Working Load Limit Question

We were taught to do the calculations like this. 1/2 weight of coil / WLL of chain = number of chains. 17250(34500/2) / 11300(wll) = 1.52 = 2. We only carry 1/2" and that coil requires 4 of them (WLL 4700). To me 7 seems excessive.

Given the way the question is worded, that was my answer as well, and as I said, I got the rest of them correct as well. The only way I saw that that made sense is if the chain was connected to the same side of the trailer, which would cut the WLL in half. But even then I would think it would be 8, not 7.

Ok grumpy I’m going to try to help here with explaining their math. It appears in the example they have left a lot of details out of the question. Also don’t kick yourself too hard over this.

In the example it’s asking how many 1/2” grade 70 chains are needed to secure a 34500 lbs load keeping WLL to 50% of rating. Just to keep it simple we know 1/2” grade 70 has a WLL of 11,300 lbs. 34500/11300=3.05 if you stop and round to 4 here the chains are at just over 75% of their WLL in a ideal world. So we multiply the 3.05*2 to get 6.10 and round to 7 for again in a perfect world using 44% of the WLL on the chains. All the other variables mentioned by others are definitely something to be remembered but the test wanted the simplest answer possible.

Are you saying I should only rely on 50% of the chain's actual WLL?

As in, each chain which has a WLL of 11,500 should be treated as if it will only secure 5,650 lbs?

Oklahoma DOT requires a four point tie down on vehicles towed on a flatbed wrecker. Common practice is to use either 4 grade 70 chains with a 3700lb WLL or 4 2” wheel straps with a 3500lb WLL. Not a single vehicle I towed came close to a 14000 gvwr let alone actually weighing that much. So seven chains rated at 11,300 for a 34,500 coil doesn’t sound out of place to me. Just an observation based on my past experience as a non-CDL tow driver though.

Ok grumpy I’m going to try to help here with explaining their math. It appears in the example they have left a lot of details out of the question. Also don’t kick yourself too hard over this.

In the example it’s asking how many 1/2” grade 70 chains are needed to secure a 34500 lbs load keeping WLL to 50% of rating. Just to keep it simple we know 1/2” grade 70 has a WLL of 11,300 lbs. 34500/11300=3.05 if you stop and round to 4 here the chains are at just over 75% of their WLL in a ideal world. So we multiply the 3.05*2 to get 6.10 and round to 7 for again in a perfect world using 44% of the WLL on the chains. All the other variables mentioned by others are definitely something to be remembered but the test wanted the simplest answer possible.

Are you saying I should only rely on 50% of the chain's actual WLL?

As in, each chain which has a WLL of 11,500 should be treated as if it will only secure 5,650 lbs?

That seems to be the way the question is worded. Even if done so in a trick question sort of way.

Ok grumpy I’m going to try to help here with explaining their math. It appears in the example they have left a lot of details out of the question. Also don’t kick yourself too hard over this.

In the example it’s asking how many 1/2” grade 70 chains are needed to secure a 34500 lbs load keeping WLL to 50% of rating. Just to keep it simple we know 1/2” grade 70 has a WLL of 11,300 lbs. 34500/11300=3.05 if you stop and round to 4 here the chains are at just over 75% of their WLL in a ideal world. So we multiply the 3.05*2 to get 6.10 and round to 7 for again in a perfect world using 44% of the WLL on the chains. All the other variables mentioned by others are definitely something to be remembered but the test wanted the simplest answer possible.

Are you saying I should only rely on 50% of the chain's actual WLL?

As in, each chain which has a WLL of 11,500 should be treated as if it will only secure 5,650 lbs?

That seems to be the way the question is worded. Even if done so in a trick question sort of way.

Except that all the other questions did not. It was strictly the aggregate of all chains WLL to secure at least 50% of the load's weight in all the other questions. I have searched the NY handbook, the Federal handbook, and none reference the multiply by two shown on that single question. I thought when I got the question on review, it would explain it, but it only shows the formula, no explanation

Haven’t made my way to the ny coils material yet so I don’t have the foggiest of ideas what the requirements are. Comparing the math in the question to my experience in the towing industry in the state of OK is why it makes sense to me. Although, I imagine there are major differences between hauling coils and hauling a vehicle on a flatbed wrecker. But the question does seem to want you to calculate the number of chains based on half the WLL rating. Just out of curiosity are you familiar with how securement devices (chains, straps, cable, etc.) are rated?

Haven’t made my way to the ny coils material yet so I don’t have the foggiest of ideas what the requirements are. Comparing the math in the question to my experience in the towing industry in the state of OK is why it makes sense to me. Although, I imagine there are major differences between hauling coils and hauling a vehicle on a flatbed wrecker. But the question does seem to want you to calculate the number of chains based on half the WLL rating. Just out of curiosity are you familiar with how securement devices (chains, straps, cable, etc.) are rated?

Yes, they provide a chart showing the WLL of rope, chains, straps, and wire rope (cable).