Thesis presentation

Objectives Literature Survey Assumptions Kinematics Stress and Strain Measures Order analysis Derivation of governing equations FEM formulation Numerical Implementation Coupling with heat equation Results Appendix

Finite Element Study of Elastic Rods underThermal Loading

Jaspreet SinghDepartment of Mechanical EngineeringIndian Institute of Technology, Kanpur

16/07/20151 / 57


Overview

1 Objectives

2 Literature Survey

3 Assumptions

4 Kinematics

5 Stress and Strain Measures

6 Order analysis

7 Derivation of governing equations

8 FEM formulation

9 Numerical Implementation

10 Coupling with heat equation

11 Results

12 Appendix

2 / 57


Objectives

To determine the displacement, strains, stresses in an elasticrod subjected to thermal loads.

Explore the possibility of buckling under thermal loads.

Study the effect of temperature on buckling in response tomechanical loads.

3 / 57


Literature Survey

Theoretical aspects: Green, Dill and Antman.

Derivation of the equations of rod theory from 3-D elasticity.Order analysis.

Numerical implementation: Simo, Argyris and Bathe.

Snaps throughs, saddle points and discontinuities of derivativeslead to failure of standard algorithms.Non-commutative finite rotations.Non-conservative external loading.

Recent work

Holmes and Cisternas: 2-D model, uniaxial temperature field,calculation of buckling temperature.Coffin and Bloom: Analytical solution for simply supportedrod.Goriely et al: Finite growth strains, applications to arteries,growth of stems and bones etc.

4 / 57


Assumptions

Large deformations, small strains.

Linearly isotropic elastic deformations.

Plane CS remains plane, warping is disallowed.

Dimensions of CS Length of rod and radii of curvature.

Additive decomposition of total strain into elastic and thermalcomponents.

5 / 57


Kinematics

6 / 57


Any point in reference configuration:r(S , ζ1, ζ2, 0) = φ

0(S , 0) + ζ1E 1 + ζ2E 2.

Any point in current configurationr(S , ζ1, ζ2, t) = φ

0(S , t) + ζ1(1 + ε1)t1 + ζ2(1 + ε2)t2.

Evolution of centroidal axis:φ

0(S , t) = φ

0(S , 0) + u1e1 + u2e2 + u3e3.

Evolution of orthonormal frame:Λ(S , t) = t1 ⊗ E 1 + t2 ⊗ E 2 + t3 ⊗ E 3

Curvature:dE idS = ω0 × E i and

dt idS = ω × t i

7 / 57


Stress and Strain Measures

P is the first Piola Kirchoff stress tensor such that

P = T 1 ⊗ E 1 + T 2 ⊗ E 2 + T 3 ⊗ E 3.

The spatial description of the stress measures is as follows:

Stress Strain

n =∫CS T 3 dA γ =

dφ0

dS − t3

= s1t1 + s2t2 + et3

m =∫CS(r − φ

0)× T 3 dA β = ω − Λω0

= k1t1 + k2t2 + k3t3

Material Analogs of n,m, γ and β are N,M, Γ and κ, respectively.

8 / 57


Order analysis

Let h, a, b and L denote the dimension of CS, radii of curvature inundeformed and current configuration and characteristic length ofthe rod.

ε = greatest(h/L, h/a, h/b, e, s1, s2, ε1, ε2)ζ1, ζ2, T i ∼ O(ε)

Theory is valid when ε 1.For additive decomposition to hold true it has been assumed

αT (S , ζ1, ζ2) ∼ O(ε)

9 / 57


Inclusion of Temperature

Note that, the temperature field is allowed to have mostgeneral form of T (S , ζ1, ζ2).

Up to first order, deformation gradient is written as

F = (1 + ε1)t1 ⊗ E 1 + (s1 − ζ2k3)t1 ⊗ E 3

(1 + ε2)t2 ⊗ E 2 + (s2 + ζ1k3)t2 ⊗ E 3

(1 + e + ζ2k1 − ζ1k2)t3 ⊗ E 3

= FiI t i ⊗ E I

F e = F (F θ)−1 =1

1 + αTF ≈ (1− αT )F

= (1− αT )(t i ⊗ E i + f ) where f ∼ O(ε)

= (1− αT )t i ⊗ E i + f

(1)

10 / 57


T 3 = G (s1− ζ1k3)t1 +G (s2 + ζ2k3)t2 +E (e− ζ1k2 + ζ2k1−αT)t3

(2)Furthermore,

∫CS ζ1dA = 0,

∫CS ζ2dA = 0 and

∫CS ζ1ζ2dA = 0

n =

∫CS

T 3dA = GA1s1t1 + GA2s2t2 + (EAe − PT)t3

where PT (S) = E

∫CSαT (ζ1, ζ2,S) dA

(3)

m =

∫CS

(r − φ0)× T 3 dA

= [EI1k1 −MT1]t1 + [EI2k2 + MT2]t2 + GJk3t3

where MT1 = E

∫CSζ2αT dA and MT2 = E

∫CSζ1αT dA

(4)

Temperature field is, thus, completely decomposed into a bodyforce and body moment.

11 / 57


Using∫CS ζ1dA =

∫CS ζ2dA =

∫CS ζ1ζ2dA = 0 and∫

VolP : F dV =

∫ S

0

∫CS

S : E dAdS

=

∫ S

0[GAs1s1 + GAs2s2 + EAee

+ EI1k1k1 + EI2k2k2 + GJk3k3

−MT1k1 + MT2k2 − PTe]dS

=

∫ S

0[N.Γ + M.κ]dS

(5)

where∫CS ζ

21 dA = I2,

∫CS ζ

22 dA = I1 and

∫CS(ζ2

1 + ζ22 )dA = J.

12 / 57


Linearization of strain measures

Set of all possible configurations of the rod is given asC = (φ

0(S , t),Λ(S , t)) : φ

0∈ R3,Λ ∈ SO(3).

Φ = (φ0,Λ) ∈ C is perturbed to Φε = (φ

ε,Λε) ∈ C , according the

following scheme.

φε

= φ0

+ εη0

Λε = exp (εΘ)Λ

where η0

= ηie i and Θ = θiI e i ⊗ e I =

0 −ϑ3 ϑ2

ϑ3 0 −ϑ1

−ϑ2 ϑ1 0

ϑ = axial(Θ) = ϑie i

(6)

η = (η0, ϑ) ∈ TΦC , tangent space of C at Φ.

13 / 57


Under the described scheme the linearization of any quantity Q(Φ)is defined as

δQ(Φ, η) =d

dεQ(φ

ε,Λε)|ε=0 (7)

Thus the linearization of strain measures is

Γε = ΛTε φε − E 3

δΓ(Φ, η) =d

dεΓε|ε=0

= ΛT (dη

0

dS− ϑ×

dφ0

dS)

(8)

δκ(Φ, η) = ΛTϑ (9)

14 / 57


Governing equations

For a rod with length L0 loaded under distributed loads, n and mand contact loads N and M on the ends.

δWint =

∫ L0

0(N.δΓ + M.δκ)dS

=

∫ L0

0(n.(

dη0

dS− ϑ×

dφ0

dS) + m.

dϑ

dS)dS

(10)

δWext =

∫ L0

0(n.η

0+ m.ϑ)dS + [N.η

0+ M.ϑ]0,L0 (11)

15 / 57


Governing equations are as follows:

δWint − δWext = 0 (12)

dn

dS+ n = 0 (13)

dm

dS+

dφ0

dS× n + m = 0 (14)

Drichlet Boundary Conditions Neumann Boundary Conditions

η0

= 0 n = N

ϑ = 0 m = M

n = GAs1t1 + GAs2t2 + (EAe − PT)t3

m = (EI1k1 −MT1)t1 + (EI2k2 + MT2)t2 + GJk3t3.

16 / 57


FEM formulation

6 dof/node i.e. p =[u1 u2 u3 ϑ1 ϑ2 ϑ3

]Number of elements =nel

G =nel∑e=1

Ge (15)

Number of nodes per element=n=2

η = (η0, ϑ) is the virtual displacement such that η

0= 0 and

ϑ = 0 at S = 0, L0.

η0

=n∑

i=1

Niηi

ϑ =n∑

i=1

Niϑi

(16)

17 / 57


From Eq[10] and Eq[11]

G (Φ, η) =nel∑e=1

Ge

Ge(Φ, η) = δWint − δWext

=

∫ L0

0(n.(

dη0

dS− ϑ×

dφ0

dS) + m.

dϑ

dS)dS −

∫ L0

0(n.η

0+ m.ϑ)dS

(17)

18 / 57


Ge =

∫ L0

0(n.(

n∑i=1

N ′i ηi +n∑

i=1

Niϑi ×dφ

0

dS) + m.

n∑i=1

N ′iϑi )dS

−∫ L0

0(n.

n∑i=1

Niηi + m.n∑

i=1

Niϑi )dS

=n∑

i=1

Gei =n∑

i=1

Pei .[ηi ϑi ]T

(18)

where

Pei =

∫ L0

0(

[Ni I 0

−Ni [φ′0×] Ni I

] [nm

]−[Ni I 00 Ni I

] [nm

])dS

Pe =[

Pe1 Pe2 . . . Pei . . . Pen

]T (19)

Pei is the residual force vector corresponding to i th node of eth

element.19 / 57


Note that

[nm

]=

[Λ 00 Λ

]

GAs1

GAs2

EAe − PT

EI1k1 −MT1

EI2k2 + MT2

GJk3

(20)

Let

Ξ =

[Ni I 0

−Ni [φ′0×] Ni I

]C = diag[GA GA EA EI1 EI2 GJ]

Π =

[Λ 00 Λ

]NT =

[0 0 PT

]TMT =

[MT1 −MT2 0

]T(21)

20 / 57


Residual given by Eq[19] can be written in the conventional form as

Pei = f int − f ext

f int =

∫ L0

0(ΞΠ

GAs1

GAs2

EAeEI1k1

EI2k2

GJk3

)dS =

∫ L0

0(Ξ Π C

[Γκ

])dS

f ext =

∫ L0

0(Ξ Π

[NT

MT

]+

[Ni I 00 Ni I

] [nm

])dS

(22)

21 / 57


To derive the stiffness matrix, variation of Eq[19] is taken.Contrary to virtual displacement herein perturbation u = (u0, ψ) isa physical perturbation.

∆n = ∆(ΛN) = [ψ×]n + ΛC1ΛT (u′0 − ψ × φ′0) (23)

∆m = ∆(ΛM) = [ψ×]m + ΛC2ΛTψ′ (24)

∆(φ′0× n) = −[n×]u′0 + [(n.φ′

0)I + n ⊗ φ′

0]ψ+

[φ′0×]ΛC1ΛTu′0 + [φ′

0×]ΛC1ΛT [φ′

0×]Tψ

(25)

22 / 57


∆Pei =

∫ L0

0

( [ N ′i ∆nN ′i ∆m − Ni∆(φ′

0× n)

] )dS

=

∫ L0

0

n∑j=1

(Seij + Teij

).uj

(26)

Putting in the values from Eq[23], Eq[24] and Eq[25] yields therequisite stiffness matrix.

23 / 57


Numerical implementation

Problem: f int(u)− λf f ext(u) = 0

1: Define R = f int(un)− λnf ext(un) and use arclength method totrace the curve. n denotes the loadstep.

2: Break at nth loadstep, if λn > λf .3: Define R = f int(u)− λf f ext(u). Use un−1 as the initial guess

and use NR procedure to converge to the exact solutioncorresponding to λn = λf .

4: If NR iterations in step[3] do not converge, go back to step[1]and change the step size in the arclength method.

24 / 57


Coupling with heat equation

In the domain shown following problem for temperature field (T) issolved using FEM

∇2T = h in Ω

T = T 0 on ΓT

k∂T

∂n= q on Γq

(27)

Knowing the dimensions of the CS (say [−a, a]× [0, b] andζ ∈ [−a, a]), numerical integration is carried out to calculatePTm =

∫CS EαTmdA = b

∫ a−a EαTmdζ and

MTm = b∫ a−a EαTmζdζ for each iteration in every loadstep.

25 / 57


Validation of code

CS 1× 1 cm2, length 1m, E=200× 109 Pa and G=1012 Pa

error= 1N

√N∑

n=1(‖x fn − xen‖2 + ‖y fn − y en‖2)

N 2 3 5 7 10 25 50

3D 0.2741 0.1820 0.1290 0.1066 0.0882 0.0552 0.0390

2D 0.2740 0.1821 0.1291 0.1067 0.0882 0.0553 0.0390

26 / 57


Figure: Error v/s Number of elements27 / 57


Thermal buckling of a beam

A simply supported beam on both ends with E=200× 109,length=1, G=1014, area=10−4 and moment of inertia=10−8,α = 10−5 is considered. The hypothesis postulated is

e ∼ s ∼ O(ε) ∼ αT (28)

Hence, the formulation should give correct results upto∆T ∼ ε

α ≈ 1000K . The magnitude of e and s gives an idea aboutthe total strain.

28 / 57


Figure: Initial and Final Shapes of Rod29 / 57


Figure: Midpoint deflection v/s Temperature30 / 57


Figure: Shear Strain v/s Temperature31 / 57


Figure: Total Axial strain (e) v/s Temperature32 / 57


Figure: Elastic Axial Strain (e − αT ) v/s Temperature33 / 57


Figure: Thermal Strain (αT ) v/s Temperature34 / 57


Effect of temperature field on buckling loads

Euler Column, fixed at(A) and subjected to a compressive load (P)at (B) is considered. The temperature profile is given asT (S , ζ) = T0ζ/b where ζ ∈ [−b, b] is the coordinate on the CS([−b, b]× [0, a]) measured from the neutral axis, 2b is thethickness of the cross section, a is the out of the plane dimensionof CS. α = 10−5, E=200× 109 Pa, G= 1014 Pa, length= 1m,area= 10−4 m2 and moment of inertia = 10−8 m4. T0 is theparameter whose effect on buckling load is considered.

35 / 57


Figure: Buckling of Euler Column36 / 57


Angular beam subjected to constant temperature field

A curved rod fixed at both ends with length=2 m, α = 10−5,thedimensions of CS 10−2 × 10−3 m2, E = 200× 109Pa, G = E/2subjected to a constant temperature rise (T (S , ζ1, ζ2) = T ) isconsidered. The objective of the problem is to explore the effect ofthe radius of curvature on buckling temperature. The problem issolved for φ = π/24, π/12, π/6, π/3, π/2, 2π/3 and π.

37 / 57


Figure: Thermal buckling of a curved rod38 / 57


Figure: Thermal buckling of a curved rod39 / 57


Non-uniform temperature distribution

Here, α = 10−5, Tmax = 200, E = 200× 109 Pa, G = 1015 Pa,area A = 10−4 m2 and moment of inertia I = 10−8 m2. Thetemperature boundary conditions are specified on the two ends(AC, BD) of the beam while the lateral surfaces (AB, CD) areconsidered to be insulated. At every loadstep, the temperature’matrix’ is λT/Tmax∀λ ∈ [0,Tmax ]. Here, Tmax = 200.

40 / 57


Figure: Temperature profile in the beam41 / 57


Figure: Initial and final shapes of the rod42 / 57


Figure: Midpoint deflection v/s load parameter (λ)43 / 57


Thank You

44 / 57


Kinematics

The complete description of rod parametrized by S ∈ [0, L0]requires specification of

Locus of the centroids (Centroidal Axis),φ

0(S , t) : S ∈ [0, L0], t ∈ R+ 7−→ R3.

Orthonormal frame attached to each cross-section (CS)[t1, t2, t3], t i (S , t)1 : S ∈ [0, L0], t ∈ R+ 7−→ R3 .

t = 0 denotes the reference configuration wherein rod isassumed to be smooth arbitrary curve.

[e1, e2, e3] is the fixed spatial frame.

1 i, I = 1, 2, 3

45 / 57


Reference Configuration

Curvilinear coordinates: (ζ1, ζ2, S)

Centroidal axis: φ0(S , 0)

Orthonormal frame at each CS:[t1(S , 0), t2(S , 0), t3(S , 0)] = [E 1(S), E 2(S), E 3(S)]

Any point on rod in reference configuration

r(S , ζ1, ζ2, 0) = φ0(S , 0) + ζ1E 1 + ζ2E 2

φ0(S , 0) =

∫ S

0E 3dS

(29)

46 / 57


Curvature tensor Ω0(S) : S ∈ [0, L0] 7→ so(3) 2

dE i

dS= Ω0E i = ω0 × E i (30)

such that, ω0 = χ01E 1 + χ0

2E 2 + χ03E 3 is the axial vector of

Ω0 = ΩiIE i ⊗ E I .

2so(3) is the space of all skew symmetric tensors.

47 / 57


Current configuration

Centroidal axis φ0(S , t)

φ0(S , t) = φ

0(S , 0) + u1e1 + u2e2 + u3e3 (31)

Orthonormal frame at each CS [t1, t2, t3] such that

t I = Λ(S , t)E I

Λ(S , t) = t1 ⊗ E 1 + t2 ⊗ E 2 + t3 ⊗ E 3

= ΛiI e i ⊗ E I

(32)

such that Λ(S , t) : S ∈ [0, L0], t ∈ R+ 7→ SO(3)3

3SO(3) is the space of all orthonormal tensors.

48 / 57


Any point on the rod

r(S , ζ1, ζ2, t) = φ0(S , t) + ζ1(1 + ε1)t1 + ζ2(1 + ε2)t2 (33)

Curvature tensor Ω(S) : S ∈ [0, L0] 7→ so(3)

dt idS

= Ωt i = ω × t i (34)

such that, ω = χ1t1 + χ2t2 + χ3t3 is the axial vector ofΩ = ΩiI t i ⊗ t I .

From Eq[4]dΛ

dS= Ω Λ (35)

49 / 57


Notation

For any vector a ∈ R3, a = a1e1 + a2e2 + a3e3 is the spatialdescription and a = ΛTa = b1E 1 + b2E 2 + b3E 3 is thecorresponding material description. For a tensor A,A = ΛTA Λ.

For any tenor A ∈ so(3) ∃ a ∈ R3 such that Ah = a× h,known as axial vector, denote by a = axial(A).

50 / 57


F e =

1 + ε1 − αT 0 00 1 + ε2 − αT 0

s1 − ζ2k3 s2 + ζ1k3 1 + e + ζ2k1 − ζ1k2

= F e

iI t i ⊗ E I

(36)

The elastic Green-Langrange strain tensor is written as:

E e =1

2((F e)TF e − I) (37)

Following constitutive relation is used

S = 2µE e + λtrE eI (38)

where S is second Piola Kirchhoff stress.

S = σkmG k ⊗ Gm ≈ σkmE k ⊗ Em (39)

51 / 57


Assume σ11 = σ22 = 0. This gives

σ31 = σ13 = G (s1 − ζ1k3)

σ32 = σ23 = G (s2 + ζ2k3)

σ33 = E (e − ζ1k2 + ζ2k1 − αT )

(40)

P = F S = (t i ⊗ E i + O(ε))S = σkmtk ⊗ Em + O(ε2) (41)

This gives

T 3 = G (s1− ζ1k3)t1 +G (s2 + ζ2k3)t2 +E (e− ζ1k2 + ζ2k1−αT )t3

(42)

52 / 57


Strain Measures

From Eq[18], the strain measures obtained are as follows:

Γ = ΛTγ = ΛTdφ

0

dS− E 3 = s1E 1 + s2E 2 + eE 3

κ = ΛTβ = k1E 1 + k2E 2 + k3E 3

(43)

Spatial counterparts can be obtained as follows

γ = ΛΓ = s1t1 + s2t2 + et3 =dφ

0

dS− t3

β = Λκ = k1t1 + k2t2 + k3t3

(44)

53 / 57


Stress measures

The material counterparts of the same are given as

N = GAs1E 1 + GAs2E 2 + (EAe − PT)E 3

M = (EI1k1 −MT1)E 1 + (EI2k2 + MT2)E 2 + GJk3E 3

(45)

The spatial description of the stress measures is given for reference.

n = Λ N = GAs1t1 + GAs2t2 + (EAe − PT)t3

m = Λ M = (EI1k1 −MT1)t1 + (EI2k2 + MT2)t2 + GJk3t3

(46)

54 / 57


Simplifications in 2D

Continuous functions β(S) : S ∈ [0, L0] 7→ R andθ(S) : S ∈ [0, L0] 7→ R are enough to describe the orthonormalframe and curvature of the rod in reference and currentconfigurations, respectively.t1 is tangent to the centroidal axis.

E 1 = cosβe1 + sinβe2

E 2 = − sinβe1 + cosβe2

t1 = cos(β + θ)e1 + sin(β + θ)e2

t2 = − sin(β + θ)e1 + cos(β + θ)e2

ω0 = βSe3 ω = (βs + θS)e3

κ = θSe3∫ S2

S1

ωdS = (θ + β)|S2S1

e3

(47)

55 / 57


Derivation of governing equations for rods withtemperature field

1: Calculate contravariant base vectors G 1,G 2,G 3 from Eq[1].2: Calculate covariant base vector g

1, g

2, g

3from Eq[5].

3: Calculate deformation gradient F = gi⊗ G i .

4: Calculate F e = F (F θ)−1 where F θ = (1 + αT (S , ζ1, ζ2))I5: Retain only first order terms and neglect higher order terms.6: Use the constitutive relation to calculate P.7: Calculate the stress measures as described.

56 / 57


Bibliography

E. H. Dill. Kirchhoff’s Theory of Rods. Archive for history of Exact Sciences 21.V, Vol 44, Issue 1, pp 1-23,

1992

Bruno A. Boley, and Jerome H. Weiner. Theory of thermal stresses. John Wiley and Sons, Inc, 1962.

J.C. Simo and L. Vu-Quoc. A three dimensional finite strain rod model. Part II: Computational aspects .

Computer Methods in Applied Mechanics and Engineering, 116:58-79, 1986.

J. Cisternas and P. Holmes. Buckling of extensible thermoelastic rods . Mathematical and Computer

Modelling, 36:233-243, 2002.

S S Antman. Nonlinear problems of elasticity (2nd edn). Springer, 1989.

A.E. Green, F.R.S., N. Laws A general theory of rods. Proc. R. Soc. Lond. A 1966 293, 145-155.

J.H. Argyris and Sp Symeonidis. Nonlinear finite element analysis of elastic systems under nonconservative

loading- Natural formulation. Part 1 Quasistatic problems . Computer Methods in Applied Mechanics andEngineering, 26:75-123, 1981.

K. J. Bathe, S. Bolourchi. Large displacement analysis of 3-D beam structures . Internation Journal for

Numerical Methods in Engineering, 14:961-986,1979.

D. W. Coffin and F. Bloom Elastica solution for the hygrothermal buckling of a beam. International Journal

of Non-Linear Mechanics, 34: 935-947, 1999.

Crisfield M.A. A fast incremental/iterative solution procedure that handles snap-through. Computer and

Structures, 13(1):5562, 1981.

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