OSHA Guardrail 3

[cal91]

I'm checking a 3/8" Ø aircraft cable to be used for guardrail and lifeline during construction. Everything is checking out, except for the 3" max deflection requirement for guardrail found under OSHA 1910.29(b)(4)

"When the 200-pound (890-N) test load is applied in a downward direction, the top rail of the guardrail system must not deflect to a height of less than 39 inches (99 cm) above the walking-working surface."

The guardrail is at 42" so 42"-39" = 3" deflection.

With the stretch in the line accounted for I'm getting that I can't have more than an 8 foot length of cable between anchorages, which is impractical. I also know that this cable is used all the time as guardrail for up to 120' between anchorages. I am trying to get this to work for up to 18'0" spans and up to 120'-0" between end anchorage. But that case gives me 12" of deflection even though strength works out.

Just wanted to pick the brains of you fellow eng-tippers (is that what we're called?) about this situation. Do I not even need to apply this requirement since it's only for construction and not permanent?

 

[cal91]

CANPRO, I appreciate your time in this discussion and respect your thoughts, but pretension plays a larger role than what you are giving credit for.

If I understand your process correctly (please correct me if I'm wrong), pretension's benefit is limited to the magnitude of the initial sag, because that is all it eliminates. I.E. if there was a safety cable in space with 0 gravity when a 200lb point load is applied, then pretension would make no difference in the deflection of the cable, because there was no initial sag. I would disagree and say that pretension would result in a lesser deflection. Here's an analogy to explain my point.

Take a guitar and loosen one of its strings (but not to the point that it starts deflecting). You'll find it takes little force to deflect the string 1" transversely. Tighten it up a bunch and apply the same amount of force, and it'll barely budge. In both cases there was no initial deflection, yet the pretension provides a drastic change in transverse stiffness. Pretension's role is not solely limited to eliminating the initial deflection. I would say in fact that unless the wire weighs the same magnitude as the load being applied to it, initial deflection effectively plays no role at all.

The mathematical explanation for the guitar string analogy, which we can apply to our safety cables is captured by equilibrium and constitutive equations. We agree on the equilibrium equation, and we agree on part of the constitutive equation. The part of the constitutive equation we don't agree on is where I account for a pretensioned cable being SHORTER than the length between it's anchors. That's what pretensioning is, taking a cable that is shorter than its anchorage length, and stretching it to be the anchorage length. Also in the constitutive equation is the additional stretch associated with the point load applying even more tension in the cable, which is the part we agree on.

I apologize if I have misinterpreted or misrepresented your process, feel free to correct if so.

 

[dik]

I'm not sure if I understand the disagreement, if there is one. When a cable is tensioned at the ends and a horizontal load is applied at some point between vertical support posts, the tension in the cable is increased. This increase in tension extends in both directions for the full length of the cable and the entire cable contributes to the elongation. I think cal91's calculation reflects this.

Dik

 

[cal91]

Dik, I'm not entirely sure there is a disagreement either, unless CANPRO is indeed analyzing the affects of pretension in the cable as I stated above.

If I understand your process correctly (please correct me if I'm wrong), pretension's benefit is limited to the magnitude of the initial sag, because that is all it eliminates.

 

[BridgeSmith]

If you ignore the weight of the wire (which is reasonable for this application), you can calculate the required pretension directly by considering the cable deflected to its limiting value of 3" under a 200lb load. The force vector triangle is a similar triangle to the triangle created by the deflected shape, so tension is equal to the length of the cable from the point of support to the point of deflection, assumed to be at the middle of the 10' span (60.07" in this case), divided by deflection, and multiplied by the force going to that side, which is half the total (think of it as two 100lb forces applied a fraction of an inch apart at the middle of the span). In this case, 60.07" / 3" * 100lbs = 2002lbs. From there, subtract the tension required to change the length from 60ft (undeflected) to 60.0125ft (deflected shape) ---> 3018psi --> 215lbs, and the required pretension is 1786lbs.

 

[CANPRO]

My view isn't that it only removes sag - my view is it limits the amount of cable that is available to form the fully deflected cable, which reduces the loaded deflection. However, there comes a point that increasing your pretension no longer reduces the deflection in any meaningful way. My calculations reflect this, and so do cal91's - our difference is in how fast we get to the limiting deflection and what that limit is - my method trends towards a higher deflection at a faster rate than cal91's method. Compared to cal, I have more deflection but less tension.

Cal, the one line out of your calculations that I don't agree with is where you write L = L[sub]0[/sub] + stretch. The stretch due to the point load should be added to the actual cable length immediately before the load is applied, in this case 60ft.

Consider this - I think we both agree that eventually there will be a limit to how much loaded deflection you can reduce regardless of increasing pretension - since the additional tension in the cable from the point load is a function of the geometry of the cable, that means the additional cable tension is limited as well, which coming full circle means that your additional stretch will eventually be limited too. With that said, what happens to your calculations as your pretension increases (and L[sub]0[/sub] decreases) to the point where L[sub]0[/sub] + cable stretch is < 60 ft?

If L0 continues to decrease as pretension increases but the cable stretch eventually reaches a limiting value, then eventually the hypotenuse of your loaded triangle is going to be less than 5ft at which point your iterations will crash. I re-created your calcs in excel and that's what happened. At about 1555 lbs of pretension, (L[sub]0[/sub] + cable stretch) was less than 60' and excel returned an error. With a pretension of 1200lbs, the solution appeared to converge on 0.3' fairly quick, but returned an error after additional iterations.

I think your results that you posted on April 6 also indicate there is an error in your method - your results suggest that when PT = 0, the sag due to a 200 lbs point load is 6.32"... Forget about all the math and debate over initial lengths for a second and picture there is a 3/8" cable laid out in front of you 60 ft long. It is laying on the ground, and someone is standing on each end of the cable - it is anchored 60' apart and there is no tension in the cable. If you picked that cable up in the middle, do you really think it would take 200 lbs of force to move that cable 6.32"? You would move that cable significantly farther than that with much less force.

The link that dik posted on April 7 has an example you can use to check your work against. See page 3 of the PDF - their calculated loaded sag is 1.66 ft. With your method, I get within 5% BUT if you input too high of a pretension, the calculation fails. My spreadsheet returns a value of 1.66ft. Our results are much closer in this example because the cable length is 200' and the difference between initial lengths becomes relatively minor.

 

[dik]

Can anyone post a table of cable properties, for example, 7-19 cable, sizes, yield, area of steel and modulus of elasticity... I can only find tables with safe loading.

My experience with pre-tension is that it is often difficult to find something to anchor the cable ends to.

Thanks, Dik

 

[CANPRO]

dik, cal91 posted the effective properties of his cable on April 6.

cal91, just wanted to add a comment on my last post - I did that late at night (I've been obsessing over this problem a bit)...now that I've slept on it, I realized that you posted results from pretensions in excess of 10,000lbs previously...when I tried to re-create your procedure based on your calcs from April 10 I returned an error at higher pretensions based on my description above, so I may not be interpreting your method correctly.

Second thing - in my thought experiment with the cable on the ground and the 200 lbs vs 6.32" deflection...that wasn't really a fair comparison because your 6.32" deflection occurred in a 10ft span where in my example it occurred in a 60ft span. However, I still think you could reach that deflection value with much less effort than 200 lbs.

I probably shouldn't be trying to figure this out late at night...but I can't let this consume my time at work either. I would very much like to reconcile our results. I've started to look at this problem from a work and energy perspective and I believe that both of our current methods contain an error. When I have time I will expand on this.

 

[dik]

cal91 posted the effective properties of his cable on April 6.

Looking for a *.pdf file with properties dia, As, fy, fu, E, etc. for a 'bunch' of different cables, not just the one. Do you have such a table? or anyone else? I can find a lot of info on 'breaking strength' or 'working strength', but, nothing with the info I seek. In some instances, the published E value is approx half that of a solid bar.

thanks, Dik

 

[BridgeSmith]

I added the weight of the cable into the Excel spreadsheet I developed for span-wire signal structures, applied the 200lb load and limited the deflection to 3". The total cable tension was 2010lbs. The change in length of the cable to deflect 3" is 012493'. Using a A = 0.0714 sq in and E = 14,500,000 psi from Cal91, the change in tension stress due to the deflection is 3019 psi (60.012493' / 60' * 14500000), yielding a tension force of about 216 lbs. The required pretension is about 1800 lbs, assuming rigid end anchorages. If the cable anchorages have some give to them, this reduces the allowable change in tension below the 216 lbs, so the required pretension gets closer to the 2010 lbs unadjusted value the more "give" there is in the end anchorages.

 

[dik]

I'm writing a small program in SMath and want to include an array with this information so it can automatically be included with the data just by specifying an index no.

Dik

 

[CANPRO]

Dik, see the link below - this document seems to provide good info on effective area and modulus of elasticity for a variety of wire rope types. It also covers the topic of prestretched rope. The good stuff is on page 16 of the PDF (page number 24/25 in the document).

http://www.cimafbrasil.com.br/down/espec_rev3_220413demos.pdf

Cal, I think to really iron out the fundamental difference to our approach, we should simplify the problem. Lets consider this: we have a vertical wire rope anchored over head with zero initial tension and initial length of L[sub]0[/sub]. A load P[sub]1[/sub] is applied to the cable and it stretches a value of Δ[sub]1[/sub], and the new cable length is L[sub]1[/sub]. The load P[sub]1[/sub] is equivalent to our pretension. Once the cable has stretched, an additional load P[sub]2[/sub] is applied, and the cable stretches Δ[sub]2[/sub] to a final length of L[sub]2[/sub]. P[sub]2[/sub] being our 200lbs point load.

The below calculations map out the solution to this problem using work and energy. First thing to note is that a strong argument could be made that the 200lbs load is applied suddenly, in which case we're both off by a significant margin.

I am unsure if in step 2 you would use length L[sub]1[/sub] or L[sub]0[/sub]. I think at this step you're using L[sub]0[/sub] and I'm using L[sub]1[/sub]. I'd have to dig deeper into my calculations to really see how that translates to my original method. If someone would comment on this, that would be great.

I think the biggest difference in our methods is this - when you apply the load P[sub]2[/sub] you are adding the additional stretch Δ[sub]2[/sub] to L[sub]0[/sub], which mathematically means that you're considering the load P[sub]2[/sub] to travel a distance Δ[sub]1[/sub] and Δ[sub]2[/sub] instead of just Δ[sub]2[/sub]. Basically, I think your method returns results assuming that P[sub]1[/sub] and P[sub]2[/sub] are applied at the same to the cables initial state, which is not the same thing as P[sub]1[/sub] being applied and then P[sub]2[/sub].

Your statement that I'm underestimating the effect of pretension may very well be true, as I'm starting to think that I'm neglecting the fact that P[sub]1[/sub] also travels a distance Δ[sub]2[/sub]. Since I didn't approach my original solution this way, I'm not 100% sure about that, I'll have to take a closer look through my original solution. However, I'm comforted by the fact that my spreadsheet matches the results from the PTI link that dik provided above. That said, I don't like to be right by accident. I was very confident about my solution previous to this conversation, but not so much now.

 

[dik]

Just what I was looking for... thank you very much.

Dik