Page 64 - Final Report - KAUSC Team

KAUSC Team

−

1 ((4( )2 (4( )2 1
+ 3( )2)0.5 )2)0.5 3
= 16 3 + + 3( ) (18)
( )

3 = 32 (( ) + √3 ( )) (19)
2

3 = 32 2.2 104.5 √3 3 15.04
(( 200 106 ) + 2 (400 106))

= 4, = 37
= 3, = 34
For optimum shaft design n=2

D = 30 mm
Second run
The second is by applying the material that is found in local market. (AISI 4130)

Tensile strength = 1050 MPa
Yield strength = 985 MPa
Elongation = 14.1 %

= ′ (20)

= 0.922, = 0.862, ′ = 1000

= 0.922 0.862 1000

= 794.764

3 = 32 2.2 104.5 √3 3 15.04
((794.764 106) + 2 (1050 106))

For optimum shaft design d = 35 mm
n = 13

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