If the permissible crushing stress for the material of a key is double the permissible shear stress, then the sunk key will be equally strong in shearing and crushing if the key is a

Concept:

When a key is used in transmitting torque from a shaft to a rotor or hub the following two types of forces act on the key:

1. Forces due to its fit in a keyway. These forces produce compressive stresses in the key which are difficult to determine.

2. Forces due to the torque transmitted by the shaft. These forces say F₁ and F₂ produce crushing (compressive) and shearing stresses in the key respectively.

Calculation:

Let us consider a rectangular key of length ‘l’, width ‘w’, and thickness ‘t’.

Crushing area \(= \frac{t}{2} \times l\)

F₁ =  Crushing stress Crushing area

F₁ = σ_c \(\left( {\frac{t}{2} \times l} \right)\)

Shearing area = w \( \times \,l\)

F₂ = shearing stress × shearing area

F₂ \(= {\rm{\tau }} \times \left( {W \times l} \right)\)

The key will be equally strong in shearing and crushing if torque transmitted is the same in both the crushing and shearing.

\(\therefore {F_1} \times \frac{d}{2} = {F_2} \times \frac{d}{2}\)

where d = diameter of the shaft

\({\sigma _c} \times \left( {\frac{t}{2} \times l} \right) = \tau \times \left( {w \times l} \right)\)

In case when: σ_c  = 2τ

Then, w = t (i.e. square key)

Therefore, option 3) is correct.