wrinkles as a relaxation of compressive stresses in an annular ......wrinkles as a relaxation of...

Wrinkles as a relaxation of compressive stresses in anannular thin film

Peter BellaCourant Institute, NYU

PIRE and OxMOS workshopPattern formation and multiscale phenomena in materials

09/26/2011

joint work with Robert V. Kohn

Peter Bella (Courant Institute) Wrinkles in an annular thin film Oxford, 09/26/2011 1 / 15

Motivation

We are interested in wrinklingobserved in thin elastic sheets undertension.

Figure: Cerda and Mahadevan, PRL,2/2003.

lots of wrinkles inside, no wrinkles near edges

free boundary between them (where is free bdry?)

more accessible – annulus-shaped sheet loaded uniformly –Davidovitch et al. (PNAS 2011)

approach: identification of the scaling law for minimum of the energy


Setting

annular sheet rather than arectangle

uniform loads in the radialdirection at the boundaries

if inner loads are sufficientlylarge we observe wrinkles nearthe inner boundary

Tin

Tout


Energy

Average energy per unit thickness h

E 3Dh (u3Dh ) =

1

h

∫Ω×(−h/2,h/2)

W3D(∇u3Dh )dx

− Tinh

∫|x |=Rin

u3Dh · ν −Touth

∫|x |=Rout

u3Dh · ν,

where Ω = {Rin < |x | < Rout} ⊂ R2 is annulus.Consider simpler 2D energy:

Eh(uh) =

∫ΩW (Duh) + h

2|D2uh,3|2 dx

− Tin∫|x |=Rin

uh · ν − Tout∫|x |=Rout

uh · ν

Here uh : Ω ⊂ R2 → R3 and Duh ∈ R3×2.Reduced 2D energy is easier to understand and still captures the mainfeatures.


Main result

We are interested in the scaling law:

Theorem

Under suitable assumptions on data (i.e. on loads and stored energyfunctions W3D ,W ) there exist E∗ and 0 < C0 < C1 such that

E∗ + C0h ≤ minE 3Dh (u3Dh ) ≤ E∗ + C1h

and similarlyE∗ + C0h ≤ minEh(uh) ≤ E∗ + C1h.

E∗ is energy due to in-plane tension = minimum of the relaxed problem.

The second term, linear in h, comes from the wrinkling.


The big picture

2D problem:

simpler, easier to understand – morefocus in this talk

proof uses interpolation at its heart

3D problem:

more general hypotheses on the energydensity (slower growth at ∞)partially based on the proof for 2D

in addition uses rigidity estimates(Friesecke-James-Müller)

Figure: Davidovitch et al.,PNAS 2011


The relaxed problem

The relaxed energy

E0(u) :=

∫ΩQW (Du)dx + loads,

where QW is quasiconvexification of W .

Formally, one can obtain E0 from Eh by letting h→ 0.

Properties:

E0 is convex (but not strictly convex)

E0 has a unique minimizer u0 (up to translations)

the relaxed solution u0 is radially symmetric and planar

u0 is “compressed” in the hoop direction near the inner boundary

arc length needed to waste to avoid compression growths linearly


Upper and lower bounds

To prove the scaling law we need to show the upper and lower bound:

1 There exists C1 s.t. for each h > 0 there exists uh satisfying

Eh(uh) ≤ E∗ + C1h.

2 There exists C0 > 0 s.t. for each h > 0 and any uh we have

Eh(uh) ≥ E∗ + C0h.

Here E∗ = minE0(u) = E0(u0).


Upper bound - Naive construction

A naive idea:Superimpose wrinkles on u0 to avoid compression.

Optimal period of wrinkles ≈ h1/2, amplitudeuh,3 ≈ h1/2ρ1/2 (ρ = dist from the free boundary).

Fixable problem: Infinite bending energy.Back-of-the-envelope calculation: ∂rruh,3 ≈ h1/2ρ−3/2 implies

h2bending ≥ h2∫ R

0|∂rruh,3|2 dρ ≈ h3

∫ R0ρ−3 dρ =∞.

Solution: Cut-off in a boundary layer of size h near the free boundary toobtain bending h3

∫ Rh ρ−3 dρ ≈ h.


Upper bound - Successful construction

More serious problem: membrane term diverges.

uh,3 ≈ h1/2ρ1/2 implies ∂ruh,3 ≈ h1/2ρ−1/2, and so

stretching &∫ R

0|∂ruh,3|2 dρ ≈ h

∫ R0ρ−1 dρ

boundary layer produces only unsatisfactory energy scaling h| log h|essence of problem: uh,3 grows too fast near the free boundary

Solution: decrease the amplitude of wrinkles by decreasing their period(but also increasing the bending energy)

construction with cascade of wrinkles produces the right energy scaling

we believe that to obtain the optimal scaling a construction similar tocascade of wrinkles is always necessary


Lower bound

Want to show

minEh ≥ E∗ + C0h

for some C0 > 0. Assume the contrary, i.e. for smallδ > 0 and h > 0 there exists uh such that

Eh(uh) ≤ E∗ + δh.

Three parts of the proof:

1 uh,3 (out-of-plane displacement) is small – think rubber bands

2 interpolation implies the planar projection of uh has also small energy

3 compare areas of images of u0 and projected uh to get a contradiction


Lower bound argument - part 1

We know

E0(uh)− E0(u0) ≤ δh.

Strict convexity of QW implies

||Duh − Du0||2L2 ≤ Cδh

in the non-relaxed region. In relaxed region:

||(Duh − Du0) · n||2L2 ≤ Cδh

relaxed

non− relaxed

Poincaré inequality implies

||uh − u0||2L2 ≤ Cδh,

in particular the out-of-plane displacement is small:

||uh,3||2L2 ≤ Cδh.



1 small energy impliesh2||D2uh,3||2L2 ≤ δh

2 interpolation with ||uh,3||2L2 ≤ Cδh (from previous slide)

||Duh,3||2L2 ≤ Cδ

(i.e. uh almost planar)

3 uh,12 = projection of uh into x1x2-plane; uh,12 has no bending

|Eh(uh)− Eh(uh,12)| ≤ Cδ

4 triangle inequality implies

Eh(uh,12)− E0(u0) ≤ Cδ

5 observe: h doesn’t play any role in the last relation



We have obtained:

For any δ > 0 small there exists a planar functionuh,12 : Ω→ R2 such that∫

ΩW (Duh,12)− QW (Du0) + loads ≤ δ.

Since u0 is compressed in some region (and uh,12 should avoid compressionto keep its energy small), the area of the image of uh,12 should be muchlarger than the area of the image of u0. At the same time, their closenessin L2 implies those areas should be almost equal – a contradiction.


Final remarks

Comment: our analysis rests crucially on uniqueness and “partial”strict convexity of the relaxed problem – typical for problems withuniaxial tension (buckling due to 2D compression is different)

3D setting

- This is the real problem

Open questions:

- pointwise results about wrinkling pattern?- less symmetric problems (like Cerda and Mahadevan)?

This is only one of a whole family of problems which involve cascades


wrinkles as a relaxation of compressive stresses in an annular ......wrinkles as a relaxation of...

Documents