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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