100%(1)1 out of 1 people found this document helpful
This preview shows page 8 - 11 out of 14 pages.
Tighten to Yield Method.Most fasteners are provided with guaranteed yield strength; therefore, it takes a predictable
CHAPTER NINETEENFasteners 637where Fe=externally applied loadP=initial clamping load [as used in Equation (19–3)]Fb=final force in boltFc=final force on clamped memberskb=stiffness of boltkc=stiffness of clamped membersand the clamped members according to their relative stiff-nesses as follows:Fb=P+kbkb+kcFe(19–8)Fc=P-kckb+kcFe(19–9)Example Problem19–2Assume that the joint described in Example Problem 19–1 was subjected to an additional external load of 3000 lb after the initial clamping load of 4000 lb was applied. Also assume that the stiff-ness of the clamped members is three times that of the bolt. Compute the force in the bolt, the force in the clamped members, and the final stress in the bolt after the external load is applied.SolutionWe will first use Equations (19–8) and (19–9) with P=4000 lb, Fe=3000 lb, and kc=3kb:Fb=P+kbkb+kcFe=P+kbkb+3kbFe=P+kb4kbFeFb=P+Fe>4=4000+3000/4=4750 lbFc=P-kckb+kcFe=P-3kbkb+3kbFe=P-3kb4kbFeFc=P-3Fe>4=4000-3(3000)>4=1750 lbBecause Fcis still greater than zero, the joint is still tight. Now the stress in the bolt can be found. For the 3/8–16 bolt, the tensile stress area is 0.0775 in2. Thus,σ=PAt=4750lb0.0775in2=61 300 psiThe proof strength of the Grade 5 material is 85 000 psi, and this stress is approximately 72% of the proof strength. Therefore, the selected bolt is still safe. But consider what would happen with a relatively “soft” joint, discussed in Example Problem 19–3.Example Problem19–3Solve Example Problem 19–2 again, but assume that the joint has a flexible elastomeric gasket separating the clamping members and that the stiffness of the bolt is then 10 times that of the joint.SolutionThe procedure will be the same as that used previously, but now kb=10kc. Thus,Fb=P+kbkb+kcFe=P+10kc10kc+kcFe=P+10kc11kcFeFb=P+10Fe>11=4000+10(3000)>11=6727 lbFc=P-kckb+kcFe=P-kc10kc+kcFe=P-kc11kcFeFc=P-Fe>11=4000-3000>11=3727 lbThe stress in the bolt would beσ=6727 lb0.0775 in2=86 800 psiThis exceeds the proof strength of the Grade 5 material and is dangerously close to the yield strength.
638PART THREEDesign Details and Other Machine Elementsrelatively low strength. The required length of engagement to develop at least the full strength of the bolt isLe=sutB(2 AtB)sutNπODBmin[0.5+0.57735 n(ODBmin-PDNmax)](19–12)wheresutB=ultimate tensile strength of the bolt materialsutN=ultimate tensile strength of the nut materialODBmin=minimum outside diameter of the bolt threadsPDNmax=maximum pitch diameter of the nut threadsThe shear area of the root of the threads of the nut isAsN=πLeODBmin[0.5+0.57735n(ODBmin-PDNmax)](19–13)Equal Strength for Nut and Bolt Material.For this case failure is predicted as shear of either part at the nominal pitch diameter, PDnom. The required length of engagement to develop at least the full strength of the bolt isLe=4 AtBπPDnom(19–14)