Vehicle Cable Barriers - Short Lengths 1

[Boiler106]

Im designing cable barriers for my concrete commercial building, and am running across an issue.

It seems when you have short lengths of cable, the required number of cables increases beyond reason. I wanted to check and see if anyone had ideas or similar experiences.

Currently, i have a single 12ft run that requires 8 cables to be struck using a 5000 lbs vehicle at 5mph per PTI Tech Note 14. That's 8 cables over a 1 ft area which seems unreasonable.

Similarly, i have a single 20 ft run that requires 5 cables to be struck using the same setup.

Anyone with experience have thoughts on this?

 

[Ingenuity]

Boiler106 and MacGruber22

I have used the GRABB-IT splice hardware (not in barrier cable applications but in structural repair of severed prestressed strands to precast concrete floor products) and the hardware requires a calibrated torque wrench to correlate to a magnitude of tension as per the following table:

We usually use a 36" or 48" long calibrated torque wrench with a crow-foot attachment. The threads of the GRABB-IT must be lubricated with anti-seize compound.

 

[Boiler106]

MacGruber, thanks so much for the help. I think im having trouble understanding backstressing and how this is performed.

If i follow the tech note #14 guide example no.1 on page 6 with all of the variables for that particular example (not my own), i determine that the total load to the structure is 50,305 lbs, with (3) cables imparting 8,768 lbs from the vehicle arrest on the column, and (8) cables imparting 3,000 lbs from the pretensioning force.

Next, i determine the jacking force, Fpj = 3000lbs (pretension) + 757 lbs (jacking) = 3,757 lbs. Does this additional force get imparted on my column so that 3,757 lbs x (11) cables of force gets transmitted to my structure?

The guide isnt so clear about that and you seem to hint at it in your response above. am i applying 50,305lbs to my column or 11x3757lbs plus my arresting force?

 

[MacGruber22]

Boiler,

I am a bit confused by your nomenclature. Should be F_e+F_seatloss=F_pj ---> (effective cable force + expected seat loss force < required jacking force)

757# can't be your seating loss for the 12' cable length. It should be much closer to 11 kips because it takes much more P to elongate L to the estimated seating loss (3/8"). Is this a different cable run?

Stages of loading
1. Cable #1 is jacked to the required F_pj. There is now 1*F_pj kips in your support.
2. Cable #1 is backstressed to 80% of MUTS. They grab the cable with the ram on the other side of the support and jack to that total force. Only additional force imparted to the support is bearing of the ram - if it is a concrete column, they can use a slotted steel plate similar to the bearing plates used when re-stressing a cable to full strength in a slab (4" square slotted steel plate). Once that has been done, F_seatloss has been removed from that cable. You now have 1*(F_pj - F_seatloss) kips in your support. The purpose of backstressing is that barrel anchors need are designed to MECHANICALLY anchor, and not by FRICTION. Thus, they need to be seated fully (80% of MUTS) so that when the cable is hit, the tendon doesn't pull thru the wedge assembly and release. Yikes.
3. Repeat 1 through 2 to the remainder of cables. If you don't backstress after each cable, then you have all that extra jacking force still in the cables, which could be a problem for your support. After this stage you have all cables loaded to F_e.
4. Car hits cables as designed (say three cables). Resultant force on your support is 3*(F_e+F_impact) + 8*F_e. Is 3-kips your required effective force?


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