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vwmeche94

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PF,

I need to design a cart that will move (horizontally/linearly) a load of ~40,000 lbs including its own weight (about 6500 lbs) at a total distance of 70 ft on a set of rails.

Known data:

1. Cart will need to move at a distance of 1 ft/s

2. Diameter of the cart wheels (4) is 15 in.

3. Coefficient of friction, COF (steel-on-steel) = 0.8

4. The shaft diameter (4140 steel) is 2 3/16"

5. Electric motor of 2.5 HP

I have calculated the diameter to yield 15.27 rpm and FLT (Full Load Torque) of a 2.5 HP motor to be 859.85 lb-ft using that rpm.

Based on an equation shared with me from a Machine Design Book,

W (weight) x mu (COF) / r (radius) = 40,000 lb x 0.8 ÷ 7.5 in = 4266 lb/in - load at each wheel.

What I need to determine is the shaft torque under load and the type of gearing (gear box) I should use to accelerate to and maintain a speed of 1 ft/s with a 40,000 lb load. I also need to determine if the 2.5 HP motor is adequate enough to handle repeated moving of the load.

Thank you in advance,

I need to design a cart that will move (horizontally/linearly) a load of ~40,000 lbs including its own weight (about 6500 lbs) at a total distance of 70 ft on a set of rails.

Known data:

1. Cart will need to move at a distance of 1 ft/s

2. Diameter of the cart wheels (4) is 15 in.

3. Coefficient of friction, COF (steel-on-steel) = 0.8

4. The shaft diameter (4140 steel) is 2 3/16"

5. Electric motor of 2.5 HP

I have calculated the diameter to yield 15.27 rpm and FLT (Full Load Torque) of a 2.5 HP motor to be 859.85 lb-ft using that rpm.

Based on an equation shared with me from a Machine Design Book,

W (weight) x mu (COF) / r (radius) = 40,000 lb x 0.8 ÷ 7.5 in = 4266 lb/in - load at each wheel.

What I need to determine is the shaft torque under load and the type of gearing (gear box) I should use to accelerate to and maintain a speed of 1 ft/s with a 40,000 lb load. I also need to determine if the 2.5 HP motor is adequate enough to handle repeated moving of the load.

Thank you in advance,

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