Question

For a bolted assembly with six bolts, the stiffness of each bolt is kb = 2 Mlbf/in and the stiffness of the members is km = 14 Mlbf/in per bolt. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in-13 UNC grade 8 bolts with rolled threads. It is desired to find the range of torque that a mechanic could apply to initially preload the bolts without expecting failure once the joint is loaded. Assume a torque coefficient of K = 0. 2.


A. Determine the yielding factor of safety.

B. Determine the overload factor of safety

Answer 1

The calculated value is nP = 4.9 nL = 1.50. the range of torque that a mechanic could apply to initially preload the bolts without expecting failure once the joint is loaded.

DUE TO DATA

applied external load (p) = 85 kips

bolt stiffness (Kb) is equal to 3(106) Ibf/in.

Member stiffness (Km) is equal to 12(106) Ibf per inch.

Bolts' diameter (d) is 1/2 in. and are UNC grade 8.

There are six bolts.

assumptions

UNC and UNF are unified screw threads.

Area under Tensile Stress (A) = 0.1419 in2

SAE requirements for grade 8 steel bolts

There are

120 kpsi is the minimum proff strength (Sp)

Tensile strength (St) minimum: 150 Kpsi

External load / number of bolts = 85 / 6 = 14.17 kips; load bolt (p)

Considering the ensuing values

12.771 kip is equal to Fi = 75%* Sp*At (0.75*120*0.1419)

Preload tension

i=0.75Sp=0.75*120=90 kpsi

stability of rigidity

C = = = 0.2

Yielding Safety Factor

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

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