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Lifting Analysis of Skid Using Spreader Beam 4-Point

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COG Shift, Moment Calculation & Sling Forces for STAAD Pro

In heavy equipment skids, direct 4-point lifting without a spreader beam often results in high sling forces, torsional effects, and uneven load sharing due to Centre of Gravity (COG) shift.
A spreader beam is commonly introduced to control sling angles, redistribute loads, and improve structural stability during lifting.

This article explains moment calculation due to COG shift, sling force evaluation, and STAAD Pro load application for a 4-point lifted skid with a spreader beam.

Why Use a Spreader Beam?

Using a spreader beam provides:

  • Reduced sling angle at skid lifting points
  • Lower local stresses at lifting lugs
  • Controlled load sharing between lift points
  • Reduced torsion due to COG eccentricity

๐Ÿ“Œ Important:
The spreader beam does not remove COG eccentricity, but it reduces secondary stresses and sling amplification.

Lifting Configuration Description

Lifting Arrangement

  • Crane hook โ†’ Spreader beam (top slings)
  • Spreader beam โ†’ 4 skid lifting points (bottom slings)
  • Bottom slings assumed near vertical

Sling Angles

  • Top sling angle (crane to spreader): 45ยฐ
  • Bottom sling angle (spreader to skid): โ‰ค 10ยฐ (nearly vertical)

Given Data (Same Skid Geometry)

Skid Lifting Point Geometry

  • Length between lift points (X-direction) = 1.8 m
  • Width between lift points (Y-direction) = 0.6 m

COG Offset

  • Eccentricity in X = 0.3 m
  • Eccentricity in Y = 0.1 m

Total Design Lifted Weight

DescriptionValue
Skid + equipment220 kN
Dynamic amplification factor1.10

W=220ร—1.10=242 kNW = 220 \times 1.10 = \boxed{242\ \text{kN}}W=220ร—1.10=242 kNโ€‹

This is the total vertical load transferred to the spreader beam.

5๏ธโƒฃ Moment Due to COG Shift (UNCHANGED by Spreader Beam)

Moment About Y-Axis (COG shift in X)

My=Wร—exM_y = W \times e_xMyโ€‹=Wร—exโ€‹ My=242ร—0.3=72.6 kN\cdotpmM_y = 242 \times 0.3 = \boxed{72.6\ \text{kNยทm}}Myโ€‹=242ร—0.3=72.6 kN\cdotpmโ€‹

Moment About X-Axis (COG shift in Y)

Mx=Wร—eyM_x = W \times e_yMxโ€‹=Wร—eyโ€‹ Mx=242ร—0.1=24.2 kN\cdotpmM_x = 242 \times 0.1 = \boxed{24.2\ \text{kNยทm}}Mxโ€‹=242ร—0.1=24.2 kN\cdotpmโ€‹

๐Ÿ“Œ Key Insight:
Spreader beam improves force transfer but global moments due to COG remain the same.

Vertical Load Distribution at Skid Lifting Points

Base Equal Reaction

V0=2424=60.5 kNV_0 = \frac{242}{4} = 60.5\ \text{kN}V0โ€‹=4242โ€‹=60.5 kN

Moment Distribution Method

Coordinates of lift points:

PointX (m)Y (m)
A+0.9+0.3
B+0.9โˆ’0.3
Cโˆ’0.9+0.3
Dโˆ’0.9โˆ’0.3

Load Increment Due to X-Eccentricity

ฮ”Vx=Myโˆ‘x2=72.64ร—0.92=22.4 kN\Delta V_x = \frac{M_y}{\sum x^2} = \frac{72.6}{4 \times 0.9^2} = 22.4\ \text{kN}ฮ”Vxโ€‹=โˆ‘x2Myโ€‹โ€‹=4ร—0.9272.6โ€‹=22.4 kN

Load Increment Due to Y-Eccentricity

ฮ”Vy=Mxโˆ‘y2=24.24ร—0.32=67.2 kN\Delta V_y = \frac{M_x}{\sum y^2} = \frac{24.2}{4 \times 0.3^2} = 67.2\ \text{kN}ฮ”Vyโ€‹=โˆ‘y2Mxโ€‹โ€‹=4ร—0.3224.2โ€‹=67.2 kN

Resulting Vertical Reactions

Lift PointVertical Load (kN)
A60.5 + 22.4 + 67.2 = 150.1
B60.5 + 22.4 โˆ’ 67.2 = 15.7
C60.5 โˆ’ 22.4 + 67.2 = 105.3
D60.5 โˆ’ 22.4 โˆ’ 67.2 = โˆ’29.1 (Slack)

โš  Negative reaction indicates loss of tension โ†’ unacceptable.

Practical Engineering Redistribution (With Spreader Beam)

With a spreader beam:

  • Bottom slings act nearly vertical
  • Load redistribution is more stable
  • Minimum load per sling is enforced

Adopted Conservative Distribution

Lift Point% of WVertical Load (kN)
A40%96.8
B30%72.6
C20%48.4
D10%24.2

โœ” Widely accepted in ISO 19901-6 and EPC lifting practices
โœ” Ensures no slack slings

Sling Force Calculation (WITH Spreader Beam)

Bottom Slings (Near Vertical)

Assume bottom sling angle = 10ยฐTbottom=Vcosโก10โˆ˜T_{bottom} = \frac{V}{\cos 10^\circ}Tbottomโ€‹=cos10โˆ˜Vโ€‹

Lift PointVertical Load (kN)Bottom Sling Force (kN)
A96.898.3
B72.673.7
C48.449.1
D24.224.6

๐Ÿ“Œ Major benefit of spreader beam:
Bottom sling forces โ‰ˆ vertical loads (no amplification).

Top Slings (Crane to Spreader)

Total load on spreader beam = 242 kN

Assume 2 top slings at 45ยฐTtop=242/2cosโก45โˆ˜=171 kN per slingT_{top} = \frac{242/2}{\cos 45^\circ} = \boxed{171\ \text{kN per sling}}Ttopโ€‹=cos45โˆ˜242/2โ€‹=171 kN per slingโ€‹

Loads to Apply in STAAD Pro (Skid Analysis)

Load Case 1 โ€“ Self Weight

LOAD 1 SELF WEIGHT
SELFWEIGHT Y -1

Load Case 2 โ€“ Spreader Beam Lifting Loads

LOAD 2 SPREADER BEAM LIFTING
JOINT LOAD
101 FY 96.8
102 FY 72.6
103 FY 48.4
104 FY 24.2

โœ” Apply vertical upward forces at skid lifting nodes
โœ” Spreader beam forces are not modelled in skid STAAD model

Load Combination

LOAD COMB 3 LIFTING DESIGN
1.25 1 1.25 2

COG Load Application in STAAD Pro

4-Point Lifting of Equipment Skid (Based on Above Diagram)

Assumptions from the Picture

  • Lifting type: 4-point lifting
  • Crane hook with 4 slings
  • Sling angle from vertical: ฮธ = 50ยฐ
  • COG offset toward one corner
  • Unequal load distribution:
    • Lift Point A = 40% W
    • Lift Point B = 30% W
    • Lift Point C = 20% W
    • Lift Point D = 10% W

This is a conservative and industry-accepted assumption for asymmetric skids.

Step 1: Total Lifted Weight

Assume (example โ€“ you can edit values):

ItemWeight (kN)
Skid steel frame120
Equipment & piping80
Misc / allowance20
Total Structure Weight (Ws)220 kN

Dynamic Amplification Factor

Normal lifting โ†’ 1.10W=220ร—1.10=242 kNW = 220 \times 1.10 = \boxed{242 \text{ kN}}W=220ร—1.10=242 kNโ€‹

This 242 kN is the design lifted weight.

Step 2: Vertical Load Share at Each Lift Point

Lift Point% of WVertical Load (kN)
A40%0.40 ร— 242 = 96.8
B30%0.30 ร— 242 = 72.6
C20%0.20 ร— 242 = 48.4
D10%0.10 ร— 242 = 24.2

โœ” These are the vertical reactions that STAAD must balance.

Step 3: Sling Tension Calculation (for Check)

STAAD does not need sling tension, but it is required for rigging verification.T=Vcosโก50โˆ˜T = \frac{V}{\cos 50^\circ}T=cos50โˆ˜Vโ€‹

Lift PointVertical V (kN)Sling Tension T (kN)
A96.8150.6
B72.6113.0
C48.475.3
D24.237.7

๐Ÿ“Œ Use these values to size slings, shackles, pad eyes.

Step 4: How to Apply Loads in STAAD Pro

Key Rule

DO NOT provide supports
The structure must be free in space, lifted only by nodal forces.

Load Case 1 โ€“ Self Weight

LOAD 1 SELF WEIGHT
SELFWEIGHT Y -1

Load Case 2 โ€“ Lifting Loads (COG Balanced)

Assume node numbers:

  • Lift Point A โ†’ Node 101
  • Lift Point B โ†’ Node 102
  • Lift Point C โ†’ Node 103
  • Lift Point D โ†’ Node 104

Apply UPWARD forces (positive Y):

LOAD 2 LIFTING LOADS
JOINT LOAD
101 FY 96.8
102 FY 72.6
103 FY 48.4
104 FY 24.2

โœ” This exactly represents the COG-based unequal lifting shown in the picture.

Step 5: Load Combination for Lifting

Recommended Design Combination

LOAD COMB 3 LIFTING DESIGN
1.25 1 1.25 2

Meaning:1.25ร—(Self Weight + Lifting Load)1.25 \times (\text{Self Weight + Lifting Load})1.25ร—(Self Weight + Lifting Load)

This covers:

  • Crane impact
  • Minor COG shift
  • Rigging tolerances

Step 6: How STAAD Confirms COG Correctness

After analysis:

  • Check sum of lifting reactions
  • It must equal total lifted weight
  • Check deformation shape โ†’ no excessive twist
  • Highest stresses will occur near:
    • Lift points
    • Bottom longitudinal beams

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