Working Load Limit Question

The question seems a little funny if that is all the information given. Even the length of the item being secured could affect the proper answer. Like I said, I will look into this one and get back to ya later.

That is the entire question, as far as I can see. I was getting tired during that chapter, so maybe I missed something in the lesson, but the question is above. The weird part is I got the rest of the questions correct, so I think I do understand the principles involved.

For instance I had to figure out the WLL of 4 different chains, etc. And actually, even though I got it right, it was one 70 grade, and 3 of another, lower grade chain. Wouldn't that mean the 70 grade chain should be considered the same strength as the lower grade chains, since they would be the weakest link? Of course, it doesn't say how the chains were used, maybe the 70 grade was on one end, and the 3 weaker chains on the other, which would be OK, if I understand it correctly.

If I have to do math in my head for the test, I will be screwed, I need paper. I have never been good at calculations in my head

It is totally OK to have different chains with different WLL ratings on the same load. What OS meant with the weakest point statement, is regarding each individual piece of securement.

On each piece, you'll have your securement device (chain), 2 anchor points, and at least one tension device (binder). So if your chain is rated at 11,300 lbs, but your anchor point or binder is only 5,600 lbs, then the lower WLL is your legal number for that specific piece of securement.

You can use a 1" strap, 2" strap, 4" strap, 3/8" chain, 5/8" chain all on the same load, and the WLL is only relative to each individual securement device.

Back to the original question, I can definitely see how that is confusing without having the full context of the actual load. I'm not really sure why it is that way.

I've been scratching my head over this one too. Something is either wrong or missing.

As for NY Coil, I have the endorsement. You won't have to do any hard ciphering in your head. It's been over a year and a half since I took the test, but if I remember correctly the test is fairly basic. Nothing too difficult and they give you a paper securement chart, so you can scratch your numbers on that.

I've been scratching my head over this one too. Something is either wrong or missing.

As for NY Coil, I have the endorsement. You won't have to do any hard ciphering in your head. It's been over a year and a half since I took the test, but if I remember correctly the test is fairly basic. Nothing too difficult and they give you a paper securement chart, so you can scratch your numbers on that.

Thanks for the effort, I have been searching for an answer as well, but so far, nothing. The only thing I found related to a sling, as in from a crane, and each leg is only 50%. And in case anyone runs into it, 4 legs are treated as 3, because there is no reliable way to get equal distance in the 4 legs, they said. :)

As long as I have some scratch paper I'm fine. My problem is, I forget the number I have in my head while I am figuring out the second part of the equation. I've always been like that for some reason.

Where I took the metal coil endorsement, they gave a sheet of scrap paper, a pen, and the sheet with the ratings for all the different size and types of securement (that you were not allowed to write on.)

Where I took the metal coil endorsement, they gave a sheet of scrap paper, a pen, and the sheet with the ratings for all the different size and types of securement (that you were not allowed to write on.)

That is great! I am not good at memorizing, either. No way I would have been able to memorize the chain sizes, grades, and weight limits.

Thanks for the info.

I have searched and searched, and can't find an explanation for this.

Brett, do you have any insight? I know the answer now, but I would like to know how to figure it out on my own.

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.

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.

I got all the other questions correct, so I just decided it was an error in the question. As written, it makes no sense.

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.