Question

A solid shaft is designed to twist no more than an angle of 2º when subjected to a torque of 800 Nm. The shaft is 1 m long and made from a material that has a shear modulus, G = 70 GN/m². Select an appropriate minimal diameter for the design specifications if

1. Solid shaft was used.
2. Hollow shaft of wall thickness 5mm was used.
3. If the shaft is to rotate at 2000 rpm, determine the maximum power that can be transmitted by the HOLLOW shaft.

Answer 1

Given data:

Max angle of twist,

Torque, T = 800 N m

Length of the shaft, L = 1 m

Shear modulus, G = 70 GN/m²

Solution:

The angle of twist in radians is,

a. when the solid shaft was used,

The angle of twist formula is,

......Eq. (1)

Where

= angle of twist in radians

L = length of the shaft in m

T = torque in N-m

G = shear modulus of rigidity in N/m²

D is the solid diameter of the shaft

Substitute , L = 1 m, T = 800 N-m, G = 70*10⁹ N/m² in Eq. (1)

D = 42.73 mm

b. When hollow shaft of wall thickness 5 mm was used.

The angle of twist formula is,

......Eq. (2)

Where

= angle of twist in radians

L = length of the shaft in m

T = torque in N-m

G = shear modulus of rigidity in N/m²

D = the outside diameter of the shaft

d = inside diameter of the shaft.

Since wall thickness is 5 mm

d = D-2*5 = D-10 mm = (D - 0.01) m

The equation 2 becomes

.......Eq. (3)

Substitute , L = 1 m, T = 800 N-m, G = 70*10⁹ N/m² in Eq. (3)

D = 0.04849 m

D = 48.49 mm

d = 48.49 - 10 = 38.49 mm

C. If the shaft is to rotate 2000 rpm,

The hollow shaft is designed to carry a torque of 800 N-m.

The relation between torque, rpm and power is,

P = 0.105*rpm*T ......Eq. (4)

Where

P = power in W

rpm = revolutions per minute

T = torque in N-m

Substitute rpm = 2000, T = 800 N-m in #q. (4)

P = 0.105*2000*800

P = 168000 W

P = 168 kW

Therefore,

The maximum power that can be transmitted by the hollow shaft is 168 kW.