Problem 10.18. Circular plate subjected to concentrated load (elastically supported and built-in)

Radius, R [m]=

Step 1.  Determine the maximum moment of a plate on Winkler type foundation subjected to concentrated load.

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See Figure 10.43.

$$\begin{gathered} M_{\max }=\frac{\left(1+\nu \right)P}{4\pi }\left\{\ln \left(\frac{1}{\lambda b}\right)+0.27\right\}=\frac{\left(1+0.3\right)P}{4\pi }\left\{\ln \left(\frac{1}{0.983\times 0.15}\right)+0.27\right\}=0.226P \\ \mathrm{where} \\ D=\frac{E{h}^3}{12\left(1–{\nu }^2\right)}=\frac{18.3\times {200}^3}{12\left(1–0.3^2\right)\times {10}^3}=13.4\times {10}^3 \mathrm{kNm} \\ \mathrm{and} \\ \lambda =\sqrt[4]{\frac{c}{4D_s}}=\sqrt[4]{\frac{50000}{4\times 13.4\times {10}^3}}=0.983\frac{1}{\mathrm{m}} \end{gathered}$$

Step 2.  Write the maximum positive moment of a circular plate with built-in edges subjected to concentrated load.

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See Table 10.12 4^th row

$M_{\max }=\frac{P}{4\pi }(1+\nu )\ln \left(\frac{R}{b}\right)=\frac{P}{4\pi }(1+0.3)\ln \left(\frac{R}{0.15}\right)$

Step 3.  Compare the above solutions. Express the radius from their equality.

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See Table 10.12 third row

$$\begin{gathered} 0.226P=\frac{P}{4\pi }(1+0.3)\ln \left(\frac{R}{0.15}\right) \\ \to \mathit{R}=0.15e^{\frac{0.226\times 4\mathrm{\pi }}{(1+0.3)}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{ }\mathbf{m} \end{gathered}$$

The maximum moment of a plate on Winkler type foundation subjected to concentrated load is

See Figure 10.43.

$$\begin{gathered} M_{\max }=\frac{\left(1+\nu \right)P}{4\pi }\left\{\ln \left(\frac{1}{\lambda b}\right)+0.27\right\}=\frac{\left(1+0.3\right)P}{4\pi }\left\{\ln \left(\frac{1}{0.983\times 0.15}\right)+0.27\right\}=0.226P \\ \mathrm{where} \\ D=\frac{E{h}^3}{12\left(1–{\nu }^2\right)}=\frac{18.3\times {200}^3}{12\left(1–0.3^2\right)\times {10}^3}=13.4\times {10}^3 \mathrm{kNm} \\ \mathrm{and} \\ \lambda =\sqrt[4]{\frac{c}{4D_s}}=\sqrt[4]{\frac{50000}{4\times 13.4\times {10}^3}}=0.983\frac{1}{\mathrm{m}} \end{gathered}$$

The maximum positive moment of a circular plate with built-in edges subjected to concentrated load is

See Table 10.12 4^th row

$M_{\max }=\frac{P}{4\pi }(1+\nu )\ln \left(\frac{R}{b}\right)=\frac{P}{4\pi }(1+0.3)\ln \left(\frac{R}{0.15}\right)$

Comparing the above solutions the unknown radius can be expressed:

See Table 10.12 third row

$$\begin{gathered} 0.226P=\frac{P}{4\pi }(1+0.3)\ln \left(\frac{R}{0.15}\right) \\ \to \mathit{R}=0.15e^{\frac{0.226\times 4\mathrm{\pi }}{(1+0.3)}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{ }\mathbf{m} \end{gathered}$$