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Stability of discretized nonlinear elastic systems
Arnaud Lazarus, Corrado Maurini, Sébastien Neukirch
To cite this version:
Arnaud Lazarus, Corrado Maurini, Sébastien Neukirch. Stability of discretized nonlinear elastic sys- tems. Extremely Deformable Structures, 562, Springer-Verlag Wien, 2015, CISM International Centre for Mechanical Sciences, 978-3-7091-1876-4. �10.1007/978-3-7091-1877-1�. �hal-01447329�
Stability of discretized nonlinear elastic systems
Arnaud Lazarus*†, Corrado Maurini*†, and S´ebastien Neukirch*†
* Sorbonne Universit´es, UPMC Univ Paris 06, UMR 7190, Institut Jean Le Rond d’Alembert, F-75005, Paris, France
† Centre National de la Recherche Scientifique, UMR 7190, Institut Jean Le Rond d’Alembert, F-75005 Paris, France Abstract. The goal of these notes is to give a short introduction to the methods for the study of stability of elastic structures. We con- sider only the finite-dimensional case, where the state of the system is represented by a discrete set of parameters. The core of the ex- position focuses on the illustration of energetic methods where the equilibrium and stability are found by studying the point of station- arity and minima of a scalar function of the state parameters. After three introductory sections presenting the links between stability and energy minimisations (Section 2), potential energy (Section 3) and discretisation methods (Section 4), we detail the mathemati- cal methods required to minimise a function of n variables (Section 5-8). We include the theory and recipes to deal with equality and inequality constraints, providing several examples of applications to simple structures. Hence, we show how to classify regular and sin- gular points (bifurcations) in force-displacement diagrams (Section 9) and give a fully worked example with several degrees of freedom (Section 10). Section 11 presents through an example the dynamical theory of stability including Floquet theory for systems with peri- odic coefficients. Finally, Section 12 shows how energetic methods can be applied to the study of material instabilities, by considering the case of springs with irreversible damage.
Figure 1. Dynamics of a simple pendulum. Associated potential energy V (θ) and phase plane (θ,θ).˙
1 Introduction
We study the equilibrium and stability of nonlinear systems. Nonlinearity has two main consequences: (i) closed-form solutions are the exception and numerical methods have to be used; and (ii) unicity is lost, that is several solutions may exist for the same loading. While the ubiquity of computers reduces the importance of the first consequence, the second raises questions such as: which solution is to be considered? Which ones are stable? etc.
We restrict to the conservative case, where for example there is no fric- tion, no viscous force (apart from Section 11 when dynamical stability will be introduced). Hence internal and external forces derivate from poten- tial energies, for example 1/2L
0 κ2ds (curvature strain energy in a beam), mgz (gravitational energy), −F ·OA (work of external load). Hence we do not consider follower or gyroscopic loads (see for example Ziegler (1977) or Bigoni (2012)).
2 Energy minimization and dynamical systems
We follow the principle stating that in conservative systems a stable equi- librium corresponds to a local minimum of the potential energy. In this
section we relate this energy-based stability criterion to the study of stabil- ity through the dynamics of the system.
We consider a simple pendulum, a punctual mass M fixed at the end of a massless rigid bar of length L, see Figure 1. The dynamics of the system obeys
d2θ
dt2 = − g
L sinθ (1)
where g is the acceleration of gravity. Using the dimensionless time τ :=
t
L/g, Eq. (1) writes ¨θ + sinθ = 0, where ˙( ) := d/dτ. This equation is integrated once to yield
1 2
θ˙2 + 1−cosθ = E (2)
where the integration constant E is the total mechanical energy of the system, the sum of the kinetic energy T = 12θ˙2 and the potential energy V = 1−cosθ, that is E = T +V. In figure 1 the stable (respectively unsta- ble) equilibrium position A (resp. B) clearly appears as the local minimum (resp. maximum) of the potential energy V . Trajectories corresponding to the dynamics of the system for all possible initial conditions are drawn in the phase space (θ,θ). There, equilibrium are points and the stable equi-˙ librium A has closed trajectories in its neighborhood: perturbation of the equilibrium A leads to small vibrations around the position θA = 0. The unstable equilibrium B has escaping trajectories in its neighborhood: some perturbations lead to evolution of the system far from B.
3 Potential energy for an elastic beam
In order to illustrate the theory, we focus on a simple elastic system : an elastic beam bent in the (x, y) plane, see Figure 2. The potential energy is then given by the functional (Audoly and Pomeau, 2010)
V [θ] = 1 2EI
L 0
θ(s)2ds+P L
0
cosθ(s) ds (3)
the first term being the internal strain energy and the second term the work done by the external applied load P. For the two cases in Figure 2, in the minimization process we have to take boundary (4) and/or integral conditions (5) into account:
θ(0) = 0 (4)
y(L) = 0 that is L
0
sinθ(s) ds = 0 (5)
Figure 2. Elastic beams buckling in the plane. (Left) Clamped-free bound- ary conditions. (Right) Pinned-pinned boundary conditions.
Figure 3. (Left) Solution θ(s) for the configuration in Figure 2-Left.
(Right) A system of rigid bars linked by spiral springs.
4 Three ways to discretize a continuous system
4.1 Finite differences
Using si = ih, with i ∈ (0, n + 1), and sn+1 = L, we apply a finite differences scheme for the derivative of θ(s), θ(s) (θi+1 − θi)/h. We then apply the rectangle rule for integration and the energy functional (3) becomes:
V Vfd = 1 2EI
n−1 i=0
θi+1 −θi h
2
h+P n
i=0
cosθih (6) We introduce dimensionless variables:
v = 1 n+ 1
VfdL EI = 1
2
n−1 i=0
(θi+1 −θi)2 + p (n+ 1)2
n i=0
cosθi (7) where p = P LEI2.
Figure 4.(Right) Piecewise linear approximation of the solution θ(s) shown in Figure 3-Left. (Left) A triangle base function, and the partitioning of the interval (0, L) into n+ 1 subintervals (or elements).
This is equivalent to the energy of a system of n+ 1 rigid bars linked by spiral springs (see Figure 3):
Vbars = 1 2k
n−1
i=0
(θi+1 −θi)2 +P h n
i=0
cosθi (8)
where the spiral spring stiffness is k and the rigid bars have length h. The equivalence Vbars = Vfd is obtained using EI = k h and L = (n+ 1)h.
4.2 Finite elements
In the former section, the solutionθ(s) was approximated by a discontin- uous piecewise constant function. Here we look for a continuous piecewise linear approximation. We therefore introduce triangle functions that take the value 1 at their center and have a compact support, see Figure 4. The interval (0, L) is split into n+ 1 elements of size h. The ith element corre- sponds to s ∈ (si, si+1), with si = i h. The triangle functions ϕi(s) span over elements i−1 and i, with ϕi(si) = 1. Consequently, in the ith element the two basis functions that are non zero are:
ϕi(s) = si+1 −s
h (9a)
ϕi+1(s) = s−si
h (9b)
And in this element the approximation to the function θ(s) is given by θfe(s) = ciϕi(s) + ci+1ϕi+1(s) (10) where the boundary condition θ(L) = 0 imposes cn+1 = cn. Putting this ansatz in the functional (3) we obtain
v = 1 n+ 1
V L EI
= 1 2
n−1 i=0
(ci+1 −ci)2 + p (n+ 1)2
n−1 i=0
sinci+1 −sinci
ci+1 −ci
+ p
(n+ 1)2 coscn (11) At the discretization gets finer, the second term tends toward cosci and
v h−→→0 1 2
n−1 i=0
(ci+1 −ci)2 + p (n+ 1)2
n i=0
cosci (12)
which is the same result as (7) with θi = ci. 4.3 Spectral decomposition
A third way to discretize the system in Figure 2-Left is to use the first n buckling modes as a functional basis and decompose the solution in this basis, see Figure 5. Here we use n = 2, the buckling modes are
θ1(s) = sin π
2s (13a)
θ2(s) = sin 3π
2 s (13b)
The approximate solution is then written as θ(s) = α sin π
2s+β sin 3π
2 s (14)
We inject this ansatz into the functional (3) and obtain:
v(α, β) = 1 2
1 0
α π
2 cos π
2s+β 3π
2 cos 3π 2 s
2
ds +P
1 0
cos α sin π
2s+β sin 3π 2 s
ds (15)
Figure 5. Spectral discretization based on the first two buckling modes.
Figure 6. A saddle point (S), a minimum (A), and a maximum (B)
The first integral is easily evaluated but not the second one. Gauss-Legendre integration is then used, with 2 points. We obtain
v(α, β) = π2
16(α2 + 9β2) + P
2 [cos (0.95α−0.54β) + cos (0.33α−0.84β)]
(16)
5 Minimum, maximum, extremum
Once the discretization is achieved, we are left with a function V (x), where V ∈ R and x ∈ Rn, and we look for the point(s) for which V is locally minimum. Examples are:
• V(x1, x2) = x21 +x22 has a global minimum at (x1, x2) = (0,0).
• V(x1, x2) = x21 − x22 has no global or local minimum. The point (x1, x2) = (0,0) is a saddle point.
• V(x) = x2 −x4 has a local minimum at x = 0.
Stated in this way the problem of the equilibrium and stability of an elastic structure becomes a problem of nonlinear optimization. We therefore use methods of nonlinear programming, see for example Luenberger and Ye (2008).
5.1 Values of a function in one direction
We are given a function V (x), where V ∈ R and x ∈ Rn. We are interested to know if the point x0 is a local minimum. Instead of looking at the values of V all around x0, we restrict to the values taken by V in the direction d. That is we study the one-variable function
V(x0)(d) := V(x0 + d) = V( ) for ∈ R (17) Example: For the function V(x1, x2) = x21 −x22, the point (x1, x2) = (0,0) is a saddle point:
• If we study V (x0)(d) with x0 = (0,0) and d = (1,0). We have V(x0)(d) = V( ) = 2. In this direction we have a local minimum.
• If we study V (x0)(d) with x0 = (0,0) and d = (0,1). We have V(x0)(d) = V( ) = − 2. In this direction we have a local maximum.
5.2 Derivative of a function in one direction
We now want to study the derivative with regard to of V( ), that is V ( ). This is the directional derivative of V(x) in the direction d, at the point x0:
V (x0)(d) := d
d V(x0 + d) =0
(18) Example : for the function V(x1, x2) = x21 −x22,
• we compute the directional derivative at the point (x1, x2) = (0,0) in the direction d = (1,0). We have V( ) = V(x0+ d) = 2, V( ) = 2 , and finally V(x0)(d) = V( )|=0 = 0.
• we compute the directional derivative at the point (x1, x2) = (1,0) in the direction d = (1,2). We have V( ) = V (x0 + d) = −3 2+ 2 + 1, and V(x0)(d) = V( )|=0 = 2.
5.3 Directional derivative and gradient
We consider V(x0+ d) with x0 = (x01, x02) and d = (d1, d2). The Taylor expansion around x0 for small writes:
V (x01 + d1, x02 + d2) = V (x01, x02) + ∂V
∂x1
x=x0
d1 + ∂V
∂x2
x=x0
d2 +O( 2)
= V (x01, x02) + ∇V|x0 ·d+O( 2) (19) where ∇ := ∂∂x.
5.4 Taylor expansion up to order two The Taylor expansion up to second order is:
V (x01 + d1, x02 + d2) = V (x01, x02) + ∇V|x0 · d+ 1 2
2d21 ∂2V
∂x21
x=x0
+ 2d1d2
∂2V
∂x1∂x2
x=x0
+ 1 2
2d22 ∂2V
∂x22
x=x0
(20) The second order term can be written as:
1 2
2d· H0 ·d (21)
where H = H(x) is the Hessian matrix, Hij = ∂x∂2V
i∂xj, and H0 means H when evaluated on x =x0.
6 Minimization without constraints
6.1 First derivative
We are given a function V(x), where V ∈ R and x ∈ Rn and we require the point x0 to be a local minimum. That is we want
V (x) ≥V (x0) ∀x near x0 (22) Hence we write x = x0+ d and consider small and all possible directions d (but not d =0). Using (19), the condition (22) yields
∂V
∂x
x0
·d ≥ 0 ∀ and ∀d (23) As positive and negative can be considered this implies ∂V∂x
x0 ·d = 0 ∀d, and finally
∂V
∂x
x0
= 0 (24)
This is a necessary condition in order to have a minimum at x0. If V is a potential energy of a system, then these n equations for the n unknowns (x1, x2, . . . , xn) are the equilibrium equations.
6.2 Second derivative
Having solved the (nonlinear) equilibrium equations (24) and found one or more solutions xA, xB, . . . , we want to test whether these solutions are
stable, that is whether they correspond to local minimum of the energy.
As for these solutions the first derivative vanishes, we need to look at the second derivative term in (20). The requirement (22) for a minimum now yields:
d·H0 ·d ≥ 0 ∀d (25) A sufficient condition is
d·H0 ·d > 0 ∀d (26) In the case where there is a d for which d·H0 ·d = 0, we cannot conclude and would have to compute higher orders. In the case there is one (or more) d for which d·H0 ·d < 0, then the point x0 is not a minimum.
Example: Minimize the function V(x1, x2) = x21 −x22 +x42 in R2. 6.3 Recipe
As the matrix H0 is symmetric, we can write it in its eigenspace. It is then diagonal with real entries, its eigenvalues λi with i ∈ (1, n). In this representation, the condition (26) writes:
d21λ1 +d22λ2 +. . .+d2nλn > 0 ∀di (27) which means that we must have λi > 0 ∀i, that is all eigenvalues must be strictly positive.
Here is the procedure to follow when we have the potential energy V(x) of a system to minimize:
• compute the equilibria, that is the solution(s) of ∂V∂x = 0 then, for each equilibrium solution:
• compute the matrix Hij = ∂x∂2V
i∂xj, and evaluate it on the equilibrium solution, then compute its eigenvalues λi
• if λi > 0 ∀i then the equilibrium is stable
• if there is one (or more) i such that λi < 0, then the equilibrium is unstable
• if the λi are all positive, but there is one (or more) i such that λi = 0, we cannot at this stage conclude on the stability.
6.4 Discrete vs. Continuous
If on the one hand we consider the continuous energy (3) and write Euler-Lagrange equations for it, we obtain
EI θ(s) +P sinθ(s) = 0 (28)
Figure 7. System of 4 bars linked by spiral springs. (Left) Ends are hori- zontal. (Right) Ends are horizontal and aligned
If now on the other hand we consider the discretized energy (6) and write the first derivative condition (24) we obtain
−EI θi+1 −2θi +θi−1
h −P hsinθi = 0 (29)
and we see that (29) is the finite differences version of (28), showing that the discretization and minimization procedures are commutative operations:
V[θ(s)] Euler-Lagragne
−−−−−−−−−→ EIθ +P sinθ = 0
⏐⏐
⏐⏐
finite diff.
⏐⏐
⏐⏐
finite diff.
V(θi) First derivative
−−−−−−−−−→ EI θi+1−2θh2i+θi−1 +P sinθi = 0 6.5 Example: The clamped beam
We here compute the buckling load of the two degrees of freedom system of Figure 7-Left. The energy of the system is
V = 1
2k(θ1 −θ0)2 + 1
2k(θ2 −θ1)2 + 1
2k(θ3 −θ2)2
+P h(cosθ0 + cosθ1 + cosθ2 + cosθ3) (30) and the boundary conditions reads: θ0 = 0 = θ3. The variables are then (θ1, θ2). We introduce dimensionless variables v = V /k and p = 16P h/k in order to follow notations of equation (7). Equilibrium equations are
∂v
∂θ1
= 2θ1 −θ2 − p
16 sinθ1 = 0 (31a)
∂v
∂θ2
= −θ1 + 2θ2 − p
16 sinθ2 = 0 (31b)
One trivial solution is θ1 = 0 = θ2. We study its stability. The Hessian matrix is
H0 =
2−p/16 −1
−1 2−p/16
(32) and has λ1 = 1 − p/16 and λ2 = 3 − p/16 as eigenvalues. We conclude the trivial solution is stable for p < 16 and unstable for p > 16. Buckling therefore happens at p = 16.
7 Minimization with equality constraints
Given a function V (x), where V ∈ R and x ∈ Rn, we look for the point(s) for which V is locally minimum, but only for the point(s) that satisfy one (or more) constraints g(x) = 0.
Example: Minimize V (x1, x2) = x21+x22 in the subset where g(x1, x2) = x1 −1 = 0.
7.1 First derivative
We are given a functionV(x) that is to be minimized in the subset where g(x) = 0. We require the point x0 to be a local minimum by stating that for small
V(x = x0 + d) ≥ V(x0) ∀x such that g(x0 + d) = 0 (33) This means that not all d directions will be tested. Only the directions d fulfilling
g(x0 + d) = g(x0) + ∇g|x0 ·d+O( 2) = 0 (34) will be tested. As in the unconstrained case, positive and negative can be considered. Therefore we need to have
∇V|x0 · d = 0 ∀d such that ∇g|x0 ·d = 0 (35) This implies that ∇V and ∇g are colinear, that is there is a real number λ such that
∇V |x0 = λ∇g|x0 (36) The function g is given and ∇g defines a direction in space. The vectors d perpendicular to the gradient of g are in the tangent (hyper-)plane of the surface g = 0 at point x0, see Figure 8. The equilibrium equations are the n equations (36) plus g(x) = 0, and the unknowns are the n components of x plus the real number λ.
Figure 8. Tangent planes in 2D and 3D
A way to memorize Eq. (36) is through the introduction of the augmented energy W = V −λ g and solve for ∇W = 0. In such a view λ is called a Lagrange multiplier.
Example: We minimizeV(x1, x2) = x21+x22 under the constraintg(x1, x2) = (x1 −1)2+ (x2 −2)2−1 = 0. The first order condition ∇V = λ∇g has two solutions. The first one is pointA such that xA1 = 1−1/√
5, xA2 = 2−2/√ 5, λA = 1 − √
5, and the second one is point B such that xB1 = 1 + 1/√ 5, xB2 = 2 + 2/√
5, λB = 1 +√ 5.
7.2 Second derivative
For the second derivative, we need to consider a perturbation to the equi- librium solutionx0 such that xstays on the surfaceg = 0. Contrary to what was done in the unconstrained case we need to develop the perturbation to second order:
x =x0 + d+ 1 2
2e+O( 3) (37)
where | | 1 and where g(x) = O( 3). Now d and e cannot be freely chosen, as they are related by the condition
g(x)−g(x0) = 0 = ∇g|x0 ·d+ 1 2
2∇g|x0 ·e+ 1 2
2d·G0·d+O( 3) (38) where G = G(x) is the matrix with second derivative of g: Gij = ∂x∂2g
i∂xj, and G0 means G when evaluated on x = x0. The Taylor expansion of V
to second order now writes:
V(x)−V(x0) = ∇V|x0 · d+ 1 2
2∇V|x0 ·e+ 1 2
2d·H0 · d+O( 3) (39) Note that in the unconstrained case, the term involving e vanishes because we had ∇V|x0 = 0, but this is no longer the case here. Using (36), (38), and (39), the condition (33) writes:
V(x)−V(x0) = 1 2
2d·(H0−λG0)·d ≥ 0 ∀d such that ∇g·d = 0 (40) We note ˆH := H −λG the augmented Hessian matrix. As in the uncon- strained case, a sufficient condition is obtained by requiring the matrix to be positive definite in the tangent plane of the surface g = 0 at point x0 (that is the ≥ sign is replaced by a > sign in (40)).
Example: We study the potential energy V(x1, x2) = x21 +x22 under the constraint g(x1, x2) = (x1 −1)2 + (x2 −2)2 −1 = 0. Equilibrium are point A, such that xA1 = 1− 1/√
5, xA2 = 2 −2/√
5, λA = 1−√
5, and B, such that xB1 = 1 + 1/√
5, xB2 = 2 + 2/√
5, λB = 1 +√
5. The augmented Hessian matrix is
HˆA = HA −λAGA =
+2√
5 0
0 +2√
5
(41a) HˆB = HB −λB GB =
−2√
5 0
0 −2√ 5
(41b) We first note that this matrix is positive definite for point A and nega- tive definite for point B. We now study them in the tangent plane of the surface g = 0 at points A and B. At point A, ∇V = −2/√
5 (1,2).
The perpendicular directions d write α(−2,1) with α ∈ R. Consequently d · HˆA · d = α210√
5 > 0. Point A is a local minimum, a stable equilib- rium. At point B, the perpendicular directions d also write α(−2,1) and d·HˆB·d =−α210√
5 > 0. Point B is not a local minimum, it corresponds to an unstable equilibrium.
We note that the restriction ∇g ·d = 0 did not change anything to the result: in this example stability could have been correctly inferred from (41).
Example: We study the potential energy V(x1, x2) = x21 −x22 under the constraint g(x1, x2) = x2 + 1 = 0. The equilibrium point is A with xA1 = 0, xA2 = −1, λA = 2. The augmented hessian matrix is:
HˆA = HA −λAGA =
2 0 0 −2
(42)
