  ### Container Lifting Support

This section covers aspects of container lifting that are relevant but are not required in the main subject body. It covers aspects of container geometry and loading that may be of general interest to anyone with related problems.
This information is to be used as a guide only and Product Developments cannot accept liability for loss or damage arising from the use of this information. Please see disclaimer

#### Centre of Gravity for containers eccentrically loaded. Limiting payload distribution
The current offset is for longitudinal eccentricity only and as stated in BS ISO 3874, allows 60% and 40% of cargo load to be distributed over half the length of container as shown here. To establish position of Centre of Gravity these two distributed payloads are replaced with point loads at their respective centroids.

#### Position of Centre of Gravity Xc Equivalent point load positions
The C of G for total payload is given as Xc the longitudinal distance from centre of container, found by equating moments about the centroid of unit resulting with the expression below.(Since moments are a product of force and distance, g should be included but is ignored since it is common to all terms,) Tare weight is assumed to be central and as such does not affect the position of the C of G and, therefore, omitted.
P Xc = (0.6P x 0.25L) - (0.4P x 0.25L)
From the above Xc = .05L or 5%L #### Level side lifting eccentrically loaded containers.

Certain lifting arrangements hold the container horizontal during a lift. When end lifting with Side Loaders the lifting arms can be operated independantly holding the container level. When bottom side lifting, maintaining an eccentrically loaded container level can only be achieved using a horizontal top-frame with equal length strops, that may be vertical or angled depending on specific top-frame in use,eg fixed length top-frame Straddle Carriers.
When single point top lifting, the C of G ALWAYS aligns itself with the top lifting point that inevitably results with rotation of unit being lifted. The amount of rotation is evaluated below at Tilt angle induced by offset loading Equation 8 and resulting scenarios,depending on lifting mode, are covered in detail in Eccentric Loading.

#### Vertical corner forces when level lifting. Forces for eccentric loading.
Since the maximum vertical end forces only are required, the simplest way to find Vmax at the 'heavy end', is from moments about the aperture centre point O at the right hand corner fitting.

Maximum vertical corner force at the heavy end is:

Vmax =
GW
4
[1 + 0.1(L/S)]
From this expression the term
[S + 0.1L]
4 S
Load Factors for Series 1 containers are :
Series 1
E
A
B
C
D
Length
45ft
40ft
30ft
20ft
10ft
.2786
.2754
.2756
.2759
.2768
Offset Load Factors when level lifting in this manner is Average .276 or 27.6% (For E Type 27.8%)
The vertical sling forces at the 'heavy end' carry higher proportion of the payload for eccentric loading thereby reducing the GW capacity.
Vmax = 0.276 GW

When C of G is central ie, uniformly loaded, Load Factor is 0.25.

#### Lifting Sling Forces when level lifting Level lifting forces for offset loading.

The sling force F is determined by both its angle α and a proportion of the payload by the expresion V= F sinα.
The proportion of the GW is expressed in terms of the maximum vertical corner force component V developed above.

When the sling force is limited to WLL the above equations give the Limiting Gross Weight as GW =
WLL sinα
0.276
The greater the Load Factor the lower the GW

#### Vertical corner forces when END lifting. End lifting forces

The maximum vertical end lifting forces are also considered since the end sling forces are further apart, by almost 280mm. When moments are taken about right hand lug load line, the maximum vertical corner force at the 'heavy end' is taken as: Vmax=27.5%Rg
Any longitudinal offset inbalance can be controlled by independant lifting arms but any transverse eccentricity may result with sideways tilting as C of G re-aligns itself with top lifting point that could be a problem.

Vmax =
GW (0.55L + 38)
2(L + 76 )

Again the maximum vertical end forces at the heavy end is found from moments about the right hand lug hole centre. The offset load factor never exceeding 0.275 or 27.5%. Lifting sling angle α.
When the sling force is limited to WLL the Limiting Gross Weight is same as equation 6. GW =
WLL sinα
0.275
This expression was used to form the GW range for End Lifting eccentrically loaded containers with side loaders.
GAP 5 Positions of C of G's for eccentric loading.

### Geometry

The geometry of eccentrically loaded container when single point lifting with slings of equal lengths is shown here with 0.05L longitudinal offset of C of G. Since the height of C of G is the great unknown, vertical positions have been chosen at 25%, 50% and 75% to see what effect it has on the tilt angle. Lifting is only possible when the C of G is aligned with top lifting point P, and the unit has rotated through the tilt angle, shaded in red, for the 3 cases considered.

The tilt angle θ is found from the arctan ratio of x/y ie

θ=tan-1 [x/y]

The x value is simply the 5%L offset whilst the value of y is distance between C of G and lifting point P ie headroom + position of C of G from the TOP of container. ie

25% way up   y = (S/2) tanα+70 - 0.25H
Half way up   y = (S/2) tanα+70 - 0.5H
75% way up   y = (S/2) tanα+70 - 0.75H

The lug pivots about the position fixed collar 70mm from bottom of corner fitting hence the expression from which the tilt angle is given when C of G is half way up is:

θ = tan-1[(0.1L)/(S tanα +140 - H)] The sling and container assembly rotates about the top lifting point P through the Tilt angle θ whilst the lug pivots about the position fixed collar in the corner fitting aperture thus eliminating abraision between lug and corner fitting when container is swinging.

When all three C of G positions are considered, resulting tilt angles for 20ft and 40ft units are shown below over sling angle range. Tilt angles over sling lifting angle range when lifting
eccentrically loaded 20ft and 40ft containers
From the above Tilt Angle distributions:
these are excessive for short slings and vary with position of C of G, consequently it should always be kept as low as possible.
GAP 6

Lifting via Bottom Lift Sling methods can only take place when C of G is aligned with top lifting point P. The slings of equal length at a specific angle α now includes the tilt angle θ. The GW is formulated to include this angle with slings loaded to WLL of lugs and were found to be approximately 10% lower than those for uniform loading, only the deadload defined in Lifting Cycle is considered.
( Additional live loads may arise when using luffing cranes where load moves in a circular path, eg ship to shore etc, these are transient self equilibriating forces resulting from centripetal force and reactive centrifugal force. Although impractical to quantify such forces, it would be wise to be aware of it's significance when using this lifting mode.) This arrangement provides a direct method of evaluating limiting gross weights for offset container loading when C of G is aligned with top lifting point. The same three positions of C of G are considered, the highest at 75%H, gives the lowest lifting force for this state. Rather than analyse the corner fitting forces it is simpler to consider the forces at the top lifting point since the sling tension for each leg is constant providing it clears the side of container when spreaders are used. At the lifting point P two slings are connected whilst their bottom ends are attached to lifting lugs and coupled with the side corner fittings. At the 'Heavy End', on the left, the sling force FL cannot exceed the WLL of lugs and, therefore, will dictate the maximum gross weight for combined sling and tilt angles.

The coplaner forces shown at the lifting point mirror those at the corner fittings and for equilibrium the two horizontal forces must be the same for zero motion during a lift. Similarly, the total resolved vertical forces due to total sling tension are limited by WLL of the lugs and these determine the gross weight that can be lifted.

For lifting sling force F, the vertical forces V, and horizontal forces H, are respectively:

V= F sin(α ± θ)
H= F cos(α ± θ)
and

From the condition that the horizontal forces are equal, the sling force FR in terms of FL is:

FR = FLcos(α + θ)/cos(α - θ)

The lifting force Rg is the sum of all the vertical force components hence:

Rg =2FLsin(α + θ) + 2FRsin(α - θ)

After substituting for FR and simplifying

Rg = 2FL [cos(α + θ) tan(α - θ) + sin(α + θ)]

FL is the maximum tensile force in left hand sling and must not exceed Working Load Limit (WLL) of lug hence substituting FL with WLL gives the limiting gross weight that can be lifted in this case as: GW = 2WLL [cos(α + θ) tan(α - θ) + sin(α + θ) ]

#### Alternative method to evaluate Limiting GW when Tilted This alternative method uses equilibrium of moments about 'O' at the RH corner due to forces acting on the container ie the body force mg or GW, and the resolved H and V components at the LH lug due to sling force F. The magnitude of these forces requires their horizontal distances from point 'O', lifting sling angle and tilt angle. In this case the C of G is taken at 0.75 height of 9ft 6in 20ft container for worst case scenario.

Again the two forces at LH corner are
V= F sin(α + θ)
H= F cos(α + θ)
and
at distances of
Scos(α+θ) for V
Ssin(α+θ) for H
and
The weight GW acts at the C of G at Xg cosθ
(S/2 + .05L + .75h tanθ)cosθ
After equating clockwise to anticlockwise moments and dividing by cosθ gives:
GW Xg=VS-HS tanθ
Finally substituting WLL for sling force F in force units gives limiting gross weight, in this case at 75% height. GW= FS
[sin(α + θ) - cos(α + θ) tanθ]
(.5S + .05L + .75h tanθ )
Since lifting force Rg is equal to GW, when WLL is substituted this gives the same results as the above method but from different equations. 