Problem 10.10. Vibration of a RC slab

Problem a)

Fundamental frequency with hinged supports, f_n [Hz]=

Problem b)

Fundamental frequency with supporting beam, f_n [Hz]=

Step 1.  Calculate the mass and stiffness data.

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Problem a)

Step 2.  Give the natural frequency assuming hinged edges.

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See Table 10.6 first row or Eq.(10-100)

$$\begin{gathered} f^2=f_x^2+f_y^2+f_t^2=\frac{{\pi }^{2}D}{4m{L}_x^4}+\frac{{\pi }^{2}D}{4m{L}_y^4}+\frac{2{\pi }^{2}D}{4m{L}_x^2{L}_y^2}=\frac{{\pi }^{2}D}{m}{\left(\frac{1}{L_x^2}+\frac{1}{L_y^2}\right)}^2= \\ \frac{{\pi }^2\times 1.067\times {10}^7}{407.7}{\left(\frac{1}{6.0^2}+\frac{1}{5.0^2}\right)}^2=296.52 \frac{1}{{\sec }^2} \\ \mathit{f}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{22}\mathbf{ }\mathbf{Hz} \end{gathered}$$

which is much higher than the limits of vibrations induced by walking

f >>3 Hz 

the slab satisfies the frequency limits.

Problem b)

Step 2.  Give the natural frequency assuming one edge supported by the given beam.

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See Table 10.8 first row.

$$\begin{gathered} f^2={\left(\frac{1}{f_x^2+f_{\mathrm{t}}^2-0.25f_{66}^2}+\frac{1}{f_{EI}^2}\right)}^{-1}+f_y^2+0.25f_{66}^2 \\ \mathrm{where} \\ f_{EI}^2=3\frac{{\pi }^{2}EI}{4m{L}_x{L}_y^4}=3\frac{{\mathrm{\pi }}^2\times 9.375\times {10}^7}{4\times 407.7\times 6.0\times 5.0^4}=453.85 \frac{1}{{\sec }^2} \\ {f}_x^2=\frac{{\pi }^{2}D}{4m{L}_x^4}=\frac{{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 6.0^4}=49.8\frac{1}{{\sec }^2} \\ {f}_y^2=\frac{{\pi }^{2}D}{4m{L}_y^4}=\frac{{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 5.0^4}=103.2\frac{1}{{\sec }^2} \\ {f}_{66}^2=\frac{4{\pi }^{2}D{\displaystyle \frac{\left(1-\nu \right)}{2}}}{4m{L}_x^2{L}_y^2}=\frac{2{\pi }^2\times 1.067\times {10}^7\left(1-0.2\right)}{4\times 407.7\times 6.0^2\times 5.0^2}=114.75\frac{1}{{\sec }^2} \\ {f}_{\mathrm{t}}^2=\frac{2{\pi }^{2}D}{4m{L}_x^2{L}_y^2}=\frac{2{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 6^4}=143.4\frac{1}{{\sec }^2} \end{gathered}$$

$$\begin{gathered} \to \mathit{f}=\sqrt{{\left(\frac{1}{f_x^2+f_{\mathrm{t}}^2-0.25f_{66}^2}+\frac{1}{f_{EI}^2}\right)}^{-1}+f_y^2+0.25f_{66}^2} =\mathbf{ } \\ =\sqrt{{\left(\frac{1}{49.8+143.4-0.25\times 114.7}+\frac{1}{453.8}\right)}^{-1}+103.3+0.25\times 114.7}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{Hz} \end{gathered}$$

The mass and stiffness data of the slab are the following:

$$\begin{gathered} m=\frac{\rho }{g}h=\frac{25\times {10}^3}{9.81}\times 0.16=407.7 \mathrm{kg}/{\mathrm{m}}^2 \\ D\mathit{=}\frac{E{t}^3}{12(1-{\nu }^2)}\mathit{=}\frac{30000\times {160}^3}{12(1-0.2^2)\times {10}^3}=1.067\times {10}^7 \mathrm{Nm} \\ EI=30000\frac{300\times {500}^3}{12\times {10}^6}=9.375\times {10}^7 {\mathrm{Nm}}^2 \end{gathered}$$

Problem a)

The natural frequency assuming hinged edges is

See Table 10.6 first row or Eq.(10-100)

$$\begin{gathered} f^2=f_x^2+f_y^2+f_t^2=\frac{{\pi }^{2}D}{4m{L}_x^4}+\frac{{\pi }^{2}D}{4m{L}_y^4}+\frac{2{\pi }^{2}D}{4m{L}_x^2{L}_y^2}=\frac{{\pi }^{2}D}{m}{\left(\frac{1}{L_x^2}+\frac{1}{L_y^2}\right)}^2= \\ \frac{{\pi }^2\times 1.067\times {10}^7}{407.7}{\left(\frac{1}{6.0^2}+\frac{1}{5.0^2}\right)}^2=296.52 \frac{1}{{\sec }^2} \\ \mathit{f}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{22}\mathbf{ }\mathbf{Hz} \end{gathered}$$

which is much higher than the limits of vibrations induced by walking

f >>3 Hz 

the slab satisfies the frequency limits.

Problem b)

The natural frequency when one edge is supported by the given beam:

See Table 10.8 first row.

$$\begin{gathered} f^2={\left(\frac{1}{f_x^2+f_{\mathrm{t}}^2-0.25f_{66}^2}+\frac{1}{f_{EI}^2}\right)}^{-1}+f_y^2+0.25f_{66}^2 \\ \mathrm{where} \\ f_{EI}^2=3\frac{{\pi }^{2}EI}{4m{L}_x{L}_y^4}=3\frac{{\mathrm{\pi }}^2\times 9.375\times {10}^7}{4\times 407.7\times 6.0\times 5.0^4}=453.85 \frac{1}{{\sec }^2} \\ {f}_x^2=\frac{{\pi }^{2}D}{4m{L}_x^4}=\frac{{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 6.0^4}=49.8\frac{1}{{\sec }^2} \\ {f}_y^2=\frac{{\pi }^{2}D}{4m{L}_y^4}=\frac{{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 5.0^4}=103.2\frac{1}{{\sec }^2} \\ {f}_{66}^2=\frac{4{\pi }^{2}D{\displaystyle \frac{\left(1-\nu \right)}{2}}}{4m{L}_x^2{L}_y^2}=\frac{2{\pi }^2\times 1.067\times {10}^7\left(1-0.2\right)}{4\times 407.7\times 6.0^2\times 5.0^2}=114.75\frac{1}{{\sec }^2} \\ {f}_{\mathrm{t}}^2=\frac{2{\pi }^{2}D}{4m{L}_x^2{L}_y^2}=\frac{2{\pi }^2\times 1.067\times {10}^7}{4\times 407.7\times 6^4}=143.4\frac{1}{{\sec }^2} \end{gathered}$$

$$\begin{gathered} \to \mathit{f}=\sqrt{{\left(\frac{1}{f_x^2+f_{\mathrm{t}}^2-0.25f_{66}^2}+\frac{1}{f_{EI}^2}\right)}^{-1}+f_y^2+0.25f_{66}^2} =\mathbf{ } \\ =\sqrt{{\left(\frac{1}{49.8+143.4-0.25\times 114.7}+\frac{1}{453.8}\right)}^{-1}+103.3+0.25\times 114.7}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{Hz} \end{gathered}$$