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
<u>a) Determine the yielding factor of safety</u>
------ ( 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 ) =
= 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
<u>b) Determine the overload factor of safety</u>
------- ( 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 :
= 0.9191![n_{L} = 0.9191](https://tex.z-dn.net/?f=n_%7BL%7D%20%20%3D%200.9191)
<u>C) Determine the factor of safety based on joint separation</u>
![n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }](https://tex.z-dn.net/?f=n_%7B0%7D%20%3D%20%5Cfrac%7BF_%7Bi%7D%20%7D%7BP_%7Bmax%7D%281%20-%20C%20%29%20%7D)
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence
= 1.056
<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>
nf = 0.849
attached below is the detailed solution .