Problem 10.11. Vibration of a steel-concrete composite floor – neglecting 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.

Show deflectionsHide deflection

See Table 3.5.

$$\begin{gathered} w_{\mathrm{b}}=\frac{5}{384}\frac{mgb{L}_x^4}{EI}=\frac{5}{384}\frac{540\times 9.81\times 2.4\times 8.0^4}{6.42\times {10}^8}=0.001056 \mathrm{m}=1.056 \mathrm{mm} \\ {w}_{\mathrm{s}}=\frac{1}{384}\frac{mg{b}^4}{D}=\frac{5}{384}\frac{540\times 9.81\times 2.4^4}{9.3\times {10}^6}=4.92\times {10}^{–6} \mathrm{m}=0.0492 \mathrm{mm} \end{gathered}$$

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

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

Show eigenfrequencyHide eigenfrequency

Eq.(8-57).

$$\begin{gathered} w=w_{\mathrm{b}}+w_{\mathrm{s}}=1.056+0.0492=1.105 \mathrm{mm} \\ {\mathit{f}}_{\mathbf{n}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{1.105}}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{Hz} \end{gathered}$$

Problem b)

Step 1. Calculate eigenfrequency of the beam and that of the slab separately.

Show eigenfrequencies of beam and slabHide eigenfrequencies

Eigenfrequency of the hinged beam:

Eq.(10-102).

$f_x^2=\frac{{\mathrm{\pi }}^{2}EI}{4mb{L}_x^4}=\frac{{\mathrm{\pi }}^2\times 6.42\times {10}^8}{4\times 540\times 2.4\times 8.0^4}=298.4 \frac{1}{{\sec }^2}$

Eigenfrequency of the hinged slab:

Eq.(10-103).

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

Step 2. Draw primary vibration mode. Calculate the eigenfrequency.

Show eigenfrequencyHide eigenfrequency

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.

Eq.(10-104).

$$\begin{gathered} f_{\mathrm{I}}^2={\left(\frac{1}{f_x^2}+\frac{1}{5.1f_{\mathrm{s}}^2}\right)}^{–1}={\left(\frac{1}{298.4}+\frac{1}{5.1\times 1281}\right)}^{–1}=242.0\frac{1}{{\sec }^2} \\ {\mathit{f}}_{\mathbf{I}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{89}\mathbf{ }\mathbf{Hz} \end{gathered}$$

Step 3. Draw second vibration mode. Calculate the eigenfrequency.

Show eigenfrequencyHide eigenfrequency

See Example 10.7.

Problem a)

The deflections of the beam and the slab are

See Table 3.5.

$$\begin{gathered} w_{\mathrm{b}}=\frac{5}{384}\frac{mgb{L}_x^4}{EI}=\frac{5}{384}\frac{540\times 9.81\times 2.4\times 8.0^4}{6.42\times {10}^8}=0.001056 \mathrm{m}=1.056 \mathrm{mm} \\ {w}_{\mathrm{s}}=\frac{1}{384}\frac{mg{b}^4}{D}=\frac{5}{384}\frac{540\times 9.81\times 2.4^4}{9.3\times {10}^6}=4.92\times {10}^{–6} \mathrm{m}=0.0492 \mathrm{mm} \end{gathered}$$

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

The eigenfrequency of the slab is approximated by the superposition of the above deflections.

Eq.(8-57).

$$\begin{gathered} w=w_{\mathrm{b}}+w_{\mathrm{s}}=1.056+0.0492=1.105 \mathrm{mm} \\ {\mathit{f}}_{\mathbf{n}}=\frac{18}{\sqrt{w}}=\frac{18}{\sqrt{1.105}}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{Hz} \end{gathered}$$

Problem b)

First the eigenfrequency of the beam and that of the slab are calculated separately.

Eigenfrequency of the hinged beam:

Eq.(10-102).

$f_x^2=\frac{{\mathrm{\pi }}^{2}EI}{4mb{L}_x^4}=\frac{{\mathrm{\pi }}^2\times 6.42\times {10}^8}{4\times 540\times 2.4\times 8.0^4}=298.4 \frac{1}{{\sec }^2}$

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 primary 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.

Eq.(10-104).

$$\begin{gathered} f_{\mathrm{I}}^2={\left(\frac{1}{f_x^2}+\frac{1}{5.1f_{\mathrm{s}}^2}\right)}^{–1}={\left(\frac{1}{298.4}+\frac{1}{5.1\times 1281}\right)}^{–1}=242.0\frac{1}{{\sec }^2} \\ {\mathit{f}}_{\mathbf{I}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{89}\mathbf{ }\mathbf{Hz} \end{gathered}$$

In the second vibration mode only the slab vibrates between the beam, the beams remain straight as it is shown in the Figure.

The vibration shape corresponds to hinged supports at the ends of the slab.

${\mathit{f}}_{\mathbf{II}}=f_{\mathrm{s}}^{}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{79}\mathbf{ }\mathbf{Hz}$

which is mush higher than the eigenfrequency of the primary mode.