Design the keys - Shear Keys:
A solid circular shaft is to transmit 200 kN at 200 rpm. If the maximum shear stress is not to exceed 80 n/mm^{2} design the shaft. Design the keys and discover number of 16 mm diameter bolt needed to connect coupling along with a bolt circle of 240 mm diameter.
Permissible stress in keys = 100 N/mm^{2}
Permissible stress in bolts = 100 N/mm^{2}
Solution
Length of the key = 200 mm
P = 200 kN N = 200 rpm Shaft
d = ?
P = 2π NT / 60
⇒200 × 10^{3} = (2π × 200 × T) /60
∴ T = 9.5 × 10^{3} N-m
T /J = τ_{max} /R ∴ τ _{max } = 16T/ π d ^{3}
⇒ 80 × 10^{6} = 16 × (9.5 × 10 ) / π d^{ 3}
∴ d = 0.085 m ≈ 85 mm
Shear Keys
T = 9.5 × 10^{3 }N-m
τ_{1} = 100 N/mm^{2}
L = 200 mm = 0.2 m
T = τ_{1} L b R
⇒ 9.5 × 10^{3} = (100 × 10^{6} ) (0.2) b ( 0.085/2)
∴ b = 0.011 m ≈ 11 mm
Bolts
τ_{2} = 100 N/mm^{2}
d = 16 mm
F = τ_{2} (π/4) d ^{2}
= 100 × (π /4 )× 16^{2} = 20106.2 N
R_{1} = 240 /2 = 120 mm = 0.12 m
T = η . F_{2} . R_{1}
⇒ 9.5 × 10^{3} = η × 20106.2 × 0.12
∴ η = 3.9 ≈ 4