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Zorluk: OrtaCargo Securement: Tie-Down Rules and WLL

A driver is transporting an unblocked crate containing industrial pump machinery on an open flatbed trailer. The cargo measures 22 feet22\text{ feet} in length and weighs 24000 pounds24{}000\text{ pounds}. If indirect synthetic web straps with a Working Load Limit (WLL) of 4000 pounds4{}000\text{ pounds} each are used across the top of the cargo, what is the minimum number of straps required to comply with federal cargo securement regulations?

  1. A
    3 straps3\text{ straps}
  2. 4 straps4\text{ straps}Cevap
  3. C
    6 straps6\text{ straps}
  4. D
    2 straps2\text{ straps}

Cevap

4 straps4\text{ straps}
Federal Motor Carrier Safety Administration (FMCSA) cargo securement standards require meeting two separate criteria: aggregate Working Load Limit (WLL) and tie-down count based on length. For weight, the aggregate WLL must equal at least 50%50\% of the cargo weight (12000 pounds12{}000\text{ pounds} for a 24000-pound24{}000\text{-pound} load), which requires 3 straps3\text{ straps} rated at 4000 pounds4{}000\text{ pounds} each. For length, when cargo is not blocked against a bulkhead or header board, regulations require 2 tie-downs2\text{ tie-downs} for the first 10 feet10\text{ feet}, plus 1 tie-down1\text{ tie-down} for each additional 10 feet10\text{ feet} or fraction thereof. A 22-foot22\text{-foot} article requires 2+1+1=4 tie-downs2 + 1 + 1 = 4\text{ tie-downs}. Because both rules must be satisfied, the driver must use at least 4 straps4\text{ straps}.

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1
Calculate the minimum aggregate Working Load Limit (WLL) required by weight.
Aggregate WLL must equal at least 50%50\% of the cargo weight: 0.50×24000 pounds=12000 pounds0.50 \times 24{}000\text{ pounds} = 12{}000\text{ pounds}.
Federal regulations mandate that the total WLL of all securement devices must equal at least half the weight of the cargo.
2
Determine the number of straps required to satisfy the weight requirement.
12000 pounds÷4000 pounds/strap=3 straps12{}000\text{ pounds} \div 4{}000\text{ pounds/strap} = 3\text{ straps}.
Each indirect strap provides its full rated WLL of 4000 pounds4{}000\text{ pounds} toward the aggregate requirement.
3
Calculate the minimum number of tie-downs required by cargo length for unblocked freight.
First 10 feet=2 tie-downs10\text{ feet} = 2\text{ tie-downs}; next 10 feet(10–20 ft)=1 tie-down10\text{ feet} (10\text{–}20\text{ ft}) = 1\text{ tie-down}; remaining 2 feet(20–22 ft)=1 tie-down2\text{ feet} (20\text{–}22\text{ ft}) = 1\text{ tie-down}. Total by length =2+1+1=4 tie-downs= 2 + 1 + 1 = 4\text{ tie-downs}.
Articles not blocked against a front-end structure require two tie-downs for the first 10 feet10\text{ feet} of length, plus one tie-down for every additional 10 feet10\text{ feet} or fraction thereof.
4
Compare the requirements and select the governing minimum.
The cargo requires max⁡(3,4)=4 straps\max(3, 4) = 4\text{ straps}.
Both the weight-based WLL rule and the length-based count rule must be satisfied simultaneously.

Anahtar Kavram

Cargo Securement: Aggregate Working Load Limit (WLL) and Unblocked Length-Based Tie-Down Rules
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Cargo Securement: Tie-Down Rules and WLL

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