Solutions Manual Continuum Mechanics Lai 4th Edittion

Lai et al, Introduction to Continuum Mechanics

CHAPTER 2, PART A 2.1

Given ⎡1 0 2 ⎤ ⎡1 ⎤ ⎢ ⎥ ⎡⎣ Sij ⎤⎦ = 0 1 2 and [ ai ] = ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 0 3 ⎥⎦ ⎢⎣ 3 ⎥⎦ Evaluate (a) Sii , (b) Sij Sij , (c) S ji S ji , (d) S jk Skj (e) am am , (f) Smn am an , (g) Snm am an

------------------------------------------------------------------------------Ans. (a) Sii = S11 + S22 + S33 = 1 + 1 + 3 = 5 . 2 2 2 2 2 2 2 2 2 (b) Sij Sij = S11 + S12 + S13 + S21 + S22 + S23 + S31 + S32 + S33 =

1 + 0 + 4 + 0 + 1 + 4 + 9 + 0 + 9 = 28 . (c) S ji S ji = Sij Sij =28. (d) S jk S kj = S1k S k1 + S2 k S k 2 + S3k S k 3

= S11S11 + S12 S21 + S13 S31 + S21S12 + S22 S22 + S23 S32 + S31S13 + S32 S23 + S33 S33

= (1)(1) + ( 0 )( 0 ) + ( 2 )( 3) + ( 0 )( 0 ) + (1)(1) + ( 2 )( 0 ) + ( 3)( 2 ) + ( 0 )( 2 ) + ( 3)( 3) = 23 .

(e) am am = a12 + a22 + a32 = 1 + 4 + 9 = 14 . (f) Smn am an = S1n a1an + S2 n a2 an + S3n a3 an = S11a1a1 + S12 a1a2 + S13 a1a3 + S21a2 a1 + S22 a2 a2 + S23a2 a3 + S31a3a1 + S32 a3a2 + S33a3a3

= (1)(1)(1) + ( 0 )(1)( 2 ) + ( 2 )(1)( 3) + ( 0 )( 2 )(1) + (1)( 2 )( 2 ) + ( 2 )( 2 )( 3) + ( 3)( 3)(1) + ( 0 )( 3)( 2 ) + ( 3)( 3)( 3) = 1 + 0 + 6 + 0 + 4 + 12 + 9 + 0 + 27 = 59.

(g) Snm am an = Smn am an =59. __________________________________________________________________ Determine which of these equations have an identical meaning with ai = Qij a 'j . 2.2 (a) a p = Q pm am' , (b) a p = Qqp aq' , (c) am = an' Qm n . ------------------------------------------------------------------------------Ans. (a) and (c) __________________________________________________________________ Given the following matrices 2.3 ⎡1 ⎤ ⎡ 2 3 0⎤ ⎢ ⎥ [ ai ] = ⎢0 ⎥ , ⎡⎣ Bij ⎤⎦ = ⎢⎢0 5 1 ⎥⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 0 2 1 ⎥⎦ Demonstrate the equivalence of the subscripted equations and corresponding matrix equations in the following two problems. (a) bi = Bij a j and [b ] = [ B ][ a ] , (b) s = Bij ai a j and s = [ a ]

T

[ B ][ a ]

------------------------------------------------------------------------------Ans. (a) bi = Bij a j → b1 = B1 j a j = B11a1 + B12 a2 + B13a3 = ( 2 )(1) + ( 3)( 0 ) + ( 0 )( 2 ) = 2 b2 = B2 j a j = B21a1 + B22 a2 + B23a3 = 2, b3 = B3 j a j = B31a1 + B32 a2 + B33 a3 = 2 .

Copyright 2010, Elsevier Inc 2-1

Lai et al, Introduction to Continuum Mechanics ⎡ 2 3 0⎤ ⎡1 ⎤ ⎡ 2⎤ [b] = [ B ][ a ] = ⎢⎢0 5 1 ⎥⎥ ⎢⎢0 ⎥⎥ = ⎢⎢ 2⎥⎥ . Thus, bi = Bij a j gives the same results as [b] = [ B ][ a ] ⎢⎣ 0 2 1 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ (b) s = Bij ai a j = B11a1a1 + B12 a1a2 + B13 a1a3 + + B21a2 a1 + B22 a2 a2 + B23 a2 a3 + B31a3 a1 + B32 a3a2 + B33a3a3 = ( 2 ) (1)(1) + ( 3) (1)(0) + ( 0 ) (1)(2) + ( 0 ) (0)(1) + ( 5 ) (0)(0) + (1) (0)(2) + ( 0 ) (2)(1) + ( 2 ) (2)(0) + (1) (2)(2) = 2 + 4 = 6. ⎡ 2 3 0⎤ ⎡1 ⎤ ⎡2⎤ ⎢ ⎥ ⎢ ⎥ and s = [ a ] [ B ][ a ] = [1 0 2] ⎢ 0 5 1 ⎥ ⎢ 0 ⎥ = [1 0 2] ⎢⎢ 2 ⎥⎥ = 2 + 4 = 6 . ⎢⎣ 0 2 1 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ __________________________________________________________________ T

2.4

Write in indicial notation the matrix equation (a) [ A] = [ B ][C ] , (b) [ D ] = [ B ] [C ] and (c) T

[ E ] = [ B ]T [C ][ F ] . ------------------------------------------------------------------------------Ans. (a) [ A] = [ B ][C ] → Aij = Bi m Cm j , (b) [ D ] = [ B ] [C ] → Aij = Bm i Cmj . T

(c) [ E ] = [ B ] [C ][ F ] → Eij = Bmi Cmk Fkj . T

__________________________________________________________________ ∂ 2φ ∂ 2φ ∂ 2φ Write in indicial notation the equation (a) s = A12 + A22 + A32 and (b) 2 + 2 + 2 = 0 . 2.5 ∂x1 ∂x2 ∂x3 ------------------------------------------------------------------------------∂ 2φ ∂ 2φ ∂ 2φ ∂ 2φ Ans. (a) s = A12 + A22 + A32 = Ai Ai . (b) 2 + 2 + 2 = 0 → =0. ∂xi ∂xi ∂x1 ∂x2 ∂x3 __________________________________________________________________ 2.6 Given that Si j =ai a j and Sij′ =ai′a′j , where ai′ =Qmi am and a′j =Qn j an , and Qik Q jk = δ ij . Show that Sii′ =Sii . ------------------------------------------------------------------------------Ans. Sij′ =Qmi am Qn j an =Qmi Qn j am an → Sii′ =Qmi Qni am an =δ mn am an =am am = S mm = Sii . __________________________________________________________________ ∂v ∂v Write ai = i + v j i in long form. 2.7 ∂t ∂x j ------------------------------------------------------------------------------Ans. ∂v ∂v ∂v ∂v ∂v ∂v i = 1 → a1 = 1 + v j 1 = 1 + v1 1 + v2 1 + v3 1 . ∂t ∂x j ∂t ∂x1 ∂x2 ∂x3

i = 2 → a2 =

∂v2 ∂v ∂v ∂v ∂v ∂v + v j 2 = 2 + v1 2 + v2 2 + v3 2 . ∂t ∂x j ∂t ∂x1 ∂x2 ∂x3

i = 3 → a3 =

∂v3 ∂v ∂v ∂v ∂v ∂v + v j 3 = 3 + v1 3 + v2 3 + v3 3 . ∂t ∂x j ∂t ∂x1 ∂x2 ∂x3

__________________________________________________________________ Copyright 2010, Elsevier Inc 2-2


2.8

Given that Tij = 2 μ Eij + λ Ekk δ ij , show that

(a) Tij Eij = 2μ Eij Eij + λ ( Ekk ) and (b) TijTij = 4μ 2 Eij Eij + ( Ekk ) (4μλ + 3λ 2 ) 2

2

------------------------------------------------------------------------------Ans. (a) Tij Eij = (2μ Eij + λ Ekk δ ij ) Eij = 2μ Eij Eij + λ Ekk δ ij Eij = 2 μ Eij Eij + λ Ekk Eii = 2μ Eij Eij + λ ( Ekk )2 (b) TijTij = (2μ Eij + λ Ekk δ ij )(2μ Eij + λ Ekk δ ij ) = 4μ 2 Eij Eij + 2μλ Eij Ekk δ ij + 2 μλ Ekk δ ij Eij +λ 2 ( Ekk ) δ ijδ ij = 4μ 2 Eij Eij + 2 μλ Eii Ekk + 2μλ Ekk Eii + λ 2 ( Ekk ) δ ii 2

2

= 4 μ 2 Eij Eij + ( Ekk ) (4μλ + 3λ 2 ). 2

__________________________________________________________________ ′ . Given that ai =Tij b j , and ai′ =Tij′b′j , where ai =Qim am′ and Tij =Qim Q jnTmn 2.9 ′ bn′ = Qim Q jnTmn ′ b j and (b) if Qik Qim =δ km , then Tkn ′ (bn′ − Q jn b j ) = 0 . (a) Show that QimTmn

------------------------------------------------------------------------------′ and Tij =Qim Q jnTmn ′ , therefore, ai =Tij b j → . Ans. (a) Since ai =Qim am ′ b j (1), Now, ai′ =Tij′b′j → am′ =Tmj ′ b′j = Tmn ′ bn′ , therefore, Eq. (1) becomes Qim am′ = Qim Q jnTmn Qi mTm′ n bn′ = Qi m Q j nTm′ n b j . (2)

(b) To remove Qim from Eq. (2), we make use of Qik Qim =δ km by multiplying the above equation, Eq.(2) with Qik . That is,

Qik QimTm′ nbn′ = Qik QimQ jnTm′ nb j → δ kmTm′ nbn′ = δ kmQ jnTm′ nb j → Tk′nbn′ = Q jnTk′nb j → Tk′n (bn′ − Q jnb j ) = 0 . __________________________________________________________________

2.10

⎡1 ⎤ ⎡0⎤ ⎢ ⎥ Given [ ai ] = ⎢ 2 ⎥ and [bi ] = ⎢⎢ 2 ⎥⎥ Evaluate [di ] , if d k = ε ijk ai b j and show that this result is ⎣⎢ 0 ⎦⎥ ⎣⎢ 3 ⎦⎥

the same as d k = ( a × b ) ⋅ e k . ------------------------------------------------------------------------------Ans. d k = ε ijk ai b j →

d1 = ε ij1ai b j = ε 231a2b3 + ε 321a3b2 = a2b3 − a3b2 = (2)(3) − (0)(2) = 6 d 2 = ε ij 2 ai b j = ε 312 a3b1 + ε132 a1b3 = a3b1 − a1b3 = (0)(0) − (1)(3) = −3 d3 = ε ij 3 ai b j = ε123 a1b2 + ε 213a2b1 = a1b2 − a2b1 = (1)(2) − (2)(0) = 2

( a × b ) = ( e1 + 2e2 ) × ( 2e2 + 3e3 ) = 6e1 − 3e2 + 2e3 . d1 = ( a × b ) ⋅ e1 = 6, d 2 = ( a × b ) ⋅ e 2 = −3, d3 = ( a × b ) ⋅ e3 = 2 .

Next,

__________________________________________________________________ 2.11 (a) If ε ijk Tij = 0 , show that Tij = T ji , and (b) show that δ ij ε ijk =0 -------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. (a) for k = 1, ε ij1Tij = 0 → ε 231T23 + ε 321T32 = 0 → T23 − T32 → T23 = T32 . for k = 2, ε ij 2Tij = 0 → ε 312T31 + ε132T13 = 0 → T31 − T13 → T31 = T13 . for k = 3, ε ij 3Tij = 0 → ε123T12 + ε 213T21 = 0 → T12 − T21 → T12 = T21 .

(b) δ ij ε ijk = δ11ε11k + δ 22ε 22k + δ 33ε 33k = (1)( 0 ) + (1)( 0 ) + (1)( 0 ) = 0 . __________________________________________________________________ 2.12 Verify the following equation: ε ijmε klm = δ ik δ jl − δ il δ jk . (Hint): there are 6 cases to be considered (i) i = j , (2) i = k , (3) i = l , (4) j = k , (5) j = l , and (6) k =l . ------------------------------------------------------------------------------Ans. There are 4 free indices in the equation. Therefore, there are the following 6 cases to consider: (i) i = j , (2) i = k , (3) i = l , (4) j = k , (5) j = l , and (6) k = l . We consider each case below where we use LS for left side, RS for right side and repeated indices with parenthesis are not sum: (1) For i = j , LS=ε (i )(i ) mε klm = 0, RS = δ (i ) k δ (i )l − δ (i )l δ (i ) k = 0.

(2) For i = k , LS=ε (i ) j1ε (i )l1 + ε (i ) j 2ε (i )l 2 + ε (i ) j 3ε (i )l 3 ,

RS = δ (i )(i )δ jl − δ (i )l δ j (i )

if j ≠ l ⎧0 ⎪ LS=RS = ⎨ 0 if j = l = i . ⎪ 1 if j = l ≠ i ⎩ (3) For i = l , LS=ε (i ) jmε k (i ) m ,

RS = δ (i ) k δ j (i ) − δ (i )(i )δ jk

if j ≠ k ⎧0 ⎪ LS=RS = ⎨ 0 if j = k = i ⎪ −1 if j = k ≠ i ⎩ (4) For j = k , LS=ε i ( j ) mε ( j )lm ,

RS = δ i ( j )δ ( j )l − δ il δ ( j )( j )

if i ≠ l ⎧0 ⎪ LS=RS = ⎨ 0 if i = l = j ⎪ −1 if i = l ≠ j ⎩ (5) For j = l , LS=ε i ( j ) mε k ( j ) m ,

RS = δ ik δ ( j )( j ) − δ i ( j )δ ( j ) k

if i ≠ k ⎧0 ⎪ LS=RS = ⎨0 if i = k = j ⎪ 1 if i = k ≠ j ⎩ (6) For k = l , LS=ε ijmε ( k )( k ) m =0,

RS = δ i ( k )δ j ( k ) − δ i ( k )δ j ( k ) = 0

__________________________________________________________________ 2.13 Use the identity ε ijmε klm = δ ik δ jl − δ il δ jk as a short cut to obtain the following results: (a) ε ilmε jlm = 2δ i j and (b) ε ijk ε ijk = 6 . ------------------------------------------------------------------------------Ans. (a) ε ilmε jlm = δ i jδ ll − δ il δ lj = 3δ i j − δ i j = 2δ i j . (b) ε ijk ε ijk = δ iiδ jj − δ i jδ ji = (3)(3) − δ ii = 9 − 3 = 6 . __________________________________________________________________ 2.14 Use the identity ε ijmε klm = δ ik δ jl − δ il δ jk to show that a × (b × c) = (a ⋅ c)b − (a ⋅ b)c .


Lai et al, Introduction to Continuum Mechanics ------------------------------------------------------------------------------Ans. a × (b × c) = am e m × (ε ijk b j ck ei ) = ε ijk am b j ck (e m × ei ) = ε ijk am b j ck (ε nmi e n ) = ε ijk ε nmi amb j ck e n = ε jkiε nmi amb j ck e n = (δ jnδ km − δ jmδ kn ) am b j ck e n = δ jnδ km am b j ck e n − δ jmδ kn am b j ck e n = ak bn ck e n − a j b j cne n = (a ⋅ c)b − (a ⋅ b)c .

__________________________________________________________________ 2.15 (a) Show that if Tij = −T ji , Tij ai a j = 0 and (b) if Tij = −T ji , and Sij = S ji , then Tij Sij = 0 ------------------------------------------------------------------------------Ans. Since Tij ai a j = T ji a j ai (switching the original dummy index i to j and the original index

j to i ), therefore Tij ai a j = T ji a j ai = −Tij a j ai = −Tij ai a j → 2Tij ai a j = 0 → Tij ai a j = 0 . (b) Tij Sij = T ji S ji (switching the original dummy index i to j and the original index j to i ), therefore, Tij Sij = T ji S ji = −Tij S ji = −Tij Sij → 2Tij Sij = 0 → Tij Sij = 0 . __________________________________________________________________

(

)

(

)

Let Tij = Sij + S ji /2 and Rij = Sij − S ji / 2 , show that Tij = T ji , Rij = − R ji ,

2.16

and Sij = Tij + R ij . -------------------------------------------------------------------------------

(

)

(

)

Ans. Tij = Sij + S ji / 2 → T ji = S ji + Sij / 2 = Tij .

(

)

(

)

(

)

Rij = Sij − S ji / 2 → R ji = S ji − Sij / 2 = − Sij − S ji / 2 = − Rij .

(

) (

)

Tij +Rij = Sij + S ji /2+ Sij − S ji / 2 = Sij . __________________________________________________________________ 2.17 Let f ( x1 , x2 , x3 ) be a function of x1 , x2 ,and x3 and vi ( x1 , x2 , x3 ) be three functions of x1 , x2 ,and x3 . Express the total differential df and dvi in indicial notation. ------------------------------------------------------------------------------∂f ∂f ∂f ∂f dx1 + dx2 + dx3 = dxi . Ans. df = ∂x1 ∂x2 ∂x3 ∂xi ∂v ∂v ∂v ∂v dvi = i dx1 + i dx2 + i dx3 = i dxm . ∂x1 ∂x2 ∂x3 ∂xm __________________________________________________________________ 2.18 Let Aij denote that determinant of the matrix ⎡⎣ Aij ⎤⎦ . Show that Aij = ε ijk Ai1 A j 2 Ak 3 ------------------------------------------------------------------------------Ans. ε ijk Ai1 A j 2 Ak 3 = ε1 jk A11 A j 2 Ak 3 + ε 2 jk A21 A j 2 Ak 3 + ε 3 jk A31 A j 2 Ak 3

= ε123 A11 A22 A33 + ε132 A11 A32 A23 + ε 231 A21 A32 A13 + ε 213 A21 A12 A33 + ε 312 A31 A12 A23 + ε 321 A31 A22 A13 = A11 A22 A33 − A11 A32 A23 + A21 A32 A13 − A21 A12 A33 + A31 A12 A23 − A31 A22 A13 A11 A12 A13 = A21 A22 A23 A31 A32 A33 __________________________________________________________________



CHAPTER 2, PART B 2.19 A transformation T operate on any vector a to give Ta = a / a , where a is the magnitude of a . Show that T is not a linear transformation. ------------------------------------------------------------------------------a a+b a b Ans. Since Ta = for any a , therefore T(a + b) = . Now Ta + Tb = + a a+b a b therefore T(a + b) ≠ Ta + Tb and T is not a linear transformation. _________________________________________________________________ 2.20 (a) A tensor T transforms every vector a into a vector Ta = m × a where m is a specified vector. Show that T is a linear transformation and (b) If m = e1 + e2 , find the matrix of the tensor T . ------------------------------------------------------------------------------Ans. (a) T(α a + β b) = m × (α a + β b) = m × α a + m × β b = α m × a + β m × b = α Ta + β Tb. Thus, the given T is a linear transformation. (b) Te1 = m × e1 = (e1 + e2 ) × e1 = −e3 , Te2 = m × e2 = (e1 + e2 ) × e2 = e3 , Te3 = m × e3 = (e1 + e2 ) × e3 = −e2 + e1 . Thus,

⎡0 0 1⎤ [T] = ⎢⎢ 0 0 −1⎥⎥ . ⎢⎣ −1 1 0 ⎥⎦ _________________________________________________________________ 2.21 A tensor T transforms the base vectors e1 and e2 such that Te1 = e1 + e2 and Te2 = e1 − e2 . If a = 2e1 + 3e2 and b = 3e1 + 2e2 , use the linear property of T to find (a) Ta ,(b) Tb , and (c) T(a + b) . -----------------------------------------------------------------------------Ans. (a)Ta = T(2e1 + 3e2 ) = 2Te1 + 3Te2 = 2 ( e1 + e2 ) + 3 ( e1 − e2 ) = 5e1 − e2 . (b)Tb = T(3e1 + 2e2 ) = 3Te1 + 2Te2 =3 ( e1 + e2 ) + 2 ( e1 − e2 ) =5e1 + e2 .

(c)T(a + b) = Ta + Tb = ( 5e1 − e2 ) + ( 5e1 + e2 ) = 10e1 . _________________________________________________________________ 2.22 Obtain the matrix for the tensor T which transforms the base vectors as follows: Te1 = 2e1 + e3 , Te2 = e2 + 3e3 , Te3 = −e1 + 3e2 . -----------------------------------------------------------------------------⎡ 2 0 −1⎤ Ans. [ T] = ⎢⎢ 0 1 3 ⎥⎥ . ⎢⎣1 3 0 ⎥⎦ _________________________________________________________________ 2.23 Find the matrix of the tensor T which transforms any vector a into a vector b = m(a ⋅ n) where m =

(

)

2 / 2 ( e1 + e 2 ) and n =

(

)

2 / 2 ( −e1 + e3 ) .

-----------------------------------------------------------------------------Ans. Te1 = m ( e1 ⋅ n ) = n1m = − 2 / 2 ⎡ 2 / 2 ( e1 + e 2 ) ⎤ = − ( e1 + e2 ) / 2 . ⎣ ⎦

(

)(

)


Lai et al, Introduction to Continuum Mechanics Te2 = m ( e 2 ⋅ n ) = n2m = 0m = 0 . Te3 = m ( e3 ⋅ n ) = n3m =

(

)(

)

2 / 2 ⎡ 2 / 2 ( e1 + e 2 ) ⎤ = ( e1 + e2 ) / 2 . ⎣ ⎦ ⎡ −1 / 2 0 1 / 2 ⎤ Thus, [ T] = ⎢⎢ −1 / 2 0 1 / 2 ⎥⎥ . ⎢⎣ 0 0 0 ⎥⎦ _________________________________________________________________ 2.24 (a) A tensor T transforms every vector into its mirror image with respect to the plane whose normal is e 2 . Find the matrix of T . (b) Do part (a) if the plane has a normal in the e3 direction. -----------------------------------------------------------------------------⎡1 0 0 ⎤ Ans. (a) Te1 = e1 , Te 2 = −e 2 , Te3 = e3 , thus, [ T] = ⎢⎢0 −1 0 ⎥⎥ . ⎢⎣0 0 1 ⎥⎦ ⎡1 0 0 ⎤ (b) Te1 = e1 , Te 2 = e 2 , Te3 = −e3 , thus, [ T] = ⎢⎢0 1 0 ⎥⎥ . ⎢⎣0 0 −1⎥⎦ _________________________________________________________________ 2.25 (a) Let R correspond to a right-hand rotation of angle θ about the x1 -axis. Find the matrix of R . (b) do part (a) if the rotation is about the x2 -axis. The coordinates are right-handed. -----------------------------------------------------------------------------Ans.(a) Re1 = e1 , Re 2 = 0e1 + cosθ e 2 + sinθ e3 , Re3 = 0e1 − sinθ e 2 + cosθ e3 . Thus, 0 0 ⎤ ⎡1 ⎢ [ R ] = ⎢0 cosθ − sin θ ⎥⎥ . ⎢⎣0 sin θ cos θ ⎥⎦ (b) Re1 = −sinθ e3 + cosθ e1 , Re2 = e2 , Re3 = cosθ e3 + sinθ e1 . Thus, ⎡ cos θ 0 sin θ ⎤ [ R ] = ⎢⎢ 0 1 0 ⎥⎥ . ⎢⎣ − sin θ 0 cos θ ⎥⎦ _________________________________________________________________ 2.26 Consider a plane of reflection which passes through the origin. Let n be a unit normal vector to the plane and let r be the position vector for a point in space. (a) Show that the reflected vector for r is given by Tr = r − 2(r ⋅ n)n , where T is the transformation that corresponds to the reflection. (b) Let n = (e1 + e 2 + e3 ) / 3 , find the matrix of T . (c) Use this linear transformation to find the mirror image of the vector a = e1 + 2e 2 + 3e3 . -----------------------------------------------------------------------------Ans. (a) Let the vector r be decomposed into two vectors rn and rt , where rn is in the direction of n and rt is in a direction perpendicular to n . That is, rn is normal to the plane of reflection and rt is on the plane of reflection and r = rt + rn . In the reflection given by T , we have, Trn = −rn and Trt = rt , so that Tr = Trt + Trn = rt − rn = ( r − rn ) − rn = r − 2rn = r − 2(r ⋅ n)n .

(b) n = (e1 + e 2 + e3 ) / 3 → e1 ⋅ n = e2 ⋅ n = e3 ⋅ n = 1 / 3 . Copyright 2010, Elsevier Inc 2-7


(

)

Te1 = e1 − 2(e1 ⋅ n)n = e1 − 2 1 / 3 ⎡⎣ (e1 + e 2 + e3 ) / Te 2 = e 2 − 2(e 2 ⋅ n)n = e 2 − 2 1 / 3 ⎡⎣ (e1 + e 2 + e3 ) / Te3 = e3 − 2(e3 ⋅ n)n = e3 − 2 1 / 3 ⎡⎣(e1 + e 2 + e3 ) / ⎡ 1 −2 −2 ⎤ 1⎢ [T] = ⎢ −2 1 −2⎥⎥ . 3 ⎢⎣ −2 −2 1 ⎥⎦

( (

) )

3 ⎤⎦ = ( e1 − 2e2 − 2e3 ) / 3 . 3 ⎤⎦ = ( −2e1 + e 2 − 2e3 ) / 3 .

3 ⎤⎦ = ( −2e1 − 2e 2 + e3 ) / 3 .

⎡ 1 −2 −2 ⎤ ⎡ 1 ⎤ ⎡ −3⎤ 1⎢ (c) [ T][a ] = ⎢ −2 1 −2 ⎥⎥ ⎢⎢ 2 ⎥⎥ = ⎢⎢ −2 ⎥⎥ → Ta = − ( 3e1 + 2e 2 + e3 ) . 3 ⎢⎣ −2 −2 1 ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣ −1⎥⎦ _________________________________________________________________ 2.27 Knowing that the reflected vector for r is given by Tr = r − 2(r ⋅ n)n (see the previous problem), where T is the transformation that corresponds to the reflection and n is the normal to the mirror, show that in dyadic notation, the reflection tensor is given by T = I − 2nn and find the matrix of T if the normal of the mirror is given by n = (e1 + e 2 + e3 ) / 3 , -----------------------------------------------------------------------------Ans. From the definition of dyadic product, we have , Tr = r − 2(r ⋅ n)n = r − 2(nn)r = (Ir − 2(nn)r ) = (I − 2nn)r → T = I − 2nn . ⎡1⎤ ⎡1 1 1⎤ 2⎢ ⎥ 2⎢ For n = (e1 + e 2 + e3 ) / 3 → [2nn] = ⎢1⎥ [1 1 1] = ⎢1 1 1⎥⎥ . 3 3 ⎢⎣1⎥⎦ ⎢⎣1 1 1⎥⎦ ⎡ 1 −2 −2 ⎤ 1⎢ → [T] = [I ] − [2nn] = ⎢ −2 1 −2 ⎥⎥ . 3 ⎢⎣ −2 −2 1 ⎥⎦ _________________________________________________________________ 2.28 A rotation tensor R is defined by the relation Re1 = e 2 , Re 2 = e3 , Re3 = e1 (a) Find the matrix of R and verify that R T R = I and det R = 1 and (b) find a unit vector in the direction of the axis of rotation that could have been used to effect this particular rotation. -----------------------------------------------------------------------------0 0 1 ⎡0 0 1 ⎤ ⎡ 0 1 0 ⎤ ⎡0 0 1 ⎤ ⎡1 0 0 ⎤ T ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ Ans. (a) [ R ] = ⎢1 0 0 ⎥ → [ R ] [ R ] = ⎢ 0 0 1 ⎥ ⎢1 0 0 ⎥ = ⎢ 0 1 0 ⎥ , det [ R ] = 1 0 0 = 1 . ⎢⎣0 1 0 ⎥⎦ ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ 0 1 0 (b) Let the axis of rotation be n = α1e1 + α 2e2 + α 3e3 , then ⎡ −1 0 1 ⎤ ⎡ α1 ⎤ ⎡0 ⎤ Rn = n → [ R − I ][n ] = [ 0] → ⎢⎢ 1 −1 0 ⎥⎥ ⎢⎢α 2 ⎥⎥ = ⎢⎢0 ⎥⎥ → −α1 + α 3 = 0, α1 − α 2 = 0, α 2 − α 3 =0 . ⎢⎣ 0 1 −1⎥⎦ ⎢⎣α 3 ⎥⎦ ⎢⎣0 ⎥⎦ Thus, α1 = α 2 = α 3 , so that a unit vector in the direction of the axis of rotation is n = (e1 + e 2 + e3 ) / 3 . _________________________________________________________________ Copyright 2010, Elsevier Inc 2-8

Lai et al, Introduction to Continuum Mechanics 2.29 A rigid body undergoes a right hand rotation of angle θ about an axis which is in the direction of the unit vector m . Let the origin of the coordinates be on the axis of rotation and r be the position vector for a typical point in the body. (a) show that the rotated vector of r is given by: Rr = (1 − cos θ )( m ⋅ r ) m + cos θ r + sin θ ( m × r ) , where R is the rotation tensor. (b) Let m = (e1 + e 2 + e3 ) / 3 , find the matrix for R . -----------------------------------------------------------------------------Ans. (a) Let the vector r be decomposed into two vectors rm and r p , where rm is in the direction of m and r p is in a direction perpendicular to m , that is, r = r p + rm . Let p ≡ rp / rp be the unit vector in the direction of r p , and let q ≡ m × p . Then, ( m , p, q) forms an orthonormal set of vectors which rotates an angle of θ about the unit vector m . Thus, Rrm = rm and Rrp = r p ( cos θ p + sin θ q ) . From r = rp + rm , we have,

{

}

Rr = Rrp + Rrm = r p ( cosθ p + sin θ q ) + rm = cosθ r p p + sin θ rp ( m × p ) + rm

{

(

= cos θ r p + sin θ m × rp

)} + rm = {cosθ ( r − rm ) + sin θ ( m × ( r − rm ) )} + rm

= r cos θ + rm (1 − cos θ ) + sin θ m × ( r − rm ) = r cosθ + rm (1 − cosθ ) + sin θ m × r We note that rm = (r ⋅ m)m , so that Rr = r cos θ + (r ⋅ m)m (1 − cos θ ) + sin θ m × r .

(b) Use the result of (a), that is, Rr = r cos θ + (r ⋅ m) (1 − cos θ ) + sin θ m × r , we have, Re1 = e1 cos θ + (e1 ⋅ m)m (1 − cos θ ) + sin θ m × e1 ,

Re2 = e 2 cos θ + (e 2 ⋅ m)m (1 − cos θ ) + sin θ m × e 2 ,

Re3 = e3 cos θ + (e3 ⋅ m )m (1 − cos θ ) + sin θ m × e3 .

Now, m = (e1 + e 2 + e3 ) / 3 , therefore, m ⋅ e1 = m ⋅ e 2 = m ⋅ e3 = 1 / 3

(

)

(

)

(

)

m × e1 = 1 / 3 (−e3 + e 2 ), m × e 2 = 1 / 3 (e3 − e1 ), m × e3 = 1 / 3 ( −e2 + e1 ) . Thus,

Re1 = e1 cos θ + (e1 ⋅ m)m (1 − cosθ ) + sin θ m × e1

(

)

= e1 cosθ + (1 / 3) (e1 + e2 + e3 ) (1 − cos θ ) + sin θ 1 / 3 (−e3 + e2 )

{

(

)}

{

(

= (1 / 3){1 + 2cos θ } e1 + e 2 (1 / 3)(1 − cosθ ) + sin θ 1 / 3 + e3 (1 / 3)(1 − cosθ ) − sin θ 1 / 3 Re2 = e 2 cos θ + (e 2 ⋅ m)m (1 − cosθ ) + sin θ m × e 2

(

)

)}

= e 2 cosθ + (1 / 3) (e1 + e 2 + e3 ) (1 − cosθ ) + sin θ 1 / 3 (e3 - e1 )

{

(

)

}

{

(

)}

= (1 / 3)(1 − cos θ ) − 1 / 3 sin θ e1 + (1 / 3)(1 + 2cosθ ) e 2 + (1 / 3)(1 − cosθ ) + sin θ 1 / 3 e3

Re3 = e3 cos θ + (e3 ⋅ m)m (1 − cosθ ) + sin θ m × e3

(

)

= e3 cosθ + (1 / 3) (e1 + e2 + e3 ) (1 − cosθ ) + sin θ 1 / 3 (-e2 + e1 )

{

(

)

} {

(

)}

= (1 / 3)(1 − cos θ ) + 1 / 3 sin θ e1 + (1 / 3)(1 − cosθ ) − sin θ 1 / 3 e2 + (1 / 3)(1 + 2cosθ ) e3 Thus,


Lai et al, Introduction to Continuum Mechanics ⎡ 1 + 2cos θ (1 − cosθ ) − 3 sin θ (1 − cosθ ) + 3 sin θ ⎤ ⎥ 1⎢ (1 + 2cosθ ) (1 − cosθ ) − 3 sin θ ⎥ . [T] = ⎢(1 − cosθ ) + 3 sin θ 3⎢ ⎥ (1 + 2cosθ ) ⎥⎦ ⎢⎣(1 − cos θ ) − 3 sin θ (1 − cosθ ) + 3 sin θ _________________________________________________________________ 2.30 For the rotation about an arbitrary axis m by an angle θ , (a) show that the rotation tensor is given by R = (1 − cosθ )(mm ) + cosθ I + sin θ E , where mm denotes that dyadic product of m and E is the antisymmetric tensor whose dual vector (or axial vector) is m , (b) find the R A , the antisymmetric part of R and (c) show that the dual vector for R A is given by (sin θ )m . Hint, Rr = (1 − cos θ )( m ⋅ r ) m + cosθ r + sin θ ( m × r ) (see previous problem). -----------------------------------------------------------------------------Ans. (a) We have, from the previous problem, Rr = (1 − cos θ )( m ⋅ r ) m + cosθ r + sin θ ( m × r ) . Now, by the definition of dyadic product, we have ( m ⋅ r ) m = (mm)r , and by the definition of dual vector we have, m × r = Er , thus Rr = (1 − cos θ ) (mm)r + cos θ r + sin θ Er

= {(1 − cos θ ) (mm) + cos θ I + sin θ E} r , from which, R = (1 − cos θ ) (mm) + cos θ I + sin θ E .

(b) R A = ( R − R T ) / 2 →

{

}

2R A = {(1 − cosθ ) (mm) + cosθ I + sin θ E} − (1 − cosθ ) (mm)T + cos θ I + sin θ ET . Now

[mm ] = ⎡⎣ mi m j ⎤⎦ = ⎡⎣ m j mi ⎤⎦ = [mm ]

T

, and the tensor E , being antisymmetric, E = −ET , therefore,

2R A = 2sin θ E , that is, R A = sin θ E .

(c) dual vector of R A = (sin θ )(dual vector of E) = sin θ m . _________________________________________________________________ 2.31 (a) Given a mirror whose normal is in the direction of e 2 . Find the matrix of the tensor S which first transforms every vector into its mirror image and then transforms them by a 45o righthand rotation about the e1 -axis. (b) Find the matrix of the tensor T which first transforms every vector by a 45o right-hand rotation about the e1 -axis, and then transforms them by a reflection with respect to the mirror (whose normal is e2 ). (c) Consider the vector a = (e1 + 2e2 + 3e3 ) , find the transformed vector by using the transformation S . (d) For the same vector a = (e1 + 2e2 + 3e3 ) , find the transformed vector by using the transformation T . -----------------------------------------------------------------------------Ans. Let T1 and T2 correspond to the reflection and the rotation respectively. We have ⎡1 0 0 ⎤ T1e1 = e1 , T1e2 = −e2 , T1e3 = e3 → [ T1 ] = ⎢⎢0 −1 0 ⎥⎥ . ⎢⎣0 0 1 ⎥⎦

0 ⎡1 ⎢ T2e1 = e1 , T2e 2 = ( e2 + e3 ) , T2e3 = ( −e2 + e3 ) → [T2 ] = ⎢0 1 / 2 2 2 ⎢0 1 / 2 ⎣ 1

1


⎤ ⎥ −1 / 2 ⎥ . 1 / 2 ⎥⎦ 0

Lai et al, Introduction to Continuum Mechanics 0 ⎡1 ⎢ (a) [S ] = [ T2 ][ T1 ] = ⎢0 1 / 2 ⎢0 1 / 2 ⎣

0 0 ⎤ ⎤ ⎡1 0 0 ⎤ ⎡1 ⎥⎢ ⎢ ⎥ ⎥ −1 / 2 ⎥ ⎢0 −1 0 ⎥ = ⎢0 −1 / 2 −1 / 2 ⎥ . 1 / 2 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 −1 / 2 1 / 2 ⎥⎦ 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) [ T] = [ T1 ][ T2 ] = ⎢0 −1 0 ⎥ ⎢0 1 / 2 −1 / 2 ⎥ = ⎢ 0 −1 / 2 1 / 2 ⎥ . ⎢⎣0 0 1 ⎥⎦ ⎢0 1 / 2 1 / 2 ⎥ ⎢ 0 1 / 2 1 / 2 ⎥ ⎣ ⎦ ⎣ ⎦ 0 0 ⎤ ⎡1 ⎤ ⎡ 1 ⎤ ⎡1 ⎢ ⎥ ⎢ ⎥ (c) [b ] = [S ][a ] = ⎢0 −1 / 2 −1 / 2 ⎥ ⎢⎢ 2 ⎥⎥ = ⎢ −5 / 2 ⎥ . ⎢0 −1 / 2 1 / 2 ⎥ ⎢ 3 ⎥ ⎢ 1 / 2 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ 0 0 ⎤ ⎡1 ⎤ ⎡ 1 ⎤ ⎡1 ⎢ ⎥ ⎢ ⎥ (d) [c ] = [ T ][a ] = ⎢0 −1 / 2 1 / 2 ⎥ ⎢⎢ 2 ⎥⎥ = ⎢1 / 2 ⎥ . ⎢ 0 1 / 2 1 / 2 ⎥ ⎢ 3 ⎥ ⎢5 / 2 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ _________________________________________________________________ 2.32 Let R correspond to a right-hand rotation of angle θ about the x3 -axis (a) find the matrix 0

of R 2 . (b) Show that R 2 corresponds to a rotation of angle 2θ about the same axis (c) Find the matrix of R n for any integer n . ------------------------------------------------------------------------------⎡cos θ − sin θ 0 ⎤ Ans. (a) [ R ] = ⎢⎢ sin θ cosθ 0 ⎥⎥ . ⎢⎣ 0 0 1 ⎥⎦ ⎡cosθ 2 → ⎡ R ⎤ = ⎢⎢ sin θ ⎣ ⎦ ⎢⎣ 0

− sin θ cos θ 0

0 ⎤ ⎡cosθ 0 ⎥⎥ ⎢⎢ sin θ 1 ⎥⎦ ⎢⎣ 0

− sin θ cosθ 0

2 2 0 ⎤ ⎡cos θ − sin θ ⎢ 0⎥⎥ = ⎢ 2sin θ cosθ ⎢ 1 ⎥⎦ ⎢ 0 ⎣

−2sin θ cosθ cos 2 θ − sin 2 θ 0

0⎤ ⎥ 0⎥ . ⎥ 1⎥ ⎦

(b) ⎡ cos 2 θ − sin 2 θ ⎢ ⎡ R 2 ⎤ = ⎢ 2sin θ cosθ ⎣ ⎦ ⎢ 0 ⎢⎣

−2sin θ cosθ 2

2

cos θ − sin θ 0

0 ⎤ ⎡ cos 2θ ⎥ 0 ⎥ = ⎢⎢ sin 2θ ⎥ 1 ⎥ ⎢⎣ 0 ⎦

− sin 2θ cos 2θ 0

0⎤ 0 ⎥⎥ . 1 ⎥⎦

Thus, R 2 corresponds to a rotation of angle 2θ about the same axis ⎡cos nθ − sin nθ 0 ⎤ n (c) ⎡ R ⎤ = ⎢⎢ sin nθ cos nθ 0 ⎥⎥ . ⎣ ⎦ ⎢⎣ 0 0 1 ⎥⎦ _________________________________________________________________ 2.33 Rigid body rotations that are small can be described by an orthogonal transformation R = I + ε R* where ε → 0 as the rotation angle approaches zero. Consider two successive small rotations R1 and R 2 , show that the final result does not depend on the order of rotations. ------------------------------------------------------------------------------



(

)(

)

(

)

Ans. R 2 R1 = I + ε R*2 I + ε R1* = I + ε R*2 + ε R1* + ε 2 R*2 R1* = I + ε R*2 + R1* + ε 2 R*2 R1* .

(

)

As ε → 0 , R 2 R1 ≈ I + ε R*2 + R1* = R1R 2 . _________________________________________________________________ 2.34 Let T and S be any two tensors. Show that (a) TT is a tensor, (b) TT + S T = (T + S)T and (c) (TS)T = S T TT . ------------------------------------------------------------------------------Ans. Let a , b , c be three arbitrary vectors and α β be any two scalars, then (a) a ⋅ TT (α b + β c) = (α b + β c) ⋅ Ta = α b ⋅ Ta + β c ⋅ Ta = α a ⋅ TT b + β a ⋅ TT c

(

)

(

)

= a ⋅ α TT b + β TT c → TT (α b + β c) = α TT b + β TT c . Thus, TT is a linear transformation, i.e.,

tensor. (b) a ⋅ (T + S)T b = b ⋅ (T + S)a = b ⋅ Ta + b ⋅ Sa = a ⋅ TT b + a ⋅ S T b = a ⋅ (TT + S T )b → (T + S)T = TT + S T .

(c) a ⋅ (TS)T b = b ⋅ (TS)a = b ⋅ T(Sa) = (Sa) ⋅ TT b = a ⋅ S T TT b → (TS)T = S T TT . _________________________________________________________________ 2.35 For arbitrary tensors T and S , without relying on the component form, prove that (a) (T−1 )T = (TT )−1 and (b) (TS)−1 = S −1T−1 -------------------------------------------------------------------------------

Ans. (a) TT−1 = I → (TT−1 )T = I → (T−1 )T TT = I → (T−1 )T = (TT )−1 . (b) (TS)(S −1T−1 ) = T(SS −1 )T−1 = TT−1 = I , thus, (TS)−1 = S −1T−1 . _________________________________________________________________ 2.36 Let {ei } and {e′i } be two Rectangular Cartesian base vectors. (a) Show that if e′i = Qmi em , then ei = Qim e′m and (b) verify Qmi Qmj = δ ij = Qim Q jm . ------------------------------------------------------------------------------Ans. (a) e′i = Qmi e m → e′i ⋅ e j = Qmi e m ⋅ e j = Qmiδ mj = Q ji → e j = Q jm e′m → ei = Qim e′m . (b) We have, e′i ⋅ e′j = δ ij = ei ⋅ e j , thus,

δ ij = e′i ⋅ e′j = Qmi e m ⋅ Qnj e n = Qmi Qnj em ⋅ en = Qmi Qnjδ mn = Qmi Qmj . And δ ij = ei ⋅ e j = Qimem ⋅ Q jn en = QimQ jn em ⋅ en = QimQ jnδ mn = QimQ jm . _________________________________________________________________ 2.37 The basis {e′i } is obtained by a 30o counterclockwise rotation of the {ei } basis about the e3

axis. (a) Find the transformation matrix [Q ] relating the two sets of basis, (b) by using the vector

transformation law, find the components of a = 3e1 + e 2 in the primed basis, i.e., find ai′ and (c) do part (b) geometrically. -----------------------------------------------------------------------------Ans. (a) e1′ = cos30o e1 + sin 30o e2 , e′2 = − sin 30o e1 + cos30o e2 , e′3 = e3 . Thus,

[Q ]ei

⎡ cos30o ⎢ = ⎢ sin 30o ⎢ ⎢⎣ 0

− sin 30o

cos30o 0

0⎤ ⎥ 0⎥ . ⎥ 1⎥ ⎦ Copyright 2010, Elsevier Inc

2-12

Lai et al, Introduction to Continuum Mechanics ⎡ a1' ⎤ ⎡ 3 / 2 ⎢ ⎥ ⎢ → ⎢ a2' ⎥ = ⎢ −1 / 2 ⎢ ⎥ ⎢ ⎢⎣ a3' ⎥⎦ ⎢⎣ 0

0⎤ ⎡ 3 ⎤ ⎡ 2⎤ ⎥⎢ ⎥ (b) [a ]e' = [Q ] [a ]e 3 / 2 0 ⎥ ⎢ 1 ⎥ = ⎢⎢ 0 ⎥⎥ → a = 2e1′ i i 0 1 ⎥⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎦ (c) Clearly a = 3e1 + e 2 is a vector in the same direction as e1′ and has a length of 2. See figure below T

1/ 2

_________________________________________________________________ 2.38 Do the previous problem with the {e′i } basis obtained by a 30o clockwise rotation of the

{ei } basis about the e3 axis.

------------------------------------------------------------------------------Ans. (a) e1′ = cos30o e1 − sin 30o e 2 , e′2 = sin 30o e1 + cos30o e 2 , e′3 = e3 . Thus, ⎡ cos30o ⎢ = ⎢ − sin 30o ⎢ 0 ⎢⎣

0⎤ ⎥ cos30o 0 ⎥ . [Q ]ei ⎥ 0 1⎥ ⎦ ⎡ a1' ⎤ ⎡ 3 / 2 ⎢ ⎥ ⎢ T (b) [a ]e' = [Q ] [a ]e → ⎢ a2' ⎥ = ⎢ 1 / 2 i i ⎢ ⎥ ⎢ ⎢⎣ a3' ⎥⎦ ⎢⎣ 0 (c) See figure below sin 30o

0⎤ ⎡ 3 ⎤ ⎡ 1 ⎤ ⎥⎢ ⎥ ⎢ ⎥ 3 / 2 0 ⎥ ⎢ 1 ⎥ = ⎢ 3 ⎥ → a = e1′ + 3e′2 0 1 ⎥⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎦

−1 / 2

_________________________________________________________________ 2.39 The matrix of a tensor T with respect to the basis {ei } is Copyright 2010, Elsevier Inc 2-13

Lai et al, Introduction to Continuum Mechanics ⎡ 1 5 −5⎤ [T] = ⎢⎢ 5 0 0 ⎥⎥ ⎢⎣ −5 0 1 ⎥⎦ ′ with respect to a right-handed basis {e′i } where e1′ is in the direction of Find T11′ , T12′ and T31 −e 2 + 2e3 and e′2 is in the direction of e1 . -----------------------------------------------------------------------------Ans. The basis {e′i } is given by: e1′ = (−e 2 + 2e3 ) / 5, e′2 = e1 , e′3 = e1′ × e′2 = (2e2 + e3 ) / 5 . ⎡ 1 5 −5⎤ ⎡ 0 ⎤ ⎢ ⎥ T11′ = e1′ ⋅ Te1′ = ⎡⎣0 −1 / 5 2 / 5 ⎤⎦ ⎢⎢ 5 0 0 ⎥⎥ ⎢ −1 / 5 ⎥ = 4 / 5 . ⎢⎣ −5 0 1 ⎥⎦ ⎢ 2 / 5 ⎥ ⎣ ⎦ ⎡ 1 5 −5⎤ ⎡1 ⎤ T12′ = e1′ ⋅ Te′2 = ⎡⎣ 0 −1 / 5 2 / 5 ⎤⎦ ⎢⎢ 5 0 0 ⎥⎥ ⎢⎢0 ⎥⎥ = −15 / 5 . ⎢⎣ −5 0 1 ⎥⎦ ⎢⎣0 ⎥⎦ ⎡ 1 5 −5⎤ ⎡ 0 ⎤ ⎢ ⎥ ′ = e′3 ⋅ Te1′ = ⎡0 2 / 5 1 / 5 ⎤ ⎢ 5 0 0 ⎥ ⎢ −1 / 5 ⎥ = 2 / 5 . T31 ⎣ ⎦⎢ ⎥ ⎢⎣ −5 0 1 ⎥⎦ ⎢ 2 / 5 ⎥ ⎣ ⎦ _________________________________________________________________ 2.40 (a) For the tensor of the previous problem, find ⎡⎣Tij′ ⎤⎦ , i.e., [ T]e' if {e′i } is obtained by a i

o

90 right hand rotation about the e3 axis and (b) obtain Tii′ and the determinant Tij′ and compare them with Tii and Tij . -----------------------------------------------------------------------------⎡ 0 −1 0 ⎤ Ans. (a) e1′ = e 2 , e′2 = −e1 , e′3 = e3 → [Q ] = ⎢⎢1 0 0 ⎥⎥ . ⎢⎣ 0 0 1 ⎥⎦ ⎡ 0 1 0 ⎤ ⎡ 1 5 −5⎤ ⎡0 −1 0 ⎤ ⎡ 0 −5 0 ⎤ T ′ ⎡⎣Tij′ ⎤⎦ = [ T] = [Q ] [ T][Q ] = ⎢ −1 0 0 ⎥ ⎢ 5 0 0 ⎥ ⎢1 0 0 ⎥ = ⎢ −5 1 5 ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ −5 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0 5 1 ⎥⎦ ′ + T33 ′ = 0 + 1 + 1 = 2, Tij′ = −25 . (b) Tii′ = T11′ + T22

Tii = T11 + T22 + T33 = 1 + 0 + 1 = 2,

Tij = −25 .

_________________________________________________________________ 2.41 The dot product of two vectors a = ai ei and b = bi ei is equal to ai bi . Show that the dot product is a scalar invariant with respect to orthogonal transformations of coordinates. ------------------------------------------------------------------------------Ans. From ai′ = Qmi am and bi' = Qmi bm , we have,

ai′bi' = Qmi am Qni bn = Qmi Qni am bn = δ mn am bn = am bm = ai bi . __________________________________________________________________ Copyright 2010, Elsevier Inc 2-14

Lai et al, Introduction to Continuum Mechanics 2.42

If Tij are the components of a tensor (a) show that TijTij is a scalar invariant with respect to

orthogonal transformations of coordinates, (b) evaluate TijTij with respect to the basis {ei } for ⎡1 0 0 ⎤ ⎡0 0 1 ⎤ ⎢ ⎥ ′ [T] = ⎢1 2 5⎥ , (c) find [T] , if e′i = Qei , where [Q ] = ⎢⎢1 0 0⎥⎥ and ⎢⎣ 0 1 0 ⎥⎦ e ⎢⎣1 2 3⎥⎦ e i i

(d) verify for the above [ T] and [ T]′ that Tij′Tij′ = TijTij . -----------------------------------------------------------------------------Ans. (a) Since Tij are the components of a tensor, Tij′ = Qmi QnjTmn . Thus,

Tij′Tij′ = Qmi Qnj Tmn (Q pi QqjT pq ) = (Qmi Q pi )(Qnj Qqj )TmnT pq = δ mpδ nqTmnTpq = TmnTmn 2 2 2 + T22 + T23 + T312 + T322 + T332 = 1 + 1 + 4 + 25 + 1 + 4 + 9 = 45 . (b) TijTij = T112 + T122 + T132 + T21

⎡ 0 1 0 ⎤ ⎡1 0 0 ⎤ ⎡0 0 1 ⎤ ⎡ 0 1 0⎤ ⎡ 0 0 1⎤ ⎡ 2 5 1⎤ (c) [ T]′ = [Q ] [ T][Q ] = ⎢⎢ 0 0 1 ⎥⎥ ⎢⎢1 2 5 ⎥⎥ ⎢⎢1 0 0 ⎥⎥ = ⎢⎢ 0 0 1 ⎥⎥ ⎢⎢ 2 5 1⎥⎥ = ⎢⎢ 2 3 1⎥⎥ ⎢⎣1 0 0 ⎥⎦ ⎢⎣1 2 3⎥⎦ ⎢⎣0 1 0 ⎥⎦ ⎢⎣1 0 0⎥⎦ ⎢⎣ 2 3 1⎥⎦ ⎢⎣ 0 0 1⎥⎦ (d) Tij′Tij′ = 4 + 25 + 1 + 4 + 9 + 1 + 1 = 45 . T

_________________________________________________________________ 2.43 Let [ T] and [ T]′ be two matrices of the same tensor T , show that det [ T] =det [ T]′ . -----------------------------------------------------------------------------T T Ans. ⎡T ⎤′ = ⎡Q ⎤ ⎡T ⎤ ⎡Q ⎤ → det ⎡T ⎤′ = det ⎡Q ⎤ det ⎡Q ⎤ det ⎡ T⎤ = (±1)(±1) det ⎡ T⎤ = det ⎡ T⎤ . ⎣ ⎦

⎣ ⎦ ⎣ ⎦⎣ ⎦

⎣ ⎦

⎣ ⎦

⎣ ⎦

⎣ ⎦

⎣ ⎦

⎣ ⎦

_________________________________________________________________ 2.44 (a) If the components of a third order tensor are Rijk , show that Riik are components of a vector, (b) if the components of a fourth order tensor are Rijkl , show that Riikl are components of a second order tensor and (c) what are components of Riik ... , if Rijk ... are components of a tensor of nth order? ------------------------------------------------------------------------------Ans. (a) Since Rijk are components of a third order tensor, therefore,

′ = Qmi Qnj Q pk Rmnp → Riik ′ = Qmi Qni Q pk Rmnp = δ mn Q pk Rmnp = Q pk Rnnp , therefore, Riik are Rijk

components of a vector. (b) Consider a 4th order tensor Rijkl , we have, ′ = Qmi Qnj Q pk Qql Rmnpq → Riikl ′ = Qmi Qni Q pk Qql Rmnpq = δ m n Q pk Qql Rm npq = Q pk Qql Rnnpq , Rijkl

therefore, Riikl are components of a second order tensor. (c) Riik ... are components of a tensor of the (n − 2)th order. _________________________________________________________________ 2.45 The components of an arbitrary vector a and an arbitrary second tensor T are related by a triply subscripted quantity Rijk in the manner ai = Rijk T jk for any rectangular Cartesian basis {ei } . Prove that Rijk are the components of a third-order tensor. ------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics ′ T jk′ ; Since a is a vector, therefore, Ans. Since ai = Rijk T jk is true for any basis, therefore, ai′ = Rijk

ai′ = Qmi am and since T is a second order tensor, therefore, Tij′ = Qmi QnjTmn . Thus, ′ T jk′ = Qmi ( Rmjk T jk ) . Multiply the last equation with Qsi and noting that ai′ = Qmi am → Rijk

Qsi Qmi = δ sm , we have, ' T ' = Q Q ( R T ) → Q R' T ' = δ R T → Q R' T ' = R T Qsi Rijk jk si mi mjk jk si ijk jk sm mjk jk si ijk jk sjk jk ' Q Q T ' → Qsi Rijk mj nk mn = Rsjk T jk → Qsi Rijk Qmj Qnk Tmn = RsmnTmn . Thus,

(R

smn

)

' − Qsi Qmj Qnk Rijk Tmn = 0 . Since this last equation is to be true for all Tmn , therefore,

′ , which is the transformation law for components of a third order tensor. Rsmn = Qsi Qmj Qnk Rijk

_________________________________________________________________ 2.46

For any vector a and any tensor T , show that (a) a ⋅ TA a = 0 and (b) a ⋅ Ta = a ⋅ TSa ,

where TA and TS are antisymmetric and symmetric part of T respectively. ------------------------------------------------------------------------------

Ans. (a) TA is antisymmetric, therefore, (TA )T = −TA , thus, a ⋅ TA a = a ⋅ (TA )T a = −a ⋅ TA a → 2a ⋅ TA a = 0 → a ⋅ T A a = 0 .

(b) Since T = TS + TA , therefore, a ⋅ Ta = a ⋅ (TS + TA )a = a ⋅ TSa + a ⋅ TA a = a ⋅ TSa . __________________________________________________________________ 2.47 Any tensor can be decomposed into a symmetric part and an antisymmetric part, that is T = TS + TA . Prove that the decomposition is unique. (Hint, assume that it is not true and show contradiction). ------------------------------------------------------------------------------Ans. Suppose that the decomposition is not unique, then ,we have, T = TS + TA = SS + S A → (TS − SS ) + (TA − S A ) = 0 . Let a be any arbitrary vector, we have,

a ⋅ (TS − SS )a + a ⋅ (TA − S A )a = 0 → a ⋅ TSa − a ⋅ SSa + a ⋅ TA a − a ⋅ S A a = 0 . But a ⋅ T A a = a ⋅ S A a = 0 (see the previous problem). Therefore,

a ⋅ TSa − a ⋅ SSa = 0 → a ⋅ (TS − SS )a = 0 → TS − SS = 0 → TS = SS . It also follows from (TS − SS ) + (TA − S A ) = 0 that TA = S A . Thus , the decomposition is unique. _________________________________________________________________ ⎡1 2 3⎤ 2.48 Given that a tensor T has the matrix [ T] = ⎢⎢ 4 5 6 ⎥⎥ , (a) find the symmetric part and the ⎢⎣ 7 8 9 ⎥⎦ anti-symmetric part of T and (b) find the dual vector (or axial vector) of the anti-symmetric part of T. ------------------------------------------------------------------------------⎧ ⎡1 2 3⎤ ⎡1 4 7 ⎤ ⎫ ⎡ 2 6 10 ⎤ ⎡1 3 5 ⎤ 1 1 ⎪⎢ ⎪ 1⎢ T ⎥ ⎢ ⎥ S⎤ ⎡ Ans. (a) T = [ T] + [ T] = ⎨ ⎢ 4 5 6 ⎥ + ⎢ 2 5 8 ⎥ ⎬ = ⎢ 6 10 14 ⎥⎥ = ⎢⎢3 5 7 ⎥⎥ . ⎣ ⎦ 2 2⎪ ⎪ 2 ⎢10 14 18⎥ ⎢5 7 9 ⎥ ⎣ ⎦ ⎣ ⎦ ⎩ ⎣⎢7 8 9 ⎦⎥ ⎣⎢ 3 6 9 ⎦⎥ ⎭

{

}



⎧ ⎡ 1 2 3⎤ ⎡ 1 4 7 ⎤ ⎫ ⎡ 0 −2 −4 ⎤ ⎡ 0 −1 −2⎤ ⎡ TA ⎤ = 1 [ T] − [ T]T = 1 ⎪⎨ ⎢ 4 5 6 ⎥ − ⎢ 2 5 8 ⎥ ⎪⎬ = 1 ⎢ 2 0 −2 ⎥ = ⎢ 1 0 −1⎥ . ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 2 ⎪⎢ ⎪ 2 ⎢ 4 2 18 ⎥ ⎢ 2 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ 7 8 9 3 6 9 ⎦ ⎣ ⎦⎭ ⎣ ⎦ ⎣ ⎦ ⎩⎣

{

}

A (b) t A = −(T23 e1 + T31A e2 + T12A e3 ) = −(−1e1 + 2e 2 − 1e3 ) = e1 − 2e2 + e3 . _________________________________________________________________ 2.49 Prove that the only possible real eigenvalues of an orthogonal tensor Q are λ = ±1 . Explain the direction of the eigenvectors corresponding to them for a proper orthogonal(rotation) tensor and for an improper orthogonal (reflection) tensor. -----------------------------------------------------------------------------Ans. Since Q is orthogonal, therefore, for any vector n , we have, Qn ⋅ Qn = n ⋅ n . Let n be an eigenvector, then Qn = λ n , so that Qn ⋅ Qn = n ⋅ n →

λ 2 (n ⋅ n) = (n ⋅ n) → (λ 2 − 1)(n ⋅ n) = 0 → λ 2 − 1 = 0 → λ = ±1 . The eigenvalue λ = 1 ( Qn = n ) corresponds to an eigenvector parallel to the axis of rotation for a proper orthogonal tensor (rotation tensor); Or, it corresponds to an eigenvector parallel to the plane of reflection for an improper orthogonal tensor (reflection tensor). The eigenvalue λ = −1 , ( Qn = −n ) corresponds to an eigenvector perpendicular to the axis of rotation for an 180o rotation; or, it corresponds to an eigenvector perpendicular to the plane of reflection. _________________________________________________________________ ⎡ 1 −2 −2 ⎤ 1⎢ 2.50 Given the improper orthogonal tensor [Q ] = ⎢ −2 1 −2 ⎥⎥ . (a) Verify that det [Q ] = −1 . 3 ⎢⎣ −2 −2 1 ⎥⎦ (b) Verify that the eigenvalues are λ = 1 and − 1 (c) Find the normal to the plane of reflection (i.e., eigenvectors corresponding to λ = −1 ) and (d) find the eigenvectors corresponding λ = 1 (vectors parallel to the plane of reflection). -------------------------------------------------------------------------------

Ans. (a) det [Q ] = (1 / 3) (1 − 8 − 8 − 4 − 4 − 4) = (−27) / 27 = −1 . 3

(b) I1 = 3 / 3 = 1, I 2 = (1 / 3) {(1 − 4) + (1 − 4) + (1 − 4)} = −1, 2

I 3 = −1 →

λ 3 − λ 2 − λ + 1 = 0 → (λ − 1)(λ 2 − 1) = 0 → λ = 1,1, − 1 (c) For λ = −1 , 2 2 2 2 2 2 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎜ + 1⎟ α1 − α 2 − α 3 = 0, − α1 + ⎜ + 1⎟ α 2 − α 3 = 0, − α1 − α 2 + ⎜ + 1⎟ α 3 = 0 . That 3 3 3 3 3 3 ⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠ is, 2α1 − α 2 − α 3 = 0, − α1 + 2α 2 − α 3 = 0, − α1 − α 2 + 2α 3 = 0 , thus, α1 = α 2 = α 3 , therefore,

n = ± (e1 + e 2 + e3 ) / 3 , this is the normal to the plane of reflection. (d) For λ = 1 , 2 2 2 2 2 2 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎜ − 1⎟ α1 − α 2 − α 3 = 0, − α1 + ⎜ − 1⎟ α 2 − α 3 = 0, − α1 − α 2 + ⎜ − 1⎟ α 3 = 0 3 3 3 3 3 3 3 3 ⎝ ⎠ ⎝ ⎠ ⎝3 ⎠ All three equations lead to α1 + α 2 + α 3 = 0 → α 3 = −α1 − α 2 . Thus, 1 1 [α1e1 + α 2e 2 − (α1 + α 2 )e3 ] , e.g., n = n= (e1 + e2 − 2e3 ) etc. these vectors are all 6 α12 + α 22 + α 32

perpendicular to n = ±(e1 + e2 + e3 ) / 3 and thus parallel to the plane of reflection. _________________________________________________________________ Copyright 2010, Elsevier Inc 2-17

Lai et al, Introduction to Continuum Mechanics 2.51 Given that tensors R and S have the same eigenvector n and corresponding eigenvalue r1 and s1 respectively. Find an eigenvalue and the corresponding eigenvector for the tensor T = RS . -----------------------------------------------------------------------------Ans. We have, Rn = r1n and Sn = s1n , thus, Tn = RSn = Rs1n = s1Rn = r1s1n . Thus, an eigenvalue for T = RS is r1s1 with eigenvector n . _________________________________________________________________ 2.52 Show that if n is a real eigenvector of an antisymmetric tensor T , then the corresponding eigenvalue vanishes. -----------------------------------------------------------------------------Ans. Tn = λ n → n ⋅ Tn = λ (n ⋅ n) . Now, from the definition of transpose, we have n ⋅ Tn = n ⋅ TT n . But, since T is antisymmetric, i.e., TT = −T , therefore, n ⋅ TT n = −n ⋅ Tn . Thus, n ⋅ Tn = −n ⋅ Tn → 2n ⋅ Tn = 0 → n ⋅ Tn = 0 . Thus, λ (n ⋅ n) = 0 → λ = 0 . _________________________________________________________________ 2.53 (a) Show that a is an eigenvector for the dyadic product ab of vectors a and b with eigenvalue a ⋅ b , (b) find the first principal scalar invariant of the dyadic product ab and (c) show that the second and the third principal scalar invariants of the dyadic product ab vanish, and that zero is a double eigenvalue of ab . -----------------------------------------------------------------------------Ans. (a) From the definition of dyadic product, we have, (ab )a = a(b ⋅ a) , thus a is an eigenvector for the dyadic product ab with eigenvalue a ⋅ b . (b) Let T ≡ ab , then Tij = ai b j and the first scalar invariant of ab is Tii = ai bi = a ⋅ b . (c) I 2 =

a1b1 a1b2 a2b2 + a2b1 a2b2 a3b2

a1b1

a1b2

I 3 = a2b1 a2b2 a3b1 a3b2

a1b3

a2b3 a1b1 a1b3 + =0+0+0=0. a3b3 a3b1 a3b3 b1 b2

a2b3 = a1a2 a3 b1 b2 a3b3 b1 b2

b3 b3 = 0 . b3

Thus, the characteristic equation is λ 3 − I1λ 2 = 0 → (λ − I1 )λ 2 = 0 → λ1 = I1 , λ2 = λ3 = 0 . _________________________________________________________________ 2.54 For any rotation tensor, a set of basis {e′i } may be chosen with e′3 along the axis of rotation so that Re1′ = cos θ e1′ + sin θ e′2 , Re′2 = − sin θ e1′ + cosθ e′2 , Re′3 = e′3 , where θ is the angle of right hand rotation. (a) Find the antisymmetric part of R with respect to the basis {e′i } , i.e., find [R A ]e′i . (b) Show that the dual vector of R A is given by t A = sin θ e′3 and (c) show that the first scalar invariant of R is given by 1 + 2cosθ . That is, for any given rotation tensor R , its axis of rotation and the angle of rotation can be obtained from the dual vector of R A and the first scalar invariant of R. -----------------------------------------------------------------------------Ans. (a) From Re1′ = cosθ e1′ + sin θ e′2 , Re′2 = − sin θ e1′ + cosθ e′2 , Re′3 = e′3 , we have,

[ R ]e'i

⎡cosθ = ⎢⎢ sin θ ⎢⎣ 0

− sin θ cos θ 0

0⎤ ⎡ 0 ⎥ A 0 ⎥ → ⎡ R ⎤ = ⎢⎢sin θ ⎣ ⎦ e'i ⎢⎣ 0 1 ⎥⎦ e' i A

− sin θ 0 0

0⎤ 0 ⎥⎥ 0 ⎥⎦ e'

i

(b) the dual vector (or axial vector) of R is given by Copyright 2010, Elsevier Inc 2-18

Lai et al, Introduction to Continuum Mechanics ′ e1′ + T31 ′ e′2 + T12′ e′3 ) = −(0e1′ + 0e′2 − sin θ e′3 ) = sin θ e′3 . t A = −(T23 (c) The first scalar invariant of R is I1 = cosθ + cosθ + 1 = 1 + 2cosθ . __________________________________________________________________ 2.55 The rotation of a rigid body is described by Re1 = e2 , Re2 = e3 , Re3 = e1 . Find the axis of rotation and the angle of rotation. Use the result of the previous problem. -----------------------------------------------------------------------------Ans From the result of the previous problem, we have, the dual vector of R A is given by t A = sin θ e3' , where e3' is in the direction of axis of rotation and θ is the angle of rotation. Thus, we can obtain the direction of axis of rotation and the angle of rotation θ by obtaining the dual vector of R A . From Re1 = e2 , Re2 = e3 , Re3 = e1 , we have, ⎡0 0 1 ⎤ ⎡ 0 −1 1 ⎤ 1 ⎢ ⎥ A⎤ 1⎢ ⎡ [ R ] = ⎢1 0 0⎥ → ⎣ R ⎦ = ⎢ 1 0 −1⎥⎥ → t A = ( e1 + e2 + e3 ) . Thus , 2 2 ⎣⎢0 1 0 ⎥⎦ ⎣⎢ −1 1 0 ⎦⎥ 3 ( e1 + e2 + e3 ) 3 ' 1 tA = = e3 , where e3' = ( e1 + e2 + e3 ) is in the direction of the axis of 2 2 3 3 rotation and the angle of rotation is given by sin θ = 3 / 2 , which gives θ = 60o or 120o . On the other hand, the first scalar invariant of R is 0. Thus, from the result in (c) of the previous problem, we have, I1 = 1 + 2cos θ = 0 , so that cosθ = −1 / 2 which gives θ = 120o or 240o . We therefore conclude that θ = 120o . _________________________________________________________________ ⎡ −1 0 0 ⎤ 2.56 Given the tensor [Q ] = ⎢⎢ 0 −1 0 ⎥⎥ . (a) Show that the given tensor is a rotation tensor. (b) ⎢⎣ 0 0 1 ⎥⎦ Verify that the eigenvalues are λ = 1 and − 1 . (c) Find the direction for the axis of rotation (i.e., eigenvectors corresponding to λ = 1 ). (d) Find the eigenvectors corresponding λ = −1 and (e) obtain the angle of rotation using the formula I1 = 1 + 2cos θ (see Prob. 2.54), where I1 is the first scalar invariant of the rotation tensor. -------------------------------------------------------------------------------

Ans. (a) det [Q ] = +1 , and [Q ][Q ] = [ I ] therefore it is a rotation tensor. I

(b) The principal scalar invariants are: I1 = −1, I 2 = −1, I 3 = 1 → characteristic equation is

(

)

λ 3 + λ 2 − λ − 1 = ( λ + 1) λ 2 − 1 = 0 → the eigenvalues are: λ = −1,1,1 . (c) For λ = 1 , clearly, the eigenvector are: n = ±e3 , which gives the axis of rotation. (d) For λ = −1 , with eigenvector n = α1e1 + α 2e2 + α 3e3 , we have 0α1 = 0, 0α 2 = 0, 2α 3 = 0 . Thus, α1 = arbitrary,α 2 = arbitrary, α 3 = 0 . The eigenvectors are:

n = α1e1 + α 2e2 , α12 + α 22 = 1 . That is, all vectors perpendicular to the axis of rotation are eigenvectors. (e) The first scalar invariant of Q is I1 = −1 . Thus, 1 + 2cosθ = −1 → cos θ = −1 → θ = π . ( We note that for this problem, the antisymmetric part of Q = 0 , so that t A = 0 = sin θ n , of which θ = π is a solution). _________________________________________________________________


Lai et al, Introduction to Continuum Mechanics 2.57 Let F be an arbitrary tensor. (a) Show that F T F and FF T are both symmetric tensors. (b) If F = QU = VQ , where Q is orthogonal and U and V are symmetric, show that U 2 = F T F and V 2 = FF T (c) If λ and n are eigenvalue and the corresponding eigenvector for U , find the eigenvalue and eigenvector for V . [note corrections for text] -------------------------------------------------------------------------------

Ans. (a) (F T F )T = F T (F T )T = F T F , thus F T F is symmetric. Also (FF T )T = (F T )T F T = FF T ,

therefore, FF T is also symmetric. (b) F = QU → F T = U T QT → F T F = U T QT QU = U T U → F T F = U 2 . F = VQ → F T = QT V T → FF T = VQQT V T = VV T → FF T = V 2 . (c) Since F = QU = VQ , and Un = λn , therefore, VQn = QUn = Q(λ n) → V (Qn) = λ (Qn) , therefore, Qn is an eigenvector for V with the eigenvalue λ . _________________________________________________________________ 2.58 Verify that the second principal scalar invariant of a tensor T can be written: I 2 = TiiT jj − TijT ji / 2 .

(

)

-----------------------------------------------------------------------------2 Ans. TiiT jj = (T11 + T22 + T33 )2 = T112 + T22 + T332 + 2T11T22 + 2T22T33 + 2T33T11 . 2 TijT ji = T1 j T j1 + T2 jT j 2 + T3 jT j 3 = T112 + T12T21 + T13T31 + T21T12 + T22 + T23T32 + T31T13 + T32T23 + T332 . 2 + T332 + 2T11T22 + 2T22T33 + 2T33T11 ) Thus, TiiT jj − Tij T ji = (T112 + T22 2 −(T112 + T22 + T332 + 2T12T21 + 2T13T31 + 2T23T32 ) = 2(T11T22 − T12T21 + T22T33 − T23T32 + T33T11 − T13T31 ) . Thus, TiiT jj − TijT ji / 2 = (T11T22 − T12T21 + T22T33 − T23T32 + T33T11 − T13T31 )

(

=

)

T11

T12

T21 T22

+

T22 T23 T32

T33

+

T11 T13 T31 T33

= I2 .

_________________________________________________________________ 2.59 A tensor has a matrix [ T] given below. (a) Write the characteristic equation and find the principal values and their corresponding principal directions. (b) Find the principal scalar invariants. (c) If n1 , n 2 , n3 are the principal directions, write [ T]n . (d) Could the following matrix

[S ] represent the same tensor T with respect to some basis.

i

⎡5 4 0⎤ ⎡7 2 0 ⎤ ⎢ ⎥ [T] = ⎢ 4 −1 0⎥ , [S ] = ⎢⎢ 2 1 0 ⎥⎥ . ⎢⎣ 0 0 3⎥⎦ ⎣⎢ 0 0 −1⎦⎥ -----------------------------------------------------------------------------Ans. (a) The characteristic equation is: 5−λ 4 0 4 −1 − λ 0 = 0 → (3 − λ ) [ (5 − λ )(−1 − λ ) − 16] = (3 − λ )(λ 2 − 4λ − 21) = (3 − λ )(λ + 3)(λ − 7) = 0 0 0 3−λ Thus, λ1 = 3, λ2 = −3, λ3 = 7 . For λ1 = 3, clearly, n1 = ±e3 . Copyright 2010, Elsevier Inc 2-20

Lai et al, Introduction to Continuum Mechanics For λ2 = −3 (5 + 3)α1 + 4α 2 = 0, 4α1 + (−1 + 3)α 2 = 0, (3 + 3)α 3 = 0, α12 + α 22 + α 32 = 1 → 8α1 + 4α 2 = 0, 4α1 + 2α 2 = 0, 6α 3 = 0 → α 2 = −2α1 , α 3 = 0. → n 2 = ± (e1 − 2e 2 ) / 5. For λ3 = 7 (5 − 7)α1 + 4α 2 = 0, 4α1 + (−1 − 7)α 2 = 0, (3 − 7)α 3 = 0, α12 + α 22 + α 32 = 1 → −2α1 + 4α 2 = 0, 4α1 − 8α 2 = 0, − 4α 3 = 0 → α1 = 2α 2 , α 3 = 0. → n3 = ± (2e1 + e 2 ) / 5. (b) The principal scalar invariants are: I1 = 5 − 1 + 3 = 7, I 2 = (−5 − 16) + (−3 − 0) + (15 − 0) = −9, I 3 = −15 − 48 = −63 . We note that

λ 3 − 7λ 2 − 9λ + 63 = 0 → ( λ − 7 ) λ 2 − 9 ( λ − 7 ) = 0 → ( λ − 7 ) (λ 2 − 9) = 0 , same as obtained in (a) ⎡3 0 0⎤ ⎡7 2 0 ⎤ ⎢ ⎥ . (d) det ⎢⎢ 2 1 0 ⎥⎥ = −3 ≠ −63 , therefore, the answer is NO. Or, (c) [ T]n = ⎢0 −3 0 ⎥ i ⎢⎣0 0 7 ⎥⎦ n ,n ,n ⎢⎣ 0 0 −1⎥⎦ { 1 2 3} clearly one of the eigenvalue for [S ] is −1 , which is not an eigenvalue for [ T] , therefore the answer is NO. _________________________________________________________________ 2.60 Do the previous problem for the following matrix: ⎡3 0 0⎤ [T] = ⎢⎢0 0 4⎥⎥ ⎢⎣0 4 0 ⎥⎦ -----------------------------------------------------------------------------Ans. (a) The characteristic equation is: 3−λ 0 0 0−λ 4 = 0 → (3 − λ )(λ 2 − 16) = (3 − λ )(λ − 4)(λ + 4) = 0 4 0−λ

0 0

Thus, λ1 = 3, λ2 = 4, λ3 = −4 . For λ1 = 3, clearly, n1 = ±e1 , because Te1 = 3e1 . For λ2 = 4 (3 − 4)α1 = 0, (0 − 4)α 2 + 4α 3 = 0, 4α 2 + (0 − 4)α 3 = 0,

α12 + α 22 + α 32 = 1

−α1 = 0, − 4α 2 + 4α 3 = 0, 4α 2 − 4α 3 = 0 → α1 = 0, α 2 = α 3 , → n 2 = ± (e 2 + e3 ) / 2. For λ3 = −4 (3 + 4)α1 = 0, (0 + 4)α 2 + 4α 3 = 0, 4α 2 + (0 + 4)α 3 = 0,

α12 + α 22 + α 32 = 1

→ 7α1 = 0, 4α 2 + 4α 3 = 0, 4α 2 + 4α 3 = 0 → α1 = 0, α 2 = −α 3 , → n3 = ± (e 2 − e3 ) / 2 (b) I1 = 3, I 2 = (0 − 0) + (0 − 16) + (0 − 0) = −16, I 3 = −48 .

λ 3 − 3λ 2 − 16λ + 48 = 0 → ( λ − 3) λ 2 − 16 ( λ − 3) = 0 → ( λ − 3) (λ 2 − 16) = 0 , same as in (a) . (c) [ T]n

i

⎡3 0 0 ⎤ = ⎢⎢0 4 0 ⎥⎥ ⎢⎣0 0 −4 ⎥⎦ n ,n ,n { 1 2 3}



⎡7 2 0 ⎤ (d) det [S ] = det ⎢⎢ 2 1 0 ⎥⎥ = −(7 − 4) = −3 ≠ −48 , therefore, the answer is NO. ⎢⎣ 0 0 −1⎥⎦ Or, clearly one of the eigenvalue for [S ] is −1 , which is not an eigenvalue for [ T] , therefore the answer is NO. _________________________________________________________________ 2.61 A tensor T has a matrix given below. Find the principal values and three mutually ⎡1 1 0 ⎤ perpendicular principal directions: [ T] = ⎢⎢1 1 0 ⎥⎥ ⎢⎣ 0 0 2 ⎥⎦ -----------------------------------------------------------------------------Ans. The characteristic equation is: 1− λ 1 0 2 1 1− λ 0 = 0 → ( 2 − λ ) ⎡(1 − λ )2 − 1⎤ = ( 2 − λ ) (−2λ + λ 2 ) = −λ ( 2 − λ ) = 0 . ⎣ ⎦ 0 0 2−λ Thus, λ1 = 0, λ2 = λ3 = 2 . That is, there is a double root λ2 = λ3 = 2 . For λ1 = 0 , (1 − 0)α1 + α 2 = 0, α1 + (1 − 0)α 2 = 0, (2 − 0)α 3 = 0,

α12 + α 22 + α 32 = 1

→ α1 + α 2 = 0, 2α 3 = 0 → α1 = −α 2 , α 3 = 0, → n1 = ± (e1 − e 2 ) / 2. For λ2 = λ3 = 2 , one eigenvector is clearly n3 . There are infinitely many others all lie on the plane

whose normal is n1 = ± (e1 − e 2 ) / 2 . In fact, we have, (1 − 2)α1 + α 2 = 0, α1 + (1 − 2)α 2 = 0, (2 − 2)α 3 = 0,

α12 + α 22 + α 32 = 1

→ −α1 + α 2 = 0, 0α 3 = 0 → α1 = α 2 = α , α 3 = 1 − 2α 2 → n = ± (α e1 + α e 2 + α 3e3 ),

which include the case where α = 0, α 3 = ±1 → n = ±e3 . _________________________________________________________________

CHAPTER 2, PART C d dT dS ( T + S ) = + , using the definition of derivative of a tensor. dt dt dt -----------------------------------------------------------------------------Ans. d {T(t + Δt ) + S(t + Δt )} − {T(t ) + S(t )} {T(t + Δt ) − T(t )} + {S(t + Δt ) − S(t )} = lim ( T + S ) = lim Δt →0 Δt →0 Δt Δt dt _ {T(t + Δt ) − T(t )} {{S(t + Δt ) − S(t )} dT dS lim + lim = + . Δt → 0 Δt →0 Δt Δt dt dt _________________________________________________________________ d dS dT 2.63 Prove the identity ( TS ) = T + S using the definition of derivative of a tensor. dt dt dt ------------------------------------------------------------------------------

2.62

Prove the identity


Lai et al, Introduction to Continuum Mechanics d T(t + Δt )S(t + Δt ) − T(t )S(t ) ( TS ) = lim Δt → 0 dt Δt T(t + Δt )S(t + Δt ) − T(t + Δt )S(t ) + T(t + Δt )S(t ) − T(t )S(t ) = lim Δt →0 Δt T(t + Δt ){S(t + Δt ) − S(t )} {T(t + Δt ) − T(t )}S(t ) = lim + lim Δt →0 Δ t → 0 Δt Δt {S(t + Δt ) − S(t )} {T(t + Δt ) − T(t )} dS dT S(t ) = T S. = T(t ) lim + lim + Δt → 0 Δt →0 Δt Δt dt dt _________________________________________________________________ Ans.

T

dT T ⎛ dT ⎞ T =⎜ ⎟ by differentiating the definition a ⋅ Tb = b ⋅ T a , where dt ⎝ dt ⎠ a and b are constant arbitrary vectors. -----------------------------------------------------------------------------2.64

Prove that

Ans. a ⋅ Tb = b ⋅ TT a → a ⋅ ( dT / dt )b = b ⋅ (dTT / dt )a . Now, the definition of transpose also gives a ⋅ (dT / dt )b = b ⋅ ( dT / dt ) a . Thus, b ⋅ ( dT / dt ) a = b ⋅ (dTT / dt )a . T

T

T

⎛ dT ⎞ ⎛ dT ⎞ Since a and b arbitrary vectors, therefore, ⎜ ⎟⎟ . ⎟ = ⎜⎜ ⎝ dt ⎠ ⎝ dt ⎠ _________________________________________________________________ T

2.65 Consider the scalar field φ = x12 + 3x1 x2 + 2 x3 . (a) Find the unit vector normal to the surface of constant φ at the origin (0,0,0) and at (1,0,1) . (b) what is the maximum value of the directional derivative of φ at the origin? At (1,0,1)? (c) Evaluate dφ / dr at the origin if dr = ds (e1 + e3 ) . -----------------------------------------------------------------------------Ans. (a) ∇φ = (2 x1 + 3 x2 )e1 + 3 x1e 2 + 2e3 ,

at (0,0,0), ∇φ = 2e3 → n = e3 , at (1,0,1), ∇φ = 2e1 + 3e2 + 2e3 , → n = ( 2e1 + 3e2 + 2e3 ) / 17 .

(b) At (0,0,0), (dφ / dr ) max = ∇φ = 2 in the direction of n =e3 . At (1,0,1) , (dφ / dr ) max = ∇φ = 17 . (c) At (0,0,0), dφ / dr = (∇φ )o ⋅ dr / dr = 2e3 ⋅ (e1 + e3 ) / 2 = 2 . _________________________________________________________________ 2.66 Consider the ellipsoidal surface defined by the equation x 2 / a 2 + y 2 / b 2 + z 2 / c 2 = 1 . Find the unit vector normal to the surface at a given point ( x, y, z ) . -----------------------------------------------------------------------------x2 y 2 z 2 Ans. Let f ( x, y , z ) = 2 + 2 + 2 − 1 , then a b c ∂f 2 x ∂f 2 y ∂f 2 z 2x 2y 2z = 2, = 2, = 2 → ∇f = 2 e1 + 2 e 2 + 2 e3 , thus, ∂x a ∂y b ∂z c a b c −1/2

2 2 2 ∇f ⎡ ⎛ 2 x ⎞ ⎛ 2 y ⎞ ⎛ 2 z ⎞ ⎤ 2y 2z ⎞ ⎛ 2x n= = ⎢⎜ 2 ⎟ + ⎜ 2 ⎟ + ⎜ 2 ⎟ ⎥ ⎜ 2 e1 + 2 e 2 + 2 e3 ⎟ . ∇f ⎢⎣⎝ a ⎠ ⎝ b ⎠ ⎝ c ⎠ ⎥⎦ b c ⎝a ⎠ _________________________________________________________________


Lai et al, Introduction to Continuum Mechanics 2.67 Consider the temperature field given by: Θ = 3x1 x2 . (a) Find the heat flux at the point A(1,1,1) , if q = − k ∇Θ . (b) Find the heat flux at the same point if q = −K∇Θ , where ⎡k 0 0 ⎤ [K ] = ⎢⎢ 0 2k 0 ⎥⎥ ⎢⎣ 0 0 3k ⎥⎦ -----------------------------------------------------------------------------Ans. Θ = 3x1 x2 → ∇Θ = 3( x2e1 + x1e2 ) → (∇Θ) A = 3(e1 + e2 ) . (a) q = − k ∇Θ = −3k (e1 + e2 ) . ⎡ k 0 0 ⎤ ⎡ 3⎤ ⎡ 3k ⎤ ⎢ ⎥ ⎢ ⎥ (b) [q ] = − [ K∇Θ ] = − ⎢ 0 2k 0 ⎥ ⎢ 3⎥ = − ⎢⎢ 6k ⎥⎥ → q = −(3ke1 + 6ke 2 ) . ⎢⎣ 0 0 3k ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0 ⎥⎦ _________________________________________________________________ 2.68 Let φ ( x1 , x2 , x3 ) and ψ ( x1 , x2 , x3 ) be scalar fields, and let v( x1 , x2 , x3 ) and w ( x1 , x2 , x3 ) be vector fields. By writing the subscripted components form, verify the following identities. ∂ (φ + ψ ) ∂φ ∂ψ (a) ∇(φ + ψ ) = ∇φ + ∇ψ , sample solution: [∇(φ + ψ )]i = = + = ∇φ + ∇ψ . ∂xi ∂xi ∂xi (b) div( v + w ) = divv + divw , (c) div(φ v ) = (∇φ ) ⋅ v + φ (divv ) and (d) div(curlv ) = 0 . -----------------------------------------------------------------------------∂ (vi + wi ) ∂vi ∂wi Ans. (b) div( v + w ) = = + = divv + divw . ∂xi ∂xi ∂xi ∂ (φ vi ) ∂v ∂φ vi = φ (divv ) + (∇φ ) ⋅ v . =φ i + (c) div(φ v ) = ∂xi ∂xi ∂xi (d) curlv = −ε ijk

∂v j ∂xk

ei = ε ijk

∂vk ∂v ∂ ∂ ∂vk ei → div(curlv) = . ε ijk k = ε ijk ∂x j ∂xi ∂x j ∂xi ∂x j

By changing the dummy indices, ( i → j , j → i ) we have, ε ijk

∂ ∂vk ∂ ∂vk . Thus, = ε jik ∂xi ∂x j ∂x j ∂xi

⎞ ⎟ = 0 . Thus, div(curlv ) = 0 . ⎟ ⎠ _________________________________________________________________

ε ijk

∂v ∂ ∂vk ∂ ∂vk ∂ ∂vk ∂ ⎛ = −ε ijk → 2ε ijk =0→ ⎜ ε ijk k ∂xi ∂x j ∂xi ⎜⎝ ∂x j ∂xi ∂x j ∂x j ∂xi

2.69

Consider the vector field v = x12e1 + x32e 2 + x22e3 . For the point (1,1,0) , find (a) ∇v , (b)

(∇v) v , (c) div v and curl v and (d) the differential dv for dr = ds (e1 + e 2 + e3 ) / 3 . -----------------------------------------------------------------------------0 ⎤ ⎡ 2 x1 0 ⎡ 2 0 0⎤ ⎢ ⎥ Ans.(a) [∇v ] = ⎢ 0 0 2 x3 ⎥ → [∇v ](1,1,0 ) = ⎢⎢ 0 0 0 ⎥⎥ . ⎢⎣ 0 2 x2 ⎢⎣ 0 2 0 ⎥⎦ 0 ⎥⎦ ⎡ 2 0 0 ⎤ ⎡1 ⎤ ⎡ 2 ⎤ (b) [ (∇v) v ] = ⎢⎢ 0 0 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → (∇v) v = 2e1 . ⎢⎣ 0 2 0 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 0 ⎥⎦ (c) div v = 2 x1 + 0 + 0 = 2 x1 → at (1,1,0), div v = 2 . Copyright 2010, Elsevier Inc 2-24

Lai et al, Introduction to Continuum Mechanics ⎛ ∂v ⎛ ∂v ∂v ⎞ ⎛ ∂v ∂v ⎞ ∂v ⎞ curlv = ⎜ 3 − 2 ⎟ e1 + ⎜ 1 − 3 ⎟ e 2 + ⎜ 2 − 1 ⎟ e3 = 2 ( x2 − x3 ) e1 . ⎝ ∂x1 ∂x2 ⎠ ⎝ ∂x2 ∂x3 ⎠ ⎝ ∂x3 ∂x1 ⎠ At (1,1,0) , curlv = 2 (1 − 0 ) e1 = 2e1 . (d) ⎡ 2 0 0 ⎤ ⎡ ds / 3 ⎤ ⎡ 2ds / 3 ⎤ ⎢ ⎥ ⎢ ⎥ At (1,1,0 ) , dv = ( ∇v ) dr → [ dv ] = [∇v ][ dr ] = ⎢⎢ 0 0 0 ⎥⎥ ⎢ ds / 3 ⎥ = ⎢ 0 ⎥ . ⎢⎣ 0 2 0 ⎥⎦ ⎢ ds / 3 ⎥ ⎢ 2ds / 3 ⎥ ⎦ ⎣⎢ ⎦⎥ ⎣ → dv = 2ds (e1 + e3 ) / 3 _________________________________________________________________

CHAPTER 2, PART D 2.70

Calculate div u for the following vector field in cylindrical coordinates:

(a) ur = uθ = 0, u z = A + Br 2 . (b) ur = sin θ / r , uθ = u z = 0 , and (c) ur = r 2 sin θ / 2, uθ =r 2 cos θ / 2, u z = 0 . -----------------------------------------------------------------------------∂u 1 ∂uθ ur ∂u z Ans.(a) ur = uθ = 0, u z = A + Br 2 → div u = r + + + =0+0+0+0=0. r ∂r r ∂θ ∂z ∂u 1 ∂uθ ur ∂u z (b) ur = sin θ / r , uθ = u z = 0 → div u = r + + + = − sin θ / r 2 + 0 + sin θ / r 2 + 0 = 0 r ∂r r ∂θ ∂z (c) ur = r 2 sin θ / 2, uθ =r 2 cos θ / 2, u z = 0 ∂u 1 ∂uθ ur ∂u z → div u = r + + + = r sin θ − r sin θ / 2 + r sin θ / 2 + 0 = r sin θ . ∂r r ∂θ r ∂z _________________________________________________________________ 2.71 Calculate ∇u for the following vector field in cylindrical coordinate: ur = A / r , uθ = Br , u z = 0 . -----------------------------------------------------------------------------⎡ ∂ur 1 ⎛ ∂ur ⎞ ∂ur ⎤ ⎡ A ⎤ ⎢ ∂r r ⎜ ∂θ − uθ ⎟ ∂z ⎥ ⎢ − 2 − B 0 ⎥ ⎝ ⎠ ⎢ ⎥ ⎢ r ⎥ ⎢ ∂uθ 1 ⎛ ∂uθ A ⎞ ∂uθ ⎥ ⎢ Ans. [∇u ] = ⎢ + ur ⎟ 0 ⎥⎥ . ⎥=⎢ B ⎜ 2 r ⎠ ∂z ⎥ ⎢ ⎢ ∂r r ⎝ ∂θ ⎥ ⎢ ∂u z ⎥ 0 0⎥ ∂u z 1 ∂u z ⎢ 0 ⎢ ⎥ ⎥⎦ ∂z ⎥⎦ ⎢⎣ r ∂θ ⎢⎣ ∂r

_________________________________________________________________ 2.72 Calculate div u for the following vector field in spherical coordinates ur = Ar + B / r 2 , uθ = uφ = 0 -----------------------------------------------------------------------------1 ∂ ( r 2 ur ) 1 ∂ (uθ sin θ ) 1 ∂uφ Ans. → div u = 2 + + r sin θ r sin θ ∂φ ∂r ∂θ r Copyright 2010, Elsevier Inc 2-25


(

)

1 ∂ ⎧ 2⎛ B ⎞⎫ 1 ∂ r Ar + 2 ⎟ ⎬ = 2 Ar 3 + B = 3 A. 2 ∂r ⎨ ⎜ r r ⎠ ⎭ r ∂r ⎩ ⎝ _________________________________________________________________ 2.73 Calculate ∇u for the following vector field in spherical coordinates: =

ur = Ar + B / r 2 , uθ = uφ = 0 . ------------------------------------------------------------------------------

⎡ ∂ur ⎢ ⎢ ∂r ⎢ ∂u Ans. [∇u ] = ⎢ θ ⎢ ∂r ⎢ ⎢ ∂uφ ⎢ ∂r ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎞⎥ ⎟⎥ ⎠⎦ ⎤ 0 ⎥ 0 ⎥. 3⎥ A+ B / r ⎦

⎛ 1 ∂ur uφ ⎞ − ⎟ ⎜ ⎝ r sin θ ∂φ r ⎠ ⎛ 1 ∂uθ uφ cot θ ⎞ ⎛ 1 ∂uθ ur ⎞ − ⎜ ⎟ ⎜ r ∂θ + r ⎟ r ⎝ ⎠ ⎝ r sin θ ∂φ ⎠ ⎛ 1 ∂uφ ur uθ cot θ 1 ∂uφ + + ⎜ r ∂θ r r ⎝ r sin θ ∂φ ⎛ 1 ∂ur uθ ⎞ ⎜ r ∂θ − r ⎟ ⎝ ⎠

0 0 ⎤ ⎡ A − 2B / r 3 ⎡∂ur / ∂r ⎢ = ⎢⎢ 0 ur / r 0 ⎥⎥ = ⎢ 0 ⎢⎣ 0 0 ur / r ⎥⎦ ⎢⎣ 0

0 A + B / r3 0

_________________________________________________________________ 2.74 From the definition of the Laplacian of a vector, ∇ 2 v = ∇ ( div v ) − curlcurl v , derive the following results in cylindrical coordinates: ⎛ ∂ 2v 1 ∂ 2 vr ∂ 2 vr 2 ∂vθ 1 ∂vr vr ⎞ ∇ 2 v = ⎜ 2r + 2 + − + − ⎟ and ⎜ ∂r r r ∂θ 2 ∂z 2 r 2 ∂θ r ∂r r 2 ⎟⎠ ⎝

(

)

∂ 2 vθ

1 ∂ 2 vθ

∂ 2 vθ

1 ∂vθ 2 ∂v v + 2 r − θ2 . r ∂r r ∂θ r ∂r r ∂θ ∂z --------------------------------------------------------------------------Ans. Let v ( r ) be a vector field. The Laplacian of v is ∇ 2 v = ∇ ( div v ) − curlcurl v . Now,

(

∇2 v

)θ

=

2

+

2

2

+

2

+

∂vr 1 ∂vθ vr ∂vz + + + , so that ∂r r ∂θ r ∂z ∂ ⎛ ∂v 1 ∂vθ vr ∂vz ⎞ 1 ∂ ⎛ ∂vr 1 ∂vθ vr ∂vz ∇ ( div v ) = ⎜ r + + + + + + er + ⎟ ∂r ⎝ ∂r r ∂θ r ∂z ⎠ r ∂θ ⎝⎜ ∂r r ∂θ r ∂z

div v =

⎞ ⎟ eθ ⎠

⎛ ∂ 2 vr 1 ∂ 2 vθ 1 ∂vθ 1 ∂vr vr ∂ 2 vz ⎞ ⎞ e = + − + − 2+ ⎟⎟ e r ⎟ z ⎜⎜ 2 r ∂r ∂θ 2 ∂θ r ∂ r ∂ r ∂ z ∂ r r r ⎠ ⎝ ⎠ 2 2 2 2 2 2 ⎛ 1 ∂ vr ⎛ ∂ vr 1 ∂ vθ 1 ∂vr ∂ vz ⎞ 1 ∂ vθ 1 ∂vr 1 ∂ vz ⎞ + + + + 2 ⎟ ez . e + +⎜ + + ⎜ r ∂θ∂r r 2 ∂θ 2 r 2 ∂θ r ∂θ∂z ⎟⎟ θ ⎜⎜ ∂z∂r r ∂z∂θ r ∂z ∂z ⎟⎠ ⎝ ⎠ ⎝ Next, v 1 ∂vr ⎞ ⎛ 1 ∂vz ∂vθ ⎞ ⎛ ∂v ⎛ ∂v ∂v ⎞ e r + ⎜ r − z ⎟ eθ + ⎜ θ + θ − ez , so that − curl v = ⎜ ⎟ ∂z ⎠ ∂r ⎠ r r ∂θ ⎠⎟ ⎝ ∂z ⎝ r ∂θ ⎝ ∂r +

∂ ⎛ ∂vr 1 ∂vθ vr ∂vz + + + ∂z ⎝⎜ ∂r r ∂θ r ∂z



( curlcurl v )r =

1 ∂ ⎛ ∂vθ vθ 1 ∂vr + − r ∂θ ⎝⎜ ∂r r r ∂θ

⎞ ∂ ⎛ ∂vr ∂vz ⎞ ⎟ − ∂z ⎜ ∂z − ∂r ⎟ ⎝ ⎠ ⎠

⎛ 1 ∂ 2 vθ 1 ∂v 1 ∂ 2 vr ⎞ ⎛ ∂ 2 vr ∂ 2 vz ⎞ =⎜ + 2 θ − 2 − − , ⎜ r ∂θ∂r r ∂θ r ∂θ 2 ⎟⎟ ⎜⎜ ∂z 2 ∂z∂r ⎟⎟ ⎠ ⎝ ⎠ ⎝ ∂v v ∂ 1 ∂v ∂ ∂v 1 ∂v ( curlcurl v )θ = ⎛⎜ z − θ ⎞⎟ − ⎛⎜ θ + θ − r ⎞⎟ ∂z ⎝ r ∂θ ∂z ⎠ ∂r ⎝ ∂r r r ∂θ ⎠ ∂ ∂v ∂v 1 ∂v ∂v 1 ∂ ⎛ 1 ∂vz ∂vθ ⎞ − ( curlcurl v ) z = ⎛⎜ r − z ⎞⎟ + ⎛⎜ r − z ⎞⎟ − ∂r ⎝ ∂z ∂r ⎠ r ⎝ ∂z ∂r ⎠ r ∂θ ⎜⎝ r ∂θ ∂z ⎟⎠ Thus, ⎛ ∂ 2 v 1 ∂ 2 vθ 1 ∂v 1 ∂vr vr ∂ 2 vz ⎞ ∇ 2 v = ⎜ 2r + − 2 θ + − + ⎟ ⎜ r r ∂r ∂θ r ∂θ r ∂r r 2 ∂r ∂z ⎠⎟ ⎝ ∂r ⎛ 1 ∂ 2 vθ 1 ∂v 1 ∂ 2 vr ⎞ ⎛ ∂ 2 vr ∂ 2 vz ⎞ ⎛ ∂ 2 vr 1 ∂ 2 vr ∂ 2 vr 2 ∂vθ 1 ∂vr vr ⎞ . −⎜ + 2 θ − 2 + − = + + − + − ⎜ r ∂θ∂r r ∂θ r ∂θ 2 ⎟⎟ ⎜⎜ ∂z 2 ∂z∂r ⎟⎟ ⎜⎜ ∂r 2 r 2 ∂θ 2 ∂z 2 r 2 ∂θ r ∂r r 2 ⎟⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2 2 2 ⎛ 1 ∂ vr 1 ∂ vθ 1 ∂v 1 ∂ vz ⎞ ∇2 v = ⎜ + 2 + 2 r + ⎟ 2 ⎜ θ r ∂θ r ∂θ∂z ⎟⎠ ⎝ r ∂θ∂r r ∂θ ∂ ⎛ 1 ∂vz ∂vθ ⎞ ∂ ⎛ ∂vθ vθ 1 ∂vr ⎞ ⎛ 1 ∂ 2 vr 1 ∂ 2 vθ 1 ∂vr 1 ∂ 2 vz ⎞ − ⎜ − + + − = + + + ⎜ ⎟ ∂z ⎝ r ∂θ ∂z ⎟⎠ ∂r ⎜⎝ ∂r r r ∂θ ⎟⎠ ⎜⎝ r ∂θ∂r r 2 ∂θ 2 r 2 ∂θ r ∂θ∂z ⎟⎠ ⎛ 1 ∂ 2 vz ∂ 2 vθ ⎞ ⎛ ∂ 2 vθ 1 ∂vθ vθ 1 ∂ 2 vr 1 ∂vr ⎞ +⎜ − + + + − − + ⎜ r ∂z∂θ ∂z 2 ⎟⎟ ⎜⎜ ∂r 2 r ∂r r 2 r ∂r ∂θ r 2 ∂θ ⎟⎟ ⎝ ⎠ ⎝ ⎠

(

)

(

)

∂ 2 vθ

1 ∂ 2 vθ

∂ 2 vθ

1 ∂vθ 2 ∂v v + 2 r − θ2 . r ∂r r ∂θ r r ∂θ ∂r ∂z _________________________________________________________________ =

2

+

2

2

+

2

+

2.75 From the definition of the Laplacian of a vector, ∇ 2 v = ∇ ( div v ) − curlcurl v , derive the following result in spherical coordinates:

(

∇2 v

)

r

⎛ 1 ∂ 2 r 2 vr 2 ∂r 2 vr 1 ∂ 2 vr cot θ ∂vr ∂ 2 vr 1 2 ∂vθ sin θ 2 ∂vφ ⎞ =⎜ 2 − + + + − 2 − 2 ⎟ 3 ∂r 2 2 2 ∂θ 2 2 2 ⎜ r ∂r 2 ∂θ r r ∂θ r r sin θ ∂φ r sin θ r sin θ ∂φ ⎟⎠ ⎝

-----------------------------------------------------------------------------Ans.

(

)

2 1 ∂ r vr 1 ∂ ( vθ sin θ ) 1 ∂vφ From 2 + + , we have, r sin θ r sin θ ∂φ ∂r ∂θ r

∇ ( divv ) =

2 ⎛ 1 ∂r 2 vr 1 ∂vθ sin θ 1 ∂vφ ⎞ 1 ∂ ⎛ 1 ∂r vr 1 ∂vθ sin θ 1 ∂vφ ⎞ + + + + ⎜ 2 ⎟ er + ⎜ 2 ⎟ eθ ∂r ⎝ r ∂r ∂θ ∂r ∂θ r sin θ r sin θ ∂φ ⎠ r ∂θ ⎝ r r sin θ r sin θ ∂φ ⎠

∂

⎛ 1 ∂r 2 vr 1 ∂vθ sin θ 1 ∂vφ ⎞ ++ + ⎜ 2 ⎟ eφ , that is ∂r ∂θ r sin θ ∂φ ⎝ r r sin θ r sin θ ∂φ ⎠ 2 2 2 ⎛ 1 ∂ 2 r 2 vr ∂vφ ⎞ ∂vθ sin θ 2 ∂ r vr 1 ∂ vθ sin θ 1 1 ∂ vφ 1 ⎟ er ∇ ( divv ) = ⎜ − + − + − 3 2 ⎜ r 2 ∂r 2 ∂r r sin θ ∂r ∂θ ∂θ r sin θ ∂r ∂φ r 2 sin θ ∂φ ⎟ r r sin θ ⎝ ⎠ +

1

∂

( )

( )



( )

2 2 ⎛ 1 ∂ 2 r 2 vr ∂ ( vθ sin θ ) ∂ vθ sin θ 1 1 1 ⎛ cos θ ⎞ ∂vθ sin θ 1 ∂ 1 ∂vφ ⎞ ⎟ eθ + + − + ⎜ ⎟ 3 2 2 2 2 2 2 ⎜r ∂θ∂r r ∂θ r sin θ ∂φ ⎟ r sin θ ∂θ r sin θ ∂θ r ⎝ sin θ ⎠ ∂θ ⎝ ⎠ 2 ⎛ ∂ 2 vφ ⎟⎞ ∂ 2 ( vθ sin θ ) 1 1 1 ∂ ∂ r vr +⎜ 3 + 2 2 + 2 2 eφ . Also, ⎜ r sin θ ∂φ ∂r ∂φ∂θ r sin θ r sin θ ∂ 2φ ⎟ ⎝ ⎠ ∂ v sin θ ∂ rv ⎛ 1 ⎛ 1 ∂vr 1 φ ⎞ 1 ∂vθ ⎞ ⎛ 1 ∂rvθ 1 ∂vr ⎞ φ curl v = ⎜ − − − eφ ⎟ er + ⎜ ⎟ eθ + ⎜ r sin θ ∂φ ⎠ r ∂θ ⎟⎠ ∂θ ⎝ r ∂r ⎝ r sin θ ⎝ r sin θ ∂φ r ∂r ⎠

+⎜

(

)

so that

curlcurl v =

1 r sin θ

⎧⎪ ∂ ⎛ 1 ∂rvθ 1 ∂vr − ⎨ ⎜ r ∂θ ⎪⎩ ∂θ ⎝ r ∂r

∂ ⎛ 1 ∂vr 1 ∂rvφ ⎞ ⎟ sin θ − ∂φ ⎜ r sin θ ∂φ − r ∂r ⎠ ⎝

1 ∂vθ ⎞ 1 ∂ ⎛ 1 ∂rvθ 1 ∂vr ∂ ⎛ 1 ∂vφ sin θ ⎪⎧ 1 r +⎨ − − ⎜ ⎟− r sin θ ∂φ ⎠ r ∂r ⎜⎝ r ∂r r ∂θ ∂θ ⎩⎪ r sin θ ∂φ ⎝ r sin θ

⎞ ⎫⎪ ⎟ ⎬ er ⎠ ⎪⎭

⎞ ⎪⎫ ⎟ ⎬ eθ ⎠ ⎭⎪

⎧⎪ 1 ∂ ⎛ 1 ∂vr 1 ∂rvφ ⎞ 1 ∂ ⎛ 1 ∂vφ sin θ 1 ∂vθ ⎞ ⎫⎪ r⎜ − +⎨ − ⎟− ⎜ ⎟ ⎬ eφ ∂θ r sin θ ∂φ ⎠ ⎪⎭ ⎪⎩ r ∂r ⎝ r sin θ ∂φ r ∂r ⎠ r ∂θ ⎝ r sin θ i.e., curlcurl v = 2 ⎧⎪ 1 ⎛ ∂ 2 rv ∂ 2 vr ⎞ cot θ ⎛ 1 ∂rvθ 1 ∂vr ⎞ ⎛ ∂ 2 vr 1 1 ∂ rvφ ⎞ ⎫⎪ θ ⎟ ⎬ er − + − −⎜ − ⎨ 2 ⎜⎜ 2 ⎟ ⎟ r ⎜⎝ r ∂r r ∂θ ⎟⎠ ⎜ r 2 sin 2 θ ∂φ 2 r 2 sin θ ∂φ∂r ⎟ ⎪ ⎪⎩ r ⎝ ∂θ∂r ∂θ ⎠ ⎝ ⎠⎭ ⎛ ∂ 2 vφ sin θ ∂ 2 v ⎞ 1 ⎛ ∂rv ∂vr ⎞ ⎛ 1 ∂ 2 rvθ 1 1 ∂rvθ 1 ∂ 2 vr 1 ∂vr ⎞ ⎪⎫ ⎪⎧ θ ⎟ θ +⎨ 2 2 ⎜ − − − − − − + ⎜ ⎟ ⎬ eθ ⎜ ∂θ ⎠⎟ ⎝⎜ r ∂r 2 r ∂r ∂θ r 2 ∂θ ⎠⎟ ⎪ ∂φ 2 ⎠⎟ r 2 ⎝ ∂r r 2 ∂r ⎪⎩ r sin θ ⎝⎜ ∂φ∂θ ⎭ 2 ⎧⎛ 1 ∂vr 1 1 ∂vr 1 ∂ rvφ 1 ∂rvφ ⎞ 1 ⎛ 1 ∂vr ∂rvφ ⎞ ⎫ ⎪⎜ ⎟+ ⎜ − 2 − + 2 − ⎟⎪ ∂r ⎠ ⎪ ⎪⎪⎝⎜ r sin θ ∂r ∂φ r sin θ ∂φ r ∂r 2 r ∂r ⎟⎠ r 2 ⎝ sin θ ∂φ ⎪ +⎨ ⎬ eφ ∂ 2 vφ ∂ 2 vθ ⎞ ⎪ 1 1 ⎛ cos θ ⎛ ∂vφ sin θ ∂vθ ⎞ ⎪ ⎜ −vφ sin θ + sin θ ⎟+ − − ⎜ ⎟ ⎪ ⎪− 2 ∂φ ⎠ ⎪ ∂θ 2 ∂θ∂φ ⎟⎠ r 2 sin 2 θ ⎝ ∂θ ⎪⎩ r sin θ ⎜⎝ ⎭

Thus, ∇ 2 v = ∇ ( divv ) − curlcurl v gives:

(

)

(

)

2 2 ⎛ ∂ 2 r 2v ⎞ 2 r 1 2 ∂ r vr 1 ∂ ( vθ sin θ ) 1 ∂ ( vθ sin θ ) 1 ∂ vφ 1 ∂vφ ⎟ ⎜ ∇ v = 2 − 3 + − 2 + − ⎜r r r sin θ r sin θ ∂r ∂φ r 2 sin θ ∂φ ⎟ ∂r ∂r ∂θ ∂θ r r sin θ ∂r 2 ⎝ ⎠ 2 2 2 ∂ 2 vr 1 1 ∂ rvφ ⎞ ⎪⎫ ⎪⎧ 1 ⎛ ∂ rvθ ∂ vr ⎞ cot θ ⎛ 1 ∂rvθ 1 ∂vr ⎞ ⎛ ⎜ ⎟⎬ −⎨ 2 ⎜ − + − − − ⎟ 2 ⎟ ⎜ r ⎝⎜ r ∂r r ∂θ ⎟⎠ ⎜ r 2 sin 2 θ ∂φ 2 r 2 sin θ ∂φ∂r ⎟ ⎪ ⎪⎩ r ⎝ ∂θ∂r ∂θ ⎠ ⎝ ⎠⎭ i.e.,

(

2

)

(∇ v ) 2

r

⎛ 1 ∂ 2 r 2 vr 2 ∂r 2 vr 1 ∂ 2 vr cot θ ∂vr ∂ 2 vr 1 2 ∂vθ sin θ 2 ∂vφ ⎞ =⎜ 2 − 3 + 2 + 2 + 2 2 − 2 − 2 ⎟ 2 2 2 ⎜ r ∂r ∂θ r ∂r r ∂θ r ∂θ r sin θ ∂φ r sin θ r sin θ ∂φ ⎟⎠ ⎝

_________________________________________________________________ Copyright 2010, Elsevier Inc 2-28

Lai et al, Introduction to Continuum Mechanics From the equation (divT) ⋅ a = div(TT a) − tr(TT ∇a) [See Eq. 2.29.3)] verify that in polar ∂T 1 ∂Tθθ Trθ + Tθ r + . coordinates, the θ -component of the vector (divT) is: (divT)θ = θ r + ∂r r ∂θ r -----------------------------------------------------------------------------Ans. (divT) ⋅ a = div(TT a) − tr(TT ∇a) → (divT) ⋅ eθ = div(TT eθ ) − tr(TT ∇eθ ) Now, 2.76

Te r = Trr e r + Tθ r eθ , Teθ = Trθ e r + Tθθ eθ → e r ⋅ TT eθ = eθ ⋅ Te r = Tθ r , eθ ⋅ TT eθ = eθ ⋅ Teθ = Tθθ ∂T 1 ∂Tθθ Tθ r i.e., TT eθ = Tθ r e r + Tθθ eθ → div(TT eθ ) = θ r + + . Also ∂r r ∂θ r Tθ r ⎤ ⎡0 −1 / r ⎤ ⎡ 0 −Trr / r ⎤ ⎡T ⎡ 0 −1 / r ⎤ ⎡ T eθ = 0e r + 1eθ → [∇eθ ] = ⎢ → T ∇eθ ⎤ = ⎢ rr = ⎥⎢ ⎥ ⎣ ⎦ 0 ⎦ 0 ⎥⎦ ⎢⎣ 0 −Trθ / r ⎥⎦ ⎣0 ⎣Trθ Tθθ ⎦ ⎣0 Thus,

∂Tθ r 1 ∂Tθθ Tθ r ∂T 1 ∂Tθθ Tθ r + Trθ + + − (0 − Trθ / r ) = θ r + + . ∂r r ∂θ r ∂r r ∂θ r _________________________________________________________________ 2.77 Calculate div T for the following tensor field in cylindrical coordinates; B B Trr = A + 2 , Tθθ = A − 2 , Tzz = constant, Trθ = Tθ r = Trz = Tzr = Tθ z = Tzθ = 0 r r -----------------------------------------------------------------------------∂T 1 ∂Trθ Trr − Tθθ ∂Trz 2B 2B Ans. (divT)r = rr + + + = − 3 + 3 = 0. ∂r r ∂θ r ∂z r r ∂T 1 ∂Tθθ Trθ + Tθ r ∂Tθ z + + = 0. (divT)θ = θ r + ∂r ∂z r ∂θ r ∂T 1 ∂Tzθ ∂Tzz Tzr + + =0. (divT) z = zr + ∂r ∂z r ∂θ r _________________________________________________________________ 2.78 Calculate divT for the following tensor field in cylindrical coordinates; ⎛ Az 3Bz 3 ⎞ ⎛ Ar 3Brz 2 ⎞ Az 3Br 2 z Az Trr = 3 − , T , T , T T = = − + = = − ⎜ ⎟ ⎜⎜ 3 + ⎟ θθ zz rz zr ⎜ R3 R R5 R3 R5 ⎟⎠ R5 ⎟⎠ ⎝ ⎝R (divT)θ = div(TT eθ ) − tr(TT ∇eθ ) =

Trθ = Tθ r = Tθ z = Tzθ = 0, R 2 = r 2 + z 2 . -----------------------------------------------------------------------------Ans. ∂T 1 ∂Trθ Trr − Tθθ ∂Trz ∂ ⎛ Az 3Br 2 z ⎞ 3Brz ∂ ⎛ Ar 3Brz 2 ⎞ (divT) r = rr + + + = ⎜ 3− − ⎜ 3+ ⎟− ⎟ r ∂θ r ∂r ∂z ∂r ⎜⎝ R ∂z ⎜⎝ R R5 ⎟⎠ R5 R 5 ⎟⎠ ∂ 1 ∂ 1 3Bz ∂ 2 ⎞ 3Brz ⎛ ∂ 1 3Br ∂ 2 ∂ 1 ⎞ ⎛ = ⎜ Az − 3Br 2 z − 5 + 5 r ⎟ − 5 − ⎜ Ar z + 3Brz 2 ⎟ 3 5 3 ∂r R ∂z R5 ⎠ R ∂r ⎠ R R ∂z ⎝ ∂r R ⎝ ∂z R ⎛ 3 Az ∂R 15Br 2 z ∂R 6 Brz ⎞ 3Brz ⎛ 3 Ar ∂R 6 Bzr 15Brz 2 ∂R ⎞ = ⎜− 4 + − 5 ⎟ − 5 −⎜− 4 + 5 − ⎟ ⎜ R ∂r ⎜ R ∂z R 6 ∂r R ⎟⎠ R R R 6 ∂z ⎟⎠ ⎝ ⎝ ⎛ 3 Arz 15Br 3 z 6 Brz ⎞ 3Brz ⎛ 3 Arz 6 Bzr 15Brz 3 ⎞ = ⎜− 5 + − 5 ⎟− 5 +⎜ 5 − 5 + ⎟ ⎜ R ⎜ R R7 R ⎟⎠ R R R 7 ⎟⎠ ⎝ ⎝


Lai et al, Introduction to Continuum Mechanics ⎛ 15r 3 z 15rz 15rz 3 ⎞ 15rz ⎞ ⎛ 15rz 15rz ⎞ ⎛ 15rz = B ⎜ 7 − 5 + 7 ⎟ = B ⎜ 7 r2 + z2 − 5 ⎟ = ⎜ 5 − 5 ⎟ = 0 . ⎜ R ⎟ R R ⎠ R ⎠ ⎝ R R ⎠ ⎝ R ⎝

(

)

∂Tθ r 1 ∂Tθθ Trθ + Tθ r ∂Tθ z + + + =0+0+0+0=0 ∂r ∂z r ∂θ r ∂T 1 ∂Tzθ ∂Tzz Tzr ∂ ⎛ A 3Brz 2 ⎞ ∂ ⎛ Az 3Bz 3 ⎞ ⎛ A 3Bz 2 ⎞ (divT) z = zr + =− ⎜ 3 + + 5 ⎟−⎜ 3 + 5 ⎟ + + ⎟− ⎜ ∂r r ∂θ ∂z r ∂r ⎜⎝ R R5 ⎟⎠ ∂z ⎜⎝ R3 R ⎟⎠ ⎜⎝ R R ⎟⎠ (divT)θ =

⎛ A 3 Ar 2 3Bz 2 15 Br 2 z 2 ⎞ ⎛ A 3 Az 2 9 Bz 2 15Bz 4 ⎞ ⎛ A 3Bz 2 ⎞ = −⎜ 3 − 5 + 5 − − 5 + 5 − + 5 ⎟ ⎟−⎜ ⎟−⎜ ⎜R R R R 7 ⎟⎠ ⎜⎝ R 3 R R R 7 ⎟⎠ ⎜⎝ R 3 R ⎟⎠ ⎝ ⎛ 3A 3A ⎛ 3 A 3 A 15 Bz 2 15Bz 2 ⎞ 15 Bz 2 15 Bz 2 2 2 ⎞ r z = ⎜ − 3 + 5 r2 + z2 − + + == − + − + ⎟ ⎟=0. ⎜ R ⎟ ⎜⎜ R3 R3 R R5 R7 R5 R 5 ⎟⎠ ⎝ ⎠ ⎝ _________________________________________________________________ 2.79 Calculate div T for the following tensor field in spherical coordinates; 2B B Trr = A − 3 , Tθθ = Tφφ = A + 3 , Trθ = Tθ r = Tθφ = Tφθ = Trφ = Tφ r = 0 r r ------------------------------------------------------------------------------

(

)

(

)

2 1 ∂ ( r Trr ) 1 ∂ (Trθ sin θ ) 1 ∂Trφ Tθθ + Tφφ + + Ans. (divT) r = 2 r ∂r r sin θ ∂θ r sin θ ∂φ r

(

)

2 T +T 1 ∂ r Trr 1 ∂ ⎛ 2 2B ⎞ ⎛ A B ⎞ − θθ φφ = 2 = 2 ⎜ Ar − ⎟ − 2⎜ + ⎟ r ∂r r r ∂r ⎝ r ⎠ ⎝ r r4 ⎠ 1⎛ 2B ⎞ ⎛ A B ⎞ ⎛ 2 A 2B ⎞ ⎛ A B ⎞ = 2 ⎜ 2 Ar + 2 ⎟ − 2 ⎜ + 4 ⎟ = ⎜ + 4 ⎟ − 2 ⎜ + 4 ⎟ = 0. r ⎝ r ⎠ ⎝r r ⎠ ⎝ r r ⎠ ⎝r r ⎠ 3 1 ∂ ( r Tθ r ) 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ Trθ − Tθ r − Tφφ cot θ + + (divT)θ = 3 + r ∂r r sin θ ∂θ r sin θ ∂φ r T cot θ −Tφφ cot θ = θθ + = 0. r r 3 1 ∂ ( r Tφ r ) 1 ∂ (Tφθ sin θ ) 1 ∂Tφφ Trφ − Tφ r + Tθφ cot θ (divT)φ = 3 + = 0. + + r ∂r r sin θ ∂θ r sin θ ∂φ r

_________________________________________________________________ 2.80 From the equation (divT) ⋅ a = div(TT a) − tr(TT ∇a) [See Eq. 2.29.3)] verify that in spherical coordinates, the θ -component of the vector (divT) is: 1 ∂ (r 3Tθ r ) 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ Trθ − Tθ r − Tφφ cot θ + ++ + . 3 ∂r r sin θ ∂θ r sin θ ∂φ r r -----------------------------------------------------------------------------Ans. (divT) ⋅ a = div(TT a) − tr(TT ∇a) → (divT) ⋅ eθ = div(TT eθ ) − tr(TT ∇eθ ) . Now, (divT)θ =


Lai et al, Introduction to Continuum Mechanics 1 ∂ (r 2Tθ r ) 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ + + . Also, 2 ∂r ∂θ r sin θ r sin θ ∂φ r 0 ⎤ ⎡ 0 −1 / r ⎢ eθ = 0e r + 1eθ + 0eφ → [∇eθ ] = ⎢ 0 0 0 ⎥⎥ ⎢⎣ 0 0 cot θ / r ⎥⎦ ⎡ Trr Tθ r Tφ r ⎤ ⎡0 −1 / r 0 ⎤ ⎡ 0 −Trr / r Tφ r cot θ / r ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 ⎥⎥ = ⎢ 0 −Trθ / r Tφθ cot θ / r ⎥ → ⎡TT ∇eθ ⎤ = ⎢Trθ Tθθ Tφθ ⎥ ⎢0 ⎣ ⎦ ⎢ ⎥ ⎢0 0 cot θ / r ⎦⎥ ⎢⎢ 0 −Trφ / r Tφφ cot θ / r ⎥⎥ ⎣⎢Trφ Tθφ Tφφ ⎦⎥ ⎣ ⎣ ⎦ Tφφ cot θ T → tr ⎡ TT ∇eθ ⎤ = − rθ + . Thus, ⎣ ⎦ r r (divT)θ = div(TT eθ ) − tr(TT ∇eθ ) TT eθ = Tθ r e r + Tθθ eθ + Tθφ eφ → div(TT eθ ) =

=

1 ∂ (r 2Tθ r ) 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ Trθ Tφφ cot θ + + + − 2 r sin θ r sin θ ∂φ r r ∂r ∂θ r

1 ∂ (r 3Tθ r ) Tθ r 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ Trθ Tφφ cot θ − + + + − . ∂r ∂θ r r sin θ r sin θ ∂φ r r r3 _________________________________________________________________ =



CHAPTER 3 3.1 Consider the motion: x1 = (1 + kt ) X1 / (1 + kto ), x2 = X 2 , x3 = X 3 .

(a) Show that reference time is t = to . (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 = X 2 , x3 = X 3 . ----------------------------------------------------------------------------------Ans. (a) At t = to , x1 = X1 , x2 = X 2 , x3 = X 3 . Thus, t = to is the reference time. (b) In material description, v1 = kX1 / (1 + kto ), v2 = v3 = 0 . Now, from x1 = (1 + kt ) X1 / (1 + kto ) , → X1 = (1 + kto ) x1 / (1 + kt ) , therefore, → v1 = kX1 = kx1 / (1 + kt ), v2 = v3 = 0 . (c) For x1 = (1 + kt ) X1 , x2 = X 2 , x3 = X 3 , → 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 = X 2 , x3 = X 3 , where the material coordinates X i designate the position of a particle at t = 0 . (a) Determine the velocity and acceleration of a particle in both a material and a spatial description. (b) If the temperature field in spatical description is given by θ = Ax1 , what is its material description? Find the material derivative of θ , using both descriptions of the temperature. (c) Do part (b) if the temperature field is θ = Bx2 ----------------------------------------------------------------------------------Ans. (a) Material description: v1 = Dx1 / Dt = ( ∂x1 / ∂t ) X −fixed = α , v2 = v3 = 0 ,

a1 = Dv1 / Dt = ( ∂v1 / ∂t ) X

i

i − fixed

= 0, a2 = a3 = 0 .

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 Dθ ∂θ = + v1 + v2 + v3 = 0 + α ( A) + (0)(0) + (0)(0) = Aα . ∂x1 ∂x2 ∂x3 Dt ∂t (c) Using the material description: θ = BX 2 → Dθ / Dt = (∂ / ∂t )( BX 2 ) = 0 . Using the spatical description: θ = Bx2 → ∂θ ∂θ ∂θ Dθ ∂θ = + v1 + v2 + v3 = 0 + α (0) + ( 0 )( B ) + ( 0 )( 0 ) = 0 . ∂x1 ∂x2 ∂x3 Dt ∂t _______________________________________________________________________

3.3 Consider the motion

x1 = X1 , x2 = β X12t 2 + X 2 , x3 = X 3 , where X i are the material coordinates. (a) at t = 0 , the corners of a unit square are at A(0,0,0), B(0,1,0), C (1,1,0) and D(1,0,0) . Determine the position 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. -----------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. For the material line AB, ( X1 , X 2 , X 3 ) = ( 0, X 2 ,0 ) ; at t = 1 , ( x1 , x2 , x3 ) = ( 0, X 2 ,0 )

( ) ) = ( X ,0,0 ) ; at t = 1 , ( x , x , x ) = ( X , β X ,0 )

For the material line BC , ( X1 , X 2 , X 3 ) = ( X1 ,1,0 ) ; at t = 1 , ( x1 , x2 , x3 ) = X1 , β X12 + 1,0 For the material line AD, ( X1 , X 2 , X 3

1

1

2

3

1

2 1

For the material line CD, ( X1 , X 2 , X 3 ) = (1, X 2 ,0 ) ; at t = 1 , ( x1 , x2 , x3 ) = (1, β + X 2 ,0 ) The shape of the material square at t = 1 is shown in the figure.

⎛ ∂x ⎞ ⎛ ∂v ⎞ → v1 = v3 = 0, v2 = 2β X12t ; a1 = a3 = 0, a2 = 2β X12 (b) vi = ⎜ i ⎟ , ai = ⎜ i ⎟ ⎝ ∂t ⎠ X i −fixed ⎝ ∂t ⎠ X i −fixed (c) Since x1 = X1 , in spatial descrip. v1 = v3 = 0, v2 = 2 β x12t ; a1 = a3 = 0, a2 = 2β x12 __________________________________________________________________ 3.4 Consider the motion: x1 = β X 22t 2 + X1 , x2 = kX 2t + X 2 , x3 = X 3 (a) At t = 0 , the corners of a unit square are at A(0,0,0), B(0,1,0), C (1,1,0) and D(1,0,0) . Sketch 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 , X 2 , X 3 ) = ( 0, X 2 ,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = 4 β X 22 , 2kX 2 + X 2 ,0 . For material line BC , ( X1 , X 2 , X 3 ) = ( X1 ,1,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( 4 β + X1 , 2k + 1,0 ) . For material line AD, ( X1 , X 2 , X 3 ) = ( X1 ,0,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( X1 ,0,0 ) .

(

)

For mat. line CD, ( X1 , X 2 , X 3 ) = (1, X 2 ,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = 4β X 22 + 1, 2kX 2 + X 2 ,0 . The shape of the material square at t = 2 is shown in the figure. x2

4 B’ 2k

B

A

C’ C

D

x1


Lai et al, Introduction to Continuum Mechanics ∂x ⎛ ∂v ⎞ 2 2 → , v1 = 2 β X 2 t , v2 = kX 2 , v3 = 0; a1 = 2 β X 2 , a2 = a3 = 0 . b) vi = ⎛⎜ i ⎞⎟ , ai = ⎜ i ⎟ ⎝ ∂t ⎠ X − fixed ⎝ ∂t ⎠ X − fixed i

i

(c) x2 = ( kt + 1) X 2 → , v1 =

2 β x22t

(1 + kt )

, v2 = 2

2β x22 kx2 , v3 = 0; a1 = , a2 = a3 = 0 . (1 + kt ) (1 + kt )2

_________________________________________________________________ 3.5 Consider the motion: x1 = k ( s + X1 ) t + X1 , x2 = X 2 , x3 = X 3 . (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 , X 2 , X 3 ) = ( 0, X 2 ,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2ks, X 2 ,0 ) .

For material line BC , ( X1 , X 2 , X 3 ) = ( X1 ,1,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2ks + 2kX1 + X1 ,1,0 ) .

For material line AD, ( X1 , X 2 , X 3 ) = ( X1 ,0,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2ks + 2kX1 + X1 ,0,0 ) .

For material line CD, ( X1 , X 2 , X 3 ) = (1, X 2 ,0 ) ; at t = 2 , ( x1 , x2 , x3 ) = ( 2ks + 2k + 1, X 2 ,0 ) . The shape of the material square at t = 2 is shown in the figure.

x2 2ks B

2k(s+1)

B’

C’

C

1 A

A’

D’

D

x1

∂x ⎛ ∂v ⎞ (b) vi = ⎛⎜ i ⎞⎟ → , v1 = k ( s + X 1 ) , v2 = 0, v3 = 0; a1 = a2 = a3 = 0 . and ai = ⎜ i ⎟ ⎝ ∂t ⎠ X − fixed ⎝ ∂t ⎠ X − fixed i

i

Thus, for the particle ( X1 , X 2 , X 3 ) = ( 0,0,0 ) , v1 = ks, v2 = 0, v3 = 0 and a1 = 0, a2 = 0, a3 = 0

(c) x1 = k ( s + X 1 ) t + X 1 → x1 = kst + ( kt + 1) X 1 → X 1 = ( x1 − kst ) / (1 + kt ) , thus, in spatial descriptions, x − kst ⎪⎫ k ( s + x1 ) ⎪⎧ v1 = k ⎨ s + 1 , v = 0, v3 = 0 and a1 = 0, a2 = 0, a3 = 0 . ⎬= (1 + kt ) ⎭⎪ (1 + kt ) 2 ⎩⎪

At the position ( x1 , x2 , x3 ) = ( 0,0,0 ) , v1 = ks / (1 + kt ), v2 = 0, v3 = 0 and a1 = 0, a2 = 0, a3 = 0 . _________________________________________________________________ 3.6 The position at time t of a particle initially at ( X 1 , X 2 , X 3 ) is given by x1 = X 1 − 2 β X 22t 2 , x2 = X 2 − kX 3t , x3 = X 3 , where β = 1 and k = 1 .

(a) Sketch the deformed shape, at time t = 1 of the material line OA which was a straight line at t = 0 with the point O at ( 0,0,0 ) and the point A at ( 0,1,0 ) . (b) Find the velocity at t = 2 , of the particle which was at (1,3,1) at t = 0 . (c) Find the velocity of the particle which is at (1,3,1) at t=2.


Lai et al, Introduction to Continuum Mechanics --------------------------------------------------------------------------------------------Ans. With β = 1 and k = 1 , x1 = X 1 − 2 X 22t 2 , x2 = X 2 − X 3t , x3 = X 3 For the material line OA : ( X 1 , X 2 , X 3 ) = (0, X 2 , 0) : at t = 1 , x1 = −2 X 22 , x2 = X 2 , x3 = 0 . Thus, the deformed shape of the material line at t = 1 is a parabola given in the figure shown.

x2 A

A’

x1

O

(b) v1 = Dx1 / Dt = −4 X 22t , v2 = Dx2 / Dt = − X 3 , v3 = Dx3 / Dt = 0 For the particle ( X 1 , X 2 , X 3 ) = (1,3,1) , at t = 2 , v1 = −4(3) 2 (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 = X 1 − 8 X 22 , 3 = X 2 − 2 X 3 , 1 = X 3 → X 1 = 201, X 2 = 5, X 3 = 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 ( X 1 , X 2 , X 3 ) is given by: x1 = X 1 + k ( X 1 + X 2 ) t , x2 = X 2 + k ( X 1 + X 2 ) t , x3 = X 3 ,

(a) Find the velocity at t = 2 , of the particle which was at (1,1,0) at the reference time t = 0 . (b) Find the velocity of the particle which is at (1,1,0) at t = 2 . ----------------------------------------------------------------------------------------Ans. (a) v1 = Dx1 / Dt = k ( X 1 + X 2 ) , v2 = Dx2 / Dt = k ( X 1 + X 2 ) , v3 = Dx3 / Dt = 0. For the particle ( X 1 , X 2 , X 3 ) = (1,1, 0) , at t = 2 , v1 = k (1 + 1) = 2k , v2 = k (1 + 1) = 2k , 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 = X 1 + 2k ( X 1 + X 2 ) , 1 = X 2 + 2k ( X 1 + X 2 ) , 0 = X 3 . → X1 =

1 1 + 4k

, X2 =

1 1 + 4k

, X3 = 0 ,

→ v1 = v2 = k ( X 1 + X 2 ) =

2k 1 + 4k

, v3 = 0

_________________________________________________________________ 3.8 The position at time t of a particle initially at ( X1 , X 2 , X 3 ) is given by x1 = X1 + β X 22t 2 , x2 = X 2 + kX 2t , x3 = X 3 , 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 . -------------------------------------------------------------------------------------------Ans. With β = 1 and k = 1 , x1 = X1 + X 22t 2 , x2 = X 2 + X 2t , x3 = X 3


Lai et al, Introduction to Continuum Mechanics (a) t = 0 → ( x1 , x2 , x3 ) = ( X1 , X 2 , X 3 ) = (1,1,0) ,

t = 1 → ( x1 , x2 , x3 ) = ( X1 + X 22 , X 2 + X 2 , X 3 ) = (2, 2,0) t = 2 → ( x1 , x2 , x3 ) = ( X1 + 4 X 22 , X 2 + 2 X 2 , X 3 ) = (5,3,0) (b) x1 = X1 + X 22t 2 x2 = X 2 + X 2t , x3 = X 3 , at t = 2 → 1 = X1 + 4 X 22 , 3 = 3 X 2 , 2 = X 3 → X1 = −3, X 2 = 1, X 3 = 2 . (c) x1 = X1 + X 22t 2 x2 = X 2 + X 2t , x3 = X 3 → v1 = 2 X 22t , v2 = X 2 , v3 = 0 . → a1 = 2 X 22 , a2 = 0, a3 = 0 . For ( X1 , X 2 , X 3 ) = (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, X 2 = 1, X 3 = 2 . 2

Thus, → a1 = 2 X 22 = 2(1) 2 = 2, a2 = 0, a3 = 0 . _________________________________________________________________ 3.9 (a) Show that the velocity field vi = kxi / (1 + kt ) corresponds to the motion xi = X i (1 + kt ) and (b) find the acceleration of this motion in material description. ----------------------------------------------------------------------------------------Ans. (a) From xi = X i (1 + kt ) and X i = xi / (1 + kt ) → vi = kX i =kxi / (1 + kt ) . (b) vi = kX i → ai = 0 , or kx j kδ ij ∂v k 2 xi k 2 xi kxi k ⎛ ∂v ⎞ + vj i = − + = − + =0. ai = ⎜ i ⎟ 2 ∂ ∂ + + + + + 1 1 1 1 1 t x kt kt kt kt )( ) ( ) ( ) ( kt ) ⎝ ⎠ xi − fixed j (1 + kt ) ( _________________________________________________________________ 3.10 Given the two dimensional velocity field: vx = −2 y, v y = 2 x . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------∂v ∂v ∂v Ans. (a) ax = x + vx x + v y x = 0 + ( −2 y ) (0) + ( 2 x )( −2 ) = −4 x , ∂t ∂x ∂y ∂v y

∂v y

∂v y

= 0 + ( −2 y ) (2) + ( 2 x )( 0 ) = −4 y , i.e., a = −4 xe x − 4 ye y ∂t ∂x ∂y dx dy dy x (b) = −2 y and = 2x → = − → xdx + ydy = 0 , → x 2 + y 2 = constant=X 2 + Y 2 , dt dt dx y

ay =

+ vx

+ vy

dx dy d2x dy d 2x = −2 y and = 2x → = −2 = − 2 ( 2 x ) → + 4x = 0 dt dt dt dt dt → x = A sin 2t + B cos 2t and y = − A cos 2t + B sin 2t , where A = −Y , B = X . _________________________________________________________________

Or,

3.11 Given the two dimensional velocity field: vx = kx, v y = − ky . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------∂v ∂v ∂v Ans. (a) ax = x + vx x + v y x = 0 + ( kx ) ( k ) + ( − ky )( 0 ) = k 2 x ∂t ∂x ∂y



ay =

∂v y ∂t

+ vx

∂v y ∂x x

(b)

+ vy

∂v y ∂y

(

= 0 + ( kx ) (0) + ( − ky )( − k ) = k 2 y , That is, a = k 2 xe x + ye y

)

t

dx dx x = kx → ∫ = ∫ kdt → ln x − ln X = kt → ln = kt → x = Xe kt . dt x 0 X X

Similarly derivation gives → y = Ye− kt . Or, xy = XY where ( X , Y ) are material coordinates. _________________________________________________________________ 3.12 Given the two dimensional velocity field: vx = k ( x 2 − y 2 ), v y = −2kxy . Obtain the acceleration field. ----------------------------------------------------------------------------------------∂v ∂v ∂v Ans. ax = x + vx x + v y x = 0 + k x 2 − y 2 (2kx) + ( −2kxy )( −2ky ) = 2 xk 2 ( x 2 + y 2 ) . ∂t ∂x ∂y ∂v y ∂v y ∂v y ay = + vx + vy = 0 + k x 2 − y 2 ( −2ky ) − 2kxy ( −2kx ) = 2 yk 2 ( x 2 + y 2 ) . ∂t ∂x ∂y

(

(

(

That is, a = 2k 2 x 2 + y 2

) ( xe

x

+ ye y

)

)

)

_________________________________________________________________ Dv ∂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 ≠ a A+B , where

3.13 In a spatial description, the equation to evaluate the acceleration

a A and a B denote respectively the acceleration fields corresponding to the velocity fields

v A and v B each existing alone, a A+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 x1e2 , v B = 2 x2e1 − 2 x1e2 -------------------------------------------------------------------------------------------Dv ∂v Ans. From = + ( ∇v ) v Dt ∂t 0 0 −2 ⎤ ⎡ −2x2 ⎤ ⎡ −4 x1 ⎤ ⎡ 0 ⎤ ⎡ 0 2 ⎤ ⎡ 2x2 ⎤ ⎡ −4 x1 ⎤ ⎡a A ⎤ = ⎡⎢ ⎤⎥ + ⎡⎢ , ⎡a B ⎤ = = + = ⎣ ⎦ ⎣0 ⎦ ⎣ 2 0 ⎦⎥ ⎢ 2x1 ⎥ ⎢ −4 x2 ⎥ ⎣ ⎦ ⎣⎢ 0 ⎦⎥ ⎣⎢ −2 0 ⎦⎥ ⎢ −2x1 ⎥ ⎢ −4 x2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

→ a A = −4 x1e1 − 4 x2e 2 , a B = −4 x1e1 − 4 x2e 2 → a A + a B = −8 x1e1 − 8 x2e 2 . On the other hand, v A +v B =0, so that a A+B = 0 . Thus, a A + a B ≠ a A+B _________________________________________________________________ 3.14 Consider the motion: x1 = X1 , x2 = X 2 + ( sin π t )( sin π X1 ) , x3 = X 3 (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.


Lai et al, Introduction to Continuum Mechanics ----------------------------------------------------------------------------------------Ans. (a) Since x1 = X1 and x3 = X 3 , 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 = X 2 − 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 ,

a1 = 0, a2 = −π 2 ( sin π t )( sin π X1 ) , a3 = 0

Since x1 = X1 , the spatial descriptions are of the same form as above except that X1 is replaced with x1 . _________________________________________________________________ 3.15 Consider the following velocity and temperature fields: v = α ( x1e1 + x2e 2 ) / ( x12 + x22 ), 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. ----------------------------------------------------------------------------------------Ans. (a) In polar coordinates, x1e1 + x2e2 = rer , where r 2 = x12 + x22 and er is the unit vector in the r direction, so that v =

α

er , Θ =kr 2 . 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.

(b) From vr =

α

and vθ = 0 , and Eq. (3.4.12) r v ∂v v2 ∂v ∂v α2 ⎛ α ⎞⎛ α ⎞ ar = r + vr r + θ r − θ = 0 + ⎜ ⎟⎜ − 2 ⎟ + 0 + 0 = − 3 . r ∂θ r ∂t ∂r r ⎝ r ⎠⎝ r ⎠ ∂vθ ∂vθ vθ ∂vθ vr vθ aθ = + vr + + =0. ∂t ∂r r ∂θ r That is, a = −α 2 / r 3e r . At the point A(1,1,0), r = 2 , a = −α 2 / ( 2)3 e r = −α 2 2 / 4e r .

DΘ ∂Θ ∂Θ vθ ∂Θ ⎛α ⎞ = + vr + = 0 + ⎜ ⎟ ( 2kr ) = 2α k . Dt r ∂θ ∂t ∂r ⎝r⎠ _________________________________________________________________ 3.16 Do the previous problem for the following velocity and temperature fields:



v=

α ( − x2e1 + x1e2 ) x12

x22

(

, Θ=k x12 + x22

)

+ -----------------------------------------------------------------------------------------Ans. With x1 = r cos θ , x2 = r sin θ and x12 + x22 = r 2 , we have v=

α ( - x2e1 + x1e2 ) x12

x22

=

α r ( − sin θ e1 + cosθ e2 ) 2

=

α

eθ and Θ=kr 2

+ r Particles move in concentric circles with their speed inversely proportional to r . Isotherms are circles. (b) With vr = 0, vθ =

α r

, we have, from Eq.(3.4.12). 2

ar =

r

v ∂v v2 ∂vr ∂v α2 ⎛α ⎞ 1 + vr r + θ r − θ = − ⎜ ⎟ = − 3 , r ∂θ r ∂t ∂r r ⎝r⎠ r

aθ =

∂vθ ∂v v ∂v vv + vr θ + θ θ + r θ = 0 ∂t ∂r r ∂θ r

i.e., a = −α 2 / r 3e r . At the point A, r = 2 , therefore, a = − 2α 2 / 4e r

DΘ ∂Θ ⎛α ⎞ = + v ⋅ ∇Θ = 0 + ⎜ eθ ⎟ ⋅ 2kre r = 0 . Dt ∂t ⎝r ⎠ _________________________________________________________________

(

1 3.17 Consider : x = X + X1ke1 . let dX( ) = dS1 / 2

) (e1 +e2 ) & dX( 2) = ( dS2 / 2 ) ( −e1 +e2 )

be

differential material elements in the undeformed configuration. (a) Find the deformed elements 1 2 dx( ) and dx( ) . (b) Evaluate the stretches of these elements ds / dS and ds / dS and the change 1 −2

1

2

2

in the angle between them. (c) Do part (b) for k = 1 and k = 10 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 = X 2 , x3 = X 3 → [ F ] = ⎢⎢ 0 1 0 ⎥⎥ , dx = FdX → ⎢⎣ 0 0 1 ⎥⎦

⎡1 + k 0 0 ⎤ ⎡1 ⎤ dS ⎛ ⎞ ⎛ dS ⎞ 1 1 ⎢ ( ) 1 ⎡ dx ⎤ = 0 1 0 ⎥⎥ ⎢⎢1 ⎥⎥ → dx( ) = ⎜ 1 ⎟ ⎡⎣(1 + k ) e1 + e2 ⎤⎦ . ⎣⎢ ⎦⎥ ⎜⎝ 2 ⎟⎠ ⎢ ⎝ 2⎠ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡1 + k 0 0 ⎤ ⎡ −1⎤ ⎥ ⎢ 1 ⎥ → dx( 2 ) = ⎛ dS2 ⎞ ⎡ − 1 + k e + e ⎤ . ⎡ dx( 2 ) ⎤ = ⎛ dS2 ⎞ ⎢ 0 1 0 ) 1 2⎦ ⎜ ⎟ ⎜ ⎟⎣ ( ⎥⎢ ⎥ ⎣⎢ ⎦⎥ ⎝ 2 ⎠ ⎢ 2⎠ ⎝ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ (b)

ds1 ds2 ⎛ 1 ⎞ = =⎜ ⎟ dS1 dS2 ⎝ 2 ⎠

(1 + k )2 + 1 .

Let γ be the decrease in angle (from 90o ), then (π / 2 ) − γ is the angle between the two deformed differential elements. Thus,


Lai et al, Introduction to Continuum Mechanics 2 (1) ( 2) 1 ⎛ dS1 ⎞⎛ dS2 ⎞ ⎡ ⎛π ⎞ dx ⋅ dx 2 ⎤ = − (1 + k ) + 1 → cos ⎜ − γ ⎟ = = − 1 + k + 1 ( ) ⎜ ⎟⎜ ⎟ ⎦⎥ (1 + k )2 + 1 ds1ds2 ds1ds2 ⎝ 2 ⎠⎝ 2 ⎠ ⎣⎢ ⎝2 ⎠

− (1 + k ) + 1 2

sin γ =

(1 + k )2 + 1

.

(c) For k = 1 ,

ds1 ds2 5 3 , sin γ = − . = = dS1 dS2 2 5

For k = 10−2 ,

ds1 ds2 ⎛ 1 ⎞ = =⎜ ⎟ dS1 dS2 ⎝ 2 ⎠

− (1 + k ) + 1 2

sin γ =

(1 + k )

2

+1

≈

⎛ 1 ⎞ ⎟ 2 + 2k = 1 + k = 1.01 = 1.005 . ⎝ 2⎠

(1 + k )2 + 1 ≈ ⎜

−2k −k −0.01 = = → γ = −0.0099 radian ( − sign indicates increase in 2 + 2k 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 ) → E11 = [1 1 0] ⎢⎢ 0 0 0⎥⎥ ⎢⎢1 ⎥⎥ = [1 1 2 2 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

0 0⎤ ⎡ k 0 0⎤ ⎥ 0 0 ⎥ → [ E] = ⎢⎢ 0 0 0 ⎥⎥ , ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎥⎦ ⎡k ⎤ k 0] ⎢⎢ 0 ⎥⎥ = , 2 ⎢⎣ 0 ⎥⎦

ds − dS k ds k = → = 1 + = 1.005 , same as the result of part (c). 2 2 dS dS Also with ⎡ k 0 0 ⎤ ⎡ −1⎤ ⎡−k ⎤ k 1 1 1 ⎢ ⎥ ⎢ ⎥ e′2 = ( -e1 + e2 ) → E12′ = [1 1 0] ⎢ 0 0 0⎥ ⎢ 1 ⎥ = [1 1 0] ⎢⎢ 0 ⎥⎥ = − → 2 E12′ = −k 2 2 2 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ′ = E11

Thus, the decrease in angle = −k , or the increase in angle is 0.01 ≈ 0.0099 . _________________________________________________________________ 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 , X 2 , X 3 ) is defined by:

x1 = X1 + kX 3 , x2 = X 2 + kX 2 , x3 = X 3 , k = 10−5 . (a) Find the components of the strain tensor and (b) find the unit elongation of an element initially in the direction of e1 + e2 .


Lai et al, Introduction to Continuum Mechanics ----------------------------------------------------------------------------------------Ans. (a) u1 = x1 − X1 = kX 3 , u2 = x2 − X 2 = kX 2 , u3 = x3 − X 3 = 0 0 k / 2⎤ ⎡0 0 k ⎤ ⎡ 0 T ∇u ] + [ ∇u ] [ ⎢ ⎥ ⎢ → [ ∇u ] = ⎢ 0 k 0 ⎥ → [ E ] = =⎢ 0 0 ⎥⎥ k 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ k / 2 0 0 ⎥⎦ 0 k / 2 ⎤ ⎡1 ⎤ ⎡ 0 −5 1 1 ⎢ ⎥ ⎢1 ⎥ = k = 10 ′ ′ ′ ′ → = E k (b) Let e1′ = 1 1 0 0 0 e + e → E = e ⋅ Ee ( 1 2 ) 11 1 1 11 [ ]⎢ ⎥⎢ ⎥ 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( ) = dX e and dX( ) = dX e that emanate from a particle designated by X = e + e . (b) 1 1

2 2

1

Sketch deformed positions of these two elements. ----------------------------------------------------------------------------------------⎡ 4kX1 + kX 2 kX1 0 ⎤ 0 2kX 2 0 ⎥⎥ , Ans. (a) [∇u ] = ⎢⎢ ⎢⎣ 0 0 0 ⎥⎦

⎡ 5k k 0 ⎤ ⎡ 5k k / 2 0 ⎤ ⎢ ⎥ At ( X1 , X 2 , X 3 ) = (1,1,0 ) , [∇u ] = ⎢ 0 2k 0 ⎥ → [ E] = ⎢⎢ k / 2 2k 0 ⎥⎥ . ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ 1 2 Unit elong. for dX( ) = dX1e1 is E11 = 5k = 5 × 10−4 , unit elong. for dX( ) = dX 2e 2 is E22 = 2k = 2 × 10−4 . Decrease in angle between them is 2 E12 = k = 10−4 radian . 1 1 1 1 (b) For dX( ) = dX1e1 , dx( ) = dX( ) + ( ∇u ) dX( ) = dX1e1 + 5kdX1e1 = (1 + 5k ) dX1e1 , 2 For dX( ) = dX e , 2 2

2 2 2 dx( ) = dX( ) + ( ∇u ) dX( ) = dX 2e 2 + (kdX 2e1 + 2kdX 2e 2 ) = kdX 2e1 + (1 + 2k )dX 2e 2

The deformed positions of these two elements are shown below:

(1 + 2 k) dX 2

k dX 2

P’

P

u2=k

dX 2

(1 + 5 k) dX1

dX1

u1= 3k


2

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⎤ 1 k 10−4 ′ = e1′ ⋅ Ee1′ = [1 1 1] ⎢ 0 0 0 ⎥ ⎢1⎥ = = . direction is E11 ⎢ ⎥⎢ ⎥ 3 3 3 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣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 OA ' , (see Figure), where OA = 3 and

OA ' = (1 + k ) 2 + 1 + 1 = 3 + 2k + k 2 = 3(1 + 2k / 3 + k 2 / 3) → OA '− OA = 3(1 + 2k / 3 + k 2 / 3)1/2 − 3 .

(

Using binomial theorem, 1 + 2k / 3 + k 2 / 3

)

1/2

= 1 + (1 / 2)(2k / 3) + ... ≈ 1 + k / 3

Thus, OA '− OA = 3(1 + k / 3) − 3 = 3k / 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 ⎤ given by the matrix [ E ] = ⎢⎢ 3 4 −1⎥⎥ × 10−4 . (a) What is the unit elongation in the direction of ⎢⎣ 0 −1 2 ⎥⎦ 2e1 + 2e 2 + 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:


Lai et al, Introduction to Continuum Mechanics ⎡5 3 0 ⎤ ⎡ 2⎤ 1 58 ′ = e1′ ⋅ Ee1′ = [ 2 2 1] ⎢ 3 4 −1⎥ ⎢ 2 ⎥ × 10−4 = × 10−4 . E11 ⎢ ⎥ ⎢ ⎥ 9 9 ⎢⎣ 0 −1 2 ⎥⎦ ⎢⎣ 1 ⎥⎦ 1 Let e′2 = ( 3e1 − 6e3 ) , then the decrease in angle between the two elements is: 45 ⎡5 3 0 ⎤ ⎡ 3 ⎤ 2 32 ′ = 2e1′ ⋅ Ee′2 = × 10−4 rad . 2 E12 [ 2 2 1] ⎢⎢3 4 −1⎥⎥ ⎢⎢ 0 ⎥⎥ × 10−4 = 3 45 45 ⎢⎣0 −1 2 ⎥⎦ ⎢⎣ −6 ⎥⎦ _________________________________________________________________ 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 37 ⎛1⎞ ′ = e1′ ⋅ Ee1′ = ⎜ ⎟ [3 −4 0] ⎢ 3 4 −1⎥ ⎢ −4 ⎥ × 10−4 = × 10−4 = 1.48 × 10−4 E11 ⎢ ⎥ ⎢ ⎥ 5 25 ⎝ ⎠ ⎢⎣ 0 −1 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ (b) Let e1'' =

1 1 ( 3e1 − 4e3 ) and e''2 = ( 4e1 + 3e3 ) , then the decrease in angle between these two 5 5

elements is:

⎡5 3 0 ⎤ ⎡ 4⎤ 2 1 72 ⎛ ⎞ = 2e1'' ⋅ Ee''2 = 2 ⎜ ⎟ [3 0 −4] ⎢⎢ 3 4 −1⎥⎥ ⎢⎢ 0 ⎥⎥ × 10−4 = × 10−4 = 2.88 × 10−4 rad . 25 ⎝5⎠ ⎢⎣ 0 −1 2 ⎥⎦ ⎢⎣ 3 ⎥⎦ _________________________________________________________________ ''' 2 E12

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 ⎤ ⎡3 0 0⎤ ⎢ ⎥ −4 ⎢ [E] = ⎢3 4 −1⎥ × 10 , ⎢0 6 0 ⎥⎥ × 10−4 ⎢⎣ 0 −1 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ----------------------------------------------------------------------------------------Ans. (a) I1 = ( 5 + 4 + 2 ) × 10−4 = 11 × 10−4 , I2 =

5 3 3 4

× 10−8 +

4

−1

−1

2

× 10−8 +

5 0 0 2

× 10−8 = 28 × 10−8

5 3 0 I 3 = 3 4 −1 × 10−12 = 17 × 10−12 0 −1 2


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) ⎤ ⎢ ( ⎡ Ans. For ⎢ T ⎥ = ⎢τ 0 0 ⎥⎥ , I1 = 0, I 2 = −τ 2 , I 3 = 0 . ⎣ ⎦ ⎢⎣0 0 0 ⎥⎦ {ei }

⎡0 2) ⎤ ⎢ ( ⎡ For ⎢ T ⎥ = ⎢ −τ ⎣ ⎦ ⎢⎣ 0

−τ 0 0

0⎤ 0 ⎥⎥ 0 ⎥⎦

I1 = 0, I 2 = −τ 2 , I 3 = 0

{ei }

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 (e + e ) / 2 , whereas the eigenvector for T( ) is 1

2

(e1 − e 2 ) / 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 X 3 + X12 , k = 10−6 , find the maximum unit elongation for an element that is initially at (1,0,0 ) . ----------------------------------------------------------------------------------------⎡ 2kX1 0 0 ⎤ ⎢ ⎥ 0 kX 3 kX 2 ⎥ , thus, for ( X1 , X 2 , X 3 ) = (1,0,0 ) , Ans. [∇u ] = ⎢ ⎢ k ( 2 X 3 + 2 X1 ) 0 2kX1 ⎥ ⎣ ⎦ 2 k 0 0 2 k 0 k ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ [∇u ] = ⎢ 0 0 0 ⎥ → [E] = ⎢ 0 0 0 ⎥⎥ , the characteristic equation for this tensor is: ⎢⎣ 2k 0 2k ⎥⎦ ⎢⎣ k 0 2k ⎥⎦ 2k − λ

0

0 k

0−λ 0

k

2 0 = 0 → ( −λ ) ⎡⎢( 2k − λ ) − k 2 ⎤⎥ = 0 → λ1 = 0, λ2 = 3k , λ3 = k . ⎣ ⎦ 2k − λ


Lai et al, Introduction to Continuum Mechanics Thus, the maximum unit elongation at (1,0,0 ) is λ2 = 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 k2 be so that no element changes its volume? ----------------------------------------------------------------------------------------Δ ( dV ) = E11 + E22 + E33 = ( k1 − 2k2 ) X 2 = 0 . Thus, the particles which were on the plane Ans. (a) dV X 2 = 0 do not suffer any change of volume. (b) If ( k1 − 2k2 ) = 0, ie., k1 = 2k2 , then no element changes its volume. _________________________________________________________________ 3.28 The displacement components for a body are:

u1 = k ( X12 + X 2 ), u2 = k (4 X 32 − 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 0 ⎤ ⎡ 2kX1 ⎡ 2kX1 k ⎢ ⎢ ⎥ 0 4kX 3 ⎥⎥ Ans. (a) [∇u ] = ⎢ − k 0 8kX 3 ⎥ → [ E] = ⎢ 0 ⎢⎣ 0 ⎢⎣ 0 0 0 ⎥⎦ 4kX 3 0 ⎥⎦ ⎡ 2k (b) At (1, 2,1) , [ E] = ⎢⎢ 0 ⎢⎣ 0 for

e1'

=

1 2

( e1 + e2 ) ,

E11'

0 0 4k

0⎤ 4k ⎥⎥ , 0 ⎥⎦ ⎡ 2k 1 = [1 1 0] ⎢⎢ 0 2 ⎢⎣ 0 2k − λ 0 0

= e1' ⋅ Ee1'

(c) The characteristic equation is

−λ 4k

0 0

0 0 4k

0 ⎤ ⎡1 ⎤ 4k ⎥⎥ ⎢⎢1 ⎥⎥ = k 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

2 4k = 0 → ( 2k − λ ) ⎢⎡λ 2 − ( 4k ) ⎥⎤ = 0 ⎣ ⎦ −λ

→ λ1 = 2k , λ2 = 4k , λ3 = −4k . The maximum elongation is 4k . (d) Change of volume per unit volume = Eii = 2kX1 , which is a function of X1 . Thus, 1

1

ΔV = ∫ 2kX1dV = 2∫ kX1 (1) dX1 = kX12 = k . o

o

_________________________________________________________________


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 + Ekk ) = ρo where ρo denote the initial density and ρ , the current density. (b) Use the smallness of Ekk to show that the current density is given by ρ = ρo (1 − Ekk ) . ----------------------------------------------------------------------------------------⎛ ΔdV ⎞ dV + ΔdV ρ dV =ρ o = ρ ⎜1 + Ans. (a) ρo dVo = ρ dV → ρ o = ⎟, dVo dVo dVo ⎠ ⎝ ΔdV For small deformation, = Ekk → ρo = ρ (1 + Ekk ) . dVo (b) From bionomial theorem, for small Ekk , (1+Ekk )

−1

≈ 1 − Ekk , thus,

−1

ρ = ρo (1 + Ekk ) = ρo (1 − Ekk ) . _________________________________________________________________ 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 = 3k , 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 k 0 ⎥ → k k − λ 0 = 0, → ( 3k − λ ) ⎢⎡⎣( k − λ )2 − k 2 ⎥⎤⎦ = 0 ⎢⎣ 0 0 3k ⎥⎦ 0 0 3k − λ

(

)

→ ( 3k − λ ) −2λ k + λ 2 = 0 → λ1 = 3k , λ2 = 0, λ3 = 2k . 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 ?


Lai et al, Introduction to Continuum Mechanics ----------------------------------------------------------------------------------------Ans.

With e1′ =

1 2

( e1 + e2 ) , we have,

⎡ E11 E12 E13 ⎤ ⎡1 ⎤ 1 1 ′ = e1′ ⋅ Ee1′ = [1 1 0] ⎢ E21 E22 E23 ⎥ ⎢1 ⎥ = ( E11 + E12 + E21 + E22 ) , with E12 = E21 , E11 ⎢ ⎥ ⎢ ⎥ 2 2 ⎢⎣ E31 E32 E33 ⎥⎦ ⎢⎣ 0 ⎥⎦ ( E + E22 ) 1 ′ = ( E11 + 2 E12 + E22 ) → E12 = E11 ′ − 11 E11 . Thus, the strain components are: 2 2 (a + c) E11 = a, E22 = c, E12 = b − . 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 e 2 respectively. (b) Find the principal directions, assuming

E31 = E32 = E33 = 0 . (c) How will the result of part b be altered if E33 ≠ 0 . ----------------------------------------------------------------------------------------′ = 50 × 10−6 and E22 = 100 × 10−6 , we have, from the results Ans. (a) With E11 = 200 × 10−6 , E11 ′ − of the previous problem, E12 = E11

E11 − λ

E12

E12 0

E22 − λ 0

(b)

E11 + E22 ⎛ 200 + 100 ⎞ −6 −6 = ⎜ 50 − ⎟ × 10 = −100 × 10 2 2 ⎝ ⎠

0 2 ⎤ 0 = 0 → λ ⎡ ( E11 − λ )( E22 − λ ) − E12 =0 ⎣ ⎦ −λ

(

)

2 ⎤ → λ ⎡ + λ 2 − λ ( E11 + E22 ) + E11E22 − E12 =0, ⎣ ⎦

λ1,2 =

( E11 + E22 ) ± ( E11 − E22 )2 + 4 E122 2

, λ3 = 0 ,

thus,

⎡ ( 200 + 100 ) ± ( 200 − 100 )2 + 4 ( −100 )2 ⎤⎥ −6 261.8 × 10−6 ⎢ × 10 = , λ3 = 0 λ1,2 = ⎢ ⎥ 2 38.2 × 10−6 ⎣⎢ ⎦⎥ The principal direction for λ3 is e3 . The principal directions corresponding to the other two eigenvalues lie on the plane of e1 and e2 . Let n = α1e1 + α 2e 2 ≡ cosθ e1 + sin θ e 2 , then ( E11 − λ )α1 + E12α 2 = 0 ,



→

( λ − E11 ) α2 = tan θ = , α1 E12

For λ1 = 261.8 × 10−6 , tanθ = Or, n = 0.851e1 − 0.525e 2 For λ2 = 38.2 × 10−6 , tanθ =

λ1 − E11 E12

λ2 − E11 E12

=

261.8 − 200 61.8 = = −0.618 → θ = −31.7o , −100 −100

=

38.2 − 200 = 1.618 → θ = 58.3o −100

Or, n = 0.525e1 + 0.851e 2 . (c) If E33 ≠ 0 , then the principal strain corresponding to the direction e3 is E33 instead of zero. Nothing else changes. _________________________________________________________________ ′ = E22 = 1000 × 10−6 . 3.34 Repeat the previous problem with E11 = E11 ----------------------------------------------------------------------------------------E + E22 ⎛ 2000 ⎞ ′ − 11 Ans. (a) From the results of Problem 3.32, E12 = E11 = ⎜ 1000 − × 10−6 = 0 , 2 2 ⎟⎠ ⎝ ⎡10−3 0 0 ⎤ ⎢ ⎥ (b) and (c) [ E ] = ⎢ 0 10−3 0 ⎥ , the principal strains are 10−3 in any directions lying on the ⎢ ⎥ 0 E33 ⎥ ⎢⎣ 0 ⎦ plane of e1 and e 2 and the principal strain E33 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 + 3e 2 and 2 2 respectively, what are the strain components E11 , E22 and E12 ?

(

)

(

)

----------------------------------------------------------------------------------------Ans.

With e1' = (e1 + 3e 2 ) / 2, e1'' = ( −e1 + 3e2 ) / 2 , we have,



E11'

= e1' ⋅ Ee1'

1 = ⎣⎡1 4

"' E11

= e1" ⋅ Ee1"

1 = ⎣⎡ −1 4

⎡ E11 3 0 ⎦⎤ ⎢⎢ E21 ⎢⎣ E31 ⎡ E11 3 0 ⎦⎤ ⎢⎢ E21 ⎢⎣ E31

E12 E22 E32 E12 E22 E32

E13 ⎤ ⎡ 1 ⎤ ⎢ ⎥ 1 E23 ⎥⎥ ⎢ 3 ⎥ = E11 + 2 3E12 + 3E22 (i) 4 E33 ⎥⎦ ⎢⎣ 0 ⎥⎦ E13 ⎤ ⎡ −1 ⎤ ⎢ ⎥ 1 E23 ⎥⎥ ⎢ 3 ⎥ = E11 − 2 3E12 + 3E22 4 E33 ⎥⎦ ⎢⎣ 0 ⎥⎦

(

)

(

)

(ii)

'' E ' − E11 1 1 b−c ''' , E11 = a . − E11 ⎤ = [ 2b + 2c − a ] , E12 = 11 (i) & (ii), → E22 = ⎡ 2 E11' + 2 E11 = ⎣ ⎦ 3 3 3 3 _________________________________________________________________

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 E22 = [ 2b + 2c − a ] = ⎡⎣( 2 )(1) + ( 2 )(1.5 ) − 2 ⎤⎦ × 10−6 = 1 × 10−6 , 3 3 b−c 1 E12 = =− × 10−6 , E11 = 2 × 10−6 . 3 2 3 _________________________________________________________________ 3.37 Repeat the previous problem for the case a = b= c = 2000 × 10−6 . ----------------------------------------------------------------------------------------1 1 Ans. E22 = [ 2b + 2c − a ] = ⎡⎣( 2 )( 2000 ) + ( 2 )( 2000 ) − 2000 ⎤⎦ × 10−6 = 2 × 10−3 , 3 3 b−c E12 = = 0 , E11 = 2 × 10−3 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 + e2 ) / 2 at x = 5e1 + 3e 2 . ----------------------------------------------------------------------------------------Ans. v1 = kx22 , v2 = v3 = 0 , ⎡ 0 2kx2 0 → [∇v ] = ⎢⎢ 0 ⎢⎣ 0 0

0⎤ ⎡ 0 S ⎥ 0 ⎥ → [ D] = [∇v ] = ⎢⎢ kx2 ⎢⎣ 0 0 ⎥⎦ (b) At the position x = 5e1 + 3e 2 , ⎡ 0 3k [ D] = ⎢⎢3k 0 ⎢⎣ 0 0

0⎤ 0 ⎥⎥ , 0 ⎥⎦

kx2 0 0

0⎤ 0 ⎥⎥ , 0 ⎥⎦

[ W ] = [∇v ]

⎡ 0 3k [ W ] = ⎢⎢ −3k 0 ⎢⎣ 0 0


A

0⎤ 0 ⎥⎥ 0 ⎥⎦

⎡ 0 = ⎢⎢ − kx2 ⎢⎣ 0

kx2 0 0

0⎤ 0 ⎥⎥ 0 ⎥⎦

Lai et al, Introduction to Continuum Mechanics For the element dx = dsn with n = (e1 + e2 ) / 2 , the rate of extension is: ⎡ 0 3k 0 ⎤ ⎡1 ⎤ 1 D( n )( n ) = n ⋅ Dn = [1 1 0] ⎢⎢3k 0 0 ⎥⎥ ⎢⎢1 ⎥⎥ = 3k . 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ _________________________________________________________________ ⎛ t+k ⎞ 3.39 For the velocity field: v = α ⎜ ⎟ e1 , find the rates of extension for the following material ⎝ 1 + x1 ⎠ 1 2 elements: dx( ) = ds e and dx( ) = ds / 2 e + e at the origin at time t = 1 .

(

1 1

)( 1

2

2

)

----------------------------------------------------------------------------------------⎡ −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 ⎤ ⎢ ⎥ 0 0 0⎥ . At t = 1 and at ( x1 , x2 , x3 ) = ( 0,0,0 ) , [ D] = ⎢ ⎢ 0 0 0 ⎥⎦ ⎣ 1 2 Rate of extension for dx( ) = ds1e1 is D11 = −α (1 + k ) ; for dx( ) = ds2 / 2 ( e1 + e 2 ) , it is:

(

)

⎡ −α (1 + k ) 0 0 ⎤ ⎡1 ⎤ 1 1 ⎢ ⎥ = [1 1 0] ⎢ 0 0 0 ⎥ ⎢⎢1 ⎥⎥ = − α (1 + k ) 2 2 ⎢ 0 0 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎣ _________________________________________________________________ D11'

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( ) = ds e , dx( ) = ds e and dx( ) = ds / 2 e + e . 1 1

2 2

(

3

)( 1

2

)

----------------------------------------------------------------------------------------Ans. (a) With v1 = 0, v2 = ( cos t )( sin π x1 ) , v3 = 0 , ⎡ 0 0 0⎤ 0 ⎡ ⎢ ⎢ ⎥ [∇v ] = ⎢π cos t cos π x1 0 0⎥ → [ D] = ⎢(π cos t cos π x1 ) / 2 ⎢ ⎢⎣ 0 0 0 ⎥⎦ 0 ⎣

⎡ − (π cos t cos π x1 ) / 2 0 ⎢ 0 [ W ] = ⎢(π cos t cos π x1 ) / 2 ⎢ 0 0 ⎣ ⎡ 0 (b) At t = 0 and ( x1 , x2 , x3 ) = ( 0,0,0 ) , [ D] = ⎢⎢π / 2 ⎢⎣ 0

(π cos t cos π x1 ) / 2 0 0

0⎤ ⎥ 0⎥ . 0 ⎥⎦

π / 2 0⎤ 0 0

0 ⎥⎥ . 0 ⎥⎦


0⎤ ⎥ 0⎥ , 0 ⎥⎦

Lai et al, Introduction to Continuum Mechanics 1 2 For dx( ) = ds1e1 , rate of extension is D11 =0, for dx( ) = ds2e 2 , D22 = 0 and

π / 2 0 ⎤ ⎡1 ⎤ ⎡ 0 π 1 3) ⎢ ( ' 0 0 ⎥⎥ ⎢⎢1 ⎥⎥ = for dx = ds3 / 2 ( e1 + e 2 ) , D11 = [1 1 0] ⎢π / 2 2 2 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ _________________________________________________________________

(

)

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 ⎤ ⎢ ⎥ Ans. [∇v ] = ⎢ −1 0 1 ⎥ → [ D] = ⎢⎢ 0 0 0 ⎥⎥ ⎢⎣ 1 −1 0 ⎥⎦ ⎢⎣0 0 0 ⎥⎦ Therefore, the velocity field is a rigid body motion.. _________________________________________________________________ 1 3.42 Given the velocity field v = e r , (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 1 ⎛ ∂vr ⎞ ∂vr ⎤ ⎡ 1 ⎤ − vθ ⎟ ⎢ ⎥ ⎢− 2 0 0⎥ ⎜ θ r r z ∂ ∂ ∂ ⎝ ⎠ ⎢ ⎥ ⎢ r ⎥ ⎢ ∂vθ 1 ⎛ ∂vθ 1 ⎞ ∂vθ ⎥ ⎢ ⎥= D , W = 0 . ∇ v v = + = 0 0 [ ] ⎢ ⎥ ⎢ r⎟ ⎜ 2 ⎥ [ ] [ ] [ ] r ⎠ ∂z ⎥ ⎢ ⎢ ∂r r ⎝ ∂θ ⎥ ⎢ ∂v 0 0⎥ ∂vz ⎥ ⎢ 0 1 ∂vz ⎢ z ⎥ ⎢ ⎥⎦ ∂z ⎥⎦ ⎣ r ∂θ ⎢⎣ ∂r

(b) The rate of extension for a radial element is Drr = −

1

. r2 _________________________________________________________________ 3.43 Given the two-dimensional velocity field in polar coordinates:

4 r (a) Find the acceleration at r = 2 and (b) find the rate of deformation tensor at r = 2 . ----------------------------------------------------------------------------------------vr = 0, vθ = 2r +

2 (v ) v ⎛ ∂v ∂vr ∂v 1⎛ 4⎞ ⎞ + vr r + θ ⎜ r − vθ ⎟ = − θ = − ⎜ 2r + ⎟ , r ⎝ ∂θ r r⎝ r⎠ ∂t ∂r ⎠ ∂v ∂v v ⎛ ∂v ⎞ aθ = θ + vr θ + θ ⎜ θ + vr ⎟ = 0 . At r = 2 , ar = −(6) 2 / 2 = −18 , aθ = 0 . ∂t ∂r r ⎝ ∂θ ⎠ 2

Ans. (a) Using Eq. (3.4.12), ar =


Lai et al, Introduction to Continuum Mechanics ⎡ ∂vr ⎢ ∂r ⎢ (b) Eq. (2.34.5) → [∇v ] = ⎢ ∂vθ ⎢ ⎣ ∂r ⎡

⎞⎤ ⎡ 0 ⎟⎥ ⎠⎥ ⎢ =⎢ 1 ⎛ ∂vθ ⎞ ⎥ ⎢ ∂vθ + vr ⎟ ⎥ ⎢ r ⎜⎝ ∂θ ⎠ ⎦ ⎣ ∂r

1 ⎛ ∂vr − vθ r ⎝⎜ ∂θ

−

4 ⎞⎤ ⎛ vθ ⎤ ⎡ 0 − ⎜ 2 + 2 ⎟⎥ ⎢ ⎥ r ⎠⎥ ⎝ r . ⎥=⎢ ⎢⎛ ⎥ 4 ⎞ ⎥ 0 0 ⎥ ⎥⎦ ⎢⎜⎝ 2 − r 2 ⎟⎠ ⎣ ⎦

−4 / r 2 ⎤ ⎡ 0 −1⎤ ⎥ , at r = 2 , [ D] = ⎢ ⎥. ⎢⎣ −4 / r 2 0 ⎥⎦ ⎣ −1 0 ⎦ _________________________________________________________________

[ D] = [∇v ]S = ⎢

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), 2 vφ ⎛ ∂vr vφ2 vθ ⎛ ∂vr B ⎞ 1⎛ ⎞ ⎞ ar = + vr + ⎜ − vθ ⎟ + − vφ sin θ ⎟ = − = − ⎜ Ar + 2 ⎟ sin 2 θ ⎜ r ⎝ ∂θ r r⎝ ∂t ∂r ⎠ r sin θ ⎝ ∂φ r ⎠ ⎠

∂vr

∂vr

2 vφ ⎛ ∂vθ vφ2 vθ ⎛ ∂vθ cos θ sin θ ⎛ B ⎞ ⎞ ⎞ + vr + ⎜ + vr ⎟ + − vφ cos θ ⎟ = − cot θ = − aθ = ⎜ Ar + 2 ⎟ ⎜ ∂t ∂r r ⎝ ∂θ r r ⎝ ⎠ r sin θ ⎝ ∂φ r ⎠ ⎠

∂vθ

aφ =

∂vφ ∂t

∂vθ

+ vr

∂vφ ∂r

+

vθ ∂vφ r ∂θ

+

(b) Eq. (2.35.25) → ⎡ ∂vr ⎛ 1 ∂vr vθ − ⎢ ⎜ r ⎢ ∂r ⎝ r ∂θ ⎢ ∂v v 1 ∂v [∇v ] = ⎢⎢ ∂rθ ⎛⎜ r ∂θθ + rr ⎝ ⎢ ⎢ ∂vφ 1 ∂vφ ⎢ r ∂θ ⎢⎣ ∂r

⎛ ∂vφ ⎞ + vr sin θ + vθ cos θ ⎟ = 0 ⎜ r sin θ ⎝ ∂φ ⎠ vφ

⎤ vφ ⎤ ⎥ ⎡ 0 0 − ⎢ ⎥ ⎥ r ⎥ ⎢ ⎥ vφ cot θ ⎥ ⎥ =⎢ 0 0 − ⎢ ⎥ , thus ⎥ r ⎥ ⎥ ⎢ ∂v ∂vφ ⎢ ⎥ 1 φ ⎥ ⎞ , 0 ⎢ ⎥ ⎟ ⎥ ⎣ ∂r r ∂θ ⎦ ⎠ ⎥⎦ the nonzero components of rate of deformation tensor are: 1 ⎛ vφ ∂vφ ⎞ 3B Drφ = ⎜ − + ⎟ = − 3 sin θ , ∂r ⎠ 2⎝ r 2r 1 ⎛ vφ cot θ 1 ∂vφ ⎞ 1 ⎡ ⎛ B⎞ ⎛ B ⎞⎤ Dθφ = ⎜ − + ⎟ = ⎢ − ⎜ A + 3 ⎟ + ⎜ A + 3 ⎟ ⎥ cos θ = 0 . 2⎝ r r ∂θ ⎠ 2 ⎣ ⎝ r ⎠ ⎝ r ⎠⎦ _________________________________________________________________ ⎛ 1 ∂vr vφ ⎞ − ⎟ ⎜ r ⎠ ⎝ r sin θ ∂φ ⎛ 1 ∂vθ vφ cot θ ⎞ ⎞ − ⎜ ⎟ ⎟ r ⎠ ⎠ ⎝ r sin θ ∂φ ⎛ 1 ∂vφ vr vθ cot θ + + ⎜ r r ⎝ r sin θ ∂φ

⎞ ⎟ ⎠

3.45 A motion is said to be irrotational if the spin tensor vanishes. Show that the following velocity field is irrotational: −x e + x e v = 2 2 2 1 2 , r 2 = x12 + x22 r -----------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics ⎡ ∂v1 ⎢ ∂x x2 x1 2 1 2 2 Ans. v1 = − 2 , v2 = 2 , r = x1 + x2 , → [∇v ] = ⎢ ⎢ ∂v2 r r ⎢ ⎣ ∂x1

r 2 = x12 + x22 → 2r

∂v1 ⎤ ⎡ 2 x2 ∂r ∂x2 ⎥ ⎢ r 3 ∂x1 ⎥=⎢ ∂v2 ⎥ ⎢ 1 2 x1 ∂r ⎥ ⎢ − ∂x2 ⎦ ⎣ r 2 r 3 ∂x1

−

2 x2 ∂r ⎤ ⎥ r r 3 ∂x2 ⎥ , ⎥ 2 x ∂r − 31 ⎥ r ∂x2 ⎦ 1

2

+

x ∂r ∂r x1 ∂r = 2 x1 → = , also, = 2, r ∂x1 ∂x1 r ∂x2

⎡ 2 x1 x2 x22 − x12 ⎤ ⎢ ⎥ 4 S r4 ⎥ = [∇v ] → [ W ] = 0. [∇v ] = ⎢⎢ 2 r 2 ⎥ 2x x x −x ⎢ 2 4 1 − 14 2 ⎥ r ⎦ ⎣ r _________________________________________________________________ 1 2 3.46 Let dx( ) = ds1n and dx( ) = ds2m 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 ⎢ ⎥ cosθ − sin θ ds2 Dt ⎦ Dt ⎣ ds1 Dt where θ is the angle between m and n . 1 2 (b) Consider the case of dx( ) = dx( ) , what does the above formula reduce to?

(c) Consider the case where θ =

π

1 2 , i.e., dx( ) and dx( ) are perpendicular to each other, where

2 does the above formula reduces to? ----------------------------------------------------------------------------------------Ans. ⎛ (1) Ddx( 2 ) ⎞ D ⎛ D (1) ⎞ 1) 2) 2) ( ( ( ⎟ = ( ∇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( ) = 2dx( ) ⋅ Ddx( ) .

)

(

(

)

(

)

{(

)

(

)}

1 2 With dx( ) = ds1n and dx( ) = ds2m , the above formula give, D D ( ds1ds2n ⋅ m ) = 2ds1ds2 ( n ⋅ Dm ) → ( ds1ds2 cosθ ) = 2ds1ds2 ( n ⋅ Dm ) . Thus, Dt Dt Dds1 Dds2 D cos θ ( ds2 cosθ ) + ( ds1 cosθ ) + ( ds1ds2 ) = 2ds1ds2 ( n ⋅ Dm ) , Dt Dt Dt ⎧⎪ 1 D ( ds1 ) 1 D ( ds2 ) ⎫⎪ Dθ →⎨ + = 2 ( n ⋅ Dm ) = 2 ( m ⋅ Dn ) . ⎬ cos θ − sin θ ds Dt ds Dt Dt 2 ⎩⎪ 1 ⎭⎪

⎧⎪ 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 dx( ) perpendicular to dx( ) , θ = 90o , we have, −

Dθ = 2 ( n ⋅ Dm ) = 2 Dnm . Dt


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( ) = ds e , dx( ) = ds e and dx( ) = ds e be 1 1

{

2 2

3 3

}

1 2 3 three material elements. Consider the material derivative ( D / Dt ) dx( ) ⋅ dx( ) × dx( ) and show

1 D ( dV ) = D1 + D2 + D3 , where dV = ds1ds2 ds3 . dV Dt ----------------------------------------------------------------------------------------Ans. Since the principal directions are (or can always be chosen to be) mutually perpendicular, 1 2 3 therefore, dx( ) ⋅ dx( ) × dx( ) = ds ds ds = dV . that

→

D ( dV )

=

D ( ds1ds2 ds3 )

1

= ds2 ds3

2

3

D ( ds1 )

+ ds1ds3

D ( ds2 )

+ ds1ds2

D ( ds3 )

, Dt Dt Dt 1 D ( ds2 ) 1 D ( ds3 ) → = + + = D11 + D22 + D33 . dV Dt ds1 Dt ds2 Dt ds3 Dt _________________________________________________________________ Dt 1 D ( dV )

Dt 1 D ( ds1 )

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 = ω × n Dt ----------------------------------------------------------------------------------------D Dn Dds 1 Dds ⎞ ⎛ Dn ⎛ Dn ⎞ Ans. (a) = ds ⎜ +n = ds ⎜ + (n ⋅ Dn)n ⎟ . [see ( dsn ) = ds + n ⎟ Dt Dt Dt ds Dt ⎠ ⎝ Dt ⎝ Dt ⎠ D D Eq.(3.13.12) ]. We also have, ( dsn ) = ( dx ) = ( ∇v ) dx = ds ( ∇v ) n , therefore, Dt Dt Dn ⎛ Dn ⎞ ( ∇v ) n = ⎜ + n ( n ⋅ Dn ) ⎟ → = ( ∇v ) n − n ( n ⋅ Dn ) = ( D + W ) n − n ( n ⋅ Dn ) . Dt ⎝ Dt ⎠ (b) If n is an eigenvector of D , then Dn = λn , therefore, Dn Dn = Wn . = ( D + W ) n − n ( n ⋅ Dn ) = λ n + Wn − nλ = Wn . That is, Dt Dt Since W is antisymmetric → Wn = ω × n , where ω is the dual vector for W . Thus Dn = Wn = ω × n . Dt That is, the principal axes of D rotates with an angular velocity given by the dual vector of the spin tensor. _________________________________________________________________

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. ----------------------------------------------------------------------------------------2


Lai et al, Introduction to Continuum Mechanics ∂v1 ∂v2 ∂v3 + + = 0 → 0 − x1 + kx1 = 0 → k = 1 ∂x1 ∂x2 ∂x3 _________________________________________________________________ Ans.

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 v ∂v v ⎛ 1 ⎞ (k cos θ ) k cos θ (k cos θ ) ⎛ 1⎞ 1 vr = → r = ( k cos θ ) ⎜ − ⎟ 3/2 , r = → r + r =⎜ ⎟ . 3/2 ∂r ∂r r ⎝ 2 ⎠ r 3/2 r r r ⎝ 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 f ( r ) = 0 . Thus, vθ = − ⎜ ⎟ . ⎝2⎠ r _________________________________________________________________ 3.52 Are the following two velocity fields isochoric (i.e., no change of volume)? xe +x e −x e + x e (i) v = 1 1 2 2 2 , r 2 = x12 + x22 and (ii) v = 2 1 2 1 2 , r 2 = x12 + x22 r r ----------------------------------------------------------------------------------------Ans. (i) With v1 = x1 / r 2 , v2 = x2 / r 2 , r 2 = x12 + x22 ,

∂v1 1 2 x1 ∂r 1 2 x2 = 2− 3 = 2 − 41 . ∂x1 r r ∂x1 r r

⎛ 2 ⎞ ∂r ∂r 2 2 = 2 x1 , 2r = 2 x2 ⎟ . ⎜ r = x1 + x2 → 2r ∂x1 ∂x2 ⎝ ⎠

∂v2 1 2 x2 ∂r 1 2 x2 = 2− 3 = 2 − 42 , ∂x2 r r ∂x2 r r

∂v1 ∂v2 2 2 x2 2 x2 2 2 + = 2 − 41 − 42 = 2 − 2 = 0. ∂x1 ∂x2 r r r r r

(ii) v1 = − x2 / r 2 , v2 = x1 / r 2 , r 2 = x12 + x22


Lai et al, Introduction to Continuum Mechanics ⎛ 2 ⎞ ∂r 2 2 = 2 x1 ⎟ ⎜ r = x1 + x2 → 2r ∂x1 ⎝ ⎠ ∂v2 2 x1 ∂r 2 x1 x2 ∂v1 ∂v2 2 x2 x1 2 x1 x2 =− 3 =− 4 , + = 4 − 4 = 0. ∂x2 ∂x1 ∂x2 r ∂x2 r r r _________________________________________________________________ ∂v1 2 x2 ∂r 2 x2 x1 = 3 = 4 ∂x1 r ∂x1 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, Dρ ∂ρ ∂ρ ∂ρ =0→ + v1 + v2 = 0 → 0 + v1 ( 0 ) + v2 k = 0 → v2 = 0. Dt ∂t ∂x1 ∂x2 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. ∂x1 ∂x2 ∂x1 _________________________________________________________________

α x1 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 ρ ( 0 ) = ρo . (b) Find the density if it depends only on x1 , i.e., ρ = ρˆ ( x1 ) , with ρˆ ( xo ) = ρ * . 3.54 Consider the velocity field: v =

----------------------------------------------------------------------------------------Ans.(a) Equation of conservation of mass is ⎛ ∂v ∂v ∂v ⎞ α x1 ∂ρ ∂ρ ∂ρ ∂ρ , v2 = v3 = 0 , + v1 + v2 + v3 + ρ ⎜ 1 + 2 + 3 ⎟ = 0 . With v1 = 1 + kt ∂t ∂x1 ∂x2 ∂x3 ⎝ ∂x1 ∂x2 ∂x3 ⎠ ρ

→

dρ α dρ dt ρ α ρ −α / k +ρ =0→ ∫ = −α ∫ → ln = − ln (1 + kt ) → = (1 + kt ) . dt k 1 + kt 1 + kt ρ ρo ρo 0 ρ t

o

(b) with ρ = ρ ( x1 ) and v1 =

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 ⎠ ρ

→ x1

x

1 x dx x ρ dρ dρ + ρ = 0, → ∫ = − ∫ 1 → ln = − ln 1 → ρ = ρ * o ρ ρ* dx1 x1 xo x1 ρ* x o

where ρo is the density at x1 = xo . _________________________________________________________________ 3.55 Given the velocity field: v = α ( x1te1 + x2te 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. -----------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. Equation of conservation of mass is ⎛ ∂v ∂v ∂v ⎞ ∂ρ ∂ρ ∂ρ ∂ρ + v1 + v2 + v3 + ρ ⎜ 1 + 2 + 3 ⎟ = 0 . With v1 = α x1t , v2 = α x2t , v3 = 0 , ∂t ∂x1 ∂x2 ∂x3 ⎝ ∂x1 ∂x2 ∂x3 ⎠ ρ

2 dρ dρ ρ + ρ (α t + α t ) = 0 → ∫ = −2α ∫ tdt → ln = −α t 2 → ρ = ρo e−α t . dt ρ ρo ρ 0

t

o

_________________________________________________________________ ∂Wim ∂Eik ∂Ekm ∂u ⎞ 1 ⎛ ∂u = − , where Eim = ⎜ i + m ⎟ is the strain tensor and 2 ⎝ ∂X m ∂X i ⎠ ∂X k ∂X m ∂X i ∂u ⎞ 1 ⎛ ∂u Wim = ⎜ i − m ⎟ is the rotation tensor. 2 ⎝ ∂X m ∂X i ⎠ -----------------------------------------------------------------------------------------Ans. ∂Wim ∂ 2 um ⎞ ∂ 1 ⎛ ∂ui ∂um ⎞ 1 ⎛ ∂ 2ui = − = − ⎜ ⎟= ⎜ ⎟ ∂X k ∂X k 2 ⎝ ∂X m ∂X i ⎠ 2 ⎜⎝ ∂X m ∂X k ∂X i ∂X k ⎟⎠ ∂ 2 uk ∂ 2 uk ∂ 2 um ⎞ 1 ⎛ ∂ 2ui + − − ⎜⎜ ⎟= 2 ⎝ ∂X m ∂X k ∂X m ∂X i ∂X m ∂X i ∂X i ∂X k ⎟⎠ 1 ⎛ ∂ ⎛ ∂ui ∂uk ⎞ ∂ ⎛ ∂uk ∂um ⎞ ⎞ ∂Eik ∂Ekm + + − ⎜ ⎜ ⎟− ⎜ ⎟⎟ = ∂xi 2 ⎜⎝ ∂X m ⎝ ∂X k ∂X i ⎠ ∂X i ⎝ ∂X m ∂X k ⎠ ⎟⎠ ∂xm _________________________________________________________________ 3.56 Show that

3.57 Check whether or not the following distribution of the state of strain satisfies the compatibility conditions: X1 X2 ⎤ ⎡ X1 + X 2 , ⎢ X2 + X3 X 3 ⎥⎥ , k = 10−4 [ E] = k ⎢ X1 ⎢⎣ X 2 X3 X1 + X 3 ⎥⎦ ----------------------------------------------------------------------------------------Ans. Yes. We note that the given Eij are linear in X1 , X 2 and X 3 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 X 22 + X 32 X1 X 3 ⎤ ⎢ ⎥ 0 X1 ⎥ , k = 10−4 [E] = k ⎢ X 22 + X 32 ⎢ ⎥ X1 X 22 ⎥ ⎢⎣ X1 X 3 ⎦ ----------------------------------------------------------------------------------------Ans.


Lai et al, Introduction to Continuum Mechanics ∂ 2 E11 ∂X 22 ∂ 2 E22

+

∂ 2 E22

+

∂ 2 E33

∂X12

=2

∂ 2 E12 → 0 + 0 = 0,OK ∂X1 X 2

=2

∂ 2 E23 → 0 + 2k ≠ 0, not satisfied ∂X 2 X 3

∂X 32 ∂X 22 The given strain components are not compatible. _________________________________________________________________ 3.59 Does the displacement field: u1 = sin X1 , u2 = X13 X 2 , u3 = cos X 3 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 X 2 , k = 10−4 and all other Eij = 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 E11 ∂ 2 E22 ∂ 2 E12 Ans. (a) 2 + = → 0 + 0 ≠ 2k . This compatibility condition is not satisfied. ∂X1 X 2 ∂X 22 ∂X12 (b) E11 = 0 → Now, 2 E12 =

∂u1 ∂u = 0 → u1 = u1 ( X 2 , X 3 ) . Also, E22 = 0 → 2 = 0 → u2 = u2 ( X1 , X 3 ) . ∂X1 ∂X 2

∂u ( X , X ) ∂u ( X , X ) ∂u1 ∂u2 + → 2kX1 X 2 = 1 2 3 + 2 1 3 = f ( X 2 , X 3 ) + g ( X1 , X 3 ) , ∂X 2 ∂X1 ∂X 2 ∂X1

That is, 2kX1 X 2 = f ( X 2 , X 3 ) + g ( X1 , X 3 ) . 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

α

f ( X 2 , X 3 ) , E22 = E33 = −

ν f ( X 2 , X 3 ) , E12 = E13 = E23 = 0 . α

Show that for the strains to be compatible, f ( X 2 , X 3 ) must be linear in X 2 and X 3 . ----------------------------------------------------------------------------------------Ans 2 2 ∂ 2 E13 ∂ 2 E11 ∂ 2 E22 ∂ 2 E12 ∂ 2 E11 ∂ 2 E33 1 ∂ f ( X2, X3 ) 1 ∂ f ( X2, X3 ) + = → = + = → =0 2 0, 2 ∂X1 X 2 ∂X1 X 3 α α ∂X 22 ∂X12 ∂X 22 ∂X 32 ∂X12 ∂X 32 , 2 ∂ 2 E11 ∂ ⎛ ∂E23 ∂E31 ∂E12 ⎞ 1 ∂ f ( X2, X3 ) = − + + → = 0 , Thus, ⎜ ⎟ ∂X 2 ∂X 3 ∂X1 ⎝ ∂X1 ∂X 2 ∂X 3 ⎠ α ∂X 2 ∂X 3


Lai et al, Introduction to Continuum Mechanics ∂2 f ( X 2 , X 3 )

= 0,

∂2 f ( X 2 , X 3 )

∂X 22 ∂X 32 X 2 and X 3 . We note also ∂ 2 E22 ∂X 32

+

∂ 2 E33 ∂X 22

=−

= 0,

∂2 f ( X 2 , X 3 ) ∂X 2 ∂X 3

= 0 . → f ( X 2 , X 3 ) is a linear function of

∂ 2 E23 ν ⎛ ∂2 f ∂2 f ⎞ + = = 0 2 , ⎜⎜ ⎟ ∂X1 X 3 α ⎝ ∂X 32 ∂X 22 ⎠⎟

∂ 2 E22 ∂ ⎛ ∂E31 ∂E12 ∂E23 ⎞ ν ∂2 f =− =0= + + ⎜− ⎟, ∂X 3∂X1 ∂X 2 ⎝ ∂X 2 ∂X 3 ∂X1 ⎠ α ∂X 3∂X1 ∂ 2 E33 ∂ ⎛ ∂E12 ∂E23 ∂E31 ⎞ ν ∂2 f =− =0= + + ⎜− ⎟. ∂X1∂X 2 ∂X 3 ⎝ ∂X 3 ∂X1 ∂X 2 ⎠ α ∂X1∂X 2

Thus, if f ( X 2 , X 3 ) 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 + dr ( r + dr ) dθ dz . Thus ,

the net mass flux into dV through the pair of faces r = r and r = r + dr is ( ρ vr )r =r ( rdθ ) dz − ( ρ vr )r =r + dr ( r + dr ) dθ dz = ⎡⎣( ρ vr )r =r − ( ρ vr )r =r + dr ⎤⎦ rdθ dz − ( ρ vr )r = r + dr drdθ dz .

⎡ ∂ ( ρ vr ) ⎤ Now, ⎡⎣( ρ vr )r = r − ( ρ vr )r = r + dr ⎤⎦ rdθ dz = − ⎢ ⎥ dr ( rdθ dz ) and ∂ r ⎣ ⎦ − ( ρ vr )r = r + dr drdθ dz = − ⎡⎣( ρ vr ) + d ( ρ vr ) ⎤⎦ drdθ dz = − ( ρ vr ) drdθ dz , 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 vr ) ⎬ drdθ dz . Similarly, ⎨−r ⎜ ⎟ ⎩ ⎝ ∂r ⎠ ⎭ the net mass flux into dV through the pair of faces θ = θ and θ = θ + dθ is ⎛ ∂ρ vθ ⎞ −⎜ ⎟ dθ ( drdz ) , ⎝ ∂θ ⎠ and the net mass flux into dV through the pair of faces z = z and z = z + dz is


Lai et al, Introduction to Continuum Mechanics ⎛ ∂ρ vz ⎞ −⎜ ⎟ dz ⎡⎣ dr ( rdθ ) ⎤⎦ ⎝ ∂z ⎠ Thus, the total influx of mass through these three pairs of faces is: ⎪⎧⎛ ∂ρ vr ⎞ ( ρ vr ) 1 ⎛ ∂ρ vθ ⎞ ⎛ ∂ρ vz ⎞ ⎪⎫ − ⎨⎜ ⎟ + r + r ⎜ ∂θ ⎟ + ⎜ ∂z ⎟ ⎬ dr ( rdθ ) dz ⎠ ⎭⎪ ⎝ ⎠ ⎝ ⎩⎪⎝ ∂r ⎠ ∂ ∂ρ On the other hand, the rate of increase of mass inside dV is ( ρ rdθ drdz ) = rdθ drdz . ∂t ∂t Therefore, the conservation of mass principle gives, ∂ρ ⎪⎧⎛ ∂ρ vr ⎞ ( ρ vr ) 1 ⎛ ∂ρ vθ ⎞ ⎛ ∂ρ vz ⎞ ⎪⎫ − ⎨⎜ + + ⎜ +⎜ rdθ drdz , That is: ⎬ dr ( rdθ ) dz = ⎟ ⎟ ⎟ ∂t r r ⎝ ∂θ ⎠ ⎝ ∂z ⎠ ⎭⎪ ⎩⎪⎝ ∂r ⎠ ∂ρ ∂ρ vr ρ vr 1 ⎛ ∂ρ vθ + + + ⎜ ∂t ∂r r r ⎝ ∂θ

⎞ ⎛ ∂ρ vz ⎟ + ⎜ ∂z ⎠ ⎝

⎞ ⎟ = 0 , Or, ⎠

⎧ ∂ρ ∂ρ vθ ⎛ ∂ρ ⎞ ∂ρ ⎫ ⎛ ∂vr vr 1 ∂vθ ∂vz ⎞ + ⎜ + vz + + + = 0 . This is the same as ⎨ + vr ⎬+ ρ⎜ ⎟ ∂r r ⎝ ∂θ ⎠ ∂z ⎭ r r ∂θ ∂z ⎟⎠ ⎝ ∂r ⎩ ∂t Eq.(3.15.7). _________________________________________________________________ 3.63 Given the following deformation in rectangular Cartesian coordinates: x1 = 3 X 3 , x2 = − X1 , x3 = −2 X 2 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 −1 0 ⎤ ⎡ 0 0 3⎤ ⎡1 0 0 ⎤ ⎡ 0 0 3⎤ T ⎢ ⎥ Ans. (a) [ F ] = ⎢ −1 0 0 ⎥ , (b) [C] = [ F ] [ F ] = ⎢⎢ 0 0 −2 ⎥⎥ ⎢⎢ −1 0 0 ⎥⎥ = ⎢⎢ 0 4 0 ⎥⎥ , ⎢⎣ 3 0 0 ⎥⎦ ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣ 0 0 9 ⎥⎦ ⎢⎣ 0 −2 0 ⎥⎦ ⎡1 0 0 ⎤ [ 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 ⎥⎦ 1/2

−1

(d) [ R ] = [ F ][ U ]

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 ⎥⎦


Lai et al, Introduction to Continuum Mechanics 0⎤ ⎡4 / 9 0 0 ⎤ ⎡0 0 0 ⎤ ⎡0 0 1 1 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 0 8 ⎥⎦ ⎢⎣ 0 0 4 ⎥⎦ 0 3 / 8⎥⎦ ΔV (g) = det B = ( 9 )(1)( 4 ) = 6 , ΔVo *⎤

⎡ 0 −6 0 ⎤ 1⎢ n o , dAo = 1, det F = 6, [ F ] = ⎢ 0 0 −3⎥⎥ , no = e2 , (h) dA = dAo ( det F ) F 6 ⎢⎣ 2 0 0 ⎥⎦ ⎡ 0 0 2 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ T 1⎢ ⎡ ⎤ −1 [ dA ] = dAo ⎢( det F ) F no ⎥ = (1)( 6 ) 6 ⎢ −6 0 0 ⎥⎥ ⎢⎢1 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dA = −3e3 . ⎣ ⎦ ⎢⎣ 0 −3 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −3⎥⎦ _________________________________________________________________

( ) −1

T

−1

( )

3.64 Do the previous problem for the following deformation: x1 = 2 X 2 , x2 = 3 X 3 , x3 = X1 . ----------------------------------------------------------------------------------------⎡ 0 0 1 ⎤ ⎡0 2 0 ⎤ ⎡1 0 0 ⎤ ⎡0 2 0⎤ T ⎢ ⎥ Ans. (a) [ F ] = ⎢ 0 0 3⎥ . (b) [C] = [ F ] [ F ] = ⎢⎢ 2 0 0 ⎥⎥ ⎢⎢0 0 3⎥⎥ = ⎢⎢0 4 0 ⎥⎥ . ⎢⎣ 0 3 0 ⎥⎦ ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 0 9 ⎥⎦ ⎢⎣1 0 0 ⎥⎦ ⎡1 0 0 ⎤ 1/2 [ U ] = [C] = ⎢⎢0 2 0⎥⎥ . (The only positive definite root). ⎢⎣0 0 3⎥⎦ ⎡0 2 0⎤ ⎡0 (c) [ B ] = [ F ][ F ] = ⎢⎢ 0 0 3⎥⎥ ⎢⎢ 2 ⎢⎣1 0 0 ⎥⎦ ⎢⎣ 0 ⎡0 2 0 ⎤ ⎡1 −1 ⎢ (d) [ R ] = [ F ][ U ] = ⎢0 0 3⎥⎥ ⎢⎢0 ⎢⎣1 0 0 ⎥⎦ ⎢⎣0 T

0 1 ⎤ ⎡ 4 0 0⎤ 0 0 ⎥⎥ = ⎢⎢ 0 9 0 ⎥⎥ . 3 0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ 0 0 ⎤ ⎡0 1 0⎤ 1 / 2 0 ⎥⎥ = ⎢⎢ 0 0 1 ⎥⎥ . 0 1 / 3⎥⎦ ⎢⎣1 0 0 ⎥⎦

0⎤ 0⎤ ⎡3 / 8 0 ⎡0 0 0⎤ ⎡0 0 1 1 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 0 8 ⎥⎦ ⎢⎣ 0 0 4 ⎥⎦ 0 0 ⎥⎦ ΔV (g) = det B = ( 4 )( 9 )(1) = 6 . ΔVo *⎤

( )

(h) dA = dAo ( det F ) F

−1

T

−1

n o , dAo = 1, det F = 6, [ F ]

⎡0 0 6⎤ 1⎢ = ⎢ 3 0 0 ⎥⎥ , no = e 2 6 ⎢⎣ 0 2 0 ⎥⎦


Lai et al, Introduction to Continuum Mechanics ⎡0 3 0 ⎤ ⎡ 0⎤ ⎡ 3⎤ 1⎢ ⎤ n o ⎥ = (1)( 6 ) ⎢ 0 0 2 ⎥⎥ ⎢⎢1 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dA = 3e1 6 ⎦ ⎢⎣ 6 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ _________________________________________________________________

( )

[ dA ] = dAo ⎡⎢( det F ) F −1 ⎣

T

3.65 Do Prob. 3.63 for the following deformation: x1 = X1 , x2 = 3 X 3 , x3 = −2 X 2 ----------------------------------------------------------------------------------------⎡1 0 0 ⎤ Ans. (a) [ F ] = ⎢⎢ 0 0 3⎥⎥ . ⎢⎣ 0 −2 0 ⎥⎦ ⎡1 0 0 ⎤ ⎡1 (b) [C] = [ F ] [ F ] = ⎢⎢0 0 −2 ⎥⎥ ⎢⎢0 ⎢⎣0 3 0 ⎥⎦ ⎢⎣0 ⎡1 0 0 ⎤ 1/2 is [ U ] = [C] = ⎢⎢0 2 0 ⎥⎥ . ⎢⎣0 0 3⎥⎦ ⎡1 0 0 ⎤ ⎡1 T (c) [ B ] = [ F ][ F ] = ⎢⎢ 0 0 3⎥⎥ ⎢⎢0 ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣0 ⎡1 0 0 ⎤ ⎡1 −1 ⎢ (d) [ R ] = [ F ][ U ] = ⎢ 0 0 3⎥⎥ ⎢⎢ 0 ⎢⎣ 0 −2 0 ⎥⎦ ⎢⎣ 0 T

0 ⎤ ⎡1 0 0 ⎤ 0 3⎥⎥ = ⎢⎢0 4 0 ⎥⎥ , The only positive definite root −2 0 ⎥⎦ ⎢⎣0 0 9 ⎥⎦ 0

0 ⎤ ⎡1 0 0 ⎤ 0 −2 ⎥⎥ = ⎢⎢0 9 0 ⎥⎥ . 3 0 ⎥⎦ ⎢⎣0 0 4 ⎥⎦ 0 0 ⎤ ⎡1 0 0 ⎤ 1 / 2 0 ⎥⎥ = ⎢⎢ 0 0 1 ⎥⎥ , 0 1 / 3⎥⎦ ⎢⎣ 0 −1 0 ⎥⎦ 0

0⎤ 0 0 ⎤ ⎡0 ⎡0 0 1 1 ⎢ ⎢ ⎥ − 1 (e) ⎡ E = [C − I ] = ⎢ 0 3 / 2 0 ⎥ , (f) ⎡e* ⎤ = ⎡I − B ⎤ = ⎢ 0 4 / 9 0 ⎥⎥ . ⎣ ⎦ 2⎣ ⎦ ⎣ ⎦ 2 ⎢⎣ 0 ⎢⎣ 0 0 4 ⎥⎦ 0 3 / 8⎥⎦ *⎤

(g)

ΔV = det B = ΔVo

(1)( 9 )( 4 ) = 6 .

⎡6 0 0 ⎤ 1⎢ n o , dAo = 1, det F = 6, [ F ] = ⎢ 0 0 −3⎥⎥ , n o = e 2 (h) dA = dAo ( det F ) F 6 ⎢⎣ 0 2 0 ⎥⎦ ⎡6 0 0 ⎤ ⎡0 ⎤ ⎡ 0 ⎤ T 1⎢ ⎡ ⎤ −1 [ dA ] = dAo ⎢( det F ) F no ⎥ = (1)( 6 ) 6 ⎢0 0 2⎥⎥ ⎢⎢1 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dA = −3e3 ⎣ ⎦ ⎢⎣ 0 −3 0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ −3⎥⎦

( ) −1

T

−1

( )

_________________________________________________________________ 3.66 Do Prob. 3.63 for the following deformation:


Lai et al, Introduction to Continuum Mechanics x1 = 2 X 2 , x2 = − X1 , x3 = 3 X 3 ----------------------------------------------------------------------------------------⎡ 0 −1 0 ⎤ ⎡ 0 2 0 ⎤ ⎡1 0 0 ⎤ ⎡ 0 2 0⎤ T ⎢ ⎥ Ans. (a) [ F ] = ⎢ −1 0 0 ⎥ . (b) [C] = [ F ] [ F ] = ⎢⎢ 2 0 0 ⎥⎥ ⎢⎢ −1 0 0 ⎥⎥ = ⎢⎢ 0 4 0 ⎥⎥ , ⎢⎣ 0 0 3⎥⎦ ⎢⎣ 0 0 3⎥⎦ ⎢⎣ 0 0 9 ⎥⎦ ⎢⎣ 0 0 3⎥⎦ ⎡1 0 0 ⎤ 1/2 [ U ] = [C] = ⎢⎢0 2 0⎥⎥ . (The only positive definite root). ⎢⎣0 0 3⎥⎦ ⎡0 (c) [ B ] = [ F ][ F ] = ⎢⎢ −1 ⎢⎣ 0 ⎡0 −1 ⎢ (d) [ R ] = [ F ][ U ] = ⎢ −1 ⎢⎣ 0 T

2 0⎤ ⎡0 0 0 ⎥⎥ ⎢⎢ 2 0 3⎥⎦ ⎢⎣ 0 2 0 ⎤ ⎡1 0 0 ⎥⎥ ⎢⎢ 0 0 3⎥⎦ ⎢⎣ 0

−1 0 ⎤ ⎡ 4 0 0 ⎤ 0 0 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥ . 0 3⎥⎦ ⎢⎣ 0 0 9 ⎥⎦ 0 0 ⎤ ⎡ 0 1 0⎤ 1 / 2 0 ⎥⎥ = ⎢⎢ −1 0 0 ⎥⎥ . 0 1 / 3⎥⎦ ⎢⎣ 0 0 1 ⎥⎦

0⎤ ⎡0 0 1 ⎢ *⎤ ⎡ (e) E = [C − I ] = ⎢ 0 3 / 2 0 ⎥⎥ , ⎣ ⎦ 2 ⎢⎣ 0 0 4 ⎥⎦ (g)

ΔV = det B = ΔVo

0 ⎤ ⎡3 / 8 0 1⎡ −1 ⎤ ⎢ * ⎡ ⎤ (f) e = I − B 0 ⎥⎥ = 0 0 ⎣ ⎦ 2⎣ ⎦ ⎢ ⎢⎣ 0 0 4 / 9 ⎥⎦

( 4 )(1)( 9 ) = 6 .

⎡ 0 −6 0 ⎤ 1⎢ n o , dAo = 1, det F = 6, [ F ] = ⎢ 3 0 0 ⎥⎥ , n o = e 2 (h) dA = dAo ( det F ) F 6 ⎢⎣ 0 0 2 ⎥⎦ ⎡ 0 3 0 ⎤ ⎡ 0⎤ ⎡ 3⎤ 1⎢ −1 T [dA] = dAo (det F)[F ] [n o ] = (1)(6) ⎢ −6 0 0 ⎥⎥ ⎢⎢1 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dA = 3e1 6 ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣0 ⎥⎦

( ) −1

T

−1

_________________________________________________________________ 3.67 Given x1 = X1 + 3 X 2 , x2 = X 2 , x3 = X 3 . 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 0 ⎤ T ⎢ ⎥ Ans. (a) [ F ] = ⎢ 0 1 0 ⎥ , [C] = [ F ] [ F ] = ⎢⎢ 3 1 0 ⎥⎥ ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ 3 10 0 ⎥⎥ . ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ (b) the characteristic equation is


Lai et al, Introduction to Continuum Mechanics 1− λ 3 0

3

0

(

)

10 − λ 0 = 0 → (1 − λ ) λ 2 − 11λ + 1 = 0, 0 1− λ

11 ± 121 − 4 → λ1 = 10.908326, λ2 =0.0916735, λ3 = 1 2 For λ1 = 10.908326 ,

λ1,2 =

(1 − λ1 )α1 + 3α 2 = 0 → α 2 = − (1 − λ1 )α1 / 3 = 3.302775α1 ,

1 ( e1 + 3.302775e2 ) = 0.289785e1 + 0.957093e2 . 3.450843 For λ2 = 0.0916735 , n1 =

(1 − λ2 )α1 + 3α 2 = 0 → α 2 = − (1 − λ2 )α1 / 3 = −0.3027755α1 ,

1 ( e1 − 0.3027755e2 ) = 0.957093e1 − 0.289784e2 . 1.044832 For λ3 = 1, n3 = e3 , (c) The matrices with respect to the principal axes are as follows 0 0⎤ 0 0⎤ ⎡10.9083 ⎡3.30277 ⎢ ⎢ ⎥ 0.0916735 0 ⎥ → [ U ]n = ⎢ 0 0.302774 0 ⎥⎥ . [C]ni = ⎢ 0 i ⎢⎣ 0 ⎢⎣ 0 0 1 ⎥⎦ 0 1 ⎥⎦ 0 0⎤ ⎡0.302774 ⎢ − 1 ⎡U ⎤ = 0 3.302772 0 ⎥⎥ . ⎣ ⎦ ni ⎢ ⎢⎣ 0 0 1 ⎥⎦ n2 =

The matrices with respect to the ei -basis are given by the formula [ U ]{e } = [Q ] [ U ]{n } [Q ] : T

i

[ U ]e

i

i

0 0 ⎤ ⎡ 0.289785 0.957093 0 ⎤ ⎡ 0.289785 0.957093 0 ⎤ ⎡3.30277 ⎢ ⎥ ⎢ = ⎢ 0.957093 −0.289785 0 ⎥ ⎢ 0 0.302774 0 ⎥⎥ ⎢⎢ 0.957093 −0.289785 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 ⎡ U −1 ⎤ = ⎢ 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 ⎥⎦


Lai et al, Introduction to Continuum Mechanics ⎡1 3 0 ⎤ ⎡ 3.050852 −0.832052 0 ⎤ ⎡ 0.55470 0.83205 0 ⎤ (d) [ R ]e = [ F ] ⎡ U = ⎢ 0 1 0 ⎥⎥ ⎢⎢ −0.832052 0.554701 0 ⎥⎥ = ⎢⎢ −0.83205 0.55470 0⎥⎥ . ⎣ ⎦ ⎢ i ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ _________________________________________________________________ −1 ⎤

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 + kX 2 , x2 = X 2 , x3 = X 3 , are given by: With f = (1 + k 2 / 4)−1/2 , kf / 2 0⎤ ⎡ f kf / 2 0 ⎤ ⎡ f ⎢ ⎥ ⎢ 2 [ U ] = ⎢ kf / 2 1 + k / 2 f 0⎥ , [ R ] = ⎢−kf / 2 f 0⎥⎥ . ⎢ ⎥ ⎢⎣ 0 0 1 ⎥⎦ 0 1 ⎦⎥ ⎣⎢ 0 ----------------------------------------------------------------------------------------kf / 2 0⎤ kf / 2 0 ⎤ ⎡ f ⎡ f ⎢ ⎥ f Ans. [ RU ] = ⎢⎢ − kf / 2 0 ⎥⎥ ⎢ kf / 2 1 + k 2 / 2 f 0 ⎥ ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎢ 0 0 1 ⎦⎥ ⎣

(

)

(

)

( (

) )

⎡ f 2 + ( kf / 2 )( kf / 2 ) f ( kf / 2 ) + ( kf / 2 ) 1 + k 2 / 2 f 0 ⎤ ⎢ ⎥ ⎢ ⎥ 2 = ⎢( − kf / 2 ) f + f ( kf / 2 ) ( − kf / 2 )( kf / 2 ) + f 1 + k / 2 f 0 ⎥ ⎢ ⎥ 0 0 1⎥ ⎢ ⎣ ⎦ 2 2 2 2 ⎡ f 1 + k / 4 kf 1 + k / 4 0 ⎤ ⎢ ⎥ ⎡1 k 0 ⎤ ⎢ ⎥ 2 2 0 =⎢ f 1 + k / 4 0 ⎥ = ⎢⎢0 1 0 ⎥⎥ = the given [ F ] ⎢ ⎥ 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( ) =ds m and dx( ) = ds n , 1

Cαβ Nα(1) N β(2)

2

ds ds 1 2 , where N( ) = Nα(1)eα , N( ) = Nα(2)eα , λ1 = 1 and λ2 = 2 . λ1λ2 dS1 dS2 ----------------------------------------------------------------------------------------1 2 1 2 1 2 1 2 Ans. dx( ) ⋅ dx( ) = FdX( ) ⋅ FdX( ) = dX( ) ⋅ F T FdX( ) = dX( ) ⋅ CdX( ) , then, cos θ =

1 2 1 ( 2) → ds1ds2 cosθ = dS1dS2 N( ) ⋅ CN( ) = dS1dS2 ( Nα( )eα ) ⋅ C( N β e β ),

→ cosθ =

1 ( 2) Cαβ Nα( ) N β dS1dS2 (1) ( 2 ) . Nα N β eα ⋅ Ce β = λ1λ2 ds1ds2


Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 3.70 Given the following right Cauchy-Green deformation tensor at a point 0 ⎤ ⎡9 0 ⎢ [C] = ⎢0 4 0 ⎥⎥ ⎣⎢0 0 0.36 ⎦⎥ (a) Find the stretch for the material elements which were in the direction of e1 , e 2 ,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 dx( ) where dX( ) = dS e and dX( ) = dS e deform to 1 1

1 2 dx( ) = ds1m and dx( ) = ds2n . -----------------------------------------------------------------------------------------

2 1

Ans. (a) For the elements which were in e1 , e 2 ,and e3 direction, the stretches are C11 , C22 , C33 , that is, 3, 2 and 0.6 respectively. 0 ⎤ ⎡1 ⎤ ⎡9 0 ⎡9 ⎤ 1 1 1 13 ⎢ ⎥ ⎢ ⎥ ' (b) Let e1' = ( e1 + e2 ) → C11 = [1 1 0] ⎢0 4 0 ⎥ ⎢1 ⎥ = [1 1 0] ⎢⎢ 4⎥⎥ = . 2 2 2 2 ⎢⎣0 0 0.36 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0 ⎥⎦ That is, the stretch for dX = dSe1' is

( ds / dS ) =

' C11 = 13 / 2 .

(c) C12 = 0 → cosθ = 0 → θ = 90o . There is no change in angle. (note, {e1 , e 2 , e3} are principal axes for C . _________________________________________________________________ 3.71 Given the following large shear deformation: x1 = X1 + X 2 , x2 = X 2 , x3 = X 3 . (a) Find the stretch tensor U (Hint: use the formula given in problem 3.68) and verify that U 2 = 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 + e 2 . (d) What is the angle between the deformed elements of dS1e1 and dS2e2 ?. ----------------------------------------------------------------------------------------Ans. (a) For x1 = X1 + kX 2 , x2 = X 2 , x3 = X 3 , from Prob. 3.68, we have

⎡ f ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎣⎢ 0

(1 + k / 2) f

⎡ f [ U ] = ⎢⎢ f / 2 ⎢⎣ 0

f /2 (3 / 2) f 0

kf / 2 2

0

0⎤ ⎥ ⎛ k2 0 ⎥ where f = ⎜1 + ⎜ 4 ⎥ ⎝ 1 ⎦⎥

⎞ ⎟⎟ ⎠

−

1 2

. Thus, with k = 1 , f = 2 / 5

⎡ 1 1/ 2 0⎤ 0 ⎤ ⎡ 1 1/ 2 ⎢ 2 0 ⎥⎥ = f ⎢⎢1 / 2 3 / 2 0 ⎥⎥ = ⎢1 / 2 3 / 2 5⎢ ⎢⎣ 0 1 ⎥⎦ 0 1 / f ⎥⎦ 0 ⎣ 0


⎤ ⎥ ⎥. 5 / 2 ⎥⎦ 0 0

Lai et al, Introduction to Continuum Mechanics ⎡ 1 4⎢ [ U ][ U ] = ⎢1 / 2 5⎢ ⎣ 0

1/ 2 (3 / 2) 0

⎤⎡ 1 ⎥⎢ ⎥ ⎢1 / 2 5 / 2 ⎥⎦ ⎢⎣ 0 0 0

1/ 2 (3 / 2) 0

⎤ ⎡1 1 0 ⎤ ⎥ ⎢ ⎥ ⎥ = ⎢ 1 2 0 ⎥ = [ C] . 5 / 2 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ 0 0

(b) The stretch for the element which was in the direction e 2 is

C22 = 2 .

(c) Let e1' = ( e1 + e2 ) / 2 , ⎡1 1 0 ⎤ ⎡1 ⎤ ⎡ 2⎤ 1 1 5 ds ⎢ ⎥ ⎢ ⎥ = [1 1 0] ⎢1 2 0 ⎥ ⎢1 ⎥ = [1 1 0] ⎢⎢ 3 ⎥⎥ = → = 5/ 2 . 2 2 2 dS ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎛ ds ⎞⎛ ds ⎞ 1 → θ = 45o . (d) ⎜ 1 ⎟⎜ 2 ⎟ cosθ = C12 → (1) 2 cosθ = 1 → cosθ = 2 ⎝ dS1 ⎠⎝ dS2 ⎠ _________________________________________________________________ ' C11

( )

3.72 Given the following large shear deformation: x1 = X1 + 2 X 2 , x2 = X 2 , x3 = X 3 (a) Find the stretch tensor U (Hint: use the formula given in problem 3.68) and verify that U 2 = 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 dS2e 2 . ----------------------------------------------------------------------------------------Ans. For x1 = X1 + kX 2 , x2 = X 2 , x3 = X 3 , from Prob. 3.68, we have ⎡ f ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0 ⎡ f ⎢ [ U ] = ⎢ kf / 2 ⎢ ⎢⎣ 0

0⎤ ⎥ 1 + k 2 / 2 f 0 ⎥ where f ⎥ 0 1 ⎥⎦ kf / 2 0⎤ ⎡1 ⎥ ⎢ 1 1 + k 2 / 2 f 0⎥ = ⎢1 2⎢ ⎥ 0 1 ⎥⎦ ⎣0 ⎡1 1 kf / 2 0 ⎤ ⎡ f 1 ⎢ ⎢ ⎥ [ R ] = ⎢ −kf / 2 f 0⎥ = ⎢ −1 1 2⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎣0 0

(

(

kf / 2

)

)

⎛ k2 = ⎜1 + ⎜ 4 ⎝ 1 3 0

⎞ ⎟⎟ ⎠

−

1 2

. Thus, with k = 2 , f = 1 / 2

0 ⎤ ⎥ 0 ⎥. 2 ⎥⎦

0 ⎤ ⎥ 0 ⎥. 2 ⎥⎦

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 e2 is C22 = 5 . (c) Let e1′ = ( e1 + e2 ) / 2 ,


Lai et al, Introduction to Continuum Mechanics ⎡ 1 2 0 ⎤ ⎡1 ⎤ ⎡ 3⎤ 1 1 ds ⎢ ⎥ ⎢ ⎥ ′ = [1 1 0] 2 5 0 1 = [1 1 0] ⎢7 ⎥ = 5 → C11 = 5 = 2.236 . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2 2 dS ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎛ ds ⎞⎛ ds ⎞ 2 (d) ⎜ 1 ⎟⎜ 2 ⎟ cosθ = C12 → (1) 5 cosθ = 2 → cos θ = . dS dS 5 ⎝ 1 ⎠⎝ 2 ⎠

( )

_________________________________________________________________ 3.73 Show that for any tensor A ( X1 , X 2 , X 3 ) ,

( )

∂ det A = ( det A ) A −1 ∂X m

∂A jn nj

∂X m

----------------------------------------------------------------------------------------Ans

A11 A = A21

A12 A22

A31

A32

∂A11 ∂X m

A13 ∂A = A21 A23 → ∂X m A33 A31

∂A12 ∂X m

∂A13 ∂X m

A22

A23

A32

A33

A11

A12

∂A + 21 ∂X m

∂A22 ∂X m

A31

A32

A13

A11 ∂A23 + A21 ∂X m ∂A31 A33 ∂X m

A12 A22

A13 A23

∂A32 ∂X m

∂A33 ∂X m

. c Let Aijc denote the cofactor of Aij , i.e., A11 =

Then,

∂A ∂X m

That is,

=

∂A ∂X m

∂A

A22 A32

A23 c A , A12 = − 21 A33 A31

A23 etc. A33

∂A11 c ∂A12 c ∂A13 c ∂A21 c ∂A22 c A11 + A12 + A13 + A21 + A22 + ... ∂X m ∂X m ∂X m ∂X m ∂X m =

∂Aij ∂X m

Aijc

.

( )

= det A A −1

( )

On the other hand, A ∂Aij

−1

ij

=

Acji

det A

( )

→ Acji = det A A −1

ij

.

∂Anj

( )

= det A A −1

. ji ∂X jn ∂X ∂X m m m _________________________________________________________________ Thus,

3.74 Show that if TU = 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 λ2T12 λ3T13 ⎤ [TU ] = [T][ U ] = ⎢⎢T21 T22 T23 ⎥⎥ ⎢⎢ 0 λ2 0 ⎥⎥ = ⎢⎢λ1T21 λ2T22 λ3T23 ⎥⎥ ⎣⎢T31 T32 T33 ⎦⎥ ⎣⎢ 0 0 λ3 ⎦⎥ ⎣⎢ λ1T31 λ2T32 λ3T33 ⎦⎥ Thus, TU = 0 gives, all Tij = 0 , that is, T = 0 . _________________________________________________________________ 2

2

2

⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ 3.75 Derive Eq. (3.29.21), that is, Bθθ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ -----------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. Bθθ = eθ ⋅ Beθ = eθ ⋅ FF T eθ . From Eq. (3.29.15). we have, r ∂θ o r ∂θ o r ∂θ o F T eθ = er + eθ + ez , thus, ∂ro ∂zo ro ∂θ o

⎛ r ∂θ o r ∂θ o r ∂θ o ⎞ r ∂θ r ∂θ r ∂θ er + eθ + ez ⎟ = eθ ⋅ Feor + eθ ⋅ Feθo + eθ ⋅ Feoz . Bθθ = eθ ⋅ F ⎜ ro ∂θ o ro ∂θ o ∂zo ⎠ ∂ro ∂zo ⎝ ∂ro r ∂θ r ∂θ r ∂θ , eθ ⋅ Feθo = , eθ ⋅ Feoz = , [See Eq. (3.29.10)], therefore, Since, eθ ⋅ Feor = ∂ro ∂zo ro ∂θ o 2

2

2

⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ Bθθ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ . ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ _________________________________________________________________ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ 3.76 Derive Eq. (3.29.23), i.e., Brz = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠ ----------------------------------------------------------------------------------------Ans. Brz = er ⋅ Bez = er ⋅ FF T ez , from Eq. (3.29.16), we have, ∂z o ∂z o ∂z o FTez = er + eθ + e z , thus, ∂ro ro ∂θ o ∂zo ⎛ ∂z o ∂z o ∂z o ⎞ ∂z ∂z ∂z er + eθ + ez , ⎟ = er ⋅ Feor + er ⋅ Feθo + er ⋅ Feoz . Brz = er ⋅ F ⎜ r r z r r z ∂ ∂ ∂ ∂ ∂ ∂ θ θ o o o o o o o ⎝ o ⎠ ∂r ∂r ∂r , er ⋅ Feθo = , eθ ⋅ Feoz = , thus, From Eq. (3.29.9), er ⋅ Feor = ∂ro ∂zo ro ∂θ o ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ Brz = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠ _________________________________________________________________ 3.77 From ro = ro ( r ,θ , z , t ) , θ o = θ o ( r ,θ , z , t ) , zo = zo ( r ,θ , z , t ) ,derive the components of B −1 with respect to the basis at x . ----------------------------------------------------------------------------------------Ans. From dX = F −1dx , where dx = drer + rdθ eθ + dzez and dX = dro eor + ro dθ o eθo + dzo eoz , we have, dro eor + ro dθ o 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 dr + o dθ + o dz = dr eor ⋅ F −1er + rdθ eor ⋅ F −1eθ + dz eor ⋅ F −1ez ∂r ∂θ ∂z ∂ro ∂r ∂r o −1 → er ⋅ F er = , eor ⋅ F −1eθ = o , eor ⋅ F −1e z = o . ∂r ∂z r ∂θ

→


)

Lai et al, Introduction to Continuum Mechanics Similarly, ro ∂θ o r ∂θ r ∂θ , eθo ⋅ F −1eθ = o o , eθo ⋅ F −1ez = o o . ∂r ∂z r ∂θ ∂z ∂z ∂z eoz ⋅ F −1er = o , eoz ⋅ F −1eθ = o , eoz ⋅ F −1ez = o . ∂r ∂z r ∂θ Thus, ∂r r ∂θ ∂z ∂r r ∂θ ∂z F −1er = o eor + o o eθo + o eoz , F −1eθ = o eor + o o eθo + o eoz ∂r ∂r ∂r r ∂θ r ∂θ r ∂θ ∂r ∂z r ∂θ F −1ez = o eor + o o eθo + o eoz . ∂z ∂z ∂z Also, we have, T T ∂r ∂r er ⋅ F −1 eor = eor ⋅ F −1er = o , eθ ⋅ F −1 eor = eor ⋅ F −1eθ = o , ∂r r ∂θ T ∂ r ez ⋅ F −1 eor = eor ⋅ F −1 ez = o . ∂z Thus, T T ∂r ∂r ∂r r ∂θ r ∂θ r ∂θ F −1 eor = o er + o eθ + o ez , F −1 eθo = o o er + o o eθ + o o ez ∂r ∂z ∂r ∂z r ∂θ r ∂θ T ∂ ∂ ∂ z z z F −1 eoz = o er + o eθ + o ez . ∂r ∂z r ∂θ

eθo ⋅ F −1er =

( ) ( )

( )

( )

( ) ( )

( )

The components of B −1 with respect to the basis at x are:

(

Brr−1 = er ⋅ B −1er = er ⋅ FF T

( )

⎛ ∂r = ⎜ o er ⋅ F −1 ⎝ ∂r

T

)

( )

(

( )

∂ro eθ ⋅ F −1 r ∂θ

T

eor +

2

( ) (F e )

e r = e r ⋅ F −1

ro ∂θ o er ⋅ F −1 ∂r

eor +

−1 Bθθ = eθ ⋅ B −1eθ = eθ ⋅ FF T

=

−1

)

−1

T

T

−1

r

( )

∂zo e r ⋅ F −1 ∂r

eθo +

T

2

2

2

2

2

⎞ ⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ eoz ⎟ = ⎜ o ⎟ + ⎜ o o ⎟ + ⎜ o ⎟ . ⎠ ⎝ ∂r ⎠ ⎝ ∂r ⎠ ⎝ ∂r ⎠

( ) (F e ) T

e θ = e θ ⋅ F −1

ro ∂θ o eθ ⋅ F −1 r ∂θ

( )

2

T

−1

θ

( )

∂zo e θ ⋅ F −1 r ∂θ

eθo +

T

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ eoz = ⎜ o ⎟ + ⎜ o o ⎟ + ⎜ o ⎟ . ⎝ r ∂θ ⎠ ⎝ r ∂θ ⎠ ⎝ r ∂θ ⎠

2

⎛ r ∂θ ⎞ ⎛ ∂z ⎞ −1 ⎛ ∂ro ⎞ Bzz =⎜ +⎜ o o ⎟ +⎜ o ⎟ . ⎟ ⎝ ∂z ⎠ ⎝ ∂z ⎠ ⎝ ∂z ⎠

(

Br−θ1 = er ⋅ B −1eθ = er ⋅ FF T =

∂ro r ∂θ

( )

e r ⋅ F −1

T

eor +

ro ∂θ o r ∂θ

(

)

−1

( ) (F e )

e θ = e r ⋅ F −1

( )

e r ⋅ F −1

Brz−1 = er ⋅ B −1ez = er ⋅ FF T

)

T

eθo +

T

∂zo r ∂θ

−1

θ

( )

e r ⋅ F −1

T

⎛ ∂ro ⎞⎛ ∂ro ⎟⎜ ⎝ r ∂θ ⎠⎝ ∂r

eoz = ⎜

( ) (F e ) ∂z e ⋅ ( F ) eθ + e ⋅ (F ) ∂z −1

ez = er ⋅ F −1

T

−1

⎞ ⎛ ro ∂θ o ⎞⎛ ro ∂θ o ⎟+⎜ ⎟⎜ ⎠ ⎝ r ∂θ ⎠⎝ ∂r

z

T T T r ∂θ ∂ro ⎛ ∂r ⎞⎛ ∂r −1 −1 o o er ⋅ F −1 eor + o o r eoz = ⎜ o ⎟⎜ o r ∂z ∂z ⎝ ∂r ⎠⎝ ∂z ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ Bθ−z1 = ⎜ o ⎟⎜ o ⎟ + ⎜ o o ⎟⎜ o o ⎟ + ⎜ o ⎟⎜ o ⎟ . ⎝ r ∂θ ⎠⎝ ∂z ⎠ ⎝ r ∂θ ⎠⎝ ∂z ⎠ ⎝ r ∂θ ⎠⎝ ∂z ⎠

=

( )


⎞ ⎛ ∂zo ⎞⎛ ∂zo ⎞ ⎟+⎜ ⎟⎜ ⎟ ⎠ ⎝ r ∂θ ⎠⎝ ∂z ⎠

⎞ ⎛ ro ∂θ o ⎞⎛ ro ∂θ o ⎟ + ⎜ ∂r ⎟⎜ ∂z ⎠ ⎝ ⎠⎝

⎞ ⎛ ∂zo ⎞⎛ ∂zo ⎞ ⎟ + ⎜ ∂r ⎟⎜ ∂z ⎟ ⎠ ⎝ ⎠⎝ ⎠

Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 2

2

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ 3.78 Derive Eq. (3.29.47), that is, Cθoθo = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ ----------------------------------------------------------------------------------------r ∂θ ∂r ∂z er + eθ + e z [Eq.3.29.3] , Ans. Cθoθo = eθo ⋅ Ceθo = eθo ⋅ F T Feθo . Now, Feθo = ro ∂θ o ro ∂θ o ro ∂θ o therefore, ⎛ ∂r ⎞ r ∂θ r ∂θ o T ∂z ∂r o T er + eθ + ez ⎟ = eθ ⋅ F e r + eθ ⋅ F eθ Cθoθo = eθo ⋅ F T ⎜ ro ∂θ o ro ∂θ o ⎠ ro ∂θ o ro ∂θ o ⎝ ro ∂θ o ∂z o T eθ ⋅ F e z . Now, from Eqs. (3.29.14) (3.29.15) and (3.29.16), ro ∂θ o ∂r r ∂θ ∂z eθo ⋅ F T er = , eθo ⋅ F T eθ = , eθo ⋅ F T ez = , thus. ro ∂θ o ro ∂θ o ro ∂θ o +

2

2

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ Cθoθo = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ _________________________________________________________________ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ 3.79 Derive Eq. (3.29.49), Croθo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ -----------------------------------------------------------------------------------------r ∂θ ∂r ∂z er + eθ + e z [Eq.(3.29.3)] , Ans. Croθo = eor ⋅ Ceθo = eor ⋅ F T Feθo . Now, Feθo = ro ∂θ o ro ∂θ o ro ∂θ o ⎛ ∂r ⎞ r ∂θ ∂r o T r ∂θ o T ∂z er + eθ + ez ⎟ = er ⋅ F er + e r ⋅ F eθ Croθo = eor ⋅ F T ⎜ ro ∂θ o ro ∂θ o ⎠ ro ∂θ o ro ∂θ o ⎝ ro ∂θ o ∂z o T + er ⋅ F ez . From Eqs. (3.29.14), (3.29.15) and (3.29.16) ro ∂θ o ∂r o T r ∂θ o T ∂z eor ⋅ F T er = , e r ⋅ F eθ = , er ⋅ F ez = , ∂ro ∂ro ∂ro ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ Thus, Croθo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ _________________________________________________________________ 3.80 Derive the components of C−1 with respect to the bases at X . ----------------------------------------------------------------------------------------Ans. −1 T ∂r ∂r ∂r Cr−o1ro =eor ⋅ F T F eor =eor ⋅ F −1 F −1 eor = o eor ⋅ F −1er + o eor ⋅ F −1eθ + o eor ⋅ F −1ez r ∂θ ∂r ∂z

(

)

( )


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 −1 T r ∂θ r ∂θ r ∂θ Cr−oθ1o =eor ⋅ F T F eθo =eor ⋅ F −1 F −1 eθo == o o eor ⋅ F −1er + o o eor ⋅ F −1eθ + o o eor ⋅ F −1ez ∂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 + dYe Y + dZe Z and r = r ( X , Y , Z , t ), θ = θ ( X , Y , Z , t ), z=z( X , Y , Z , t ) , we have, dx = FdX → dx = dre r + rdθ eθ + dzez = dXFeX + dYFeY + dZFe Z ∂r ∂r r ∂θ r ∂θ ⎛ ∂r ⎞ ⎛ r ∂θ ⎞ →⎜ dX + dY + dZ ⎟ er + ⎜ dX + dY + dZ ⎟ eθ ∂Y ∂Z ∂Y ∂Z ⎝ ∂X ⎠ ⎝ ∂X ⎠ ∂z ∂z ⎛ ∂z ⎞ +⎜ dX + dY + dZ ⎟ e z = dXFe X + dYFeY + dZFe Z ∂ X ∂ Y ∂ Z ⎝ ⎠ ∂r r ∂θ ∂z ∂r r ∂θ ∂z → Fe X = er + eθ + ez , FeY = er + eθ + ez , ∂X ∂X ∂X ∂Y ∂Y ∂Y ∂r r ∂θ ∂z Fe Z = er + eθ + e z , and ∂Z ∂Z ∂Z ∂r ∂r , eY ⋅ F T er = er ⋅ FeY = , etc. e X ⋅ F T er = er ⋅ FeX = ∂X ∂Y ∂r ∂r ∂r r ∂θ r ∂θ r ∂θ F T er = eX + eY + e Z , F T eθ = eX + eY + eZ , ∂X ∂Y ∂Z ∂X ∂Y ∂Z ∂z ∂z ∂z FTe z = eX + eY + eZ . ∂X ∂Y ∂Z The components of B are: 2

2

2

∂r ∂r ∂r ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ e r ⋅ Fe X + e r ⋅ Fe Y + er ⋅ Fe Z = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ . ∂X ∂Y ∂Z ⎝ ∂X ⎠ ⎝ ∂Y ⎠ ⎝ ∂Z ⎠ r ∂θ r ∂θ r ∂θ er ⋅ FeX + er ⋅ FeY + er ⋅ Fe Z Brθ = e r ⋅ FF T eθ = ∂X ∂Y ∂Z ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ ⎞ =⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ⎝ ∂X ⎠⎝ ∂X ⎠ ⎝ ∂Y ⎠⎝ ∂Y ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎠ _________________________________________________________________ Brr = e r ⋅ FF T e r =

3.82 Derive the components of B −1 with respect to the basis {e r ,eθ ,e z } 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 + dYe Y + dZe Z and


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, dX = F −1dx → dXe X + dYeY + dZe Z = drF −1er + rdθ F −1eθ + dzF −1ez ∂X ∂X ⎞ ∂Y ∂Y ⎞ ⎛ ∂X ⎛ ∂Y →⎜ dr + dθ + dz ⎟ e X + ⎜ dr + dθ + dz ⎟ e Y ∂θ ∂z ⎠ ∂θ ∂z ⎠ ⎝ ∂r ⎝ ∂r ∂Z ∂Z ⎞ ⎛ ∂Z +⎜ dr + dθ + dz ⎟ e Z = drF −1e r + rdθ F −1eθ + dzF −1ez . ∂ ∂ ∂ θ r z ⎝ ⎠ Thus, ∂X ∂Y ∂Z ∂X ∂Y ∂Z F −1e r = eX + eY + e Z , F −1eθ = eX + eY + eZ ∂r ∂r ∂r r ∂θ r ∂θ r ∂θ ∂X ∂Y ∂Z F −1e z = eX + eY + eZ . ∂z ∂z ∂z and T T ∂X ∂Y , er ⋅ F −1 eY = eY ⋅ F −1er = , e r ⋅ F −1 e X = eX ⋅ F −1er = ∂r ∂r T ∂Z , etc. that is, e r ⋅ F −1 e Z = e Z ⋅ F −1er = ∂r T T ∂X ∂X ∂X ∂Y ∂Y ∂Y F −1 e X = er + eθ + e z , F −1 e Y = er + eθ + ez ∂r r ∂θ ∂z ∂r r ∂θ ∂z T ∂Z ∂Z ∂Z F −1 e Z = er + eθ + ez . r ∂θ ∂r ∂z Thus, −1 T T T ∂X ∂Y e r ⋅ F −1 eX + er ⋅ F −1 eY Brr−1 = e r ⋅ FF T e r = e r ⋅ F −1 F −1e r = ∂r ∂r

( ) ( ) ( ) ( )

( )

( )

(

+

)

( )

∂Z e r ⋅ F −1 ∂r

r

−1

2

( )

2

( )

2

⎛ ∂X ⎞ ⎛ ∂Y ⎞ ⎛ ∂Z ⎞ eZ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ . ⎝ ∂r ⎠ ⎝ ∂r ⎠ ⎝ ∂r ⎠ T T T ∂X ∂Y F −1eθ = e r ⋅ F −1 e X + e r ⋅ F −1 e Y r ∂θ r ∂θ T ∂ X ∂ X ∂ Y ∂ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ Y ⎞ ⎛ ∂Z ⎞⎛ ∂Z eZ = ⎜ +⎜ ⎟⎜ ⎟ ⎟⎜ ⎟+⎜ ⎟⎜ ⎝ ∂r ⎠⎝ r ∂θ ⎠ ⎝ ∂r ⎠⎝ r ∂θ ⎠ ⎝ ∂r ⎠⎝ r ∂θ

T

( ) ∂Z + e ⋅ (F ) r ∂θ Br−θ1 = e r ⋅ F −1

( )

( )

( )

⎞ ⎟ ., ⎠

etc. _________________________________________________________________ 3.83 Verify that (a) the components of B with respect to {e r ,eθ ,e z } can be obtained from ⎡ FF T ⎤ ⎣ ⎦ o o o T and (b) the component of C , with respect to e r ,eθ , e z can be obtained from ⎡ F 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) →

{

}


Lai et al, Introduction to Continuum Mechanics ⎡ ∂r ⎢ ⎢ ∂ro ⎢ ⎡ FF T ⎤ = ⎢ r ∂θ ⎣ ⎦ ⎢ ∂ro ⎢ ∂z ⎢ ⎢⎣ ∂ro 2

∂r ⎤ ⎡ ∂r ⎥⎢ ∂zo ⎥ ⎢ ∂ro r ∂θ ⎥ ⎢ ∂r ⎥⎢ ∂zo ⎥ ⎢ ro ∂θ o ∂z ⎥ ⎢ ∂r ⎥⎢ ∂zo ⎥⎦ ⎢⎣ ∂zo

∂r ro ∂θ o r ∂θ ro ∂θ o ∂z ro ∂θ o 2

r ∂θ ro ∂θ o r ∂θ ∂zo

2

⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ Brr = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ , Brθ ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ (b) ⎡ ∂r ∂z ⎤ ⎡ ∂r r ∂θ ⎢ ⎥⎢ ∂ro ∂ro ⎥ ⎢ ∂ro ⎢ ∂ro ⎢ ∂z ⎥ ⎢ r ∂θ r ∂θ ⎡ F T F ⎤ = ⎢ ∂r ⎥⎢ ⎣ ⎦ ⎢ ro ∂θ o ro ∂θ o ro ∂θ o ⎥ ⎢ ∂ro ⎢ ∂r ∂z ⎥ ⎢ ∂z r ∂θ ⎢ ⎥⎢ ∂zo ∂zo ⎦⎥ ⎣⎢ ∂ro ⎣⎢ ∂zo 2

r ∂θ ∂ro

2

=

∂z ⎤ ⎥ ∂ro ⎥ ∂z ⎥ ⎥ → ro ∂θ o ⎥ ∂z ⎥ ⎥ ∂zo ⎥⎦

∂r r ∂θ ∂r r ∂θ ∂r r ∂θ etc. + + ∂ro ∂ro ro ∂θ o ro ∂θ o ∂zo ∂zo

∂r ro ∂θ o r ∂θ ro ∂θ o ∂z ro ∂θ o

∂r ⎤ ⎥ ∂zo ⎥ r ∂θ ⎥ ⎥→ ∂zo ⎥ ∂z ⎥ ⎥ ∂zo ⎦⎥

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ Cro ro = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ , Croθo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟. ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠⎝ ro ∂θ o ⎠ ⎝ ∂ro ⎠⎝ ro ∂θ o ⎠ ⎝ ∂ro ⎠⎝ ro ∂θ 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

Bθθ

2

2

2

⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ⎞ 2 2 =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = ⎜ ⎟ + ( kr ) = 1 + ( kr ) , ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠ ⎝ ro ⎠ 2

2

2

⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎛ ∂z ⎞ Bzz = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1, ⎝ ∂ro ⎠ ⎝ ro ∂θo ⎠ ⎝ ∂zo ⎠ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ Brθ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠ ⎛ ∂r ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ ∂z ⎞ Brz = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟=0, r r r r z z ∂ ∂ ∂ ∂ ∂ ∂ θ θ ⎝ o ⎠⎝ o ⎠ ⎝ o o ⎠⎝ o o ⎠ ⎝ o ⎠⎝ o ⎠ ⎛ ∂z ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ r ∂θ ⎞ Bzθ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = rk . ⎝ ∂ro ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ∂zo ⎠


Lai et al, Introduction to Continuum Mechanics . 0 0⎤ ⎡1 ⎢ ⎥ 2 Thus, [ B ] = ⎢ 0 1 + ( rk ) rk ⎥ . ⎢0 1 ⎥⎥ rk ⎢⎣ ⎦ (b) Using Eqs.(3. 29.43) to (3. 29.51), we have, 2

Cro ro

2

2

2

2

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ =⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1, Cθoθo = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1, ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ ⎝ ro ∂θ o ⎠ 2

2

2

⎛ ∂r ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ 2 Cz o z o = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = 1 + ( rk ) , ⎝ ∂zo ⎠ ⎝ ∂z o ⎠ ⎝ ∂z o ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ Croθo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎝ ro ∂θ o ⎠⎝ ∂ro ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ Cro zo = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ⎝ ∂ro ⎠⎝ ∂zo ⎠ ⎝ ∂ro ⎠⎝ ∂zo ⎠ ⎝ ∂ro ⎠⎝ ∂zo ⎠ ⎛ ∂r ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ C zo θ o = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = rk . ⎝ ∂zo ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ro ∂θ o ⎠ ⎝ ∂zo ⎠⎝ ro ∂θ o ⎠ ⎡1 0 ⎤ 0 ⎢ ⎥ Thus, [C] = ⎢ 0 1 rk ⎥ . ⎢ 2⎥ ⎣⎢ 0 rk 1 + ( rk ) ⎦⎥

_________________________________________________________________ 3.85 Given r = ( 2aX + b )

1/2

, 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 ⎞ Bzz = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ =1, ⎝ ∂X ⎠ ⎝ ∂Y ⎠ ⎝ ∂Z ⎠ ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ Brθ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎝ ∂X ⎠⎝ ∂X ⎠ ⎝ ∂Y ⎠⎝ ∂Y ⎠ ⎝ ∂Z ⎠⎝ ∂Z

⎞ ⎟=0, ⎠

⎛ ∂r ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ ∂z ⎞ ⎛ ∂r ⎞ ⎛ ∂z ⎞ Brz = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0, ⎝ ∂X ⎠ ⎝ ∂X ⎠ ⎝ ∂Y ⎠ ⎝ ∂Y ⎠ ⎝ ∂Z ⎠ ⎝ ∂Z ⎠ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ ⎛ r ∂θ ⎞⎛ ∂z ⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ Bθ z = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0. ⎝ ∂X ⎠ ⎝ ∂X ⎠ ⎝ ∂Y ⎠⎝ ∂Y ⎠ ⎝ ∂Z ⎠ ⎝ ∂Z ⎠ Copyright 2010, Elsevier Inc 3-44

2

Lai et al, Introduction to Continuum Mechanics ⎡( a / r ) 2 0 0⎤ ⎢ ⎥ 2 ⎢ Thus, [ B ] = 0 ( r / a ) 0⎥⎥ . ⎢ 0 1⎥ ⎢ 0 ⎣ ⎦ (b) det B = 1 =1, thus, there is no change of volume. _________________________________________________________________ 3.86 Given r = r ( X ), θ = g (Y ), z = h( Z ) , where ( r ,θ , z ) and ( X , Y , Z ) are cylindrical and rectangular Cartesian coordinate with respect to the current and the reference configuration respectively. Obtain the components of the right Cauchy-Green Tensor C with respect to the basis at the reference configuration. ----------------------------------------------------------------------------------------Ans. Using Eqs.(3.29.68) etc. we have, CXX

⎛ ∂r =⎜ ⎝ ∂X ⎛ ∂r =⎜ ⎝ ∂X

2

2

2

⎞ ⎛ r ∂θ ⎞ ⎛ ∂z ⎞ ⎟ + ⎜ ∂X ⎟ + ⎜ ∂X ⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ r ∂θ ⎟⎜ ∂Y ⎟ + ⎜ ∂X ⎟⎜ ∂Y ⎠⎝ ⎠ ⎝ ⎠⎝

= ( r ′( X ) ) , CYY = ( rg ′(Y ) ) , CZZ = ( h′( Z ) ) 2

2

2

⎞ ⎛ ∂z ⎞⎛ ∂z ⎞ ⎟ + ⎜ ∂X ⎟⎜ ∂Y ⎟ = 0, CYZ = 0, CXZ = 0 . ⎠ ⎝ ⎠⎝ ⎠ 2 ⎡( r ′( X ) ) 0 0 ⎤ ⎢ ⎥ 2 ⎢ 0 ⎥ , where r ′( X ) ≡ dr / dX , etc., ( g ′(Y ) ) [ C] = ⎢ 0 ⎥ 2 ⎢ 0 0 h′( Z ) ) ⎥ ( ⎣ ⎦ _________________________________________________________________ CXY


Lai et al, Introduction to Continuum Mechanics CHARTER 4

4.1

⎡1 2 3⎤ The state of stress at a certain point in a body is given by: [ T ] = ⎢⎢ 2 4 5 ⎥⎥ MPa. . ⎢⎣ 3 5 0 ⎥⎦ ei

On each of the coordinate planes (with normal in e1 , e2 , e3 directions), (a) what is the normal stress and (b) what is the total shearing stress -----------------------------------------------------------------------------Ans. (a) The normal stress on the e1 plane (i.e., the plane whose normal is in the direction e1 ) is 1 MPa. , on the e2 plane is 4 MPa. , and on the e3 plane is 0 MPa.

(b) The total shearing stress on the e1 plane is

22 + 32 = 13 =3.61 MPa. On the e 2 plane is

22 + 52 = 29 =5.39 MPa. , and on the e3 plane is 32 + 52 = 34 = 5.83 MPa . _________________________________________________________________

4.2

⎡ 2 −1 3 ⎤ The state of stress at a certain point in a body is given by: [ T] = ⎢⎢ −1 4 0 ⎥⎥ MPa. ⎢⎣ 3 0 −1⎥⎦ ei

(a) Find the stress vector at a point on the plane whose normal is in the direction of 2e1 + 2e2 + e3 . (b) Determine the magnitude of the normal and shearing stresses on this plane. ------------------------------------------------------------------------------Ans. (a) The stress vector on the plane is t = Tn , where n = (2e1 + 2e2 + e3 ) / 3 . Thus ⎡ 2 −1 3 ⎤ ⎡ 2 ⎤ ⎡5⎤ 1⎢ 1⎢ ⎥ ⎥ ⎢ ⎥ [t ] = 3 ⎢ −1 4 0 ⎥ ⎢ 2⎥ = 3 ⎢6⎥ , → t = (5e1 + 6e2 + 5e3 ) / 3 MPa. ⎢⎣ 3 0 −1⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 5 ⎥⎦

(b) Normal stress Tn = n ⋅ t = (1 / 9) ( 2e1 + 2e2 + e3 ) ⋅ ( 5e1 + 6e2 + 5e3 ) = 3 MPa. Magnitude of shearing stress Ts =

2

t − Tn2 = 86 / 9 − 9 = 0.745 MPa. Or,

Ts = t - Tnn = ( 5e1 + 6e2 + 5e3 ) / 3 − 3 (1 / 3) ( 2e1 + 2e2 + e3 ) = ( −e1 + 2e3 ) / 3 → Ts = 5 / 3 = 0.745 __________________________________________________________________ 4.3 Do the previous problem for a plane passing through the point and parallel to the plane x1 − 2 x2 + 3 x3 = 4 . ------------------------------------------------------------------------------Ans. (a) The normal to the plane is n = (e1 − 2e2 + 3e3 ) / 14 . [ t ] = [ T][n ] → ⎡ 2 −1 3 ⎤ ⎡ 1 ⎤ ⎡13 ⎤ 1 ⎢ 1 ⎢ ⎥ ⎥ ⎢ ⎥ → [t ] = −1 4 0 ⎥ ⎢ −2 ⎥ = −9 , → t = (1 / 14)(13e1 − 9e2 ) = 3.47e1 − 2.41e2 MPa. 14 ⎢ 14 ⎢ ⎥ ⎢⎣ 3 0 −1⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ 13 ⎤ 1 (b) Normal stress Tn = n ⋅ t = [1 −2 3] ⎢⎢ −9 ⎥⎥ = 31 / 14 = 2.21 MPa. 14 ⎢⎣ 0 ⎥⎦

4-1

Copyright 2010, Elsevier Inc

Lai et al, Introduction to Continuum Mechanics Magnitude of shearing stress Ts =

t − Tn2 = 250 / 14 − ( 2.21) = 3.60 MPa. Or, 2

2

Ts = t − Tn n = (3.47e1 − 2.41e2 ) − (2.21 / 14)(e1 − 2e2 + 3e3 ) = 2.88e1 − 1.23e2 − 1.77e3 , → Ts = 3.60 MPa. _________________________________________________________________ The stress distribution in a certain body is given by 100 x1 −100 x2 ⎤ ⎡ 0 ⎢ 0 0 ⎥⎥ MPa. [T] = ⎢ 100 x1 ⎢⎣ −100 x2 0 0 ⎥⎦

4.4

(

Find the stress vector acting on a plane which passes through the point 1 / 2,

)

3 / 2, 3 and is

tangent to the circular cylindrical surface x12 + x22 = 1 at that point. -----------------------------------------------------------------------------Ans. Let f = x12 + x22 , then the unit normal to the circle f = 1 at a point ( x1 , x2 ) is given by n=

2 x e + 2 x2e2 ∇f = 11 = x1e1 + x2e2 . At the point 1 / 2, ∇f 4 x12 + 4 x22

(

)

3 / 2, 3 , n =

(

)

1 e1 + 3e 2 . 2

⎡ 0 50 −50 3 ⎤ ⎢ ⎥ and [ T ] = ⎢ 50 0 0 ⎥ , thus, ⎢ ⎥ 0 ⎦ ⎣ −50 3 0 ⎡ 0 50 −50 3 ⎤ ⎡ 1 / 2 ⎤ ⎡ 25 3 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 ⎥ ⎢ 3 / 2 ⎥ = ⎢ 25 ⎥ → t = 25 3e1 + 25e 2 − 25 3e3 MPa. [t ] = ⎢ 50 0 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 ⎦ ⎣ 0 ⎦ ⎣ −25 3 ⎦ ⎣ −50 3 0 _________________________________________________________________ 4.5

Given T11 = 1 MPa., T22 = −1 MPa. , and all other Tij = 0 at a point in a continuum.

(a)

Show that the only plane on which the stress vector is zero is the plane with normal in the e3 direction. (b) Give three planes on which there is no normal stress acting. -----------------------------------------------------------------------------⎡1 0 0 ⎤ ⎡ n1 ⎤ ⎡ n1 ⎤ Ans. (a) [ t ] = ⎢⎢ 0 −1 0 ⎥⎥ ⎢⎢ n2 ⎥⎥ = ⎢⎢ − n2 ⎥⎥ → t = n1e1 − n2e 2 . t = 0 → n1 = n2 = 0 → t = e3 . ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ n3 ⎥⎦ ⎢⎣ 0 ⎥⎦ (b) Tn = n ⋅ t = n12 − n22 . Thus, the plane with n12 − n22 = 0 has no normal stress. These include

n = e3 , n = (e1 + e2 ) / 2, n = (e1 − e2 ) / 2 etc. _________________________________________________________________

4.6

⎡ 10 50 −50 ⎤ 0 ⎥⎥ MPa. , find T11′ and T13′ where e1′ For the following state of stress [ T ] = ⎢⎢ 50 0 ⎢⎣ −50 0 0 ⎥⎦

is in the direction of e1 + 2e2 + 3e3 and e′2 is in the direction of e1 + e2 − e3 .

4-2


Lai et al, Introduction to Continuum Mechanics -----------------------------------------------------------------------------Ans. e1′ = (e1 + 2e2 + 3e3 ) / 14, e′2 = (e1 + e2 − e3 ) / 3 , thus ⎡ 10 50 −50 ⎤ ⎡ 1 ⎤ ⎡ −40 ⎤ 1 1 ⎢ ⎥ ⎢ ⎥ ' T11 = [1 2 3] ⎢ 50 0 0 ⎥ ⎢ 2 ⎥ = [1 2 3] ⎢⎢ 50 ⎥⎥ = −90 / 14 = −6.43 MPa. 14 14 ⎢⎣ −50 0 ⎢⎣ −50 ⎥⎦ 0 ⎥⎦ ⎢⎣ 3 ⎥⎦ e3' = e1' × e2' = (−5e1 + 4e2 − e3 ) / 42 , therefore, ⎡ 10 50 −50 ⎤ ⎡ −5⎤ 1 ' T13 = [1 2 3] ⎢⎢ 50 0 0 ⎥⎥ ⎢⎢ 4 ⎥⎥ = 450 / 588 =18.6 MPa . 588 ⎢⎣ −50 0 0 ⎥⎦ ⎢⎣ −1⎥⎦ _________________________________________________________________ ⎡α x2 β 0 ⎤ 0 0 ⎥⎥ where α and β are Consider the following stress distribution [ T ] = ⎢⎢ β 4.7 ⎢⎣ 0 0 0 ⎥⎦ constants. (a) Determine and sketch the distribution of the stress vector acting on the square in the x1 = 0 plane with vertices located at ( 0,1,1) , ( 0, −1,1) , ( 0,1, −1) and ( 0, −1, −1) . (b) Find the total resultant force and moment about the origin of the stress vectors acting on the square of part (a). -----------------------------------------------------------------------------Ans. (a) The normal to the plane x1 = 0 is e1 , thus, t e1 = α x2e1 + β e2 . On the plane, there is a

constant shearing stress β in the e 2 direction and a linear distribution of normal stress α x2 , (see figure).

(b) 1 1

FR = ∫ tdA = ∫

∫ (α x2e1 + β e2 )dx2 dx3 = 0e1 + 4β e2 .

−1 −1

1 1

M o = ∫ x × tdA = ∫

∫ ( x2e2 + x3e3 ) × (α x2e1 + β e2 )dx2 dx3

−1 −1

1

⎡ x3 ⎤ 4α =∫ ∫ e3 dx2 dx3 = 0e1 + 0e2 − α ( 2 ) ⎢ 2 ⎥ e3 = − 3 ⎢⎣ 3 ⎥⎦ −1 −1 −1 _________________________________________________________________ 1 1

4-3

(

− β x3e1 + α x2 x3e 2 − α x22e3

)


Lai et al, Introduction to Continuum Mechanics 4.8 Do the previous problem if the stress distribution is given by T11 = α x22 and all other Ti j = 0 . -----------------------------------------------------------------------------Ans. (a) The normal to the plane x1 = 0 is e1 , thus, t e1 = α x22e1 . On the plane, there is a parabolic distribution of normal stress α x22 , (see figure). (0,-1,1)

(0,-1,-1)

α

x3

(0,1,-1)

x2

x1

α

(0,1,1)

4α e1 . 3

(b) FR = ∫ tdA = ∫

∫ (α x2 e1 )dx2 dx3 =

M o = ∫ x × tdA = ∫

∫ ( x2e2 + x3e3 ) × (α x2 e1 )dx2 dx3

1 1

2

−1 −1 1 1

2

−1 −1

=

∫ ∫ (α x2 x3e2 - α x2e3 )dx2 dx3 = 0e2 + 0e3 1 1

2

3

−1 −1

_________________________________________________________________ 4.9

Do problem 4.7 for the stress distribution: T11 = α , T12 = T21 = α x3 and all other Ti j = 0 .

-----------------------------------------------------------------------------Ans. (a) The normal to the plane x1 = 0 is e1 , thus, t e1 = α e1 + α x3e 2 . On the plane, there is a constant normal stress of α and a linear distribution of shearing stress α x3e2 , (see figure). (0,-1,1)

(0,-1,-1)

α

x3

(0,1,-1)

x2

(0,1,1)

x1

α

1 1

(b) FR = ∫ tdA = ∫

∫ (α e1 + α x3e2 )dx2 dx3 = 4α e1 .

−1 −1

4-4


Lai et al, Introduction to Continuum Mechanics 1 1

M o = ∫ x × tdA = ∫

∫ ( x2e2 + x3e3 ) × (α e1 + α x3e2 )dx2 dx3

−1 −1

=

∫ ∫ ( −α x3 e1 + α x3e2 - α x2e3 )dx2 dx3 = − 1 1

2

−1 −1

4α e1 3

_________________________________________________________________ 4.10

Consider the following stress distribution for a circular cylindrical bar:

−α x3 α x2 ⎤ ⎡ 0 ⎢ 0 ⎥⎥ [T] = ⎢ −α x3 0 ⎢⎣ α x2 0 0 ⎥⎦ (a) What is the distribution of the stress vector on the surfaces defined by (i) the lateral surface x22 + x32 = 4 , (ii) the end face x1 = 0 , and (iii) the end face x1 = l ? (b) Find the total resultant force and moment on the end face x1 = l . -----------------------------------------------------------------------------Ans. (a) The outward unit normal vector to the lateral surface x22 + x32 = 4 is given by x e + x3e3 n1 = 2 2 . The outward unit normal vector to x1 = 0 is n 2 = −e1 and that to x1 = l is 2 n3 = e1 . Thus, −α x3 α x2 ⎤ ⎡ 0 ⎤ ⎡0 ⎤ ⎡ 0 1⎢ ⎡ t n ⎤ = −α x3 0 0 ⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢⎢0 ⎥⎥ → t n1 = 0 . ⎣ 1⎦ 2⎢ ⎢⎣ α x2 0 0 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0 ⎥⎦ t n 2 = −Te1 = − ( −α x3e 2 + α x2e3 ) = α x3e2 − α x2e3 t n 3 = Te1 = ( −α x3e 2 + α x2e3 ) = −α x3e2 + α x2e3

(b) On the end face x1 = l , t n 3 = Te1 = −α x3e 2 + α x2e3

FR = ∫ t n3 dA = ∫ ( −α x3e 2 + α x2e3 ) dA = ( −α e2 ) ∫ x3dA + (α e3 ) ∫ x2 dA = 0 .

[note: the axes are axes of symmetry, the integrals are clearly zero]. M o = ∫ x × tdA = ∫ ( x2e 2 + x3e3 ) × ( −α x3e 2 + α x2e3 ) dA

(

)

2

2

0

0

= ∫ α x22 + α x32 dAe1 = α ∫ r 2 2π rdre1 = 2πα ∫ r 3 dre1 = 8πα e1 _________________________________________________________________ 4.11

An elliptical bar with lateral surface defined by x22 + 2 x32 = 1 has the following stress

−2 x3 x2 ⎤ ⎡ 0 ⎢ distribution: [ T ] = ⎢ −2 x3 0 0 ⎥⎥ . (a) Show that the stress vector at any point ( x1 , x2 , x3 ) on ⎢⎣ x2 0 0 ⎥⎦ the lateral surface is zero. (b) Find the resultant force, and resultant moment, about the origin O , of the stress vector on the left end face x1 = 0 .

4-5



∫ x2 dA = 4 2

π

and ∫ x32 dA =

π

. 2 8 2 -----------------------------------------------------------------------------Ans. (a) The outward unit normal vector to the lateral surface x22 + 2 x32 = 1 is given by: −2 x3 x2 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎡0 ⎤ x2e 2 + 2 x3e3 1 ⎢ ⎥ ⎢ ⎥ 0 0 ⎥ ⎢ x2 ⎥ , thus, [ t ] = [ T][n1 ] = ⎢ −2 x3 = ⎢⎢0 ⎥⎥ n1 = 2 2 2 2 x2 + 4 x3 ⎢⎣ x2 0 0 ⎥⎦ ⎢⎣ 2 x3 ⎥⎦ x2 + 4 x3 ⎢⎣0 ⎥⎦

Note:

(b) On the left end face x1 = 0 , n = −e1 , the stress vector is t = 2x3e 2 − x2e3 , FR = ∫ tdA = ∫ ( 2 x3e 2 − x2e3 )dA = ( 2e 2 ) ∫ x3dA − ( e3 ) ∫ x2 dA = 0 .

[note: the axes are axes of symmetry, the integrals are clearly zero]] M o = ∫ x × tdA = ∫ ( x2e2 + x3e3 ) × ( 2 x3e2 − x2e3 )dA

2π ⎫ π ⎧ π = − ∫ x22 + 2 x32 e1dA = − ⎨ + e1. ⎬ e1 = − 2 2 ⎩4 2 8 2 ⎭ _________________________________________________________________

(

4.12

)

For any stress state T , we define the deviatoric stress S to be S = T − (Tkk / 3) I , where

Tkk is the first invariant of the stress tensor T . (a) Show that the first invariant of the deviatoric ⎡ 6 5 −2 ⎤ stress vanishes. (b) Given the stress tensor [ T ] = 100 ⎢⎢ 5 3 4 ⎥⎥ kPa. , evaluate S . (c) Show ⎢⎣ −2 4 9 ⎥⎦ that the principal directions of the stress tensor coincide with those of the deviatoric stress tensor. -----------------------------------------------------------------------------Ans. (a) From S = T − (Tkk / 3) I , we have, trS = trT − (Tkk / 3) trI = Tkk − (Tkk / 3) ( 3) = 0 . 500 −200 ⎤ ⎡ 6 5 −2 ⎤ ⎡ 0 ⎢ ⎥ ⎢ (b) [S ] = 100 ⎢ 5 3 4 ⎥ − (1800 / 3) [ I ] = ⎢ 500 −300 400 ⎥⎥ kPa. ⎢⎣ −2 4 9 ⎥⎦ ⎢⎣ −200 400 300 ⎥⎦ (c) Let n be an eigenvector of T , then Tn = λ n . Now Sn = Tn − (Tkk / 3) In = λn − (Tkk / 3) n , that is Sn = λ ′n where λ ′ = λ − (Tkk / 3) . Thus, n is also an eigenvector of S with eigenvalue

λ − (Tkk / 3) .

_________________________________________________________________ 4.13 An octahedral stress plane is one whose normal makes equal angles with each of the principal axes of stress. (a) How many independent octahedral planes are there at each point? (b) Show that the normal stress on an octahedral plane is given by one-third the first stress invariant. (c) Show that the shearing stress on the octahedral plane is given by 1 2 2 1/2 2 Ts = ⎡⎢(T1 − T2 ) + (T2 − T3 ) + (T3 − T1 ) ⎤⎥ , where T1 , T2 , T3 are principal values of the stress ⎦ 3⎣ tensor. -----------------------------------------------------------------------------Ans. (a) There are four independent octahedral planes. They are given by the following unit normal vectors:

4-6


Lai et al, Introduction to Continuum Mechanics n1 =

e1 + e2 + e3

3

We note that −

, n2 =

e1 + e2 − e3

3

, n3 =

e1 − e2 + e3

3

,n 4 =

e1 − e2 − e3

3

e1 + e2 + e3

gives the same plane as n1 , etc. 3 (b) Using the principal directions as the orthonormal basis, the matrix of T is diagonal, i.e., ⎡T1 0 0 ⎤ e ±e ±e [ T] = ⎢⎢ 0 T2 0 ⎥⎥ . The normal to an octahedral plane is 1 2 3 , thus, 3 ⎢⎣ 0 0 T3 ⎥⎦ ⎡T1 0 0 ⎤ ⎡ 1 ⎤ 1 ⎢ Tn = n ⋅ Tn = [1 ±1 ±1] ⎢ 0 T2 0 ⎥⎥ ⎢⎢ ±1⎥⎥ where in this equation, the row matrix and column 3 ⎢⎣ 0 0 T3 ⎥⎦ ⎢⎣ ±1⎥⎦ matrix of n have the same elements, that is if the row matrix is [1 −1 1] then the column matrix ⎡T1 0 0 ⎤ ⎡ 1 ⎤ ⎡1⎤ 1 1 ⎢ ⎥ is ⎢ −1⎥ . Thus, Tn = [1 ±1 ±1] ⎢⎢ 0 T2 0 ⎥⎥ ⎢⎢ ±1⎥⎥ = (T1 + T2 + T3 ) . 3 3 ⎢⎣ 0 0 T3 ⎥⎦ ⎢⎣ ±1⎥⎦ ⎢⎣ 1 ⎥⎦ 1 1 2 (c) Ts2 = t n − Tn2 = T12 + T22 + T33 − T12 + T22 + T33 + 2T1T2 + 2T1T3 + 2T2T3 3 9 2 2 1 2 2 2 2 3 = T1 + T2 + T3 − T1T2 − T1T3 − T2T3 = ⎡⎢(T1 − T2 ) + (T2 − T3 ) + (T3 − T1 ) ⎤⎥ ⎣ ⎦ 9 9 1 2 2 1/2 2 That is, Ts = ⎡⎢(T1 − T2 ) + (T2 − T3 ) + (T3 − T1 ) ⎤⎥ ⎦ 3⎣ _________________________________________________________________

(

(

) ( )

)

4.14 (a) Let m and n be two unit vectors that define two planes M and N that pass through a point P . For an arbitrary state of stress defined at the point P , show that the component of the stress vector t m in the n -direction is equal to the component of the stress vector t n in the m direction. (b) If m = e1and n = e2 , what does the results of (a) reduce to? -----------------------------------------------------------------------------Ans. (a) The component of the stress vector t m in the n -direction is n ⋅ t m = n ⋅ Tm and the component of the stress vector t n in the m direction is m ⋅ t n = m ⋅ Tn = n ⋅ TT m . Since T is symmetric, therefore, n ⋅ TT m = n ⋅ Tm , therefore, n ⋅ t m = m ⋅ t n . (b) If m = e1 and n = e2 , then e1 ⋅ t e2 = e2 ⋅ t e1 → T12 = T21 . _________________________________________________________________ 4.15 Let m be a unit vector that defines a plane M passing through a point P . Show that the stress vector on any plane that contains the stress traction t m , lies in the M plane. -----------------------------------------------------------------------------Ans. Referring to the figure below, where m is perpendicular to the plane M , and t m is the stress vector for the plane. Let N be any plane which contains the vector t m and let n be the unit vector perpendicular to the plane N . Then t n = Tn . We wish to show that t n is perpendicular to m .

4-7



Now, t n ⋅ m = Tn ⋅ m = n ⋅ TT m = n ⋅ Tm = n ⋅ t m = 0, because t m is on the N plane. Thus, t n ⋅ m = 0, so that t n lies on the M plane. _________________________________________________________________ 4.16 Let t m and t n be stress vectors on planes defined by the unit vector m and n respectively and pass through the point P . Show that if k is a unit vector that determines a plane that contains t m and t n , then t k is perpendicular to m and n . -----------------------------------------------------------------------------Ans. Since k is a unit vector that determines a plane that contains t m and t n , therefore, t ×t k = m n . Since t k = Tk , t m = Tm, and t n = Tn , therefore, tm × tn m ⋅ t k = m ⋅ Tk = k ⋅ TT m = k ⋅ Tm = k ⋅ t m =

tm ⋅ tm × tn = 0 , similarly, tm × tn

tn ⋅ tm × tn =0. tm × tn _________________________________________________________________ n ⋅ t k = n ⋅ Tk = k ⋅ TT n = k ⋅ Tn = k ⋅ t n =

4.17 Given the function f ( x, y ) = 4 − x 2 − y 2 , find the maximum value of f subjected to the constraint that x + y = 2 . -----------------------------------------------------------------------------Ans. Let g ( x, y ) = 4 − x 2 − y 2 + λ ( x + y − 2) , then we have the following three equations to solve for x, y and λ : ∂g ∂g = −2 y + λ = 0 and x + y = 2 . = −2 x + λ = 0 , ∂x ∂y Thus, −2 x + λ = 0 → λ = 2 x, − 2 y + λ = 0 → λ = 2 y , therefore, x = y → x + y = 2 → 2 x = 2 → x = 1 = y . That is, f max occurs at x = y = 1 . That is,

f max = 4 − (1) 2 − (1) 2 = 2 _________________________________________________________________ 4.18 True or false: (i) Symmetry of stress tensor is not valid if the body has an angular acceleration. (ii) On the plane of maximum normal stress, the shearing stress is always zero. -----------------------------------------------------------------------------Ans. (i) False. (ii) True. _________________________________________________________________

4-8


Lai et al, Introduction to Continuum Mechanics 4.19 True or false: (i) On the plane of maximum shearing stress, the normal stress is always zero. (ii) A plane with its normal in the direction of e1 + 2e2 − 2e3 has a stress vector t = 50e1 + 100e2 − 100e3 MPa. It is a principal plane. -----------------------------------------------------------------------------Ans. (i) Not true in general. Maybe true in some special cases. (ii) True. We note that t = 50e1 + 100e2 − 100e3 = 50 ( e1 + 2e2 − 2e3 ) . Therefore, t is normal to the plane, so that there is no shearing stress on the plane. That is, it is a principal plane. _________________________________________________________________ 4.20

Why can the following two matrices not represent the same stress tensor?

⎡100 200 40 ⎤ ⎡ 40 100 60 ⎤ ⎢ 200 0 ⎥ 0 ⎥ MPa., ⎢⎢100 100 0 ⎥⎥ MPa. . ⎢ ⎢⎣ 40 ⎢⎣ 60 0 −50 ⎥⎦ 0 20 ⎥⎦ -----------------------------------------------------------------------------Ans. The first scalar invariant for the first matrix is 50 MPa. The first scalar invariant for the second matrix is 160 MPa. They are not the same, therefore, they can not represent the same stress tensor. _________________________________________________________________ ⎡ 0 100 0 ⎤ 4.21 Given [ T] = ⎢⎢100 0 0 ⎥⎥ MPa. (a) Find the magnitude of shearing stress on the plane ⎢⎣ 0 0 0 ⎥⎦ whose normal is in the direction of e1 + e2 . (b) Find the maximum and minimum normal stresses and the planes on which they act. (c) Find the maximum shearing stress and the plane on which it acts. -----------------------------------------------------------------------------⎡ 0 100 0 ⎤ ⎡1 ⎤ ⎡1 ⎤ 1 1 ⎢ 100 ⎢ ⎥ ⎥ ⎢ ⎥ Ans. (a) Let n = ( e1 + e2 ) . Then tn = ⎢100 0 0⎥ ⎢1 ⎥ = ⎢1 ⎥ i.e., 2 2 2 ⎢⎣ 0 ⎢⎣0 ⎥⎦ 0 0 ⎥⎦ ⎢⎣0 ⎥⎦

t n = 100n → shearing stress Ts = 0 . 0 − λ 100 (b) The characteristic equation is 100 0 − λ 0 0

0 0 = 0 → −λ λ 2 − 1002 = 0 −λ

(

)

→ λ1 = 100 MPa., λ2 = −100MPa., λ3 = 0 . The maximum normal stress is 100 MPa. and the minimum normal stress is −100MPa . For λ1 = 100MPa., − 100α1 + 100α 2 = 0, so that α1 = α 2 , n1 = (e1 + e2 ) / 2 .

For λ2 = −100MPa., 100α1 + 100α 2 = 0 → α1 = −α 2 , n 2 = (e1 − e2 ) / 2 . (c) (Ts )max =

(Tn )max − (Tn )min

100 − ( −100 )

= 100 MPa. The maximum shearing stress acts on 2 2 the planes n = (n1 ± n 2 ) / 2 , i.e., on the planes e1 and e2 . _________________________________________________________________

4-9

=


Lai et al, Introduction to Continuum Mechanics 4.22 Tn =

Show the equation for the normal stress on the plane of maximum shearing stress is

(Tn )max + (Tn )min

. 2 -----------------------------------------------------------------------------Ans. Let {n1 , n 2 , n3 } be the principal axes of the stress tensor with principal values T1 > T2 > T3 , ⎡T1 0 0 ⎤ then [ T] = ⎢⎢ 0 T2 0 ⎥⎥ . On the plane n = ( n1 ± n3 ) / 2 , the shearing stress is a maximum. On ⎢⎣ 0 0 T3 ⎥⎦ this plane, the normal stress is: ⎡T1 0 0 ⎤ ⎡ 1 ⎤ (Tn )max + (Tn )min T +T 1 ⎢ Tn = n ⋅ Tn → Tn = [1 0 ±1] ⎢ 0 T2 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = 1 3 = 2 2 2 ⎢⎣ 0 0 T3 ⎥⎦ ⎢⎣ ±1⎥⎦ _________________________________________________________________ 4.23 The stress components at a point are given by: T11 = 100MPa., T22 = 300MPa., T33 = 400MPa . T12 = T13 = T23 = 0 . (a) Find the maximum shearing stress and the planes on which they act. (b) Find the normal stress on these planes. (c) Are there any plane/planes on which the normal stress is 500 MPa. ? -----------------------------------------------------------------------------Ans. (a) The maximum normal stress is clearly T33 = 400MPa ., acting on the e3 plane and the minimum normal stress is clearly T11 = 100 MPa. , acting on the e1 plane. Thus, the maximum 400 − 100 1 shearing stress is (Ts )max = = 150 MPa. , acting on the plane n = ( e1 ± e3 ) . 2 2 0 ⎤⎡ 1 ⎤ ⎡100 0 ⎡ 100 ⎤ 1 1 ⎢ ⎥ ⎢ ⎥ (b) Tn = [1 0 ±1] ⎢ 0 300 0 ⎥ ⎢ 0 ⎥ = [1 0 ±1] ⎢⎢ 0 ⎥⎥ = 250 MPa. 2 2 ⎢⎣ 0 ⎢⎣ ±400⎥⎦ 0 400 ⎥⎦ ⎢⎣ ±1⎥⎦ Note: We can also use the result of Prob. 4.22 to obtain Tn =

Tmax + Tmin 400 + 100 = = 250 MPa. 2 2

(c) No, because Tmax = 400MPa _________________________________________________________________ 4.24

The principal values of a stress tensor T are T1 = 10MPa., T2 = −10 MPa. and

⎡T11 0 0 ⎤ T3 = 30MPa. If the matrix of the stress is given by: [ T] = ⎢⎢ 0 1 2 ⎥⎥ × 10 MPa. , find the ⎢⎣ 0 2 T33 ⎥⎦ values of T11 and T33 . -----------------------------------------------------------------------------Ans. I1 = 10 − 10 + 30 = 10(T11 + 1 + T33 ) → 2 = (T11 + T33 ) → T11 =2 − T33 (i) I 3 = (10 )( −10 )( 30 ) = (10 ) (T11T33 − 4T11 ) → −3 = (T11T33 − 4T11 ) (ii) 3

(i) and (ii) → −3 = ( 2 − T33 )(T33 − 4 ) → T332 − 6T33 + 5 = 0. Thus,

4-10


Lai et al, Introduction to Continuum Mechanics T33 = [6 ± 36 − 20] / 2 = 3 ± 2. Thus, T33 is either 5 or 1. To determine which is the correct value, we check I 2 = ( −10 )( 30 ) + (10 )( −10 ) + (10 )( 30 ) = 102 ⎡⎣(T33 − 4 ) + T11T33 + T11 ⎤⎦ → T33 + T11T33 + T11 − 3 = 0 . (iii) Try T33 = 1 first, from (i), T11 = 1 , so that (iii) is clearly satisfied. Next try T33 = 5 , eq (i) gives T11 = 2 − 5 = −3 , then left side of (iii) becomes 5 + (−3)(5) − 3 − 3 = −16 ≠ 0 . Thus, T33 = 1 and T11 = 1 . _________________________________________________________________

4.25

0 0 ⎤ ⎡300 ⎢ If the state of stress at a point is: [ T] = ⎢ 0 −200 0 ⎥⎥ kPa. , find (a) the magnitude of ⎢⎣ 0 0 400 ⎥⎦

the shearing stress on the plane whose normal is in the direction of ( 2e1 + 2e2 + e3 ) and (b) the maximum shearing stress. -----------------------------------------------------------------------------1 Ans. (a) Let n = ( 2e1 + 2e2 + e3 ) , then 3 0 0 ⎤ ⎡ 2⎤ ⎡300 ⎡ 600 ⎤ 1⎢ 1⎢ 100 ⎥ ⎢ ⎥ [t n ] = 3 ⎢ 0 −200 0 ⎥ ⎢ 2⎥ = 3 ⎢ −400⎥⎥ → t n = 3 ( 6e1 − 4e2 + 4e3 ) ⎢⎣ 0 ⎢⎣ 400 ⎥⎦ 0 400 ⎥⎦ ⎢⎣ 1 ⎥⎦

Tn = n ⋅ t n =

100 800 = 88.89kPa. (12 − 8 + 4 ) = 9 9

104 × 68 104 × 64 − = 104 × 6.76 → Ts = 260 kPa. 9 81 400 − ( −200 ) = 300 kPa. (b) (Ts )max = 2 _________________________________________________________________ 2

Ts2 = t n − Tn2 =

⎡1 4 0⎤ 4.26 Given [ T] = ⎢⎢ 4 1 0 ⎥⎥ MPa. (a) Find the stress vector on the plane whose normal is in ⎢⎣ 0 0 1 ⎥⎦ the direction of e1 + e2 . (b) Find the normal stress on the same plane. (c) Find the magnitude of the shearing stress on the same plane. (d) Find the maximum shearing stress and the planes on which this maximum shearing stress acts. -----------------------------------------------------------------------------⎡ 1 4 0 ⎤ ⎡1 ⎤ ⎡5⎤ 1 1 ⎢ 1 ⎢ ⎥ 5 ⎥ ⎢ ⎥ Ans. (a) Let n = ( e1 + e2 ) , then [t n ] = ⎢ 4 1 0⎥ ⎢1 ⎥ = ⎢5⎥ → tn = ( e1 + e2 ) . 2 2 2 2 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣0 ⎥⎦ (b) Tn = n ⋅ t n =

4-11

1 ( 5 + 5) = 5 MPa. (c) Ts2 = tn 2 − Tn2 = 25 − 25 = 0 2


Lai et al, Introduction to Continuum Mechanics 2 (d) The characteristic equation is (1 − λ ) ⎢⎡(1 − λ ) − 42 ⎥⎤ = 0 → λ1 = 5, λ2 = −3, λ3 = 1 . Thus, ⎣ ⎦ (Tn )max = 5 MPa. and (Tn )min = −3 MPa.

(

)

For (Tn )max = 5 MPa. (1 − 5 ) α1 + 4α 2 = 0 → α1 = α 2 → n1 = 1 / 2 ( e1 + e 2 ) .

(

)

For (Tn )min = −3 MPa. (1+3) α1 + 4α 2 = 0 → α1 = −α 2 → n 2 = 1 / 2 ( e1 − e 2 ) . Thus, (Ts )max =

5 − ( −3)

= 4 MPa. , acting on the plane whose normal is 2 n = 1 / 2 ( n1 ± n 2 ) → n = e1 and n = e 2 .

(

)

_________________________________________________________________ 4.27 The stress state in which the only non-vanishing stress components are a single pair of shearing stresses is called simple shear. Take T12 = T21 = τ and all other Ti j = 0 . (a) Find the principal values and principal directions of this stress state. (b) Find the maximum shearing stress and planes on which it acts. -----------------------------------------------------------------------------⎡0 τ 0⎤ Ans. (a) With [ T ] = ⎢⎢τ 0 0 ⎥⎥ , the characteristic equation is ⎢⎣ 0 0 0 ⎥⎦

(

)

λ λ 2 − τ 2 = 0 → λ1 = τ , λ2 = −τ , λ3 = 0.

(

)

For λ1 = τ , ( 0 − τ )α1 + τα 2 = 0 → α1 = α 2 → n1 = 1 / 2 ( e1 + e 2 ) .

(

)

For λ2 = −τ → ( 0+τ )α1 + τα 2 = 0 → α1 = −α 2 → n 2 = 1 / 2 ( e1 − e 2 ) . For λ3 = 0 → n3 = e3 . (b) (Ts )max =

τ − ( −τ )

= τ , acting on the plane whose normal is 2 n = 1 / 2 ( n1 ± n 2 ) → n = e1 and n = e 2 .

(

)

_________________________________________________________________ 4.28 The stress state in which only the three normal stress components do not vanish is called a tri-axial state of stress. Take T11 = σ 1 , T22 = σ 2 , T33 = σ 3 with σ 1 > σ 2 > σ 3 and all other Ti j = 0 . Find the maximum shearing stress and the plane on which it acts. -----------------------------------------------------------------------------σ −σ3 1 , n= Ans. (Ts )max = 1 ( e1 ± e3 ) . 2 2 _________________________________________________________________ 4.29 Show that the symmetry of the stress tensor is not valid if there are body moments per unit volume, as in the case of a polarized anisotropic dielectric solid. ------------------------------------------------------------------------------

4-12


Lai et al, Introduction to Continuum Mechanics Ans. Let M* = M1*e1 + M 2*e 2 + M 3*e3 be the body moments per unit volume. Then referring to the figure shown below, the total moments of all the surface forces and the body force and body moment about the axis which passes through the center point A and parallel to the x3 axis is : ∑ ( M c )3 = T21 ( Δx2 Δx3 ) ( Δx1 / 2 ) + (T21 + ΔT21 ) ( Δx2 Δx3 ) ( Δx1 / 2 ) − T12 ( Δx1Δx3 ) ( Δx2 / 2 ) − (T12 + ΔT12 ) ( Δx1Δx3 ) ( Δx2 / 2 ) + M 3* ( Δx1Δx2 Δx3 ) 2 2 = (1 / 12 ) (density) ( Δx1Δx2 Δx3 ) ⎡⎢( Δx1 ) + ( Δx2 ) ⎤⎥ α 3 ⎣ ⎦ where α 3 is the x3 components of the angular acceleration of the element. We now let Δx1 → 0, Δx2 → 0, Δx3 → 0 and drop all terms of small quantities of higher order than

( Δx1Δx2 Δx3 ) , we obtain, T21 ( Δx1Δx2 Δx3 ) − T12 ( Δx1Δx2 Δx3 ) + M 3* ( Δx1Δx2 Δx3 ) = 0 → T12 − T21 = M 3* ,

Similarly, one can show that T13 − T31 = M 2* and T23 − T32 = M1* .

_________________________________________________________________

⎡ x1 + x2 T12 ( x1 , x2 ) 0 ⎤ ⎢ ⎥ 0 ⎥ , find T12 so 4.30 Given the following stress distribution: [ T ] = ⎢T12 ( x1 , x2 ) x1 − 2 x2 ⎢ 0 0 x2 ⎥⎦ ⎣ that the stress distribution is in equilibrium with zero body force and so that the stress vector on the plane x1 = 1 is given by t = (1 + x2 ) e1 + ( 5 − x2 ) e 2 . -----------------------------------------------------------------------------∂Ti j Ans. The equations of equilibrium are + ρ Bi = 0 . Now with Bi = 0 , we have, ∂x j ∂T11 ∂T12 ∂T13 ∂T + + = 1 + 12 = 0 → T12 = − x2 + f ( x1 ) . ∂x1 ∂x2 ∂x3 ∂x2 ∂T21 ∂T22 ∂T23 ∂T12 + + = − 2 = 0 → T12 = 2 x1 + g ( x2 ) , thus, T12 = 2 x1 − x2 + C . ∂x1 ∂x2 ∂x3 ∂x1 ∂T31 ∂T32 ∂T33 + + = 0 → 0 = 0. ∂x1 ∂x2 ∂x3 To determine C, we have, the stress vector on the plane x1 = 1 is t = Te1 = T11e1 + T21e 2 + T31e3 = ⎡⎣( x1 + x2 ) e1 + ( 2 x1 − x2 + C ) e2 ⎤⎦ . Thus, x1 =1

(1 + x2 ) e1 + ( 2 − x2 + C ) e2 = (1 + x2 ) e1 + ( 5 − x2 ) e2 → C = 3 → T12 = 2 x1 − x2 + 3 . _________________________________________________________________

4-13



4.31

⎡ x2 Consider the following stress tensor: [ T] = α ⎢⎢ − x3 ⎢⎣ 0

− x3 0 − x2

0 ⎤ − x2 ⎥⎥ . Find an expression T33 ⎥⎦

for T33 such that the stress tensor satisfies the equations of equilibrium in the presence of body force vector B = − ge3 , where g is a constant. -----------------------------------------------------------------------------∂Ti j + ρ Bi = 0 . With B1 = B2 = 0, B3 = − g , we have, Ans. The equations of equilibrium are ∂x j ∂T ∂T11 ∂T12 ∂T13 ∂T ∂T + + + ρ B1 = 0 + 0 + 0 + 0 = 0, 21 + 22 + 23 + ρ B2 = 0 + 0 + 0 + 0 = 0, ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3

⎛ ∂T31 ∂T32 ∂T33 ∂T ⎞ ∂T ρg + + + ρ B3 = α ⎜ −1 + 33 ⎟ − ρ g = 0 → 33 = +1 α x x ∂x1 ∂x2 ∂x3 ∂ ∂ 3 ⎠ 3 ⎝ ⎛ ρg ⎞ → T33 = ⎜ + 1⎟ x3 + f ( x1 , x2 ) ⎝ α ⎠ _________________________________________________________________ 4.32

In the absence of body forces, the equilibrium stress distribution for a certain body is

T11 = Ax2 , T12 = T21 = x1 , T22 = Bx1 + Cx2 , T33 = (T11 + T22 ) / 2, all other Ti j = 0 . Also, the boundary plane x1 − x2 = 0 for the body is free of stress. (a) Find the value of C and (b) determine the value of A and B . -----------------------------------------------------------------------------∂Ti j + ρ Bi = 0 . With Bi = 0, , we have, Ans. (a) The equations of equilibrium are ∂x j ∂T11 ∂T12 ∂T13 + + + ρ B1 = 0 + 0 + 0 + 0 = 0, ∂x1 ∂x2 ∂x3 ∂T21 ∂T22 ∂T23 + + + ρ B2 = 1 + C + 0 + 0 = 0 → C = −1 , ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + + ρ B3 = 0 + 0 + 0 + 0 = 0 . ∂x1 ∂x2 ∂x3

(b) The unit normal to the boundary plane x1 − x2 = 0 is n = ( e1 − e 2 ) / 2 . Thus, on this plane (note x1 = x2 ), we have, ⎡ Ax1 ⎤⎡ 1 ⎤ x1 0 ⎡ Ax1 − x1 ⎤ ⎡0 ⎤ 1 ⎢ ⎥⎢ ⎥ 1 ⎢ x1 − Bx1 + x1 ⎥⎥ = ⎢⎢0 ⎥⎥ , thus, −1⎥ = 0 [t ] = ⎢ x1 Bx1 − x1 ⎥ ⎢ ⎢ 2⎢ 2 ⎢⎣ ⎥⎦ ⎢⎣0 ⎥⎦ 0 0 (T11 + T22 ) / 2⎥⎦ ⎢⎣ 0 ⎥⎦ ⎣ 0 Ax1 − x1 = 0 → A = 1 and x1 − Bx1 + x1 = 0 → B = 2 . _________________________________________________________________ 4.33 In the absence of body forces, do the following stress components satisfy the equations of equilibrium:

4-14



(

)

(

)

(

)

T11 = α ⎡ x22 + ν x12 − x22 ⎤ , T22 = α ⎡ x12 + ν x22 − x12 ⎤ , T33 = αν x12 + x22 , ⎣ ⎦ ⎣ ⎦ T12 = T21 = −2αν x1 x2 , T13 = T31 = 0, T23 = T32 = 0. -----------------------------------------------------------------------------∂Ti j + ρ Bi = 0 . With Bi = 0, , we have, Ans. The equations of equilibrium are ∂x j ∂T11 ∂T12 ∂T13 + + + ρ B1 = 2αν x1 − 2αν x1 + 0 + 0 = 0, ∂x1 ∂x2 ∂x3 ∂T21 ∂T22 ∂T23 + + + ρ B2 = −2αν x2 + 2αν x2 + 0 + 0 = 0 ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + + ρ B3 = 0 + 0 + 0 + 0 = 0 . Yes, the equations of equilibrium are all satisfied. ∂x1 ∂x2 ∂x3 _________________________________________________________________ 4.34

Repeat the previous problem for the stress distribution

⎡ x1 + x2 2 x1 − x2 0 ⎤ [ T] = α ⎢⎢ 2 x1 − x2 x1 − 3x2 0 ⎥⎥ ⎢⎣ 0 x1 ⎥⎦ 0 -----------------------------------------------------------------------------∂Ti j + ρ Bi = 0 . With Bi = 0, , we have, Ans. The equations of equilibrium are ∂x j ∂T11 ∂T12 ∂T13 + + + ρ B1 = 0 → α (1 − 1 + 0 ) = 0 → 0 = 0, ∂x1 ∂x2 ∂x3 ∂T21 ∂T22 ∂T23 + + + ρ B2 = α (2 − 3 + 0) ≠ 0 ∂x1 ∂x2 ∂x3 No, the second equation of equilibrium is not satisfied. _________________________________________________________________ 4.35

Suppose that the stress distribution has the form (called a plane stress state)

⎡T11 ( x1 , x2 ) T12 ( x1 , x2 ) 0 ⎤ [T] = ⎢⎢T12 ( x1 , x2 ) T22 ( x1 , x2 ) 0⎥⎥ ⎢ 0 0 0 ⎥⎦ ⎣ (a) If the state of stress is in equilibrium, can the body forces be dependent on x3 ? (b) If we introduce a function ϕ ( x1 , x2 ) such that T11 =

∂ 2ϕ ∂x22

, T22 =

∂ 2ϕ ∂x12

and T12 = −

∂ 2ϕ , What should ∂x1∂x2

be the function ϕ ( x1 , x2 ) for the equilibrium equations to be satisfied in the absence of body forces? -----------------------------------------------------------------------------∂T ( x , x ) ∂T ( x , x ) ∂T ∂T ∂T Ans. (a) 11 + 12 + 13 + ρ B1 = 11 1 2 + 12 1 2 + ρ B1 = 0 . ∂x1 ∂x2 ∂x3 ∂x1 ∂x2

4-15


Lai et al, Introduction to Continuum Mechanics ∂T ( x , x ) ∂T ( x , x ) ∂T21 ∂T22 ∂T23 + + + ρ B2 = 21 1 2 + 22 1 2 + ρ B2 = 0 . ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 Thus, B1 and B2 must be independent of x3 . ⎞ ∂ ⎛ ∂ 2ϕ ⎞ ∂ ⎛ ∂ 2ϕ ⎞ ⎜⎜ 2 ⎟⎟ − ⎜⎜ 2 ⎟⎟ = 0 . ⎟⎟ + 0 + 0 = ∂ ∂ x x ∂ x 1 1 ⎠ ⎝ 2 ⎠ ⎝ ∂x2 ⎠ ∂T21 ∂T22 ∂T23 ∂ ⎛ ∂ 2ϕ ⎞ ∂ ⎛ ∂ 2ϕ ⎞ ∂ ⎛ ∂ 2ϕ ⎞ ∂ ⎛ ∂ 2ϕ ⎞ + + + ρ B2 = − ⎜⎜ 2 ⎟⎟ + 0 + 0 = − ⎜ ⎟+ ⎜ ⎟ = 0. ⎜⎜ ⎟⎟ + ∂x1 ∂x2 ∂x3 ∂x1 ⎝ ∂x1∂x2 ⎠ ∂x2 ⎝ ∂x1 ⎠ ∂x2 ⎜⎝ ∂x12 ⎟⎠ ∂x2 ⎜⎝ ∂x12 ⎟⎠ Thus, the equations of equilibrium are satisfied for any function ϕ ( x1 , x2 ) which is continuous up to the third derivatives. _________________________________________________________________

(b)

∂T11 ∂T12 ∂T13 ∂ ⎛ ∂ 2ϕ ⎞ ∂ ⎛ ∂ 2ϕ + + + ρ B1 → ⎜ ⎟− ⎜ ∂x1 ∂x2 ∂x3 ∂x1 ⎜⎝ ∂x22 ⎟⎠ ∂x2 ⎜⎝ ∂x1∂x2

4.36 In cylindrical coordinates ( r ,θ , z ) , consider a differential volume of material 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. Derive the r and θ equations of motion in cylindrical coordinates and compare the equations with those given in Section 4.8. -----------------------------------------------------------------------------Ans.

From the free body diagram above, we have, ∑ Fr = −Trr ( rdθ ) dz + (Trr + dTrr ) ( ( r + dr ) dθ ) dz − Trθ drdz cos ( dθ / 2 )

+ (Trθ + dTrθ ) drdz cos ( dθ / 2 ) − Tθθ drdz sin ( dθ / 2 ) − (Tθθ + dTθθ ) drdz sin ( dθ / 2 )

−Trz ( rdθ ) dr + (Trz + dTrz )( rdθ ) dr + ρ Br ( rdθ ) drdz = ⎡⎣ ρ ( rdθ ) drdz ⎤⎦ ar . 2

3

1 ⎛ dθ ⎞ ⎛ dθ ⎞ ⎛ d θ ⎞ ⎛ dθ ⎞ 1 ⎛ d θ ⎞ Now, cos ⎜ =1+ ⎜ + ...and sin ⎜ ⎟ ⎟ ⎟=⎜ ⎟− ⎜ ⎟ + ... and keeping only terms 2! ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 3! ⎝ 2 ⎠ involving products of three differentials (i.e., terms involving product of 4 or more differentials drop out in the limit when these differentials approach zero), we have, Trr drdθ dz + ( dTrr ) rdθ dz + dTrθ drdz − 2Tθθ drdz ( dθ / 2 )

+ ( dTrz )( rdθ ) dr + ρ Br ( rdθ ) drdz = ⎡⎣ ρ ( rdθ ) drdz ⎤⎦ ar

Dividing the equation by rdθ drdz , we get, Trr ∂Trr ∂Trθ Tθθ ∂Trz + + − + + ρ Br = ρ ar . This is Eq. (4.8.1) ∂r ∂z r rdθ r Next,

4-16



∑ Fθ = −Tθ r ( rdθ ) dz + (Tθ r + dTθ r ) ( ( r + dr ) dθ ) dz + Trθ drdz sin ( dθ / 2 )

+ (Trθ + dTrθ ) drdz sin ( dθ / 2 ) − Tθθ drdz cos ( dθ / 2 ) + (Tθθ + dTθθ ) drdz cos ( dθ / 2 ) −Tθ z ( rdθ ) dr + (Tθ z + dTθ z ) ( rdθ ) dr + ρ Bθ ( rdθ ) drdz = ⎡⎣ ρ ( rdθ ) drdz ⎤⎦ aθ . 2

3

1 ⎛ dθ ⎞ ⎛ dθ ⎞ ⎛ d θ ⎞ ⎛ dθ ⎞ 1 ⎛ dθ ⎞ Again, cos ⎜ =1+ ⎜ + ...and sin ⎜ ⎟ ⎟ ⎟=⎜ ⎟− ⎜ ⎟ + ... and keeping only terms 2! ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 3! ⎝ 2 ⎠ involving products of three differentials, we have, (Tθ r ) drdθ dz + ( dTθ r ) rdθ dz + 2Trθ drdz ( dθ / 2 ) + ( dTθθ ) drdz + ( dTθ z ) rdθ dr + ρ Bθ ( rdθ ) drdz = ⎡⎣ ρ ( rdθ ) drdz ⎤⎦ aθ

Dividing the equation by rdθ drdz , we get, Tθ r ∂Tθ r Trθ 1 ∂Tθθ ∂Tθ z + + + + + ρ Bθ = ρ aθ , this is Eq. (4.8.2). ∂r ∂z r r r ∂θ _________________________________________________________________ 4.37 Verify that the following stress field satisfies the z-equation of equilibrium in the absence of body forces: ⎛ z 3r 2 z ⎞ ⎛ z 3z3 ⎞ ⎛ r 3rz 2 Az Trr = A ⎜ 3 − 5 ⎟ , Tθθ = 3 , Tzz = − A ⎜ 3 + 5 ⎟ , Trz = − A ⎜ 3 + 5 ⎜R ⎜R ⎜R R ⎟⎠ R R ⎟⎠ R ⎝ ⎝ ⎝

⎞ ⎟⎟ , Trθ = Tzθ = 0 ⎠

R2 = r 2 + z 2 -----------------------------------------------------------------------------Ans. The z equation of equilibrium in cylindrical coordinate is: ∂Tzr 1 ∂Tzθ Tzr ∂Tzz ∂R r ∂R z + + + + ρ Bz = 0 . Now = , = so that ∂r r ∂θ r ∂z ∂r R ∂z R ⎛ ∂ r ⎛ 1 ∂Tzr ∂ 3rz 2 ⎞ 3r ∂R 3z 2 15rz 2 ∂R ⎞ = − A⎜ + = − − + − 6 A ⎟ ⎜ ⎟ ⎜ ∂r R3 ∂r R5 ⎟ ⎜ R3 R 4 ∂r R5 ∂r R ∂r ⎟⎠ ⎝ ⎠ ⎝ ⎛ 1 3r 2 3 z 2 15r 2 z 2 ⎞ ⎛ 1 3r 2 3z 2 15r 2 z 2 ⎞ ∂Tzθ = − A⎜ 3 − 5 + 5 − = − − + 5 − =0 A ⎟ ⎜ ⎟, ⎜ ⎜ R3 R5 ∂θ R R R 7 ⎠⎟ R R 7 ⎟⎠ ⎝R ⎝ ⎛ 1 3z 2 ⎞ Trz = − A⎜ 3 + 5 ⎟ , ⎜R r R ⎟⎠ ⎝ ⎛ 1 ⎛ 1 3z 2 9 z 2 15 z 4 ⎞ ∂Tzz 3z ∂R 9 z 2 15 z 3 ∂R ⎞ = − A⎜ 3 − 4 + 5 − 6 = − A⎜ 3 − 5 + 5 − 7 ⎟ ⎟ ⎜R ⎜R ∂z R ∂z R R ∂z ⎟⎠ R R R ⎟⎠ ⎝ ⎝ ⎛ 3 3r 2 3 z 2 3z 2 3z 2 9 z 2 15r 2 z 2 15 z 4 ⎞ ∂Tzr T ∂T + rz + zz = − A ⎜ 3 − 5 − 5 + 5 + 5 + 5 − − 7 ⎟ Thus, ⎜R ∂r r ∂z R R R R R R7 R ⎟⎠ ⎝ ⎛ 3 3r 2 3 z 2 3z 2 3z 2 9 z 2 15r 2 z 2 15 z 4 ⎞ = − A⎜ 3 − 5 − 5 + 5 + 5 + 5 − − 7 ⎟ ⎜R R R R R R R7 R ⎟⎠ ⎝ 2 2 2 2 2 ⎞ ⎛ ⎛ 3 3R 2 15 z 2 15 z 2 R 2 ⎞ 3 3 r +z 15 z 2 15 z r + z ⎟ ⎜ = −A 3 − + − = − A − + 5 − ⎜ ⎟=0. ⎜ R3 R5 ⎜⎜ R ⎟⎟ R5 R5 R7 R R 7 ⎟⎠ ⎝ ⎝ ⎠

(

)

(

)

_________________________________________________________________

4-17


Lai et al, Introduction to Continuum Mechanics 4.38 Trr = −

Given the following stress field in cylindrical coordinates: 3Pzr 2

, Tzz = −

3Pz 3

, Trz = − 5

3Pz 2 r

, Tθθ = Trθ = Tzθ = 0, R 2 = r 2 + z 2

2π R 2π R 2π R Verify that the state of stress satisfies the equations of equilibrium in the absence of body forces. -----------------------------------------------------------------------------Ans. ∂Trr ∂Trθ Trr − Tθθ ∂Trz ∂ ⎛ 3Pzr 2 ⎞ 1 ⎛ 3Pzr 2 ⎞ ∂ ⎛ 3Pz 2 r ⎞ + + + = ⎜− ⎟ + ⎜− ⎟ + ⎜− ⎟ rdθ r ∂r ∂z ∂r ⎜⎝ 2π R 5 ⎟⎠ r ⎜⎝ 2π R5 ⎟⎠ ∂z ⎜⎝ 2π R5 ⎟⎠ 3Pz ∂ 2 ⎛ 3Pzr 2 ⎞ ∂ ⎛ 1 ⎞ 1 ⎛ 3Pzr 2 ⎞ ⎛ 3Pr ⎞ ∂ 2 ⎛ 3Pz 2 r ⎞ ∂ ⎛ 1 ⎞ r +⎜− z + ⎜− =− + − + − ⎟ ⎜ ⎟ ⎜ ⎟⎟ ∂r ⎝⎜ R 5 ⎠⎟ r ⎜⎜ 2π R5 ⎟⎟ ⎝⎜ 2π R5 ⎠⎟ ∂z ⎜ π 2 2π ⎠⎟ ∂z ⎝ R5 ⎠ 2π R5 ∂r ⎝ ⎠ ⎝ ⎠ ⎝ 3Pzr ⎛ 15 Pzr 2 ⎞ ∂R ⎛ 3Pzr ⎞ ⎛ 3P zr ⎞ ⎛ 15 Pz 2 r ⎞ ∂R =− +⎜ −⎜ ⎟ ⎟ ⎟ + ⎜− ⎟+⎜ π R5 ⎜⎝ 2π R 6 ⎟⎠ ∂r ⎝ 2π R5 ⎠ ⎝ π R5 ⎠ ⎜⎝ 2π R 6 ⎟⎠ ∂z 2 2 ⎞ ⎛ 15 Pzr ⎛ 15Pzr 3 ⎞ ⎛ 15Pz 3 r ⎞ 15 Pzr ⎜ 15 Pzr r + z ⎟ =− +⎜ + =0. ⎟+⎜ ⎟=− ⎟⎟ 2π R 5 ⎜⎝ 2π R 7 ⎟⎠ ⎜⎝ 2π R 7 ⎟⎠ 2π R 5 ⎜⎜ 2π R 7 ⎝ ⎠ ∂Tθ r 2Trθ 1 ∂Tθθ ∂Tθ z + + + =0+0+0+0=0 ∂r r r ∂θ ∂z ∂Tzr 1 ∂Tzθ Tzr ∂Tzz ∂ ⎛ 3Pz 2 r ⎞ 1 ⎛ 3Pz 2 r ⎞ ∂ ⎛ 3Pz 3 ⎞ + + + + ρ Bz = ⎜ − ⎟ + ⎜− ⎟ + ⎜− ⎟ ∂r ∂z ∂r ⎜⎝ 2π R5 ⎟⎠ r ⎜⎝ 2π R5 ⎟⎠ ∂z ⎜⎝ 2π R 5 ⎟⎠ r ∂θ r ⎛ 3Pz 2 ⎞ ⎛ 3Pz 2 r ⎞ ∂ ⎛ 1 ⎞ 3Pz 2 9 Pz 2 ⎛ 3Pz 3 ⎞ ∂ ⎛ 1 ⎞ = ⎜− + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟− ⎟ ⎜ 2π R5 ⎟⎟ ⎜⎜ 2π ⎟⎠ ∂r ⎝ R5 ⎠ 2π R5 2π R5 ⎜⎝ 2π ⎟⎠ ∂z ⎝ R 5 ⎠ ⎝ ⎠ ⎝ ⎛ 3Pz 2 ⎞ ⎛ 15Pz 2 r ⎞ ∂R 3Pz 2 9 Pz 2 ⎛ 15Pz 3 ⎞ ∂R = ⎜− + − − + ⎜ 2π R5 ⎟⎟ ⎜⎜ 2π R 6 ⎟⎟ ∂r 2π R5 2π R5 ⎜⎜ 2π R 6 ⎟⎟ ∂z ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 ⎛ 3Pz 2 ⎞ ⎛ 15Pz 2 r 2 ⎞ 3Pz 2 9 Pz 2 ⎛ 15 Pz 4 ⎞ ⎛ 15 Pz 2 ⎞ 2⎛r + z ⎞ 15 = ⎜− + − − + = − + =0 Pz ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ 2π R5 ⎟ ⎜ 2π R 7 ⎟ 2π R5 2π R 5 ⎜ 2π R 7 ⎟ ⎜ 2π R5 ⎟ ⎜ 2π R 7 ⎟⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ _________________________________________________________________ 5

5

( )

(

)

4.39 For the stress field given in Example 4.9.1, determine the constants A and B if the inner cylindrical wall is subjected to a uniform pressure pi and the outer cylindrical wall is subjected to a uniform pressure po . -----------------------------------------------------------------------------Ans. The given stress field is: B B Trr = A + 2 , Tθθ = A − 2 , Tzz = constant and Trθ = Trz = Tθ z = 0 . r r On the outer wall, r = ro , and Trr = − po , and on the inner wall, r = ri , and Trr = − pi , therefore, B B we have, − po = A + 2 (i) and − pi = A + 2 (ii). ro ri → po − pi =

4-18

B ri2

−

B ro2

→

( p − pi ) ri2 ro2 , B= o

(r

2 o

− ri2

)

(pr A= (r

2 i i 2 o

− po ro2 − ri2

)

).


Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 4.40 Verify that Eq. (4.8.4) to (4.8.6) are satisfied by the stress field given in Example 4.9.2 in the absence of body forces. -----------------------------------------------------------------------------2B B Ans. The given stress field is: Trr = A − 3 , Tθθ = Tφφ = A + 3 , Trθ = Trφ = Tθφ = 0 . r r

(

)

2 1 ∂ r Trr 1 ∂ (Trθ sin θ ) 1 ∂Trφ Tθθ + Tφφ + + 2 ∂r r sin θ ∂θ r sin θ ∂φ r r

=

B ⎞ 2⎛ B⎞ B⎞ 1 ∂ ⎛ 2 2B ⎞ 2⎛ 2⎛ r A− ⎟ + 0 + 0 − ⎜ A + 3 ⎟ = ⎜ A + 3 ⎟ + 0 + 0 − ⎜ A + 3 ⎟ = 0. 2 ∂r ⎜ r ⎠ r⎝ r⎝ r r ⎠ r⎝ r ⎠ r ⎠ ⎝

(

)

(

)

3 1 ∂ r Tθ r 1 ∂ (Tθθ sin θ ) 1 ∂Tθφ Trθ − Tθ r − Tφφ cot θ + + + 3 ∂r ∂θ r sin θ r sin θ ∂φ r r T cot θ Tφφ cot θ = 0 + 0 + 0+ θθ − = 0. r r 3 1 ∂ r Tφ r 1 ∂ Tφθ sin θ 1 ∂Tφφ Trφ − Tφ r + Tθφ cot θ + + = 0 + 0 + 0+ 0 = 0 . + ∂r r sin θ ∂θ r sin θ ∂φ r r3 _________________________________________________________________

(

)

4.41 In Example 4.9.2, if the spherical shell is subjected to an inner pressure pi and an outer pressure po , determine the constant A and B . -----------------------------------------------------------------------------2B 2B 2B Ans. From the example, we have, Trr = A − 3 , thus, − po = A − 3 and − pi = A − 3 r ro ri → A=−

po ro3 − pi ri3

(r

3 o

− ri3

)

and B = −

( po − pi ) ro3ri3 .

(

2 ro3 − ri3

)

_________________________________________________________________ 4.42

The equilibrium configuration of a body is described by:

1 1 x1 = 16 X1 , x2 = − X 2 , x3 = − X 3 . If the Cauchy stress tensor is given by: 4 4 T11 = 1000 MPa.,and all other Ti j = 0 , (a) calculate the first Piola –Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is e1 plane and (b) calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the same plane. -----------------------------------------------------------------------------1 1 Ans. From x1 = 16 X1 , x2 = − X 2 , x3 = − X 3 , we obtain the deformation gradient F and its 4 4 inverse as:

4-19


Lai et al, Introduction to Continuum Mechanics 0 0 ⎤ ⎡16 ⎡1 / 16 0 0 ⎤ ⎢ ⎥ − 1 [F ] = ⎢ 0 −1 / 4 0 ⎥ , ⎣⎡F ⎦⎤ = ⎢⎢ 0 −4 0 ⎥⎥ and detF = 1 . ⎢⎣ 0 ⎢⎣ 0 0 0 −4 ⎥⎦ −1 / 4 ⎥⎦

( )

(a) The first Piola-Kirchhoff stress tensor is, from To = ( det F ) T F −1

T

:

⎡1000 0 0 ⎤ ⎡1 / 16 0 0 ⎤ ⎡1000 / 16 0 0⎤ ⎢ −4 0 ⎥⎥ = ⎢⎢ 0 0 ⎥⎥ ⎢⎢ 0 0 0 0 ⎥⎥ MPa. ⎥⎦ = (1) ⎢ 0 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ For a unit area in the deformed state in the e1 direction, its undeformed area is 0 0 ⎤ ⎡1 ⎤ ⎡16 ⎤ ⎡16 1 ⎢ T F n = (1) ⎢ 0 −1 / 4 0 ⎥⎥ ⎢⎢0 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → dAo n o = 16e1 . ( dAono ) = det F ⎢⎣ 0 0 −1 / 4 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 0 ⎥⎦

( )

[To ] = ( det F ) [T] ⎡⎢ F −1 ⎣

T⎤

That is, its undeformed plane is also e1 plane. The corresponding pseudo stress vector is given by ⎡1000 / 16 0 0 ⎤ ⎡1 ⎤ t o = To no , where no = e1 . Thus t o = To no → [ t o ] = ⎢⎢ 0 0 0 ⎥⎥ ⎢⎢0 ⎥⎥ → t o = (1000 / 16 ) e1 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣0 ⎥⎦ We note that the pseudo stress vector is in the same direction as the Cauchy stress vector and the intensity of the pseudo stress vector is 1/16 of the Cauchy stress vector simply because the undeformed area is 16 times the deformed area and both areas have the same normal direction.

( )

% = ( det F ) F −1T F −1 (b)The second Piola-Kirchhoff stress tensor is, from T

T

⎡1 / 16 0 0 ⎤ ⎡1000 0 0 ⎤ ⎡1 / 16 0 0 ⎤ ⎢ 0 0 ⎥⎥ ⎢⎢ 0 −4 0 ⎥⎥ ⎢⎢ 0 −4 0 ⎥⎥ ⎥⎦ = (1) ⎢ 0 ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎡1 / 16 0 0 ⎤ ⎡1000 / 16 0 0 ⎤ ⎡1000 / 256 0 0 ⎤ ⎢ 0 −4 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎥ = ⎢⎢ 0 0 0 ⎥⎥ MPa. ⎢ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ % , where n = e . Thus, The corresponding pseudo stress vector is given t% = Tn

( )

⎡ −1 ⎡ ⎡%⎤ ⎣ T ⎦ = ( det F ) ⎣ F ⎦ [ T] ⎢⎣ F −1 ⎤

T⎤

o

o

1

% → t% = (1000 / 256 ) e . t% = Tn o 1 _________________________________________________________________ 4.43 Can the following equations represent a physically acceptable deformation of a body? Give reason. 1 1 x1 = − X1 , x2 = X 3 , x3 = −4 X 2 . 2 2 -----------------------------------------------------------------------------0 ⎤ ⎡ −1 / 2 0 ⎢ 0 1 / 2 ⎥⎥ → det F = −1 . The given equations are not acceptable as a Ans. [ F ] = ⎢ 0 ⎢⎣ 0 −4 0 ⎥⎦ physically acceptable deformation because it gives a negative ratio of deformed volume to the undeformed volume.

4-20


Lai et al, Introduction to Continuum Mechanics _________________________________________________________________ 4.44

The deformation of a body is described by:

x1 = 4 X1 , x2 = − (1 / 4 ) X 2 , x3 = − (1 / 4 ) X 3 . (a) For a unit cube with sides along the coordinate axes what is its deformed volume? What is the deformed area of the e1 face of the cube? (b) If the Cauchy stress tensor is given by: T11 = 100 MPa.,and all other Ti j = 0 , calculate the first Piola –Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is e1 plane. (c) Calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is e1 plane. Also, calculate the pseudo differential force for the same plane. -----------------------------------------------------------------------------Ans. From x1 = 4 X1 , x2 = − (1 / 4 ) X 2 , x3 = − (1 / 4 ) X 3 , we have (a) 0 0 ⎤ ⎡4 ⎢ [F ] = ⎢0 −1 / 4 0 ⎥⎥ → det F = 1 / 4 , thus dV = ( det F ) dVo → dV = (1 / 4 ) (1) = 1 / 4 . ⎢⎣ 0 0 −1 / 4 ⎥⎦ ⎡1 / 4 0 0 ⎤ ⎡1 ⎤ ⎡1 / 4 ⎤ ⎢ ⎥ ⎢ ⎥ dA = dAo ( det F ) F n o = (1)(1 / 4 ) ⎢ 0 −4 0 ⎥ ⎢ 0 ⎥ = (1 / 4 ) ⎢⎢ 0 ⎥⎥ → dA = (1 / 16 ) e1 ⎢⎣ 0 ⎢⎣ 0 ⎥⎦ 0 −4 ⎥⎦ ⎢⎣ 0 ⎥⎦ That is, the deformed volume is 1/4 of its original volume and the e1 face of unit area deformed into an area 1/16 of it original area and remain in the same direction. These results are quite obvious from the geometry of the deformation.

( ) −1

T

(b) The first PK stress tensor is: ⎡100 0 ⎛ 1 ⎞⎢ ⎥⎦ = ⎜ 4 ⎟ ⎢ 0 0 ⎝ ⎠ ⎢⎣ 0 0 The corresponding pseudo stress vector for

( )

[To ] = ( det F ) [T] ⎡⎢ F −1 ⎣

T⎤

0 ⎤ ⎡1 / 4 0 0 ⎤ ⎡100 / 16 0 0 ⎤ 0 ⎥⎥ ⎢⎢ 0 −4 0 ⎥⎥ = ⎢⎢ 0 0 0 ⎥⎥ MPa. 0 ⎥⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ e1 -plane in the deformed state, whose undeformed

plane is also e1 -plane, is given by t o = To no , where no = e1 , that is t o = (100 / 16 ) e1MPa. The Cauchy stress vector on the e1 face in the deformed state is t = 100e1MPa. Clearly the Cauchy stress vector has a larger magnitude because the area in the deformed state is 1/16 of the undeformed area. (c) The second PK stress tensor is: ⎡1 / 4 0 0 ⎤ ⎡100 / 16 0 0 ⎤ ⎡100 / 64 0 0 ⎤ −1 ⎤ % ⎡ ⎡ T ⎤ = F [ To ] = ⎢ 0 −4 0 ⎥ ⎢ 0 0 0 ⎥⎥ = ⎢⎢ 0 0 0 ⎥⎥ MPa. ⎣ ⎦ ⎣ ⎢ ⎥⎢ ⎦ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ The corresponding pseudo stress vector for the e1 -plane in the deformed state, whose undeformed % , where n = e . Thus, t% = (100 / 64 ) e MPa. The plane is also e1 -plane, is given by t% = Tn o o 1 1

4-21


Lai et al, Introduction to Continuum Mechanics pseudo force df% is related to the force df (= t o dAo = 100 / 16e1 for dAo = 1) by the formula ⎡1 / 4 0 0 ⎤ ⎡100 / 16 ⎤ ⎛ 100 ⎞ -1 -1 % % df = F df , Thus, ⎡⎣ df ⎤⎦ = ⎡ F ⎤ [ df ] = ⎢⎢ 0 −4 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ → df% = ⎜ ⎟ e1 ⎣ ⎦ ⎝ 64 ⎠ ⎢⎣ 0 0 −4 ⎥⎦ ⎢⎣ 0 ⎥⎦ ______________________________________________________________ 4.45 The deformation of a body is described by: x1 = X1 + kX 2 , x2 = X 2 , x3 = X 3 . (a) For a unit cube with sides along the coordinate axes what is its deformed volume? What is the deformed area of the e1 face of the cube? (b) If the Cauchy stress tensor is given by: T12 = T21 = 100 MPa.,and all other Ti j = 0 , calculate the first Piola – Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is e1 plane and compare it with the Cauchy stress vector in the deformed state. (c) Calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is e1 plane. Also, calculate the pseudo differential force for the same plane. -----------------------------------------------------------------------------Ans. From x1 = X1 + kX 2 , x2 = X 2 , x3 = X 3 , we have (a) ⎡1 k 0 ⎤ [ F ] = ⎢⎢0 1 0⎥⎥ → det F = 1 , thus dV = ( det F ) dVo → dV = dVo = 1 . ⎢⎣ 0 0 1 ⎥⎦ ⎡ 1 0 0 ⎤ ⎡1 ⎤ ⎡ 1 ⎤ dA = dAo ( det F ) F n o = (1)(1) ⎢⎢ − k 1 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = ⎢⎢ − k ⎥⎥ → dA = e1 − ke 2 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ That is, the deformed volume is the same as its original volume and the e1 face of unit area

( ) −1

T

deformed into an area 1 + k 2 of it original area and whose normal is in the direction of e1 − ke 2 . These results are quite obvious from the geometry of the deformation. (b) The first PK stress tensor is: ⎡ 0 100 0 ⎤ ⎡ 1 0 0 ⎤ ⎡ −100k 100 0 ⎤ ⎢ ⎥⎢ ⎥ ⎢ 0 0 ⎥⎥ MPa. ⎥⎦ = ⎢100 0 0 ⎥ ⎢ − k 1 0 ⎥ = ⎢ 100 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ The corresponding pseudo stress vector for the (e1 − ke 2 ) plane, whose undeformed plane is the

( )

[To ] = ( det F ) [T] ⎡⎢ F −1 ⎣

T⎤

e1 plane, is given by t o = To no , where no = e1 . Thus, t o = 100 ( − ke1 + e 2 ) MPa. The Cauchy

stress vector on the (e1 − ke 2 ) face in the deformed configuration is ⎡ 0 100 0 ⎤ ⎡ 1 ⎤ 1 ⎢100 0 0 ⎥ ⎢ − k ⎥ → t = 100 − ke + e ( 1 2) [t ] = [ T][n ] = ⎥⎢ ⎥ 2 ⎢ 2 1+ k ⎢ 0 1 + k 0 0 ⎦⎥ ⎣⎢ 0 ⎦⎥ ⎣ The Cauchy stress vector has a smaller magnitude because the deformed area is 1 + k 2 times the undeformed area.

4-22


Lai et al, Introduction to Continuum Mechanics (c) The second PK stress tensor is: ⎡1 − k 0 ⎤ ⎡ −100k 100 0 ⎤ ⎡ −200k 100 0 ⎤ % ⎤ = ⎡ F −1 ⎤ [ T ] = ⎢ 0 1 0 ⎥ ⎢ 100 ⎡T 0 0 ⎥⎥ = ⎢⎢ 100 0 0 ⎥⎥ MPa. ⎣ ⎦ ⎣ ⎥⎢ ⎦ o ⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ The corresponding pseudo stress vector for the e1 − ke 2 plane, whose undeformed plane is the % , where n = e . Thus, t% = 100 ( −2ke + e ) MPa. The pseudo force e1 plane, is given by t% = Tn o o 1 1 2 df% is related to the force df ( = t o dAo = 100 ( − ke1 + e 2 ) for dAo = 1) by the formula df% = F -1df ,

⎡1 − k Thus, ⎡⎣ df% ⎤⎦ = ⎡F [ df ] = 100 ⎢⎢0 1 ⎣ ⎦ ⎢⎣0 0 -1 ⎤

0⎤ ⎡−k ⎤ 0 ⎥⎥ ⎢⎢ 1 ⎥⎥ → df% = 100 ( −2ke1 + e 2 ) . 1 ⎥⎦ ⎢⎣ 0 ⎥⎦

_________________________________________________________________ 4.46 The deformation of a body is described by: x1 = 2 X1 , x2 = 2 X 2 , x3 = 2 X 3 . (a) For a unit cube with sides along the coordinate axes, what is its deformed volume? What is the deformed area of the e1 face of the cube? (b) If the Cauchy 0 ⎤ ⎡100 0 ⎢ stress tensor is given by: ⎢ 0 100 0 ⎥⎥ Mpa. , calculate the first Piola –Kirchhoff stress tensor ⎢⎣ 0 0 100 ⎥⎦ and the corresponding pseudo stress vector for the plane whose undeformed plane is the e1 plane and compare it with the Cauchy stress vector on its deformed plane, (c) calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is the e1 plane. Also, calculate the pseudo differential force for the same plane. -----------------------------------------------------------------------------Ans. From, x1 = 2 X1 , x2 = 2 X 2 , x3 = 2 X 3 we have (a) ⎡2 0 0⎤ [F ] = ⎢⎢0 2 0 ⎥⎥ → det F = 8 , thus dV = ( det F ) dVo → dV = 8dVo = 8 . ⎢⎣ 0 0 2 ⎥⎦ 0 ⎤ ⎡1 ⎤ ⎡1 / 2 0 ⎡1 / 2⎤ ⎢ ⎥ ⎢ ⎥ dA = dAo ( det F ) F n o = (1)( 8 ) ⎢ 0 1 / 2 0 ⎥ ⎢ 0 ⎥ = ( 8 ) ⎢⎢ 0 ⎥⎥ → dA = 4e1 ⎢⎣ 0 ⎢⎣ 0 ⎥⎦ 0 1 / 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ (b) The first PK stress tensor is: 0 ⎤ ⎡1 / 2 0 0 ⎤ ⎡ 400 0 0 ⎤ ⎡100 0 ⎡ −1 T ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ [To ] = ( det F ) [T] ⎢ F ⎥ = (8) ⎢ 0 100 0 ⎥ ⎢ 0 1 / 2 0 ⎥ = ⎢ 0 400 0 ⎥⎥ MPa. ⎣ ⎦ ⎢⎣ 0 0 100 ⎥⎦ ⎢⎣ 0 0 1 / 2 ⎥⎦ ⎢⎣ 0 0 400 ⎥⎦ The corresponding pseudo stress vector for the e1 plane in the deformed state, whose undeformed plane is also e1 plane, is t o = 400e1 MPa. The Cauchy stress vector on the e1 plane is t = 100e1 MPa. The Cauchy stress vector has a smaller magnitude because the area is four times larger.

( ) −1

T

( )

(c) The second PK stress tensor is:

4-23


Lai et al, Introduction to Continuum Mechanics 0 ⎤ ⎡ 400 0 0 ⎤ ⎡ 200 0 0 ⎤ ⎡1 / 2 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ %⎤ ⎡ ⎡T ⎣ ⎦ = ⎣ F ⎦ [ To ] = ⎢ 0 1 / 2 0 ⎥ ⎢ 0 400 0 ⎥ = ⎢ 0 200 0 ⎥ MPa. ⎢⎣ 0 0 1 / 2 ⎥⎦ ⎢⎣ 0 0 400 ⎥⎦ ⎢⎣ 0 0 200 ⎥⎦ −1 ⎤

The corresponding pseudo stress vector for the e1 plane in the deformed state, whose undeformed plane is also e1 plane, is t% = 200e1MPa. The pseudo force df% is related to the force df ( = t o dAo = 400e1 for dAo = 1) by the formula df% = F -1df . Thus, 0 ⎤ ⎡1 ⎤ ⎡1 / 2 0 ⎡ df% ⎤ = ⎡F -1 ⎤ [ df ] = 400 ⎢ 0 1 / 2 0 ⎥ ⎢ 0 ⎥ → df% = 200e1 . ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢⎣ 0 0 1 / 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ _________________________________________________________________

4-24



CHAPTER 5, PART A 5.1 Show that the null vector is the only isotropic vector. (Hint: Assume that a is an isotropic vector, and use a simple change of basis to equate the primed and unprimed components). -------------------------------------------------------------------------------Ans. For an isotropic a , by definition, [a ]e = [a ]e′ , where {ei } and {e′i } are any two orthonormal i

bases. That is [a ]e′ = [Q ]

T ei

i

[a]e

i

→ [a ]e = [Q ]e [a ]e for all [Q ]e . T

i

i

i

i

i

⎡ a1 ⎤ ⎡ −1 0 0 ⎤ ⎡ a1 ⎤ ⎡ −1 0 0 ⎤ ⎢ ⎥ Method I. Choose [Q ] = ⎢ 0 −1 0 ⎥ , then ⎢⎢ a2 ⎥⎥ = ⎢⎢ 0 −1 0 ⎥⎥ ⎢⎢ a2 ⎥⎥ gives ⎢⎣ a3 ⎥⎦ ⎢⎣ 0 0 −1⎥⎦ ⎢⎣ a3 ⎥⎦ ⎣⎢ 0 0 −1⎦⎥ a1 = − a1 = 0, a2 = − a2 = 0, a3 = − a3 = 0 . In other words, the only isotropic vector is the null vector.

Method II. The matrix equation [a ]e = [Q ]e [a ]e , with the same basis for each matrix, is T

i

i

i

equivalent to the equation a = Q a . That is, a is an eigenvector for QT for any orthogonal tensor Q . But clearly, there is no non-zero vector which is an eigenvector for all orthogonal tensors. T

________________________________________________________________________

Show that the most general isotropic second-order tensor is of the form of α I , where α is a scalar and I is the identity tensor. ------------------------------------------------------------------------------Ans. For an isotropic T , by definition, [ T ]e = [ T]e′ , where {ei } and {e′i } are any two orthonormal 5.2

i

⎡T11 T12 ⎢T ⎢ 21 T22 ⎣⎢T31 T32

i

⎡ −1 0 0 ⎤ bases. Choose [Q ] = ⎢⎢ 0 1 0 ⎥⎥ , then [ T]e i ⎢⎣ 0 0 1 ⎥⎦ T13 ⎤ ⎡ −1 0 0 ⎤ ⎡T11 T12 T13 ⎤ ⎡ −1 0 T23 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥ ⎢⎢T21 T22 T23 ⎥⎥ ⎢⎢ 0 1 T33 ⎦⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣T31 T32 T33 ⎥⎦ ⎢⎣ 0 0

⎡ T11 −T12 = ⎢⎢ −T21 T22 ⎢⎣ −T31 T32

= [Q ]e [ T]e [Q ]e gives T

i

i

0 ⎤ ⎡ −1 0 0 ⎤ ⎡ −T11 T12 T13 ⎤ 0 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥ ⎢⎢ −T21 T22 T23 ⎥⎥ 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ −T31 T32 T33 ⎥⎦

−T13 ⎤ T23 ⎥⎥ → T12 = −T12 = 0, T21 = −T21 = 0, T13 = −T13 = 0 T31 = −T31 = 0. T33 ⎥⎦ ⎡1 0 0 ⎤ Next, the choice of [Q ] = ⎢⎢0 −1 0 ⎥⎥ gives, ⎢⎣0 0 1 ⎥⎦


5-1

i

Lai et al, Introduction to Continuum Mechanics 0 ⎤ ⎡1 0 0 ⎤ ⎡T11 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 0 ⎡T11 0 ⎢0 T ⎥ ⎢ ⎢ ⎥ T22 T23 ⎥⎥ ⎢⎢ 0 −1 0 ⎥⎥ = ⎢⎢ 0 −1 22 T23 ⎥ = ⎢ 0 −1 0 ⎥ ⎢ 0 ⎢ ⎣⎢ 0 T32 T33 ⎥⎦ ⎣⎢0 0 1 ⎦⎥ ⎢⎣ 0 T32 T33 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 0 ⎤ ⎡T11 ⎢ = ⎢ 0 T22 −T23 ⎥⎥ → T23 = −T23 = 0, T32 = −T32 ⎢⎣ 0 −T32 T33 ⎥⎦

0 ⎤ ⎡T11 0 0 ⎤ ⎢ ⎥ 0 ⎥ ⎢ 0 −T22 T23 ⎥⎥ 1 ⎥⎦ ⎢⎣ 0 −T32 T33 ⎥⎦ = 0.

⎡0 1 0 ⎤ Next, the choice of [Q ] = ⎢⎢1 0 0 ⎥⎥ gives, ⎢⎣0 0 1 ⎥⎦ 0 ⎤ ⎡0 1 0 ⎤ ⎡T11 0 0 ⎤ ⎡0 1 0 ⎤ ⎡T22 0 0 ⎤ ⎡T11 0 ⎢0 T ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ 0 ⎥ = ⎢1 0 0 ⎥ ⎢ 0 T22 0 ⎥ ⎢1 0 0 ⎥ = ⎢ 0 T11 0 ⎥⎥ → T11 = T22 22 ⎢ ⎢⎣ 0 0 T33 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0 0 T33 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0 0 T33 ⎥⎦ 0 ⎤ ⎡T11 0 0⎤ ⎡T11 0 ⎡1 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ Finally, the choice of [Q ] = ⎢ 0 0 1 ⎥ gives ⎢ 0 T11 0 ⎥ = ⎢ 0 T33 0 ⎥⎥ → T11 = T33 . ⎢⎣ 0 ⎢⎣ 0 1 0 ⎥⎦ 0 T33 ⎥⎦ ⎢⎣ 0 0 T11 ⎥⎦ Thus, T11 = T22 = T33 =α and T12 = T21 = T13 = T31 = T23 = T32 = 0 → [ T] = α [ I ] ________________________________________________________________________

5.3

For an isotropic linearly elastic body, (a) verify the μ = μ ( λ , EY ) as given in Table 5.1. (b) Obtain the value of μ as EY / λ → 0 ------------------------------------------------------------------------------μ ( EY − 2μ ) Ans. From Table 5.1, λ = → 2 μ 2 + ( 3λ − EY ) μ − EY λ = 0 3μ − EY

→μ= (b) μ =

− ( 3λ − EY ) +

( 3λ − EY )2 + 8EY λ 4

.

− ( 3λ − EY ) + ( 3λ − EY ) 1 + 8 ( EY / λ ) / ( 3 − EY / λ )

2

4

As EY / λ → 0 , 1 + 8 ( EY / λ ) / ( 3 − EY / λ ) → 1 + 4 ( EY / λ ) / 9 , where we have used the 2

binomial theorem. Thus,

μ → {−(3λ − EY ) + (3λ − EY )[1 + 4( EY / λ ) / 9]} / 4 = (3λ − EY )( EY / λ ) / 9 = (3 − EY / λ )( EY ) / 9 Thus, as EY / λ → 0 , μ → EY / 3 . ________________________________________________________________________

5.4

From λ =

λ (1 + ν ) ν EY 2 μν ,λ = and k = obtain μ = μ ( EY ,ν ) and 3ν (1 + ν )(1 − 2ν ) (1 − 2ν ) k = k ( μ ,ν )


5-2

Lai et al, Introduction to Continuum Mechanics ------------------------------------------------------------------------------ν EY EY 2 μν = →μ= . Ans. λ = 2 (1 + ν ) (1 + ν )(1 − 2ν ) (1 − 2ν )

λ=

2μ (1 + ν ) 2 μν 3ν k = →k = . 3 (1 − 2ν ) (1 − 2ν ) 1 + ν

________________________________________________________________________

5.5

Show that for an incompressible material (ν → 1 / 2 ) that (a) μ = EY / 3, λ → ∞, k → ∞, but k − λ = 2μ / 3

(b) T = 2 μ E + (Tkk / 3)I where Tkk is constitutively indeterminate. -----------------------------------------------------------------------------Ans. (a) From Table 5.1, we have EY EY E ν EY 2 2 μ= = = Y , λ= → ∞, λ + μ = k → k − λ = μ . 2(1 + ν ) 2(1 + 1 / 2) 3 (1 + ν )(1 − 2ν ) 3 3 (b) In general, T = λ eI + 2μ E . Now, from Eq.(5.4.2), we have Tkk T T e= . As ν → 1 / 2 , λ → ∞ , and λ e → kk so that T = kk I + 2 μ E . 3 (2 μ + 3λ ) 3 We note that because of incompressibility, Tkk will be constitutively indeterminate. It becomes determinate when the boundary condition(s) is (are) taken into account. ________________________________________________________________________

5.6

Given Aijkl = δ ijδ kl and Bijkl = δ ik δ jl . (a) Obtain A11 jk and B11 jk . (b) Identity those A11 jk that are different from B11 jk . ------------------------------------------------------------------------------Ans. (a) A11kl = δ11δ kl = δ kl , B11kl = δ1k δ1l . (b) A1111 = A1122 = A1133 = 1, all other A11kl = 0, B1111 = 1, all other B11kl = 0 . A1122 ≠ B1122 , A1133 ≠ B1133 .

________________________________________________________________________

5.7

Show that for an anisotropic linear elastic material, the principal directions of stress and strain are in general not coincident. ------------------------------------------------------------------------------Ans. We have, Tij = Cijkl Ekl . Let ei be the principal basis for E , then [ E ]e is diagonal. Thus, i

T12 = C12 kl Ekl = C1211E11 + C1222 E22 + C1233 E33 . This equation shows that in general, T12 ≠ 0 . Similarly, in general T13 ≠ 0 and T23 ≠ 0 . Thus the matrix of T is not diagonal with respect to the principal basis of E . ________________________________________________________________________


5-3


If the Lamé Constants for a material are:

λ = 119.2GPa (17.3 × 106 psi), μ = 79.2GPa (11.5 × 106 psi ) Find Young's modulus, Poisson's ratio and the bulk modulus. ------------------------------------------------------------------------------Ans. From Table 5.1, we have, μ ( 3λ + 2 μ ) 79.2 ⎡⎣3 (119.2 ) + 2 ( 79.2 ) ⎤⎦ = = 206 GPa 30 × 106 psi EY = λ+μ 119.2 + 79.2 λ 119.2 ν= = = 0.3 , 2(λ + μ ) 2(119.2 + 79.2)

(

)

k = λ + 2μ / 3 = 119.2 + 2 ( 79.2 ) / 3 = 172 GPa (25 × 106 psi ) . ________________________________________________________________________

5.9

Given Young's modulus EY = 103 GPa and Poisson's ratioν = 0.34 . Find the Lamé constants λ and μ . Also find the bulk modulus. ------------------------------------------------------------------------------Ans. 0.34 (103) ν EY λ= = = 81.7 GPa 11.8 × 106 psi (1 + ν )(1 − 2ν ) (1.34 )( 0.32 )

(

μ=

(

EY 103 = = 38.4 GPa 5.56 × 106 psi 2 (1 + ν ) 2 × 1.34

)

)

(

k = λ + 2μ / 3 = 81.7 + 2 ( 38.4 ) / 3 = 107.3 GPa 15.6 × 106 psi

)

________________________________________________________________________

Given Young's modulus EY = 193 GPa ., shear modulus μ = 76 GPa . Find Poisson's ratioν , Lamé constant λ and the bulk modulus k . ------------------------------------------------------------------------------2 ( 76 )( 0.27 ) EY 193 2 μν Ans. ν = −1 = − 1 = 0.27, λ = = = 89.1 GPa 12.9 × 106 psi 2μ 2 ( 76 ) 1 − 2ν 1 − 0.54

5.10

(

(

k = λ + 2 μ / 3 = 89.1 + 2 ( 76 ) / 3 = 140 GPa 20.3 × 106 psi

)

)

________________________________________________________________________

5.11

The components of strain at a point of structural steel are:

E11 = 36 × 10−6 , E22 = 40 × 10−6 , E33 = 25 × 10−6 E12 = 12 × 10−6 , E23 = 0, E13 = 30 × 10−6

(

)

(

Find the stress components. λ = 119.2 GPa 17.3 × 106 psi , μ = 79.2 GPa 11.5 × 106 psi ------------------------------------------------------------------------------Ans. From T = λ eI + 2μ E , we have, with e = ( 36 + 40 + 25 ) × 10−6 = 101 × 10−6


5-4

)

Lai et al, Introduction to Continuum Mechanics ⎡1 0 0 ⎤ ⎡36 12 30 ⎤ ⎡17.7 1.9 4.75⎤ ⎢ ⎥ ⎢ ⎥ −6 −6 [T] = (119.2 )(101) ⎢0 1 0⎥ × 10 + 2 ( 79.2 ) ⎢12 40 0 ⎥ × 10 = ⎢⎢ 1.9 18.4 0 ⎥⎥ MPa. ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣30 0 25⎥⎦ ⎢⎣ 4.75 0 16.0 ⎥⎦ ________________________________________________________________________

5.12

Do the previous problem if the strain components are:

E11 = 100 × 10−6 , E22 = −200 × 10−6 , E33 = 100 × 10−6 E12 = −100 × 10−6 , E23 = 0, E13 = 0 ------------------------------------------------------------------------------Ans. From T = λ eI + 2μ E , we have, with e = (100 − 200 + 100 ) × 10−6 = 0 ⎡ 100 −100 0 ⎤ ⎡ 15.8 −15.8 0 ⎤ ⎢ ⎥ −6 [T] = 2 ( 79.2 ) ⎢ −100 −200 0 ⎥ × 10 = ⎢⎢−15.8 −31.7 0 ⎥⎥ MPa . ⎢⎣ 0 ⎢⎣ 0 0 100 ⎥⎦ 0 15.8⎥⎦ ________________________________________________________________________

5.13

An isotropic elastic body ( EY = 207GPa, μ = 79.2GPa ) has a uniform state of stress 60 ⎤ ⎡100 40 ⎢ given by: [ T ] = ⎢ 40 −200 0 ⎥⎥ MPa . ⎢⎣ 60 0 200 ⎥⎦

(a) What are the strain components? (b) What is the total change of volume for a five centimeter cube of the material? ------------------------------------------------------------------------------1 ⎡ (1 + ν )Tij − ν (Tkk )δ ij ⎤ , therefore, we first Ans. (a) We would like to use the equation Eij = ⎦ EY ⎣ E 207 − 1 = 0.306 , then obtain Tkk = T11 + T22 + T33 = 100MPa . Thus, obtain ν = Y − 1 = 2μ 2 ( 79.2 ) ⎧ 60 ⎤ ⎡100 40 ⎡1 0 0 ⎤ ⎫ 1 ⎪ ⎢ ⎥ [ E] = E ⎨1.31 ⎢ 40 −200 0 ⎥ − ( 0.306 ) (100) ⎢⎢0 1 0⎥⎥ ⎬⎪ Y ⎪ ⎢⎣ 60 ⎢⎣0 0 1 ⎥⎦ ⎪⎭ 0 200 ⎥⎦ ⎩ ⎡ 100 52.4 78.6 ⎤ ⎡ 0.483 0.253 0.380 ⎤ 1 ⎢52.4 −292 = 0 ⎥⎥ = ⎢⎢ 0.253 −1.41 0 ⎥⎥ × 10−3 3 ⎢ 207 × 10 ⎢⎣ 78.6 0 231 ⎥⎦ ⎢⎣0.380 0 1.12 ⎥⎦ (b) Dilatation e = Ekk = ( 0.483 − 1.41 + 1.12 ) × 10−3 = 0.193 × 10−3 .

( )(

)

Total change of volume = ΔV = (V )( e ) = 53 0.193 × 10−3 = 24.1 × 10−3 cm3 . ________________________________________________________________________


5-5


5.14

An isotropic elastic sphere ( EY = 207GPa, μ = 79.2GPa ) of 5 cm radius is under the uniform stress field ⎡6 2 0⎤ [T] = ⎢⎢ 2 −3 0⎥⎥ MPa ⎢⎣ 0 0 0 ⎥⎦ Find the change of volume for the sphere. -----------------------------------------------------------------------------E 207 1 ⎡ (1 + ν )Tij − ν (Tkk )δ ij ⎤ gives Ans. ν = Y − 1 = − 1 = 0.306 , Eij = ⎦ 2μ 2 ( 79.2 ) EY ⎣ ⎧ 0 ⎤ ⎡6 2 0⎤ ⎡1 0 0 ⎤ ⎫ ⎡3.35 1.26 ⎪ ⎪ ⎢ ⎢ ⎥ ⎢ ⎥ 0 ⎥⎥ × 10−5 ⎨1.31 ⎢ 2 −3 0 ⎥ − ( 0.306 ) (3) ⎢0 1 0 ⎥ ⎬ = ⎢1.26 −2.34 ⎪ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎪⎭ ⎢⎣ 0 −0.443⎥⎦ 0 ⎩ ⎛ 4π 53 ⎞ Thus, e = 0.567 × 10−5 → ΔV = ( 0.567 ) ⎜ × 10−5 = 2.96 × 10−3 ⎜ 3 ⎟⎟ ⎝ ⎠

1 [E] = EY

________________________________________________________________________

5.15 x1 = X1 + k ( X1 + X 2 ) ,

Given a motion

x2 = X 2 + k ( X1 − X 2 ) , show that for a function f (a, b) = ab ∂f ( x1 , x2 )

=

∂f ( X1 , X 2 )

+ O( k ) , ∂x1 ∂X1 where O(k ) → 0 as k → 0 ------------------------------------------------------------------------------Ans. (a) f ( x1 , x2 ) = x1 x2 = ⎡⎣ X1 + k ( X1 + X 2 ) ⎤⎦ ⎡⎣ X 2 + k ( X1 − X 2 ) ⎤⎦

(a) f ( x1 , x2 ) = f ( X1 , X 2 ) + O(k ) ,

= X1 X 2 + k { X1 ( X1 − X 2 ) + X 2 ( X1 + X 2 )} + k 2 ( X1 + X 2 )( X1 − X 2 ) . That is, f ( x1 , x2 ) = X1 X 2 + O ( k ) , where O(k ) → 0 as k → 0 , i.e.

f ( x1 , x2 ) = f ( X1 , X 2 ) + O(k ) → f ( x1 , x2 ) ≈ f ( X1 , X 2 ) as k → 0 . ∂f = x2 = X 2 + k ( X1 − X 2 ) = X 2 + O ( k ) , and ∂x1 ∂f ∂f ∂f f ( X1 , X 2 ) = X1 X 2 → = X 2 . Thus, ≈ as k → 0 . ∂X1 ∂x1 ∂X1

(b) f ( x1 , x2 ) = x1 x2 →

________________________________________________________________________

5.16

Do the previous problem for f ( a, b) = a 2 + b 2

------------------------------------------------------------------------------Ans. (a) f ( x1 , x2 ) = x12 + x22 = ⎡⎣ X1 + k ( X1 + X 2 ) ⎤⎦ + ⎡⎣ X 2 + k ( X1 − X 2 ) ⎤⎦ 2

= X12 + X 22 + 2k { X1 ( X1 + X 2 ) + X 2 ( X1 − X 2 )} + k 2

{( X + X )


5-6

1

2

2

2

+ ( X1 − X 2 )

2

}.

Lai et al, Introduction to Continuum Mechanics That is, f ( x1 , x2 ) = X12 + X 22 + O ( k ) , where O(k ) → 0 as k → 0 , i.e. f ( x1 , x2 ) = f ( X1 , X 2 ) + O(k ) → f ( x1 , x2 ) ≈ f ( X1 X 2 ) as k → 0 .

∂f = 2 x1 = 2 X1 + 2k ( X1 +X 2 ) =2 X1 + O ( k ) and ∂x1 ∂f ∂f ∂f as k → 0 . f ( X1 , X 2 ) = X12 + X 22 → ≈ = 2 X1 . Thus, ∂x1 ∂X1 ∂X1

(b) f ( x1 , x2 ) = x12 + x22 →

________________________________________________________________________

5.17

Given the following displacement field in an isotropic linearly elastic solid:

(

)

u1 = kX 3 X 2 , u2 = kX 3 X1 , u3 = k X12 − X 22 , k = 10−4 (a) Find the stress components and (b) in the absence of body forces, is the state of stress a possible equilibrium stress field? -------------------------------------------------------------------------------Ans. ⎡ kX 3 kX 2 ⎤ 0 2X3 ⎡ 0 ( 2 X1 + X 2 ) ⎤ k⎢ ⎢ ⎥ 0 kX1 ⎥ → [ E] = ⎢ 2 X 3 0 (a) [∇u ] = ⎢ kX 3 ( X1 − 2 X 2 ) ⎥⎥ 2 ⎢( 2 X 1 + X 2 ) ( X 1 − 2 X 2 ) ⎥ ⎢⎣ 2kX1 −2kX 2 0 ⎥⎦ 0 ⎣ ⎦

⎡ 0 2X3 ( 2 X1 + X 2 ) ⎤ ⎢ 0 Thus, Ekk = 0 → [ T] = 2μ [ E] = μ k ⎢ 2 X 3 ( X1 − 2 X 2 ) ⎥⎥ ⎢( 2 X 1 + X 2 ) ( X 1 − 2 X 2 ) ⎥ 0 ⎣ ⎦ Since the displacement components are small (of the order of k ), therefore, xi ≈ X i , so that

T11 = T22 = T33 = 0, T12 = T21 = 2 μ kx3 , T13 = T31 = μ k ( 2 x1 + x2 ) , T23 = T32 = μ k ( x1 − 2 x2 ) . (b) Substituting the above stress components into the equations of equilibrium, we have, ∂T ∂T11 ∂T12 ∂T13 ∂T ∂T + + = 0 → 0 + 0 + 0 = 0 , 21 + 22 + 23 = 0 → 0 + 0 + 0 = 0 and ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + = 0 → 2μ k − 2μ k + 0 = 0 . Thus, all equations of equilibrium are satisfied. ∂x1 ∂x2 ∂x3 Since the stress field is obtained from a given displacement field, therefore, the state of stress is a possible equilibrium stress field. ________________________________________________________________________

5.18


u1 = kX 2 X 3 , u2 = kX1 X 3 , u3 = kX1 X 2 , k = 10−4 (a) Find the stress components and (b) in the absence of body forces, is the state of stress a possible equilibrium stress field? ------------------------------------------------------------------------------Ans.


5-7

Lai et al, Introduction to Continuum Mechanics kX 2 ⎤ ⎡ 0 X3 X2 ⎤ ⎥ 0 kX1 ⎥ = [ E] , Thus, Ekk = 0 → [ T] = 2 μ [ E ] = 2μ k ⎢⎢ X 3 0 X1 ⎥⎥ ⎢⎣ X 2 X1 0 ⎥⎦ kX1 0 ⎥⎦ Since the displacement components are small (of the order of k ), therefore, xi ≈ X i , so that T11 = T22 = T33 = 0, T12 = T21 = 2μ kx3 , T13 = T31 = 2μ kx2 , T23 = T32 = 2μ kx1 . (b) Substituting the above stress components into the equations of equilibrium, we have, ∂T ∂T11 ∂T12 ∂T13 ∂T ∂T + + = 0 → 0 + 0 + 0 = 0 , 21 + 22 + 23 = 0 → 0 + 0 + 0 = 0 and ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + = 0 → 0 + 0 + 0 = 0 . Thus, all equations of equilibrium are satisfied. Since the ∂x1 ∂x2 ∂x3 stress field is obtained from a given displacement field, therefore, the state of stress is a possible equilibrium stress field. ⎡ 0 (a) [∇u ] = ⎢⎢ kX 3 ⎢⎣ kX 2

kX 3

________________________________________________________________________

5.19


(

)

u1 = kX 2 X 3 , u2 = kX1 X 3 , u3 = k X1 X 2 + X 32 , k = 10−4 (a) Find the stress components and (b) in the absence of body forces, is the state of stress a possible equilibrium stress field? -------------------------------------------------------------------------------Ans. kX 3 kX 2 ⎤ ⎡ 0 0 kX1 ⎥⎥ = [ E] , Thus, Ekk = 2kX 3 → [ T] = λ ( 2kX 3 ) I + 2μ [ E ] (a) [∇u ] = ⎢⎢ kX 3 ⎢⎣ kX 2 kX1 2kX 3 ⎥⎦ Since the displacement components are small (of the order of k ), therefore, xi ≈ X i , so that ⎡ λ x3 [ T] = 2k ⎢⎢ μ x3 ⎢ μ x2 ⎣

μ x3 λ x3 μ x1

⎤ μ x2 ⎥ μ x1 ⎥. ( λ + 2μ ) x3 ⎥⎦

(b) Substituting the above stress components into the equations of equilibrium, we have, ∂T ∂T11 ∂T12 ∂T13 ∂T ∂T + + = 0 → 0 + 0 + 0 = 0 , 21 + 22 + 23 = 0 → 0 + 0 + 0 = 0 and ∂x1 ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + = 0 → 0 + 0 + ( λ + 2 μ ) ≠ 0 . Thus, the stress field is not an equilibrium stress ∂x1 ∂x2 ∂x3 field in the absence of body forces. The given state of stress is not a possible equilibrium stress field. ________________________________________________________________________


5-8


Show that for any function f ( s ) , the displacement u1 = f ( s ) where s = x1 ± cL t satisfies the wave equation

∂ 2u1 ∂t 2

= cL2

∂ 2u1 ∂x12

------------------------------------------------------------------------------Ans. ∂u1 df ∂s df ∂ 2u d 2 f ∂s d2 f = = ( ± cL ) → 21 = ( ± cL ) 2 = cL2 2 , ∂t ds ∂t ds ∂t ds ∂t ds

∂u1 df ∂s df ∂ 2u d 2 f ∂s d 2 f = = → 21 = 2 = . ∂x1 ds ∂x1 ds ∂x1 ds ∂x1 ds 2 Thus,

∂ 2u1 ∂t2

= cL2

d2 f ds 2

= cL2

∂ 2u1 ∂x12

________________________________________________________________________

5.21

Calculate the ratio of the phase velocities cL / cT for Poisson 's ratio equal 1 / 3, 0.49 and 0.499 .

----------------------------------------------------------------------------------------------------------2μ (1 − ν ) μ 2 μν 1 − 2ν Ans. From Table 5.1, we have λ = , → λ + 2μ = → = λ + 2 μ 2 (1 − ν ) 1 − 2ν 1 − 2ν Thus,

cL λ + 2μ 2(1 − ν ) = = . Thus, for 1 − 2ν μ cT

ν = 1 / 3, cL / cT = 2(2 / 3) / (1 − 2 / 3) =

( 4 / 3) / (1 / 3) = 2.

ν = 0.49, cL / cT = 2(0.51) / (1 − 0.98 ) = 1.02 / 0.02 = 7.14. ν = 0.499, cL / cT = 2(0.501) / (1 − 0.998 ) = 1.002 / 0.002 = 22.4. __________________________________________________________________ 5.22

Assume a displacement which depends only x2 and t , i.e., ui = ui ( x2 , t ) , i = 1, 2,3 .

Obtain the differential equations which ui ( x2 , t ) must satisfy in order to be a possible motion in the absence of body forces. ------------------------------------------------------------------------------Ans. From the Navier equations, we have,

e = ∂u2 / ∂x2 → ∂ e / ∂x1 = 0, ∂e / ∂x2 = ∂ 2u2 / ∂x22 , ∂e / ∂x3 = 0. Thus,

ρo (∂u12 / ∂t 2 ) = μ (∂ 2u1 / ∂x22 ) → (∂u12 / ∂t 2 ) = cT2 (∂ 2u1 / ∂x22 ), ρo (∂u22 / ∂t 2 ) = (λ + μ )(∂ 2u2 / ∂x22 ) + μ (∂ 2u2 / ∂x22 ),

(

→ ρo (∂u22 / ∂t 2 ) = (λ + 2μ )(∂ 2u2 / ∂x22 ) → (∂u22 / ∂t 2 ) = cL2 ∂ 2u2 / ∂x22

ρo (∂u32 / ∂t 2 ) = μ (∂ 2u3 / ∂x22 ) → (∂u32 / ∂t 2 ) = cT2 (∂ 2u3 / ∂x22 ). Copyright 2010, Elsevier Inc

5-9

)


Consider a linear elastic medium. Assume the following form for the displacement field: u1 = ε ⎡⎣sin β ( x3 − ct ) + α sin β ( x3 + ct ) ⎤⎦ , u2 = u3 = 0

(a) What is the nature of this elastic wave (longitudinal, transverse, direction of propagation?) (b) Find the strains, stresses and determine under what condition(s), the equations of motion are satisfied in the absence of body forces. (c)Suppose that there is boundary at x3 = 0 that is traction free. Under what condition(s) will the above motion satisfy this boundary condition for all time. (d) Suppose that there is boundary at x3 = l that is also traction free. What further conditions will be imposed on the above motion to satisfy this boundary condition for all time. ------------------------------------------------------------------------------Ans. (a) Transverse wave, propagating in the e3 direction. (b)The only nonzero strain components are: E13 = E31 = (1 / 2 ) ( ∂u1 / ∂x3 ) = ( εβ / 2 ) ⎡⎣ cos β ( x3 − ct ) + α cos β ( x3 + ct ) ⎤⎦ . The only nonzero stress components are: T13 = T31 = μ ( ∂u1 / ∂x3 ) = ( εμβ ) ⎡⎣ cos β ( x3 − ct ) + α cos β ( x3 + ct ) ⎤⎦ ,

(

)

x1 equation of motion is: ρ 0 ∂ 2u1 / ∂t 2 = ∂T13 / ∂x3 → − ρ 0 β 2 c 2u1 = − β 2 μ u1 → c 2 = μ / ρ 0 . The other two equations are 0=0. (c) The boundary condition on x3 = 0 is:

T ( −e3 ) = 0 → T13 ( 0, t ) = 0 → [ cos β ct + α cos β ct ] = 0 → α = −1 .

(d) The boundary condition on x3 = l is,

T ( e3 ) = 0 → T13 ( l, t ) = 0 → ⎡⎣cos β ( l − ct ) − cos β ( l + ct ) ⎤⎦ = 0,[note α = −1].

→ 2sin β l sin β ct = 0 → sin β l = 0 → β = nπ / l, n = 1, 2,3... _________________________________________________________________ 5.24

Do the previous problem (Prob. 5.23) if the boundary x3 = 0 is fixed (no motion) and x3 = l is still traction free.

------------------------------------------------------------------------------Ans. (a) and (b) are the same as in the previous problem. (c) The boundary condition on x3 = 0 is: u1 ( 0, t ) = 0 → u1 = ε [ − sin β ct + α sin β ct ] = 0 → α = 1 . (d) The boundary condition on x3 = l is:

T ( e3 ) = 0 → T13 ( l, t ) = 0 → ⎡⎣ cos β ( l − ct ) + cos β ( l + ct ) ⎤⎦ = 0

→ 2cos β l cos β ct = 0 → cos β l = 0 → β = nπ / ( 2l ) , n = 1,3,5... _________________________________________________________________ 5.25

Do Problem 5.23 if the boundary x3 = 0 and x3 = l are both rigidly fixed (no motion) -------------------------------------------------------------------------------Ans. (a) and (b) are the same as in the previous problem 5.23. Copyright 2010, Elsevier Inc

5-10

Lai et al, Introduction to Continuum Mechanics (c) The boundary condition on x3 = 0 is: u1 ( 0, t ) = 0 → u1 = ε [ − sin β ct + α sin β ct ] = 0 → α = 1 (d) The boundary condition on x3 = l is:

u1 ( l, t ) = 0 → u1 = ε ⎡⎣sin β ( l − ct ) + sin β ( l + ct ) ⎤⎦ = 0 → sin β l cos β ct → 0 → β = nπ / l, n = 1, 2,3... __________________________________________________________________ 5.26

Do Problem 5.23, if the assumed displacement field is of the form:

u3 = ε ⎡⎣sin β ( x3 − ct ) + α sin β ( x3 + ct ) ⎤⎦ , u1 = u2 = 0 -------------------------------------------------------------------------------Ans. (a) Longitudinal, propagating in the e3 direction. (b)The only nonzero strain components are: E33 = ( ∂u3 / ∂x3 ) = εβ ⎡⎣ cos β ( x3 − ct ) + α cos β ( x3 + ct ) ⎤⎦ .

The nonzero stress components are: T11 = T22 = λ ( ∂u3 / ∂x3 ) , T33 = λ ( ∂u3 / ∂x3 ) + 2μ ( ∂u3 / ∂x3 ) = ( λ + 2μ ) ( ∂u3 / ∂x3 ) , where ( ∂u3 / ∂x3 ) = ( εβ ) ⎡⎣cos β ( x3 − ct ) + α cos β ( x3 + ct ) ⎤⎦ .

(

)

x3 equation of motion is:

ρ0 ∂ 2u3 / ∂t 2 = ∂T33 / ∂x3 → − ρ0 β 2 c 2u3 = − β 2 ( λ + 2μ ) u3 → c 2 = ( λ + 2μ ) / ρ 0 . The other two equations are 0=0. (c) The boundary condition on x3 = 0 is:

T ( −e3 ) = 0 → T33 ( 0, t ) = 0 → ( λ + 2μ )( εβ ) ⎡⎣ cos β ( ct ) + α cos β ( ct ) ⎤⎦ = 0 → α = −1 .

(d) The boundary condition on x3 = l is:

T ( e3 ) = 0 → T33 ( l, t ) = 0 → ⎡⎣ cos β ( l − ct ) − cos β ( l + ct ) ⎤⎦ = 0.[Note α = −1],

→ 2sin β l sin β ct = 0 → sin β l = 0 → β = nπ / l, n = 1, 2,3... __________________________________________________________________ Do the previous problem, Problem 5.26, if the boundary x3 = 0 is fixed (no motion) and x3 = l is traction free ( t = 0 ).

5.27

------------------------------------------------------------------------------Ans. (a) and (b) are the same as the previous problem, problem 5.26. (c) The boundary condition on x3 = 0 is: u3 ( 0, t ) = 0 → u3 = ε [ − sin β ct + α sin β ct ] = 0 → α = 1 . (d) The boundary condition on x3 = l is, with α = 1 ,

T ( e3 ) = 0 → T33 ( l, t ) = 0 → ⎡⎣ cos β ( l − ct ) + cos β ( l + ct ) ⎤⎦ = 0

→ 2cos β l cos β ct = 0 → cos β l = 0 → β = nπ / ( 2l ) , n = 1,3,5... __________________________________________________________________ 5.28

Do Problem 5.26, if the boundary x3 = 0 and boundary x3 = l are both rigidly fixed. ------------------------------------------------------------------------------Copyright 2010, Elsevier Inc

5-11

Lai et al, Introduction to Continuum Mechanics Ans. (a) (b) and (c) are the same as in Prob. 5.27, with α = 1 . (d) The boundary condition on x3 = l is u3 ( l, t ) = 0 → u3 = ε ⎡⎣sin β ( l − ct ) + sin β ( l + ct ) ⎤⎦ = 0

→ sin β l cos β ct = 0 → sin β l = 0 → β = nπ / l, n = 1, 2,3... __________________________________________________________________ 5.29

Consider the displacement field: ui = ui ( x1 , x2 , x3 , t ) . In the absence of body forces,

(a) obtain the governing equation for ui for the case where the motion is equivoluminal and (b) obtain the governing equation for the dilatation e for the case where the motion is irrotational

( ∂ui / ∂x j = ∂u j / ∂xi ) .

------------------------------------------------------------------------------Ans. From the Navier equations of motion, Eq. (5.6.4) ∂ 2u ∂ 2ui ∂e +μ , we have ρo 2i = ρo Bi + ( λ + μ ) ∂xi ∂x j ∂x j ∂t (a) with e = 0 and Bi =0 , ρo (b) For irrotational motion

ρo

∂ 2ui ∂t 2

∂t 2

=μ

∂ 2ui . ∂x j ∂x j

∂ui ∂u j ∂ 2ui ∂ ∂u j ∂ ∂u j ∂e = → = = = . Thus, ∂x j ∂xi ∂x j ∂x j ∂x j ∂xi ∂xi ∂x j ∂xi

∂e ∂e ∂e ∂ 2 ∂u ∂ 2e +μ = ( λ + 2μ ) → ρo 2 i = ( λ + 2μ ) ∂xi ∂xi ∂xi ∂xi ∂xi ∂t ∂xi

( λ + 2μ )

∂ 2e . ∂xi ∂xi ρo ∂t 2 __________________________________________________________________ →

∂ 2e

= (λ + μ )

∂ 2ui

=

5.30 (a) Write a displacement field for an infinite train of longitudinal waves propagating in the direction of 3e1 + 4e 2 . (b) Write a displacement field for an infinite train of transverse waves propagating in the direction of 3e1 + 4e 2 and polarized in the x1 x2 plane. ------------------------------------------------------------------------------Ans. Let e n = (1 / 5 )( 3e1 + 4e 2 ) , then x ⋅ e n = (1 / 5 )( 3x1 + 4 x2 ) . Also, e t = ± (1 / 5 )( 4e1 − 3e 2 ) ⎛ 2π ⎞ (a) Equation 5.10.8 of Example 5.10.3 gives u = ε sin ⎜ x ⋅ e n − cL t − η ⎟ e n . Thus, l ⎝ ⎠ ⎡ 2π ⎛ 3x 4 x ⎡ 2π ⎛ 3x 4 x 3ε 4ε ⎞⎤ ⎞⎤ u1 = sin ⎢ ⎜ 1 + 2 − cL t − η ⎟ ⎥ , u2 = sin ⎢ ⎜ 1 + 2 − cL t − η ⎟ ⎥ , u3 = 0 5 5 5 5 5 5 l l ⎠⎦ ⎠⎦ ⎣ ⎝ ⎣ ⎝ ⎛ 2π ⎞ (b) Equation 5.10.10 of Example 5.10.3 gives u = ε sin ⎜ x ⋅ e n − cT t − η ⎟ e t . Thus, ⎝ l ⎠ ⎡ 2π ⎛ 3x 4 x ⎡ 2π ⎛ 3x 4 x 4ε 3ε ⎞⎤ ⎞⎤ u1 = ± sin ⎢ ⎜ 1 + 2 − cT t − η ⎟ ⎥ , u2 = m sin ⎢ ⎜ 1 + 2 − cT t − η ⎟ ⎥ , u3 = 0 5 5 5 5 ⎠⎦ ⎠⎦ ⎣ l ⎝ 5 ⎣ l ⎝ 5


5-12

Lai et al, Introduction to Continuum Mechanics __________________________________________________________________ 5.31

Solve for ε 2 and ε 3 in terms of ε1 from the following two algebra equations: 1 n ------------------------------------------------------------------------------Ans. cos 2α1 n(sin 2α 3 ) 1 2 Δ= = − ⎡⎢( cos 2α1 ) + n 2 (sin 2α 3 )sin 2α1 ⎤⎥ 1 ⎣ ⎦ n sin 2α1 − (cos 2α1 ) n Thus, ⎡ 1 ⎤ ⎡ε 2 ⎤ 1 ⎢ − (cos 2α1 ) − n(sin 2α 3 ) ⎥ ⎡ ε1 cos 2α1 ⎤ n ⎢ ⎥= ⎥ ⎢ −ε sin 2α1 ⎥⎦ ⎣ε 3 ⎦ Δ ⎢⎢ − sin 2α cos 2α1 ⎦⎥ ⎣ 1 1 ⎣

ε 2 (cos 2α1 ) + ε 3n(sin 2α 3 ) = ε1 cos 2α1 (i) ε 2 sin 2α1 − ε 3 (cos 2α1 ) = −ε1 sin 2α1 , (ii)

=−

ε1 ⎡ (cos 2α1 )2 − n 2 (sin 2α 3 ) ( sin 2α1 ) ⎤ ⎢ ⎥ nΔ ⎢⎣ n sin 4α1 ⎥⎦ That is,

ε 2 = ε1

(cos 2α1 ) − n (sin 2α 3 ) ( sin 2α1 ) 2

( cos 2α1 )

2

, ε 3 = ε1

n sin 4α1

+ n (sin 2α 3 )sin 2α1 ( cos 2α1 ) + n2 (sin 2α 3 )sin 2α1 __________________________________________________________________ 2

2

2

5.32 A transverse elastic wave of amplitude ε1 incidents on a traction free plane boundary. If the Poisson's ratio ν = 1 / 3 , determine the amplitudes and angles of reflection of the reflected waves for the following two incident angles (a) α1 = 0 and (b) α1 = 15o . -------------------------------------------------------------------------------Ans. From Eq. (5.11.14), we have, for ν = 1 / 3

(1 − 2ν ) / 2 (1 −ν ) = (1 − 2 / 3) / 2 (1 − 1 / 3) = (1 / 3) / 2 ( 2 / 3) = 1 / 2 . Thus, (1 / 2 ) sin α 3 = sin α1 . Using this equation, and Equations (cos 2α1 )2 − n 2 (sin 2α 3 ) ( sin 2α1 ) n sin 4α1 ε 2 = ε1 , ε 3 = ε1 2 2 ( cos 2α1 ) + n 2 (sin 2α 3 )sin 2α1 ( cos 2α1 ) + n2 (sin 2α 3 )sin 2α1

n = cT / cL =

we have, (a) α1 = 0 → α 2 = 0 and (1 / 2 ) sin α 3 = sin α1 = 0 → α 3 = 0. Also, the above equations give

ε 2 = ε1 , and ε 3 = 0 ,. That is, there is no reflected longitudinal wave. There is only a reflected transverse wave of the same amplitude which completely cancels out the incident transverse wave. (b) α1 = 15o → α 2 = 15o and sin α 3 = 2sin15o = 0.5176 → α 3 = 31.17o ,


5-13


ε 2 = ε1 ε 3 = ε1

( ) = 0.742ε ( cos30 ) + (1 / 4) (sin 62.34 ) (sin 30 ) (cos30o )2 − (1 / 4 ) (sin 62.34o ) sin 30o 2

o

o

o

sin 600 / 2

(

cos30o

)

2

(

+ (1 / 4 ) (sin 62.34o ) sin 30o

)

1

= 0.503ε1

That, the reflected transverse wave has an amplitude ε 2 = 0.742ε1 , with a reflected angle of

α 2 = α1 = 15o . The reflected longitudinal wave has an amplitude ε 3 = 0.503ε1 ,with a reflected angle of α 3 = 31.17o . __________________________________________________________________ 5.33

Referring to Figure 5.11.1 (Section 5.11), consider a transverse elastic wave incident on a traction-free plane surface ( x2 = 0 ) with an angle of incident α1 with the x2 axis and

polarized normal to x1 x2 , the plane of incidence. Show that the boundary condition at x2 = 0 can be satisfied with only a reflected transverse wave that is similarly polarized. What is the relation of the amplitudes, wavelengths, and direction of propagation of the incident and reflected wave? -------------------------------------------------------------------------------Ans. Let the plane of incidence be x1 x2 plane with the angle of incidence of the transverse wave be α1 . That is, e n1 = sin α1e1 − cos α1e 2 . The waves are polarized normal to the plane of incidence, therefore, u1 = u2 = 0 , and u3 = ε1 sin ϕ1 + ε 2 sin ϕ2 , with 2π 2π ϕ1 = ( x1 sin α1 − x2 cos α1 − cT t − η1 ), ϕ2 = ( x1 sin α 2 + x2 cos α 2 − cT t − η 2 ) l1 l2 The nonzero stress components are: T13 = T31 = μ∂u3 / ∂x1 = 2πμ ⎡⎣( ε1 / l1 ) cos ϕ1 sin α1 + ( ε 2 / l 2 ) cos ϕ2 sin α 2 ⎤⎦ , T23 = T32 = μ∂u3 / ∂x2 =2πμ ⎡⎣ − ( ε1 / l1 ) cos ϕ1 cos α1 + ( ε 2 / l 2 ) cos ϕ 2 cos α 2 ⎤⎦ .

(

)

The x3 equation of motion ρo ∂ 2u3 / ∂t 2 = ∂T31 / ∂x1 + ∂T32 / ∂x2 gives:

(

)

(

)

(

)

(

)

2 2 ρo ( 2π cT ) ⎡ ε1 / l12 sin ϕ1 + ε 2 / l 22 sin ϕ 2 ⎤ = ( 2π ) μ ⎡ ε1 / l12 sin ϕ1 + ε 2 / l 22 sin ϕ 2 ⎤

⎣

⎦

⎣

⎦

→ ρo cT2 = μ → cT2 = μ / ρo . The traction free boundary at x2 = 0 requires that T12 = T22 = T32 = 0 on the surface, thus, ⎡⎣ − ( ε1 / l1 ) cos ϕ1 cos α1 + ( ε 2 / l 2 ) cos ϕ 2 cos α 2 ⎤⎦ = 0 , where x2 = 0

2π 2π ( x1 sin α1 − cT t − η1 ), ϕ 2 = ( x1 sin α 2 − cT t − η 2 ) . l1 l2 Thus, the boundary condition is satisfied if α1 = α 2 , ε1 = ε 2 , l1 = l 2 , η1 = η2 . __________________________________________________________________

ϕ1 =


5-14


Do the problem of Section 5.11.(Reflection of Plane Elastic Waves, Figure 5.11-1) for the case where the boundary x2 = 0 is fixed. ------------------------------------------------------------------------------Ans. As in Section 5.11, we assume u1 = ( cos α1 ) ε1 sin ϕ1 + ( cos α 2 ) ε 2 sin ϕ 2 + ( sin α 3 ) ε 3 sin ϕ3

u2 = ( sin α1 ) ε1 sin ϕ1 − ( sin α 2 ) ε 2 sin ϕ2 + ( cos α 3 ) ε 3 sin ϕ3 , u3 = 0 where 2π 2π ϕ1 = ( x1 sin α1 − x2 cos α1 − cT t − η1 ), ϕ 2 = ( x1 sin α 2 + x2 cos α 2 − cT t − η 2 ) l1 l2

ϕ3 =

2π ( x1 sin α 3 + x2 cos α 3 − cL t − η3 ) l3

The equations of motion are satisfied with ( cL ) = ( λ + 2 μ ) / ρ0 , ( cT ) = μ / ρ0 2

2

Now, at x2 = 0 ,

⎡⎣( cos α1 ) ε1 sin ϕ1 + ( cos α 2 ) ε 2 sin ϕ2 + ( sin α 3 ) ε 3 sin ϕ3 ⎤⎦ =0 x2 =0 ⎡⎣( sin α1 ) ε1 sin ϕ1 − ( sin α 2 ) ε 2 sin ϕ2 + ( cos α 3 ) ε 3 sin ϕ3 ⎤⎦ =0 x2 =0 Thus, at x2 = 0 , sin ϕ1 = sin ϕ2 = sin ϕ3 , so that 2π 2π 2π ϕ1 = ( x1 sin α1 − cT t − η1 ) = ( x1 sin α 2 − cT t − η 2′ ) = ( x1 sin α 3 − cL t − η3′ ), l1 l2 l3

η2′ = η2 − ( ± pl 2 ) ,η3′ = η3 − ( ± ql 2 )

Thus, as in Section 5.11, we have, with n = cT / cL , l 2 = l1 , nl 3 = l1 , α1 = α 2 , n sin α 3 = sin α1 , η2′ = η1 , nη3′ = η1 . However, the relations between the amplitudes are different. In fact, from ( cos α1 ) ε 2 + ( sin α 3 ) ε 3 = − ( cos α1 ) ε1 ,

( sin α1 ) ε 2 − ( cos α 3 ) ε 3 = ( sin α1 ) ε1. we can obtain, sin α sin α − cos α cos α1 ) ( 3 )( 1 ) ( − sin 2α1 3 )( ε2 = , ε3 = . ( sin α1 ) ( sin α 3 ) + ( cosα1 ) ( cos α 3 ) ( sin α1 ) ( sin α 3 ) + ( cos α1 ) ( cos α 3 ) ________________________________________________________________ 5.35 A longitudinal elastic wave is incident on a fixed boundary x2 = 0 with an incident angle of α1 with the x2 axis (similar to Fig. 5.11.1 of Section 5.11). (a) Show that in general, there are two reflected waves, one longitudinal and the other transverse (also polarized in the incident plane x1 x2 ). (b)Find the amplitude ratio of reflected to incident elastic waves. -----------------------------------------------------------------------Ans. (a) Let u1 = ( sin α1 ) ε1 sin ϕ1 + ( sin α 2 ) ε 2 sin ϕ2 + ( cos α 3 ) ε 3 sin ϕ3 u2 = ( − cos α1 ) ε1 sin ϕ1 + ( cos α 2 ) ε 2 sin ϕ2 − ( sin α 3 ) ε 3 sin ϕ3 , u3 = 0, where Copyright 2010, Elsevier Inc

5-15


ϕ1 = ϕ3 =

2π 2π ( x1 sin α1 − x2 cos α1 − cL t − η1 ), ϕ2 = ( x1 sin α 2 + x2 cos α 2 − cL t − η 2 ) l1 l2 2π ( x1 sin α 3 + x2 cos α 3 − cT t − η3 ) l3

The equations of motion are satisfied with ( cL ) = ( λ + 2μ ) / ρ0 , ( cT ) = μ / ρ0 . 2

2

Now, at x2 = 0 ,

⎡⎣( sin α1 ) ε1 sin ϕ1 + ( sin α 2 ) ε 2 sin ϕ 2 + ( cos α 3 ) ε 3 sin ϕ3 ⎤⎦ =0 x2 = 0 ⎡⎣( − cos α1 ) ε1 sin ϕ1 + ( cos α 2 ) ε 2 sin ϕ 2 − ( sin α 3 ) ε 3 sin ϕ3 ⎤⎦ =0 x2 = 0 Thus, at x2 = 0 , sin ϕ1 = sin ϕ2 = sin ϕ3 , so that ϕ1 = (2π / l1 )( x1 sin α1 − cL t − η1 ) = (2π / l 2 )( x1 sin α 2 − cL t − η 2′ ) = (2π / l 3 )( x1 sin α 3 − cT t − η3′ ),

η2′ = η2 − ( ± pl 2 ) ,η3′ = η3 − ( ± ql 3 ) .

Thus, we have,

ϕ1 = (2π / l1 )( x1 sin α1 − cL t − η1 ) = (2π / l 2 )( x1 sin α 2 − cL t − η 2′ ) = (2π / l 3 )( x1 sin α 3 − cT t − η3′ ),

η2′ = η2 − ( ± pl 2 ) ,η3′ = η3 − ( ± ql 2 ) .

Thus, sin α1 sin α 2 sin α 3 cL cL cT η1 η 2′ η3′ , = = = = , = = l1 l2 l3 l1 l 2 l 3 l1 l 2 l 3 So that α1 = α 2 , l 2 = l1 , l 3 = nl1 , sin α 3 = n sin α1 , η 2′ = η1 , η3′ = nη1 where n = cT / cL . We note that unlike the problem in Sect. 5.11, here l 3 = nl1 , sin α 3 = n sin α1 ( instead of nl 3 = l1 , n sin α 3 = sin α1 ). With sin ϕ1 = sin ϕ2 = sin ϕ3 , we have

( sin α1 ) ε 2 + ( cos α 3 ) ε 3 = − ( sin α1 ) ε1 , ( cos α1 ) ε 2 − ( sin α 3 ) ε 3 = ( cosα1 ) ε1

Thus, ε 3 / ε1 = − sin 2α1 / cos (α1 − α 3 ) , ε 2 / ε1 = cos (α1 + α 3 ) / cos (α1 − α 3 ) __________________________________________________________________ 5.36

Do the previous problem (Prob. 5.35) for the case where x2 = 0 is a traction free boundary ------------------------------------------------------------------------------Ans. Let u1 = ( sin α1 ) ε1 sin ϕ1 + ( sin α 2 ) ε 2 sin ϕ2 + ( cos α 3 ) ε 3 sin ϕ3

u2 = ( − cos α1 ) ε1 sin ϕ1 + ( cos α 2 ) ε 2 sin ϕ2 − ( sin α 3 ) ε 3 sin ϕ3 , u3 = 0, where

ϕ1 = ϕ3 =

2π 2π ( x1 sin α1 − x2 cos α1 − cL t − η1 ), ϕ2 = ( x1 sin α 2 + x2 cos α 2 − cL t − η 2 ) l1 l2 2π ( x1 sin α 3 + x2 cos α 3 − cT t − η3 ) l3

The equations of motion are satisfied with ( cL ) = ( λ + 2μ ) / ρ0 , ( cT ) = μ / ρ0 . 2

At x2 = 0 , T21 = T22 = T23 = 0 . Thus,


5-16

2

Lai et al, Introduction to Continuum Mechanics ∂u1 / ∂x2 + ∂u2 / ∂x1 = 0 and ( λ +2μ )( ∂u2 / ∂x2 ) + λ ( ∂u1 / ∂x1 ) = 0 . That is, (ε1 / l1 ) cos ϕ1 ( − sin 2α1 ) + (ε 2 / l 2 ) cos ϕ2 ( sin 2α 2 ) + (ε 3 / l 3 ) cosϕ3 ( cos 2α 3 ) = 0 and 2 (ε1 / l1 ) ⎡⎣⎢λ + 2μ ( cos α1 ) ⎤⎦⎥ cos ϕ1 + (ε 2 / l 2 ) ⎡⎣⎢λ + 2μ ( cosα 2 )2 ⎤⎦⎥ cos ϕ2

(i)

(ii)

+ ( ε 3 / l 3 ) ( 2μ ) ⎡⎣( − sin α 3 cos α 3 ) ⎤⎦ cos ϕ3 = 0

In order for the above two equations to be satisfied for all x1 and t , we must have, at x2 = 0 cos ϕ1 = cos ϕ2 = cos ϕ3 , which gives sin α1 sin α 2 sin α 3 cL cL cT η1 η 2 m pl 2 η3 m ql 3 , = = = = , = = l1 l2 l3 l1 l 2 l 3 l 1 l2 l3 Thus, α1 = α 2 , l 2 = l1 , l 3 = nl1 , sin α 3 = n sin α1 , η 2 m pl 2 = η1 , η3 m ql 3 = nη1 , where n = cT / cL . (i) and (ii) now gives (iii) (ε1 / l1 )( − sin 2α1 ) + (ε 2 / l 2 )( sin 2α1 ) + (ε 3 / l 3 )( cos 2α 3 ) = 0 (ε1 / l1 )(λ + 2μ cos 2 α1 ) + (ε 2 / l1 )(λ + 2μ cos 2 α1 ) − (ε 3 / l 3 )(2μ )sin α 3 cos α 3 = 0 ⎛ λ + 2μ

λ + 2μ cos 2 α1 = λ + 2 μ − 2μ sin 2 α1 = μ ⎜

μ

⎝

(ε 2 )

μ

(1 − 2n n

2

)

⎞ − 2sin 2 α1 ⎟ = ⎠

sin 2 α1 − (ε 3 /)(2μ )sin α 3 cos α 3 = −ε1

μ

(1 − 2n n

2

(iv) sin 2 α1

)

Since ⎛ λ + 2μ ⎞ μ λ + 2μ cos 2 α1 = λ + 2μ − 2μ sin 2 α1 = μ ⎜ − 2sin 2 α1 ⎟ = 2 1 − 2n 2 sin 2 α1 , ⎝ μ ⎠ n and l 3 = nl1 , therefore, (iii) and (iv) become

(

( n sin 2α1 ) ε 2 + ( cos 2α 3 ) ε 3 = ε1n sin 2α1

(1 − 2n

2

)

)

(v)

(

)

sin 2 α1 ε 2 − 2n sin α 3 cos α 3ε 3 = − 1 − 2n 2 sin 2 α1 ε1

(vi)

(v) and (vi) give 2n sin 2α1 (1 − 2n 2 sin 2 α1 )

ε3 = , ε1 n 2 sin 2α1 sin 2α 3 + (1 − 2n 2 sin 2 α1 ) cos 2α 3 ε 2 n 2 sin 2α1 sin 2α 3 − (1 − 2n 2 sin 2 α1 ) cos 2α 3 = ε1 n 2 sin 2α1 sin 2α 3 + (1 − 2n 2 sin 2 α1 )cos 2α 3 __________________________________________________________________ 5.37

Verify that the thickness stretch vibration given by Eq.(5.12.3), i.e., u1 = ( A cos kx1 + B sin kx1 )(C cos cL kt + D sin cL kt )

(

does satisfy the longitudinal wave equation ∂ 2u1 / ∂t 2 = ( cL ) ∂ 2u1 / ∂x12 2


5-17

)

Lai et al, Introduction to Continuum Mechanics --------------------------------------------------------------------------------Ans. u1 = ( A cos kx1 + B sin kx1 )(C cos cL kt + D sin cL kt ) , ∂ 2u1 / ∂x12 = − k 2 ( A cos kx1 + B cos kx1 )(C cos cL kt + D sin cL kt ) ∂ 2u1 / ∂t 2 = −(kcL ) 2 ( A cos kx1 + B cos kx1 )(C cos cL kt + D sin cL kt )

that is, ∂ 2u1 / ∂x12 = − k 2u1 and ∂ 2u1 / ∂t 2 = − ( cL k ) u1 . Thus, cL2 ∂ 2u1 / ∂x12 = ∂ 2u1 / ∂t 2 __________________________________________________________________ 2

5.38 (a) Find the thickness-stretch vibration of a plate, where the left face ( x1 = 0 ) is subjected to a forced displacement u = (α cosωt )e1 and the right face x1 = l is free. (b)Determine the values of ω that give resonance. --------------------------------------------------------------------------------Ans. Let (a) u1 = ( A cos kx1 + B sin kx1 )(C cos cL kt + D sin cL kt ) . Using the boundary condition u(0, t ) = (α cosωt )e1 , we have , α cos ωt = u1 (0, t ) = AC cos cL kt + AD sin cL kt Thus, AC = α , k = ω / cL , D = 0 → u1 = (α cos kx1 + BC sin kx1 ) cos ωt .

At x1 = l , T11 = T12 = T13 = 0 . Now, T11 = ( λ + 2 μ )( ∂u1 / ∂x1 ) , thus ( ∂u1 / ∂x1 ) x =l = 0 , i.e., 1

k (−α sin k l + BC cos k l) cos ωt = 0 → BC = α tan k l , → u1 = α [cos(ω x1 / cL ) + tan(ωl / cL )sin(ω x1 / cL )]cos ωt . (b) Resonance occurs at : ω l / cL = nπ / 2, n = 1,3,5... __________________________________________________________________

(a) Find the thickness stretch vibration if the x1 = 0 face is being forced by a traction

5.39

t = ( β cos ωt ) e1 and the right hand face x1 = l is fixed. (b) Find the resonance frequencies.

-------------------------------------------------------------------------------Ans. (a) u1 = ( A cos kx1 + B sin kx1 )(C cos cL kt + D sin cL kt ) . At x1 = 0, n = −e1 , t = −Te1 = −(T11e1 + T21e2 + T31e3 ) = β cos ωte1 → T11 = − β cos ωt , T21 = T31 = 0 Since T11 = ( λ + 2 μ )( ∂u1 / ∂x1 ) , therefore, the boundary condition at x1 = 0 is:

( λ + 2μ )( ∂u1 / ∂x1 ) x1 =0 = − β cosωt , → D = 0, cL k = ω , BC = −

u1 = ( AC cos k l − AC =

β

k ( λ + 2μ )

β

→ ( λ + 2μ ) k ( B )(C cos cL kt + D sin cL kt ) = − β cosωt ,

k ( λ + 2μ )

, → u1 = ( AC cos kx1 −

At x1 = l , sin k l) cos ωt = 0 → AC cos k l −

k ( λ + 2μ )

β

k ( λ + 2μ )

sin kx1 ) cos ωt

sin k l = 0 →

⎡ β cL ωx β cL ωx ⎤ ωl tan k l → u1 = ⎢ tan cos 1 − sin 1 ⎥ cos ωt k ( λ + 2μ ) cL cL ω ( λ + 2 μ ) cL ⎥⎦ ⎢⎣ ω ( λ + 2μ ) (b) Resonance occurs at

β


5-18

β


ω = nπ cL / ( 2l ) , n = 1,3,5... __________________________________________________________________ 5.40 (a) Find the thickness-shear vibration if the left hand face x1 = 0 has a forced displacement u = (α cosωt )e3 and the right-hand face x1 = l is fixed. (b) Find the resonance frequencies. -------------------------------------------------------------------------------Ans. (a) Let u3 = ( A cos kx1 + B sin kx1 )(C cos cT kt + D sin cT kt ), u1 = u2 = 0 In the absence of body forces, the x3 Navier equation of motion (5.6.7) gives:

ρo

∂ 2u3 ∂t 2

= (λ + μ )

⎛ ∂2 ∂e ∂2 ∂2 + μ⎜ 2 + 2 + 2 ⎜ ∂x ∂x3 ⎝ 1 ∂x21 ∂x3

⎞ ∂ 2u3 ∂ 2u3 ⎟ u3 → ρ o = μ ⎟ ∂t 2 ∂x12 ⎠

leads to − ρo ( cT k ) u3 = − μ k 2u3 → ( cT ) = μ / ρo . 2

2

The boundary condition at x1 = 0 ,

u3 ( 0, t ) = (α cosωt ) → u3 = ( A)(C cos cT kt + D sin cT kt ) = α cos ωt → D = 0, AC = α , k = ω / cT . → u3 = (α cos kx1 + BC sin kx1 ) cos ωt . The boundary condition at x1 = l , u3 (l, t ) = 0 → u3 = (α cos k l + BC sin k l) cos ωt = 0 → BC = −α cot k l . → u3 = α [ cos(ω x1 / cT ) − cot(ω l / cT )sin(ω x1 / cT ) ] cos ωt .

(b) Resonance occurs at ω = nπ cT / l, n = 1, 2,3... __________________________________________________________________ 5.41 (a) Find the thickness-shear vibration if the left hand face x1 = 0 has a forced displacement u = α (cosωte 2 + sin ωte3 ) and the right-hand face x1 = l is fixed. (b) Find the resonance frequencies. ------------------------------------------------------------------------------Ans. (a) If the left hand face x1 = 0 has a forced displacement u = α cosωte 2 and the right-hand face x1 = l is fixed, it is clear from the result of the previous problem, u2 = α ⎡⎣ cos (ω x1 / cT ) − cot (ω l / cT ) sin (ω x1 / cT ) ⎤⎦ cos ωt , u1 = u3 = 0 .

If the left hand face x1 = 0 has a forced displacement u = α sin ωte3 and the right-hand face x1 = l is fixed, the displacement field is clearly given by u3 = α ⎡⎣cos (ω x1 / cT ) − cot (ω l / cT ) sin (ω x1 / cT ) ⎤⎦ sin ωt , u1 = u2 = 0

Thus, the solution to the present problem can be obtained by superposition to be u1 = 0 , u2 = α ⎡⎣ cos (ω x1 / cT ) − cot (ω l / cT ) sin (ω x1 / cT ) ⎤⎦ cos ωt , u3 = α ⎡⎣ cos (ω x1 / cT ) − cot (ω l / cT ) sin (ω x1 / cT ) ⎤⎦ sin ωt .

(b) Resonance occurs at ω = nπ cT / l, n = 1, 2,3... Copyright 2010, Elsevier Inc

5-19

Lai et al, Introduction to Continuum Mechanics __________________________________________________________________ 5.42 A cast iron bar, 200 cm long and 4 cm in diameter, is pulled by equal and opposite axial force P at its ends. (a) Find the maximum normal and shearing stresses if P=90,000N. (b) Find the total elongation and lateral contraction. ( EY = 103GPa.,ν = 0.3) -------------------------------------------------------------------------------Ans. A = π (4 × 10−2 )2 / 4 = 12.6 × 10−4 m 2 .

(

)

(a) (Tn )max = P / A = 90,000 / 12.6 × 10−4 = 71.4 × 106 N ,

(Ts )max = P / ( 2 A) = 35.7 × 106 N .

(b) δ l = ( P / A)(l / EY ) = (71.4 × 106 ) × 2 / (103 × 109 ) = 1.39 × 10−3 m,

δ d = −ν ( P / A)(d / EY ) = (0.3)(71.4 × 106 ) × (4 × 10−2 ) / (103 × 109 ) = −0.832 × 10−5 m. ____________________________________________________________________ 5.43 A composite bar, formed by welding two slender bars of equal length and equal crosssectional area, is loaded by an axial load P as shown in Figure below. If Young's moduli of the two portions are EY(1) and EY(2) , find how the applied force is distributed between the two halves. ------------------------------------------------------------------------------Ans. Taking the whole bar as a free body, let P1 be the compressive reactional force from the right wall to the bar and P2 be the compressive reactional force from the left wall to the bar, then the equation of static equilibrium requires

P = P1 − P2 .

(i) There is no net elongation of the composite bar, therefore,

P1l AEY(1)

+

P2 l AEY(2)

=0

(ii)

Combining Eq. (i) and (ii), we obtain −P . , P2 = P1 = (2) (1) (1) 1 + ( EY / EY ) 1 + ( EY / EY(2) ) _________________________________________________________________

P

(iii)

5.44 A bar of cross-sectional area A is stretched by a tensile force P at each end. (a) Determine the normal and shearing stresses on a plane with a normal vector which makes an angle α with the axis of the bar. (b) For what value of α are the normal and shearing stresses equal? (c) If the load carrying capacity of the bar is based on the shearing stress on the plane defined by α = α o to be less than τ o what is the maximum allowable load P ? -------------------------------------------------------------------------------


5-20


⎡σ Ans. [ T ] = ⎢⎢ 0 ⎢⎣ 0

0 0⎤ 0 0 ⎥⎥ , n = cos α e1 + sin α e 2 , 0 0 ⎥⎦

(a) For the plane with a normal given by n = cos α e1 + sin α e 2 , we have,

⎡σ [t ] = [T][n ] = ⎢⎢ 0 ⎢⎣ 0

0 0 ⎤ ⎡cos α ⎤ ⎡σ cos α ⎤ 0 0 ⎥⎥ ⎢⎢ sin α ⎥⎥ = ⎢⎢ 0 ⎥⎥ → t = σ cos α e1 → Tn = t ⋅ n = σ cos 2 α , 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

(

2

)

Ts2 = t − Tn2 = σ 2 cos 2 α − σ 2 cos 4 α = σ 2 cos 2 α 1 − cos 2 α = σ 2 cos 2 α sin 2 α

(b)

σ sin 2α

→ Ts = σ sin 2α / 2 . = σ cos 2 α → cos α sin α = cos 2 α → cos α ( sin α − cos α ) = 0 .

2 Thus, (i) cos α = 0 → α = π / 2 → Ts = Tn = 0, and (ii) sin α = cos α → α = π / 4 → Ts = Tn = σ / 2 . σ sin 2α o 2τ o 2τ o (c) ≤ τo → σ ≤ . Max allowable P ≤ A 2 sin 2α o sin 2α o _________________________________________________________________

5.45 A cylindrical bar, whose lateral surface is constrained so that there can be no lateral expansion, is then loaded with an axial compressive stress T11 = −σ . (a) Find T22 and T33 in terms of σ and the Poisson's ratio ν , (b) show that the effective Young's modulus

( EY )eff

≡ T11 / E11 is given by ( EY )eff = (1 − ν ) / (1 − ν − 2ν 2 ) . [note misprint in text].

------------------------------------------------------------------------------Ans. (a) E22 = 0 → T22 − ν (T33 + T11 ) = 0 , E33 = 0 → T33 − ν (T11 + T22 ) = 0 . Thus, T22 − ν T33 = −νσ and T33 − ν T22 = −νσ . From these two equations, we have, T22 = T33 = −νσ / (1 − ν ) .

(b) E11 =

2 2 ⎞ ⎤ −σ ⎡ ⎛ 2ν ⎞ ⎤ −σ ⎛ 1 − ν − 2ν ⎞ ⎟ ⎥ = E ⎢1 − ⎜⎜ 1 − ν ⎟⎟ ⎥ = E ⎜⎜ 1 − ν ⎟⎟ ⎠⎦ Y ⎢ Y ⎝ ⎠ ⎥⎦ ⎠ ⎣ ⎝ E (1 − ν ) EY (1 − ν ) −σ . = = Y = 2 E11 1 − ν − 2ν (1 − 2ν )(1 + ν )

1 1 ⎡ ⎛ νσ ⎡T11 − ν (T22 + T33 ) ⎦⎤ = −σ + 2ν ⎜ ⎢ ⎣ EY EY ⎣ ⎝ 1 −ν Thus, ( EY )eff ≡

T11 → ( EY )eff E11

__________________________________________________________________ 5.46

Let the state of stress in a tension specimen be given by T11 = σ and all other Tij =0 . (a)

Find the components of the deviatoric stress defined by To = T − (1 / 3) Tkk I . (b) Find the principal scalar invariants of To ------------------------------------------------------------------------------Ans. (a) Tkk = T11 + T22 + T33 = σ → T11o = T11 − Tkk / 3 = σ − σ / 3 = 2σ / 3 .


5-21

Lai et al, Introduction to Continuum Mechanics o o o T22 = T22 − Tkk / 3 = −σ / 3 = T33 , T12o = T13o = T23 = 0. (b) o o o I1 = T11 + T22 + T33 = 2σ / 3 − σ / 3 − σ / 3 = 0. o o o o + T22 = ( 2σ / 3)( −σ / 3) + ( −σ / 3)( −σ / 3) + ( 2σ / 3)( −σ / 3) = −σ 2 / 3. I 2 = T11oT22 T33 + T11oT33 o o I 3 = T11oT22 T33 = ( 2σ / 3)( −σ / 3)( −σ / 3) = 2σ 3 / 27.

_________________________________________________________________ 5.47 A circular cylindrical bar of length l hangs vertically under gravity force from the ceiling. Let x1 axis coincides with the axis of the bar and points downward and let the

point ( x1 , x2 , x3 ) = ( 0,0,0 ) be fixed at the ceiling. (a) Verify that the following stress field

satisfies the equations of equilibrium in the presence of the gravity force: T11 = ρ g ( l − x1 ) , all other Tij = 0 and (b) verify that the boundary conditions of zero surface traction on the lateral face and the lower end face are satisfied and (c) obtained the resultant force of the surface traction at the upper face. ------------------------------------------------------------------------------Ans. (a) The body force per unit volume is given by ρ B = ρ ge1 . Thus, with T11 = ρ g ( l − x1 ) , we have, ∂T11 ∂T12 ∂T13 + + + ρ g = − ρ g + 0 + 0 + ρ g = 0 and the other two equations are trivially ∂x1 ∂x2 ∂x3 satisfied. (b) On the bottom end face x1 = l , n = e1 , t = Te1 = T11 x =l e1 = ρ g ( l − l ) e1 = 0 . 1

⎡T11 0 0 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ On the lateral face, n = n2e 2 + n3e3 , [ t ] = ⎢⎢ 0 0 0 ⎥⎥ ⎢⎢ n2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ → t = 0 . ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ n3 ⎥⎦ ⎢⎣ 0 ⎥⎦ (c) On the top face at x1 = 0 , n = −e1 , t = −Te1 = −T11 x =0 e1 = − ρ g ( l − 0 ) e1 = − ρ g le1 1

Let the area of the face be A , then the resultant force is tA = − ρ g lAe1 = −We1 where W = ρ gAl is the weight of the bar and the minus sign indicates that the resultant force at the ceiling is upward which balances the weight of the bar. __________________________________________________________________ 5.48 A circular steel shaft is subjected to twisting couples of 2700 Nm . The allowable tensile stress is 0.124 GPa . If the allowable shearing stress is 0.6 times the allowable tensile stress, what is the minimum allowable diameter? ------------------------------------------------------------------------------Ma 2M t Ans. (Tn )max = (Ts )max = t = M t a / π a 4 / 2 = . Thus Ip π a3

(


5-22

)

Lai et al, Introduction to Continuum Mechanics 2 ( 2700 )

πa

3

(

)

≤ ( 0.6 ) 0.124 × 109 → a3 ≥

2 ( 2700 )

(

π ( 0.6 ) 0.124 × 10

9

)

= 23.1 × 10−6 m3

→ a ≥ 2.85 × 10−2 m = 2.85 cm → d ≥ 5.7 cm. _________________________________________________________________ 5.49

In Figure 5P.2, a twisting torque M t is applied to the rigid disc A . Find the twisting moments transmitted to the circular shafts on either side of the disc.

Figure 5P.2 ------------------------------------------------------------------------------Ans. Let M1 and M 2 be the twisting moments transmitted to the left and the right shaft respectively. Then equilibrium of the disc demands that M1 + M 2 = M t (i) In addition, the disc is rigid, therefore, the angle of twist of the left shaft at the disc relative to the left wall must equal the angle of twist of the right shaft at the disc relative to the right wall, i.e., M1l1 M 2 l 2 = → M1l1 = M 2 l 2 (ii) μI p μI p Thus,

⎛ l2 ⎞ ⎛ l1 ⎞ M1 = ⎜ ⎟ Mt , M2 = ⎜ ⎟ Mt ⎝ l1 + l 2 ⎠ ⎝ l1 + l 2 ⎠ for l1 = l 2 , M1 = M 2 = M t / 2. _________________________________________________________________

(iii)

5.50 What needs to be changed in the solution for torsion of a solid circular bar obtained in Section 5.14 for it to be valid for torsion of a hollow circular bar with inner radius a and outer radius b ? ------------------------------------------------------------------------------Ans. The hollow circular bar differs from the solid circular bar in that there is an inner lateral surface which is also traction free. However, the normal to the inner lateral surface differs from that to the outer surface only by a sign so that the zero surface traction in the inner surface is also satisfied since that for the outer surface is satisfied. However, in calculating the resultant force and resultant moment due to the surface traction on the end faces, the integrals are now to be integrated over the circular ring area between by r = a and r = b rather than the whole solid


5-23

Lai et al, Introduction to Continuum Mechanics circular area of radius b . Thus, the only change that needs to be made is that the polar area second moment I p is now given by I p =

π

(

)

b4 − a4 . 2 __________________________________________________________________

5.51 A circular bar of radius ro is under the action of an axial tensile load P and a twisting couple of M t . (a) Determine the stress throughout the bar. (b) Find the maximum normal and shearing stress ------------------------------------------------------------------------------M P Ans. Superpose the solutions for tension and for torsion, we have, with σ = , β ≡ t A Ip (a) T11 = σ , T12 = T21 = − β x3 , T13 = T31 = β x2 , all otherTij = 0 . (b) The characteristic equation is

σ − λ − β x3 β x2 2 2 − β x3 −λ 0 = 0 → σλ 2 − λ 3 + λ ( β x2 ) + λ ( β x3 ) = 0 → λ λ 2 − σλ − β 2 r 2 = 0, where r 2 = x22 + x32 β x2 −λ 0

(

Thus, λ1,2 =

σ ± σ 2 + 4β 2 r 2 2

σ + σ 2 + 4β 2 r 2

)

, λ3 = 0 . Thus

1 σ 2 + 4β 2 r 2 . 2 2 __________________________________________________________________

(Tn )max =

, Ts =

5.52 Compare the twisting torque which can be transmitted by a shaft with an elliptical crosssection having a major diameter equal to twice the minor diameter with a shaft of circular cross-section having a diameter equal to the major diameter of the elliptical shaft. Both shafts are of the same material. Also compare the unit twist (i.e., twist angle per unit length) under the same twisting moment. Assume that the maximum twisting moment which can be transmitted is controlled by the maximum shearing stress. ---------------------------------------------------------------------------------Ans. (a) For an elliptical shaft with major diameter 2b and minor diameter 2a (i.e., b > a ), 2 ( M t )ell . (Ts )max = π a 2b 2 ( M t )cir For a circular shaft with radius b , (Ts )max = , thus π b3 2 ( M t )ell 2 ( M t )cir ( M t )ell ⎛ a ⎞2 ⎛ a ⎞2 1 , = → = = = . (Ts )max = ( M t )cir ⎜⎝ b ⎟⎠ ⎜⎝ 2a ⎟⎠ 4 π a 2b π b3 (b) α 'ell

⎛ a2 + b2 ⎞ M , =⎜ ⎜ μπ a3b3 ⎟⎟ t ⎝ ⎠

α 'cir

⎛ 2 = ⎜⎜ 4 ⎝ μπ b


5-24

(

)

( )

2 2 ( 2 a ) 5a 2 ⎞ α 'ell b a + b = =5 . = ⎟⎟ M t , thus, α 'cir 2a 3 2a 3 ⎠

Lai et al, Introduction to Continuum Mechanics __________________________________________________________________ 5.53

Repeat the previous problem except that the circular shaft has a diameter equal to the minor diameter of the elliptical shaft. -------------------------------------------------------------------------------Ans. (a) For an elliptical shaft with major diameter 2b and minor diameter 2a (i.e., b > a ), 2 ( M t )ell . (Ts )max = π a 2b 2 ( M t )cir For a circular shaft with radius a , (Ts )max = , thus, π a3 2 ( M t )ell 2 ( M t )cir ( M t )ell ⎛ b ⎞ ⎛ 2a ⎞ = → = = = 2. , (Ts )max = 2 3 ( M t )cir ⎜⎝ a ⎟⎠ ⎜⎝ a ⎟⎠ πa b πa

⎛ a2 + b2 ⎞ ⎛ 2 ⎞ α 'ell ⎛ 5a 2 ⎞⎛ a 4 ⎞ 5 (b) α 'ell = ⎜ , ' M M = → =⎜ α ⎟ ⎟⎜ ⎟⎟ = . ⎜⎜ cir 4 ⎟ ⎟ t ⎜ μπ a3b3 ⎟ t α 'cir ⎜⎝ 8a 6 ⎟⎜ ⎝ μπ a ⎠ ⎝ ⎠ ⎠⎝ 2 ⎠ 16 __________________________________________________________________ 5.54

Consider torsion of a cylindrical bar with an equilateral triangular cross-section as

(

)

shown in Fig. P.5.3. (a) Show that a warping function ϕ = C 3x22 x3 − x33 generate an equilibrium stress field. (b) Determine the constant C , so as to satisfy the traction free boundary condition on the lateral surface x2 = a . With C so obtained, verify that the other two lateral surfaces are also traction free. (c) Evaluate the shear stress at the corners and along the line x3 = 0 .(d) Along the line x3 = 0 where does the greatest shear stress occur? -------------------------------------------------------------------------------. ∂ϕ ∂ 2ϕ ∂ 2ϕ ∂ 2ϕ = = − + = 0 , so that equations of equilibrium are Ans. (a) 6 Cx , 6 Cx , thus 3 3 ∂x22 ∂x32 ∂x22 ∂x32 satisfied. 2

x3

(−2a,0)

(a,0)

x2

(b) For the lateral surface x2 = a , n =e2 , t = Te 2 = T12e1 = ⎣⎡ − μ x3α '+ μ ( ∂ϕ / ∂x2 ) ⎦⎤

→ x3α ' = [ 6Cx2 x3 ]x

(

)

2 =a

e x2 = a 1

=0

→ x3α ' = 6Cax3 → C = α '/ 6a .

(

On the lateral surface x3 = 1 / 3 ( x2 + 2a ) → 3 x3 − x2 = 2a → n = (1 / 2 ) −e 2 + 3e3

(

)

(

)

(

)

t = Tn = (1 / 2 ) −Te 2 + 3Te3 = (1 / 2 ) −T12 + 3T13 e1 . Now, for ϕ = (α '/ 6a ) 3 x22 x3 − x33


5-25

)

Lai et al, Introduction to Continuum Mechanics T12 = − μ x3α '+ μ (∂ϕ / ∂x2 ) = − μ x3α '+ μ (α '/ a )( x2 x3 ),

, T13 = μ x2α '+ μ (∂ϕ / ∂x3 ) = μ x2α '+ μ (α '/ 2a)( x22 − x32 ). μα ' ⎡ Therefore, −T12 + 3T13 = 2ax3 − 2 x2 x3 + 2 3ax2 + 3 x22 − x32 ⎤ . With x2 = 3 x3 − 2a , ⎦ 2a ⎣

(

−T12 + 3T13 =

μα ' ⎡

μα ' ⎡

2ax3 − 2 2a ⎢⎣

(

(

)

3 x3 − 2a x3 + 2 3a

) (

(

)

)

)

⎛ 3 x3 − 2a + 3 ⎜ ⎝

(

3 x3 − 2a

(

)

2

⎞⎤ − x32 ⎟ ⎥ ⎠⎦

)

2ax3 + −2 3 x32 + 4ax3 + 6ax3 − 4 3a 2 + 3 3 x32 − 4 3ax3 + 4a 2 − x32 ⎤ ⎦ 2a ⎣ μα ' ⎡ 2ax3 + 4ax3 + 6ax3 − 12ax3 − 4 3a 2 + 4 3a 2 + −2 3 x32 + 3 3 x32 − 3 x32 ⎤ = 0. = ⎦ 2a ⎣

=

(

) (

(

)

)

That is, on x3 = 1 / 3 ( x2 + 2a ) , t = 0 . Clearly, for the lateral surface

(

)

x3 = − 1 / 3 ( x2 + 2a ) , t = 0 .

(c) at the corner ( −2a,0 ) , T12 = − μ x3α '+ μ (α '/ a ) ( x2 x3 ) = 0 and

(

)

( ) At the corners ( a, ± 3a ) , T = μα ' ( x / a ) ( x − a ) = μα ' ( ± 3 ) ( a − a ) = 0 and T = μα ' ⎡ x + ( x − x ) / 2a ⎤ = μα ' ⎡ a + ( a − 3a ) / 2a ⎤ = 0 . ⎣ ⎦ ⎣ ⎦ T13 = μ x2α '+ μ (α '/ 2a ) x22 − x32 = ( −2a ) μα '+ μ (α '/ 2a ) 4a 2 = 0 . 12

13

2

2 2

3

2

2 3

2

2

That is, the shear stress at all three corners are zero. Along x3 = 0 , T12 = − μ x3α '+ μ (α '/ a ) ( x2 x3 ) = 0 ,

(

)

(

)

T13 = μα ' ⎡ x2 + x22 − x32 / 2a ⎤ = ( μα '/ 2a ) 2ax2 + x22 . ⎣ ⎦

(d) dT13 / dx2 = ( μα '/ 2a )( 2a + 2 x2 ) = 0 → x2 = − a → (T13 ) x

(

2 =− a

)

= μα ' a / 2 .

( )

But at ( x2 , x3 ) = ( a,0 ) , T13 = μα ' ⎡ x2 + x22 − x32 / 2a ⎤ = μα ' ⎡ a + a 2 / 2a ⎤ = ( 3a / 2 ) μα ' . ⎣ ⎦ ⎣ ⎦ Thus, along x3 = 0 , the greatest shear stress occurs at ( x2 , x3 ) = ( a,0 ) with Ts = ( 3a / 2 ) μα ' . __________________________________________________________________ 5.55

Show from the compatibility equations that the Prandtl's stress function ψ ( x2 , x3 ) for torsion problem must satisfy the equation

∂ 2ψ ∂x33

+

∂ 2ψ ∂x22

= constant

------------------------------------------------------------------------------∂ψ ∂ψ , T13 = − , and all other Tij = 0 , we have, the nonzero strain components Ans. With T12 = ∂x3 ∂x2 1 ∂ψ 1 ∂ψ , E13 = − , all other Eij = 0 . All equations of compatibility are are: E12 = 2 μ ∂x3 2 μ ∂x2 identically satisfied except the following two:


5-26

Lai et al, Introduction to Continuum Mechanics ∂ 2 E33 ∂ 2 E22 ∂ ⎛ ∂E31 ∂E12 ∂E23 ⎞ ∂ ⎛ ∂E12 ∂E23 ∂E31 ⎞ = + + = + + ⎜− ⎟, ⎜− ⎟ ∂x3∂x1 ∂x2 ⎝ ∂x2 ∂x3 ∂x1 ⎠ ∂x1∂x2 ∂x3 ⎝ ∂x3 ∂x1 ∂x2 ⎠ which leads to ∂ ⎛∂ ψ ∂ψ ⎞ ∂ ⎛ ∂ 2ψ ∂ 2ψ + ⎜ ⎜⎜ 2 + 3 ⎟⎟ = 0 , ∂x3 ⎜⎝ ∂x33 ∂x22 ∂x2 ⎝ ∂x2 ∂x3 ⎠ 2

3

⎞ ∂ 2ψ ∂ 2ψ ⎟⎟ = 0 → 3 + 2 = constant ∂x3 ∂x2 ⎠

________________________________________________________________________

5.56

Given that the Prandtl' stress function for a rectangular bar in torsion is given by

⎛ 32 μα ' a 2 ⎞ ∞ 1 nπ x2 ⎪ cosh ( nπ x3 / 2a ) ⎫⎪ ( n −1) /2 ⎧ ⎟⎟ ∑ 3 ( −1) ⎨1 − ⎬ cos 3 cosh ( nπ b / 2a ) ⎭⎪ 2a ⎝ π ⎠ n =1,3,5 n ⎩⎪ The cross section is defined by − a ≤ x2 ≤ a and − b ≤ x3 ≤ b . Assume b > a , (a) Find the maximum shearing stress. (b) Find the maximum normal stress and the plane it acts. --------------------------------------------------------------------------------

ψ = ⎜⎜

Ans. We know that when a rectangular membrane, fixed on its side, is subjected to a uniform pressure on one side of the membrane, the deformed surface has a maximum slope at the mid point of the longer side. Thus, based on the membrane analogy discussed in Example 5.17.3, on any plane x1 = constant , the maximum shearing stress occurs on the mid point of the longer side. That is at the point x2 = a and x3 = 0 . From the given functionψ ( x2 , x3 ) , we obtain the stress components as 2 ∞ ⎛ 32μα ' a ⎞ nπ x2 ∂ψ 1 ⎪ sinh ( nπ x3 / 2a ) ⎫⎪ ( n −1) /2 ⎛ nπ ⎞ ⎧ = −⎜ −1) T12 = cos . ( ⎟ ⎨ ⎬ ∑ ⎜ ⎟ 3 3 ⎜ π ⎟ ∂x3 2a ⎝ 2a ⎠ ⎩⎪ cosh ( nπ b / 2a ) ⎭⎪ ⎝ ⎠ n =1,3,5 n T13 = −

⎞ ∞ 1 ⎪ cosh ( nπ x3 / 2a ) ⎫⎪ nπ x2 ( n −1) /2 nπ ⎧ . ⎟⎟ ∑ 3 ( −1) ⎨1 − ⎬ sin 2a ⎩⎪ cosh ( nπ b / 2a ) ⎭⎪ 2a ⎠ n =1,3,5 n ( n +3) /2 At x2 = a, x3 = 0, note sin ( nπ / 2 ) = ( −1) , n = 1,3,5..

∂ψ ⎛ 32 μα ' a 2 =⎜ ∂x2 ⎜⎝ π 3

(

)

1 ⎪⎫ ⎛ 16μα ' a ⎞ ∞ 1 ⎧⎪ T13 = ⎜ ⎟ ∑ 2 ⎨1 − cosh nπ b / 2a ⎬, T12 = 0 2 ( ) ⎭⎪ ⎝ π ⎠ n =1,3,5 n ⎩⎪ That is ∞ ⎫⎪ 16 μα ' a 1 ⎧ 1 (Ts )max = ⎛⎜ 2 ⎞⎟ ∑ 2 ⎪⎨1 − ⎬. ⎝ π ⎠ n =1,3,5 n ⎩⎪ cosh ( nπ b / 2a ) ⎭⎪ ∞

Or, since

∑

1

n =1,3,5 n

2

=

⎛ 16μα ' a ⎞ ∞ 1 , therefore (Ts )max = 2 μα ' a − ⎜ ⎟ ∑ 8 ⎝ π 2 ⎠ n =1,3,5 n 2

π2

⎧⎪ ⎫⎪ 1 ⎨ ⎬ ⎪⎩ cosh ( nπ b / 2a ) ⎪⎭

Since at this point, the only nonzero stress components are T13 and T31 ( = T13 ) , therefore, the characteristic equation is −λ 3 + λT132 = 0 so that the maximum normal stress is

(Tn )max = T13 = (Ts )max , which acts on plane whose normal is in the direction (1 / Copyright 2010, Elsevier Inc

5-27

)

2 ( e1 ± e3 ) .


Obtain the relationship between the twisting moment M t and the twist angle per unit length α ' for a rectangular bar under torsion. Note: 1 +

1

+

1

+ ... =

π4

96 34 54 ------------------------------------------------------------------------------Ans. We have, [see Eq. (5.18.10)], 2 ⎛ 32 μα ' a ⎞ ∞ 1 nπ x2 ⎪ cosh ( nπ x3 / 2a ) ⎫⎪ ( n −1) /2 ⎧ . ψ = ⎜⎜ ⎟⎟ ∑ 3 ( −1) ⎨1 − ⎬ cos 3 cosh ( nπ b / 2a ) ⎭⎪ 2a ⎝ π ⎠ n =1,3,5 n ⎩⎪

⎛ 32μα 'a 2 Thus, if let A = ⎜ ⎜ π3 ⎝

.

⎞ cosh ( nπ x3 / 2a ) , then, we have, ⎟⎟ and F ( x3 ) = cosh ( nπ b / 2a ) ⎠

M t = 2∫ψ dA = ∞ ∞ ⎛ ⎞ ⎛ ⎞ a nπ x2 nπ x2 ⎡ b 1 1 ( n −1) /2 ( n −1) /2 a 2b ) ∫ cos dx2 ⎟ − ⎜ 2 A ∑ 3 ( −1) cos F ( x3 )dx3 ⎤ dx2 ⎟ ⎜ 2 A ∑ 3 ( −1) ( ∫ ∫ −a −a ⎜ n =1,3,5 n ⎟ ⎜ n =1,3,5 n ⎟ ⎦⎥ 2a 2a ⎣⎢ −b ⎝ ⎠ ⎝ ⎠ ≡ M − N.

nπ x2 nπ ⎛ 4a ⎞ ⎛ 2a ⎞ =⎜ dx2 = ⎜ 2sin ( −1)( n+3)/2 , n = 1,3,5 , therefore, ⎟ ⎟ 2a 2 ⎝ nπ ⎠ ⎝ nπ ⎠ ∞ ∞ ∞ a nπ x 1 8a 1 32μα ' 1 ( n −1) /2 M = 2 A ∑ 3 ( −1) ( 2b ) ∫− a cos 2 dx2 = A ⎛⎜ ⎞⎟ ( 2b ) ∑ 4 = ⎛⎜ 4 ⎞⎟ ( 2a )3 ( 2b ) ∑ 4 2a ⎝π ⎠ ⎝ π ⎠ n =1,3,5 n n =1,3,5 n n =1,3,5 n Now,

Next,

a

∫−a cos

b

b

∫−b F ( x3 )dx3 = ∫−b

. nπ b ⎛ 2a ⎞ 2sinh ( nπ b / 2a ) ⎛ 4a ⎞ =⎜ , so that dx3 = ⎜ ⎟ ⎟ tanh π π π cosh ( nπ b / 2a ) n cosh n b / 2 a n 2a ( ) ⎝ ⎠ ⎝ ⎠

cosh ( nπ x3 / 2a )

⎛ 64μα ' ( 2a )4 ⎞ ∞ 1 nπ x2 ⎡ b nπ b ⎤ ⎟ ∑ N = 2 A ∑ 3 ( −1) cos F ( x3 )dx3 dx2 = ⎜ tanh ∫ ∫ 5 5 ⎢ ⎥ −a ⎜ ⎟ n =1,3,5 n ⎦ 2a ⎣ −b 2a π n =1,3,5 n ⎝ ⎠ Thus, 4 ∞ 1 ⎛ 64μα ' ( 2a ) ⎞ ∞ 1 nπ b ⎛ 32 μα ' ⎞ 3 ⎜ ⎟ ∑ Mt ≡ M − N = ⎜ tanh ( 2a ) ( 2b ) ∑ 4 − 4 ⎟ 5 5 ⎜ ⎟ n =1,3,5 n 2a π ⎝ π ⎠ n =1,3,5 n ⎝ ⎠ ∞

1

( n −1) /2

a

4 ⎛ π 4 ⎞ ⎛ 64μα ' ( 2a ) 3 ⎛ 32 μα ' ⎞ ⎜ 2 a 2 b =⎜ − ⎟ ( ) ( ) ⎜⎜ 96 ⎟⎟ ⎜ π5 ⎝ π4 ⎠ ⎝ ⎠ ⎝

⎞ ∞ 1 nπ b ⎟ ∑ tanh . ⎟ n =1,3,5 n5 2a ⎠

Or, ∞ ⎡ 192 a 1 nπ b ⎤ ⎛ μα ' ⎞ Mt = ⎜ ( 2a )3 ( 2b ) ⎢1 − 5 ⎛⎜ ⎞⎟ ∑ 5 tanh ⎥ ⎟ 2a ⎥⎦ ⎝ 3 ⎠ ⎢⎣ π ⎝ b ⎠ n =1,3,5 n _________________________________________________________________


5-28


In pure bending of a bar, let M L = M 2e 2 + M 3e3 = −M R , where e 2 and e3 are not along the principal axes, show that the flexural stress T11 is given by T11 = −

M 2 I 23 + M 3 I 22

(I

33 I 22

− I 23

2

)

x2 +

M 2 I 33 + M 3 I 23

(I

33 I 22

− I 232

)

x3

------------------------------------------------------------------------------Ans. Refer to Section 5.19, we had [see Eq.(5.19.4)(5.19.6) and (5.19.7)] T11 = β x2 + γ x3 , where M 2 = β I 23 + γ I 22 , M 3 = − β I 33 − γ I 23 Solving the above two equations for β and γ , in terms of M 2 and M 3 , we obtain M I + M 3 I 22 M I + M 3 I 23 β = − 2 23 and γ = 2 33 . 2 2 I 33 I 22 − I 23 I 33 I 22 − I 23

(

)

Thus, T11 = −

(

M 2 I 23 + M 3 I 22

(I

33 I 22

− I 232

)

x2 +

)

M 2 I 33 + M 3 I 23

(I

33 I 22

− I 232

)

x3 .

_________________________________________________________________ 5.59 From the strain components for pure bending M x ν M 2 x3 E11 = 2 3 , E22 = E33 = − , E12 = E13 = E23 = 0 I 22 EY I 22 EY Obtain the displacement field ------------------------------------------------------------------------------Ans. Integration of ∂u1 / ∂x1 = Ax3 , ∂u2 / ∂x2 = −ν Ax3 , ∂u3 / ∂x3 = −ν Ax3 , whereA ≡

M2 gives I 22 EY

u1 = Ax3 x1 + f1 ( x2 , x3 ) , u2 = −ν Ax3 x2 + f 2 ( x1 , x3 ) , u3 = −ν Ax32 / 2 + f3 ( x1 , x2 )

(i)

where f1 ( x2 , x3 ) , f 2 ( x1 , x3 ) and f3 ( x1 , x2 ) are integration functions. Substituting (i) into

∂u1 / ∂x2 + ∂u2 / ∂x1 = 0, ∂u1 / ∂x3 + ∂u3 / ∂x1 = 0 and ∂u2 / ∂x3 + ∂u3 / ∂x2 = 0 , we obtain ∂f1 ( x2 , x3 ) / ∂x2 = −∂f 2 ( x1 , x3 ) / ∂x1 = g1 ( x3 )

∂f1 ( x2 , x3 ) / ∂x3 = −∂f3 ( x1 , x2 ) / ∂x1 = g 2 ( x2 )

(ii)

∂f 2 ( x1 , x3 ) / ∂x3 = −∂f3 ( x1 , x2 ) / ∂x2 = g3 ( x1 )

where g ( x1 ) , g ( x2 ) , g ( x3 ) are integration functions. Integrations of (ii) give, f1 = g1 ( x3 ) x2 + g 4 ( x3 ) and f1 = g 2 ( x2 ) x3 + g 6 ( x2 )

− f 2 = g1 ( x3 ) x1 + g5 ( x3 ), and f 2 = g3 ( x1 ) x3 + g8 ( x1 )

− f3 = g 2 ( x2 ) x1 + g 7 ( x2 ) and − f3 = g3 ( x1 ) x2 + g9 ( x1 )

From (iii), g1 ( x3 ) = a1 x3 + b1 , g 2 ( x2 ) = a1 x2 + b2 , g 4 ( x3 ) = b2 x3 + c2 , g6 ( x2 ) = b1 x2 + c2 From (iv) and (vi), g3 ( x1 ) = − a1 x1 + b3 , g8 ( x1 ) = −b1 x1 + c3 , − g5 ( x3 ) = b3 x3 + c3

From (v) (vi),(vii), a1 = 0, g9 ( x1 ) = b2 x1 + c4 , g 7 ( x2 ) = b3 x2 + c4 Thus, f1 = b1 x2 + b2 x3 + c2 , f 2 = b3 x3 − b1 x1 + c3 , f3 = −b2 x1 − b3 x2 − c4


5-29

(iii) (iv) (v) (vi) (vii) (viii) (ix)

Lai et al, Introduction to Continuum Mechanics So that, M2 ν M2 u1 = x3 x1 + b1 x2 + b2 x3 + c2 , u2 = − x3 x2 + b3 x3 − b1 x1 + c3 , I 22 EY I 22 EY u3 = −

ν M2

(x)

− b2 x1 − b3 x2 − c4 . 2 I 22 EY _________________________________________________________________ x32

5.60 In pure bending of a bar, let M L = M 2e2 + M 3e3 = −M R , where e2 and e3 are along the principal axes, show that the neutral axis, (that is, the axis on the cross section where the flexural stress T11 is zero) is, in general, not parallel to the couple vectors. -------------------------------------------------------------------------------M x M x Ans. From Eq.(5.19.10), we have, T11 = 2 3 − 3 2 , thus the neutral axis is given by: I 22 I 33 M 2 x3 M 3 x2 x ⎛I ⎞M − = 0 . That is, the neutral axis is given by 3 = ⎜ 22 ⎟ 3 . Thus, only when I 22 I 33 x2 ⎝ I 33 ⎠ M 2 I 22 = I 33 is the neutral axis parallel to the couple vector M L = M 2e2 + M 3e3 = −M R . __________________________________________________________________ 5.61

For plane strain problem, derive the bi-harmonic equation for the Airy stress function ------------------------------------------------------------------------------Ans. We have [Eq.(5.20.7)}

E11 =

1 ⎡ EY ⎣

E12 = − EY

∂ 2 E11 ∂x22

(

2 ⎢ 1 −ν

1 EY

) ∂x

∂ 2ϕ

− ν (1 + ν )

2 2

(1 + ν )

∂ 2ϕ ∂x1∂x2

∂ 2ϕ ⎤

1 ⎡

∂x12 ⎦

EY ⎣

⎥ , E22 =

(

2 ⎢ 1 −ν

) ∂x

∂ 2ϕ 2 1

− ν (1 + ν )

∂ 2ϕ ⎤

⎥,

∂x22 ⎦

E13 = E23 = E33 = 0.

,

4 ⎡ ∂ 2 E22 ⎡ ∂ 4ϕ ∂ 4ϕ ⎤ ∂ 4ϕ ⎤ 2 ∂ ϕ ν ν ν ν ν E 1 , 1 1 = ⎢ 1 −ν 2 − + = − − + ( ) ( ) ⎥ ⎢ ⎥, Y ∂x24 ∂x22 ∂x12 ⎦⎥ ∂x12 ∂x14 ∂x12 ∂x22 ⎦⎥ ⎢⎣ ⎣⎢

(

)

2 EY

(

∂ 2 E12 ∂x22 ∂x12

= −2(1 + ν )

∂ 2ϕ ∂x12 ∂x22

. Thus, the compatibility equation

⎛ ∂ 2 E11 ∂ 2 E22 ∂ 2 E12 + − 2 ⎜⎜ 2 ∂x22 ∂x22 ∂x12 ⎝ ∂x2

⎡ → ⎢ 1 −ν 2 ⎢⎣

(

⎧

)

⎞ ⎟⎟ = 0 → ⎠

⎫

4

⎭

2 2

⎤ ⎥ = 0, 2 1 ⎥ ⎦

) ⎪⎨⎪ ∂∂xϕ + ∂∂xϕ ⎪⎬⎪ + {2 (1 +ν ) − 2ν (1 +ν )} ∂x∂ ∂ϕx ⎩

4

4

4 2

4 1

⎛ ∂ 4ϕ ∂ 4ϕ ⎛ ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ ⎞ ∂ 4ϕ ⎞ → 1 −ν 2 ⎜ 4 + 4 + 2 2 2 ⎟ = 0 → ⎜ 4 + 4 + 2 2 2 ⎟ = 0 ⎜ ∂x ⎜ ∂x ∂x2 ∂x1 ⎟⎠ ∂x2 ∂x1 ⎟⎠ ⎝ 2 ∂x1 ⎝ 2 ∂x1 __________________________________________________________________

(

)


5-30


For plane stress problem, derive the bi-harmonic equation for the Airy stress function ------------------------------------------------------------------------------Ans. 2 2 1 ⎛∂ ϕ 1 ⎛ ∂ 2ϕ (1 + ν ) ∂ 2ϕ ∂ ϕ⎞ ∂ 2ϕ ⎞ From E11 = , [Eq. (5.22.3)] ⎜⎜ 2 − ν 2 ⎟⎟ , E22 = ⎜⎜ 2 − ν 2 ⎟⎟ , E12 = − EY ⎝ ∂x2 EY ⎝ ∂x1 EY ∂x1∂x2 ∂x1 ⎠ ∂x2 ⎠ we get ⎛ ∂ 2 E11 ∂ 2 E22 EY ⎜ + ⎜ ∂x 2 ∂x12 2 ⎝

⎞ ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ ⎟⎟ = 4 + 4 − 2ν 2 2 , ∂x1 ∂x2 ⎠ ∂x1 ∂x2

The compatibility equation →

∂ 4ϕ

∂ 4ϕ

− 2ν

∂ 4ϕ

∂ 2ϕ

∂ 2 E11 ∂x22 − 2ν

+

2 EY

∂ 2 E22 ∂x12

∂ 2ϕ

∂ 2 E12 ∂ 2ϕ ∂ 2ϕ = −2 2 2 − 2ν 2 2 ∂x1∂x2 ∂x1 ∂x2 ∂x1 ∂x2 =2

∂ 2 E12 then gives ∂x1∂x2 ∂ 4ϕ

= −2

→ Thus,

+

∂ 4ϕ

+2

∂ 2ϕ

=0 ∂x14 ∂x24 ∂x12 ∂x22 ∂x14 ∂x24 ∂x12 ∂x22 ∂x12 ∂x22 ∂x12 ∂x22 __________________________________________________________________ +

5.63 Consider the Airy stress function ϕ = α1 x12 + α 2 x1 x2 + α 3 x22 . (a) Verify that it satisfies the bi-harmonic equation. (b) Determine the in-plane stresses T11 , T12 and T22 . (c) Determine and sketch the tractions on the four rectangular boundaries x1 = 0, x1 = b, x2 = 0, x2 = c .(d) As a plane strain solution, determine T13 , T23 , T33 and all the strain components . (e) As a plane stress solution, determine T13 , T23 , T33 , and all the strain components . ------------------------------------------------------------------------------Ans. (a) ∂ ϕ / ∂x14 = 0, ∂ 4ϕ / ∂x24 = 0, ∂ 4ϕ / ∂x12 ∂x22 = 0 , thus ϕ = α1 x12 + α 2 x1 x2 + α 3 x22 satisfies the bi-harmonic equation. 4

(b) T11 = ∂ 2ϕ / ∂x22 = 2α 3, T12 = − ∂ 2ϕ / ∂x1∂x2 = −α 2, T22 = ∂ 2ϕ / ∂x12 = 2α1 (c) On x1 = 0, t = −Te1 = − (T11e1 + T21e2 ) = −2α 3e1 + α 2e2 , on x1 = b, t = Te1 = T11e1 + T21e2 = 2α 3e1 − α 2e2 , on x2 = 0, t = −Te 2 = − (T12e1 + T22e2 ) = α 2e1 − 2α1e2 , on x1 = c, t = Te2 = T12e1 + T22e2 = −α 2e1 + 2α1e2 .


5-31


2 α1

α2

α2 2 α3

2 α3

c

α2

b

α2

2 α1 (d) As a plane strain solution, T33 = ν (T11 + T22 ) = 2ν (α 3 + α1 ) , T13 = T23 = 0, E13 = E23 = E33 = 0, E11 = (1 / EY ) ⎡(1 − ν 2 )T11 − ν (1 + ν ) T22 ⎤ = 2 (1 / EY ) ⎡ (1 − ν 2 )α 3 − ν (1 + ν )α1 ⎤ , ⎣ ⎦ ⎣ ⎦ E22 = (1 / EY ) ⎡(1 − ν 2 )T22 − ν (1 + ν ) T11 ⎤ = 2 (1 / EY ) ⎡ (1 − ν 2 )α1 − ν (1 + ν )α 3 ⎤ , ⎣ ⎦ ⎣ ⎦ E12 = (1 / EY )(1 + ν ) T12 = − (1 / EY )(1 + ν )α 2,

T33 = T13 = T23 = 0, E13 = E23

(e) As a plane stress solution, = 0, E11 = (1 / EY )(T11 − ν T22 ) = 2 (1 / EY ) (α 3 − να1 ) ,

E22 = (1 / EY )(T22 − ν T11 ) = 2 (1 / EY ) (α1 − να 3 ) , E12 = (1 / EY )(1 + ν ) T12 = −α 2 (1 + ν ) / EY = −α 2 / 2μ . E33 = (1 / EY ) ⎡⎣ −ν (T11 + T22 ) ⎤⎦ = −2 (ν / EY ) (α 3 + α1 ) . Note, for this problem, since T11 + T22 is a linear function of x1 and x2 , in fact, a constant, therefore, all the compatibility equations are satisfied so that E33 is meaningful and u3 does exist. _________________________________________________________________ 5.64 Consider the Airy stress function ϕ = α x12 x2 . (a) Verify that it satisfies the bi-harmonic equation. (b) Determine the in-plane stresses T11 , T12 and T22 . (c) Determine and sketch the tractions on the four rectangular boundaries x1 = 0, x1 = b, x2 = 0, x2 = c . (d) As a plane strain solution, determine T13 , T23 , T33 and all the strain components. (e) As a plane stress solution, determine T13 , T23 , T33 . and all the strain components . ------------------------------------------------------------------------------Ans. (a) ∂ ϕ / ∂x14 = 0, ∂ 4ϕ / ∂x24 = 0, ∂ 4ϕ / ∂x12 ∂x22 = 0 , thus ϕ = α x12 x2 satisfies the bi-harmonic equation. 2 2 2 (b) T11 = ∂ ϕ / ∂x2 = 0, T12 = − ∂ ϕ / ∂x1∂x2 = −2α x1 , T22 = ∂ 2ϕ / ∂x12 = 2α x2 4


5-32

Lai et al, Introduction to Continuum Mechanics (c) On x1 = 0, t = −Te1 = − (T11e1 + T21e2 ) = 2α x1e2 = 0, On x1 = b, t = Te1 = T11e1 + T21e2 = −2α be2 . On x2 = 0, t = −Te2 = − (T12e1 + T22e2 ) = 2α x1e1 − 2α x2e2 = 2α x1e1. On x2 = c, t = Te 2 = T12e1 + T22e2 = −2α x1e1 + 2α x2e2 = −2α x1e1 + 2α ce2 .

x2 2 αc 2 αx1

c o

2 αb

b

x1

2 α x1

(d) As a plane strain solution, T33 = ν (T11 + T22 ) = 2να x2 , T13 = T23 = 0. E13 = E23 = E33 = 0, E11 = (1 / EY ) ⎡(1 − ν 2 )T11 − ν (1 + ν ) T22 ⎤ = − ⎡⎣ 2αν (1 + ν ) / EY ⎤⎦ x2 , ⎣ ⎦ E22 = (1 / EY ) ⎡(1 − ν 2 )T22 − ν (1 + ν ) T11 ⎤ = ⎡ 2α (1 − ν 2 ) / EY ⎤ x2 , ⎣ ⎦ ⎣ ⎦ E12 = (1 / EY )(1 + ν ) T12 = − ⎡⎣ 2α (1 + ν ) / EY ⎤⎦ x1. (d) As a plane stress solution, T33 = T13 = T23 = 0, E13 = E23 = 0, E33 = (1 / EY ) ⎡⎣ −ν (T11 + T22 ) ⎤⎦ = −2α (ν / EY ) x2 .

E11 = (1 / EY )(T11 − ν T22 ) = − ( 2αν / EY ) x2 , E22 = (1 / EY )(T22 − ν T11 ) = ( 2α / EY ) x2 .

E12 = (1 / EY )(1 + ν ) T12 = − ⎣⎡ 2α (1 + ν ) / EY ⎦⎤ x1 = −α x1 / μ . Note, for this problem, since T11 + T22 is a linear function of x1 and x2 , therefore, all the compatibility equations are satisfied so that E33 is meaningful and u3 does exist. _________________________________________________________________ 5.65

(

)

Consider the Airy stress function ϕ = α x14 − x24 . (a) Verify that it satisfies the bi-

harmonic equation. (b) Determine the in-plane stresses T11 , T12 and T22 . (c) Determine and sketch the tractions on the four rectangular boundaries x1 = 0, x1 = b, x2 = 0, x2 = c . (d) As a plane strain solution, determine T13 , T23 , T33 and all the strain components . (e) As a plane stress solution, determine T13 , T23 , T33 and all the strain components . --------------------------------------------------------------------------------


5-33


(

)

Ans. (a) ∂ 4ϕ / ∂x14 = 24α , ∂ 4ϕ / ∂x24 = −24α , ∂ 4ϕ / ∂x12 ∂x22 = 0 , thus ϕ = α x14 − x24 satisfies the (b) T11 = ∂ ϕ 2

/ ∂x22

=

−12α x22 ,

bi-harmonic equation. T22 = ∂ 2ϕ / ∂x12 = 12α x12 , T12 = −∂ 2ϕ / ∂x1∂x2 = 0 .

(c) On x1 = 0, t = −Te1 = − (T11e1 + T21e2 ) = 12α x22e1 , On x1 = b, t = Te1 = T11e1 + T21e2 = −12α x22e1. On x2 = 0, t = −Te2 = − (T12e1 + T22e2 ) = −12α x12e2 , On x2 = c, t = Te2 = T12e1 + T22e2 = 12α x12e2 .

x2

12 αx1 2

12 αx22

c b

o

12 αx22

x1

12 αx

2 1

T33 = ν (T11 + T22 ) = 12αν

(

(d) As a plane strain solution, x12

−

x22

), T

13

= T23 = 0, E13 = E23 = E33 = 0, E12 = (1 / EY )(1 + ν ) T12 = 0.

E11 = (1 / EY ) ⎡(1 − ν 2 )T11 − ν (1 + ν ) T22 ⎤ = −12α (1 / EY ) ⎡ x22 (1 − ν 2 ) + ν (1 + ν ) x12 ⎤ . ⎣ ⎦ ⎣ ⎦ 2 2 2 2 E22 = (1 / EY ) ⎡(1 − ν )T22 − ν (1 + ν ) T11 ⎤ = 12α (1 / EY ) ⎡(1 − ν ) x1 + x2ν (1 + ν ) ⎤ . ⎣ ⎦ ⎣ ⎦

(d) As a plane stress solution,

(

)

T33 = T13 = T23 = 0, E13 = E23 = 0, E11 = (1 / EY )(T11 − ν T22 ) = −12α (1 / EY ) x22 + ν x12 ,

(

)

E22 = (1 / EY )(T22 −ν T11 ) = 12α (1 / EY ) x12 + ν x22 , E12 = (1 / EY )(1 + ν ) T12 = 0.

(

)

.E33 = (1 / EY ) ⎡⎣ −ν (T11 + T22 ) ⎤⎦ = 12αν (1 / EY ) x22 − x12 . Since T11 + T22 is not a linear function of x1 and x2 , E33 is meaningless, because u3 does not exist. __________________________________________________________________ 5.66 Consider the Airy's stress function ϕ = α x1 x22 + x1 x23 . (a) Verify that it satisfies the biharmonic equation. (b) Determine the in-plane stresses T11 , T12 and T22 . (c) Determine the condition necessary for the traction at x2 = c to vanish and (d) determine the tractions on the remaining boundaries x1 = 0, x1 = b and x2 = 0 . ------------------------------------------------------------------------------Ans. (a) ∂ ϕ / ∂x14 = 0, ∂ 4ϕ / ∂x24 = 0, ∂ 4ϕ / ∂x12 ∂x22 = 0 , thus ϕ = α x1 x22 + x1 x23 satisfies the biharmonic equation. 4

(b) T11 = ∂ 2ϕ / ∂x22 = 2α x1 + 6 x1 x2 , T22 = ∂ 2ϕ / ∂x12 = 0, T12 = −∂ 2ϕ / ∂x1∂x2 = −2α x2 − 3x22 . Copyright 2010, Elsevier Inc

5-34

Lai et al, Introduction to Continuum Mechanics (c)On x2 = c, t = Te2 = T12e1 + T22e2 = (−2α c − 3c 2 )e1 , → −2α c − 3c 2 = 0 → 2α c = −3c 2 → α = −3c / 2 .

(

)

(d) On x1 = 0, t = −Te1 = − (T11e1 + T21e2 ) = 2α x2 + 3 x22 e2 = 3x2 ( x2 − c ) e2 . On x1 = b, t = Te1 = T11e1 + T21e2 = 3b ( 2 x2 − c ) e1 − 3 x2 ( x2 − c ) e2 . On x2 = 0, t = −Te 2 = − (T12e1 + T22e2 ) = 0. __________________________________________________________________ 5.67

Obtain the in-plane displacement components for the plane stress solution for the cantilever beam from the following strain strain-displacement relations.

⎞ ⎛ P ⎞ ⎛ h2 ∂u1 Px1 x2 ∂u ν Px1 x2 , E22 = 2 = − , E12 = ⎜ = − x22 ⎟ . ⎟ ⎜⎜ ⎟ EY I EY I ∂x1 ∂x2 ⎝ 4μ I ⎠ ⎝ 4 ⎠ ------------------------------------------------------------------------------Ans. 2 ∂u1 Px1 x2 Px x ∂u2 ν Px1 x2 ν Px1 x22 = → u1 = 1 2 + f1 ( x2 ) , =− → u2 = − + f 2 ( x1 ) , EY I EY I 2 EY I 2 EY I ∂x1 ∂x2 E11 =

⎛ P ∂u1 ∂u2 + = 2⎜ ∂x2 ∂x1 ⎝ 4μ I

⎞ ⎞ ⎛ h2 Px12 df ν Px22 df 2 ⎛ P − x22 ⎟ → + 1 − + =⎜ ⎟ ⎜⎜ ⎟ 2 E I dx EY I dx1 ⎝ 2μ I ⎠⎝ 4 Y 2 ⎠

⎞ ⎞ ⎛ h2 − x22 ⎟ , ⎟ ⎜⎜ ⎟ ⎠⎝ 4 ⎠

⎞ ⎛ P ⎞ ⎛ h2 ⎞ ⎟+⎜ ⎟ ⎜ ⎟. ⎠ ⎝ 2 μ I ⎠ ⎜⎝ 4 ⎟⎠ In the above equation, the left side is a function of x1 only, right side is a function of x2 only, thus both sides must equal to the same constant, say c1 . That is, →

⎛ P Px12 df 2 df ν Px22 − x22 ⎜ + =− 1 + dx2 2 EY I dx1 EY I ⎝ 2μ I

Px12 df 2 df Px12 Px13 + = c1 → 2 = c1 − → f 2 = c1 x1 − + c2 . dx1 2 EY I dx1 2 EY I 6 EY I ⎛ P df1 ν Px22 = − x22 ⎜ dx2 EY I ⎝ 2μ I

⎞ ⎛ P ⎟+⎜ ⎠ ⎝ 2μ I

⎞ ⎛ h2 ⎟ ⎜⎜ ⎠⎝ 4

⎞ ν Px23 x23 ⎛ P − → = − ⎜ c f ⎟⎟ 1 1 E I 6 3 ⎝ 2μ I Y ⎠

Thus, ⎛ ⎞ ⎛ P u1 = + −⎜ ⎟+⎜ 2 EY I 6 EY I ⎜⎝ 6 μ I ⎟⎠ ⎝ 2 μ I Px12 x2

ν Px23

ν Px1 x22

Px23

⎞ ⎛ P ⎞ ⎛ h2 ⎟+⎜ ⎟⎜ ⎠ ⎝ 2μ I ⎠ ⎜⎝ 4

⎞ ⎟⎟ x2 − c1 x2 + c3 ⎠

2

⎞⎛ h ⎞ ⎟⎜ ⎟ x2 − c1 x2 + c3 , ⎠⎝ 2 ⎠

Px13 + c1 x1 + c2 . 2 EY I 6 EY I _________________________________________________________________ u2 = −

5.68

−

(a) Let the Airy stress function be of the form ϕ = f ( x2 )cos general form of


5-35

mπ x1 . Show that the most l


f ( x2 ) is f ( x2 ) = C1 cosh λm x2 + C2 sinh λm x2 + C3 x2 cosh λm x2 + C4 x2 sinh λm x2 . (b) Is the mπ x1 ? l ------------------------------------------------------------------------------Ans. (a) The function ϕ ( x1 , x2 ) must satisfy the bi-harmonic equation. Now, answer the same if ϕ = f ( x2 )sin

mπ x1 ⎛ mπ ⎞ f ( x2 ) cos , = −⎜ ⎟ 2 l ∂x1 ⎝ l ⎠

∂ 2ϕ

2

mπ x1 ⎛ mπ ⎞ f ( x2 )cos , =⎜ ⎟ 4 l ∂x1 ⎝ l ⎠

∂ 4ϕ

4

2 mπ x1 mπ x1 ∂ 4ϕ d 4 f ⎛ mπ ⎞ d f cos , , = − = 4 cos ⎜ ⎟ 2 2 2 4 l l dx2 ∂x1 ∂x2 ∂x2 ⎝ l ⎠ dx2 Thus, 4 2 2 4 mπ x1 ⎡⎛ mπ ⎞ ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ ⎛ mπ ⎞ d f d f ⎤ 4 f ( x ) 2 ∇ ϕ = 4 + 2 2 2 + 4 = cos − + ⎢⎜ ⎥=0. 2 ⎟ ⎜ ⎟ l ⎣⎢⎝ l ⎠ dx24 ⎥⎦ ∂x1 ∂x1 ∂x2 ∂x2 ⎝ l ⎠ dx22 2 d4 f mπ 2 d f . Therefore, − 2 λ + λm4 f = 0, where λm ≡ m 4 2 l dx2 dx2 2

∂ 4ϕ

The characteristic equation for the above ODE is D 4 − 2λm2 D 2 + λm4 = 0 . The roots of this equation consists of two sets of double roots. They are: D = λm , λm , −λm , −λm . Thus, f ( x2 ) = C1 cosh λm x2 + C2 sinh λm x2 + C3 x2 cosh λm x2 + C4 x2 sinh λm x2 .

(b) Yes, the same __________________________________________________________________ 5.69 Consider a rectangular bar defined by −l ≤ x1 ≤ l, − c ≤ x2 ≤ c, − b ≤ x3 ≤ b , where b / l is very small. At the boundaries x2 = ±c , the bar is acted on by equal and opposite cosine normal stress Am cos λm x1 , where λm = mπ / l (per unit length in x3 direction). (a) Obtain the in-plane stresses inside the bar. (b) Find the surface tractions at x1 = ±l . Under what conditions can these surface tractions be removed without affecting T22 and T12 (except near x1 = ± l )? How would T11 be affected by the removal. Hint: Assume

ϕ = f ( x2 ) cos λm x1 , where λm = mπ / l and use the results of the previous problem ------------------------------------------------------------------------------Ans. (a) Boundary conditions are (T12 ) x =± c = 0 , (T22 ) x =± c = Am cos λm x1 2

2

Let ϕ = f ( x2 ) cos λm x1 , where λm = mπ / l . Then (see previous problem) , f ( x2 ) = C1 cosh λm x2 + C2 sinh λm x2 + C3 x2 cosh λm x2 + C4 x2 sinh λm x2 . The in-plane stresses are:


5-36


T22 = T12 = −

∂ 2ϕ ∂x12

= − ( λm ) f ( x2 ) cos λm x1 , T11 = 2

∂ 2ϕ ∂x22

(

)

= d 2 f / dx22 cos λm x1 ,

∂ ∂ϕ = λm ( df / dx2 ) sin λm x1 . Now, applying the boundary condition: ∂x1 ∂x2

(T22 ) x2 =± c = Am cos λm x1 → − ( λm )2 f ( ±c ) cos λm x1 = Am cos λm x1 → f ( ±c ) = − Am / ( λm )2 . From f ( ±c ) = − Am / ( λm ) , f ( +c ) = f ( −c ) , so that C2 = C3 = 0 and 2

f ( ±c ) = C1 cosh λm c + C4 c sinh λm c = − Am / λm2 Applying the other boundary condition: (T12 ) x =± c = 0 → ( df / dx2 ) x =± c = 0 → 2

(i)

2

C1λm sinh λm c + C4 ( sinh λm c + λm c cosh λm c ) = 0 (i) and (ii) give 2 A ⎡ ( λ c ) cosh λm c + sinh λm c ⎤ 2 Am ⎡ λm sinh λm c ⎤ C1 = − 2m ⎢ m ⎥ , C4 = 2 ⎢ ⎥. sinh 2λm c + 2λm c λm ⎣ λm ⎣ sinh 2λm c + 2λm c ⎦ ⎦ With f ( x2 ) = C1 cosh λm x2 + C4 x2 sinh λm x2 , we have ,

(ii)

T22 = − ( λm ) f ( x2 ) cos λm x1 = − ( λm ) {C1 cosh λm x2 + C4 λm x2 sinh λm x2 } cos λm x1 2

2

⎡ {( λm c ) cosh λm c + sinh λm c} cosh λm x2 − {λm x2 sinh λm x2 sinh λm c} cos λm x1 ⎤ ⎥. = 2 Am ⎢ sinh 2λm c + 2λm c ⎢⎣ ⎥⎦ T12 = λm ( df / dx2 ) sin λm x1 = ⎡C1λm2 sinh λm x2 + C4 λm ( sinh λm x2 + λm x2 cosh λm x2 ) ⎤ sin λm x1 ⎣ ⎦ ⎡ {− ( λm c ) cosh λm c} sinh λm x2 + sinh λm c ( λm x2 cosh λm x2 ) ⎤ ⎥ sin λm x1. = 2 Am ⎢ sinh 2λm c + 2λm c ⎢⎣ ⎥⎦ T11 =

∂ 2ϕ ∂x22

(

)

= d 2 f / dx22 cos λm x1 =

⎡ − ( λm c ) cosh λm c cosh λm x2 + sinh λm c ( λm x2 sinh λm x2 + cosh λm x2 ) ⎤ 2 Am ⎢ ⎥ cos λm x1. sinh 2λm c + 2λm c ⎣ ⎦ (b) Surface tractions at x1 = ± l are: T12 ( ± l, x2 ) = []sin mπ = 0 .

⎡ − ( λm c ) cosh λm c cosh λm x2 + sinh λm c ( λm x2 sinh λm x2 + cosh λm x2 ) ⎤ T11 ( ± l, x2 ) = 2 Am ⎢ ⎥ cos mπ sinh 2λm c + 2λm c ⎣ ⎦ At x1 = ± l , T11 is an even function of x2 , which gives rise to equal and opposite resultant force of magnitude FR at the two ends. Removal of these resultants will have little effects on T12 and T22 ,


5-37

Lai et al, Introduction to Continuum Mechanics if l / c is very large. However, T11 will need to be modified by subtracting the normal stress ( FR /Area) caused by the resultant forces. _________________________________________________________________ Verify that the equations of equilibrium in polar coordinates are satisfied by 1 ∂ϕ 1 ∂ 2ϕ ∂ 2ϕ ∂ ⎛ 1 ∂ϕ ⎞ + 2 , = , Trθ = − ⎜ Trr = T θθ ⎟. 2 2 ∂r ⎝ r ∂θ ⎠ r ∂r r ∂θ ∂r -------------------------------------------------------------------------------Ans. 2 ⎞ ⎛ 2 ⎛ Tθθ 1 ∂ (rTrr ) 1 ∂ ∂ϕ 1 ∂ ϕ 1 ∂ ϕ 1 ∂ 3ϕ 1 ∂ 2ϕ ⎞ 1 ∂ 2ϕ , = + = + − − = − ⎜⎜ ⎟ ⎜ ⎟ r ∂r r ∂r ⎝ ∂r r ∂θ 2 ⎟⎠ ⎜⎝ r ∂r 2 r 2 ∂r ∂θ 2 r 3 ∂θ 2 ⎟⎠ r r ∂r 2 5.70

⎡ 1 ⎛ ∂ 3ϕ ⎞ 1 ∂ 2ϕ ⎤ ⎡ ∂ ⎛ 1 ∂ϕ ⎞ ⎤ 1 ∂ ⎡ 1 ⎛ ∂ 2ϕ ⎞ 1 ∂ϕ ⎤ = − − = − − 3 ⎢ 2 ⎜⎜ ⎥ ⎢ ⎥ ⎜ ⎟ ⎟⎥ ⎢ ⎜ 2 ⎟ 2 r ∂θ ⎢⎣ r ⎜⎝ ∂r ∂θ ⎟⎠ r 2 ∂θ ⎥⎦ ⎣ ∂r ⎝ r ∂θ ⎠ ⎦ ⎢⎣ r ⎝ ∂r ∂θ ⎟⎠ r ∂θ ⎥⎦ Thus, [See Eq.(4.8.1), 1 ∂ (rTrr ) 1 ∂Trθ Tθθ + − r ∂r r ∂θ r ⎛ 1 ∂ 2ϕ 1 ∂ 3ϕ 1 ∂ 2ϕ ⎞ ⎡ 1 ⎛ ∂ 3ϕ ⎞ 1 ∂ 2ϕ ⎤ 1 ∂ 2ϕ =⎜ + − − − − =0 ⎜ r ∂r 2 r 2 ∂r ∂θ 2 r 3 ∂θ 2 ⎟⎟ ⎢⎢ r 2 ⎜⎜ ∂r ∂θ 2 ⎟⎟ r 3 ∂θ 2 ⎥⎥ r ∂r 2 ⎝ ⎠ ⎣ ⎝ ⎠ ⎦ Similarly, 2 1 ∂ (r Tθ r ) 1 ∂ ⎡ 2 ∂ ⎛ 1 ∂ϕ ⎞ ⎤ 1 ∂ ⎡ ⎛ ∂ 2ϕ ⎞ ∂ϕ ⎤ 1 ⎛ ∂ 3ϕ ⎞ r r = − = − − = − ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ r ⎝⎜ ∂r 2 ∂θ ⎠⎟ ∂r r2 r 2 ∂r ⎣ ∂r ⎝ r ∂θ ⎠ ⎦ r 2 ∂r ⎢⎣ ⎝⎜ ∂r ∂θ ⎠⎟ ∂θ ⎥⎦

1 ∂Trθ 1 ∂ =− r ∂θ r ∂θ

1 ∂Tθθ 1 ∂ ∂ 2ϕ 1 ∂ 3ϕ , = = r ∂θ r ∂θ ∂r 2 r ∂θ∂r 2 1 ∂ (r 2Tθ r ) 1 ∂Tθθ + = 0. r ∂θ ∂r r2 __________________________________________________________________ Thus, [See Eq.(4.8.2)]

5.71

⎡T From the transformation law : ⎢ rr ⎣Tθ r

Trθ ⎤ ⎡ cos θ = Tθθ ⎦⎥ ⎢⎣ − sin θ

sin θ ⎤ ⎡T11 T12 ⎤ ⎡cos θ cos θ ⎥⎦ ⎣⎢T21 T22 ⎦⎥ ⎢⎣ sin θ

and ∂ 2ϕ 1 ∂ϕ ⎛ 1 ⎞ ∂ 2ϕ T = +⎜ ⎟ , obtain rr ∂x1∂x2 r ∂r ⎝ r 2 ⎠ ∂θ 2 ∂x22 ∂x12 ------------------------------------------------------------------------------Ans.

T11 =

∂ϕ 2

, T22 =

∂ ϕ 2

and T12 = −


5-38

− sin θ ⎤ cosθ ⎥⎦

Lai et al, Introduction to Continuum Mechanics Trθ ⎤ ⎡ cos θ = Tθθ ⎥⎦ ⎢⎣ − sin θ

⎡ Trr ⎢T ⎣ θr

sin θ ⎤ ⎡T11 T12 ⎤ ⎡ cosθ cos θ ⎥⎦ ⎢⎣T21 T22 ⎥⎦ ⎢⎣ sin θ

⎡ T cos 2 θ + 2T sin θ cos θ + T sin 2 θ 11 12 22 =⎢ ⎢ 2 2 ⎢⎣(T22 − T11 ) cos θ sin θ + T12 cos θ − sin θ

(

− sin θ ⎤ cosθ ⎥⎦

(T22 − T11 ) sin θ cosθ + T12 ( cos2 θ − sin 2 θ )⎤⎥

) (

T11 sin 2 θ + T22 cos 2 θ − 2T12 sin θ cos θ

)

⎥ ⎥⎦

,

r 2 = x12 + x22 → ∂r / ∂x1 = x1 / r = cos θ , ∂r / ∂x2 = x2 / r = sin θ , −x x2 x sin θ cos θ → ∂θ / ∂x1 = 2 2 2 = − , ∂θ / ∂x2 = 2 1 2 = , x1 r r x1 + x2 x1 + x2

θ = tan −1

∂ϕ ∂ϕ ∂r ∂ϕ ∂θ ∂ϕ 1 ∂ϕ = + = sin θ + cosθ , ∂x2 ∂r ∂x2 ∂θ ∂x2 ∂r r ∂θ T11 = = =

∂ 2ϕ ∂x22

=

∂ ⎛ ∂ϕ 1 ∂ϕ ∂ ⎛ ∂ϕ 1 ∂ϕ ⎞ cos θ ⎞ sin θ + cos θ ⎟ sin θ + cos θ ⎟ sin θ + ⎜ ⎜ ∂θ ⎝ ∂r r ∂θ ∂r ⎝ ∂r r ∂θ ⎠ r ⎠

⎛ 1 ∂ 2ϕ ⎞ cos θ 1 ∂ϕ ⎞ ⎛ ∂ 2ϕ ∂ϕ 1 ∂ 2ϕ 1 ∂ϕ 2 sin + cos sin − + sin + cos + cos − sin θ θ θ θ θ θ θ ⎜ ⎟ ⎜ ⎟⎟ 2 ⎜ ⎟ ⎜ ∂r r ∂θ 2 r ∂θ ∂r 2 ⎝ r ∂r ∂θ r ∂θ ⎠ ⎝ ∂r ∂θ ⎠ r

∂ 2ϕ ∂ 2ϕ ∂r 2

sin 2 θ +

cos 2 θ ∂ϕ cos 2 θ ∂ 2ϕ 2cosθ sin θ ∂ 2ϕ 2cosθ sin θ ∂ϕ + + − , ∂r ∂θ ∂θ r ∂r r r2 r 2 ∂θ 2

∂ϕ ∂ϕ ∂r ∂ϕ ∂θ ∂ϕ ∂ϕ sin θ = + = cos θ − , ∂x1 ∂r ∂x1 ∂θ ∂x1 ∂r ∂θ r T22 =

∂ 2ϕ ∂x12

=

∂ ⎛ ∂ϕ ∂ϕ sin θ ⎞ ∂ ⎛ ∂ϕ ∂ϕ sin θ ⎞ sin θ cosθ − cosθ − cosθ − ⎜ ⎟ ⎜ ∂r ⎝ ∂r ∂θ r ⎠ ∂θ ⎝ ∂r ∂θ r ⎟⎠ r

⎛ ∂ 2ϕ sin 2 θ ∂ϕ sin 2 θ ∂ 2ϕ 2sin θ cos θ ∂ 2ϕ 2sin θ cosθ ∂ϕ ⎞ = ⎜ 2 cos 2 θ + + 2 − + ⎟. ⎜ ∂r ∂θ ∂θ ⎟⎠ r ∂r r ∂θ 2 r r2 ⎝ ∂r ∂ ∂ϕ ∂ ⎛ ∂ϕ ∂ϕ sin θ ⎞ ∂ ⎛ ∂ϕ ∂ϕ sin θ ⎞ cosθ −T12 = = ⎜ cos θ − cosθ − ⎟ sin θ + ⎜ ⎟ ∂x2 ∂x1 ∂r ⎝ ∂r ∂θ r ⎠ ∂θ ⎝ ∂r ∂θ r ⎠ r =

∂ 2ϕ ∂r 2

cos θ sin θ −

sin θ cos θ ∂ 2ϕ r2

∂θ 2

+

sin 2 θ ∂ϕ cos 2 θ ∂ϕ cos 2 θ ∂ 2ϕ sin 2 θ ∂ 2ϕ sin θ cosθ ∂ϕ − + − − . r ∂θ∂r r ∂r ∂θ r ∂r r 2 ∂θ r 2 ∂θ Thus,


5-39

Lai et al, Introduction to Continuum Mechanics Trr = T11 cos 2 θ + 2T12 sin θ cosθ + T22 sin 2 θ = ⎛ ∂ 2ϕ 2 cos 2 θ ∂ϕ cos 2 θ ∂ 2ϕ 2cosθ sin θ ∂ 2ϕ 2cosθ sin θ ∂ϕ ⎞ 2 + + − ⎜⎜ 2 sin θ + ⎟ cos θ r ∂r r ∂r ∂θ ∂θ ⎟⎠ r 2 ∂θ 2 r2 ⎝ ∂r ⎛ ∂ 2ϕ sin θ cosθ ∂ 2ϕ sin 2 θ ∂ϕ cos 2 θ ∂ϕ cos 2 θ ∂ 2ϕ ⎞ + 2 − + ⎜ 2 cos θ sin θ − ⎟ 2 2 2 r ∂θ∂r ⎟ ∂θ ∂θ r r r r ∂ ∂ θ ⎜ sin θ cos θ −2 ⎜ sin 2 θ ∂ 2ϕ sin θ cosθ ∂ϕ ⎟ ⎜− ⎟ − ∂r r ∂r ∂θ r ⎝ ⎠ ⎛ ∂ 2ϕ sin 2 θ ∂ϕ sin 2 θ ∂ 2ϕ 2sin θ cosθ ∂ 2ϕ 2sin θ cosθ ∂ϕ ⎞ + sin 2 θ ⎜ 2 cos 2 θ + + 2 − + ⎟. ⎜ ∂r r ∂r r ∂r ∂θ ∂θ ⎟⎠ r r2 ∂θ 2 ⎝ ⎛ ∂ 2ϕ 1 ∂ϕ ∂ 2ϕ ∂ 2ϕ ⎞ Trr = sin 2 θ cos 2 θ ⎜ 2 − 2 2 + 2 ⎟ + cos 4 θ + sin 4 θ + 2sin 2 θ cos 2 θ ⎜ ∂r ⎟ r ∂r r r ∂ ∂ ⎝ ⎠

(

(

+ cos 4 θ + sin 4 θ + 2sin 2 θ cos 2 θ

(

)

) r1 ∂∂θϕ − r2 ∂∂ϕθ (sin θ cosθ − sin θ cos θ + cos θ sin θ − sin θ cosθ ) 2

3

2

2

3

3

3

2

sin θ cos3 θ − cos3 θ sin θ + sin 3 θ cos θ − sin 3 θ cos θ

)

2 ∂ 2ϕ . r ∂r ∂θ

That is, 1 ∂ϕ ⎛ 1 ⎞ ∂ 2ϕ +⎜ ⎟ . Trr = r ∂r ⎝ r 2 ⎠ ∂θ 2 __________________________________________________________________ 5.72

Obtain the displacement field for the plane strain solution of the axis-symmetric stress distribution from that for the plane stress solution obtained in Section 5.28. ------------------------------------------------------------------------------Ans. From Section 5. 29, we have, for plane stress solution, [See Eq.(5.29.15) and .(5.29.16) and note EY = 2μ (1 + ν ) ] ⎡ A ⎤ − (1 + ν ) + 2 B (1 − ν ) r ln r − (1 + ν ) Br + 2C (1 − ν ) r ⎥ + H sin θ + G cos θ , ⎢ 2 μ (1 + ν ) ⎣ r ⎦ 2 Brθ + H cos θ − G sin θ + Fr . uθ = μ (1 + ν ) To obtain the corresponding displacement field for the plane strain solution, we replace the Poisson ratio ν with ν / (1 − ν ) in the above equation [see Section. 5.26]. That is, ur =

1

(1 + ν ) → ⎛⎜1 + ⎝

ν ⎞ 1 = , ⎟ 1 −ν ⎠ 1 −ν

(1 −ν ) → ⎛⎜1 −

⎝ Thus, for plane strain:


5-40

ν ⎞ 1 − 2ν . = 1 − ν ⎟⎠ 1 − ν

Lai et al, Introduction to Continuum Mechanics 1 A 1 (1 − 2ν ) (1 − 2ν ) ⎤ r ln r − B r + 2C r + H sin θ + G cos θ + 2B ⎢− 2 μ ⎣ r (1 − ν ) (1 − ν ) (1 − ν ) ⎥⎦ (1 − ν ) 1 ⎡ A ⎤ − + 2 B (1 − 2ν ) r ln r − Br + 2C (1 − 2ν ) r ⎥ + H sin θ + G cos θ = 2μ ⎢⎣ r ⎦

ur =

(1 − ν ) ⎡

(1 + ν ) ⎡ A

⎤ ⎢⎣ − r + 2 B (1 − 2ν ) r ln r − Br + 2C (1 − 2ν ) r ⎥⎦ + H sin θ + G cos θ . and 2 Brθ (1 − ν ) 4 Brθ (1 − ν )(1 + ν ) uθ = + H cos θ − G sin θ + Fr = + H cosθ − G sin θ + Fr . EY μ =

EY

_________________________________________________________________ 5.73

Let the Airy stress function be ϕ = f (r )sin nθ , find the differential equation for f (r ) . Is this the same ODE for f (r ) if ϕ = f (r )cos nθ ?

--------------------------------------------------------------------------------Ans. 2 ∂ϕ ∂ ϕ ∂ϕ ∂ 2ϕ ϕ = f (r )sin nθ → = f 'sin nθ → 2 = f ''sin nθ ; = nf cos nθ → 2 = − n 2 f sin nθ . ∂r ∂θ ∂r ∂θ Thus, ∂2 ⎞ 1 ∂ϕ 1 ∂ 2ϕ ∂ 2ϕ ⎛ 1 ∂ 1 ∂2 ⎛ f' 1 2 ⎞ + 2 + = + + ⎜ ⎟ f (r )sin nθ = ⎜ − 2 n f + f '' ⎟ sin nθ ≡ g ( r ) sin nθ r ∂r r ∂θ 2 ∂r 2 ⎜⎝ r ∂r r 2 ∂θ 2 ∂r 2 ⎟⎠ r r ⎝ ⎠ , 2 1 d n2 ⎞ ⎛ f' 1 ⎞ ⎛ d where g ( r ) ≡ ⎜ − 2 n 2 f + f '' ⎟ = ⎜ 2 + − ⎟f . r dr r 2 ⎟⎠ ⎝ r r ⎠ ⎜⎝ dr Now, 2 2 2 ⎛1 ∂ 1 ∂ 1 ∂2 ∂ ⎞⎛ 1 ∂ϕ 1 ∂ ϕ ∂ 2ϕ ⎞ ⎛ 1 ∂ ∂2 ⎞ + 2 + + + = + + ⎜⎜ ⎟⎜ ⎟ ⎜ ⎟ g ( r ) sin nθ 2 2 2 ∂r 2 ⎟⎜ ∂r 2 ⎟⎠ ⎜⎝ r ∂r r 2 ∂θ 2 ∂r 2 ⎟⎠ ⎝ r ∂r r ∂θ ⎠⎝ r ∂r r ∂θ ⎛ d 2 1 d n2 ⎞ ⎛ d 2 1 d n 2 ⎞⎛ d 2 1 d n 2 ⎞ =⎜ 2 + − 2 ⎟ g ( r ) sin nθ = ⎜ 2 + − ⎟⎜ 2 + − ⎟ f ( r ) sin nθ = 0. ⎜ dr ⎜ dr r dr r ⎟⎠ r dr r 2 ⎟⎜ r dr r 2 ⎟⎠ ⎝ ⎝ ⎠⎝ dr Therefore, 2 2 ⎛ d 1 d n ⎞⎛ d 2 1 d n 2 ⎞ − ⎟⎜ 2 + − ⎟ f (r ) = 0 . ⎜⎜ 2 + r dr r 2 ⎟⎜ r dr r 2 ⎟⎠ ⎝ dr ⎠⎝ dr The same equation will be obtained if ϕ = f (r )cos nθ _________________________________________________________________ 5.74

Obtain the four independent solutions for the following equation ⎛ d 2 1 d n 2 ⎞⎛ d 2 f 1 df n 2 ⎞ − ⎟⎜ 2 + − f ⎟=0 ⎜⎜ 2 + r dr r ⎟⎜ r dr r ⎟⎠ ⎝ dr ⎠⎝ dr


5-41

Lai et al, Introduction to Continuum Mechanics --------------------------------------------------------------------------------Ans. Let f = r m .

⎛ d 2 f 1 df n 2 ⎞ − 2 f ⎟ = ⎡ m ( m − 1) + m − n 2 ⎤ r m − 2 = m 2 − n 2 r m − 2 ⎜⎜ 2 + ⎦ ⎟ ⎣ r dr dr r ⎝ ⎠ ⎛ d 2 1 d n 2 ⎞⎛ d 2 f 1 df n 2 ⎞ − 2 ⎟⎜ 2 + − 2 f ⎟ = m2 − n 2 ⎡( m − 3)( m − 2 ) + ( m − 2 ) − n 2 ⎤ r m − 4 = 0 ⎜⎜ 2 + ⎣ ⎦ ⎟⎜ ⎟ r dr r ⎠⎝ dr r dr r ⎝ dr ⎠ 2 → m 2 − n 2 ⎡⎢( m − 2 ) − n 2 ⎤⎥ = 0. ⎣ ⎦ Thus, m1 = + n, m2 = − n, m3 = 2 + n, m4 = 2 − n .

(

(

(

)

)

)

For n ≠ 0 and n ≠ 1 , the four independent solutions for f are: r + n , r − n , r + n + 2 and r − n+ 2 . For n = 0 , m1 = m2 = 0, m3 = m4 = 2 . Two independent solutions for f are given by C and r 2 . Additional solutions are given by ⎛ d n⎞ ⎛ d ⎞ = r n ln r = ln r , and ⎜ r n + 2 ⎟ = ⎡ r n + 2 ln r ⎤ = r 2 ln r . ⎜ dn r ⎟ ⎣ ⎦ n → 0 n = 0 ⎝ ⎠ n →0 ⎝ dn ⎠ n →0

(

)

The four independent solutions are: C , r 2 ,ln r and r 2 ln r . ⎛ d ⎞ = r n ln r For n = 1 , m1 = m4 = 1 , in addition to r , r −1 , r 3 , we have, ⎜ r n ⎟ dn ⎝ ⎠ n→1

(

)

n =1

= r ln r

Thus, the four independent solutions are: r , r −1 , r 3and r ln r . __________________________________________________________________ 5.75

(

)

(

)

⎡d n ⎤ ⎡d n ⎤ r cos nθ ⎥ , ⎢ r sin nθ ⎥ , Evaluate ⎢ ⎣ dn ⎦ n =0 ⎣ dn ⎦ n =0 ⎡ d − n+2 ⎤ ⎡d n ⎤ r cos nθ ⎥ cos nθ ⎥ and ⎢ ⎢ dn r ⎣ ⎦ n =1 ⎣ dn ⎦ n =1

(

)

(

)

-------------------------------------------------------------------------------Ans. ⎡d n ⎤ = ⎡ r n ln r cos nθ − r nθ sin nθ ⎤ = ln r. ⎢ dn r cos nθ ⎥ ⎦ n =0 ⎣ ⎦ n =0 ⎣

(

)

(

)

(

)

(

)

⎡d n ⎤ = ⎡ r n ln r sin nθ + r nθ cos nθ ⎤ = θ. ⎢ dn r sin nθ ⎥ ⎦ n =0 ⎣ ⎦ n =0 ⎣

(

)

⎡ d − n+ 2 ⎤ cos nθ ⎥ = ⎡ − r − n + 2 ln r cos nθ − r − n + 2θ sin nθ ⎤ = − r ln r cosθ − rθ sin θ ⎢ dn r ⎦ n =1 ⎣ ⎦ n =1 ⎣

(

)

⎡d n ⎤ n ⎡ n ⎤ ⎢ dn r cos nθ ⎥ = ⎣ r ln r cos nθ − r θ sin nθ ⎦ n =1 = r ln r cosθ − rθ sin θ ⎣ ⎦ n =1 ________________________________________________________________


5-42

Lai et al, Introduction to Continuum Mechanics 5.76 In the Flamont Problem (Sect. 5.37), if the concentrated line load F , acting at the origin on the surface of a 2D half-space (defined by −π / 2 ≤ θ ≤ π / 2 ), is tangent to the surface ⎛ 2 F ⎞ sin θ and in the direction of θ = 90o , show that: Trr = − ⎜ , Tθθ = Trθ = 0 ⎟ ⎝ π ⎠ r

. ------------------------------------------------------------------------------Ans. The boundary conditions are: Tθθ = Trθ = 0 at θ = ±π / 2, r ≠ 0 . π /2

(i),

π /2

∫ (Trr cosθ − Trθ sin θ ) rdθ = 0

(ii),

−π /2

∫ (Trr sin θ + Trθ cosθ ) rdθ = − F .

(iii)

−π /2

From the stress field obtained in Sect. 5.37, Trr = r −1 ( 2 B5 cosθ − 2 B5 sin θ ) , Tθθ = 0, Trθ = 0 , π /2

∫

−π /2

π /2

∫

−π /2

(iv)

we obtain, from Eqs.(ii) and (iv) :

( 2B cos θ − B sin 2θ ) dθ = 0 → 2B 2

5

5

5

π /2

∫

cos 2 θ dθ = 0 → B5 = 0 .

−π /2

From Eqs.(iii) and (iv)

(

)

π /2

B5 sin 2θ − 2 B5 sin 2 θ dθ = − F → −2 B5

∫

sin 2 θ dθ = − F → 2 B5

−π /2

π 2

= F → 2 B5 =

Thus sin θ 2 F ⎛ ⎞ ⎛ sin θ ⎞ , Tθθ = 0, Trθ = 0 . Trr = −2 B5 ⎜ =− ⎟ π ⎜⎝ r ⎟⎠ ⎝ r ⎠ _________________________________________________________________ 5.77

2F

π (v)

Verify that the displacement field for the Flamont Problem under a normal force P is given by P P ur = − (1 −ν )θ sin θ + 2ln r cosθ } , uθ = { {(1 + ν ) sin θ + 2ln r sin θ − (1 −ν )θ cosθ } , π EY π EY The 2D half space is defined by −π / 2 ≤ θ ≤ π / 2 . ------------------------------------------------------------------------------Ans. From the given displacement field, we have, Trr 2 P ⎛ cos θ ⎞ . Err = ∂ur / ∂r = − ⎜ ⎟ , i.e., Err = EY π ⎝ r ⎠ EY ∂uθ P + ur = π EY ∂θ −

P π EY

{(1 + ν ) cosθ + 2ln r cosθ − (1 −ν ) cosθ + (1 −ν )θ sin θ }

{(1 −ν )θ sin θ + 2ln r cosθ } = π PE {(1 + ν ) cosθ − (1 −ν ) cosθ } = 2νπPEcosθ . Y


5-43

Y


νT 1 ⎛ ∂u ⎞ 2ν P cosθ , i.e., Eθθ = − rr . That is, Eθθ = ⎜ θ + ur ⎟ = r ⎝ ∂θ EY ⎠ π EY r Next, P {(1 − ν ) sin θ + (1 + ν ) sin θ } 2 P sin θ 1 ⎛ ∂u ⎞ ∂u 2 Erθ = ⎜ r − uθ ⎟ + θ = − + r ⎝ ∂θ π EY r π EY r ⎠ ∂r 2 P sin θ 2 P sin θ =− + = 0, i.e., 2μ Erθ = 0 = Trθ . π EY r π EY r _________________________________________________________________ 5.78

Show that Eq. (5.38.6), i.e., u = Ψ −

1 ∇(x ⋅ Ψ + Φ) 4(1 − ν )

can also be written as: 2μ u = −4 (1 − ν )ψ + ∇ ( x ⋅ψ + φ ) where Ψ = −ψ

2 (1 − ν )

μ

, Φ = −φ

2 (1 − ν )

μ

------------------------------------------------------------------------------2 (1 − ν ) 2 (1 − ν ) 2 (1 − ν ) , we have, x ⋅ Ψ = − x ⋅ψ → . Ans. With Ψ = −ψ , Φ = −φ

μ

→u =Ψ−

μ

μ

2 (1 − ν ) 1 1 ∇(x ⋅ Ψ + Φ) = − ∇ ( x ⋅ψ + φ ) . ψ+ μ 4(1 − ν ) 2μ That is,

2μ u = −4 (1 − ν )ψ + ∇ ( x ⋅ψ + φ ) . ________________________________________________________________________

5.79 ui = Ψ i −

Show that with

1 ∂ ( xn Ψ n + Φ ) , the Navier Equations become : 4(1 − ν ) ∂xi

⎛ ∂∇ 2 Ψ n ∂∇ 2 Φ ⎞ − (1 − 4ν ) ∇ 2 Ψ i + ⎜⎜ xn ⎟ + Bi = 0 2 (1 − 2ν ) ⎝ ∂xi ∂xi ⎟⎠ ------------------------------------------------------------------------------1 ∂ 1 ⎛ ∂Ψ n ∂Φ ⎞ Ans. ui = Ψ i − + Ψi + ( xn Ψ n + Φ ) = Ψ i − ⎜ xn ⎟ 4(1 − ν ) ∂xi 4(1 − ν ) ⎝ ∂xi ∂xi ⎠ −

μ


5-44


→e=

∂ui ∂Ψ m ∂2Ψn ∂Ψ m ∂ 2Φ 1 ⎛ = − +2 + ⎜⎜ xn ∂xi ∂xm 4(1 − ν ) ⎝ ∂xm ∂xm ∂xm ∂xm ∂xm

∂e ∂ ∂Ψ m 1 ⎛ ∂ ⎜ = − ∂xi ∂xi ∂xm 4(1 − ν ) ⎜⎝ ∂xi =

⎞ ⎟⎟ ⎠

∂ 2 Ψ n ⎪⎫ ∂ ∂Ψ m ∂ ∂ 2Φ ⎪⎧ + ⎨ xn ⎬+2 ∂xi ∂xm ∂xi ∂xm ∂xm ⎩⎪ ∂xm ∂xm ⎭⎪

⎞ ⎟ ⎟ ⎠

1 ⎛ ∂ ∂Ψ m (1 − 2ν ) ∂ ∂2Ψ n ∂ 2 ⎞ − + ∇2Ψi + ∇ Φ⎟ ⎜⎜ xn ⎟ ∂xi ∂xm 2(1 − ν ) 4(1 − ν ) ⎝ ∂xi ∂xm ∂xm ∂xi ⎠

⎛ μ μ ∂e ∂ ∂Ψ m ∂ ∂2Ψ n ∂ 2 ⎞ = − + ∇2Ψi + ∇ Φ⎟ ⎜⎜ xn ⎟ ∂xi 1 − 2ν ∂xi 2(1 − ν ) ∂xi ∂xm 4 (1 − 2ν ) (1 − ν ) ⎝ ∂xi ∂xm ∂xm ⎠

μ

Also,

μ

∂ ui ∂ 2 ∂ ∂Ψ j ∂ 2 ⎞ μ ⎛ = μ∇ 2 Ψ i − ∇ Ψn + 2 + ∇2Ψi + ∇ Φ⎟ ⎜ xn ⎟ ∂x j ∂x j ∂x j ∂xi ∂xi 4(1 − ν ) ⎜⎝ ∂xi ⎠ 2

∂2Ψ j

⎛ ∂ 2 ∂ 2 ⎞ ∇ Ψ n + ∇ 2 Ψ i [1 − 4(1 − ν )] + ∇ Φ⎟ ⎜ xn ∂xi 2(1 − ν ) ∂x j ∂xi 4(1 − ν ) ⎝ ∂xi ⎠ Thus, ⎛ ∂ 2ui ∂ 2 ∂ 2 ⎞ μ ∂e μ + =− ∇ Ψ n − 2 (1 − 4ν )(1 − ν ) ∇ 2 Ψ i + ( 2 − 2ν ) ∇ Φ⎟ μ ⎜ ( 2 − 2ν ) xn 4(1 − ν ) (1 − 2ν ) ⎝ ∂x j ∂x j 1 − 2ν ∂xi ∂xi ∂xi ⎠

μ

=−

=−

−

μ

⎛ ∂ 2 ∂ 2 ⎞ ∇ Ψ n − (1 − 4ν ) ∇ 2 Ψ i + ∇ Φ⎟ ⎜ xn 2 (1 − 2ν ) ⎝ ∂xi ∂xi ⎠

μ

i.e., ⎛ ∂ ui ∂ 2 ∂ 2 ⎞ μ ∂e μ + =− ∇ Ψ n − (1 − 4ν ) ∇ 2 Ψ i + ∇ Φ⎟ , μ ⎜ xn ∂x j ∂x j 1 − 2ν ∂xi ∂xi 2 (1 − 2ν ) ⎝ ∂xi ⎠ 2

so that the Navier Equations become: ⎛ ∂∇ 2 Ψ n ∂∇ 2 Φ ⎞ μ − − (1 − 4ν ) ∇ 2 Ψ i + ⎜⎜ xn ⎟ + Bi = 0 . ∂xi ∂xi ⎟⎠ 2 (1 − 2ν ) ⎝ _________________________________________________________________ 5.80

Consider the potential function given in Eq. (5.38.32) [See Example 5.38.5], i.e.,

ψ = ψ ( R ) eR , φ = φ ( R ) , where ⎛ d 2ψ 2 dψ 2ψ ⎞ 2 dφ = and + − = 0. 0 ⎜ ⎜ dR 2 R dR R 2 ⎟⎟ dR 2 R dR ⎝ ⎠ Show that these functions generate the following displacements, dilatation and stresses as given in Eq.(5.38.5) to (5.38.38): ∇ 2φ =

d 2φ

+


5-45

Lai et al, Introduction to Continuum Mechanics dφ ⎞ ⎛ dψ (a) Displacements: 2 μ u R = ⎜ R , uθ = uβ = 0 + (−3 + 4ν )ψ + dR ⎟⎠ ⎝ dR (1 − 2ν ) ⎧ dψ + 2ψ ⎫ (b) Dilation: e = − ⎨ ⎬ μ ⎩ dR R ⎭

dψ ψ d 2φ dψ 3ψ 1 dφ ⎫ ⎧ + ( 2 − 4ν ) + 2 , Tββ = Tθθ = − ⎨( 2ν − 1) + − ⎬ dR R R dR ⎭ dR R dR ⎩ ------------------------------------------------------------------------------Ans. With x = Re R , we have, x ⋅ψ = Rψ , thus 2μ u = −4 (1 − ν )ψ + ∇ ( x ⋅ψ + φ ) →

(c) Stresses: TRR = ( 2ν − 4 )

d dψ dφ ( Rψ + φ ) eR = ( −3 + 4ν )ψ eR + ⎛⎜ R + ⎞⎟ eR , dR ⎝ dR dR ⎠ dψ dφ + , i.e., 2μ u R = ( −3 + 4ν )ψ + R dR dR (b) The non zero strain components are: ∂u dψ ⎛ d 2ψ dψ d 2φ ⎞ dψ d 2ψ d 2φ +⎜R 2 + + 2 ⎟ = ( −2 + 4ν ) +R 2 + 2 . 2μ ERR = 2 μ R = ( −3 + 4ν ) ∂R dR ⎜⎝ dR dR dR ⎟⎠ dR dR dR 2 2 d ψ 2 dψ 2ψ d ψ 2ψ 2 dψ + − 2 =0→ = − But 2 R dR dR R dR 2 R 2 R dR dψ d 2ψ d 2φ ∂ψ 2ψ d 2φ . → 2 μ ERR = ( −2 + 4ν ) + R 2 + 2 = −4 (1 − ν ) + + ∂R dR R dR 2 dR dR 2μuR ψ dψ 1 dφ = ( −3 + 4ν ) + + . 2μ Eββ = 2 μ Eθθ = R R dR R dR Therefore, 2 ψ 1 dφ ⎫ dψ 2ψ d φ ⎧ dψ + + 2 + ⎨2 ( −3 + 4ν ) + 2 +2 2μ e =→ 2 μ e = −4 (1 − ν ) ⎬ dR R dR R dR R dR ⎭ ⎩ 2μ u = −4 (1 − ν )ψ e R +

= ( −2 + 4ν )

2ψ ⎪⎧ d 2φ 2 dφ ⎪⎫ dψ ⎛ dψ 2ψ + ( −2 + 4ν ) +⎨ 2 + + ⎬ = −2 (1 − 2ν ) ⎜ dR R ⎪⎩ dR R dR ⎪⎭ R ⎝ dR (1 − 2ν ) ⎛ dψ + 2ψ ⎞ . →e=− μ ⎜⎝ dR R ⎟⎠ (c) the stresses are: 2μν dψ ⎛ dψ 2ψ ⎞ + + TRR = e + 2μ ERR = −2ν ⎜ ⎟ − 4 (1 − ν ) 1 − 2ν R ⎠ dR ⎝ dR

⎞ ⎟. ⎠

2ψ d 2φ + R dR 2

dψ ψ d 2φ + ( 2 − 4ν ) + 2 . dR R dR 2μν ψ dψ 1 dφ ⎛ dψ 2ψ ⎞ + + Tββ = Tθθ = e + 2μ Eθθ = −2ν ⎜ ⎟ + ( −3 + 4ν ) + 1 − 2ν R ⎠ R dR R dR ⎝ dR dψ 3ψ 1 dφ = (1 − 2ν ) − + . dR R R dR = ( 2ν − 4 )

________________________________________________________________________


5-46


Consider the following potential functions for axis-symmetric problems: ψ = 0, φ = φ ( r , z ) = φˆ ( R, β ) , ∇ 2φ = ∇ 2φˆ = 0 , where (r ,θ , z ) and ( R,θ , β ) are cylindrical and spherical coordinates respectively with z as the axis of symmetry, θ the longitudinal angle and β the angle between z axis and e R (the azimuthal angle). Shows that these functions generate the following displacements, dilatation and stresses: Cylindrical coordinates ∂φ ∂φ (a) Displacements: 2 μ ur = , uθ = 0, 2 μ u z = ∂r ∂z (b) Dilation: e = 0 2 μν ∂ 2φ 2μν 1 ∂φ e + 2μ Err = 2 , Tθθ = e + 2 μ Eθθ = (c) Trr = ν r ∂r 1 − 2ν 1 − 2 ∂r 2 2 μν ∂φ ∂ 2φ Tzz = e + 2 μ E zz = 2 , Erθ = 0, Eθ z = 0, Trz = 2 μ Erz = 1 − 2ν ∂r ∂z ∂z Spherical coordinates: ˆ 1 ∂φˆ ∂φ , uθ = 0, 2μ uβ = (d) Displacements: 2μ u R = R ∂β ∂R (e) Dilation: e = 0 (f) Stresses: 2μν 2μν 1 ∂ 2φˆ 1 ∂φˆ TRR = e + 2 μ ERR = 2 , Tββ = e + 2μ Eββ = 2 + 1 − 2ν 1 − 2ν R ∂β 2 R ∂R ∂R 2μν 1 ∂φˆ cot β ∂φˆ Tθθ = e + 2 μ Eθθ = , TRθ = 0, Tθβ = 0 + 2 R ∂R 1 − 2ν R ∂β 1 ∂ 2φ 1 ∂φ TRβ = 2μ ERβ = − 2 R ∂β∂R R ∂β ------------------------------------------------------------------------------Ans. With x = rer + zez = ReR , ψ = 0, φ = φ ( r , z ) = φˆ ( R, β ) we have,[see Eqs.(2.34.4) and (2.35.15)] 1 ∂φˆ ∂φ ∂φ ∂φˆ er + ez = eR + eβ 2μ u = −4 (1 − ν )ψ + ∇ ( x ⋅ψ + φ ) → 2μ u = R ∂β ∂r ∂z ∂R That is, in cylindrical coordinates ∂φ ∂φ 2 μ ur = , uθ = 0, 2 μ u z = ∂r ∂z and in spherical coordinates; 1 ∂φˆ ∂φˆ 2μ uR = , uθ = 0, 2μ uβ = ∂R R ∂β (b) The non zero strain components are: ∂ φˆ 2

In cylindrical coordinates: [See Eqs.(3.7.20)] Copyright 2010, Elsevier Inc

5-47


2μ Err = 2 μ

∂ur ∂ 2φ 2 μ ur 1 ∂φ ∂u ∂ 2φ = 2 , 2μ Eθθ = = , 2μ Ezz = 2μ z = 2 ∂r ∂r r r ∂r ∂z ∂z

∂u ⎞ ∂ 2φ ⎛ ∂u Erθ = 0, Eθ z = 0, 2μ Erz = μ ⎜ r + z ⎟ = ∂r ⎠ ∂r ∂z ⎝ ∂z ∂ 2φ 1 ∂φ ∂ 2φ 2μ e = 2μ ( Err + Eθθ + E zz ) = 2 + + 2 = ∇ 2φ = 0 r r ∂ ∂r ∂z In spherical coordinates: [see Eqs.(3.7.21)] ∂u ∂ 2φˆ 2 μ ∂uβ 2 μ u R 1 ∂ 2φˆ 1 ∂φˆ = 2 + 2μ ERR = 2 μ R = 2 , 2μ Eββ = + ∂R ∂R R ∂β R R ∂β 2 R ∂R 2μ u R 2 μ uβ cot β 1 ∂φˆ cot β ∂φˆ + = + 2 2μ Eθθ = , ERθ = 0, Eθβ = 0 R R R ∂R R ∂β 2μ ⎛ 1 ∂u R uβ ∂uβ ⎞ 1 ⎛ 1 ∂ 2φˆ 1 ∂φˆ 1 ∂ 2φˆ 1 ∂φˆ ⎞ − + = − + − 2 μ ER β = ⎜⎜ ⎟ ⎜ ⎟ R ∂R ⎠ 2 ⎝ R ∂β∂R R 2 ∂β R ∂R∂β R 2 ∂β ⎟⎠ 2 ⎝ R ∂β 1 ⎛ 2 ∂ 2φˆ 2 ∂φˆ ⎞ ⎛ 1 ∂ 2φˆ 1 ∂φˆ ⎞ = ⎜ − 2 =⎜ − 2 ⎟ ⎟ 2 ⎜⎝ R ∂β ∂R R ∂β ⎟⎠ ⎜⎝ R ∂β ∂R R ∂β ⎟⎠ ∂ 2φˆ 1 ∂ 2φˆ 1 ∂φˆ 1 ∂φˆ cot β ∂φˆ + + + 2 2μ e = 2μ ERR + Eββ + Eθθ = 2 + 2 ∂R R ∂β 2 R ∂R R ∂R R ∂β 2ˆ 2ˆ 1 ∂ φ 2 ∂φˆ cot β ∂φˆ ∂φ + + + 2 = ∇ 2φ = 0 [see Eq.2.35.37)] 2 2 2 R ∂R R ∂β R ∂β ∂R

(

)

(c) the stresses are: In cylindrical coordinates: 2 μν ∂ 2φ 2μν 1 ∂φ Trr = e + 2 μ Err = 2 , Tθθ = e + 2 μ Eθθ = 1 − 2ν 1 − 2ν r ∂r ∂r 2 2μν ∂φ ∂ 2φ Tzz = e + 2 μ E zz = 2 , Erθ = 0, Eθ z = 0, Trz = 2 μ Erz = 1 − 2ν ∂r ∂z ∂z

TRR

In spherical coordinates: 2μν 2μν 1 ∂ 2φˆ 1 ∂φˆ ∂ 2φˆ = + e + 2 μ ERR = 2 , Tββ = e + 2μ Eββ = 2 1 − 2ν 1 − 2ν ∂R R ∂β 2 R ∂R 2μν 1 ∂φˆ cot β ∂φˆ Tθθ = e + 2 μ Eθθ = , TRθ = 0, Tθβ = 0 + 2 R ∂R 1 − 2ν R ∂β

1 ∂ 2φ 1 ∂φ − 2 R ∂β∂R R ∂β These are the formulas given in Example 5.38.6. __________________________________________________________________ TRβ = 2μ ERβ =


5-48


For the potential functions given in Eq.(5.38.46), [see Example 5.38.7)], i.e., :

ψ = ψ ( R, β ) ez , φ = 0 , where ∇ 2ψ = 0 , shows that these functions generate the following displacements ui , dilatation e and the stresses Tij (in spherical coordinates) as given in Eq.(5.38.47) to (5.38.50)::

TRR Tββ

(a) Displacements: ⎧ ∂ψ ⎫ ∂ψ ⎫ ⎧ 2μ u R = − ⎨(3 − 4ν )ψ − R ⎬ cos β , 2 μ uβ = ⎨(3 − 4ν )ψ sin β + cos β ⎬ , uθ = 0 . ∂R ⎭ ∂β ⎭ ⎩ ⎩ ⎛ ∂ψ sin β ∂ψ ⎞ − (b) Dilation: 2μ e = − ( 2 − 4ν ) ⎜ cos β ⎟. ∂R R ∂β ⎠ ⎝ (c) Stresses 2 ⎛ ∂ψ ∂ ψ 2ν sin β ∂ψ ⎞ = − ⎜ 2(1 − ν ) cos β − R cos β − ⎟⎟ , 2 ⎜ β R R ∂ ∂ R ∂ ⎝ ⎠ ⎛ sin β ∂ψ cos β ∂ 2ψ ⎞ ∂ψ = − ⎜ ( 2ν − 1) cos β − (2 − 2ν ) − ⎟, ⎜ R ∂β R ∂β 2 ⎠⎟ ∂R ⎝ ⎛ ⎛ ∂ψ 1 ⎞ ∂ψ ⎞ Tθθ = − ⎜ ( 2ν − 1) cos β − ⎜ ( 2ν − 1) sin β + ⎟, ⎟ ∂R sin β ⎠ R∂β ⎠ ⎝ ⎝ ⎡ ∂ψ ∂ 2ψ ∂ψ ⎤ 1 TRβ = − ⎢ 2(1 − ν ) cos β − cos β − sin β (1 − 2ν ) ⎥ . TRθ = Tθβ = 0 R ∂β ∂β∂R ∂R ⎦⎥ ⎣⎢ ------------------------------------------------------------------------------Ans (a) with x = R eR , eR ez = cos β and x ⋅ψ = R eR ⋅ψ e z = ψ R cos β , e z = cos β eR − sin β e β , 2μ u = −4 (1 − ν )ψ + ∇ ( x ⋅ψ + φ ) = −4 (1 − ν )ψ e z + ∇ ( Rψ cos β ) →

(

)

2μ u = −4 (1 − ν )ψ cos β e R − sin β e β +

∂ 1 ∂ (ψ R cos β ) eR + (ψ R cos β ) e β ∂R R ∂β

⎡ ⎤ ∂ψ ∂ψ eR + cos β e β − ψ sin β e β ⎥ = −4 (1 − ν )ψ cos β e R − sin β e β + ⎢ψ cos β eR + R cos β ∂R ∂β ⎣ ⎦ ⎡ ∂ψ ⎤ ∂ψ ⎤ ⎡ = − cos β ⎢( 3 − 4ν )ψ − R eR + ⎢( 3 − 4ν )ψ sin β + cos β eβ . ⎥ ∂R ⎦ ∂β ⎥⎦ ⎣ ⎣ (b) The strain components are: ⎧⎪ ⎧⎪ ∂u ∂ψ ∂ 2ψ ∂ψ ⎫⎪ ∂ψ ∂ 2ψ ⎫⎪ −R 2 − = − − − 2μ ERR = 2 μ R = − ⎨(3 − 4ν ) cos β (2 4 ν ) R ⎬ ⎨ ⎬ cos β ∂R ∂R ∂R ⎭⎪ ∂R ∂R ∂R 2 ⎭⎪ ⎩⎪ ⎩⎪

(

)

⎛ 1 ∂uβ u R ⎞ 1 ∂ ⎧ ∂ψ ⎫ 1 ⎧ ∂ψ ⎫ 2μ Eββ = 2 μ ⎜ + ⎟= ⎨(3 − 4ν )ψ sin β + cos β ⎬ − ⎨(3 − 4ν )ψ − R ⎬ cos β R ⎠ R ∂β ⎩ ∂β ⎭ R ⎩ ∂R ⎭ ⎝ R ∂β ⎧⎪ ∂ψ ⎫ sin β ∂ψ ψ cos β ∂ 2ψ ⎫⎪ 1 ⎧ = ⎨(2 − 4ν ) + (3 − 4ν ) cos β + ⎬ − ⎨(3 − 4ν )ψ − R ⎬ cos β 2 ∂R ⎭ R ∂β R R ∂β ⎪⎭ R ⎩ ⎪⎩


5-49


= (2 − 4ν ) uβ cot β ⎛u 2μ Eθθ = 2 μ ⎜ R + R ⎝ R

(

2 μ 2 ERβ

)

sin β ∂ψ cos β ∂ 2ψ ∂ψ . + + cos β 2 R ∂β R ∂β ∂R

⎞ ∂ψ ⎫ 1⎧ cot β ⎬ cos β + ⎟ = − ⎨(3 − 4ν )ψ − R R R R ∂ ⎩ ⎭ ⎠

⎧ ∂ψ ⎫ ⎨(3 − 4ν )ψ sin β + cos β ⎬ ∂β ⎭ ⎩

⎛ ∂ψ cot β ∂ψ ⎞ =⎜ + ⎟ cos β . R ∂β ⎠ ⎝ ∂R ⎛ 1 ∂u R uβ ∂uβ ⎞ ∂ 2ψ ⎪⎫ ⎧ cos β ∂ψ ψ ∂ψ ⎫ ⎪⎧ = 2μ ⎜ − + − cos β ⎬ + ⎨(3 − 4ν ) − ⎬ sin β ⎟ = − ⎨(3 − 4ν ) ∂R ⎠ ∂β ∂R ⎪⎭ ⎩ R R ∂β R ∂R ⎭ ⎝ R ∂β ⎩⎪ ⎧ sin β cos β ∂ψ ⎫ ⎧⎪ ∂ψ ∂ 2ψ ⎫⎪ − ⎨(3 − 4ν )ψ + + − + (3 4 ν )sin β cos β ⎬ ⎨ ⎬ ∂R ∂R∂β ⎭⎪ R R ∂β ⎭ ⎪⎩ ⎩ cos β ∂ψ ∂ 2ψ ∂ψ . = −4(1 − ν ) + 2cos β + 2(1 − 2ν )sin β R ∂β ∂R∂β ∂R cos β ∂ψ ∂ 2ψ ∂ψ . + cos β + (1 − 2ν )sin β R ∂β ∂R∂β ∂R ⎧⎪ ⎛ ∂ψ cot β ∂ψ ⎞ ∂ψ ∂ 2ψ ⎪⎫ = − ⎨(2 − 4ν ) − R 2 ⎬ cos β + ⎜ + ⎟ cos β ∂R R ∂β ⎠ ∂R ⎭⎪ ⎝ ∂R ⎩⎪

i.e., 2μ ERβ = −2(1 − ν )

(

2μ ERR + Eθθ + Eββ

)

+(2 − 4ν )

sin β ∂ψ cos β ∂ 2ψ ∂ψ + + cos β 2 R ∂β R ∂β ∂R

∂ 2ψ ⎪⎫ cos β ∂ 2ψ 1 ∂ψ ⎪⎧ ∂ψ = ⎨4ν + R 2 ⎬ cos β + + cos 2 β + (2 − 4ν )sin 2 β 2 R ∂β R sin β ∂β ∂R ⎭⎪ ⎩⎪ ∂R

(

)

⎛ ∂ 2ψ 1 ∂ 2ψ cot β ∂ψ ⎞ sin β ∂ψ ∂ψ = R cos β ⎜ 2 + 2 + 2 + (2 − 4ν ) ⎟⎟ + 4ν cos β 2 ⎜ ∂R R ∂β ∂R R ∂β R ∂β ⎠ ⎝ ⎛ sin β ∂ψ ∂ψ ∂ψ sin β ∂ψ ⎞ ⎛ 2 ∂ψ ⎞ = R cos β ⎜ − + (2 − 4ν ) = − ( 2 − 4ν ) ⎜ cos β − ⎟. ⎟ + 4ν cos β R ∂β R ∂β ⎠ ∂R ∂R ⎝ R ∂R ⎠ ⎝ 1 ⎛ ∂ 2ψ ⎞ 2 ∂ψ cot β ∂ψ ∂ 2ψ where we have used, the relation: + + 2 = 0. ⎜ ⎟+ ∂R 2 R 2 ⎜⎝ ∂β 2 ⎟⎠ R ∂R R ∂β ⎛ ∂ψ sin β ∂ψ ⎞ − Thus, 2μ e = − ( 2 − 4ν ) ⎜ cos β ⎟. R ∂β ⎠ ∂R ⎝ (c) The stresses are: ⎛ ∂ψ sin β ∂ψ ⎞ 2 μν − with e = −2ν ⎜ cos β ⎟ , we have, ∂R 1 − 2ν R ∂β ⎠ ⎝


5-50


TRR =

⎛ 2μν ∂ψ sin β ∂ψ ⎞ ⎧⎪ ∂ψ ∂ 2ψ ⎪⎫ e + 2 μ ERR = −2ν ⎜ cos β ν (2 4 ) R − − − − ⎨ ⎬ cos β ⎟ R ∂β ⎠ ⎪⎩ 1 − 2ν ∂R ∂R ∂R 2 ⎪⎭ ⎝ ∂ψ 2ν sin β ∂ψ ∂ 2ψ + + R cos β 2 . ∂R ∂β R ∂R Similarly, 2μν ∂ψ sin β ∂ψ cos β ∂ 2ψ = e + 2 μ Eββ = − ( 2ν − 1) cos β + (2 − 2ν ) + . 1 − 2ν ∂R R ∂β R ∂β 2

= (−2 + 2ν ) cos β

Tββ

⎛ 2μν ∂ψ 1 ⎞ ∂ψ e + 2μ Eθθ = − ( 2ν − 1) cos β + ⎜ ( 2ν − 1) sin β + . ⎟ 1 − 2ν ∂R sin β ⎠ R∂β ⎝ cos β ∂ψ ∂ 2ψ ∂ψ . + cos β + sin β (1 − 2ν ) TRβ = 2 μ ERβ = −2(1 − ν ) TRθ = Tθβ = 0 R ∂β ∂β∂R ∂R _________________________________________________________________ Tθθ =

5.83

Show that (1 / R ) is a harmonic function (i.e., it satisfies the Laplace

Equation ∇ 2 (1 / R ) = 0 ), where R is the radial distance from the origin. ------------------------------------------------------------------------------∂R x1 ∂R x2 ∂R x3 = , = , = so that Ans (a) R 2 = x12 + x22 + x32 , therefore, ∂x1 R ∂x2 R ∂x3 R x1 ∂ ⎛1⎞ ∂2 ⎛ 1 ⎞ 1 ∂R 1 3x1 ⎛ x1 ⎞ 1 3x12 = − = − = − + = − + and . ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂x1 ⎝ R ⎠ R 2 ∂x1 R3 ∂x12 ⎝ R ⎠ R3 R 4 ⎝ R ⎠ R3 R5 ∂2 ⎛ 1 ⎞ ∂2 ⎛ 1 ⎞ 1 3 x22 1 3 x32 = − + = − + , ⎜ ⎟ ⎜ ⎟ ∂x22 ⎝ R ⎠ R3 R5 ∂x32 ⎝ R ⎠ R3 R5 Thus, 2 2 2 2 ∂ ∂ ∂ ⎞ ⎛ 1 ⎞ ⎛ 1 3 x12 ⎞ ⎛ 1 3 x22 ⎞ ⎛ 1 3x32 ⎞ ⎛1⎞ ⎛ ∂ = + + + + + ⎜ ⎟⎜ ⎟ = ⎜ − ⎟ ⎟ + ⎜− ⎟ + ⎜− ⎜ ⎟ ∂x j ∂x j ⎝ R ⎠ ⎜⎝ ∂x12 ∂x22 ∂x32 ⎟⎠ ⎝ R ⎠ ⎜⎝ R3 R5 ⎟⎠ ⎜⎝ R3 R 5 ⎟⎠ ⎜⎝ R 3 R5 ⎟⎠ Similarly,

=−

3 R

3

+

3( x12 + x22 + x32 ) R

5

=−

3 R

3

+

3 R3

= 0.

________________________________________________________________________

In Kelvin's problem, we used the potential function Ψ = ψ ez where in cylindrical A coordinates: ψ = , R 2 = r 2 + z 2 .Using the results in Example 5.38.6, obtain the stresses. R ------------------------------------------------------------------------------1 ∂ψ z ∂ 2ψ 3 z 2 1 Ans. ψ = , =− 3, = 5 − 3, 2 R ∂z R ∂z R R 2 2 2 ∂ψ r ∂ ψ 3r 1 ∂ψ ∂ ⎛ r ⎞ 3rz =− 3 → 2 = 5 − 3, = ⎜− 3 ⎟ = 5 , ∂r ∂z∂r ∂z ⎝ R ⎠ R R R R ∂r 5.84


5-51

Lai et al, Introduction to Continuum Mechanics 2 ∂ψ ∂ 2ψ z ⎞ z 3r 2 z ⎛ z ⎞ ⎛ 3r z + z 2 = −2ν ⎜ − 3 ⎟ + ⎜ 5 − 3 ⎟ = − (1 − 2ν ) 3 + 5 . ∂z ∂r R ⎠⎟ R R ⎝ R ⎠ ⎝⎜ R ∂ψ z ∂ψ ⎛ z ⎞ z⎛ r ⎞ ⎛ z ⎞ Tθθ = −2ν + = −2ν ⎜ − 3 ⎟ + ⎜ − 3 ⎟ = − (1 − 2ν ) ⎜ 3 ⎟ . ∂z r ∂r ⎝ R ⎠ r⎝ R ⎠ ⎝R ⎠ 2 2 ∂ψ ∂ψ ⎛ r ⎞ ⎛ 3rz ⎞ ⎛ r ⎞ ⎛ 3rz ⎞ +z = − (1 − 2ν ) ⎜ − 3 ⎟ + z ⎜ 5 ⎟ = (1 − 2ν ) ⎜ 3 ⎟ + ⎜ 5 ⎟ . Trz = − (1 − 2ν ) ∂r ∂r ∂z ⎝ R ⎠ ⎝R ⎠ ⎝ R ⎠ ⎝⎜ R ⎠⎟ ⎛ 3z 2 ∂ 2ψ ∂ψ 1 ⎞ z 3z 3 z = z ⎜ 5 − 3 ⎟ + 2(1 − ν ) 3 = 5 + (1 − 2ν ) 3 . Tzz = z 2 − 2(1 − ν ) ⎜ R ∂z ∂z R ⎟⎠ R R R ⎝ _________________________________________________________________

Trr = −2ν

5.85

Show that for ϕ = C ln( R + z ), R 2 = r 2 + z 2 , ∂ 2ϕ

⎧ z ⎫ 1 =C⎨ 3 − ⎬. R( R + z ) ⎭ ∂r ⎩R ------------------------------------------------------------------------------∂ϕ C r 1 ∂ϕ C = , = Ans. . ∂r ( R + z ) R r ∂r R ( R + z ) 2

⎧⎪ ⎫⎪ ⎧r ∂ 1 1 C r2 r2 ∂ r⎫ C 1 = + = − − + ⎨ ⎬ ⎨ ⎬ 2 ∂r 2 ⎩ R ∂r ( R + z ) ( R + z ) ∂r R ⎭ R( R + z ) ⎪⎩ R ( R + z ) R ⎭⎪ ⎧⎪ ⎧⎪ R 2 − r 2 + zR r 2 ⎫⎪ C r2 r 2 R ( R + z ) ⎫⎪ C = − 2+ − 2⎬ ⎨− ⎬= ⎨ R ( R + z ) ⎩⎪ R ( R + z ) R R ( R + z ) ⎭⎪ R ( R + z ) ⎩⎪ R ( R + z ) R ⎭⎪ ∂ 2ϕ

=

⎧⎪ z R 2 − z 2 ⎫⎪ ⎧⎪ Rz + z 2 − R 2 ⎪⎫ ⎧ z ⎫ 1 C C = − C . − = ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ 3 R( R + z ) ⎭ R ( R + z ) ⎩⎪ R R 2 ⎭⎪ R ( R + z ) ⎩⎪ R2 ⎩R ⎭⎪

______________________________________________________________________________________

5.86

Given the following potential functions:

ψ = (∂ϕ / ∂z )ez , φ = (1 − ν )ϕ , where ϕ = C ln( R + z ), R 2 = r 2 + z 2 . From the results of Example 5.38.4, and Eqs (i), (ii) (iii) of Section 5.40, obtain

{

}

Trr = C 3r 2 z / R5 − (1 − 2ν ) / [ R( R + z )] ,

{

Tzz = C (3z 3 / R5 ) , Trz = C (3rz 2 ) / R5 . -------------------------------------------------------------------------------Ans. 3 2 ∂ϕ ⎛ 1 ∂ϕ ⎞ ∂ ϕ Trr = z 2 + 2ν ⎜ ⎟+ ∂r ∂z ⎝ r ∂r ⎠ ∂r 2 2 ⎧⎪⎛ 3r 2 z ⎞ 1 − 2ν ⎫⎪ ⎛ z ⎞ ⎫⎪ 1 ⎞ 2ν 1 ⎪⎧ ⎛ 3r C = C ⎨z ⎜ 5 − 3 ⎟ + +⎜ 3 − = ⎨⎜⎜ 5 ⎟⎟ − ⎬, ⎟⎬ ⎜ R( R + z ) ⎠ ⎪⎭ R ⎠⎟ R ( R + z ) ⎝ R ⎪⎩ ⎝ R ⎪⎩⎝ R ⎠ R ( R + z ) ⎪⎭


5-52

}

Tθθ = C (1 − 2ν ) − z / R3 + 1 / [ R( R + z )] .

Lai et al, Introduction to Continuum Mechanics ∂ 2ϕ z ∂ 2ϕ 1 ∂ϕ ⎪⎫ ⎪⎧ + (1 − 2ν ) Tθθ = ⎨−2ν 2 + ⎬ r ∂r ∂z r ∂r ⎭⎪ ∂z ⎩⎪ C Cz C ⎛ Cz ⎞ z ⎛ Cr ⎞ , = −2ν ⎜ − 3 ⎟ + ⎜ − 3 ⎟ + (1 − 2ν ) = − 3 (1 − 2ν ) + (1 − 2ν ) R( R + z R( R + z R ⎝ R ⎠ r⎝ R ⎠ ⎧ z ⎫ 1 = C (1 − 2ν ) ⎨− 3 + ⎬ R( R + z ) ⎭ ⎩ R ⎧⎪ ⎛ z 3 z 3 ⎞ z ⎫⎪ ⎛ ∂ 3ϕ ⎞ ∂3ϕ ∂ 2ϕ 3z 3 3rz 2 = , Tzz = z 3 − 2 = C ⎨− ⎜ 3 − 5 ⎟ + 3 ⎬ = C 5 , Trz = ⎜ z C ⎟ ⎜ ∂r ∂z 2 ⎟ ⎜ ∂z ∂z R ⎟⎠ R ⎭⎪ R R5 ⎝ ⎠ ⎩⎪ ⎝ R Trθ = Tθ z = 0 . _________________________________________________________________ 5.87

The stresses in Boussinesq problem in cylindrical coordinate are given by:

Trr = −

Fz ⎧⎪ 3r 2 z (1 − 2ν ) ⎫⎪ − ⎨ ⎬, R ( R + z ) ⎭⎪ 2π ⎩⎪ R5

Tθθ = −

Fz (1 − 2ν ) ⎧ z ⎫ 1 ⎨− 3 + ⎬, R( R + z ) ⎭ 2π ⎩ R

Fz 3z 3 Fz 3rz 2 T , = − , Trθ = Tθ z = 0 . rz 2π R5 2π R5 Obtain the stresses in rectangular Cartesian coordinates. -------------------------------------------------------------------------------Ans. ⎡Txx Txy Txz ⎤ ⎡ cos θ − sin θ 0 ⎤ ⎡T 0 Trz ⎤ ⎡ cosθ sin θ 0⎤ rr ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢Tyx Tyy Tyz ⎥ = ⎢ sin θ cos θ 0 ⎥ ⎢ 0 Tθθ 0 ⎥ ⎢ − sin θ cosθ 0⎥ ⎢ ⎥ ⎢ 0 0 1 ⎥⎦ ⎢⎣Tzr 0 Tzz ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎣⎢Tzx Tyz Tzz ⎦⎥ ⎣ Tzz = −

⎡Trr cos 2 θ + Tθθ sin 2 θ ⎢ ⎢ (Trr − Tθθ ) sin θ cos θ ⎢ Trz cos θ ⎢⎣

(Trr − Tθθ ) sin θ cosθ Trr sin 2 θ + Tθθ cos 2 θ Trz sin θ

Trz cosθ ⎤ ⎥ Trz sin θ ⎥ . Thus, ⎥ 1 ⎥ ⎦

Txx = Trr cos 2 θ + Tθθ sin 2 θ =−

Fz (1 − 2ν ) ⎧ z ⎫ 2 Fz ⎧⎪ 3r 2 z (1 − 2ν ) ⎪⎫ 2 1 ⎨ 5 − ⎬ cos θ − ⎨− 3 + ⎬ sin θ 2π ⎩⎪ R 2π R ( R + z ) ⎭⎪ R( R + z ) ⎭ ⎩ R

=−

Fz 2π

⎡ 3 x 2 z (1 − 2ν ) (1 − 2ν ) z sin 2 θ + (1 − 2ν ) sin 2 θ ⎤ cos 2 θ − ⎢ 5 − ⎥ R( R + z ) R( R + z ) R3 ⎣⎢ R ⎦⎥

⎡ 3 x 2 z (1 − 2ν ) z (1 − 2ν ) ⎧ z ⎫⎤ 2 2 ( R z ) cos θ 2cos θ 1 + + − + ⎢ 5 − ⎨ ⎬⎥ R( R + z ) ⎩ R 2 R3 ⎭⎥⎦ ⎢⎣ R (1 − 2ν ) ⎧ z F ⎡ 3x 2 z (1 − 2ν ) z (1 − 2ν ) 2 2 ⎫⎤ =− z ⎢ 5 − + + ⎨ 2 ( R + z )cos θ − 2cos θ ⎬⎥ 3 2π ⎣⎢ R R( R + z ) R( R + z ) ⎩ R R ⎭⎦⎥ =−

Fz 2π


5-53


=−

(1 − 2ν ) ⎧⎪ R( z − R) + z 2 − R 2 ⎫⎪ 2 ⎤ Fz ⎡ 3 x 2 z (1 − 2ν ) z (1 − 2ν ) + + ⎢ 5 − ⎨ ⎬ cos θ ⎥ 2π ⎣⎢ R R ( R + z ) R ( R + z ) ⎩⎪ R3 R2 ⎭⎪ ⎦⎥

⎡ 3 x 2 z (1 − 2ν ) z (1 − 2ν ) (1 − 2ν ) ⎧⎪ R( z − R) + z 2 − R 2 ⎫⎪ 2 ⎤ + + ⎢ 5 − ⎨ ⎬ cos θ ⎥ R ( R + z ) R ( R + z ) ⎩⎪ R3 R2 ⎭⎪ ⎣⎢ R ⎦⎥ Now, x2 x2 r cos θ = x → cosθ = x / r , r 2 + z 2 = R 2 , cos 2 θ = 2 = R − z 2 ( R + z )( R − z )

=−

Fz 2π

Therefore, ⎤ (1 − 2ν ) ⎧⎪ R( z − R) + z 2 − R 2 ⎫⎪ F ⎡ 3x 2 z (1 − 2ν ) z (1 − 2ν ) x2 Txx = − z ⎢ 5 − + + ⎥ ⎨ ⎬ R ( R + z ) R( R + z ) ⎪⎩ 2π ⎢⎣ R R3 R2 ⎪⎭ ( R + z )( R − z ) ⎥⎦ =−

⎡ 3 x 2 z (1 − 2ν ) z (1 − 2ν ) (1 − 2ν ) ⎧⎪ 1 x 2 x 2 ⎫⎪⎤ + − + ⎢ 5 − ⎨ ⎬⎥ . R ( R + z ) R ( R + z ) ⎪⎩ R ( R + z ) R 2 ⎪⎭⎥⎦ R3 ⎢⎣ R _________________________________________________________________

Fz 2π

5.88 Obtain the variation of Tzz along the z axis for the case where the normal load on the surface of an elastic half-space is uniform with intensity qo , and the loaded area is a circle of radius ro with its center at the origin. -------------------------------------------------------------------------------Ans. Using Eq.(5.41.3), we have, ro r ′dr ′ 3 z 3 qo 2π r ′dr ′ , where = −3 z 3qo ∫ Tzz = − ∫ 5 r ′= o R '5 2π R' Ro dR ′ R′2 = r ′2 + z 2 → R′dR′ = r ′dr ′ → Tzz = −3 z 3qo ∫ z R ′4 R

qo z 3 qo z 3 ⎡ 1 ⎤ o → Tzz = −3z 3qo ⎢ − = − = − qo . q o ⎥ (ro2 + z 2 )3/2 Ro3 ⎣ 3R '3 ⎦ R '= z __________________________________________________________________ 5.89

For the potential function ψ = D1R −2 cos β ez , where ( R, β ,θ ) are the spherical coordinates with β as the azimuthal angle. Find TRR and Tβ R .

------------------------------------------------------------------------------∂ψ ∂ 2ψ ∂ψ = −2 D1R −3 cos β → 2 = 6 D1R −4 cos β , = − D1R −2 sin β . Ans. ψ = D1R −2 cos β → ∂R ∂ β ∂R From Example 5.38.7,


5-54


TRR = −2(1 − ν ) cos β

∂ψ ∂ 2ψ 2ν sin β ∂ψ + R cos β + ∂R ∂β R ∂R 2

2ν sin β ⎡ ⎤ = D1 ⎢ −2(1 − ν ) cos β −2 R −3 cos β + R cos β 6 R −4 cos β + − R −2 sin β ⎥ R ⎣ ⎦

(

)

(

)

(

)

= D1 ⎡ 2 ( 5 − ν ) cos 2 β − 2ν ⎤ R −3 . ⎣ ⎦ 2(1 − ν ) ∂ψ ∂ 2ψ ∂ψ cos β + cos β + sin β (1 − 2ν ) TRβ = − ∂β ∂β∂R ∂R R ⎡ 2(1 − ν ) ⎤ = D1 ⎢ − cos β − R −2 sin β + cos β 2 R −3 sin β + sin β (1 − 2ν ) −2 R −3 cos β ⎥ R ⎣ ⎦

(

)

(

)

(

)

= D1 ⎡ 2(1 − ν ) R −3 cos β sin β + 2 R −3 sin β cos β − 2 R −3 cos β sin β (1 − 2ν ) ⎤ ⎣ ⎦ = D1R −3 cos β sin β [ 2(1 −ν ) + 2 − 2(1 − 2ν )] = 2 D1R −3 (1 + ν ) cos β sin β . __________________________________________________________________ 5.90

(

)

For the potential function, φ% = φ% ( R, β ) = C1[ R −3 3cos 2 β − 1 / 2] + C2 R −1 ,

where ( R, β ,θ ) are the spherical coordinates with β as the azimuthal angle, obtain TRR and Tβ R . -------------------------------------------------------------------------------Ans. With ∂φ% = ( C1 / 2 ) −3R −4 3cos 2 β − 1 + ( C2 ) − R −2 φ% ( R, β ) = ( C1 / 2 ) R −3 3cos 2 β − 1 + C2 R −1 → ∂R ∂ 2φ% → 2 = 6C1R −5 3cos 2 β − 1 + 2C2 R −3 ∂R ∂φ% ∂ ∂φ% = −3C1R −3 sin β cos β → = 9C1R −4 sin β cos β . and ∂β ∂R ∂β From Example 5.38.6, we have ∂ 2φˆ TRR = 2 = 6C1R −5 3cos 2 β − 1 + 2C2 R −3 . ∂R 2 ˆ ˆ 1⎛ ∂ φ 1 ∂φ ⎞ 1 ⎡ 1 ⎤ −4 − −3C1R −3 sin β cos β ⎥ = 12C1R −5 sin β cos β TRβ = ⎜ ⎟⎟ = ⎢9C1R sin β cos β − ⎜ R ⎝ ∂β∂R R ∂β ⎠ R ⎣ R ⎦

(

(

)

(

)(

)

(

)

(

)

(

)

. TRθ = Tθβ = 0 . _________________________________________________________________

CHARTER 5, PART B 5.91

Demonstrate that if only E2 and E3 are nonzero, then Eq.(5.46.4) becomes


5-55

)

Lai et al, Introduction to Continuum Mechanics C23 ⎤ ⎡ E2 ⎤ ⎡C E3 ] ⎢ 22 ⎥⎢ ⎥. ⎣C32 C33 ⎦ ⎣ E3 ⎦ ------------------------------------------------------------------------------Ans. Eq. .(5.46.4) gives ⎡ C12 E2 + C13 E3 ⎤ ⎢C E + C E ⎥ 23 3 ⎥ ⎢ 22 2 ⎢ C E + C33 E3 ⎥ 2U = [ 0 E2 E3 0 0 0] ⎢ 23 2 ⎥ = ⎡⎣ E2 ( C22 E2 + C23 E3 ) + E3 ( C23 E2 + C33 E3 ) ⎤⎦ . ⎢C24 E2 + C34 E3 ⎥ ⎢C25 E2 + C35 E3 ⎥ ⎢ ⎥ ⎢⎣C26 E2 + C36 E3 ⎥⎦ 2U = [ E2

This is the same as C23 ⎤ ⎡ E2 ⎤ ⎡C E3 ] ⎢ 22 ⎥ ⎢ ⎥ = ⎡⎣ E2 ( C22 E2 + C23 E3 ) + E3 ( C23 E2 + C33 E3 ) ⎤⎦ . ⎣C23 C33 ⎦ ⎣ E3 ⎦ _________________________________________________________________

[ E2

5.92

Demonstrate that if only E1 and E3 are nonzero, then Eq.(5.46.4) becomes C13 ⎤ ⎡ E1 ⎤ ⎡C E3 ] ⎢ 11 ⎥⎢ ⎥ ⎣C31 C33 ⎦ ⎣ E3 ⎦ -------------------------------------------------------------------------------Ans. ⎡ C11 C12 C13 C14 C15 C16 ⎤ ⎡ E1 ⎤ ⎢C ⎥⎢ ⎥ ⎢ 12 C22 C23 C24 C25 C26 ⎥ ⎢ 0 ⎥ ⎢C C23 C33 C34 C35 C36 ⎥ ⎢ E3 ⎥ 2U = [ E1 0 E3 0 0 0] ⎢ 13 ⎥⎢ ⎥ ⎢C14 C24 C34 C44 C45 C46 ⎥ ⎢ 0 ⎥ ⎢C15 C25 C35 C45 C55 C56 ⎥ ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎣⎢C16 C26 C36 C46 C56 C66 ⎦⎥ ⎣⎢ 0 ⎦⎥ 2U = [ E1

= ⎡⎣ E1 ( C11E1 + C13 E3 ) + E3 ( C13 E1 + C33 E3 ) ⎤⎦ This is the same as ⎡ C11 C13 ⎤ ⎡ E1 ⎤ 1 [ E1 E3 ] ⎢C C ⎥ ⎢ E ⎥ = 2 ⎡⎣ E1 ( E1C11 + C13 E3 ) + E3 ( C13 E1 + C33 E3 )⎤⎦ = 2U 33 ⎦ ⎣ 3 ⎦ ⎣ 13 __________________________________________________________________ 5.93

Write stress strain laws for a monoclinic elastic solid in contracted notation, whose plane of symmetry is the x1 x2 plane.

-------------------------------------------------------------------------------Ans. All Cijkl = 0 where the indices ijkl contain an odd number of 3. Therefore,


5-56

Lai et al, Introduction to Continuum Mechanics ⎡ C11 C12 ⎢C ⎢ 12 C22 ⎢C C23 [C ] = ⎢ 13 0 ⎢ 0 ⎢ 0 0 ⎢ ⎢⎣C16 C26

C13 C23

0 0

0 0

C33 0

0 C44

0 C45

0 C36

C45 0

C55 0

C16 ⎤ C26 ⎥⎥ C36 ⎥ ⎥. 0 ⎥ 0 ⎥ ⎥ C66 ⎥⎦

________________________________________________________________________

5.94

Write stress strain laws for a monoclinic elastic solid in contracted notation, whose plane of symmetry is the x1 x3 plane.

-------------------------------------------------------------------------------Ans. All Cijkl = 0 where the indices ijkl contain an odd number of 2. Therefore, ⎡ C11 C12 ⎢C ⎢ 12 C22 ⎢C C23 [C ] = ⎢ 13 0 ⎢ 0 ⎢C15 C25 ⎢ 0 ⎢⎣ 0

C13 C23 C33

0 0 0

C15 C25 C35

0 C35 0

C44 0 C46

0 C55 0

⎤ ⎥ ⎥ ⎥ ⎥ C46 ⎥ 0 ⎥ ⎥ C66 ⎥⎦ 0 0 0

________________________________________________________________________

5.95

For transversely isotropic solid with e3 as the axis of transversely isotropy, show from ′ = Qmi Qnj Qrk Qsl Cmnrs that C1113 ′ = 0 (See Sect.5.50) the transformation law Cijkl ------------------------------------------------------------------------------Ans. Since Q33 = 1, Q13 = Q23 = Q31 = Q32 = 0 , therefore,

′ = Qm1Qn1Qr1Qs 3Cmnrs = Qm1Qn1Qr1Q33Cmnr 3 = Qm1Qn1Qr1Cmnr 3 = Q11Qn1Qr1C1nr 3 + Q21Qn1Qr1C2 nr 3 C1113 = Q11Q11Qr1C11r 3 + Q11Q21Qr1C12 r 3 + Q21Q11Qr1C21r 3 + Q21Q21Qr1C22 r 3 . Now, all Cijkl with odd number of either 1 or 2 are zero because e1 plane and e 2 -plane are planes ′ =0 of material symmetry. Thus, Qr1C11r 3 = Qr1C12 r 3 = Qr1C21r 3 = Qr1C22 r 3 = 0 . Thus, C1113 _________________________________________________________________________

5.96

Show that for a transversely isotropic elastic material with e3 as the axis of transverse isotropy, C1133 = C2233 (see Sect.5.50) .

--------------------------------------------------------------------------------Ans. e1′ = cos β e1 + sin β e2 , e′2 = − sin β e1 + cos β e2 , e′3 = e3 Q11 = cos β , Q12 = − sin β , Q21 = sin β , Q22 = cos β , Q33 = 1, Q31 = Q13 = Q23 = Q32 = 0 . Thus, ′ = Qm1Qn 2Qr 3Qs 3Cmnrs = Qm1Qn 2Q33Q33Cmn33 = Qm1Qn 2Cmn33 = Q11Qn 2C1n33 + Q21Qn 2C2 n33 C1233 = Q11Q12C1133 + Q11Q21C1233 + Q21Q12C2133 + Q21Q22C2233 .


5-57

Lai et al, Introduction to Continuum Mechanics Now C1233 = C2133 = 0 because e1 plane (as well as e2 -plane) is a plane of material symmetry. Thus, ′ = Q11Q12C1133 + Q21Q22C2233 = − cos β sin β C1133 + sin β cos β C2233 C1233 = ( −C1133 + C2233 ) cos β sin β .

′ = 0 , because e1′ is also a plane of symmetry. Thus C1133 = C2233 . Again, C1233 ________________________________________________________________________

5.97

Show that for a transversely isotropic elastic material with e3 as the axis of transverse isotropy(see Sect.5.50)

(sin β ) 2 C1111 + ⎡(cos β ) 2 − (sin β ) 2 ⎤ C1122 + 2 ⎡(cos β )2 − (sin β )2 ⎤ C1212 − (cos β )2 C2222 = 0 . ⎣ ⎦ ⎣ ⎦ -------------------------------------------------------------------------------Ans. Since Q13 = Q31 = Q23 = Q32 = 0 and Cijkl = 0 when the indices ijkl contain an odd number ′ = Qm1Qn 2Qr 2Qs 2Cmnrs C1222

of either 1 or 2, therefore, = Q11Qn 2Qr 2Qs 2C1nrs + Q21Qn 2Qr 2Qs 2C2 nrs

= Q11Q12Qr 2Qs 2C11rs + Q11Q22Qr 2Qs 2C12 rs + Q21Q12Qr 2Qs 2C21rs + Q21Q22Qr 2Qs 2C22 rs = Q11Q12Q12Q12C1111 + Q11Q12Q22Q22C1122 + Q11Q22Q12Q22C1212 + Q11Q22Q22Q12C1221 +Q21Q12Q12Q22C2112 + Q21Q12Q22Q12C2121 + Q21Q22Q12Q12C2211 + Q21Q22Q22Q22C2222 . Thus, 3 3 ′ = − cos β (sin β ) C1111 − sin β (cos β ) C1122 − (cos β )3 sin β C1212 − (cos β )3 sin β C1221 C1222 + (sin β )3 cos β C2112 + (sin β )3 cos β C2121 + (sin β )3 cos β C2211 + (cos β )3 sin β C2222 = − cos β sin β ⎡sin 2 β C1111 + (cos 2 β − sin 2 β )C1122 + 2(cos 2 β − sin 2 β )C1212 − cos 2 β C2222 ⎤ . ⎣ ⎦ where we have used Cijkl = C jikl , Cijkl = C jilk and Cijkl = Cklij . ′ = 0 because e1′ is also a plane of symmetry, therefore, Now, C1222 sin β C1111 + (cos 2 β − sin 2 β )C1122 + 2(cos 2 β − sin 2 β )C1212 − cos 2 β C2222 = 0 . 2

________________________________________________________________________

5.98 In Section 5.50, we obtained the reduction in the elastic coefficients for a transversely isotropic elastic solid by demanding that each S β plane is a plane of material symmetry. We ′ be the same for all β . Use this can also obtain the same reduction by demanding the Cijkl procedure to obtain the result: C1133 = C2233 . ------------------------------------------------------------------------------Ans. Since Q31 = Q13 = Q32 = Q23 = 0, Q33 = 1 , therefore, ′ = Qm1Qn1Qr 3Qs 3Cmnrs = Qm1Qn1Q33Q33Cmn33 = Qm1Qn1Cmn33 . C1133 Now, Cijkl = 0 when the indices contain an odd number of either 1 or 2, therefore, ′ = Q11Q11C1133 + Q21Q21C2233 = cos 2 β C1133 + sin 2 β C2233 . C1133

′ = C1133 for all β , therefore, Now, C1133 C1133 = cos 2 β C1133 + sin 2 β C2233 → C1133 sin 2 β = sin 2 β C2233 . Copyright 2010, Elsevier Inc

5-58

Lai et al, Introduction to Continuum Mechanics Thus, C1133 = C2233 . ________________________________________________________________________

5.99 Invert the compliance matrix for a transversely isotropic elastic solid to obtain the relationship between Cij and the engineering constants. That is, verify Eq. (5.53.2) and (5.53.3) by inverting the following matrix: −ν 21 / E1 −ν 31 / E3 ⎤ ⎡ 1 / E1 ⎢ [ A] = ⎢ −ν 21 / E1 1 / E1 −ν 31 / E3 ⎥⎥ ⎢⎣ −ν13 / E1 −ν 13 / E1 1 / E3 ⎥⎦ ------------------------------------------------------------------------------−ν 21 / E1 −ν 31 / E3 ⎤ ⎡ 1 / E1 ⎢ −ν 31 / E3 ⎥⎥ 1 / E1 Ans. [ A] = ⎢ −ν 21 / E1 ⎢⎣ −ν13 / E1 −ν 13 / E1 1 / E3 ⎥⎦ Δ = det [ A] =

1 E12 E3 Δ=

(1 − 2ν 21ν13ν 31 − 2ν13ν 31 −ν 21ν 21 ) =

1 E12 E3

(1 + ν 21 ) (1 −ν 21 − 2ν13ν 31 ) .

Now, ν 31 / E3 = ν13 / E1 →

1 E12 E3

(1 + ν 21 ) (1 −ν 21 − 2ν 312 ( E1 / E3 ) ) =

(

1 E12 E3

(1 + ν 21 ) D

)

2 where D = 1 − ν 21 − 2ν 31 ( E1 / E3 ) . 2 ⎡1 − ν 31 ( E1 / E3 ) ⎤ −ν 31 / E3 E12 E3 E1 1 1 / E1 1 ⎣ ⎦ C11 = 1 ν ν = − = ( 31 13 ) / E 1 / E ν − D (1 + ν 21 ) E1E3 D Δ 13 1 (1 + ν 21 ) 3

C22 = C11

C33

2 ⎡1 − ν 21 ( E3 / E1 ) ⎤ −ν 21 / E1 E1 1 1 / E1 1 ⎣ ⎦ 2 1 − ν 21 ( E3 / E1 ) = = = 1 / E3 D Δ −ν 21 / E1 ΔE1E3 (1 + ν 21 )

(

)

(

2 E1 ν 21 + ν 31 E1 / E3 1 −ν 21 / E1 −ν 31 / E3 1 C12 = − = ν 21 + ν 31ν 13 ) = ( 1 / E3 Δ −ν 13 / E1 ΔE1E3 (1 + ν 21 ) D

C13 =

ν E 1 −ν 21 / E1 −ν 31 / E3 1 = ν 21ν 31 + ν 31 ) = 31 1 ( −ν 31 / E3 ΔE1E3 D Δ 1 / E1

C23 = −

−ν 31 / E3 ν E 1 1 / E1 1 ν 31 + ν 21ν 31 ) = 31 1 = ( D Δ −ν 21 / E1 −ν 31 / E3 ΔE1E3

________________________________________________________________________

5.100 Obtain Eq.(5.53.6) from Eq. (5.53.2) and (5.53.3). ------------------------------------------------------------------------------Ans. From


5-59

)


(

2 2 ⎡1 − ν 31 E1 ν 21 + ν 31 E1 / E3 ( E1 / E3 ) ⎤ E1 ⎣ ⎦ C11 = and C12 = D (1 + ν 21 ) (1 + ν 21 ) D

(

2 2 E1 / E3 E1 ⎡1 − ν 31 ( E1 / E3 ) ⎤ E1 ν 21 + ν 31 ⎣ ⎦ C11 − C12 = − (1 + ν 21 ) D (1 + ν 21 ) D

=

E1

(1 + ν 21

1 −ν )D{

21

}

2 − 2ν 31 ( E1 / E3 ) =

E1

(1 + ν 21 ) D

Thus, [see Eq.5.53.5], 2G12 =

{D} =

)

)

E1 (1 + ν 21 )

E1 . (1 + ν 21 )

________________________________________________________________________

5.101

Invert the compliance matrix for an orthotropic elastic solid to obtain the relationship between Cij and the engineering constants. ------------------------------------------------------------------------------−1

⎡ 1 ν ⎤ ν − 21 − 31 ⎥ ⎢ E2 E3 ⎥ ⎢ E1 ⎡ C11 C12 C13 ⎤ ⎢ ν 32 ⎥ ν12 1 −1 ⎢ ⎥ − Ans. Let [ A] = ⎢ − ⎥ = ⎢C12 C22 C23 ⎥ E2 E3 ⎥ ⎢ E1 ⎢⎣C13 C23 C33 ⎥⎦ ⎢ ν ν 23 1 ⎥ 13 ⎢− ⎥ − E2 E3 ⎥⎦ ⎢⎣ E1 [1 −ν12ν 23ν 31 −ν13ν 21ν 32 −ν13ν 31 −ν 23ν 32 −ν 21ν 12 ] . Since det [ A] ≡ Δ = E1E2 E3 ⎛ E1ν 21 ⎞ ⎛ E2ν 32 ⎞ ⎛ E3ν 13 ⎞ ⎟⎜ ⎟⎜ ⎟ = ν 21ν 32ν 13 , therefore, ⎝ E2 ⎠ ⎝ E3 ⎠ ⎝ E1 ⎠ [1 − 2ν13ν 21ν 32 −ν13ν 31 −ν 23ν 32 −ν 21ν12 ] . Δ= E1E2 E3 Next

ν12ν 23ν 31 = ⎜

1 1 E2 C11 = Δ ν 23 − E2 − C12 = −

1 Δ

−

ν 21 E2

ν 23 E2

−

−

ν 32 E3 1 E3

=

1 ⎛ 1 1 ν 32 ν 23 ⎞ 1 − (1 −ν 32ν 23 ) , etc., ⎜ ⎟= Δ ⎝ E2 E3 E3 E2 ⎠ ΔE2 E3

ν 31 E3

1 E3

− =

1 1 (ν 21 + ν 31ν 23 ) , C13 = ΔE2 E3 Δ

ν 21 E2

1 E2


5-60

− −

ν 31 E3

ν 32 E3

=

1 (ν 31 + ν 21ν 32 ) ΔE2 E3


C23 = −

1 Δ

1 E1 −

ν 12 E1

− −

ν 31 E3

ν 32

=

1 (ν 32 + ν 31ν12 ) ΔE1E3

E3

_________________________________________________________________ 5.102 Obtain the restriction given in Eq.(5.54.8) for engineering constants for an orthotropic elastic solid -------------------------------------------------------------------------------Ans. ⎡ 1 ν ⎤ ν − 21 − 31 ⎥ ⎢ E2 E3 ⎥ ν ⎤ ⎡ 1 ⎢ E1 − 21 ⎥ ⎢ ⎢ ν 12 E E2 ν ⎥ 1 ⎥ = 1 (1 − ν 21ν 12 ) → 1 − ν 21ν 12 > 0 , − 32 ⎥ → det ⎢ 1 ⎢− ⎢ ν 12 1 ⎥ E1E2 E2 E3 ⎥ ⎢ E1 − ⎢ ⎥ ⎢ ν E2 ⎦ ν 1 ⎥ ⎣ E1 ⎢ − 13 − 23 ⎥ E2 E3 ⎦⎥ ⎣⎢ E1 ν ν ν ν E E 2 But, 12 = 21 → 1 − ν 21ν 12 = 1 − 21 21 1 > 0 → 2 > ν 21 . Also, E1 E2 E2 E1 1 − ν 21ν 12 = 1 − ⎡ 1 ⎢ E 2 Next, det ⎢ ⎢ ν 23 ⎢− ⎣ E2 But,

ν 32 E3

=

−

ν 23 E2

−

E1

>0→

E1 2 > ν 12 E2

ν 32 ⎤

E3 ⎥ ⎥ = 1 (1 − ν ν ) → 1 − ν ν > 0 . 32 23 32 23 1 ⎥ E2 E3 ⎥ E3 ⎦ E3 2 E2 = 1 − ν 32 , E2 E3 E 2 and 3 > ν 32 . E2

2 → 1 − ν 32ν 23 = 1 − ν 23

1 − ν 32ν 23 > 0 →

⎡ 1 ⎢ E 1 det ⎢ ⎢ ν 13 ⎢− ⎣ E1

2 ν12 E2

E2 2 > ν 23 E3 Also,

ν 31 ⎤

E3 ⎥ ⎥ = 1 (1 − ν ν ) = 1 ⎛⎜1 − ν 2 E3 ⎞⎟ = 1 ⎛⎜ 1 − ν 2 E1 ⎞⎟ 31 13 13 31 1 ⎥ E1E3 E1E3 ⎝ E1 ⎠ E1E3 ⎝ E3 ⎠ ⎥ E3 ⎦

(1 −ν 31ν13 ) > 0 → ν132 <

E E1 2 and ν 31 < 3 . E3 E1

________________________________________________________________________

5.103 Write down all the restrictions for the engineering constants for a monoclinic solid in determinant form (no need to expand the determinant). Copyright 2010, Elsevier Inc

5-61

Lai et al, Introduction to Continuum Mechanics -------------------------------------------------------------------------------−ν 21 / E2 −ν 31 / E3 η 41 / G4 0 0 ⎤ ⎡ T11 ⎤ ⎡ E11 ⎤ ⎡ 1 / E1 ⎢ E ⎥ ⎢ −ν / E −ν 32 / E3 η 42 / G4 1 / E2 0 0 ⎥⎥ ⎢⎢T22 ⎥⎥ ⎢ 22 ⎥ ⎢ 12 1 ⎢ E ⎥ ⎢ −ν / E −ν 23 / E2 1 / E3 η43 / G4 0 0 ⎥ ⎢T33 ⎥ Ans. ⎢ 33 ⎥ = ⎢ 13 1 ⎥⎢ ⎥ 1 / G4 0 0 ⎥ ⎢T23 ⎥ ⎢ 2 E23 ⎥ ⎢ η14 / E1 η24 / E2 η34 / E3 ⎢ 2 E31 ⎥ ⎢ μ65 / G6 ⎥ ⎢T31 ⎥ 0 0 0 0 1 / G5 ⎥⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 0 μ56 / G5 1 / G6 ⎦⎥ ⎣⎢T12 ⎦⎥ ⎢⎣ 2 E12 ⎥⎦ ⎣⎢ (i) E1 > 0, E2 > 0, E3 > 0, G4 > 0, G5 > 0, G6 > 0 (ii )

1 / E1 −ν 21 / E2 1 / E2 > 0, −ν 12 / E1 −ν 23 / E2 −ν 23 / E2

1 / G5 μ56 / E2

μ65 / G6

−ν 32 / E3 1 / E3 η 43 / G4 > 0, >0 1 / E3 η34 / E3 1 / G4

1 / E1 −ν 12 / E1 −ν 13 / E1

1 / E1 −ν 31 / E3 1 / E2 η42 / G4 1 / E1 η41 / G4 > 0, > 0, >0 1 / E3 −ν 13 / E1 η24 / E2 1 / G4 η14 / E1 1 / G4 (iii) −ν 21 / E2 −ν 31 / E3 1 / E2 −ν 32 / E3 η 42 / G4 1 / E2 −ν 32 / E3 > 0, −ν 23 / E2 1 / E3 η 43 / G4 > 0 −ν 23 / E2 1 / E3 η24 / E2 η34 / E3 1 / G4

1 / E1 −ν 13 / E η14 / E1

−ν 31 / E3 η41 / G4 1 / E1 −ν 21 / E2 η 41 / G4 1 / E3 η43 / G4 > 0, −ν12 / E1 1 / E2 η42 / G4 > 0 1 / G4 η34 / E3 1 / G4 η14 / E1 η24 / E2

1 / G6

> 0,

(iv) 1 / E1 −ν 21 / E2 −ν 31 / E3 η41 / G4 −ν 12 / E1 1 / E2 −ν 32 / E3 η42 / G4 >0 −ν 13 / E1 −ν 23 / E2 1 / E3 η43 / G4 η14 / E1 η24 / E2 η34 / E3 1 / G4 _________________________________________________________________

CHAPRTR 5, PART C 5.104 Show that if a tensor is objective, then its inverse is also objective . ------------------------------------------------------------------------------Ans. Let T be an objective tensor, then in a change of frame: x* = c(t ) + Q(t )(x − xo )

T* = Q(t )TQ(t )T . Taking the inverse of this equation, we get, since Q −1 = QT .

(

T−1* = Q(t )TQ(t )T

)

−1

= Q(t )T−1Q(t )T .

Thus, T−1 is objective.

________________________________________________________________________


5-62

Lai et al, Introduction to Continuum Mechanics 5.105 Show that the rate of deformation tensor D = [∇v + ( ∇v ) ] / 2 is objective. [See Example 5.56.2)]. ------------------------------------------------------------------------------& T . Thus, Ans. From Eq.(5.56.13), we have ∇ * v* = Q(t ) ( ∇v ) QT (t ) + QQ T

( )

T & (∇ * v*)T = Q(t ) ( ∇v ) Q T (t ) + Q Q

T

. Now,

( d / dt ) (QQT ) = 0 → Q& QT + QQ& T → Q& QT = −QQ& T . Thus, & T + Q(t ) ( ∇ v )T Q T (t ) − QQ & T = Q (t ) ⎡ ( ∇ v ) + ( ∇ v ) T ⎤ Q T (t ) ∇ * v * +(∇ * v*)T = Q(t ) ( ∇v ) Q T (t ) + QQ ⎢⎣ ⎥⎦ → D* = Q(t )DQT (t ) . ________________________________________________________________________

5.106 Show that in a change of frame, the spin tensor W = [∇v − ( ∇v ) ] / 2 transforms in T

& Q T . [See Example 5.56.2)]. accordance with the equation W* = Q(t ) WQ T (t ) + Q ------------------------------------------------------------------------------& T . Thus Ans. From Eq.(5.56.13), we have ∇ * v* = Q(t ) ( ∇v ) Q T (t ) + QQ

( )

T & (∇ * v*)T = Q(t ) ( ∇v ) Q T (t ) + Q Q

( d / dt ) ( QQT ) = 0 = Q& QT + QQ& T → QQ& T = −Q& QT ,

T

. Now,

T & QT . → (∇ * v*)T = Q(t ) ( ∇v ) Q T (t ) − Q

T & T Thus, ( ∇ * v * ) − (∇ * v*)T = Q(t )[( ∇v ) − ( ∇v ) ]Q T (t ) + 2QQ

& QT . → W* = Q(t ) WQT (t ) + Q ____________________________________________________________ 5.107 Show that in a change of frame, the material derivative of an objective tensor T & &T, & * = QTQ & T (t ) + Q(t )TQ T transforms in accordance with the equation T (t ) + Q(t )TQ where a super-dot indicates material derivative. Thus the material derivative of an objective tensor T is not objective. ------------------------------------------------------------------------------Ans. Since T is objective, therefore, in a change of frame, T* = Q(t )TQ T (t ) . Taking the material derivative of this equation and noting that t* = t , we have, T & & T . Since T & * ≠ Q(t )TQ & T (t ) , therefore, DT is non-objective. & * = QTQ & T + QTQ T + QTQ Dt ________________________________________________________________________

5.108 The second Rivlin-Ericksen tensor is defined by: & + A ( ∇v ) + ( ∇v )T A , where A & ≡ D A , where A = 2D = ∇v + ( ∇v )T . Show A2 = A 1 1 1 1 1 1 Dt that A 2 is objective. [See Prob.Error! Reference source not found. and Error! Reference source not found.].


5-63

Lai et al, Introduction to Continuum Mechanics ------------------------------------------------------------------------------Ans. From Prob. 5.105, we had, D* = Q(t )DQ T (t ) → A1* = Q(t ) A1Q T (t ) .

& *=Q & (t ) A QT (t ) + Q(t ) A & QT (t ) + Q(t ) A Q & T (t ) . →A 1 1 1 1

(i)

& We also have, from Eq.(5.56.13), ∇ * v* = Q(t ) ( ∇v ) Q (t ) + QQ . T

T

Thus A1 * ∇ * v * + ( ∇ * v * ) A1 * T

& T ] + [Q ( ∇v ) QT + QQ & T ]T QA QT = QA1Q T [Q ( ∇v ) QT + QQ 1

(

)

& T ] + [Q ( ∇v )T A QT + QQ & T QA QT ] . = [QA1 ( ∇v ) Q T + QA1Q T QQ 1 1 & T + QQ & T = 0 → QQ & T = −QQ & T , therefore, Since ( D / Dt )QQ T = 0 → QQ A1 * ∇ * v * + ( ∇ * v * ) A1 * = T

(

)

& T ] + [Q ( ∇v )T A Q T − QQ & T QA Q T ] [QA1 ( ∇v ) Q T − QA1Q T QQ 1 1 i.e.,

& T − QA & QT A1 * ∇ * v * + ( ∇ * v * ) A1 * = [QA1 ( ∇v ) Q + Q ( ∇v ) A1Q T ] − QA1Q 1 (i) and (ii) give & * +A * ∇ * v * + ( ∇ * v * )T A * →A 1 1 1 T

T

T

(ii)

& Q T + QA & Q T + QA Q & T + [QA ( ∇v ) QT + Q ( ∇v )T A QT ] − QA Q & T − QA & QT = QA 1 1 1 1 1 1 1 T & Q T + QA ( ∇v ) Q T + Q ( ∇v )T A Q T = Q ⎡ A & ⎤ T = QA 1 1 1 ⎢⎣ 1 + A1 ( ∇v ) + ( ∇ v ) A1 ⎥⎦ Q . & + A ( ∇v ) + ( ∇v )T A is objective. Thus, A

1

1

1

_________________________________________________________________

& + TW − WT , where 5.109 The Jaumann Derivative of a second order objective tensor T is : T W is the spin tensor. Show that the Jaumann derivative of T is objective. [See Prob. 5.106 and Prob. 5.107] -------------------------------------------------------------------------------Ans. We have, since T is objective, therefore, in a change of frame, T* = QTQ T . & QT and in Prob. 5.107, we had In Prob.5.106, we had W* = Q(t ) WQT (t ) + Q T & & T . Also, & * = QTQ & T (t ) + Q(t )TQ T (t ) + Q(t )TQ & T + QQ & T → QQ & T = −QQ &T ( D / Dt ) QQT = 0 → QQ

Thus,

(

)

& T = QTWQ T − QT Q & T, T * W* = QTQ QWQ + QTQ T −QQ T

T

& Q T QTQ T = QWTQT + Q & TQT . W * T* = QWQ T QTQ T + Q & T −Q & TQ T . → T * W * − W * T* = Q ( TW − WT ) Q T − QT Q Thus,


5-64

Lai et al, Introduction to Continuum Mechanics T & & T + Q ( TW − WT ) QT − QT Q & T −Q & TQT & * +T * W * − W * T* = QTQ & T + QTQ + QTQ T

& T + Q ( TW − WT ) QT . = QTQ

(

)

& * +T * W * − W * T* = Q T & + TW − WT QT . That is, T

(

)

& + TW − WT is objective. Therefore, the Jaumann derivative of T , that is , T ________________________________________________________________________

% is related to the first Piola-Kirchhoff stress 5.110 The second Piola Kirchhoff stress tensor T − 1 % = F T , or to the Cauchy stress tensor T by tensor T by the formula T −1

o −1 T

o

% = (detF)F T(F ) Show that, in a change of frame, T %*=T % . [See Example 5.56.3 and T Example 5.57.1] -------------------------------------------------------------------------------Ans. In Example 5.56.3 and Example 5.57.1, we obtained that in a change of frame, F* = Q(t )F and To * = QTo . Thus, % * = F *−1 T * = (Q(t )F) −1 QT = F −1Q −1QT = F −1T . That is, T %*=T %. T o

o

o

o

________________________________________________________________________

5.111 Starting from the constitutive assumption that T = H (F ) and T* = H (F*) , where T is Cauchy stress and F is deformation gradient, show that in order that the assumption be independent of observers, H (F ) must transform in accordance with the equation

QTQ T = H (QF) . (b) Choose Q = R T to obtain T = RH (U )R T , where R is the rotation % = h(U ) , where tensor associated with F and U is the right stretch tensor. (c) Show that T h = (det U) U −1H (U)U −1 . Since C = U 2 , therefore, we may write T = f (C) . ------------------------------------------------------------------------------Ans. (a) In a change of frame, T* = QTQT and F* = QF , therefore, T* = H (F*) → QTQT = H (QF ) . (b) From QTQ T = H (QF) , with Q = R T → R T TR = H (R T F ) . But R T F = R T ( RU) = U where

U is the right stretch tensor. Therefore, R T TR = H (R T F ) → T = RH (U) R T . (c) F = RU → R = FU −1 → R T = U −1F T , thus,

T = RH (U)R T → T = FU −1H (U)U −1F T → F −1T(F T )−1 = U −1H (U)U −1 . Now since J = det F = det U , we can write JF −1T(F T ) −1 = (det U) U −1H (U)U −1 . % and the right side The left side of the above equation is the second Piola-Kirchhoff stress tensor T % = h(U) , or since U 2 = C , one can write is a function of the right stretch tensor U . Thus, T

% = h(C) . T ________________________________________________________________________


5-65

Lai et al, Introduction to Continuum Mechanics 5.112 From r = ( 2α X + β )

1/2

θ =cY ,

z = Z , α =1 / c , obtain the right Cauchy-Green deformation tensor B . -------------------------------------------------------------------------------

,

Ans., we have, with r = ( 2α X + β )

1/2

θ =cY , z = Z , α ∂r ∂r −1/2

,

∂r 1 −1/2 = ( 2α X + β ) = =0 ( 2α ) = α ( 2α X + β ) = , ∂X 2 r ∂Y ∂Z ∂θ ∂θ ∂θ ∂z ∂z ∂z = 0, = c, = 0; = 0, = 0, =1 ∂X ∂Y ∂Z ∂X ∂Y ∂Z Thus, Using Eq.3.29.59 to 3.29.64, 2

2

2

2

2

2

2

2 ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ α ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ Brr = ⎜ +⎜ +⎜ = ⎜ ⎟ , Bθθ = ⎜ ⎟ ⎟ ⎟ ⎟ + ⎜ ∂Y ⎟ + ⎜ ∂Z ⎟ = ( rc ) X Y Z r X ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2

2

2

⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎛ ∂z ⎞ Bzz = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ =1, ⎝ ∂X ⎠ ⎝ ∂Y ⎠ ⎝ ∂Z ⎠ ⎛ ∂r ⎞⎛ ∂z Brz = ⎜ ⎟⎜ ⎝ ∂X ⎠⎝ ∂X ⎛ r ∂θ ⎞⎛ ∂z Bθ z = ⎜ ⎟⎜ ⎝ ∂X ⎠⎝ ∂X

⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ ⎞ ⎛ ∂r ⎞⎛ r ∂θ Brθ = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎝ ∂X ⎠⎝ ∂X ⎠ ⎝ ∂Y ⎠⎝ ∂Y ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎟ + ⎜ ∂Y ⎟⎜ ∂Y ⎟ + ⎜ ∂Z ⎟⎜ ∂Z ⎟ = 0 , ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎞ ⎛ r ∂θ ⎞⎛ ∂z ⎞ ⎛ r ∂θ ⎞⎛ ∂z ⎞ ⎟ + ⎜ ∂Y ⎟⎜ ∂Y ⎟ + ⎜ ∂Z ⎟⎜ ∂Z ⎟ = 0 ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

⎞ ⎟=0 ⎠

________________________________________________________________________

5.113 From r = λ1R,

θ = Θ + KZ ,

z = λ3 Z , λ12 λ3 = 1 , obtain the right Cauchy-Green deformation tensor B . -------------------------------------------------------------------------------Ans. With r = λ1R, θ = Θ + KZ , z = λ3 Z , λ12 λ3 = 1 , we have, r = λ1R,

θ = Θ + KZ ,

z = λ3 Z ,

λ12 λ3 = 1, ∂θ ∂θ

∂r ∂r ∂r ∂θ = λ1 , = 0, = 0, = 0, = 1, = K, ∂R ∂Θ ∂Z ∂R ∂Θ ∂Z ∂z ∂z ∂z = 0, = 0, = λ3 . ∂R ∂Θ ∂Z Using Eq. (3.29.19) to Eq. (3.29.24) and noting that ro ≡ R, θo ≡ Θ, zo ≡ Z , 2

Brθ

2

2

⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎛ ∂r ⎞ Brr = ⎜ +⎜ +⎜ = (λ1 ) 2 . ⎟ ⎟ ⎟ ⎝ ∂R ⎠ ⎝ R∂Θ ⎠ ⎝ ∂Z ⎠ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ ⎛ r ∂θ ⎞⎛ ∂r ⎞ =⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0 = Bθ r . ⎝ ∂R ⎠⎝ ∂R ⎠ ⎝ R∂Θ ⎠⎝ R∂Θ ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎠ 2

2

2

2

⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ∂θ ⎞ ⎛ r ⎞ 2 2 2 +⎜ +⎜ = ⎜ ⎟ + ( rK ) = ( λ1 ) + ( rK ) . Bθθ = ⎜ ⎟ ⎟ ⎟ ⎝ ∂R ⎠ ⎝ R∂Θ ⎠ ⎝ ∂Z ⎠ ⎝ R ⎠ 2

2

2

⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎛ ∂z ⎞ 2 Bzz = ⎜ ⎟ + ⎜ R∂Θ ⎟ + ⎜ ∂Z ⎟ = (λ3 ) . R ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠


5-66


Bzθ

⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ ⎛ ∂r ⎞⎛ ∂z ⎞ Brz = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ = 0 = Bzr . ⎝ ∂R ⎠⎝ ∂R ⎠ ⎝ R∂Θ ⎠⎝ R∂Θ ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎠ ⎛ ∂z ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ r ∂θ ⎞ ⎛ ∂z ⎞⎛ r ∂θ ⎞ =⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟ = λ3rK = Bθ z . ⎝ ∂R ⎠⎝ ∂R ⎠ ⎝ R∂Θ ⎠⎝ R∂Θ ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎠ ⎝ ∂Z ⎠⎝ ∂Z ⎠

________________________________________________________________________


5-67


CHAPTER 6 6.1 In Figure P 6-1, the gate AB is rectangular with width b = 60 cm and length L= 4 m. The gate is hinged at the upper edge A . Neglect the weight of the gate, find the reactional force at B . Take the specific weight of water to be 9800 N / m3 and neglect frictions.

30 o

3m

hinge

A

so

4m

B

s Figure P 6-1 ------------------------------------------------------------------------------Ans. Take the gate AB as a free body. With so measured from the water surface along the inclined plane to point A, s measured from point A along the length of the plate (AB) and α = 30o as shown in the figure, we have, dF = pdA = ⎡⎣ ρ g ( so + s ) sin α ⎤⎦ (bds ), thus , L ⎧⎪ L2 L3 ⎫⎪ = → = = + = + ⎬ . Therefore, M R L sdF b g s s s ds b g 0 ρ sin α ρ sin α ⎨ so ∑ A B ∫ ∫ (o ) 2 3 ⎭⎪ ⎪ ⎩ 0 A ⎧⎪ 3 4 ( 0.5 ) ⎫⎪ ⎧ s sin α L sin α ⎫ 4 + RB = bL ρ g ⎨ o ⎬ = ( 0.6 )( 4 )( 9800 ) ⎨ + ⎬ = 5.1× 10 N . 3 ⎭ 3 ⎪⎭ ⎩ 2 ⎪⎩ 2 _________________________________________________________________

6.2 The gate AB in Figure P 6-2 is 5 m long and 3 m wide. Neglect the weight of the gate, compute the water level h for which the gate will start to fall. Take the specific weight of water to be 9800 N / m3 . 20,000 N

20,000 N

A

A

W

∇

60

o

B

h

5m

5m

h

∇

60

o

F

B

Figure P 6-2 -------------------------------------------------------------------------------Ans. Consider the gate plus the triangular region of water above the gate as the free body diagram. Then, Horizontal force from water to gate: F = ρ g ( h / 2)(bh) = ρ gbh 2 / 2 acting at 1/3 from base.


Lai et al, Introduction to Continuum Mechanics Weight of water on gate: W=ρ gb(1 / 2)[h(h tan 30o )] = ρ gbh 2 / (2 3) .

∑ M B = 0 → W (1 / 3)(h / h3 =

3) + Fh / 3 = P ( AB) →

9 P( AB ) 9 ( 20000 ) (5) = = 15.31 → h = 2.48m. 2 ρ gb 2 ( 9800 )( 3)

__________________________________________________________________ 6.3 The liquids in the U-tube shown in Figure P 6-3 is in equilibrium. Find h2 as a function of ρ1 , ρ 2 , ρ3 , h1 and h3 . The liquids are immiscible. ρ3 ρ1

h3

h1 h2 ρ2

1

2

Figure P 6-3 -------------------------------------------------------------------------------Ans. p1 = ρ1 gh1 , p2 = ρ3 gh3 + ρ 2 gh2 , p1 = p2 → ρ1 gh1 = ρ3 gh3 + ρ 2 gh2 → h2 = ( ρ1h1 − ρ3h3 ) / ρ 2 . _________________________________________________________________ 6.4 In Figure P 6-4, ,the weight WR is supported by the weight WL , via the liquids in the container. The area under WR is twice that under WL . Find WR in terms of WL , ρ1 , ρ 2 , AL , and h

( ρ2 < ρ1and assume no mixing ) .

WR

WL

1

AL

h

AR

4

ρ

2

2

3 ρ1

Figure P 6-4 ------------------------------------------------------------------------------Ans. p3 = p2 = p1 + ρ1 gh , p4 = p3 − ρ 2 gh = p1 + ( ρ1 − ρ 2 ) gh WR = p4 AR = p1 AR + ( ρ1 − ρ 2 ) ghAR , i.e.,

WR = 2 p1 AL + 2 ( ρ1 − ρ 2 ) ghAL = 2WL + 2 ( ρ1 − ρ 2 ) ghAL _________________________________________________________________


Lai et al, Introduction to Continuum Mechanics 6.5 Referring to Figure P 6-5, the radius and length of the cylinder are r and L respectively, The specific weight of the liquid is γ . (a) Find the buoyancy force on the cylinder and (b) Find the resultant force on the cylindrical surface due to the water pressure. The centroid of a semi-circular area is 4r / 3π from the diameter.

Figure P 6-5 ------------------------------------------------------------------------------Ans. (a) Buoyancy force is the net upward force due to the water pressure on the left half of the boundary of the cylinder which is submerged in the water. It is therefore equal to the weight of the water displaced by this left half. That is, Buoyancy force = γ (π r 2 / 2) L .

(

)

(b) Horizontal water force: Fx = γ (2r / 2) ( 2rL ) = γ 2r 2 L . The line of action of Fx is 2r / 3 above the ground. The line of action of Fy (the buoyancy force) passes through the centroid of the semi-circular area, i.e., 4r / 3π left of the diameter. _________________________________________________________________ 6.6 A glass of water moves vertically upward with a constant acceleration a . Find the pressure at a point whose depth from the surface of the water is h . Take the atmospheric pressure to be pa . ------------------------------------------------------------------------------Ans. Let z axis be appointing vertically upward, then dp dp − − ρ g = ρa → − = ρ ( g + a) → p = −ρ ( g + a) z + C . dz dz At the instant of interest, let the origin be at the free surface, then C = pa , the atmospheric

pressure. Thus, p − pa = − ρ ( g + a ) z . At a point which is at z = − h, p − pa = ρ ( g + a ) h . __________________________________________________________________ 6.7 A glass of water moves with a constant acceleration a in the direction shown in Figure P 6-6. (a) Show that the free surface is a plane and find its angle of inclination and (b) find the pressure at the point A . Take the atmospheric pressure to be pa .


Lai et al, Introduction to Continuum Mechanics y x

a

θ

g ro

h

A

Figure P 6-6 ------------------------------------------------------------------------------Ans. (a) With respect to the coordinates shown, the governing equations are: ∂p ∂p ∂p = ρ a cosθ , (ii) − − ρ g = ρ a sin θ , (iii) − = 0 , thus (i) − ∂x ∂y ∂z ∂p df = . (iii) → p = p ( x, y ), (i) → p = − ( ρ a cos θ ) x + f ( y ) → ∂y dy df (ii) → = − ρ ( g + a sin θ ) → f = − ρ ( g + a sin θ ) y + C → p = − ( ρ a cosθ ) x − ρ ( g + a sin θ ) y + C dy . At the instant of interest, let the origin be at the center of the surface, then C = pa .and

p = − ( ρ a cos θ ) x − ρ ( g + a sin θ ) y + pa . On every point on the free surface, p = pa ,

therefore, − ( ρ a cos θ ) x − ρ ( g + a sin θ ) y = 0 . Thus, the free surface is a plane. The angle of inclinations is given by tan β =

dy a cos θ =− . dx g + a sin θ

(b) At the point A, x = −ro , y = −h . Thus, p = ( ρ a cos θ ) ro + ρ ( g + a sin θ ) h + pa _________________________________________________________________ 6.8 The slender U-tube shown in Figure P 6-7 moves horizontally to the right with an acceleration a . Determine the relation between a, l and h . h a

z

Figure P 6-7 -----------------------------------------------------------------------a a h al Ans. The slope of the free surface is given by − . Thus − = − → h = . g g l g _________________________________________________________________



A liquid in a container rotates with a constant angular velocity ω about a vertical axis.

Show that the free surface is a paraboloid given by z = r 2ω 2 / (2 g ) where the origin is on the axis of rotation and z is measured upward from the lowest point of the free surface. ------------------------------------------------------------------------------Ans. Let z be pointing vertically upward with the origin at the lowest point of the free surface. We have, ∂p ∂p ∂p df = ρ − rω 2 and (ii) − − ρ g = 0 → p = − ρ gz + f (r ) → = (i) − . ∂r ∂z ∂r dr ∂p df ρ r 2ω 2 ρ r 2ω 2 (i) → = = ρ rω 2 → f = + C , → p = − ρ gz + +C. 2 2 ∂r dr ρ r 2ω 2 At (r , z ) = (0,0) , p = pa , therefore, p = − ρ gz + + pa . The free surface is characterized 2 r 2ω 2 . by p = pa , therefore, the equation of the surface is: z = 2g _________________________________________________________________

(

)

6.10 The slender U-tube rotates with an angular velocity ω about the vertical axis shown in Figure P 6-8. Find the relation between δ h( ≡ h1 − h2 ), ω , r1 and r2 .

ω

h1 r1

o

r2

h2

Figure P 6-8 ------------------------------------------------------------------------------Ans. The equation for the free surface is given by (see the previous problem) z = r 2ω 2 / (2 g ) , where the origin is on the axis of rotation and z is measured upward from the lowest point of the free surface. Thus, we have, r 2ω 2 r 2ω 2 ω2 z1 = 1 and z2 = 2 → z1 − z2 = r12 -r22 , but, z1 − z2 = h1 − h2 2g 2g 2g

(

)

→ h1 − h2 = ( r12 -r22 )ω 2 / (2 g ) . ____________________________________________________________ 6.11 For minor altitude differences, the atmosphere can be assumed to have constant temperature. Find the pressure and density distribution for this case. The pressure p , density ρ and absolute temperature Θ are related by the ideal gas law p = ρ RΘ . -------------------------------------------------------------------------------Ans.: Let gravity be in the negative x3 direction, then we have ∂p / ∂x1 = 0, ∂p / ∂x2 = 0, ∂p / ∂x3 = − ρ g


(i)

Lai et al, Introduction to Continuum Mechanics Thus, p depends only on x3 . Let po denote the pressure at x = 0 , then, we have dp dp g g − g / RΘ ) x3 dx3 → ln p = − x3 + ln po → p = po e ( = −ρ g → =− dx3 p RΘ RΘ − g / RΘ ) x3 . If ρ is the density at x = 0 , then ρ = ρ e ( o

(ii)

o

3

____________________________________________________________ 6.12 In astrophysical applications, an atmosphere having the relation between the density ρ and the pressure p given by p / po = ( ρ / ρo ) , where po and ρo are some reference pressure and density, is known as a polytropic atmosphere. Find the distribution of pressure and density in a polytropic atmosphere. -----------------------------------------------------------------------n

Ans. Let z axis point upward, then dp / dz = − ρ g . From p / po = ( ρ / ρo ) , we have, n

ρ = Cp1/ n , where C = ρo po−1/ n . Thus, dp / dz = −Cp1/ n g → p −1/ n dp = −Cgdz →∫

p po

z

p −1/ n dp = − ∫ Cgdz . Thus, zo

(A) for n ≠ 1 , [n / (n − 1)] p(

n −1) / n

p po

p(

n −1) / n

z n −1 / n n −1 / n = − Cgz z → p( ) − po( ) = −[(n − 1) / n]Cg ( z − zo ) → o

(

)

n −1 / n = po( ) − [(n − 1) / n] ρo po−1/ n g ( z − zo ) = po−1/ n ⎡⎣ po − {(n − 1) / n}ρo g ( z − zo ) ⎤⎦

n / ( n −1) n −1 ⎤ −1/ n −1 ⎡ . p = po ( ) ⎢ po − ρ o g ( z − zo ) ⎥ n ⎣ ⎦ (B) for n = 1 , p dp z ∫ po p = − ∫zo Cgdz → ln( p / po ) = −Cg ( z − zo ) → p = po exp ⎡⎣−Cg ( z − zo )⎤⎦

→ p = po exp ⎡ − ρo po−1/ n g ( z − zo ) ⎤ ⎣ ⎦ __________________________________________________________________ 6.13 Given the following velocity field for a Newtonian liquid with viscosity

(

)

μ =0.982 mPa.s 2.05 × 10−5 lb×s/ft 2 : v1 = −c ( x1 + x2 ) , v2 = c ( x2 − x1 ) , v3 = 0, c = 1 s −1

For a plane whose normal is in the e1 direction, (a) find the excess of the total normal compressive stress over the pressure p, and (b) find the magnitude of the shearing stress. -----------------------------------------------------------------------------∂v Ans. (a) T11 = − p + 2μ D11 → ( −T11 ) − p = −2μ D11 , where D11 = 1 = −c = −1 s −1 , ∂x1 Thus, ( −T11 ) − p = −2 ( 0.982 )( −1) = 1.96 mPa.

⎛ ∂v ∂v ⎞ (b) T12 = 2μ D12 = μ ⎜ 1 + 2 ⎟ = −2c μ = −1.96 mPa. ⎝ ∂x2 ∂x1 ⎠ Thus, the magnitude of shearing stress = 1.96 mPa .

⎛ ∂v ∂v ⎞ T13 = 2μ D13 = μ ⎜ 1 + 3 ⎟ = 0 . ⎝ ∂x3 ∂x1 ⎠


Lai et al, Introduction to Continuum Mechanics __________________________________________________________________ 6.14 For a steady parallel flow of an incompressible linearly viscous fluid, if we take the flow direction to be e3 , (a) show that the velocity field is of the form v1 = 0, v2 = 0 and v3 = v( x1 , x2 ) (b) If v( x1 , x2 ) = kx2 , find the normal and shear stresses on the plane whose normal is in the direction of e2 + e3 in terms of viscosity μ and pressure p . (c) On what planes are the total normal stresses given by p . ------------------------------------------------------------------------------∂v ∂v ∂v ∂v Ans. (a) From the equation of continuity 1 + 2 + 3 = 0 , we get, 3 = 0 , thus v3 is ∂x1 ∂x2 ∂x3 ∂x3 independent of x3 i.e., v3 = v( x1 , x2 ) . (b) with v1 = 0, v2 = 0 and v3 = kx2 , we have, 0 0 ⎤ 0 ⎤ ⎡0 0 0⎤ ⎡0 ⎡− p 0 ⎢ ⎥ ⎢ ⎥ ⎢ [∇v ] = ⎢0 0 0⎥ → [ D] = ⎢0 0 k / 2⎥ → [ T] = − p [I ] + 2μ [ D] = ⎢ 0 − p μ k ⎥⎥ ⎢⎣ 0 k 0 ⎥⎦ ⎢⎣0 k / 2 ⎢⎣ 0 μ k − p ⎥⎦ 0 ⎥⎦ On the plane with n = (e 2 + e3 ) / 2 , 0 ⎤ ⎡0⎤ ⎡− p 0 ⎡ 0 ⎤ 1 ⎢ 1 ⎢ ⎥ ⎢ ⎥ [t ] = [T][n ] = ⎢ 0 − p μ k ⎥ ⎢1 ⎥ = ⎢ μ k − p ⎥⎥ → Tn = t ⋅ n = μ k − p . 2 2 ⎢⎣ 0 μ k − p ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ μ k − p ⎥⎦ 1 2 2 2 2 2 Ts2 = t − (Tn ) = ⎡⎢( μ k − p ) + ( μ k − p ) ⎤⎥ − ( μ k − p ) = 0 . ⎦ 2⎣ (c) 0 ⎤ ⎡ n1 ⎤ ⎡ − pn1 ⎤ ⎡− p 0 ⎢ ⎥ ⎢ ⎢ ⎥ [ t ] = ⎢ 0 − p μ k ⎥ ⎢ n2 ⎥ = ⎢ − pn2 + μ kn3 ⎥⎥ , n12 + n22 + n32 = 1 ⎢⎣ 0 μ k − p ⎥⎦ ⎢⎣ n3 ⎥⎦ ⎢⎣ μ kn2 − pn3 ⎥⎦

(

) (

)

→ Tn = t ⋅ n = − pn12 + − pn22 + μ kn3 n2 + μ kn2 n3 − pn32 = − p + 2μ kn3 n2 . Thus, − p + 2μ kn3 n2 = − p → n2 = 0 and/or n3 = 0 That is, on any plane ( n1 ,0, n3 ) and ( n1 , n2 ,0 ) , where n12 + n22 + n32 = 1 , the normal component of stress is − p , these include the three coordinate planes (1,0,0 ) , ( 0,1,0 ) and ( 0,0,1) . __________________________________________________________________

6.15 Given the following velocity field for a Newtonian incompressible fluid with a viscosity μ = 0.96 mPa.s :

(

)

v1 = k x12 − x22 , v2 = −2kx1 x2 , v3 = 0, k = 1 s −1m −1 . At the point (1,2,1)m and on the plane whose normal is in the direction of e1 , (a) find the excess of the total normal compressive stress over the pressure p and (b) find the magnitude of the shearing stress. -----------------------------------------------------------------------------Ans. (a)


Lai et al, Introduction to Continuum Mechanics 0⎤ ⎡ − p + 4μ kx1 ⎥ −2kx1 0 ⎥ = [ D] , [ T] = ⎢⎢ −4μ kx2 ⎢⎣ 0 0 ⎥⎦ 0 −8μ ⎡ − p + 4μ ⎢ At (1, 2,1) and for k = 1, [ T] = − ⎢ −8μ − p − 4μ ⎢⎣ 0 0 ⎡ 2kx1 [∇v ] = ⎢⎢ −2kx2 ⎢⎣ 0

−2kx2

On e1 -plane ( −T11 ) − p = −4μ = −3.84 mPa.

−4μ kx2 − p − 4 μ kx1 0

0 ⎤ 0 ⎥⎥ . − p ⎥⎦

0 ⎤ 0 ⎥⎥ . − p ⎥⎦

(b) on the same plane, Ts = 8μ = 7.68 mPa. _________________________________________________________________ 6.16 Do Problem 6.15 except that the plane has a normal in the direction 3e1 + 4e2 . ------------------------------------------------------------------------------Ans. −4 μ kx2 0 ⎤ ⎡ 2kx1 −2kx2 0 ⎤ ⎡ − p + 4μ kx1 ⎢ ⎥ ⎢ [∇v ] = ⎢ −2kx2 −2kx1 0⎥ = [ D] → [ T] = ⎢ −4μ kx2 − p − 4μ kx1 0 ⎥⎥ . − p ⎦⎥ 0 0 ⎦⎥ 0 0 ⎣⎢ 0 ⎣⎢ −8μ 0 ⎤ ⎡ − p + 4μ ⎢ − p − 4 μ 0 ⎥⎥ , At (1, 2,1) and for k = 1, [ T] = − ⎢ −8μ ⎢⎣ 0 − p ⎥⎦ 0 −8μ 0 ⎤ ⎡3 / 5⎤ ⎡ − p + 4μ ⎡ −3 p − 20μ ⎤ 1⎢ ⎢ ⎥ ⎢ ⎥ [t ] = [T][n ] = ⎢ −8μ − p − 4μ 0 ⎥ ⎢ 4 / 5⎥ = 5 ⎢ −4 p − 40μ ⎥⎥ . ⎢⎣ 0 ⎢⎣ ⎥⎦ − p ⎥⎦ ⎢⎣ 0 ⎥⎦ 0 0 1 44 μ 44 μ ⎡⎣( −3 p − 20 μ ) 3 + ( −4 p − 40 μ ) 4 ⎤⎦ = − p − Tn = t ⋅ n = . (−Tn ) − p = . 25 5 5 1 ⎡ 1 2 2 2⎤ − − + − − = μ μ 3 p 20 4 p 40 25 p 2 + 440 p μ + 2000μ 2 (b) t = ( ) ( ) ⎦⎥ 25 25 ⎣⎢ 88μ p 1936 2 64 8μ + μ . (Tn )2 = p 2 + (Ts )2 = t 2 − (Tn )2 = μ 2 → Ts = . 5 25 25 5 __________________________________________________________________

(

)

6.17 Use the results of Sect. 2.34., chapter 2 and the constitutive equations for the Newtonian viscous fluid, verify the Navier Stokes Equation in the r -direction for cylindrical coordinates, i.e., Eq. (6.8.1). ------------------------------------------------------------------------------Ans. For a Newtonian fluid, the stress tensor in cylindrical coordinates is given by:


Lai et al, Introduction to Continuum Mechanics ⎡ ∂vr ∂v ⎞ ⎤ ⎛ 1 ∂vr vθ ∂vθ ⎞ ⎛ ∂v μ⎜ μ⎜ r + z ⎟ ⎥ − + ⎢ − p + 2μ ⎟ r ∂r ∂r ⎠ ∂r ⎠ ⎥ ⎝ ∂z ⎝ r ∂θ ⎢ ⎢ 1 ∂vz ⎞ ⎥ ⎛ 1 ∂vθ vr ⎞ ⎛ ∂v − p + 2μ ⎜ + ⎟ μ⎜ θ + [T] = ⎢ T21 ⎟⎥ r ⎠ ⎝ r ∂θ ⎝ ∂z r ∂θ ⎠ ⎥ ⎢ ⎢ ∂v ⎥ T31 T32 − p + 2μ z ⎥ ⎢ ∂z ⎥⎦ ⎢⎣ The Equations of motion in terms of the stress components in the r-direction is [see Eq.(4.8.1)]: ∂Trr 1 ∂Trθ Trr − Tθθ ∂Trz + + + + ρ Br = ρ ar ∂r ∂z r ∂θ r We also have the equation of continuity [see Eq.(2.34.6) or Eq.(6.8.4)]: ∂vr ⎛ 1 ∂vθ vr ⎞ ∂vz + + ⎟+ =0 ∂r ⎜⎝ r ∂θ r ⎠ ∂z ∂vr ∂T ∂ 2v ∂p → rr = − + 2 μ 2r ∂r ∂r ∂r ∂r ⎛ 1 ∂ 2 vr 1 ∂vθ 1 ∂ 2 vθ ⎛ 1 ∂vr vθ ∂vθ ⎞ 1 ∂Trθ − + → = − 2 + Trθ = μ ⎜ μ ⎜⎜ 2 2 ∂r ⎟⎠ r ∂θ ∂ θ r r ∂r ∂θ ∂ θ r r ⎝ r ∂θ ⎝ Trr − Tθθ ⎛ 1 ∂vr 1 ∂vθ vr ⎞ = 2μ ⎜ − − ⎟ r ⎝ r ∂r r 2 ∂θ r 2 ⎠

Now, Trr = − p + 2 μ

⎞ ⎟⎟ ⎠

⎛ ∂ 2v ∂v ⎞ ∂T ∂ 2 vz ⎞ ⎛ ∂v Trz = μ ⎜ r + z ⎟ → rz = μ ⎜ 2r + ⎟ ⎜ ∂z ∂r ⎠ ∂z ∂r ∂z ⎟⎠ ⎝ ∂z ⎝ Thus, ∂T ∂ 2v ∂p 1 ∂Trθ Trr − Tθθ ∂Trz + + = − + 2 μ 2r (divT)r = rr + ∂r r ∂θ r ∂z ∂r ∂r ⎛ 1 ∂ 2 vr 1 ∂vθ 1 ∂ 2 vθ ⎞ ⎛ 1 ∂vr 1 ∂vθ vr +μ ⎜ 2 − 2 + + 2μ ⎜ − − ⎟ 2 ⎜ r ∂θ r ∂r ∂θ ⎠⎟ ⎝ r ∂r r 2 ∂θ r 2 ⎝ r ∂θ ⎡ ∂ 2v ∂p 1 ∂ 2 vr ∂ 2 vr 1 ∂vr 2 ∂vθ vr ⎤ = − + μ ⎢ 2r + 2 + 2 + − − ⎥ ∂r r ∂r r 2 ∂θ r 2 ⎥⎦ ∂z r ∂θ 2 ⎢⎣ ∂r

⎛ ∂ 2 vr ∂ 2 vz ⎞ ⎞ + + μ ⎜ ⎟ ⎜ 2 ∂r ∂z ⎟⎟ ⎠ ⎝ ∂z ⎠

⎛ ∂ 2v 1 ∂v 1 ∂ 2 vθ vr 1 ∂vr ∂ 2 vz ⎞ + μ ⎜ 2r − 2 θ + − + + ⎟ ⎜ ∂r r ∂θ r ∂r ∂θ r 2 r ∂r ∂r ∂z ⎟⎠ ⎝ But using the equation of continuity, we have, ⎛ ∂ 2 vr 1 ∂vθ 1 ∂ 2 vθ vr 1 ∂vr ∂ 2 vz ⎞ ∂ ⎛ ∂vr 1 ∂vθ vr ∂vz + − + + + + + ⎜⎜ 2 − 2 ⎟= ⎜ ∂z r r ∂θ r ∂r ∂θ r 2 r ∂r ∂r ∂z ⎠⎟ ∂r ⎝ ∂r r ∂θ ⎝ ∂r Thus, (divT)r +ρ Br =ρ ar →

⎞ ⎟=0 ⎠

⎡ ∂ 2v ∂p 1 ∂ 2 vr ∂ 2 vr 1 ∂vr 2 ∂vθ vr ⎤ + μ ⎢ 2r + 2 + 2 + − − ⎥ + ρ Br =ρ ar ∂r r ∂r r 2 ∂θ r 2 ⎥⎦ ∂z r ∂θ 2 ⎢⎣ ∂r __________________________________________________________________ −


Lai et al, Introduction to Continuum Mechanics 6.18 Use the results of Sect. 2.35, chapter 2 and the constitutive equations for the Newtonian viscous fluid, verify Navier-Stokes Equation in the r-direction in spherical coordinates, i.e., Eqs. (6.8.5). -------------------------------------------------------------------------------Ans. For a Newtonian fluid, the stress tensor in spherical coordinates is given by: ⎡ ⎤ ⎡⎛ 1 ∂vr vφ ⎞ ∂vφ ⎤ ⎡⎛ 1 ∂vr vθ ⎞ ∂vθ ⎤ ∂v − ⎟+ − ⎟+ μ ⎢⎜ ⎢ − p + 2 μ r μ ⎢⎜ ⎥ ⎥ ⎥ r ⎠ ∂r ⎦ r ⎠ ∂r ⎥⎦ ∂r ⎢ ⎢⎣⎝ r sin θ ∂φ ⎥ ⎣⎝ r ∂θ ⎢ ⎥ ⎡⎛ 1 ∂vθ vφ cot θ ⎞ 1 ∂vφ ⎤ ⎥ ⎛ 1 ∂vθ vr ⎞ ⎢ − p + 2μ ⎜ + ⎟ − μ ⎢⎜ [T] = ⎢ Tθφ ⎥ ⎟+ r ∂θ r ⎠ r sin θ ∂φ r ⎠ r ∂θ ⎦⎥ ⎥ ⎝ ⎢ ⎝ ⎣ ⎢ ⎥ ⎢ v ∂ ⎛ 1 v cot θ ⎞ ⎥⎥ v φ ⎢ − p + 2μ ⎜ + r + θ Tφ r Tφθ ⎟ r r ⎢ ⎝ r sin θ ∂φ ⎠ ⎦⎥ ⎣ The equation of motion in the r-direction is given by [see Eq.(4.8.4)]

(

)

(

)

2 1 ∂ r Trr 1 ∂ (Trθ sin θ ) 1 ∂Trφ Tθθ + Tφφ + + + ρ Br = ρ ar . r sin θ r sin θ ∂φ r ∂r ∂θ r2 We also have the equation of continuity [see Eq.(2.35.26) or Eq.(6.8.8)] ∂vr ⎛ 1 ∂vθ 2vr ⎞ ⎛ 1 ∂vφ vθ cot θ ⎞ + + +⎜ + ⎟ =0. ∂r ⎜⎝ r ∂θ r ⎟⎠ ⎝ r sin θ ∂φ r ⎠ Now, 2 ⎡ ∂ 2 vr 2 ∂vr ⎤ 1 ∂ r Trr ∂p 2 p μ 2 = − − + ⎢ 2 + ⎥, r r ∂r ⎦⎥ ∂r ∂r r2 ⎣⎢ ∂r

1 ∂ (Trθ sin θ ) r sin θ ∂θ

⎡⎛ 1 ∂v v = μ ⎢⎜ 2 r − θ2 ⎣⎝ r ∂θ r

⎞ 1 ∂vθ ⎟+ ⎠ r ∂r

⎡⎛ 1 ∂ 2 v ⎤ 1 ∂vθ ⎞ 1 ∂ 2 vθ ⎤ r + − θ μ cot ⎢ ⎥. ⎜ ⎟+ ⎥ 2 ⎜ 2 r 2 ∂θ ⎠⎟ r ∂r ∂θ ⎦⎥ ⎦ ⎣⎢⎝ r ∂θ

2 ⎡⎛ ∂vφ ⎞ ∂ 2 vr 1 ∂Trφ 1 1 1 ∂ vφ ⎤ ⎥ = μ ⎢⎜ 2 2 − + ⎟ r sin θ ∂φ ⎢⎣⎜⎝ r sin θ ∂φ 2 r 2 sin θ ∂φ ⎟⎠ r sin θ ∂r ∂φ ⎥⎦ Tθθ + Tφφ ∂vφ vθ cot θ ⎞ ⎛ 1 2v ⎞ 2p ⎛ 1 ∂v − =+ − 2μ ⎜ 2 θ + 2r ⎟ − 2 μ ⎜ 2 + ⎟. r r r ⎠ r2 ⎠ ⎝ r ∂θ ⎝ r sin θ ∂φ Thus.



(

)

2 Tθθ + Tφφ 1 ∂ r Trr 1 ∂ (Trθ sin θ ) 1 ∂Trφ + + − = 2 ∂r r sin θ ∂θ r sin θ ∂φ r r

−

⎡ ∂ 2v ⎡⎛ 1 ∂vr vθ 2 ∂vr ⎤ ∂p 2 p − + 2 μ ⎢ 2r + − 2 ⎥ + μ ⎢⎜ 2 r r ∂r ⎥⎦ ∂r ⎢⎣ ∂r ⎣⎝ r ∂θ r

⎞ 1 ∂vθ ⎤ ⎟ + r ∂r ⎥ cot θ ⎠ ⎦

2 ⎡⎛ ⎡⎛ 1 ∂ 2 v ∂vφ ⎞ ∂ 2 vr 1 ∂vθ ⎞ 1 ∂ 2 vθ ⎤ 1 1 1 ∂ vφ ⎤ r ⎥ + μ ⎢⎜ 2 − 2 − ⎥ + μ ⎢⎜⎜ 2 2 ⎟+ ⎟+ 2 r ∂θ ⎟⎠ r ∂r ∂θ ⎥⎦ ⎢⎣⎝ r sin θ ∂φ 2 r 2 sin θ ∂φ ⎟⎠ r sin θ ∂r ∂φ ⎥⎦ ⎢⎣⎜⎝ r ∂θ ⎛ 1 ∂vφ vθ cot θ ⎞ 2v ⎞ 2p ⎛ 1 ∂v + − 2 μ ⎜ 2 θ + 2r ⎟ − 2μ ⎜ 2 + ⎟ r r ⎠ r2 ⎠ ⎝ r ∂θ ⎝ r sin θ ∂φ ⎛ ∂ 2v ∂ 2 vr ⎞ 2 ∂vr 2vr 1 ∂ 2 vr 1 ∂vr 1 ∂p cot θ = − + μ ⎜ 2r + − 2 + 2 + + ⎟ ⎜ ∂r r ∂r ∂r r r ∂θ 2 r 2 ∂θ r 2 sin 2 θ ∂φ 2 ⎟⎠ ⎝ ⎛ 2v cot θ ⎞ 2 ∂vφ 2 ∂v +μ ⎜ − 2 − 2 θ − θ 2 ⎟ r ⎝ r sin θ ∂φ r ∂θ ⎠ 2 ⎡ ∂ 2v ∂vφ vθ 2 ∂vr 2vr 1 ∂ 2 vθ 1 ∂vθ 1 ∂vθ 1 1 ∂ vφ ⎤ r ⎥ cot θ − 2 cot θ − 2 +μ ⎢ 2 + − 2 + − + + r ∂r r ∂r ∂θ r 2 ∂θ r ∂r r r r sin θ ∂φ r sin θ ∂r ∂φ ⎥⎦ ⎢⎣ ∂r Now, differentiate the equation of continuity with respect to r, we have, 1 ∂vφ vθ cot θ ⎞ ∂ ⎛ ∂vr 2vr 1 ∂vθ + + + + ⎜ ⎟ = 0 , that is, ∂r ⎝ ∂r r r ∂θ r sin θ ∂φ r ⎠

2 2 ∂vr 2vr 1 ∂ 2 vθ 1 ∂vθ 1 ∂ vφ 1 ∂vφ − + − + − 2 + 2 2 2 r ∂r r ∂r ∂θ r ∂θ r sin θ ∂r ∂φ r sin θ ∂φ ∂r r cot θ ∂vθ vθ cot θ − =0 r ∂r r2

∂ 2 vr

+

(

)

2 Tθθ + Tφφ 1 ∂ r Trr 1 ∂ (Trθ sin θ ) 1 ∂Trφ + + − Thus, 2 ∂r r sin θ ∂θ r sin θ ∂φ r r

=−

⎛ ∂ 2v ∂ 2 vr ∂p 2 ∂vr 2vr 1 ∂ 2 vr 1 ∂vr 1 + μ ⎜ 2r + − 2 + 2 + + θ cot ⎜ ∂r ∂r r ∂r r r ∂θ 2 r 2 ∂θ r 2 sin 2 θ ∂φ 2 ⎝

⎛ 2v cot θ ⎞ 2 ∂vφ 2 ∂v +μ ⎜ − 2 − 2 θ − θ 2 ⎟ r ⎝ r sin θ ∂φ r ∂θ ⎠ ⎡∂ 1 ∂ 2 ∂vr ⎞ ∂ 2 vr ⎤ ∂p ∂ ⎛ 1 1 =− +μ⎢ + + θ r v sin ⎥ r ⎜ ⎟ 2 ∂r ∂θ ⎠ r 2 sin 2 θ ∂φ 2 ⎥⎦ r 2 sin θ ∂θ ⎝ ⎢⎣ ∂r r ∂r ⎡ ⎤ 2 ∂vφ 2 ∂ + μ ⎢− 2 − 2 ( vθ sin θ ) ⎥ ⎣ r sin θ ∂φ r sin θ ∂θ ⎦ Finally, the Navier-Stokes equation in r direction is:

(

)


⎞ ⎟⎟ ⎠

Lai et al, Introduction to Continuum Mechanics ⎡∂ 1 ∂ 2 ∂vr ⎞ ∂ 2 vr ⎤ ∂p ∂ ⎛ 1 1 +μ⎢ + + θ r v sin ⎥ r ⎜ 2 ∂r ∂θ ⎟⎠ r 2 sin 2 θ ∂φ 2 ⎥⎦ r 2 sin θ ∂θ ⎝ ⎣⎢ ∂r r ∂r ⎡ ⎤ 2 ∂vφ 2 ∂ + μ ⎢− 2 − 2 ( vθ sin θ ) ⎥ + ρ Br = ρ ar ⎣ r sin θ ∂φ r sin θ ∂θ ⎦ __________________________________________________________________

(

=−

)

6.19 Show that for a steady flow, the streamline containing a point P coincides with the pathline for a particle which passes through the point P at some time t . ------------------------------------------------------------------------------Ans. For a steady flow, the velocity at every point on a streamline does not change with time. Therefore, any particle, which is at a point P on the streamline at a given time t, will move along the streamline at all time. That is, its pathline coincides with the streamline containing the point P. We can also demonstrate this mathematically as follows: For steady flow, the velocity field is independent of time, that is, v = v (x) . Let x = x(t ) be the pathline, then, the differential system for the pathline is: dx = v {x ( t )} , subjected to the condition x ( to ) = xo (1) dt Let x = x( s ) be the parametric equation for the streamline passing though xo , then the differential system for the streamline is: dx = v {x ( s )} , subjected to the condition x ( so ) = xo (2) ds The two differential systems are identical. They determine the same curve. _________________________________________________________________ 6.20 Given the two dimensional velocity field kx x v1 = 1 2 , v2 = 0 1 + kx2t

(a) Find the streamline at time t , which passes through the spatial point (α1 , α 2 ) and,

(b) find the pathline equation x = x(t ) for a particle which is at ( X1 , X 2 ) at time to ------------------------------------------------------------------------------Ans. (a) Since the flow is in e1 direction only, therefore, both the streamline and the pathline are straight line in the e1 direction. The streamline equation which passes through the spatial point

(α1 ,α 2 )

is simply x2 = α 2 .

(b) The pathline for a particle which is at ( X1 , X 2 ) at time to is simply x2 = X 2 . To find the time history of the particle along the pathline, i.e., to find x = x(t ) with X = x(to ) ,we have, x

t

1 dx1 kx X dx kX 2 x 1 + kX 2t = 1 2 → ∫ 1=∫ dt → ln 1 = ln , dt 1 + kX 2t x1 t 1 + kX 2t X1 1 + kX 2to X 1

o

1 + kX 2t → x1 = X1 and x2 = X 2 . 1 + kX 2to _________________________________________________________________ 6.21 Given the two dimensional flow: v1 = kx2 , v2 = 0


Lai et al, Introduction to Continuum Mechanics (a) Obtain the streamline passing through the point (α1 , α 2 ) . (b) Obtain the pathline for the

particle which is at ( X1 , X 2 ) at t = 0 , including the time history of the particle along the pathline -----------------------------------------------------------------------------Ans. (a) The streamline is clearly x2 = α 2 .

(b) The pathline for the particle which is at ( X1 , X 2 ) at time 0 is simply x2 = X 2 . To find the time history of the particle along the pathline, i.e., to find x = x(t ) with X = x(0) , we have, dx2 dx = 0 → x2 = X 2 , → 1 = kX 2 → x1 = X1 + kX 2t , 0 ≤ t ≤ ∞ . dt dt _________________________________________________________________ 6.22 Do Prob. 6.21 for the following velocity field :

v1 = ω x2 , v2 = −ω x1 .

------------------------------------------------------------------------------dx1 dx2 dx1 x Ans. (a) From = ω x2 , = −ω x1 → = − 2 → x1dx1 + x2 dx2 = 0 , ds ds dx2 x1

→ x12 + x22 = C → x12 + x22 = α12 + α 22 . The streamline is a circle. (b) Since the flow is steady, clearly, the pathline is also a circle given by x12 + x22 = X12 + X 22 . To find the time history of the particle along the pathline, i.e., to find x = x(t ) with X = x(0) , we have, dx1 dx2 d 2x dx d2x = ω x2 , = −ω x1 → 21 = ω 2 = −ω 2 x1 → 21 + ω 2 x1 = 0 , dt dt dt dt dt 1 dx1 → x1 = A sin ωt + B cos ωt , → x2 = = A cos ωt − B sin ωt . ω dt at t = 0, x1 = X1 , x2 = X 2 , thus, x1 = X 2 sin ωt + X1 cos ωt , → x2 = X 2 cos ωt − X1 sin ωt _________________________________________________________________ 6.23 Given the following velocity field in polar coordinates ( r ,θ ) : vr =

Q , vθ = 0 . 2π r

(a) Obtain the streamline passing through the point ( ro ,θ o ) ,

(b) Obtain the pathline for the particle which is at ( R, Θ ) at t = 0 , including the time history of the particle along the pathline ------------------------------------------------------------------------------Ans. Both the streamline and the path lines are radial lines with θ =constant. (a) the streamline passing through the point r = ro ,θ = θo is θ = θo .

(b) the pathline for the particle which is at ( R, Θ ) at t = 0 is θ = Θ . To find the time history of the particle, we have, r t Q dθ dr Q Q = 0 → θ = Θ, = → ∫ rdr = ∫ dt → r 2 = R 2 + t . 0 2π R π dt dt 2π r _________________________________________________________________ 6.24 Do Prob. 6.23 for the following velocity field in polar coordinates ( r ,θ ) :

vr = 0, vθ = C / r . ------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. Both the streamline and the path lines are circles r = constant . (a) the streamline passing through the point r = r ,θ = θo is r = ro

(b) the pathline for the particle which is at ( R, Θ ) at t = 0 is r = R . To find the time history of

dr dθ C dθ C C =0→r = R→R = → = 2 →θ = 2 t + Θ . dt dt R dt R R _________________________________________________________________ the particle, we have,

6.25 From the Navier-stokes equations, obtain Eq. (6.11.2) for the velocity distribution of the plane Couette flow. ------------------------------------------------------------------------------Ans. With x2 axis pointing vertically upward, we have

v1 = v ( x2 ) , v2 = 0, v3 = 0 and a1 = a2 = a3 = 0 , thus, with p = p ( x2 ) , the Navier-Stoke’s equation

in the x1 direction become 0=μ

d 2v dx22

→ v = C1 x2 + C2 . At x2 = 0, v = 0, → C2 = 0. At x2 = d , v = vo → C1 =

vo v → v = o x2 . d d

_________________________________________________________________ 6.26 For the plane Couette flow, if in addition to the movement of the upper plate, there is also an applied negative pressure gradient ∂p / ∂x1 , obtain the velocity distribution. Also obtain the volume flow rate per unit width. ------------------------------------------------------------------------------Ans. With x2 axis pointing vertically upward, we have v1 = v ( x2 ) , v2 = 0, v3 = 0 and a1 = a2 = a3 = 0 , thus, the Navier-Stoke’s Equations become,

0=−

d 2v ∂p ∂ ⎛ ∂p ⎞ +μ 2 → ⎜ ⎟=0 ∂x1 ∂x1 ⎝ ∂x1 ⎠ dx2

∂p ∂ ⎛ ∂p ⎞ ∂ ⎛ ∂p ⎞ → ⎜ ⎟= ⎜ ⎟=0 ∂x2 ∂x1 ⎝ ∂x2 ⎠ ∂x2 ⎝ ∂x1 ⎠ ∂p ∂ ⎛ ∂p ⎞ ∂ ⎛ ∂p ⎞ 0=− → ⎜ ⎟= ⎜ ⎟=0 ∂x3 ∂x1 ⎝ ∂x3 ⎠ ∂x3 ⎝ ∂x1 ⎠ 0=−

Thus d 2v dx22

∂p = constant ≡ −α ∂x1 =−

⎛ α ⎞ x2 α → v = − ⎜ ⎟ 2 + C1 x2 + C2 , At x2 = 0, v = 0 → C2 = 0. At x2 = d , v = vo → μ ⎝μ⎠ 2

v ⎛α ⎞ d ⎛α ⎞ d2 vo = − ⎜ ⎟ + C1d → C1 = o + ⎜ ⎟ → d ⎝μ⎠2 ⎝μ⎠ 2 vo ⎛ α ⎞ x 2 ⎪⎧ v ⎛ α ⎞ d ⎪⎫ ⎛ α ⎞ 2 v = − ⎜ ⎟ 2 + ⎨ o + ⎜ ⎟ ⎬ x2 = ⎜ ⎟ x2 d − x2 + x2 . d ⎝ μ ⎠ 2 ⎪⎩ d ⎝ μ ⎠ 2 ⎪⎭ ⎝ 2μ ⎠ The volume flow rate per unit width is given by

(

)


Lai et al, Introduction to Continuum Mechanics d d ⎡ ⎛ α ⎞ x2 ⎧ v ⎛ α ⎞ d ⎫ ⎤ Q = ∫ v ( x2 )dx2 = ∫ ⎢ − ⎜ ⎟ 2 + ⎨ o + ⎜ ⎟ ⎬ x2 ⎥dx2 0 0 ⎢ ⎣ ⎝ μ ⎠ 2 ⎩ d ⎝ μ ⎠ 2 ⎭ ⎥⎦

⎛ α ⎞ d 3 ⎧ v ⎛ α ⎞ d ⎫ d 2 ⎛ α ⎞ ⎛ d 3 ⎞ ⎧ vo ⎫ d 2 = −⎜ ⎟ + ⎨ o + ⎜ ⎟ ⎬ = ⎜ ⎟⎜ ⎟ + ⎨ ⎬ ⎝ μ ⎠ 6 ⎩ d ⎝ μ ⎠ 2 ⎭ 2 ⎝ μ ⎠ ⎝ 12 ⎠ ⎩ d ⎭ 2 _________________________________________________________________ 6.27 Obtain the steady uni-directional flow of an incompressible viscous fluid layer of uniform depth d flowing down an inclined plane which makes an angle θ with the horizontal. ------------------------------------------------------------------------------Ans. With x2 axis normal to the flow and pointing away from the fluid and x1 axis in the flow

direction, we are looking for the velocity field in the following form: v1 = v ( x2 ) , v2 = 0, v3 = 0 , which clearly satisfies the continuity equation. Now the N-S equations give d 2v ∂p ∂ ⎛ ∂p ⎞ 0=− + ρ g sin θ + μ 2 → ⎜ ⎟=0. ∂x1 ∂x1 ⎝ ∂x1 ⎠ dx2 ∂p ∂ ∂p ∂ ∂p − ρ g cosθ → = =0 ∂x2 ∂x1 ∂x2 ∂x2 ∂x1 ∂p ∂ ∂p ∂ ∂p 0=− =0→ = = 0 . Thus ∂x3 ∂x1 ∂x3 ∂x3 ∂x1 ∂p =C. ∂x1 The constant C can be determined from the pressure condition on the free surface ( x2 = d ), where pressure p = pa , the atmospheric pressure which is independent of x1 , thus ∂p / ∂x1 = 0 → C = 0 so that ∂p / ∂x1 = 0 for the whole flow field. Thus, d 2v d 2v dv 0 = ρ g sin θ + μ 2 → μ 2 = − ρ g sin θ → μ = − ρ g sin θ x2 + C1 dx2 dx2 dx2 0=−

x22 + C1 x2 + C2 . At x2 = 0, v = 0 (non slip condition) → C2 = 0 . 2 At x2 = d , shear stress T12 = 0 → μ dv /dx2 =0 → C1 =ρ gdsinθ . → μ v = − ρ g sin θ

x ⎞ ⎛ → μ v = ρ gsinθ ⎜ d − 2 ⎟ x2 . 2 ⎠ ⎝ __________________________________________________________________

6.28 A layer of water ( ρ g = 62.4lb / ft 3 ) flows down an inclined plane ( θ = 30o ) with a uniform thickness of 0.1 ft . Assuming the flow to be laminar, what is the pressure at any point on the inclined plane. Take the atmospheric pressure to be zero. -----------------------------------------------------------------------------Ans. With flow in the x1 direction, the N-S equation in the x2 direction (pointing away from the inclined plane) gives, [note: p is independent of x1 and x3 , see Prob. 6.27] −∂p / ∂x2 − ρ g cosθ = 0 → p = − ( ρ g cosθ ) x2 + C .


Lai et al, Introduction to Continuum Mechanics At x2 = d , p = pa = 0 → ( ρ g cosθ ) d = C , → p = ( ρ g cosθ )( d − x2 ) .

(

)

At x2 = 0 → p = ( ρ g cos θ )( d ) = 62.4cos30o ( 0.1) = 5.40 lb / ft 2 . We can also obtain the same result by using the fact that the piezometric head p / ( ρ g ) + z =constant for any points on the same plane perpendicular to the direction of the flow p p (see example 6.7.2), therefore a + za = b + zb → pb = ρ g ( za − zb ) = ρ gd cos θ . ρg ρg __________________________________________________________________ 6.29 Two layers of liquids with viscosities μ1 and μ2 , densities ρ1 and ρ 2 respectively, and with equal depths b , flow steadily between two fixed horizontal parallel plates. Find the velocity distribution for this steady uni-directional flow. Neglect body forces. ------------------------------------------------------------------------------Ans. We are looking for velocity fields in the two layers in the following form corresponding to the uni-directional steady laminar flows: t (t ) (t ) (t ) For the top layer: v1 = v( ) ( x2 ) , v2 = v3 = 0 . b (b) (b) (b ) For the bottom layer: v = v( ) x , v = v = 0 .

( 2)

1

2

3

From the N-S equations for the top layer, we have t t t 0 = −∂p( ) / ∂x1 + μ1d 2 v( ) / dx22 → (∂ / ∂x1 )(∂p( ) / ∂x1 ) = 0 t t t 0 = −∂p( ) / ∂x2 → (∂ / ∂x1 )(∂p( ) / ∂x2 ) = (∂ / ∂x2 )(∂p( ) / ∂x1 ) = 0 t t t 0 = −∂p( ) / ∂x → (∂ / ∂x )(∂p( ) / ∂x ) = (∂ / ∂x )(∂p( ) / ∂x ) = 0 3

1

3

3

1

t Thus, ∂p ( ) / ∂x1 = −α1 (a constant) . Now, μ1d 2 v( t ) / dx22 = −α1 → μ1dv( t ) / dx2 = −α1 x2 + A1 → μ1v( t ) = −(α1 x22 / 2) + A1 x2 + B1. Similarly from the N-S equations for the bottom layer, we have, b ∂p( ) / ∂x = −α (a constant) .

1

2

μ2 dv( b ) / dx2 = −α 2 x2 + A2 → μ2 v( b ) = −(α 2 x22 / 2) + A2 x2 + B2 . The constants A1 , B1 , A2 , B2 will be determined from the boundary and the interface conditions: t At x = b (the top plate), v( ) = 0 , 0 = −(α b 2 / 2) + A b + B (1) 2

1

1

1

b At x2 = −b (the bottom plate), v( ) = 0 , 0 = −(α 2b 2 / 2) − A2b + B2

(2) (t )

(b )

At x2 = 0 (the interface), there is no slip between the two layers of flow, i.e., v = v → B1 / μ1 = B2 / μ 2 (3) rd Also, according to Newton’s 3 law, the action and reaction at the interface between the two fluid must be equal and opposite, that is, both the shear stress and the normal stress must by continuous t t (t ) (b) = μ dv( ) / dx =A, T = μ dv( ) / dx =A at x = 0 . Since T 2

12

x2 =0

1

(

2

Therefore, A1 = A2 , t (b) (t ) and T22 = − p( ) ( x1 ,0), T22 x2 =0

)

x2 =0

1

12

2

x2 =0

(

2

)

x2 =0

2

(4) x2 =0

b = − p( ) ( x1 ,0),

(t ) T22

x2 = 0

=

(b) T22

x2 = 0

→

t b t b p( ) ( x1 ,0) = p( ) ( x1 ,0) → (∂p( ) / ∂x1 )( x1 ,0) = (∂p( ) / ∂x1 )( x1 ,0) → α1 = α 2 ≡ ∂p / ∂x1. Now, Eqs.(1)(2)(3)(4) determine the four constants A1 , B1 , A2 , B2 as a function of α = −∂p / ∂x1 :


Lai et al, Introduction to Continuum Mechanics μ1 ⎞ μ2 ⎞ ⎪⎧ μ − μ1 ⎪⎫ 2 ⎛ 2 ⎛ A1 = A2 = α b ⎨ 2 ⎬ , B1 = b α ⎜ ⎟ , B2 = b α ⎜ ⎟ . Thus, ⎝ μ1 + μ2 ⎠ ⎝ μ1 + μ2 ⎠ ⎩⎪ 2 ( μ1 + μ2 ) ⎭⎪ ⎡ x22

μ1 ⎞ ⎤ b ⎛ μ − μ1 ⎞ 2⎛ − ⎜ 2 ⎟ x2 − b ⎜ ⎟⎥ , ⎝ μ1 + μ2 ⎠ ⎦⎥ ⎣⎢ 2 2 ⎝ μ1 + μ2 ⎠

μ1v( t ) = −α ⎢

⎡x ⎛ μ2 ⎞ ⎤ b⎛ μ −μ ⎞ b μ2 v( ) = −α ⎢ 2 − ⎜ 2 1 ⎟ x2 − b 2 ⎜ ⎟⎥ . ⎝ μ1 + μ2 ⎠ ⎦⎥ ⎣⎢ 2 2 ⎝ μ1 + μ2 ⎠ 2

__________________________________________________________________ 6.30 For the Couette flow of Section 6.15, (a) obtain the shear stress at any point inside the fluid (b) obtain the shear stress on the outer and inner cylinder (c) obtain the torque which must be applied to the cylinders to maintain the flow. ------------------------------------------------------------------------------Ans. (a) Eq.(6.15.4) and (6.15.7), give vθ = Ar + B / r , where B = r12 r22 ( Ω1 − Ω2 ) / (r22 − r12 ) . Thus, Trθ = Tθ r = 2μ Drθ = μ r

2μ r12 r22 ( Ω1 − Ω2 ) 1 d ⎛ vθ ⎞ 2μ B = − = − dr ⎜⎝ r ⎟⎠ r2 r22 − r12 r2

(b) On the outer wall: r = r2 , the shear stress is Trθ = Tθ r = 2μ r12 (Ω2 − Ω1 ) / (r22 − r12 ) , On the inner wall, r = r1 , the shear stress is Trθ = Tθ r = 2 μ r22 ( Ω2 − Ω1 ) / (r22 − r12 ) . (c) On the outer wall, per unit height, the torque is given by 2μ r 2 ( Ω − Ω ) 4πμ r12 r22 ( Ω2 − Ω1 ) eθ ( M )r2 = ⎡⎢⎣(Trθ )r =r2 ( 2π r2 ) (1) ⎤⎥⎦ r2eθ = 1 2 2 2 1 2π r22 eθ = r −r r2 − r2

(

2

)

1

2

1

The torque on the inner wall is equal and opposite to that on the outer wall. __________________________________________________________________ 6.31 Verify the equation β 2 = ρω / 2μ for the oscillating plane problem of Section 6.16. -------------------------------------------------------------------------------Ans. With v = α e − β x2 cos(ωt − β x2 + ε ) , ∂v / ∂t = −ωα e− β x2 sin(ωt − β x2 + ε ) , ∂v / ∂x2 = − βα e − β x2 cos(ωt − β x2 + ε ) + βα e− β x2 sin(ωt − β x2 + ε ) and

∂ 2 v / ∂x22 = β 2α e− β x2 cos(ωt − β x2 + ε ) − β 2α e− β x2 sin(ωt − β x2 + ε ) − β 2α e− β x2 sin(ωt − β x2 + ε ) − β 2α e− β x2 cos(ωt − β x2 + ε ) = −2 β 2α e− β x2 sin(ωt − β x2 + ε ) Thus ρ∂v / ∂t = μ∂ 2 v / ∂x22 → − ρωα e− β x2 sin(ωt − β x2 + ε ) = − μ 2β 2α e− β x2 sin(ωt − β x2 + ε ) → ρω = μ 2β 2 → β 2 = ρω / (2 μ ). __________________________________________________________________ 6.32 Consider the flow of an incompressible viscous fluid through the annular space between two concentric horizontal cylinders. The radii are a and b . (a) Find the flow field if there is no variation of pressure in the axial direction and if the inner and the outer cylinders have axial


Lai et al, Introduction to Continuum Mechanics velocities va and vb respectively and (b) find the flow field if there is a pressure gradient in the axial direction and both cylinders are fixed. Take body forces to be zero. -------------------------------------------------------------------------------Ans. (a) We look for the following form of velocity field in cylindrical coordinates: vr = 0, vθ = 0, vz = v(r ) and ∂p / ∂z = 0 . The N-S equations give, in the absence of body forces 0=−

⎛ d 2 v 1 dv ⎞ ∂p 1 ∂p , 0=− , 0 = μ⎜ 2 + ⎟⎟ . ⎜ dr r ∂θ r dr ∂r ⎝ ⎠

The first two equations together with ∂p / ∂z = 0 give, p = constant .

⎛ d 2 v 1 dv ⎞ 1 d ⎛ dv ⎞ dv dv C 0 = μ⎜ 2 + =C → = → v = C ln r + D ⎟⎟ → ⎜r ⎟ = 0 → r ⎜ dr r dr ⎠ r dr ⎝ dr ⎠ dr dr r ⎝ At r = a, v = va and at r = b, v = vb , thus, va = C ln a + D and vb = C ln b + D → va − vb = C ln(a / b) . v −v v ln b − vb ln a → C = a b and D = a . So that, ln ( a / b ) ln ( b / a ) v=

va − vb v ln b − vb ln a ln r + a . ln ( a / b ) ln ( b / a )

(b) 0 = −∂p /∂r and 0 = −(1 / r )∂p / ∂θ → p = p ( z ) , 0=−

⎛ d 2 v 1 dv ⎞ d 2 p dp dp + μ⎜ 2 + → 2 =0→ = constant ≡ −α , ⎟ ⎜ dr dz r dr ⎟⎠ dz dz ⎝

⎛ d 2 v 1 dv ⎞ μ d ⎛ dv ⎞ d ⎛ dv ⎞ αr → μ⎜ 2 + → ⎟⎟ = −α → ⎜ r ⎟ = −α → ⎜ r ⎟ = − ⎜ dr r dr r dr dr dr dr μ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 dv αr dv α r C αr2 →r =− +C → =− + →v=− + C ln r + D . μ 2 μ2 r 4μ dr dr The boundary conditions are: v( a ) = v(b) = 0 → 0=−

α a2 α b2 + C ln a + D and 0 = − + C ln b + D → 4μ 4μ

(

)

(

)

(

2 2 2 2 2 2 1 dp a − b 1 dp a ln b − b ln a α a −b =− , D=− C= 4μ ln ( a / b ) 4 μ dz ln ( a / b ) 4 μ dz ln ( b / a )

(

)

(

2 2 b 2 ln a − a 2 ln b 1 dp ⎡ 2 a − b ⎢r + v= ln r + 4 μ dz ⎢ ln ( b / a ) ln ( b / a ) ⎣

)

) ⎤⎥ . ⎥⎦

__________________________________________________________________ 6.33 Show that for the velocity field : vx = v( y, z ), v y = vz = 0 , the Navier-Stokes equations, with ρ B =0, reduces to

∂ 2v

+

∂ 2v

=

∂y ∂z ------------------------------------------------------------------------------Ans. With vx = v( y, z ), v y = vz = 0 , we have, 2

2


1 dp

μ dx

= β = constant .

Lai et al, Introduction to Continuum Mechanics ∂vx ∂v ∂v ∂v + vx x + v y x + vz x = 0 + 0 + 0 + 0 = 0, and a y = az = 0 , thus, the N-S equations in ∂t ∂x ∂y ∂z the absence of body forces are: ⎛ ∂ 2v ∂ 2v ⎞ ∂p ∂p ∂p 0 = − + μ ⎜ 2x + 2x ⎟ , 0 = − , 0 = − . Thus, p = p ( x) , and ⎜ ∂y ∂x ∂y ∂z ∂z ⎟⎠ ⎝ ⎛ ∂ 2v ∂ 2 v ⎞ dp ∂ 2v ⎞ d 2 p d2 p ∂ ⎛ ∂ 2v μ ⎜ 2x + 2x ⎟ = → μ ⎜ 2x + 2x ⎟ = 2 → 0 = 2 → , ⎜ ∂x ⎜⎝ ∂y dx ∂z ⎟⎠ dx ∂z ⎟⎠ dx ⎝ ∂y ax =

∂ 2v ∂ 2v dp 1 dp = constant → 2x + 2x = ≡β. dx μ dx ∂y ∂z __________________________________________________________________ 6.34 Given the velocity field in the form of

(

)

vx = v = A y 2 / a 2 + z 2 / b 2 + B , v y = vz = 0 . Find A and B for the steady laminar flow of a Newtonian fluid in a pipe having an elliptical cross section given by y 2 / a 2 + z 2 / b 2 = 1 . Assume no body forces and use the governing equation obtained in the previous problem. ------------------------------------------------------------------------------Ans. The governing equation is [see the previous problem] : ∂ 2 vx ∂ 2 vx 1 dp + 2 = ≡ β . Now, vx = v = A y 2 / a 2 + z 2 / b 2 + B → μ dx ∂y 2 ∂z

(

∂ 2 vx ∂y 2

+

∂ 2 vx

2 ⎛ 2 = A⎜ 2 + 2 b ⎝a

∂z 2

⎛ a2 + b2 ⎞ 2 = A ⎜ ⎟ ⎜ a 2b 2 ⎠ ⎝

⎛ a2 + b2 ⎞ ⎟⎟ → 2 A ⎜⎜ 2 2 ⎝ a b ⎠

)

⎞ a 2b 2 ⎟⎟ = β → A = β 2 a 2 + b2 ⎠

(

)

On the boundary y 2 / a 2 + z 2 / b 2 = 1 , no slip condition requires that vx = 0 , therefore,

A (1) + B = 0 → B = − A , thus,

⎡⎛ y 2 z 2 ⎞ ⎤ a 2b 2 β vx = A ⎢⎜ 2 + 2 ⎟ − 1⎥ = b ⎠⎟ ⎥⎦ 2 a 2 + b 2 ⎢⎣⎝⎜ a

(

)

⎡⎛ y 2 z 2 ⎞ ⎤ ⎢⎜ 2 + 2 ⎟⎟ − 1⎥ . b ⎠ ⎥⎦ ⎢⎣⎝⎜ a

__________________________________________________________________ 6.35 Given the velocity field in the form of

⎛ b ⎞⎛ b ⎞⎛ b ⎞ vx = A ⎜ z + ⎟⎜ z + 3 y − ⎟⎜ z − 3 y − ⎟ + B , v y = vz = 0 2 3 ⎠⎝ 3 ⎠⎝ 3⎠ ⎝ Find A and B for the steady laminar flow of a Newtonian fluid in a pipe having an equilateral triangular cross-section defined by the planes: b b b z+ = 0, z + 3 y − = 0, z − 3 y − =0. 2 3 3 3 Assume no body forces and use the governing equation obtained in Prob. 6.33. ------------------------------------------------------------------------------∂ 2 vx ∂ 2 vx 1 dp Ans. The governing equation is + 2 = ≡ β . [See problem 6.33] μ dx ∂y 2 ∂z

(

)(

)(

)

With vx = A z + b / (2 3) z + 3 y − b / 3 z − 3 y − b / 3 + B ,



(

∂vx / ∂y = A z + b / (2 3)

(

){ 3 ( z −

)

(

3y − b / 3 − 3 z + 3y − b / 3

) ∂ 2 vx / ∂y 2 = −6 A ( z + b / (2 3) ) . Let f ( z ) = ( z + b / (2 3) ) , g ( y, z ) = ( z +

)}

= A z + b / (2 3) ( −6 y ) →

)

(

)

3 y − b / 3 , h( y , z ) = z − 3 y − b / 3 ,

then vx = Af ( z ) g ( y, z )h( y, z ) + B → ∂vx / ∂z = Ag ( y , z )h( y, z ) + Af ( z ) {h( y , z ) + g ( y, z )}

(

) (

) (

) 3 − b 2 / 3)

Now, h( y , z ) + g ( y, z ) = z + 3 y − b / 3 + z − 3 y − b / 3 = 2 z − b / 3

( )( 2 3 ) − 3 y 2 = z 2 − 2bz /

) (

f ( z ) {h( y, z ) + g ( y, z )} = 2 z − b / 3 z + b / (2 3) = 2 z 2 − zb /

(

g ( y , z ) h( y , z ) = z − b /

3 + b 2 / 3 − 3 y 2 . Thus,

∂vx = Ag ( y, z )h( y, z ) + Af ( z ) {h( y , z ) + g ( y , z )} ∂z

(

) ( ) ( ) / ∂z = A ( 6 z − 3b / 3 ) = 6 A ( z − b / (2 3) ) . Thus, ∂ v 1 dp + = −6 A ( z + b / (2 3) ) + 6 A ( z − b / (2 3) ) = −6 A ( b / 3 ) = ≡β, dx μ ∂z

= A z 2 − 2bz / 3 + b 2 / 3 − 3 y 2 + A 2 z 2 − zb / 3 − b 2 / 3 = A 3z 2 − 3bz / 3 − 3 y 2 ∂ vx 2

∂ 2 vx ∂y

2

2

2

x 2

from which, A = − β / (2 3b) . The non-slip condition on the boundary requires B = 0 . _________________________________________________________________ 6.36 For the steady-state, time dependent parallel flow of water ( density ρ = 103 Kg / m3 , viscosity, μ = 10−3 Ns / m2 ) near an oscillating plate, calculate the wave length for ω = 2 cps . ------------------------------------------------------------------------------Ans. v = ae − β x2 ( cos ωt − β x2 + ε ) , the wave length is given by 2π / β , where

β=

ρω . Here we have, ρ = 103 Kg / m3 , ω = 4π rad / s , μ = 10−3 Ns / m 2 , thus 2μ

ρω = β= 2μ

(10 ) ( 4π ) =10 2 (10 ) 3

3

−3

2π m −1 → wave length=

2π

β

=

2π 3

10

= 2.51× 10−3 m

_________________________________________________________________ 6.37 The space between two concentric spherical shells is filled with an incompressible Newtonian fluid. The inner shell (radius ri ) is fixed; the outer shell (radius ro ) rotates with an angular velocity Ω about a diameter. Find the velocity distribution. Assume the flow to be laminar without secondary flow. -------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. We look for solution in the form of vr = 0, vθ = 0, vφ = f ( r ) sin θ . This velocity field clearly satisfies the continuity equation [see Section 6.8, Eq.(6.8.8)]: ∂vr 2vr 1 ∂vθ vθ cot θ 1 ∂vφ + + + + =0. ∂r r r ∂θ r r sin θ ∂φ The N-S equations in spherical coordinates give[see Section 6.8]: vφ2 cot θ vφ2 1 ∂p 1 ∂p (1), − (2), =− − =− ρ ∂r ρ r ∂θ r r 0=−

∂p

1

ρ r sin θ ∂φ

+

μ ⎡ 1 ∂ ⎛ 2 ∂vφ ⎞ 1 ∂ ⎛ 1 ∂ ⎞⎤ vφ sin θ ) ⎟ ⎥ ⎢ 2 ⎜r ( ⎟+ 2 ⎜ ρ ⎣⎢ r ∂r ⎝ ∂r ⎠ r ∂θ ⎝ sin θ ∂θ ⎠ ⎦⎥

(3).

∂2 p ∂2 p ∂2 p = 0, eq.(2)→ = 0 and eq.(3)→ =0. ∂φ∂r ∂φ∂θ ∂φ∂φ Thus, ∂p / ∂φ = constant . The constant must be zero, otherwise p will not be single-valued. Eq.(1)→

Eq. (3) now becomes, with ∂p / ∂φ = 0 and vφ = f ( r ) sin θ , 0=

2 d ⎛ 2 df ⎞ df sin θ d ⎛ 2 df ⎞ 2 f ( r ) 2 d f − . That is, θ r − 2 f = 0 → r +2r −2f =0. r sin ⎟ ⎜ ⎟ 2 dr ⎜ 2 2 dr ⎝ dr ⎠ dr dr r r ⎝ dr ⎠

(

)

The general solution of this equation is: f = Ar + B / r 2 . Thus, vφ = Ar + B / r 2 sin θ . The inner shell (radius ri ) is fixed; therefore, at r = ri , vφ = 0 →

(

0 = Ari + B / ri 2

The outer shell (radius ro ) rotates with an angular velocity Ω , therefore, at

(

)

(

r = ro , vφ = ( ro sin θ ) Ω → ( ro sin θ ) Ω = Aro + B / ro2 sin θ → ro Ω = Aro + B / ro2

)

)

(4)

(5)

Equations (4) and (5) are two equations for the two unknowns A and B : Ωro3 r 3r 3 B⎞ ⎛ A= , B = − o i Ω , and vφ = ⎜ Ar + 2 ⎟ sin θ . r ⎠ ⎝ ro3 − ri3 ro3 − ri3

(

)

(

)

_________________________________________________________________ 6.38 Consider the following velocity field in cylindrical coordinates for an incompressible fluid: vr = v(r ), vθ = 0, vz = 0 A where A is a constant so that the equation of conservation of mass is r satisfied. (b) If the rate of mass flow through the circular cylindrical surface of radius r and unit length (in z direction) is Qm , determine the constant A in terms of Qm . ------------------------------------------------------------------------------Ans. (a) The equation of continuity is ∂v 1 ∂ 1 ∂v 1 d A ( rvr ) + θ + z = 0 . Thus, ( rvr ) = 0 → rvr = A → vr = . r ∂r r ∂θ ∂z r dr r Qm Qm A Qm → = → A= . (b) vr ( 2π r )(1) = Qm → vr = 2π r 2π r 2π r _________________________________________________________________ (a) Show that vr =

6.39 Given the following velocity field in cylindrical coordinates for an incompressible fluid: vr = v(r ,θ ), vθ = 0, vz = 0


Lai et al, Introduction to Continuum Mechanics (a) Show that vr = f (θ ) / r , where f (θ ) is any function of θ . (b) In the absence of body forces, show that d2 f f kμ ρf2 + 4 f + + k = 0 , p = 2μ 2 + 2 + C , where k and C are constants. 2 μ dθ r 2r ------------------------------------------------------------------------------Ans. (a). The equation of continuity is ∂v 1 ∂ 1 ∂v 1 ∂ ( rvr ) + θ + z = 0 . Thus, ( rvr ) = 0 → rvr = f (θ ) → vr = f (θ ) / r r ∂r r ∂r r ∂θ ∂z (b).The N-S equation in the r direction gives v ⎛ ∂v ∂vr ∂v ∂v 1 ∂p ⎞ + vr r + θ ⎜ r − vθ ⎟ + vz r = − + Br r ⎝ ∂θ ∂t ∂r ∂z ρ ∂r ⎠ +

μ ⎡ ∂ 2 vr 1 ∂ 2 vr ∂ 2 vr 1 ∂vr 2 ∂vθ vr ⎤ f2 1 ∂p μ ⎛ 1 d 2 f ⎞ + + − − → − = − + ⎜ ⎟ (1) ⎢ 2 + 2 ⎥ ρ ⎢⎣ ∂r ρ ∂r ρ ⎜⎝ r 3 dθ 2 ⎟⎠ ∂z 2 r ∂r r 2 ∂θ r 2 ⎥⎦ r ∂θ 2 r3

The N-S equation in the z direction gives −∂p / ∂z = 0 → p is independent of z. The N-S equation in the θ direction gives ∂p 2 μ ⎛ df ⎞ 1 ∂p μ ⎛ 2 ∂vr ⎞ 1 ∂p μ 2 ⎛ df ⎞ =− + ⎜ 2 →− + =0→ = ⎜ ⎟ ⎟ 3 ∂θ r 2 ⎜⎝ dθ ⎟⎠ ρ r ∂θ ρ ⎝ r ∂θ ⎠ ρ r ∂θ ρ r ⎝ dθ ⎠ 2μ f ∂p 4 μ f dg → p = 2 + g (r ) → =− 3 + . Thus, Eq.(1) gives: ∂ r dr r r ⎛ d2 f ⎞ r 3 ⎛ dg ⎞ d 2 f ρf2 ρ f 2 r 3 ⎛ dg ⎞ → + + = ⎜ ⎟. − =− ⎜ ⎟+ + 4 f 4 f ⎜ ⎟ ⎜ dθ 2 ⎟ μ μ ⎝ dr ⎠ dθ 2 μ μ ⎝ dr ⎠ ⎝ ⎠ The left side of the above equation is a function of θ , the right side is a function of r , therefore, they must be equal to a constant, say, −k , i.e., ⎛ d2 f ⎞ ρ f 2 r 3 ⎛ dg ⎞ μk + + = ⎜ ⎟ = − k → g = 2 + C . Therefore, 4 f ⎜⎜ 2 ⎟⎟ μ μ ⎝ dr ⎠ 2r ⎝ dθ ⎠ ⎛ d2 f ⎞ ρf2 2μ f k μ + k = 0, p = 2 + 2 + C . ⎜⎜ 2 ⎟⎟ + 4 f + μ 2r r ⎝ dθ ⎠ _________________________________________________________________ 6.40 Consider the steady two dimensional channel flow of an incompressible Newtonian fluid under the action of an applied negative pressure gradient ∂p / ∂x1 , as well as the movement of the top plate with velocity vo in its own plane.[See Prob. 6.26]. Determine the temperature distribution for this flow due to viscous dissipation when both plates are maintained at the same fixed temperature θ o . Assume constant physical properties. -------------------------------------------------------------------------------

(

)

Ans. From the result of Prob. 6.26, we have v1 = vo x2 / d + (α / 2μ ) x2 d − x22 , v2 = v3 = 0. Let the temperature distribution be denoted by Θ = Θ ( x2 ) . From Eq. (6.18.3), we have,

ρc

(

)

DΘ ∂ 2Θ 2 2 2 2 2 2 + D22 + D33 + 2 D12 + 2 D13 + 2 D23 , represents = Φ inc + κ , where Φinc = 2μ D11 ∂x j ∂x j Dt

the heat generated through viscous forces. For this problem, only D12 is nonzero, thus,


Lai et al, Introduction to Continuum Mechanics 2

2

2 ⎤ ⎡ vo ⎛ α ⎞ ⎤ ⎛ 1 ⎞ ⎡v ⎛ α ⎞ 2 Φinc = 4μ D12 = ( 4μ ) ⎜ ⎟ ⎢ o + ⎜ ⎟ ( d − 2 x2 ) ⎥ = μ ⎢ + ⎜ ⎟ ( d − 2 x2 ) ⎥ . ⎝ 2 ⎠ ⎣ d ⎝ 2μ ⎠ ⎦ ⎣ d ⎝ 2μ ⎠ ⎦ 2

⎤ μ ⎡ vo ⎛ α ⎞ μ2 ∂Θ d 2 x = − + − → = ( ) ⎢ ⎜ ⎟ 2 ⎥ κ ⎣ d ⎝ 2μ ⎠ ∂x2 3κα ∂ 2 x2 ⎦ ∂ 2Θ

μ3 Θ=− 12κα 2

3

⎡ vo ⎛ α ⎞ ⎤ ⎢ +⎜ ⎟ ( d − 2 x2 ) ⎥ + C → ⎣ d ⎝ 2μ ⎠ ⎦

4

⎡ vo ⎛ α ⎞ ⎤ ⎢ +⎜ ⎟ ( d − 2 x2 ) ⎥ + Cx2 + D . ⎣ d ⎝ 2μ ⎠ ⎦ 4

At x2 = 0, Θ = θ o → θ o = −

μ 3 ⎛ vo α d ⎞ ⎜ + ⎟ + D. 12κα 2 ⎝ d 2 μ ⎠ 4

μ 3 ⎛ vo α d ⎞ At x2 = d , Θ = θ o → θ o = − ⎜ − ⎟ + Cd + D. 12κα 2 ⎝ d 2 μ ⎠ Thus, 4

μ 3 ⎛ vo α d ⎞ μ3 + D = θo + and 0 = ⎜ ⎟ 12κα 2 12κα 2 ⎝ d 2μ ⎠

μ 3 ⎧⎪⎛ vo α d ⎞ ⎛ vo α d ⎞ ⎫⎪ ⎨⎜ + ⎟ −⎜ − ⎟ ⎬. 12κα 2 d ⎪⎩⎝ d 2 μ ⎠ ⎝ d 2μ ⎠ ⎪⎭ 4

→C =−

4 4 ⎧⎪⎛ v α d ⎞ ⎛ vo α d ⎞ ⎫⎪ o ⎨⎜ + ⎟ −⎜ − ⎟ ⎬ + Cd d 2μ ⎠ ⎝ d 2μ ⎠ ⎪ ⎩⎪⎝ ⎭ 4

_________________________________________________________________ 6.41 Determine the temperature distribution in the plane Poiseuille flow where the bottom plate is kept at a constant temperature Θ1 and the top plate Θ2 . Include the heat generated by viscous dissipation. ------------------------------------------------------------------------------Ans. For the plane Poiseuille flow [see Eq.(6.12.9)],

(

)

v1 = (α / 2 μ ) b 2 − x22 , v2 = v3 = 0,α ≡ −∂p / ∂x1 > 0 Let the temperature distribution be denoted by Θ = Θ ( x2 ) . From Eq. (6.18.3), we have,

ρc

(

)

DΘ ∂ 2Θ 2 2 2 2 2 2 + D22 + D33 + 2 D12 + 2 D13 + 2 D23 , represents = Φ inc + κ , where Φinc = 2μ D11 ∂x j ∂x j Dt

the heat generated through viscous forces. For this problem, only D12 is nonzero, 2

D12 =

⎛ 1α ⎞ 1 ∂v1 1α α2 2 2 =− = 4μ ⎜ − x2 → Φ inc = 4 μ D12 x2 ⎟ = x . Thus, 2 ∂x2 2μ μ 2 ⎝ 2μ ⎠ 2

DΘ ∂ 2Θ α2 2 d 2Θ d 2Θ 1 ⎛ ∂p ⎞ 2 =− ⎜ ρc = Φinc + κ →0= x2 + κ 2 → ⎟ x 2 Dt ∂x j ∂x j μ κμ ⎝ ∂x1 ⎠ 2 dx2 dx2 2

→

2

dΘ 1 ⎛ ∂p ⎞ x23 1 ⎛ ∂p ⎞ x24 =− ⎜ + → Θ = − + Cx2 + D . C ⎟ ⎜ ⎟ dx2 κμ ⎝ ∂x1 ⎠ 3 κμ ⎝ ∂x1 ⎠ 12



At x2 = +b, Θ = Θ 2 → Θ 2 = −

1 ⎛ ∂p ⎞ b 4 + Cb + D ⎜ ⎟ κμ ⎝ ∂x1 ⎠ 12 2

1 ⎛ ∂p ⎞ b 4 At x2 = −b, Θ = Θ1 → Θ1 = − − Cb + D . Thus, ⎜ ⎟ κμ ⎝ ∂x1 ⎠ 12 2

Θ1 + Θ2 Θ2 − Θ1 1 ⎛ ∂p ⎞ 4 + . ⎜ ⎟ b . C= 2b 2 12κμ ⎝ ∂x1 ⎠ __________________________________________________________________ D=

6.42 Determine the temperature distribution in the steady laminar flow between two coaxial cylinders (Couette flow) if the temperatures at the inner and the outer cylinders are kept at the same fixed temperature θ o . ------------------------------------------------------------------------------B Ans. For Couette flow, we have, vr = 0, vθ = Ar + , vz = 0 , where r 2 2 2 2 r r (Ω − Ω ) Ω r − Ω1r1 A = 2 22 , B = 1 2 2 1 2 2 . The only nonzero rate of deformation is 2 r2 − r1 r2 − r1 Drθ =

1 ⎧ 1 ⎛ ∂vr − vθ ⎨ 2 ⎩ r ⎜⎝ ∂θ

B ⎞ ∂vθ ⎫ 1 ⎧ vθ ∂vθ ⎫ ⎟ + ∂r ⎬ = 2 ⎨− r + ∂r ⎬ = − 2 . r ⎠ ⎩ ⎭ ⎭ 2

4μ B 2 DΘ ∂ 2Θ ⎛ B⎞ Φinc = 2μ 2 Dr2θ = 4 μ ⎜ − 2 ⎟ = 4 . ρ c = Φinc + κ → ∂x j ∂x j Dt r ⎝ r ⎠

(

→0=

)

4μ B 2 r4

→Θ=−

+κ

1 d ⎛ dΘ ⎞ d ⎛ dΘ ⎞ 4μ B 2 d Θ 2μ B 2 C → = − → = + r r ⎜ ⎟ ⎜ ⎟ r dr ⎝ dr ⎠ dr ⎝ dr ⎠ dr r κ r3 κ r3

μ B2 + C ln r + D . κ r2

At r = ri ,

Θ = Θo → Θo = −

μ B2 + C ln ri + D. κ ri2

At r = ro ,

Θ = Θo → Θo = −

μ B2 + C ln ro + D. κ ro2

Thus, ⎛ r 2 − r 2 ⎞⎛ μ B 2 ⎞ ⎛ ri ⎞ ⎡ ⎤ ro μ B 2 2 r 2 , D r r r r C = ⎜ o 2 2i ⎟⎜ / ln ln ln ln / ln o . = Θ + − ⎟ ⎜ ⎟ ⎢ ⎥ o o o i i 2 2 ⎜ r r ⎟⎜ κ ⎟ ri κ ri ro ri ⎠ ⎝ ro ⎠ ⎣ ⎦ ⎝ i o ⎠⎝ _________________________________________________________________

{

}

6.43 Show that the dissipation function for a compressible fluid can be written as that given in Eq. (6.17.10). ------------------------------------------------------------------------------Ans. From

(

)

2 2 2 2 2 2 Φ = λ ( D11 + D22 + D33 ) + 2 μ D11 + D22 + D33 + 2 D12 + 2 D13 + 2 D23 , we get, 2



(

)

(

2 2 2 2 2 2 Φ = ( λ + 2 μ ) D11 + D22 + D33 + 2λ ( D11D22 + D11D33 + D22 D33 ) + 4 μ D12 + D13 + D23

)

We now verify that this is the same as Φ = ( λ + 2 μ / 3) ( D11 + D22 + D33 ) + 2

(

2 2 2 +4μ D12 + D13 + D23

)

2 ⎡ 2 2 2 μ ⎢( D11 − D22 ) + ( D11 − D33 ) + ( D22 − D33 ) ⎤⎥ ⎦ 3 ⎣

Expanding the above equation, we have,

(

2 2 2 Φ = ( λ + 2 μ / 3) D11 + D22 + D33 + 2 D11D22 + 2 D11D33 + 2 D22 D33

(

)

) (

)

2 2 2 2 2 2 + ( 2μ / 3) ⎡ 2 D11 + D22 + D33 − ( 2 D11D22 + 2 D11D33 + 2 D22 D33 ) ⎤ + 4μ D12 + D13 + D23 . ⎣ ⎦ i.e., 2 ⎞ ⎛ 2 2 2 Φ = ( λ + 2 μ / 3) D11 + D22 + D33 + ⎜ λ + μ ⎟ ( 2 D11D22 + 2 D11D33 + 2 D22 D33 ) 3 ⎠ ⎝

(

(

)

)

(

2 2 2 2 2 2 + ( 4μ / 3) D11 + D22 + D33 − ( 2 μ / 3) ( 2 D11D22 + 2 D11D33 + 2 D22 D33 ) + 4μ D12 + D13 + D23

(

)

)

2 2 2 = D11 + D22 + D33 ( λ + 2μ / 3 + 4μ / 3) + 2 ( D11D22 + D11D33 + D22 D33 ) ( λ + 2μ / 3 − 2μ / 3)

(

)

2 2 2 +4μ D12 + D13 + D23 .

That is

(

)

(

2 2 2 Φ = D11 + D22 + D33 ( λ + 2μ ) + 2λ ( D11D22 + D11D33 + D22 D33 ) + 4μ D122 + D132 + D232

)

Thus, Φ = Φ __________________________________________________________________ 6.44 Given the velocity field of a linearly viscous fluid v1 = kx1 , v2 = − kx2 , v3 = 0 (a) Show that the velocity field is irrotational. (b) Find the stress tensor. (c) Find the acceleration field. (d) Show that the velocity field satisfies the Navier-Stokes equations by finding the pressure distribution directly from the equations. Neglect body forces. Take p = po at the origin. (e) Use the Bernoulli equation to find the pressure distribution. (f) Find the rate of dissipation of mechanical energy into heat. (g) If x2 = 0 is a fixed boundary, what condition is not satisfied by the velocity field. ------------------------------------------------------------------------------⎡ k 0 0⎤ ⎡k 0 0⎤ ⎡0 0 0⎤ ⎢ ⎥ ⎢ ⎥ Ans. (a) [∇v ] = ⎢ 0 − k 0 ⎥ → [ D] = ⎢ 0 − k 0 ⎥ , [ W ] = ⎢⎢0 0 0 ⎥⎥ ⎣⎢ 0 0 0 ⎦⎥ ⎣⎢ 0 0 0 ⎦⎥ ⎣⎢0 0 0 ⎦⎥ therefore, the flow is irrotational. (b) T11 = − p + 2 μ k , T22 = − p − 2 μ k , T33 = − p, T12 = T13 = T23 = 0 . ⎡ a1 ⎤ ⎡ k 0 ⎢ (c) ⎢ a2 ⎥⎥ = ⎢⎢ 0 − k ⎢⎣ a3 ⎥⎦ ⎢⎣ 0 0

2 0 ⎤ ⎡ kx1 ⎤ ⎡ k x1 ⎤ ⎢ ⎥ 0 ⎥⎥ ⎢⎢ − kx2 ⎥⎥ = ⎢ k 2 x2 ⎥ . ⎢ ⎥ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢ 0 ⎥ ⎣ ⎦

(d)



(

)

(

)

⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ∂p ∂p + μ ⎜ 21 + 21 + 21 ⎟ → ρ k 2 x1 = − ⎜ ⎟ ∂x1 ∂x1 ∂x2 ∂x3 ⎠ ⎝ ∂x1 ⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ∂p ∂p and =− + μ ⎜ 22 + 22 + 22 ⎟ → ρ k 2 x2 = − ⎜ ⎟ x ∂x2 ∂ x x x ∂ ∂ ∂ 2 2 3 ⎠ ⎝ 1

ρ k 2 x1 = − ρ k 2 x2 0=−

∂p → p is independent of x3 . ∂x3

Thus,

ρ k 2 x1 = − → f =−

ρ k 2 x12 df ( x2 ) ∂p ∂p df ( x2 ) → p=− + f ( x2 ) → = → − ρ k 2 x2 = ∂x1 ∂x2 2 dx2 dx2

ρ k 2 x22 2

+C → p =−

(

)(

ρk 2 2

(x

2 1

)

+ x22 + C. At x1 = x2 = 0, p = po → C = po

)

(

)

That is, p = − ρ k 2 / 2 x12 + x22 + po → p = − ( ρ / 2 ) v12 + v22 + po . (e) From the Bernoulli Equation, we have p p v2 p v 2 + v22 ⎡ p v12 + v22 ⎤ + = constant → + 1 =⎢ + = o +0 ⎥ 2 2 ⎦⎥ ρ 2 ρ ρ ⎣⎢ ρ at origin

(

)

→ p = po − ρ k 2 x12 + x22 / 2. (f)

(

)

(

)

2 2 2 2 2 2 Φ = λ ( D11 + D22 + D33 ) + 2 μ D11 + D22 + D33 + 2 D12 + 2 D13 + 2 D23 = 2μ k 2 + k 2 = 4μ k 2 2

(h) if x2 = 0 is a fixed boundary, then v must be zero there. But v = k ( x1e1 − x2e2 ) = kx1e1 ≠ 0

at x2 = 0 , therefore the non slip boundary condition at x2 = 0 is not satisfied for a viscous fluid. _________________________________________________________________

(

)

6.45 Do Problem 6.44 for the following velocity field: v1 = k x12 − x22 , v2 = −2kx1 x2 , v3 = 0 . ------------------------------------------------------------------------------Ans. (a) ⎡ 2kx1 −2kx2 0 ⎤ ⎡ x1 − x2 0 ⎤ ⎡0 0 0⎤ ⎢ ⎥ ⎢ ⎥ [∇v ] = ⎢ −2kx2 −2kx1 0⎥ → [ D] = 2k ⎢ − x2 − x1 0⎥ , [ W ] = ⎢⎢0 0 0⎥⎥ ⎢⎣ 0 ⎢⎣ 0 ⎢⎣0 0 0 ⎥⎦ 0 0 ⎥⎦ 0 0 ⎥⎦ therefore, the flow is irrotational T11 = − p + 4 μ kx1 , T22 = − p − 4 μ kx1 , T33 = − p, T12 = T13 = T23 = 0 (b) (c) 2 2 ⎤ ⎡ 2k 2 x x 2 + x 2 ⎤ ⎡ 1 1 2 ⎥ ⎡ a1 ⎤ ⎡ 2kx1 −2kx2 0 ⎤ ⎢ k x1 − x2 ⎥ ⎢ ⎢ ⎢ a ⎥ = ⎢ −2kx −2kx 0 ⎥ ⎢ −2kx x ⎥ = 2k 2 x x 2 + x 2 ⎥ 2 1 1 2 2 1 2 ⎥ ⎢ 2⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢⎣ a3 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 ⎥ 0 ⎥ ⎣ ⎦ ⎢⎣ ⎦ (d)

(

)


( (

) )

Lai et al, Introduction to Continuum Mechanics ⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ∂p ∂p + μ ⎜ 21 + 21 + 21 ⎟ → ρ ⎡ 2k 2 x1 x12 + x22 ⎤ = − ρ ⎡ 2k 2 x1 x12 + x22 ⎤ = −

(

⎣

ρ ⎡ 2k 2 x2 ⎣

0=−

(

)⎦

(

)

⎜ ∂x ⎣ ⎦ ∂x1 ∂x2 ∂x3 ⎟⎠ ⎝ 1 ⎛ ∂ 2v ∂ 2v ∂ 2v ⎞ ∂p ∂p x12 + x22 ⎤ = − + μ ⎜ 22 + 22 + 22 ⎟ → ρ ⎡ 2k 2 x2 x12 + x22 ⎤ = − ⎜ ⎟ ⎦ ⎣ ⎦ ∂x2 ∂x2 ∂x2 ∂x3 ⎠ ⎝ ∂x1 ∂x1

)

(

)

∂p → p is independent of x3 ∂x3

Thus, ⎛ x4 ⎞ df ∂p ∂p ρ ⎡ 2k 2 x1 x12 + x22 ⎤ = − → p = − ρ k 2 ⎜ 1 + x12 x22 ⎟ + f ( x2 ) → = −2 ρ k 2 x12 x2 + ⎣

and

(

)⎦

∂x1

(

)⎦

∂x2

⎜ ⎝ 2

⎟ ⎠

(

∂x2

)⎦

dx2

∂p df ρ ⎡ 2k 2 x2 x12 + x22 ⎤ = − → − ρ ⎡ 2k 2 x2 x12 + x22 ⎤ = −2 ρ k 2 x12 x2 + ⎣

⎣

dx2

ρ k 2 x24 df → f =− + C. dx2 2

→ −2 ρ k 2 x23 =

Since p = po at origin, therefore, C = po , → p=−

ρk 2 2

(

)

x14 + 2 x12 x22 + x24 + po → p = −

ρk2 2

(x

2 1

+ x22

)

2

+ po .

Or, since

(

)

(

⎡ v1 = k x12 − x22 , v2 = −2kx1 x2 , v12 + v22 = k 2 ⎢ x12 − x22 ⎣

(

)

)

2

(

⎤ + 4 x12 x22 ⎥ = k 2 x12 + x22 ⎦

)

2

→ p = − ρ v12 + v22 / 2 + po . (e) From the Bernoulli Equation, we have p p v2 p v 2 + v22 ⎡ p v12 + v22 ⎤ + = constant → + 1 =⎢ + = o +0 ⎥ 2 2 ⎦⎥ ρ 2 ρ ρ ⎣⎢ ρ at origin

(

→ p = po − ρ k 2 x12 + x22

)

2

/ 2.

(f)

(

)

(

)

(

)

2 2 2 Φ = λ ( D11 + D22 + D33 ) + 2 μ D11 + D22 + 2 D12 = 0 + 2μ 8k 2 x12 + 8k 2 x22 = 16 μ k 2 x12 + x22 (h) 2

if x2 = 0 is a fixed boundary, then v must be zero there. But v = kx12e1 ≠ 0 at x2 = 0 , therefore the non slip boundary condition at x2 = 0 is not satisfied for a viscous fluid. _________________________________________________________________ 6.46 Obtain the vorticity vector for the plane Poiseuille flow. -------------------------------------------------------------------------------

(

)

Ans. With v1 = v ( x2 ) = (α / 2 μ ) b 2 − x22 , where α = −∂p / ∂x1 and v2 = v3 = 0 , the spin tensor is

[ W ] = [ ∇v ]

A

0 ⎡ 1⎢ = ⎢ −∂v1 / ∂x2 2 ⎢⎣ 0

∂v1 / ∂x2 0 0

0⎤ 0 ⎥⎥ 0 ⎥⎦


Lai et al, Introduction to Continuum Mechanics The vorticity tensor is 2W and the vorticity vector is twice the axial vector ⎛ 1 ∂v1 ⎞ αx ∂v1 1 ∂p xe . ς = 2ω = 2 (W32e1 + W13e2 + W21e3 ) = 2 ⎜ − e3 = 2 e3 = − ⎟ e3 = − μ μ ∂x1 2 3 ∂x2 ⎝ 2 ∂x2 ⎠ _________________________________________________________________ 6.47 Obtain the vorticity vector for the Hagen-Poiseuille flow. ------------------------------------------------------------------------------⎞ α ⎛ d2 ∂p Ans. With vz = − r 2 ⎟ , vr =vθ = 0, α = − , the spin tensor is ⎜⎜ ⎟ ∂z 4μ ⎝ 4 ⎠ ⎡ 1 ⎛ 1 ∂vr vθ ∂vθ ⎞ 1 ⎛ ∂vr ∂vz ⎞ ⎤ − − − 1 ⎛ ∂v ⎞ ⎤ ⎢ 0 ⎥ ⎡ 0 − ⎜ z ⎟⎥ ∂r ⎟⎠ 2 ⎜⎝ ∂z ∂r ⎟⎠ ⎥ ⎢ 0 2 ⎜⎝ r ∂θ r ⎢ 2 ⎝ ∂r ⎠ ⎥ ⎢ 1 ⎛ ∂vθ 1 ∂vz ⎞ ⎥ ⎢ A ⎥ The 0 0 0 − [ W ] = [∇v ] = ⎢W21 ⎥=⎢ 0 2 ⎜⎝ ∂z r ∂θ ⎟⎠ ⎥ ⎢ ⎥ ⎢ 1 ⎛ ∂vz ⎞ ⎢ ⎥ ⎢W ⎥ 0 0 W32 0 ⎥ ⎢ 31 ⎥ ⎢⎣ 2 ⎜⎝ ∂r ⎟⎠ ⎦ ⎣⎢ ⎦⎥ vorticity tensor is 2W and the vorticity vector is twice the axial vector αr 1 ∂p ⎛ ∂v ⎞ reθ . ς = 2ω = 2 (Wzθ e r + Wrz eθ + Wθ r ez ) = 2Wrz eθ = − ⎜ z ⎟ eθ = eθ = − 2μ 2 μ ∂z ⎝ ∂r ⎠ _________________________________________________________________ 6.48 For a two-dimensional flow of an incompressible fluid, we can express the velocity components in terms of a scalar function ψ (known as the Lagrange stream function) by the ∂ψ ∂ψ relations vx = . (a) Show that the equation of conservation of mass is , vy = − ∂y ∂x automatically satisfied for any ψ ( x, y ) which has continuous second partial derivatives. (b) Show that for two-dimensional flow of an incompressible fluid, ψ =constants are streamlines. ∂ϕ (c) If the velocity field is irrotational, then vi = − where ϕ is known as the velocity potential. ∂xi Show that the curves of constant velocity potential ϕ = constant and the streamline ψ =constant are orthogonal to each other. (d) Obtain the only nonzero vorticity component in terms ofψ . ------------------------------------------------------------------------------∂v y ∂v ∂ ∂ψ ∂ ∂ψ ∂ψ ∂ψ Ans. (a) With vx = , we have, x + and v y = − = − = 0. ∂y ∂x ∂x ∂y ∂x ∂y ∂y ∂x (b) From ψ ( x, y ) = C , we have, vy ∂ψ ∂ψ ∂ψ / ∂x ⎛ dy ⎞ ⎛ dy ⎞ dψ = dx + dy = 0 → ⎜ ⎟ =− →⎜ ⎟ = ∂x ∂y ∂ψ / ∂y ⎝ dx ⎠ψ =constant vx ⎝ dx ⎠ψ =constant Thus, ψ ( x, y ) = C are streamlines. (c) From v ∂ϕ ∂ϕ ∂ϕ / ∂x ⎛ dy ⎞ ⎛ dy ⎞ ϕ ( x, y ) = C → d ϕ = dx + dy = 0 → ⎜ ⎟ =− →⎜ ⎟ =− x vy ∂x ∂y ∂ϕ / ∂y ⎝ dx ⎠ϕ =constant ⎝ dx ⎠ϕ =constant


Lai et al, Introduction to Continuum Mechanics ⎛ dy ⎞ ⎛ dy ⎞ Thus, → ⎜ ⎟ = −1 . ⎜ ⎟ ⎝ dx ⎠ϕ =constant ⎝ dx ⎠ψ = constant (d) ⎛ ∂ 2ψ ∂ 2ψ ⎞ ⎛ ∂v y ∂vx ⎞ − ς = 2ω = 2 Wzy e x + Wxz e y + Wyx ez = ⎜ ⎟ e z = − ⎜⎜ 2 + 2 ⎟⎟ e z . ∂y ⎠ ∂x ⎠ ⎝ ∂x ⎝ ∂y _________________________________________________________________

(

)

⎛ a2 ⎞ 6.49 Show that ψ = Vo y ⎜1 − 2 ⎟ represents a two-dimensional irrotational flow of an ⎜ x + y 2 ⎟⎠ ⎝ inviscid fluid. ------------------------------------------------------------------------------⎛ a2 ⎞ , we have, Ans. With ψ = Vo y ⎜1 − 2 2 ⎟ ⎜ ⎟ x y + ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ( −2 )( 2 x )2 a 2 ⎟ ∂ψ 2a 2 ⎜ 2 xa 2 ⎟ ∂ 2ψ ⎜ = Vo y ⎜ → 2 = Vo y ⎜ + 2 ⎟ 2 3 ⎟ ∂x ∂x ⎜ x2 + y 2 ⎟ ⎜ x2 + y 2 ⎟ x2 + y 2 ⎝ ⎠ ⎝ ⎠

(

)

(

⎛ 1 ⎜ = Vo y 2a ⎜ 2 ∂x ⎜ x2 + y 2 ⎝

)

(

)

⎞ ⎛ 8a 2 yx 2 ⎟ ⎜ 2a 2 yo − = − V o⎜ 2 3⎟ 2 ⎜ x2 + y2 x2 + y 2 ⎟ x2 + y 2 ⎠ ⎝ ⎛ ⎞ ⎛ a2 ⎞ ∂ψ ⎜ 2 y2a2 ⎟ = Vo ⎜1 − 2 +V 2 ⎟ 2 ⎟ ⎜ ⎟ o⎜ 2 ∂y x y + ⎝ ⎠ ⎜ x + y2 ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ 2 2 y 2 a 2 ( 2 )( 2 y ) ⎟ ∂ 2ψ ⎜ 2 ya 2 ⎟ ⎜ 2(2 y ) a = Vo ⎜ + Vo ⎜ − ⎟ 2 ⎟ 2 3 ∂y 2 ⎜ x2 + y 2 ⎟ ⎜ x2 + y 2 ⎟ x2 + y 2 ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ∂ 2ψ 8 y3a 2 ⎟ ⎜ 2 ya 2 ⎟ ⎜ 4 ya 2 = Vo ⎜ + Vo ⎜ − . Thus, 2 ⎟ 2 3⎟ ∂y 2 ⎜ x2 + y 2 ⎟ ⎜ x2 + y 2 x2 + y 2 ⎟ ⎝ ⎠ ⎝ ⎠ ∂ 2ψ

( ) 2

(

4 x2

) (

)

(

)

(

)

(

)

(

) (

∂ 2ψ

(

(

)

)

)

⎞ ⎛ 8a 2 yx 2 ⎟ ⎜ 2 ya 2 − + V o⎜ 3⎟ ⎜ x2 + y 2 x2 + y 2 ⎟ ⎠ ⎝

) ( 2

) (

)

(

⎛ ⎜ 2a 2 yo + = V o⎜ ∂y 2 ∂x 2 ⎜ x2 + y 2 ⎝

∂ 2ψ

(

⎞ ⎟ 3⎟ ⎟ ⎠

)

(

)

⎞ ⎛ ⎟ ⎜ 4 ya 2 + V ⎟ o⎜ 2 ⎟ ⎜ x + y2 ⎠ ⎝

) ( ) ⎛ ⎞ ⎞ ⎛ ⎞ ⎜ 8a y ( x + y ) ⎟ ⎟ ⎜ 8a y ⎟ ⎟=0 ⎟ =V ⎜ ⎟ −V ⎜ ⎜ ⎟ ⎜ ⎟ x + y x y + ) ⎟⎠ ) ⎠ ⎝( )⎠ ⎝ ( 2

(

⎞ ⎟ − 3⎟ x2 + y 2 ⎟ ⎠ 8 y 3a 2

2

⎛ ⎞ ⎛ 2 2 2 2 8 y 3a 2 ⎜ 8a 2 y ⎟ ⎜ 8a 2 yx 2 Vo ⎜ − Vo ⎜ + o o 2 ⎟ 3 3 2 3 2 2 2 2 ⎜ x2 + y 2 ⎟ ⎜ x2 + y 2 x2 + y 2 ⎝ ⎠ ⎝ Therefore, the given stream function ψ represents a two-dimensional irrotational flow of an inviscid fluid. __________________________________________________________________

(

)

(

) (


Lai et al, Introduction to Continuum Mechanics 6.50 Referring to Figure P 6-9, compute the maximum possible flow of water. Take the atmospheric pressure to be 93.1 kPa. , the specific weight of water 9810 N / m3 , and the vapor pressure 17.2 kPa. Assume the fluid to be inviscid. Find the length l for this rate of discharge. 10cm dia

3m

Δ

5m

Figure P 6-9 -----------------------------------------------------------------------------Ans. The Bernoulli equation gives, with point 1 at the reservoir top and point 2 at the highest point inside the tube, we have, p1 / ρ + v12 / 2 + g (0) = p2 / ρ + v22 / 2 + g (3) . Thus, assuming v1 to be very small and negligible, we have, with p1 = 93,100 Pa., p2 = 17, 200 Pa., ρ = 1000 kg / m3 and g = 9.81m / s 2 .

v22 / 2 = ( p1 − p2 ) / ρ − 3g = ( 93,100 − 17, 200 ) / 1000 − 3 ( 9.81) = 46.47

(

)

→ v2 = 9.64 m / s. Qmax = v2 A = 9.64 π ( 0.1) / 4 = 0.0757 m3 / s. 2

With point 3 at the exit, we have, p2 / ρ + v22 / 2 + g (0) = p3 / ρ + v32 / 2 + g (−l) now, v2 = v3 , p2 = pv (the vapor pressure), p3 = pa (atm.pressure)

l = ( pa − pv ) / ( ρ g ) = ( 93,100 − 17, 200 ) / 9810 = 7.74 m. _________________________________________________________________

6.51 Water flows upward through a vertical pipeline which tapers from cross sectional area A1 to area A2 in a distance of h . If the pressure at the beginning and end of the constriction are p1 and p2 respectively. Determine the flow rate Q in terms of ρ , A1 , A2 , p1 , p2 and h . Assume the fluid to be inviscid. ------------------------------------------------------------------------------Ans. Let the lower point be denoted as point 1, and the upper point denoted as point 2, we have

(

)

p1 / ρ + v12 / 2 + g (0) = p2 / ρ + v22 / 2 + g (h) → ( p1 − p2 ) / ρ − g (h) = v22 − v12 / 2 Let Q be the flow rate, then Q = A1v1 = A2 v2 and p1 − p2

ρ

2 2 2 ⎡⎣( p1 − p2 ) − ρ gh ⎤⎦ A22 A12 1 ⎡⎛ Q ⎞ ⎛ Q ⎞ ⎤ 2 ⎢ ⎥ − g ( h) = ⎜ ⎟ −⎜ ⎟ →Q = 2 ⎢⎝ A2 ⎠ ⎝ A1 ⎠ ⎥ ρ A12 − A22 ⎣ ⎦

→ Q = A1 A2

(

)

2 ⎡⎣( p1 − p2 ) − ρ g (h) ⎤⎦

(

ρ A12 − A22

)

_________________________________________________________________


Lai et al, Introduction to Continuum Mechanics 6.52 Verify that the equation of conservation of mass is automatically satisfied if the velocity components in cylindrical coordinates are given by 1 ∂ψ 1 ∂ψ , vz = , vθ = 0 ρ r ∂z ρ r ∂r where the density ρ is a constant and ψ is any function of r and z having continuous second partial derivatives. ------------------------------------------------------------------------------Ans. The equation of continuity is ∂v 1 ∂ 1 ∂v 1 ∂ψ 1 ∂ψ , vz = , vθ = 0 , we have, ( rvr ) + θ + z = 0 . With vr = − r ∂r r ∂θ ∂z ρ r ∂z ρ r ∂r vr = −

1 ∂ 1 ∂ ⎛ 1 ∂ψ ⎞ 1 ⎛ ∂ 2ψ ⎞ ∂vz ∂ ⎛ 1 ∂ψ ⎞ 1 ⎛ ∂ 2ψ ⎞ = ⎜ ( rvr ) = − ⎟, ⎜ ⎟ ⎜r ⎟ = − ⎜⎜ ⎟= ∂z ∂z ⎝ ρ r ∂r ⎠ ρ r ⎜⎝ ∂z∂r ⎟⎠ r ∂r r ∂r ⎝ ρ r ∂z ⎠ ρ r ⎝ ∂r ∂z ⎟⎠ Thus, the equation of continuity is automatically satisfied for any functionψ ( x, y ) . _________________________________________________________________ 6.53 From the constitutive equation for a compressible fluid Tij = − pδ ij − (2μ / 3)Δδ ij + 2 μ Dij + k Δδ ij , Δ = ∂v j / ∂x j , derive the equation Dvi ∂ 2 vi ∂p μ ∂ ⎛ ∂v j ⎞ ∂ ⎛ ∂v j ⎞ = ρ Bi − + +k ⎜ ⎟+μ ⎜ ⎟ ∂xi 3 ∂xi ⎜⎝ ∂x j ⎟⎠ ∂x j ∂x j ∂xi ⎜⎝ ∂x j ⎟⎠ Dt -----------------------------------------------------------------------------Ans. ∂Tij ∂p ∂ 1 ⎛ ∂vi ∂v j ⎞ ∂Δ 2 ∂Δ =− + δ ij − μ δ ij + 2μ δ ij ⎜ ⎟+k ⎜ ⎟ ∂x j ∂x j ∂x j 2 ⎝ ∂x j ∂xi ⎠ ∂x j 3 ∂x j

ρ

⎛ ∂ 2 vi ∂p 2 ∂Δ ∂Δ ⎞ ∂Δ − μ + μ⎜ + ⎟+k ⎜ ⎟ ∂xi 3 ∂xi ∂xi ⎝ ∂x j ∂x j ∂xi ⎠ That is, ∂Tij ∂Tij ∂ 2 vi Dv ∂p μ ∂Δ ∂Δ =− + +μ +k . Thus, ρ i = ρ Bi + → ∂x j ∂xi 3 ∂xi ∂x j ∂x j ∂xi Dt ∂x j =−

ρ

Dvi ∂ 2 vi ∂p μ ∂Δ ∂Δ = ρ Bi − + +μ +k Dt ∂xi 3 ∂xi ∂x j ∂x j ∂xi

_________________________________________________________________ 6.54 Show that for a one-dimensional, steady, adiabatic flow of an ideal gas, the ratio of temperature Θ1 / Θ2 at sections 1 and 2 is given by 1 1 + ( γ − 1) M12 Θ1 2 = Θ2 1 + 1 γ − 1 M 2 ( ) 2 2 where γ is the ratio of specific heat, M1 and M 2 are local Mach number at section 1 and section 2 respectively. -------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics p ⎛ γ ⎞ v12 p2 ⎛ γ ⎞ v22 p ⎛ γ ⎞ v2 = constant , we have 1 ⎜ = . ⎜ ⎟+ ⎟+ ⎜ ⎟+ ρ1 ⎝ γ − 1 ⎠ 2 ρ 2 ⎝ γ − 1 ⎠ 2 ρ ⎝ γ −1 ⎠ 2 In terms of the Mach numbers M1 = v1 / c1 and M 2 = v2 / c2 , we have, Ans.

p1 ⎛ γ ⎞ c12 M12 p2 ⎛ γ ⎞ c22 M 22 = . ⎜ ⎟+ ⎜ ⎟+ 2 2 ρ1 ⎝ γ − 1 ⎠ ρ2 ⎝ γ − 1 ⎠ γp For an ideal gas, p = ρ RΘ , and c 2 = = γ RΘ , therefore,

ρ

ρ1RΘ1 ⎛ γ ⎞ ⎜ ⎟+ ρ1 ⎝ γ − 1 ⎠

γ RΘ1M12 2

=

ρ 2 RΘ2 ⎛ γ ⎞ γ RΘ2 M 22 → ⎜ ⎟+ ρ2 ⎝ γ − 1 ⎠ 2

⎛ γ ⎞ ⎛ γ ⎞ γΘ2 M 22 γΘ1M12 1 1 − → Θ1 + ( γ − 1) Θ1M12 = Θ 2 + ( γ − 1) Θ 2 M 22 → ⎜ ⎟ Θ1 − ⎜ ⎟ Θ2 = 2 2 2 2 ⎝ γ −1 ⎠ ⎝ γ −1⎠ Θ 1 + M 2 ( γ − 1) / 2 ⎡ 1 ⎤ ⎡ 1 ⎤ . Θ1 ⎢1 + ( γ − 1) M12 ⎥ = Θ 2 ⎢1 + ( γ − 1) M 22 ⎥ → 1 = Θ2 1 + M12 ( γ − 1) / 2 ⎣ 2 ⎦ ⎣ 2 ⎦ _________________________________________________________________ 2

6.55 Show that for a compressible fluid in isothermal flow with no external work, dM 2

dv , where M is the Mach number. (Assume perfect gas). v M ------------------------------------------------------------------------------2

=2

Ans. Since M 2 ≡ v 2 /c 2 and c 2 = γ RΘ for ideal gas, therefore, M 2 ≡ v 2 / (γ RΘ) For isothermal flow, Θ =constant , therefore, v2 2vdv dM 2 2vdv γ RΘ 2dv M2 ≡ → dM 2 = → = = . v γ RΘ γ RΘ M 2 γ RΘ v 2 _________________________________________________________________ 6.56 Show that for a perfect gas flowing through a duct of constant cross sectional area at dp 1 dM 2 constant temperature =− . [Use the results of the last problem]. p 2 M2 ------------------------------------------------------------------------------Ans. We have, from ρ Av = constant, ( d ρ ) v + ρ ( dv ) = 0 → d ρ / ρ = − dv / v Since Θ = constant , therefore, p = ρ RΘ → dp = RΘd ρ →

dp RΘd ρ d ρ = = p ρ RΘ ρ

Thus, dp / p = − dv / v . From the results of last problem, we have, dM 2 / M 2 = 2dv / v , therefore,

dp / p = −(1 / 2)(dM 2 / M 2 ) . _________________________________________________________________ 6.57 For the flow of a compressible inviscid fluid around a thin body in a uniform stream of speed Vm in the x1 direction, we let the velocity potential be ϕ = −Vo ( x1 + ϕ1 ) , where ϕ1 is


Lai et al, Introduction to Continuum Mechanics assumed to be very small. Show that for steady flow the equation governing ϕ1 is, with M o = Vo / co ,

(1 − M ) ∂∂xϕ 2 o

2

1 2 1

+

∂ 2ϕ1 ∂x22

+

∂ 2ϕ1 ∂x32

=0.

-----------------------------------------------------------------------∂v ∂ρ + ρ i = 0 , in terms of the potential Ans. For steady flow, the equation of continuity is vi ∂xi ∂xi function ϕ , we have, → −

∂ϕ ∂ρ ∂ 2ϕ −ρ =0. ∂xi ∂xi ∂xi ∂xi

(i).

The equation of motion is: ∂v ⎤ ∂p 1 ∂p 1 ∂p ∂ρ 1 ∂ρ ⎡ vj i = − =− = − c2 = local sound speed ⎥ , ⎢ note : c 2 = ∂x j ∂ρ ρ ∂xi ρ ∂ρ ∂xi ρ ∂xi ⎣ ⎦ which becomes, ∂ϕ ∂ 2ϕ 1 ∂ρ ∂ρ ρ ∂ϕ ∂ 2ϕ = − c2 → =− 2 (ii) ∂x j ∂x j ∂xi ρ ∂xi ∂xi c ∂x j ∂x j ∂xi ∂ϕ ⎛ ∂ϕ ∂ 2ϕ ⎞ 2 ∂ 2ϕ = 0 (iii). Now, with ϕ = −Vo ( x1 + ϕ1 ) , we have, ⎜ ⎟−c ∂xi ⎜⎝ ∂x j ∂x j ∂xi ⎟⎠ ∂xi ∂xi ⎛ ⎛ ∂ϕ ⎞ ∂ϕ ⎞ ⎡ ⎛ ∂ϕ ⎞ ∂ 2ϕ1 ⎤ ∂ϕ ∂ϕ ⎛ ∂ϕ ∂ 2ϕ ⎞ ⎥ = −Vo ⎜ δ i1 + 1 ⎟ , → ⎜ ⎟ = −Vo ⎜ δ i1 + 1 ⎟ ⎢Vo ⎜ δ j1 + 1 ⎟ Vo ∂xi ⎝⎜ ∂x j ∂x j ∂xi ⎠⎟ ∂xi ⎠ ⎢ ⎝⎜ ∂x j ⎠⎟ ∂x j ∂xi ⎥ ∂xi ∂xi ⎠ ⎝ ⎝ ⎣ ⎦ ⎛ ∂ϕ ⎞ ⎡ ∂ 2ϕ1 ⎤ ∂ 2ϕ1 ∂ 2ϕ1 3 . ≈ −Vo3 ⎜ δ i1 + 1 ⎟ ⎢δ j1 = −Vo3 ⎥ ≈ −Vo δ i1δ j1 ∂xi ⎠ ⎣⎢ ∂x j ∂xi ⎦⎥ ∂x j ∂xi ∂x1∂x1 ⎝

(ii) into (i) →

⎛ ∂ 2ϕ1 ∂ 2ϕ1 ∂ 2ϕ1 ⎞ 2 c V + + 2 + 2 ⎟=0, o⎜ ⎜ ∂x 2 ∂x12 ∂x2 ∂x3 ⎟⎠ ⎝ 1 ∂ 2ϕ1 ⎛ ∂ 2ϕ1 ∂ 2ϕ1 ⎞ +⎜ + 2 ⎟=0. ∂x12 ⎜⎝ ∂x22 ∂x3 ⎟⎠

Thus, Eq.(iii) → −Vo3

(

→→ 1 − M o2

)

∂ 2ϕ

_________________________________________________________________ 6.58 For a one dimensional steady flow of a compressible fluid through a convergent channel, obtain (a) the critical pressure and (b) the corresponding velocity. That is, verify equation (6.30.7) and Eq. (6.30.8) -------------------------------------------------------------------------------Ans. (a) From Eq. (6.30.6), 1

γ +1 ⎫ ⎤ 2 2 ⎡ ⎧ γ ⎛ ⎞ ⎛ ⎞ ⎢ 2γ p γ ⎪⎥ dm ⎪ p = A2 ⎢ p1ρ1 ⎨⎜ 2 ⎟ − ⎜ 2 ⎟ ⎬⎥ , dt ⎝ p1 ⎠ ⎪⎥ ⎪⎝ p1 ⎠ ⎢γ −1 ⎩ ⎭⎦ ⎣ we have,


(i)

Lai et al, Introduction to Continuum Mechanics d ( dm / dt ) d ( p2 / p1 )

Thus,

1 ⎛ dm ⎞ = ⎜ ⎟ 2 ⎝ dt ⎠

d ( dm / dt ) d ( p2 / p1 )

2 ⎛ p2 ⎞ ⎜ ⎟ γ ⎝ p1 ⎠ That is,

2−γ

γ

−

1 2

2 −γ 1⎫ ⎧ 2γ γ + 1 ⎛ p2 ⎞ γ ⎪ ⎪ 2 ⎛ p2 ⎞ γ − pρ ⎨ ⎜ ⎟ ⎜ ⎟ ⎬. γ − 1 1 1 ⎪ γ ⎝ p1 ⎠ γ ⎝ p1 ⎠ ⎪ ⎩ ⎭

(ii)

= 0 gives, 1

1

⎡

⎛ p2 ⎞ γ ⎢ 2 ⎛ p2 ⎞ γ + 1 ⎛ p2 ⎞ γ − ⎜ ⎟ = 0, or ⎜ ⎟ ⎢ ⎜ ⎟ γ ⎝ p1 ⎠ ⎝ p1 ⎠ ⎢ γ ⎝ p1 ⎠

1−γ

γ

(iii)

⎥ ⎦

⎣

1−γ

⎤

γ + 1⎥ − =0. γ ⎥

γ

⎛p ⎞ ⎛ 2 ⎞ γ −1 2 ⎛ p2 ⎞ γ γ +1 , therefore ⎜ 2 ⎟ = =⎜ ⎜ ⎟ ⎟ . γ ⎝ p1 ⎠ γ ⎝ p1 ⎠critical ⎝ γ + 1 ⎠ (b) Substituting this critical pressure into Eq. (6.30.4) for the velocity, we get γ −1 ⎞ ⎛ −1 ⎛ ⎞ p2 γ ⎟ 2γ p1 ⎛ ⎛ γ + 1 ⎞ ⎞ 2γ p1 ⎛ γ − 1 ⎞ p1 ⎛ 2γ ⎞ 2γ p1 ⎜ ⎜ ⎟= v22 = 1 1 − = − ⎜ ⎟ ⎜ ⎟= ⎜ ⎟. γ − 1 ρ1 ⎜⎜ ⎝ p1 ⎠crit ⎟⎟ γ − 1 ρ1 ⎝⎜ ⎜⎝ 2 ⎟⎠ ⎠⎟ γ − 1 ρ1 ⎝ γ + 1 ⎠ ρ1 ⎝ γ + 1 ⎠ ⎝ ⎠ p ⎛ρ ⎞ From 1 = ⎜ 2 ⎟ p2 ⎝ ρ1 ⎠

−γ

⎛ p1 ⎞⎛ ρ1 ⎞⎛ ρ 2 ⎞ ⎛ ρ 2 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟ ⎝ ρ1 ⎠⎝ ρ 2 ⎠⎝ p2 ⎠ ⎝ ρ1 ⎠

→

1

1−γ

⎛ ρ 2 ⎞ ⎛ p2 ⎞ γ ⎛ ρ2 ⎞ ⎜ ⎟=⎜ ⎟ →⎜ ⎟ ⎝ ρ1 ⎠ ⎝ p1 ⎠ ⎝ ρ1 ⎠

⎛p ⎞ =⎜ 2 ⎟ ⎝ p1 ⎠

1−γ

γ

−γ

1−γ

⎛ p ⎞ ⎛ρ ⎞ →⎜ 1 ⎟=⎜ 2 ⎟ ⎝ ρ1 ⎠ ⎝ ρ1 ⎠

⎛p ⎞ ⎛p ⎞ →⎜ 1 ⎟=⎜ 2 ⎟ ⎝ ρ1 ⎠ ⎝ p1 ⎠

1−γ

γ

⎛ p2 ⎞ ⎜ ⎟ . But ⎝ ρ2 ⎠

⎛ p2 ⎞ ⎜ ⎟ . Now, at ⎝ ρ2 ⎠

γ

⎛ p2 ⎞ ⎛ p1 ⎞ γ + 1 ⎛ p2 ⎞ ⎛ γ + 1 ⎞1−γ =⎜ ⎜ ⎟ ⎜ ⎟ . Thus, ⎟ , we have, ⎜ ⎟ = 2 ⎝ ρ2 ⎠ ⎝ p1 ⎠critical ⎝ 2 ⎠ ⎝ ρ1 ⎠ ⎛p ⎞ v22 = γ ⎜ 2 ⎟ =speed of sound at section (2) . ⎝ ρ2 ⎠ __________________________________________________________________


(iv)


CHAPTER 7

7.1

Verify the divergence theorem

∫S v ⋅ ndS = ∫V div v dV for the vector field

v = 2xe1 + ze2 ,

by considering the region bounded by x = 0, x = 2, . y = 0, y = 2, z = 0, z = 2 . --------------------------------------------------------------------------------------Ans. With v = 2xe1 + ze2 , we have For the face x = 0, n = −e1 , v ⋅ n = −2x = 0, For the face x = 2 , n = +e1 , v ⋅ n = +2x = 4,

For the face y = 0, n = −e 2 , v ⋅ n = −z,

∫ v ⋅ ndS = 0 . ∫ v ⋅ ndS = 4 A = 4(4) = 16 . 2

2

2 ∫ v ⋅ ndS = − ∫0 z (2dz ) = −2 ⎡⎣ z / 2⎤⎦ 0 = −4 . 2

∫ v ⋅ ndS = ∫0 z (2dz ) = +4 . For the face z = 0, n = −e3 , v ⋅ n = 0, ∫ v ⋅ ndS = 0 . For the face z = 2 , n = +e3 , v ⋅ n = 0, ∫ v ⋅ ndS = 0 . Thus, ∫ v ⋅ ndS =16 − 4 + 4 = 16 and ∫ ( div v ) dV = ∫ 2dV = 2 ( 2 × 2 × 2 ) = 16. S So, ∫ v ⋅ ndS = ∫ div vdV . S V For the face y = 2, n = +e2 , v ⋅ n = z,

________________________________________________________________________ 7.2


∫S v ⋅ ndS = ∫V div v dV for the vector field, which in

cylindrical coordinates, is v = 2rer + zez , by considering the region bounded by r = 2 , z = 0 and z = 4. --------------------------------------------------------------------------------------Ans. For the cylindrical surface r = 2 , n = e r , → v ⋅ n = 2r = 2(2) = 4, → ∫ v ⋅ ndS = 4∫ dS = 4 S = 4 ⎡⎣( 2π 2 ) (4) ⎤⎦ = 64π . For the end face z = 0 , n = −e z , → v ⋅ n = − z z =0 = 0, → ∫ v ⋅ ndS = 0 . For the end face z = 4 , n = e z , → v ⋅ n = z z = 4 = 4, → ∫ v ⋅ ndS = 4S = 4π 22 = 16π . Therefore,

div v =

∫S v ⋅ ndS = 64π + 16π = 80π .

∂vr vr ∂vz ∂ ( 2r ) 2r ∂z + + = + + = 2 + 2 +1= 5. ∂r r ∂z ∂r r ∂z

2 ∫V div vdV =5 (π 2 ) ( 4 ) = 80π , thus, ∫S v ⋅ ndS = ∫V divv dV .

________________________________________________________________________ 7.3


∫S v ⋅ ndS = ∫V divv dV for the vector field, which in

spherical coordinates is v = 2rer , by considering the region bounded by the spherical surface r = 2. ---------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. For the spherical surface at r = 2 , n = er , → v ⋅ n = 2r = 2(2) = 4, →

∫ v ⋅ ndS = ∫ 4dS =4S = 4 ( 4π 2

2

) = 64π

On the other hand [see Eq.(2.35.26)],

(

)

( )

2 3 ⎛ 4π 23 ⎞ 1 d r vr 1 d 2r = 2 = 6, ∫ div v dV = 6 ⎜ = 64π . div v = 2 ⎜ 3 ⎟⎟ dr dr r r ⎝ ⎠ ________________________________________________________________________

7.4

Show that

∫S x ⋅ n dS = 3V . where x is the position vector and V is the volume enclosed by the boundary S . --------------------------------------------------------------------------------------∂x ∂x ∂x Ans. x = x1e1 + x2e 2 + x3e3 , div x = 1 + 2 + 3 = 1 + 1 + 1 = 3 . ∂x1 ∂x2 ∂x3 Thus

∫S x ⋅ ndS = ∫V div x dV = 3V .

________________________________________________________________________ 7.5 (a) Consider the vector field v = ϕ a , where ϕ is a given scalar field and a is an arbitrary constant vector (independent of position). Using the divergence theorem, prove that

∫V ∇ϕ dV = ∫S ϕn dS . (b) Show that for any closed surface S that

∫S ndS = 0 . --------------------------------------------------------------------------------------Ans. (a) With v = ϕ a , v ⋅ n = ϕ a ⋅ n → ∫ v ⋅ ndS = a ⋅ ∫ ϕ ndS , div v = div (ϕ a ) =

∂ϕ ai ∂ϕ = ai = a ⋅ ∇ϕ . ∂xi ∂xi

∫S v ⋅ ndS = ∫S div vdV → a ⋅ ∫S ϕ ndS = a ⋅ ∫V ∇ϕ dV . Since a is arbitrary, therefore, ∫S ϕ ndS = ∫V ∇ϕ dV . (b) Take ϕ = 1 in the results of part (a), we have ∫ ndS = 0 . S

Thus,

________________________________________________________________________ 7.6 A stress field T is in equilibrium with a body force ρ B . Using the divergence theorem, show that for any volume V and boundary surface S , that

∫S tdS + ∫V ρ BdV = 0 . where t is the stress vector. That is, the total resultant force is equipollent to zero. --------------------------------------------------------------------------------------Ans. The stress vector t is related to the stress tensor T by t = Tn , therefore, ∫ tdS = ∫ TndS = ∫ div TdV , thus, ∫ tdS + ∫ ρ BdV = ∫ ( divT + ρ B ) dV . S

S

V

S

V

V


Lai et al, Introduction to Continuum Mechanics But in equilibrium, ( divT + ρ B ) = 0 , therefore,

∫S tdS + ∫V ρ BdV = 0 .

________________________________________________________________________ 1 T Let u* define an infinitesimal strain field E* = ⎡⎢∇u* + ( ∇u* ) ⎤⎥ and let T** be the ⎣ ⎦ 2 symmetric stress tensor in static equilibrium with a body force ρ B ** and a surface traction t ** . Using the divergence theorem, verify the following identity (theory of virtual work).

7.7

∫S t ** ⋅ u* dS + ∫V ( ρ B** ) ⋅ u* dV = ∫V Tij

** * Eij dV

.

--------------------------------------------------------------------------------------Ans. T T ∫S t ** ⋅ u* dS = ∫S ( T** n ) ⋅ u* dS = ∫S n ⋅ ( T** ) u* dS = ∫V div ⎡⎣⎢( T** ) u* dV ⎤⎥⎦ . = ∫ div(T** u* )dV V

Now, div ( T** u* ) =

∂ (Tij**u*j ) ∂xi

=

∂Tij** ∂xi

u*j + Tij**

∂u*j ∂xi

= divT** ⋅ u* +Tij**

∫S t ** ⋅ u* dS + ∫V ρ B** ⋅ u* dV = ∫ ⎡⎣( divT** + ρ B** ) ⋅ u* +Tij

**

∂u*j ∂xi

, therefore,

∂u*j / ∂xi ⎤dV = ∫ Tij**∂u*j / ∂xi dV . ⎦

Now, since Tij** = T ji** , therefore, ⎛ ∂u* ∂u*j 1 Tij** Eij* = Tij** ⎜ i + ⎜ ∂x j ∂xi 2 ⎝

⎞ 1 ∂u* 1 ∂u* ∂u* 1 ∂u* 1 ⎟ = Tij** i + Tij** j = Tij** i + T ji** j ⎟ 2 ∂x j 2 ∂xi 2 ∂x j 2 ∂xi ⎠

∂u j ∂ui* ∂u* = T ji** i = Tij** . ∂x j ∂x j ∂xi *

= Tij** Thus,

∫S t ** ⋅ u* dS + ∫V ( ρ B* ) ⋅ u* dV = ∫V Tij

** * Eij dV

________________________________________________________________________ Using the equations of motion and the divergence theorem, verify the following rate of 7.8 work identity. Assume the stress tensor to be symmetric.

D ⎛ v2 ⎞ ⎟⎟ dV + ∫V Tij Dij dV ⎝ 2 ⎠

∫S t ⋅ vdS + ∫V ρ B ⋅ vdV = ∫V ρ Dt ⎜⎜

--------------------------------------------------------------------------------------Ans. ∫ t ⋅ vdS = ∫ Tn ⋅ vdS = ∫ n ⋅ TT vdS = ∫ div(TT v ) dV = ∫ div(Tv )dV S

Now, div(Tv ) =

S

S

∂Tij v j ∂xi

=

∂Tij ∂xi

V

v j + Tij

∂v j ∂xi

= divT ⋅ v + Tij

V

∂v j ∂xi

, therefore,

∫S t ⋅ vdS + ∫V ρ B ⋅ vdV = ∫V [( divT + ρ B ) ⋅ v + Tij ∂v j / ∂xi ]dV = ∫V ρ ( Dv / Dt ) ⋅ vdV + Tij ∂v j / ∂xi . 1 D( v ⋅ v) D ⎛ v2 ⎞ 1 ⎛ ∂v ∂v j ⎛ Dv ⎞ Now, ⎜ ρ ρ ρ = ⋅ = v ⎜⎜ ⎟⎟ and Tij Dij = Tij ⎜⎜ i + ⎟ 2 ⎝ ∂x j ∂xi 2 Dt Dt ⎝ 2 ⎠ ⎝ Dt ⎠


⎞ ∂v j ∂v . ⎟ = Tij i = Tij ⎟ ∂x j ∂xi ⎠

Lai et al, Introduction to Continuum Mechanics ⎞ ⎟⎟ dV + ∫V Tij Dij dV . ⎠ ________________________________________________________________________

Therefore,

∫S

t ⋅ vdS + ∫ ρ B ⋅ vdV = ∫ ρ V

V

D ⎛ v2 ⎜ Dt ⎜⎝ 2

Consider the velocity and density fields1

7.9

−α t −t v = α x1e1 , ρ = ρo e ( o ) (a) Check the equation of mass conservation. (b) Compute the mass and rate of increase of mass in the cylindrical control volume of crosssection A and bounded by x1 = 0 and x1 = 3 . (c) Compute the net mass inflow into the control volume of part (b). Does the net mass inflow equal the rate of mass increase inside the control volume? --------------------------------------------------------------------------------------Dρ −α t −t −α t −t + ρ div v = −αρo e ( o ) + ρo e ( o ) (α ) = 0 . That is, the conservation of mass Ans. (a). Dt equation is satisfied. (b) Inside the control volume, 3 −α t −t −α t −t −α t −t m = ρ dV = ρ e ( o ) Adx = 3ρ e ( o ) A, and dm / dt = −3αρ e ( o ) A . That is, the

∫

∫0

o

1

o

o

−α t −t mass inside the volume is decreasing at the rate of 3αρo e ( o ) A . (c). Rate of inflow from the face x1 = 0 is zero because at x1 = 0 , v1 = 0. −α t −t Rate of outflow from the face x = 3 is given by ρ v A = 3αρ e ( o ) A . There is no flow

1

1 x =3 1

o

across the cylindrical surface because flow is only in the x1 direction. Thus, the rate of outflow exactly equals the rate of decrease of mass inside the volume. ________________________________________________________________________ 7.10

(a) Check that the motion

α t −t x1 = X1e ( o ) , x2 = X 2 , x3 = X 3 corresponds to the velocity field v = α x1e1 . −α t −t (b) For a density field ρ = ρ e ( o ) , verify that the mass contained in the material volume that o

was coincident with the control volume of Prob. 7.9 at time to , remains a constant at all times, as it should (conservation of mass). (c) Compute the total linear momentum for the material volume of part (b). (d) Compute the force acting on the material volume --------------------------------------------------------------------------------------∂x ∂x ∂x α t −t Ans. (a) v1 = 1 = α X1e ( o ) = α x1 , v2 = 2 = 0, v3 = 3 = 0 , i.e., v = α x1e1 . ∂t ∂t ∂t (b) The particles which are at x1 = 0 at time to have the material coordinates X1 = 0 . These particles remain at x1 = 0 at all time. The particles which are at x1 = 3 at time to have the material 1

It should be remarked that, for a real fluid, to achieve the given velocity and density fields in this and some other problems may require body force distributions and/or a pressure density relationship that are not realistic.


Lai et al, Introduction to Continuum Mechanics α t −t coordinates X1 = 3 . These particles move in such a way that x1 = 3e ( o ) . Thus, to find the mass inside this material volume as a function of time, we have

M=

α t −t 3e ( o )

∫ 0

−α t −t −α t −t α t −t ρo e ( o ) Adx1 = ρo e ( o ) A ⎡⎢3e ( o ) ⎤⎥ = 3ρo A .

⎣

⎦

(c) Linear momentum in the material volume

P=

α t −t 3e ( o )

∫

ρo e

−α ( t −to )

v1 Adx1e1 =

0

α t −t 3e ( o )

∫

ρo e−α ( t −to )α x1 Adx1e1

0 α ( t − to )

⎡ 9e 2α ( t −to ) ⎤ 9 −α ( t −to ) α t −t ⎢ ⎥ e1 = Aρoα e ( o )e1. x dx e ρ α e A = ∫ 1 11 o 2 2 ⎢⎣ ⎥⎦ 0 (d) Force acting on the material volume dP 9 α t −t F= = Aρoα 2 e ( o )e1 . dt 2 We see that both the linear momentum and the force increase exponentially with time. This is due to the given data of density and velocity fields, which describe the space occupied by the fixed material increases exponentially with time, ⎡ 0 ≤ x1 ≤ 3eα (t −to ) ⎤ ,while the density decreases ⎣ ⎦ t t − α − ( ) o ⎤ exponentially ⎡⎢ ρo e to conserve the mass. We note also that at t = to , the materials occupy ⎣ ⎦⎥ 3e

−α t −t = ρoα e ( o ) A

9 Aρoα 2e1 . 2 ________________________________________________________________________

the space between x1 = 0 and x1 = 3 , and F =

7.11 Do Problem 7.9 for the velocity field v = α x1e1 and the density field ρ = k ρo / x1 and for the cylindrical control volume bounded by x1 = 1 and x1 = 3 . --------------------------------------------------------------------------------------⎛ ρ ⎞ ρ ∂v ∂ρ Dρ Ans. (a). + ρ div v = v1 + ρ 1 = α x1 ⎜ − k 2o ⎟ + k o (α ) = 0 . That is, the conservation of ⎜ ∂x1 ∂x1 Dt x1 x1 ⎟⎠ ⎝ mass equation is satisfied. 3 ρ dm =0. (b) Inside the control volume, m = ∫ ρ dV = ∫ k o Adx1 = kAρo ln 3, and 1 x1 dt

⎡ ρ ⎤ (c). Rate of inflow from the face x1 = 1 is [ ρ v1 A]x = ⎢ k o α x1 A⎥ = kαρ o A . 1 ⎣ x1 ⎦ x1 =1 ⎡ ρ ⎤ Rate of outflow from the face x1 = 3 is given by [ ρ v1 A]x =3 = ⎢ k o α x1 A⎥ = kαρo A . There is 1 ⎣ x1 ⎦ x1 =3 no flow across the cylindrical surface because flow is only in the x1 direction. The net mass inflow is 0 , which is equal to the rate of increase of mass inside the control volume. ________________________________________________________________________ 7.12

The center of mass xc.m of a material volume is defined by the equation mxc.m = ∫ xρ dV , where m = ∫ ρ dV Vm

Vm


Lai et al, Introduction to Continuum Mechanics Demonstrate that the linear momentum principle may be written in the form ∫ tdS + ∫ ρ BdV = mac.m S

V

where ac.m is the acceleration of the mass center. --------------------------------------------------------------------------------------Ans. We have from the principle of linear momentum:

D

∫S tdS + ∫V ρ BdV = Dt ∫V

ρ vdV

m

D D D D ( ρ dV ) = 0 , therefore, ∫ ρ vdV = ∫ ( ρ xdV ) = ∫ ( ρ xdV ) = mxc.m = mv c.m . Dt Dt Dt Dt D Therefore, ∫ tdS + ∫ ρ BdV = mv c.m = mac.m . Dt S V

Now, since

________________________________________________________________________ 7.13

Consider the following velocity field and density field ρ α x1 v= e1 , ρ = o 1 + αt 1+ αt (a) Compute the total linear momentum and rate of increase of linear momentum in a cylindrical control volume of cross-sectional area A and bounded by the planes x1 = 1 and x1 = 3 . (b) Compute the net rate of outflow of linear momentum from the control volume of (a) (c) Compute the total force on the material in the control volume. (d) Compute the total kinetic energy and rate of increase of kinetic energy for the control volume of (a). (e) Compute the net rate of outflow of kinetic energy from the control volume. --------------------------------------------------------------------------------------Ans. (a) Linear momentum is 3 4 ρo Aα ρ ρ Aα 3 α x1 ⎛ 9 1 ⎞ ρ Aα P = ∫ ρ vdV = ∫ o e . Adx1e1 = o 2 ∫ x1dx1e1 = ⎜ − ⎟ o 2 e1 = 2 1 2 2 α α 1 1 + + t t ⎝ ⎠ α + + 1 t 1 t α α 1 + t ( ) ( ) ( ) 1 1 Rate of increase of linear momentum inside the control volume is 8 ρ Aα 2 dP = − o 3 e1 . dt (1 + α t ) (b) Net rate of outflow of linear momentum in e1 direction

(

= ρ Av12

)

x1 =3

(

− ρ Av12

)

x1 =1

=

ρo A ⎛⎜ 9α 2 α 2 ⎞⎟ 8α 2 ρo A − = . 1 + α t ⎜ (1 + α t )2 (1 + α t )2 ⎟ (1 + α t )3 ⎝ ⎠

(c) Total force = Rate of inc. of P inside control volume + net outflux of P 8 ρ Aα 2 8α 2 ρo A = − o 3 e1 + e1 =0. (1 + α t ) (1 + α t )3 (d) Total kinetic energy inside the control volume 1 1 ρ ⎛ α x1 ⎞ 1 ρo Aα 2 13 ρ o Aα 2 2 K .E. = ∫ ρ v 2 dV = ∫ o ⎜ Adx x dx = = 1 1 1 2 2 1 1 + α t ⎝ 1 + α t ⎟⎠ 2 (1 + α t )3 ∫1 3 (1 + α t )3 3

2

3

13ρo Aα 3 d . K .E = − dt (1 + α t )4 (e) Net rate of outflow of kinetic energy from the control volume=


Lai et al, Introduction to Continuum Mechanics 1 ρo A ⎛ 27α 3 α 3 ⎞⎟ 13α 3 ρo A ⎛1 ⎛1 3⎞ 3⎞ ⎜ . − = − = Av Av ⎜ ρ 1⎟ ⎜ ρ 1⎟ ⎝2 ⎠ x1 =3 ⎝ 2 ⎠ x1 =1 2 1 + α t ⎝⎜ (1 + α t )3 (1 + α t )3 ⎠⎟ (1 + α t )4 ________________________________________________________________________ 7.14

Consider the velocity and density fields

−α t −t v = α x1e1 , ρ = ρo e ( o ) For an arbitrary time t , consider the material contained in the cylindrical control volume of crosssectional area A , bounded by x1 = 0 and x1 = 3 . (a) Determine the linear momentum and rate of increase of linear momentum in this control volume. (b) Determine the outflux of linear momentum. (c) Determine the net resultant force that is acting on the material contained in the control volume. --------------------------------------------------------------------------------------Ans. (a) Linear momentum inside the control volume:

3

3

−α t −t −α t −t P = ∫ ρ vdV = ∫ ρo e ( o ) v1 Adx1e1 = ∫ ρo e ( o )α x1 Adx1e1

0

0 3

9 ⎡9⎤ −α t −t −α t −t −α t −t = ρoα e ( o ) A∫ x1dx1e1 = ρ oα e ( o ) A ⎢ ⎥ e1 = Aρoα e ( o )e1. 2 2 ⎣ ⎦ 0 Rate of increase of linear momentum inside the control volume dP 9 −α t −t = − α 2 Aρ o e ( o )e1 . dt 2 (b) Out flux of linear momentum from the control volume in the e1 direction: −α t −t ρ Av 2 − ρ Av 2 = ρ Aα 2 x 2 − ρ Aα 2 x 2 = 9 Aρ α 2 e ( o ) .

(

1

)

x1 =3

(

1

)

x1 = 0

(

1

)

x1 =3

(

1

)

x1 = 0

o

(c) The total force = rate of increase of linear momentum inside the control volume + net momentum outflux from the control volume. Thus, 9 9 −α t −t −α t −t −α t −t F = − α 2 Aρo e ( o )e1 + 9 Aρoα 2 e ( o )e1 = Aρoα 2 e ( o )e1 2 2 We see that the force exerted on the material within the control volume decreases exponentially with time. This is due to the given data of density field and velocity field, which states that within the fixed space defined by 0 ≤ x1 ≤ 3 , the density decreases exponentially with time while speed at 9 each spatial point is independent of time. We also note that at t = to , F = Aρoα 2e1 , the same 2 results was obtained in Problem 7.10. ________________________________________________________________________ 7.15

Do Problem 7.14 for the same velocity field, v = α x1e1 but with ρ = k

cylindrical control volume bounded by x1 = 1 and x1 = 3 . --------------------------------------------------------------------------------------Ans. (a) Linear momentum inside the control volume:


ρo x1

and the


P = ∫ ρ vdV = ∫ ( k ρ o / x1 ) v1 Adx1e1 = ∫ ( k ρo / x1 ) (α x1 ) Adx1e1 = ρo kα A∫ dx1e1 = 2 ρo kα Ae1 3

3

1

1

1

Rate of increase of linear momentum inside the control volume dP = 0. dt (b) Out flux of linear momentum from the control volume in the e1 direction:

( ρ Av ) 2 1

x1 =3

(

− ρ Av12

)

x1 =1

(

= k ( ρ o / x1 ) Aα 2 x12

)

x1 =3

(

− k ( ρo / x1 ) Aα 2 x12

)

x1 =1

= 2k ρo Aα 2 .

(c) The total force = rate of increase of linear momentum inside the control volume + net momentum outflux from the control volume F = 0 + 2k ρo Aα 2e1 = 2k ρo Aα 2e1 ________________________________________________________________________

7.16 Consider the flow field v = k ( xe1 − ye2 ) with ρ =constant. For a control volume defined by x = 0, x = 2, y = 0, y = 2, z = 0, z = 2 , determine the net resultant force and moment about the origin (note misprint in text) that are acting on the material contained in this volume. --------------------------------------------------------------------------------------Ans. Since the flow is steady, the resultant force = net linear momentum outflux through the three pairs of faces: (i) through x = 0 and x = 2 , 2 2 ∫ ρ v1vdA x=2 − ∫ ρ v1vdA x=0 = k ⎡⎣ ∫ ∫ ρ x ( xe1 − ye2 ) dydz ⎤⎦ x=2 − k ⎡⎣ ∫ ∫ ρ x ( xe1 − ye2 ) dydz ⎤⎦ x=0

(

= k2ρ

)

(

)

⎡ 2 ⎤ ⎡ 2 ⎤ 2 ⎢ ⎥ − = − e y e dz dy k ρ e y e dy 4 2 2 4 2 ⎢ ⎥ ( ) ( ) 2 2 ∫ ⎢∫ 1 ∫ 1 ⎢ ⎥ ⎥ y =0 ⎣ z =0 ⎦ ⎣ y =0 ⎦ 2

i

= 2k 2 ρ ( 8e1 − 4e 2 ) = k 2 ρ (16e1 − 8e2 ) . (ii) through y = 0 and y = 2 ,

( ∫ ρ v2 vdA) y=2 − ( ∫ ρ v2 vdA) y=0 = k 2 ⎡⎣ ∫ ∫ ρ ( − y )( xe1 − ye2 ) dxdz ⎤⎦ y=2 − k 2 ⎡⎣ ∫ ∫ ρ ( − y )( xe1 − ye2 ) dxdz ⎤⎦ y=0 = k2ρ ∫

2

⎡

2 ⎤ ∫ ( −2 xe1 + 4e2 ) dz ⎥⎦ dx = 2k ρ ∫x=0 ( −2 xe1 + 4e2 ) dx 2

2

x =0 ⎢ ⎣ z =0

= 2k 2 ρ ⎡ − x 2e1 + 4 xe 2 ⎤ = 2k 2 ρ ( −4e1 + 8e 2 ) = k 2 ρ ( −8e1 + 16e2 ) . ⎣ ⎦ x =2 (iii) through z = 0 and z = 2 ,

( ∫ ρ v3 vdA) z =2 − ( ∫ ρ v3 vdA)z =0 = 0, (v3 =0) .

Thus, the total net force F = k 2 ρ (16e1 − 8e 2 ) + k 2 ρ ( −8e1 + 16e2 ) = k 2 ρ ( 8e1 + 8e2 ) . The flow is steady, the resultant moment about a point = net moment of momentum outflux about the same point. Take the point to be the origin, then


Lai et al, Introduction to Continuum Mechanics (i) through x = 0 and x = 2 ,

( ∫ ρ v1 (r × v ) dA)x=2 − ( ∫ ρ v1 (r × v ) dA)x=0 = k 2 ⎡⎣ ∫ ∫ ρ x ( xe1 + ye2 + ze3 ) × ( xe1 − ye2 ) dydz ⎤⎦ x=2 2 2 = k 2 ρ ⎡ ∫ ∫ ( xyze1 + x 2 ze 2 - 2 x 2 ye3 ) dydz ⎤ = k 2 ρ ∫ ⎡ ∫ ( 2 yze1 + 4 ze 2 - 8 ye3 ) dz ⎤ dy (i) ⎢ z =0 y =0 ⎣ ⎣ ⎦ x=2 ⎦⎥ = k2ρ ∫

2 y =0

( 4 ye1 + 8e2 - 16e3 )dy = k 2 ρ (8e1 + 16e2 - 32e3 ) .

through y = 0 and y = 2 ,

( ∫ ρ v2 (r × v ) dA) y=2 − ( ∫ ρ v2 (r × v ) dA) y=0 = k 2 ⎡⎣ ∫ ∫ ρ ( − y ) ( xe1 + ye2 + ze3 ) × ( xe1 − ye2 ) dxdz ⎤⎦ y=2 2 2 = k 2 ρ ⎡ ∫ ∫ ( − y 2 ze1 − xyze 2 + 2 xy 2e3 ) dxdz ⎤ = k 2 ρ ∫ ⎡ ∫ ( −4 ze1 − 2 xze 2 + 8 xe3 ) dz ⎤dx ⎥⎦ x =0 ⎢ z ⎣ ⎦ y =2 ⎣ =0 = k2ρ ∫

2 y =0

( −8e1 − 4 xe2 + 16 xe3 ) dx = k 2 ρ ( −16e1 − 8e2 + 32e3 ) .

(iii) through z = 0 and z = 2 ,

( ∫ ρ v3 (r × v ) dA)z =2 − ( ∫ ρ v3 (r × v ) dA) z =0 = 0

Thus, M o = k 2 ρ ( 8e1 + 16e 2 − 32e3 ) + k 2 ρ ( −16e1 − 8e2 + 32e3 ) = k ρ ( −8e1 + 8e2 ) . ________________________________________________________________________ 7.17

(

)

For Hagen-Poiseuille flow in a pipe, v = C ro2 − r 2 e1 . Calculate the momentum flux

across a cross-section. For the same flow rate, if the velocity is assumed to be uniform, what is the momentum flux across a cross section? Compare the two results. --------------------------------------------------------------------------------------Ans. Momentum flux across a cross section = ∫ ρ v12 dAe1 = ρ C 2 ∫

ro

o

(r

2 o

− r2

)

2

( 2π r ) dre1 = ( ρ C 2π ro6 / 3) e1

Volume flow rate is Q = ∫ v1dA = C ∫

ro

o

(r

2 o

)

(

)

− r 2 ( 2π r )dr = C π ro4 / 2 .

The uniform flow which has the same flow rate Q is given by : v = (Q / π ro2 )e1 = (Cro2 / 2)e1 . The momentum flux across a cross section for this uniform flow is given by

ρ Qv1e1 = (πρ C 2 ro6 / 4)e1 . 4 that of the uniform flow. 3 ________________________________________________________________________

Thus, the momentum flux for the Hagen-Poiseuille flow is

7.18 Consider a steady flow of an incompressible viscous fluid of density ρ , flowing up a vertical pipe of radius R . At the lower section of the pipe, the flow is uniform with a speed vl and a pressure pl .After flowing upward through a distance l , the flow becomes fully developed with a parabolic velocity distribution at the upper section, where the pressure is pu . Obtain an expression for the fluid pressure drop pl − pu between the two sections in terms of ρ , R and the frictional force F f , exerted on the fluid column from the wall though viscosity. ---------------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. Let the control volume encloses the fluid between the two sections. The linear momentum theorem states that: For steady flow, Force on the fluid= Momentum outflux – momentum influx. The force on the fluid in the control volume is given by: ( pl − pu ) A − ρ g ( Al ) − F f .

(

)

The momentum influx through the lower section = ρ vl2 π R 2 . The momentum outflux through the upper section

∫ ρ vu ( 2π rdr ) , where vu = C ( R 2

constant C can be obtain as follows R

R

0

0

(

2

− r 2 ) . The

)

Q = ∫ vu 2π rdr = 2π C ∫ ( R 2 − r 2 )rdr = π CR 4 / 2 = vl π R 2 → C = 2vl / R 2 . Thus, vu =

2vl R

2

(R2 − r 2 ) .

Momentum outflux

(

= 8 ρπ vl2 / R 4

)∫

R

o

R

∫o

(

ρ vu2 ( 2π rdr ) = ρ 2π 2vl / R 2

(

( R 2 − r 2 ) 2 ( rdr ) = 8ρπ vl2 / R 4

)

2

R

∫o ( R

2

− r 2 )2 ( rdr )

)( R / 6) = 4ρπ v R 6

2 l

2

/ 3.

Thus, ( pl − pu ) A − ρ g ( Al ) − F f = 4 ρπ vl2 R 2 / 3 − ρ vl2 (π R 2 ) = ρ vl2 (π R 2 ) / 3 → ( pl − pu ) (π R 2 ) = ρ vl2 (π R 2 ) / 3 + F f + ρ g (π R 2 ) ( l ) . That is,

( pl − pu ) = ρ vl2 / 3 + F f

/ (π R 2 ) + ρ g l .

________________________________________________________________________ 7.19 A pile of chain on a table falls through a hole from the table under the action of gravity. Derive the differential equation governing the hanging length x . [Assume the pile is large compared with the hanging portion] --------------------------------------------------------------------------------------Ans. Using a control volume (Vc )2 [see Fig. 7.6-1 in Section7.6] enclosing the hanging down portion x of the chain, we can obtain the same equation as that given in Eq. (iv) of Section 7.6, i.e., with μ denoting m / l , mass per unit length:

μ gx − T = μ xd 2 x / dt 2

(1) where T is the tension on the chain at the hole. Next, using a control volume enclosing the pile above the table, then, since the particles of the chain pile stay essentially at rest at any given instant (except those near the hole), we can assume that the rate of change of momentum inside the control volume is zero (quasi-static approximation). Further, we assume that the net force acting at the pile is the tension T at the hole (the reaction of the supporting table exactly balances the weight of the pile). Then, the momentum principle gives:

T = μ ( dx / dt )

2

(2).

Equations (1) and (2) give d2x

2

⎛ dx ⎞ (3) +⎜ ⎟ 2 dt ⎝ dt ⎠ We note that this equation is a good approximation when the length of the pile is large compared with the hanging portion x . Eventually, when the pile reduces to essentially a flat straight segment on the table, Eq. (vi) of Section 7.6 becomes a better approximation. ________________________________________________________________________ gx = x



A water jet of 5 cm. diameter moves at 12 m / sec , impinges on a curved vane which

deflects it 60o from its original direction. Neglect the weight, obtain the force exerted by the liquid on the vane. (see Fig. 7.6-2 of Example 7.6.2). --------------------------------------------------------------------------------------Ans. Referring to Fig. 7.6-2, we have, vo = 12 m / s, θ =60o , volume flow rate

(

)

Q = vo (π d 2 / 4) = 12 π (5 × 10−2 ) 2 / 4 = 235.6 × 10−4 m3 , ρ Qvo = ( 998 ) (235.6 × 10−4 ) (12 ) = 282 N Thus, force on the jet =

(

)

− ρ Qvo 1 − cos 60o e1 + ρ Qvo sin 60o e 2 = −282 ( 0.5 ) e1 + 282 ( 0.866 ) e 2 = −141e1 + 244e 2 N . Force on the vane from the jet is 141e1 − 244e 2 N . ________________________________________________________________________ 7.21 A horizontal pipeline of 10 cm. diameter bends through 90o , and while bending, changes its diameter to 5 cm. The pressure in the 10 cm. pipe is 140 kPa. Estimate the resultant force on the bends when 0.005 m3 / sec .of water is flowing in the pipeline. --------------------------------------------------------------------------------------Ans. Let ( vu , pu , Au ) and ( vd , pd , Ad ) denote upstream and downstream (speed, pressure and cross-sectional area) respectively and Q the volume discharge. We have, Q = 0.005 m3 / s,

(

)

(

)

vu = 0.005 / π ( 0.1) / 4 = 0.6366 m / s, vd = 0.005 / π ( 0.05 ) / 4 = 2.546 m / s 2

2

Upstream pressure pu = 140,000 Pa . Down stream pressure can be obtained from Bernoulli vu2 pd vd2 . Thus, = + ρ 2 ρ 2 ρ 2 2 998 pd = pu + vu − vd = 140,000 + 0.63662 − 2.5462 = 137,000 . 2 2 Let e1 be the direction of the incoming flow and e 2 be the direction after the 90o bend, then, we have, Momentum outflux = ρ Qvd = ( 998 )(.005 )( 2.546 ) e 2 = 12.7e 2 . Equation:

pu

(

+

)

(

)

Momentum influx = ρ Qvu = ( 998 )(.005 )( 0.6366 ) = 3.18e1 .

Momentum principle gives: pu Au e1 − pd Ad e 2 + Fw = ρ Qvd e 2 − ρ Qvu e1 . Force on water Fw = − ( ρ Qvu + pu Au ) e1 + ( ρ Qvd + pd Ad ) e 2

(

) (

)

= − 3.177 + 140,000(π ( 0.1) / 4) e1 + 12.7 + 137000π ( 0.05 ) / 4 e 2 = −1100e1 + 282e 2 N . 2

2

Thus, the force from water to the bend is −Fw = 1100e1 − 282e 2 N . ________________________________________________________________________ 7.22 Figure P7.1 shows a steady water jet of area A impinging onto the flat wall. Find the force exerted on the wall. Neglect weight and viscosity of water.


Lai et al, Introduction to Continuum Mechanics vo

vo

vo

--------------------------------------------------------------------------------------Ans. Let the control volume be coincident with the outline of the flow shown in the figure. Force on the liquid FL = momentum outflux-momentum influx = 0 − ρ Qvo e1 Force on the wall = ρ Qvo e1 = ρ Avo2 e1 . ________________________________________________________________________ 7.23 Frequently in open channel flow, a high speed flow “jumps” to a low speed flow with an abrupt rise in the water surface. This is known as a hydraulic jump. Referring to Fig. p7.2, if the flow rate is Q per unit width, show that when the jump occurs, the relation between y1 and y2 .is given by y2 = −

y1 1 8v 2 + y1 1 + 1 gy1 2 2

Assume the flow before and after the jump is uniform and the pressure distribution is hydrostatic.

v1

y2

y1

--------------------------------------------------------------------------------------Ans. Use a control volume enclosing the water with an upstream section before the jump and a downstream section after the jump. According to the momentum principle, the force on the fluid per unit width is given by (neglect friction from the ground and air)

(

)

Fx = ρ gy12 / 2 − ρ gy22 / 2 = ρ Qv2 − ρ Qv1 , thus , ( g / 2) y12 − y22 = Q ( v2 − v1 ) = v1 y1 ( v2 − v1 ) . Conservation of mass gives: v1 y1 = v2 y2 → v2 − v1 = v1 y1 / y2 − v1 = v1 ( y1 − y2 ) / y2 . Therefore,

(

)

we have, ( gy2 / 2) y12 − y22 = v12 y1 ( y1 − y2 ) . The above equation shows that y1 = y2 is a root for the equation. This solution corresponds to a

flow without a jump. To look for the jump solution, we eliminate the factor ( y1 − y2 ) and obtain

gy2 ( y1 + y2 ) / 2 = v12 y1 → y22 + y1 y2 − (2v12 / g ) y1 = 0

(

)

1 y2 = (1 / 2 ) ⎡ − y1 + y12 + (8v12 y1 / g ) ⎤ = − y1 / 2 + y1 1 + 8v12 / gy1 ⎣⎢ ⎦⎥ 2 ________________________________________________________________________


Lai et al, Introduction to Continuum Mechanics 7.24 If the curved vane of Example 7.6.2 moves with a velocity v < vo in the same direction as the oncoming jet, find the resultant force exerted on the vane by the jet. --------------------------------------------------------------------------------------Ans. Fig. 7.6-2 of Example 7.6.2 is reproduced below. vo

e2

B

e1 vo

A

Let the control volume surrounding the jet moves with the vane, then the flow is steady with respect to the moving control volume. Momentum outflux relative to the control volume = ρ Q(vo − v) ( cosθ e1 + sin θ e 2 ) = ρ A(vo − v)2 ( cosθ e1 + sin θ e2 ) Momentum influx relative to the control volume is

ρ Q(vo − v)e1 = ρ A(vo − v) 2 e1 Thus, since the control volume moves with a constant speed, there is no extra term to be added to the momentum equation for the fixed control volume case. Thus, force acting on the jet is Fjet = ρ A(vo − v) 2 ( cos θ e1 + sin θ e 2 ) − ρ A(vo − v) 2 e1 = ρ A(vo − v) 2 ⎡⎣( cos θ − 1) e1 + sin θ e 2 ⎤⎦ and the force on the vane is Fvane = ρ A(vo − v) 2 ⎡⎣(1 − cos θ ) e1 − sin θ e 2 ⎤⎦ . ________________________________________________________________________ 7.25

For the half-arm sprinkler shown in Fig. P7.3, find the angular speed if

Q = 0.566m3 / sec. Neglect friction.

d=2.54 cm 1.83 m --------------------------------------------------------------------------------------Ans. Let the control volume Vc rotate with the arm. Then, relative to the control volume, the outflux of moment of momentum about an axis passing through O and perpendicular to the plane of the paper is ρ Q ( Q / A ) ro e3 , where ro is the length of the arm. There is no influx of moment of momentum about the same axis since the inflow is parallel to it. Since the control volume is rotating with an angular velocity ω about the same axis, we need to add terms to the left hand side of Eq. (7.9.8) , the moment of momentum principle. The terms that need to be added are given in Eq. (7.9.9). With ω = ωe3 and x = xe1 , we have, ω × x = xωe2 , ω × (ω × x ) = − xω 2e1 so that x × ω × (ω × x ) = 0 . We also have, ao = 0 and ω& = 0 , therefore, the only non-zero term is −2∫ x × (ω × v ) dm = −2 ∫ xe1 × (ωe3 × ( Q / A ) e1 ) ρ Adx = −2 ρωQ ∫ xdxe3 = − ρωQro2e3 . ro

0

Adding this term to Eq. (7.9.8), whose left hand side is zero (because frictional torque is neglected) and whose right hand side is the net moment of momentum outflux, we have,


Lai et al, Introduction to Continuum Mechanics − ρωQro2e3 = ρ Q (Q / A)ro e3 → ω = −Q / ( Aro ) .

Now, A = π (2.54 × 10−2 )2 / 4 = 5.067 × 10−4 m 2 ,

therefore, ω = −Q / ( Aro ) = −0.566 / [(5.067 × 10−4 )(1.83)] = −610.4 rad / s . The minus sign means the rotation is clockwise looking from the top. ________________________________________________________________________ 7.26

The tank car shown in fig. P7.4 contains water and compressed air which is regulated to

force a water jet out of the nozzle at a constant rate of Q m3 / sec. The diameter of the jet is d cm. , the initial total mass of the tank car is M o . Neglecting frictional forces, find the velocity of the car as a function of time.

d

--------------------------------------------------------------------------------------Ans. Let the control volume Vc encloses the whole tank car and moves with the car. Then relative to the control volume, the momentum outflux is

(

)

− ρ Q 4Q / π d 2 e1 = −4 ρ Q 2 / (π d 2 )e1 . There is no momentum influx. Since the control volume moves with the car which has an acceleration ao e1 , therefore, the momentum principle in the e1 direction takes the form [see Eq.(7.8.20}: (with al frictional/resistance force neglected): − ( M o − ρ Qt ) (dv / dt ) = −4 ρ Q 2 / (π d 2 ) . → dv / dt = [4 ρ Q 2 / (π d 2 )] / ( M o − ρ Qt ) . Integrating, we have, v = −[4Q / (π d 2 )]ln ( M o − ρ Qt ) + C . If the initial velocity is zero then we have v = [4Q / (π d 2 )] ⎡⎣ − ln ( M o − ρ Qt ) + ln M o ⎤⎦ . ________________________________________________________________________ 7.27

For the one dimensional problem discussed in Section 7.10,

(a) from the continuity equation ρ1v1 = ρ 2 v2 and the momentum equation p1 − p2 = ρ 2 v22 − ρ1v12 , obtain ⎞ v2 1 ⎛ p2 =1− − 1⎟ ⎜ 2 v1 γ M1 ⎝ p1 ⎠ (b) From the energy equation

γ

1 γ p2 1 2 + v12 = + v2 , obtain γ − 1 ρ1 2 γ − 1 ρ2 2

p1

⎛ v2 ⎞ γ − 1 v12 ⎛ v22 ⎞ ⎜ ⎟ ⎜ ⎟+ 2 a12 2 a12 ⎜⎝ v12 ⎟⎠ p1 ⎝ v1 ⎠ (c) From the results of (a) and (b), obtain 2

1+

γ − 1 v1

2

=

p2

⎛ p2 ⎞ ⎛p ⎞ 2 2 ⎛ γ −1 ⎞ 1 + γ M12 ⎜ 2 ⎟ − − γ M12 ⎟ = 0 . ⎜ ⎟ − ⎜ ⎠ ⎝ p1 ⎠ γ + 1 ⎝ p1 ⎠ γ + 1 ⎝ 2 ---------------------------------------------------------------------------------------

(

)


Lai et al, Introduction to Continuum Mechanics Ans. (a) Using ρ1v1 = ρ 2 v2 , p1 − p2 = ρ 2 v22 − ρ1v12 → p1 − p2 = ρ1v1 ( v2 − v1 ) → 1−

→

⎞ 1 − p2 / p1 p2 ρ1v1 ( v2 − v1 ) ⎛ ρ1v12 ⎞ ( v2 − v1 ) ⎛ ρ1v12 ⎞ ⎛ v2 v = =⎜ =⎜ +1= 2 ⎟⎟ ⎟⎟ ⎜ − 1⎟ → 2 ⎜ ⎜ p1 p1 v1 ρ1v1 / p1 ⎠ ⎝ p1 ⎠ v1 ⎝ p1 ⎠ ⎝ v1 1 − p2 / p1

γ v12 ( ρ1 / γ p1 )

+1=

⎞ v2 1 − p2 / p1 v2 v2 1 ⎛ p2 → + = − 1 . That is, 1 . = 1 − ⎜ ⎟ v1 v1 v1 γ v12 / a12 γ M12 ⎝ p1 ⎠

(b) γ p1 1 2 γ p2 1 2 + v1 = + v2 γ − 1 ρ1 2 γ − 1 ρ2 2

→1+

2

γ − 1 v1 2

a12

=

⎛ ρ1 ⎜ p1 ⎝ ρ 2

p2

→1+ γ − 1 2γ

ρ1

p1

v12 =

p2 ρ1

ρ 2 p1

+

γ − 1 ρ1 2 v2 2γ p1

2 ⎞ γ − 1 v22 p ⎛v γ − 1 v1 , 1 + + = 2⎜ 2 →→ ⎟ 2 2 2 a1 2 a1 p1 ⎝ v1 ⎠

⎞ γ − 1 v12 ⎛ v22 ⎜ ⎟+ 2 a12 ⎜⎝ v12 ⎠

⎡ note γ p1 / ρ1 = a12 and ρ1v1 =ρ 2 v2 ⎤ . ⎣ ⎦ (c) v 1 ⎛ p2 ⎞ γ − 1 v12 p2 ⎛ v2 1 1 + = Using 2 = 1 − − , the equation ⎜ ⎟ ⎜ p1 ⎝ v1 v1 2 a12 γ M12 ⎜⎝ p1 ⎟⎠ 1+

γ −1 2

M12

p2 ⎡

⎞ γ − 1 v12 ⎛ v22 ⎜ ⎟+ 2 a12 ⎜⎝ v12 ⎠

1 ⎛ p2 ⎞ ⎤ γ − 1 2 ⎡ 1 ⎛ p2 ⎞ ⎤ 1 1 = ⎢1 − − − − 1⎟⎟ ⎥ + M ⎥ ⎢ ⎜ ⎟ 1 2⎜ 2 ⎜⎜ ⎟ p1 ⎢ 2 ⎠ ⎥⎦ ⎠ ⎥⎦ ⎣ γ M1 ⎝ p1 ⎣⎢ γ M1 ⎝ p1

⎡ ⎛ 1 ⎜ ⎛ =⎢ − ⎢ γ M 2 ⎜ ⎜⎜ 1 ⎢⎣ ⎝ ⎝

⎞ ⎟⎟ ⎠.

⎞ ⎟⎟ → ⎠

2

⎤

⎤ p2 ⎞ p2 ⎞⎟ p2 ⎥ γ − 1 2 ⎡⎢ 2 ⎛ p2 ⎞ 1 ⎛ p2 ⎞ ⎥ + + M1 1 − − + − 1 1 ⎟⎟ + ⎜ ⎟ ⎜ ⎟ ⎟ γ 2M 4 ⎜ p ⎟ ⎥ ⎢ γ M12 ⎜⎝ p1 p1 ⎠ p1 ⎟ p1 ⎥ 2 ⎠ ⎠ 1 ⎝ 1 2

2

⎥⎦

⎠

⎣

⎦

2

γ − 1 ⎤ p2 1 ⎡ γ M12 + 1 ⎤ γ − 1 1 ⎡ 2 4 1 ⎛ p2 ⎞ ⎡ = − + γ M1 + 2γ M12 + 1⎤⎦ . 1 ⎢ ⎥+ ⎜ ⎟ ⎢ ⎥+ 2 2⎜ p ⎟ 2 ⎣ 2γ ⎦ p1 γ M1 ⎣⎢ γ γ M1 ⎝ 1 ⎠ ⎣ ⎦⎥ 2γ γ M1 Thus, 1+

γ −1 2

1 ⎛ p2 ⎞ ⎡ − ( γ + 1) ⎤ = ⎥+ ⎜ ⎟ ⎢ γ M12 ⎜⎝ p1 ⎟⎠ ⎢⎣ 2γ ⎥⎦ 2

M12

p2 1 ⎡ γ M12 + 1 ⎤ γ − 1 1 ⎡ 2 4 γ M1 + 2γ M12 + 1⎤⎦ . ⎢ ⎥+ 2 ⎣ 2 γ p1 γ M12 ⎢⎣ γ γ M ⎥⎦ 1

Rearranging, ⎡ − ( γ + 1) ⎤ p 1 ⎡ γ M12 + 1 ⎤ ⎢ ⎥+ 2 ⎥ 2 ⎢ 2 ⎢⎣ 2γ ⎥⎦ p1 γ M1 ⎣⎢ γ ⎦⎥ 2γ 2 M12 ⎤ γ −1 2 γ −1 1 ⎡ 2 2 γ 1 + + − M1 + M ⎢ ⎥ 1 2 2γ γ M12 ⎣⎢ (γ − 1) ⎦⎥

γ −1

M12

1 ⎛ p2 ⎞ = ⎜ ⎟ γ M12 ⎜⎝ p1 ⎟⎠

2

1 ⎛ p2 ⎞ ⎡ − ( γ + 1) ⎤ →0= ⎥+ ⎜ ⎟ ⎢ γ M12 ⎝⎜ p1 ⎠⎟ ⎢⎣ 2γ ⎦⎥ 2

(

)

p2 1 ⎡ γ M12 + 1 ⎤ γ − 1 1 ⎡ 2γ M12 ⎤ 1− ⎢ ⎥+ ⎥ 2 ⎢ p1 γ M12 ⎣⎢ γ ⎦⎥ 2γ γ M1 ⎣⎢ ( γ − 1) ⎦⎥

2 2 γ M12 + 1 p ⎛ p2 ⎞ ⎞ 2 ⎛ ( γ − 1) 2 − That is, ⎜⎜ − γ M12 ⎟⎟ = 0 . ⎟⎟ − ⎜⎜ (γ + 1) p1 (γ + 1) ⎝ 2 ⎠ ⎝ p1 ⎠ The above equation has two solutions:


Lai et al, Introduction to Continuum Mechanics (1) p2 = p1 (2) p2 =

1 ⎡ 1 ⎡ 2 ρ1v12 − ( γ − 1) p1 ⎤ , or p2 = 2γ M12 − ( γ − 1) ⎤ p1 ⎦ ⎦ γ +1 ⎣ γ +1 ⎣



CHAPTER 8 8.1 Show that for an incompressible Newtonian fluid in Couette flow, the pressure at the outer cylinder ( r = Ro ) is always larger than that at the inner cylinder. That is, obtain

⎡⎣ −Trr ( Ro ) ⎤⎦ − ⎡⎣ −Trr ( Ri ) ⎤⎦ = ρ ∫ o rω 2 ( r ) dr Ri R

------------------------------------------------------------------------------B Ans. In Couette flow, vr = vz = 0 and vθ = Ar + , [see Eq. (6.15.4) and (6.15.7)]. Thus, r v ⎞ ⎛ dv Trr = Tθθ = Tzz = − p , Trz = Tθ z = 0 , and Trθ = μ ⎜ θ − θ ⎟ = function of r only . r ⎠ ⎝ dr ∂T 1 ∂Trθ Trr − Tθθ ∂Trz + + = − ρ rω 2 becomes: Thus, the r-equation of motion rr + ∂r r ∂θ r ∂z Ro ∂Trr Ro ∂Trr = − ρ rω 2 . Now, ∫ dr = − ρ ∫ rω 2 ( r ) dr . Thus,. Ri ∂r Ri ∂r

⎡⎣ −Trr ( Ro ) ⎤⎦ − ⎡⎣ −Trr ( Ri ) ⎤⎦ = ρ

Ro

2 ∫ rω ( r ) dr . The right hand side of this last equation is always

Ri

positive. ____________________________________________________________ 8.2

Show that the constitutive equation

τ = τ 1 + τ 2 + τ 3 , with τ n + λn ∂τ n / ∂t = 2μn D, n = 1, 2,3 is equivalent to τ + a1∂τ / ∂t + a2 ∂ 2τ / ∂t 2 + a3∂ 3τ / ∂t 3 = bo D + b1∂D / ∂t + b2∂ 2 D / ∂t 2 where a1 = ( λ1 + λ2 + λ3 ) , a2 = ( λ1λ2 + λ2λ3 + λ3λ1 ) , a3 = λ1λ2λ3 bo = 2 ( μ1 + μ2 + μ3 ) , b1 = 2 ⎡⎣ μ1 ( λ2 + λ3 ) + μ2 ( λ1 + λ3 ) + μ3 ( λ2 + λ1 ) ⎤⎦

b2 = 2 ( μ1λ2 λ3 + μ2 λ1λ3 + μ3λ1λ2 ) ------------------------------------------------------------------------------Ans. ⎞ ⎛ 3 ⎞ ⎛ 3 ∂τ j ⎞ 3 3 ∂τ j ⎛ 3 ⎞ ∂τ ⎛ 3 ⎞ ∂ ⎛ 3 ⎟ = ∑∑ λi ⎜ ∑ λi ⎟ = ⎜ ∑ λi ⎟ ⎜⎜ ∑ τ j ⎟⎟ = ⎜ ∑ λi ⎟ ⎜⎜ ∑ ⎝ i =1 ⎠ ∂t ⎝ i =1 ⎠ ∂t ⎝ j =1 ⎠ ⎝ i =1 ⎠ ⎝ j =1 ∂t ⎟⎠ i =1 j =1 ∂t 3 3 3 3 3 3 ∂τ j ∂τ j ∂τ = ∑ λi i + ∑∑ λi = ∑ ( 2μi D − τ i ) + ∑∑ λi i =1

∂t

i =1 j =1 j ≠i

∂t

i =1

i =1 j =1 j ≠i

∂t

3 3 3 3 3 ∂τ j ∂τ j ⎛ 3 ⎞ 3 = 2D ⎜⎜ ∑ μi ⎟⎟ − ∑τ i + ∑∑ λi = 2D∑ μi − τ + ∑∑ λi ∂t ∂t i =1 j =1 i =1 i =1 j =1 ⎝ i =1 ⎠ i =1 j ≠i

j ≠i

That is,


Lai et al, Introduction to Continuum Mechanics ⎛ 3 ⎞ ⎛ 3 ⎞ ∂τ ∂τ ⎞ ⎛ ∂τ ∂τ ⎞ ⎛ ∂τ ∂τ ⎞ ⎛ ∂τ 2 + λ1 3 ⎟ + ⎜ λ2 1 + λ2 3 ⎟ + ⎜ λ3 1 + λ3 2 ⎟ ⎜ ∑ λi ⎟ = 2D ⎜⎜ ∑ μi ⎟⎟ − τ + ⎜ λ1 ∂t ⎠ ⎝ ∂t ∂t ⎠ ⎝ ∂t ∂t ⎠ ⎝ ∂t ⎝ i =1 ⎠ ∂t ⎝ i =1 ⎠

(i)

Next, we have, ( λ1λ2 + λ2λ3 + λ3λ1 ) ∂ 2τ / ∂t 2 = ( λ1λ2 + λ2λ3 + λ3λ1 ) ∂ 2τ 1 / ∂t 2 + ( λ1λ2 + λ2λ3 + λ3λ1 ) ∂ 2τ 2 / ∂t 2 + ( λ1λ2 + λ2 λ3 + λ3λ1 ) ∂ 2τ 3 / ∂t 2 = ( λ2 λ1 + λ3λ1 ) ∂ 2τ 1 / ∂t 2 + λ2 λ3∂ 2τ 1 / ∂t 2 + ( λ1λ2 + λ2λ3 ) ∂ 2τ 2 / ∂t 2 + λ3λ1∂ 2τ 2 / ∂t 2 + ( λ2 λ3 + λ3λ1 ) ∂ 2τ 3 / ∂t 2 + λ1λ2 ∂ 2τ 3 / ∂t 2 . i.e.,

( λ1λ2 + λ2λ3 + λ3λ1 ) ∂ 2τ / ∂t 2 = ⎡⎣ 2μ1 ( λ2 + λ3 ) + 2μ2 ( λ1 + λ3 ) + 2μ3 ( λ2 + λ1 )⎤⎦ ∂D / ∂t − ( λ2 + λ3 ) ∂τ 1 / ∂t − ( λ1 + λ3 ) ∂τ 2 / ∂t − ( λ2 + λ1 ) ∂τ 3 / ∂t + λ2 λ3∂ 2τ 1 / ∂t 2 + λ3λ1∂ 2τ 2 / ∂t 2

(ii)

+ λ1λ2 ∂ 2τ 3 / ∂t 2 . Finally, we have, ∂3τ ∂ 3τ ∂ 3τ ∂ 3τ λ1λ2 λ3 3 = λ1λ2λ3 31 + λ1λ2 λ3 32 + λ1λ2λ3 33 ∂t ∂t ∂t ∂t 2 2 2 2 ⎛ ⎛ ⎛ ∂ D ∂τ ⎞ ∂ D ∂τ ⎞ ∂ 2 D ∂ 2τ ⎞ = λ2 λ3 ⎜ 2μ1 2 − 21 ⎟ + λ1λ3 ⎜ 2μ 2 2 − 22 ⎟ + λ1λ2 ⎜ 2μ3 2 − 23 ⎟ ⎜ ⎜ ⎜ ∂t ∂t ⎠⎟ ∂t ∂t ⎠⎟ ∂t ∂t ⎠⎟ ⎝ ⎝ ⎝ = 2λ2 λ3 μ1

∂2D ∂t 2

that is, λ1λ2 λ3

+ 2λ1λ3 μ2

∂ 3τ ∂t 3

∂2D ∂t 2

= 2λ2 λ3 μ1

+ 2λ1λ2 μ3

∂2D ∂t 2

+ 2λ1λ3 μ 2

∂2D ∂t 2 ∂2D ∂t 2

− λ2λ3

∂ 2τ 1 ∂t 2

+ 2λ1λ2 μ3

− λ1λ3

∂2D ∂t 2

∂ 2τ 2 ∂t 2

− λ2 λ3

− λ1λ2

∂ 2τ 1 ∂t 2

∂ 2τ 3

− λ1λ3

∂t 2 ∂ 2τ 2 ∂t 2

− λ1λ2

∂ 2τ 3 ∂t 2

(iii)

Thus, (i) + (ii)+ (iii) gives τ + ( λ1 + λ2 + λ3 ) ∂τ / ∂t + ( λ1λ2 + λ2 λ3 + λ3λ1 ) ∂ 2τ / ∂t 2 + λ1λ2λ3∂ 3τ / ∂t 3 = 2D ( μ1 + μ2 + μ3 ) + 2 ⎡⎣ μ1 ( λ2 + λ3 ) + μ 2 ( λ1 + λ3 ) + μ3 ( λ2 + λ1 ) ⎤⎦ ∂D / ∂t +2 ( μ1λ2 λ3 + μ2 λ1λ3 + μ3λ1λ2 ) ∂ 2 D / ∂t 2 . That is, ∂τ ∂ 2τ ∂ 3τ ∂D ∂2D τ + a1 + a2 2 + a3 3 = bo D + b1 + b2 2 . ∂t ∂t ∂t ∂t ∂t where a1 = ( λ1 + λ2 + λ3 ) , a2 = ( λ1λ2 + λ2λ3 + λ3λ1 ) , a3 = λ1λ2λ3

bo = 2 ( μ1 + μ2 + μ3 ) , b1 = 2 ⎡⎣ μ1 ( λ2 + λ3 ) + μ2 ( λ1 + λ3 ) + μ3 ( λ2 + λ1 ) ⎤⎦

b2 = 2 ( μ1λ2 λ3 + μ2 λ1λ3 + μ3λ1λ2 ) ____________________________________________________________ 8.3 Obtain the force-displacement relationship for the Kelvin-Voigt solid, which consists of a dashpot (with damping coefficient η ) and a spring (with spring constant G ) connected in parallel. Also, obtain its relaxation function. ------------------------------------------------------------------------------Ans. Since the spring and the dashpot are connected in parallel, therefore, the total force is given by: S = S sp + Sdash and the total displacement ε is given by ε = ε sp = ε dash . Now,


Lai et al, Introduction to Continuum Mechanics dε dε , therefore, S = Gε + η . To find the relaxation function, we let dt dt ε = ε o H (t ) , where H (t ) is the Heaviside function. Then S = Gε o H (t ) + ηε oδ (t ) . Thus, the relaxation function is S / ε o = GH (t ) + ηδ (t ) . ____________________________________________________________ S sp = Gε and Sdash =η

(a) Obtain the force-displacement relationship for a dashpot (damping coefficient ηo ) 8.4 and a Kelvin-Voigt solid (damping coefficient η and spring constant G, see the previous problem) connected in series. (b) Obtain its relaxation function. ------------------------------------------------------------------------------Ans. (a) Let Skv and Sd be the force transmitted by the Kelvin-Voight element and the dashpot respectively and let ε kv and ε d be the elongation of the Kelvin-Voight element and the dashpot respectively. Then we have, the total force is given by S = Sd = Skv (i) and the total displacement dε dε is given by ε = ε d + ε kv (ii), where Sd = ηo d = S = Skv = Gε kv + η kv (iii). From (ii) and dt dt d ε d ε d d ε kv S 1 S S G = + = + ( S − Gε kv ) = + − ( ε − ε d ) (iv). Thus, (iii) we have, ηo η ηo η η dt dt dt

⎛ η + ηo ⎞ dS G d ε G d ε d ηηo d 2ε (η + ηo ) dS dε , or, = − + = − ηo +S. ⎜ ⎟ 2 2 dt dt dt G G dt dt ηη η η dt dt o ⎠ ⎝ Thus, the force-displacement relationship is given by: (η + ηo ) dS = η d ε + ηηo d 2ε . S+ o G dt dt G dt 2 (b) Let ε = ε o H (t ) , where H (t ) is Heaviside function. Then Eq. (v) gives εηG ε ηη d δ dS G + , S = o o δ (t ) + o o dt (η + ηo ) (η + ηo ) (η + ηo ) dt

d 2ε

where δ ( t ) is the Dirac function. The integration factor for this ODE is exp ⎡⎣Gt / (η + ηo ) ⎤⎦ . Gt Gt ⎡ Gt ⎤ ε oηηo η +ηo d δ d ⎢ η +ηo ⎥ ε oηo G η +ηo = and Se e e δ (t ) + Thus, ⎥ (η + ηo ) dt ⎢ dt (η + ηo ) ⎣ ⎦ Gt

Se =

η +ηo

Gt

Gt

t εηG t ε ηη dδ = o o ∫ eη +ηo δ (t )dt + o o ∫ eη +ηo dt dt (η + ηo ) t =−∞ (η + ηo ) −∞

t ε oηo G ε ηη ⎧ G + o o ⎨ ⎡eG t /(η +ηo )δ (t ) ⎤ − ⎦ −∞ η + η (η + ηo ) (η + ηo ) ⎩ ⎣ o Gt

Se

η +ηo

⎫ dt ⎬ . That is, ⎭ G t Gt ⎧ ⎫ ⎧ ⎫ ε oηo G ε oηηo ⎪ η +ηo ε oηo2G G ⎪ ⎪ ε oηηo η +ηo ⎪ e δ t e δ t ( ) ( ) = + − = + ⎨ ⎬ ⎨ ⎬. 2 η + ηo ⎪ (η + ηo ) ⎪ (η + ηo ) (η + ηo ) (η + ηo ) ⎪ ⎪⎭ ⎩ ⎭ ⎩

Thus, the relaxation function is

S

εo

=

−G t

ηo2G

(η + ηo )

t

∫−∞ δ e

2

e

η +ηo

+

G t /(η +ηo )

ηηo δ (t ) . (η + ηo )

____________________________________________________________


(v)

(vi)

Lai et al, Introduction to Continuum Mechanics 8.5 A linear Maxwell fluid, defined by Eq. (8.1.2), is between two parallel plates which are one unit apart. Staring from rest, at time t = 0 , the top plate is given a displacement u = vo t while the bottom plate remains fixed. Neglect inertia effects, obtain the shear stress history. -----------------------------------------------------------------------Ans. The velocity field for the fluid in this motion is given by (inertia neglected) v1 = vo H (t ) x2 , v2 = v3 = 0 , where H ( t ) is the Heaviside Function. The only non-zero rate of

deformation component is D12 = vo H ( t ) / 2 . Thus, from the constitutive equation for the linear

μv dS12 d ⎡ S12 et / λ ⎤ = o et / λ H ( t ) . That is, = μ vo H ( t ) . Thus, ⎣ ⎦ dt dt λ μ vo t t / λ μ vo t t / λ μ vo ⎡ t / λ ⎤ t t /λ = e H ( t ) dt = e dt = λ e ⎦ = μ vo e − 1 . Thus, the shear 0 λ ∫−∞ λ ∫0 λ ⎣

Maxwell fluid, we obtain, S12 + λ S12 et / λ

(

(

)

)

stress history is: S12 = μ vo 1 − e −t / λ . ____________________________________________________________ Obtain Eq. (8.3.1) i.e., S = 2 ∫ φ (t − t ' )D(t ' )dt ' , where φ (t ) = ( μ / λ ) e −t / λ , by solving t

8.6

−∞

the linear non-homogeneous ordinary differential equation S + λ

dS = 2μ D . dt

------------------------------------------------------------------------Ans. The integration factor for this ODE is exp [t / λ ] . That is the equation can be written as; t d 2μ t / λ e D . Thus, Set / λ = ( 2 μ / λ ) ∫ et / λ D ( t ) dt → Se t / λ = −∞ dt λ

(

)

→ S = e −t / λ ( 2 μ / λ ) ∫

t

−∞

et ′/ λ D ( t ′ ) dt ′ . That is,

t μ −( t −t ' ) / λ e D t ' dt ' ≡ 2∫ φ t − t ' D t ' dt ' . −∞ λ −∞

S = 2∫

t

()

(

) ()

____________________________________________________________ 8.7

Show that for the linear Maxwell fluid, defined by Eq. (8.1.2),

∫−∞ φ (t − t ' ) J ( t ' ) dt ' = t , t

where φ ( t ) is the relaxation function and J ( t ) is the creep compliance function.

------------------------------------------------------------------------------Ans. Let S12 = So H ( t ) be applied to the top plate of a channel of unit depth in which is the linear

Maxwell fluid. [ H ( t ) is the unit step function, i.e., Heaviside function]. Neglecting inertia, the velocity field is v ( x2 ) = vo x2 , where vo is the velocity of the top plate. Then from the

constitutive equation S + λ dS / dt = 2μ D , we obtain So + λ Soδ ( t ) = 2μ D12 = 2μ ( vo / 2 ) = μ duo / dt ,

duo So λ = + S δ ( t ) , we obtain μ μ o dt t duo t So So So λ t λ ∫o dt dt = ∫o μ dt + μ ∫o Soδ ( t ) → uo = μ t + μ So = μ (t + λ ) . Thus, the creep compliance function is: J (t ) = uo / So = (t + λ ) / μ . where uo ( t ) is the displacement of the top plate. From

Since the relaxation function is φ (t ) = ( μ / λ )e−t / λ , therefore,



∫−∞ φ ( t − t ′) J ( t ′) dt ′ = ∫−∞ (μ / λ )e t

t

=∫

t

−∞

(t ′ + λ ) / μ dt ′ = (1 / λ ) ∫

t

−∞

− t −t ′ / λ e ( ) (t ′ + λ )dt ′

t − t −t ′ / λ − t −t ′ / λ e ( ) dt ′ + (1 / λ ) ∫ e ( ) (t ′)dt ′. −∞

t

−( t −t ′ ) / λ

Now

∫−∞ e

t

−( t −t ′ ) / λ

∫−∞ e

( )

− t −t ' / λ

dt ′ = [λ e

( )

− t −t ' / λ t ]t ′=−∞

= λ and

( t ′) dt ' = ∫−∞ (λ de−( t −t′)/ λ )(t ′)dt ′ = ⎡⎣⎢λ e−(t −t′)/ λ t ′⎤⎦⎥ t

∫−∞ φ ( t − t ' )J ( t ' ) dt ' = λ + λ ( λt − λ 1

t

Thus,

2

t t ′=−∞

− λ∫

t

−∞

− t −t ′ / λ e ( ) dt ′ = λ t − λ 2 .

) = λ + (t − λ ) = t .

____________________________________________________________ Obtain the storage modulus and loss modulus for the linear Maxwell fluid with a ∞ H ( λ ) −t / λ continuous relaxation spectrum defined by Eq. (8.4.1), i.e., φ ( t ) = ∫ e dλ .

8.8

o

λ

-------------------------------------------------------------------------------

Ans. Let the shear strain be: γ 12 = γ o eiω t . For this strain history, the rate of deformation history is dγ given by 2 D12 = 12 = iωγ o eiω t . Thus, from the constitutive equation, dt S = 2 ∫ φ ( t − t ′ )D ( t ′ ) dt ′ , we have S12 = 2 ∫ φ ( t − t ′ )D12 ( t ′ ) dt ′ = iωγ o ∫ φ ( t − t ′ ) eiω t ′ dt ′ . t

−∞

∞

With φ ( t ) = ∫ ⎡⎣ H ( λ ) / λ ⎤⎦ e o

S12 = iωγ o ∫

t

∫

∞

−∞ o

t

∫t′=−∞ e

Now, S12

γo

= iω eiω t

e

∞

H (λ )

λ =o

1+ iλω )t ′ / λ

(1 + λ 2ω 2 )

∞

λ =o

λ

t =−∞

et ′/ λ ei ω t ′ dt ′d λ

λ 1+ iλω )t / λ e( . Thus, (1 + iλω ) ∞

H (λ )

∫ (1 + iλω )d λ λ =o

is the complex modulus.

(1 − iλω ) H ( λ ) dλ (1 + iλω )(1 − iλω )

∞ λω 2 H (λ ) ω H (λ ) d + i dλ λ ∫ λ =o (1 + λ 2ω 2 ) λ =o (1 + λ 2ω 2 )

dλ = ∫

∞

λ 2ω 2 H (λ ) dλ, λ =o λ (1 + λ 2ω 2 )

Thus, G ′ = ∫

dt ′ =

d λ ≡ G*eiω t , where G* = iω

d λ = iω ∫

(iω + λω 2 ) H (λ )

∞

−∞

d λ , we have,

e(

t

t ′=−∞

(1 + iλω )

−∞

λ =o

dt ′ = ∫

H (λ )

t

−t / λ ∞ H (λ ) e t − t −t ′ / λ e ( ) eiω t ′ d λdt ′ = iωγ o ∫ ∫'

∫ 1 + iλω ) λ =o (

∞

λ =o

=∫

λ

t ′ / λ iωt ′

Now, G* = iω ∫

H (λ )

−t / λ

t

∞

λω H (λ ) dλ λ =o λ (1 + λ 2ω 2 )

G ′′ = ∫

∞

____________________________________________________________ Show that the viscosity μ of a linear Maxwell fluid, define by S = 2 ∫ φ ( t − t ′ )D ( t ′ ) dt ′ , t

8.9

−∞

is related to the relaxation function φ ( t ) and the memory function f ( s ) by the relation ∞

∞

o

o

μ = ∫ φ ( s ) ds = − ∫ sf ( s ) ds .


Lai et al, Introduction to Continuum Mechanics -----------------------------------------------------------------------Ans. S12 = 2∫

t

t ′ = −∞

φ ( t − t ′ ) D12 ( t ′ ) dt ′ = −2∫

0

s =∞

φ ( s )D12 ( t − s ) ds = 2 ∫

∞

s =0

φ ( s )D12 ( t − s ) ds .

For simple shearing flow, v1 = kx2 , v2 = v3 = 0 , 2D12 = k , so that S12 = k ∫

∞

s =0

φ ( s )ds → μ = S12 / k = ∫

∞

s =0

φ ( s )ds . Now, the memory function f ( s ) is related to the

relaxation function φ ( s ) by the relation dφ ( s ) / ds = f ( s ) . Thus, ∞ dφ ( s ) ds = − ∫ sf ( s ) ds . 0 0 0 ds ____________________________________________________________ ∞

∞

∞

μ = ∫ φ ( s ) ds = ⎡⎣ sφ ( s ) ⎤⎦ 0 − ∫ s

8.10 Show that the relaxation function for the Jeffrey model [Eq. (8.2.7)] with a2 = 0 is given by [note: Reference to Eq.(8.2.7) is missing in the problem statement in the text]

φ (t ) =

S12

γo

=

⎤ bo ⎡⎛ b1 ⎞ −t / a1 b1 + δ (t ) ⎥ , δ (t ) = Dirac Function . ⎢⎜ 1 − ⎟e bo 2a1 ⎢⎣⎝ bo a1 ⎠ ⎥⎦

-----------------------------------------------------------------------Ans. Let the shear strain γ 12 be given by γ 12 = γ o H (t ) . Then 2 D12 = d γ 12 / dt = γ oδ (t ) , where δ (t ) is Dirac function. From the constitutive equation, we have,

γ γ ∂δ γ ⎛b ∂S12 ∂S b ∂δ ⎞ 1 = bo o δ + b1 o → 12 + S12 = o ⎜ o δ + 1 ⎟ ∂t ∂t a1 a1 ∂t ⎠ 2 2 ∂t 2 ⎝ a1

S12 + a1

⎛b b ∂δ ⎞ ∂ S12 et / a1 = et / a1 γ o ⎜ o δ + 1 ⎟ ∂t a1 ∂t ⎠ ⎝ a1 b t S b t d δ (t ) dt → 2 12 et / a1 = o ∫ et / a1δ (t )dt + 1 ∫ et / a1 −∞ −∞ a1 a1 dt γo

→2

(

)

t bo b1 ⎡ t / a1 b b t 1 b b e δ (t ) ⎤ − 1 ∫ δ (t )et / a1 dt = o + 1 et / a1δ (t ) − 12 + ⎣ ⎦ −∞ a1 a1 a1 −∞ a1 a1 a1 a1 Thus, the relaxation function is:

=

φ (t ) ≡

S12

γo

=

⎤ bo ⎛ b ⎡⎛ b1 ⎞ − t / a1 b1 b ⎞ b δ ( t ) = o ⎢⎜1 − 1 ⎟ e− t / a1 + 1 δ ( t ) ⎥ + ⎜1 − ⎟e 2a1 ⎝ a1bo ⎠ 2a1 2a1 ⎢⎣⎝ a1bo ⎠ bo ⎦⎥

____________________________________________________________ Given the following velocity field: v1 = 0, v2 = v ( x1 ) , v3 = 0 . Obtain (a) the particle 8.11 pathline equations using the current time as the reference time, (b) the relative right CauchyGreen deformation tensor and (c) the Rivlin-Ericksen tensors using the equation

Ct = I + (τ - t ) A1 + (τ - t ) A 2 / 2 + .. (d) the Rivlin-Ericksen tensor A 2 using the recursive 2

equation, [ A 2 ] = [ DA1 / Dt ] + [ A1 ][∇v ] + [∇v ]

T

[ A1 ] etc.

------------------------------------------------------------------------------Ans. (a) Let x′ = xi′ei be the position at time τ of the particle which is at x = xi ei at time t . Then xi′ = xi′ ( x1 , x2 , x3 ,τ ) gives the pathline equation. Thus, dx1′ = v1 = 0 (i) , dτ

dx3′ dx2′ = v2 = v ( x1′ ) (ii) , = 0 (iii) dτ dτ


Lai et al, Introduction to Continuum Mechanics with the initial conditions: xi = xi′ ( x1 , x2 , x3 , t ) . Eq (i) gives x1′ = f ( x1 , x2 , x3 ) = x1 , Eq. (iii) dx′ gives x3′ = g ( x1 , x2 , x3 ) = x3 . Eq. (ii) becomes, 2 = v ( x1 ) → x2′ = v ( x1 )τ + h ( x1 , x2 , x3 ) , dτ → x2 = v ( x1 ) t + h ( x1 , x2 , x3 ) → h ( x1 , x2 , x3 ) = x2 − v ( x1 ) t →→ x2' = x2 + v ( x1 )(τ − t ) . Thus, x1′ = x1 , x2′ = x2 + v ( x1 )(τ − t ) , x3′ = x3 ,

1 0 0⎤ ⎡ 1 0 0⎤ ⎡ ⎢ ⎥ ⎢ (b) [ Ft ] = [∇x′t ] = ⎢( dv / dx1 )(τ − t ) 1 0 ⎥ = ⎢ k (τ − t ) 1 0 ⎥⎥ , k ≡ dv / dx1 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ 2 ⎡1 k (τ − t ) 0 ⎤ ⎡ 1 0 0 ⎤ ⎡1 + k 2 (τ − t ) k (τ − t ) 0 ⎤ ⎢ ⎥ ⎥ 1 0 ⎥ ⎢⎢ k (τ − t ) 1 0 ⎥⎥ = ⎢ k (τ − t ) 1 0⎥ [Ct ] = [ Ft ]T [Ft ] = ⎢⎢0 ⎢0 0 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎢ 0 0 1 ⎥⎥ ⎣ ⎣ ⎦ ⎡ 2k 2 ⎡1 0 0 ⎤ ⎡ 0 k 0 ⎤ ⎢ = ⎢⎢ 0 1 0 ⎥⎥ + ⎢⎢ k 0 0 ⎥⎥ (τ − t ) + ⎢ 0 ⎢ 0 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎣ ⎡ 0 k 0⎤ (c) [ A1 ] = ⎢⎢ k 0 0 ⎥⎥ , ⎢⎣ 0 0 0 ⎥⎦

⎡ 2k 2 ⎢ [ A2 ] = ⎢ 0 ⎢ 0 ⎣

0 0⎤ 2 ⎥ (τ − t ) 0 0⎥ 2 0 0⎥ ⎦ 0 0⎤ ⎥ dv 0 0⎥ , k = dx1 0 0⎥ ⎦

⎡ 0 k 0⎤ ⎡ DA1 ⎤ T ⎢ ⎥ (d) [ A 2 ] = ⎢ ⎥ + [ A1 ][∇v ] + [∇v ] [ A1 ] , where [ A1 ] = ⎢ k 0 0 ⎥ . Thus, Dt ⎣ ⎦ ⎢⎣ 0 0 0 ⎥⎦ ∂ ( A1 )ij ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ = + ∇ A v → = + vk = 0 + 0 = 0 . [ ][ ] 1 ⎢ Dt ⎥ ⎢ ∂t ⎥ ⎢ Dt ⎥ ⎢ ⎥ ∂xk ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ij ⎣ ∂t ⎦ ij 2 ⎡ 0 k 0⎤ ⎡ 0 0 0⎤ ⎡ k ⎢ [ A1 ][∇v ] = ⎢⎢ k 0 0⎥⎥ ⎢⎢ k 0 0⎥⎥ = ⎢ 0 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 ⎣

0 0⎤ ⎥ 0 0⎥ , 0 0⎥ ⎦

⎡k 2 0 0⎤ ⎢ ⎥ [∇v ]T [ A1 ] = ⎢ 0 0 0⎥ . ⎢ 0 0 0⎥ ⎣ ⎦ 2 ⎡ 2k 0 0⎤ ⎢ ⎥ T Thus, [ A 2 ] = [ A1 ][∇v ] + [∇v ] [ A1 ] = ⎢ 0 0 0 ⎥ . ⎢ 0 0 0⎥ ⎣ ⎦ ____________________________________________________________ Given the following velocity field: v1 = −kx1 , v2 = kx2 , v3 = 0 . Obtain (a) the particle 8.12 pathline equations using the current time as the reference time, (b) the relative right CauchyGreen deformation tensor and (c) the Rivlin-Ericksen tensors using the equation Ct = I + (τ - t ) A1 + (τ - t ) A 2 / 2 + .. (d) the Rivlin-Ericksen tensor A 2 and A 3 using the recursive 2


T

[ A1 ] etc.

-------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. (a) (a) Let x′ = xi′ei be the position at time τ of the particle which is at x = xi ei at time t . Then xi′ = xi′ ( x1 , x2 , x3 ,τ ) gives the pathline equation. Thus, dx3′ dx1′ dx2′ = v1 = − kx1′ (i) , = v2 = kx2′ (ii) , = 0 (iii) dτ dτ dτ with the initial conditions: xi = xi′ ( x1 , x2 , x3 , t ) . Now, dx1′ = − kx1′ → ln x1' = − kτ + g ( x1 , x2 , x3 ) → g ( x1 , x2 , x3 ) = ln x1 + kt dτ − k τ −t → ln x′ = − kτ + ln x + kt → ln x′ − ln x = − k τ − t → x′ = x e ( ) . 1

1

1

1

(

)

1

1

dx′ k τ −t Similarly, 2 = kx2′ → x2′ = x2 e ( ) and x3′ = f ( x1 , x2 , x3 ) = x3 dτ − k τ −t k τ −t Thus, x1′ = x1e ( ) , x2′ = x2 e ( ) , x3′ = x3 , ⎡e − k (τ −t ) ⎡e−2 k (τ −t ) 0 0⎤ 0 0⎤ ⎢ ⎥ ⎢ ⎥ T 2 k τ −t k τ −t (b) [ Ft ] = [∇x′t ] = ⎢ 0 e ( ) 0 ⎥ , [Ct ] = [ Ft ] [ Ft ] = ⎢ 0 e ( ) 0⎥ ⎢ ⎥ ⎢ ⎥ 0 1⎥ 0 1⎥ ⎢ 0 ⎢ 0 ⎣ ⎦ ⎣ ⎦ 2 3 4k 8k m2 k τ −t Since e ( ) = 1 m 2k (τ − t ) + (τ − t )2 m (τ − t )2 + ... , therefore, 2 3! 2 ⎡ −8k 3 0 0 ⎤ 0 0⎤ ⎡ −2k 0 0 ⎤ ⎢⎡ 4k 2 ⎥ ⎥ (τ − t )3 τ − t) ⎢ ( ⎢ ⎥ 2 3 [Ct ] = [ I ] + (τ − t ) ⎢ 0 2k 0⎥ + ⎢ 0 4k 0⎥ 2 + ⎢ 0 8k 0⎥ 3! + ... ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 0 0⎥ 0 0⎥ ⎢⎣ 0 ⎣ ⎦ ⎦ 2 3 ⎡ 4k ⎡ −8k 0 0⎤ 0 0⎤ ⎡ −2k 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ (c) [ A1 ] = ⎢⎢ 0 2k 0 ⎥⎥ , [ A 2 ] = ⎢ 0 4k 2 0 ⎥ , [ A3 ] = ⎢ 0 8k 3 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 0⎥ 0 0 0⎥ ⎢⎣ 0 ⎢ ⎦ ⎣ ⎦ − k 0 0 − 2 k 0 0⎤ ⎡ ⎤ ⎡ ⎢ ⎥ ⎢ (d) with v1 = − kx1 , v2 = kx2 , v3 = 0 , [∇v ] = ⎢ 0 k 0 ⎥ → [ A1 ] = ⎢ 0 2k 0 ⎥⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦

∂ ( A1 )i j ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ = + ∇ → = + vk = 0 + 0 = 0 A v [ ][ ] 1 ⎢ Dt ⎥ ⎢ ∂t ⎥ ⎢ Dt ⎥ ⎢ ⎥ ∂xk ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ i j ⎣ ∂t ⎦ i j

[ A 2 ] = [ A1 ][∇v ] + [∇v ]T [ A1 ] ⎡ −2k = ⎢⎢ 0 ⎢⎣ 0

0 2k 0

0⎤ ⎡ −k 0 ⎥⎥ ⎢⎢ 0 0 ⎥⎦ ⎢⎣ 0

0 0⎤ ⎡ −k k 0 ⎥⎥ + ⎢⎢ 0 0 0 ⎥⎦ ⎢⎣ 0

0 0 ⎤ ⎡ −2 k k 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎦ ⎢⎣ 0

0 2k 0

2 0 ⎤ ⎡ 4k ⎢ 0⎥⎥ = ⎢ 0 ⎢ 0⎥⎦ ⎢ 0 ⎣

Next,


0 4k 2 0

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎦

Lai et al, Introduction to Continuum Mechanics ⎡ 4k 2 ⎢ [ A 2 ][∇v ] = ⎢ 0 ⎢ ⎢⎣ 0

0⎤ ⎡ −k ⎥ 0 ⎥ ⎢⎢ 0 ⎥ 0 ⎥ ⎢⎣ 0 ⎦

3 0 0⎤ 0 0 ⎤ ⎡ −4k ⎢ ⎥ 4k 2 k 0 ⎥⎥ = ⎢ 0 4k 3 0 ⎥ . ⎢ ⎥ 0 0 0 ⎥⎦ ⎢ 0 0 0⎥ ⎣ ⎦ 3 ⎡ −8k 0 0⎤ ⎢ ⎥ T Thus, [ A3 ] = [ 0] + [ A 2 ][∇v ] + [∇v ] [ A 2 ] = ⎢ 0 8k 3 0 ⎥ . ⎢ ⎥ 0 0⎥ ⎢⎣ 0 ⎦ ____________________________________________________________

0

Given the following velocity field: v1 = kx1 , v2 = kx2 , v3 = −2kx3 . Obtain (a) the 8.13 particle pathline equations using the current time as the reference time, (b) the relative right Cauchy-Green deformation tensor and (c) the Rivlin-Ericksen tensors using the equation Ct = I + (τ - t ) A1 + (τ - t ) A 2 / 2 + .. (d) the Rivlin-Ericksen tensor A 2 and A3 using the recursive 2


T

[ A1 ] etc.

------------------------------------------------------------------------------Ans. (a) Let x′ = xi′ei be the position at time τ of the particle which is at x = xi ei at time t . Then xi′ ( x1 , x2 , x3 ,τ ) gives the pathline equation. Thus, dx3′ dx1′ dx2′ = v1 = kx1′ (i) , = v2 = kx2′ (ii) , = −2kx3′ dτ dτ dτ with the initial conditions: xi = xi′ ( x1 , x2 , x3 , t ) . Now, dx1′ = kx1′ → ln x1′ = kτ + g ( x1 , x2 , x3 ) → g ( x1 , x2 , x3 ) = ln x1 − kt dτ k τ −t → ln x′ = kτ + ln x − kt → ln x′ − ln x = k τ − t → x′ = x e ( ) . 1

1

1

1

(

)

1

1

dx′ k τ −t −2 k τ − t Similarly, 2 = kx2′ → x2′ = x2 e ( ) and x3′ = x3e ( ) . Thus, dτ k (τ −t ) k τ −t −2 k τ − t x1′ = x1e , x2′ = x2 e ( ) , x3′ = x3e ( ) (b) ⎡ e k (τ −t ) ⎡e 2 k (τ −t ) 0 0 ⎤ ⎢ ⎥ ⎢ T 0 ⎥ , [Ct ] = [ Ft ] [ Ft ] = ⎢ 0 [Ft ] = ⎡⎣∇xt' ⎤⎦ = ⎢⎢ 0 ek (τ −t ) ⎥ ⎢ −2 k τ − t ⎢ 0 ⎢ 0 0 e ( )⎥ ⎣ ⎦ ⎣ Since 4k 2 8k 3 2 k τ −t e ( ) = 1 + 2k (τ − t ) + (τ − t )2 + (τ − t )3 + ..., 2 3!

e

−4 k (τ −t )

= 1 − 4k (τ − t ) +

therefore, [Ct ] = [ I ]

16k 2 64k 3 (τ − t )2 − (τ − t )3 + ..., 2 3!


(iii)

0 2 k τ −t e ( ) 0

⎤ ⎥ 0 ⎥ ⎥ −4 k τ − t e ( )⎥ ⎦ 0

Lai et al, Introduction to Continuum Mechanics ⎡8 k 3 0 0 ⎤ 0 ⎤ ⎥ (τ − t )2 ⎢ ⎥ (τ − t )3 2 3 + ⎢ 0 8k + ... 2k 4k 0 ⎥ 0 ⎥ 2 3! ⎥ ⎢ ⎥ 0 0 16k 2 ⎥ 0 −64k 3 ⎥ ⎢⎣ 0 ⎦ ⎦ 2 3 ⎡ 4k ⎡8k 0 0 ⎤ 0 0 ⎤ 0 ⎤ ⎡ 2k 0 ⎢ ⎥ ⎢ ⎥ (c) [ A1 ] = ⎢⎢ 0 2k 0 ⎥⎥ , [ A 2 ] = ⎢ 0 4k 2 0 ⎥ , [ A 3 ] = ⎢ 0 8k 3 0 ⎥ ⎢ ⎥ ⎢ ⎥ 0 16k 2 ⎥ 0 0 −64k 3 ⎥ ⎣⎢ 0 0 −4k ⎦⎥ ⎢⎣ 0 ⎢ ⎦ ⎣ ⎦ k 0 0 2 k 0 0 ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ (d) with v1 = kx1 , v2 = kx2 , v3 = −2kx3 , [∇v ] = ⎢ 0 k 0 ⎥ → [ A1 ] = ⎢ 0 2k 0 ⎥⎥ ⎢⎣ 0 0 −2k ⎥⎦ ⎢⎣ 0 0 −4k ⎥⎦ ⎡ 2k + (τ − t ) ⎢⎢ 0 ⎢⎣ 0

0

2 0 ⎤ ⎡ 4k ⎢ 0 ⎥⎥ + ⎢ 0 ⎢ −4k ⎥⎦ ⎢ 0 ⎣

0

∂ ( A1 )ij ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ = + ∇ → = + vk = 0 + 0 = 0 A v [ ][ ] 1 ⎢ Dt ⎥ ⎢ ∂t ⎥ ⎢ Dt ⎥ ⎢ ⎥ ∂xk ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ij ⎣ ∂t ⎦ ij

[ A 2 ] = [ A1 ][∇v ] + [∇v ]T [ A1 ] ⎡ 2k = ⎢⎢ 0 ⎢⎣ 0

0 2k 0

0 ⎤ ⎡ k 0 0 ⎤ ⎡ k 0 0 ⎤ ⎡ 2k 0 ⎥⎥ ⎢⎢ 0 k 0 ⎥⎥ + ⎢⎢ 0 k 0 ⎥⎥ ⎢⎢ 0 −4k ⎥⎦ ⎢⎣ 0 0 −2k ⎥⎦ ⎢⎣ 0 0 −2k ⎥⎦ ⎢⎣ 0

0 2k 0

0 ⎤ ⎡ 4k ⎢ 0 ⎥⎥ = ⎢ 0 ⎢ −4k ⎥⎦ ⎢ 0 ⎣

2

0 4k 2 0

0 ⎤ ⎥ 0 ⎥. ⎥ 16k 2 ⎥ ⎦

Next, ⎡ 4k 2 ⎢ [ A 2 ][∇v ] = ⎢ 0 ⎢ ⎢⎣ 0

3 0 ⎤ ⎡k 0 0 0 ⎤ 0 ⎤ ⎡ 4k ⎥⎢ ⎢ ⎥ 4k 2 0 ⎥ ⎢0 k 0 ⎥⎥ = ⎢ 0 4k 3 0 ⎥. ⎥ ⎢ ⎥ 0 16k 2 ⎥ ⎢⎣ 0 0 −2k ⎥⎦ ⎢ 0 0 −32k 3 ⎥ ⎦ ⎣ ⎦ 3 ⎡8k 0 0 ⎤ ⎢ ⎥ T Thus, [ A3 ] = [ 0] + [ A 2 ][∇v ] + [∇v ] [ A 2 ] = ⎢ 0 8k 3 0 ⎥ Etc. ⎢ ⎥ 0 −64k 3 ⎥ ⎢⎣ 0 ⎦ ____________________________________________________________

0

8.14 Given the following velocity field: v1 = kx2 , v2 = kx1 , v3 = 0 . Obtain (a) the particle pathline equations using the current time as the reference time, (b) the relative right CauchyGreen deformation tensor and (c) the Rivlin-Ericksen tensors using the equation

Ct = I + (τ - t ) A1 + (τ - t ) A 2 / 2 + .. (d) the Rivlin-Ericksen tensor A 2 and A3 using the recursive 2


T

[ A1 ] etc.

------------------------------------------------------------------------------Ans. (a) Let x′ = xi′ei be the position at time τ of the particle which is at x = xi ei at time t . Then xi′ = xi′ ( x1 , x2 , x3 ,τ ) gives the pathline equation. Thus, dx1′ = v1 = kx2′ (i) , dτ

dx2′ = v2 = kx1′ (ii) , dτ

dx3′ = 0 (iii) dτ


Lai et al, Introduction to Continuum Mechanics with the initial conditions: xi = xi′ ( x1 , x2 , x3 , t ) . Now, dx1′ d 2 x1′ dx2′ d 2 x1′ 2 ′1 → k k x = kx2′ → = = − k 2 x1′ = 0 → 2 2 dτ d τ dτ dτ x1′ = A sinh kτ + B cosh kτ → x1 = A sinh kt + B cosh kt (iv) 1 dx1′ = A cosh kτ + B sinh kτ → x2 = A cosh kt + B sinh kt (v) (ii) → x2′ = k dτ (iv) and (v) gives A = − x1 sinh kt + x2 cosh kt , B = x1 cosh kt − x2 sinh kt x1′ = x1 ( cosh kt cosh kτ − sinh kt sinh kτ ) + x2 ( cosh kt sinh kτ − sinh kt cosh kτ ) (i) →

x2′ = x1 ( cosh kt sinh kτ − sinh kt cosh kτ ) + x2 ( cosh kt cosh kτ − sinh kt sinh kτ ) That is, x1′ = x1 cosh k (τ − t ) + x2 sinh k (τ − t ) , x2′ = x1 sinh k (τ − t ) + x2 cosh k (τ − t ) , x3′ = x3

⎡cosh k (τ − t ) sinh k (τ − t ) 0 ⎤ [Ft ] = [∇x′t ] = ⎢⎢ sinh k (τ − t ) cosh k (τ − t ) 0⎥⎥ , ⎢ 0 0 1 ⎥⎦ ⎣

(b)

⎡cosh 2 x + sinh 2 x 2cosh x sinh x 0⎤ ⎢ ⎥ T [Ct ] = [ Ft ] [Ft ] = ⎢ 2cosh x sinh x sinh 2 x + cosh 2 x 0⎥ , x ≡ k (τ − t ) ⎢ ⎥ 0 0 1⎥ ⎢⎣ ⎦ Since x2 x3 + O( x 4 ), sinh x = x + + O( x5 ) cosh x = 1 + 2 6

cosh 2 x = 1 + x 2 + O( x 4 ), sinh 2 x = x 2 + O( x 4 ), sinh x cosh x = x +

2 x3 + O( x5 ) 3

4 3 ⎡ ⎤ 2 ⎢ 1 + 2 x + .. 2 x + 3 x + ... 0 ⎥ ⎡ 0 2k 0 ⎤ ⎢ ⎥ 4 3 2 ⎢ ⎥ [Ct ] = ⎢ 2 x + 3 x + ... 1 + 2 x + ... 0⎥ = [I ] + ⎢⎢ 2k 0 0⎥⎥ (τ − t ) ⎢⎣ 0 0 0 ⎥⎦ ⎢ ⎥ 0 0 1⎥ ⎢ ⎥⎦ ⎣⎢ ⎡ 4k 2 ⎡ 0 8k 3 0 ⎤ 0 0⎤ ⎢ ⎥ (τ − t )2 ⎢ ⎥ (τ − t )3 2 3 0⎥ 0 0⎥ + ⎢ 0 4k + ⎢8k + ... ⎢ ⎥ 2 ⎢ ⎥ 6 0 0⎥ 0 0⎥ ⎢⎣ 0 ⎢⎣ 0 ⎦ ⎦ Thus, (c) ⎡ 4k 2 ⎡ 0 8k 3 0 0⎤ ⎡ 0 2k 0 ⎤ ⎢ ⎥ ⎢ [ A1 ] = ⎢⎢ 2k 0 0⎥⎥ , [ A 2 ] = ⎢ 0 4k 2 0⎥ , [ A3 ] = ⎢8k 3 0 ⎢ ⎥ ⎢ ⎢⎣ 0 0 0 ⎥⎦ 0 0⎥ 0 ⎢⎣ 0 ⎢⎣ 0 ⎦


0⎤ ⎥ 0 ⎥ etc. ⎥ 0⎥ ⎦

Lai et al, Introduction to Continuum Mechanics ⎡ 0 k 0⎤ ⎡ 0 2k ⎢ ⎥ (d) with v1 = kx2 , v2 = kx1 , v3 = 0 , [∇v ] = ⎢ k 0 0 ⎥ → [ A1 ] = ⎢⎢ 2k 0 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 ∂ ( A1 )ij ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ ⎡ DA1 ⎤ ⎡ ∂A1 ⎤ vk = 0 + 0 = 0 . = + ∇ A v → = + [ ][ ] 1 ⎢ Dt ⎥ ⎢ ∂t ⎥ ⎢ Dt ⎥ ⎢ ⎥ ∂xk ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ij ⎣ ∂t ⎦ ij

0⎤ 0 ⎥⎥ . 0 ⎥⎦

[ A 2 ] = [ A1 ][∇v ] + [∇v ]T [ A1 ] ⎡0 [ A1 ][∇v ] = ⎢⎢ 2k ⎢⎣ 0

2 0 ⎤ ⎡ 0 k 0 ⎤ ⎡ 2k ⎢ 0 ⎥⎥ ⎢⎢ k 0 0 ⎥⎥ = ⎢ 0 ⎢ 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 ⎣

2k 0 0

0 2k 2 0

⎡ 4k 2 0⎤ ⎥ ⎢ 0⎥ → [ A2 ] = ⎢ 0 ⎥ ⎢ 0⎥ ⎢⎣ 0 ⎦

0 4k 2 0

0⎤ ⎥ 0⎥ . ⎥ 0⎥ ⎦

Next, ⎡ 4k 2 ⎢ [ A 2 ][∇v ] = ⎢ 0 ⎢ ⎢⎣ 0

0 ⎤ ⎡ 0 k 0 ⎤ ⎡ 0 4k 3 0 ⎤ ⎥ ⎢ ⎥ 4k 2 0 ⎥ ⎢⎢ k 0 0 ⎥⎥ = ⎢ 4k 3 0 0⎥ . ⎥ ⎢ ⎥ 0 0 ⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 0 0⎥ ⎦ ⎣ ⎦ 3 ⎡ 0 8k 0 ⎤ ⎢ ⎥ T Thus, [ A3 ] = [ 0] + [ A 2 ][∇v ] + [∇v ] [ A 2 ] = ⎢8k 3 0 0 ⎥ . ⎢ ⎥ 0 0⎥ ⎢⎣ 0 ⎦ ____________________________________________________________ 8.15

0

Given the velocity field in cylindrical coordinates: vr = 0, vθ = 0, vz = v ( r ) , obtain the

second Rivlin-Ericksen tensors A N , N = 2,3,... using the recursive formula. -----------------------------------------------------------------------⎡ 0 0 0⎤ ⎡0 0 k ⎤ dv T ⎢ ⎥ Ans. [∇v ] = ⎢ 0 0 0 ⎥ , k = , [ A1 ] = [∇v ] + [∇v ] = ⎢⎢ 0 0 0 ⎥⎥ dr ⎢⎣ k 0 0 ⎥⎦ ⎢⎣ k 0 0 ⎥⎦ Since ( A1 )ij =constant, independent of time and space, therefore ⎡⎛ DA ⎞ ⎤ ⎡⎛ ∂A ⎞ ⎤ 1 ⎢⎜ ⎥ = ⎢⎜ 1 ⎟ + ( ∇A1 )ijk vk ⎥ = [ 0] . ⎟ ⎢⎣⎝ Dt ⎠ij ⎥⎦ ⎢⎣⎝ ∂t ⎠ij ⎥⎦



[ A 2 ] = ⎡⎣⎢ A1 ( ∇v ) + ( ∇v )T A1 ⎤⎦⎥ 2 ⎡ 0 0 k ⎤ ⎡ 0 0 0 ⎤ ⎡ 0 0 k ⎤ ⎡ 0 0 k ⎤ ⎡ 2k ⎢ = ⎢⎢ 0 0 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎥ + ⎢⎢ 0 0 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎥ = ⎢ 0 ⎢⎣ k 0 0 ⎥⎦ ⎢⎣ k 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ k 0 0 ⎥⎦ ⎢ 0 ⎣

0 0⎤ ⎥ 0 0⎥ . 0 0⎥ ⎦

[ A3 ] = ⎡⎢⎣ A 2 ( ∇v ) + ( ∇v )T A 2 ⎤⎥⎦ ⎡ 2k 2 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎡ 0 0 k ⎤ ⎡ 2k 2 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ 0 0 0 ⎥ ⎢⎢ 0 0 0 ⎥⎥ + ⎢⎢0 0 0 ⎥⎥ ⎢ 0 0 0 ⎥ = ⎢⎢ 0 0 0 ⎥⎥ . ⎢ 0 0 0⎥ ⎢ k 0 0 ⎥ ⎢0 0 0 ⎥ ⎢ 0 0 0⎥ ⎢0 0 0⎥ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ Thus, A N = 0, N = 3, 4... ____________________________________________________________ Using the equations given in Appendix 8.1 for cylindrical coordinates, verify that the 8.16 rrθ component of the third order tensor ∇T is given by:

⎡ 1 ∂Trr Tθ r + Trθ ⎤ =⎢ − ⎥ r ⎣ r ∂θ ⎦ -----------------------------------------------------------------------------Ans. From the equations ∂T ( ∇T )ijm hm = ij + Tqj Γ qmi + Tiq Γ qmj no sum on m, sum on q, ∂xm

( ∇T )rrθ

and hr = 1, hθ = r , hz = 1; Γ rθθ = 1, Γθθ r = −1 , all other Γijk = 0 we have,

∂Trr ∂T + Tqr Γ qθ r + Trq Γ qθ r = rr + Tθ r Γθθ r + Trθ Γθθ r , thus, ∂θ ∂θ ∂Trr 1 ∂Trr Tθ r + Trθ + Tθ r ( −1) + Trθ ( −1) → ( ∇T )rrθ = − . ( ∇T )rrθ r = ∂θ r ∂θ r ____________________________________________________________

( ∇T )rrθ hθ

=

Using the equations given in Appendix 8.1 for cylindrical coordinates, verify that the 8.17 rθθ component of the third order tensor ∇T is given by: 1 ∂Trθ Trr − Tθθ + r ∂θ r -----------------------------------------------------------------------------Ans. From the equations ∂T ( ∇T )ijm hm = ij + Tqj Γ qmi + Tiq Γ qmj no sum on m, sum on q ∂xm

( ∇T )rθθ

=

and hr = 1, hθ = r , hz = 1; Γ rθθ = 1, Γθθ r = −1 , all other Γijk = 0. we have,


Lai et al, Introduction to Continuum Mechanics ∂Trθ ∂T + Tqθ Γ qθ r + Trq Γ qθθ → ( ∇T )rθθ hθ = rθ + Tθθ Γθθ r + Trr Γ rθθ ∂θ ∂θ ∂T 1 ∂Trθ Trr − Tθθ → ( ∇T )rθθ r = rθ + Tθθ ( −1) + Trr (1) → ( ∇T )rθθ = + . ∂θ r ∂θ r ____________________________________________________________

( ∇T )rθθ hθ

=

Using the equations given in Appendix 8.1 for spherical coordinates, verify that the 8.18 rrφ component of the third order tensor ∇T is given by:

(

Tφ r + Trφ 1 ∂Trr − r sin θ ∂φ r -----------------------------------------------------------------------Ans. From the equations ∂T ( ∇T )ijm hm = ij + Tqj Γ qmi + Tiq Γ qmj no sum on m, sum on q ∂xm

( ∇T )rrφ =

)

and hr = 1, hθ = r , hφ = r sin θ ; Γ rθθ = 1, Γ rφφ = sin θ , Γφφ r = − sin θ , Γφφθ = − cosθ , Γθθ r = −1, Γθφφ = cos θ all other Γijk = 0 we have, ∂Trr ∂T + Tqr Γ qφ r + Trq Γ qφ r → ( ∇T )rrφ hφ = rr + Tφ r Γφφ r + Trφ Γφφ r ∂φ ∂φ ∂T → ( ∇T )rrφ ( r sin θ ) = rr + Tφ r ( − sin θ ) + Trφ ( − sin θ ) → ∂φ 1 ∂Trr Tφ r + Trφ → ( ∇T )rrφ = − . r sin θ ∂φ r ____________________________________________________________

( ∇T )rrφ hφ

8.19

=

Using the equations given in Appendix 8.1 for spherical coordinates, verify that the

φφφ component of the third order tensor ∇T is given by:

(

) (

)

Trφ + Tφ r Tθφ + Tφθ cot θ 1 ∂Tφφ + + r sin θ ∂φ r r ------------------------------------------------------------------------------Ans. From the equations ∂T ( ∇T )ijm hm = ij + Tqj Γ qmi + Tiq Γ qmj no sum on m, sum on q ∂xm hr = 1, hθ = r , hφ = r sin θ ; Γ rθθ = 1, Γ rφφ = sin θ , Γφφ r = − sin θ , Γφφθ = − cosθ , Γθθ r = −1, Γθφφ = cosθ all other Γijk = 0 we have,



( ∇T )φφφ hφ

∂Tφφ

+ Tqφ Γ qφφ + Tφ q Γ qφφ ∂φ ∂Tφφ → ( ∇T )φφφ hφ = + Trφ Γ rφφ + Tθφ Γθφφ + Tφ r Γ rφφ + Tφθ Γθφφ ∂φ ∂Tφφ → ( ∇T )φφφ hφ = + Trφ + Tφ r Γ rφφ + Tθφ + Tφθ Γθφφ ∂φ ∂Tφφ = + Trφ + Tφ r ( sin θ ) + Tθφ + Tφθ cosθ ∂φ =

(

(

)

)

(

)

(

)

(

) (

)

Trφ + Tφ r Tθφ + Tφθ cot θ 1 ∂Tφφ + + r sin θ ∂φ r r ____________________________________________________________ → ( ∇T )φφφ =

8.20

Given the velocity field in cylindrical coordinates: vr = 0, vθ = v ( r ) , vz = 0 , obtain (a)

the first Rivlin-Ericksen tensor A1 (b) ∇A1 (c) the second Rivlin-Ericksen tensors A 2 , using the recursive formula.. -----------------------------------------------------------------------⎡ ∂vr 1 ⎛ ∂vr ⎞ ∂vr ⎤ ⎡ v( r ) ⎤ − vθ ⎟ ⎢ ⎥ ⎢0 − ⎜ 0⎥ r r z θ ∂ ∂ ∂ ⎝ ⎠ r ⎢ ⎥ ⎢ ⎥ ⎢ ∂vθ 1 ⎛ ∂vθ ⎥ ∂ v dv ⎞ ⎢ ⎥ θ 0 0 Ans. [∇v ] = ⎢ v + = ⎥ ⎢ r⎟ ⎜ ∂θ ⎥ r r z dr ∂ ∂ ⎝ ⎠ ⎢ ⎥ ⎢ ⎥ ⎢ ∂v 0 0⎥ ∂vz ⎥ ⎢ 0 1 ∂vz z ⎢ ⎥ ⎦⎥ r ∂θ ∂z ⎦⎥ ⎣⎢ ⎣⎢ ∂r ⎡ 0 k ( r ) 0⎤ dv v( r ) ⎞ T ⎢ [ A1 ] = [∇v ] + [∇v ] = ⎢ k ( r ) 0 0⎥⎥ , k = ⎛⎜ − ⎟ r ⎠ ⎝ dr ⎢ 0 ⎥ 0 0⎦ ⎣ ⎡⎛ DA ⎞ ⎤ ⎡⎛ ∂A ⎞ ⎤ 1 ⎢⎜ ⎥ = ⎢⎜ 1 ⎟ + ( ∇A1 )ijk vk ⎥ ⎟ ⎢⎣⎝ Dt ⎠ij ⎥⎦ ⎢⎣⎝ ∂t ⎠ij ⎥⎦

⎡ ( ∇A1 ) vθ ( ∇A1 ) vθ ( ∇A1 ) vθ ⎤ rrθ rθθ rzθ ⎢ ⎥ ⎡ ⎤ = ( ∇A1 )ijk vk = ⎢( ∇A1 )θ rθ vθ ( ∇A1 )θθθ vθ ( ∇A1 )θ zθ vθ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢⎣ ( ∇A1 ) zrθ vθ ( ∇A1 ) zθθ vθ ( ∇A1 ) zzθ vθ ⎥⎦ The components of the third order tensor ( ∇A1 ) can be obtained from Appendix 8.1 as: 2k 1 ∂Arθ Arr − Aθθ ⎡ 1 ∂Arr Aθ r + Arθ ⎤ =⎢ − = − , ( ∇A1 )rθθ = + =0 ⎥ r r r ∂θ r ⎣ r ∂θ ⎦ 1 ∂Arz Aθ z − =0 ( ∇A1 )rzθ = r ∂θ r A −A A 1 ∂A 2k 1 ∂A ( ∇A1 )θ rθ = θ r + rr θθ = 0 , ( ∇A1 )θθθ = , ( ∇A1 )θ zθ = θ z + rz = 0 r ∂θ r r ∂θ r r

( ∇A1 )rrθ


Lai et al, Introduction to Continuum Mechanics 1 ∂Azθ Azr 1 ∂Azz + = 0 , ( ∇A1 ) zzθ = =0 r ∂θ r r ∂θ 0 ⎤ ⎡ −2kv / r 0 0⎤ ⎥ ⎢ 0⎥ = ⎢ 0 2kv / r 0 ⎥⎥ , 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ −v(r ) / r 0 ⎤ ⎡ kdv / dr 0 0⎤ ⎥ ⎢ 0 0⎥ = ⎢ 0 − kv / r 0 ⎥⎥ , 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦

1 ∂Azr Azθ − = 0 , ( ∇A1 ) zθθ r ∂θ r 0 ⎡ −2k / r ⎡⎛ DA ⎞ ⎤ ⎢ 1 2k / r Thus, ⎢⎜ ⎟ ⎥ = vθ ⎢ 0 ⎢⎣⎝ Dt ⎠ij ⎥⎦ ⎢⎣ 0 0

( ∇A1 ) zrθ

=

=

⎡ 0 k ( r ) 0⎤ ⎡ 0 ⎢ [ A1 ][∇v ] = ⎢ k ( r ) 0 0⎥⎥ ⎢⎢ dv / dr ⎢ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎣ dv / dr 0 ⎤ ⎡ 0 k ( r ) 0 ⎤ ⎡ kdv / dr ⎡ 0 T ⎢ ⎢ ⎥ [∇v ] [ A1 ] = ⎢ −v(r ) / r 0 0⎥ ⎢ k ( r ) 0 0⎥⎥ = ⎢⎢ 0 ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0

0 0⎤ − kv / r 0 ⎥⎥ , 0 0 ⎥⎦

⎡ k ( dv / dr − v / r ) 0 0 ⎤ ⎡ 2k 2 ⎡ DA1 ⎤ T ⎢ ⎥ ⎢ A2 = ⎢ + [ A1 ][∇v ] + [∇v ] [ A1 ] = 2 ⎢ 0 0 0⎥ = ⎢ 0 ⎥ ⎣ Dt ⎦ ⎢ 0 0 0 ⎥⎦ ⎢ 0 ⎣ ⎣ ____________________________________________________________

0 0⎤ ⎥ 0 0⎥ . 0 0⎥ ⎦

DA N T + A N ( ∇v ) + ( ∇v ) A N . Dt -----------------------------------------------------------------------Ans. We had {see Eq. (8.11.7)}, DA N DN D N +1 Ddx Ddx 2 = ⋅ → ⋅ A N dx + dx ⋅ . That is, x A x ds d d ds 2 = dx + dx ⋅ A N N N N +1 Dt Dt Dt Dt Dt

Derive Eq. (8.11.3), i.e., A N +1 =

8.21

( )

D N +1

( ds ) == (∇v ) dx ⋅ A

( )

DA N dx + dx ⋅ A N ( ∇v ) dx Dt Dt DA N T = dx ⋅ ( ∇v ) A N dx + dx ⋅ A N ( ∇v ) dx + dx ⋅ dx Dt DA N ⎤ ⎡ T = dx ⋅ ⎢ ( ∇ v ) A N + A N ( ∇ v ) + dx = dx ⋅ A N +1dx. Dt ⎥⎦ ⎣ DA N T Thus, A N +1 = + A N ( ∇v ) + ( ∇v ) A N Dt ____________________________________________________________ N +1

8.22

2

N dx + dx ⋅

Let S ≡ DT / Dt + TW − WT , where T is an objective tensor and W is the spin tensor,

show that S is objective, i.e., S* = Q ( t ) SQ T ( t ) . ------------------------------------------------------------------------Ans. Since T is objective, therefore T* = Q ( t ) TQ T ( t ) and from Eq. (8.13.13),

W* = ( dQ / dt ) QT ( t ) + Q ( t ) WQT ( t ) , therefore,



S* ≡

DT* + T* W* - W*T* Dt

dQ dQT ⎡ dQ T ⎡ ⎛ DT ⎞ T + + QTQT ⎤ TQ T + Q ⎜ Q QT Q + QTQT QWQT ⎤ ⎟ ⎣ ⎦ ⎣ ⎦ dt dt dt ⎝ Dt ⎠ dQ T ⎡ − Q QTQ T ⎤ − ⎡QWQT Q ( t ) TQT ( t ) ⎤ . ⎣ ⎦ ⎣ ⎦ dt dQ T dQ T Now, Q ( t ) Q T ( t ) = I → Q ( t ) = −Q ( t ) , therefore, the above equation becomes dt dt dQ dQ T ⎡ dQT ⎤ ⎡ ⎛ DT ⎞ T T S* = TQT + Q ⎜ − ⎢QT ⎥ + ⎣QTWQ ⎤⎦ ⎟ Q + QT dt dt dt ⎥⎦ ⎝ Dt ⎠ ⎢⎣ ⎡ dQ ⎤ −⎢ TQT ⎥ − ⎡QWTQT ( t ) ⎤ . ⎦ ⎣ dt ⎦ ⎣ That is, ⎛ DT ⎞ S* = Q ( t ) ⎜ + TW − WT ⎟ Q T ( t ) = Q ( t ) SQ T ( t ) . ⎝ Dt ⎠ ____________________________________________________________ =

8.23

Obtain the viscosity function and the two normal stress function for the nonlinear ∞

f 2 ( s ) ⎡I − Ct−1 (t − s ) ⎤ ds ⎣ ⎦ -----------------------------------------------------------------------------Ans. For v1 = kx2 , v2 = v3 = 0 , we have [see Section 8.9, Eq.(8.9.12)]

viscoelastic fluid defined by S = ∫

0

k (τ − t ) ⎡ 1 ⎢ 2 [Ct (τ )] = ⎢ k (τ − t ) k (τ − t )2 + 1 ⎢ 0 0 ⎢⎣ ⎡1 + k 2 s 2 ks 0 ⎤ ⎢ ⎥ ⎡Ct−1 (t − s ) ⎤ = ⎢ ks 1 0⎥ . ⎣ ⎦ ⎢ 0 0 1⎥ ⎣ ⎦ ∞

S12 = − k ∫ sf 2 ( s )ds → μ ≡ S11 = − k

0 ∞ 2 2

∫s

0⎤

⎡1 + k 2 (τ − t )2 − k (τ − t ) 0 ⎤ ⎥ ⎢ ⎥ 0 ⎥ , ⎡Ct−1 (τ ) ⎤ = ⎢ − k (τ − t ) 1 0⎥ ⎣ ⎦ ⎥ ⎢ 0 0 1 ⎥⎥ 1 ⎥⎦ ⎣⎢ ⎦ ⎡ − k 2 s 2 − ks 0 ⎤ ⎢ ⎥ Thus, ⎡ I − Ct−1 (t − s ) ⎤ = ⎢ − ks 0 0⎥ ⎣ ⎦ ⎢ 0 0 0⎥ ⎣ ⎦ ∞

S12 = − ∫ sf 2 ( s )ds,. k 0

f 2 ( s )ds, S22 = 0, S33 = 0 ,

0 ∞

σ1 = S11 − S22 = − k 2 ∫ s 2 f 2 ( s )ds, σ 2 = S22 − S33 = 0. 0

____________________________________________________________ 8.24 Derive the following transformation laws [Eqs.(8.13.8) and Eq. (8.13.12)] under a change of frame. Vt* = Q (τ ) Vt QT (τ ) and R*t = Q (τ ) R t QT ( t ) -------------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics Ans. Since Ft = Vt R t and Ft* = Vt*R*t , therefore, from Ft* (τ ) = Q (τ ) Ft (τ ) QT ( t ) , we get Vt*R*t = Q (τ ) Vt R t QT ( t ) = ⎡Q (τ ) Vt QT (τ ) ⎤ ⎡Q (τ ) R t QT ( t ) ⎤ , where ⎡Q (τ ) Vt QT (τ ) ⎤ is a ⎣ ⎦⎣ ⎦ ⎣ ⎦ T symmetry tensor and ⎡Q (τ ) R t Q ( t ) ⎤ is an orthogonal tensor. Therefore, the uniqueness of the ⎣ ⎦ polar decomposition leads to Vt* = Q (τ ) Vt Q T (τ ) and R*t = Q (τ ) R t QT ( t ) . ____________________________________________________________

8.25

( ⎡ DJ (τ ) ⎤ ⎡ DF (τ ) ⎤ From T ≡ ⎢ L and ⎢ t ⎥ ⎥ = ∇v , show that ⎣ Dτ ⎦τ =t ⎣ Dτ ⎦τ =t

( o T = T + TD + DT . [note misprint in the problem in text] -----------------------------------------------------------------------------Ans. From J L (τ ) = FtT (τ ) T (τ ) Ft (τ ) , we have,

DJ L (τ ) Dτ

=

DFtT (τ )

T (τ ) Ft (τ ) + FtT (τ )

DT (τ )

Ft (τ ) + FtT (τ ) T (τ )

DFt (τ )

Dτ Dτ Dτ ⎡ DJ L (τ ) ⎤ DT T + T ( t ) ∇v [Note Ft ( t ) = FtT ( t ) = I ] Thus, ⎢ ⎥ = ( ∇v ) T ( t ) + Dt ⎣ Dτ ⎦τ =t Now, ∇v = D + W , therefore, ⎡ DJ L (τ ) ⎤ DT DT T + (D + W) T + T(D + W) = + DT + TD + W T T + TW ⎢ ⎥ = D τ Dt Dt ⎣ ⎦τ =t o DT = + DT + TD + ( TW − WT ) = T + DT + TD Dt ____________________________________________________________

(

)

Consider J U (τ ) = Ft−1 (τ ) T (τ ) Ft−1T (τ ) . Show that (a) ⎡⎣ DJ U (τ ) / Dτ ⎤⎦ is objective τ =t o DT T − T ( ∇v ) − ( ∇v ) T = T − ( TD + DT ) . and (b) ⎡⎣ DJ U (τ ) / Dτ ⎤⎦ = τ =t Dτ ------------------------------------------------------------------------------8.26

( )

Ans. (a) Given J U (τ ) = Ft−1 (τ ) T (τ ) Ft−1

T

(τ ) , and

(

J*U (τ ) = F* t−1 (τ ) T* (τ ) F* t−1

In a change of frame (see Section 8.13. Eq.(8.13.6), Ft * (τ ) = Q (τ ) Ft (τ ) Q

(

F* t−1 (τ ) = Q ( t ) Ft−1 (τ ) QT (τ ) and F* t−1

)

T

T

)

T

(τ ) .

( t ) , so that

(τ ) = Q (τ ) ( Ft−1 ) (τ ) QT ( t ) . T

Also, T* (t ) = Q(t )T(t )QT (t ) . Thus

( )

J*U (τ ) = Q ( t ) Ft−1 (τ ) QT (τ ) Q(τ )T(τ )QT (τ )Q (τ ) Ft−1

( )

= Q ( t ) Ft−1 (τ ) T(τ ) Ft−1

T

(τ ) QT ( t ) .

T

(τ ) QT ( t )

That is,

(

)

J*U (τ ) = Q ( t ) JU (τ ) QT ( t ) and D N J*U (τ ) / Dτ N = Q ( t ) D N JU (τ ) / Dτ N QT ( t ) . Thus,


Lai et al, Introduction to Continuum Mechanics ⎡ D N J*U (τ ) ⎢ N ⎢⎣ Dτ

⎡⎛ D N J (τ ) ⎤ U ⎥ = Q ( t ) ⎢⎜ ⎜ ⎢ Dτ N ⎥⎦τ =t ⎣⎝

⎞⎤ ⎟ ⎥ QT ( t ) . ⎟⎥ ⎠ ⎦τ =t

( )

⎡ (b) ⎡⎣ DJ U (τ ) / Dτ ⎤⎦ = ⎡ DFt−1 / Dτ ⎤ T ( t ) + [ DT / Dτ ]τ =t + T ( t ) ⎢ D Ft−1 ⎣ ⎦τ =t τ =t ⎣

T

⎤ / Dτ ⎥ . ⎦τ =t

Now, Ft (τ ) Ft −1 (τ ) = I → [ DFt / Dτ ] Ft −1 (τ ) + Ft (τ ) ⎡ DFt −1 / Dτ ⎤ = 0 → ⎣ ⎦ − 1 − 1 − 1 ⎡ DFt (τ ) / Dτ ⎤ = −Ft ( t ) [ DFt / Dτ ] Ft ( t ) = − [ DFt / Dτ ] = − ⎡( ∇v ) Ft (τ ) ⎤ ⎣ ⎦τ =t τ =t τ =t ⎣ ⎦τ =t = − ( ∇v ) . Thus,

DT DT T ⎡⎣ DJ U (τ ) / Dτ ⎤⎦ = − T ( ∇v ) − ( ∇v ) T = − T(D − W) − (D + W)T . τ =t Dt Dt That is, the upper convected derivative of T can be written: o ˆ ≡ ⎡⎢ DJ U (τ ) ⎤⎥ = DT + TW − WT − ( TD + DT ) = T − ( TD + DT ) . T ⎣ Dτ ⎦τ =t Dt ____________________________________________________________

Given the velocity field of a plane Couette flow: v1 = 0, v2 = kx1 . (a) For a Newtonian

8.27

o fluid, find the stress field [ T ] and the co-rotational stress rate ⎡ T ⎤ . (b) Consider a change of ⎣ ⎦ frame (change of observer) described by:

⎡ x1* ⎤ ⎡cos ωt − sin ωt ⎤ ⎡ x1 ⎤ ⎡cos ωt − sin ωt ⎤ , [Q ] = ⎢ ⎢ ⎥=⎢ ⎢ ⎥ ⎥ ⎥ ⎢⎣ x2* ⎥⎦ ⎣ sin ωt cos ωt ⎦ ⎣ x2 ⎦ ⎣ sin ωt cos ωt ⎦ Find ⎡ v* ⎤ , ⎡∇* v* ⎤ , ⎡ D* ⎤ and ⎡ W* ⎤ . ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (c) Find the co-rotational stress rate for the starred frame (d) Verify that the two stress rates are related by the objective tensorial relation. -----------------------------------------------------------------------------Ans. k / 2⎤ −k / 2⎤ ⎡ 0 0⎤ ⎡ 0 ⎡ 0 , [ D] = ⎢ , [ W] = ⎢ . Thus, stress tensor is (a) [∇v ] = ⎢ ⎥ ⎥ 0 ⎦ 0 ⎥⎦ ⎣ k 0⎦ ⎣k / 2 ⎣k / 2

⎡− p

[T] = ⎢ 0 ⎣

0 ⎤ k / 2⎤ ⎡ − p μ k ⎤ ⎡ 0 + 2μ ⎢ = . ⎥ − p⎦ 0 ⎥⎦ ⎢⎣ μ k − p ⎥⎦ ⎣k / 2 ⎡− p

[TW ] − [ WT] = ⎢ μ k ⎣

−k / 2⎤ ⎡ 0 −k / 2⎤ ⎡ − p μ k ⎤ ⎡ μ k 2 − =⎢ − p ⎥⎦ ⎢⎣ k / 2 0 ⎥⎦ ⎢⎣ k / 2 0 ⎥⎦ ⎢⎣ μ k − p ⎥⎦ ⎢⎣ 0

μk ⎤ ⎡ 0

0 ⎤ ⎥ − μ k 2 ⎥⎦

2 0 ⎤ ⎡μk 2 0 ⎤ o ⎡ DT ⎤ ⎡ μ k + =⎢ Co-rotational stress rate is: ⎡ T ⎤ = ⎢ ⎢ ⎥ ⎥. ⎥ ⎣ ⎦ ⎣ Dt ⎦ ⎢ 0 2 2 ⎥ ⎢ ⎥⎦ μ 0 μ − − k k ⎣ ⎦ ⎣ (b) From Eq. (5.56.12) of Chapter 5, we have, [v * (x*)] = [Qv ] + ⎡⎣( dQ / dt ) x ⎤⎦ = [Qv ] + ⎡⎣( dQ / dt ) QT x *⎤⎦ . Thus, * ⎡ v1* ⎤ ⎡ cos ωt − sin ωt ⎤ ⎡ 0 ⎤ ⎡ − sin ωt − cos ωt ⎤ ⎡ cos ωt sin ωt ⎤ ⎡ x1 ⎤ + ω ⎢ ⎥=⎢ ⎢ ⎥⎢ ⎥ ⎢ cos ωt − sin ωt ⎥ ⎢ − sin ωt cos ωt ⎥ * ⎥ . Since, ⎢⎣ v2* ⎥⎦ ⎣ sin ωt cos ωt ⎦ ⎣v2 ⎦ ⎣ ⎦⎣ ⎦ ⎢⎣ x2 ⎥⎦


Lai et al, Introduction to Continuum Mechanics * ⎡ x1 ⎤ ⎡ cos ωt sin ωt ⎤ ⎡ x1 ⎤ * * * * = ⎢ ⎢ x ⎥ ⎢ − sin ωt cos ωt ⎥ * ⎥ → x1 = cos ωtx1 + sin ωtx2 → v2 = k cos ωtx1 + sin ωtx2 ⎦ ⎢⎣ x2 ⎥⎦ ⎣ 2⎦ ⎣ Therefore, ⎡ − cos ωt sin ωtx* + sin 2 ωtx* ⎤ * ⎡ v1* ⎤ ⎡ −v2 sin ωt ⎤ ⎡ − x2* ⎤ 1 2 ⎡ 0 −1⎤ ⎡ x1 ⎤ ⎢ ⎥ +ω⎢ ⎢ ⎥=⎢ ⎥ ⎥ + ω ⎢1 0 ⎥ ⎢ * ⎥ = k ⎢ * 2 * * ⎥ ⎢⎣ v2* ⎥⎦ ⎣ v2 cos ωt ⎦ ⎢ ⎥⎦ x ⎣ ⎦ ⎢⎣ x2 ⎥⎦ cos sin cos + ω tx ω t ω tx 1 ⎣ 1 2 ⎣⎢ ⎦⎥ from which, we get, ⎡ − ( sin 2ωt ) / 2 − sin 2 ωt ⎤ ⎡ 0 −1⎤ ⎡⎣( ∇ * v * ) ⎤⎦ = k ⎢ ⎥ +ω⎢ ⎥, 2 ⎢⎣ cos ωt ( sin 2ωt ) / 2⎦⎥ ⎣1 0 ⎦ ⎡ − ( sin 2ωt ) / 2 ( cos 2ωt ) / 2 ⎤ ⎡ 0 −1 / 2 ⎤ ⎡ 0 −1⎤ [ D *] = k ⎢ cos 2ωt / 2 sin 2ωt / 2 ⎥ , [ W *] = k ⎢1 / 2 0 ⎥ + ω ⎢1 0 ⎥ . ) ( ) ⎦ ⎣ ⎦ ⎣ ⎦ ⎣(

(

(

(

)

) )

(c) For the Newtonian fluid, the stress field in the starred-frame is: ⎡ − p − μ k ( sin 2ωt ) μ k cos 2ωt ⎤ , [T *] = ⎢ μ k cos 2ωt ( ) − p + μ k ( sin 2ωt )⎥⎦ ⎣ where the indeterminate pressure p is time independent. Thus, ⎡ − ( cos 2ωt ) − ( sin 2ωt ) ⎤ ⎡ DT * ⎤ ⎢ Dt ⎥ = 2 μω k ⎢ − ( sin 2ωt ) ( cos 2ωt ) ⎥ , and ⎣ ⎦ ⎣ ⎦ p + μ k sin 2ωt ⎤ p + μ k sin 2ωt ⎤ ⎡ μ k cos 2ωt ⎡ μ k cos 2ωt ⎡ T* W* ⎤ = k ⎢ +ω⎢ ⎥ ⎣ ⎦ 2 − p + μ k ( sin 2ωt ) − μ k cos 2ωt ⎥ ⎣ − p + μ k sin 2ωt − μ k cos 2ωt ⎦ ⎣ ⎦ − μ k cos 2ωt p − μ k sin 2ωt ⎤ p − μ k sin 2ωt ⎤ ⎡ − μ k cos 2ωt ⎡ W*T* ⎤ = k ⎡⎢ +ω⎢ ⎥ ⎣ ⎦ 2 ⎣ − p − μ k sin 2ωt μ k cos 2ωt ⎦ μ k cos 2ωt ⎦⎥ ⎣ − p − μ k sin 2ωt Thus, μ k cos 2ωt μ k sin 2ωt ⎤ ⎡cos 2ωt sin 2ωt ⎤ ⎡ T* W* ⎤ − ⎡ W*T* ⎤ = k ⎢⎡ + 2μω k ⎢ ⎥ ⎥. ⎣ ⎦ ⎣ ⎦ ⎣ μ k sin 2ωt − μ k cos 2ωt ⎦ ⎣ sin 2ωt − cos 2ωt ⎦ o ⎡ cos 2ωt sin 2ωt ⎤ Thus, T * = ⎡ DT* / Dt ⎤ + ⎡ T* W* − W*T* ⎤ = μ k 2 ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ sin 2ωt − cos 2ωt ⎦ o T (d) [Q ] ⎡T ⎤ [Q ] ⎣ ⎦ 2 ⎡ cos ωt − sin ωt ⎤ ⎡ μ k =⎢ ⎢ ⎥ ⎣ sin ωt cos ωt ⎦ ⎣⎢ 0 Thus, we have

0 ⎤ ⎡ cos ωt sin ωt ⎤ ⎡ cos 2ωt sin 2ωt ⎤ = μk 2 ⎢ ⎥ ⎥ 2 ⎢ − sin ω t cos ω t ⎥ − μ k ⎦⎥ ⎣ ⎦ ⎣ sin 2ωt − cos 2ωt ⎦

o ⎡ T* ⎤ = [Q ] ⎡To ⎤ [Q ]T . ⎣ ⎦ ⎣ ⎦ ____________________________________________________________

8.28 Given the velocity field: v1 = −kx1 , v2 = kx2 , v3 = 0 . Obtain (a) the stress field for a second-order fluid (b) the co-rotational derivative of the stress tensor ------------------------------------------------------------------------


Lai et al, Introduction to Continuum Mechanics ⎡ −k Ans. (a) [∇v ] = ⎢⎢ 0 ⎢⎣ 0

0 0⎤ k 0 ⎥⎥ = [ D] , 0 0 ⎥⎦

⎡ −2k ⎢ [ A1 ] = [ 2D] = ⎢ 0 ⎢⎣ 0

0

[ W ] = [ 0] ,

0⎤ ⎡ −2k 2 ⎥ ⎢ 0 ⎥ → [ A1 ] = ⎢ 0 ⎢⎣ 0 0 ⎥⎦

2k 0

0 ⎤ ⎡ −2k 0 ⎥⎥ ⎢⎢ 0 0 ⎥⎦ ⎢⎣ 0

0 2k 0

2 0 ⎤ ⎡ 4k ⎢ 0 ⎥⎥ = ⎢ 0 ⎢ 0 ⎥⎦ ⎢ 0 ⎣

0 2k 0

0 4k 2 0

[ A 2 ] = ⎡⎣⎢ DA1 / Dt + A1 ( ∇v ) + ( ∇v )T A1 ⎤⎦⎥ = ⎡⎣⎢ A1 ( ∇v ) + ( ∇v )T A1 ⎤⎦⎥ ⎡ −2k = ⎢⎢ 0 ⎢⎣ 0

0 2k 0

0⎤ ⎡ −k 0 ⎥⎥ ⎢⎢ 0 0 ⎥⎦ ⎢⎣ 0

0 0⎤ ⎡ −k k 0 ⎥⎥ + ⎢⎢ 0 0 0 ⎥⎦ ⎢⎣ 0

0 0 ⎤ ⎡ −2 k k 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎦ ⎢⎣ 0

0 2k 0

2 0 ⎤ ⎡ 4k ⎢ 0⎥⎥ = ⎢ 0 ⎢ 0⎥⎦ ⎢ 0 ⎣

0 4k 2 0

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎦

The second-order fluid is defined by Eq.(8.18.6): T = − pI + μ1A1 + μ2 A12 + μ3 A 2 → ⎡ −2k [T] = − p [I ] + μ1 ⎢⎢ 0 ⎢⎣ 0

0 2k 0

⎡ 4k 2 0⎤ ⎢ 0 ⎥⎥ + μ2 ⎢ 0 ⎢ 0 ⎥⎦ ⎢⎣ 0

0 4k 2 0

⎡ 4k 2 0⎤ ⎥ ⎢ 0 ⎥ + μ3 ⎢ 0 ⎥ ⎢ 0⎥ ⎢⎣ 0 ⎦

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎦

0 4k 2 0

T11 = − p − 2μ1k + 4( μ2 + μ3 )k 2 , T22 = − p + 2μ1k + 4( μ2 + μ3 )k 2 , T33 = − p . To obtain the pressure p, we first calculate the acceleration: 2 ⎡ − k 0 0 ⎤ ⎡ − kx1 ⎤ ⎡⎢ k x1 ⎤⎥ [a] = [∂v / ∂t ] + [∇v ][ v ] = ⎢⎢ 0 k 0⎥⎥ ⎢⎢ kx2 ⎥⎥ = ⎢ k 2 x2 ⎥ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎢ 0 ⎥⎥ ⎣ ⎦ ∂Tij Equations of motion = ρ ai then give ∂x j −

∂p = ρ k 2 x1 , ∂x1

−

∂p = ρ k 2 x2 , ∂x2

−

∂p = 0, thus, ∂x2

p = − ρ k 2 ( x12 + x22 ) / 2 +C . o

(b) The co-rotational derivative of T : T = DT / Dt + TW - WT . Since W = 0 , ⎡1 0 0 ⎤ ∂Tij ⎤ ⎛ ∂p ∂p ⎞ ⎢ ⎡ o ⎤ ⎡⎛ DT ⎞ ⎤ ⎡ ∂Tij ⎥ T ⎥ = ⎢⎜ ⎟ ⎥ = ⎢ ∂x v1 + ∂x v2 ⎥ = ⎜ − ∂x v1 − ∂x v2 ⎟ ⎢ 0 1 0 ⎥ ij ⎦ ⎢⎝ Dt ⎠ ⎥ ⎣⎢ 2 1 2 ⎠ ⎢0 0 1 ⎥ ij ⎦ ⎣ 1 ⎦ ⎝ ⎣ ⎣ ⎦ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎛ ∂p ∂p ⎞ ⎢ ⎥ 2 2 2 2 ⎢ = k⎜ x1 − x2 ⎟ ⎢ 0 1 0 ⎥ = ρ k − k x1 + k x2 ⎢0 1 0 ⎥⎥ = ρ k v22 − v12 [ I ] ∂x2 ⎠ ⎝ ∂x1 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦

( )

(

)

(

____________________________________________________________


)

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎦

Lai et al, Introduction to Continuum Mechanics ( Show that the Lower Convected derivative of A1 is A 2 , i.e., A1 = A 2 .

8.29

------------------------------------------------------------------------Ans. From Eq.(8.19.22), ( o A1 = A1 + A1D + DA1 = ( DA1 / Dt + A1W − WA1 ) + A1D + DA1 = DA1 / Dt + A1 ( W + D ) + ( D − W ) A1 = DA1 / Dt + A1 ( ∇v ) + ( ∇v ) A1 = A 2 . T

____________________________________________________________ 8.30

The Reiner-Rivlin fluid is defined by the constitutive equation: T = − pI + S, S = φ1 ( I 2 , I 3 ) D + φ2 ( I 2 , I 3 ) D2

where Ii are the scalar invariants of D . Obtain the stress components for this fluid in a simple shearing flow. ------------------------------------------------------------------------------Ans. In a simple shearing flow, v1 = kx2 , v2 = v3 = 0 , ⎡k 2 / 4 0 0⎤ k / 2 0⎤ ⎡ 0 ⎢ ⎥ k2 [ D] = ⎢⎢ k / 2 0 0⎥⎥ , ⎡⎣ D2 ⎤⎦ = ⎢ 0 k 2 / 4 0⎥ , I 2 = − 4 , I3 = 0 ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 0⎥ ⎢⎣ 0 ⎦ ⎡k 2 / 4 0 0⎤ k / 2 0⎤ ⎡ 0 ⎢ ⎥ [T] = − p [I ] + φ1 k 2 / 4,0 ⎢⎢ k / 2 0 0⎥⎥ + φ2 k 2 / 4,0 ⎢ 0 k 2 / 4 0⎥ . ⎢ ⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 0⎥ ⎢⎣ 0 ⎦ ____________________________________________________________

(

)

(

)

N

8.31

The exponential of a tensor A is defined as: exp [ A ] = I + ∑

tensor, is exp [ A ] also objective?

1

1 n A . If A is an objective n!

------------------------------------------------------------------------------Ans. Yes. Because

( )

A* = Q ( t ) AQ T ( t ) → A*

( )

→ A*

N

2

= Q ( t ) AQ T ( t ) Q ( t ) AQ T ( t ) = Q ( t ) A 2QT ( t )

= Q ( t ) A N QT ( t )

( )

That is, A*

N

is objective for all N . As a consequence, exp [ A ] is objective.

____________________________________________________________ Why is it that the following constitutive equation is not acceptable: 8.32 T = − pI + S, S = α ( ∇v ) , where v is velocity and α is a constant ------------------------------------------------------------------------Ans. Because ∇v is not objective. ____________________________________________________________


Lai et al, Introduction to Continuum Mechanics 8.33 Let da and dA denote the differential area vectors at time τ and time t respectively. For an incompressible fluid, show that ⎡ D N da 2 / Dτ N ⎤ = dA ⋅ ⎡ D N Ct−1 / Dτ N ⎤ dA ≡ − dA ⋅ M N dA ⎣ ⎦τ =t ⎣ ⎦τ =t where da is the magnitude of da and the tensors M N are known as the White-Metzner tensors. -------------------------------------------------------------------------------Ans. From Eq. (3.27.12), we have, [note here dA is the reference area and da is the area at the

( ) dA . For an incompressible fluid, ( det F ) = 1 , so that da ⋅ da = ( F ) dA ⋅ ( F ) dA = dA ⋅ ( F )( F ) dA = dA ⋅ ( F F ) dA .

running time τ ], da = ( det F ) F −1

( )

da = F −1

T

dA ,

−1

T

T

−1

T

−1

−1

T

T

−1

That is, da 2 = dA ⋅ Ct−1dA . Thus, ⎡ D N Ct−1 ⎤ ⎡ D N Ct−1 ⎤ ⎡ D N da 2 ⎤ = ⋅ ≡ − ⋅ = − , where . d d d d A A A M A M ⎢ ⎢ ⎢ N N N ⎥ N ⎥ N ⎥ ⎣⎢ Dτ ⎦⎥τ =t ⎣⎢ Dτ ⎦⎥τ =t ⎣⎢ Dτ ⎦⎥τ =t ____________________________________________________________

8.34 (a) Verify that Oldroyd's lower convected derivatives of the identity tensor I are the Rivlin-Ericksen tensor A N . (b) Verify that Oldroyd upper derivatives of the identity tensor are the negative White-Metzner tensors [see Prob. 8.33 for the definition of White-Metzner tensor]. ------------------------------------------------------------------------------Ans. (a) The Nth lower convected derivative of T is given by ⎡ D N J L / Dτ N ⎤ , where J L (τ ) = FtT (τ ) T (τ ) Ft (τ ) . For T = I , ⎣ ⎦τ =t ⎡ D N Ct (τ ) ⎤ ⎡ DN JL ⎤ = J L (τ ) = FtT (τ ) Ft (τ ) = Ct (τ ) . Thus, ⎢ ⎢ ⎥ = AN ⎥ N N ⎣⎢ Dτ ⎦⎥τ =t ⎢⎣ Dτ ⎦⎥τ =t (b) The Nth upper convected derivative of T is given by

( )

⎡ D N JU / Dτ N ⎤ , where JU (τ ) = Ft−1 (τ ) T (τ ) Ft−1 ⎣ ⎦τ =t

T

(τ ) .

For T = I ,

⎡ D N Ct−1 (τ ) ⎤ ⎡ D N JU ⎤ = ⎢ ⎥ = −M N . N ⎥ N ⎥⎦τ =t ⎢⎣ Dτ ⎥⎦τ =t ⎢⎣ Dτ ____________________________________________________________

( )

JU (τ ) = Ft−1 (τ ) Ft−1

T

(τ ) = Ct−1 (τ ) . Thus, ⎢

( ( T Obtain the equation T = DT / Dt + T∇v + ( ∇v ) T , where T is the lower convected 8.35 derivative of T. ------------------------------------------------------------------------------Ans. By definition, the lower convected derivative is ⎡⎣ DJ L (τ ) / Dτ ⎤⎦ , where τ =t

J L (τ )

= FtT

(τ ) T (τ ) Ft (τ ) . Thus,

⎡⎣ DJ L (τ ) / Dτ ⎤⎦ = τ =t

⎡ DFtT ⎣

/ Dτ ⎤ T ( t ) Ft ( t ) ⎦τ =t

+ FtT ( t ) [ DT / Dτ ]τ =t Ft ( t ) + FtT ( t ) T ( t ) ⎡ DF T / Dτ ⎤ . ⎣ ⎦τ =t T T Now, ⎡ DFtT (τ ) / Dτ ⎤ = DFtT / Dt = ( DFt / Dt ) = ( ∇v ) [see Eq.(8.12.3)] and Ft ( t ) = I , ⎣ ⎦τ =t therefore,


Lai et al, Introduction to Continuum Mechanics ( ⎡ DJ (τ ) ⎤ DT T + T∇ v + ( ∇ v ) T . T=⎢ L ⎥ = ⎣ Dτ ⎦τ =t Dt ____________________________________________________________

8.36

Consider the following constitutive equation:

S + λ ( D*S / Dt ) = 2μ D, where ( D*S / Dt ) ≡ S + α ( DS + SD ) and S is co-rotational derivative of S . Obtain the shear stress function and the two normal stress functions for this fluid. ------------------------------------------------------------------------------Ans. With v1 = kx2 , v2 = v3 = 0 , the rate of deformation tensor and spin tensor are: k / 2 0⎤ k / 2 0⎤ ⎡ 0 ⎡ 0 ∂S ⎢ ⎥ ⎢ [ D] = ⎢ k / 2 0 0⎥ , [ W ] = ⎢ −k / 2 0 0⎥⎥ . Since the flow is steady, ∂t = 0 . ⎢⎣ 0 ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎥⎦ o

o

The co-rotational derivative is, for symmetric S: S = SW − WS = SW + ( SW ) . Now, o

⎡ − S12 S11 0 ⎤ ⎡ − S12 − S22 − S32 ⎤ k⎢ k⎢ T ⎥ [SW ] = 2 ⎢− S22 S21 0⎥ , [SW ] = 2 ⎢ S11 S21 S31 ⎥⎥ , ⎢⎣ − S32 S31 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ S11 − S22 − S32 ⎤ ⎡ −2S12 ⎡ 2S12 o k k ⎢ ⎥ ⎡S ⎤ = S − S22 2S21 S31 ⎥ , [SD] + [ DS ] = ⎢⎢ S11 + S22 ⎣ ⎦ 2 ⎢ 11 2 ⎢⎣ − S32 ⎢⎣ S32 0 ⎥⎦ S31

T

S11 + S22 2S21 S31

S32 ⎤ S31 ⎥⎥ . 0 ⎥⎦

Thus, D*S / Dt ≡ S + α ( DS + SD ) gives o

⎡ 2 (α − 1) S12 ⎡ D*S ⎤ k ⎢ ⎢ Dt ⎥ = 2 ⎢(1 + α ) S11 + (α − 1) S22 ⎣ ⎦ ⎢ (α − 1) S32 ⎣ Therefore, S + λ

(1 + α ) S11 + (α − 1) S22 (α − 1) S32 ⎤ 2 (1 + α ) S21 (1 + α ) S31 ⎥⎥ ⎥ 0 (1 + α ) S31 ⎦

D*S = 2μ D → Dt

⎛ λk ⎞ S11 + λ k (α − 1) S12 = 0 (i) , S12 + ⎜ ⎟ ⎣⎡(1 + α ) S11 + (α − 1) S22 ⎦⎤ = μ k (ii) , ⎝ 2 ⎠ λk S13 + (α − 1) S23 = 0 (iii), S22 + λ k (1 + α ) S12 = 0 (iv), 2 λk S23 + (1 + α ) S13 = 0 (v), S33 =0 (vi) . 2 Now, (iii) , (v) and (vi) give S13 = S23 = S33 = 0 . Eq. (i) gives S11 = λ k (1 − α ) S12 , Eq.(iv) gives S22 = −λ k (1 + α ) S12 , thus, with

(

)

A(k ) ≡ 1 + 1 − α 2 ( λ k ) , we have, 2

S12 = μ k / A(k ) , S11 = λμ k 2 (1 − α ) / A( k ) , S22 = −λμ k 2 (1 + α ) / A(k ) The shear stress function is S12 = μ k / A(k ) . The normal stress functions are:

σ1 ≡ S11 − S22 = 2λμ k 2 / A(k ), σ 2 = S22 − S33 = −λμ k 2 (1 + α ) / A( k ) .


Lai et al, Introduction to Continuum Mechanics ____________________________________________________________ Obtain the apparent viscosity and the normal stress functions for the Oldroyd 3-constant 8.37 fluid [see (C) of Section 8.20]. ------------------------------------------------------------------------------Ans. For the simple shearing flow, k / 2 0⎤ k / 2 0⎤ ⎡ 0 ⎡ 0 ⎢ ⎥ ⎢ [ D] = ⎢ k / 2 0 0⎥ , [ W ] = ⎢ −k / 2 0 0⎥⎥ . ⎢⎣ 0 ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎥⎦ S11 − S22 − S32 ⎤ ⎡ −2S12 ⎢ 2S21 S31 ⎥⎥ , S = [ 0] + [SW ] − [ WS ] = ( k / 2 ) ⎢ S11 − S22 ⎢⎣ − S32 S31 0 ⎥⎦ S11 + S22 S23 ⎤ ⎡ 2S12 ⎢ [SD] + [ DS ] = ( k / 2 ) ⎢ S11 + S22 2S12 S13 ⎥⎥ , ⎢⎣ S23 S13 0 ⎥⎦ ⎡ −4S12 −2S22 −2 S23 ⎤ o Sˆ = S − ( SD + DS ) = ( k / 2 ) ⎢⎢ −2S22 0 0 ⎥⎥ , ⎢⎣ −2S23 0 0 ⎥⎦ o

⎡ −k 2 / 2 0 0⎤ ⎢ ⎥ o k2 D = [ 0] + [ DW ] − [ WD] = ⎢ 0 k 2 / 2 0 ⎥ , DD = 4 ⎢ ⎥ 0 0⎥ ⎢⎣ 0 ⎦ 2 ⎡ −k ⎡ −2λ2 μ k 2 0 0⎤ o ⎢ ⎥ ˆ = D − 2D 2 = 0 0 0 , 2 μ D + λ D ˆ = ⎢ μk D ⎢ ⎥ ⎢ 2 ⎢ ⎢ 0 0 0⎥ 0 ⎣ ⎦ ⎣ ⎡ S11 − 2k λ1S12 S12 − k λ1S22 S13 − k λ1S23 ⎤ ⎥ S + λ1Sˆ = ⎢⎢ S12 − k λ1S22 S22 S23 ⎥ ⎢⎣ S13 − k λ1S23 ⎥⎦ S23 S33

(

(

)

⎡1 0 0 ⎤ ⎢0 1 0 ⎥ ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦

μ k 0⎤ 0 0

⎥ 0⎥ 0⎥ ⎦

)

ˆ → S + λ1Sˆ = 2 μ D + λ2 D ⎡ S11 − 2k λ1S12 ⎢ S − kλ S 1 22 ⎢ 12 ⎢⎣ S13 − k λ1S23 Thus,

S12 − k λ1S22 S22 S23

S13 − k λ1S23 ⎤ ⎡ −2λ2 μ k 2 ⎥ = ⎢ μk S23 ⎥ ⎢ ⎥⎦ ⎢ S33 0 ⎣

μ k 0⎤ 0 0

⎥ 0⎥ 0⎥ ⎦

S22 = S23 = S33 = S13 = 0, S12 = μ k , S11 − 2k λ1S12 = −2λ2 μ k 2 , so that, we have, S12 = μ k , S11 = 2μ k 2 ( λ1 − λ2 ) , all other Sij = 0 . The apparent viscosity is

η ( k ) = S12 / k = μ , σ1 =T11 − T22 = 2 μ k 2 ( λ1 − λ2 ) , σ 1 =T22 − T33 = 0 . ____________________________________________________________


Lai et al, Introduction to Continuum Mechanics 8.38 Obtain the apparent viscosity and the normal stress functions for the Oldroyd 4-constant fluid [see (D) of Section 8.20] ------------------------------------------------------------------------------Ans. For the simple shearing flow k / 2 0⎤ k / 2 0⎤ S11 − S22 − S23 ⎤ ⎡ −2S12 ⎡ 0 ⎡ 0 o ⎛ k ⎞⎢ ⎢ ⎥ ⎢ ⎥ S13 ⎥⎥ , [ D] = ⎢ k / 2 0 0⎥ , [ W ] = ⎢ −k / 2 0 0⎥ , S = ⎜ 2 ⎟ ⎢ S11 − S22 2S12 ⎝ ⎠ ⎢⎣ − S32 ⎢⎣ 0 ⎢⎣ 0 0 0 ⎥⎦ 0 0 ⎥⎦ S31 0 ⎥⎦ ⎡ −4S12 ˆS = So − ( SD + DS ) = ( k / 2 ) ⎢ −2S 22 ⎢ ⎢⎣ −2S23

−2S22 0 0

⎡ −k 2 / 2 0 ⎢ o k2 / 2 D = [ 0] + [ DW ] − [ WD] = ⎢ 0 ⎢ 0 ⎢⎣ 0 ⎡ −1 0 0 ⎤ 2 k2 ⎢ ⎥− k 2 ˆ D = D − 2D = 0 1 0 ⎥ 2 2 ⎢ ⎢⎣ 0 0 0 ⎥⎦ o

⎡1 0 ⎢0 1 ⎢ ⎢⎣0 0

−2 S23 ⎤ 0 ⎥⎥ , 0 ⎥⎦ 0⎤ ⎥ 0 ⎥ , DD = ⎥ 0⎥ ⎦ 0⎤ ⎡ −k 2 ⎢ 0 ⎥⎥ = ⎢ 0 0 ⎥⎦ ⎢ 0 ⎣

⎡1 0 0 ⎤ k2 ⎢ 0 1 0 ⎥⎥ , 4 ⎢ ⎢⎣0 0 0 ⎥⎦ 0 0⎤ ⎥ 0 0⎥ , 0 0⎥ ⎦

k / 2 0⎤ ⎡ 0 ⎢ μo ( trS ) [ D] = μo ( S11 + S22 + S33 ) ⎢ k / 2 0 0 ⎥⎥ . ⎢⎣ 0 0 0 ⎥⎦

(

)

ˆ → S + λ1Sˆ + μo ( trS ) D = 2μ D + λ2 D

⎡ S11 − 2k λ1S12 S12 − k λ1S22 + μo k ( S11 + S22 + S33 ) / 2 S13 − k λ1S23 ⎤ ⎢ ⎥ S22 S23 ⎢ S12 − k λ1S22 + μo k ( S11 + S22 + S33 ) / 2 ⎥ ⎢ ⎥ S13 − k λ1S23 S23 S33 ⎣ ⎦ ⎡ −2λ2 μ k 2 ⎢ = ⎢ μk ⎢ 0 ⎣

μ k 0⎤ 0 0

⎥ 0⎥ . 0⎥ ⎦

hus, S22 = S23 = S33 = S13 = 0, S11 − 2k λ1S12 = − 2λ2 μ k 2 , S12 + μo kS11 / 2 = μ k From which, we get, with B (k ) ≡ (1 + λ1μo k 2 ) ,

(

)

S11 =2μ k 2 ( λ1 − λ2 ) / B (k ), S12 = μ k 1 + λ2 μo k 2 / B (k ) . Thus, the apparent viscosity is: η ( k ) = S12 / k = μ (1 + λ2 μo k 2 ) / B (k ) . Normal stress functions are: σ1 =T11 − T22 = 2μ k 2 (λ1 − λ2 ) / B( k ), σ 2 =T22 − T33 = 0 . ____________________________________________________________


T

Lai et al, Introduction to Continuum Mechanics ⎡ −1 0 0 ⎤ ⎡0 1 0⎤ ⎢ ⎥ Given [Q ] = ⎢ 0 1 0 ⎥ and [ N ] = ⎢⎢ 0 0 0 ⎥⎥ and ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ {ni } {ni }

8.39

(

)

A1 = k N + N T and A 2 = 2k 2 N T N . (a) Verify that QA1QT = − A1 and QA 2QT = A 2 . (b) From

(

)

T = − pI + f ( A1 , A 2 ) and Qf ( A1 , A 2 ) QT = f QA1QT , QA 2QT , show that QT ( k ) Q T = T ( − k ) and (c) From the results of part (b), show that the viscometric functions have the properties: S ( k ) = −S ( − k ) , σ 1 ( k ) = σ 1 ( − k ) , σ 2 ( k ) = σ 2 ( − k ) . ------------------------------------------------------------------------------⎡ 0 k 0⎤ ⎡0 0 0⎤ ⎡0 1 0⎤ ⎡0 0 0⎤ ⎢ ⎥ ⎢ ⎥ T 2 Ans. (a) [ A1 ] = k ⎡ N + N ⎤ = ⎢ k 0 0 ⎥ and A 2 = 2k ⎢⎢1 0 0 ⎥⎥ ⎢⎢ 0 0 0 ⎥⎥ = ⎢ 0 2k 2 0 ⎥ ⎣ ⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢ 0 0 0 ⎥ ⎣ ⎦ ⎡ −1 0 0 ⎤ ⎡ 0 k 0 ⎤ ⎡ −1 0 0 ⎤ ⎡ 0 k 0⎤ ⎡QA1QT ⎤ = ⎢ 0 1 0 ⎥ ⎢ k 0 0 ⎥ ⎢ 0 1 0 ⎥ = − ⎢ k 0 0 ⎥ = − [ A1 ] ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦

(

)

⎡ −1 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎡ −1 0 0 ⎤ ⎡ 0 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎡QA 2Q = ⎢ 0 1 0 ⎥ ⎢ 0 2k 2 0 ⎥ ⎢ 0 1 0 ⎥ = ⎢ 0 2k 2 0 ⎥ = [ A 2 ] ⎥ ⎢ ⎥ ⎣ ⎦ ⎢ ⎢⎣ 0 0 1 ⎥⎦ ⎢ 0 0 0 ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢ 0 0 0 ⎥ ⎣ ⎦ ⎣ ⎦ (b) QT ( k ) QT = Q ⎡⎣ − pI + f ( A1 ,A 2 ) ⎤⎦ QT = − pI + Qf ( A1 ,A 2 ) QT T⎤

(

)

= − pI + f QA1QT ,QA 2QT = − pI + f ( − A1 ( k ) ,A 2 ( k ) ) . Now, A1 ( −k ) = − A1 ( k ) and A 2 ( − k ) = A 2 ( k ) , thus,.

QT ( k ) QT = − pI + f ( A1 ( − k ) ,A 2 ( − k ) ) and QT ( − k ) QT = − pI + f ( A1 ( k ) ,A 2 ( k ) ) .

That is, QT ( −k ) QT = T ( k ) and QT ( k ) QT = T ( − k ) ⎡ −1 0 0 ⎤ ⎡T11 T12 T13 ⎤ ⎡ −1 0 0 ⎤ ⎡ T11 −T12 T⎤ ⎢ ⎡ (c) [Q ] ⎡⎣T ( k ) ⎤⎦ Q = ⎢ 0 1 0 ⎥⎥ ⎢⎢T21 T22 T23 ⎥⎥ ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ −T21 T22 ⎣ ⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣T31 T32 T33 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ −T31 T32

−T13 ⎤ T23 ⎥⎥ T33 ⎥⎦

QT ( − k ) Q T = T ( k ) → T11 ( − k ) = T11 ( k ) , T22 ( − k ) = T22 ( k ) , T33 ( − k ) = T33 ( k )

−T12 ( − k )12 = T12 ( k ) , − T13 ( − k ) = T13 ( k ) , − T23 ( − k ) = T23 ( k ) . Thus,

σ1 ( k ) = σ1 ( −k ) , σ 2 ( k ) = σ 2 ( − k ) , S ( k ) = − S ( − k ) . [Note, in viscometric flow, T13 = T23 = 0 ]. ____________________________________________________________ 8.40

For the velocity field given in example 8.21.2, i.e., vr = 0, vθ = 0, vz = v ( r ) , (a) obtain

the stress components in terms of the shear stress function S ( k ) and the normal stress functions

σ1 ( k ) and σ 2 ( k ) , where k = dv / dr , (b) obtain the following velocity distribution for the


Lai et al, Introduction to Continuum Mechanics Poiseuille flow under a pressure gradient of ( − f ): v ( r ) = ∫ γ ( fr / 2 )dr , where γ is the inverse R

r

(

)

shear stress function, and (c)obtain the relation γ ( Rf / 2 ) = ⎡1 / (π R3 f 2 ) ⎤ ∂ f 3Q / ∂f . ⎣ ⎦ ------------------------------------------------------------------------------Ans. (a) In example 8.21.2, we see that the velocity field vr = 0, vθ = 0, vz = v ( r ) describes a viscometric flow with the nonzero Rivlin-Ericksen tensors given by 0 0⎤ ⎡0 ⎡ 0 k ( r ) 0⎤ ⎢ ⎥ ⎢ ⎥ [ A1 ] = ⎢ k ( r ) 0 0⎥ , [ A 2 ] = ⎢0 2k 2 ( r ) 0⎥ , ⎢0 ⎢ 0 0 0 ⎥⎦ 0 0⎥ ⎣ ⎣ ⎦ ni ni

where n1 = e z , n 2 = er , n3 = eθ and k ( r ) = dv / dr (see Example 8.10.2, but, note the differences

in the order of bases). Thus the stress components with respect to the basis {ni } are given by (See section 8.22): S z r = τ (k ), S z z − Sr r = σ1 ( k ) , Sr r − Sθ θ = σ 2 ( k ) , S z θ = Sr θ = 0 . (b) With Sij depending only on r , the equations of motion become:

∂S rr S rr − Sθθ ∂p ∂p ∂p 1 ∂ + − = 0 (i ), = 0 (ii) , ( rSrz ) − = 0 (iii) ∂r ∂r ∂θ ∂z r r ∂r ∂ ⎛ ∂p ⎞ ∂ ⎛ ∂p ⎞ ∂ ⎛ ∂p ⎞ Eq. (i) gives = 0 , Eq. (ii) gives =0 = 0 and Eq. (iii) gives ⎜ ⎟ ⎜ ⎟ ∂r ⎝ ∂z ⎠ ∂z ⎜⎝ ∂z ⎟⎠ ∂θ ⎝ ∂z ⎠ Thus, ∂p / ∂z = a constant ≡ − f . Eq. (iii) becomes 1 ∂ ∂ fr C ( rSrz ) = − f → ( rSrz ) = − fr → Srz = − + . Since Srz must be finite at r = 0 , thus, r ∂r ∂r 2 r C = 0 and Srz = − fr / 2 . Now, Srz = τ (k ) where τ ( k ) is the shear stress function and k = dv / dr . −1

Thus, τ ( k ) = − fr / 2 . Inverting this equation, we have, k = τ ( − fr / 2 ) ≡ γ ( − fr / 2 ) . Since τ (k ) is an odd function of k , therefore, γ is also an odd function of k , so that k = −γ ( fr / 2 ) → dv / dr = −γ ( fr / 2 ) → dv = −γ ( fr / 2 ) dr .

Thus, v ( R ) − v ( r ) = − ∫ γ ( fr / 2 )dr . Since v ( R ) = 0 , therefore, v ( r ) = ∫ γ ( fr / 2 )dr . R

R

r

r

(c) The volume discharge is given by Q = ∫ v ( r ) 2π rdr . Therefore, R

0

R R R R R dv ⎫ dv ⎧ Q = π ∫ v ( r ) dr 2 = π ⎨ ⎡v ( r ) r 2 ⎤ − ∫ r 2 dr ⎬ = −π ∫ r 2 dr = π ∫ r 2γ ( fr / 2 )dr ⎣ ⎦ 0 0 0 0 0 dr ⎭ dr ⎩

Thus, Q / π = ∫ r 2γ ( fr / 2 )dr . Let fr / 2 ≡ s → dr ≡ 2ds / f and r 2 = 4s 2 / f 2 , then , R

0

Q/π = ∫

R

r =0

r γ ( fr / 2 ) dr = ∫ 2

R f /2

s =0

( 8s

2

)

/ f 3 γ ( s )ds → f 3Q / π = ∫

R f /2

s =0

8s 2γ ( s )ds .

Differentiating the last equation with respect to f , we obtain

(

)

3 ∂ ⎛ Rf 1∂ f Q = ⎜ ∂f ⎝ 2 π ∂f

⎞ ⎧⎪ ⎛ Rf ⎞ ⎛ Rf ⎟ ⎨8 ⎜ 2 ⎟ γ ⎜ 2 ⎠ ⎩⎪ ⎝ ⎠ ⎝ 2

(

)

⎞ ⎫⎪ ⎛ R ⎞ ⎛ Rf ⎞ ⎛ Rf ⎞ 3 2 ⎛ Rf ⎞ ⎟⎬ = ⎜ 2 ⎟ 8 ⎜ 2 ⎟ γ ⎜ 2 ⎟ = R f γ ⎜ 2 ⎟ . ⎠ ⎭⎪ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2

3 1 ∂ f Q ⎛ Rf ⎞ Thus, γ ⎜ . ⎟= ⎝ 2 ⎠ π R3 f 2 ∂f


Lai et al, Introduction to Continuum Mechanics ____________________________________________________________


Solutions Manual Continuum Mechanics Lai 4th Edittion

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