A four–column semi–submersible floater is located offshore. The diameter of each column is $5~\text{m}$. Consider the total displaced weight of seawater of the semi–submersible as $4000~\text{tonnes}$. Assume added mass contribution as $50\%$ of the semi–submersible weight, and seawater density as $1025~\text{kg/m}^3$. (Acceleration due to gravity $=9.81~\text{m/s}^2$.) The natural period of oscillation of the floater in vertical mode is __________ seconds (rounded to one decimal place).
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Solution and Explanation
Four circular columns of diameter $d=5~\text{m}$ $\Rightarrow$ area per column $=\dfrac{\pi d^2}{4}=6.25\pi~\text{m}^2$.
Total $A_{wp}=4\times 6.25\pi=78.54~\text{m}^2$.
Step 2: Heave restoring coefficient.
$C_z=\rho g A_{wp}=1025\times 9.81\times 78.54=7.897\times 10^5~\text{N/m}$.
Step 3: Effective mass (structure + added mass).
Displaced weight $=4000~\text{tonnes}\Rightarrow$ structural mass $m=4.0\times 10^6~\text{kg}$.
Added mass $=0.5\,m=2.0\times 10^6~\text{kg}$.
$m_{\text{eff}}=m+m_a=6.0\times 10^6~\text{kg}$.
Step 4: Natural period in heave.
\[ T=2\pi\sqrt{\frac{m_{\text{eff}}}{C_z}} =2\pi\sqrt{\frac{6.0\times 10^6}{7.897\times 10^5}} =17.32~\text{s}\approx 17.3~\text{s}. \]
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