A bundle of structural steel tubing measuring 19\text{ feet} in length and weighing 14{}00… — Commercial Driver's License (CDL) General Knowledge Exam

A bundle of structural steel tubing measuring $19\text{ feet}$ in length and weighing $14{}000\text{ pounds}$ is transported on a flatbed trailer and blocked against forward movement by a certified front-end header board. Securing this cargo with $2\text{ synthetic webbing straps}$ , each having a Working Load Limit ( $\text{WLL}$ ) of $4{}000\text{ pounds}$ , meets both the minimum tie-down count and aggregate $\text{WLL}$ requirements under federal cargo securement regulations.

Answer: Answer

Answer

True, because the load satisfies both the length-based tie-down requirement for blocked cargo and the

50\%

aggregate Working Load Limit rule.

The statement is correct because the cargo is blocked against forward movement, making

2\text{ tie-downs}

sufficient for

19\text{ feet}

of length (

1

per

10\text{ feet}

or fraction thereof), and the

8{}000\text{ pounds}

of combined strap rating exceeds the required

50\%

weight threshold of

7{}000\text{ pounds}

Step-by-Step Solution

Determine the minimum number of tie-downs required based on cargo length and forward blocking status.

For cargo blocked against forward movement, regulations mandate

1\text{ tie-down}

per

10\text{ feet}

of article length or fraction thereof. For a

19\text{ foot}

article:

\lceil 19 / 10 \rceil = 2\text{ tie-downs}

. The driver used

2\text{ straps}

, fulfilling the count requirement.

FMCSA regulations permit fewer tie-downs when cargo is positively prevented from forward movement by a header board or bulkhead compared to unblocked freight.

Calculate the required minimum aggregate Working Load Limit (

\text{WLL}

) based on total cargo weight.

\text{Required WLL} = 14{}000\text{ pounds} \times 0.50 = 7{}000\text{ pounds}

Federal standards require the aggregate

\text{WLL}

of all securement devices to equal at least one-half (

50\%

) of the total cargo weight.

Calculate the provided aggregate

\text{WLL}

and compare it against the required threshold.

\text{Provided WLL} = 2 \times 4{}000\text{ pounds} = 8{}000\text{ pounds}

. Since

8{}000\text{ pounds} \ge 7{}000\text{ pounds}

, the aggregate strength requirement is satisfied.

Both individual regulatory criteria (minimum number of tie-downs and aggregate strength) must be met simultaneously for lawful transport.

Key Concept

Cargo Securement: Tie-Down Rules and WLL for Blocked Cargo

Estimated Time:1m 0s

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