Machinery's Handbook, 31st Edition
TORQUE AND TENSION IN FASTENERS 1665 With the coefficient of friction μ 2 between a nut or bolt-head pressure face and a compo nent face, as in Fig. 7 , the friction load is equal to μ 2 P B . Assuming the force is applied mid way between the nominal (bolt) diameter d and the pressure-face diameter b , the torque T 3 to overcome the nut or bolt underhead friction is: (21) The total torque, T , required to develop axial bolt load, P B , is equal to the sum of the torques T 1 , T 2 , and T 3 as follows: (22) For a fastener system with 60 ° threads, α = 30 ° and d 2 is approximately 0.92 d . If no loose washer is used under the rotated nut or bolt head, b is approximately 1.5 d and Equation (22) reduces to: (23) In addition to the conditions of Equation (23), if the thread and bearing friction coeffi cients, μ 1 and μ 2 , are equal (which is not necessarily so), then μ 1 = μ 2 = μ , and the previous equation reduces to: (24) Example: Estimate the torque required to tighten a UNC 1 ∕ 2 -13 grade 8 steel bolt to a pre load equivalent to 55 percent of the minimum tensile bolt strength. Assume that the bolt is unplated and both the thread and bearing friction coefficients equal 0.15. l d 0159 1156 B µ = + ^ h Solution: The minimum tensile strength for SAE grade 8 bolt material is 150,000 psi (from page 1666 ). To use Equation (24), find the stress area of the bolt using Equation (9) with P = 1 ∕ 13, d m = d − 1.2990 P , and d p = d − 0.6495 P , and then calculate the necessary preload, P B , and the applied torque, T . . . . . . . . . . . A P A T 4 2 0 4500 0 4001 0 1419 0 55 150000 0 1419 11707 11707 13 0 159 1 156 0 15 0 500 1158 965 in. lb lb-in. lb-ft s B s 2 2 allow f # # # # # π σ = + = = = = = + = = a a k k T d b P 4 B 3 2 = + µ cos T P l = c d d b + 2 2 π + 4 B 2 1 µ 2 h m µ α + ^ . T P l d 0159 0531 0625 B 1 2 # µ µ = + + ^ h 6 @ . . . . T P
Contact surface of bolt (pressure surface) u 2 P B
d /2 ( d + b )/4
b /2
P B
Fig. 7. Nut or Bolt Head Friction Force
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