8.4 Center of Gravity Off-Center & Asymmetric Load Calculations

In real-world industrial rigging, loads are rarely perfectly symmetrical or uniformly dense. Heavy internal components—such as electric motors, gearboxes, refractory linings, or liquid pockets—shift the Center of Gravity (CG) away from the geometric center of the object. Lifting an asymmetric load without calculating off-center CG distribution leads to severe load tilting, unequal sling leg tension, unexpected load swings, and potential rigging failure.

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Physics of Center of Gravity (CG) and Center of Lift (CL)

The Center of Gravity (CG) is the theoretical point through which the entire weight of a body acts in three-dimensional space. The Center of Lift (CL) is the point in space located directly below the crane hook apex where the sling forces converge.

The Fundamental Law of Static Suspension

LAW OF SUSPENSION: When any body is suspended freely from a single crane hook, it will automatically rotate and tilt until its Center of Gravity sits on a vertical line running directly beneath the crane hook.

If the pick points are placed symmetrically around the geometric center of an asymmetric load, the heavier end will sink and the lighter end will rise until the CG aligns vertically under the hook. This rotation causes the sling leg attached to the heavy end to become much steeper, drastically increasing tension in that leg while the light-end leg goes slack.

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2-Point Load Sharing Formula (Static Equilibrium)

To calculate how total load weight ($W$) is shared between two lifting points (Point A and Point B) when the Center of Gravity is off-center, we apply static moment equilibrium about each pick point.

Let:

• $W = \text{Total gross load weight}$
• $L = d_A + d_B = \text{Total horizontal distance between Pick Points A and B}$
• $d_A = \text{Horizontal distance from Pick Point A to the Center of Gravity}$
• $d_B = \text{Horizontal distance from Pick Point B to the Center of Gravity}$
• $W_A = \text{Vertical load share carried by Pick Point A}$
• $W_B = \text{Vertical load share carried by Pick Point B}$

Derivation from Moment Equilibrium

Taking moments about Pick Point B ($\sum M_B = 0$):

$W_A \cdot (d_A + d_B) - W \cdot d_B = 0 \implies W_A = W \cdot \left(\frac{d_B}{d_A + d_B}\right)$

Taking moments about Pick Point A ($\sum M_A = 0$):

$W_B = W \cdot \left(\frac{d_A}{d_A + d_B}\right)$

Key Principle of Asymmetric Distribution

Notice the inverse horizontal relationship: The pick point closest to the Center of Gravity carries the larger share of the weight. As $d_A \to 0$, $W_A \to W$.

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Calculating Leg Tension for Off-Center CG

Once the vertical load shares ($W_A$ and $W_B$) are determined, the actual tension in each sling leg ($T_A$ and $T_B$) must be calculated by incorporating the sling leg angles ($\theta_A$ and $\theta_B$ measured to the vertical):

$T_A = \frac{W_A}{\cos\theta_A} = \frac{W \cdot d_B}{(d_A + d_B) \cdot \cos\theta_A}$

$T_B = \frac{W_B}{\cos\theta_B} = \frac{W \cdot d_A}{(d_A + d_B) \cdot \cos\theta_B}$

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Engineering Methods for Managing Asymmetric Lifts

When handling asymmetric loads, riggers employ three standard engineering solutions:

1. Adjusting Sling Leg Lengths (Leg Shorteners / Turnbuckles)

To keep an asymmetric load perfectly level during lifting, the crane hook must be positioned directly above the CG before hoisting. Because the CG is closer to Point A than Point B, the distance from Point A to the hook is shorter than from Point B to the hook. Riggers use chain shortening clutches or turnbuckles to shorten Leg A and lengthen Leg B. This aligns the hook over the CG while holding the load structure perfectly horizontal.

2. De-rating Sling Assemblies or Selecting Asymmetric Capacities

If equal-length sling legs are used and the load is allowed to tilt, or if a multi-leg (3 or 4 leg) sling is used on an asymmetric load, the load distribution must be assumed non-uniform. Standard LEEA practice mandates:

• Calculate maximum tension in the heavily loaded leg.
• Select a sling size where each individual leg WLL is capable of supporting the maximum tension ($T_{max}$), effectively ignoring assistance from lighter legs.

3. Multi-Leg (3 & 4 Leg) Asymmetric Rule

For 3-leg and 4-leg slings on asymmetric loads, statics shows that two legs closest to the CG may carry up to 80% to 100% of the entire load weight. The rigger must evaluate load distribution in two orthogonal planes ($X$ and $Y$ axes) to calculate individual leg loads.

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Tailing Lifts with Two Cranes (Upending Operations)

A tailing lift is a complex, high-risk operation where two cranes (a Main Crane and a Tailing Crane) work in tandem to tilt or upend a heavy vessel, column, or vessel from a horizontal orientation to vertical (or vice-versa).

Dynamic Load Distribution during Tilting

As the vessel transitions from horizontal ($0^\circ$) to vertical ($90^\circ$), the distribution of load weight between the Main Crane ($W_{main}$) and Tailing Crane ($W_{tail}$) shifts dynamically as a function of tilt angle ($\phi$):

1. Horizontal Position ($\phi = 0^\circ$): Static 2-point load sharing applies based on CG distance: $W_{main} = W \cdot \left(\frac{d_{tail}}{L}\right), \quad W_{tail} = W \cdot \left(\frac{d_{head}}{L}\right)$

2. Tilting Transition ($0^\circ < \phi < 90^\circ$): As the Main Crane lifts the head lug upward, the Center of Gravity shifts horizontally toward the main head lug. The load share on the tailing crane steadily decreases while the load share on the main crane increases.

3. Vertical Position ($\phi = 90^\circ$): When the vessel reaches true vertical, the Center of Gravity is suspended directly beneath the Main Crane hook. $W_{main} = 100\% \text{ of Total Weight } W, \quad W_{tail} = 0$

CRITICAL TAILING SAFETY: Riggers and crane operators must continuously monitor load indicator displays during upending. The main crane must be rated for $100%$ of the gross vessel weight, while the tailing crane and tailing lug must be rated for the maximum horizontal load share plus impact factors. Swivel hoist rings or trunnions must be used to prevent side-loading crane blocks during rotation.

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Step-by-Step Worked Engineering Examples

Worked Example 1: Asymmetric Load Sharing & Leg Tension

Problem: A skid containing a heavy compressor motor has a total mass of $12.0\text{ tonnes}$ and a total length of $6.0\text{ metres}$. The Center of Gravity is located $1.8\text{ metres}$ from Pick Point A and $4.2\text{ metres}$ from Pick Point B. The load is lifted using a 2-leg chain sling with legs adjusted so that each leg forms an angle of $30^\circ$ to the vertical while keeping the skid level. Calculate:

1. Vertical load share at Pick Point A ($W_A$) and Pick Point B ($W_B$).
2. Actual leg tension in Leg A ($T_A$) and Leg B ($T_B$).

Solution:

1. Calculate Vertical Load Shares: $L = d_A + d_B = 1.8\text{ m} + 4.2\text{ m} = 6.0\text{ metres}$ $W_A = W \cdot \left(\frac{d_B}{L}\right) = 12.0\text{ t} \times \left(\frac{4.2}{6.0}\right) = 12.0 \times 0.70 = 8.40\text{ tonnes}$ $W_B = W \cdot \left(\frac{d_A}{L}\right) = 12.0\text{ t} \times \left(\frac{1.8}{6.0}\right) = 12.0 \times 0.30 = 3.60\text{ tonnes}$ Check: $W_A + W_B = 8.40 + 3.60 = 12.00\text{ tonnes}$.

2. Calculate Sling Leg Tensions ($\theta = 30^\circ, \cos(30^\circ) = 0.8660$): $T_A = \frac{W_A}{\cos(30^\circ)} = \frac{8.40\text{ t}}{0.8660} = 9.6998 \approx 9.70\text{ tonnes}$ $T_B = \frac{W_B}{\cos(30^\circ)} = \frac{3.60\text{ t}}{0.8660} = 4.1570 \approx 4.16\text{ tonnes}$

3. Engineering Conclusion: Leg A experiences $9.70\text{ tonnes}$ of tension—more than double the tension in Leg B ($4.16\text{ t}$). The rigger must select a chain size for Leg A rated for at least $9.70\text{ tonnes}$ WLL (e.g., 16mm Grade 80 chain, $WLL = 8.0\text{ t}$ is insufficient; 20mm Grade 80, $WLL = 12.5\text{ t}$ is required).

Worked Example 2: Two-Crane Tailing Lift Upending Load Allocation

Problem: A $40.0\text{-tonne}$ distillation column $20.0\text{ metres}$ in length is horizontal. Its CG is located $8.0\text{ metres}$ from the Head Lug (Main Crane) and $12.0\text{ metres}$ from the Tail Lug (Tailing Crane).

1. Calculate initial crane load sharing in the horizontal position.
2. Determine the required capacity of the Main Crane when upending is complete.

Solution:

1. Horizontal Load Sharing ($\phi = 0^\circ$): $W_{main} = W \cdot \left(\frac{d_{tail}}{L}\right) = 40.0\text{ t} \times \left(\frac{12.0\text{ m}}{20.0\text{ m}}\right) = 40.0 \times 0.60 = 24.0\text{ tonnes}$ $W_{tail} = W \cdot \left(\frac{d_{head}}{L}\right) = 40.0\text{ t} \times \left(\frac{8.0\text{ m}}{20.0\text{ m}}\right) = 40.0 \times 0.40 = 16.0\text{ tonnes}$

2. Upending Completion ($\phi = 90^\circ$): When fully vertical, the tailing crane is disconnected and the Main Crane carries $100%$ of the load.

   • Main Crane Requirement: Must be rated for at least $40.0\text{ tonnes}$ plus dynamic safety factor.