Stability of Structures

STABILITY

OF

STRUCTURES

GENERAL CONCEPTS OF STABILITY

Collapse: In a structure we can have different types of collapse:

• MATERIAL LEVEL: It can be distinguished on the basis of the behavior of the structure
  • BRITTLE BEHAVIOUR ⇒ local failure
  • DUCTILE BEHAVIOUR ⇒ global failure
  the elasto-plastic structure reach a global failure limit under a given COLLAPSE LOAD

oss.: In many cases response is given by a TIME-INDEPENDENT behaviour. But failure can happen also with damage accumulation induced by TIME-VARIABLE actions.

• GEOMETRICAL EFFECTS can be also responsible of collapse of a structure

⇒ In fact there can be failure conditions with stress and/or damage condition far from reach LIMIT CONDITIONS.

Consider the case:

beam with an imperfection that cause an initial deformation e_o

Applying a compressive force P, the presence of imperfection has an effect of destabilization and induces bending MOMENT that cannot be resisted ⇒ UNSTABLE CONDITION

At the opposite, in the case of tensile action we have a STABILIZING MOMENT tending to straight geometry ⇒ STABLE CONFIGURATION

Based on experience, in many situations, when AXIAL COMPRESSIONS reach CRITICAL VALUE, any small disturbance can induce UNSTABLE CONFIGURATIONS ⇒ STABILITY PROBLEMS

usually treated with dynamic formulations but in many cases they can be treated also by an EQUIVALENT STATIC APPROACH

The classical Theory of Structures is generally based on the SMALL DISPLACEMENTS HYPOTHESIS, which stands upon two independent assumptions:

• (a) It is possible to rely over the (linearized) small strain definition (ε = ε^l = e+nl)
• (b) displacements are small enough to allow the equilibrium to be fixed in the reference (undeformed) configuration l (t)
• (b) For ε = χ ...^t

Simple stability examples:

1 DOF SYSTEMS ⇒ LAGRANGIAN COORDINATE θ

a) We try to define EQUILIBRIUM in the undeformed configuration.

In this way, force P and not auto write moment equilibrium

LINEAR RELATION with θ

b) We can go through LARGE DISPLACEMENTS and ROTATIONS description. We have only the measured force αP:

P= ...

if we bring back the system to x_c

with a different path we can define:

*EXTERNALLY CONSERVATIVE SYSTEM ∮ W_e= ∮ F_e(x)dx=0 the work done by the applied forces is null along any closed cycle.

*INTERNALLY CONSERVATIVE SYSTEM ∮ W_i= ∮ F_i(x)dx=0 the work done by the reacting forces / stress is null over any closed cycle (case of hyper elastic law).

=>CONSERVATIVE SYSTEM if it is both internally and externally conservative

we can define some quantities:

• W_e(x_c,x_a)= -ΔΠ(x_c,x_a) — called potential energy of the external forces
• W_i(x_c,x_a)= -ΔU(x_c,x_a) — called deformation energy

A conservative system can always be associated to the quantity TOTAL POTENTIAL ENERGY

ΔV(x_c,x_a)= ΔΠ(x_c,x_a)+ ΔU(x_c,x_a)

© in a generic continuum, considering small displacement HP:

u=u(x)

V(u)= ∫_v w (E(u))dV + ∫ F_eu dV+ ∫ g_σudS -TOTAL POTENTIAL ENERGY

Note: Classical Mechanics states the

• Kinetic Energy Law: work
• external and internal, in the variation ΔT, that is ΔT(x_c,x_a) :
• If the system is cons

In general, when internal dissipation happens D(x_c,x_a)= ∮ W_d ≥0

If external forces are non-conservative D(x_c,x_a) <0, energy of the system increases (left the problem)

* What types of forces are CONSERVATIVES?

Consider the following simple problem considering the state variation:

x_o (0,0), θ_a (0+2) -> x_c (π/2,π)

and the work computed along two different path β₁:

β₂

θ_c=θ₂

= ₁/² kΘ² - ₁/² PLΘ² - αPLΘ

V₂(Θ) = -αPLΘ + ₁/² (k - PL) Θ²

Now again we apply stationarity and the condition on II^nd order variation in order to check equilibrium and stability.

* EQUILIBRIUM

δV₂ = 0 → -αPL + (K - PL)Θ = 0

(k-PL)Θ = αPL

Θ = ^αPL/_{k - PL}

this is the equilibrium that we've already got by directly introducing the approx. up to the I^st order

Add second order approximation in the energy product we get the same equation of equilibrium that we find with the first order approximations in equilibrium

* STABILITY CHECK

∂²V/∂Θ² = k - PL

δ ₁/² δ[² V = ₁/² ∂²V/∂Θ²Θ² > 0

₁/² (k - PL) Θ² > 0

(k - PL) > 0

P _L < K or p_L < 1

V₂(Θ) = -αPLΘ + ₁/² (k - PL) Θ²

we have the II^nd variation without make any computation we have just to draw when this quantity is >> 0

The limit P_L/k/l doesn't depend upon the configuration of equilibrium. It is just an overall stability condition for the problem regardless the current value of the result

We can find any solution depending upon the load, α, or (k - PL), but at the same time we find an answer to the stability. We can find a solution depending on P_L/k/l, but for when upon P_L ≠ P_L/k/l = 1 the II^nd variation is zero and the system become unstable.

The level of external loading that we find corresponding to the lack of stability of the system for P_L/k/l→1 means the same level of loading for which the equilibrium solution become singular

Example: Tensile Force

Δ₁ = L(1 - cosΘ)

Δ₂ = L senΘ tanΘ

Δ = Δ₂ - Δ₁ = L[senΘ tanΘ - (1 - cosΘ)]

V(Θ) = 1/2 k Θ² - TΔ = 1/2 k Θ² - TL[senΘ tanΘ - (1 - cosΘ)]

V₂(Θ) = 1/2 k Θ² - TL[Θ²/2 + 1 - Θ²/2]

V₂(Θ) = 1/2 k Θ² - TLΘ²/2 = 1/2 (k - TL) Θ² = 1/2 (1 - t) Θ²

V₂(Θ) = 0 → (1 - t) = 0, t = 1

Even if T is a tensile force

Continuos Systems

1-Axially Inextensible Euler-Bernoulli Beam

HP: * perfect constraints * perfectly-elastic behaviour

1 DOF Θ or Δ

rotation deflection

We focus now on Δ = displacement performed on the right extremity of the beam

To find out Δ we can try to sum up the dl or dlu that are the difference between the end in undeformed configuration of a dx-element that we are considering minus the projection of deformed configuration

dl / dx = dx cosΘ

Due to inextensibility property dx doesn't change

We rewrite B.C. in terms of :

• (0)=0
• (0)=0
• (0)=0
• (0)=0

We can write it in a matrix form:

IF the determinant of the matrix is NON ZERO

the only unique solution is X = 0

When det(M)=0, we get a solution, but not unique

X=0 det(M)=0

oss: When we determined n(x) dependent on a constant, that is a

discretization into the solution and it is like mass balance

discrete equation of equilibrium and we final out that

there & constants don't be ≠ form zero only in one

one circumstance - det M = 0

det(M) =

CONDITION OF

SINGULARITY of M

=kN*

A=0;B=0;C;D=0