Strength of Taper Pins Machinery's Handbook, 31st Edition
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(28) The decrease in the horizontal diameter will be about 92 percent of the increase in the vertical diameter given by Formula (28). In the above formulas, P = load on ring in pounds; D = mean diameter of ring in inches; S = tensile stress in psi, I = moment of inertia of sec- tion in inches 4 ; and E = modulus of elasticity of material in psi. Strength of Taper Pins.— The mean diameter of taper pin required to safely transmit a known torque may be found from the formulas: (29) and (30a) where T = torque in inch-pounds; S = safe unit stress in psi; HP = horsepower transmitted; N = number of revolutions per minute; and d and D denote dimensions shown in the figure. Formula (29) can be used with metric SI units where d and D denote dimensions shown in the figure in millimeters; T = torque in newton-millimeters (N·mm); and S = safe unit stress in newtons per millimeter 2 (N/mm 2 ). Formula (30a) is replaced by: (30b) where d and D denote dimensions shown in the figure in millimeters; S = safe unit stress in N/mm 2 ; N = number of revolutions per minute, and Power = power transmitted in watts. Example 6(a): A lever secured to a 2-inch round shaft by a steel tapered pin (dimension d = 3 ∕ 8 inch) has a pull of 50 pounds at a 30-inch radius from shaft center. Find S , the unit working stress on the pin. By rearranging Formula (29): . . S Dd T 127 2 8 3 1 27 50 30 6770 psi 2 2 # # # ≅ = = a k . EI PD 00186 Increase in verticaldiameter inches 3 = . DS T 113 = d d NDS = 283 HP . 1103 = d NDS Power
6770 psi is a safe unit working stress for machine steel in shear. Let P = 50 pounds, R = 30 inches, D = 2 inches, and S = 6000 pounds unit working stress. Using Formula (29) to find d : . . . . d DS T 113 113 2 6000 50 30 113 8 1 04inch # # = = = =
d
D
Example 6(b) : A similar example using SI units is as follows: A lever secured to a 50 mm round shaft by a steel tapered pin ( d = 10 mm) has a pull of 200 newtons at a radius of 800 mm. Find S , the working stress on the pin. By rearranging Formula (29): . 40 6 megapascals = = . . S Dd T 127 50 10 127 200 800 2 2 # # # = = 40.6 N / mm 2 If a shaft of 50 mm diameter is to transmit power of 12 kilowatts at a speed of 500 rpm, find the mean diameter of the pin for a material having a safe unit stress of 40 N/mm 2 . Using Equation (30b):
, 500 50 40 12 000 # #
NDS Power
= =
=
. 1103
then d
. 1103
d
=
. 110 3 0 1096 12 09 mm . . #
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