The individual containers are standard containers, i.e. no high cubes. The centre of gravity within each container is assumed to be only 25% within each container.


This “stack” of a given height would tip over at a certain angle of inclination. In this case, that is only 7.8°.


The only securing counteracting this is the securing of the base of the stack.


Or in Diagram C: An upright tower on which transverse lateral forces act would also tilt if the tilting moment is not counteracted by a sufficient retaining moment.


Or in a second example:


An upright tower on which transverse lateral forces act would also tilt if the tilting moment is not counteracted by a sufficient retaining moment.


Tipping moment MT Retaining moment MR


MT < MR = ax x m x heightCOG = az x m x half tower width


If the prevailing tipping moments exceed the effective retaining moments, then again only the securing at the base of the stack prevents the stack from tipping over.


HeightCOG


= 7.89m (2.43 m x 3.25; 3 standard DV containers, 8.0’ high, plus a quarter container height)


Half width = 1.22m ax


The tipping moment MT is calculated as follows: MT = 0.5 x 9.81 x 7 x 10 mt x·7.89 m = 2,709 kNm This is compared to a retaining moment of: MR = 1.0 x 9.81 x 7 x 10 mt x 1.22 m = 837.7 kNm.


az


Or in a second example: =


=


The tipping moment MT is calculated as follows: MT


0.5 x 9.81 x 7 x 10 mt x∙7.89 m


Based on the above calculation there is a difference (tipping moment surp 1,871.3 kNm. What does this mean for the lowest layer of the twist locks?


An upright tower on which transverse lateral forces act would also tilt if the tilting moment is not counteracted by a sufficient retaining moment.


This is compared to a retaining moment of: MR


1.0 x 9.81 x 7 x 10 mt x 1.22 m =


Tipping moment MT = ax x m x heightCOG Retaining moment MR = az x m x half tower width MT < MR


Based on the above calculation there is a difference (tipping moment surplus) of 1,871.3 kNm.


What does this mean for the lowest layer of the twist locks?


Diagram D: If we look at the situation at the base of a stack of 7 containers compared to other containers below, it becomes clear that the tilting motion results in the acting of a vertical pulling force on the twist locks.


Diagram C 58 | The Report • June 2021 • Issue 96


If the prevailing tipping moments exceed the effective retaining moments, then again only the securing at the base of the stack prevents the stack from tipping over.


Diagram D


If we look at the situation at the base of a stack of 7 containers compared to other containers below, it becomes clear that the tilting motion results in the acting of a vertical pulling force on the twist locks.


If we look at the situation at of a stack of 7 containers co other containers below, it be clear that the tilting motion the acting of a vertical pullin the twist locks.


When using a simple numerical example to calculate the forces during tilting in a curve without vertical acceleration by way of example, a very practical problem


In order to estimate/calculate the magnitude of this pulling force, the excess moment is to be divided by the total lever arm length.


=


= 0.5 x g (comparable to the accelerations whilst negotiating a curve in a truck or a train)


= 1 x g (no vertical acceleration, i.e. no consideration made for wave motion)


Tipping moment MT


Retaining moment MR MT


= = <


ax az


x m x heightCOG x m x half tower width MR


If the prevailing tipping moments exceed the effective retaining moments, then again only the securing at the base of the stack prevents the stack from tipping over.


When using a simple numerical example to calculate the forces during tilting in a curve without vertical acceleration by way of example, a very practical problem becomes clear:


The following parameters are assumed as the basis for the calculation: mContainer = 10mt


2,709 kNm 837.7 kNm.


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