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0.75 0.375 F = 8000 lbs 1.0 1.0 Problem 6: A 5/8 inch Grade 8 bolt (NC) is used to connect two steel plates, each 1.5 inch thick. The bol is rightened to 75 percent of the proof load. A force pulling the two plates apart fluctuates from zero to 12,000 lbs. Compute the following: (a) The proper torque to apply to the bolt. (o) The mean factor of safery, load factor and joint opening factor (np, n and no) for the maximum applied load of 12,000 lbs. (c) The fatigue life of the joint, assuming the load computed in part (b) fluctuates from zero to maximum (R 0) (use the Smith-Watson-Topper approach). (d) The fatigue life of the joint, assuming someone forgot to tighten the bolt. (The tightened such that Fi0- use the Smith-Watson-Topper approach). bolt was finger-
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Proof Stress, sY Utilization factor, X Nominal Diameter of Bolt, D Thread per inch Pitch of thread Thread Friction Coefficient, K Tensile Stress Area (from bolt table), As Proof Load of the Bolt, P1 Tightening Torque, T 120000 [lb/in 2] 75% [%] 0.625 in] 11.0 TPI] 0.091 [in] 0.2 1 0.226 [in 2] 20340 [lbf] 2543 [Ib-in] 212 [lb-ft aries as per material grade ASTM A307 standard Nominal diamter Varies for fine and coarse thread series 1/TPI Varies with surface treatment and lubrication As 0.7854 * [D-0.9743/TPIj2 Proof stress* As Minimum Load on bolted joint, F1 Maximum Load on bolted joint, F2 0 [lbf 15000 [Ibf Assumption There is no gasket between the clamped members Pb Share of external load on bolt Pc Share of external load on clamped part P2- External load HenceP2 = Pb + Pc , Till parts are not separated Hence: Pb P2 * Kb /(Kb+Kc) P2* K For metalic contacts, joint without gasket, K0.0-0.10 РЫКЬ Pc/Kc (elongation in bolt-compression in clamoed parts) Hence, total load on bolt P3 P1+0.1*F2- 21840.2 Since, there is no bending or shear forces. Load Factor = Yield Stress of the Bolt Material / Stress at Pre-load Thus, load factor P1/Utilization Factor/P3 1.242

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