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Zorluk: ZorWeight Definitions and Limits (GVWR, GCWR, Axle Weights)

A commercial two-axle flatbed straight truck is being loaded for an interstate delivery. The vehicle certification label lists a Gross Vehicle Weight Rating (GVWR) of 33,000 lbs33,000\text{ lbs}, a front steer Gross Axle Weight Rating (GAWR) of 12,000 lbs12,000\text{ lbs}, and a rear drive GAWR of 21,000 lbs21,000\text{ lbs}. The unladen tare weight of the empty vehicle is 14,500 lbs14,500\text{ lbs} (7,000 lbs7,000\text{ lbs} on the steer axle and 7,500 lbs7,500\text{ lbs} on the drive axle). On federal interstate highways, the statutory single-axle weight limit is 20,000 lbs20,000\text{ lbs}. Assuming cargo weight can be distributed optimally across both axles, what is the maximum payload weight that this vehicle can legally transport without exceeding any manufacturer rating or federal statutory limit?

  1. A
    18,500 lbs18,500\text{ lbs}
  2. 17,500 lbs17,500\text{ lbs}Cevap
  3. C
    25,500 lbs25,500\text{ lbs}
  4. D
    12,500 lbs12,500\text{ lbs}

Cevap

17,500 lbs17,500\text{ lbs}
Commercial vehicle weight compliance requires adhering to the most restrictive limitation among manufacturer ratings (GVWR, GAWR) and legal statutory limits. The front steer axle is limited by its manufacturer GAWR of 12,000 lbs12,000\text{ lbs}, while the rear drive axle is limited by the federal single-axle limit of 20,000 lbs20,000\text{ lbs} (which is lower than its 21,000 lbs21,000\text{ lbs} GAWR). Summing these maximum allowable axle weights yields a total allowable vehicle operating weight of 32,000 lbs32,000\text{ lbs}, which is below the vehicle's 33,000 lbs33,000\text{ lbs} GVWR. Subtracting the tare weight of 14,500 lbs14,500\text{ lbs} results in a maximum legal cargo payload of 17,500 lbs17,500\text{ lbs}.

Adım Adım Çözüm

1
Determine the maximum allowable weight for each axle by comparing manufacturer GAWR against statutory federal limits
Steer axle maximum = min⁡(12,000 lbs GAWR,20,000 lbs statutory)=12,000 lbs\min(12,000\text{ lbs GAWR}, 20,000\text{ lbs statutory}) = 12,000\text{ lbs}; Drive axle maximum = min⁡(21,000 lbs GAWR,20,000 lbs statutory)=20,000 lbs\min(21,000\text{ lbs GAWR}, 20,000\text{ lbs statutory}) = 20,000\text{ lbs}.
A commercial motor vehicle must never exceed either the manufacturer's structural axle rating or statutory highway weight limits.
2
Calculate the effective maximum legal Gross Vehicle Weight (GVW)
Effective maximum legal GVW = min⁡(33,000 lbs GVWR,12,000 lbs+20,000 lbs)=32,000 lbs\min(33,000\text{ lbs GVWR}, 12,000\text{ lbs} + 20,000\text{ lbs}) = 32,000\text{ lbs}.
Although the chassis GVWR is 33,000 lbs33,000\text{ lbs}, the vehicle cannot legally exceed 32,000 lbs32,000\text{ lbs} total because the drive axle cannot exceed the 20,000 lbs20,000\text{ lbs} legal statutory limit and the steer axle cannot exceed its 12,000 lbs12,000\text{ lbs} rating.
3
Subtract the unladen vehicle tare weight from the effective maximum legal GVW
Maximum legal cargo payload = 32,000 lbs−14,500 lbs=17,500 lbs32,000\text{ lbs} - 14,500\text{ lbs} = 17,500\text{ lbs}.
Payload capacity is the remaining allowable weight after accounting for empty vehicle weight.

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Weight Ratings vs. Legal Statutory Limits (GVWR, GAWR, and Axle Limits)
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Weight Definitions and Limits (GVWR, GCWR, Axle Weights)

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