Question 21.P.15: A pin-ended column of length L has its central portion reinf...
A pin-ended column of length L has its central portion reinforced, the second moment of its area being I2 while that of the end portions, each of length a, is I_{1} . Use the Rayleigh-Ritz method to determine the critical load of the column assuming that its centreline deflects into the parabola v = kx(L – x) and taking the more accurate of the two expressions for bending moment.
In the case where I_{2} = 1.6I_{1} and a = 0.2 L find the percentage increase in strength due to the reinforcement.
The bending moment at any section of the column is given by
M=P_{\mathrm{CR}} v=P_{\mathrm{CR}} k x(L-x)
Also
\frac{\mathrm{d} v}{\mathrm{~d} x}=k(L-2 x)
Substituting in Eq. (21.65) U+V=\int_{0}^{L} \frac{M^{2}}{2 E I} \mathrm{~d} x-\frac{P_{\mathrm{CR}}}{2} \int_{0}^{L}\left(\frac{\mathrm{d} v}{\mathrm{~d} x}\right)^{2} \mathrm{~d} x
\begin{aligned}U+V=&\left(\frac{P_{\mathrm{CR}}^{2} k^{2}}{2 E}\right)\left[\left(\frac{1}{I_{1}}\right) \int_{0}^{a}\left(L x-x^{2}\right)^{2} \mathrm{~d} x+\left(\frac{1}{I_{2}}\right) \int_{a}^{L-a}\left(L x-x^{2}\right)^{2} \mathrm{~d} x\right.\\&\left.+\left(\frac{1}{I_{1}}\right) \int_{L-a}^{L}\left(L x-x^{2}\right)^{2} \mathrm{~d} x\right]-\left(\frac{P_{\mathrm{CR}} k^{2}}{2}\right) \int_{0}^{L}(L-2 x)^{2} \mathrm{~d} x\end{aligned}
i.e.
\begin{aligned}U+V=&\left(\frac{P_{\mathrm{CR}} k^{2}}{2 E I_{2}}\right)\left\{\left(\frac{I_{2}}{I_{1}}-1\right)\left[\frac{L^{2} a^{3}}{3}-\frac{L a^{4}}{2}+\frac{a^{5}}{5}-\frac{L^{2}(L-a)^{3}}{3}+\frac{L(L-a)^{4}}{2}-\frac{(L-a)^{5}}{5}\right]\right.\\&\left.+\left(\frac{I_{2}}{I_{1}}\right)\left(\frac{L^{5}}{30}\right)\right\}-\frac{P_{\mathrm{CR}} k^{2} L^{3}}{6}\end{aligned}
From the principle of the stationary value of the total potential energy
\frac{\partial(U+V)}{\partial k}=0
Then, since I_{2}=1.6 I_{1} and a = 0.2L this gives
P_{\mathrm{CR}}=\frac{14.96 E I_{1}}{L^{2}}
Without the reinforcement
P_{\mathrm{CR}}=\frac{\pi^{2} E I_{1}}{L^{2}}
The ratio of the two is 14.96/π² = 1.52
Therefore the increase in strength is 52%.