Problem 8.2. Additional mass – frequency of beam with hinged ends

Problem a)

Additional mass, M [kg]=

Problem a)

Additional mass, M [kg]=

Problem a)

Step 1.  Calculate the maximum deflection of the beam with distributed mass only. Approximate natural frequency with the aid of the deflection. 

Show frequency, fm Hide frequency, fm

See Table 3.5. and Eq.(8-57)

$$\begin{gathered} w_m=\frac{5mg{L}^4}{384EI}=\frac{5\times 60\times 10\times 6.0^4}{384\times 6000\times {10}^3}{10}^3=1.69 \mathrm{mm} \\ {f}_m^{}=\frac{18}{\sqrt{w_m}}=\frac{18}{\sqrt{1.69}}=13.8 \mathrm{Hz} \end{gathered}$$

Step 2.  Applying superposition give the maximum deflection of the beam with the distributed and the additional concentrated masses. Approximate natural frequency with the aid of the deflection. 

Show frequency, fm+MHide frequency, fm+M

See Table 3.5. and Eq.(8-57)

$$\begin{gathered} w_{m+M}=w_m+\frac{Mg{L}^3}{48 EI} \\ {f}_{m+M}^{}=\frac{18}{\sqrt{w_{m+M}}}=\frac{18}{\sqrt{w_m+\frac{Mg{L}^3}{48 EI}}} \end{gathered}$$

Step 3.  Express additional mass, M from the equality f_m+M = f_m/2.

Show mass valueHide mass value

Problem b)

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

Show frequency, fm Hide frequency, fm

Eq.(8-49)

$f_m^{}=\sqrt{\frac{{\pi }^{2}EI}{4m{L}^4}}=\sqrt{\frac{{\pi }^2\times 6000\times {10}^3}{4\times 60\times 6.0^4}}=13.8 \mathrm{Hz}$

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

Show frequency, fMHide frequency, fM

Eq.(8-15)

$f_M^{}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{M}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{48EI}{M{L}^3}} \mathrm{where} k=\frac{P}{u}=\frac{P}{{\displaystyle \frac{P{L}^3}{48EI}}}=\frac{48EI}{L^3}$

Step 3.  Approximate fundamental frequency with Dunkerley’s summation theorem.

Show frequency, fM+mHide frequency, fM+m

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

$\frac{1}{f_{M+m}^2}=\frac{{\displaystyle 1}}{{\displaystyle f_m^2}}+\frac{{\displaystyle 1}}{{\displaystyle f_M^2}}$

Step 4.  Express additional mass, M from the equality f_m+M = f_m/2.

Show mass valueHide mass value

Problem a)

Maximum deflection of the beam with distributed mass only is 

See Table 3.5.

Its natural frequency can be approximated as:

Eq.(8-57)

$f_m^{}=\frac{18}{\sqrt{w_m}}=\frac{18}{\sqrt{1.69}}=13.8 \mathrm{Hz}$

Applying superposition the maximum deflection of the beam with the distributed and the additional concentrated masses is also calculated. 

See Table 3.5.

$w_{m+M}=w_m+\frac{Mg{L}^3}{48 EI}$

The approximate natural frequency is:

$f_{m+M}^{}=\frac{18}{\sqrt{w_{m+M}}}=\frac{18}{\sqrt{w_m+\frac{Mg{L}^3}{48 EI}}}$

Additional mass, M is expressed from the equality f_m+M = f_m/2:

$$\begin{gathered} \frac{f_m}{2}= f_{m+M}^{}=\frac{18}{\sqrt{w_m+{\displaystyle \frac{Mg{L}^3}{48EI}}}} \\ \to \mathit{M}=\frac{48EI}{g{L}^3}\left({\left(\frac{2\times 18}{f_m}\right)}^2-w_m\right)=\frac{48\times 6000}{10\times 6.0^3}\left({\left(\frac{2\times 18}{13.85}\right)}^2-1.69\right)\mathbf{=}\mathbf{675}\mathbf{ }\mathbf{kg} \end{gathered}$$

Problem b)

Natural frequency of the beam with uniform mass only is the following:

Eq.(8-49)

Natural frequency of the beam with a concentrated mass only is

Eq.(8-15)

$f_M^{}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{M}}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{48EI}{M{L}^3}} \mathrm{where} k=\frac{P}{u}=\frac{P}{{\displaystyle \frac{P{L}^3}{48EI}}}=\frac{48EI}{L^3}$

The fundamental frequency is approximated by Dunkerley’s summation theorem.

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

Additional mass, M is expressed from the equality f_m+M = f_m/2.

$$\begin{gathered} \frac{f_m}{2}=f_{m+M}^{}=\left(\frac{1}{f_m^2}+\frac{1}{f_M^2}\right){ }^{-0.5} \to \frac{4}{f_m^2}=\frac{1}{f_m^2}+\frac{1}{f_M^2} \to f_M^2=\frac{f_m^2}{3}=\frac{1}{4{\pi }^2}\frac{48EI}{M{L}^3} \\ \to \mathit{M}=\frac{48EI}{4{\mathrm{\pi }}^2{L}^3}\frac{3}{f_m^2}=\frac{48\times 6000\times {10}^3}{10\times 6.0^3}\frac{3}{13.8^2}\mathbf{=}\mathbf{532}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kg} \end{gathered}$$

Note that the second solution gives more accurate result.