8 Components of tooth forces tan\u03b1 BC BD P sP twhere Ps separating component N \u03b1

# 8 components of tooth forces tanα bc bd p sp twhere

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The figure2.8 Components of tooth forces tanα = BC BD = P s P t where, P s = separating component (N) α = pressure angle (degrees) Table 2.2 forces on bevel gear tooth where, T = torque transmitted by gears (N-mm) r m = radius of the pinion at the midpoint along the face width (mm), and given by the expression r m = D p 2 bsinγ 2 Example 2.3 A pair of bevel gears transmitting 7.5 kW at 300 rpm is shown in the figure The pressure angle is 20°. Determine the components of the resultant gear tooth force. Solution Given kW = 7.5, n p = 300 rpm, α = 20°, D p =150 mm, D g = 200 mm ,b = 40 mm T = 60 × 10 6 × ( kW ) 2 π n p = 60 × 10 6 × ( 7.5 ) 2 π × 300 ¿ 238732.41 Nmm tanγ = z p z g = D p D g = 150 200 = 0.75 Or γ = 36.87˚ r m = D p 2 bsinγ 2 = 150 2 40sin ( 36.87 ) 2 = 63 mm Force Expression Tangential force ( P t ) P t = T r m Radial force ( P r ) P r = P t tanαcosγ Axial or thrust force ( Pa ) P a = P t t anαsinγ P t = T r m = 238732.41 63 = 3789.40 N P r = P t tanαcosγ = 3789.40 × tan ( 20 ) cos ( 36.87 ) = 1103.38 N P a = P t t anαsinγ = 3789.40 × tan ( 20 ) sin ( 36.87 ) = 827.54 N Beam Strength of Bevel Gear In order to determine the beam strength of the tooth of a bevel gear, it is considered to be equivalent to a formative spur gear in a plane perpendicular to the tooth element. Consider an elemental section of the tooth at a distance x from the apex O and having a width dx. Applying the Lewis equation to a formative spur gear at a distance x from the apex, The figure2.9 Beam Strength of the bevel gear tooth δ ( S b )= m x b x σ b Y δ(S b ) = beam strength of the elemental section (N) m x = module of the section (mm) b x = face width of elemental section (mm) Y = Lewis form factor based on virtual number of teeth From the figure2.9 r x R = x A 0 At elemental section m x = 2 r x z = 2 Rx zA 0 At the large end of the tooth m = 2 R z We get m x = m x A 0 And b x = dx [ r x δ ( S b ) ] = ( b YR A 0 2 ) x 2 dx T = ( m σ b YR A 0 2 ) ( A 0 b ) A 0 x 2 dx Upon integration we get, T = mbσ b YR [ 1 b A 0 + b 2 3 A 0 2 ] Assuming beam strength (S b ) as the tangential force at the large end of the tooth, T = S b R Hence, S b = mbσ b Y [ 1 b A 0 + b 2 3 A 0 2 ] The face width of the bevel gear is limited to one-third of the cone distance. Therefore, the last term in the bracket will never be more than (1/27). Neglecting the last term, S b = mbσ b Y [ 1 b A 0 ] where, S b = beam strength of the tooth (N) m = module at the large end of the tooth (mm) b = face width (mm) σ b = permissible bending stress (Sut/3) (N/mm 2 ) Y = Lewis form factor based on formative number of teeth A 0 = cone distance (mm) [ 1 b A 0 ] is the bevel factor. Wear strength of helical gears The contact between two meshing teeth of straight bevel gears is a line contact, which is similar to that of spur gears. In order to determine the wear strength, the bevel gear is considered to be equivalent to a formative spur gear in a plane which is perpendicular to the tooth at the large end. Applying Buckingham’s equation to these formative gears, S w = bQd p ' K where, b = face width of gears (mm) Q = ratio factor d p ’= pitch circle diameter of a formative pinion (mm) K = material constant (N/mm 2 ) And d p ' = D p cosγ where D p is the pitch circle diameter of the pinion at the large end of the tooth.  #### You've reached the end of your free preview.

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