Offshore crane shock absorber

Offshore lifting capacity is governed by classification society rules. The fundamental rule is that the design load should not be exceeded. The design load is given by:
${\displaystyle F_{m}=\psi m_{s}g}$

Where:
${\displaystyle F_{m}}$ - Crane design load
${\displaystyle \psi }$ - Dynamic coefficient
${\displaystyle m_{s}}$ - Safe working load
${\displaystyle g}$ - Acceleration of gravity

The classification societies require that ${\displaystyle \psi }$ ≥1.3 for offshore cranes. This means that ${\displaystyle m_{s}}$ cannot exceed ${\displaystyle {\frac {F_{m}}{\psi g}}}$ even for platform lifts.^[2][3][4]

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To calculate the dynamic load the classification societies use energy equations. The kinetic energy of the load is expressed as:
${\displaystyle E_{k}={\frac {1}{2}}m_{s}v_{r}^{2}}$, or alternatively as potential energy ${\displaystyle E_{p}=m_{s}gh}$.

Where:
${\displaystyle E_{k}}$ - Kinetic energy of the load
${\displaystyle v_{r}}$ - Relative velocity between crane hook and load

${\displaystyle h}$ - Free fall distance of payload (typically due to slack slings)

Energy absorbed by the crane is given by:
${\displaystyle E_{c}={\frac {1}{2}}ky^{2}}$

Where:
${\displaystyle E_{c}}$ - Spring energy stored in crane structure
${\displaystyle k}$ - Stiffness of offshore crane
${\displaystyle y}$ - Vertical displacement of crane hook

The spring force is given by:
${\displaystyle F_{c}=ky}$

Combining the two previous equations yields an alternative description of the energy stored in the crane structure:
${\displaystyle E_{c}={\frac {1}{2}}k({\frac {F_{c}}{k}})^{2}={\frac {F_{c}^{2}}{2k}}}$

During an offshore lift the crane does not start to absorb energy from the load before the force in the crane wire exceeds the static weight of the load, which means that we can write the energy absorption of the crane as:
${\displaystyle E_{c}={\frac {(m_{s}\psi g-m_{s}g)^{2}}{2k}}}$

Setting ${\displaystyle E_{c}}$ equal to ${\displaystyle E_{k}}$: ${\displaystyle {\frac {(m_{s}\psi g-m_{s}g)^{2}}{2k}}={\frac {1}{2}}m_{s}v_{r}^{2}}$

And solving for ${\displaystyle \psi }$:
${\displaystyle \psi =1+{\frac {v_{r}}{g}}{\sqrt {\frac {k}{m}}}}$

The dynamic factor will always be larger than 1 because there will always be a velocity difference between the crane hook and the load during offshore lifts. The value of ${\displaystyle \psi }$ determines the lifting capacity as a function of ${\displaystyle v_{r}}$ which is dependent on the significant wave height. If the offshore crane has a shock absorber mounted it will absorb energy according to:^[1]
${\displaystyle E_{a}=\eta \mu Sm_{s}g(\psi -1)}$

Where:
${\displaystyle E_{a}}$ - Energy absorbed by shock absorber
${\displaystyle \mu }$ - Stroke utilization (safety factor), standard value 0.9
${\displaystyle \eta }$ - Efficiency of shock absorber, typically 0.5 to 0.7 and depending on payload speed, and 0.9 and beyond for pure shock absorbers
${\displaystyle S}$ - Stroke length

Adding this to the energy balance yields:

${\displaystyle {\frac {(m_{s}\psi g-m_{s}g)^{2}}{2k}}+\eta \mu Sm_{s}g(\psi -1)={\frac {1}{2}}m_{s}v_{r}^{2}}$

Again solving for ψ:
${\displaystyle \psi =1+{\frac {{\sqrt {(\mu \eta Sk)^{2}+m_{s}kv_{r}^{2}}}-\mu \eta Sk}{m_{s}g}}}$

The plot to the right shows the effect of the shock absorber stroke length.