Q&A forum: Wire Rope Calculator
(including Warrington and Seale)

The attached image shows the construction of a typical 6x19 → 7x19
There are 163 filaments (not 133).
The (two) filament diameters appear to be 1.3mm & 2.418mm (my estimate from the image)

As you know, the wire rope calculator allows for just one filament size, so you need to work out an equivalent diameter for bending stiffness (EI) and the average axial stiffness (EA) to find the expected bend radius and axial growth.

The Young’s modulus of the material of the two filaments are assumed to be the same, so I can ignore 'E'.
The square root of the area (axial stiffness) of the 2.418mm filament is 0.9357 of the 1.3mm filament.
This means that the equivalent [single] filament size is 2.3mm (taking into account an estimated helix angle of 80°).
This gives the following result:
<INPUT DATA>
T [%],19.5
Ø,32
d,2.3
n,163
SMYS,1770
ρ,7.85E-6
Eᶠ,2.07E+5
<OUTPUT DATA>
Fb,1,198,688.625
Aᶠ,677.22522
ρˡ,0.005316
A,804.247742
I,51,471.855469
E,182,433.328125
ρᵖ,84.206045
Iᵀ,49,493.523438
Eᵀ,185,122.828125
ρᵖᵀ,85.872483
Rᵀ,2,060.103027
δØᵀ,-0.975049
δLᵀ,1.978997
for an equivalent stretch calculation.

Given that I have estimated the filament diameters from a photograph (attached) and I don’t know the lay angle, I may be a little out.
But given that this filament diameter (2.3mm) gives an extension of 2%, and 2.37mm makes the extension go negative, it cannot be far out.

As a result of one of our customer's problems disseminating the output data in the wire rope calculator, we have decided to rewrite it with a view to making it easier to use.

This calculator was reissued (version 1.1) on
the 17th of September 2016

1) The properties of a wire rope are defined by the filament diameter, strength and packing density (wire rope diameter), not the construction 'type'. The packing density is only achievable with particular construction types. Therefore the construction types are unnecessary to define the loaded wire rope.

2) The bending stiffness provided is for a wire rope with 50% breaking load applied. Under such conditions the filament Young's modulus has negligible effect on the bending stiffness of the steelcord. However, as your customer is the second person to raise this question I have modified the calculator to provide [include] bending stiffness and radius for the wire rope in unloaded condition.

The Wire Rope calculator includes no 'modelling' in its truest sense.

Over many years dealing with multi-strand wire I have generated numerous models for the properties of various wire rope constructions (steelcord, hosecord, tyrecord, OTR, structural, etc.) in an attempt to find an all-encompassing model that can be successfully applied across the board. I am afraid that to-date I have not yet discovered one.

In doing so, however, I have been able to establish a set of factors that provide an acceptable level of accuracy for most constructions in most applications and that require minimal input data. It is these factors that have been used in the Wire Rope calculator.

The Wire Rope calculator is not a ‘right-or-wrong’ facility in the same way that most of CalQlata’s other products are, in that they provide one mathematically correct result. If I was to provide a mathematically correct model capable of extreme accuracy for any steelcord construction, the input data necessary to ensure an accurate result would need to be extensive and in some cases difficult to acquire as the information is copyright protected by the manufacturer. It would also only apply to that piece of cord, the same properties of another section of cord extracted from the same length would most probably not be so accurately predicted using the same model.

By way of assistance; structural wire ropes such as hosecord, mooring, lifting and bridge building constructions are normally (but not always) relatively consistent in their properties and generally work well with the Wire Rope calculator. Tyrecord is also fairly predictable but OTR can be wildly variable if manufactured with ‘less-than-best’ quality and even my ‘universal’ constants would struggle to predict the properties of some of these cords within an acceptable degree of accuracy. That said most of the complex models I have generated in the past would also have failed to produce reliable results for many OTR constructions.

Note: When I have required accurate data in the past I always tested the wire rope concerned, but frequently discovered that even the properties of pieces that were cut from the same manufactured lengths varied in some cases by more than the accuracy I would expect from my ‘universal’ factors.

Dear Sir
Your query concerning the Wire Rope calculator ...
It looks to me that filament diameter of 1.3 is too large for the rope diameter
If you change this to 1.2 or increase the diameter to 16.8 the calculation works perfectly well
How sure are you of the filament and/or wire diameters?

Postscript:
You should remember that all our calculators follow the GiGo principle
In most cases the wire rope calculator will know if you are trying to squeeze a quart into a pint pot
Its calculations rely on actual dimensions rather than design or catalogue dimensions, i.e.;
The 'Ø' dimension represents a hole that the cord will pass through freely.
Most catalogue dimensions are design-target integer values.

Wire rope is not as predictable as homogeneous metals, that much everybody understands.

Your customer is very observant in his suggestion that friction is taken into account in our calculation of E in the calculator, albeit not exactly:
a) The formula for E is for the bending plane (normal to the axis of the wire), it does not apply to the tensile modulus used for axial or torsional calculations (see explanation below)
b) The formula used in the calculator for bending stiffness has been established over many years of testing and applies to the cord at (or close to) 50% breaking strain
c) This formula applies to bend radii at or close to minimum bend radius

Whilst the above may not be much comfort, I can recommend a rule of thumb for his calculation, which appears to work at relatively low tensions (note: E for wire rope increases with tension), i.e. less than 50% breaking load:
EA = E.(½A).ρᵖ.Cos(average lay angle)
Where:
E = tensile modulus of the material from which the filaments are made
½A = half cross-sectional area of all the filaments (less than half the filaments will generate more than half the resistance at low tensions)
ρᵖ = packing density of the filaments
average lay angle = the average lay angle of all the filaments and the strands in the cord. Regular lay 7x7 is approximately 35° (strand angle - filament angle), lang lay will be approximately 55° (strand angle + filament angle)

It has been my experience that EA for most wire ropes (not OTR as this will be noticeably less) will be about 10% that expected for a solid piece of the filament wire material the same diameter of the wire rope. But the above calculation is more accurate. However, your customer’s cable configuration has only one strand and very large filament diameters, so I would expect his EA value to be nearer to 25% than 10%

E for torsional calculation of stiffness will be different for both directions of twist and will vary similarly to the values for E in EA but the angle you use in the calculation will be different.

Moreover, I must admit I have never studied the behaviour of Titanium in wire rope. It has a much lower tensile modulus that steel but if the filaments are kept within their elastic deformation range the calculations should work reasonably well. My only word of caution is that under significant compression (e.g. between two adjacent filaments) the coefficient of friction will be different, resulting in [slightly] higher EI values. I suggest that these should be increased by 5% for titanium over those predicted in the wire rope calculator.

All of that said, your customer has quite rightly pointed out that the technical help web page for this calculator does not make this sufficiently clear.