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I was debating with some of my co workers the other day about strut loading and I came up with some questions that nobody could answer. Here it is!

Let's say we have a 10 ton load on a strut to make it easy. The strut is at a 60 degree angle and is anchored to the load at the base by a 10 ton ratchet strap that has a working load of 3300lbs. How much of the 10ton load is exerted on on the strap? The reason I am curious is that you always hear that that using ratchets is stronger that just driving pickets in the base and that chain is better that using a ratchet strap. Why is this and how can I calculate what load is being applied to that strap/train at different angles. I realize there is probably a lot of variables that I am not considering but any thoughts would be greatly appreciated.

2. First let me qualify my answer, my degree included some engineering courses in the curriculum, that being said that was years ago and I didn't use it much in my career. So it's kind of like a dull knife it's still a knife, but it doesn't cut anything.

My calculation gave me 5 tons of force in your strap, over the working load. Of course you are already 2000 pounds over the working load of the highest configuration possible for a rescue 42 composite strut.

Also, just a note if you increase the angle to 70 degrees (the steepest allowed by rescue 42) you have 3.4 tons of tension in the strap.

I should note here that I neglected friction of the base plate which is going to take some force out of the strap.

Of course there is the chance that I don't remember what the heck I'm doing so if someone knows better set me straight, you won't hurt my feelings.

I attached a PDF I made explaining how I calculated this, so you can play with the numbers to.

3. Originally Posted by sfdtruck1medic
Let's say we have a 10 ton load on a strut to make it easy. The strut is at a 60 degree angle and is anchored to the load at the base by a 10 ton ratchet strap that has a working load of 3300lbs.
Maybe I'm not picturing it right, or relying on the previous PDF for a mental image. How is your 10 ton load supported by a single strut at 60 degrees?

There are two typos in the PDF, though the math is still correct.
y = cosine (30°) / 10 tons
should be
y = cosine (30°) * 10 tons
Same with the sine formula

As stated it's for a frictionless environment. I think the friction between the ground and base plate is going to be a bigger influence that can't be ignored.

4. I dont see how you could support a load with two struts, let alone just one. If you let mother earth carry part of the load, I would use thirds as a rule of thumb. One third on the ground, two thirds supported by struts. If you support the entire load with struts, you would need a minimum of four struts. Either way, that would put you within the limits of a R42.

5. Ok I'm going to attempt to read the mind of the original poster.

I believe his intent was to explore the relationship between the strut load and the tension in the strap. The theoretical object he is trying to stabilize weighs over 10 tons because he specifically states "Let's say we have a 10 ton load on a strut to make it easy". Key words Load on a strut. Example 60 ton object fully supported by 6 struts, yes we are overloading the struts, and I can't imagine trying to fully support a 60 ton object, but that is not the point.

I believe he choose 10 tons because it is a nice round number to explore the relationship of forces throughout the system, not that he is wanting to load a strut to 10 tons (I hope).

Voyager- the more I thought about the problem I agree that friction plays a significant role.

6. Originally Posted by ellwood1
Ok I'm going to attempt to read the mind of the original poster.
I concur and realized after my first post that it is more of a thought experiment rather then a actual real-life scenario. I would love to see a 10-ton load magically balanced on a single strut at 60 degrees though.. from a distance.

Originally Posted by ellwood1
Voyager- the more I thought about the problem I agree that friction plays a significant role.
I'm not a civil engineer, and it's been 10+ years since I took physics. A little googling:Friction Force (Ff) == friction coefficient(us) * Normal Force (Fn). A standard friction coefficient of steel on gravel is about 0.4. Figuring out the normal force should be easy using your PDF..

7. Originally Posted by ellwood1
Voyager- the more I thought about the problem I agree that friction plays a significant role.
Speaking as an engineer, friction would play a part, however when doing the calculations you would always want to assume that there is no friction at all.

This would be a worst case scenario but certainly something you would want to plan for. That way any friction which you do have is only going to make things better.

To determine the actual coefficient of friction would be very difficult if as I assume, the bottom of the strut has some kind of teeth or ridges which will bite into the ground.

8. Originally Posted by KB1OEV
To determine the actual coefficient of friction would be very difficult if as I assume, the bottom of the strut has some kind of teeth or ridges which will bite into the ground.
The Resque42's that we have don't have teeth or anything, just a base plate. It does have holes for spikes that we pound through if the situation warrants.

9. Originally Posted by voyager9
The Resque42's that we have don't have teeth or anything, just a base plate. It does have holes for spikes that we pound through if the situation warrants.
Ok, so without spikes I would definitely go with the no-friction model. I'm sure the spikes give you a lot of hold, but would still be very difficult to quantify.

10. My original post was just theoretical ideas of the relationship of the weight on the strut compared to the amount of force in relation that would be applied to the strap. I was looking for a relationship such as 1/3 of the force would be applied to the strap. We often hear things in regards to strut loading like at 60 degrees will increase the force on the strut to a 1:1.15 ratio. So if I understand this correctly then a 10 ton load on a strut would actually have 11.5 tons applied to it based on the angle in which it is set. This lead me to the question of how much of the force is being put on the strap, chain,etc....that is holding the base.

A couple of other examples I have heard:

45 degree strut angle = 1:1.4 so 10 ton load would be 14ton
75 degree strut angle= 1.1.035 so 10 ton load would be 10.4 tons

And obviously at 90 degree would be a 1:1 ratio so 10 tons would be 10 tons.

Hope this clarified some of the questions! Thanks again for any input.

11. Originally Posted by KB1OEV
Ok, so without spikes I would definitely go with the no-friction model. I'm sure the spikes give you a lot of hold, but would still be very difficult to quantify.
You Wanna change that? Put your strut base on a rubber pad. That will change your anchorability by QUITE a bit. T.C.

12. Your thought process is valid but as the practical apps go you form a structure with a Device that is Underrated. Should you load beyond the specified WLL? NO, but the strut will take MORE than that. Like wire rope, that is WLL at 10 ton. It will take more than 10 ton to break it but it is VERY unwise to operate over the WLL. T.C.

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