Question 8.16: A uniform column (Fig. P.8.16), of length l and bending stif......
A uniform column (Fig. P.8.16), of length l and bending stiffness EI, is rigidly built in at the end z = 0 and simply supported at the end z = l. The column is also attached to an elastic foundation of constant stiffness k/unit length. Representing the deflected shape of the column by a polynomial,
\nu=\sum\limits_{n=0}^{p}{a_{n}\eta ^{n},} where \eta =z/l
determine the form of this function by choosing a minimum number of terms p such that all the kinematic (geometric) and static boundary conditions are satisfied, allowing for one arbitrary constant only. Using the result thus obtained, find an approximation to the lowest flexural buckling load P_{CR} by the Rayleigh–Ritz method.
Answer: P_{\mathrm{CR}}=21.05E I/l^{2}+0.09k l^{2}
There are four boundary conditions to be satisfied, namely υ = 0 at z = 0 and z=l, dυ/dz = 0 at z = 0 and d²υ/dz² (i.e., bending moment)=0 at z=l. Thus, since only one arbitrary constant may be allowed for, there cannot be more than five terms in the polynomial. Suppose
\upsilon=a_{0}+a_{1}\left(\frac{z}{l}\right)+a_{2}\left(\frac{z}{l}\right)^{2}+a_{3}\left(\frac{z}{l}\right)^{3}+a_{4}\left(\frac{z}{l}\right)^{4} (i)
Then, since υ = 0 at z = 0, a_{0}=0. Also, since dυ/dz = 0 at z = 0, a_{1}=0. Hence, Eq. (i) becomes
\upsilon=a_{2}\biggl(\frac{z}{l}\biggr)^{2}+a_{3}\biggl(\frac{z}{l}\biggr)^{3}+a_{4}\biggl(\frac{z}{l}\biggr)^{4} (ii)
When z=l, υ = 0, thus
0=a_{2}+a_{3}+a_{4} (iii)
When z=l, d²υ/dz² = 0, thus
0=a_{2}+3a_{3}+6a_{4} (iv)
Subtracting Eq. (iv) from Eq. (iii),
0=-2a_{3}-5a_{4}from which a_{3}=-\,5a_{4}/2.
Substituting for a_{3} in Eq. (iii) gives a_{4}=2a_{2}/3 so that a_{3}=-5a_{2}/3. Eq. (ii) then becomes
\upsilon=a_{2}\biggl(\frac{z}{l}\biggr)^{2}-\frac{5a_{2}}{3}\biggl(\frac{z}{l}\biggr)^{3}+\frac{2a_{2}}{3}\biggl(\frac{z}{l}\biggr)^{4} (v)
Then
{\frac{\mathrm{d}v}{\mathrm{d}z}}=2a_{2}{\frac{z}{l^{2}}}-5a_{2}{\frac{z^{2}}{l^{3}}}+{\frac{8a_{2}z^{3}}{3\ l^{4}}} (vi)
and
{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}={\frac{2a_{2}}{l^{2}}}-10a_{2}{\frac{z}{l^{3}}}+8a_{2}{\frac{z^{2}}{l^{4}}} (vii)
The total strain energy of the column will be the sum of the strain energy due to bending and the strain energy due to the resistance of the elastic foundation. For the latter, consider an element, δz, of the column. The force on the element when subjected to a small displacement, υ, is kδzv. Thus, the strain energy of the element is {\textstyle{\frac{1}{2}}}k\nu^{2}\delta z and the strain energy of the column due to the resistance of the elastic foundation is
\int_{0}^{l}{{\frac{1}{2}}k v^{2}\mathrm{d}z}Substituting for υ from Eq. (v),
U(elastic foundation) {=\frac{1}{2}k\frac{a_{2}^{2}}{l^{4}}\int_{0}^{l}\left(z^{4}-\frac{10z^{5}}{3l}+\frac{37z^{6}}{9l^{2}}-\frac{20z^{7}}{9l^{3}}+\frac{4z^{8}}{9l^{4}}\right)\mathrm{d}z}
i.e., U(elastic foundation) =0.071k a_{2}^{2}l.
Now substituting for d²υ/dz² and dυ/dz in Eq. (8.48) and adding U (elastic foundation) gives
U+V={\frac{E I}{2}}\int_{0}^{l}{\left({\frac{\mathrm{d}^{2}\nu}{\mathrm{d}z^{2}}}\right)^{2}\mathrm{d}z-{\frac{P_{\mathrm{CR}}}{2}}} \int_{0}^{l}{\left({\frac{\mathrm{d}\nu}{\mathrm{d}z}}\right)^{2}\mathrm{d}z} (8.48)
Eq. (viii) simplifies to
U+V={\frac{0.4E I}{l^{3}}}a_{2}^{2}+0.0017k a_{2}^{2}l-{\frac{0.019a_{2}^{2}P_{\mathrm{CR}}}{l}}From the principle of the stationary value of the total potential energy
{\frac{\partial(U+V)}{\partial a_{2}}}={\frac{0.8E I}{l^{3}}}a_{2}+0.0034k a_{2}I-{\frac{0.038a_{2}P_{\mathrm{CR}}}{l}}=0whence,
P_{\mathrm{CR}}={\frac{21.05E I}{l^{2}}}+0.09k l^{2}