For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.6 MN/mm per bolt. The bolts are preloaded to 75 percent of proof strength. Assume the external load is equally distributed to all the bolts. The bolts are M6 × 1 class 5.8 with rolled threads. A fluctuating external load is applied to the entire joint with Pmax = 60 kN and Pmin = 20 kN.(a) Determine the yielding factor of safety.(b) Determine the overload factor of safety.(c) Determine the factor of safety based on joint separation.(d) Determine the fatigue factor of safety using the Goodman criterion.

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

a) Determine the yielding factor of safety

[tex]n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }[/tex]  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = [tex]\frac{380*20.1}{0.277*7500*5728.5}[/tex] = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

[tex]n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}[/tex]  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : [tex]n_{L}[/tex] = 0.9191[tex]n_{L} = 0.9191[/tex]

C)  Determine the factor of safety based on joint separation

[tex]n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }[/tex]

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence [tex]n_{0}[/tex] = 1.056

D)  Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

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