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Difficulty: MediumCargo Securement: Tie-Down Rules and WLL

A driver is securing a rigid crate containing automated manufacturing equipment on a flatbed trailer. The crate measures 23 feet23\text{ feet} in length, weighs 28,000 pounds28,000\text{ pounds}, and is loaded without front-end blocking or contact against a bulkhead. If the driver uses tie-downs rated at a Working Load Limit (WLL) of 3,000 pounds3,000\text{ pounds} each, what is the minimum number of tie-downs required to meet federal cargo securement standards?

  1. A
    3 tie-downs3\text{ tie-downs}
  2. B
    2 tie-downs2\text{ tie-downs}
  3. 5 tie-downs5\text{ tie-downs}Answer
  4. D
    10 tie-downs10\text{ tie-downs}

Answer

A minimum of 5 tie-downs5\text{ tie-downs} is required.
To secure cargo safely and legally, the securement system must satisfy both length and weight mandates: (1) For length, an unblocked 23-foot23\text{-foot} item requires one tie-down per 10 feet10\text{ feet} or fraction thereof, resulting in 3 tie-downs. (2) For weight, the aggregate WLL must equal at least 50%50\% of the 28,000-pound28,000\text{-pound} load, which is 14,000 pounds14,000\text{ pounds}. Dividing 14,000 lbs14,000\text{ lbs} by the 3,000-pound3,000\text{-pound} rating of each strap gives 4.674.67, requiring 5 straps. The driver must apply the stricter of the two requirements, meaning at least 5 tie-downs are necessary.

Step-by-Step Solution

1
Calculate the minimum number of tie-downs required by cargo length.
At least 3 tie-downs3\text{ tie-downs} based on length.
Federal regulations mandate at least one tie-down for every 10 feet10\text{ feet} of cargo length or fraction thereof. An unblocked article measuring 23 feet23\text{ feet} requires 1 for the first 10 ft+1 for the next 10 ft+1 for the remaining 3 ft=3 tie-downs1\text{ for the first } 10\text{ ft} + 1\text{ for the next } 10\text{ ft} + 1\text{ for the remaining } 3\text{ ft} = 3\text{ tie-downs}.
2
Calculate the minimum aggregate Working Load Limit (WLL) required by cargo weight.
Required aggregate WLL=14,000 pounds\text{WLL} = 14,000\text{ pounds}.
The aggregate WLL of all securement devices must equal at least 50%50\% (one-half) of the total weight of the cargo: 28,000 lbs2=14,000 pounds\frac{28,000\text{ lbs}}{2} = 14,000\text{ pounds}.
3
Determine the number of 3,000-pound3,000\text{-pound} rated tie-downs needed to satisfy the aggregate WLL and compare with the length requirement.
14,000 lbs3,000 lbs/strap=4.67→5 tie-downs\frac{14,000\text{ lbs}}{3,000\text{ lbs/strap}} = 4.67 \rightarrow 5\text{ tie-downs}.
Because fractional tie-downs cannot be used, the count must be rounded up to 5 (5×3,000 lbs=15,000 lbs≥14,000 lbs5 \times 3,000\text{ lbs} = 15,000\text{ lbs} \ge 14,000\text{ lbs}). Taking the higher of the length requirement (3) and the WLL requirement (5) yields 5 tie-downs.

Key Concept

Cargo securement requires satisfying both length-based tie-down minimums (one per 10 feet or fraction thereof) and weight-based aggregate Working Load Limit minimums (at least 50% of total cargo weight).
Estimated Time:1m 15s
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Topics

Cargo Securement: Tie-Down Rules and WLL

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