How to calculate weight distribution in shipping containers

How to calculate weight distribution in shipping containers

A cargo container is designed to carry cargo. Most of us see a container as an extremely strong steel box designed to carry weights of over 20 tons. Why should we worry about where or how we place the cargo on the floor?

Probably the reason is simply that we generally never see the underside of a container, which consists of pieces of plywood screwed on to a number of cross members down the length of the container. In between the cross members it’s only plywood, which is not strong at all and the floor is therefore not designed for heavy selective loads.

The container may have a maximum payload of e.g. 28 tons but this is based on being spread over the entire area of the container floor! This is easy to accomplish with light weight palletised loads. Yet, for cargos with a large mass concentrated in a small area like steel coils or heavy machinery, it is extremely important to first ask the question whether the container floor will be able to withstand this concentrated weight without e.g. timber supports on the floor.

Cargo weight should be evenly distributed over the entire length of the container floor

  • a 20ft container floor, build accordingly ISO 668 standard, may be loaded with 4.5 ton/meter
  • a 40ft container floor, build accordingly ISO 668 standard, may be loaded with 3 ton/meter

If the ballast exceeds these numbers per meter, more beams should be placed lengthwise. These beams cannot stick out from underneath both sides of the cargo by more than 1 meter. If this still is not sufficient you need to find other means of transportation, e.g. a flat rack.

When driving a forklift inside a container you should not exceed the maximum floor ballast. According ISO standard 1496/1. a forklift can drive inside a cargo container if:

  • the weight of the front axle (forklift including load) does not exceed 5.460 kg
  • the contact surface meets a minimum of 142cm2 per tire
  • the minimum width of the tires is 18 cm
  • the distance between the wheels on 1 axle is 76 cm

Simple trick for calculating floor support

Here is a simple trick to calculate floor support/weight distribution when you are working in the field if floor support is required.

1. Count the number of rows of screws (see picture) from back to front in the container floor e.g. 18 rows for this 20’ container

2. The number of rows of screws in the container floor will tell you the number of container cross members under the plywood floor e.g. in this example 18

3. Divide the max. payload of the container by the number of rows of screws. This provides the max weight that one cross member can carry i.e. in this example 28 ÷ 18 = 1.56 tons

4. If your cargo is a steel coil of 5 tons it must be spread over at least 5 ÷ 1.56 = 3.2 or, rounded up, 4 cross members

5. This will require the coil being bedded on timber beams that spread the load along these 4 container cross members

“Cracking ice does not break”

There is a saying that cracking ice does not break. However…when the ice starts to crack, the breaking point is very close. Please be careful that your cargo “doesn’t fall through the ice” and make sure your cargo weight is correctly distributed on the loading floor.

Don’t forget to secure and/or block and brace your cargo

Of course, securing your cargo against movement is also key to successful, damage free transport. We offer a wide variety of container securing solutions, including our Cordstrap composite lashing and dunnage bags. These are globally the most commonly used products to safely secure cargo of any weight, size and dimensions. All our container securing solutions are CTU Code Compliant, AAR approved and DNV/GL Certified. Our team will be happy to meet with you for an site consultation; and ofcourse we are always happy to assist with complimentary 3D loading diagrams to make sure the weight in your container is evenly distributed!

Very interesting, but pls permitt a question: regarding the outriggers in the hc, we can make the same calc? If we need to add the forklift wheigt, can we use the Euclid's second level formula?

回复

要查看或添加评论,请登录

Cargo Restraint Systems Pty Ltd的更多文章

社区洞察

其他会员也浏览了