Putting the stress tensor (and traction vector) into explicit vector form. http://sites.google.com/site/peeterjoot2/math2012/continuumstressTensorVectorForm.pdf Peeter Joot — [email protected] Revision https://github.com/peeterjoot/physicsplay commit 1fa4a520eb388b1d15e856ab9c00172ebb6a7582 Apr/8/2012 continuumstressTensorVectorForm.tex Keywords: PHY454H1S, stress tensor, strain tensor, consitutative relationship, curl, wedge product, geometric algebra, traction vector

Contents 1

Motivation.

1

2

Verifying the relationship.

2

3

Cylindrical strain tensor. 3.1 With nˆ = rˆ . . . . . . . ˆ . . . . . 3.2 With nˆ = φ. 3.3 With nˆ = zˆ . . . . . . . 3.4 Summary. . . . . . .

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Spherical strain tensor. 4.1 With nˆ = rˆ . . . . . . ˆ . . . . 4.2 With nˆ = θ. ˆ . . . . 4.3 With nˆ = φ. 4.4 Summary . . . . . .

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6 . 7 . 8 . 10 . 11

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3 3 4 5 6

1. Motivation. Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as t = − pnˆ + µ(2(nˆ · ∇)u + nˆ × (∇ × u)).

(1)

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write t = − pnˆ + µ(2(nˆ · ∇)u + (∇ ∧ u) · nˆ ).

(2)

(∇ ∧ u) · nˆ = nˆ × (∇ × u) = ∇0 (nˆ · u0 ) − (nˆ · ∇)u

(3)

In either case we have

(where the primes indicate the scope of the gradient, showing here that we are operating only ˆ on u, and not n). 1

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements. We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating ˆ 2enˆ = 2(nˆ · ∇)u + nˆ × (∇ × u) = 2(nˆ · ∇)u + (∇ ∧ u) · n.

(4)

2. Verifying the relationship. Let’s start with the the plain old cross product version

(nˆ × (∇ × u) + 2(nˆ · ∇)u)i = n a (∇ × u)b eabi + 2n a ∂ a ui = n a ∂r us ersb eabi + 2n a ∂ a ui [rs]

= n a ∂r us δia + 2n a ∂ a ui = n a (∂i u a − ∂ a ui ) + 2n a ∂ a ui = n a ∂i u a + n a ∂ a ui = n a ( ∂i u a + ∂ a ui ) = σia n a We can also put the double cross product in wedge product form nˆ × (∇ × u) = − I nˆ ∧ (∇ × u) I = − (nˆ (∇ × u) − (∇ × u)nˆ ) 2 I = − (− I nˆ (∇ ∧ u) + I (∇ ∧ u)nˆ ) 2 I2 = − (−nˆ (∇ ∧ u) + (∇ ∧ u)nˆ ) 2 = (∇ ∧ u) · nˆ Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship a · (c ∧ d ∧ e ∧ · · · ) = (a · c)(d ∧ e ∧ · · · ) − (a · d)(c ∧ e ∧ · · · )+ so we find

((∇ ∧ u) · nˆ + 2(nˆ · ∇)u)i = (∇0 (u0 · nˆ ) − (nˆ · ∇)u + 2(nˆ · ∇)u)i = ∂i u a n a + n a ∂ a ui = σia n a .

2

(5)

3. Cylindrical strain tensor. Let’s now compute the strain tensor (and implicitly the traction vector) in cylindrical coordinates. Our gradient in cylindrical coordinates is the familiar

∇ = rˆ

1 ∂ ∂ ∂ ˆ +φ + zˆ , ∂r r ∂φ ∂z

(6)

and our cylindrical velocity is ˆ φ + zˆ uz . u = rˆ ur + φu

(7)

Our curl is then 
∂ 1 ∂ ∂ ˆ ˆ φ + zˆ uz ∇ ∧ u = rˆ + φ + zˆ ∧ rˆ ur + φu ∂r r ∂φ ∂z    
1 1 1 ˆ ˆ ∧ (∂φ rˆ )ur + (∂φ φ ˆ )uφ ˆ ∂φ uz − ∂z uφ + zˆ ∧ rˆ (∂z ur − ∂r uz ) + φ = rˆ ∧ φ ∂r uφ − ∂φ ur + φ ∧ zˆ r r r 

ˆ = −rˆ , we have only one cross term and our curl is Since ∂φ rˆ = θˆ and ∂φ φ u 1 ˆ ∂r u φ − ∂ φ ur + φ ∇ ∧ u = rˆ ∧ φ r r 



ˆ ∧ zˆ +φ



1 ∂φ uz − ∂z uφ r



+ zˆ ∧ rˆ (∂z ur − ∂r uz ) .

(8)

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term. 3.1. With nˆ = rˆ . Our directional derivative component for a rˆ normal direction doesn’t have any cross terms
ˆ φ + zˆ uz 2(rˆ · ∇)u = 2∂r rˆ ur + φu
ˆ r uφ + zˆ ∂r uz . = 2 rˆ ∂r ur + φ∂ Projecting our curl bivector onto the rˆ direction we have     uφ 1 1 ˆ ˆ + (φ ∧ zˆ ) · rˆ ∂φ uz − ∂z uφ + (zˆ ∧ rˆ ) · rˆ (∂z ur − ∂r uz ) (∇ ∧ u) · rˆ = (rˆ ∧ φ) · rˆ ∂r uφ − ∂φ ur + r r r   u 1 ˆ ∂r uφ − ∂φ ur + φ + zˆ (∂z ur − ∂r uz ) . = −φ r r Putting things together we have

3

 uφ 1 ˆ r uφ + zˆ ∂r uz − φ ˆ ∂r u φ − ∂ φ ur + 2erˆ = 2 rˆ ∂r ur + φ∂ + zˆ (∂z ur − ∂r uz ) r r   uφ 1 ˆ = rˆ (2∂r ur ) + φ 2∂r uφ − ∂r uφ + ∂φ ur − + zˆ (2∂r uz + ∂z ur − ∂r uz ) . r r




For our stress tensor σ rˆ = − pˆr + 2µerˆ ,

(9)

we can now read off our components by taking dot products to yield ∂ur σrr = − p + 2µ ∂r   uφ ∂uφ 1 ∂ur + − σrφ = µ ∂r r ∂φ r   ∂ur ∂uz + . σrz = µ ∂r ∂z

(10a) (10b) (10c)

ˆ 3.2. With nˆ = φ. ˆ normal direction will have some cross terms Our directional derivative component for a φ ˆ are functions of φ since both rˆ and φ
2 ˆ φ + zˆ uz ∂φ rˆ ur + φu r
2 ˆ φ uφ + zˆ ∂φ uz + (∂φ rˆ )ur + (∂φ φ ˆ )uφ = rˆ ∂φ ur + φ∂ r
2 ˆ (∂φ uφ + ur ) + zˆ ∂φ uz rˆ (∂φ ur − uφ ) + φ = r

ˆ · ∇)u = 2( φ

ˆ direction we have Projecting our curl bivector onto the φ    uφ 1 1 ˆ = (rˆ ∧ φ ˆ)·φ ˆ ∂r u φ − ∂ φ ur + ˆ ∧ zˆ ) · φ ˆ ˆ ( ∂ z ur − ∂r u z ) (∇ ∧ u) · φ + (φ ∂φ uz − ∂z uφ + (zˆ ∧ rˆ ) · φ r r r     uφ 1 1 = rˆ ∂r uφ − ∂φ ur + − zˆ ∂φ uz − ∂z uφ r r r 

Putting things together we have    
uφ 2 1 1 ˆ 2eφˆ = rˆ (∂φ ur − uφ ) + φ(∂φ uφ + ur ) + zˆ ∂φ uz + rˆ ∂r uφ − ∂φ ur + − zˆ ∂φ uz − ∂z uφ r r r r    
uφ 1 1 2 ˆ ∂φ uφ + ur + zˆ = rˆ ∂ φ ur − + ∂r u φ + φ ∂φ uz + ∂z uφ . r r r r 4

For our stress tensor ˆ + 2µeφˆ , σ φˆ = − pφ we can now read off our components by taking dot products to yield   1 ∂uφ ur σφφ = − p + 2µ + r ∂φ r   ∂uφ 1 ∂uz + σφz = µ r ∂φ ∂z   uφ ∂uφ 1 ∂ur σφr = µ − + . r ∂φ r ∂r

(11)

(12a) (12b) (12c)

3.3. With nˆ = zˆ . Like the rˆ normal direction, our directional derivative component for a zˆ normal direction will not have any cross terms
ˆ φ + zˆ uz 2(zˆ · ∇)u = ∂z rˆ ur + φu ˆ z uφ + zˆ ∂z uz = rˆ ∂z ur + φ∂ Projecting our curl bivector onto the zˆ direction we have     u 1 1 ˆ ∧ zˆ ) · zˆ ˆ = (rˆ ∧ φ ˆ ) · zˆ ∂r uφ − ∂φ ur + φ + (φ ∂φ uz − ∂z uφ + (zˆ ∧ rˆ ) · zˆ (∂z ur − ∂r uz ) (∇ ∧ u) · φ r r r   1 ˆ ∂φ uz − ∂z uφ − rˆ (∂z ur − ∂r uz ) =φ r Putting things together we have   1 ˆ ˆ 2ezˆ = 2ˆr∂z ur + 2φ∂z uφ + 2ˆz∂z uz + φ ∂φ uz − ∂z uφ − rˆ (∂z ur − ∂r uz ) r   1 ˆ 2∂z uφ + ∂φ uz − ∂z uφ + zˆ (2∂z uz ) = rˆ (2∂z ur − ∂z ur + ∂r uz ) + φ r   1 ˆ ∂z uφ + ∂φ uz + zˆ (2∂z uz ) . = rˆ (∂z ur + ∂r uz ) + φ r For our stress tensor σ zˆ = − pzˆ + 2µezˆ ,

(13)

we can now read off our components by taking dot products to yield σzz = − p + 2µ

5

∂uz ∂z

(14a)

 σzr = µ  σzφ = µ

∂ur ∂uz + ∂z ∂r



∂uφ 1 ∂uz + ∂z r ∂φ

(14b)  .

(14c)

3.4. Summary.

σφφ

∂ur σrr = − p + 2µ ∂r   1 ∂uφ ur = − p + 2µ + r ∂φ r

∂uz σzz = − p + 2µ ∂z   ∂uφ 1 ∂ur uφ σrφ = µ + − ∂r r ∂φ r   ∂uφ 1 ∂uz σφz = µ + r ∂φ ∂z   ∂uz ∂ur σzr = µ + ∂z ∂r

(15a) (15b) (15c) (15d) (15e) (15f)

4. Spherical strain tensor. Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier? We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was dr ˆ (r sin θ φ˙ ), = rˆ r˙ + θˆ (r θ˙ ) + φ dt and our gradient mirrors this structure

∇ = rˆ

1 ∂ 1 ∂ ∂ ˆ + θˆ +φ . ∂r r ∂θ r sin θ ∂φ

(16)

(17)

We also previously calculated [2] the unit vector differentials ˆ ˆ sin θdφ + θdθ dˆr = φ

(18a)

ˆ cos θdφ − rˆ dθ dθˆ = φ

(18b)

ˆ = −(rˆ sin θ + θˆ cos θ )dφ, dφ

(18c)

and can use those to read off the partials of all the unit vectors

6

∂ˆr ˆ φ ˆ sin θ } = {0, θ, ∂{r, θ, φ} ∂θˆ ˆ cos θ } = {0, −rˆ , φ ∂{r, θ, φ} ˆ ∂φ = {0, 0, −rˆ sin θ − θˆ cos θ }. ∂{r, θ, φ}

(19) (20) (21)

Finally, our velocity in spherical coordinates is just ˆ θ + φu ˆ φ, u = rˆ ur + θu

(22)

from which we can now compute the curl, and the directional derivative. Starting with the curl we have 


∂ 1 ∂ 1 ∂ ˆ θ + φu ˆ ˆ φ ∇ ∧ u = rˆ + θˆ ∧ rˆ ur + θu +φ ∂r r ∂θ r sin θ ∂φ   1 ˆ = rˆ ∧ θ ∂r uθ − ∂θ ur r   1 1 ˆ + θˆ ∧ φ ∂θ uφ − ∂φ uθ r r sin θ   1 ˆ ∂ φ ur − ∂r u φ + φ ∧ rˆ r sin θ   1ˆ  ˆ + θ ∧ uθ ∂θ θˆ +uφ ∂θ φ |{z} |{z} r −rˆ 0  

+

 1 ˆ ∧ uθ ∂φ θˆ +uφ φ  |{z} r sin θ ˆ cos θ φ

ˆ ∂φ φ |{z}

 . 

−rˆ sin θ −θˆ cos θ

So we have  u 1 θ ∇ ∧ u = rˆ ∧ θˆ ∂r uθ − ∂θ ur + r r   uφ cot θ 1 1 ˆ ˆ +θ∧φ ∂θ uφ − ∂φ uθ + r r sin θ r   uφ 1 ˆ ∧ rˆ +φ ∂ φ ur − ∂r u φ − . r sin θ r 

4.1. With nˆ = rˆ . The directional derivative portion of our strain is ˆ θ + φu ˆ φ) 2(rˆ · ∇)u = 2∂r (rˆ ur + θu ˆ r uθ + φ∂ ˆ r u φ ). = 2(rˆ ∂r ur + θ∂ 7

(23)

The other portion of our strain tensor is   uθ 1 ˆ (∇ ∧ u) · rˆ = (rˆ ∧ θ) · rˆ ∂r uθ − ∂θ ur + r r   uφ cot θ 1 1 ˆ ) · rˆ + (θˆ ∧ φ ∂θ uφ − ∂φ uθ + r r sin θ r   uφ 1 ˆ ∧ rˆ ) · rˆ + (φ ∂ φ ur − ∂r u φ − r sin θ r   u 1 = −θˆ ∂r uθ − ∂θ ur + θ r r   uφ 1 ˆ ∂ φ ur − ∂r u φ − +φ . r sin θ r Putting these together we find 2erˆ = 2(rˆ · ∇)u + (∇ ∧ u) · rˆ  ˆ ˆ ˆ = 2(rˆ ∂r ur + θ∂r uθ + φ∂r uφ ) − θ ∂r uθ −  1 = rˆ (2∂r ur ) + θˆ 2∂r uθ − ∂r uθ + ∂θ ur − r

   uφ 1 uθ 1 ˆ ∂ θ ur + +φ ∂ φ ur − ∂r u φ − r r r sin θ r    uφ uθ 1 ˆ 2∂r uφ + +φ ∂ φ ur − ∂r u φ − . r r sin θ r

Which gives     uφ 1 uθ 1 ˆ ∂r u φ + 2erˆ = rˆ (2∂r ur ) + θˆ ∂r uθ + ∂θ ur − +φ ∂ φ ur − r r r sin θ r

(24)

For our stress tensor σ rˆ = − pˆr + 2µerˆ ,

(25)

we can now read off our components by taking dot products ∂ur σrr = − p + 2µ ∂r   ∂uθ 1 ∂ur uθ σrθ = µ + − ∂r r ∂θ r   ∂uφ uφ 1 ∂ur σrφ = µ + − . ∂r r sin θ ∂φ r

(26a) (26b) (26c)

This is consistent with (15.20) from [3] (after adjusting for minor notational differences). ˆ 4.2. With nˆ = θ. Now let’s do the θˆ direction. The directional derivative portion of our strain will be a bit more work to compute because we have θ variation of the unit vectors

8

1 ˆ θ + φu ˆ φ) (θˆ · ∇)u = ∂θ (rˆ ur + θu r
1 ˆ θ uθ + φ∂ ˆ θ uφ + = rˆ ∂θ ur + θ∂ r
1 ˆ θ uθ + φ∂ ˆ θ uφ + = rˆ ∂θ ur + θ∂ r


1 ˆ )uφ (∂θ rˆ )ur + (∂θ θˆ )uθ + (∂θ φ r
1 ˆ θur − rˆ uθ . r

So we have 2 2 2 ˆ θ uφ , rˆ (∂θ ur − uθ ) + θˆ (∂θ uθ + ur ) + φ∂ (27) r r r and can move on to projecting our curl bivector onto the θˆ direction. That portion of our strain tensor is 2(θˆ · ∇)u =

  u 1 (∇ ∧ u) · θˆ = (rˆ ∧ θˆ ) · θˆ ∂r uθ − ∂θ ur + θ r r   uφ cot θ 1 1 ˆ ˆ ˆ + (θ ∧ φ) · θ ∂θ uφ − ∂φ uθ + r r sin θ r   u 1 φ ˆ ∧ rˆ ) · θˆ + (φ ∂ φ ur − ∂r u φ − r sin θ r     uφ cot θ uθ 1 1 1 ˆ −φ ∂θ uφ − ∂φ uθ + . = rˆ ∂r uθ − ∂θ ur + r r r r sin θ r Putting these together we find 2eθˆ = 2(θˆ · ∇)u + (∇ ∧ u) · θˆ 2 2 2 ˆ θ uφ = rˆ (∂θ ur − uθ ) + θˆ (∂θ uθ + ur ) + φ∂ r r  r   uφ cot θ 1 uθ 1 1 ˆ + rˆ ∂r uθ − ∂θ ur + −φ ∂θ uφ − ∂φ uθ + . r r r r sin θ r Which gives  2eθˆ = rˆ

1 uθ ∂ θ ur + ∂r u θ − r r



+ θˆ



2 2 ∂ θ u θ + ur r r



ˆ +φ



uφ cot θ 1 1 ∂θ uφ + ∂φ uθ − r r sin θ r

 . (28)

For our stress tensor σ θˆ = − pθˆ + 2µeθˆ , we can now read off our components by taking dot products   2 ∂uθ 2 σθθ = − p + µ + ur r ∂θ r

9

(29)

(30a)

uφ cot θ 1 ∂uθ 1 ∂uφ + − =µ r ∂θ r sin θ ∂φ r   1 ∂ur ∂u u + θ− θ . σθr = µ r ∂θ ∂r r 

σθφ

 (30b) (30c)

This again is consistent with (15.20) from [3]. ˆ 4.3. With nˆ = φ. ˆ direction. This directional derivative portion of our strain will also be a Finally, let’s do the φ ˆ variation of the unit vectors bit more work to compute because we have φ 1 ˆ θ + φu ˆ φ) ∂φ (rˆ ur + θu r sin θ 1 ˆ φ uθ + φ∂ ˆ φ uφ + (∂φ rˆ )ur + (∂φ θˆ )uθ + (∂φ φ ˆ )uφ ) = (rˆ ∂φ ur + θ∂ r sin θ 1 ˆ φ uθ + φ∂ ˆ φ uφ + φ ˆ sin θur + φ ˆ cos θuθ − (rˆ sin θ + θˆ cos θ )uφ ) = (rˆ ∂φ ur + θ∂ r sin θ

ˆ · ∇)u = (φ

So we have 

uφ 1 ∂ φ ur − r sin θ r





   1 1 1 1 1 ˆ · ∇)u = 2ˆr ˆ 2( φ + 2θˆ ∂φ uθ − cot θuφ + 2φ ∂φ uφ + ur + cot θuθ , r sin θ r r sin θ r r (31) ˆ direction. That portion of our strain and can move on to projecting our curl bivector onto the φ tensor is   1 uθ ˆ ˆ ˆ (∇ ∧ u) · φ = (rˆ ∧ θ) · φ ∂r uθ − ∂θ ur + r r   uφ cot θ 1 1 ˆ)·φ ˆ + (θˆ ∧ φ ∂θ uφ − ∂φ uθ + r r sin θ r   uφ 1 ˆ ∧ rˆ ) · φ ˆ + (φ ∂ φ ur − ∂r u φ − r sin θ r   u cot θ 1 1 φ ∂θ uφ − ∂φ uθ + = θˆ r r sin θ r   uφ 1 − rˆ ∂ φ ur − ∂r u φ − . r sin θ r Putting these together we find ˆ · ∇)u + (∇ ∧ u) · φ ˆ 2eθˆ = 2(φ       uφ 1 1 1 1 1 1 ˆ ˆ = 2ˆr ∂ φ ur − + 2θ ∂φ uθ − cot θuφ + 2φ ∂φ uφ + ur + cot θuθ r sin θ r r sin θ r r sin θ r r     uφ cot θ uφ 1 1 1 + θˆ ∂θ uφ − ∂φ uθ + − rˆ ∂ φ ur − ∂r u φ − . r r sin θ r r sin θ r 10

Which gives  2eφˆ = rˆ

uφ ∂ φ ur − + ∂r u φ r sin θ r



+ θˆ



uφ cot θ ∂θ uφ ∂φ uθ − + r sin θ r r



ˆ + 2φ



∂φ uφ ur cot θuθ + + r sin θ r r

 . (32)

For our stress tensor ˆ + 2µeφˆ , σ φˆ = − pφ we can now read off our components by taking dot products   ur cot θuθ 1 ∂uφ + + σφφ = − p + 2µ r sin θ ∂φ r r   uφ ∂uφ 1 ∂ur − + σφr = µ r sin θ ∂φ r ∂r   uφ cot θ 1 ∂uφ 1 ∂uθ σφθ = µ − + . r sin θ ∂φ r r ∂θ

(33)

(34a) (34b) (34c)

This again is consistent with (15.20) from [3]. 4.4. Summary ∂ur σrr = − p + 2µ ∂r   ur 1 ∂uθ σθθ = − p + 2µ + r ∂θ r   ur cot θuθ 1 ∂uφ + + σφφ = − p + 2µ r sin θ ∂φ r r   ∂uθ 1 ∂ur u σrθ = µ + − θ ∂r r ∂θ r   uφ cot θ 1 ∂uφ 1 ∂uθ σθφ = µ − + . r sin θ ∂φ r r ∂θ   uφ ∂uφ 1 ∂ur − + σφr = µ r sin θ ∂φ r ∂r

(35a) (35b) (35c) (35d) (35e) (35f)

References [1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990. 1 [2] Peeter Joot. Continuum mechanics., chapter Introduction and strain tensor. Available from: http://sites.google.com/site/peeterjoot2/math2012/phy454.pdf. 4 [3] L.D. Landau and E.M. Lifshitz. A Course in Theoretical Physics-Fluid Mechanics. Pergamon Press Ltd., 1987. 4.1, 4.2, 4.3

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