Elongation of Wire Rope 1

[nutte]

I wonder if anybody can help me determine how much elongation you get from a wire rope under an applied load. I downloaded the USS Wire Rope Engineering Hand Book from SlideRuleEra's web site (

http://www.slideruleera.net/USSWireRopeEngrHandBook.zip

- Thanks for hosting it). It has some information on stretch of rope, but I'm concerned it might be dated, being from 1968. Page 29 of the Hand Book has a table for approximate metallic areas of wire rope based on your diameter and rope type. Page 30 has a table for approximate moduli of elasticity, based on your material.

This sounds easy enough, but it's not like pulling "A36" from a table where you know you have a match. There are decent variations even between the closest two options.

I have seen the following equation in another engineer's work, but I don't know the source of it:

AE = 7,140,000*D^2 (where D=wire rope diameter)

For a 1/2" rope, this gives you AE=1,785,000 #.

Perhaps not coincidentally, if you used an effective diameter of 56%*D, and an E of 29,000,000 psi, you get the following:

De = 0.56*0.5" = 0.28"
Ae = pi/4*De^2 = 0.06158 in^2
AE = Ae*E = 1,785,681 #, virtually the same answer. This makes more sense to me, but who knows if it's correct.

Does anybody here know where the 7,140,000*D^2 equation comes from? Is there any validity to the "effective diameter" method? How do others here go about calcualting the stretch of a wire rope?

 

[RockEngineer]

I use the Wire Rope User's Manual 4th Edition which is put out by the Wire Rope Technical Board. It's the standard for the industry. Chapter 6 is Physical Properties and has information on stretch which varies depending on your wire rope construction. There are tables for "approximate" E based on the type of rope and whether it is above or below 20% loading. The 20% comes from using a FS=5 in wire rope design. The E varies from 8,100,000 psi to 13,500,000 psi depending on type of construction.

There is also a table for the approximate metalic area for 1" wire rope based on construction type. For other than 1" diameter rope you multiply the 1" number by your ropes diameter squared. Then they calculate stretch by PL/AE. The metallic area for 1" rope varies from 0.23 sq in to 0.505 sq in depending on type of construction.

Example
6x19 1" Seale IWRC E=12,000,000 psi A=0.470 sq in

stretch=P*L/(12,000,000*0.470)

 

[nutte]

Thanks for the information. It sounds like I need to get hold of that manual.

In the meantime, could I talk you out of the E and A values for 1/2" 6x19 EIPS IWRC?

 

[RockEngineer]

You need to know the wire rope makeup in the 6x19 IWRC class. The manual list for 6x19 class: 6x19 Seale IWRC, 6x25 Filler Wire IWRC and 6x26 Warrington Seale IWRC. You could get any of these is you only specify 6x19 EIPS IWRC. Some companies have their own special makeup but using these numbers should get you close. There isn't a huge variation within this class of wire rope.

I gave you the more common Seale 6x19 IWRC. The E for all three in 6x19 class is the same. This is the E for less than 20% of breaking strength.
E= 12,000,000 psi

6x25 FW A=0.483 sq in
6X26 WS A=0.476 sq in

 

[nutte]

Are those for 1/2"? They look closer to the example for 1" you posted above.

 

[RockEngineer]

Yes I gave you the values for 1" diameter wire rope.

As stated in the first reply "For other than 1" diameter rope you multiply the 1" number by your ropes diameter squared." So for 1/2" wire rope you multiply the approximate metalic area of 1" diameter wire rope by 1/2 squared or 0.25.

 

[nutte]

Ahh, I see. Thanks again, I sure do appreciate it.

 

[Denial]

Nutte,

The rough-as-guts formula you ask about (7140000*D^2) is very similar to the approach I take when confronted with a need to guesstimate the stretch characteristics of a wire rope about which I know nothing more that its notional diameter. I use:
E[sub]effective[/sub] = 0.5*E[sub]material[/sub]
to allow for the twisted strands,
and A[sub]effective[/sub] = 0.6*A[sub]gross[/sub]
to allow for the "circles within circles aspect of the cross section.

This leads to
(AE)[sub]effective[/sub] = 0.3*A[sub]gross[/sub]*E[sub]material)[/sub]
where your formula corresponds to
(AE)[sub]effective[/sub] = 0.303*A[sub]gross[/sub]*E[sub]material)[/sub]

Note that this approach really is as rough as guts, and you should always make every effort to ascertain the correct extensibility.