Question 21.P.16: A tubular column of length L is tapered in wall thickness so...
A tubular column of length L is tapered in wall thickness so that the area and the second moment of area of its cross section decrease uniformly from A_{1} and I_{1} at its centre to 0.2A_{1} and 0.2I_{1} at its ends, respectively.
Assuming a deflected centreline of parabolic form and taking the more correct form for the bending moment, use the Rayleigh-Ritz method to estimate its critical load; the ends of the column may be taken as pinned. Hence show that the saving in weight by using such a column instead of one having the same radius of gyration and constant thickness is about 15%.
Assume the equation of the deflected centre line of the column is
v=\left(\frac{4 \delta}{L^{2}}\right) x^{2}
in which δ is the deflection of the ends of the column relative to its centre and the origin for x is at the centre of the column. The second moment of area varies in accordance with the relationship
I=I_{1}\left[1-1.6\left(\frac{x}{L}\right)\right]
The bending moment at any section of the column is given by
M=P_{\mathrm{CR}}(\delta-v)=P_{\mathrm{CR}} \delta\left[1-4\left(\frac{x^{2}}{L^{2}}\right)\right]
Also, from the above
\frac{\mathrm{d} v}{\mathrm{~d} x}=\left(\frac{8 \delta}{L^{2}}\right) 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
U+V=\left(\frac{P_{\mathrm{CR}} \delta^{2}}{E I_{1} L^{3}}\right) \int_{0}^{\frac{L}{2}}\left[\frac{\left(L^{2}-4 x^{2}\right)^{2}}{L-1.6 x}\right] \mathrm{d} x-\left(\frac{64 P_{\mathrm{CR}} \delta^{2}}{L^{4}}\right) \int_{0}^{\frac{L}{2}} x^{2} \mathrm{~d} x
i.e.
U+V=\frac{0.3803 P_{\mathrm{CR}}^{2} \delta^{2} L}{E I_{1}}-\frac{8 P_{\mathrm{CR}} \delta^{2}}{3 L}
From the principle of the stationary value of the total potential energy
\frac{\partial(U+V)}{\partial \delta}=\frac{0.7606 P_{\mathrm{CR}}^{2} \delta L}{E I_{1}}-\frac{16 P_{\mathrm{CR}} \delta}{3 L}=0
which gives
P_{\mathrm{CR}}=\frac{7.01 E I_{1}}{L^{2}}
For a column of constant thickness and second moment of area I_{2}
P_{\mathrm{CR}}=\frac{\pi^{2} E I_{2}}{L^{2}}
For the columns to have the same buckling load
\frac{\pi^{2} E I_{2}}{L^{2}}=\frac{7.01 E I_{1}}{L^{2}}
so that
I_{2}=0.7 I_{1}
Therefore, since the radii of gyration are the same, A_{2}=0.7 A_{1}
The weight of the constant thickness column is equal to \rho A_{2} L=0.7 \rho A_{1} L.
The weight of the tapered column is equal to ρ × average thickness × L=\rho \times 0.6 A_{1} L so that the saving in weight is 0.1 \rho A_{1} L , i.e.
\text { saving in weight }=\frac{0.1 \rho A_{1} L}{0.7 \rho A_{1} L}=0.143
i.e. about 15%.