A driver secures an unblocked industrial transformer measuring 26\text{ feet} in length an… — Commercial Driver's License (CDL) General Knowledge Exam

A driver secures an unblocked industrial transformer measuring $26\text{ feet}$ in length and weighing $30{}000\text{ pounds}$ on a flatbed trailer. The driver uses exactly four tie-downs: two indirect tie-downs routed side-to-side over the cargo, each with a working load limit (WLL) of $5{}400\text{ pounds}$ , and two direct tie-downs attached from the trailer to the cargo's anchor points, each with a WLL of $4{}200\text{ pounds}$ . This securement configuration satisfies federal regulations for both the minimum tie-down quantity and the minimum aggregate WLL.

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The statement is true; the securement precisely meets both the minimum quantity requirement and the aggregate WLL requirement.

The securement is perfectly compliant. The unblocked

26\text{-foot}

transformer requires exactly 4 tie-downs (2 for the first

10\text{ ft}

, 2 for the remaining

16\text{ ft}

). The required WLL is half the weight (

15{}000\text{ lbs}

). The indirect tie-downs provide 100% of their WLL (

10{}800\text{ lbs}

), while the direct tie-downs provide 50% of their WLL (

4{}200\text{ lbs}

), totaling exactly

15{}000\text{ lbs}

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Calculate the minimum required number of tie-downs based on length.

2 tie-downs for the first

10\text{ feet}

, plus 2 tie-downs for the remaining

16\text{ feet}

(two fractions of

10\text{ feet}

). Total = 4 tie-downs.

Federal rules require unblocked cargo longer than

10\text{ feet}

to have 2 tie-downs for the first

10\text{ feet}

and 1 additional tie-down for every

10\text{ feet}

or fraction thereof.

Calculate the required minimum aggregate Working Load Limit (WLL).

30{}000 \times 0.5 = 15{}000\text{ pounds}

The aggregate WLL must be at least 50% of the total cargo weight.

Calculate the provided WLL from the indirect tie-downs.

2 \times 5{}400 = 10{}800\text{ pounds}

Tie-downs that pass over the cargo from one side of the vehicle to the other count for 100% of their rated WLL.

Calculate the provided WLL from the direct tie-downs.

2 \times (4{}200 \times 0.5) = 4{}200\text{ pounds}

Tie-downs that attach from the vehicle directly to an anchor point on the cargo only count for 50% of their rated WLL.

Sum the total provided WLL and evaluate compliance.

10{}800 + 4{}200 = 15{}000\text{ pounds}

. The securement is fully compliant.

The provided WLL (

15{}000\text{ pounds}

) exactly meets the required minimum (

15{}000\text{ pounds}

), and the driver used the correct quantity (4).

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Cargo Securement: Tie-Down Rules and WLL

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