Problem 8.3. Additional mass – frequency of beam with built-in ends

Problem a)

Eigenfrequency, f [Hz]=

Problem b)

Eigenfrequency, f [Hz]=

Problem a)

Step 1.  Applying superposition give the maximum deflection of the beam with the distributed and the additional concentrated masses.

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See Table 3.5.

$w=\frac{1}{384}\frac{mg{L}^4}{EI}+\frac{1}{192}\frac{Mg{L}^3}{EI}=\frac{1\times 42.2\times 9.81\times 7.0^4}{384\times 18640\times {10}^3}{10}^3+\frac{1}{192}\frac{150\times 9.81\times 7.0^3}{18640\times {10}^3}{10}^3=0.280 \mathrm{mm}$

Step 2.  Approximate eigenfrequency with the aid of the deflection. 

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Eq.(8-57)

${\mathit{f}}_{\mathbf{m}}^{}=\frac{18}{\sqrt{w_m}}=\frac{18}{\sqrt{0.280}}\mathbf{=}\mathbf{34}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{Hz}$

Problem b)

Step 1.  Calculate natural frequency of the beam with uniform mass only.

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Eq.(8-49)

$f_m^{\mathrm{bb}}=\sqrt{5.1}{f}_m^{\mathrm{ss}}=\sqrt{5.1}\sqrt{\frac{{\pi }^{2}EI}{4m{L}^4}}=\sqrt{5.1}\sqrt{\frac{{\pi }^2\times 18640\times {10}^3}{4\times 42.2\times 7.0^4}}=48.11 \mathrm{Hz}$

Step 2.  Determine natural frequency of the beam with a concentrated mass only. 

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Equation (8-15)

$$\begin{gathered} f_M^{}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{M}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{192EI}{M{L}^3}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{192\times 18640\times {10}^3}{150\times 7.0^3}}=41.98 \mathrm{Hz} \\ \mathrm{where} k=\frac{P}{u}=\frac{P}{{\displaystyle \frac{P{L}^3}{192EI}}}=\frac{192EI}{L^3} \end{gathered}$$

Step 3.  Approximate fundamental frequency with Dunkerley’s expression.

Show frequency, fM+mHide frequency, fM+m

See Table 8.3 and Eq.(8-58).

$\frac{1}{f_{M+m}^2}=\frac{{\displaystyle 1}}{{\displaystyle f_m^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_M^2}} \to {\mathit{f}}_{\mathbf{M}\mathbf{+}\mathbf{m}}^{}={\left(\frac{1}{48.{11}^2}+\frac{1}{41.{98}^2}\right)}^{–0.5}\mathbf{=}\mathbf{31}\mathbf{.}\mathbf{63}\mathbf{ }\mathbf{Hz}$

Note that the second solution gives more accurate result.

Problem a)

Maximum deflection of the beam with the distributed and the additional concentrated masses is calculated by superposition.

See Table 3.5.

Eigenfrequency is approximated with the aid of the deflection:

Equation (8-57)

${\mathit{f}}_{\mathbf{m}}^{}=\frac{18}{\sqrt{w_m}}=\frac{18}{\sqrt{0.280}}\mathbf{=}\mathbf{34}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{Hz}$

Problem b)

First the natural frequency of the beam with uniform mass only is determined.

Equation (8-49)

Second natural frequency of the beam with a concentrated mass is calculated.

Equation (8-15)

$$\begin{gathered} f_M^{}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{M}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{192EI}{M{L}^3}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{192\times 18640\times {10}^3}{150\times 7.0^3}}=41.98 \mathrm{Hz} \\ \mathrm{where} k=\frac{P}{u}=\frac{P}{{\displaystyle \frac{P{L}^3}{192EI}}}=\frac{192EI}{L^3} \end{gathered}$$

Finally theeigenfrequency with both masses is approximated with Dunkerley’s expression.

See Table 8.3 and Equation (8-58).

Note that the second solution gives more accurate result.