Minimum Tiedown Count and Aggregate WLL
The minimum number of tiedowns a piece of cargo needs and the aggregate working load limit it needs.
Example
You enter
- Article length (ft) 24
- Article weight (lb) 12000
- Tiedowns planned 4
- Working load limit per tiedown (lb) 5400
- Each tiedown attached and secured at both ends true
You get
- Min tiedowns 4
- Required wll (lb) 6000
- Provided wll (lb) 21600
Details, formula, and sources
Which are two independent rules that both have to be satisfied. The count rule is about the article's LENGTH: an article of 5 ft or less weighing 1,100 lb or less takes one tiedown and a heavier short article takes two; anything over 5 ft and up to 10 ft takes two regardless of weight; and past 10 ft the requirement is two plus one more for each additional 10 ft or fraction. Long cargo needs more attachment points so it cannot rotate or shift within the securement, and the count keeps climbing every ten feet no matter how light the piece is. The aggregate working load limit rule is about WEIGHT: the sum of the tiedowns' working load limits must be at least half the cargo weight, on the reasoning that a tiedown restrains in more than one direction. A tiedown that passes over the load and is secured at both ends counts its full working load limit, while one anchored at only one end counts half. The two rules govern in different situations, which is why both are reported and the controlling one named. A 24 ft, 12,000 lb steel beam needs 2 + ceil(14/10) = 4 tiedowns by count while needing only 6,000 lb of aggregate working load limit, so the count governs and four chains is exactly the minimum. Reverse it -- a 4 ft, 14,000 lb block -- and the count rule asks for two while the working load limit rule asks for 7,000 lb, so the chain rating decides. A crew that has internalized only one of the two rules will be wrong about half the time. A screen; 49 CFR 393 in full, the working load limits marked on the actual hardware, and the driver's own inspection govern.
min_tiedowns = 1 if (length <= 5 ft and weight <= 1,100 lb), 2 if (length <= 5 ft and weight > 1,100 lb), 2 if (5 ft < length <= 10 ft), otherwise 2 + ceil((length - 10) / 10); required_wll_lb = 0.5 x weight_lb; provided_wll_lb = tiedowns x (full WLL if secured at both ends, half otherwise).
49 CFR 393.110 minimum tiedown count by article length and 49 CFR 393.106 aggregate working load limit at half the cargo weight, cited by section and not reproduced; both rules apply and the controlling one is named. 49 CFR 393 in full, the working load limits marked on the actual hardware, and the driver's inspection govern.
The federal securement rules are public law, cited by section rather than mirrored. The article's length and weight and the hardware's marked working load limits are the driver's own values.
Estimate. AHJ and licensed professional govern.
Field names used by the API: length_ft, weight_lb, tiedowns, wll_per_tiedown_lb, secured_both_ends, min_tiedowns, required_wll_lb, provided_wll_lb
- Count rule by article LENGTH: 1 or 2 up to 5 ft by weight, 2 through 10 ft, then 2 + one per additional 10 ft or fraction49 CFR 393.110
- Aggregate WLL rule the sum of tiedown working load limits must be at least half the cargo weight49 CFR 393.106
- Both ends a tiedown secured at both ends counts full WLL; anchored at one end it counts half49 CFR 393.106