Subject: Machine Design -I

Topic: Shaft, Keys and Coupling

Difficulty: Medium

The forces acting on a flat key, with width as b and height h are show in figure

The exact location of the force P on the surface AC is unknown. In order to simplify the analysis, it is assumed that the force P is tangential to the shaft diameter. Therefore,

$P=\frac{M_T}{d/2}=\frac{2M_T}{d}$ (a)

where $M_T$=transmitted torque

d=shaft diameter

P=force on key

The shear stress $\tau$ in the plane AB is given by,

$\tau=\frac{P}{area of plane}=\frac{P}{bl}$ (b)

where, b=width of key

l=length of key

From (a) and (b),

$\tau=\frac{2M-T}{dbl}$ (1)

The failure due to compressive stress will occur on surfaces AC or DB. It is assumed that,

$AC=BD=\frac{h}{2}$

where, h=height of key

The compressive stress in the key is given by,

$\sigma_c=\frac{P}{area of surface AC}=\frac{P}{hl/2}=\frac{2P}{hl} (c)$

From (a) and (c),

$\sigma_c=\frac{4M_T}{dhl}$ (2)

From square key:

h=b

Substituting the above relationship in (1) and (2),

$\tau=\frac{2M_T}{dbl}$ (a)

$\sigma_c=\frac{4M_T}{dbl}$ (b)

From (a) and (b),

$\sigma_c=2\tau$

Therefore, the compressive stress induced in a square key due to the transmitted torque is twice the shear stress.