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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/mm2 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/mm2
Permissible stress in bolts = 100 N/mm2
Solution
Length of the key = 200 mm
P = 200 kN N = 200 rpm Shaft
d = ?
P = 2π NT / 60
⇒200 × 103 = (2π × 200 × T) /60
∴ T = 9.5 × 103 N-m
T /J = τmax /R ∴ τ max = 16T/ π d 3
⇒ 80 × 106 = 16 × (9.5 × 10 ) / π d 3
∴ d = 0.085 m ≈ 85 mm
Shear Keys
T = 9.5 × 103 N-m
τ1 = 100 N/mm2
L = 200 mm = 0.2 m
T = τ1 L b R
⇒ 9.5 × 103 = (100 × 106 ) (0.2) b ( 0.085/2)
∴ b = 0.011 m ≈ 11 mm
Bolts
τ2 = 100 N/mm2
d = 16 mm
F = τ2 (π/4) d 2
= 100 × (π /4 )× 162 = 20106.2 N
R1 = 240 /2 = 120 mm = 0.12 m
T = η . F2 . R1
⇒ 9.5 × 103 = η × 20106.2 × 0.12
∴ η = 3.9 ≈ 4
Equilibrium of body - force acted at any angle: Magnitude of force ' p ' which is required to move the body down the plane. When ' p ' is acted with an angle of φ.
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