Solution Manual For Applied Continuum Mechanics for Thermo-Fluids, 1st Edition By Jafar Ghazanfaria

Jafar Ghazanfarian

Solution manual: Applied Continuum Mechanics for Thermo-fluids

Contents Chapter 1 ■ Cartesian Tensor Formalism and Dyadic Algebra 1.1

EXERCISES

Chapter 2 ■ Preliminary Concepts 2.1

EXERCISES

1 1

7 7

iii

CHAPTER

1

Cartesian Tensor Formalism and Dyadic Algebra

1.1

EXERCISES

1.1 Prove that in three-dimensional space, the double-dot product of a symmetric tensor and an anti-symmetric tensor equals 0.

Solution: Similar to the two-dimensional case explained in the text, in three-dimensional space, αij βji

= α11 β11 + α12 β21 + α13 β31 + α21 β12 + α22 β22 + α23 β32 + α31 β13 + α32 β23 + α33 β33 =

0 + α12 β21 − α12 β21 + 0

+ α13 β31 − α13 β31 + α23 β32 − α23 β32 + 0 =

0

1.2 Obtain the following relation for the gradient of the dot product of two vectors. ∇(A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇)B + (B · ∇)A Solution: The first term in the RHS in the ith direction (Ii ) is

∂Bm ∂Bm Ii = ϵijk Aj (∇ × B)k = ϵijk Aj ϵklm = ϵijk ϵklm Aj ∂xl ∂xl

1

2 ■ Solution manual: Applied Continuum Mechanics for Thermo-fluids Using the identity ϵijk ϵklm = δil δjm − δim δjl , Ii = δil δjm Aj

∂Bm ∂Bm ∂Bj ∂Bi ∂Bj −δim δjl Aj = Aj −Aj = Aj −(A·∇)Bi ∂xl ∂xl ∂xi ∂xj ∂xi

Similarly for the second term in the RHS (IIi ) IIi = Bj

∂Aj − (B · ∇)Ai ∂xi

The sum of two terms yields Ii + IIi = Aj

∂Bj ∂Aj + Bj − (A · ∇)Bi − (B · ∇)Ai ∂xi ∂xi

Finally, the ith component of the RHS equals the ith component of the LHS, ∂(Aj Bj ) = ∇(A · B) (RHS)i = ∂xi 1.3 Obtain the following relation for the curl of the product of a scalar and the gradient of another scalar. ∇ × (f ∇g) = ∇f × ∇g Solution: First we need to prove that ∇ × (f ∇g) = ∇f × ∇g + f ∇ × ∇g The first term in the RHS equals I = ϵijk

∂f ∂g ∂xj ∂xk

On the other hand, the LHS can be simplified to

∂g ∂2g ∂ ∂f ∂g LHS = ϵijk f = ϵijk + ϵijk f ∂xj ∂xk ∂xj ∂xk ∂xj ∂xk The last term vanishes because it is the product of two antisymmetric 2 g and symmetric tensors. ∂x∂j ∂x is symmetric referring to the Schwarz’ k theorem.

So we proved that LHS = I + 0 = ∇f × ∇g 1.4 If F is a second-order anti-symmetric tensor, and M is a vector equal to 0.5ϵijk Fjk , find the product ϵpqi Mi .

Solution: The given statement reads ϵpqi Mi = 0.5ϵijk ϵpqi Fjk = 0.5 (δjp δkq − δjq δpk ) Fjk Using the identity of product of two permutation symbols, ϵpqi Mi = 0.5(Fpq − Fqp ) = 0.5(Fpq + Fpq ) = Fpq

Cartesian Tensor Formalism and Dyadic Algebra ■ 3 1.5 Write down the velocity gradient tensor and obtain its dual vector. Repeat this procedure to obtain the dual vector of the anti-symmetric part of the velocity gradient tensor. Show that both results are identical.

Solution: Start from the two-dimensional case. The third component of the dual vector is d3 = ϵ312 L21 + ϵ321 L12 or anti d3 = ϵ312 Lanti 21 + ϵ321 L12

which is equal to the z-component of the vorticity vector ωz =

∂u ∂v − ∂x ∂y

1.6 Use the conclusion of the previous exercise to prove ω × u = 2Lanti · u, where u is velocity (a vector), ω is the curl of u (vorticity), and ω × u is the Lamb vector.

Solution: The Lamb vector can be simplified to (ω × u)i = ϵijk ωj uk in which the vorticity vector equals the dual-vector of the rotation tensor (ω)j = ϵjmn wnm Combining two relations (ω × u)i = ϵijk ϵjmn wnm uk Using the identity of product of two permutation symbols (ω × u)i = (δkm δin − δkn δim )wnm uk = wik uk − wki uk w is anti-symmetric. Hence (ω × u)i = wik uk + wik uk = 2wik uk = (2w · u)i So, ω × u = 2w · u ⇒ ω × u = 2Lanti · u 1.7 If A is a second-order constant tensor and x is a vector in threedimensional space, compute the result of ∇ · (Axx).

Solution: using the tensor notation ∇ · (Axx) =

∂ ∂xk ∂xj (Aij xj xk ) = Aij xj + Aij xk ∂xk ∂xk ∂xk ∂x

j k We use two relations ∂x ∂xk = δkk = 3 and ∂xk = δjk ,

∇ · (Axx) = Aij xj × 3 + Aij xj = 4Aij xj

4 ■ Solution manual: Applied Continuum Mechanics for Thermo-fluids 1.8 If A is a second-order symmetric constant tensor and x is a vector, compute the result of (Aij xi xj ),k .

Solution: (Aij xi xj ),k = Aij xj

∂xi ∂xj + Aij xi ∂xk ∂xk

∂x

∂xi = δik and ∂xkj = δjk , We use two relations ∂x k

(Aij xi xj ),k = Akj xj + Aik xi = Akj xj + Aki xi = 2Akj xj 1.9 Compute the determinant of dyadic product of two vectors |a ⊗ b|.

Solution: Using the definition of determinant of a matrix det(Aij ) = ϵijk Aai A2j A3k for the dyadic product |a ⊗ b| = ϵijk (a1 a2 a3 )(bi bj bk ) = 0 The last expression is zero since it is the product of an anti-symmetric (permutation tensor) and a symmetric tensor (bi bj bk ).

H 1.10 Compute S xi nj dS where nj is the normal unit vector of S which bounds the volume ∀.

Solution: Using the divergence theorem, I Z Z ∂xi xi nj dS = ∇ ⊗ xd∀ = d∀ = δij ∀ ∂xj S ∂A

δij if Aij = xi xj .

1.11 Compute ∂xij k Solution: Replace Aij ∂Aij ∂Aii δij = = 2xk ∂xk ∂xk 1.12 If Sij + 3Skk δim δjm = Tij , then obtain a relation for Sij .

Solution: First we need to find Sii , i = j ⇒ Sii + 3Skk δii = Tii ⇒ Tii = 10Sii Substitute in the main relation Sij = Tij − 3Skk δij ⇒ Sij = Tij −

3 δij Tkk 10

Cartesian Tensor Formalism and Dyadic Algebra ■ 5 1.13 If Aij is an anti-symmetric constant tensor and ci = ϵijk Ajk is a vector, then compute Z ϵpqi ci d∀ Solution: Substitute in the integral and use the identity of product of two permutation symbols Z Z ϵpqi ϵijk Ajk d∀ = (δpj δqk − δpk δqj )Ajk d∀ hence Z

Z ϵpqi ϵijk Ajk d∀ =

Z (Apq − Aqp )d∀ =

2Apq ∀ = 2Apq ∀

CHAPTER

2

Preliminary Concepts

2.1

EXERCISES

2.1 If eigenvalues and eigenvectors of a given transformation (A) is known, compute the eigenvalues and eigenvectors of a new transformation which is obtained by applying the transformation five successive times A5 .

Solution: When applying the transformation we should use the relation Ax = λx (AAAAA)x

=

(AAAA)λx

=

(AAA)λ2 x = (AA)λ3 x

= Aλ4 x = λ5 x On the other hand, during each transformation the input vector is mapped on a same direction and just the length of the output vector changes. So the eigenvalues equal λ5 and eigenvectors remain unchanged.

2.2 Prove that the advection part of the material derivative can be written as the sum of the Lamb vector and half of the gradient of u · u.

Solution: By adding and subtracting uj,i uj in the advection part of the material derivative (u · ∇u)i = uj ui,j − uj,i uj + uj,i uj = uj (ui,j − uj,i ) + uj uj,i Using the relation derived in exercise problem 1.6, (u · ∇u)i = uj (2wij ) + 0.5(uj uj ),i So u · ∇u = 2w · u + 0.5∇(u · u) = ω × u + 0.5∇(u · u) 2.3 Find the traction vector and the normal stress at the point p located on

7

8 ■ Solution manual: Applied Continuum Mechanics for Thermo-fluids the plane 2x1 − 2x2 − x3 = 0, if the stress tensor at that point is   1 2 3  2 4 6  3 6 1 Solution: First, we need to compute the unit normal vector of the plane which is 1 √ (2, −2, −1)T 9 The traction vector is the product of the stress tensor and the unit normal vector. Hence 1 T = √ (−5, −10, −7)T 9 The normal stress can be obtained by the dot product of the traction vector and the unit normal vector σn =

17 1 (−10 + 20 + 7) = 9 9

2.4 Compute the curl of a shear flow with the velocity field (ky, 0, 0)T .

Then write down the given velocity as a sum of k2 (y, x)T and k2 (y, −x)T .

Compute the gradients of two new velocity fields by decomposing into symmetric and anti-symmetric parts.

Solution: The curl of the velocity field is ω = ∇ × u = −kez By decomposing the given vector field u = u1 + u2 =

k k (y, x)T + (y, −x)T 2 2

Gradients of u1 (no symmetric part) and u2 (no symmetric part) respectively are

0 k2 +0 k 0 2

0+

0 − k2

k 2

0

2.5 If axes of coordinates are principle directions of the stress tensor, compute the normal stress on a plain with unit normal vector √13 (1, 1, −1)T as a function of the first invariant of the stress tensor.

Preliminary Concepts ■ 9 Solution: The stress tensor in this case is   σ1 0 0  0 σ2 0  0 0 σ3 The first invariant of the tensor is σ1 + σ2 + σ3 . The traction vector on the given plain is 1 √ (σ1 , σ2 , −σ3 )T 3 The normal stress corresponding to this traction vector becomes σn =

I1 3

2.6 If the velocity field in the Lagrangian viewpoint is ui = α(Xi + 1) where α is a positive constant, compute the velocity components in Eulerian description.

Solution: Start from the definition ui =

dxi = α(Xi + 1) ⇒ xi = α(Xi + 1)t + c dt

The constant can be computed using xi = Xi at t = 0. xi = (αt + 1)Xi + αt ⇒ Xi =

xi − αt αt + 1

Using the given velocity field

xi − αt α(xi + 1) ui = α +1 = αt + 1 αt + 1 2.7 If the stress tensor equals 

x1 −x2  −x2 x3 f (x1 ) 0

 f (x1 ) 0  2x3

and the body is in equilibrium without any external body force, compute the function f (x1 ).

Solution: The divergence of the stress tensor should be zero, ∇ · σ = (1 − 1 + 0, 0 + 0 + 0,

∂f (x1 ) + 2)T = 0 ∂x1

Solving the differential equation yields f (x1 ) = −2x1 + c

10 ■ Solution manual: Applied Continuum Mechanics for Thermo-fluids 2.8 If the product of eigenvalues of the following tensor equals 120, find the value or values of x.   6 x x  x 8 4  x 4 9 Solution: The determinant of the matrix equals the product of eigenvalues, √ 6(72 − 16) + x(4x − 9x) + x(4x − 8x) = 120 ⇒ x = ± 24 2.9 Obtain the time-rate of change of temperature field T = x1 + tx2 when the displacement in Lagrangian viewpoint is x1

= X1 + AtX2

x2

= X2 − AtX1

x3

= X3

Solution: First obtain the velocity components u1

= AX2

u2

= −AX1

u3

=

0

Then use the definition of the material derivative and replace x2 = X2 − AtX1 DT = x2 + (AX2 ) × 1 + (−AX1 )t + 0 = X1 (−2At) + X2 (1 + A) Dt 2.10 Obtain magnitudes of σ1 and σ2 , if the principle stresses of the following stress tensor are 4, 3, -2 MPa.   σ1 −3 0  −3 σ2 0  M P a 0 0 3 Solution: Two equations with two unknowns will be obtained. The trace of the matrix equals the sum of eigenvalues σ1 + σ2 + 3 = 4 + 3 − 2 The determinant of the matrix should be equal to the product of eigenvalues 3(σ1 σ2 − 9) = 4 × 3 × −2 So, σ1 = 1M pa and σ2 = 1M P a.

Preliminary Concepts ■ 11 2.11 A displacement field in Lagrangian viewpoint is given. x1

= λ1 X1

x2

= −λ3 X3

x3

= λ2 X2

Compute the Jacobian of the transformation.

Solution: The Jacobian matrix is  λ1 0 0 J= 0 0 λ2

 0 −λ3  0

The determinant of the matrix is λ1 λ2 λ3 .

2.12 Compute unknowns a, b, c, if the stress vector components on the octahedral plane for the following stress tensor are zero.   1 a b  a 1 c  b c 1 Solution: The unit normal vector of the octahedral plane is √13 (1, 1, 1), so 1+a+b

=

0

1+a+c =

0

1+b+c =

0

After solving the system of equations a = b = c = − 21 .

2.13 Principle stresses of a stress field are 10, 20, 30 MPa. Compute the equivalent stress.

Solution: The first and the second invariants of the stress field are I1 = 10 + 20 + 30 = 60M P a I2 = 10 × 20 + 20 × 30 + 10 × 30 = 1100M P a The second invariant of the deviatoric part of the stress tensor is I2D = 1100 − Then σE =

√

−3I2D =

√

3600 = −100M P a 3

300 MPa.