ 2 2 k 0 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ 1 2 Unit elong. for dX( ) = dX 1e1 is E11 = 5k = 5 × 10 −4 , unit elong. for dX( ) = dX 2 e2 is 4 E22 = 2 k = 2 × 10 − . 4 E= = k 10− radian Decrease in angle between them is 2 12 . 1 1 1 1 (b) For dX( ) = dX 1e1 , dx( ) = dX( ) + ( ∇u ) dX( ) = dX1e1 + 5kdX1e1 = (1 + 5 k ) dX1e1 ,
( 2) For dX = dX 2 e2 , ( 2) (2 ) (2) dx = dX + ( ∇u ) dX = dX 2e2 + ( kdX 2 e1 + 2 kdX 2 e2 ) = kdX 2 e1 + (1 + 2 k) dX2 e 2
The deformed positions of these two elements are shown below: k d X 2 2
X d ) k
2
+
1 (
P’
d X 2
k
=
2
P
u
(1 + 5 k) d X 1
d X
1 u 1= 3k
Copyright 2010, Elsevier Inc 3-10
Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 3.21 Given displacement field: u1 = kX1 , u2 = u3 = 0, k = 10−4 . Determine the increase in length for the diagonal element OA of the unit cube (see figure below) in the direction of e1 + e2 + e3 (a) by using the strain tensor and (b) by geometry. ----------------------------------------------------------------------------------------⎡ k 0 0⎤ 1 ⎢ ⎥ Ans. (a) [u ] = 0 0 0 = [E ] . Let e1′ = ( e1 + e2 + e3 ) , then the unit elongation in the e1' ⎢ ⎥ 3 ⎢⎣ 0 0 0 ⎥⎦
⎡ k 0 0⎤ ⎡1⎤ 4 k 10− ′ = e1′ ⋅ Ee1′ = [1 1 1] ⎢ 0 0 0 ⎥ ⎢1⎥ = = direction is E 11 . ⎢ ⎥⎢ ⎥ 3 3 3 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣1⎥⎦ 1
(b) From the given displacement field, we see that the unit cube becomes longer in the x1 direction by an amount of k , while the other two sides remain the same. The diagonal OA
becomes ' Figure), where OA = 3 and OA , (see 2
2
2
OA ' = (1 + k ) + 1 + 1 = 3 + 2k + k = 3(1 + 2k / 3 + k / 3)
→ OA '− OA = 3(1 + 2 k / 3 + k 2 / 3)1/2 − 3 .
(
Using binomial theorem, 1 + 2k / 3 + k 2 / 3
1/2
)
= 1 + (1 / 2)(2 k / 3) + ... ≈ 1 + k / 3
Thus, OA '− OA = 3(1 + k / 3) − 3 = 3 k / 3 → ( OA' − OA) / OA = k / 3 , same as that obtained in part (a). _________________________________________________________________ 3.22 With reference to a rectangular Cartesian coordinate system, the state of strain at a point is ⎡5 3 0 ⎤
⎢ ⎥ 4 −1 × 10 −4 . (a) What is the unit elongation in the direction of ⎢ ⎥ ⎢⎣0 −1 2 ⎥⎦
given by the matrix [E ] = 3
2e1 + 2e2 + e3 ? (b) What is the change in angle between two perpendicular lines (in the undeformed state) emanating from the point and in the directions of 2e1 + 2e2 + e3 and 3e1 − 6e3 ? ---------------------------------------------------------------------------------------- Ans. Let e1′ = (2e1 + 2e2 + e3 ) / 3 , the unit elongation in this direction is:
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Lai et al, Introduction to Continuum Mechanics
⎡ 5 3 0 ⎤ ⎡ 2⎤ 58 ′ = e1′ ⋅ Ee1′ = [ 2 2 1] ⎢3 4 −1⎥ ⎢2 ⎥ ×10 −4 = ×10 −4 . E 11 ⎢ ⎥⎢ ⎥ 9 9 ⎢⎣0 −1 2 ⎥⎦ ⎢⎣ 1⎥⎦ 1
1
Let e′2 =
( 3e1 − 6e3 ) , then the decrease in angle between the two elements is:
45
⎡5 3 0 ⎤ ⎡ 3 ⎤ 32 ′ = 2e1′ ⋅ Ee′2 = 2 12 ×10 −4 [2 E2 1] ⎢⎢3 4 −1⎥⎥ ⎢⎢ 0 ⎥⎥ ×10 −4 = 3 45 45 ⎢⎣0 −1 2 ⎥⎦ ⎢⎣ −6⎥⎦ 2
rad
.
_________________________________________________________________ 3.23 For the strain tensor given in the previous problem, (a) find the unit elongation in the direction of 3e1 − 4e2 and (b) find the change in angle between two elements in the dir. of 3e1 − 4e3 and 4e1 + 3e3 . ----------------------------------------------------------------------------------------1 Ans. (a) Let e1′ = ( 3e1 − 4e2 ) , the unit elongation in this direction is: 5 ⎡5 3 0 ⎤ ⎡ 3 ⎤ 2 1⎞ 37 ⎛ ′ = e1′ ⋅ Ee1′ = ⎜ ⎟ [3 −4 0] ⎢3 4 −1⎥ ⎢ −4 ⎥ ×10 −4 = ×10 −4 = 1.48 ×10 −4 E 11 ⎢ ⎥⎢ ⎥ 25 ⎝5⎠ ⎢⎣0 −1 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ (b) Let e1'' =
1 5
( 3e1 − 4e3 )
and e''2 =
1 5
( 4e1 + 3e3 ) , then the decrease in angle between these two
elements is: 2
''' 12
⎡ 5 3 0 ⎤ ⎡ 4⎤ 2 1 72 ⎛ ⎞ = 2e1'' ⋅ Ee''2 = 2 ⎜ ⎟ [3 0 E−4 ] ⎢⎢3 4 −1⎥⎥ ⎢⎢0 ⎥⎥ ×10 −4 = ×10 −4 = 2.88 ×10 −4 25 ⎝5⎠ ⎢⎣0 −1 2 ⎥⎦ ⎢⎣ 3⎥⎦
.
_________________________________________________________________ 3.24 (a) Determine the principal scalar invariants for the strain tensor given below at the left and (b) show that the matrix given below at the right can not represent the same state of strain.
⎡5 3 0 ⎤ [E] = ⎢⎢3 4 −1⎥⎥ ×10 −4 , ⎢⎣0 −1 2 ⎥⎦
⎡3 0 0 ⎤ ⎢0 6 0 ⎥ ×10 −4 ⎢ ⎥ ⎢⎣0 0 2 ⎥⎦
----------------------------------------------------------------------------------------−4 −4 Ans. (a) I 1 = ( 5 + 4 + 2 ) × 10 = 11× 10 ,
I 2 =
5 3 3 4
× 10−8 +
4
−1
−1
2
×10 −8 +
5
3
0
I 3 = 3
4
−1 × 10 −12 = 17 ×10 −12
−1
2
0
5
0
0
2
×10 −8 = 28 ×10 −8
Copyright 2010, Elsevier Inc 3-12
ra
Lai et al, Introduction to Continuum Mechanics
⎡3 0 0 ⎤ ⎢ ⎥ (b) For 0 6 0 ×10 −4 , I 3 = 36 ×10 −12 , which is different from the I 3 in (a), therefore, the ⎢ ⎥ ⎢⎣0 0 2 ⎥⎦ two matrices can not represent the same tensor. _________________________________________________________________ 3.25 Calculate the principal scalar invariants for the following two tensors. What can you say about the results?
⎡0 τ 0 ⎤ ⎡ 0 −τ 0⎤ ⎡T(1) ⎤ = ⎢τ 0 0⎥ and ⎡ T( 2 ) ⎤ = ⎢ −τ 0 0⎥ . ⎥ ⎥ ⎢⎣ ⎥⎦ ⎢ ⎢⎣ ⎥⎦ ⎢ ⎢⎣0 0 0 ⎥⎦ ⎢⎣ 0 0 0⎥⎦ ----------------------------------------------------------------------------------------⎡0 τ 0⎤ 1 ⎢ ⎥ , I = 0, I = −τ 2 , I = 0 . ( ) Ans. For ⎡ T ⎤ = τ 0 0 1 2 3
⎥ ⎦⎥ ⎢ ⎢⎣0 0 0 ⎥⎦{e } i
⎣⎢
⎡ 0 −τ 0⎤ ⎢ ⎥ (2) 2 For ⎡ T ⎤ = −τ 0 0 1I = 0, 2I = −τ , 3I = 0 ⎥ ⎢⎣ ⎥⎦ ⎢ ⎢⎣ 0 0 0 ⎥⎦{e } i We see that these two tensors have the same principal scalar invariants. This result demonstrates that two different tensors can have the same three principal scalar invariants and therefore the same eigenvalues (in fact, λ1 = τ , λ2 = −τ , λ3 = 0 ). However, corresponding to the same (1) (2) eigenvalue τ , the eigenvector for T is (e1 + e2 ) / 2 , whereas the eigenvector for T is (e1 − e2 ) / 2 . We see from this example that having the same principal scalar invariants is a necessary but not sufficient condition for the two tensors to be the same. _________________________________________________________________
(
)
3.26 For the displacement field: u1 = kX12 , u2 = kX 2 X 3 , u3 = k 2 X1 X3 + X12 , k = 10−6 , find the maximum unit elongation for an element that is initially at (1, 0, 0) . ----------------------------------------------------------------------------------------⎡ 2kX 1 0 0 ⎤
⎢
Ans. [∇u ] = ⎢
0
kX 3
⎥
kX 2 ⎥ , thus, for ( X1 , X2 , X3 ) = (1, 0, 0) , 2 kX1 ⎥⎦
⎢ k ( 2 X 3 + 2 X1 ) 0 ⎣ ⎡ 2k 0 0 ⎤ ⎡ 2 k 0 k ⎤ ⎢ ⎥ [∇u ] = ⎢ 0 0 0 ⎥ → [E ] = ⎢⎢ 0 0 0 ⎥⎥ , the characteristic equation for this tensor is: ⎢⎣ 2k 0 2 k ⎥⎦ ⎢⎣ k 0 2 k ⎥⎦ 2k − λ
0
k
0
0−λ
0
k
0
2 k − λ
2 = 0 → ( −λ ) ⎡⎢( 2 k − λ ) − k 2 ⎤⎥ = 0 → λ1 = 0, λ2 = 3k, λ3 = k. ⎣ ⎦
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Lai et al, Introduction to Continuum Mechanics Thus, the maximum unit elongation at (1, 0, 0 ) is λ 2 = 3k = 3 × 10 −6 . _________________________________________________________________ 3.27 Given the matrix of an infinitesimal strain tensor as
⎡ k1 X 2 [E] = ⎢⎢ 0 ⎢⎣ 0
0
⎤ ⎥. 0 ⎥ −k2 X 2 ⎥⎦ 0
− k2 X 2 0
(a) Find the location of the particle that does not undergo any volume change. (b) What should the relation between k1 and k 2 be so that no element changes its volume? ----------------------------------------------------------------------------------------Δ ( dV ) = E11 + E22 + E33 = ( k1 − 2 k2 ) X2 = Ans. (a) 0 . Thus, the particles which were on the plane dV X 2 = 0 do not suffer any change of volume. (b) If ( k1 − 2k2 ) = 0, ie., k1 = 2 k2 , then no element changes its volume. _________________________________________________________________ 3.28 The displacement components for a body are: 2
2
u1 = k ( X1 + X 2 ), u2 = k (4 X3 − X1 ), u3 = 0,
k =10 − . 4
(a) Find the strain tensor. (b) Find the change of length per unit length for an element which was at (1, 2,1) and in the direction of e1 + e2 . (c) What is the maximum unit elongation at the same point (1, 2,1) ? (d) What is the change of volume for the unit cube with a corner at the origin and with three of its edges along the positive coordinate axes? ----------------------------------------------------------------------------------------0 ⎤ 0 ⎤ ⎡ 2kX1 k ⎡2 kX1 0 Ans. (a) [∇u ] =
⎢ −k ⎢ ⎢⎣ 0
⎥ ⎥ 0 ⎥⎦
0 8kX 3 → [E ] = 0
⎢ 0 ⎢ ⎢⎣ 0
0 4 kX 3
⎥ ⎥ 0 ⎥⎦
4 kX 3
⎡ 2k 0 0 ⎤ ⎢ ⎥ (b) At (1, 2,1) , [ E] = 0 0 4k , ⎢ ⎥ ⎢⎣ 0 4k 0 ⎥⎦ ⎡ 2k 0 0 ⎤ ⎡1 ⎤ ⎢ ⎥ ⎢k1 ⎥ = for e1' = 0 4 E= e1' ⋅ Ee1' = [1 1 0] 0 k ( e1 + e2 ) , ⎢ ⎥⎢ ⎥ 2 2 ⎢⎣ 0 4k 0 ⎥⎦ ⎢⎣0 ⎥⎦ 2k − λ 0 0 2 −λ 4k = 0 → ( 2 k − λ ) ⎡⎢λ 2 − ( 4 k ) ⎤⎥ = 0 (c) The characteristic equation is 0 ⎣ ⎦ 0 4k −λ 1
1
' 11
→ λ1 = 2k , λ2 = 4k , λ 3 = −4 k . The maximum elongation is 4k . (d) Change of volume per unit volume = iiE = 2 kX 1 , which is a function of X 1 . Thus, 1
1
ΔV = ∫ 2kX1dV = 2∫ kX1 (1) dX1 = kX12 = k . o
o
_________________________________________________________________
Copyright 2010, Elsevier Inc 3-14
Lai et al, Introduction to Continuum Mechanics 3.29 For any motion, the mass of a particle (material volume) remains a constant (conservation of mass principle). Consider the mass to be the product of its volume and its mass density and show that (a) for infinitesimal deformation ρ (1 + E kk ) = ρ o where ρ o denote the initial density and ρ , the current density. (b) Use the smallness of E kk to show that the current density is given by ρ = ρ o (1 − E kk ) . ---------------------------------------------------------------------------------------- Ans. (a) ρo dVo = ρdV → ρo =
For small deformation,
ΔdV dV o
ρ dV dVo
−1
dVo
⎞ ⎟, ⎠
= E kk → ρo = ρ (1 + E kk ) .
(b) From bionomial theorem, for small ρ= oρ(1 + Ekk )
⎛ ΔdV = ρ⎜1 + dVo ⎝
dVo + ΔdV
= ρ
E kk , (1+ E kk )
−1
≈ 1− E kk , thus,
E kk ) . = ρ o (1 −
_________________________________________________________________ 3.30 True or false: At any point in a body, there always exist two mutually perpendicular material elements which do not suffer any change of angle in an arbitrary small deformation of the body. Give reason(s). ---------------------------------------------------------------------------------------- Ans. True. The strain tensor E is a real symmetric tensor, for which there always exists three principal directions, with respect to which, the matrix of E is diagonal. That is, the non-diagonal elements, which give one-half of the change of angle between the elements which were along the principal directions, are zero. _________________________________________________________________ 3.31 Given the following strain components at a point in a continuum: E11 = E12 = E22 = k,
E33 = 3 k,
E13 = E23 = 0,
k = 10 −
6
Does there exist a material element at the point which decreases in length under the deformation? Explain your answer. ---------------------------------------------------------------------------------------- Ans. k − λ k 0 ⎡k k 0 ⎤
[E] = ⎢⎢ k
⎥ ⎥ ⎢⎣ 0 0 3k ⎥⎦ k
(
0 →
k
k−λ
0
0
0
3k − λ
)
2 = 0, → ( 3 k − λ ) ⎢⎡( k − λ ) − k 2 ⎥⎤ = 0 ⎣ ⎦
→ ( 3k − λ ) −2λ k + λ 2 = 0 → λ1 = 3k, λ2 = 0, λ3 = 2 k.
Thus, the minimum unit elongation is 0 . Therefore, there does not exist any element at the point which has a negative unit elongation (i.e., decreases in length). _________________________________________________________________ 3.32 The unit elongation at a certain point on the surface of a body are measured experimentally by means of strain gages that are arranged 45o apart (called the 45o strain rosette) in the direction 1 of e1 , ( e1 + e2 ) and e2 . If these unit elongation are designated by a, b, c respectively, what are 2 the strain components E11 , E22 and E12 ?
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Lai et al, Introduction to Continuum Mechanics ---------------------------------------------------------------------------------------- Ans.
With e1′ =
1 2
( e1 + e2 ) , we have,
⎡ E11 ′ = e1′ ⋅ Ee1′ = [1 1 0 ] ⎢ E21 E11 ⎢ 2 ⎢⎣ E31
E12
E13 ⎤ ⎡1 ⎤
E22
E23
⎥ ⎢1 ⎥ = 1 E + E + E + E , with E = E , 21 22 ) 12 21 ⎥ ⎢ ⎥ 2 ( 11 12 E32 E33 ⎥⎦ ⎢⎣ 0 ⎥⎦ ( E + E 22 ) 1 ′ = ( E11 + 2 E12 + E22 ) → E12 = E11 ′ − 11 . Thus, the strain components are: E11 1
2
2
E11 = a, E22 = c, E12 = b−
( a + c)
. 2 _________________________________________________________________ 3.33 (a) Do the previous problem, if the measured strains are 200 × 10 −6 , 50 × 10 −6 and 100 × 10 −6 in the direction e1 , e1′ and e2 respectively. (b) Find the principal directions, assuming E31 = E32 = E33 = 0 . (c) How will the result of part b be altered if E 33 ≠ 0 .
----------------------------------------------------------------------------------------6 ′ = 50 × 10−6 and E 22 = 100 × 10−6 , we have, from the results Ans. (a) With E 11 = 200 × 10− , E 11
′ − of the previous problem, E12 = E 11 E11 − λ
E 12
E 12
E 22 − λ
0
0
(b)
E11 + E 22
2
200 + 100 ⎞ ⎛ = ⎜ 50 − × 10−6 = −100 × 10 −6 ⎟ 2 ⎝ ⎠
0 2 ⎤ 0 = 0 → λ ⎡( E 11 − λ )( E 22 − λ ) − E 12 = 0
⎣
−λ
⎦
2 ⎤ → λ ⎡ +λ 2 − λ ( E11 + E22 ) + E11 E22 − E12 =0, ⎣ ⎦
(
λ1,2 =
( E11 +
E22 ) ±
( E11 − 2
)
2
2 E22 ) + 4 E12 , λ 3 = 0 ,
thus, λ1,2
2 2 ⎡ + ± − + − 200 100 200 100 4 100 ( ) ( ) ( ) =⎢ ⎢ 2 ⎢⎣
⎤ −6 ⎥ × 10−6 = 261.8 ×10 , λ = 0 3 ⎥ 38.2 × 10 −6 ⎥⎦
The principal direction for λ 3 is e3 . The principal directions corresponding to the other two eigenvalues lie on the plane of e1 and e2 . Let n = α1e1 + α 2 e2 ≡ cos θ e1 + sin θ e2 , then ( E11 − λ )α1 + E 12α2 = 0 ,
Copyright 2010, Elsevier Inc 3-16
Lai et al, Introduction to Continuum Mechanics
→
α 2 α 1
= tan θ =
( λ − E11 ) E 12
,
For λ1 = 261.8 × 10 −6 , tanθ =
λ 1 − E 11 E 12
261.8 − 200
=
−100
=
61.8
−100
= −0.618 → θ = −31.7 o ,
Or, n = 0.851e1 − 0.525e2 For λ2 = 38.2 × 10 −6 , tanθ =
λ 2 − E 11 E 12
=
38.2 − 200
−100
= 1.618 → θ = 58.3o
Or, n = 0.525e1 + 0.851e2 . (c) If E 33 ≠ 0 , then the principal strain corresponding to the direction e3 is E 33 instead of zero. Nothing else changes. _________________________________________________________________
′ = E22 = 1000 × 10 −6 . 3.34 Repeat the previous problem with E11 = E11 ---------------------------------------------------------------------------------------- E + E 22 ⎛ 2000 ⎞ ′ − 11 = ⎜ 1000 − × 10 −6 = 0 , Ans. (a) From the results of Problem 3.32, E12 = E 11 ⎟ 2 2 ⎠ ⎝
⎡10−3 0 ⎢ (b) and (c) [E ] = ⎢ 0 10−3 ⎢ 0 ⎢⎣ 0
0 ⎤
⎥
0 ⎥ , the principal strains are 10−3 in any directions lying on the
⎥ ⎦
E 33 ⎥
plane of e1 and e2 and the principal strain E 33 is in e3 direction. _________________________________________________________________ 3.35 The unit elongation at a certain point on the surface of a body are measured experimentally by means of strain gages that are arranged 60o apart (called the 60o strain rosette) in the direction 1 1 −e1 + 3e2 . If these unit elongation are designated by a, b, c of e1 , e1 + 3e2 and 2 2 respectively, what are the strain components E11 , E22 and E12 ?
(
)
(
)
---------------------------------------------------------------------------------------- Ans.
With e1' = (e1 + 3e2 ) / 2,
e1'' = (− e1 + 3e2 ) / 2 , we have,
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Lai et al, Introduction to Continuum Mechanics
'
E11 = e1' ⋅ Ee1' =
"' E11
=
" " e1 ⋅ Ee1
=
⎡ E11 ⎢ 3 0 ⎤ E21 ⎦⎢ ⎢⎣ E31
1
⎡1 4⎣
1
⎡ −1
4⎣
⎡ E11 ⎢ 3 0 ⎤ E21 ⎦⎢ ⎢⎣ E31
E12 E22 E32 E12 E22 E32
E13 ⎤ ⎡ 1 ⎤
⎥⎢ ⎥ 1 E23 ⎢ 3 ⎥ = ⎥ 4 E33 ⎥⎦ ⎢ 0 ⎥
( E11 + 2
3 E 12 + 3 E 22
)
⎦ E13 ⎤ ⎡ −1 ⎤ ⎥⎢ ⎥ 1 E −2 3 E +3 E E23 ⎢ 3 ⎥ = 11 12 22 ⎥ 4 ⎢ ⎥ ⎥ E33 ⎦ 0 ⎣ ⎦
(i)
⎣
(
)
(ii)
' '' − E 11 E11 1 b−c ' ''' ⎡ ⎤ = (i) & (ii), → E22 = 2 E11 + 2 E11 − E11 = [ 2 b+ 2 c− a] , E 12 = , E11 = a . ⎦ 3 3⎣ 3 3
1
_________________________________________________________________ 3.36 If the 60o strain rosette measurements give a = 2 × 10−6 , b = 1 × 10 −6 , c = 1.5 × 10 −6 , obtain E11 , E12 and E22 . Use the formulas obtained in the previous problem. ---------------------------------------------------------------------------------------- Ans. Using the formulas drived in the previous problem, we have, 1 1 −6 −6 E22 = [ 2 b+ 2 c− a] = ⎡⎣( 2 )(1) + ( 2 )(1.5) − 2 ⎤⎦ × 10 = 1× 10 , 3 3 1 b−c =− × 10−6 , E 11 = 2 × 10−6 . E 12 = 3 2 3 _________________________________________________________________ 3.37 Repeat the previous problem for the case a = b= c = 2000 ×10 −6 . ----------------------------------------------------------------------------------------1 1 6 3 Ans. E22 = [ 2 b+ 2 c− a] = ⎡⎣( 2 )( 2000 ) + ( 2 )( 2000 ) − 2000 ⎤⎦ ×10 − = 2 × 10 − , 3 3 b−c = 0 , E 11 = 2 × 10−3 E 12 = 3 _______________________________________________________________________ 3.38 For the velocity field: v = kx22e1 , (a) find the rate of deformation and spin tensors. (b) Find the rate of extension of a material element dx = dsn where n = ( e1 + e 2 ) / 2 at x = 5e1 + 3e2 . ----------------------------------------------------------------------------------------2 Ans. v1 = kx2 , v2 = v3 = 0 ,
⎡ 0 2kx2 0 ⎤ ⎡ 0 S ⎥ ⎢ 0 0 → [ D] = [∇ v ] = kx2 → [∇v ] = ⎢⎢ 0 ⎥ ⎢ ⎢⎣ 0 ⎢⎣ 0 0 0 ⎥⎦ (b) At the position x = 5e1 + 3e2 , ⎡ 0 3k 0 ⎤ [ D] = ⎢⎢3k 0 0 ⎥⎥ , ⎢⎣ 0 0 0 ⎥⎦
kx2
0 0
0⎤
⎥ ⎥ 0 ⎥⎦
0 ,
[W ] = [∇ v ]
⎡ 0 3k 0 ⎤ [ W ] = ⎢⎢−3k 0 0 ⎥⎥ ⎢⎣ 0 0 0 ⎥⎦
Copyright 2010, Elsevier Inc 3-18
A
⎡ 0 = ⎢⎢− kx2 ⎢⎣ 0
kx2
0 0
0⎤
⎥ ⎥ 0 ⎥⎦ 0
Lai et al, Introduction to Continuum Mechanics For the element dx = dsn with n = (e1 + e 2 ) / 2 , the rate of extension is:
⎡0 ⎢ ( n)D (n ) = n ⋅ Dn = 2 [1 1 0 ] ⎢3 ⎢⎣ 0
3k 0 ⎤ ⎡1 ⎤
1
k0
0
⎥ ⎢ 1⎥ = 3 . k ⎥⎢ ⎥ 0⎥⎦ ⎢⎣0 ⎥⎦
0
_________________________________________________________________
⎛ t + k ⎞ ⎟ e1 , find the rates of extension for the following material ⎝ 1 + x1 ⎠
3.39 For the velocity field: v = α ⎜ elements: dx
(1)
= ds1e1 and dx( 2 ) = ds2 / 2 ( e1 + e2 ) at the origin at time t = 1 .
(
)
-----------------------------------------------------------------------------------------
⎡ −α ( t + k ) / (1 + x ) 2 0 0 ⎤ 1 ⎢ ⎥ ⎛ t + k ⎞ 0 0 0 ⎥ = [ D] . Ans. v1 = α ⎜ ⎟ , v2 = v3 = 0 → [∇v ] = ⎢ ⎝ 1 + x1 ⎠ ⎢ 0 0 0⎥ ⎢⎣ ⎥⎦ ⎡ −α (1 + k ) 0 0 ⎤ ⎢ ⎥ At t = 1 and at ( x1 , x2 , x3 ) = ( 0,0, 0 ) , [ D] = ⎢ 0 0 0⎥ . ⎢ 0 0 0 ⎥⎦ ⎣ 1 ( 2) Rate of extension for dx( ) = ds1e1 is D11 = −α (1 + k ) ; for dx = ds2 /
(
2
) ( e1 + e2 ) , it is:
⎡ −α (1 + k ) 0 0 ⎤ ⎡1 ⎤ 1 ⎢ ⎥⎢ ⎥ ' 0 0 0 ⎥ 1 = − α (1 + k ) D11 = [1 1 0 ] ⎢ ⎢ ⎥ 2 2 ⎢ ⎥ ⎢⎣0 ⎥⎦ 0 0 0 ⎣ ⎦ 1
_________________________________________________________________ 3.40 For the velocity field v = ( cos t )( sin π x1 ) e 2 (a) find the rate of deformation and spin tensors, and (b) find the rate of extension at t = 0 for the following elements at the origin: (1) (2 ) (3 ) dx = ds1e1 , dx = ds2 e2 and dx = ds3 / 2 ( e1 + e2 ) .
(
)
---------------------------------------------------------------------------------------- Ans. (a) With v1 = 0, v2 = ( cos t )( sin π x1 ) , v3 = 0 ,
⎡ 0 (π cos t cos π x1 ) / 2 0 ⎤ ⎢ ⎥ ⎥ 0 → [ D] = ⎢(π cos t cos π x1 ) / 2 0 0⎥ , ⎥ ⎢ 0 ⎥⎦ 0 0 0 ⎥⎦ ⎣ − (π cos t cos π x1 ) / 2 0 ⎤ ⎥ 0 0⎥ . 0 0 ⎥⎦ π / 2 0 ⎤ ⎡ 0 ⎢ ⎥ (b) At t = 0 and ( x1 , x2 , x3 ) = ( 0,0, 0) , [ D] = π / 2 0 0 . ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎡ [∇v ] = ⎢⎢π cos t cos π x1 0 ⎢⎣ 0 0 ⎡ 0 ⎢ [ W ] = ⎢(π cos t cos π x1 ) / 2 ⎢ 0 ⎣
0⎤
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Lai et al, Introduction to Continuum Mechanics
For dx for dx
(1)
( 3)
2 = ds1e1 , rate of extension is D11 =0, for dx( ) = ds2e2 , D22 = 0 and
(
)
= ds3 / 2 ( e1 + e2 )
' , D11
⎡ 0 π / 2 0 ⎤ ⎡1 ⎤ ⎥ ⎢ ⎥ π = [1 1 0 ] ⎢⎢π / 2 0 0 1 = ⎥⎢ ⎥ 2 2 ⎢⎣ 0 0 0⎥⎦ ⎢⎣0 ⎥⎦ 1
_________________________________________________________________ 3.41 Show that the following velocity components correspond to a rigid body motion. v1 = x2 − x3 , v2 = − x1 + x3 , v3 = x1 − x2
---------------------------------------------------------------------------------------⎡ 0 1 −1⎤ ⎡0 0 0 ⎤
⎢ ⎢ ⎢⎣ 1
Ans. [∇v ] = −1
0
−1
⎥ ⎥ 0 ⎥⎦
⎢ ⎥ ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦
1 → [D] = 0 0 0
Therefore, the velocity field is a rigid body motion.. _________________________________________________________________ 3.42 Given the velocity field v =
1
er , (a) find the rate of deformation tensor and the spin tensor r and (b) find the rate of extension of a radial material line element.
----------------------------------------------------------------------------------------1 Ans. With vr = , vθ = vz = 0 , we have, using Eq. (2.34.5) r
[
⎡ ∂vr ⎢ ∂r ⎢ ⎢ ∂v v] = ⎢ θ ⎢ ∂r ⎢ ∂v ⎢ ⎣⎢ ∂r
⎞ ∂vr ⎤ ⎡ 1 v − θ ⎟ ⎥ − ⎜ r ⎝ ∂θ ⎠ ∂z ⎥ ⎢ r 2 ⎢ 1 ⎛ ∂vθ ⎞ ∂vθ ⎥ ⎢ + vr ⎟ ⎥= 0 ⎜ r ⎝ ∂θ ⎠ ∂z ⎥ ⎢⎢ ∂v ⎥z ⎢ 0 1 ∂v z ⎥ ∂z ⎦⎥ ⎢⎣ r ∂θ
1 ⎛ ∂vr
z
0 1 2
r 0
⎤
0⎥
⎥ ⎥ 0 ⎥ = [ D] , ⎥ 0⎥ ⎥⎦
(b) The rate of extension for a radial element is Drr = −
[ W ] = [0 ] .
1
. 2 r _________________________________________________________________ 3.43 Given the two-dimensional velocity field in polar coordinates: vr = 0, vθ = 2 r +
4
r (a) Find the acceleration at r = 2 and (b) find the rate of deformation tensor at r = 2 . ----------------------------------------------------------------------------------------2
2
( vθ ) ∂v ∂v v ⎛ ∂v 1⎛ 4⎞ ⎞ = − ⎜ 2r + ⎟ , Ans. (a) Using Eq. (3.4.12), ar = r + vr r + θ ⎜ r − vθ ⎟ = − ∂t ∂r r ⎝ ∂θ r r⎝ r⎠ ⎠ ∂v ∂v v ⎛ ∂v ⎞ 2 aθ = θ + vr θ + θ ⎜ θ + vr ⎟ = 0 . At r = 2 , ar = −(6) / 2 = −18 , aθ = 0 . ∂t ∂r r ⎝ ∂θ ⎠
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Lai et al, Introduction to Continuum Mechanics
(b) Eq. (2.34.5) → [
⎡ ∂vr ⎢ ∂r v] = ⎢ ⎢ ∂vθ ⎢ ⎣ ∂r
⎡ 0 [ D] = [∇v ] = ⎢ ⎢⎣ −4 / r 2 S
⎞⎤ − vθ ⎟ ⎥ ⎡ 0 ⎜ r ⎝ ∂θ ⎠⎥ ⎢ =⎢ 1 ⎛ ∂vθ ⎞ ⎥ ⎢ ∂vθ + vr ⎟ ⎥ ⎢ ⎜ r ⎝ ∂θ ⎠ ⎦ ⎣ ∂r
1 ⎛ ∂vr
−
⎡ 4 ⎞⎤ ⎛ − ⎜ 2 + 2 ⎟⎥ 0 ⎢ r ⎠ ⎥ ⎝ r ⎥ ⎢ . ⎥= ⎢ ⎥ 4 ⎛ ⎞ 0 ⎥ ⎢⎜ 2 − 0 ⎥ ⎥⎦ ⎝ 2 ⎟ r ⎠ ⎣ ⎦ vθ ⎤
−4 / r 2 ⎤ ⎡ 0 −1⎤ ⎥ , at r = 2 , [ D] = ⎢ ⎥. 0 ⎥⎦ ⎣ −1 0 ⎦
_________________________________________________________________ 3.44 Given the velocity field in spherical coordinates: B ⎞ ⎛ vr = 0, vθ = 0, vφ = ⎜ Ar + 2 ⎟ sin θ r ⎠ ⎝ (a) Determine the acceration field and (b) find the rate of deformation tensor. ---------------------------------------------------------------------------------------- Ans. (a) From Eq. (3.4.16), ar = aθ = aφ =
∂vr ∂t ∂vθ ∂t ∂vφ
+ vr + vr + vr
∂vr ∂r ∂vθ ∂r ∂vφ
+
vθ r vθ
+
r
+
vθ
r ∂t ∂r (b) Eq. (2.35.25) →
[
⎡ ∂vr ⎢ ⎢ ∂r ⎢ ∂v v] = ⎢ θ ⎢ ∂r ⎢ ⎢ ∂vφ ⎢ ⎣⎢ ∂r
2
2
vφ ⎛ ∂vr vφ 1⎛ B ⎞ ⎞ ⎛ ∂vr ⎞ v v sin θ − + − = − = − ⎜ Ar + 2 ⎟ sin 2 θ ⎜ θ ⎟ φ ⎜ ⎟ r r⎝ r ⎠ ⎝ ∂θ ⎠ r sin θ ⎝ ∂φ ⎠ 2
vφ ⎛ ∂vθ vφ cosθ sin θ ⎛ B ⎞ ⎞ ⎛ ∂vθ ⎞ v v cos θ cot θ = − + + − = − ⎜ Ar + 2 ⎟ φ ⎜ r ⎟ ⎜ ⎟ r r ⎝ ⎝ ∂θ ⎠ r sin θ ⎝ ∂φ r ⎠ ⎠ vφ ⎛ ∂vφ ∂vφ ⎞ + + vr sin θ + vθ cosθ ⎟ = 0 ⎜ ∂θ r sin θ ⎝ ∂φ ⎠
⎤ ⎛ 1 ∂vr vφ ⎞ − ⎥ ⎡ 0 ⎜ ⎟ 0 ⎝ r sin θ ∂φ r ⎠ ⎥ ⎢ ⎥ ⎢ ⎛ 1 ∂vθ vφ cot θ ⎞ ⎥ ⎢ ⎛ 1 ∂vθ vr ⎞ − 0 ⎜ ⎟ ⎥ =⎢ 0 ⎜ r ∂θ + r ⎟ sin θ φ ∂ r r ⎝ ⎠ ⎝ ⎠ ⎥ ⎢ ∂ ∂ ⎛ 1 ∂vφ vr vθ cot θ ⎞ ⎥ ⎢ vφ , 1 vφ 1 ∂vφ + + ⎜ ⎟ ⎥ ⎢⎣ ∂r r ∂θ sin θ φ ∂ r ∂θ r r r ⎝ ⎠ ⎦⎥ ⎛ 1 ∂vr vθ ⎞ ⎜ r ∂θ − r ⎟ ⎝ ⎠
1 ⎛ vφ cot θ
⎜− 2⎝
r
+
vφ
1 ∂vφ ⎞
B⎞ ⎛ B ⎞⎤ 1⎡ ⎛ ⎟ = ⎢− ⎜ + A3 ⎟ + ⎜ + A3 ⎟ ⎥ cos θ = 0 . r ∂θ ⎠ 2 ⎣ ⎝ r ⎠ ⎝ r ⎠ ⎦
_________________________________________________________________ 3.45 A motion is said to be irrotational if the spin tensor vanishes. Show that the following velocity field is irrotational: v=
− x2e2 + x1e2 2
, r 2 = x12 + x22
r -----------------------------------------------------------------------------------------
Copyright 2010, Elsevier Inc 3-21
⎤ ⎥ r ⎥ vφ cot θ ⎥ − ⎥ , thus r ⎥ ⎥ 0 ⎥ ⎦ −
the nonzero components of rate of deformation tensor are: 1 ⎛ vφ ∂vφ ⎞ 3B + Dr φ = ⎜ − ⎟ = − 3 sin θ , ∂r ⎠ 2⎝ r 2r
= θφ D
2
Lai et al, Introduction to Continuum Mechanics 1 2 x ∂r ⎤ ⎡ ∂v1 ∂v1 ⎤ ⎡ 2 x2 ∂r − 2 + 32 ⎢ ∂ x ∂ x⎥ ⎢ ⎥ 3 r ∂ 1x r r ∂ 2x⎥ x2 x1 1 2⎥ 2 2 2 ⎢ ⎢ = , Ans. v1 = − 2 , v2 = 2 , r = x1 + x2 , → [∇v ] = ⎢ ∂v2 ∂v2 ⎥ ⎢ 1 2 x1 ∂r 2 x1 ∂r ⎥ r r − 3 ⎢∂ ⎥ ⎢ 2− 3 ∂ ⎥ r r ∂ 2x ⎦ 1x ⎣ 1x ∂ 2x⎦ ⎣ r ∂r ∂r x1 ∂r x2 2 2 2 = 2 x1 → = , also, = , r = x1 + x2 → 2 r ∂ x1 ∂ x1 ∂ x2 r r
⎡ 2 x1 x2 ⎢ 4 r ⎢ ∇ v = [ ] ⎢ 2 2 x − x ⎢ 2 4 1 ⎣ r
2 2 x2 − x1 ⎤
⎥ ⎥ = [∇v ]S → [W ] = 0. 2xx ⎥ − 14 2 ⎥ r ⎦ 4
r
_________________________________________________________________ 1 2 3.46 Let dx( ) = ds1n and dx ( ) = ds2 m be two material elements that emanate from a particle (1) (2 ) P which at present has a rate of deformation D . (a) Consdier ( D/ Dt) ( dx ⋅ dx ) to show that
⎡ 1 D( ds1 ) 1 D( ds2 ) ⎤ Dθ + = 2m ⋅ Dn ⎢ ⎥ cos θ − sin θ ds Dt ds Dt Dt 2 ⎣ 1 ⎦ where θ is the angle between m and n . 1 2 (b) Consider the case of dx( ) = d x ( ) , what does the above formula reduce to? π (1) ( 2) (c) Consider the case where θ = , i.e., d x and d x are perpendicular to each other, where 2 does the above formula reduces to? ---------------------------------------------------------------------------------------- Ans. (2) ⎛ x Dd⎞ D(1) D ⎛ (2 ) (1) ⎞ ( 2 ) ⎜ (1 ) ⎟ = (∇v ) dx(1) dx (2 ) + dx (1 ) (∇ v ) dx(2 ) = ⎜ dx ⎟ dx + dx dx ⋅ dx ⎜ ⎟ Dt Dt Dt ⎝ ⎠
)
(
⎝
⎠
{
}
T T 1 2 1 2 1 2 1 2 = dx( ) ( ∇v ) dx( ) + dx( ) (∇ v ) dx( ) = dx( ) (∇ v ) + ( ∇ v) dx( ) = 2 dx( ) ⋅ Ddx( ) .
With dx D
(1)
2 = ds1n and dx( ) = ds2m , the above formula give,
( ds1ds2n ⋅ m ) = 2ds1ds2 ( n ⋅ Dm ) → Dt
1
Dds
( ds2 cosθ ) + Dt
2
Dds
( ds1 cos θ ) + Dt
D
( ds1ds2 cos θ ) = 2 ds1ds2 ( n ⋅ Dm) . Thus,
D cosθ
Dt
( ds1ds2 ) = 2 ds1 ds2 ( n ⋅ Dm ) , Dt
⎧⎪ 1 D( ds1 ) 1 D( ds2 ) ⎫⎪ Dθ →⎨ + = 2 ( n ⋅ Dm ) = 2 ( m ⋅ Dn ) . ⎬ cos θ − sin θ ds2 Dt ⎪ Dt ⎪⎩ ds1 Dt ⎭ ⎧⎪ 1 D( ds) ⎫⎪ 1 2 (b) For , dx( ) = dx( ) = dsn the above formula → ⎨ ⎬ = ( n ⋅ Dn ) = D( n) (n ) , no sum on n . ⎪⎩ ds Dt ⎪⎭ 1 2 (c) For d x( ) perpendicular to d x( ) , θ = 90o , we have, Dθ − = 2 ( n ⋅ Dm ) = 2 Dnm . Dt
Copyright 2010, Elsevier Inc 3-22
Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 3.47 Let e1 , e2 ,e3 and D1 , D2 , D3 be the principal directions and corresponding principal values of (1) (2 ) (3 ) a rate of deformation tensor D . Further, let dx = ds1e1 , dx = ds2 e2 and dx = ds3 e3 be three material elements. Consider the material derivative that
1 D ( dV ) dV
Dt
( D/
{
Dt) dx
(1)
dx
(2 )
}
(3 ) dx and show
= D1 + D2 + D3 , where dV = ds1ds2 ds3 .
---------------------------------------------------------------------------------------- Ans. Since the principal directions are (or can always be chosen to be) mutually perpendicular, 1 2 3 therefore, dx( ) ⋅ dx ( ) × dx ( ) = ds1 ds2 ds3 = dV .
→ →
D ( dV)
=
1
Dt D( dV)
dV
Dt
(D =
)
1ds 2ds 3ds
D ( ds 1)
= ds2 ds3
1
Dt D( ds1 )
ds1
Dt
+
D ( ds 2)
+ ds1ds3
Dt
1
D( ds2 )
ds2
Dt
+
(D 3ds)
, Dt
Dt
1 D ( ds3 ) ds3
+ ds1ds2
Dt
= D11 + D22 + D33 .
_________________________________________________________________ 3.48 Consider a material element dx = dsn (a) Show that
( D/
Dt ) n = Dn + Wn − ( n ⋅ Dn ) n ,
where D is rate of deformation tensor and W is the spin tensor. (b) Show that if n is an eigenvector of D , then, Dn
= Wn = n Dt ----------------------------------------------------------------------------------------1 Dds n D D Dds ⎛ n D ⎞ ⎛ nD ⎞ = ds ⎜ +n = ds ⎜ + (n Dn )n ⎟ . [see Ans. (a) ( dsn ) = ds + n ⎟ Dt Dt Dt ds Dt⎠ ⎝ Dt ⎝ Dt ⎠ Eq.(3.13.12) ]. We also have,
⎛ Dn ⎞ + n ( n Dn ) ⎟ → Dt ⎝ ⎠
( ∇v ) n = ⎜
D
( dsn ) =
Dt Dn
D
( dx ) = ( ∇v ) dx = ds ( ∇ v) n , Dt
therefore,
= ( ∇v ) n − n ( n Dn ) = ( D + W ) n − n ( n Dn ) .
Dt (b) If n is an eigenvector of D , then Dn = λ n , therefore, Dn Dn = ( D + W ) n − n ( n ⋅ Dn ) = λ n + Wn − nλ = Wn . That is, = Wn . Dt Dt Since W is antisymmetric → Wn = is the dual vector for W . Thus n , where Dn = Wn = n. Dt That is, the principal axes of D rotates with an angular velocity given by the dual vector of the spin tensor. _________________________________________________________________ 2
3.49 Given the following velocity field: v1 = k ( x2 − 2 ) x3 , v2 = − x1 x2 , v3 = kx1 x3 for an incompressible fluid, determine the value of k , such that the equation of mass conservation is satisfied. -----------------------------------------------------------------------------------------
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Lai et al, Introduction to Continuum Mechanics
Ans.
∂v1 ∂v2 ∂v3 + + = 0 → 0 − 1x+ kx 1 = 0 → k= 1 ∂ x1 ∂ x2 ∂ x3
_________________________________________________________________ 3.50 Given the velocity field in cylindrical coordinates: vr = f ( r, θ ), vθ = vz = 0 . For an incompressible material, from the conservation of mass principle, obtain the most general form of the function f ( r , θ ) . ---------------------------------------------------------------------------------------- Ans. The equation of continuity for an incompressible material is [see Eq.(3.15.11)]: ∂vr 1 ∂vθ vr ∂vz ∂f f 1 ∂ + + + = 0→ + = 0→ ( fr) = 0,→ fr= g(θ ) . ∂r r ∂θ r ∂z ∂r r r ∂r Therefore, f = g(θ ) / r, where g (θ ) is an arbitrary function of θ . _________________________________________________________________ 3.51 An incompressible fluid undergoes a two-dimensional motion with vr = k cos θ /
r . From
the consideration of the principle of conservation of mass, find vθ , subject to the condition that vθ = 0 at θ = 0 .
---------------------------------------------------------------------------------------- Ans. ∂v ∂vr vr ⎛ 1 ⎞ (k cos θ ) v k cos θ (k cos θ ) ⎛ 1⎞ 1 vr = → r = ( k cosθ ) ⎜ − ⎟ 3/2 , r = → + = ⎜ ⎟ 3/2 . ∂r ∂r r ⎝ 2 ⎠ r r r r 3/2 ⎝ 2 ⎠ r The equation of continuity for an incompressible fluid is [see ∂v 1 ∂vθ vr ∂vz + + = 0 . Thus, Eq.(3.15.11)]: r + ∂r r ∂θ r ∂z ∂vθ ⎛ k ⎞ cos θ ⎛ k ⎞ sin θ = −⎜ ⎟ → vθ = − ⎜ ⎟ + f ( r ) . Since vθ = 0 at θ = 0 , Therefore, ∂θ ⎝2⎠ r ⎝ 2 ⎠ r
⎛ k ⎞ sin θ . ⎟ ⎝ 2 ⎠ r
f ( r ) = 0 . Thus, vθ = − ⎜
_________________________________________________________________ 3.52 Are the following two velocity fields isochoric (i.e., no change of volume)? (i) v =
x1e1 + x2 e2 2
, r 2 = x12 + x22 and (ii) v =
− x2e1 + x1e2 2
, r 2 = x12 + x22
r r ----------------------------------------------------------------------------------------2 2 2 2 2 Ans. (i) With v1 = x1 / r , v2 = x2 / r , r = x1 + x2 , 2 ⎛ ⎞ ∂v1 1 2 x1 ∂r ∂r ∂r 1 2x = 2− 3 = 2 − 41 . ⎜ r 2 = x12 + x22 → 2 r = 2 x1 , 2 r = 2 x2 ⎟ . ∂1xr ∂1x ∂2x r ∂1 x r r ⎝ ⎠
∂v2 1 2 x2 ∂r 1 2x22 = − = − , ∂ 2x r 2 r 3 ∂ 2x r 2 r 4
∂v1 ∂v2 2 2 x12 2 x22 2 2 + = − − 4 = 2 − 2 = 0. ∂ 1x ∂ 2x r 2 r 4 r r r
(ii) v1 = − x2 / r 2 , v2 = x1 / r 2 , r 2 = x12 + x22
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⎞ ∂v1 2x2 ∂r 2 x2 x1 ⎛ 2 ∂r 2 2 = 3 = 2 x1 ⎟ = 4 ⎜ r = x1 + x2 → 2 r ∂ 1 x r ∂ 1 x r ∂1 x ⎠ ⎝ 2 x ∂r 2x x ∂v2 ∂v1 ∂v2 2 x2 x1 2 x1 x2 = − 31 = − 14 2 , + = 4 − 4 = 0. ∂ 2x ∂ 1x ∂ 2x r ∂ 2x r r r _________________________________________________________________ 3.53 Given that an incompressible and inhomogeneous fluid has a density field given by ρ = kx2 . From the consideration of the principle of conservation of mass, find the permissible form of velocity field for a two dimensional flow ( v3 = 0 ) . ---------------------------------------------------------------------------------------- Ans. Since the fluid is incompressible, therefore, D ρ ∂ρ ∂ρ ∂ρ =0→ + v1 + v2 = 0 → 0 + v1 ( 0 ) + v2 k = 0 → v2 = 0. Dt ∂ t ∂ 1x ∂ 2x The conservation of mass equation of an incompressible fluid in two dimensional flow is ∂v1 ∂v2 ∂v + = 0 → 1 = 0 → v1 = f ( x2 ) , v2 = 0. ∂ 1x ∂ 2x ∂ 1x _________________________________________________________________ α x1
3.54 Consider the velocity field: v =
e1 . From the consideration of the principle of 1 + kt conservation of mass, (a) Find the density if it depends only on time t , i.e., ρ = ρ (t ) , with
ˆ ( x1 ) , with ρˆ ( xo ) = ρ * . ρ ( 0 ) = ρ o . (b) Find the density if it depends only on x1 , i.e., ρ = ρ ---------------------------------------------------------------------------------------- Ans.(a) Equation of conservation of mass is α x1 ∂ρ ∂ρ ∂ρ ∂ ρ ⎛ ∂v1 ∂v2 ∂v3 ⎞ + v1 + v2 + v3 + ρ ⎜ + + , v2 = v3 = 0 , ⎟ = 0 . With v1 = 1 + kt ∂t ∂x1 ∂x2 ∂x3 ⎝ ∂ x1 ∂ x2 ∂ x3 ⎠
→
d ρ dt
+ ρ
α 1 + kt
ρ
=0→
∫
d ρ
t
= −α ∫
1 + kt 0
ρ
ρ o
(b) with ρ = ρ ( x1 ) and v1 =
dt
→ ln
ρ o
=− ρ
α ln (1 + kt ) → k
ρ
−α / k
= (1 + kt )
o
ρ
.
α x1
, v2 = v3 = 0 1 + kt
⎛ ∂v ∂v ∂v ⎞ α x1 d ρ ∂ρ ∂ρ ∂ρ ∂ρ α + v1 + v2 + v3 + ρ ⎜ 1 + 2 + 3 ⎟ = 0 → + ρ =0 ∂t ∂x1 ∂x2 ∂x3 1 + kt dx1 1 + kt ⎝ ∂x1 ∂ x2 ∂ x3 ⎠ → x1
d dx1
ρ
+
ρ
= ρ0, →
∫
ρ *
d
ρ
ρ
x1
= − ∫ xo
dx1 x1
→ ln
ρ ρ *
= − ln
x1 xo
→
x
= ρ * ρo x1
where ρ o is the density at x1 = xo . _________________________________________________________________ 3.55 Given the velocity field: v = α ( x1 te1 + x2 te 2 ) . From the consideration of the principle of conservation of mass, determine how the fluid density varies with time, if in a spatial description, it is a function of time only. -----------------------------------------------------------------------------------------
Copyright 2010, Elsevier Inc 3-25
Lai et al, Introduction to Continuum Mechanics Ans. Equation of conservation of mass is
∂ρ ∂ρ ∂ρ ∂ ρ ⎛ ∂v1 ∂v2 ∂v3 ⎞ + v1 + v2 + v3 + ρ ⎜ + + ⎟ = 0 . With v1 = α x1t, v2 = α x2 t, v3 = 0 , ∂t ∂x1 ∂x2 ∂x3 ∂ ∂ ∂ x x x 2 3 ⎠ ⎝ 1 d dt
ρ + ρ( αt + αt ) = 0 →
ρ
∫
ρ o
ρ
d
ρ
t
= −2 α∫ tdt → ln 0
ρ
ρ o
2
= − αt 2 → ρ = ρo e−α t .
_________________________________________________________________
∂Wim ∂Eik ∂E km ∂u 1 ⎛ ∂ui = − + m , where E im = ⎜ ∂ Xk ∂ Xm ∂ Xi 2 ⎝ ∂ X m ∂X i ∂u ⎞ 1 ⎛ ∂ui − m ⎟ is the rotation tensor. W im = ⎜ 2 ⎝ ∂ X m ∂X i ⎠
3.56 Show that
⎞ ⎟ is the strain tensor and ⎠
----------------------------------------------------------------------------------------- Ans.
∂Wim ∂ 1 ⎛ ∂ui ∂um = − ⎜ ∂ Xk ∂ Xk 2 ⎝ ∂ Xm ∂ Xi
⎞ ⎟= ⎠
1⎛
∂ 2 ui ∂ 2 um − ⎜ 2 ⎜⎝ ∂ Xm ∂ Xk ∂ Xi∂ Xk
⎞ ⎟⎟ = ⎠
1⎛
∂ 2ui ∂2 uk ∂2 u k ∂2 um ⎞ + − − ⎜ ⎟= 2 ⎜⎝ ∂ Xm ∂ Xk ∂ Xm ∂ Xi ∂ Xm ∂ Xi ∂ Xi ∂ Xk ⎟⎠
1 ⎛ ∂ ⎛ ∂ui
∂u ⎞ ∂ + k ⎟− ⎜⎜ ⎜ 2 ⎝ ∂ Xm ⎝ ∂ Xk ∂ Xi ⎠ ∂ Xi
⎛ ∂uk ∂um + ⎜ ⎝ ∂ Xm ∂ Xk
⎞ ⎞ ∂Eik ∂Ekm − ⎟ ⎟⎟ = ∂ xi ⎠ ⎠ ∂ xm
_________________________________________________________________ 3.57 Check whether or not the following distribution of the state of strain satisfies the compatibility conditions: X1 ⎡ X1 + X2 , X 2 + X3 [E] = k ⎢⎢ X1 ⎢⎣ X2 X3
X2 ⎤
⎥ , k = 10 4 − ⎥ X1 + X3 ⎥⎦ X3
---------------------------------------------------------------------------------------- Ans. Yes. We note that the given E ij are linear in X1 , X2 and X3 and the terms in the compatibility conditions all involve second derivatives with respect to X i , therefore these conditions are obviously satisfied by the given strain components. _________________________________________________________________ 3.58 Check whether or not the following distribution of the state of strain satisfies the compatibility conditions:
⎡ X12 ⎢ E = k [ ] ⎢ X 22 + X 32 ⎢ ⎢⎣ X1 X3
2 2 X2 + X3
X1 X3 ⎤
⎥
0
X1 ⎥ , k = 10 −4
X1
X2 ⎥
2
⎥ ⎦
---------------------------------------------------------------------------------------- Ans.
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Lai et al, Introduction to Continuum Mechanics
∂2 E ∂ 2 E22 ∂ 2 12 E 11 + = → 0 + 0 = 0,OK 2 ∂ X1 X 2 ∂ X 22 ∂X 12
∂ 2 E 23 ∂ 2 E 22 ∂ 2 E33 + =2 → 0 + 2k ≠ 0, not satisfied ∂ X 2 X 3 ∂ X 32 ∂X 22 The given strain components are not compatible. _________________________________________________________________ 3 3.59 Does the displacement field: u1 = sin X1 , u2 = X1 X 2 , u3 = cos X3 correspond to a
compatible strain field? ---------------------------------------------------------------------------------------- Ans. Yes. The displacement field obviously exists. In fact, the displacement field is given. There is no need to check the compatibility conditions. Whenever a displacement field is given, there is never any problem of compatibility of strain components. _________________________________________________________________ 3.60 Given the strain field: E12 = E21 = kX1 X2 , k = 10−4 and all other E ij = 0 . (a) Check the equations of compatibility for this strain field and (b) by attempting to integrate the strain field, show that there does not exist a continuous displacement field for this strain field. -----------------------------------------------------------------------------------------
∂2 E ∂2 E ∂ 2 12 E 11 22 + =2 → 0 + 0 ≠ 2k . This compatibility condition is not satisfied. Ans. (a) 2 2 ∂ X1 X 2 ∂ X 2 ∂X 1 ∂u ∂u2 = 0 → u2 = u2 ( X1 , X3 ) . (b) E11 = 0 → 1 = 0 → u1 = u1 ( X2 , X3 ) . Also, E22 = 0 → ∂ X1 ∂X 2 ∂u ( X , X ) ∂u ( X , X ) ∂u1 ∂u2 + → 2 kX1 X2 = 1 2 3 + 2 1 3 = f( X2 , X3 ) + g( X1 , X3 ) , Now, 2 E12 = ∂ X ∂ 1X ∂ X ∂ 1X 2 2 That is, 2kX1 X 2 = f ( X 2 , X 3 ) + g ( X1 , X3 ) . Clearly, there is no way this equation can be satisfied, because the right side can not have terms of the form of X1 X 2 . _________________________________________________________________ 3.61 Given the following strain components: E11 =
1
f ( X 2 , X3 ) , E22 = E33 = −
ν
f ( X2 , X3 ) , E12 = E13 = E23 = 0 . α α Show that for the strains to be compatible, f ( X2 , X3 ) must be linear in X 2 and X 3 .
---------------------------------------------------------------------------------------- Ans 2 ∂ 2 11E ∂ 2 22 ∂ 2 12E E 1 ∂ f( X2 , X3 ) + =2 → = 0, 2 2 ∂ α X X ∂ X22 ∂ X ∂ X 1 2 1 2
2 ∂ 2 E 13 ∂ 2 11E ∂ 2 E 33 1 ∂ f( X2 , X3 ) + =2 → =0 2 2 ∂ α X X ∂ X32 ∂ X ∂ X 1 3 1 3
, 2 ∂ 2 E11 ∂ ⎛ ∂ E23 ∂E 31 ∂E 12 ⎞ 1 ∂ f ( X2 , X3 ) = + + = 0 , Thus, ⎜− ⎟→ ∂ X2 ∂ X3 ∂ X1 ⎝ ∂ X1 ∂ X2 ∂ X3 ⎠ α ∂ X2 ∂ X3
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∂ 2 f ( X2 , X3 ) ∂ 2 f ( X2 , X3 ) ∂ 2 f ( X2 , X3 ) = 0, = 0, = 0 . → f ( X2 , X3 ) is a linear function of ∂ X 2 ∂X 3 ∂ X 22 ∂X 32 X 2 and X 3 . We note also
∂ 2 E 23 ∂ 2 E 22 ∂ 2 E33 ν ⎛ ∂ 2 f ∂ 2 f ⎞ + =− ⎜ 2 + , ⎟= 0= 2 ∂ X1 X 3 α ⎜⎝ ∂ X3 ∂ X22 ⎟⎠ ∂ X32 ∂ X22 ∂ 2 E22 ∂ ⎛ ∂E31 ∂E 12 ∂E 23 ⎞ ν ∂ 2 f =− =0= + + ⎜− ⎟, ∂ X3∂ X1 ∂ X2 ⎝ ∂ X2 ∂ X3 ∂ X1 ⎠ α ∂ X3∂ X1 ⎞ ∂ 2 33 E ν ∂ 2 f ∂ ⎛ ∂ E 12 ∂ 23 E ∂ 31 E =− =0= + + ⎜− ⎟. α ∂ X1∂ X2 ∂ X1∂ X2 ∂ X3 ⎝ ∂ X3 ∂ X1 ∂ X2 ⎠
Thus, if f ( X2 , X3 ) is a linear function of X 2 and X 3 , then all compatibility equations are satisfied. _________________________________________________________________ 3.62 In cylindrical coordinates ( r ,θ , z ) , consider a differential volume bounded by the three pairs of faces: r = r and r = r + dr; θ =θ and θ =θ + dθ ; z = z and z = z + dz. The rate at which mass is flowing into the volume across the face r = r is given by ρ vr ( rdθ )( dz ) and similar expressions for the other faces. By demanding that the net rate of inflow of mass must be equal to the rate of increase of mass inside the differential volume, obtain the equation of conservation of mass in cylindrical coordinates. Check your answer with Eq. (3.15.7 ). ---------------------------------------------------------------------------------------- Ans. Mass flux across the face r = r into the differential volume dV is ( ρ vr )( rdθ ) dz . That across the face r = r + dr out of the volume is ( ρ vr ) r =r +d r ( r + dr ) dθ dz . Thus , the net mass flux into dV through the pair of faces r = r and r = r + dr is
(
ρvr )r =r ( rd θ) dz − ( ρvr ) r =r + dr ( r + dr) d θdz = ⎡( ρvr ) r =r − ( ρvr ) r =r +dr ⎤ rd θdz
⎣
⎦
− ( ρ vr )r =r +d r drdθ dz . ⎡ ∂ ( ρ vr ) ⎤ θ ) and ⎥ dr ( rd dz ⎣ ∂r ⎦
Now, ⎡⎣( vρr )r =r − ( vρr ) r =r + dr ⎤⎦ rd dz θ =−⎢
− ( ρvr )r =r +d r drd θdz = − ⎡⎣( ρvr ) + d ( ρvr ) ⎤⎦ drd θdz = − ( ρvr ) drd θdz , where we have dropped the higher order term involving ⎡⎣ d ( ρ vr ) ⎤⎦ drdθ dz which approaches zero in the limit compared to the terms involving only three differentials. Thus, the net mass flux into dV through the pair of faces r = r and r = r + dr is
⎧ ⎛ ∂ ρ vr ⎞ ⎫ − ρ v ( ) ⎨−r ⎜ r ⎬ drdθ dz . Similarly, ⎟ ⎩ ⎝ ∂r ⎠ ⎭ the net mass flux into dV through the pair of faces θ = θ and θ = θ + d θ is ⎛ ∂ ρ vθ ⎞ −⎜ ⎟ dθ ( drdz ) , ⎝ ∂θ ⎠ and the net mass flux into dV through the pair of faces z= zand z = z+ dzis
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Lai et al, Introduction to Continuum Mechanics
⎛ ∂ ρ v z −⎜ ⎝ ∂ z
⎞ ⎟ dz ⎡⎣dr (rd θ ) ⎤⎦ ⎠
Thus, the total influx of mass through these three pairs of faces is:
⎪⎧⎛ ∂ ρ vr − ⎨⎜ ⎪⎩⎝ ∂r
⎞ ( ρ vr ) 1 ⎛ ∂ρ vθ ⎞ ⎛ ∂ρ vz ⎟ + r + r ⎜ ∂θ ⎟ + ⎜ ∂z ⎠ ⎝ ⎠ ⎝
⎞ ⎪⎫ ⎟ ⎬ dr ( rdθ ) dz ⎠ ⎪⎭
On the other hand, the rate of increase of mass inside dV is
∂ ∂ ρ ( ρrd θdrdz ) = rd θdrdz . ∂t ∂t
Therefore, the conservation of mass principle gives,
⎧⎪⎛ ∂ ρ vr ⎞ ( ρ vr ) 1 ⎛ ∂ρ vθ ⎞ ⎛ ∂ρ vz ⎞ ⎫⎪ ∂ρ − ⎨⎜ + + ⎜ +⎜ rdθ drdz , That is: ⎬ dr ( rdθ ) dz = ⎟ ⎟ ⎟ ∂ r r ⎝ ∂θ ⎠ ⎝ ∂z ⎠ ⎪ t ⎪⎩⎝ ∂r ⎠ ⎭ ∂ ρ ∂ vρr vρ 1 ⎛ ∂ ρ vθ ⎞ ⎛ ∂ vρz ⎞ + + r+ ⎜ ⎟ = 0 , Or, ⎟+ ⎜ ∂t ∂r r r ⎝ ∂θ ⎠ ⎝ ∂z ⎠ ⎧∂ ρ ∂ ρ vθ ⎛ ∂ ρ⎞ ∂ ρ⎫ ⎛ ∂vr vr 1 ∂vθ ∂vz ⎞ + ⎜ ⎟ + vz + + + = 0 . This is the same as ⎨ + vr ⎬ + ρ ⎜ ∂r r ⎝ ∂θ ⎠ ∂z ⎭ ∂ z ⎟⎠ r r ∂θ ⎝ ∂r ⎩ ∂t Eq.(3.15.7). _________________________________________________________________ 3.63 Given the following deformation in rectangular Cartesian coordinates: x1 = 3 X3 , x2 = − X1 , x3 = − 2 X2
Determine (a) the deformation gradient F , (b) the right Cauchy-Green tensor C and the right stretch tensor U , (c) the left Cauchy-Green tensor B , (d) the rotation tensor R , (e) the Lagrangean strain tensor E* (f) the Euler strain tensor e* , (g) ratio of deformed volume to initial volume, (h) the deformed area (magnitude and its normal) for the area whose normal was in the direction of e 2 and whose magnitude was unity for the undeformed area. ----------------------------------------------------------------------------------------⎡ 0 0 3⎤ ⎡ 0 −1 0 ⎤ ⎡ 0 0 3 ⎤
⎢ ⎥ ⎢ 0 0 , (b) [C] = [F ] [F ] = 0 0 ⎢ ⎥ ⎢ ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣ 3 0 ⎡1 0 0 ⎤ 1/2 [ U] = [C ] = ⎢⎢0 2 0 ⎥⎥ . (The only positive definite root). ⎢⎣0 0 3 ⎥⎦ ⎡ 0 0 3⎤ ⎡0 −1 0 ⎤ ⎡9 0 0 ⎤ T ⎢ ⎥⎢ ⎥ ⎢ ⎥ (c) [ B] = [F ][F ] = −1 0 0 0 0 −2 = 0 1 0 . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣3 0 0 ⎥⎦ ⎢⎣0 0 4 ⎥⎦
Ans. (a) [F ] = −1
(d) [ R ] = [F ][U]
−1
T
⎡1 0 0 ⎤ −2 ⎥⎥ ⎢⎢−1 0 0 ⎥⎥ = ⎢⎢0 4 0 ⎥⎥ , 0 ⎥⎦ ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣0 0 9 ⎥⎦
0 ⎤ ⎡0 0 1⎤ ⎡ 0 0 3 ⎤ ⎡1 0 ⎢ ⎥ ⎢ ⎥ ⎢ = ⎢−1 0 0 ⎥ ⎢0 1 / 2 0 ⎥ = ⎢−1 0 0 ⎥⎥ . ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣0 0 1 / 3⎥⎦ ⎢⎣ 0 −1 0 ⎥⎦
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⎡0 0 0 ⎤ ⎡0 0 0 ⎤ ⎡4 / 9 0 0 ⎤ 1 ⎥ ⎢ ⎥ ⎢ ⎥ − 1 (e) ⎡ E ⎤ = [C - I ] = 0 3 0 = 0 3 / 2 0 , (f) ⎡e* ⎤ = ⎡I − B ⎤ = 0 0 0 . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 ⎣ ⎦ ⎣ ⎦ 2 2 ⎢⎣0 0 8 ⎥⎦ ⎢⎣0 0 4 ⎥⎦ ⎢⎣ 0 0 3 / 8 ⎥⎦ ΔV (g) = det B = ( 9 ) (1) ( 4) = 6 , ΔV o ⎡ 0 −6 0 ⎤ T 1 1 − ⎢ 0 0 −3⎥ , n = e , (h) dA = dAo ( det F ) F −1 no , dAo = 1, det F = 6, [ F ] = ⎥ o 2 6⎢ ⎢⎣ 2 0 0 ⎥⎦ 1⎢
1
*
( )
⎡ [ dA ] = dAo ⎢( det F ) F −1 ⎣
( )
T
⎡ 0 0 2 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ 1⎢ ⎤ no ⎥ = (1 )( 6 ) −6 0 0 ⎥⎥ ⎢⎢1 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dA = −3e3 . ⎢ 6 ⎦ ⎢⎣ 0 −3 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −3⎥⎦
_________________________________________________________________ 3.64 Do the previous problem for the following deformation: x1 = 2 X2 , x2 = 3 X3 , x3 = X1 .
----------------------------------------------------------------------------------------⎡0 2 0 ⎤ ⎡ 0 0 1 ⎤ ⎡ 0 2 0 ⎤ ⎡1 0 0 ⎤
⎢ ⎢ ⎢⎣1 ⎡1 = ⎢⎢0 ⎢⎣0
Ans. (a) [F ] = 0
1/2
[ U ] = [C ]
T
(c) [ B] = [F ][F ]
(d) [ R ] = [F ][U]
T ⎥ ⎢ ⎥ ⎢ 0 0 3 ⎥ = ⎢0 4 0 ⎥ . ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0⎦ ⎣ 0 3 0 ⎦ ⎣ 1 0 0 ⎦ ⎣0 0 9 ⎥⎦ 0 0⎤ ⎥ 2 0 . (The only positive definite root). ⎥ 0 3 ⎥⎦ ⎡ 0 2 0 ⎤ ⎡ 0 0 1 ⎤ ⎡4 0 0 ⎤ = ⎢⎢ 0 0 3 ⎥⎥ ⎢⎢ 2 0 0 ⎥⎥ = ⎢⎢0 9 0 ⎥⎥ . ⎢⎣1 0 0 ⎥⎦ ⎢⎣ 0 3 0 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
0
−1
3 . (b) [C] = [F ] [F ] = 2 0 0
0 ⎤ ⎡0 1 0 ⎤ ⎡0 2 0 ⎤ ⎡1 0 = ⎢⎢0 0 3⎥⎥ ⎢⎢0 1 / 2 0 ⎥⎥ = ⎢⎢0 0 1 ⎥⎥ . ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 0 1 / 3 ⎥⎦ ⎢⎣1 0 0 ⎥⎦
⎡ 0 0 0 ⎤ ⎡0 0 0 ⎤ ⎡3 / 8 0 0 ⎤ 1 ⎥ ⎢ ⎥ ⎢ ⎥ − 1 (e) ⎡ E ⎤ = [C − I ] = 0 3 0 = 0 3 / 2 0 , (f) ⎡e* ⎤ = ⎡ I − B ⎤ = 0 4/9 0 . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 ⎣ ⎦ 2⎣ ⎦ 2 ⎢⎣ 0 0 8 ⎥⎦ ⎢⎣0 0 4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ΔV (g) = det B = ( 4 ) (9) (1) = 6 . ΔV o *
1
1⎢
(h) dA = dAo ( det F ) F −1
( )
T
−1
no , dAo = 1, det F = 6, [ F ]
⎡0 0 6⎤ = ⎢ 3 0 0 ⎥⎥ , no = e2 6 ⎢⎣ 0 2 0 ⎥⎦ 1⎢
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Lai et al, Introduction to Continuum Mechanics
⎡ [ dA ] = dAo ⎢( det F ) F −1 ⎣
( )
T
⎡ 0 3 0 ⎤ ⎡0 ⎤ ⎡ 3 ⎤ 1⎢ ⎤ ⎥⎢ ⎥ ⎢ ⎥ no ⎥ = (1 )( 6 ) 0 0 2 1 = 0 → dA = 3e1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 6 ⎦ ⎢⎣ 6 0 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦
_________________________________________________________________ 3.65 Do Prob. 3.63 for the following deformation: x1 = X1 , x2 = 3 X3 , x3 = − 2 X2
----------------------------------------------------------------------------------------⎡1 0 0 ⎤
⎢ 0 ⎢ ⎢⎣0 −2 ⎡1 T ⎢ (b) [C] = [F ] [F ] = 0 ⎢ ⎢⎣0
Ans. (a) [F ] = 0
⎥ ⎥ 0 ⎥⎦
3 .
0 0 ⎤ ⎡1 0 ⎥ ⎢ ⎥ ⎢ 0 −2 0 0 3 = 0 4 ⎥⎢ ⎥ ⎢ 3 0 ⎥⎦ ⎢⎣0 −2 0 ⎥⎦ ⎢⎣0 0 ⎡1 0 0 ⎤ 1/2 ⎢ ⎥ is [ U ] = [C ] = 0 2 0 . ⎢ ⎥ ⎢⎣0 0 3 ⎥⎦ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 0 T ⎢ ⎥⎢ ⎥ ⎢ (c) [ B] = [F ][F ] = 0 0 3 0 0 −2 = 0 9 ⎢ ⎥⎢ ⎥ ⎢ ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣0 3 0 ⎥⎦ ⎢⎣0 0 0 ⎤ ⎡1 ⎡ 1 0 0 ⎤ ⎡1 0 −1 ⎢ ⎥ ⎢ ⎥ ⎢ (d) [ R ] = [F ][U] = 0 0 3 0 1 / 2 0 = 0 ⎢ ⎥⎢ ⎥ ⎢ ⎢⎣0 −2 0 ⎥⎦ ⎢⎣0 0 1 / 3 ⎥⎦ ⎢⎣0 0
0 ⎤ ⎡1
0⎤
⎥ ⎥ 9 ⎥⎦
0 , The only positive definite root
0⎤
⎥ ⎥ 4 ⎥⎦
0 . 0
0⎤
0
1 ,
⎥ ⎥ −1 0 ⎥⎦
0 ⎤ ⎡0 0 0 ⎤ ⎡0 0 1 ⎢ ⎥ ⎢ ⎥ (e) ⎡ E* ⎤ = [C − I ] = 0 3 / 2 0 , (f) ⎡e* ⎤ = ⎡I − B−1 ⎤ = 0 4 / 9 0 . ⎢ ⎥ ⎥ ⎣ ⎦ 2 ⎣ ⎦ 2⎣ ⎦ ⎢ ⎢⎣0 0 3 / 8 ⎥⎦ ⎢⎣0 0 4 ⎥⎦ 1
(g)
ΔV = det B = (1) (9) ( 4 ) = 6 . ΔV o
( )
(h) dA = dAo ( det F ) F
⎡ ⎣
[ dA ] = dAo ⎢( det F ) ( F −1 )
⎡6 0
−1
no , dAo = 1, det F = 6, [ F ]
1⎢
0⎤
= ⎢ 0 0 −3⎥⎥ , no = e2 6 ⎢⎣ 0 2 0 ⎥⎦ ⎡ 6 0 0 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ T 1 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ no ⎥ = (1 )( 6 ) 0 0 2 1 = 0 → dA = −3e3 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 6 ⎦ ⎢⎣ 0 −3 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −3⎥⎦
−1 T
_________________________________________________________________ 3.66 Do Prob. 3.63 for the following deformation:
Copyright 2010, Elsevier Inc 3-31
Lai et al, Introduction to Continuum Mechanics x1 = 2 X2 , x2 = − X1 , x3 = 3 X3
----------------------------------------------------------------------------------------⎡ 0 2 0⎤ ⎡ 0 −1 0 ⎤ ⎡ 0 2 0 ⎤ ⎡1
0 0⎤
T ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 . (b) [C] = [F ] [F ] = 2 0 0 −1 0 0 = 0 4 0 , ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 3 ⎥⎦ ⎢⎣ 0 0 3⎥⎦ ⎢⎣ 0 0 3 ⎥⎦ ⎢⎣0 0 9 ⎥⎦ ⎡1 0 0 ⎤ 1/2 [ U] = [C ] = ⎢⎢0 2 0 ⎥⎥ . (The only positive definite root). ⎢⎣0 0 3 ⎥⎦ ⎡ 0 2 0 ⎤ ⎡0 −1 0 ⎤ ⎡4 0 0 ⎤ T ⎢ ⎥⎢ ⎥ ⎢ ⎥ (c) [ B] = [F ][F ] = −1 0 0 2 0 0 = 0 1 0 . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 3 ⎥⎦ ⎢⎣0 0 3 ⎥⎦ ⎢⎣0 0 9 ⎥⎦
Ans. (a) [F ] = −1 0
(d) [ R ] = [F ][U]
−1
0 ⎤ ⎡ 0 1 0⎤ ⎡ 0 2 0 ⎤ ⎡1 0 = ⎢⎢−1 0 0 ⎥⎥ ⎢⎢0 1 / 2 0 ⎥⎥ = ⎢⎢−1 0 0 ⎥⎥ . ⎢⎣ 0 0 3 ⎥⎦ ⎢⎣ 0 0 1 / 3⎥⎦ ⎢⎣ 0 0 1 ⎥⎦
⎡0 0 0 ⎤ ⎡3 / 8 0 0 ⎤ 1 ⎢ ⎥ ⎢ ⎥ − 1 (e) ⎡ E ⎤ = [C − I ] = 0 3 / 2 0 , (f) ⎡e* ⎤ = ⎡ I − B ⎤ = 0 0 0 ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 ⎣ ⎦ 2⎣ ⎦ ⎢⎣0 0 4 ⎥⎦ ⎢⎣ 0 0 4 / 9 ⎥⎦ *
(g)
1
ΔV = det B = ( 4 ) (1) (9) = 6 . ΔV o
⎡ 0 −6 0 ⎤ ⎥ (h) dA = dAo ( det F ) F −1 no , dAo = 1, det F = 6, [ F ] = 3 0 0 , no = e2 ⎢ ⎥ 6 ⎢⎣ 0 0 2 ⎥⎦ ⎡ 0 3 0 ⎤ ⎡0 ⎤ ⎡3 ⎤ 1⎢ ⎥⎢ ⎥ ⎢ ⎥ −1 T [dA] = dAo (det F )[F ] [no ] = (1)(6) −6 0 0 1 = 0 → dA = 3e1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 6 ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦
( )
T
−1
1⎢
_________________________________________________________________ 3.67 Given x1 = X1 + 3 X2 , x2 = X2 , x3 = X3 . Obtain (a) the deformation gradient F and the right Cauchy-Green tensor C , (b) The eigenvalues and eigenvector of C , (c) the matrix of the stretch tensor U and U −1 with respect to the ei -basis and (d) the rotation tensor R with respect to the ei -basis. ----------------------------------------------------------------------------------------⎡1 3 0 ⎤ ⎡1 0 0 ⎤ ⎡1 3 0 ⎤ ⎡1 3
⎢ 1 0⎥ , ⎢ ⎥ ⎢⎣0 0 1 ⎥⎦
Ans. (a) [F ] = 0
[C ] = [F ] [F ] = ⎢⎢3 T
⎥ ⎢0 1 0 ⎥ = ⎢3 10 0 ⎥ . ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ 1
0
(b) the characteristic equation is
Copyright 2010, Elsevier Inc 3-32
0⎤
Lai et al, Introduction to Continuum Mechanics 1 − λ
3
0
3
10 − λ
0
0
0
1 − λ
λ1,2 =
11 ± 121 − 4
(
)
= 0 → (1 − λ ) λ 2 − 11λ + 1 = 0,
→ λ1 = 10.908326, λ2 =0.0916735, λ3 = 1
2 For λ 1 = 10.908326 ,
(1 − λ1 )α1 + 3α 2 = 0 → α 2 = − (1 − λ1 )α1 / 3 = 3.302775α1,
1
n1 =
(e1 + 3.302775e2 ) = 0.289785 e1 + 0.957093 e2. 3.450843 For λ 2 = 0.0916735 , (1 − λ2 )α1 + 3α 2 = 0 → α 2 = − (1 − λ2 )α1 / 3 = − 0.3027755α1,
1
n2 =
( e1 − 0.3027755e 2 ) = 0.957093e1 − 0.289784 e2. 1.044832 For λ 3 = 1, n3 = e3 , (c) The matrices with respect to the principal axes are as follows 0 0⎤ 0 ⎡10.9083 ⎡3.30277
= ⎢⎢ ⎢⎣
⎥ ⎢ 0 0.0916735 0 → [ U ]n = [C]ni ⎥ ⎢ i ⎢⎣ 0 0 1 ⎥⎦ 0 0⎤ ⎡0.302774 ⎢ ⎥ − 1 ⎡U ⎤ = 0 3.302772 0 . ⎥ ⎣ ⎦ ni ⎢ ⎢⎣ 0 0 1 ⎥⎦
0 0
0⎤
⎥ ⎥ 1 ⎥⎦
0.302774 0 . 0
T
The matrices with respect to the ei -basis are given by the formula [ U ]{e } = [Q ] i
[ U ]e
i
[U ]{n } [Q ] : i
0 0 ⎤ ⎡0.289785 0.957093 0 ⎤ ⎡ 0.289785 0.957093 0 ⎤ ⎡3.30277 ⎢ ⎥ ⎢ ⎥⎢ ⎥ 0.302774 0 0.957093 −0.289785 0 = ⎢ 0.957093 −0.289785 0 ⎥ ⎢ 0 ⎥⎢ ⎥ ⎢⎣ ⎥ ⎢ ⎥ ⎢ 0 0 1⎦ ⎣ 0 0 1⎦ ⎣ 0 0 1 ⎥⎦
⎡ 0.554704 0.832057 0 ⎤ ⎢ ⎥ = 0.832057 3.05087 0 . ⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦ 0 0 ⎤ ⎡0.289785 0.957093 0 ⎤ ⎡0.289785 0.957093 0 ⎤ ⎡0.302774 ⎢ ⎥ ⎢ ⎥⎢ ⎥ 1 − ⎡ U ⎤ = 0.957093 −0.289785 0 0 3.302772 0 0.957093 −0.289785 0 ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦ ei ⎢ ⎢⎣ ⎥ ⎢ ⎥ ⎢ 0 0 1⎦ ⎣ 0 0 1⎦ ⎣ 0 0 1 ⎥⎦ ⎡ 3.050852 −0.832052 0 ⎤ ⎢ ⎥ = −0.832052 0.554701 0 . ⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦
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Lai et al, Introduction to Continuum Mechanics
(d) [ R ]e
⎡1 3 0 ⎤ ⎡ 3.050852 −0.832052 0 ⎤ ⎡ 0.55470 0.83205 0 ⎤ = [ F ] ⎡ U = ⎢⎢0 1 0 ⎥⎥ ⎢⎢−0.832052 0.554701 0 ⎥⎥ = ⎢⎢−0.83205 0.55470 0 ⎥⎥ . ⎣ ⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ −1 ⎤
i
_________________________________________________________________ 3.68 Verify that with respect to rectangular Cartesian base vectors, the right stretch tensor U and the rotation tensor R for the simple shear deformation , x1 = X1 + kX2 , x2 = X2 , x3 = X3 , are given by: With f = (1 + k 2 / 4) −1/2 , f ⎡ ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0
0⎤
/ 2 kf
(1 + k
2
)
/2 f
0
f ⎡ ⎢ [ R ] = ⎢− kf / 2 ⎢⎣ 0
⎥ 0⎥, ⎥ 1 ⎥⎦
kf/ 2 f
0
0⎤
⎥ ⎥ 1 ⎥⎦
0 .
----------------------------------------------------------------------------------------/ 2 0⎤ f kf f kf/ 2 0 ⎤ ⎡ ⎡
⎢ ⎢ ⎢⎣
Ans. [ RU ] = − kf / 2
f
0
0
⎥⎢
0 ⎢ kf / 2 ⎥
1 ⎥⎦ ⎢⎢ 0 ⎣
(1 + k
2
)
/2 f
0
⎥
0 ⎥
⎥
1 ⎥⎦
2 ⎡ f2 + ( kf/ 2 )( kf/ 2 ) f( kf/ 2 ) + ( kf/ 2 ) 1 + k / 2 f 0 ⎤ ⎢ ⎥ ⎢ ⎥ = ⎢( − kf / 2 ) f + f ( kf / 2 ) ( − kf / 2)( kf / 2 ) + f 1 + k 2 / 2 f 0 ⎥ ⎢ ⎥ 0 0 1⎥ ⎢ ⎣ ⎦ 2 2 ⎡ f2 1 + k2 / 4 kf 1 + k / 4 0 ⎤ ⎢ ⎥ ⎡1 k 0 ⎤ ⎢ ⎥ ⎢ ⎥ 2 2 =⎢ 0 0 ⎥ = 0 1 0 = the given [ F ] f 1 + k / 4 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦ 0 0 1⎥ ⎢ ⎣ ⎦
( (
(
)
( (
) )
) )
Since the decomposition of F is unique, therefore, the given R and U are the rotation and the stretch tensor respectively. _________________________________________________________________ 1 1 2 2 3.69 Let dX( ) = dS1N ( ) , dX ( ) = dS2 N ( ) be two material elements at a point P. Show that if θ (1) (2 ) denotes the angle between their respective deformed elements dx = ds1m and dx = ds2 n , (1)
then, cosθ =
(2)
Cαβ Nα N β
λ1λ 2
ds ds 1 2 , where N( ) = Nα(1) eα , N ( ) = N α(2) eα , λ1 = 1 and λ 2 = 2 . dS1 dS2
----------------------------------------------------------------------------------------(1) (2 ) (1) (2 ) (1 ) T (2 ) (1 ) (2 ) Ans. dx ⋅ dx = FdX ⋅ FdX = dX ⋅ F F dX = dX ⋅ C dX , 2 1 2 1 → ds1ds2 cos θ = dS1dS2 N( ) ⋅ CN ( ) = dS1dS2 ( Nα( )eα ) C( N β ( )eβ ), (1) ( 2 ) Cαβ Nα N β (1) ( 2 ) . Nα N β eα ⋅ Ceβ = → cos θ = λ1λ 2 ds1ds2
dS1dS2
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Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 3.70 Given the following right Cauchy-Green deformation tensor at a point
⎡9 0 0 ⎤ [C] = ⎢⎢0 4 0 ⎥⎥ ⎢⎣0 0 0.36 ⎥⎦ (a) Find the stretch for the material elements which were in the direction of e1 , e2 ,and e3 . (b) Find the stretch for the material element which was in the direction of e1 + e2 . (c) Find cosθ , 1 2 1 2 where θ is the angle between dx( ) and d x ( ) where dX( ) = dS1e1 and dX( ) = dS2e1 deform to (1) (2 ) dx = ds1m and dx = ds2n . ---------------------------------------------------------------------------------------- Ans. (a) For the elements which were in e1 , e2 , and e3 direction, the stretches are C11 , C22 , C 33 , that is, 3, 2 and 0.6 respectively.
⎡ 9 0 0 ⎤ ⎡1 ⎤ ⎡9⎤ 1 13 (b) Let e1' = ( e1 + e2 ) → C 11' = [1 1 0 ] ⎢⎢0 4 0 ⎥⎥ ⎢⎢1 ⎥⎥ = [1 1 0 ] ⎢⎢4 ⎥⎥ = . 2 2 2 2 ⎢⎣0 0 0.36 ⎥⎦ ⎢⎣0⎥⎦ ⎢⎣0 ⎥⎦ 1
1
That is, the stretch for dX = dSe1' is
( ds / dS ) =
'
C11 = 13 / 2 .
(c) C 12 = 0 → cos θ = 0 → θ = 90o . There is no change in angle. (note, {e1 , e2 , e3 } are principal axes for C . _________________________________________________________________ 3.71 Given the following large shear deformation: x1 = X1 + X2 , x2 = X2 , x3 = X3 .
(a) Find the stretch tensor U (Hint: use the formula given in problem 3.68) and verify that 2
U = C , the right Cauchy-Green deformation tensor. (b) What is the stretch for the element which was in the direction e 2 ?
(c) Find the stretch for an element which was in the direction of e1 + e2 . (d) What is the angle between the deformed elements of dS1e1 and dS2 e2 ?. ---------------------------------------------------------------------------------------- Ans. (a) For x1 = X1 + kX2 , x2 = X2 , x3 = X3 , from Prob. 3.68, we have f ⎡ ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0
/ 2 kf
(1 + k
2
)
/2 f
0
0⎤ f / 2 ⎡ f [ U ] = ⎢⎢ f/ 2 ( 3 / 2 ) f 0 ⎥⎥ = ⎢⎣ 0 0 1 ⎥⎦
0⎤
⎥
⎛ ⎜ ⎝
0 ⎥ where f = ⎜1 +
⎥ 1 ⎥⎦
k ⎞ 2
⎟ 4 ⎟⎠
−
1 2
. Thus, with k = 1 , f = 2 / 5
⎡ 1 1/ 2 0 ⎤ ⎡ 1 1/ 2 ⎢ 2 ⎢f1 / 2 3 / 2 0 ⎥ = ⎢1 / 2 3 / 2 ⎢ ⎥ 5⎢ ⎢⎣ 0 0 1 / f ⎥⎦ 0 ⎣ 0
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⎤ ⎥ 0 ⎥. ⎥ 5 / 2⎦ 0
Lai et al, Introduction to Continuum Mechanics
⎡ 1 4⎢ [ U] [U ] = ⎢1 / 2 5⎢ ⎣ 0
1/ 2
⎤⎡ 1 1/ 2 ⎥⎢ 0 ⎥ ⎢1 / 2 (3 / 2 ) ⎥⎢ 5 / 2⎦ ⎣ 0 0
⎤ ⎡1 1 0 ⎤ ⎥ ⎢ ⎥ 0 ⎥ = 1 2 0 = [C] . ⎢ ⎥ ⎥ ⎢ ⎥⎦ 0 0 1 5 / 2⎦ ⎣
0
(3 / 2 ) 0
0
(b) The stretch for the element which was in the direction e 2 is
C 22 = 2 .
(c) Let e1' = ( e1 + e2 ) / 2 ,
⎡ 1 1 0 ⎤ ⎡1 ⎤ ⎡2 ⎤ 1 ds ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 5 ' = 5/2 . C 11 = [1 1 0 ] 1 2 0 1 = [1 1 0 ] 3 = → ⎢ ⎥⎢ ⎥ 2 ⎢ ⎥ 2 2 dS ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎛ ds ⎞⎛ ds ⎞ 1 (d) ⎜ 1 ⎟⎜ 2 ⎟ cos θ = C 12 → (1) 2 cos θ = 1 → cos θ = → θ = 45 o . 2 ⎝ dS1 ⎠⎝ dS2 ⎠ 1
( )
_________________________________________________________________ 3.72 Given the following large shear deformation: x1 = X1 + 2 X2 , x2 = X2 , x3 = X3
(a) Find the stretch tensor U (Hint: use the formula given in problem 3.68) and verify that 2
U = C , the right Cauchy-Green deformation tensor. (b) What is the stretch for the element which was in the direction e2 .
(c) Find the stretch for an element which was in the direction of e1 + e2 . (d) What is the angle between the deformed elements of dS1e1 and dS2 e2 . ---------------------------------------------------------------------------------------- Ans. For x1 = X1 + kX2 , x2 = X2 , x3 = X3 , from Prob. 3.68, we have f ⎡ ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0 f ⎡ ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0
/ 2 kf
(1 + k
2
)
/2 f
0⎤
⎥
⎛ ⎜ ⎝
0 ⎥ where f = ⎜1 +
k ⎞ 2
⎟ 4 ⎟⎠
−
1 2
. Thus, with k = 2 , f = 1 / 2
⎥ 0 1 ⎥⎦ / 2 0⎤ kf ⎡1 1 0 ⎤ ⎥ ⎥ 1 ⎢ 1 + k 2 / 2 f 0 ⎥ = ⎢1 3 0 ⎥ . 2⎢ ⎥ ⎥ 2⎦ 0 1 ⎥⎦ ⎣0 0 ⎡1 1 0 ⎤ f kf/ 2 0 ⎤ ⎡ ⎢ ⎥ 1 [ R ] = ⎢⎢ − kf / 2 f 0 ⎥⎥ = ⎢−1 1 0 ⎥ . 2⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦ 2⎦ ⎣0 0 ⎡ 0 ⎤ ⎡1 1 0 ⎤ ⎡1 2 0 ⎤ 2 1 1 ⎥⎢ ⎥ ⎛ 1 ⎞ ⎢ [ U ][U ] = ⎜ ⎟ ⎢1 3 0 ⎥ ⎢1 3 0 ⎥ = ⎢⎢2 5 0 ⎥⎥ = [C] . ⎝ 2⎠ ⎢ ⎥⎢ ⎥ 2 ⎦ ⎣0 0 2 ⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎣0 0
(
)
(b) The stretch for the element which was in the direction e 2 is (c) Let e1′ = ( e1 + e2 ) / 2 ,
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C 22 = 5 .
Lai et al, Introduction to Continuum Mechanics
⎡ 1 2 0 ⎤ ⎡1 ⎤ ⎡3⎤ 1 ds ⎢ ⎥ ⎢ ⎥ ′ = [1 1 0 ] 2 5 0 1 = [1 1 0 ] ⎢7 ⎥ = 5 → = 5 = 2.236 . C 11 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 2 dS ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎛ ds ⎞⎛ ds ⎞ 2 (d) ⎜ 1 ⎟⎜ 2 ⎟ cos θ = C 12 → (1) 5 cos θ = 2 → cos θ = . 5 ⎝ dS1 ⎠⎝ dS2 ⎠ 1
( )
_________________________________________________________________ 3.73 Show that for any tensor A ( X1 , X2 , X3 ) ,
∂ ∂ X m
det A = ( det A ) A −1
(
)
∂ Ajn nj
∂X m
---------------------------------------------------------------------------------------- Ans
A 11
A 12
A = A21
A22
A31
A32
∂ A11 ∂ Xm
A 13
∂A = A21 A23 → ∂ X m A33
∂A12 ∂ Xm A22
A31
A32
∂ A13 A 11 ∂ Xm ∂ A21 A23 + ∂ X m
∂A22 ∂ X m
A 31
A 32
A33
A 12
A 13
A 11
∂ A23 + A21 ∂ X m ∂ A31 A 33 ∂ Xm
A 12
A 13
A22
A23
∂ A32 ∂ Xm
∂ A33 ∂ Xm
. c = Let Aijc denote the cofactor of Aij , i.e., A11
Then,
A22
A23
A32
A33
c =− , A12
A21
A 23
A 31
A 33
etc.
∂ A ∂ 11 ∂ A13 c ∂ 21 ∂ 12 ∂ 22 A c A c A c A c = A A A A A + ... 11 + 12 + 13 + 21 + ∂ Xm ∂ Xm ∂ Xm ∂ Xm ∂ Xm ∂ Xm 22
c A ji ∂ A ∂ Aij c −1 = = → A jic = det A A −1 That is, Aij . On the other hand, A ij ∂ X m ∂X m det A ∂ Aij ∂Anj ∂A = det A A −1 = det A A −1 Thus, . ∂ mX ∂ mXji ∂ m Xjn
(
(
)
(
)
(
)
ij
.
)
_________________________________________________________________ 3.74 Show that if T U = 0 , where the eigenvalues of U are all positive (nonzero), then T = 0 . ---------------------------------------------------------------------------------------- Ans. Using the eigenvectors of U as basis, we have, ⎡ T11 T12 T13 ⎤ ⎡λ1 0 0 ⎤ ⎡ λ1T11 λ2 T12 λ3 T13 ⎤
[TU ] = [T ][U ] = ⎢⎢T21
T22
⎢⎣T31 T32
⎥⎢ 0 ⎥⎢ T33 ⎥⎦ ⎢⎣ 0 T23
λ2 0
⎥ ⎢
0 = λ1T21 ⎥ ⎢ λ3 ⎥⎦ ⎢⎣ λ1T31
λ2 T22 λ2 T32
⎥
λ3 T23 ⎥
λ3 T33 ⎥⎦
Thus, T U = 0 gives, all T ij = 0 , that is, T = 0 . _________________________________________________________________ 2
2
⎛ r∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ 3.75 Derive Eq. (3.29.21), that is, Bθθ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠
2
-----------------------------------------------------------------------------------------
Copyright 2010, Elsevier Inc 3-37
Lai et al, Introduction to Continuum Mechanics T Ans. Bθθ = eθ ⋅ Beθ = eθ ⋅ FF e θ . From Eq. (3.29.15). we have,
F T eθ =
r∂θ
∂ro
eor +
r ∂θ ro ∂θ o
eθo +
r ∂θ
∂zo
eoz , thus,
⎛ r∂θ o r∂θ o r∂θ o ⎞ r∂θ r∂θ r ∂θ o o o er + eθ + ez ⎟ = eθ ⋅ Fer + eθ ⋅ Feθ + eθ ⋅ Fez . ∂ ∂ ∂ ∂ ∂ ∂ θ θ r r z r r z o o o o o o o ⎝ o ⎠ r ∂θ r∂θ r ∂θ Since, eθ ⋅ Feor = , eθ ⋅ Feoz = , [See Eq. (3.29.10)], therefore, , eθ ⋅ Feoθ = ∂ro ∂zo ro ∂θ o
Bθθ = eθ ⋅ F ⎜
2
2
2
⎛ r∂θ ⎞ ⎛ r∂θ ⎞ ⎛ r ∂θ ⎞ Bθθ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ . ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ _________________________________________________________________
⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎟⎜ ⎟+⎜ ⎟⎜ ∂ ∂ r r ⎝ o ⎠⎝ o ⎠ ⎝ ro ∂θo ⎠⎝ ro ∂θ o
3.76 Derive Eq. (3.29.23), i.e., Brz = ⎜
⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎟+⎜ ⎟⎜ ⎟ ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠
----------------------------------------------------------------------------------------T
Ans. Brz = er ⋅ Bez = er ⋅ FF ez , from Eq. (3.29.16), we have, T
F ez =
∂ zo ∂ z o ∂ zo er + eθ + ez , thus, ∂ro ∂zo ro ∂θ o ⎛∂z o ∂z o ∂z o ⎞ ∂z ∂z ∂z er + eθ + ez , ⎟ = er ⋅ Feor + er ⋅ Feoθ + er ⋅ Feoz . ∂zo ∂zo ro ∂θo ro ∂θ o ⎝ ∂ro ⎠ ∂ro
Brz = e r ⋅ F ⎜
From Eq. (3.29.9), er ⋅ Feor =
∂r ∂r ∂r o , eθ ⋅ Feoz = , thus, , er ⋅ Feθ = ∂ro ∂zo ro ∂θ o
⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θo ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠
Brz = ⎜
_________________________________________________________________ 1 3.77 From ro = ro ( r ,θ , z , t ) , θo = θo ( r ,θ , z , t ) , zo = zo ( r ,θ , z , t ) ,derive the components of B −
with respect to the basis at x . ----------------------------------------------------------------------------------------1 o o o Ans. From dX = F − d x , where dx = dre r + rdθ eθ + dzez and dX = dro er + ro dθ o eθ + dzo ez , we
have, dro eor + ro dθo eθo + dzo eoz = F −1 ( drer + rdθ eθ + dzez )
→ dro = dr eor ⋅ F −1er + rdθ eor ⋅ F−1eθ + dz eor ⋅ F−1ez
(
)
(
)
(
)
∂ro ∂r ∂r o 1 o 1 o 1 dr + o dθ + o dz = dr e r ⋅ F − er + rdθ er ⋅ F − eθ + dz er ⋅ F− ez ∂r ∂θ ∂z ∂r ∂r ∂r → eor ⋅ F −1er = o , eor ⋅ F −1eθ = o , eor ⋅ F −1ez = o . ∂r ∂z r∂θ →
(
)
(
)
Copyright 2010, Elsevier Inc 3-38
(
)
Lai et al, Introduction to Continuum Mechanics Similarly, −1
o
eθ ⋅ F er =
ro ∂θ o
, eθo ⋅ F−1eθ =
ro ∂θo
r o ∂θ o
o , eθ ⋅ F−1ez =
. ∂z r∂θ ∂ z o −1 ∂ z o −1 ∂oz o −1 = . e ⋅ Fz er = o , e ⋅ Fz eθ = o , e ⋅ F e z z ∂r ∂z r ∂θ Thus, ∂r ∂z ∂r r ∂θ r ∂θ o ∂zo o − − F 1e r = o eor + o o eθo + o e zo , F 1eθ = o eor + o o eθ + ez ∂r ∂r ∂r r∂θ r∂θ r∂θ ∂r o r ∂θ o ∂z o −1 F e z = o e r + o o eθ + o e z . ∂ z ∂ z ∂ z Also, we have, ∂r ∂r −1 T o −1 −1 T o −1 o o er ⋅ F er = er ⋅ F er = o , eθ ⋅ F er = er ⋅ F eθ = o , ∂r r ∂θ ∂r − T − ez ⋅ F 1 eor = eor ⋅ F 1 ez = o . ∂ z Thus, T T ∂r ∂r ∂r r ∂θ r ∂θ r ∂θ o o −1 −1 F er = o e r + o eθ + o ez , F eθ = o o er + o o eθ + o o ez ∂r ∂z ∂r ∂z r ∂θ r∂θ ∂ z ∂ z ∂ z −1 T o F ez = o e r + o eθ + o ez . ∂r ∂z r ∂θ
∂r
( )
( )
( )
( )
( )
( )
( )
The components of B −1 with respect to the basis at x are: − = er ⋅ B− er = er ⋅ FF Brr 1
(
1
⎛ ∂r = ⎜ o e r ⋅ F −1 ⎝ ∂r
( )
T
o er
T
+
∂r = o eθ ⋅ F −1 r ∂θ
( )
−1
B zz
T
o er
+
2
T
ro ∂θo r ∂θ
r
∂z 1 eθ + o er ⋅ F− ∂r
( )
T
eθ = e θ ⋅ F −
T
2
o e z
2
∂ro r ∂θ
( −)
er ⋅ F
1
T
o
er +
ro ∂θo r ∂θ
(
)
er
−1 = er ⋅ B−1ez = er ⋅ FFT Brz
1
−1
( )
T
o eθ
1
θ
∂z + o eθ ⋅ F−1 r ∂θ
T
( )
2
o e z
2
2
⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ =⎜ o ⎟ +⎜ o o ⎟ +⎜ o ⎟ . ⎝ r∂θ ⎠ ⎝ r∂θ ⎠ ⎝ r∂θ ⎠
)
−1
T
( ) ( F− e ) ∂z e ⋅ (F ) ⋅ (F ) e + r ∂θ 1 eθ = er ⋅ F − −1 T o
θ
−1 T o
o
θ
−1
1
r
T
⎛ ∂ro ⎞⎛ ∂ro ⎞ ⎛ ro ∂θo ⎞⎛ ro ∂θ o ⎞ ⎛ ∂zo ⎞⎛ ∂zo ⎞ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ ⎝ r∂θ ⎠⎝ ∂r ⎠ ⎝ r∂θ ⎠⎝ ∂r ⎠ ⎝ r∂θ ⎠⎝ ∂z ⎠
e z = ⎜
( ) ( F− e )
1 ez = e r ⋅ F −
1
z
∂ro ∂z r ∂θ −1 T o −1 T o −1 T o ⎛ ∂ro ⎞⎛ ∂ro ⎞ ⎛ ro ∂θo ⎞⎛ ro ∂θ o er ⋅ F e r + o o er ⋅ F eθ + o er ⋅ F e z = ⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ∂z ∂z ∂z ⎝ ∂ r ⎠⎝ ∂ z ⎠ ⎝ ∂ r ⎠⎝ ∂ z ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ 1 Bθ − z = ⎜ o ⎟⎜ o ⎟ + ⎜ o o ⎟⎜ o o ⎟ + ⎜ o ⎟⎜ o ⎟ . ⎝ r∂θ ⎠⎝ ∂z ⎠ ⎝ r∂θ ⎠⎝ ∂z ⎠ ⎝ r∂θ ⎠⎝ ∂z ⎠
( )
( )
( )
Copyright 2010, Elsevier Inc 3-39
2
(
=
2
⎞ ⎛ ∂ro ⎞ ⎛ ro ∂θ o ⎞ ⎛ ∂zo ⎞ ⎟ = ⎜ ∂r ⎟ + ⎜ ∂r ⎟ + ⎜ ∂r ⎟ . ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( ) ( F− e )
eθ ⋅ F
2
1
⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ =⎜ o ⎟ +⎜ o o ⎟ +⎜ o ⎟ . ⎝ ∂ z⎠ ⎝ ∂ z ⎠ ⎝ ∂ z⎠
1 1 T Br −θ = er ⋅ B− eθ = er ⋅ FF
=
−1
1
−1 T o
( )
)
T
( − ) ( F− e )
er = er ⋅ F
er ⋅ F
∂r
(
1
−1
ro ∂θo
− = eθ ⋅ B− eθ = eθ ⋅ FF Bθθ 1
)
⎞ ⎛ ∂zo ⎞⎛ ∂zo ⎞ ⎟ + ⎜ ∂ r ⎟⎜ ∂ z ⎟ ⎠ ⎝ ⎠⎝ ⎠
Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 2
3.78 Derive Eq. (3.29.47), that is, C θo θo
2
⎛ ∂r ⎞ ⎛ r∂θ ⎞ ⎛ ∂z ⎞ =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ∂ ∂ r r θ θ o o o o ⎝ ⎠ ⎝ ⎠ ⎝ r o ∂θ o ⎠
2
----------------------------------------------------------------------------------------∂r ∂z r∂θ o o o o T o er + eθ + ez [Eq.3.29.3] , Ans. C θo θo = e θ ⋅ Ceθ = eθ ⋅ F Fe θ . Now, Feθ = ro ∂θo ro ∂θo r o ∂θ o therefore,
⎛ ∂r ⎞ ∂z ∂r o T r ∂θ r∂θ o T er + eθ + ez ⎟ = eθ ⋅ F er + eθ ⋅ F eθ ro ∂θo ro ∂θo ⎠ ro ∂θo ro ∂θo ⎝ ro ∂θo
o T C θo θo = e θ ⋅ F ⎜
∂ z o T eθ ⋅ F ez . Now, from Eqs. (3.29.14) (3.29.15) and (3.29.16), r o ∂θ o ∂r ∂z r ∂θ o T , eθo ⋅ FT eθ = , eoθ ⋅ FT ez = , thus. eθ ⋅ F er = ro ∂θo ro ∂θo r o ∂θ o +
2
C θo θo
2
⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ∂ ∂ θ θ r r ⎝ o o ⎠ ⎝ o o ⎠ ⎝ r o ∂θ o ⎠
2
_________________________________________________________________
⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ ⎝ ro ∂θo ⎠⎝ ∂ro ⎠ ⎝ ro ∂θo ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠
3.79 Derive Eq. (3.29.49), C r oθo = ⎜
-----------------------------------------------------------------------------------------∂r ∂z r ∂θ o o o o T o e r + eθ + e z [Eq.(3.29.3)] , Ans. C r oθo = er ⋅ Ceθ = er ⋅ F Feθ . Now, Feθ = ro ∂θ o ro ∂θo r o ∂θ o
⎛ ∂r ⎞ ∂z ∂r o T r ∂θ r ∂θ o T er + eθ + ez ⎟ = e r ⋅ F er + er ⋅ F eθ ∂ ∂ ∂ ∂ ∂ θ θ θ θ θ r r r r r o o o o o o ⎝ o o ⎠ o o
o T C r oθo = er ⋅ F ⎜
∂ z o T er ⋅ F ez . From Eqs. (3.29.14), (3.29.15) and (3.29.16) r o ∂θ o ∂r o T ∂z r ∂θ o T o T , er ⋅ F eθ = , er ⋅ F ez = , e r ⋅ F er = ∂ro ∂ro ∂r o +
⎛ ∂r ⎞⎛ ∂r ⎟⎜ ⎝ ro ∂θo ⎠⎝ ∂ro
Thus, C r oθo = ⎜
⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ∂ ∂ ∂ ∂ θ θ r r r r o o o o o o ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
_________________________________________________________________ 3.80 Derive the components of C−1 with respect to the bases at X . ---------------------------------------------------------------------------------------- Ans. T −1 o ∂r o −1 ∂r o −1 ∂r o −1 1 o T o 1 1 o er = o er ⋅ F er + o er ⋅ F eθ + o er ⋅ F ez C r−or o =er ⋅ F F er =er ⋅ F− F − ∂r ∂z r∂θ
(
)
( )
Copyright 2010, Elsevier Inc 3-40
Lai et al, Introduction to Continuum Mechanics
∂ro ∂ro ∂ro ∂ro ∂ro ∂ro . [See Eqs.(3.29.30), (3.29.31) and (3.29.32)]. + + ∂r ∂r r∂θ r∂θ ∂z ∂z T −1 o r ∂θ o −1 r ∂θ o −1 r ∂θ o −1 1 o T o 1 1 o eθ == o o er ⋅ F er + o o er ⋅ F eθ + o o er ⋅ F ez C r −oθ o =e r ⋅ F F eθ =er ⋅ F − F− ∂r ∂z r ∂θ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ = ⎜ o o ⎟⎜ o ⎟ + ⎜ o o ⎟⎜ o ⎟ + ⎜ o o ⎟⎜ o ⎟ . ⎝ ∂r ⎠⎝ ∂r ⎠ ⎝ r∂θ ⎠⎝ r∂θ ⎠ ⎝ ∂z ⎠⎝ ∂z ⎠ =
(
)
( )
The other components can be similarly derived. _________________________________________________________________ 3.81 Derive components of B with respect to the basis {er ,eθ ,ez } at x for the pathline equations given by r = r ( X , Y , Z , t), θ = θ ( X , Y , Z , t), z=z( X , Y , Z , t) . -------------------------------------------------------------------------------------------- Ans. From dx = drer + rdθ eθ + dzez and dX = dXeX + dYeY + dZeZ and r = r ( X , Y , Z , t ), θ = θ ( X , Y , Z , t ), z=z( X , Y , Z , t ) , we have, dx = FdX → dx = drer + rdθ eθ + dzez = dXFeX + dYFeY + dZFe Z
∂r ∂r r ∂θ r∂θ ⎞ ⎛ ∂r ⎞ ⎛ r∂θ →⎜ dX + dY + dZ ⎟ er + ⎜ dX + dY + dZ e ∂Y ∂Z ⎠ ∂Y ∂ Z ⎟⎠ θ ⎝∂X ⎝∂X ∂ z ∂ z ⎞ ⎛∂ z +⎜ dX + dY + dZ e = dXFeX + dYFeY + dZFe Z ∂ Y ∂ Z ⎟⎠ z ⎝∂ X ∂r ∂z ∂r ∂z r ∂θ r∂θ → FeX = er + eθ + ez , FeY = er + eθ + e , ∂X ∂X ∂X ∂Y ∂Y ∂Y z ∂r ∂z r ∂θ Fe Z = e r + eθ + ez , and ∂ Z ∂ Z ∂ Z ∂r ∂r T , eY ⋅ FT er = er ⋅ FeY = , etc. eX ⋅ F er = er ⋅ FeX = ∂ X ∂Y ∂r ∂r ∂r r∂θ r∂θ r∂θ F T er = eX + eY + eZ , FT eθ = eX + eY + e , ∂X ∂Y ∂Z ∂X ∂Y ∂Z Z ∂ z ∂ z ∂ z T F e z = eX + eY + eZ . ∂ X ∂ Y ∂ Z The components of B are: 2
2
2
∂r ∂r ∂r ⎛ ∂r ⎞ ⎛ ∂ r ⎞ ⎛ ∂ r ⎞ er ⋅ FeX + er ⋅ FeY + er ⋅ FeZ = ⎜ Brr = er ⋅ FF er = ⎟ +⎜ ⎟ + ⎜ ⎟ . ∂X ∂Y ∂Z ⎝ ∂ X⎠ ⎝ ∂ Y⎠ ⎝ ∂ Z⎠ r∂θ r ∂θ r ∂θ T er ⋅ FeX + er ⋅ FeY + e ⋅ Fe Br θ = er ⋅ FF eθ = ∂ X ∂ Y ∂ Zr Z ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r∂θ ⎞ ⎛ ∂r ⎞⎛ r∂θ ⎞ =⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟. ⎝ ∂ X ⎠⎝ ∂ X ⎠ ⎝ ∂ Y ⎠⎝ ∂ Y ⎠ ⎝ ∂ Z ⎠⎝ ∂ Z ⎠ T
_________________________________________________________________ 3.82 Derive the components of B −1 with respect to the basis {er ,eθ ,ez } at x for the pathline equations given by X = X ( r, θ , z, t), Y = Y ( r, θ , z, t), Z = Z( r,θ , z, t) . ---------------------------------------------------------------------------------------- Ans. From dx = drer + rdθ eθ + dzez and dX = dXeX + dYeY + dZeZ and
Copyright 2010, Elsevier Inc 3-41
Lai et al, Introduction to Continuum Mechanics X = X ( r, θ , z, t), Y = Y ( r, θ , z, t), Z = Z ( r,θ , z, t) , we have, 1 1 1 1 dX = F − dx → dXe X + dYeY + dZe Z = drF − er + rdθ F − eθ + dzF− ez
∂X ∂X ⎞ ∂Y ∂ Y ⎞ ⎛∂X ⎛∂ Y →⎜ dr + dθ + dz ⎟ eX + ⎜ dr + dθ + dz e ∂θ ∂z ⎠ ∂θ ∂z ⎟⎠ Y ⎝ ∂r ⎝ ∂r ∂ Z ∂ Z ⎞ ⎛∂ Z 1 1 1 +⎜ dr + dθ + dz ⎟ e Z = drF − er + rdθ F − eθ + dzF− ez . ∂θ ∂z ⎠ ⎝ ∂r Thus,
∂X ∂Y ∂Z ∂X ∂Y ∂Z 1 eX + eY + eZ , F− eθ = eX + eY + eZ ∂r ∂r ∂r r∂θ r∂θ r∂θ ∂ X ∂ Y ∂ Z 1 F − ez = eX + eY + eZ . ∂ z ∂ z ∂ z 1 F − er =
and T ∂ X ∂Y , er ⋅ F −1 eY = eY ⋅ F−1er = , ∂r ∂r T ∂ Z , etc. that is, er ⋅ F −1 eZ = e Z ⋅ F−1er = ∂r ∂X ∂X ∂X ∂Y ∂Y ∂Y − T − T F 1 eX = er + eθ + e z , F 1 eY = er + eθ + e ∂r ∂z ∂r ∂z z r∂θ r∂θ ∂ Z ∂ Z ∂ Z − T F 1 eZ = er + eθ + ez . ∂r ∂z r ∂θ
er ⋅ F −1
( )
T
eX = eX ⋅ F−1er =
( )
( )
( )
( )
( ) Thus,
−1
−1 T
T
∂Y − er ⋅ F 1 ∂r
)
∂ Z + e r ⋅ F −1 ∂r
⎛ ∂ X⎞ ⎛ ∂ Y⎞ ⎛ ∂ Z⎞ eZ = ⎜ +⎜ ⎟ + ⎜ ⎟ . ⎟ ⎝ ∂r ⎠ ⎝ ∂r ⎠ ⎝ ∂r ⎠ T ∂ X ∂Y −1 −1 T −1 T F eθ = er ⋅ F eX + er ⋅ F eY r ∂θ r ∂θ T ⎛ ∂ X ⎞⎛ ∂ X ⎞ ⎛ ∂ Y ⎞⎛ ∂ Y ⎞ ⎛ ∂ Z⎞⎛ ∂ Z ⎞ eZ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ ., ⎝ ∂r ⎠⎝ r∂θ ⎠ ⎝ ∂r ⎠⎝ r∂θ ⎠ ⎝ ∂r ⎠⎝ r∂θ ⎠
(
( )
T
( )
1 1 Br −θ = er ⋅ F −
+
∂Z − er ⋅ F 1 r ∂θ
( )
er = e r ⋅ F 2
−
∂ X − er ⋅ F 1 ∂r
−1 T Brr = er ⋅ FF
F 1er = 2
( )
eX +
T
( )
eY
2
( )
( )
( )
etc. _________________________________________________________________ 3.83 Verify that (a) the components of B with respect to {er ,eθ ,ez } can be obtained from ⎡FF T ⎤
⎣
{
}
⎦
and (b) the component of C , with respect to eor ,eθo , eoz can be obtained from ⎡ F T F ⎤ , where [ F ]
⎣
is the matrix of the two points deformation gradient tensor given in Eq. (3.29.12). ---------------------------------------------------------------------------------------- Ans. (a) Eq. (3.29.12) →
Copyright 2010, Elsevier Inc 3-42
⎦
Lai et al, Introduction to Continuum Mechanics
⎡ ∂r ⎢ ⎢ ∂ro ⎢ ⎡ FF T ⎤ = ⎢ r∂θ ⎣ ⎦ ⎢ ∂ro ⎢ ∂z ⎢ ⎢⎣ ∂ro
∂r ⎤ ⎡ ∂r ⎥⎢ ∂zo ⎥ ⎢ ∂ro r ∂θ ⎥ ⎢ ∂ r ⎥⎢ ∂zo ⎥ ⎢ ro ∂θo ∂z ⎥⎢ ∂ r ⎥⎢ ∂z o ⎥⎦ ⎢⎣ ∂zo
∂r ro ∂θ o r ∂θ ro∂θo ∂z ro ∂θ o
2
2
r∂θ
∂ro r∂θ ro ∂θo r∂θ ∂zo
∂z ⎤ ⎥ ∂ro ⎥ ∂z ⎥ ⎥ → ro ∂θ o ⎥ ∂z ⎥ ⎥ ∂zo ⎥⎦
2
⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ ∂r r∂θ ∂r r∂θ ∂r r∂θ + + etc. Brr = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ , Br θ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ r r z r r r r z z θ θ θ o o o o o o o o ⎝ o⎠ ⎝ o o⎠ ⎝ o⎠ (b)
⎡ ∂r ⎢ ⎢ ∂ro ⎢ ⎡ F T F ⎤ = ⎢ ∂r ⎣ ⎦ ⎢ ro ∂θo ⎢ ∂r ⎢ ⎢⎣ ∂ zo 2
Croro
∂z ⎤ ⎡ ∂r ⎥⎢ ∂ro ⎥ ⎢ ∂ro ∂ z ⎥ ⎢ r∂θ ⎥⎢ ro ∂θo ⎥ ⎢ ∂ro ∂z ⎥ ⎢ ∂z ⎥⎢ ∂ zo ⎥⎦ ⎢⎣ ∂ ro
r ∂θ
∂ro r ∂θ ro ∂θo r∂θ ∂ zo 2
∂r ro∂θ o r ∂θ ro ∂θo ∂z ro ∂θ o
∂r ⎤ ⎥ ∂z o ⎥ r ∂θ ⎥ ⎥→ ∂ zo ⎥ ∂z ⎥ ⎥ ∂ zo ⎥⎦
2
⎛ ∂r ⎞ ⎛ r∂θ ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂r =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ , C r oθ o = ⎜ ⎟⎜ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠⎝ ro ∂θo
⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎟+ ⎜ ⎟⎜ ⎠ ⎝ ∂ro ⎠⎝ ro ∂θo
⎞ ⎛ ∂ z ⎞⎛ ∂z ⎞ ⎟+⎜ ⎟⎜ ⎟. ∂ ∂ θ r r ⎠ ⎝ o ⎠⎝ o o ⎠
_________________________________________________________________ 3.84 Given r = ro , θ = θ o + kzo , z = zo . (a) Obtain the components of the Left Cauchy-Green tensor B , with respect to the basis at the current configuration ( r ,θ , z ) . (b) Obtain the components of the right Cauchy-Green tensor C with respect to the basis at the reference configuration. ---------------------------------------------------------------------------------------- Ans. (a), Using Eqs (3.29.19) to (3.29.24). we obtain 2
2
2
⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ Brr = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ =1, ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ 2
2
2
2
⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r∂θ ⎞ ⎛ r ⎞ 2 2 = ⎜ B ⎟ +⎜ ⎟ +⎜ ⎟ = ⎜ ⎟ + ( ) = 1 +kr( ) , θθ ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ ⎝ ro ⎠ 2
2
kr
2
⎛ ∂ z⎞ ⎛ ∂ z ⎞ ⎛ ∂ z⎞ B zz = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1, r r θ z ∂ ∂ ∂ ⎝ o⎠ ⎝ o o⎠ ⎝ o⎠ ⎛ r∂θ ⎞⎛ ∂r ⎞ ⎛ r∂θ ⎞⎛ ∂ r ⎞ ⎛ r∂θ ⎞⎛ ∂r ⎞ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θo ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠
Br θ = ⎜
⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟=0, ∂ ∂ ∂ ∂ ∂ ∂ r r r θ r θ z z ⎝ o ⎠⎝ o ⎠ ⎝ o o ⎠⎝ o o ⎠ ⎝ o ⎠⎝ o ⎠
Brz = ⎜
zθ
⎛ ∂ z ⎞ ⎛ r∂θ ⎞ ⎛ ∂ z ⎞⎛ r∂θ =⎜ ⎟⎜ ⎟B+ ⎜ ⎟⎜ ∂ ∂ r r ⎝ o ⎠ ⎝ o ⎠ ⎝ ro ∂θo ⎠⎝ ro ∂θ o
⎞ ⎛ ∂ z ⎞⎛ r∂θ ⎞ ⎟+⎜ ⎟⎜ ⎟= ∂ ∂ z z ⎠ ⎝ o ⎠⎝ o ⎠
.
Copyright 2010, Elsevier Inc 3-43
rk
Lai et al, Introduction to Continuum Mechanics . 0 ⎡1 ⎢ 2 Thus, [ B ] = ⎢ 0 1 + ( rk ) ⎢0 rk ⎢⎣
0⎤
⎥
rk ⎥ .
1⎥
⎥⎦
(b) Using Eqs.(3. 29.43) to (3. 29.51), we have, 2
Cror o
2
2
2
2
⎛ ∂r ⎞ ⎛ r∂θ ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ r∂θ ⎞ ⎛ ∂z =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1, C θoθ o = ⎜ ⎟ +⎜ ⎟ +⎜ ∂ ∂ ∂ ∂ ∂ r r r r r θ θ ⎝ o⎠ ⎝ o ⎠ ⎝ o⎠ ⎝ o o ⎠ ⎝ o o ⎠ ⎝ ro ∂θ o 2
2
2
⎞ ⎟ = 1, ⎠
2
⎛ ∂r ⎞ ⎛ r∂θ ⎞ ⎛ ∂z ⎞ 2 Czo zo = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1 + ( rk ) , ⎝ ∂zo ⎠ ⎝ ∂zo ⎠ ⎝ ∂zo ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ C r oθ o = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ∂ ∂ ∂ ∂ ∂ ∂ θ θ θ r r r r r r ⎝ o o ⎠⎝ o ⎠ ⎝ o o ⎠⎝ o ⎠ ⎝ o o ⎠⎝ o ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ C ro zo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ∂ ∂ ∂ ∂ ∂ ∂ r z r z r z ⎝ o ⎠⎝ o ⎠ ⎝ o ⎠⎝ o ⎠ ⎝ o ⎠⎝ o ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r∂θ ⎞⎛ r∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ∂ ∂ ∂ ∂ θ θ z r z r ⎝ o ⎠⎝ o o ⎠ ⎝ o ⎠⎝ o o ⎠ ⎝ ∂ zo ⎠⎝ ro ∂θ o ⎡1 0 ⎤ 0 ⎢ ⎥ Thus, [C] = ⎢ 0 1 rk ⎥ . ⎢ 2⎥ ⎢⎣ 0 rk 1 + ( rk ) ⎥⎦ C zoθo = ⎜
⎞ ⎟ = rk . ⎠
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3.85 Given r = ( 2aX + b )
, θ = Y / a, z = Z , where ( r ,θ , z ) are cylindrical coordinates for the
cuurent configuration and ( X , Y, Z ) are rectangular coordinates for the reference configuration. (a) Obtain the components of [ B ] with respect to the basis at the current configuration and (b) calculate the change of volume. ---------------------------------------------------------------------------------------- Ans. (a) Using Eqs.(3.29.59) to (3.29.64), we have, 2
2
2
2
2
2
2
2
2
2
⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ a ⎞ ⎛ r ∂θ ⎞ ⎛ r∂θ ⎞ ⎛ r∂θ ⎞ ⎛ r ⎞ + ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ , Bθθ = ⎜ Brr = ⎜ ⎟ ⎟ +⎜ ⎟ +⎜ ⎟ =⎜ ⎟ ⎝ ∂ X⎠ ⎝ ∂ Y⎠ ⎝ ∂ Z⎠ ⎝ r⎠ ⎝ ∂ X ⎠ ⎝ ∂ Y ⎠ ⎝ ∂ Z ⎠ ⎝ a⎠ ⎛ ∂ z⎞ ⎛ ∂ z⎞ ⎛ ∂ z⎞ B zz = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ =1, ⎝ ∂ X⎠ ⎝ ∂ Y⎠ ⎝ ∂ Z⎠ ⎛ ∂r ⎞⎛ r∂θ ⎞ ⎛ ∂r ⎞⎛ r∂θ ⎞ ⎛ ∂r ⎞⎛ r∂θ ⎞ Br θ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ =0, ⎝ ∂ X ⎠⎝ ∂ X ⎠ ⎝ ∂ Y ⎠⎝ ∂ Y ⎠ ⎝ ∂ Z ⎠⎝ ∂ Z ⎠
⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ 0, ⎟⎜ ∂ X ⎟ + ⎜ ∂ Y ⎟⎜ ∂ Y ⎟ + ⎜ ∂ Z ⎟⎜ ∂ Z ⎟ = ∂ X ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ r∂θ ⎞⎛ ∂z ⎞ ⎛ r∂θ ⎞⎛ ∂z ⎞ ⎛ r∂θ ⎞⎛ ∂z ⎞ =⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟ = 0 . ⎝ ∂ X ⎠⎝ ∂ X ⎠ ⎝ ∂ Y ⎠⎝ ∂ Y ⎠ ⎝ ∂ Z ⎠⎝ ∂ Z ⎠
Brz = ⎜ Bθ z
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