A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the minimum diameter required if the shear stress must not exceed 150 MPa and the angle of twist must not exceed 7 degree. Report both the maximum shear stress and the angle of twist at this minimum diameter.

a) minimum diameter = 16.90 mm

b) minimum diameter = 17.39 mm

Explanation:

The formula to apply here is;

Angular deformation of a solid shaft is

α=32LT/GπD⁴

where

α=angular shaft deformation in radians

L=length of shaft in meters

T=twisting moment in Nm

G=shear modulus of rigidity in Pa

D=diameter of shaft in m

Given

α=7°= 0.1222 rad

L=720 mm =0.72 m

T= 155 N

G=114×10⁹ Pa

D=diameter of shaft in m

For the angle of twist

α=32LT/GπD⁴

0.1222= 32*0.72*155/114*10⁹*3.142 *D⁴

0.1222*114*10⁹*3.142 *D⁴=32*0.72*155

D⁴=32*0.72*155 / 0.1222*114*10⁹*3.142

D=(32*0.72*155 / 0.1222*114*10⁹*3.142)^1/4

D=0.01690 m =16.90 mm

For maximum shear stress apply

D=1.72*(Tmax/tmax)^1/3

where

D=diameter of solid shaft in m

Tmax= maximum twisting moment in Nm

tmax = maximum shear stress in Pa

Given

tmax=150*10⁶ Pa

Tmax=155 Nm

D= 1.72 * ( 155/150*10⁶)^1/3

D= 1.72 * 0.0101098989

D=0.01739 m = 17.39 mm

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explanation: