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8.1: Prelude to Stability of Elastic Structures

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    21518
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    In Chapter 7 we have formulated the condition of static equilibrium of bodies and structures by studying a small change (variation) of the total potential energy. The system was said to be in equilibrium if the first variation of the total potential energy vanishes. The analysis did not say anything about the stability of equilibrium. The present lecture will give an answer to that question by looking more carefully what is happening in the vicinity of the equilibrium state.

    To illustrate the concept, consider a rigid body (a ball) siting in an axisymmetric paraboloid. shown in Figure (\(\PageIndex{1}\)).

    8.1.1.png
    Figure \(\PageIndex{1}\): Illustration of stable, neutral and unstable equilibrium.

    In the case of a rigid body the total potential energy is just the potential energy

    \[\prod = mgh = Cu^2 \]

    where u is the horizontal displacement of the ball from the resting position. Let’s calculate the first and second variation of the function \(\prod (u)\)

    \[\delta \prod = \frac{\partial\prod}{\partial u} \delta u = 2Cu \delta u \]

    \[\delta^2 \prod = \delta (\delta \prod) = 2C \delta u \delta u \]

    At the origin of the coordinate system \(u = 0\), so the first variation of \(\prod\) is zero no matter what the sign of the coefficient \(C\) is. In the expression for the second variation, the product \(\delta u \delta u = (\delta u)^2\) is always non-negative. Therefore, the sign of the second variation depends on the sign of the coefficient \(C\). From Figure \(\PageIndex{1}\) we infer that \(C > 0\) corresponds to a stable configuration. The ball displaced by a small amount \(\delta u\) will return to the original position. By contrast, for \(C < 0\), the ball, when displaced by a tiny amount \(\delta u\), will roll down and disappear. We call this an unstable behavior. The case \(C = 0\) corresponds to the neutral equilibrium.

    One can formalize the above consideration to the elastic body (structure), where the total potential energy is a function of a scalar parameter, such as a displacement amplitude \(u\). The function \(\prod (u)\) can be expanded in Taylor series around the reference point \(u_o\)

    \[\prod (u) = \prod (u_o) + \left. \frac{d\prod}{du} \right|_{u=u_o} (u − u_o) + \frac{1}{2} \left. \frac{d^2\prod}{du^2}\right|_{u=u_o} (u − u_o)^2 + \dots \]

    The incremental change of the potential energy \(\Delta \prod = \prod (u) − \prod (u_o)\) upon small variation of the argument \(\delta u = u − u_o\) is

    \[\Delta \prod = \frac{d\prod}{du} \delta u + \frac{1}{2} \frac{d^2 u}{du^2} (\delta u)^2 = \delta \prod + \delta^2 \prod + \dots \]

    For the system in equilibrium the first variation must be zero. Therefore, to the second term expansion, the sign of the increment of \(\prod\) depends on the sign of the second variation of the potential energy. We can now distinguish three cases

    \[\delta^2 \prod \begin{cases} > 0, & \text{ Positive (stable equilibrium)} \\ = 0, & \text{ Zero (neutral equilibrium)} \\ < 0, & \text{ Negative (unstable equilibrium)} \end{cases} \nonumber \]


    This page titled 8.1: Prelude to Stability of Elastic Structures is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tomasz Wierzbicki (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.