3 years ago

91430 SPS cover edited - Electronic Fasteners Inc

91430 SPS cover edited - Electronic Fasteners Inc

F 1F 1Fig. 10 Force

F 1F 1Fig. 10 Force diagrams show the effect of the loading planes of the external load on the bolt load.Fig. 11 Modified joint diagramshows nonlinear compressionof joint at low preloads.60

JOINT DIAGRAMSF eB = Φ F e must be modified to :F eB = n Φ F ewhere n equals the ratio of the length of the clampedparts due to F e to the joint length l j . The value of n canrange from 1, when F e is applied under the head and nut,to O, when F e is applies at the joint center. Consequentlythe stress amplitude:σ B = ±Φ F ebecomes2 A mσ B = ± n Φ F e2 A mGeneral Design FormulaeHitherto, construction of the joint diagram has assumedlinear resilience of both bolt and joint members. However,recent investigations have shown that this assumption isnot quite true for compressed parts.Taking these investigations into account, the jointdiagram is modified to Fig. 11. The lower portion of thejoint spring rate is nonlinear, and the length of the linearportion depends on the preload level F i . The higher F i thelonger the linear portion. By choosing a sufficiently highminimum load, F min >2F e , the non-linear range of the jointspring rate is avoided and a linear relationship betweenF eB and F e is maintained.Also from Fig. 11 this formula is derived:F i min = F J min + ( 1 – Φ) F e + ∆F iwhere ∆F i is the amount of preload loss to beexpected. For a properly designed joint, a preload loss∆F i = – (0.005 to 0.10) F i should be expected.The fluctuation in bolt load that results from tighteningis expressed by the ratio:a= F i maxF i minwhere a varies between 1.25 and 3.0 depending on thetightening method.Considering a the general design formulae are:F i nom = F J min = (1 – Φ) F eF i max = a [ F j min + (1 – Φ) F e + ∆F i ]F B max = a [ F j min + (1 – Φ) F e + ∆F i ] + ΦF eConclusionThe three requirements of concentrically loaded jointsthat must be met for an integral bolted joint are:1. The maximum bolt load FB max must be less thanthe bolt yield strength.2. If the external load is alternating, the alternatingstress must be less than the bolt endurance limit to avoidfatigue failures.3. The joint will not lose any preload due to permanentset or vibration greater than the value assumed for∆F i .SYMBOLSA Area (in. 2 )Am Area of minor thread diameter (in. 2 )A s Area of substitute cyliner (in. 2 )A x Area of bolt part 1 x (in. 2 )d Diameter of minor thread (in.)D c Outside diameter of bushing (cylinder) (in.)D H Diameter of Bolt head (in.)D h Diameter of hole (in.)D J Diameter of JointE Modulus of Elasticity (psi)F Load (lb)F e External load (lb.)F eB Additinal Bolt Load due to external load (lb)F eJ Reduced Joint load due to external load (lb)F i Preload on Bolt and Joint (lb)∆F i Preload loss (–lb)F i min Minimum preload (lb)F i max Maximum preload (lb)F j nom Nominal preload (lb)F B max Maximum Bolt load (lb)F J min Minimum Joint load (lb)K Spring rate (lb/in.)K B Spring rate of Bolt (lb/in.)K J Spring rate of Joint (lb/in.)K x Spring rate of Bolt part l x (lb/in.)l Length (in.)∆l Change in length (in.)l B Length of Bolt (in.)∆l B Bolt elongation due to F i (in.)l J Length of Joint (in.)∆l J Joint compression to F i (in.)l x Length of Bolt part x (in.)Length of clamped partsnTotal Joint Lengthα Tightening factorΦ Force ratioλ Bolt and Joint elongation due to F e (in.)Bolt stress amplitude (± psi)σ B61

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