Baseball Field Netting Cable Tensions

[structuralengr89]

If any of you have experience with calculating forces in cables, I'd appreciate your input on my attached calculations.

I have received calculations from a Contractors sub for the tensions and sag in a backstop netting system at a baseball field. The system consists of poles supporting the net at points C and D and then tied back to my structure at points E and F. There are no poles at points A and B.

He stated that the elevation of the net at points A and B will be brought up to the same elevation as points C and D. He says the maximum sag will be 3".

We have a slight discrepancy in our calculations: He says the force in the cable tied back to my structure will be 650#. I came up with 29,500#. Someone is vastly incorrect!!!

Thanks!

 

[racookpe1978]

Hmmmn. Have not checked all of the calc's - but weight of net = 50 lbs per linear foot.

Yet, over a 50 ft span, each section load is written in the square box as 450 lbs [as if 9 feet x 50 lbs/ft ?] not [40 ft x 50 lbs/ft.]

Height vs length of the cables is only 5'-6". Are they mounted up high on the building wall?

 

[structuralengr89]

The note in the RED BOX is what the Contractor said is the TENSION IN THE CABLE.

Net weight = 2 PSF
Net height = 25 FT
or 50 PLF

Thanks for your reply!

 

[Ron]

I also have not done calcs on this, but taking a quick look at the sketch, my gut feel is you both might be wrong, but the contractor is closer to right. The net weighs 5600 lbs. It is supported at 4 points. Fundamentally if hung from vertical supports, you would have a stress in each vertical cable of roughly 1400 lbf. Since you have angles involved, it then becomes an exercise in statics to resolve the cable force in each location.

 

[structuralengr89]

Ron,
Are you saying that sag has no bearing on the tension force in the cables? Is the formula sag=(weight*horizontal distance^2)/(8*horizontal tension in cable) not applicable?

The 29500# force I came up with seems extremely high, but so does the idea that the net would be hung so tight horizontally that it only has 3" of sag in 40'. Other ball parks I've seen that are hung this way appear to have way more sag in the cables.

Thanks.

 

[BAretired]

With only three inches of sag, the cable tension must be in the order of 40,000#. A simple span beam spanning 40' carrying 50 pounds per lineal foot would have a total load of 2000# and a bending moment of 2000*40/8 = 10,000'#.

Cable B would have a horizontal component of tension of 10,000/s where s is the sag. With a sag of only 3" (0.25') the cable tension is roughly 40,000#, so I agree with the OP that the contractor's calculation is seriously in error.

BA

 

[Ron]

I was not considering that a 3" sag limitation was reasonable for a baseball net....in fact, if the net itself is too taut there is more potential for rebound, which can be dangerous. I would probably opt for a more reasonable sag of about 12" in 40 feet.

Even if you use the code deflection limit for screen supports (L/60), you get 8" of sag in 40 feet.

 

[BAretired]

With a 12" sag, the tension in Cable B is in the order of 10,000#. That is twenty times the 500# suggested by the Contractor.

Also, there are other factors to be considered. If the net weighs 2psf when dry, it weighs considerably more when wet, so the cables will be stressed more during and after rainfall. Wind loads will affect cable tension as well.

All cables will strain under load and the poles at C and D will deflect under load, so points A and B will shift in a somewhat unpredictable manner making it difficult to determine the exact geometry of the system.

BA

 

[hdn32]

Structuralengr89,

I have done similar projects.

Could you provide the following, so that I can put together a couple quick models for your comparison:
- Elevation of all points (or coordinates of all points would be better - save time

- Size of cable you have in mind, if any

Are you concerning the tension in cable C only or are you responsible for all cables and connections/supports?

hdn32

 

[structuralengr89]

Ron and BA,
Thanks for your replies and thank you for taking the time on this board to help those who lack your wisdom.

The Contractor had sent me his output from a Finite Element Program he was using. I was a bit concerned because I have a similar FEM program and you cannot model cables with it. He believed because he had clicked "tension only" that the member would act as a cable. I contacted the company that sells the program and spoke with one of their "Engineers". He said you can not analyze cables with their program.

Lesson for all you Young Engineers (as was told to me by a gray haired engineer when I started)
"Anytime you purchase new software, you need to check the software results with hand calculations. Always verify that the program is correct".

 

[Rabbit12]

My initial thought when I first read this without running any numbers would be that your reactions seem much more reasonable than his. Please let us know the resolution. Sure seems like the vendor was in error at this point.

Anchor forces on cables can get very big quick if you limit the sag.

 

[structuralengr89]

hdn32,

Thank you for running this. I'd really like to check my hand calcs. By the way, what software are you using?

Coordinates (X,Y,Elevation)
Point# (Coordinates)
A (0,0,25)
B (40,0,25)
C (-26.8701,-26.8701,25)
D (66.8701,-26.8701,25)
E (-6.8227,10.9186,30.5)
F (46.8227,10.9186,30.5)

Uniform load on Cable A and Cable B = 50 plf

If your program will provide this info, it would be nice to know:
Tension in Cables A, B and C
The Reactions (X,Y,Z) @ Points D and F
Assume reactions @ C, D, E and F are pinned.

Thank you!

 

[BAretired]

Something seems wrong with the last force triangle. 1950# vertical, 29,412# horizontal and 29,476# are not consistent with the assumed slope of Cable C.

BA

 

[hdn32]

Hi Structuralengr89,

Please take a look at the attached PDF file of the sketch and results analysis using PLS Tower.
You will noticed that this is a "rough" run and will need to be refined. But it will give you a "feel" for the tension in the cables.

Yes, I could not agree with you more regarding of checking/understanding the capacity of various software .

As far as handcalc for your model, I would break the structures into 3 smaller models (actually it would be 2, since 2 of them are identical).

The first model would be for cable B with supports at A & B, this will get us the tension & reactions accordingly with the sag that you want. This cable profile could be used for more refined software runs.

The 2nd (& 3rd), will include cable C & A with loads taken from previous model's reactions.

I hope this help you.

I know you already stated in your OP that ice & wind loads being ignored, is it because you just want to simplify the model for quick handcalc. check?

Regards,

hdn32

 

[BAretired]

I think the hand calcs would have been okay if the elevation of points E and F were changed from 5'-6" to 10.25" above points A, B, C and D.

BA

 

[structuralengr89]

BA,
You are correct. Thank you for catching that.

hdn32,
Thank you for running this for me! I really appreciate it!

From your first run, you placed a 2000# point load (50plf) at Point P2P and the point displaces 4.2 ft and the cable has 5340# of tension in it.

The submittal I received from the Contractor said the maximum sag(displacement) would be 2 to 3 inches and the maximum tension would be 500#.

HUGE DIFFERENCE!!!

By the way, do you know how much the program you are using costs and is it available to purchase?
Thanks again!

 

[hdn32]

Structuralengr89,

You are welcome.

You can find additional info at this link

http://www.powline.com/products.html

Good luck with your project,

hdn32