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Difficulty: MediumCargo Securement: Tie-Down Rules and WLL

A commercial driver is preparing to transport an unblocked shipment of structural steel I-beams measuring 28 feet28\text{ feet} in length and weighing 40000 pounds40{}000\text{ pounds} on a flatbed trailer. If the driver secures the load using the minimum number of tie-downs required by length, what is the minimum Working Load Limit (WLL) that each individual tie-down must have to meet federal cargo securement standards?

  1. A
    2500 pounds2{}500\text{ pounds}
  2. B
    6667 pounds6{}667\text{ pounds}
  3. 5000 pounds5{}000\text{ pounds}Answer
  4. D
    10000 pounds10{}000\text{ pounds}

Answer

Each tie-down must have a minimum Working Load Limit of 5000 pounds5{}000\text{ pounds}.
Securing an unblocked 28-foot28\text{-foot} load requires 4 tie-downs: 2 for the first 10 feet10\text{ feet}, 1 for the second 10 feet10\text{ feet}, and 1 for the remaining 8 feet8\text{ feet}. To satisfy the federal requirement that total aggregate Working Load Limit equals at least 50%50\% of the 40000-pound40{}000\text{-pound} cargo weight (20000 pounds20{}000\text{ pounds}), dividing 20000 pounds20{}000\text{ pounds} by 4 tie-downs yields a minimum requirement of 5000 pounds5{}000\text{ pounds} per tie-down.

Step-by-Step Solution

1
Calculate the minimum number of tie-downs required by length for unblocked cargo.
4 tie-downs
Federal regulations require 2 tie-downs for the first 10 feet10\text{ feet} of cargo length, plus 1 additional tie-down for each additional 10 feet10\text{ feet} or fraction thereof (10 ft+10 ft+8 ft=2+1+1=4 tie-downs10\text{ ft} + 10\text{ ft} + 8\text{ ft} = 2 + 1 + 1 = 4\text{ tie-downs}).
2
Calculate the required aggregate Working Load Limit (WLL) for the cargo.
20000 pounds20{}000\text{ pounds}
The aggregate WLL of all securement devices must equal at least 50%50\% (1/21/2) of the total cargo weight (0.50×40000 lbs=20000 lbs0.50 \times 40{}000\text{ lbs} = 20{}000\text{ lbs}).
3
Determine the minimum WLL required per individual tie-down across the 4 required tie-downs.
5000 pounds5{}000\text{ pounds} per tie-down
Dividing the required aggregate WLL of 20000 pounds20{}000\text{ pounds} evenly across the 4 required tie-downs gives 20000 lbs4=5000 pounds\frac{20{}000\text{ lbs}}{4} = 5{}000\text{ pounds}.

Key Concept

Cargo Securement: Length-based minimum tie-down rules and aggregate Working Load Limit (WLL) requirements.
Estimated Time:1m 30s
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Topics

Cargo Securement: Tie-Down Rules and WLL

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