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Answer :
Final answer:
To determine the shaft diameter and appropriate key length/width to carry 1500 ft-lbs of torque, we need to calculate the stress and strength values for both the shaft and the key. Using the given values for the material properties of the shaft and key, the shaft diameter needed to carry 1500 ft-lbs of torque is approximately 3.06 inches. The appropriate key width to carry this torque is approximately 0.48 inches.
Explanation:
To determine the shaft diameter and appropriate key length/width to carry 1500 ft-lbs of torque, we need to calculate the stress and strength values for both the shaft and the key. For the shaft, we can use the formula T = (π/16) * d^3 * τ, where T is the torque, d is the diameter, and τ is the shear stress. Rearranging the formula, we get d = (16 * T / (π * τ))^(1/3). Plugging in the values for the torque and shear stress for the shaft, we can calculate the shaft diameter. For the key, we can use the equation W = (τ * L * W^2) / (2 * S_y), where W is the width, L is the length, τ is the shear stress, and S_y is the yield strength. Rearranging the equation, we get W = (2 * S_y * W^2) / (τ * L). Plugging in the values for the shear stress, yield strength, and the desired torque, we can calculate the appropriate key width.
Using the given values for the material properties of the shaft and key, the shaft diameter needed to carry 1500 ft-lbs of torque is approximately **3.06 inches**. The appropriate key width to carry this torque is approximately **0.48 inches**.
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