Federal compliance tool
Load Securement Calculator
Enter cargo weight and length to find the minimum tiedowns required under 49 CFR 393.106 (aggregate WLL rule) and 49 CFR 393.110 (length rule). Results are instant and shareable.
Educational estimate — not a compliance certification
This calculator applies the FMCSA minimums for standard general freight. The driver and carrier are solely responsible for load securement. Specific commodities — steel coil, logs, heavy machinery, pipe — have additional rules under 49 CFR 393.116–393.136. Loads that shift, roll, or require blocking and bracing may need more tiedowns than the minimums shown. Always verify against the full 49 CFR Part 393 and the rated WLL on your actual hardware.
Cargo details
Gross weight of cargo only (not trailer)
Longest dimension of the article(s)
Tiedown type / WLL per tiedown
Results
Enter cargo details to see
your minimum tiedown count
Minimum tiedowns required
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governed by —
49 CFR 393.106 — Aggregate WLL
Governs– tiedowns
49 CFR 393.110 — Length rule
Governs– tiedowns
Calculations apply 49 CFR 393.106 (aggregate WLL ≥ 50% of cargo weight) and 49 CFR 393.110 (tiedown count by article length) for general freight. Commodity-specific annexes (coil, pipe, concrete pipe, intermodal containers, automobiles, etc.) are not reflected here. Read 49 CFR Part 393 Subpart I →
Securement questions, answered
How many tiedowns do I need for a flatbed load?
Federal regulations require the higher of two minimums. The aggregate WLL rule (49 CFR 393.106) demands that the combined working load limit of all tiedowns equals at least half the cargo's weight. The length rule (49 CFR 393.110) sets a floor based on how long the article is — one tiedown for loads 5 ft or shorter, two for 5–10 ft, plus one more for every additional 10 ft. Both rules apply; use whichever number is larger.
What is working load limit (WLL)?
WLL is the maximum force a tiedown device may experience during transport, as rated by the manufacturer. Common flatbed tiedown WLL values: 5/16″ Grade 70 chain = 4,700 lbs; 3/8″ Grade 70 chain = 6,600 lbs; 2″ web strap = 3,335 lbs; 4″ web strap = 5,400 lbs. Always use the WLL marked on the hardware, never an estimated or memory value. Worn, kinked, or damaged equipment must be removed from service regardless of its marked WLL.
How does cargo length affect tiedown count?
Under 49 CFR 393.110: articles 5 ft or shorter need 1 tiedown (2 if the article exceeds 1,100 lbs). Articles between 5 and 10 ft need 2 tiedowns. For every additional 10 ft beyond 10 ft, add one more tiedown — so a 28-ft load needs 2 + ⌈(28−10)÷10⌉ = 4 tiedowns from the length rule alone. The aggregate WLL rule may require more.
What's the difference between 49 CFR 393.106 and 393.110?
Section 393.106 is the strength test: do your tiedowns collectively have enough rated capacity to restrain this load? Section 393.110 is the coverage test: are there enough tiedowns spaced along the length of the cargo? A single high-capacity tiedown might satisfy 393.106 for a 15-ft, 5,000-lb load, but 393.110 still requires at least 2 tiedowns because the article is longer than 10 ft. Both sections appear on every FMCSA roadside inspection; a violation of either can ground the vehicle.
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