Wire-Rope Diameter for a Required WLL
The wire rope diameter for a required working load limit.
Example
You enter
- Required working load limit (tons) 5
- Construction factor (tons/in²) 46
- Design factor (safety factor) 5
You get
- Exact diameter required 0.737 in
- Next standard size 0.75
Details, formula, and sources
d = sqrt(WLL x design factor / construction factor), then rounded up to the next standard size. A 5-ton WLL at 46 tons/in^2 and 5:1 needs 0.74 in -- pick 3/4 in (5.18 t). An ESTIMATE only; the manufacturer's certified rating governs, and unmarked rope must not be placed in service.
diameter_in = sqrt(wll_required_tons x design_factor / construction_factor), the MBS = factor x d^2 / WLL = MBS / design_factor rule-of-thumb solved for the diameter; round up to the next standard rope diameter.
Wire Rope Users Manual rule-of-thumb (by name), solved for the diameter; MBS = factor x d^2, no edition cycle.
An ESTIMATE only; the manufacturer's certified breaking strength governs. Do not place unmarked or uncertified rope in service.
Estimate. Head rigger and manufacturer working-load-limit charts govern. Inspect every piece of hardware before the show.
Field names used by the API: wll_required_tons, construction_factor, design_factor, diameter_in, selected_diameter_in
- Construction factor the default 46 is the rule-of-thumb tons/in^2 for IPS 6x19; bright IPS, EIPS, and other constructions/grades differ - edit itWire Rope Users Manual
- Design factor 5:1 is typical for general rigging; the application and the AHJ set the required factorASME B30.9
- Estimate only round up to a standard size and use the certified breaking strength for any real lift; never use unmarked or uncertified ropemanufacturer certification