Problem 7.5. Water tower

Critical load parameter=

Step 1.  Neglect distributed weight of the shaft. Calculate critical load parameter of the elastically supported cantilever loaded by the top concentrated weight only.

Check critical load parameter for MHide critical load

See Figure 7.48 and Eq.(7-90)

$$\begin{gathered} N_{\mathrm{cr},1}=\frac{{\pi }^{2}EI}{4L^2}\frac{1}{1+2.5{\displaystyle \frac{EI}{kL}}}=\frac{{\pi }^2\times 1.0\times {10}^9}{4\times {40}^2}\frac{1}{1+2.5{\displaystyle \frac{1.0\times {10}^9}{300\times {10}^6\times 40}}}=1.276\times {10}^6 \mathrm{kN} \\ {\alpha }_{\mathrm{cr},1}=\frac{N_{\mathrm{cr},1}}{N_1}=\frac{1.276\times {10}^6}{10\times {10}^3}=127.6 \end{gathered}$$

The above approximate formula was derived by applying Föppl’s theorem to a flexible cantilever with rigid foundation and an elastically supported rigid bar.

Step 2.  Remove top weight. Calculate critical load parameter of the elastically supported cantilever loaded by the distributed weight of the shaft only. 

Check critical load parameter for m Hide critical load

See Figure 7.49 and Eq.(7-92)

$$\begin{gathered} N_{\mathrm{cr},2}=7.8\frac{EI}{L^2}\frac{1}{1+3.9{\displaystyle \frac{EI}{kL}}}=7.8\frac{1.0\times {10}^9}{{40}^2}\frac{1}{1+3.9{\displaystyle \frac{1.0\times {10}^9}{300\times {10}^6\times 40}}}=3.68\times {10}^6 \mathrm{kN} \\ {\alpha }_{\mathrm{cr},2}=\frac{N_{\mathrm{cr},2}}{N_2}=\frac{3.68\times {10}^6}{250\times 40}=367.9 \end{gathered}$$

(The above approximate formula was derived by applying Föppl’s theorem to a flexible cantilever with rigid foundation and an elastically supported rigid bar.)

Step 3. Apply Dunkerley approximation to determine the critical load parameter of the cantilever subjected to both weights.

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See Table 7.4.

${\mathit{\alpha }}_{\mathbf{cr}}={\left(\frac{1}{{\alpha }_{\mathrm{cr},1}}+\frac{1}{{\alpha }_{\mathrm{cr},1}}\right)}^{–1}={\left(\frac{1}{127.6}+\frac{1}{367.9}\right)}^{–1}\mathbf{=}\mathbf{94}\mathbf{.}\mathbf{8}$

Remark. An alternative solution is that first the tower with rigid support, subjected to both loads is investigated, then the rigid structure supported elastically at the bottom. After that Föppl’s approximation is applied which gives ${\alpha }_{\mathrm{cr}}=94.9$.

In the approximation first the distributed weight of the shaft is neglected. The critical load parameter of the elastically supported cantilever loaded by the top concentrated weight is calculated.

See Figure 7.48 and Eq.(7-90)

$$\begin{gathered} N_{\mathrm{cr},1}=\frac{{\pi }^{2}EI}{4L^2}\frac{1}{1+2.5{\displaystyle \frac{EI}{kL}}}=\frac{{\pi }^2\times 1.0\times {10}^9}{4\times {40}^2}\frac{1}{1+2.5{\displaystyle \frac{1.0\times {10}^9}{300\times {10}^6\times 40}}}=1.276\times {10}^6 \mathrm{kN} \\ {\alpha }_{\mathrm{cr},1}=\frac{N_{\mathrm{cr},1}}{N_1}=\frac{1.276\times {10}^6}{10\times {10}^3}=127.6 \end{gathered}$$

Next, the top weight is removed. The critical load parameter of the elastically supported cantilever loaded by the distributed weight of the shaft is calculated.

See Figure 7.49 and Eq.(7-92)

(The above approximate formulas were derived by applying Föppl’s theorem to a flexible cantilever with rigid foundation and an elastically supported rigid bar.)

The critical load parameter of the cantilever subjected to both weights  is calculated by the Dunkerley approximation.

See Table 7.4.

Remark. An alternative solution is that first the tower with rigid support, subjected to both loads is investigated, then the rigid structure supported elastically at the bottom. After that Föppl’s approximation is applied which gives ${\alpha }_{\mathrm{cr}}=94.9$.