Question 7.23: Design a rectangular key for the following application. A sh......

From Table 7 .5 select a rectangular key of dimensions b = 18 mm and h = 11 mm corresponding to shaft diameter 65 mm.

Limiting shear stress in the key \tau_{\text {key }}=0.75 \times 67=50.25 \mathrm{~MPa}

Limiting crushing stress in the key \sigma_{c, \text { key }}=2.5 \times 50.25=125.625 \mathrm{~MPa}

Now, we have shear stress developed in the shaft \tau=\frac{16 T}{\pi d^3}

T=\frac{\pi}{16} \times \tau \times d^3=\frac{\pi}{16} \times 67 \times 65^3=3612.807 \times 10^3 \mathrm{~N} \cdot \mathrm{mm}

From shearing consideration the shear stress induced in the key is

\tau=\frac{2 T}{d b l}=\frac{2 \times 3612.807 \times 10^3}{65 \times 18 \times l}

As it is given that shear stress in the key is 75% that of the shaft, so

\frac{2 \times 3612.807 \times 10^3}{65 \times 18 \times l}=0.75 \times 67

l=\frac{2 \times 3612.807 \times 10^3}{65 \times 18 \times 0.75 \times 67}=122.9

l = 123.0 mm

From crushing failure consideration,

\sigma_c=\frac{4 T}{d h l}=\frac{4 \times 3612.807 \times 10^3}{65 \times 11 \times l}=125.625

l=\frac{4 \times 3612.807 \times 10^3}{65 \times 11 \times 125.625}=160.88 \mathrm{~mm}

Hence, the length of the key is 161.0 mm