Model A: stable symmetric bifurcation buckling

This model is shown in figure [fig10.1] and it has one coordinate \(\theta, -\pi < \theta < \pi\), to describe the configuration of the model under the deadweight load P. (An external load independent of its corresponding displacement.) The model consists of a rigid rod of length L, connected by smooth hinge to a rigid base. The rod can rotate about the hinge, but it is restrained by a linear elastic torsional spring of stiffness K (dimensional units of F-L/ radian). The restoring moment of the spring acting on the bar is zero at \(\theta = 0\). Neglect the weight of the rod with respect to the applied load P. From the free body diagram of the rod shown in figure [fig10.1], the equation of motion for rotation about the fixed hinge is \[\label{eq10.1} P L \sin \theta-K \theta=I_{0} \frac{d^{2} \theta}{d t^{2}} \quad \theta=\theta(t) \quad t>0\] where \({I_0}\) is the moment of inertia of the rod about the fixed point and \(t\) is time.

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Model B: unstable symmetric bifurcation

This model consists of a coplanar arrangement of two rigid bars of length L and a linear elastic spring of stiffness K. The bars are horizontal in the initial position and connect to a center hinge, with the opposite ends of each bar supported on roller support. The vertical spring connects to the center hinge and is not stretched when the bars are in the horizontal position. A horizontal load P acts at each roller support to subject the model to compression.

The deflected configuration of the model is symmetric with respect to the vertical line through the spring, and each bar rotates through an angle \(\theta\) with respect to the original horizontal position. The deflection of the spring is \(L \sin \theta\), and the load P is independent of the corresponding displacement \(\Delta(\theta)\). The potential energy is \[\begin{aligned} \label{eq10.40} V=\frac{1}{2} K(L \sin \theta)^{2}-2 P \underbrace{[L(1-\cos \theta)]}_{\Delta}.\end{aligned}\] The potential energy is stationary at equilibrium, or \(\frac{d V}{d \theta}=0\), which leads to \[\begin{aligned} \label{eq10.41} (K L^{2} \cos \theta-2 P L) \sin \theta=0.\end{aligned}\] The solutions of eq. ([eq10.41]) are \[\begin{aligned} \label{eq10.42} p 1: \theta=0 \text { for any } P \quad p 2: P=\frac{1}{2} K L \cos \theta.\end{aligned}\] The equilibrium paths are shown on the load-deflection plot in figure [fig10.13]. Equilibrium paths intersect at the bifurcation point \((\theta,\; P)=(0,\; \textit{KL}/2)\). By the equilibrium method the critical load is \(P_{\mathrm{cr}}=K L / 2\). However, equilibrium states in figure [fig10.13] are different than those of model A shown in figure [fig10.2]. Note that there are three equilibrium positions for \(P<P_{\mathrm{cr}}\).

The stability of the equilibrium states is assessed from the second derivative of the potential energy ([eq10.40]). The second derivative is \[\begin{aligned} \label{eq10.43} \frac{d^{2} V}{d \theta^{2}}=V_{2}(\theta)=K L^{2}(\cos ^{2} \theta-\sin ^{2} \theta)-2 P L \cos \theta=K L^{2} \cos 2 \theta-2 P L \cos \theta.\end{aligned}\] On equilibrium path p1 \[\begin{aligned} \label{eq10.44} V_{2}(0)=K L^{2}-2 P L.\end{aligned}\] Therefore, on equilibrium path p1 \[\begin{aligned} \label{eq10.45} V_{2}(0)|_{p 1}>0 \text {, stable, } P<\frac{K L}{2} \quad V_{2}(0)|_{p 1}=0, \text { critical, } P=\frac{K L}{2} \quad V_{2}(0)|_{p 1}<0 \text {, unstable, } P>\frac{K L}{2}.\end{aligned}\] On equilibrium path p2 \(P=(K L / 2) \cos \theta_{0}\). The second derivative is \[\begin{aligned} \label{eq10.46} V_{2}\left(\theta_{0}\right)=K L^{2}(\cos ^{2} \theta-\sin ^{2} \theta)-2(K L / 2) \cos ^{2} \theta=-K L^{2} \sin ^{2} \theta_{0}.\end{aligned}\] Therefore, on equilibrium path p2 \[\begin{aligned} \label{eq10.47} &V_{2}\left(\theta_{0}\right)|_{p 2}>0, \text { stable, for no } 0<|\theta_{0}| \leq\frac{\pi}{2} \quad V_{2}\left(\theta_{0}\right)|_{p 2}=0, \text { critical, } \theta_{0}=0 \nonumber\\ &V_{2}\left(\theta_{0}\right)|_{p 2}<0, \text { unstable, } 0<|\theta_{0}| \leq \frac{\pi}{2}.\end{aligned}\] At the bifurcation point \(\left(\theta_{0}, P\right)=(0, K L / 2)\) on path p2 the derivatives of the potential energy are \[\begin{aligned} \label{eq10.48} V_{2}(0)=0 \quad V_{3}(0)=0 \quad V_{4}(0)=-3 K L^{2}.\end{aligned}\] The potential energy is a relative maximum at the bifurcation point, so the bifurcation point is unstable. Model B exhibits unstable symmetric bifurcation.

Model C: asymmetric bifurcation

Model C is a coplanar arrangement of two rigid bars of length L and a linear elastic spring with stiffness K. In the initial configuration the bars are horizontal and the spring is at a \(45^{\circ}\) angle with respect to the bars. The bars and the spring are connected to a smooth central pin. The opposite end of the left bar is pinned to a fixed point, and the opposite end of the spring is connected to a fixed pin at distance L below the fixed end of the left bar. The opposite end of the right bar is pinned to a roller support free to move horizontally. A compressive force P acts at the roller support and under its action the bars can rotated through an angle \(\theta\) with respect to the original horizontal position.

The potential energy is \(V=K \Delta_{s}^{2} / 2-P \Delta\), where \(\Delta_{s}\) denotes the change in the length of the spring and \(\Delta\) denotes the shortening of the distance between supports. These changes in length are related to angle \(\theta\) by \[\begin{aligned} \label{eq10.53} \Delta_{s}=\sqrt{(L \cos \theta)^{2}+(L-L \sin \theta)^{2}}-\sqrt{2} L=\sqrt{2} L(\sqrt{1-\sin \theta}-1) \quad \Delta=2 L-2 L \cos \theta.\end{aligned}\] The total potential energy \(V(\theta)\) is given by \[\begin{aligned} \label{eq10.54} V(\theta)=K L^{2}(\sqrt{1-\sin \theta}-1)^{2}-2 P L(1-\cos \theta).\end{aligned}\] The potential energy is stationary at equilibrium which leads to \[\begin{aligned} \label{eq10.55} K L^{2} \cos \theta[(1-\sin \theta)^{-1 / 2}-1]-2 P L \sin \theta=0.\end{aligned}\] The solutions of eq. ([eq10.55]) are \[\begin{aligned} \label{eq10.56} p 1: \theta=0 \text { for any } P \quad p 2: P=\frac{K L}{2} \cot \theta[(1-\sin \theta)^{-1 / 2}-1].\end{aligned}\] On path p2 as \(\theta \rightarrow 0\), we get the indeterminate form \(\frac{K L}{2} \frac{1}{0}\left(\frac{1}{1}-1\right)=\frac{K L}{2} \frac{1}{0} \times 0\). The limit of this indeterminate form is found from l’Hôpital’s rule to be \[\begin{aligned} \label{eq10.57} P=P_{\mathrm{cr}}=K L / 4\quad \text {at } \theta=0.\end{aligned}\] The equilibrium paths are plotted on the load-deflection plane in figure [fig10.18]. Equilibrium path p2 is asymmetric about \(\theta = 0\). Stability analysis leads to path p2 being stable for \(\theta > 0\) and unstable for \(\theta < 0\). Path p1 is stable for \(0 \leq P<P_{\mathrm{cr}}\) and unstable for \(P>P_{\mathrm{cr}}\). At the bifurcation point \((\theta, P)=\left(0, P_{\mathrm{cr}}\right)\) higher derivatives of the potential energy are \(V_{2}=0\) and \(V_{3}=\left(3 K L^{2}\right) / 4\). That is, the potential energy is neither a minimum nor maximum, but has a horizontal inflection point at \((\theta, P)=\left(0, P_{\mathrm{cr}}\right)\).

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Consider a geometric imperfection of model C in which the bars are at an angle \(\delta_{0}\) with respect to the horizontal before the load is applied as is shown in figure [fig10.19]. In the unloaded configuration the spring is not stretched nor contracted. The change in spring length is \[\begin{aligned} \label{eq10.58} \Delta_{s}=\sqrt{2} L\left(\sqrt{1-\sin \theta}-\sqrt{1-\sin \delta_{0}}\hspace*{3pt}\right).\end{aligned}\] The potential energy is \[\begin{aligned} \label{eq10.59} V=K L^{2}\left(\sqrt{1-\sin \theta}-\sqrt{1-\sin \delta_{0}}\hspace*{3pt}\right)^{2}-2 P L\left(\cos \delta_{0}-\cos \theta\right).\end{aligned}\] The potential energy is stationary at equilibrium, which leads to \[\begin{aligned} \label{eq10.60} K L^{2} \cos \theta\left(\frac{\sqrt{1-\sin \theta}-\sqrt{1-\sin \delta_{0}}}{\sqrt{1-\sin \theta}}\right)-2 P L \sin \theta=0.\end{aligned}\] Solve eq. ([eq10.60]) for P and divide by \(P_{\mathrm{cr}}=K L / 4\) to get \[\begin{aligned} \label{eq10.61} \frac{P}{P_{\mathrm{cr}}}=2 \cot \theta\left(\frac{\sqrt{1-\sin \theta}-\sqrt{1-\sin \delta_{0}}}{\sqrt{1-\sin \theta}}\right).\end{aligned}\] Note that a solution of eq. ([eq10.61]) is \((\theta, P)=\left(\delta_{0}, 0\right)\). The equilibrium paths determined from eq. ([eq10.61]) are shown as dashed lines in the load-deflection plane of figure [fig10.20]. The equilibrium path beginning at the unloaded state for \(\delta_{0}>0\) in figure [fig10.20](a) is stable and the deflection increases rapidly as the load approaches the critical load of the perfect system. The equilibrium path beginning at the unloaded state for \(\delta_{0}<0\) in figure [fig10.20](b) is stable until the maximum load \(P_{\mathrm{m}}\) is encountered, which is indicated by the filled circle. There are no stable adjacent equilibrium states if P is increased from \(P_{\mathrm{m}}\) or if \(\theta\) decreases from the maximum load point. Hence, model C is imperfection sensitive for \(\delta_{0}<0\).

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A real structure exhibiting asymmetric bifurcation is a pin-supported, two-member frame. The joint connecting the members is assumed rigid. Thus, each bar rotates through the same angle at the joint as shown in figure [fig10.21]. For \(e>0\) the horizontal member is in tension, which is a stabilizing effect. For \(e<0\) the horizontal member is in compression, which is a destabilizing effect.

Discussion of models A, B, and C

We have considered three one-degree-of-freedom models (one coordinate is sufficient to describe the equilibrium configuration). The equilibrium paths were plotted on the \((\theta, P)\) plane. For the perfect system \(\theta = 0\) for any P is an equilibrium state (trivial equilibrium). Two equilibrium paths of the perfect system cross at the bifurcation point \((\theta, P)=\left(0, P_{\mathrm{cr}}\right)\). There are three basic bifurcation points: stable symmetric, unstable symmetric, and asymmetric. The unstable symmetric and asymmetric cases are imperfection sensitive. A maximum load \(P_{\mathrm{m}}\) below \(P_{\mathrm{cr}}\) is possible when the system has imperfections. This theory was originally developed in the PhD dissertation by Koiter (1945 in Dutch, English translation 1970).

Model D: snap-through instability

Model D is a coplanar arrangement of two rigid bars and a linear elastic spring in the shape of an arch as shown in figure [fig10.22]. Each bar has the same length L, and the bars connect to a central pin. The bars are at angle \(\alpha\) with respect to a horizontal line passing through the supported ends of the bars. The left end of the left bar is pin-connected to a fixed support. The right end of the right bar is pin-connected to a roller support restrained to move horizontally by a linear elastic spring with stiffness K. The model is subject to a downward, deadweight load P acting at the central pin.

The total potential energy is \(V=K\left(\Delta_{s}\right)^{2 / 2-P \Delta}\), where \(\Delta_{s}\) is the change in length of the spring and \(\Delta\) is the downward displacement corresponding to the load P. The change in length of the spring and the downward displacement are \[\begin{aligned} \label{eq10.62} \Delta_{s}=2 L(\cos \theta-\cos \alpha) \quad \Delta=L(\sin \alpha-\sin \theta).\end{aligned}\] Hence, the total potential energy is \[\begin{aligned} \label{eq10.63} V(\theta)=2 K L^{2}(\cos \theta-\cos \alpha)^{2}-P L(\sin \alpha-\sin \theta).\end{aligned}\] The total potential energy is stationary at equilibrium, which yields the equilibrium equation \[\begin{aligned} \label{eq10.64} 4 K L^{2}(\cos \theta-\cos \alpha)(-\sin \theta)+P L \cos \theta=0.\end{aligned}\] Solve eq. ([eq10.64]) for load P to get \[\begin{aligned} \label{eq10.65} P=4 K L(\cos \theta-\cos \alpha) \tan \theta.\end{aligned}\] Note that the range of \(\theta\) in eq. ([eq10.65]) is \(-\pi / 2<\theta<\pi / 2\) for finite values of the load P. On a plot of the load P as a function of \(\theta\), horizontal slopes occur at \(\frac{d P}{d \theta}=0\). The derivative of eq. ([eq10.65]) with respect to \(\theta\) is \[\begin{aligned} \label{eq10.66} \frac{d P}{d \theta}=4 K L\left[-\sin \theta \tan \theta+\frac{(\cos \theta-\cos \alpha)}{\cos ^{2} \theta}\right]=4 K L \frac{(\cos ^{3} \theta-\cos \alpha)}{\cos ^{2} \theta}.\end{aligned}\] Therefore horizontal slopes occur at \[\begin{aligned} \label{eq10.67} \theta_{\mathrm{m}}=\pm \operatorname{acos}[\sqrt[3]{\cos \alpha}],\end{aligned}\] Substitute \(\cos \theta=\cos ^{1 / 3} \alpha\) into eq. ([eq10.65]) and use trigonometric identities to find the load at the horizontal slope to be \[\begin{aligned} \label{eq10.68} P_{\mathrm{m}}=4 K L(1-\cos ^{2 / 3} \alpha)^{3 / 2}.\end{aligned}\] For \(\alpha=45^{\circ}\), \(\theta_{\mathrm{m}}=\pm 27.01^{\circ}\). At \(\theta_{\mathrm{m}}=-27.01^{\circ}\) the load \(P_{\mathrm{m}}=-0.375\,KL\) with the corresponding displacement \(\Delta_{\mathrm{m}}=1.16\,{L}\). At \(\theta_{\mathrm{m}}=27.01^{\circ}\) the load \(P_{\mathrm{m}}=0.375\,{KL}\) with the corresponding displacement \(\Delta_{\mathrm{m}}=0.253\,{L}\). The load-displacement response is plotted in figure [fig10.23] by selecting \(\theta\) and computing P from eq. ([eq10.65]) and \(\Delta\) from eq. ([eq10.62]). There is one continuous path with no bifurcation. The loads at the horizontal slopes are indicated by filled circles in figure [fig10.23].

The stability of the equilibrium states are determined from the second derivative of the potential energy. The second derivative is \[\begin{aligned} \label{eq10.69} \frac{d^{2} V}{d \theta^{2}}=4 K L^{2}[\sin ^{2} \theta-\cos \theta(-\cos \alpha+\cos \theta)]-P L \sin \theta.\end{aligned}\]

Substitute the expression for P from eq. ([eq10.65]) into eq. ([eq10.69]) to evaluate the second derivative on the equilibrium path to find \[\begin{aligned} \label{eq10.70} V_{2}(\theta)=\frac{d^{2} V}{d \theta^{2}}=4 K L^{2} \frac{\left(\cos ^{3} \theta-\cos \alpha\right)}{\cos \theta}.\end{aligned}\] For \(|\theta|<\pi / 2\), \(\cos \theta>0\). Select a value of \(\theta\) in the range \(|\theta|<\pi / 2\). Then, the value of \(\Delta\) is computed from eq. ([eq10.62]) and the value of the second derivative of the potential energy is computed from eq. ([eq10.70]). The plot of the second derivative divided by \(K L^{2}\) with respect to \(\Delta / L\) is shown in figure [fig10.24].

For \(\alpha=45^{\circ}\) the range of \(\Delta\) is \(-0.2929 L<\Delta<1.707 L\) when \(\theta\) is in the interval \(\pi / 2>\theta>-\pi / 2\). From figure [fig10.24] the stability of the equilibrium path is determined as \[\begin{gathered} \label{eq10.71} V_{2}>0, \text { stable, }-0.293<\Delta / L<0.253\, \&\, 1.16<\Delta / L<1.1707, \\\label{eq10.72} V_{2}=0 \text {, critical } \Delta / L=0.253\, \&\, \Delta / L=1.16, \text{and} \\ \label{eq10.73} V_{2}<0, \text { unstable, } 0.253<\Delta / L<1.16.\end{gathered}\] The stability of the equilibrium path is depicted in figure [fig10.25]. As the load P is increased from \(\Delta=0\) a maximum load \(P_{\mathrm{m}}\) is encountered. If the load is increased further the system snaps-through. The maximum point is called a limit point. This is a different kind of instability from the perfect systems of models A, B, and C. In models A, B, and C \(\theta = 0\) was an equilibrium state of the perfect system. Model D is said to have pre-buckling “deformations.” That is \(\theta \neq 0\) before buckling. Snap-through is a dynamic event, and the system can settle to an inverted, stable equilibrium state. If the load is decreased from the inverted state to the lower limit point the system can snap back to a shape resembling the original configuration.

Model E: a two-degree-of freedom system

Models A to D are single-degree-of-freedom systems. Only one coordinate \(\theta\) determines the position of the system. Consider a two-degree-of-freedom system consisting of rigid bar restrained by two rotational springs with stiffnesses \(\textit{K}_1\) and \(\textit{K}_2\), and subject to a vertical, deadweight load P as shown in figure [fig10.26]. This model is known as Augusti’s column. See Bazant and Cedolin (1991). The position of the bar is referenced to a right-handed Cartesian coordinate system x-y-z, with corresponding unit vectors \(\hat{i}, \hat{j}, \hat{k}\). The initial position of the bar is vertical coinciding with the \(z\)-axis shown in figure [fig10.26](a), and in the deflected position it is located by two angles \(\theta_1\) and \(\theta_2\) shown in figure [fig10.26](b) The projection of the bar into the x-z plane is at angle \(\theta_1\) with respect to the \(z\)-axis. The projection of the bar in the y-z plane is at angle \(\theta_2\) with respect to the \(z\)-axis.

The angle between the bar and the \(z\)-axis is denoted by \(\varphi\). The Cartesian coordinates at the end of the bar in its deflected position in shown in figure [fig10.26] (c) are \(\left(L \sin \theta_{1}, L \sin \theta_{2}, L \cos \varphi\right)\). By the Pythagorean theorem the square of the length of the bar in the deflected position is given by \[\begin{aligned} \label{eq10.74} L^{2}=\left(L \sin \theta_{1}\right)^{2}+\left(L \sin \theta_{2}\right)^{2}+(L \cos \varphi)^{2}.\end{aligned}\] From eq. ([eq10.74]) we find that the cosine of the angle \(\varphi\) is \[\begin{aligned} \label{eq10.75} \cos \varphi=\sqrt{1-\sin ^{2} \theta_{1}-\sin ^{2} \theta_{2}}.\end{aligned}\] The displacement corresponding to load P is \(\Delta=L(1-\cos \varphi)\). The total potential energy is \[\begin{aligned} \label{eq10.76} V\left(\theta_{1}, \theta_{2}\right)=K_{1} \theta_{1}^{2} / 2+K_{2} \theta_{1}^{2} / 2-P L\left(1-\sqrt{1-\sin ^{2} \theta_{1}-\sin ^{2} \theta_{2}}\hspace*{4pt}\right).\end{aligned}\] The series expansion of \(1-\cos \varphi\) is \[\begin{aligned} \label{eq10.77} 1-\sqrt{1-\sin ^{2} \theta_{1}-\sin ^{2} \theta_{2}}=\frac{1}{2}\left(\theta_{1}^{2}+\theta_{2}^{2}+\frac{1}{2} \theta_{1}^{2} \theta_{2}^{2}-\frac{1}{12} \theta_{1}^{4}-\frac{1}{12} \theta_{2}^{4}+O(\theta^{6})\right).\end{aligned}\] Neglect terms of order six and higher in the series expansion to get the total potential energy as \[\begin{aligned} \label{eq10.78} V\left(\theta_{1}, \theta_{2}\right)=K_{1} \theta_{1}^{2} / 2+K_{2} \theta_{1}^{2} / 2-P L\left(\theta_{1}^{2}+\theta_{2}^{2}+\frac{1}{2} \theta_{1}^{2} \theta_{2}^{2}-\frac{1}{12} \theta_{1}^{4}-\frac{1}{12} \theta_{2}^{4}\right) / 2.\end{aligned}\]

Let \(\theta_{1}=\theta_{10}\) and \(\theta_{2}=\theta_{20}\) denote the angles in an equilibrium state, and let small changes in the angles with respect to the equilibrium state be denoted by \[\begin{aligned} \label{eq10.79} h_{1}=\theta_{1}-\theta_{10} \quad \text { and } \quad h_{2}=\theta_{2}-\theta_{20}.\end{aligned}\] The Taylor series of the potential energy about the equilibrium state is \[\begin{aligned} \label{eq10.80} V\left(\theta_{1}, \theta_{2}\right)=V\left(\theta_{10}, \theta_{20}\right)+\delta V+\delta^{2} V+\delta^{3} V+\delta^{4} V+\ldots,\end{aligned}\] where \(\delta V\) is called the first variation with terms linear in \(h_1\) and \(h_2\), and \(\delta^{2} V\) is called the second variation with terms quadratic in \(h_1\) and \(h_2\), etc. The change in potential energy about the equilibrium state is \(\Delta V=V\left(\theta_{1}, \theta_{2}\right)-V\left(\theta_{10}, \theta_{20}\right)\). Thus, \[\begin{aligned} \label{eq10.81} \Delta V=\delta V+\delta^{2} V+\delta^{3} V+\delta^{4} V+\ldots.\end{aligned}\] Partial derivatives of the potential energy evaluated at the equilibrium state are represented by the notation \[\begin{aligned} \label{eq10.82} V_{0}^{(m, n)}=\left.\frac{\partial^{(m\,{+}\,n)} V}{\partial \theta_{1}^{m} \partial \theta_{2}^{n}}\right|_{\theta_{10}, \theta_{20}} \quad m, n=0,1,2, \ldots.\end{aligned}\] For example, \[\begin{aligned} \label{eq10.83} V_{0}^{(1,0)}=\left.\frac{\partial V}{\partial \theta_{1}}\right|_{\theta_{10}, \theta_{20}} \quad V_{0}^{(1,1)}=\left.\frac{\partial^{2} V}{\partial \theta_{1} \partial \theta_{2}}\right|_{\theta_{10}, \theta_{20}} \quad V_{0}^{(2,2)}=\left.\frac{\partial^{4} V}{\partial \theta_{1}^{2} \partial \theta_{2}^{2}}\right|_{\theta_{10}, \theta_{20}}.\end{aligned}\] The terms in the Taylor series expansion ([eq10.81]) are \[\begin{aligned} \label{eq10.84} \begin{gathered} \delta V=V_{0}^{(1,0)} h_{1}+V_{0}^{(0,1)} h_{2} \\ \delta^{2} V=\frac{1}{2}\left(V_{0}^{(2,0)} h_{1}^{2}+2 V_{0}^{(1,1)} h_{1} h_{2}+V_{0}^{(0,2)} h_{2}^{2}\right) \\ \delta^{3} V=\frac{1}{6}\left(V_{0}^{(3,0)} h_{1}^{3}+3 V_{0}^{(2,1)} h_{1}^{2} h_{2}+3 V_0^{(1,2)} h_{1} h_{2}^{2}+V_{0}^{(0,3)} h_{2}^{3}\right) \\ \delta^{4} V=\frac{1}{24}\left(V_{0}^{(4,0)} h_{1}^{4}+4 V_{0}^{(3,1)} h_{1}^{3} h_{2}+6 V_{0}^{(2,2)} h_{1}^{2} h_{2}^{2}+4 V_{0}^{(1,3)} h_{1} h_{2}^{3}+V_{0}^{(0,4)} h_{2}^{4}\right) \end{gathered}.\end{aligned}\]

A necessary condition for the potential energy to be a relative minimum or maximum at the equilibrium state is \(\delta V=0\) for every \(h_1\) and \(h_2\), but both not equal to zero. Thus, “coefficients” \(V_{0}^{(1,0)}=0\) and \(V_{0}^{(0,1)}=0\). The potential energy is stationary at equilibrium. Take the partial derivatives of the potential energy ([eq10.78]) to get the equilibrium equations \[\begin{aligned} \label{eq10.85} V_{0}^{(1,0)}=K_{1} \theta_{10}-L P(\theta_{10}-\theta_{10}^{3} / 6+\theta_{10} \theta_{20}^{2} / 2)=0,\ \text{and} \\ \label{eq10.86} V_{0}^{(0,1)}=K_{2} \theta_{20}-L P(\theta_{20}-\theta_{20}^{3} / 6+\theta_{10}^{2} \theta_{20} / 2)=0.\end{aligned}\] A solution to the equilibrium equations ([eq10.85]) and ([eq10.86]) is \[\begin{aligned} \label{eq10.87} p 1: \theta_{10}=\theta_{20}=0 \text { for any } P.\end{aligned}\] The next non-zero term in the expansion of \(\Delta V\) is the second variation. Evaluating the second order partial derivatives of the potential energy ([eq10.78]) followed by evaluation on equilibrium path p1 we get \[\begin{aligned} \label{eq10.88} \delta^{2} V=\frac{1}{2}\left[\left(K_{1}-L P\right) h_{1}^{2}+\left(K_{2}-L P\right) h_{2}^{2}\right].\end{aligned}\]

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Buckling loads are determined when second variation vanishes for every value of \(h_1\) and \(h_2\), but both not equal to zero. This leads to two buckling loads and associated modes \[\begin{aligned} \label{eq10.89} P_{1}=K_{1} / L \quad\left(h_{1}, h_{2}\right)=(1,0),\ \text{and}\\ \label{eq10.90} P_{2}=K_{2} / L \quad\left(h_{1}, h_{2}\right)=(0,1).\end{aligned}\] The critical loads and modes are shown in the load-deflection plane of figure [fig10.27]. Take the case of \(K_{1}<K_{2}\). Then, the critical load is \(P_{\mathrm{cr}}=P_{1}=K_{1} / L\) and the associated mode is \(\left(h_{1}, h_{2}\right)=(1,0)\).

The second variation \(\delta^{2} V\) is a quadratic form in variables \(h_1\) and \(h_2\). Examples of quadratic forms and their descriptions are listed in Table [tab10.1].

The second variation ([eq10.88]) is positive definite for \(0 \leq P<K_{1} / L\). At the critical load the second variation is \[\begin{aligned} \label{eq10.91} \delta^{2} V=\frac{1}{2}\left[0 \cdot h_{1}^{2}+\left(K_{2}-K_{1}\right) h_{2}^{2}\right].\end{aligned}\] The second variation at the critical load is said to be positive semidefinite. It is zero for all non-zero values of \(h_1\) and \(h_2 = 0\), but is positive for all non-zero values of \(h_2\) and \(h_1 = 0\). The second variation ceases to be positive definite at the critical state. The stability of equilibrium path \(p_{1}\) is determined from eq. ([eq10.88]) as follows: \[\begin{aligned} \label{eq10.92} \left.\delta^{2} V\right|_{p 1}>0 \text {, stable, } 0 \leq P<K_{1} /\left.L \quad \delta^{2} V\right|_{p 1}=0 \text {, critical, } P=K_{1} /\left.L \quad \delta^{2} V\right|_{p 1}<0 \text {, unstable, } P>K_{1} / L.\end{aligned}\] The stability of the bifurcation point \(\left(\theta_{1}, \theta_{2}, P\right)=\left(0,0, K_{1} / L\right)\) is not determined from the second variation of the potential energy.

At the critical load \(\theta_{2}=0\). This suggests we seek a solution to equilibrium equations ([eq10.85]) and ([eq10.86]) with \(\theta_{10} \neq 0\) and \(\theta_{20}=0\). Equation ([eq10.86]) is identically satisfied, and eq. ([eq10.85]) reduces to \[\begin{aligned} \label{eq10.93} K_{1} \theta_{10}-L P(\theta_{10}-\theta_{10}^{3} / 6)=0.\end{aligned}\] Solve eq. ([eq10.93]) for P to get \[\begin{aligned} \label{eq10.94} \frac{P}{P_{\mathrm{cr}}}=\frac{\theta_{10}}{\theta_{10}-\theta_{10}^{3} / 6}.\end{aligned}\] The equilibrium path described by eq. ([eq10.94]) is shown in figure [fig10.28]. The load increases in the initial post-buckling response indicating the bifurcation point is stable.

Consider the case where \(K_{1}=K_{2}=K\). The critical points \(\textit{P}_1\) and \(\textit{P}_2\) coincide on the path \(p_1\) and simultaneous buckling modes \(\left(h_{1}, h_{2}\right)=(1,0)\) and \(\left(h_{1}, h_{2}\right)=(0,1)\) interact at \(\left(\theta_{1}, \theta_{2}, P\right)=(0,0, K / L)\). In this case both \(\delta V=0\) and \(\delta^{2} V=0\) at the bifurcation point, and we have to consider the next non-zero term in the expansion of the change in potential energy ([eq10.81]). To evaluate the third variation at the bifurcation point, the third partial derivatives of the total potential energy evaluated a the bifurcation point are \[\begin{aligned} \label{eq10.95} V_{0}^{(3,0)}=V_{0}^{(2,1)}=V_{0}^{(1,2)}=V_{0}^{(0,3)}=0.\end{aligned}\] Hence, \(\delta^{3} V=0\) for all values of \(h_1\) and \(h_2\). Evaluate the fourth partial derivatives of the total potential energy at the bifurcation point to find \[\begin{aligned} \label{eq10.96} V_{0}^{(4,0)}=K_{1} \quad V_{0}^{(3,1)}=0 \quad V_{0}^{(2,2)}=-K_{1} \quad V_{0}^{(1,3)}=0 \quad V_{0}^{(0,4)}=K_{1}.\end{aligned}\] The fourth variation of the potential energy is \[\begin{aligned} \label{eq10.97} \delta^{4} V=\frac{1}{24}(K_{1} h_{1}^{4}-6 K_{1} h_{1}^{2} h_{2}^{2}+K_{1} h_{2}^{4}).\end{aligned}\] The fourth variation vanishes at \(h_{2}=\pm 0.414 h_{1}\) and \(h_{2}=\pm 2.414 h_{1}\). Regions in the \(h_1\)-\(h_2\) plane where the fourth variation is positive and negative are established by plotting the locus where it is zero as shown in figure [fig10.29]. The minimum values of the fourth variation occur along the directions \(h_{2}=\pm \sqrt{3} h_{1}\) and are \(\delta^{4} V=-K_{1} h_{1}^{4} / 3\). Since the fourth variation can be positive, zero, and negative depending on the values of \(h_1\) and \(h_2\), the fourth variation is indefinite. The bifurcation point is unstable. It is shown in Bazant and Cedolin (1991) that the condition for existence of a non-zero solution to equilibrium equations ([eq10.85]) and ([eq10.86]) is \(\theta_{10}=\theta_{20}=\theta\). The two equilibrium equations reduce to the single equation \[\begin{aligned} \label{eq10.98} K \theta-P L(\theta+\theta^{3} / 3)=0.\end{aligned}\] Solve eq. ([eq10.98]) for the load P to get \[\begin{aligned} \label{eq10.99} P / P_{\mathrm{cr}}=\frac{\theta}{\theta+\theta^{3} / 3}.\end{aligned}\] The load decreases from the critical value on the post-buckling path for \(|\theta|>0\) as shown in figure [fig10.30].

For \(\textbf{\textit{K}}_{\textbf{1}} \boldsymbol{<} \textbf{\textit{K}}_{\textbf{2}}\) the bifurcation point stable and the system is imperfection insensitive. For \(\textbf{\textit{K}}_{\textbf{1}} \boldsymbol{=} \textbf{\textit{K}}_{\textbf{2}} \boldsymbol{=} \textbf{\textit{K}}\) the bifurcation point is unstable and the system is imperfection sensitive.

[sec10.7] Bazant, Z., and L. Cedolin. Stability of Structures: Elastic, Inelastic, Fracture, and Damage Theories. New York: Oxford University Press, 1991, p.265.

Brush, D. O., and BO O. Almroth. Buckling of Bars, Plates, and Shells. New York: McGraw-Hill, 1975, p. 185.

Huseyin, K. Nonlinear theory of elastic stability. Leyden, The Netherlands: Noordhoff International Publishing, 1975.

Koiter, W. T. “The Stability of Elastic Equilibrium.” PhD diss., Teehisehe Hoog School, Delft, The Netherlands. 1945. (English translation published as Technical Report AFFDL-TR-70-25, Air Force Flight Dynamics Laboratory, Wright-Patterson Air Force Base, OH, February 1970.)

Simitses, G. An Introduction to the Elastic Stability of Structures. Englewood Cliffs, NJ: Prentice Hall, Inc., 1976, Chapter 2.

Ziegler, Hans. Principles of Structural Stability. Waltham, Massachusetts: Blaisdell Publishing Company, A Division of Ginn and Company, 1968, Chapter 1.

Practice exercises

1. A rigid, straight bar of length L is pinned a point O, restrained by a linear elastic spring with stiffness K, and subject to a downward load P. Neglect the weight of the bar. The bar is vertical in the initial configuration as shown in figure [fig10.31](a).The spring remains horizontal as the bar rotates from the vertical through angle \(\theta\) as shown in figure [fig10.31](b). Refer to the free body diagram in figure [fig10.31](c) to find the equation of motion is \[\begin{aligned} \label{eq10.100} P L \sin \theta-[K(a \sin \theta)] a \cos \theta=I_{0} \frac{d^{2} \theta}{d t^{2}} \quad \theta=\theta(t),\end{aligned}\] where \({I_0}\) is the moment of inertia of the rod about the fixed point and \(t\) is time. \(\theta>0\) clockwise.

   1. Plot the equilibrium paths on the \(P-\theta\) plane for \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\) and \({P > }0\). Note, \(\theta\) is independent of t.

   2. What is the critical load \(P_{\mathrm{cr}}\)?

   3. Let the rotation angle \(\theta(t)=\theta_{0}+\varphi(t)\) where \(\theta_{0}\) is independent of time and satisfies the equilibrium equation of part (a), and where the additional rotation about the equilibrium configuration \(\varphi(t)\) is infinitesimal. Determine \(\omega^{2}\) on the equilibrium paths, and from the dynamic criterion state the stability of the equilibrium states on each equilibrium path.

2. Determine the stability of the post-buckling path for model E given by eq. ([eq10.94]) and shown in figure [fig10.28].