A total of 5 bolt are used to clamp two member plates A tota

A total of 5 bolt are used to clamp two member plates. A total external load of 40000 Ibf is applied to the entire joint. The preload on each bolt is 10000 Ibf. The stiffness of each bolt is k_b = 4 Mlbf/in and the stiffness of the member is k_m = 12 Mlbf/in. Bolt properties: 1/2 in -13 UNC, E = 30 Mpsi, Proof strength, S_p = 100 kpsi. Determine the service load on each bolt F_b. Determine the load on the member F_m. What is the load factor n_L? What is the separation safety factor n_0?

Solution

Given Data -
Total number of bolts N = 5
Total external load = 40000 lbf
Preload on each bolt = 10000 lbf
Each bolt stiffness k_b = 4 Mlbf/in
Stiffness of the member k_{m =} 12 Mlbf/in
Bolt properties - 1/2 in 13 UNC, E = 30 Mpsi, Proof strength S_p = 100 Kpsi

4.1) Service load on each bolt F_b = Total load in operation in each bolt
   = Preload on each bolt + External load on each bolt
= Preload on each bolt + (Total external load / Number of bolts)
   = 10000 lbf + (40000 / 5)lbf
= 18000 lbf

4.2) Load on the Member F_m = Total Preload due to all bolts - Total external load
   = (Preload on each bolt x Total number of bolts) - Total external load
   = (10000 lbf x 5) - 40000 lbf
   = 10000 lbf

4.3) Load factor nL = (S_p.A_t - F_i) / [C (P_total / N)]

Where, S_p is Proof strength = 7800 lbf (can be obtained from design data book UNC bolts table)
A_t is tensile stress area = 0.159 in2
F_i is initial bolt tension = 10000 lbf
C is stiffness constant = k_b / (k_b + k_m) = 4 / ( 4 + 12 ) = 1/4
PTotal external load =40000 lbf
Total number of bolts N = 5

Thus we get, Load factor nL = (7800 x 0.159 - 10000) / [ (1/4) x (40000/5) ]
= 4.3799