A shaft of 50 mm diameter transmits a torque of 800 N-m. The width of the rectangular key used is 10 mm. The allowable shear stress of the material of the key being 40 MPa, the required length of the key would be

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, T = 800 N-m, d = 50 mm, b = 10 mm, τ_per = 40 MPa

\(T = F \times \frac{d}{2}\)

\( \Rightarrow 800 = F \times \frac{{50}}{2} \times {10^{ - 3}}\)

∴ Force on key F = 800/25 = 32 kN

Allowable shear stress \(\tau = \frac{F}{A} = \frac{F}{{b \times l}}\)

\( \Rightarrow 40 = \frac{{32000}}{{10\; \times \;l}}\)

⇒ l = 3200/40 = 80 mm