STRUCTURAL MECHANICS AND ELASTICITY

ERNEST YEUNG

Abstract. Everything about Structural Mechanics and Elasticity

Contents

1. Strain tensor 2

2. Stress 3

3. Equations of Motion 4

4. “Moments” 5

5. Free energy 5

6. Linearization 6

7. Review of Thermodynamics 6

Part 1. On MIT OCW 16.20 Fall 2002 Structural Mechanics 6

Unit 3: (Review of) Language of Stress/Strain Analysis 6

Unit 4: Equations of Elasticity 6

Unit 13 Review of Simple Beam Theory 7

Part 2. On MIT OCW 16.255 Computational Mechanics of Materials 7

8. Lecture 1: Elastic Solids; Legendre Transformation; Isotropy; Equilibrium; Compatibility; Constitutive Relations; Variational Calculus; Example of a Functional: String; Extrema - Calculus of Variations; Local Form of Stationarity Condition 7

Part 3. Implementation 10

9. FEniCS 10

10. Paraview 14

11. MeshPY 14

Date: 21 juillet 2015.

1991 Mathematics Subject Classification. Structural Mechanics.

Key words and phrases. Structural Mechanics, Elasticity.linkedin : ernestyalumni .

1

12. My Examples 14

Part 4. Dictionary 15

13. Curl 15

14. Strain 16

15. Stress 16

References 17

1. Strain tensor

I was following Chapter 1, Fundamental Equations, and Sec. 1 The strain tensor of Landau and Lifshitz [1]. But I will raise anumber of issues with the derivation of the strain tensor.

Consider a diffeomorphism Φ on manifold M , representing space, to manifold N . I will allow for the targe manifold to be Nfor generality, but I suspect that N = M , i.e. Φ should map back to M . As Φ is a diffeomorphism on M , then necessarily thedimension of M , dimM , and the dimension of N , dimN , must be equal, i.e. dimM = dimN .

Let B be a compact submanifold of M , B ⊂M s.t. dimB = dimM .

So

M N

B Φ(B)

Φ

Φ

Let a be a point in B, a ∈ B, and let a be represented by the same notation for its coordinate representation. Let the coordinaterepresentation of Φ(a) be x for its notation. Then

a x = x(a) = Φ(a)Φ

Φ ∈ Diff(M), i.e. Φ is a diffeomorphism in the group of diffeomorphisms on M .Let Φt := Φ(t) i.e. diffeomorphism Φ is parametrized by t ∈ R, so that

Φ0 = 1 · a = a andΦt(a) = x(a, t)

Consider

U(a, t) :=∂x(a, t)

∂t=

∂

∂tΦt(a)

where U(a, t) is the Lagrangian or material velocity U , which is the velocity of the particle with label a at time t.

EY : 20150723 I think this velocity U(a, t) is in TaM , i.e. U(a, t) ∈ TaM ; please correct me if I’m wrong.

Now consider the metric g ∈ Γ(T ∗M ⊗ T ∗M), a (0, 2)-rank symmetric tensor, sog ∈ Γ(T ∗M ⊗ T ∗M)g = gijda

i ⊗ daj

Consider the metric g′ ∈ Γ(T ∗N ⊗ T ∗N), a (0, 2)-rank symmetric tensor on N :g′ ∈ Γ(T ∗N ⊗ T ∗N)g′ = g′ijdx

i ⊗ dxj

We want to consider the pullback Φ∗ of g′, so that Φ∗g′ ∈ Γ(T ∗M ⊗ T ∗M), i.e.

T ∗M ⊗ T ∗M T ∗N ⊗ T ∗N

M N

Φ∗

Φ

Φ∗g′ g′

a x = Φ(a)

Φ∗

Φ

For U, V ∈ Γ(TM),

Φ∗g′(U, V ) = g′ij Φ(a)dΦa(U)⊗ dΦa(V ) = g′ij∂xi

∂xk∂xj

∂alUkV l

as dΦ = ∂xi

∂aj daj and so (with U = U j ∂

∂aj )

dΦ(U) = ∂xi

∂ajUj .

So then

(Φ∗g′ − g)(U, V ) =

(g′ij

∂xi

∂ak∂xj

∂al− gkl

)UkV l := 2EklU

kV l

where

(1) Ekl =1

2(g′ij

∂xi

∂ak∂xj

∂al− gkl)

is the Lagrangian strain-tensor, E, or Green-St. Vencent strain tensor, a (0, 2)-rank symmetric tensor. Note that

(2) E =1

2(Φ∗g′ − g) ∈ Γ(T ∗M ⊗ T ∗M)

Suppose we define ui = xi− ai. It’s very unclear how the frame field ei for TM at point a ∈ TM , an orthonormal basis, canbe related in such a way to the frame field for TN at point x = Φ(a) ∈ TN , to be completely “inline” with each other (i.e.ei = Φ(ei)), in general. Nevertheless, just suppose this is the case.

2

Then substitute ui + ai for xi into the Lagrangian strain-tensor Ekl:

Ekl =1

2

[g′ij

(∂ui

∂ak+ δik

)(∂uj

∂al+ δjl

)− gkl

]=

1

2

[g′ij

(∂ui

∂ak∂uj

∂al+∂ui

∂akδjl +

∂uj

∂alδik

)− gkl

]=

=1

2

[g′ij

(∂ui

∂ak∂uj

∂al

)+∂ul∂ak

+∂uk∂al− gkl

]=

1

2

[∂ul∂ak

+∂uk∂al

+∂uj∂ak

∂uj

∂al− gkl

]Let u ∈ Γ(T ∗M ⊗ T ∗M) be the strain tensor of rank (0, 2) s.t.

(3) ukl =1

2

[∂ul∂ak

+∂uk∂al

+∂uj∂ak

∂uj

∂al

]for a = ai ∈M .

This derivation of the strain tensor is essentially what Landau and Lifshitz is saying in Chapter 1, Sec. 1 of Landau andLifshitz [1].

However, from an entirely different perspective, consider the Lie derivative of the metric g ∈ Γ(T ∗M ⊗ T ∗M). Note that theLie derivative of the metric g should give us back an object that is also a (0, 2)-rank symmetric tensor, just like g.

For a vector field u ∈ X(M) on M , with g = gijdai ⊗ daj , then

Lug = (Lugij)dai ⊗ daj + gijLudai ⊗ daj + gijdai ⊗ Ludaj

Now

Ludai = iud(dai) + diudai = 0 +

∂ui

∂ajdaj

and, for u = ui ∂∂ai

Lugij = uk∂gij∂ak

so that

Lug = gij∂ui

∂akdak ⊗ daj + gijda

i ⊗ ∂uj

∂akdak + uk

∂gij∂ak

dai ⊗ daj =∂uj∂ak

dak ⊗ daj +∂ui∂ak

dai ⊗ dak + uk∂gij∂ak

dai ⊗ daj =

=

(∂uj∂ai

+∂ui∂aj

+ uk∂gij∂ak

)dai ⊗ daj

So the correct formula for the Lie derivative of the metric g is

Lug =

(∂ui∂aj

+∂uj∂ai

+ uk∂gij∂ak

)dai ⊗ daj

Carroll (2001) and Fitzmaurice (2011) have it wrong 1. This stackexchange question and Harmark (2008) have it right 2.

EY: 20150724 I’m not sure if I can be so causal with raising and lowering indices, since it obfuscates when the metric g isdependent upon coordinate a or not. Then what can be agreed upon is this:

(4) Lug =

(gkj

∂uk

∂ai+ gik

∂uk

∂aj+ uk

∂gij∂ak

)dai ⊗ daj

Then define the strain tensor is such:

Definition 1. The strain tensor u, u ∈ Γ(T ∗M ⊗ T ∗M), a symmetric (0, 2)-rank tensor is given by

(5) u = uijdai ⊗ daj =

1

2

(gkj

∂uk

∂ai+ gik

∂uk

∂aj+ uk

∂gij∂ak

)dai ⊗ daj

2. Stress

Given the Cauchy stress tensor T , T being a (0, 2)-rank symmetric tensor, T = T ijei ⊗ ej , with eii=1...dimM being anorthonormal frame field on tangent bundle TM , and so consider this set of transformations:

T ∗M ⊗ T ∗M Ω1(M,TM) Ωn−1(M,TM)

T = T ijei ⊗ ej = T ijej ⊗ ei T ijej ⊗ ei T ijdSj ⊗ ei

([,−) (∗,−)

([,−) (∗,−)

with

T ij

√g

(n− 1)!εji2...indx

i2 ∧ · · · ∧ dxin ⊗ ei = T ijgjkdSk ⊗ ei = T ijdSj ⊗ ei

EY : 20150724 I have a serious concern about the last step, namely in how the Levi-Civita permutation symbol transforms inthe musical isomorphism. Can we really raise and lower indices with impunity, and in the (2) cases of coordinate bases and anorthonormal frame field? How does the Levi-Civita permutation symbol transform?

Define dSk as such:

Proposition 1. dSj ∈ Ωn−1(M)

(6)

dSj = iejvoln

iejvoln =

√g

(n− 1)!εji2...indx

i2 ∧ · · · ∧ dxin

Proof.

iejvoln =

√g

n!εi1...in(δi1j dx

i2 ∧ · · · ∧ dxin + (−1)dxi1δi2j ∧ dxi3 ∧ · · · ∧ dxin + · · ·+ (−1)n+1dxi1 ∧ · · · ∧ dxin−1δinj ) =

=

√g

(n− 1)!εji2...indx

i2 ∧ · · · ∧ dxin

Nevertheless, T ijdSj could be interpreted as the force, in the direction i, on a surface element dSj normal to direction j.

Now the Cauchy stress tensor acts on the surface, or surface boundary of compact body B, ∂B, and so the total force due tothe Cauchy stress tensor on the surface of the body B, ∂B, is∫

∂B

T ijdSj ⊗ ei

with T ijdSj ⊗ ei ∈ Ωn−1(M,TM).

Now, the question I have is whether Stoke’s law could be used or not to change the integral above, because we’re not necessarilyintegrating over a n− 1-form, but a vector-valued n− 1 form, and the connection has to be taken into account.

Consider the covariant exterior differential, D, which takes into account of a connection form (the usual exterior differential,or differential, d on C∞(M), df , f ∈ C∞(M), say, doesn’t involve the presence of the connection) 3.

D : Ωn−1(M ;E)→ Ωn(M ;E)

1Sean Caroll, Lecture Notes on General Relativity, March 2001, http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll5.html; Fionn Fitzmaurice, “Differential Geometry”, 2011, https://www.maths.tcd.ie/~fionnf/dg/dg.pdf2“Confusion about Lie derivative on metric”, physics.stackexchange.com, http://physics.stackexchange.com/questions/112357/confusion-about-lie-derivative-on-metric; T. Harmark, “Notes on Lie derivatives and Killing vector fields,” 2008 http://www.nbi.dk/~harmark/Killingvectors.pdf3“Exterior covariant derivative”, wikipedia, https://en.wikipedia.org/wiki/Exterior_covariant_derivative

3

with definition of D taken out of Morita (2001) (Sec.5.3. Connections in Vector Bundles and Curvature [3]). Morita has Ddefined as

D : Ωk(M ;E)→ Ωk+1(M ;E)

D(θ ⊗ s) = dθ ⊗ s+ (−1)kθ ∧∇sθ ∈ Ωk(M)

s ∈ Γ(E)

With∇ : Γ(E)→ Ω1(M ;E)

ei ∈ Ω1(M)

ei ∈ Γ(E)

Now ∇ej = ωij ⊗ ei, where ωij ∈ Ω1(M) is a connection 1-form, s.t. ωij = Γijkek, where Γijk are Christoffel symbols.

Then applying D onto T ijdSj ⊗ ei,

D(T ijdSj ⊗ ei) = d(T ijdSj)⊗ ei + (−1)n−1T ijdSj ∧∇ei =∂T ij

∂xjvoln ⊗ ei + (−1)n−1T kjdSj ∧ ωik ⊗ ei =

∂T ij

∂xjvoln ⊗ ei + T kjωik ∧ dSj ⊗ ei =

=

(∂T ij

∂xjvoln + ωik ∧ T kjdSj

)⊗ ei = ∇(T ijdSj)⊗ ei

where ∇ : Ωn−1(M)→ ωn(M) acts on (n− 1)-forms as such.

Consider ωij = Γijkek. Then

ωij ∧ T jkdSk = ωij ∧ T jk√g

(n− 1)!εji2...indx

i2 ∧ · · · ∧ dxin = Γijkek ∧ T jl

√g

(n− 1)!εli2...indx

i2 ∧ · · · ∧ dxin =

= ΓijkTjl

√g

(n− 1)!εli2...inε

ki2...inj1j2...jn

1

ndxj1 ∧ · · · ∧ dxjn = ΓijkT

jlδkl voln = ΓijkTjkvoln

So then

(7) ∇(T ijdSj) =

(∂T ij

∂xj+ ΓijkT

jk

)voln ∈ Ωn(M)

At this point, in Appendix A “Forms in Continuum Mechanics”, Sections A.a “The Classical Cauchy Stress Tensor andEquations of Motion” and A.b “Stresses in Terms of Exterior Forms” of Frankel (2004) [2] is either very obfuscating or veryerroneous. The problem is this; there are 2 possibilities:

(8)

∫∂B

T ijdSj ⊗ ei =

∫B

d(T ijdSj)⊗ ei =

∫B

∂T ij

∂xjvoln ⊗ ei or∫

∂B

T ijdSj ⊗ ei =

∫B

D(T ijdSj ⊗ ei) =

∫B

∇(T ijdSj)⊗ ei

The problem with the first possibility is the question of whether ei ⇐⇒ ∂∂xi remains “fixed” or “doesn’t change” as we move

around ∂B in the surface integral or B in the volume integral.

The problem with the second possibility is, to my knowledge, I don’t know if Stoke’s law applies for vector-valued differentialforms. For differential forms on manifold M , the exterior derivative d on differential forms is an antiderivative of degree 1(or −1, whatever your convention is), and d2 = 0 leads to a deRham (co)homology. I don’t know if D, the covariant exteriorderivative, an antiderivative of degree 1, yields a cohomology.

3. Equations of Motion

I’m going to follow the derivation in Frankel (2003; p.s. I need the 3rd. edition, please see the note at the very end) in AppendixA, Forms in Continuum Mechanics, Sec. A.a. The Classical Cauchy Stress Tensor and Equations of Motion, [2], because Iwant to point out some issues I want to raise.

Let M be a smooth manifold equipped with a metric g, (M,O,A, g) of space. If one can consider a spacetime manifold M ,please let me know. Otherwise, assume a foliation by time t ∈ R, so that we are considering R×M .Let b ∈ X(M).

b is the force per unit mass on the (compact) body B ⊂M s.t. dimB = dimMLet m := ρvoln, voln a n-volume form on M , dimM = n, with ρ ∈ C∞(M), ρ the mass density. So m ∈ Ωn(M).Let b⊗m ∈ Ωn(M,TM), a vector-valued n-form.∫Bb⊗m is the total force on the compact body B due to the external force b acting directly on the body B.

The forces on the body B due to the Cauchy stress tensor T , is on the surface or boundary of the body B, ∂B, given by∫∂B

T ijdSj .For diffeomorphism Φ : M → N , parametrize Φ by time t ∈ R, so that B(t) := ΦtB = Φ(t)B, and Φ0B = Φ(0)B = B.∫B(t)

m =∫B(t)

ρvoln is the total mass in the body B(t) at time t.

Mass conservation is

(9)d

dt

∫B(t)

m =

∫B(t)

L ∂∂t+um = 0

Consider m⊗ u ∈ Ωn(M,TM). Then m⊗ u is the momentum for voln. P = P (t) :=∫B(t)

m⊗ is the total momentum of body

B.

Take the time derivative:

P =d

dtP (t) =

d

dt

∫B(t)

m⊗ u =

∫B(t)

L ∂∂t+um⊗ u =

∫B(t)

(L ∂∂t+um)⊗ u+

∫B(t)

m⊗ L ∂∂t+uu =

= 0 +

∫B(t)

m⊗(∂ui

∂t+ uj

∂

∂xjui)

∂

∂xi=

∫B(t)

m⊗(∂ui

∂t+ u(ui)

)∂

∂xi

For a proof of the step L ∂∂t+uv =

(∂vi

∂t + uj ∂vi

∂xj

)∂∂xi , where v ∈ X(R×M), i.e. v is a time-dependent vector field, and in this

case, we have v = u, then see my Solutions to J. Lee’s Introduction to Smooth Manifolds, where I present a proof that,to my knowledge, is original [7]. I will recap it here:

Let’s define the Lie derivative:

(10) LVW = (LVW )(t,p) =d

ds

∣∣∣∣s=0

(dθ−s)θs(t,p)(Wθs(t,p)) = lim

s→0

(dθ−s)θs(t,p)(Wθs(t,p))−Wθs(t,p)

s

Use Case 1 of the proof of Lee’s Theorem 9.38 [6], for showing LVW = [V,W ].Let open neighborhood U ⊆ J ×M , with (t, p) ∈ U . On open U , choose smooth coordinates (t, ui) on U . By Theorem 9.22 ofLee [6], that at a regular point p ∈M , ∃ (ui) coordinates s.t. Vp = ∂

∂u1 , then consider

V =∂

∂t+

∂

∂u1∈ X(R×M)

with V (t)(p) = ∂∂u1 ∈ X(M). (Remember, V (t) is a vector-field that is time-dependent, but is on M . I will use this as a

justification for using Thm. 9.22).

4

Now the flow θs takes on these forms:

θ(t,p)(s) = θ(s, (t, p)) = θs(t, p) =

= (α(s, (t, p)), β(s, (t, p))) = (s+ t, β(s, (t, p)))

Given these conditions, thatβ(0, (t, p)) = p = (u1, u2, . . . un) and

∂β

∂s(s, (t, p))

∣∣∣∣s=0

= V (t, p) =∂

∂u1=

d

dsβ(t,p)(s)

∣∣∣∣s=0

then a β that satisfies these conditions above is

β(s, (t, p)) = βs(t, p) = (u1 + s, u2 . . . un)

so that we can conclude that

θs(t, p) = (t+ s, u1 + s, u2, . . . , un)

For fixed s, then

d(θ−s)θs(t,p) = 1Tθs(t,p)

(R×M)

so that

d(θ−s)θs(t,p)(Wθs(t,p)) = d(θ−s)θs(t,p) ·W

j(t+ s, u1 + s, u2 . . . un)∂

∂uj

∣∣∣∣θs(t,p)

= W j(t+ s, u1 + s, u2 . . . un)∂

∂uj

∣∣∣∣(t,p)

=⇒ d

ds

∣∣∣∣s=0

W j(t+ s, u1 + s, u2 . . . un)∂

∂uj

∣∣∣∣(t,p)

=

(∂

∂tW j(t, u1 . . . un) +

∂

∂u1W j(t, u1 . . . un)

)∂

∂uj

∣∣∣∣(t,p)

Thus, we can conclude that

(11) LVW = L ∂∂t+VW =

(L ∂∂tV

W)

(t,p)=

((∂

∂t+ V

)W j

)∂

∂xj

∣∣∣∣(t,p)

So, so far we have

(12)

∫B(t)

m⊗(∂ui

∂t+ uj

∂

∂xjui)

∂

∂xi=

∫B(t)

m⊗ b+

∫∂B(t)

T ijdSj ⊗∂

∂xi

See Eq. 8 for the 2 possibilities I raise and the issue related to it. Then we either have 1 of 2 possibilities:

(13)

ρ

(∂ui

∂t+ uj

∂ui

∂t

)= ρbi +

∂T ij

∂xjor

m

(∂ui

∂t+ uj

∂ui

∂xj

)= mbi +∇(T ijdSj)

Nevertheless, for the equilibrium condition of P = 0, then we have either 1 of 2 possibilities:

(14)0 = ρbi +

∂T ij

∂xjor

0 = mbi +∇(T ijdSj)

For the second possibility, from Eq. 7, with ∇(T ijdSj) =(∂T ij

∂xj + ΓijkTjk)

voln, then

0 = ρvolnbi +

(∂T ij

∂xj+ ΓijkT

jk

)voln

so then

(15) 0 = ρbi +

(∂T ij

∂xj+ ΓijkT

jk

)

4. “Moments”

cf. A.c. “Symmetry of Cauchy’s Stress Tensor in Rn”, Appendix A Forms in Continuum Mechanics, of Frankel (2004) [2].

5. Free energy

Consider the strain tensor u ∈ Γ(T ∗M ⊗ T ∗M) ≡ Γ(⊗2i=1T

∗M). Then consider

λ ∈ Γ((TM ⊗ TM)⊗ (TM ⊗ TM)) ≡ Γ(⊗2i=1TM ⊗2

i=1 TM)

λ = λiklm(∂i ⊗ ∂k)⊗ (∂l ⊗ ∂m)

a rank-4 tensor, s.t. λ is symmetric in i, k and l,m, from the symmetry of the strain tensor u.

General form of free energy is

F =1

2λiklmuikulm

Then the relationship to the stress tensor T is

T ik =∂F

∂uik

Now

∂F

∂uik=

1

2(λiklmulm + λlmikulm) = λiklmulm

=⇒ T ik = λiklmulm

⊗2i=1TM ⊗⊗2

i=1TM ⊗2i=1TM

λ λiklmulm∂i ⊗ ∂k = T

(−, u)

(−, u)

5

6. Linearization

cf. pp. 241, Sec. 3, Chapter 4 “Linearization” of Marsden and Hughes [10].

Definition 2. Young’s modulus E is given by

(16) E :=µ(3λ+ 2)

λ+ µ

Definition 3. Poisson’s ratio ν is given by

(17) ν =λ

2(µ+ λ)

Then, solving for λ, µ, the Lame parameters λ, µ is given as follows (one can solve this with Sage Math, using the commands

sage : E, nu ,mu, lmbd = var ( ’E nu mu lmbd ’ )sage : s o l v e ( [ E== mu∗(3∗ lmbd +2∗mu)/( lmbd+mu) , nu == lmbd /(2∗ (mu+lmbd ) ) ] , mu, lmbd )[ [ mu == 1/2∗E/(nu + 1) , lmbd == −E∗nu/(2∗nuˆ2 + nu − 1 ) ] , [mu == 0 , lmbd == 0 ] ]

Definition 4. Lame first parameter λ is

(18) λ =Eν

(1 + ν)(1− 2ν)

Definition 5. Lame second parameter or sheer modulus µ is

(19) µ =E

2(1 + ν)

7. Review of Thermodynamics

Let Σ be the manifold of equilibrium (and non-equilibrium) states of the system.

Proposition 2 (First Law: Energy Conservation).

(20) dU = Q− dW = Q− pdV

with U, p, V ∈ C∞(Σ), and dU,Q, dW, dV ∈ Ω1(Σ), and where U is internal energy, p is pressure, V is the volume of the system.

Proposition 3 (Second Law).

(21) Q = TdS

where Q, dS ∈ Ω1(Σ), and S ∈ C∞(Σ) and with

(22) dS ≥ 0

describing irreversibility.

Definition 6 (Enthalpy).

(23) H = U + pV

where H is the enthalpy, H ∈ C∞(Σ).

Now, for M being the molar mass, and defining per unit mass quantities as we go,

dH = dU + V dp+ pdV = Q+ V dp = TdS + V dp

1/M−−−→ dH

M= dh = T

dS

M+V

Mdp = Tds+

dp

ρ

where ρ = M/V .

Now the total energy in a fluid occupying region V0 is∫V0

(1

2ρv2 + ρε

)voln

where∫V0

12ρv

2voln is the total kinetic energy of fluid and ε internal energy per unit mass.

The time rate of change of the energy is

d

dt

∫V0

(1

2ρv2 + ρε)voln =

∫V0

L ∂∂t+v(

1

2ρv2 + ρε)voln =

∫V0

∂

∂t(1

2ρv2 + ρε)voln + Lv((

1

2ρv2 + ρε)voln) =

=

∫V0

∂

∂t(1

2ρv2 + ρε)voln + div((

1

2ρv2 + ρε)voln)

Let domain be the smooth submanifold D ⊂ N , N is the spatial manifold. dima = dimN = n; spacetime manifold M = R×N .a = (ai) ∈ D is a particle label.

Part 1. On MIT OCW 16.20 Fall 2002 Structural Mechanics

I review MIT OCW 16.20[4] and add onto the material.

Unit 3: (Review of) Language of Stress/Strain Analysis

Day 4,5: “Unit 3: Language of Stress/Strain Analysis (Review) Definition of Stress and Strain; Notation; Tensor Rules; Tensorvs. Engineering Notation; Contracted Notation; Matrix Notation.”

I’m going to set the notation along the lines of wikipedia, differential geometry, and Landau and Liftshitz [1].

Unit 4: Equations of Elasticity

Day 6,7,8: Unit 4: Equations of Elasticity (Review) Equations of Elasticity (Equilibrium, Strain-Displacement, Stress-Strain);Static Determinance; Compatibility; Elasticity Tensor; Material Types and Elastic Components; Materials Axes vs. “LoadingAxes”; Compliance and its Tensor; The Formal Strain Tensor; Large Strains vs. Small Strains; Linear vs. Nonlinear Srain.

MIT OCW 16.20 Unit 4

Let M be a smooth manifold equipped with a metric g, (M,O,A, g) of space. If one can consider a spacetime manifold M ,please let me know. Otherwise, assume a foliation by time t ∈ R, so that we are considering R×M .Let b ∈ X(M).

b is the force per unit mass on the (compact) body B ⊂M s.t. dimB = dimMLet m := ρvoln, voln a n-volume form on M , dimM = n, with ρ ∈ C∞(M), ρ the mass density. So m ∈ Ωn(M).Let b⊗m ∈ Ωn(M,TM), a vector-valued n-form.∫Bb⊗m is the total force on the compact body B due to the external force b acting directly on the body B.

The forces on the body B due to the Cauchy stress tensor T , is on the surface or boundary of the body B, ∂B, given by∫∂B

T ijdSj .

6

7.1. Equilibrium. From Eq. 14, the equilibrium condition that∑Fext = 0, the sum of all external forces Fext on a body B

is 0 means either 1 of 2 possibilities:

0 = ρbi +∂T ij

∂xjor

0 = mbi +∇(T ijdSj)

If the second possibility is true, then

(24) 0 = ρbi +

(∂T ij

∂xj+ ΓijkT

jk

)which is the general form for equilibrium.

7.2. Strain-Displacement. From Eq. 4 and Def. 1, we obtain the general form for the strain tensor on the reference bodyB ⊂ M , with coordinates ai, which comes from the Lie derivative of the metric g for reference manifold M , Lug, by adeformation of B according to vector field u ∈ X(B)

(25) u = uijdai ⊗ daj =

1

2

(gkj

∂uk

∂ai+ gik

∂uk

∂aj+ uk

∂gij∂ak

)dai ⊗ daj

In Cartesian coordinates, it’s clear that Eq. 25 reproduces the strain tensor in pp. 4 of the Unit 4 lecture slide.

Unit 13 Review of Simple Beam Theory

IV. General Beam Theory. Choice of x, y, z axes, or 1, 2, 3-axes appears to motivated by attach the beam to an airplane,with z or 3 axis being “vertical” (going up to space), and y or 2 axis being the symmetry axis of the airplane and also in the“horizontal” plane; x or 1 axis is along the length of the beam or its symmetry axis. Thus, the beam’s cross section is in they, z or 2, 3 directions.

For a rocket, then the z or 3-axis is the symmetry axis of the rocket, and the x, y or 1, 2 directions are in the horizontal crosssection of the rocket.

With Lagace’s notation on the left and my notation on the right,

σ = σij∂

∂xj⊗ ∂

∂xi∈ Γ(⊗2TR3) T = T ij

∂

∂xj⊗ ∂

∂xi∈ Γ(⊗2TN)

Take no stresses in y or 2-direction

σ22, σ23, σ12 = σ21 = 0 T22, T23, T12 = T21 = 0

Assumeσ11 >> σ33

σ13 >> σ33

T33 >> T22

T32 >> T22

So only significant stresses are σ11, σ13.

Part 2. On MIT OCW 16.255 Computational Mechanics of Materials

8. Lecture 1: Elastic Solids; Legendre Transformation; Isotropy; Equilibrium; Compatibility;Constitutive Relations; Variational Calculus; Example of a Functional: String; Extrema - Calculus

of Variations; Local Form of Stationarity Condition

Elastic Solids; Legendre Transformation; Isotropy; Equilibrium; Compatibility; Constitutive Relations; Variational Calculus;Example of a Functional: String; Extrema - Calculus of Variations; Local Form of Stationarity Condition.

8.1. Elastic solids. Let deformation power be

σij εij

EY : 20150728 in my notation and Landau and Lifshitz’s [1], this would be tij uij .

Consider a cycle of deformation Γ : J ⊂ R→ B

Γ(t) ∈ Bwith time interval J = [0, T ] and so t ∈ [0, T ]. Now for elastic deformation,

we would expect εij(T ) = εij(0), i.e. the material will return back to its initial state after a deformation and bringing it backto its original configuration, so that Γ(0) = Γ(T ).

If∮

ΓT ij uijdt = 0, then T ij uijdt is an exact form.

⇐⇒ ∃W (uij) s.t. T ij = ∂W∂uij

so that∮T ij uijdt =

∮∂W∂uij

duijdt dt =

∮dW (uij) = 0

W : T ∗M ⊗ T ∗M ≡ ⊗2T ∗M → R is called the strain energy density, and takes as its argument, the strain tensor, ε. For thiselastic case, the existence of W is guaranteed and σij = ∂W

∂εijwith σij being the stress tensor, but on reference body B (EY:

20150728 please correct me if I’m wrong on this point).

8.2. Energy conservation. By the definition of power, being the time rate of change of energy (rather than energy conser-vation)

dµ

dt= T ij uij

For this elastic case, as above, thendµ

dt= T ij uij =

∂W

∂uij

duijdt

elastic

Then

µ = W (uij) + µ0

8.3. Legendre Transformation. Let’s do the Legendre transform on strain energy density W , to obtain χ, the complemen-tary strain energy density, which allows one to reexpress χ in terms of what would be the conjugate of strain tensor ε (whichis what Legendre transforms do).

Given from above that σij = ∂W∂εij

(εij),

χ = σijεij −W (ε)

dχ = dσijεij + σijdεij −∂W

∂εijdεij = dσijεij + 0

7

Now

dχ =∂χ

∂σijdσij =

∂χ

∂σij∂σij

∂εkldεkl

and

dσijεij =∂σij

∂εkldεklεij

so

εij =∂χ

∂σij

There is the question of whether σij can be related to εij through ∂σij

∂εklin that this is invertible. I’ll assume it is invertible for

this case.

8.3.1. Example: Thermoelasticity. Suppose for linear thermoelasticity (Hooke’s law), that the strain energy density is

(26) W =1

2(εij − αijT )Cijkl(εkl − αklT )

Now ε−αT ∈ Γ(T ∗M ⊗T ∗M) := Γ(⊗2T ∗M), and so we can think of tensor C = Cijkl as defining as kinetic energy metric on⊗2T ∗M

W =1

2〈ε− αT, ε− αT 〉C =

1

2‖ε− αT‖C ∈ R

for C ∈ Γ(⊗2TM ⊗2 TM).

Cijkl is given by

Cijkl =∂2W

∂εij∂εklwhich is called the elastic moduli by Radovitzky [5], or (second) elasticity tensor in Marsden and Hughes [10].

Now

σij =∂W

∂εij(εij) = Cijkl(εkl − αklT )

If C is invertible, i.e. ∃C−1, thenC−1ijklσ

ij = εkl − αklT so C−1ijklσ

ij + αklT = εkl

Nowσijεij = σij(C−1

klijσkl + αijT )

W =1

2C−1klijσ

klCijpqC−1mnpqσ

mn =1

2C−1klmnσ

klσmn

so

χ = σijεij −W (ε) =1

2C−1ijklσ

ijσkl + σijαijT

confirming the expression for the Exercise on pp.4 of Unit 1 for MIT OCW 16.225 [5].

Consider the symmetries of C. Let d be the dimension of space. Usually, d = 3.For C of “rank” 4, then there are d ∗ d ∗ d ∗ d = d4 possible values or 34 = 81.

But Cijkl is symmetric about i, j. Recall that d(d+1)2 is the number of independent values of a d× d matrix (think of an upper

triangle, including the diagonal elements. Then use the sum formula for 1 + 2 + · · · + d = d(d+1)2 ). d(d+1)

2 = 6 and so if k, ldidn’t have symmetry, k,l would contribute 3 ∗ 3 = 9 different, independent values, for 6 ∗ 9 = 54.

Cijkl is also symmetric about k, l, and so, likewise, (d(d+1)2 )2 = 6 ∗ 6 = 36.

Now C ∈ Γ(⊗2TM ⊗ ⊗2TM) so it’s symmetric about i, j and k, l, i.e. Cijkl = Cklij . Then using the same d(d+1)2 formula,

except imagining that we relabel with the 6 ij coordinates (imagine a 6× 6 symmetric matrix), then 6(6+1)2 = 21.

8.3.2. Isotropy.

Cijkl = λδijδkl + µ(δikδjl + δilδjk)

λ, µ Lame constants

thermal isotropy: αij = αδij

σij = λεkkδij + µ(εij + εji)− αT (λij3 + µ2δij)

W =1

2σijεij =

1

2[λεkkδijεij + 2µεijεij ] =

1

2λε2kk + µεijεij

8.4. Summary of field equations of linearized elasticity.

S = ∂B = S1

⋃S2, S1

⋂S2 = ∅

S1: displacement boundaryS2 traction boundary

8.4.1. Compatibility. εij = 12 ( ∂u

i

∂aj + ∂uj

∂ai ) in B

u = u on S1.

εij is the usual strain tensor that describes the strain (and directed forces) that the material inherently experiences due to adeformation (diffeomorphism) from the reference body B (with parts of it labeled by ai coordinates).

S1 was called the displacement boundary, and u = u implies that this is the extent of the displacement or deformation. EY :20150729 I think this is the boundary on which we’re pulling or pushing on; please correct me if I’m wrong.

8.4.2. Equilibrium. ∂σij

∂j + fi = 0 in B

σijnj = ti on S2.

8.4.3. Constitutive relations. σij = ∂W∂εij

εij = ∂χ∂σij

8

8.5. Variational Calculus. There are a number of references cited on page 5; I don’t have any access to them. So if you’retrying to learn about structural mechanics then Marsden and Hughes’ book is available online, on the late Prof. Marsden’swebsite [10].

Let field µ : B → N

µ(a) ∈ Nbe a smooth mapping from the compact body B to manifold N . N can be the manifold of states of the

solid, or, as Marsden and Hughes had first implicated, N is the configuration of body, N , after a deformation, represented bydiffeomorphism Φ : B ⊂M → N .

Consider a fiber bundle (E, π,B) over compact body B. By the definition of a fiber bundle, ∀ (U, ai), chart for B, then ∃homeomorphism ϕ : π−1(U)→ U ×N ; without loss of generality, suppose ϕ is a diffeomorphism. So π−1(U) is diffeomorphicto U ×N , i.e. π−1(U) ∼= U ×N .

A field is part of this section of this π−1(U) ∼= U ×N .

Here’s what I mean. Field µ : B → N maps B to N . Then for section µ ∈ Γ(π), µ(a) = (a, µ(a)) for a ∈ B. That’s whyMarsden and Hughes [10] and I will call field µ a section of fiber bundle (E, π,B), even though we’ll treat µ as µ : B → N .

Thus,

E B ×N

B

∼=

µ(−, µ)

µ(a) (a, µ(a))

a

∼=

µ(−, µ)

Consider the push forward of field µ, (µ)∗ ≡ Dµ : TB → TN . Dµ, if N is considered the configuration of the body B, is the

deformation gradient of configuration µ(B). Locally, Dµa ≡ Dµ(a) = ∂µi

∂aj (a).

Thus,

TB ⊗ TN

E B ×N

TB

B

∼=

µ(−, µ)

(A, (Dµ)aA)

µ(a) (a, µ(a))

A = Ai(a) ∂∂ai

a

∼=

µ(−, µ)

At this point, I should make a comment about the “splitting” (connection) that the fiber bundle E carries, that was mentionedin Marsden and Hughes [10]:

Tµ(a)Ea

p = µ(a) ∈ Ea

TEa

π−1(a) = Ea

TaB = kerPp

a

Dµ

∀ a ∈ B, consider vector µa ∈ Tµ(a)Ea in the tangent space of the fiber over a, Ea.

Let Ξ be the bundle over B, with a fiber over B being a triple (E, TE, T (π−1)), i.e.

Ξ := (E, TE, T (π−1))πΞ−−→ B

π−1Ξ (a) = (µa, µa, (Dµ)a).

Consider a (Lagrangian) functional density L, L : B × Ξ→ R, i.e.L : Ξ→ RL(a, µa, µa, (Dµ)a).

Let φ : E → R be a smooth function on fiber bundle E over B, i.e.φ : E → Rφ(a, µ(a)) ∈ R.

Thus, the (Lagrangian) functional L is a smooth map from tangent bundle TN to R:

(27)

L : TN → R

L(µ, µ) =

∫B

L(a, µ, µ,Dµ)dV −∫S2

φ(a, µ)dS

At this point, I will defer to the derivation as presented by Radovitzky in Unit 1 [5]. With L being denoted by Radovitzky’sF in his treatment, and η ∈ N , One should note the integration by parts step:

(28)

∫B

∂L

∂(∂µi

∂aj

) ∂ηi∂aj

daj =

∫ ∂

∂aj

∂L

∂(∂µi

∂aj

)ηi− ∂

∂aj

∂L

∂(∂µi

∂aj

) ηi

dajThe second term, ∂

∂aj

(∂L

∂(∂µi

∂aj

)) ηi adds to ∂L∂µi to give, as ηi is arbitrarily small, the usual Euler-Lagrange equation part on

the compact body B:

(29)∂L∂µi− ∂

∂aj

∂L

∂(∂µi

∂aj

) = 0 ∀ a ∈ B

9

However, the first term of Eq. 28, ∂∂aj

(∂L

∂(∂µi

∂aj

)ηi)

, is problematic.

Consider this n− 1-form on B, in Ωn−1(B), with dimB = n:

∂L

∂(∂µi

∂aj

) ηi

(n− 1)!εji2...inda

i2 ∧ · · · ∧ dain

Apply the exterior derivative d : Ωn−1(B)→ Ωn(B):

∂L

∂(∂µi

∂aj

) ηi

(n− 1)!εji2...inda

i2 ∧ · · · ∧ dain d−→

∂

∂ak

∂L

∂(∂µi

∂aj

)ηi εji2...in

(n− 1)!dak ∧ dai2 · · · ∧ dain =

∂

∂ak

∂L

∂(∂µi

∂aj

)ηi εji2...in

(n− 1)!(εki2...inj1j2...jn

1

ndaj1 ∧ · · · ∧ dajn) =

∂

∂ak

∂L

∂(∂µi

∂aj

)ηi δkj

1

n!εj1...jnda

j1 ∧ · · · ∧ dajn =1√g

∂

∂aj

∂L

∂(∂µi

∂aj

)ηi voln

The problem is this: we’d want to use Stoke’s theorem. Recall, for θ ∈ Ωn−1(B), then Stoke’s theorem is this:∫B

dθ =

∫∂B

θ

and in our case, we want to go from left to right, to the surface integral.

If the volume element dV in Marsden and Hughes [10] is the volume form, voln =√g

n! εi1...indai1∧· · ·∧dain , then, as of 20150728,

I don’t know how to apply Stoke’s Theorem, as making∫B

∂∂aj

(∂L

∂(∂µi

∂aj

)ηi)dV as differential of a n − 1 form is problematic,

with the√g factors in dV , if dV is a volume form. If it is not, or if

√g does not depend on the body reference coordinates

ai ∈ B, then we can concretely conclude that∫B

∂

∂aj

∂L

∂(∂µi

∂aj

)ηi dV =

∫∂B

∂L

∂(∂µi

∂aj

) ηi

(n− 1)!εji2...inda

i2 ∧ · · · ∧ dain

with dV = da1 ∧ · · · ∧ dan =εi1...inn! dai1 ∧ · · · ∧ dain and dV is not the volume form voln.

For the variation of∫S2φ(a, µ)dS, given on pp. 8 of Unit 1 [5], the very last term of the boxed equation,∫

S2

∂φ

∂µiηidS =

∫∂φ

∂µi〈η, dS〉i

where

〈η, dS〉i = ηkδik1

(n− 1)!εki2...inda

i2 ∧ · · · ∧ dain

If η is normalized to have length 1 and so

dSj :=1

(n− 1)!εji2...inda

i2 ∧ · · · ∧ dain

dSη = ηjdSj =⇒∫S2

∂φ

∂µiηidS =

∫S2

∂φ

∂µidSi

and ∫∂B

∂L

∂(∂µi

∂aj

) ηi

(n− 1)!εji2...inda

i2 ∧ · · · ∧ dain =

∫S2

∂L

∂(∂µi

∂aj

)ηidSjIf we can exchange ηi with the ηj implied in dSj , then

(30)∂L

∂(∂µi

∂aj

)ηj − ∂φ

∂µi= 0

Marsden and Hughes [10] defined the first Pioff-Kirchhoff stress P as such:

Definition 7 (First Pioff-Kirchhoff stress tensor). P is a (1, 1)-rank tensor (that is not symmetric)

(31) P :=∂L∂Dµ

=∂L

∂(∂µi

∂aj

)dai ⊗ ∂

∂aj∈ Γ(T ∗M ⊗ TM)

Then P jinj = ∂φ∂µi , with nj being the jth component of the normal vector on ∂B or S2.

Part 3. Implementation

9. FEniCS

9.1. On Chapter 26 “Applications in solid mechanics” by Ølgaard and Well. These are some notes/commentary andimplementations of Chapter 26 “Applications in solid mechanics” by Kristian B. Ølgaard and Garth N. Well in AutomatedSolution of Differential Equations by the Finite Element Method [11].

9.1.1. On Preliminaries for the Governing Equations; the setup. cf. 26.2.1. Preliminaries, 26.2. “Governing equations” of theFEniCS book (2011) [11].

I’m trying to understand the setup and here’s how I see it:

Let polynomial domain Ω ⊂ Rd (d = 3 usually)Let ΓD

∐ΓN = ∂Ω

Let finite interval I = (0, T ]Let Ω0 ⊂ Rd, reference domainLet triangulation of domain Ω ≡ τh

triangulation of Ω0 ≡ τ0finite element cell τ ∈ τh.

Find u ∈ V .

Given F : V × V → RF (u;w) = 0 ∀w ∈ V

, V function space.

Let F : V → V ∗

F : w 7→ F (−;w)

10

F ∈ L(V, V ∗) linear in w.

If F linear in u,

F (u;w) := a(u,w)− L(w)

where a : V × V → Ra(u,w) ∈ R

, a bilinear in u,w

L : V → RL(w) ∈ R

linear in w.

So we’d want a(u,w) = L(w) for F = 0, ∀w ∈ V .

dF (u0 + εdu;w)

dε|ε=0 =

∂F

∂uj(u0;w)duj = DFdu(u0;w) := a(du,w)

L(w) = F (u0, w) (EY: 20150804 why isn’t that L(w) = a(0, w)− F (0;w) = −F (0;w))

So as Ølgaard and Well wrote, taking (26.4),(26.5) into (26.3) [11], which is

a(du,w) := DFdu(u0;w) =dF (u0 + εdu;w)

dε

∣∣∣∣ε=0

L(w) := F (u0, w)

=⇒ a(u,w) = L(w)

Then Ølgaard and Well says to take the step u0 ← u0 − du.

a(u0 + du,w) = a(u0, w) + a(du,w) = a(u0, w) +DFdu(u0;w) = L(w) +DFdu(u0;w) = F (u0, w) +DFdu(u0;w)

EY : 20150804 My read:

When u0 → u0 + du1 → u0 + du1 + du2 → · · · → u0 +∑NI duI ≡ uf ,

s.t. F (uf , w) < ε, ε is specified tolerance,Then u is found.

9.1.2. On Balance of momentum. cf. 26.2.2. Balance of momentum, 26.2. “Governing equations” of the FEniCS book (2011)[11].

With Ølgaard and Well’s notation on the left, and mine on the right, the so-called balance of momentum condition is

ρu−∇ · σ = b in Ω× Iu = g on ΓD × I D for Dirichlet

σn = h on ΓN × I N for Neumann

u(x, 0) = u0 in Ω

u(x, 0) = v0 in Ω

ρui − ∂σij

∂xj= bi in Ω× I 3 (xi, t)

u(x, t) = g(x, t) on ΓD × Iσijnj = σij(x, t)nj(x, t) = hi(x, t) on ΓN × Iu(x, 0) = u0(x) in Ω

u(x, 0) = v0(x) in Ω

whereρ : Ω× I → R mass densityρ(x, t) ∈ Ru : Ω× I → Rd displacement field

ui(x, t) ∈ R

σ : Ω× I → Rd × Rd

σ(x, t) = σij(x, t)

symmetric Cauchy stress tensor σ = σij∂

∂xj⊗ ∂

∂xi∈ Γ(⊗2TN) (or Γ(⊗2T (R×N)))

σ(x, t) = σij(x, t)∂

∂xj⊗ ∂

∂xib : Ω× I → Rd body forcebi(x, t) ∈ Rg : Ω× I → Rd prescribed boundary displacement on ΓDgi(x, t) ∈ Rh : Ω× I → Rd prescribed boundary traction on ΓNhi(x, t) ∈ Ru0 : Ω→ Rd initial displacementui0(x) ∈ Rv0 : Ω→ Rd initial velocityvi0(x) ∈ R

Constitutive model:

cijklukul = σij with

σ = cij(u, u)∂

∂xj⊗ ∂

∂xi

c ∈ Γ(⊗2T ∗N ⊗⊗2TN)

Consider weight function w : Ω× I → R.Common practice, w : Ω→ R and apply finite difference methods to deal with time derivatives.

For time t ∈ I,assuming w = 0 on ΓD ∫

Ω

wρuidv −∫

Ω

w∂σij

∂xjdv −

∫Ω

wbidv = 0

Consider the term∫

Ωw ∂σij

∂xj dv.

9.1.3. On using Stoke’s theorem. EY : 20150805 I want to clarify the usage of the “divergence” term in the general case of aRiemannian manifold, (N, g′).

One should note that even more general is the case of a vector-valued n-form on N , with dimN = n, but it is unclear if D2 = 0for D being the exterior covariant derivative on Ωn(N,TN). But for the exterior derivative d, d2 = 0, which leads to a deRhamcohomology for p-forms on N . I will consider this case.

Now ∫Ω

wdivσivolnN :=

∫Ω

w1√g′∂σij√g′

∂xjvolnN

Consider

d(wσijdSj) := d(wσij√g′

(n− 1)!εjj2...jndx

j2 ∧ · · · ∧ dxjn) =

(∂w

∂xkσij√g′ + w

∂(σij√g′)

∂xk

)εj1j2...jn(n− 1)!

dxk ∧ dxj2 ∧ · · · ∧ dxjn =

=

(∂w

∂xkσij√g′ + w

∂(σij√g′)

∂xk

)εj1j2...jn(n− 1)!

εkj2...jni1i2...indxi1 ∧ · · · ∧ dxin =

∂w

∂xjσijvolnN + w

1√g′∂(σij

√g′)

∂xjvolnN

The dSj ∈ Ωn−1(N) n− 1-form replaces the need to ever have to define a normal vector field on ∂Ω (or, in another notation,on compact body B = B(t), ∂B(t)). dSj is indeed the surface area element with its normal in the j-direction as I’ll show.

11

Consider dimN = 3. Consider these 2-forms representing a (“rectangular” in coordinate space) surface area and (the only)corresponding nonzero value for the normal vector to that surface area element:dx1 ∧ dx2 n3 = 1dx1 ∧ dx3 n2 = −1dx2 ∧ dx3 n1 = 1

Then by induction, the usual surface area element from undergraduate (high school?) vector analysis dS = ndS, is really

dS := ndS =⇒ dSj =√g′(−1)j+1dx1 ∧ · · · ∧ dx

j∧ · · · ∧ dxn

EY : 20150805 One should note that for Riemannian manifold (N, g′), the (problematic) factor√g′ in the volume for volnN

and dSj , for dSj =√g′dx1 ∧ · · · ∧ dx

j∧ · · · ∧ dxn. This is to make it manifestly covariant, because ε Levi-Civita symbol has

to transform as a pseudo-tensor. Second, I’ll show that this dSj is the same formally as dSj =√g′

(n−1)!εjj2...jndxj2 ∧ · · · ∧ dxjn .

d(wσijdSj) =

= d(wσij√g′dx1 ∧ · · · ∧ dx

j∧ · · · ∧ dxn(−1)j+1) =

(∂w

∂xkσij + w

∂σij√g′

∂xk

)dxk ∧ dx1 ∧ · · · ∧ dx

k∧ · · · ∧ dxn(−1)j+1 =

=

(∂w

∂xjσij + w

1√g′∂(σij)

√g′

∂xj

)(−1)j+1volnN (−1)j+1 =

(∂w

∂xjσij + w

1√g′∂(σij)

√g′

∂xj

)volnN

with

dxk ∧ dx1 ∧ · · · ∧ dxj∧ · · · ∧ dxn = (−1)j+1δjkdx1 ∧ · · · ∧ dxn

and volnN =√g′dx1 ∧ · · · ∧ dxn.

So clearly

d(wσijdSj) =

(∂w

∂xjσij + wdivσi

)voln

Then the use of Stoke’s theorem (∫

Ωdω =

∫∂Ωω) makes this clear:∫

Ω

wdivσivolnN =

∫Ω

(d(wσijdSj)−

∂

∂xjσijvoln

)=

∫∂Ω

wσijdSj −∫

Ω

σijvoln

With w = 0 on ΓD,

F i :=

∫Ω

wρuidv +

∫Ω

σij∂w

∂xjdv −

∫ΓN

whijdSj −∫

Ω

wbidv = 0

9.1.4. On Potential energy minimization. cf. 26.2.3. Potential energy minization, 26.2. “Governing equations” of the FEniCSbook (2011) [11].

Consider total potential energy Π of compact body Ω0

Π = Πint + Πext

internal potential energy functional Πtextint

Πint =

∫Ω0

Ψ0(u)dV

Ψ0 ≡ stored strain energy density

Πext = −∫

Ω0

b0 · udv −∫

Γ0,N

h0 · uds

form of Ψ0 defines a particular constitutive model.

Π = minu∈V Π when u stable solution ⇐⇒ DwΠ(u) = dΠ(u+εw)dε

∣∣∣ε=0

=: F (u;w)

9.2. On Constitutive models. cf. 26.3. “Constitutive models” of the FEniCS book (2011) [11].

9.2.1. On Linearized elasticity. cf. 26.3.1. Linearized elasticity, 26.3. “Constitutive models” of the FEniCS book (2011) [11].

isotropic, homogeneous material σ = 2µε+ 2tr(ε)1

σij = 2µεij + 2tr(ε)δij

stress tensor as function of strain tensor (in this case):

εij =1

2

(∂ui∂xj

+∂uj∂xi

)µ, λ Lame parameters.

σij = Cijklεkl

Cijkl = µ(δikδjl + δilδjk) + λδijδkl

9.2.2. linelast, linelast01. At this point, one should take a look at and open up linelast.py in the github repo and see thatthe Python code corresponds exactly with the mathematical expressions for the physics of the problem. This is the case forthe Unit Cube.

For linelast01.py, I use a box mesh to try to model a strut with a very long rectangle. I’m taking cross-sectional thicknessto be about 1 in. (2.54 cm) and longitudinal, symmetry axis length of 2 ft. (0.6 m) because that was about the dimensions ofthe rocket strut in question in the second stage of Falcon 9, CRS-7 4.

I only took the Dirchlet conditions at x = 0 and at x = 0.6m, “left” and “right”, respectively, and declared it as clamp,such that u = 0, with u representing the displacement, as we “clamp” down the left and right ends of the rectangle. Theycorrespond to bcl, bcr.

Then the code correspond exactly to the previously mentioned expressions. I’ll provide an abridged dictionary here:

V a function space g = 0 on ΓD × I with ΓD = ΓD left

∐ΓD right

u ∈ V w : Ω× I → R (weight function)

b : Ω× I → Rd body force

bi(x, t) ∈ Rwith bi = (0, ρ0(2g), 0) units of Pascal/m

E = 200.0× 109 Pa , ν = 0.29 Poisson’s ratio for steel

⇐⇒4https://youtu.be/YNzhVqt4WQs

12

V = VectorFunctionSpace (mesh , ’ Lagrange ’ , 2)

clamp = Constant ( ( 0 . 0 , 0 . 0 , 0 . 0 ) )

# Create t e s t and t r i a l f unc t ions , and source term

u , w = Tria lFunct ion (V) , TestFunction (V)

b = Constant ( ( 0 . 0 , rho 0 ∗2 .∗ g 0 , 0 . 0 ) )

# E l a s t i c i t y parameters

E, nu = 200.0∗10∗∗9 , 0 .29 # 200 GPa, Gigapascal , f o r s t e e l #∗∗9

Nevertheless, the most important expressions are for the stress-tensor and the so-called F :

σij = 2µεij + λεkkδij

F i =

∫Ω

σij∂w

∂xjdv −

∫Ω

wbidv = 0

⇐⇒# St r e s s

sigma = 2∗mu∗sym( grad (u ) ) + lmbda∗ t r ( grad (u ) )∗ I d e n t i t y (w. c e l l ( ) . geometr i c d imens ion ( ) )

# Governing ba lance equat ion

F = inner ( sigma , grad (w))∗ dx − dot (b ,w)∗dx

It’s clear from this example how the use of the authors’ so-called Unified Form Language (UFL) of the FEniCS book [11] comesinto play in this example.

One could plot u, the displacement, over the rectangular mesh, after it’s been solved. I found that u was in the order ofmagnitude of 10−6m, which is small. It seems that the stress tensor values σij is more physically interesting.

I get errors when defining the so-called FacetNormal for this mesh. I’ll look further into it, why this mesh with cells oftetrahedron doesn’t have facets.

However, one could simply consider

〈u, σijgjkuk∂

∂xi〉

the inner product of the displacement u (which is a vector field) to what had already been the inner product of the displacementu with stress tensor σ, to get the pressure in the direction of the displacement u, due to the displacement u. That σ has unitsof pressure (Pascal) is clear from the units of µ.

The plotting of this is implemented by

s i gma f = 2∗mu∗sym( grad (u ) ) + lmbda∗ t r ( grad (u ) )∗ I d e n t i t y (3 )

T f = s igma f ∗uT f u = inner ( T f , u )

p l o t ( T f u /norm(u )∗∗2 , i n t e r a c t i v e=True )

Note that u was declared again as a function on function space V .

Note that one could also consider

σijgjkuk ∂

∂xi

and plot those vector values over the mesh.

From looking at the plot ofσ(u, u)/‖u‖2 = σijgikgjlu

kul/‖u‖2

I obtain a maximum value near the ends of about 3.10 pascals.

From looking at the plot ofσ(u, u) = σijgikgjlu

kul

I get about 3.10× 10−8 N at its maximum at points near the ends of the beam.

I increased the acceleration to 3g, and, from looking at a plot of σ(u, u)/‖u‖2, obtained a maximum value near the ends of about7.0 pascals. From a plot of σ(u, u), I get about 10−7 N for the force, due to stress tensor in the direction of the displacementu, due to displacement u, maximum, near the ends of the beam.

9.2.3. On Flow theory of plasticity. cf. 26.3.2. Flow theory of plasticity, 26.3. “Flow Theory of plasticity” of the FEniCS book(2011) [11].

Ølgaard and Well (2008) references Lubliner (2008)

geometrically, linear plasticity is σij = Cijklεekl. εe ∈ Γ(⊗2T ∗N) elastic part of strain tensor.

Assume ε = εe + εP

f(σ, εP , κ) := φ(σ, qkin(εP ))− qiso(κ)− σy ≤ 0

with

φ(σ, qkin(εP )) scalar effective stress measureqkin stress-like internal variable, to model kinematic hardeningqiso scalar stress-like term used to model isotropic hardeningκ scalar internal variableσy initial scalar yield stress

For von Mises (i.e. J2-flow) model, with linear isotropic hardening φ, qiso

φ(σ) =

√3

2sijsij

qiso(κ) = Hκ

where sij = σij − σkkδij/3 is deviatoric stress.H constant scalar H > 0

EY : 20150805 Then

f(σ, εP , κ) =

√3

2(σij − σkk δ

ij

3)(σij − σkk

δij3

)−Hκ− σy ≤ 0

In flow theory of plasticity,plastic strain rate εP = λ ∂g∂σwhere λ = rate of plastic multiplier

g ∈ C∞(N) plastic potentialin associative plastic flow g = f .

For isotropic strain-hardening, κ =√

23 εPij ε

Pij

13

For associative von Mises plasticity, κ =√

23 λ

2 ∂g∂σij

∂g∂σij = λ2

√23∂g∂σij

∂g∂σij

EY : 20150805 I don’t understand why

κ = λ

(cf. FEniCS book (2011) [11]) please contact me to see how this was worked out.

Since Ψ0 manifestly covariant,

Ψ0 =λ

2(Ekk)2 + µEijEij

This is a powerful statement, as ε and E are related by a diffeomorphism, then they are diffeomorphically equivalent, and somanifestly covariant relations such as Ψ0 must remain so under diffeomorphisms.

Compressible neo-Hookean model (an example of hyperelasticity):

(32) Ψ0 =µ

2(IC − 3)− µ ln J +

λ

2(ln J)2

where IC = trC and J = detF , cf. Eq. (26.33) [11], i.e.

Ψ0 =µ

2(trΦ∗g′ − dimM)− µ ln detDΦ +

λ

2(ln detDΦ)2

UFL, Unified Form Language (Chapter 17 [11] permit problems to be posed as energy minimization problems, and it won’t benecessary to compute an expression for a stress tensor, nor its linearization.

9.2.4. Time integration. cf. 26.4 Time integration [11]

Newmark methods - direct integration method, in which the equations are evaluated at discrete points in time separated by atime increment ∆t.

But it’s straightforward to extend to generalized α-methods.

Forun displacement at tnun velocity at tnun acceleration at tn

un+1 = un + ∆tun +1

2∆t2(2βun+1 + (1− 2β)un+1)

un+1 = un + ∆t(γun+1 + (1− γ)un)

Then “to solve a time dependent problem” [11],pose a governing equation at tn+1

F (un+1, w) = 0 ∀w ∈ V

9.3. FEniCS examples involving structural mechanics, solid mechanics./Applications/FEniCS.app/Contents/Resources/share/dolfin/demo/documented/hyperelasticity/python/demo_hyperelasticity.py

/Applications/FEniCS.app/Contents/Resources/share/dolfin/demo/undocumented/elasticity/python/demo_elasticity.py

/Applications/FEniCS.app/Contents/Resources/share/dolfin/demo/undocumented/elastodynamics/python/demo_elastodynamics.py

10. Paraview

After obtaining a .pvd and .vtu (vtk UnstructuredGrid), to view it, I use Paraview. Downloading and installing on a Mac OS X is easyas a binary (double-click and move and drag into the Applications Folder) 5.20150801 I don’t know how to do it in FEniCS and dolfin, from Python, itself; please let me know how to do the “In” part of File I/Oof FEniCS.

11. MeshPY

I’ll try to use MeshPy to create meshes.

12. My Examples

12.1. hyperelasticity02.py. hyperelasticity02.py is based on the (documented) demo file demo_hyperelasticity.py in/Applications/FEniCS.app/Contents/Resources/share/dolfin/demo/documented/hyperelasticity/python

3.7 meters if the diameter of Falcon 9.6 I’m just going to guess that the steel strut beam is 2 meters long. I’m going to guess that thecross section is a square of width 0.1 meters.

VectorFunctionSpace cf. http://fenicsproject.org/documentation/dolfin/dev/python/programmers-reference/functions/

functionspace/VectorFunctionSpace.html

For vector space V , VectorFunctionSpace 3 f : D ⊂M → V .

I used CompiledSubDomain to define the boundaries, but I want to keep in mind this page, making subclasses of SubDomain, 21. Poissonequation with multiple subdomains.

5Download, Paraview, http://www.paraview.org/download/6Falcon 9, SpaceX, http://www.spacex.com/falcon9

14

12.1.1. DirichletBC. EY : 20150802 DirichletBC is the Dirichlet boundary condition and I need help and feedback on what kinds of(Dirichlet) boundary conditions a steel beam or steel or whatever the metal is (stainless steel, Inconel?); would it have to deal with howthe beam or strut is bolted (and is a strut a beam?)?

Comparing the code (left) with what we’ve talked about before (right),

d = u.geometric_dimension()

I = Identity(d)

F = I + grad(u) Deformation gradient

C = FT · F

d = dimΩ0

1B

DΦ ' δij +∂ui

∂aj(a)

C := Φ∗g′

E, ν, λ, µ are the Young’s modulus, Poisson’s ratio, Lame first parameter, and Lame second parameter or sheer modulus, as given inDefinitions 2, 3, 4, 5.

Units: [ν] = 1 (dimensionless), [E], [µ], [ν] = GPa (GigaPascal)

Stored strain energy density (for the compressible neo-Hookian model), denoted ψ, is as before:

ψ =µ

2(tr(C)− 3)− µ ln (J) +

λ

2(ln J)2

where J = det(F ), which is DΦ, the pushforward of the diffeomorphism (deformation) Φ.[ψ] = [µ] = GPa

For dx, an instance of Measure, with domain everywhere on M , a “cell”, and ds, the measure of an “exterior facet”, the total potentialenergy Π, given in both ways, along with other quantities, are

Π = ψdx− (B · u)dx− (T · u)ds

F = ∇vΠ(u)

J = (∇duF )(u) =∂iF

∂uj(u)

U = ψdV −BiuidV − T iuidS

F = ∇vU(u)

JF =∂iF

∂uj(u)

[Π] = (N/m2) ∗m3 = N ∗m = J , Joules.

Consider a maximum load of 10, 000 lb. which I’ll take to be 44482.22N . The volume of the steel beam is 0.1 ∗ 0.1 ∗ 2.0 = 0.02m3.[B] = N/m2. Then take B = 0.2224111 ∗ 106N/m2, with the “longitudinal” surface being 2.0 ∗ 0.1 = 0.2m2.

For “traction” T , [T ] = N/m. I’ll take the traction to be the maximum stress considered (on the chart’s axis) in the weld-ing by gas tungsten-arc process for Inconel alloy X-750, which is 220 ksi, or 1516.9 MPa 7. Taking the 0.1 m width, thenT = 1516.9 ∗ 106 ∗ 0.1 = 1.5169 ∗ 108N/m.

12.1.2. Dirichlet boundary conditions. Given a partial differential equation (PDE), ∇2y + y = 0 = F (x, y,∇2y), the Dirichlet boundarycondition is y(x) = f(x) ∀x ∈ ∂Ω 8. From wikipedia, in mechanical engineering, the Dirichlet boundary condition applies to the end(s)of the beam that’s “held at a fixed position in space.”

12.2. demo elastodynamics.py and elastodynamics 01.py. demo_elastodynamics.py

demo_elastodynamics.py is an undocumented demo found in

/Applications/FEniCS.app/Contents/Resources/share/dolfin/demo/undocumented/elastodynamics/python.

I’ll go ahead and document it, and further (try to) modify it.

It appears to follow Chapter 27 “A Computational framework for nonlinear elasticity” by Harish Narayanan [11], which I’ll review here.

12.2.1. On “A computational framework for nonlinear elasticity”. Noting “disallowing interpenetration of matter or formation of cracks.”

(deformation) diffeomorphism Φ : Ω0 × [0, T ]→ N

In Narayanan’s notation, ϕ :Ω× [0, T ]→ R2,3

Ω := Ω⋃∂Ω

Construct ∀ a ∈ Ω0, u(a, t) = Φ(a, t)− a ∈ RdimN ; dimN = dimΩ0.

Subject body Ω0 tobody forces, B(a, t)traction forces, T (a, t) ∈ X(N , forces per unit surface area, acting on Neumann boundary of body ∂ΩN

displacement boundary conditions; Dirichlet boundary on ∂ΩD.

∂ΩN

⋂∂ΩD = ∅, ∂ΩN

⋃∂ΩD = ∂Ω

From cf. 27.1.2. The basic equation we need to solve [11],

ρ∂2u

∂t2= div(P ) +B on Ω0

ρ ∈ C∞(Ω0)first Piola-Kirchhoff stress tensor P .

12.2.2. Exporting animations into ParaView. This useful Question and Answer helped with explaining how to export the animation intoParaView cf. How to make a video of a sequence of FEM functions?. Essentially, you would take elasticity.pvd, which references allthe time steps/slices of data, and open it up in Paraview. You would also need to Apply in Paraview, “WarpByVector” to see the actualdeformation.

[ParaView] Basics of Keyframe Animation from UM3DLab’s YouTube channel was also useful as a tutorial.

Part 4. Dictionary

In this part, entitled “Dictionary,” I have further entries for a “dictionary” between the laws of physics and its application in engineering.

13. Curl

James T. Wheeler had worked out explicitly how the Levi-Civita connection is related to the curl 9

(33) ∗ dv =

√g

(n− 2)!((∇iv)[j)ε

ijj3...jn

dxj3 ∧ · · · ∧ dxjn = (curlv) ∈ Ωn−2(M)

Indeed, if n = 3, Γmil = 0, gjm = δjm,

∗dv =∂vj

∂xiεijj3dx

j3 =

(∂vj

∂xi− ∂vi

∂xj

)dxj3

while, from “vector analysis”

(∇× v)k = εijk∂vj

∂xi∂

∂xk=

(∂vj

∂xi− ∂vi

∂xj

)∂

∂xk

7http://www.specialmetals.com/documents/InconelalloyX-750.pdf8“Dirichlet boundary condition”, wikipedia, https://en.wikipedia.org/wiki/Dirichlet_boundary_condition9James T. Wheeler, Differential Forms, which is (the last) part of his bigger book for (classical) mechanics Mechanics, http://www.physics.usu.edu/Wheeler/ClassicalMechanics/MechanicsBookFall2014.pdf

15

14. Strain

What’s called u in Landau and Lifshitz [1] and here is ε in a number of engineering texts [4], [5].

(U, ai) ∈ A of (M,O,A, g)

u ∈ Γ(T ∗M ⊗ T ∗M)

u = uijdai ⊗ daj

uij =1

2(gkj

∂uk

∂ai+ gik

∂uk

∂aj+ uk ∂gij

∂ak)

(U,Xi) ∈ A of (M,O,A, g)

ε ∈ Γ(T ∗M ⊗ T ∗M)

ε = εijdXi ⊗ dXj

εij =1

2(gkj

∂uk

∂Xi+ gik

∂uk

∂Xj+ uk ∂gij

∂Xk)

If gkj = δkj ,

uij =1

2

(∂uj

∂ai+∂ui

∂aj

)εij =

1

2

(∂uj

∂Xi+

∂ui

∂Xj

)

15. Stress

What I denote as the Cauchy stress tensor T is usually denoted with σ in engineering [4], [5]:

(U, xi) ∈ A of (N,O,A, g′) T ∈ Γ(TN ⊗ TN)

T = T ijej ⊗ ei

(U, xi) ∈ A of (N,O,A, g′) σ ∈ Γ(TN ⊗ TN)

σ = σijej ⊗ ei

16

References

[1] L. D. Landau, E.M. Lifshitz, Theory of Elasticity, Second Edition: Volume 7 (Course of Theoretical Physics), Pergamon Press, 1970.

[2] T. Frankel, The Geometry of Physics, Cambridge University Press, Second Edition, 2004.[3] Shigeyuki Morita, Geometry of Differential Forms (Translations of Mathematical Monographs, Vol. 201) 2001

[4] Paul Lagace. 16.20 Structural Mechanics, Fall 2002. (Massachusetts Institute of Technology: MIT OpenCourseWare), http://ocw.mit.edu (Accessed 23 Jul, 2015). License: Creative Commons BY-NC-SA http://ocw.mit.edu/courses/aeronautics-and-astronautics/

16-20-structural-mechanics-fall-2002/

[5] Raul Radovitzky. 16.225 Computational Mechanics of Materials, Fall 2003. (Massachusetts Institute of Technology: MIT OpenCourseWare), http://ocw.mit.edu (Accessed 27 Jul, 2015). License: Creative Commons BY-NC-SA http://ocw.mit.edu/courses/aeronautics-and-astronautics/

16-225-computational-mechanics-of-materials-fall-2003/#

[6] John Lee, Introduction to Smooth Manifolds (Graduate Texts in Mathematics, Vol. 218), 2nd edition, Springer, 2012, ISBN-13: 978-1441999818[7] Ernest Yeung, “Solutions To Introduction To Smooth Manifolds by John M. Lee, 2012, Springer,” 2012

[8] Darryl D. Holm, Tanya Schmah, Cristina Stoica, Geometric Mechanics and Symmetry: From Finite to Infinite Dimensions (Oxford Texts in Applied and Engineering Mathematics) 2009, ISBN-13: 978-0199212903 ISBN-10: 0199212902

[9] Tsutomu Kambe, Geometrical Theory of Dynamical Systems and Fluid Flows, Advanced Series in Nonlinear Dynamics: Volume 23, 2009, http://www.worldscientific.com/worldscibooks/10.1142/7418 ISBN: 978-981-4282-24-6 (hardcover)[10] Jerrold E. Marsden, Thomas J. R. Hughes, Mathematical Foundations of Elasticity (Dover Civil and Mechanical Engineering), Dover Publications, 1994 Available at the late Prof. Marsden’s website: http://authors.library.caltech.edu/25074/1/Mathematical_Foundations_of_Elasticity.

pdf

[11] A. Logg, K.-A. Mardal, G. N. Wells et al. (2012). Automated Solution of Differential Equations by the Finite Element Method, Springer. [doi:10.1007/978-3-642-23099-8] http://launchpad.net/fenics-book/trunk/final/+download/fenics-book-2011-10-27-final.pdf

There is a Third Edition of T. Frankel’s The Geometry of Physics [2] and a third edition of Landau and Lifshitz’s Theory of Elasticity [1], but I don’t have the funds to purchase the book (about $ 71 US dollars, with sales tax, ahd $ 66 USD for thelatter). It would be nice to have the hardcopy text to see new updates and to use for research, as the second edition allowed me to formulate fluid mechanics and elasticity in a covariant manner. Please help me out and donate at ernestyalumni.tilt.com.

E-mail address: ernestyalumni@gmail.com

URL: http://ernestyalumni.wordpress.com

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