Question

QUESTION 1 The cross section of a key for a 41 mm diameter shaft. The Torque transmitted by shaft to hub is 636,348 N-mm. The key is made of steel having Syt - 380 MPa and the factor of safety is 2. Determine the length of the key (mm) in Shear.

Answer 1

Diameter of shaft = 41mm

Yeild Strenghth of key =380MPa

Factor Of Safety=2

Torque Transmitted = 636,348 N-mm

b= Width of Key

L= Length of Key

$T=F*\frac{D}{2}$

636348=F*(41/2)

F=31041.365 N

$\tau =Syt/2$

$\tau =$380/2 = 190 MPa

Since width of Key not given , therfore designing at optimum

b= D/4

b = 10.25 mm

$\tau =\frac{F}{bL}$

$\tau$= 31041.365 / (10.25 *L)

$\tau$= 3028.425 / L

$Factor Of Safety= \frac{\tau max}{\tau }$

$2 = \frac{190}{\frac{3028.425}{L}}$

L= 31.878 mm