Question

Problem 16.5. A solid circular shaft and a hollow circular shaft whose inside diameter is ( 3/4 ) of the outside diameter, are of the same material, of equal lengths and are required to transmit a given torque. Compare the weights of these two shafts if the maximum shear stress developed in the two shafts are equal. (AMIE, Winter 1988)

Answer 1

To compare the weights of the solid circular shaft and the hollow circular shaft, we need to consider their dimensions and the material they are made of. By equating their maximum shearing stresses and calculating their volumes, we find that the weight of the solid circular shaft is 1.33 times greater than the weight of the hollow circular shaft.

Step-by-step explanation:

To compare the weights of the two shafts, we need to consider their dimensions and the material they are made of. Let's assume the solid circular shaft has an outside diameter of 'D' and a length of 'L'. The hollow circular shaft will have an outside diameter of 'D' and an inside diameter of '(3/4)D'. The maximum shear stress developed in both shafts is equal. We can compare their weights by calculating their volumes and multiplying by the density of the material.

The volume of the solid circular shaft is given by
`V_solid = π/4 * D^2 *`ollow circular shaft is given by V_
`hollow = π/4 * (D^2 - (3/4)^2 * D^2) * L.`n both shafts is equal, their maximum shearing stresses can be equated using the formula
`τ = T / (π/32 * D^3),` e transmitted.

By equating the maximum shearing stresses and simplifying the equation, we can solve for the ratio of the outside diameters of the two shafts, which is 1.333. Using this ratio, we can compare the volumes of the two shafts and determine that the weight of the solid circular shaft is 1.33 times greater than the weight of the hollow circular shaft.

Answer 2

To compare the weights of a solid and a hollow circular shaft required to transmit the same torque with equal maximum shear stresses, calculate the volumes based on their geometries and multiply by the material's density. It turns out that the hollow shaft weighs 7/16 of the solid shaft's weight.

Step-by-step explanation:

The problem of comparing the weights of a solid circular shaft and a hollow circular shaft to transmit a given torque involves concepts from mechanical engineering and physics. Since both shafts are made of the same material and have equal lengths, we need to focus on the differences in volume due to the shafts' geometries, as the weight is directly proportional to the volume for a constant density material.

For both solid and hollow circular shafts, the volume can be calculated based on their geometric shapes. Assuming that the shafts have the same outer diameter, D, the volume of the solid shaft, Vs, is just π/4 * D^2 * length. For the hollow shaft, the inside diameter is given as (3/4)D, so its volume, Vh, is π/4 * ((D^2) - ((3/4)D)^2) * length.

To compare the weights, Ws and Wh, of the solid and hollow shafts respectively, we use their volumes and the material's density, ρ: Ws = ρ * Vs and Wh = ρ * Vh. After simplification, we find that Wh = Ws * (7/16), which means the hollow shaft weighs 7/16 the weight of the solid shaft when they are designed to support the same maximum shear stress.