Question 7.24: A solid shaft of 45 mm diameter transmits 15 kW at 750 rpm b......
A solid shaft of 45 mm diameter transmits 15 kW at 750 rpm by means of Kennedy key of size 4 × 4 mm cross section. The permissible shearing and crushing strength of the key material are 70 MPa and 120 MPa respectively. Determine the length of the keys.
Torque transmitted is
T=\frac{9550 \times \mathrm{kW}}{\mathrm{rpm}}=\frac{9550 \times 15}{750}=191.0 \mathrm{~N} \cdot \mathrm{m}
From the shear failure consideration, the induced shear stress is
\tau=\frac{T}{\sqrt{2} \times b \times l \times d}=\frac{191 \times 10^3}{\sqrt{2} \times b \times l \times 45}=\frac{3001.28}{b \times l}
Equating to shear strength, we get
\frac{3001.28}{b \times l}=70
b \times l=\frac{3001.28}{70}=42.88 (7.90)
l=\frac{42.88}{4}=10.72 \mathrm{~mm}
From crushing consideration,
\sigma_c=\frac{\sqrt{2} \times T}{d b l}=\frac{\sqrt{2} \times 191 \times 10^3}{45 \times 4 \times l}=\frac{1500.64}{l} (7.91)
Equating (7.91) to allowable crushing strength,
\frac{1500.64}{l}=120
l=\frac{1500.64}{120}=12.5 \mathrm{~mm}
Hence, the required length is 12.5 mm.