If a rectangular key of 8 mm width and 6 mm height and a shaft of diameter 32 mm are made of same material, then the necessary length of the key for equal shear strength of shaft and key will be (neglecting stress concentration on the shaft)

Concept:

For a rectangular key Allowable shear stress \(\tau = \frac{F}{A} = \frac{F}{{b\; \times \;l}}\)

The torque transmitted by the shaft is given by, \(T = F \times \frac{d}{2}\)

Calculation:

Given that, b = 8 mm, h = 6 mm, d = 32 mm

\({\tau _{shaft}} = \frac{{16T}}{{\pi {d^3}}} = {\tau _{key}}\)

Torque in the key for shear failure, is given by

\(\begin{array}{l} T = {\tau _{key}} \times l \times b \times \frac{d}{2} \end{array}\)

\(\Rightarrow T = \frac{{16T}}{{\pi {d^3}}} \times l \times b \times \frac{d}{2}\)

\(\Rightarrow l = \frac{{\pi {d^3}}}{{8 \times bd}} \Rightarrow l = \frac{{\pi \times {{\left( {32} \right)}^3}}}{{8 \times 8 \times 32}} = 50.26\ mm \)