Answer:
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
