Kinematics.

Consider a differential element dz-by-dy in the initial, undeformed column, where dz is along the centroidal axis and dy is perpendicular to the centroidal axis as shown in figure 1.2. In the z-y coordinate system, the material point at coordinates (z, 0) displaces to coordinates (z*, y*) in the deformed bar, where(z*, y*) is referenced to the same z-y system. These coordinates are related by \[\begin{aligned} \label{eq11.1} z^{*}=z+w(z) \quad y^{*}=0+v(z) ,\end{aligned}\] where \(w(z)\) is the displacement parallel to the \(z\)-axis and \(v(z)\) is the displacement parallel to the \(y\)-axis. The material points along length dz map to the differential length ds* along the centroidal axis in the deformed bar. By the Pythagorean theorem \(\left(d s^{*}\right)^{2}=\left(d z^{*}\right)^{2}+\left(d y^{*}\right)^{2}\). The differential lengths in the deformed bar are \[d z^{*}=\left(1+\frac{d w}{d z}\right) d z\,\textrm{and } d y^{*}=\left(\frac{d v}{d z}\right) d z. \label{eq11.2}\] Define the stretch ratio \(\lambda\) by \(\lambda=d s^{*}/dz\). Consequently, the stretch ratio is related to the derivatives of the displacements by

\[\begin{aligned} \label{eq11.3} \lambda=\sqrt{\left(1+\frac{d w}{d z}\right)^{2}+\left(\frac{d v}{d z}\right)^{2}}.\end{aligned}\] The clockwise rotation angle of element ds* with respect to the \(z\)-direction is denoted by \(\phi_{x}(z)\). Trigonometric functions of this rotation angle are given by \[\begin{aligned} \label{eq11.4} \sin \phi_{x}=\frac{-d y^{*}}{d s^{*}}\,\textrm{and } \cos\phi_{x}=\frac{d z^{*}}{d s^{*}}.\end{aligned}\] Using the chain rule of differentiation and the definition of the stretch ratio these trigonometric functions can be written as \[\begin{aligned} \label{eq11.5} \sin \phi_{x}=\left(\frac{-d y^{*}}{d z}\right) \frac{d z}{d s^{*}}=\frac{1}{\lambda}\left(\frac{-d v}{d z}\right),\,\textrm{ and similarly } \cos\phi_{x}=\frac{1}{\lambda}\left(1+\frac{d w}{d z}\right).\end{aligned}\] We impose the hypothesis of classical theory that cross sections normal to the centroidal axis in the undeformed bar remain rigid and normal to the centroidal axis in the deformed bar. Thus, the differential line element dy in the undeformed bar does not change length in the deformed bar and remains normal to the centroidal axis in the deformed bar. That is, line element dy also rotates clockwise through angle \(\phi_{x}\). The stretch ratio ([eq11.3]) is expanded in a binomial series to get \[\lambda=1+\frac{1}{2}\left[2 \frac{d w}{d z}+\left(\frac{d w}{d z}\right)^{2}+\left(\frac{d v}{d z}\right)^{2}\right]-\frac{1}{8}\left[2 \frac{d w}{d z}+\left(\frac{d w}{d z}\right)^{2}+\left(\frac{d v}{d z}\right)^{2}\right]^{2}+\ldots=1+\frac{d w}{d z}+\frac{1}{2}\left(\frac{d v}{d z}\right)^{2}+O\left(\frac{d w}{d z}\right)^{3}. \label{eq11.6}\] The engineering strain is defined as \(\varepsilon_{z z}=\lambda-1\). If cubic powers and higher of the displacement derivatives in the series expansion of \(\lambda\) are neglected, then the strain-displacement relation is \[\varepsilon_{z z}=\frac{d w}{d z}+\frac{1}{2}\left(\frac{d v}{d z}\right)^{2}. \label{eq11.7}\]

Equilibrium.

The free body diagram of the element of the bar in the deformed state is shown in figure 1.3(a). The force F acting on the cross section of the deformed bar is resolved in two sets of orthogonal components. Components H and \(S_{y}\) act in horizontal direction and the vertical direction, respectively. Components N and \(V_{y}\) act tangent and normal to the centroidal axis, respectively. Let \(\alpha\) denote the angle between the vertical and the line of action of force F. From figure 1.3(b) \(H=F\sin\alpha\) and \(S_{y}=F \cos \alpha\). Components N and \(V_{y}\) are given by

\[\begin{gathered} N=F \sin \left(\alpha-\phi_{x}\right)=(F \sin \alpha) \cos \phi_{x}-(F \cos \alpha) \sin \phi_{x}=H \cos \phi_{x}-S_{y} \sin \phi_{x} ,\,\textrm{and} \label{eq11.8}\\ V_{y}=F \cos \left(\alpha-\phi_{x}\right)=(F \cos \alpha) \cos \phi_{x}+(F \sin \alpha) \sin \phi_{x}=S_{y} \cos \phi_{x}+H \sin \phi_{x}. \label{eq11.9}\end{gathered}\] Equilibrium in the horizontal direction requires \(dH=0\) and equilibrium in the vertical direction requires \(d S_{y}=0\). Thus, the horizontal component H and the vertical component \(S_{y}\) are spatially uniform along the length of the bar. Since the applied compressive force P is also horizontal then \(H=-P\). The bending moment is denoted by \(M_{x}\). Moment equilibrium about the right end of the element in figure 1.3(a) leads to \[\begin{aligned} \label{eq11.10} M_{x}+d M_{x}-M_{x}-\left(-d y^{*}\right) H-\left(d z^{*}\right) S_{y}=0.\end{aligned}\] The differential functions are expanded in a series. For example \(d M_{x}=\frac{d M_{x}}{d z} d z+\frac{d^{2} M_{z}}{d z^{2}}\left(d z^{2}\right)+\ldots.\) Then moment equilibrium becomes \[\begin{aligned} \label{eq11.11} \left[\frac{d M_{x}}{d z}-\left(\frac{-d y^{*}}{d z}\right) H-\left(\frac{d z^{*}}{d z}\right) S_{y}\right] d z+O\left(d z^{2}\right)=0.\end{aligned}\] Division by dz followed by the limit as \(d z \rightarrow 0\) leads to the differential equation \[\frac{d M_{x}}{d z}-\left(\frac{d v}{d z}\right) p-\left(1+\frac{d w}{d z}\right) S_{y}=0. \label{eq11.12}\]

Hooke’s law.

The normal force component N is proportional to the axial strain \(\varepsilon_{zz}\) of the centroidal axis, and the bending moment \(M_{x}\) is proportional to the rotation gradient \(\frac{d \phi_{x}}{d z}\) of the cross section. That is, \[N=\textit{E A} \varepsilon_{z z} \quad M_{x}=E I\left(\frac{d \phi_{x}}{d z}\right), \label{eq11.13}\] where the modulus of elasticity is denoted by E, the cross-sectional area by A, and the second area moment of the cross section by I.

Pre-buckling equilibrium

The trivial equilibrium configuration of the column is straight and in compression subject to the applied axial force P. The lateral displacement \(v(z)=0\), rotation \(\phi_{x}(z)=0\), and \(M_{x}=0\) for \(0 \leq z \leq L\). From eq. ([eq11.8]) \({N=H=-P}\). From eq. ([eq11.9]) \(S_{y}=V_{y}\) but overall equilibrium requires \(S_{y}=0\). Let \(w_{0}(z)\) denote the axial displacement in pre-buckling equilibrium. Then, strain ([eq11.7]) and stretch ratio reduce to \(\varepsilon_{zz}=\frac{d w_{0}}{d z}\) and \(\lambda=1+\frac{d w_{0}}{d z}\), respectively. Hooke’s law ([eq11.13]) for the axial force is \(-P=E A\left(\frac{d w_{0}}{d z}\right)\). Integrate Hooke’s law and take the axial displacement \(w_{0}(0)=0\) to get \[w_{0}(z)=\frac{-P z}{E A}. \label{eq11.14}\]

Buckling

To assess buckling of a slightly defected column, we introduce a small, dimensionless parameter \(\xi\) such that all dependent variables equal there pre-buckling equilibrium expressions as \(\xi \rightarrow 0\). The displacements and rotation are expressed as \[w(z)=w_{0}(z)+\xi w_{1}(z) \quad v(z)=\xi v_{1}(z) \quad \phi_{x}(z)=\xi \phi_{1}(z). \label{eq11.15}\] The trigonometric functions of the rotation angle are \[\begin{aligned} \sin \phi_{x}=\xi \phi_{1}(z)+O\big(\xi^{3}\big) \quad \cos \phi_{x}=1+O\big(\xi^{2}\big).\label{eq11.16}\end{aligned}\] In the following developments terms of \(\xi^{2}\) and higher degrees are neglected. The vertical shear force \(S_{y}=\xi S_{y 1}\). The axial strain ([eq11.7]) expansion is \[\varepsilon_{zz}=\frac{d w_{0}}{d z}+\xi \frac{d w_{1}}{d z}. \label{eq11.17}\] The first expression in ([eq11.8]) is written as \(N+P \cos \phi_{x}+S_{y} \sin \phi_{x}=0\) , where \(N=E A \varepsilon_{z z}\). The expansion for this equation containing the force N is \[\begin{aligned} \xi^{0}\left[E A \frac{d w_{0}}{d z}+P\right]+\xi\left[E A \frac{d w_{1}}{d z}\right]+O(\xi^{2})=0.\label{eq11.18}\end{aligned}\] To satisfy the last equation for each value of \(\xi \neq 0\), the coefficients of \(\xi^{0}\) and \(\xi^{1}\) must vanish, which leads to \[\begin{aligned} \frac{d w_{0}}{d z}=\frac{-P}{E A} \quad \frac{d w_{1}}{d z}=0. \label{eq11.19}\end{aligned}\] On the pre-buckling equilibrium path the stretch ratio \(\lambda=1-P /(E A)\).

The expansion of the equilibrium equation for bending ([eq11.12]) combined with the moment from Hooke’s law ([eq11.13]) is \[\begin{aligned} \xi^{0} \cdot 0+\xi\left[E I \frac{d^{2} \phi_{1}}{d^{2}}-P \frac{d v_{1}}{d z}-\left(1-\frac{P}{E A}\right) S_{y 1}\right]+O\big(\xi^{2}\big)=0. \label{eq11.20}\end{aligned}\]

Therefore, the significant result from eq. ([eq11.20]) is \[\begin{aligned} E I \frac{d^{2} \phi_{1}}{d z^{2}}-p \frac{d v_{1}}{d z}-\left(1-\frac{P}{E A}\right) S_{y 1}=0. \label{eq11.21}\end{aligned}\]

The first expression in eq. ([eq11.5]) relating the rotation, the stretch ratio, and the displacement v(z) is written as \[\begin{aligned} [1-P /(E A)] \sin \phi_{x}+\frac{d v}{d z}=0. \label{eq11.22}\end{aligned}\] The expansion of eq. ([eq11.22]) becomes \[\begin{aligned} \xi^{0} \cdot 0+\xi\left[\frac{d v_{1}}{d z}+\left(1-\frac{P}{E A}\right) \phi_{1}\right]+O\big(\xi^{2}\big)=0. \label{eq11.23}\end{aligned}\] Therefore, \[\begin{aligned} \frac{d v_{1}}{d z}+\left(1-\frac{P}{E A}\right) \phi_{1}=0. \label{eq11.24}\end{aligned}\]

Solve eq. ([eq11.24]) for \(\phi_{1}(z)\) and substitute the result into eq. ([eq11.21]) to get \[\begin{aligned} -E I\left[\frac{d^{3} v_{1}}{dz^{3}}\right]-P\left(1-\frac{P}{E A}\right) \frac{d v_{1}}{d z}-\left(1-\frac{P}{E A}\right)^{2} S_{y 1}=0. \label{eq11.25}\end{aligned}\] Now differentiate eq. ([eq11.25]) with respect to \(z\) and note that \(\frac{d S_{y 1}}{d z}=0\) consistent with the equilibrium equation \(\frac{d S_{y}}{d z}=0\). The final result is \[\begin{aligned} \frac{d^{4} v_{1}}{d z^{4}}+K^{2} \frac{d^{2} v_{1}}{d z^{2}}=0 \quad v_{1}=v_{1}(z) \quad 0<z<L , \label{eq11.26}\end{aligned}\] where \[\begin{aligned} K^{2}=\frac{P}{E I}\left(1-\frac{P}{E A}\right). \label{eq11.27}\end{aligned}\] The expressions for the bending moment and vertical shear force are \[\begin{aligned} \left(1-\frac{P}{E A}\right) M_{x 1}=-E I\left(\frac{d^{2} v_{1}}{d z^{2}}\right) \quad\left(1-\frac{P}{E A}\right)^{2} S_{y 1}=-E I\left[\frac{d^{3} v_{1}}{dz^{3}}+K^{2} \frac{d v_{1}}{d z}\right]. \label{eq11.28}\end{aligned}\]

The solution of eq. ([eq11.26]) for \(v_{1}(z)\) is subject to boundary conditions at \(z= 0\) and \(z =L\). There are four so-called standard boundary conditions. These are shown in figure 1.4.

One solution to the differential equation ([eq11.26]) subject to boundary conditions A-D is \(v_{1}(z)=0\) for all values of the load P. This is the trivial solution. The general solution of eq. ([eq11.26]) for \(K^{2}>0\) is \[\begin{aligned} v_{1}(z)=A_{1} \sin (K z)+A_{2} \cos (K z)+A_{3} z+A_{4} , \label{eq11.29}\end{aligned}\] where \(\textit{A}_{1}\), \(\textit{A}_{2}\), \(\textit{A}_{3}\), and \(\textit{A}_{4}\) are arbitrary constants to be determined by boundary conditions.

[ex11.1] Consider the clamped-free boundary conditions denoted as (B). Determine the critical load \(P_{cr}\) for which the perfect column admits a non-trivial equilibrium state.

Buckling equations for negligible strain at the bifurcation point

Neglecting the axial strain with respect to unity means the stretch ratio \(\lambda=1-P /(E A) \approx 1\), and eqs. ([eq11.21]), ([eq11.24]), and ([eq11.27]) simplify to \[\begin{aligned} EI \frac{d^{2} \phi_{1}}{d z^{2}}-P \frac{d v_{1}}{d z}-S_{y 1}=0 \quad \frac{d v_{1}}{d z}+\phi_{1}=0 \quad K^{2}=\frac{P}{E I} \quad 0<z<L, \label{eq11.30}\end{aligned}\] From eq. ([eq11.26]) the differential equation governing buckling is \[\begin{aligned} \frac{d^{4} v_{1}}{d z^{4}}+k^{2} \frac{d^{2} v_{1}}{d z^{2}}=0 \quad v_{1}=v_{1}(z) \quad 0<z<L \quad k^{2}=P /(E I). \label{eq11.31}\end{aligned}\]

The critical loads for boundary conditions A through D and for \(\textit{EI} = \textrm{constant}\) are given in figure 1.6.

Summary of the nonlinear equations

The overall free body diagram of the column in a deflected configuration is shown in figure [fig11.7]. The shortening of the distance between support points is denoted by \(\Delta\), where \(\Delta=-w(L)\). If the column is cut at some point along its length, then equilibrium results in the vertical force component \(S_{y}=0\) for \(0 \leq z \leq L\).

The relation of the force N to load P is obtained from eq. ([eq11.8]) with \(H=-P\) as \[\begin{aligned} E A \varepsilon_{z z}+P \cos \phi=0 \label{eq11.32}\end{aligned}\]

R138pt

In eq. ([eq11.32]) force N was replaced by Hooke’s law ([eq11.13]), and we dropped the subscript \(x\) on the rotation \(\phi_{x}\) introduced in article 1.1 for convenience in the following developments. The strain \(\varepsilon_{z z}\) is related to the derivatives of the displacements by the nonlinear relation ([eq11.7]). Substitute Hooke’s law ([eq11.13]) for the bending moment into eq. ([eq11.12]) to get the differential equation for bending as \[\begin{aligned} \label{eq11.33} E I\left(\frac{d^{2} \phi}{d t^{2}}\right)-P\left(\frac{d v}{d z}\right)=0 \quad 0<z<L ,\end{aligned}\] From eq. ([eq11.5]) the trigonometric functions of the rotation are related to the displacements by \[\begin{aligned} \label{eq11.34} \lambda \sin \phi+\frac{d v}{d z}=0 \quad \lambda \cos \phi-\left(1+\frac{d w}{d z}\right)=0 ,\end{aligned}\] where the stretch ratio is \(\lambda=1+\varepsilon_{z z}\). The boundary conditions to be satisfied are \[\begin{aligned} \label{eq11.35} w(0)=0 \quad v(0)=\left.0 \quad \frac{d \phi}{d z}\right|_{z=0}=0 \quad v(L)=\left.0 \quad \frac{d \phi}{d z}\right|_{z=L}=0.\end{aligned}\]

The perturbation expansion

From the expressions in eq. ([eq11.30]) the differential equation governing buckling in terms of the rotation is \[\begin{aligned} E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}=0 \quad \phi_{1}=\phi_{1}(z) \quad 0<z<L, \label{eq11.36}\end{aligned}\] subject to the boundary conditions \[\begin{aligned} \frac{d \phi_{1}}{d z}=0 \quad \text { at } \quad z=0, L. \label{eq11.37}\end{aligned}\] The solution to this boundary value problem is \[\begin{aligned} \phi_{1}(z)=-\cos \left(\frac{\pi z}{L}\right) \quad P=P_{\textrm{cr}}=\frac{\pi^{2} E I}{L^{2}}. \label{eq11.38}\end{aligned}\] The lateral displacement is determined from the second equation in ([eq11.30]) as \[\begin{aligned} \label{eq11.39} v_{1}(z)=\frac{L}{\pi} \sin \left(\frac{\pi z}{L}\right).\end{aligned}\]

Consider the rotation and displacement in the differential equation ([eq11.33]) to be a function the dimensionless parameter \(\xi\) as well as independent variable \(z\) (i.e., \(\phi(\dot{z},\;\xi)\) and \(v(z,\;\xi)\)). An approximate solution to the nonlinear differential equation ([eq11.33]) and kinematic equation ([eq11.34]) is to be determined by perturbation expansions of the dependent variables in the parameter \(\xi\) for very small values of \(\xi\). To effect the procedure, the displacements and rotation are expanded in a series in \(\xi\) as \[\begin{aligned} \label{eq11.40} \begin{gathered} w(z)=w_{0}(z)+\xi w_{1}(z)+\xi^{2} w_{2}(z)+\xi^{3} w_{3}(z)+\ldots \\ v(z)=\xi v_{1}(z)+\xi^{2} v_{2}(z)+\xi^{3} v_{3}(z)+\ldots \\ \phi=\xi \phi_{1}(z)+\xi^{2} \phi_{2}(z)+\xi^{3} \phi_{3}(z)+\ldots \end{gathered}.\end{aligned}\] Functions \(w_{0}(z)\), \(\phi_{1}(z)\), and \(v_{1}(z)\) are given by eqs. ([eq11.14]), ([eq11.38]), and ([eq11.39]), respectively. The remaining functions in the expansion ([eq11.40]) are to be determined. The expansion of the sine and cosine functions are \[\begin{aligned} \begin{gathered} \sin \phi_{x}=\sin \left[\xi \phi_{1}(z)+\xi^{2} \phi_{2}(z)+\xi^{3} \phi_{3}(z)+\ldots\right]=\xi \phi_{1}+\xi^{2} \phi_{2}+\xi^{3}\left(\phi_{3}-\frac{\phi_{x 1}^{3}}{6}\right)+O\big(\xi^{4}\big) \\ \cos \phi=\cos (\xi \phi_{1}(z)+\xi^{2} \phi_{2}(z)+\xi^{3} \phi_{3}(z)+\ldots)=1-\frac{1}{2} \xi^{2} \phi_{1}(z)-\xi^{3} \phi_{1}(z) \phi_{2}(z)+O\big(\xi^{4}\big) \end{gathered}. \label{eq11.41}\end{aligned}\]

Relations between the expansion functions for the rotations and lateral displacement

With respect to the discussion in article 1.1.3 we take \(\lambda=1\), so the expansion of the first equation in ([eq11.34]) is \[\begin{aligned} \label{eq11.42} \xi\left[\phi_{1}+\frac{d v_{1}}{d z}\right]+\xi^{2}\left[\phi_{2}+\frac{d v_{2}}{d z}\right]+\xi^{3}\left[\phi_{3}+\frac{d v_{3}}{d z}-\frac{1}{6} \phi_{1}^{3}\right]+O\big(\xi^{4}\big)=0.\end{aligned}\] The previous series converges to zero for each sufficiently small value of \(\xi \neq 0\) requires that the coefficient of each power of \(\xi\) must vanish. Hence, \[\begin{aligned} \label{eq11.43} \frac{d v_{1}}{d z}=-\phi_{1} \quad \frac{d v_{2}}{d z}=-\phi_{2} \quad \frac{d v_{3}}{d z}=-\phi_{3}+\frac{1}{6} \phi_{1}^{3}.\end{aligned}\]

Perturbation expansion of the load P

We utilize the relations in eq. ([eq11.43]) to get the expansion of eq. ([eq11.33]) as \[\begin{aligned} \xi\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}\right)+\xi^{2}\left(E I \frac{d^{2} \phi_{2}}{d z^{2}}+P \phi_{2}\right)+\xi^{3}\left(E I \frac{d^{2} \phi_{3}}{d z^{2}}+P \phi_{3}-\frac{P}{6} \phi_{1}^{3}\right)+O\big(\xi^{4}\big)=0. \label{eq11.44}\end{aligned}\] To determine how the load P is a function of \(\xi\) multiply eq. ([eq11.44]) by \(\phi_{1}\) and integrate with respect to \(z\) from \(z=0\) to \(z=L\): \[\begin{aligned} &\xi\left[\int_{0}^{L}\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}\right) \phi_{1} d z\right]+\xi^{2}\left[\int_{0}^{L}\left(E I \frac{d^{2} \phi_{2}}{d z^{2}}+P \phi_{2}\right) \phi_{1} d z\right]\nonumber\\ &\quad+\xi^{3}\left[\int_{0}^{L}\left(E I \frac{d^{2} \phi_{3}}{d z^{2}}+P \phi_{3}-\frac{P}{6} \phi_{1}^{3}\right) \phi_{1} d z\right] +O\big(\xi^{4}\big)=0. \label{eq11.45}\end{aligned}\] Integrate twice by parts with respect to \(z\) of the \(O\big(\xi^{2}\big)\) term in eq. ([eq11.45]) to get \[\begin{aligned} \int_{0}^{L} E I \frac{d^{2} \phi_{2}}{dz^{2}} \phi_{1} d z=\int_{0}^{L} E I \frac{d^{2} \phi_{1}}{dz^{2}} \phi_{2} d z+\underbrace{\left.\left[E I \frac{d \phi_{2}}{d z} \phi_{1}-E I \frac{d \phi_{1}}{d z} \phi_{2}\right]\right|_{0} ^{L}}_{=\;0}=\int_{0}^{L} E I \frac{d^{2} \phi_{1}}{dz^{2}} \phi_{2} d z. \label{eq11.46}\end{aligned}\] The boundary terms in the expansion functions vanish consistent with the conditions in eq. ([eq11.35]). Also, integrate twice by parts with respect to \(z\) of the \(O\big(\xi^{3}\big)\) term. After integrating by parts, eq. ([eq11.45]) is \[\begin{aligned} &\xi\left[\int_{0}^{L}\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}\right) \phi_{1} d z\right]+\xi^{2} \int_{0}^{L}\left[\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}\right) \phi_{2} d z\right]\nonumber\\ &\quad+\xi^{3}\left[\int_{0}^{L}\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P \phi_{1}\right) \phi_{3}-\frac{P}{6} \phi_{1}^{4} d z\right] +O\big(\xi^{4}\big)=0. \label{eq11.47}\end{aligned}\] From the differential equation ([eq11.36]) at buckling we have \[\begin{aligned} E I \frac{d^{2} \phi_{1}}{d z^{2}}=-P_{\textrm{cr}} \phi_{1}. \label{eq11.48}\end{aligned}\] Substitute eq. ([eq11.48]) into eq. ([eq11.47]) to find \[\begin{aligned} \xi\left(P-P_{c r}\right)\left(\int_{0}^{L} \phi_{1}^{2} d z\right)+\xi^{2}\left(P-P_{\textrm{cr}}\right) \int_{0}^{L} \phi_{1} \phi_{2} d z+\xi^{3}\left[\left(P-P_{\textrm{cr}}\right) \int_{0}^{L} \phi_{1} \phi_{3} d z-\frac{P}{6} \int_{0}^{L} \phi_{1}^{4} d z\right]+O\big(\xi^{4}\big)=0. \label{eq11.49}\end{aligned}\] The last step is to impose the orthogonality conditions on the expansion functions, which are \[\begin{aligned} \label{eq11.50} \int_{0}^{L} \phi_{1} \phi_{2} d z=0 \quad \int_{0}^{L} \phi_{1} \phi_{3} d z=0.\end{aligned}\] Equation ([eq11.49]) reduces to \[\begin{aligned} \label{eq11.51} \xi\left(P-P_{\mathrm{cr}}\right)\left(\int_{0}^{L} \phi_{1}^{2} d z\right)-\xi^{3} \frac{P}{6} \int_{0}^{L} \phi_{1}^{4} d z+O\big(\xi^{4}\big)=0.\end{aligned}\] Divide eq. ([eq11.51]) by \(\xi P_{\mathrm{cr}}\left(\int_{0}^{L} \phi_{1}^{2} d z\right)\) and rearrange terms to the form \(\frac{P}{P_{\mathrm{cr}}}=\frac{1+O\big(\xi^{3}\big)}{\left(1-\xi^{2} b\right)}\),

where \[\begin{aligned} \label{eq11.52} b=\frac{1}{6}\left[\left(\int_{0}^{L} \phi_{1}^{4} d z\right) /\left(\int_{0}^{L} \phi_{1}^{2} d z\right)\right]=\frac{1}{8}.\end{aligned}\] The series of \(\left(1-\xi^{2} b\right)^{-1}=1+b \xi^{2}+O\big(\xi^{4}\big)\), which leads to \[\begin{aligned} \label{eq11.53} \frac{P}{P_{\mathrm{cr}}}=1+b \xi^{2}+O\big(\xi^{3}\big).\end{aligned}\] In general, the expansion of \(P/P_{\mathrm{cr}}\) is written as \[\begin{aligned} \label{eq11.54} \frac{P}{P_{\mathrm{cr}}}=1+a \xi+b \xi^{2}+\ldots.\end{aligned}\]

Solutions for the rotation and lateral displacement functions

Substitute the expansion of load \(P\) from eq. ([eq11.54]) into eq. ([eq11.44]), and arrange terms in powers of \(\xi\) to get \[\begin{aligned} &\xi\left[E I \frac{d^{2} \phi_{1}}{d z^{2}}+P_{\mathrm{cr}} \phi_{1}\right]+\xi^{2}\left[E I \frac{d^{2} \phi_{2}}{d z^{2}}+P_{\mathrm{cr}} \phi_{2}+a P_{\mathrm{cr}} \phi_{1}\right] \nonumber \\ &\quad +\xi^{3}\left[E I \frac{d^{2} \phi_{3}}{d z^{2}}+P_{\mathrm{cr}} \phi_{3}+a P_{\mathrm{cr}} \phi_{2}+b P_{\mathrm{cr}} \phi_{1}-\frac{1}{6} P_{\mathrm{cr}} \phi_{1}^{3}\right]+\ldots=0. \label{eq11.55}\end{aligned}\] For the series ([eq11.55]) to converge to zero for each sufficiently small value of \(\xi\), the coefficient of each power of \(\xi\) must vanish. The coefficient of \(\xi\) is the differential equation for buckling, eqs. ([eq11.36]) and ([eq11.38]). The coefficients of \(\xi^{2}\) and \(\xi^3\) lead to differential equations \[\begin{aligned} \label{eq11.56} E I \frac{d^{2} \phi_{k}}{d z^{2}}+P_{\mathrm{cr}} \phi_{k}=F_{k} \quad \phi_{k}=\phi_{k}(z) \quad 0<z<L \quad k=2,3,\end{aligned}\] subject to boundary conditions \[\begin{aligned} \label{eq11.57} \left.\frac{d\phi_{k}}{dz}\right|_{z=0}=\left.0 \quad \frac{d \phi_{k}}{d z}\right|_{z=L}=0.\end{aligned}\] The non-homogeneous terms in eq. ([eq11.56]) depend on previous solutions of the expansion functions. That is, \[\begin{aligned} \label{eq11.58} F_{2}=-a P_{\mathrm{cr}} \phi_{1} \quad F_{3}=-a P_{\mathrm{cr}} \phi_{2}-b P_{\mathrm{cr}} \phi_{1}+\frac{1}{6} P_{\mathrm{cr}} \phi_{1}^{3}.\end{aligned}\] Let \(k_{\textrm{cr}}=P_{\textrm{cr}} /(E I)\). The general solution to differential equation ([eq11.56]) consists of a complementary function \(c_{1 k} \cos \left(k_{\textrm{cr}} z\right)+c_{2 k} \sin \left(k_{\textrm{cr}} z\right)\) that satisfies the homogeneous equation plus a particular solution denoted by \(\phi_{k}^{*}(z)\) that satisfies the non-homogeneous equation. Then, the general solution is \[\begin{aligned} \label{eq11.59} \phi_{k}(z)=c_{1 k} \cos \left(k_{c r} z\right)+c_{2 k} \sin \left(k_{c r} z\right)+\phi_{k}^{*}(z).\end{aligned}\] Consider conditions required to solve the boundary value problem presented by eqs. ([eq11.56]) and ([eq11.57]). Multiply eq. ([eq11.56]) by \(\phi_{1}(z)\) and integrate the result from \(z=0\) to \(z=L\). The result is \[\begin{aligned} \label{eq11.60} \int_{0}^{L}\left(E I \frac{d^{2} \phi_{k}}{d z^{2}} \phi_{1}+P_{\mathrm{cr}} \phi_{k} \phi_{1}\right) d z=\int_{0}^{L} F_{k} \phi_{1} d z.\end{aligned}\] Integrate the first term on the left-hand side of eq. ([eq11.60]) by parts twice to get \[\begin{aligned} \label{eq11.61} \left.\left[\phi_{1} E I \frac{d \phi_{k}}{d z}-\frac{d \phi_{1}}{d z} E I \phi_{k}\right]\right|_{z=L}-\left.\left[\phi_{1} E I \frac{d \phi_{k}}{d z}-\frac{d \phi_{1}}{d z} E I \phi_{k}\right]\right|_{z=0}+\int_{0}^{L}\left(E I \frac{d^{2} \phi_{1}}{d z^{2}}+P_{\mathrm{cr}} \phi_{1}\right) \phi_{k} d z=\int_{0}^{L} F_{k} \phi_{1} d z.\end{aligned}\] Boundary conditions ([eq11.37]) and ([eq11.57]) result in the terms on the left-hand side of eq. ([eq11.61]) evaluated at the end points of the interval equal to zero. Also, the integrand on the left-hand side vanishes since rotation \(\phi_{1}\) satisfies eq. ([eq11.36]). We are left with the condition for the solution of the boundary value problem for \(\phi_{k}\) that \[\begin{aligned} \label{eq11.62} \int_{0}^{L} F_{k} \phi_{1} d z=0.\end{aligned}\] For \(k=2\) condition ([eq11.62]) is \[\begin{aligned} \label{eq11.63} \int_{0}^{L} F_{2} \phi_{1} d z=-a P_{\mathrm{cr}} \int_{0}^{L} \phi_{1}^{2} d z=-a P_{\mathrm{cr}}\left(\frac{L}{2}\right)=0.\end{aligned}\] The only way to satisfy the condition in eq. ([eq11.63]) is to take the post-buckling coefficient \(a=0\). The solution for \(\phi_{2}\) that satisfies the boundary conditions ([eq11.57]) is \(\phi_{2}(z)=c_{12} \cos \left(k_{c r} z\right)\). The orthogonality condition ([eq11.50]) determines \(c_{12}=0\). Thus, \(\phi_{2}(z)=0\). From eq. ([eq11.43]) and boundary condition ([eq11.35]) we find \(v_{2}(z)=0\) for \({0 \leq z \leq L}\). For \(k=3\), condition ([eq11.62]) is \[\begin{aligned} \label{eq11.64} \int_{0}^{L} F_{3} \phi_{1} d z=\int_{0}^{L}\left(-b P_{\mathrm{cr}} \phi_{1}+\frac{1}{6} P_{\mathrm{cr}} \phi_{1}^{3}\right) \phi_{1} d z=0.\end{aligned}\] Equation ([eq11.64]) determines post-buckling coefficient \(b\), and it is the same as given in eq. ([eq11.52]).

From eq. ([eq11.56]) the governing equation for \(\phi_{3}(z)\) with \(a=0\) and \(b=1/8\) is \[\begin{aligned} \label{eq11.65} E I \frac{d^{2} \phi_{3}}{d z^{2}}+P_{\mathrm{cr}} \phi_{3}=-\left(\frac{1}{8}\right) P_{\mathrm{cr}} \phi_{1}+\frac{1}{6} P_{\mathrm{cr}} \phi_{1}^{3}=P_{\mathrm{cr}}\left[\left(\frac{1}{8}\right) \cos \left(\frac{\pi z}{L}\right)-\left(\frac{1}{6}\right) \cos ^{3}\left(\frac{\pi z}{L}\right)\right].\end{aligned}\] Use the trigonometric identity \(\cos ^{3} x=(3/4) \cos x+(1/4) \cos (3 x)\) to find \[\begin{aligned} \label{eq11.66} E I \frac{d^{2} \phi_{3}}{d z^{2}}+P_{\textrm{cr}} \phi_{3}=\frac{-P_{\mathrm{cr}}}{24} \cos \left(\frac{3 \pi z}{L}\right).\end{aligned}\] The solution to differential equation ([eq11.66]) is \[\begin{aligned} \label{eq11.67} \phi_{3}(z)=c_{13} \cos \left(k_{\textrm{cr}} z\right)+c_{23} \sin \left(k_{\textrm{cr}} z\right)+\left(\frac{1}{192}\right) \cos \left(\frac{3 \pi z}{L}\right).\end{aligned}\] Boundary conditions ([eq11.57]) lead to coefficient \(c_{23}=0\). Coefficient \(c_{13}\) is determined from the orthogonality condition ([eq11.50]) \[\begin{aligned} \label{eq11.68} \int_{0}^{L} \phi_{3} \phi_{1} d z=\frac{-c_{13} L}{2}=0.\end{aligned}\] Therefore \(c_{13}=0\) and the solution for function \(\phi_{3}(z)\) is \[\begin{aligned} \label{eq11.69} \phi_{3}(z)=\left(\frac{1}{192}\right) \cos \left(\frac{3 \pi z}{L}\right).\end{aligned}\] The function \(v_{3}(z)\) can now be determined from eq. ([eq11.43]). The result that satisfies boundary conditions ([eq11.35]) is \[\begin{aligned} \label{eq11.70} v_{3}(z)=\left(\frac{-L}{8 \pi}\right) \sin \left(\frac{\pi z}{L}\right)-\left(\frac{L}{64 \pi}\right) \sin \left(\frac{3 \pi z}{L}\right).\end{aligned}\]

Solutions for the axial displacement functions

We make use of eq. ([eq11.43]) to find that the expansion of the strain \(\varepsilon_{z z}\) in ([eq11.7]) is \[\begin{aligned} \label{eq11.71} \varepsilon_{z z}=\frac{d w_{0}}{d z}+\xi \frac{d w_{1}}{d z}+\xi^{2}\left[\frac{d w_{2}}{d z}+\frac{1}{2} \phi_{1}^{2}\right]+\xi^{3}\left[\frac{d w_{3}}{d z}+\phi_{1} \phi_{2}\right]+O\big(\xi^{4}\big).\end{aligned}\] Substitute expansions of the strain from eq. ([eq11.71]), the load P from eq. ([eq11.54]), and the cosine function from eq. ([eq11.41]), into eq. ([eq11.32]) to find the expansion of the axial equilibrium as \[\begin{aligned} \label{eq11.72} \xi^{0}\left[E A \frac{d w_{0}}{d z}+P_{\mathrm{cr}}\right]+\xi\left[E A \frac{d w_{1}}{d z}+a P_{\mathrm{cr}}\right]+\xi^{2}\left[E A \frac{d w_{2}}{d z}+b P_{\mathrm{cr}}-P_{\mathrm{cr}} \phi_{1}^{2}+E A \frac{1}{2} \phi_{1}^{2}\right]+O\big(\xi^{3}\big)=0.\end{aligned}\] For eq. ([eq11.72]) to converge to zero for each sufficiently small value of \(\xi\), the coefficient of each power of \(\xi\) must vanish. Thus, we get to expressions for the derivatives of the expansion functions of displacement \(w(z)\) as \[\begin{aligned} \label{eq11.73} \frac{d w_{0}}{d z}=\frac{-P_{\mathrm{cr}}}{E A} \quad \frac{d w_{1}}{d z}=\frac{-a P_{\mathrm{cr}}}{E A} \quad \frac{d w_{2}}{d z}=\frac{-2 b P_{\mathrm{cr}}+\left(2 P_{\mathrm{cr}}-E A\right) \phi_{1}^{2}}{2 E A}.\end{aligned}\] Since \(a=0\) and \(w_{1}(0)=0\), the displacement function \(w_{1}(z)=0\) for \(0 \leq z \leq L\). The expression for the derivative of displacement function \(w_{2}(z)\) is integrated with respect to \(z\), and we set \(w_{2}(0)=0\) to find \[\begin{aligned} \label{eq11.74} w_{2}(z)=-\frac{z}{4}+\frac{7 P_{\mathrm{cr}}}{8 E A} z-\frac{L}{8 \pi} \sin \left(\frac{2 \pi z}{L}\right).\end{aligned}\]

Summary

From this initial post-buckling analysis the results for the expansions of the load, displacements, and rotation are \[\begin{aligned} \label{eq11.75} \frac{P}{P_{\mathrm{cr}}}=1+\frac{\xi^{2}}{8} \quad w(z)=-\frac{P_{\mathrm{cr}}}{E A} z+\xi^{2} w_{2}(z) \quad v(z)=\xi v_{1}(z)+\xi^{3} v_{3}(z) \quad \phi(z)=\xi \phi_{1}(z)+\xi^{3} \phi_{3}(z).\end{aligned}\] The strain of the centroidal axis is \[\begin{aligned} \label{eq11.76} \varepsilon_{z z}=-\frac{P_{\mathrm{cr}}}{E A}+\xi^{2}\left[\frac{d w_{2}}{d z}+\frac{1}{2}\left(\frac{d v_{1}}{d z}\right)^{2}\right].\end{aligned}\] Rotation functions \(\phi_{1}(z)\) and \(\phi_{3}(z)\) are given by eqs. ([eq11.38]) and ([eq11.69]), respectively. Lateral displacement functions \(v_{1}(z)\) and \(v_{3}(z)\) are given by eqs. ([eq11.39]) and ([eq11.70]), respectively, and the axial displacement function \(w_{2}(z)\) is given by eq. ([eq11.74]).

[ex11.2] Consider the column with a solid, rectangular cross section of height \(h\) and width \(b\), where \(h<b\). The radius of gyration is \(r=\sqrt{I/A}=h/\sqrt{12}\). The strain at the bifurcation point is obtained from eq. ([eq11.76]) for \(\xi \rightarrow 0\) is \(-P_{\mathrm{cr}} /(E A)\), and take this strain equal to \(-0.0006\). Since \(P_{\mathrm{cr}}=\left(\pi^{2} E I\right)/ L\), we have \[\begin{aligned} \frac{P_{\mathrm{cr}}}{E A}=\frac{\pi^{2} E I}{L^{2}}\left(\frac{1}{E A}\right)=\frac{\pi^{2}}{12}\left(\frac{h}{L}\right)^{2}=0.0006. \label{eq11.2.a}\tag{a}\end{aligned}\] Hence, the span-to-thickness ratio \(L/h=37\).

The restriction on the magnitude of the expansion parameter \(\xi\) in the initial post-buckling analysis is based on the strain at the elastic limit of 7075-T6 aluminum alloy, which is about \(0.0068\). Let \(\bar{\varepsilon}_{z z}\) denote the strain of a line element parallel to the centroidal axis. It is the sum of the strain of centroidal axis \(\varepsilon_{z z}\) plus the strain due to bending. That is, \[\begin{aligned} \bar{\varepsilon}_{z z}=\varepsilon_{z z}+y\left(\frac{d \phi}{d z}\right), \label{eq11.2.b}\tag{b}\end{aligned}\] where \((-h)/2 \leq y \leq h/2\) and \(d \phi/d z\) is the curvature of the centroidal axis. The magnitude of the maximum compressive strain in post-buckling occurs at midspan, \(z=L/2\), and \(y=-h/2\). That is, \[\begin{aligned} \bar{\varepsilon}_{z z}=\varepsilon_{z z}-\left.\frac{h}{2}\left(\frac{d \phi}{d z}\right)\right|_{z=L/2}. \label{eq11.2.c}\tag{c}\end{aligned}\] The expansion for the curvature at midspan is determined from eqs. ([eq11.75]), ([eq11.38]) and ([eq11.69]). The result is \[\begin{aligned} \left.\frac{d \phi}{d z}\right|_{z=L/2}=\left[\frac{\pi}{L} \sin \left(\frac{\pi z}{L}\right)\right] \xi-\left.\left[\frac{\pi}{64 L} \sin \left(\frac{3 \pi z}{L}\right)\right] \xi^{3}\right|_{z=L/2}=\frac{\pi}{L} \xi+\frac{\pi}{64 L} \xi^{3}. \label{eq11.2.d}\tag{d}\end{aligned}\] Substitute \(h=0.270L\) and \(P_{\mathrm{cr}}=0.0006 \mathrm{EA}\) in the expansions for the strains. The numerical evaluations of the strain in eq. ([eq11.76]) and the strain from bending are \[\begin{aligned} \varepsilon_{z z}=-0.0006-0.000075 \xi^{2} \quad-\frac{h}{2}\left(\frac{d \phi}{d z}\right)=-0.0424264 \xi-0.000662193 \xi^{3}. \label{eq11.2.e}\tag{e}\end{aligned}\] The axial strain on the concave side of the bar at midspan is set equal to the elastic limit strain of \(-0.0068\). Thus, \[\begin{aligned} \bar{\varepsilon}_{z z}=-0.0006-0.042464 \xi-0.000075 \xi^{2}-0.000662913 \xi^{3}=-0.0068.\end{aligned}\] The real root of the previous polynomial is the maximum value of parameter \(\xi\), which is \[\begin{aligned} \xi_{\max }=0.14605. \label{eq11.2.f}\tag{f}\end{aligned}\] For post-buckling coefficients \(a=0\) and \(b=1/8\), we get \(P/P_{\mathrm{cr}}=1.0027\) at \(\xi=\xi_{\max}\) from eq. ([eq11.54]). There is a very small increase in the load during post-buckling. The lateral displacement of the column is determined from eqs. ([eq11.75]), ([eq11.39]), and ([eq11.70]), and it is a maximum at midspan. Evaluation of the maximum displacement is \[\begin{aligned} v_{\max }=v(L/2)=\frac{L \xi}{\pi}-\frac{7 L \xi^{2}}{64 \pi}, \label{eq11.2.g}\tag{g}\end{aligned}\] and \(v_{\max }=0.04638\,{L}\) at \(\xi=\xi_{\max }\).

The axial displacement of the column is determined from eqs. ([eq11.75]) and ([eq11.74]). The shortening of distance between supports is \[\begin{aligned} \Delta=-w(L)=L(0.0006)+\left(\frac{L}{4}+\frac{L(0.0006)}{8}\right) \xi^{2}. \label{eq11.2.h}\tag{h}\end{aligned}\] The shortening at buckling is \(\Delta_{\mathrm{cr}}=\left.\Delta\right|_{\xi=0}=L(0.0006)\), and the normalized shortening is defined by \[\begin{aligned} \Delta/\Delta_{\mathrm{cr}}=1+416.792 \xi^{2}. \label{eq11.2.i}\tag{i}\end{aligned}\] At \(\xi=\xi_{\max }\), \(\Delta/\Delta_{\mathrm{cr}}=9.89\) on the post-buckling path. The configuration of the column at \(\xi_{\max }\) is shown in figure [fig11.8].

The pre-buckling equilibrium path is determined from eq. ([eq11.14]) where \(\Delta=-w_{0}(L)=(P L) /(E A)\), or \({P=(E A/ L) \Delta}\). Divide by the critical load to get \(P/P_{\mathrm{cr}}=\left(E A/P_{\mathrm{cr}}\right)(\Delta/L)\). From eq. ([eq11.2.a]) the factor \(E A/P_{\mathrm{cr}}=1/0.0006\). Thus, \(P/P_{\mathrm{cr}}=\Delta /(0.0006 L)=\Delta/ \Delta_{\mathrm{cr}}\) on the pre-buckling equilibrium path.

The load-deflection response is shown in figure 1.7(a), and the load-shortening response is shown in figure 1.7(b). The post-buckling behavior of the column is stable symmetric bifurcation, which is the same behavior as model A in article [sec10.1] on page . The load does not decrease in post-buckling. However, the increase in load is very small in post-buckling. From a practical point of view, the column is considered neutral in post-buckling.The structural stiffness is defined as \(d P/d \Delta\). For post-buckling the structural stiffness is computed as \[\begin{aligned} \frac{d P}{d \Delta}=\frac{d P}{d \xi} \frac{d \xi}{d \Delta}=\left(\frac{P_{\mathrm{cr}}}{4} \xi\right) \frac{1}{2(L/ 4+0.0006(L/8)) \xi}=\frac{P_{\mathrm{cr}}}{L(2+0.0006)}=\frac{E A(0.0006)}{L(2.0006)}=\frac{E A}{L}(0.0003).\end{aligned}\] The structural stiffness in pre-buckling is \((E A)/L\). The ratio of the post-buckling stiffness to the pre-buckling stiffness is 0.0003, which indicates the dramatic loss of structural stiffness due to buckling.

From eq. ([eq11.54]) the perturbation expansion of the load in initial post-buckling is \(P/P_{\mathrm{cr}}=1+a \xi+b \xi^{2}\). The post-buckling coefficients \(a=0\) and \(b<0\) correspond to unstable symmetric bifurcation behavior illustrated by model B in article [sec10.2] on page . Post-buckling coefficient \(a \neq 0\) corresponds to asymmetric bifurcation behavior illustrated by model C in article [sec10.3] on page .

Axial strain-displacement relation.

The strain-displacement relation ([eq11.7]) for each bar is \[\begin{aligned} \varepsilon_{z z}=\frac{d w}{d z}+\frac{1}{2}\left(\frac{d v}{d z}\right)^{2}, \label{eq11.3.a}\tag{a}\end{aligned}\]

L122pt

where the axial displacement is denoted by \(w(z)\) and the lateral displacement is denoted by \(v(z)\). Consider the bar on the left-hand side of the truss as shown in figure [fig11.11]. At the fixed end where \(z= 0\), \(w(0)=v(0)=0\). At the end of the bar where \(z=L\) the axial displacement and the lateral displacement are related to the downward displacement \(q\) of the movable joint by \(w(L)=-q \sin \beta\) and \(v(L)=-q \cos \beta\), respectively. The axial strain in a truss bar is uniform along its length, which means that the displacements are linear in coordinate \(z\). Linear displacement functions for each displacement satisfying the end conditions are, \[\begin{aligned} w(z)=(-q \sin \beta)(z/L) \quad v(z)=(-q \cos \beta)(z/L). \label{eq11.3.b}\tag{b}\end{aligned}\] Substitute eq. ([eq11.3.b]) for the displacement functions into eq. ([eq11.3.a]) to get the strain-displacement relation \[\begin{aligned} \varepsilon_{z z}=(\sin \beta)\left(\frac{-q}{L}\right)+\frac{1}{2}\left(\cos ^{2} \beta\right)\left(\frac{q}{L}\right)^{2}. \label{eq11.3.c}\tag{c}\end{aligned}\] Substitute \(\sin \beta=H/L\), and \(\cos \beta=\big(\sqrt{L^{2}-H^{2}}\big)/L\) into eq. ([eq11.3.c]) to get \[\begin{aligned} \varepsilon_{z z}=\frac{H}{L}\left(\frac{-q}{L}\right)+\frac{\left(L^{2}-H^{2}\right)}{2 L^{2}}\left(\frac{-q}{L}\right)^{2}. \label{eq11.3.d}\tag{d}\end{aligned}\] Numerical evaluation of eq. ([eq11.3.d]) is \[\begin{aligned} \varepsilon_{z z}=\left(-3 \times 10^{-4}/\mathrm{mm}\right) q+\left(5.51056 \times 10^{6}/\mathrm{mm}^{2}\right) q^{2}. \label{eq11.3.e}\tag{e}\end{aligned}\] The strain energy of the truss is \[\begin{aligned} U=2\left(\frac{1}{2} E A L \varepsilon_{z z}^{2}\right), \label{eq11.3.f}\tag{f}\end{aligned}\] in which the leading factor of 2 accounts for the two bars. Castigliano’s first theorem determines the force Q by \[\begin{aligned} Q=\frac{\partial U}{\partial q}=2 E A L \varepsilon_{z z} \frac{\partial \varepsilon_{z z}}{\partial q}. \label{eq11.3.g}\tag{g}\end{aligned}\] Substitute eq. ([eq11.3.d]) for the strain into eq. ([eq11.3.f]) to get \[\begin{aligned} Q=2 E A\left[\frac{H^{2}}{L^{3}} q-\frac{3 H\left(L^{2}-H^{2}\right)}{2 L^{5}} q^{2}+\frac{\left(L^{2}-H^{2}\right)^{2}}{2 L^{7}} q^{3}\right]. \label{eq11.3.h}\tag{h}\end{aligned}\] Numerical evaluation of eq. ([eq11.3.h]) is \[\begin{aligned} Q=(1{,}725.3\,\mathrm{N}/\mathrm{mm}) q-(95.0736\,\mathrm{N}/\mathrm{mm}^{2}) q^{2}+(1.16424\,\mathrm{N}/ \mathrm{mm}^{3}) q^{3}. \label{eq11.3.i}\tag{i}\end{aligned}\]

In-plane buckling of the truss bars based on linear analysis.

The expressions for the axial strain ([eq11.3.e]) and applied load ([eq11.3.h]) reduce to \[\begin{aligned} \varepsilon_{z z}=\frac{H}{L}\left(\frac{-q}{L}\right)=(-3 \times 10^{-4}\,\mathrm{mm}^{-1}) q,\,\textrm{and } Q=2 E A\left[\frac{H^{2}}{L^{3}} q\right]=(1{,}725.3\,\mathrm{N}/\mathrm{mm}) q. \label{eq11.3.j}\tag{j}\end{aligned}\] The axial force in each bar is given by \[\begin{aligned} N=E A \varepsilon_{z z}=(-9{,}585\,\mathrm{N}/\mathrm{mm}) q. \label{eq11.3.k}\tag{k}\end{aligned}\] The Euler buckling force \(P_{E}=\left(\pi^{2} E I\right)/L^{2}=94.60\,\textrm{kN}\). Set \(-N=P_{E}\) to find the displacement for in-plane buckling of the truss bars \(q=9.87\,\mathrm{mm}\). The corresponding load \(Q=17.028\,\textrm{kN}\).

Equations ([eq11.3.i]) and ([eq11.3.j]) are plotted on the graph of load Q versus displacement \(q\) in figure 1.9. As the load is increased from zero on the nonlinear path ([eq11.3.i]) a limit point load of 9.038 kN at a displacement of 11.5 mm is encountered. As discussed in article [sec10.5] a dynamic snap-through motion occurs at the limit load that eventually (with damping) settles to a displacement of 58.65 mm. The linear response path ([eq11.3.j]) is the straight line in figure 1.9, and the load causing in-plane buckling of the truss bars is 17.028 kN. Thus, the critical load for this structure is at the limit point.

Southwell plot

Rearrange eq. ([eq11.88]) as follows: \(\delta_{c}\left(1-P/P_{\textrm{cr}}\right)=\left(P/P_{\textrm{cr}}\right) a_{1}\), then \(\delta_{c}-\left(P/P_{\textrm{cr}}\right) \delta_{c}=\left(P/P_{\textrm{cr}}\right) a_{1}\). Divide the last by P to get \[\begin{aligned} \frac{\delta_{c}}{P}=\frac{\left(\delta_{c}+a_{1}\right)}{P_{\textrm{cr}}}. \label{eq11.89}\end{aligned}\] We plot \(\delta_{c}/P\) versus \(\delta_{c}\) from eq. ([eq11.89]) in figure [fig11.15], which is called the Southwell plot (Southwell, 1932).¹ The Southwell plot is very useful for determining \(P_{\textrm{cr}}\) from test data in the elastic range. As \(P \rightarrow P_{\textrm{cr}}\), \(P<P_{\textrm{cr}}\), \(\delta\) becomes large and the data (\((\delta/P) \text { vs. } \delta\)) tends to plot on a straight line. Extrapolating this straight line back to toward the ordinate axis (\(\delta/P\)) one can estimate \(a_1\) and \(P_{\textrm{cr}}\). It is more difficult to determine \(P_{\textrm{cr}}\) by the load-deflection curve obtained in experiments as illustrated in figure [fig11.16].

Inelastic buckling

The column curve equation, eq. ([eq11.93]), is valid up to the proportional limit of the material, denoted by \(\sigma_{p}\). The proportional limit is defined as the stress where the compressive stress-strain curve of the material deviates from a straight line. If the stress at the onset of buckling is greater than the proportional limit, then the column is said to be of intermediate length, and the Euler formula, eq. ([eq11.93]), cannot be used. The proportional limit is difficult to measure from test data because its definition is based on the deviation from linearity. In particular, the compressive stress-strain curves for aluminum alloys typically used in aircraft construction do not exhibit a very pronounced linear range. For aluminum alloys a material law developed by Ramberg and Osgood (1943) is often used to describe the nonlinear compressive stress-strain curve. The Ramberg-Osgood equation is a three-parameter fit to the compressive stress-strain curves of aluminum alloys. From the experimental compressive stress-strain curve the slope near the origin is the modulus of elasticity E. The stress where the secant line drawn from the origin with slope 0.85 E intersects the stress-strain curve is denoted \(\sigma_{0.85}\).The stress where a second secant line drawn from the origin with slope 0.7 E intersects the stress-strain curve is denoted by \(\sigma_{0.7}\). These data are depicted in figure [fig11.19]. Note that the compressive normal strain corresponding to the stress \(\sigma_{0.7}\) is usually about the 0.2 percent offset yield strain for the material. Hence, stress \(\sigma_{0.7}\) is close to the 0.2 percent offset yield stress of the aluminum alloy. The Ramberg-Osgood equation is

\[\begin{aligned} \label{eq11.94} \varepsilon=\frac{\sigma}{E}\left[1+\frac{3}{7}\left(\frac{\sigma}{\sigma_{0.7}}\right)^{n-1}\right],\end{aligned}\] where the shape parameter \(n\) is given by \[\begin{aligned} \label{eq11.95} n=1+\ln \left(\frac{17}{7}\right)/\ln \left(\frac{\sigma_{0.7}}{\sigma_{0.85}}\right).\end{aligned}\] Equation ([eq11.94]) is re-written as \[\begin{aligned} \label{eq11.96} \frac{E \varepsilon}{\sigma_{0.7}}=\frac{\sigma}{\sigma_{0.7}}+\frac{3}{7}\left(\frac{\sigma}{\sigma_{0.7}}\right)^{n},\end{aligned}\]

and is plotted as \(\sigma/\sigma_{0.7}\) versus \((E \varepsilon)/\sigma_{0.7}\) for various values of the shape parameter \(n\), This plot is shown in figure 1.10. Some approximate values for common aluminum alloys are listed in table [tab11.3].

From the Ramberg-Osgood equation, eq. ([eq11.94]), the local slope of the compressive stress-strain curve is determined as a function of the stress. This slope of the compressive stress-strain curve is called the tangent modulus (i.e., \(\frac{d \sigma}{d \varepsilon}=E_{t}\) where \(E_t\) is the tangent modulus). Differentiate eq. ([eq11.94]) to get \[\begin{aligned} \label{eq11.97} d \varepsilon=\frac{d \sigma}{E}+\frac{3}{7} \frac{n \sigma^{n-1}}{E \sigma_{0.7}^{n}-1} d \sigma \quad \frac{d \varepsilon}{d \sigma}=\frac{1}{E_{t}}=\frac{1}{E}+\frac{3}{7} \frac{n}{E}\left(\frac{\sigma}{\sigma_{0.7}}\right)^{n-1}.\end{aligned}\] Thus, the tangent modulus is \[\begin{aligned} \label{eq11.98} E_{t}=\frac{E}{1+\frac{3}{7} n\left(\frac{\sigma}{\sigma_{0.7}}\right)^{n-1}}.\end{aligned}\]

For intermediate length columns it has been demonstrated by extensive testing that the critical stress is reasonably well predicted using the Euler curve, eq. ([eq11.93]), with the modulus of elasticity replaced by the tangent modulus. This inelastic buckling analysis is called the tangent modulus theory. That is, \[\begin{aligned} \label{eq11.99} \sigma_{\textrm{cr}}=\pi^{2} \frac{E_{t}}{\left(\frac{K L}{r}\right)^{2}}.\end{aligned}\] Now substitute eq. ([eq11.98]) for the tangent modulus in the latter equation, noting that \(\sigma=\sigma_{\textrm{cr}}\), to get \[\begin{aligned} \label{eq11.100} \sigma_{\textrm{cr}}=\frac{\pi^{2}}{\left(\frac{K L}{r}\right)^{2}}\left[\frac{E}{1+\frac{3}{7} n\left(\frac{\sigma_{\textrm{cr}}}{\sigma_{0.7}}\right)^{n-1}}\right].\end{aligned}\] After division by \(\sigma_{0.7}\), eq. ([eq11.100]) can be written as \[\begin{aligned} \label{eq11.101} \frac{\sigma_{\textrm{cr}}}{\sigma_{0.7}}+\frac{3}{7} n\left(\frac{\sigma_{\textrm{cr}}}{\sigma_{0.7}}\right)^{n}=\frac{1}{\left[\frac{(K L/r)}{\pi \sqrt{E/ \sigma_{0.7}}}\right]^{2}}.\end{aligned}\] A plot of the column curve given by eq. ([eq11.101]) is shown in figure 1.11.

Simply supported rectangular plate

Consider a plate simply supported on all four edges and subject to uniform compressive on edges \(x=0\) and \(x=a\). In the pre-buckling equilibrium configuration the plate remains flat, \(w_{0}(x, y)=0\), with a spatially uniform compressive stress equal to the applied compressive stress \(\sigma\). From the method of adjacent equilibrium, the out-of-plane displacement of the plate at the onset of buckling is \[\begin{aligned} \label{eq11.112} w_{1}(x, y)=A_{1} \sin \left(\frac{m \pi x}{a}\right) \sin \left(\frac{n \pi y}{b}\right),\end{aligned}\] where \(m\) and \(n\) are positive integers and \(A_{1}\) is an arbitrary amplitude. Integer \(m\) corresponds to the number of half waves in the \(x\)-direction and integer \(n\) corresponds to the number of half waves in the \(y\)-direction. Thus, specific values of integers \(m\) and \(n\) in eq. ([eq11.112]) characterize a buckling mode, and for each buckling mode there is a corresponding buckling stress. Equation ([eq11.110]) is the formula for the compressive stress at buckling, with the compressive buckling coefficient given by \[\begin{aligned} \label{eq11.113} k_{c}=\left(\frac{m}{a/b}+n^{2} \frac{(a/b)}{m}\right)^{2} \quad m, n=1,2, \ldots.\end{aligned}\] The critical stress is the lowest buckling stress, which occurs for a certain choice of \(m\) and \(n\). Since \(k_{c}\) is directly proportional to powers of integer \(n\), the minimum value of \(k_{c}\) occurs for \(n=1\). Then minimum values of \(k_{c}\) are related to \(a/b\) and integer \(m\) by \[\begin{aligned} \label{eq11.114} k_{c}=\left(\frac{m}{a/b}+\frac{(a/b)}{m}\right)^{2}.\end{aligned}\]

Critical values of the compressive buckling coefficient as a function of a few aspect ratios are listed in table [tab11.4].

These critical values of the compression buckling coefficient are plotted as case C in figure 1.16 on page . The buckling modes for three integer values of the aspect ratio are depicted in figure 1.17. There is one half

wave across the width (\(n=1\)) and the number of half waves across the length, \(m\), increases with increasing aspect ratio. For integer values of the aspect ratio the critical value of the compressive buckling coefficient \(k_{c r}=4\), and it follows that the critical compressive stress is \[\begin{aligned} \label{eq11.115} \sigma_{c r}=4 \pi^{2} \frac{E}{12\left(1-v^{2}\right)}\left(\frac{t}{b}\right)^{2} \quad \frac{a}{b}=m=1,2,3, \ldots \quad n=1.\end{aligned}\] From eq. ([eq11.112]) the length of a half wave in the \(x\)-direction is \(a/m\), and the length of a half wave in the \(y\)-direction is the plate width \(b\) for \(n=1\). These half wave lengths are the same when \(a/m=b\), or \(a/b=m\). That is, the half wave lengths in the \(x\)- and \(y\)-directions are the same for integer values of the aspect ratio. Hence, for integer values of the plate aspect ratio the buckling mode consists of a sequence of square buckles.

[ex11.4] Let \(a=20\) in., \(b=10\) in., \(t=0.10\) in., \(E=10 \times 10^{6}\,\textrm{lb}./\mathrm{in}.^{2}\), and \(v=0.3\). From eq. ([eq11.110]) the critical compressive stress \[\begin{aligned} \sigma_{\mathrm{cr}}=k_{c}(903.81\,\textrm{lb}./\mathrm{in.}^{2}). \label{eq11.4.a}\tag{a}\end{aligned}\] From eq. ([eq11.114]) the compression buckling coefficient is \[\begin{aligned} k_{c}=\left(\frac{m}{2}+\frac{2}{m}\right)^{2}. \label{eq11.4.b}\tag{b}\end{aligned}\] For \(m= 1, 2, 3, k_{c}=6.25,4.0,4.694\), respectively. For larger values of \(m\), coefficient \(k_{c}\) is larger. The minimum value of \(k_{c}\) is 4 corresponding to \(m=2\). Hence, the critical stress is \[\begin{aligned} \sigma_{\mathrm{cr}}=3,615.24\,\textrm{lb}./\mathrm{in.}^{2}. \label{eq11.4.c}\tag{c}\end{aligned}\] The critical compressive load \(P_{\mathrm{cr}}=\sigma_{\mathrm{cr}} bt\). Hence, \[\begin{aligned} P_{\mathrm{cr}}=3{,}615\,\textrm{lb.}. \label{eq11.4.d}\tag{d}\end{aligned}\]

The buckling mode for \((m, n)=(2,1)\) has one half sine wave in the transverse direction and two half waves in the longitudinal direction. The load \(P_{\mathrm{cr}}=3{,}615\,\textrm{lb.}\) is the lowest load at which such a plate can lose itsstability.

Solution.

The centroid and the shear center of the cross section coincide with the center of the box beam due to symmetry. The normal stress due to bending in the upper skin is calculated from the flexure formula; i.e., \[\begin{aligned} \sigma_{z}=\frac{M_{x}\left(b_{w}/2\right)}{I_{x x}}, \label{eq11.5.a}\tag{a}\end{aligned}\] where the second area moment of the cross section about the \(x\)-axis is \[\begin{aligned} I_{x x}=b_{w}^{2} A_{f}+\frac{1}{2} b_{f} b_{w}^{2} t_{f}+\frac{1}{6} b_{w}^{3} t_{w}=1{,}461.85\,\mathrm{in}.^{4} \label{eq11.5.b}\tag{b}\end{aligned}\] Hence, the bending normal stress in the upper skin is \[\begin{aligned} \sigma_{z}=-18{,}408.2\,\textrm{lb}./\text {in.}^{2}. \label{eq11.5.c}\tag{c}\end{aligned}\]

The shear stress in the upper skin is determined from the analysis of the shear flow distribution around the contour of the cross section, which is shown in figure [fig11.34].

The shear flow in the upper skin is \[\begin{aligned} q_{3}=\frac{b_{w} t_{f}\left(b_{f}-2 s_{3}\right)}{4 I_{x x}} V_{y} \quad 0 \leq s_{3} \leq b_{f}. \label{eq11.5.d}\tag{d}\end{aligned}\] The shear stress \(\tau_{3}=q_{3}/t_{f}\), and its evaluation is \[\begin{aligned} \tau_{3}=55.5803\left(24-2 s_{3}\right)\,\textrm{lb}./\text{in.}^{2} \quad 0 \leq s_{3} \leq 24\,\textrm{in}. \label{eq11.5.e}\tag{e}\end{aligned}\] Computing the maximum magnitude and the average value of this shear stress results in \[\begin{aligned} \left|\tau_{3}\right|_{\max}=1{,}333.93\,\textrm{lb}/\text{in.}^{2} \quad\left(\tau_{3}\right)_{\mathrm{ave}}=\frac{1}{b_{f}} \int_{0}^{b_{f}} \tau_{3} d s_{3}=0. \label{eq11.5.f}\tag{f}\end{aligned}\] The maximum magnitude of the shear stress in the upper skin is 7.25 percent of the magnitude of the bending normal stress. Moreover, the average value of the shear stress is zero in the upper skin. Hence, it is reasonable to neglect the effect of the shear stress on the buckling of the upper skin.

Assume the upper skin is a simply supported rectangular plate between the stringers and ribs. Actually, the stringer and ribs provide rotational constraint to the upper skin, but the assumption of no rotational constraint is conservative with respect to design. The critical compressive stress for simple support on all four edges of the upper skin underestimates its actual value. Equation ([eq11.110]) for the top cover skin is \[\begin{aligned} \sigma_{\textrm{cr}}=k_{c} \pi^{2} \frac{E}{12\left(1-v^{2}\right)}\left(\frac{t_{f}}{b_{f}}\right)^{2}, \label{eq11.5.g}\tag{g}\end{aligned}\] and eq. ([eq11.114]) for the compression buckling coefficient is \[\begin{aligned} k_{c}=\left(\frac{m}{a/b_{f}}+\frac{a/b_{f}}{m}\right)^{2} \quad m \text { a positive integer to minimize } k_{c}. \label{eq11.5.aa}\tag{a}\end{aligned}\] The margin of safety is defined by \[\begin{aligned} \label{eq11.119} MS=\frac{\text { Excess strength }}{\text { Required strength }}=\frac{\sigma_{\mathrm{cr}}-\left|\sigma_{z}\right|}{\left|\sigma_{z}\right|}.\end{aligned}\] The margin of safety ([eq11.119]) is positive for a feasible design, otherwise the design is infeasible. It should be a small positive value for a design of least weight. The computations for the margin of safety are listed intable [tab11.6].

A rib spacing of 16 in. is a feasible design with a slightly positive margin of safety.

[sec11.10] Brush, D. O., and B. O. Almroth. Buckling of Bars, Plates, and Shells. New York: McGraw-Hill, 1975, pp.75–105.

Budiansky, B., and J. W. Hutchinson. “Dynamic Buckling of Imperfection-Sensitive Structures.” In Proceedings of the Eleventh International Congress of Applied Mechanics (Munich, Germany). 1964, pp. 639–643.

Budiansky, B. “Dynamic Buckling of Elastic Structures: Criteria and Estimates.” In Proceedings of an International Conference at Northwestern University (Evanston IL). New York and Oxford: Pergamon Press, 1966.

Gerard, G., and H. Becker. Handbook of Structural Stability: Part I, Buckling of Flat Plates. Technical Report NACA-TN 3781. Washington, DC: Office of Scientific and Technical Information, U.S. Department of Energy, 1957.

Koiter, W. T. “The Stability of Elastic Equilibrium.” PhD diss., Teehische 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.)

National Advisory Committee for Aeronautics (NACA), Technical Note 3781 (NACA-TN-3781) July 1957. https://ntrs.nasa.gov/api/citations/1930084505/downloads/19930084505.pdf.

Ramberg, W., and Osgood, W. Description of Stress-Strain Curves by Three Parameters. Technical Report NACA-TN-902. Washington DC: NASA, 1943.

Southwell, Richard V. “On the Analysis of Experimental Observations in Problems of Elastic Stability.” Proc. Roy. Soc. London A, no. 135 (April 1932): 601–616.

Timoshenko, S. P., and J. M. Gere. Theory of Elastic Stability. New York: McGraw-Hill Book Company,1961, p.33.

Ugural A. C., and S. K. Fenster. Advanced Strength and Applied Elasticity, 4th ed.,Upper Saddle River, NJ: Pearson Education, Inc., Publishing as Prentice Hall Professional Technical Reference, 2003, pp.472–490.