Question 7.22: Design a square key for fixing a gear on the shaft having 25......
Design a square key for fixing a gear on the shaft having 25 mm diameter. The gear rotates at 550 rpm and transmits 12 kW power to the meshing gear. The key is made of steel having yield strength in tension as 400 N/mm². The yield stress in compression and tension may be taken equal to each other. Assume the factor of safety as 2.5.
Allowable strength
Assuming distortion energy theory
\tau_y=0.577 \sigma_y
= 0.577 × 400 = 230.8 MPa
Allowable shear strength =\frac{230.8}{\text { fos }}=\frac{230.8}{2.5}=92.32 MPa (fos is factor °F of safety).
Allowable crushing strength \sigma_c=\frac{\sigma_y}{\text { fos }}=\frac{400}{2.5}=160 \mathrm{~MPa}
Torque due to power transmission is
T=\frac{9550 \times \mathrm{kW}}{\mathrm{rpm}}=\frac{9550 \times 12}{550}=208.36 \mathrm{~N} \cdot \mathrm{m}
Width of the key b=\frac{d}{4}=\frac{25}{4}=6.25
Height or thickness of the key h = 6.25 mm
Taking standard dimensions b × h = 6 × 6 mm
Considering shear failure of the key, length is obtained.
\tau=\frac{2 T}{d b l}=\frac{2 \times 208.36 \times 1000}{25 \times 6 \times l} (7.87)
Equating (7.87) to allowable strength, we get
\frac{2 \times 208.36 \times 1000}{25 \times 6 \times l}=92.32
Solving, l = 30.09 mm
Take l = 30 mm
From crushing failure point of view, the required length is
\sigma_c=\frac{4 T}{d h l}=\frac{4 \times 208.36 \times 1000}{25 \times 6 \times l}=160
Solving, we get
l=\frac{4 \times 208.36 \times 1000}{25 \times 6 \times 160}=34.72 \mathrm{~mm}
Hence length of the key is 35 mm.
Key dimensions are 6 × 6 × 35 mm