bending moment in addition to shear, consider uniformly distributed load along the portion of pin. Taking moment about axis XX M = [(-P/2) × (t/4)] + { (P/2) × [ (t/2)+(t1/3) ] } = P/2 [(t1/3)+(t/2)-(t/4) ] = P/2 [ (t1/3)+(t/4) ] Section modulus, Z = (π/ 32)dp3 Maximum bending stress, σb σb= M/Z = { P/2 [(t1/3)+(t/4)] } / {(π/ 32)dp3} Here, we check the pin in bending and find the value of dp (3) Design of single eye : (a) To find the outside diameter of single eye (doe) The single eye is subjected to a direct tensile stress, due to this single eye under tear. σt = P/A = P/ (doe-dp)× t (b) Due to direct tensile strength, the single eye is subjected to double shear. Resisting shearing area = 2(doe-dp)×(t/2) The direct shear stress induced is ς=P/(doe-dp)×t From this equation the outside diameter of single eye doe is obtained. (C) Failure of single eye or pin due to tensile load in crushing Resisting crushing area = dp × t σc = P/(dp×t) Form this equation crushing stress checked if fail, increase the thickness of eye (t). (4) Design of fork (double eye): (a) The tearing of the double eye at weakest section due to tension Area resisting tear = (dof – dp) × 2 t1 σt = p/ [(dof – dp) × 2 t1] From this equation, find the outside diameter of fork (dof). (b) Failure of double eye (fork) in double shear due to tensile load. Area resisting shear = 4 × [(dof – dp) ]/2 × t1 = 2 × (dof – dp) t1 The shear stress is given by, ς = p/[(dof – dp) × 2 t1] From this equation, check shear stress if less than design, increase thickness of fork t1. (c) Failure double eye in crushing (thickness of fork) Double eye may fail in crushing due to tensile load The crushing stress is given by, σc = P/( 2×dp ×t1) Check crushing stress or find t1 References Mechanical fasteners