Question

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.

Answer 1

The required tube thickness for a steel tube that will transmit a specified power at a certain rotational speed without exceeding the maximum shearing stress. This involves mechanical engineering calculations that take into account the relationship between power, torque, shear stress, and the geometrical properties of the steel tube, leading to the determination of the tube's required wall thickness.

Step-by-step explanation:

The calculation of tube thickness for a steel tube that has to transmit a specific amount of power without exceeding a given shearing stress limit. This problem is solvable using the relationship between torque, shear stress, and the polar moment of inertia of a circle, and requires knowledge from the field of mechanical engineering. We are given the following parameters: outer diameter of the steel tube (50 mm), power to be transmitted (100 kW), rotational speed (34 Hz), and maximum allowable shearing stress (60 MPa).

To find the required tube thickness, we will use the following formulae:

1. Power (P) = Torque (T) × Angular Velocity (ω)

2. Torque (T) = Shear stress (τ) × Polar Moment of Inertia (J) / Radius (r)

3. Polar Moment of Inertia (J) for a hollow cylinder = π × (Outer radius4 - Inner radius4) / 2

Let t be the required thickness of the tube, and ri be the inner radius of the tube. Then, ri = (Outer diameter / 2) - t.

Solving these equations with the given values will provide us with the tube thickness t that ensures the shearing stress does not exceed 60 MPa.

Answer 2

25 -
`\sqrt[4]{26.66*10^(-8) }` mm

Step-by-step explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2
`\pi`f = 2 *
`\pi` * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter using the relation

`(J)/(c_(2) ) = (T)/(t_(max) )` = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

`(J)/(c_(2) )` =
`(\pi )/(2c_(2) ) ( c^(4) _(2) - c^(4) _(1) )` =
`(\pi )/(0.050) [ ( 0.025^(4) - c^(4) _(1) ) ]`

therefore; 0.025^4 -
`c^(4) _(1)` = 0.050 /
`\pi` (7.8 *10^-6)

`c^(4) _(1)` = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

`c_(1) = \sqrt[4]{26.66 * 10^(-8) }` =

THE TUBE THICKNESS

`c_(2) - c_(1)` = 25 -
`\sqrt[4]{26.66*10^(-8) }` mm