8.3 Determining Center of Gravity (CG) for Regular & Irregular Loads

8.3 Determining Center of Gravity (CG) for Regular & Irregular Loads

Every object hoisted by a crane has a Center of Gravity (CG). The Center of Gravity is defined as the single theoretical point in three-dimensional space at which the entire weight of the object is concentrated and perfectly balanced in all three coordinate axes ($X, Y, Z$). If an object were supported at its exact Center of Gravity, it would balance in any orientation without rotating or tilting.

In crane rigging, understanding the exact location of the CG is vital because a freely suspended load will always rotate, tilt, and shift until its Center of Gravity lies directly on the vertical line below the crane hook. If the rigger misidentifies the CG, the load will tilt uncontrollably upon lift-off, transferring unequal tensile loads into individual sling legs, causing hardware overload, and potentially dropping the load.

1. Symmetrical vs. Irregular Loads

+-----------------------------------------------------------------------------------------+
|                                 CENTER OF GRAVITY (CG)                                  |
+-----------------------------------------------------------------------------------------+
|  SYMMETRICAL UNIFORM LOAD:                       IRREGULAR / COMPOSITE LOAD:            |
|        +-------------------------+                     +-------------+                  |
|        |            |            |                     | Heavy Motor |   Light Tank     |
|        |            X CG         |                     |   (7000 lb) |    (2000 lb)     |
|        |            |            |                     +-------------+----------------+ |
|        +-------------------------+                     |  X CG                        | |
|        |<--- L/2 --->|<--- L/2 -->|                     |<-- D_cg -->|                  |
|          CG is at Geometric Centroid                     CG is Shifted Toward Heavy End |
+-----------------------------------------------------------------------------------------+

Symmetrical Objects of Uniform Density

For objects with uniform material composition and symmetrical geometric profiles (such as solid rectangular plates, uniform steel beams, solid round shafts, and concentric cylindrical tanks), the Center of Gravity coincides exactly with the geometric center (centroid):

• Uniform Rectangular Block: $\text{CG}_X = \frac{\text{Length}}{2}$, $\text{CG}_Y = \frac{\text{Width}}{2}$, $\text{CG}_Z = \frac{\text{Height}}{2}$.
• Uniform Cylinder / Pipe: $\text{CG}_X = \frac{\text{Length}}{2}$ along the centerline axis.

Irregular & Composite Objects

When an object has varying cross-sections (e.g., stepped shafts, cones), contains heavy components clustered at one end (e.g., a pump skid with motor and gearbox), or combines multiple materials (e.g., steel skid with concrete ballast), the CG shifts toward the heavier concentration of mass. Finding the CG requires applying the Principle of Moments.

2. The Principle of Moments for Locating CG

A moment (or torque) is the rotational force created by a weight acting at a horizontal distance from a reference pivot point:

$\text{Moment} = \text{Weight} \times \text{Distance} \quad (M = W \times D)$

According to the Principle of Moments (Varignon's Theorem), the total moment of a composite object about any reference axis equals the sum of the moments of its individual components about that same axis. Therefore, the distance from the reference datum to the composite Center of Gravity ($\bar{X}_{CG}$) is calculated as:

$\mathbf{\bar{X}_{CG} = \frac{\text{Total Moment}}{\text{Total Weight}} = \frac{\sum (W_i \times D_i)}{\sum W_i} = \frac{W_1 D_1 + W_2 D_2 + \dots + W_n D_n}{W_1 + W_2 + \dots + W_n}}$

Four-Step Procedure for Calculating CG

1. Establish a Reference Datum (Reference Line): Choose a clear, accessible edge or face of the object (such as the extreme left edge, mounting flange face, or base corner) as Datum $0.0\text{ ft}$.
2. Break Down into Components: Divide the object into simple geometric parts with known weights ($W_1, W_2, \dots, W_n$).
3. Determine Individual Centroid Distances ($D_i$): Measure or calculate the horizontal distance from the reference datum to the individual center of gravity of each component.
4. Construct Moment Table & Calculate $\bar{X}_{CG}$: Multiply each component weight by its distance to find individual moments ($W_i \times D_i$), sum the total moments, sum the total weight, and divide total moment by total weight.

3. Worked Example 1: Stepped Diameter Steel Shaft

+-----------------------------------------------------------------------------------------+
|                        WORKED EXAMPLE 1: STEPPED STEEL SHAFT                            |
+-----------------------------------------------------------------------------------------+
|  DATUM (0 ft)                                                                           |
|  |                                                                                      |
|  |====> [ Section 1: Dia = 12", L = 8 ft ]                                              |
|  |      Weight = 3,079 lbs | CG1 = 4.0 ft                                               |
|  |                                                                                      |
|  |                                        [ Section 2: Dia = 6", L = 12 ft ]            |
|  |                                          Weight = 1,155 lbs | CG2 = 14.0 ft          |
|  |<------------------- Composite Length = 20 ft -------------------------------------->|
|  |                                                                                      |
|  |---------> X_cg = 6.73 ft (Shifted toward heavy 12" section)                          |
+-----------------------------------------------------------------------------------------+

Problem Statement

A solid steel stepped turbine shaft consists of two sections welded end-to-end:

• Section 1 (Large End): $12\text{ inches}$ ($1.0\text{ ft}$) in diameter, $8\text{ ft}$ long. Density $= 490\text{ lb/ft}^3$.
• Section 2 (Small End): $6\text{ inches}$ ($0.5\text{ ft}$) in diameter, $12\text{ ft}$ long. Density $= 490\text{ lb/ft}^3$.
• Calculate the horizontal location of the composite CG from the extreme left face of Section 1 (Datum = $0.0\text{ ft}$).

Step-by-Step Calculation

1. Section 1 Weight and Centroid ($W_1, D_1$):

   • Volume $V_1 = \pi \times (0.5\text{ ft})^2 \times 8\text{ ft} = 6.2832\text{ ft}^3$
   • Weight $W_1 = 6.2832\text{ ft}^3 \times 490\text{ lb/ft}^3 = 3,078.8\text{ lbs}$
   • Centroid $D_1 = \frac{8\text{ ft}}{2} = 4.0\text{ ft}$ from left datum.

2. Section 2 Weight and Centroid ($W_2, D_2$):

   • Volume $V_2 = \pi \times (0.25\text{ ft})^2 \times 12\text{ ft} = 2.3562\text{ ft}^3$
   • Weight $W_2 = 2.3562\text{ ft}^3 \times 490\text{ lb/ft}^3 = 1,154.5\text{ lbs}$
   • Centroid $D_2 = 8.0\text{ ft} + \left(\frac{12.0\text{ ft}}{2}\right) = 8.0 + 6.0 = 14.0\text{ ft}$ from left datum.

3. Moment Table:

1. Calculate Composite Center of Gravity ($\bar{X}_{CG}$): $\bar{X}_{CG} = \frac{\sum M}{\sum W} = \frac{28,478.2\text{ ft-lbs}}{4,233.3\text{ lbs}} = 6.727\text{ ft} \approx 6.73\text{ ft from left datum}$

Analysis: Note that the physical geometric midpoint of the $20\text{ ft}$ shaft is at $10.0\text{ ft}$. However, because Section 1 contains over 72% of the total weight, the Center of Gravity is pulled $3.27\text{ ft}$ to the left, locating at $6.73\text{ ft}$ from the left face.

4. Worked Example 2: Skid-Mounted Modular Compressor Package

+-----------------------------------------------------------------------------------------+
|                   WORKED EXAMPLE 2: SKID-MOUNTED COMPRESSOR PACKAGE                     |
+-----------------------------------------------------------------------------------------+
|  DATUM (0 ft)                                                                           |
|  |                                                                                      |
|  |  +--------------------+                                                              |
|  |  | Electric Motor &   |             +-----------------+        +-----------------+   |
|  |  | Recip Compressor   |             | Control Panel   |        | Buffer Air Tank |   |
|  |  | Weight = 7,000 lbs |             | Weight = 2,000lb|        | Weight = 3,000lb|   |
|  |  | CG = 4.0 ft        |             | CG = 12.0 ft    |        | CG = 18.0 ft    |   |
|  |  +--------------------+             +-----------------+        +-----------------+   |
|  |==================================================================================|   |
|  |  Structural Steel Skid Frame: Weight = 4,000 lbs | CG = 10.0 ft                  |   |
|  |<-------------------------- Total Skid Length = 20 ft --------------------------->|   |
|  |                                                                                      |
|  |-------------> Composite CG = 8.00 ft from Left Datum                                 |
+-----------------------------------------------------------------------------------------+

Problem Statement

A modular industrial skid package measuring $20\text{ ft}$ in length is being rigged for a crane pick. The package contains:

1. Structural Steel Base Skid: Total weight $= 4,000\text{ lbs}$, CG located at mid-length ($10.0\text{ ft}$ from left end).
2. Motor & Compressor Unit: Total weight $= 7,000\text{ lbs}$, CG located $4.0\text{ ft}$ from left end.
3. VFD & Electrical Enclosure: Total weight $= 2,000\text{ lbs}$, CG located $12.0\text{ ft}$ from left end.
4. Receiver Storage Tank (Empty): Total weight $= 3,000\text{ lbs}$, CG located $18.0\text{ ft}$ from left end.

Moment Table & Composite CG Calculation

$\bar{X}_{CG} = \frac{\sum M}{\sum W} = \frac{146,000\text{ ft-lbs}}{16,000\text{ lbs}} = 9.125\text{ ft from left datum}$

5. Field Methods for Determining Center of Gravity

When detailed engineering drawings and moment calculations are unavailable on-site, riggers use two proven field methods to verify CG:

+-----------------------------------------------------------------------------------------+
|                        FIELD METHOD: TWO-SCALE WEIGHING SETUP                           |
+-----------------------------------------------------------------------------------------+
|                                 <------- Span Distance (L) ------->                     |
|                                                                                         |
|                           +----------------------------------------+                    |
|                           |           LOAD OBJECT                  |                    |
|                           |                  X CG                  |                    |
|                           +----------------------------------------+                    |
|                              ^                                  ^                       |
|                           [SCALE 1]                          [SCALE 2]                  |
|                          (Reaction R1)                      (Reaction R2)               |
|                           |                                                             |
|                           +-----> Distance to CG (D_cg) = (Scale 2 x L) / (S1 + S2)     |
+-----------------------------------------------------------------------------------------+

1. Two-Scale / Load Cell Weighing Method

By placing load cells or calibrated scales under two support points separated by span distance $L$:

• Total Weight $W = \text{Scale}_1 + \text{Scale}_2$
• Distance from Scale 1 to the Center of Gravity ($D_{CG}$): $\mathbf{D_{CG} = \frac{\text{Scale}_2 \times L}{\text{Scale}_1 + \text{Scale}_2}}$ Worked Field Example: A transformer resting on two beams separated by $L = 12\text{ ft}$ registers $8,000\text{ lbs}$ on Scale 1 (left) and $16,000\text{ lbs}$ on Scale 2 (right): $\text{Total Weight} = 8,000 + 16,000 = 24,000\text{ lbs}$ $D_{CG} = \frac{16,000\text{ lbs} \times 12\text{ ft}}{24,000\text{ lbs}} = \frac{192,000}{24,000} = 8.0\text{ ft from Scale 1}$

2. The Test Pick Procedure

Before executing a full lift, the rigger signals the crane operator to hoist the load only 2 to 4 inches off the ground or dunnage blocking:

• Observe Tilt: If the load tilts, the heavy end is lower, meaning the crane hook is NOT positioned vertically over the CG.
• Correct Alignment: Lower the load back onto solid blocking, adjust sling lengths (e.g., using turnbuckles or chain adjusters), and shift the crane hook toward the low (heavy) end until the load lifts perfectly level.