Problem 10.12. Vibration of a steel-concrete composite floor -including beam’s weight

Problem a)

Approximate eigenfrequency calculated from the deflecions, f_n [Hz]=

Problem b)

Approximate eigenfrequency calculated by the summation theorems, f_n [Hz]=

See Example 10.7.

Problem a)

Step 1. Determine the deflections of the beam and the slab.

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Step 2. Approximate the eigenfrequency of the slab with the superposition of the above deflections.

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Eq.(8-57).

$$\begin{gathered} w=w_{\mathrm{b}}+w_{\mathrm{s}}=1.16+0.0492=1.207 \mathrm{mm} \\ {\mathit{f}}_{\mathbf{n}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{1.207}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{38}\mathbf{ }\mathbf{Hz} \end{gathered}$$

Problem b)

Step 1. Approximate eigenfrequency of the beam using Dunkerley’s theorem.

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Approximate eigenfrequency of the hinged beam with two masses is

Eq.(10-102) and Table 8.3.

$$\begin{gathered} f_x^2={\left(\frac{1}{f_1^2}+\frac{1}{f_2^2}\right)}^{–1}={\left(\frac{4mb{L}_x^4}{{\mathrm{\pi }}^{2}EI}+\frac{4m_b{L}_x^4}{{\mathrm{\pi }}^{2}EI}\right)}^{–1}=\frac{{\mathrm{\pi }}^{2}EI}{4\left(mb+m_b\right)L_x^4}= \\ =\frac{{\mathrm{\pi }}^2\times 6.42\times {10}^8}{4\times \left(540\times 2.4+150\right)\times 8.0^4}=272.2 \frac{1}{{\sec }^2} \end{gathered}$$

Step 2. Determine the eigenfrequency of the whole plate using Föppl’s approximation.

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Eigenfrequency of the hinged slab:

Eq.(10-103).

$f_{\mathrm{s}}^2=\frac{{\mathrm{\pi }}^{2}D}{4m{b}^4}=\frac{{\mathrm{\pi }}^2\times 9.3\times {10}^6}{4\times 540\times 2.4^4}=1281 \frac{1}{{\sec }^2}$

In the relevant vibration mode both the beams and the plate undergo vibration as it is shown in the Figure.

The eigenfrequency is approximated by Föppl’s theorem. The edges of the slab is assumed to be built-in as the deformed shape shows.

$$\begin{gathered} f_{\mathrm{I}}^2={\left(\frac{1}{f_x^2}+\frac{1}{5.1f_{\mathrm{s}}^2}\right)}^{–1}={\left(\frac{1}{272.2}+\frac{1}{5.1\times 1281}\right)}^{–1}=224.5\frac{1}{{\sec }^2} \\ {\mathit{f}}_{\mathbf{I}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{16}\mathbf{ }\mathbf{Hz} \end{gathered}$$

See Example 10.7.

Problem a)

The beam’s deflection is modified by the beam’s weight.

$$\begin{gathered} w_{\mathrm{b}}=\frac{5}{384}\frac{\left(mb+m_b\right)g{L}_x^4}{EI}=\frac{5}{384}\frac{(540\times 2.4+150)\times 9.81\times 8.0^4}{6.42\times {10}^8}\times {10}^3=1.16 \mathrm{mm} \\ {w}_{\mathrm{s}}=\frac{1}{384}\frac{mg{b}^4}{D}=\frac{5}{384}\frac{540\times 9.81\times 2.4^4}{9.3\times {10}^6}\times {10}^3=4.92\times {10}^{–6} \mathrm{m}=0.0492 \mathrm{mm} \end{gathered}$$

The multispan slab is assumed to be a 1 m wide single span beam built-in at both ends.

Approximate eigenfrequency of the slab is given using the superposition of the above deflections.

Eq.(8-57).

$$\begin{gathered} w=w_{\mathrm{b}}+w_{\mathrm{s}}=1.16+0.0492=1.207 \mathrm{mm} \\ {\mathit{f}}_{\mathbf{n}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{1.207}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{38}\mathbf{ }\mathbf{Hz} \end{gathered}$$

Problem b)

Approximate eigenfrequency of the hinged beam with two masses is

Eq.(10-102) and Table 8.3.

$$\begin{gathered} f_x^2={\left(\frac{1}{f_1^2}+\frac{1}{f_2^2}\right)}^{–1}={\left(\frac{4mb{L}_x^4}{{\mathrm{\pi }}^{2}EI}+\frac{4m_b{L}_x^4}{{\mathrm{\pi }}^{2}EI}\right)}^{–1}=\frac{{\mathrm{\pi }}^{2}EI}{4\left(mb+m_b\right)L_x^4}= \\ =\frac{{\mathrm{\pi }}^2\times 6.42\times {10}^8}{4\times \left(540\times 2.4+150\right)\times 8.0^4}=272.2 \frac{1}{{\sec }^2} \end{gathered}$$

Eigenfrequency of the hinged slab:

Eq.(10-103).

$f_{\mathrm{s}}^2=\frac{{\mathrm{\pi }}^{2}D}{4m{b}^4}=\frac{{\mathrm{\pi }}^2\times 9.3\times {10}^6}{4\times 540\times 2.4^4}=1280.8 \frac{1}{{\sec }^2}$

In the relevant vibration mode both the beams and the plate undergo vibration as it is shown in the Figure.

The eigenfrequency is approximated by Föppl’s theorem. The edges of the slab is assumed to be built-in as the deformed shape shows.

$$\begin{gathered} f_{\mathrm{I}}^2={\left(\frac{1}{f_x^2}+\frac{1}{5.1f_{\mathrm{s}}^2}\right)}^{–1}={\left(\frac{1}{272.2}+\frac{1}{5.1\times 1281}\right)}^{–1}=224.5\frac{1}{{\sec }^2} \\ {\mathit{f}}_{\mathbf{I}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{16}\mathbf{ }\mathbf{Hz} \end{gathered}$$