91430 SPS cover edited - Electronic Fasteners Inc

JOINT DIAGRAMSF e2F e212nl jI j34F e2F e2Fig. 7 Analysis of external load F e and derivation ofForce Ratio Φ.F itan α = = K B and tan β = Fi = K J∆l B∆l Jλ =F eB F= eJ FeB F = = eJ ortan α tan β K B K JF eJ = λ tan β and F eB = λ tan αSince F e = F eB + F eJF e = F eB + λ tan βSubstitutingF eBtan αfor λ produces:F e = F eB + F eB tan βtan αMultiplying both sides by tan α:F e tan α = F eB (tan α + tan β) andFF eB = e tan αtan α tan βSubstituting K B for tan α and K J for tan βAFig. 8 Joint diagram shows effect of loading planes of F eon bolt loads F eB and F B max . Black diagram shows F eBand F B max resulting from F e applied in planes 1 and 4.Orange diagram shows reduced bolt loads when Fe isapplied in planes 2 and 3.F eF eEstimated:F eB = F eDefining Φ =F BK B + K JK BK B + K JF eB= Φ F eΦ = F eBF eand it becomes obvious why Φis called force ratio.BF eF eFig. 9 When external load is applied relatively near bolthead, joint diagram shows resulting alternating stress α B(A). When same value external load is applied relativelynear joint center, lower alternating stress results (B).59