Answer:
Explanation:
We have the following data:
Mass = 2 kg
R.V = 10 Grms
g = 9.8 m/s^2
edge of cube = 10 cm = 0.1 m
Assume that the force is divided between the four bolts, the restraint force will be:
[tex]F=mg\\F= 2 \times (10 \times 9.8)\\F = 196.2 \ N[/tex]
Total Restraint Force = 196.2
Force on each bolt: [tex]\frac{196.2}{4} = 49.05 \ N[/tex]
Assume the bolt material as metric class 4.6 with:
Proof strength = 225 MPa
Tensile strength = 400 MPa
Yield strength = 240 MPa
The load can be either horizontal or vertical, thus we check for tension and shear.
Tenison:
Assume that the load is on the bolts.
Maximum and Minimum bolt load is =
[tex]F_{max} = 2 \times 49.05 = 98.1 \ N\\F_{min} = 0 \ N[/tex]
Stress:
(where stress concentration factor = 2.2)
[tex]\sigma _ {max} = 2.2 \times \frac{98.1}{\pi \frac{d^2} {4}}\\\sigma _ {max} = \frac{272}{d^2} \ Pa[/tex]
[tex]\sigma _{min} = 0 \ Pa[/tex]
Mean stress:
[tex]\sigma_{mean} = \frac{\sigma_{max} + \sigma_{min}}{2} = \frac{136}{d^2} \ Pa[/tex]
Stress amplitude:
[tex]\sigma_a = \frac{\sigma_{max} - \sigma_{min}}{2} = \frac{136}{d^2} \ Pa[/tex]
Fatigue Strength:
The rotating beam fatigue strength is:
[tex]S = 0.4 \times 400 = 200 \ MPa[/tex]
Assume machined thread, the surface condition factor will be:
[tex]f = 4.51 \times 400^{-0.265} = 0.9218[/tex]
Size factor:
[tex]s = 1.24 (1000 d)^{-0.107} = 0.5921 d^{-0.107}[/tex]
Loading factor:
[tex]l = 0.85 \times axial[/tex]
Temperature factor:
[tex]t = 1[/tex]
Reliability factor:
[tex]k = 0.814 (99\% \ reliability \ assumed\ )[/tex]
Fatigue strength for infinite cycles:
[tex]S_f = 0.9218 \times (0.5921 d ^{-0.107}) \times 0.85 \times 0.814 \times 200\\S_f = 75.53 d^-0.107 \ MPa[/tex]
Using modified Goodman relation with assumed factor of safety:
[tex]n = 4\\\\\frac{136d^{-2}}{75.53d^{-0.107}\times 10^6} + \frac{136d^{-2}}{400 \times 10^6}=\frac{1}{4}\\\\d = 2.5\ mm\\[/tex]
Shear Failure:
[tex]\xi = 2.2 \times \frac{49.05}{\pi \times 0.25 \times 0.0025^2}\\\\ \xi = 22\ MPa[/tex]
Axial load of the bolt is 49.05 N.
[tex]\sigma = 2.2 \times \frac{49.05}{\pi \times 0.25 \times 0.0025^2}\\\\ \sigma = 22\ MPa[/tex]
Principal stress are:
[tex]\sigma _1 = \frac{\sigma}{2} + \sqrt{\frac{\sigma^2}{4} + \xi^2} = 35.6 MPa\\\\\sigma _2 = \frac{\sigma}{2} - \sqrt{\frac{\sigma^2}{4} + \xi^2} = 13.6 MPa\\[/tex]
[tex]\sigma _v = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} = 44MPa\\[/tex]
We know, [tex]\sigma_{max} = 44\ MPa[/tex]
[tex]\sigma_m = 22 \ MPa\\\sigma_a = \sigma_{max} - \sigma_{m} = 22 \ MPa[/tex]
Using endurance strength values:
[tex]\frac{22}{75.53 (0.0025)^{-0.107}} + \frac{22}{400} = \frac{1}{n}\\\\n=4.8[/tex]
The bolt is safe in shear.
We will need to check:
1. Natural frequency of supporting system
2. Range of excitation frequencies
