# Wall Strength of a Shipping Container

## Main Question or Discussion Point

I'm going to bolt some hooks to the inside walls of 20ft and 40ft shipping containers. What I'm wondering is, what would be the easiest way to determine the stress on the corrugated steel walls, excerted on it by placing things on these hooks?. The hooks are 280mm apart and the force will be applied at approx 80mm from the wall. The steel is 14 gauge (2mm thick) and has a yield strength of 350 MPa.

So far I've done an FEA model in strand7. I drew the profile of the corrugated sheet using plates and then simulated the hooks by creating nodes at the approx position of the weight and used a rigid link to tie it to the plates. This gives a max stress much higher than I anticipated. Restraints are all ends fixed.

Anyone got any better ideas on how to better apply the load from the hooks? Or would it just be better to just approximate it as a simple flat plate? How would you determine the uniform load per unit area for this case?
Any other ideas on how best to determine this?

All help is appreciated.

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Anyone got any ideas? Is it hard to understand, do I need to add a diagram?

nvn
Homework Helper
Vadar2012: The way you created your corrugated sheet is good. Make the wall mesh at each hook fine (dense) enough, so that you can have an approximate circle (or square) of stiffened plate elements (i.e., E = 2e7 MPa), corresponding approximately to the outside diameter of the hook flat washer. You can also make these stiffened plate elements thicker, if you wish. Use your existing rigid link, with one end attached to the center node of this stiffened region. When you read your stress results, ignore the stress in the stiffened plate elements. Only look at stress outside of the washer outside diameter. See if this helps.

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I find this interesting. I've put hundreds of hooks, racks, and shelves in shipping containers. Sometimes we would have nearly a thousand pounds hanging from one hook. I never analyzed the stresses and never had a failure.

@nvm - Thanks for that, but it didn't really help. It was still wacky. Below is another idea. I apologise for the wall of text, but I think it needs the detail.

The hooks are inside a length of RHS and a bolt is put through it at each protrusion of the corrugated sheet at equal distances from the hooks (the hooks are located in the grooves). So would it be plausable to just apply the weight on the hook as a vertical force at the node on the wall. This is assuming that the bolt is tightened to a suitable tension as to restrict bending of course. I've done this and got a much more reasonable result. With the rigid link model from before, the stress at the mid plane (neglects bending) is still ridiculously high. It indicates that the wall would fail at 20kgs, which we know it wouldn't. I also calculated the strength of the bolts (M8, 8.8) used to attach them to the wall and got a result of 686 kg per bolt, which sounds reasonable (it'd actually be 50% more but one of the bolts will be taking more). I also get about 2 tonnes if I do a shear calculation (single, even though might be double), assuming the walls of the RHS and the wall of the container are just under tension of the weight of the hook. Calcs of this are in my next post.

@Pkruse - I've asked manufacturers of shipping containers and their related accessories if they have any guidelines in regards to maximum loads etc. and I was surprised to find that they had no idea what I was talking about. I just want to make sure that the maximum load that I rate the hooks is actually safe. I've modeled and destructively tested them, so they're fine (can take 800kg before the hook starts to straighten). It's just the wall left to test, but you can't safely destructively test it, so I'm left with theory.

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Ok so here's my calcs. Diagram is shown in the attachment. Note that Fb is the tension force on the bolt that is through the RHS and the corrugated sheet wall.

The strength of bolt neglecting shear force. Assuming tension takes it.

Sum of moments about point A = 0 = 8*P= Fb*25
therefore Fb = 3.2*P

For an M8 bolt At = 36.6 mm^2, P = 300N (a random load on hook)
Sigma = P/At = 3.2*300 N/36.6 = 26.2 MPa (required strength of bolt given a 300 N load)

Assuming 8.8 SAE class steel, Sp = 600MPa (proof strength)
600 = 3.2*P / 36.6
therefore P = 6.863 kN = 686.25 kg (maximum load on hook)

Strength of bolt under shear force, assuming setup is just RHS and sheet connection under tension

Ssy = 0.58*Sy (distortion energy theory) = 0.58*600 = 382.8 MPa (600 is the strength of 8.8 SAE class steel)

Area of bolt = pi*d^2/4 = pi*8^2/4 = 50.27 mm^2 (it's an M8 bolt, thus d=8mm)

Ssy = P/A
P = 382.8*50.27 = 19243.4 N = approx 2 tonnes. I find this soultion weird, as shear is usually much less, but I guess it's because of the assumption that the force is applied vertically to the RHS with no bending.

The bearing strength of the plate (strength of the plate at the hole from the force applied by the bolt).

Sigmab = P/Ab
where :
Ab = d*t (diameter of bolt time, thickness of plate, RHS is thicker) = 8*2 = 16 mm^2
Sigmab = allowable bearing stress = 1.5*Sty (Sty = material tensile strength)
Sigmab = 1.5*450MPa = 675MPa (450 is the tensile strength of C450L0 steel)

Therefore P = 675*16 = 10.8 kN = approx 1tonne of vertical force on the hook.

Anything else you think I should check? I can't think of anything weaker than these points. If there isn't, I guess this shows that the bolt in tension governs the strength in this situation.

Are these methods sound? I don't expect anyone to check my maths.

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nvn
Homework Helper
Vadar2012: Your calculations in post 6 look fine, except you would want to use a factor of safety.
So would it be plausible to just apply the weight on the hook as a vertical force at the node on the wall?
No, you would need to also account for the applied load offset (moment). Perhaps try the approach shown in the attached diagram. Let the applied load on the hook be P1 = 1000 N (random load on hook). Therefore, the corresponding nodal applied loads, which you should apply on each corrugation ridge, are shown in the attached file. You can stiffen the plate elements in the red box by setting E = 2e6 MPa, representing the flat washer footprint.

Scale the wall stress results by P2/P1, where P2 = hook load such that wall (and bolt) is not overstressed, using a factor of safety.

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Destructive testing is really rather easy. Apply the test load with a wire rope. Run the wire rope through a sheave mounted on the floor of the container. React the sheave load through the floor to ground. Then apply a horizontal load from outside the container. For a load device, I’ve used a large tractor with a dynamometer attached, and I’ve also used the hoist line from a mobile crane.

nvn: Thanks for your help, it is much appreciated. I always calculate the max load first and then apply a safety factor. Helps me make sense of the numbers. I just didn't bother yet.

So about the model. How did you calculate those forces? I get much larger ones for a 1000N load (just applying trig using the geometry in my previous attachment). I'd like to apply a load of 300N this way.

I applied those forces to my model and got the results shown in the attachment. As you can see they are far above the 350MPa yield strength of the material (max is 450MPa), which is weird, because we know it can take far more than 100kg. I already tried increasing the mesh density. Appying the forces further down from the top and to a larger model with more of the sheet gives the same results.

I also drew the bolt hole in the model and restrained the nodes that made up the circle to a master node in the centre. I then exended this centre node out 80mm and applied a vertical force to it. Again, got results much higher than expected, but you would given the applied stress concentration factors.

Pkruse: would you be able to confirm for me that the thickness of the container walls is 2mm? Manufacturers state only 14gauge, which is t=1.87mm, but looking at a container it seems thicker than this. Note that I can't cut one up to find out. It's welded all around.

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14 ga is common, but I've found some that are heavier.

Find a machinist's depth gauge and drill a small hole big enough for the gauge. Hold a plate on one side and measure the depth thru the wall to the plate. Quick and easy way to measure the thickness.

Thank you very much for your help so far guys.

Pkruse - That's the prob though. I can't drill a hole or cut one open. They are either on rent (not ours) or in use and can't have any random holes.

nvn - I'd still really like to know how you calculated those force values to apply at the wall for the 100kg load, so I can apply it to a 30kg load. It would help immensley. I'm not one to ask without trying first, but I'v been unable to replicate it.

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nvn
Homework Helper
Vadar2012: 14 gauge steel sheet has a nominal thickness of 1.897 mm; and 14 gauge galvanized steel sheet has a nominal thickness of 1.994 mm. (Or if you want to include the tolerance, that is 1.897 +/-0.127 mm thick, and 1.994 +/-0.127 mm thick, respectively. Or a more common tolerance would be 1.897 +/-0.095 mm, and 1.994 +/-0.100 mm, respectively.)

From summation of moment about point A, you can obtain the bolt tensile force, Fb. Fb/(9 nodes) = (3.2*P1)/(9 nodes) = 3.2(1000 N)/(9 nodes) = 355.56 N/node.

From summation of horizontal force, the force at the bottom of the rectangular tube is Fb/(14 nodes) = (3.2*P1)/(14 nodes) = 3.2(1000 N)/(14 nodes) = 228.57 N/node.

The vertical force on the wall is, P1/(9 nodes) = (1000 N)/(9 nodes) = 111.11 N/node.

To change P1 to 300 N, change P1 to 300 N, above.

When you display the stress contour, do not display the elements inside the red box in my diagram. Omit those elements from the contour display.

Does your container really have a free edge, or fixed edge, close to the bolt?

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Does your container really have a free edge, or fixed edge, close to the bolt?
The restraints I've applied are just around the corrugated steel sheet where the welds would be. Top, bottom, and left are all fixed, and the right side restraint is "Y sym" so I don't have to draw the whole damm thing. I don't know where you're going with that question, but keep in mind that picture just shows the top of the sheet, there's another set of forces half way down. Am I missing some restraints? Say to help simulate the washer/bolt on the container?

When you display the stress contour, do not display the elements inside the red box in my diagram. Omit those elements from the contour display.
This is an obvious question, but just to be sure, this is just to simulate the hole without actually drawing it right? I could easily draw it and apply 355, 111.11 N at the centre node of it? (restraining all the radial nodes to a master centre node of course) This could elimate the contour problem.

Thanks again for your help, you've been extremely helpful. I can't thank you enough.

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nvn
Homework Helper
Vadar2012: I cannot think of any restraints you are missing, at the moment. Yes, the elements in the red box simulate the bolt, bolt hole, and flat washer.

The attached files, below, show a few other methods, corresponding to P1 = 1000 N. You can scale P1, as previously discussed, if you wish; or you can scale the stress results. For these second and third diagrams, you can use E = 2e7 MPa for the plate elements in the red box. You can also make the plate elements in the red box thicker, if you wish. For the third diagram, the blue element is an interpolation element, sometimes called an RBE3 element. The pink element is a short, very stiff beam element. The pink element is optional, but helps you read the force on the centre node, to see if the force is (Fx, Fz) = (3200, 1000) N.

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If you can't drill a very small hole and patch it with epoxy, you can get a thickness measurement with ultasound. Anyone who inspects boilers, pressure vessels, or fuel tanks should have the equipment. NACE inspectors often use it, too.

Thanks again guys.

nvn - Nice methods, I'll try those too. What do you think about drawing the bolt hole and appying the forces to the centre master node?

nvn
Homework Helper
What do you think about drawing the bolt hole and applying the forces to the centre master node?
Vadar2012: Sounds OK. The centre master node needs to be connected to all nodes underneath the washer, not just the nodes on the bolt hole edge.

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I think I finally got a model that gave me some decent answers. Results are attached. The red square is a different element and all circle nodes are linked to the centre master node. This makes it easy to get rid of the contours inside the red element. The results shown are only outside the red square (square is drawn as a different element and turned off after solving).

So this means that there's a safety factor of 2.5 given this setup and a load of 30kg. I really thought it'd be more than this though.

Thanks again for your help nvn. Much appreciated.

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Somewhat late information but just be aware the the side walls of a container are not the same thickness all the way along the wall.

For example the sidewall of a 40' container are actually made up from 11 x fully corrugated sheets - continuously welded together at assembly.

The two outermost sheets ( ie the ones closest to the posts on each end ) are typcally 2mm thick and the 9 inner sheets are typically 1.6mm thick

http://containerhome.info/shipping-container-drawings.html

Just a further consideration.

That is extremely helpful, thankyou. That could have saved me alot of work. Are the sheets overlaped at all, or just welded together at the edges?

The sheets are butt welded to each other, there is no overlap.

I have a reciprocal help request

As you have already pointed out the manufacturers of containers hold ( or at least are willing to share ) very little testing data driven from computer modelling - the dominant testing paradigm is destructive testing.

My interest in the use of containers for construction elements - I am sure you are familiar with the Shipping Container Homes concept.

A part of the challenge in promoting real world discussions is the almost complete lack of "engineering data" that exists - as you have found out - at the same time container construction advocates too often run around quoting the "if they can stack them 10 high on container ships they are indestructible" type "evidence" and expect to be taken seriously.

Anyway my point is anytime I meet someone that is doing any work in this area I ask if they would be willing to donate any outcomes, models, calc's etc that can add to our body of knowledge.

Even if the outcome is nothing short of illustrations that people like Architects "see" and maybe and step back with a realization that "engineering" is a requirement of this construction methodology - that you cant just stack them like legos, cut out the bits you dont want and hey presto instant building that would be really helpful to us.

Any and all resources you would be willing to contribute are greatly appreciated.

Thanks