Question 8.14: 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 I_{2}, while that of the end portions, each of length a, is I_{1}. Use the energy method to determine the critical load of the column, assuming that its centerline deflects into the parabola v = kz(l – z) and taking the more accurate of the two expressions for the 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, and compare it with the percentage increase in weight on the basis that the radius of gyration of the section is not altered.
Answer: P_{CR} = 14.96EI_{1},l^{2}, 52 percent, 36 percent
The bending moment, M, at any section of the column is given by
M=P_{\mathrm{CR}}v=P_{\mathrm{CR}}k\left(l z-z^{2}\right) (i)
Also,
{\frac{\mathrm{d}v}{\mathrm{d}z}}=k(l-2z) (ii)
Substituting from Eqs (i) and (ii) in Eq. (8.47),
U+V=\int_{0}^{l}{\frac{M}{3EI} dz}-\frac{P_{CR}}{2} \int_{0}^{l}{\left(\frac{dv}{dz} \right)^{2} } dz (8.47)
i.e.,
U+V={\frac{P_{\mathrm{CR}}^{2}k^{2}}{2EI_2}}\left\{\left(\frac{l_{2}}{l_{1}}-1 \right)\left[\frac{l^{2}a^{3}}{3}-\frac{la^{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]+\frac{I_{2} l^5}{I_{1} 30} \right\} -\frac{P_{CR}k^{2}l^{3}}{6}From the principle of the stationary value of the total potential energy,
Hence,
P_{CR}=\frac{EI_{2}l^{3}}{3\left\{\left(\frac{I_{2}}{I_{1}}-1 \right)\left[\frac{l^{2}a^{3}}{3}-\frac{la^{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]+\frac{I_{2}}{I_{1}}\frac{l^{5}}{30} \right\} } (iii)
When I_{2}=1.6I_{1} and a = 0.2l, Eq. (iii) becomes
P_{\mathrm{CR}}={\frac{14.96E I_{1}}{l^{2}}} (iv)
Without the reinforcement,
P_{\mathrm{CR}}={\frac{\pi^{2}E I_{1}}{l^{2}}} (v)
Therefore, from Eqs (iv) and (v), the increase in strength is
{\frac{E I_{1}}{l^{2}}}{\big(}14.96-\pi^{2}{\big)}Thus the percentage increase in strength is
\left[{\frac{E I_{1}}{l^{2}}}\left(14.96-\pi^{2}\right)/{\frac{l^{2}}{\pi^{2}E I}}\right]\times100=52%
Since the radius of gyration of the cross-section of the column remains unchanged,
I_{1}=A_{1}r^{2}\ \ {\mathrm{and}}\ \ I_{2}=A_{2}r^{2}\,Hence,
{\frac{A_{2}}{A_{1}}}={\frac{I_{2}}{I_{1}}}=1.6 (vi)
The original weight of the column is lA_1ρ where ρ is the density of the material of the column. Then, the increase in weight = 0.4l A_{1}\rho+0.6l A_{2}\rho-l A_{1}\rho=0.6l\rho(A_{2}-A_{1}).
Substituting for A_{2} from Eq. (vi),
Increase in weight = 0.6l\rho(1.6A_{1}-A_{1})=0.36l A_{1}\rho
i.e., an increase of 36%.