23. schweizer cadfem ansys simulation …...frank gehry, lvmh, paris inspiration mutsuro sasaki,...

The perfect shellPatrick Steingruber

23. Schweizer CADFEM ANSYS Simulation Conference

14. Juni 2018, Rapperswil

CADFEM - ANSYS

• Inspiration

• Calculus of variations

• Minimum potential energy

• Examples

• Conclusion

Summary

All that is superfluous

displeases God and nature.

All that displeases God and

nature is evil.

(Dante Alighieri, 1300)

Inspiration

Tensairity® structures

Garage parc Montreux station WEB enclosure, Tenerife

Asymmetrical load Transparent

Frank Gehry, LVMH, Paris

Inspiration

Mutsuro Sasaki, Rolex Learning Center EPFL

Inspiration

Hanging chain of equal sized

rings and another composed of

32 small lead balls.

→arch = “upside down catenary”

→“…, the shape of the great

vault was not bad at all”.

Cupola San Pietro, Roma

Long meridian

cracks

Camorino, Heinz Isler, 1973

→shell = “upside down fabric”

Inspiration

Calculus of variations

The calculus of variations can be used to find an unknown function that

minimize or maximize a functional (𝐼 𝑦 =function of function).

𝐼 𝑦 = 𝑥1

𝑥2

[𝑝 𝑥 𝑦′ 2 + 𝑞 𝑥 𝑦2 + 2𝑓 𝑥 𝑦] 𝑑𝑥

𝑦 𝑥1 = 𝑦1 𝑦 𝑥2 = 𝑦2

Brachistochrone problem:

Given two points A and B in a vertical plane,

what is the curve traced out by a point acted

on only by gravity, which starts at A and

reaches B in the shortest time

Calculus of variations

Equation of catenary via calculus of variations (2D problem)

The equilibrium state is determined from the minimum potential energy condition.

The potential energy of the catenary is

𝐼 𝑦 = 𝜌 𝑥𝐴

𝑥𝐵

𝑦 1 + 𝑦′ 2 𝑑𝑥

The length L of the chain is constant (Additional constraint

conditions = Subsidiary conditions)

Boundary conditions

𝑥𝐴

𝑥𝐵

1 + 𝑦′ 2 𝑑𝑥 = L

𝑦 𝑥𝐴 = 𝑦𝐴 𝑦 𝑥𝐵 = 𝑦𝐵

𝐼 𝑦, 𝜆 = 𝑥𝐴

𝑥𝐵

(𝑦 + 𝜆) 1 + 𝑦′ 2 𝑑𝑥

Minimum potential energy

𝜌=constant

𝑥𝐴

𝑥𝐵

1 + 𝑦′ 2 𝑑𝑥 = L

Isoperimetric

problem using

Lagrange multiplier 𝝀

Minimum potential energy

Pacific Journal of Mathematics, April 1980

1 +𝜕𝑧

𝜕𝑥

2+𝜕𝑧

𝜕𝑦

2= 1 + |𝛻𝑧|2

𝐼 𝑧, 𝜆 = Ω

(𝑧 + 𝜆) 1 + |𝛻𝑧|2𝑑𝑥 𝑑𝑦

𝜌=constant

Ω

1 + |𝛻𝑧|2𝑑𝑥 𝑑𝑦 = 𝐴

Minimum potential energy

Equation of the two-dimensional analogue of the catenary (3D problem)

Isoperimetric

problem using

Lagrange multiplier 𝝀

The surface A of the area is constant (Additional constraint

conditions = Subsidiary conditions)

with:

𝛻𝑧 𝑥, 𝑦 = 𝑔𝑟𝑎𝑑(𝑧)=𝜕𝑧

𝜕𝑥,𝜕𝑧

𝜕𝑦

Minimum potential energy

The much more complex 3D problem is no longer possible to solve in an

analytical way and has to be calculated numerically. The non-linear

Euler-Lagrange partial differential equation has to be solved with the

use of a computer.

Equation of the two-dimensional analogue of the catenary (3D problem)

Numerical solution

AΔ = s s − l1 s − l2 s − l3

Minimum potential energy

The surface can be discretized with a triangle surface mesh

The variational problem can be solved numerically

by FEM ≠ by structural FEA analysis!!

s = l1 + l2 + l3 /2

ΠN = i=1nPΔi − k(S 𝐱N −AN) stationary

𝜕ΠN𝜕𝐱N=𝜕P

𝜕𝐱N

T

− 𝑘𝜕S

𝜕𝐱N

T

= 0

𝜕ΠN𝜕k= − S 𝐱N − AN = 0

PΔi=AΔizi: Potential energy of triangle i

with area AΔik: Lagrange multiplier

S 𝐱N : Unknown surface with the

coordinate vector 𝐱NAN: Area of the total surface

The nonlinear system of equations becomes

Minimum potential energy

𝐟(𝐱) = 𝐟(𝐱𝟎) −)𝜕𝐟(𝐱

𝜕𝐱𝚫𝐱 = 𝟎

And can be solved with the Newton Raphson method

ANSYS + Fortran Compiler: new ANSYS element UEL105

Newton Raphson

Minimization / Maximization problems

Nonlinear systems of equations

Triangular element:

Ni

Nj

Nk

10 equations, 10 unknowns:xi, yi, zi, xj, yj, zj, xk, yk, zk, k

Minimum potential energy

Minimum potential energy

ANSYS UEL105

Newton Raphson

vector and area A of

the unknown surface

ANSYS will not perform a static analysis

with UEL105!!

The solution will be an unknown

surface S of area A that has the minimal

potential energy!!

Examples

Catenary shape: FE Method versus Theoretical shape

FEM Theory

Examples

Spherical dome

“Perfect dome”

“Perfect dome” with square base

Examples

A = 13.3 A = 16.5

UX=UY=UZ=0

S1=0 S2<0 S3<0

→ Only compression stresses!!

A = 13.3 A = 15.0

Examples

UX=UY=UZ=0

S1=0 S2<0 S3<0

→ Only compression stresses!!

Conclusion

• A lightweight structure is defined by the optimal use of

material to carry loads

• Material is used optimally within a structural member if the

member is subjected to membrane force rather than

bending

• The nature will always find the optimal equilibrium position

by minimizing the energy

• In nature the structures are always minimum energy

structures

Thank you for your

attention