Effective load on Gear Tooth Tangential force on the helical gear tooth is

Effective load on gear tooth tangential force on the

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Effective load on Gear Tooth Tangential force on the helical gear tooth is given in table 2.1, which depends upon the rated power and rated speed. But during the operation a dynamic load also acts on the gear tooth which can be considered in the following two basis- (i) Based on velocity factor this method of load estimation is used in preliminary stages of gear design. The effective load in this stage is given by P eff = C s P t C v where, C s = service factor C v = velocity factor, for helical gear velocity factor, is given by the following relation C v = 5.6 5.6 + √ v v = velocity in m/sec (ii) Buckingham equation In the final stages of gear design, when gear dimensions are known, errors specified and the quality of gears determined, the dynamic load is calculated by the equation derived by Earle Buckingham and given by the following equation P eff = C s P t + P d Where P d = incremental dynamic load, given by P d = 21 v ( Ceb + P t ) 21 v + ( Ceb + P t ) v = pitch line velocity (m/s)
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C = deformation factor (N/mm2) e = sum of errors between two meshing teeth (mm) b = face width of tooth (mm) P t = tangential force due to rated torque (N) Deformation factor C is given by C = K [ 1 E p + 1 E g ] where, k = constant depending upon the form of tooth E p = modulus of elasticity of pinion material (N/mm2 ) E g = modulus of elasticity of gear material (N/mm2 ) Wear strength of helical gears The wear strength equation of the spur gear is modified to suit helical gears. For this purpose, a pair of helical gears is considered to be equivalent to a formative pinion and a formative gear in a plane perpendicular to the tooth element. In case of spur gears wear strength is given by S w = bQdK Referring to Fig. 2.5 S w = (S w ) n = wear strength perpendicular to the tooth element b = b cosψ = face width along the tooth element d = d cos 2 ψ = pitch circle diameter of the formative pinion. Using the above relations we get Fig. 2.5 S ( ¿¿ w ) n = bQdK cos 3 ψ ¿ The component of (S w ) n in the plane of rotation is denoted by S w . Therefore,
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S ( ¿¿ w ) n cosψ S w = ¿ Which gives, S w = bQdK cos 2 ψ Where Q= ratio factor, given by Q = 2 Z g Z p + Z g Z p = actual number of teeth on pinion Z g = actual number of teeth on gear K= Load stress factor, given by K = σ c 2 cos α n sin α n ( 1 / E p + 1 / E g ) 1.4 σ c = surface endurance strength (N/mm 2 ) E p , E g = moduli of elasticity of materials for pinion and gear (N/mm 2 ) To design the gear S w > P t Or S w = P t ( fs ) Where, fs = factor of safety Example 2.3 A pair of parallel helical gears consists of a 20 teeth pinion meshing with a 100 teeth gear. The pinion rotates at 720 rpm. The normal pressure angle is 20°, while the helix angle is 25°. The face width is 40 mm and the normal module is 4 mm. The pinion as well as the gear is made of steel 4OC8 (S ut = 600 N/mm2) and heat-treated to a surface hardness of 300 BHN. The service factor and the factor of safety are 1.5 and 2 respectively. Assume that the velocity factor accounts for the dynamic load and calculate the power transmitting capacity of gears.
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