static and dynamic

Question

Wind turbine blades transmit 137.5 Watt power to shaft with the rotational velocity of 175 rpm. Shaft is supported by journal bearings.

a) Determine the smallest diameter of shaft for the factor of safety 2? (τallow=60 MPa).

b) Blades are connected to shaft with four screws. Determine the required smallest screw diameter for the factor of safety 2? (τallow=40 MPa).

Answer 1

For this question, the torsional shear equation is employed.

τ/r = T/J

τ* = shear stress

r = radius

T = Torque

J = polar inertia

*The shear stress is allowable shear stress. This implies that the yield shear stress was divided by the safety of margin.

Step 1) Collect What is Given

Power (P) = 137.5 W

n = 175 rpm

τ(allowable) = 60 MPa

Step 2) Calculate Torque

P = T*ω⇒ T = P/ω = 137.5/(2π*175/60) = 7.5 N-m

Step 3) Determine Polar Inertia Equation

The shaft is circular, therefore the polar inertia equation (J) = r⁴π/2

Step 4) Transpose Shear Stress Equation

τ/r = T/J

J/r = T/τ

r⁴π/2r = T/τ

r³π/2 = T/τ

r = [(2T)/(τπ)]^1/3

Step 5) Evaluate the Equation from Step 4

r = [(2T)/(τπ)]^1/3 = [(2*7.5)/(60*10⁶*π)]^(1/3) = 0.0043m = 4.3mm

Therefore, D = 2*r = 2*4.3 = 8.6mm

For part B, we'll use the equation from Step 5 and sub in the new numbers.

τ = 40 MPa

However, since there are four bolts, the torque is divided by four. Hence: T/4.

r = [(2T)/(τπ)]^1/3 = [(2*7.5)/(4*40*10⁶*π)]^(1/3) = 0.0031m = 3.1mm

We are asked to find the bolt's diameter.

D = 2*r = 2*0.0031 = 0.0062m = 6.2mm

Answers:

A) 8.6mm

B) 6.2mm