Cargo Securement Working-Load-Limit Check
Aggregate WLL vs. the half-cargo-weight requirement, pass/fail, and the minimum tiedown count for the article length.
Example
You enter
- Cargo weight (lb) 8000
- Number of tiedowns 4
- WLL per tiedown (lb, marked) 1500
- Article length (ft) 16
You get
- Aggregate WLL 6,000 lb
- Required (1/2 cargo weight) 4,000 lb
Checked against published worked examples from: FMCSA 49 CFR 393.100-393.136.
Details, formula, and sources
ScopePer FMCSA 49 CFR 393.100-393.136; WLLs user-supplied from the marked hardware.
Formula or tableAggregate WLL = number of tiedowns x per-tiedown WLL x path credit (1.0 over the cargo to the other side, 0.5 direct or same-side, per 49 CFR 393.106(d)); required >= 0.5 x cargo weight; minimum tiedown count from the length/weight rule of 393.110(b) (<= 5 ft: 1, or 2 over 1,100 lb; over 5 to 10 ft: 2; over 10 ft: 2 plus 1 per additional 10 ft or fraction).
Source and editionFMCSA 49 CFR 393.100-393.136 cargo securement (aggregate-WLL and tiedown-count rules), by name.
Free accessFree at ecfr.gov; FMCSA enforces.
What governsMath aid for personal verification. The ELD on the truck and the carrier tariff are the legal record. State limits may be lower than federal.
API field namescargo_weight_lb, tiedown_count, wll_each_lb, cargo_length_ft, aggregate_wll_lb, required_wll_lb
- WLL the marked rating of each tiedown (not breaking strength); the lowest-rated component governs each tiedown49 CFR 393.102