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Magazine Chapter 1: Q37E (page 1)
Buckling of a Thin Vertical Column In Example 4 of Section 5.2 we saw that when a constant vertical compressive force, or load, \(P\) was applied to a thin column of uniform cross section and hinged at both ends, the deflection \(y(x)\) is a solution of the BVP:
\(El\frac{{{d^2}y}}{{d{x^2}}} + Py = 0,y(0) = 0,y(L) = 0\)
(a)If the bending stiffness factor \(El\) is proportional to \(x\), then \(El(x) = kx\), where \(k\) is a constant of proportionality. If \(El(L) = kL = M\) is the maximum stiffness factor, then \(k = M/L\) and so \(El(x) = Mx/L\). Use the information in Problem 39 to find a solution of \(M\frac{x}{L}\frac{{{d^2}y}}{{d{x^2}}} + Py = 0,y(0) = 0,y(L) = 0\) if it is known that \(\sqrt x {Y_1}(2\sqrt {\lambda x} )\) is not zero at \(x = 0\).
(b) Use Table 6.4.1 to find the Euler load \({P_1}\) for the column.
(c) Use a CAS to graph the first buckling mode \({y_1}(x)\) corresponding to the Euler load \({P_1}\). For simplicity assume that \({c_1} = 1\) and \(L = 1\).
Short Answer
(a) The general solution is \(y(x) = {c_1}\sqrt x {J_1}(2\sqrt {PLx/M} )\).
(b) The Euler Load for the column is \({P_1} = \frac{{3.6705M}}{{{L^2}}}\).
(c) The graph has been plotted.
Step by step solution
Define Bessel’s equation.
Let the Bessel equation be\({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation hastwo linearly independent solutionsfor a fixed value of\(n\).A Bessel equation of the first kind,indicated by\({J_n}(x)\), is one of these solutions that may be derived usingFrobinous approach.
\(\begin{array}{l}{y_1} = {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} = {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{array}\)
At\(x = 0\), this solution is regular. The second solution, which is singular at\(x = 0\), is represented by\({Y_n}(x)\)and is calleda Bessel function of the second kind.
\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)
Determine the general solution of the DE.
(a)
Let the given equation be\(M\frac{x}{L}\frac{{{d^2}y}}{{d{x^2}}} + Py = 0,y(0) = 0,y(L) = 0\)鈥�(1)
Since\(\sqrt x {Y_1}(2\sqrt {\lambda x} )\)is not equal to zero at\(x = 0\), then the equation (1) becomes,\(x\frac{{{d^2}y}}{{d{x^2}}} + \lambda y = 0\)鈥� (2)where\(\lambda = \frac{{PL}}{M}\).
In problem 37, the general solution of above DE is given by,
\(y(x) = {c_1}\sqrt x {J_1}(2\sqrt {\lambda x} ) + {c_2}\sqrt x {Y_1}(2\sqrt {\lambda x} )\)鈥� (3)
Hence,\({J_1}(0) = 0\)from the definition of\({J_1}(x)\). As\(\sqrt x {Y_1}(2\sqrt {\lambda x} )\)is not equal to zero at\(x = 0\), and with the boundary condition\(y(0) = 0\), the result is\({c_2} = 0\).
Now, the equation (3) becomes,
\(y(x) = {c_1}\sqrt x {J_1}(2\sqrt {\lambda x} )\)鈥� (4)
Substitute the value\(\lambda = \frac{{PL}}{M}\)in the equation yields,
\(y(x) = {c_1}\sqrt x {J_1}(2\sqrt {PLx/M} )\) 鈥� (5)
Find the solution of the BVP.
(b)
Substitute the value\(y(L) = 0\)in the equation (5).
\(\begin{array}{c}{c_1}\sqrt x {J_1}(2\sqrt {PL \times L/M} ) = 0\\{J_1}(2L\sqrt {P/M} ) = 0\end{array}\)
From table 6.4.1, the first positive zero of\({J_1}\)is\(3.8317\). Therefore,
\(\begin{array}{c}2L\sqrt {{P_1}/M} = 3.8317\\{P_1} = \frac{{3.831{7^2}M}}{{4{L^2}}}\\{P_1} = \frac{{3.6705M}}{{{L^2}}}\end{array}\)
Find the critical length of the solid steel rod.
(c)
Substitute the value\({P_1} = \frac{{3.6705M}}{{{L^2}}}\)in the equation (5) to obtain\({y_1}(x) = {c_1}\sqrt x {J_1}\left( {2\sqrt {\frac{{3.6705x}}{L}} } \right)\). For\({c_1} = L = 1\), the equation (6) becomes\({y_1}(x) = \sqrt x {J_1}(3.8317\sqrt x )\). Using GNU Octave, the code to plot the graph is,
Let the graph be,
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