Question 8.15: 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 center to 0.2A_{1} and 0.2I_{1} at its ends. Assuming a deflected center-line of parabolic form and taking the more correct form for the bending moment, use the energy method to estimate its critical load when tested between pin-centers, in terms of the preceding data and Young’s modulus E. 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 percent.
Answer: 7.01E I_{1}/l^{2}
The equation for the deflected center line of the column is
\ v={\frac{4\delta}{l^{2}}}z^{2} (i)
in which δ is the deflection at the ends of the column relative to its center and the origin for z is at the center of the column. Also, the second moment of area of its cross-section varies, from the center to its ends, in accordance with the relationship
I=I_{1}\left(1-1.6{\frac{z}{\overline{{l}}}}\right) (ii)
At any section of the column the bending moment, M, is given by
M=P_{\mathrm{CR}}(\delta-v)=P_{\mathrm{CR}}\delta\!\left(1-4{\frac{z^{2}}{l^{2}}}\right) (iii)
Also, from Eq. (i),
{\frac{\mathrm{d}\upsilon}{\mathrm{d}z}}={\frac{8\delta}{l^{2}}}z (iv)
Substituting in Eq. (8.47) for M, I, and dυ/dz,
U+V=\int_{0}^{l}\!\frac{M^{2}}{2E I}\ \mathrm{d}z-\frac{P_{\mathrm{CR}}}{2}\int_{0}^{l}\!\left(\frac{\mathrm{d}\nu}{\mathrm{d}z}\right)^{2}\!\mathrm{d}z (8.47)
U+V=2\int_{0}^{l/2}{\frac{P_{\mathrm{CR}}^{2}\delta^{2}(1-4z^{2}/l^{2})^{2}}{2E I_{1}(1-1.6z/l)}}\mathrm{d}z-{\frac{P_{\mathrm{CR}}}{2}}2\int_{0}^{l/2}{\frac{64\delta^{2}}{l^{4}}}z^{2}\mathrm{d}z
or
U+V={\frac{P_{\mathrm{CR}}^{2}\delta^{2}}{E I_{1}l^{3}}}\int_{0}^{l/2}{{\frac{(l^{2}-4z^{2})^{2}}{l-1.6z}}\mathrm{d}z-{\frac{64P_{\mathrm{CR}}\delta^{2}}{l^{4}}}} \int_{0}^{l/2}{z^{2}dz} (v)
Dividing the numerator by the denominator in the first integral in Eq. (v) gives
Hence,
i.e.,
U+V=\frac{0.3803P_{\mathrm{CR}}^{2}\delta^{2}l}{E I_{1}}-\frac{8P_{\mathrm{CR}}\delta^{2}}{3l}From the principle of the stationary value of the total potential energy,
\frac{\partial(U+V)}{\partial\delta}=\frac{0.7606P_{\mathrm{CR}}^{2}\delta l}{E I_{1}}-\frac{16P_{\mathrm{CR}}\delta}{3l}=0Hence,
P_{\mathrm{CR}}={\frac{7.01E 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}}} ((see Eq.(8.5))
P_{\mathrm{CR}}={\frac{\pi^{2}E I}{l^{2}}} (8.5)
For the columns to have the same buckling load,
{\frac{\pi^{2}E I_{2}}{l^{2}}}={\frac{7.01E I_{1}}{l^{2}}}so that
I_{2}=0.7I_{1}Thus, since the radii of gyration are the same,
A_{2}=0.7A_{1}Therefore, 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 = \rho\times\mathrm{average~thickness}\times l=\rho\times 0.64 A_{1}\;l.
Hence the saving in weight = 0.7\rho A_{1}l-0.6\rho A_{1}l=0.1~\rho A_{1}~l.
Expressed as a percentage,
\mathrm{swing~in~weight}=\frac{0.1/\rho A_{1}l}{0.7\rho A_{1}l}\times100=14.3%