Homework SourseTwo parts of a machine are held together by six equallyspace
Two parts of a machine are held together by six equally-spaced M6x1.0 grade 5.8 bolts that are uniformly tightened to provide a total initial clamping force of 36 kN. Each bolt has a proof stress of 380 MPa, and a threaded area of 20.1 mm2 . The elasticities are such that the joint constant is 0.22.
(a) A uniform static tensile load of 7200 N is now applied to the joint. Determine load factor and the factor of safety against joint separation.
(b) You have determined that assembly procedure can reliably set the initial clamping force to within ±25% of the nominal value. What do you recommend as a safe working load for the entire joint if you require a factor of safety of 2.0 against any failures?
(c) What bolt preload would you recommend to maximize the amount of load that the joint can safely sustain? Assume that the assembly procedures can perfectly achieve this preload level. Hint: This preload equalizes the load factor and factor of safety against joint separation.
(d) Using your newly-optimized preload, and acknowledging that the assembly procedure can only achieve this to within ±25% of the nominal value, what should be the new load rating for the entire joint that maintains a factor of safety of 2.0 against failure by overload or joint separation?
The solutions are as follows: (a) n0=6.41, nL=6.20; (b) 1882 N; (c) 5958 N; (d) 2615 N
I have the numerical solutions but am preparing for an exam and would like to see how these solutions were reached with relevant equations.
Solution
ANSWER FOR PART 1 OF GIVEN QUESTION:-
Given
Proof load SP = 380 mpa
Threded area At = 20.1 mm2
Joint constant is C = 0.22
Given that number of bolts are 6 and intial load is 36KN
now per each bolt preload is Fi = 36/6 = 6KN
and externally applied static load P =7200N =7.2KN
Bolt load factor nb is = (SpAt - Fi)/CP
=(380 * 20.1 - 36) / (0.22 * 7.2)
=4800
The factor of safety guarding against joint separation is
ns = Fi/[P(1 - C)]
=36/(7.2(1-0.22))
=6.41
