Nonlinear mechanics of rods, sheets and ribbons - IMA-UMN

Nonlinear mechanics of rods, sheets and ribbons –interplay between geometry and mechanics

Gert van der Heijden

Department of Civil, Environmental & Geomatic EngineeringUniversity College London

Gert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 1 / 72

Outline

Nonlinear dynamics of large-deformation structures

Geometrically-exact theory of highly-flexible slender structures

We require:

Geometry of curves and surfacesCalculus of variationsNonlinear dynamics – bifurcation theory

We consider:

RodsRectangular strips/sheetsHelical ribbonsAnnular stripsMulti-strand structures (braids)

Quasi-rod theory

(joint work with Eugene Starostin)


Examples of large deformations of 1D and 2Dstructures

(The need for nonlinear mechanics)


Pipelines and cables

Marine pipelines: (a) (b)

( ) (d)

(e) (f)

Cables hanging under gravity (catenary)

Looping and loop pop-out of cables and pipelines (with self-contact)


Mechanical modelling of DNA supercoiling

Single-molecule DNA experiments and (closed) DNA plasmids:


Textile spinning

Whirling and snarling of a moving threadline (textile yarn ballooning):


Magnetic buckling and space tethers

Magnetic buckling and electrodynamic space tethers (use earth’s magnetic fieldfor propulsion):

Post-buckling tether shapes (first and fifth mode):

z

y

x

z

y

x

Control of periodic orbits of a tether moving around the earthGert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 7 / 72

Hairstyle modelling

(Bertails et al., Eurographics, 2005)

Important effects of eccentricity, natural curliness and collisions, as well as wetting


Animal and robotic whiskers

(Towal et al., PLoS Comput. Biol., 2011) (Solomon, Hartmann, Nature, 2006)


Electrospinning

(Reneker & Yarin, Polymer, 2008)


Other examples of slender one-dimensional structures

Viscous fluid jet (corn syrup) – bacterium (bacillus subtilis) – tendril (Darwin)


Wrinkling and crumpling of sheets

Ripples in a stretched sheet – triangular buckling pattern in a twisted strip:


Helical ribbons

Bistability of cholesterol ribbons:

Zastavker et al. (1999) Smith et al. (2001)

Zinc oxide and other nanosprings (Wang, 2007; Zhang et al., 2007):

Potential use in nanoscale force probes, transducersGert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 13 / 72

Annular nanostructures

Saddling of annular nanostructures:

(from: Cho et al., Nanoletters 10, 2010)


Dynamical systems view

Perhaps simplest mechanical equation

mu + ku = 0 or u + ω2u = 0, where ω =

√k

m

Geometrical view (write as (first-order) system):

x = y

y = −ω2x

(x , y) space (phase portrait)

Convenient because initial conditions are (x(0), y(0)), i.e., position and velocity,so a point in this phase portrait.

Forward mapping of all solution curves together called a flow


Phase portrait

Phase portraits of x = f (x) with f = −U ′Gert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 16 / 72

Phase portrait of the pendulum

T =1

2ml2φ2

U = mg(l − l cosφ) = −mgl cosφ+ const.

Lagrangian: L = T − U

φ+ λ sinφ = 0, λ =g

lGert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 17 / 72

Phase portrait of the pendulum with damping


Linearisation – fixed point stability

2D: u = Au, u = (x , y)T


Change of stability in parametrised equations – bifurcations

Critical changes in the system’s behaviour as parameters are varied (change ofstability)

Possibilities (elementary bifurcations of 1D equilibria):

1) saddle-node or fold x = µ− x2

2) transcritical bifurcation x = µx − x2


Bifurcations (cont’d)

3) pitchfork bifurcation x = µx − x3

4) Hopf bifurcation (involves a periodic solution, requires 2D)


Pitchfork bifurcation theorem

A parametrised ODE

z = f (x , µ), (x , µ) ∈ R× R

with f satisfying f (−x , µ) = −f (x , µ) (hence x = 0 is a fixed point) and

∂f

∂x(0, µ0) = 0,

∂2f

∂x2(0, µ0) = 0,

∂3f

∂x3(0, µ0) 6= 0,

∂f

∂µ(0, µ0) = 0,

∂2f

∂x∂µ(0, µ0) 6= 0

has a pitchfork bifurcation at (x , µ) = (0, µ0).

x = µx − x3 x = µx + x3

(supercritical) (subcritical)Gert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 22 / 72

In second-order system

x = f (x) = −dV

dx

saddle-node:

x + x2 − µ = 0

(V (x) =

1

3x3 − µx

)[

x = yy = µ− x2 linearisation: z = Az , A =

(0 1−2x 0

)Fixed points: y = 0, x = ±√µ (µ ≥ 0)

Eigenvalues: λ2 + 2x = 0

x =√µ: λ1,2 = ±i

√2√µ (centre)

x = −√µ: λ1,2 = ±√

2√µ (saddle)


pitchfork bifurcation:

x + x3 − µx = 0

(V (x) =

1

4x4 − 1

2µx2

)[

x = yy = µx − x3 linearisation: z = Az , A =

(0 1

µ− 3x2 0

)Fixed points: y = 0, x = 0, ±√µ

Eigenvalues: λ2 − (µ− 3x2) = 0

x = 0: λ1,2 = ±√µ (centre for µ < 0, saddle for µ > 0)

x = ±√µ: λ1,2 = ±i√

2µ (µ > 0) (centre)

x = µx − x3 x = µx + x3

(supercritical) (subcritical)


2D elastic rod – Euler elastica

Euler elastica

x ′ = cos θ, y ′ = sin θ

Elastic energy: U = 12B

∫ S

0

κ2 ds, curvature κ = θ′, M = Bκ, B = EI

Work done by end load: W = −PD = −P∫ S

0

(1− cos θ)ds

Total potential energy: V = U + W =

∫ S

0

L(θ, θ′)ds, L = 12Bθ

′2 + P cos θ

Energy minimisation: Bθ′′ + P sin θ = 0 (equilibrium equation)

Static-dynamic analogy


Euler elastica (phase portrait and elastic shapes)

Centreline equations: x ′ = cos θ, y ′ = sin θ


The elastica


(Spatial) chaos

(1) Intrinsic curvature:

M = EI (κ− κ0), κ =dθ

dsFor instance, sinusoidal nonstraightness: κ0 = κ sinωs

EId2θ

ds2+ F sin θ = EI

dκ0

ds= EI κω cosωs =: α cosωs

(periodic forcing)

(2) Varying cross-section: EI = EI (s)

For instance, sinusoidally varying stiffness: EI = B0 + B1 sinωs

B0d2θ

ds2+ F sin θ = −ωB1 cosωs

dθ

ds− B1 sinωs

d2θ

ds2

(parametric forcing)


IVP vs. BVP

θ′′ + α2 sin θ = 0 (α2 = P/EI or g/L)

IVP: θ(0) = θ0, θ′(0) = 0

BVP: θ′(0) = 0 = θ′(L) (pinned-pinned boundary conditions)

α1 α2 α3

Length constraint: L =

∫ds =

∫ −θ0

θ0

dθ

θ′(integral: 1

2θ′2 − α2 cos θ = h)


Stability analysis for the pendulum – spectral stability

θ′′ + α2 sin θ = 0 (α2 = g/L)

Written as a dynamical system (x = θ, y = θ′):[x ′ = yy ′ = −α sin x

Jacobian at downward hanging equilibrium (x , y) = (0, 0):

J =

(0 1

−α cos x 0

)=

(0 1−α 0

)Eigenvalues: λ2 + α = 0 =⇒ λ = ±i

√α

i.e., (stable) centre point for all α


Solving a boundary-value problem – buckling

Column of length L with pinned ends under a compressive load P

Loss of stability of the straight solution – buckling

Need only consider linearised equation about θ = 0:

θ′′ + λθ = 0

(λ =

P

EI

), θ′(0) = 0, θ′(L) = 0

Solution: θ(s) = A sin√λs + B cos

√λs

Critical loads: λ =n2π2

L2(n = 1, 2, ...)


Calculus of variations (1)

Stationary point required of functional

J[u] =

∫ b

a

L(u(x), u′(x), x)dx (L Lagrangian)

with boundary conditions

u(a) = ua, u(b) = ub

Suppose u(x) = u(x) solves the problem and consider a variation v(x) from thissolution,

u(x) = u(x) + εv(x),

that satisfiesv(a) = 0, v(b) = 0

Define

J(ε) := J[u + εv ] =

∫ b

a

L(u + εv , u′ + εv ′, x)dx

Then

J(ε) = J(0) + εδJ +1

2ε2δ2J + O(ε3)


Calculus of variations (2)

δJ =dJ(ε)

dε

∣∣∣∣ε=0

=

∫ b

a

[Lu(u, u′, x)v + Lu′(u, u′, x)v ′]dx (first variation)

δ2J =d2J(ε)

dε2

∣∣∣∣ε=0

=

∫ b

a

[Luu(u, u′, x)v2+2Luu′(u, u′, x)vv ′+Lu′u′(u, u′, x)(v ′)2]dx

(second variation)

For a local minimum:

Stationarity:

δJ =

∫ b

a

(Lu −

d

dxLu′

)v dx = 0 (for all variations v)

=⇒ d

dx

(∂L

∂u′

)− ∂L

∂u= 0 (Euler-Lagrange equation)

Minimum ifδ2J > 0 (for all variations v 6= 0)


Stability of the unconstrained elastica (1)

Functional (total potential energy, u = θ):

V [θ] =

∫ l

0

1

2Bθ′2ds − P

(l −∫ l

0

cos θds

)

Free-free boundary conditions: θ′(0) = 0 = θ′(l)

First variation (v = δθ):

δV = −∫ l

0

[Bθ′′ + P sin θ] δθds + [θ′δθ]l0

Euler-Lagrange equation (p = PB ):

θ′′ + p sin θ = 0



Second variation:

δ2V =

∫ l

0

[(δθ′)2 − (δθ)2p cos θ

]ds

or, after integration by parts,

δ2V = −∫ l

0

[δθ′′ + δθp cos θ] δθds

Consider the Sturm-Liouville problem (linearised Euler-Lagrange equation):

φ′′ + µp cos θφ = 0, φ′(0) = 0 = φ′(l)

This has real eigenvalues µn and a basis of eigenfunctions φn (n = 1, 2, ...)satisfying the orthogonality condition∫ l

0

φn(s)φm(s)p cos θ(s)ds = 0 (n 6= m)



Thus we can write

δθ(s) =∞∑n=1

cnφn(s)

and

δ2V =

∫ l

0

∞∑n=1

(µn − 1)p cos θc2nφ

2nds

or, using the identity µn

∫ l

0

p cos θ(s)φ2nds =

∫ l

0

(φ′n(s))2ds,

δ2V =∞∑n=1

(1− 1

µn

)c2n

∫ l

0

(φ′n(s))2ds

So we havestability if µn /∈ [0, 1]

instability if µn ∈ [0, 1]

(for all n)



Stability of the straight solution (θ = 0):

Sturm-Liouville problem:

φ′′ + µpφ = 0, φ′(0) = 0 = φ′(l)

Nontrivial solutions (eigenfunctions):

φn(s) = cosnπs

l, µn =

n2π2

pl2=

Pcrn

P, Pcr

n =n2π2B

l2

Conclusion:

The straight configuration is stable for P < Pcr0 = 0 and unstable for P > 0

(µ = 0 is an eigenvalue with corresponding rigid-body rotation mode φ = const.)

The solution θ = const. for P = 0 is marginally stable

Rigid-body mode (constant variation) δθ = a (θ = 0):

δ2V = −∫ l

0[δθ′′ + p cos θδθ] δθds = −

∫ l

0pa2ds = −pla2 < 0

Indeed, taking θ = θ0: V (θ0) =∫ l

0( 1

2θ′20 + p cos θ0)ds = pl cos θ0 < pl = V (0)


Stability of the constrained elastica – pinned (1)

Functional (total potential energy + constraint, λ Lagrange multiplier):

V [θ] =

∫ l

0

1

2Bθ′2ds − P

(l −∫ l

0

cos θds

)− λ

∫ l

0

sin θds

Pinned-pinned boundary conditions: θ′(0) = 0 = θ′(l), ∆y =

∫ l

0

sin θds = 0

First variation:

δV = −∫ l

0

[Bθ′′ + P sin θ + λ cos θ] δθds + [θ′δθ]l0

Euler-Lagrange equation (p = PB , λ = λ

B ): θ′′ + p sin θ + λ cos θ = 0

Integrate: λ = 0 (no shear force, unless ends coincide)

Constraint: C [θ] =

∫ l

0

sin θds = 0

So variations δθ satisfy the linearised constraint

δC =

∫ l

0

δθ cos θds = 0



Second variation (noting λ = 0):

δ2V =

∫ l

0

[(δθ′)2 − (δθ)2p cos θ

]ds

or, after integration by parts,

δ2V = −∫ l

0

[δθ′′ + pδθ cos θ] δθds

Consider the (inhomogeneous) Sturm-Liouville problem (linearised Euler-Lagrangeequation, c free parameter):

φ′′ + µp cos θφ = c cos θ, φ′(0) = 0 = φ′(l)

subject to

∫ l

0

φ cos θds = 0

Integrate: c = 0 (no (linearised) shear force, unless ends coincide)

This has real eigenvalues µn and a basis of eigenfunctions φn (n = 1, 2, ...) with∫ l

0

φn(s)φm(s)p cos θ(s)ds = 0 (n 6= m)



Thus we can write

δθ(s) =∞∑n=1

cnφn(s)

and

δ2V =

∫ l

0

∞∑n=1

(µn − 1)p cos θc2nφ

2nds

or, using the identity µn

∫ l

0

p cos θ(s)φ2nds =

∫ l

0

(φ′n(s))2ds,

δ2V =∞∑n=1

(1− 1

µn

)c2n

∫ l

0

(φ′n(s))2ds

So we havestability if µn /∈ [0, 1]

instability if µn ∈ [0, 1]

(for all n)



Stability of the straight solution (θ = 0):

Sturm-Liouville problem (c = 0):

φ′′ + µpφ = 0, φ′(0) = 0 = φ′(l),

∫ l

0

φ ds = 0

Nontrivial solutions (eigenfunctions):

φn(s) = cosnπs

l, µn =

n2π2

pl2=

Pcrn

P, Pcr

n =n2π2B

l2

Conclusion:

The straight configuration is stable for P < Pcr1 and unstable for P > Pcr

1

(Rigid-body mode, and hence zero eigenvalue, eliminated by the constraint)


Eigenvalues of the free-free elastica

1st mode 2nd mode

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4

µ

θ(L)

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4

µ

θ(L)

3rd mode

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4

µ

θ(L)

Coincident ends: K (k) = 2E (k), k = 0.9089, p = 21.5491 (θ(l) = 130.71◦)


Eigenvalue continuation – solving an IVP

θ′′ + p sin θ = 0, θ′(0) = 0, θ′(l) = 0

φ′′ + µp cos θ φ = 0, φ′(0) = 0, φ′(l) = 0 (−→ φ(0) = 1)

First three (non-rigid-body) modes of the free-free rod:

-10

-5

0

5

10

0 0.5 1 1.5 2

φ’(

l)

µ

1st mode: p = 10.7064, 2nd mode: p = 41.3133, 3rd mode: p = 92.9550


Pinned elastica – secondary instability


Bifurcation diagrams for a twisted rod

varying end rotation(R = 0 (a), π (b), 2π (c),3π (d), 4π (e) and 8.9527

(f))(solid curves stable under

rigid loading) zero end rotation (R = 0), varying Poisson’s ratio(γ = C

B = 1.7, 1.8, 3.2, 12.0)


Large-deformation (Cosserat) rod theory

Rod configuration specified by centreline r(s) and frame of directors{d1(s),d2(s),d3(s)} (s arclength)

Assumptions (Kirchhoff rod):

thin rod with uniform cross-section

intrinsically straight and untwisted

no extension (|r′| = 1) and no shear (r′ · d1 = r′ · d2 = 0): r′ = d3

linear stress-strain relations (nonlinearities purely geometrical)

Frame evolution equations:

d′i = u× di (i = 1, 2, 3)

u = κ1 d1 + κ2 d2 + τ d3 (strains: curvatures and twist)


Kinematics - frame evolution equation

Orthonormality:d i · d i = 1 =⇒ d

′i · d i = 0

d i · d j = 0 =⇒ d′i · d j = −d ′j · d i

Thus we can write (for some functions κ1, κ2, τ):

d′1 = τd 2 − κ2d 3

d′2 = −τd 1 + κ1d 3

d′3 = κ2d 1 − κ1d 2

ord′i = u × d i (i = 1, 2, 3),

whereu = κ1d 1 + κ2d 2 + τd 3 (strain vector)

κ1 and κ2 are the curvatures (about d 1 and d 2) and τ is the twist (about d 3)

Note the analogy with angular velocity in rigid-body dynamics!


Force and moment balance

Force balance:

F (s + ∆s)− F (s) +

∫ s+∆s

s

f (s) ds = 0

⇐⇒∫ s+∆s

s

F′(s)ds +

∫ s+∆s

s

f (s) ds

=

∫ s+∆s

s

[F ′(s) + f (s)] ds = 0

ThusF′ + f = 0

Moment balance:

M(s + ∆s)−M(s) + r(s + ∆s)× F (s + ∆s)− r(s)× F (s) +

∫ s+∆s

s

r × f ds = 0

⇐⇒∫ s+∆s

s

[M ′ + (r × F )′ + r × f ] ds = 0

Thus M′ + r ′ × F = 0


3D deformation – elastic (Cosserat) rod equations

−60−40

−200

−40−20

020

−300

−250

−200

−150

−100

−50

0

Equilibrium equations (F force, M moment): F′ = 0, M

′ + r ′ × F = 0

Frame evolution equation: d′i = u × d i (i=1,2,3)

Inextensible rod: r ′ = d 3 (unit tangent)

Expressed in the moving (body) frame {d 1,d 2,d 3} (F = (F1,F2,F3), Fi = F ·d i ):

F′ + u× F = 0, M′ + u×M + d3 × F = 0, d3 = (0, 0, 1)

u = (κ1, κ2, τ) (strains; κ1, κ2 curvatures, τ twist)

Linear elasticity (Hooke’s law): M1 = B1κ1, M2 = B2κ2, M3 = Cτ

B1, B2 bending stiffnesses, C torsional stiffnessGert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 49 / 72

Dynamic analogy – spinning top

Equations of motion (L angular momentum, v vector from fixed point to centre ofmass): L

′ = mgk × v (torque due to gravity, k vertical)

Expressed relative to the rotating (body) frame (principal axes), including k′ = 0:

k′ + ω × k = 0, L′ + ω × L + mgv × k = 0, v = (0, 0, l)

ω angular velocity (assume centre of mass lies on one of the principal axes)

L1 = I1ω1, L2 = I2ω2, L3 = I3ω3

I1, I2, I3 moments of inertia

Kirchhoff dynamic analogy


Reduction of the equilibrium equations in the isotropic case

d′i = u × d i for i = 1, 2, 3

ui =1

2εijk d

′j · d k

κ1 = θ′ sinφ− ψ′ sin θ cosφ

κ2 = θ′ cosφ+ ψ′ sin θ sinφ

τ = φ′ + ψ′ cos θ

Choice of fixed frame:

F′ = 0 −→ F (s) = T e3

Lagrangian:

L =1

2B1κ

21 +

1

2B2κ

22 +

1

2Cτ 2 − T e3 · d 3 = L(θ, ψ, φ, θ′, ψ′, φ′)

Isotropic rod (symmetric cross-section): B1 = B2 = B


Euler angles to parametrise the body frame

The body co-ordinate system {d 1,d 2,d 3} can be related to a fixed orthonormalsystem {e1, e2, e3} via the Euler angles θ, ψ, φ:

d 1 = (− sinψ sinφ+ cosψ cosφ cos θ) e1 +

(cosψ sinφ+ sinψ cosφ cos θ) e2 − cosφ sin θ e3,

d 2 = (− sinψ cosφ− cosψ sinφ cos θ) e1 +

(cosψ cosφ− sinψ sinφ cos θ) e2 + sinφ sin θ e3,

d 3 = cosψ sin θ e1 + sinψ sin θ e2 + cos θ e3.


Conserved quantities:

d

ds(M · e3) = 0 (moment about force, ψ cyclic)

d

ds(M · d 3) = 0 (twisting moment, φ cyclic)

d

ds

(1

2B1κ

21 +

1

2B2κ

22 +

1

2Cτ 2 + T d 3 · e3

)= 0 (‘energy’)

(Linear) constitutive relations: M1 = B1κ1, M2 = B2κ2, M3 = Cτ

β = M · d 3 = M3 = Cτ

α = M · e3 =3∑

i=1

Mi d i · e3 = Bκ1 d 1 · e3 + Bκ2 d 2 · e3 + Cτ d 3 · e3

= −Bκ1 cosφ sin θ + Bκ2 sinφ sin θ + β cos θ

ψ′ =α− β cos θ

B sin2 θ, φ′ =

β

C− ψ′ cos θ


Planar system (for isotropic rod or Lagrange top):

Bθ′′ +dV

dθ= 0, V (θ) =

(α− β cos θ)2

2B sin2 θ+ T cos θ

For a straight rod (or upright top) we need to take α = β = M0

(d 3 and e3 aligned, M0 applied twisting moment)

V (θ) =M2

2B

1− cos θ

1 + cos θ+ T cos θ

Dimensionless:

1

2θ′

2+ V (θ) = h, V (θ) =

1

2

1− cos θ

1 + cos θ+

cos θ

m2, m =

M0√BT

Special case: M0 = 0 −→ α = β = 0 −→ Euler elastica


Buckling of an isotropic rod – falling over of Lagrange top

Reduced 1-DOF equation (θ is angle between tangent and F ):

θ′′ +dV (θ)

dθ= 0, V (θ) =

1

2

1− cos θ

1 + cos θ+

cos θ

m2

(m =

M0√BT

)(same as for Lagrange, i.e., symmetric, top)

Typical phase portraits:

m > 2 m < 2


Rod-top analogy


Localised buckling of isotropic rod

Fixed points: dV /dθ = 0 (nonzero θ: helical rod, precessing top)

Eigenvalues of trivial fixed point θ = 0: λ1,2 = ±(1/2m)√

4−m2

θ′′ = −dV

dθ=

− sin θ

(1 + cos θ)2+

sin θ

m2=

(1

m2− 1

4

)θ −

(1

12+

1

6m2

)θ3 + O

(θ5)

Supercritical pitchfork bifurcation at m = 2

Homoclinic orbit (localised solution):

cos θ(s) = u0 + (1− u0) tanh2

(√1− u0

m√

2s

), u0 =

1

2m2 − 1

z

y

0 15 30 45 60 75 90 105 120 135 150-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0


3D bifurcation diagram

Equivalent oscillator for the twisted circular rod:

Note inversion of stability!Gert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 58 / 72

Localised solutions

Shape obtained by integrating the centreline equation r ′ = d 3

z

x

0 15 30 45 60 75 90 105 120 135 150-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

z

y

0 15 30 45 60 75 90 105 120 135 150-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

z

x0 15 30 45 60 75 90 105 120 135 150

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

z

y

0 15 30 45 60 75 90 105 120 135 150-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

z

x

0 15 30 45 60 75 90 105 120 135 150-1.80

-1.44

-1.08

-0.72

-0.36

0.00

0.36

0.72

1.08

1.44

1.80

z0 15 30 45 60 75 90 105 120 135 150

-1.80

-1.44

-1.08

-0.72

-0.36

0.00

0.36

0.72

1.08

1.44

1.80Gert van der Heijden (UCL) Nonlinear mechanics of rods, sheets and ribbons 59 / 72

Localised solutions (true 3D views)


Nonlinear mechanics of rods, sheets and ribbons - IMA-UMN

Documents

Transcript of Nonlinear mechanics of rods, sheets and ribbons - IMA-UMN