Page 328 - Intro to Tensor Calculus

322

I 22. In generalized orthogonal coordinates, show that the physical components of the rate of deformation
stress can be written, for i 6= j

h i ∂ v(i) h j ∂ v(j)
σ(ij)= µ ∗ + , no summation,
h j ∂x j h i h i ∂x i h j
and for i 6= j 6= k

1 ∂v(i) 1 ∂h i 1 ∂h i
σ(ii)= −p +2µ ∗ + v(j) + v(k)
h i ∂x i h i h j ∂x j h i h k ∂x k

λ ∗ ∂ ∂ ∂
+ {h 2 h 3 v(1)} + {h 1 h 3 v(2)} + {h 1h 2 v(3)} , no summation
h 1 h 2 h 3 ∂x 1 ∂x 2 ∂x 3
I 23. Find the physical components for the rate of deformation stress in Cartesian coordinates. Hint: See
problem 22.

I 24. Find the physical components for the rate of deformations stress in cylindrical coordinates. Hint: See
problem 22.
1
I 25. Verify the Navier-Stokes equations for an incompressible fluid can be written ˙v i = − p ,i +νv i,mm +b i
%
where ν = µ ∗ is called the kinematic viscosity.
%
I 26. Verify the Navier-Stokes equations for a compressible fluid with zero bulk viscosity can be written
1 ν µ ∗
˙ v i = − p ,i + v m,mi + νv i,mm + b i with ν = % the kinematic viscosity.
% 3
I 27. The constitutive equation for a certain non-Newtonian Stokesian fluid is σ ij = −pδ ij +βD ij +γD ik D kj .
Assume that β and γ are constants (a) Verify that σ ij,j = −p ,i + βD ij,j + γ(D ik D kj,j + D ik,j D kj )
(b) Write out the Cauchy equations of motion in Cartesian coordinates. (See page 236).
I 28. Let the constitutive equations relating stress and strain for a solid material take into account thermal
1+ ν ν
stresses due to a temperature T . The constitutive equations have the form e ij = σ ij − σ kk δ ij +αT δ ij
E E
where α is a coefficient of linear expansion for the material and T is the absolute temperature. Solve for the
stress in terms of strains.

I 29. Derive equation (2.5.53) and then show that when the bulk coefficient of viscosity is zero, the Navier-
Stokes equations, in Cartesian coordinates, can be written in the conservation form
2
∂(%u) ∂(%u + p − τ xx ) ∂(%uv − τ xy ) ∂(%uw − τ xz )
+ + + = %b x
∂t ∂x ∂y ∂z
2
∂(%v) ∂(%uv − τ xy ) ∂(%v + p − τ yy ) ∂(%vw − τ yz )
+ + + = %b y
∂t ∂x ∂y ∂z
2
∂(%w) ∂(%uw − τ xz ) ∂(%vw − τ yz ) ∂(%w + p − τ zz )
+ + + = %b z
∂t ∂x ∂y ∂z
2
∗
where v 1 = u,v 2 = v,v 3 = w and τ ij = µ (v i,j + v j,i − δ ij v k,k ). Hint: Alternatively, consider 2.5.29 and use
3
the continuity equation.