3.4 Symmetrical vs. Non-Symmetrical Bridle Leg Tension & Center of Gravity Alignment

Symmetrical vs. Non-Symmetrical Bridle Leg Tension & Center of Gravity Alignment

Core Principle of Moments: In any non-symmetrical rigging arrangement, the sling leg closest to the Center of Gravity (CG) carries the majority of the weight. To calculate the vertical load share on any single pick point, multiply the total load weight by the distance from the CG to the opposite pick point, divided by the total span between pick points.

In standard symmetrical rigging, the Center of Gravity lies equidistant between identical pick points, resulting in perfectly equal load sharing (W / N). However, real-world industrial loads—such as pumps with heavy electric motors, gearboxes with offset drives, transformers with internal core offsets, and fabricated machinery skids—frequently have an off-center Center of Gravity.

When rigging an off-center load, assuming equal load sharing is a critical error that can result in immediate sling failure, violent load shifting, or structural tipping.

---

The Principle of Moments & Vertical Share Derivation

Static equilibrium requires two conditions to be met simultaneously when a load is suspended:

1. Σ Fy = 0 (The sum of upward vertical forces must equal the total downward weight).
2. Σ M = 0 (The sum of clockwise moments must equal the sum of counterclockwise moments around any reference point).

                     Hook (Must be Plumb over CG)
                                 /|
                                / |
                               /  | H
                     Leg 1 (L1)/  |  \ Leg 2 (L2)
                              /   |   \
                             / θ1 | θ2 \
       [Pick Point 1] ------+-----+-----+------ [Pick Point 2]
                            |< D1 >|< D2 >|
                            |<---- Span (S) --->|
                                   [CG]
                                Total Weight (W)

Mathematical Formulation

Let:

• W = Total weight of the load
• S = Total span between Pick Point 1 and Pick Point 2 (S = D1 + D2)
• D1 = Horizontal distance from Pick Point 1 to the Center of Gravity
• D2 = Horizontal distance from Pick Point 2 to the Center of Gravity
• V1 = Vertical load share supported by Pick Point 1
• V2 = Vertical load share supported by Pick Point 2

Taking moments about Pick Point 2 (setting sum of moments to zero):

• V1 * S = W * D2 -> V1 = W * (D2 / S)

Taking moments about Pick Point 1:

• V2 * S = W * D1 -> V2 = W * (D1 / S)

[!NOTE] Notice the inverse relationship: Pick Point 1's share is proportional to distance D2, and Pick Point 2's share is proportional to distance D1. The closer a pick point is to the CG, the larger its share of the load.

---

The Two-Stage Non-Symmetrical Tension Calculation Procedure

Calculating sling tension in a non-symmetrical lift requires two distinct, sequential stages:

1. Stage 1: Calculate the Vertical Share (V) for each pick point using the moment equations:
   • V1 = W * (D2 / S)
   • V2 = W * (D1 / S)
2. Stage 2: Calculate the Leg Tension (T) for each leg independently by multiplying that leg's vertical share by its specific Load Angle Factor:
   • T1 = V1 * LAF1 = V1 * (L1 / H1) = V1 / sin(θ1)
   • T2 = V2 * LAF2 = V2 * (L2 / H2) = V2 / sin(θ2)

---

Comprehensive Worked Numerical Example: Off-Center Industrial Generator

Let us analyze a real-world complex rigging problem with complete step-by-step numbers.

Lift Specifications

• Total Load Weight (W): 18,000 lbs
• Total Span Between Lugs (S): 20.0 ft
• Center of Gravity Location: Located 6.0 ft from Lug 1 (D1 = 6.0 ft) and 14.0 ft from Lug 2 (D2 = 14.0 ft).
• Crane Hook Vertical Height (H): The crane hook is positioned directly above the CG at a vertical height of H = 8.0 ft above the pick-point plane.

                            Crane Hook
                                /|
                               / |
                              /  | H = 8.0 ft
                    L1 = 10.0'/  |  \ L2 = 16.12'
                             /   |   \
                            / θ1 | θ2 \
            [Lug 1] -------+-----+-----+------- [Lug 2]
                           | 6.0'| 14.0'|
                           |<-- Span = 20.0' -->|
                                 [CG]
                               18,000 lbs

Step 1: Calculate Vertical Load Shares (V1 and V2)

• V1 = W * (D2 / S) = 18,000 lbs * (14.0 ft / 20.0 ft) = 18,000 * 0.70 = 12,600 lbs
• V2 = W * (D1 / S) = 18,000 lbs * (6.0 ft / 20.0 ft) = 18,000 * 0.30 = 5,400 lbs
• Check: V1 + V2 = 12,600 lbs + 5,400 lbs = 18,000 lbs (Matches total weight)

Step 2: Determine Sling Leg Geometry (L1, L2, θ1, θ2)

Because the hook is centered over the CG (H = 8.0 ft):

• Leg 1 Geometry:

  • Horizontal run = D1 = 6.0 ft
  • Vertical rise = H = 8.0 ft
  • Sling Length L1 = √(6.0² + 8.0²) = √(36 + 64) = √100 = 10.0 ft
  • LAF1 = L1 / H = 10.0 ft / 8.0 ft = 1.250
  • Horizontal Angle θ1 = arctan(8.0 / 6.0) = 53.13°

• Leg 2 Geometry:

  • Horizontal run = D2 = 14.0 ft
  • Vertical rise = H = 8.0 ft
  • Sling Length L2 = √(14.0² + 8.0²) = √(196 + 64) = √260 ≈ 16.12 ft
  • LAF2 = L2 / H = 16.12 ft / 8.0 ft = 2.015
  • Horizontal Angle θ2 = arctan(8.0 / 14.0) = 29.74° ≈ 30°

Step 3: Calculate Actual Sling Leg Tensions (T1 and T2)

• T1 = V1 * LAF1 = 12,600 lbs * 1.250 = 15,750 lbs
• T2 = V2 * LAF2 = 5,400 lbs * 2.015 = 10,881 lbs

Critical Engineering Insights from Example

1. Leg 1 (Close to CG): Carries 70% of the vertical weight (12,600 lbs) and has a steep angle (53.1°), resulting in 15,750 lbs of tension.
2. Leg 2 (Far from CG): Carries only 30% of the vertical weight (5,400 lbs), but because its horizontal angle is shallow (29.7°), its Load Angle Factor is 2.015, doubling its tension to 10,881 lbs.
3. Rigging Selection: Sling 1 must have a WLL >= 16,000 lbs (8 tons), while Sling 2 must have a WLL >= 11,000 lbs (5.5 tons). Note that Leg 2 is right at the 30° prohibition threshold; to improve safety, the hook height H should be increased by lengthening both slings.

---

Center of Gravity Hook Alignment & Load Stability

Before hoisting any load, the crane hook MUST be positioned directly over the Center of Gravity.

  HOOK ALIGNED OVER CG (STABLE)        HOOK OFF-CENTER FROM CG (UNSTABLE)
              Hook                                 Hook
               |                                     |
              / \                                   / \
             /   \                                 /   \
            /  *  \                               /     \     *
       ----+-- CG -+----                     ----+-------+-- CG --
        Load lifts level                      Load violently tilts and swings
                                              upon leaving ground!

Consequences of Off-Center Hook Placement

If the crane hook is positioned over the physical center of the load rather than the Center of Gravity:

1. Violent Tilting on Liftoff: As soon as the load leaves the ground, it will instantly rotate until the Center of Gravity settles directly plumb beneath the crane hook apex.
2. Load Swing and Dynamic Shock: The lateral shift creates a swinging pendulum motion that introduces dynamic shock loading to the crane boom and rigging hardware.
3. Sling Slippage or Unhooking: Rapid rotation can cause loose slings to slip off unshouldered lugs or roll out of basket hitches.

---

Hardware for Leveling and Load Equalization

To rig non-symmetrical loads safely while keeping the load level and the crane hook centered over the CG, riggers use specialized adjustable rigging hardware: