![]() Let us assume a pin diameter 1/2" less than the hole diameter. Often the pin diameter is much less than the hole diameter. ![]() Therefore, using a factor of safety of 5, the working load is: Pw1 :=įailure Mode 2: This failure mode involves bearing failure at the pin/lifting lug interface. Therefore, the ultimate tensile load is given by: Pu := 2 ⋅ a ⋅ t ⋅ Fu Pu = 163.12 kipĪ factor of safety of 5 is common for lifting components. Rule_2 := if ( e ≥ 0.67 ⋅ d, "OK", "NG" )Įvaluation based on Failure Modes: Failure Mode 1: This failure mode involves tension failure on both sides of the hole. Thus, e > 0.67 * d For this example, e = 1.125" and since it is greater than 0.67*d which is 0.8375". ![]() Rule 2: The dimension "e" must be greater than or equal to 0.67 times the hole diameter, d. a > 1/2 * d For this example, a = 1.125" and since it is greater than 1/2*d which is 0.625". Rule 1: The dimension "a" must be greater than or equal to half the hole diameter, d. Geometric Guidelines: There are some geometric guidelines to be considered as recommended in Reference 1. The following provides a systematic method.Įxample of an Overhead Lifting Lug Embed plate Therefore, design engineers are left without adequate technical guidance on this subject. Introduction There is very little published information available on the subject of the design/analysis of lifting lugs. Anyone making use of the information set forth herein does so at their own risk and assumes any and all resulting liability arising therefrom.ĭesign/Evaluation of Overhead Lifting Lugs by Clement Rajendra, PE Project Engineer, CP&L, Southport, NC 28461 e-mail: Application of this information to a specific project should be reviewed by registered professional engineer. They are not a substitute for competent professional advice. The materials are for general information only. ![]() DISCLAIMER: The materials contained in this MathCad file are not intended as a representation or warranty on the part of or any other person named herein. ![]()
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