Question

A satellite component is in the shape of a cube with 10cm edges, has an evenly distributed mass of 2 kg and one bolt hole at each of four corners. If the maximum predicted random vibration environment is 10 Grms, what bolt size would be a good choice for restraint? What other factors might drive bolt size besides force/stress?

Answer 1

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:

$F=mg\\F= 2 \times (10 \times 9.8)\\F = 196.2 \ N$

Total Restraint Force = 196.2

Force on each bolt: $\frac{196.2}{4} = 49.05 \ N$

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 =

$F_{max} = 2 \times 49.05 = 98.1 \ N\\F_{min} = 0 \ N$

Stress:

(where stress concentration factor = 2.2)

$\sigma _ {max} = 2.2 \times \frac{98.1}{\pi \frac{d^2} {4}}\\\sigma _ {max} = \frac{272}{d^2} \ Pa$

$\sigma _{min} = 0 \ Pa$

Mean stress:

$\sigma_{mean} = \frac{\sigma_{max} + \sigma_{min}}{2} = \frac{136}{d^2} \ Pa$

Stress amplitude:

$\sigma_a = \frac{\sigma_{max} - \sigma_{min}}{2} = \frac{136}{d^2} \ Pa$

Fatigue Strength:

The rotating beam fatigue strength is:

$S = 0.4 \times 400 = 200 \ MPa$

Assume machined thread, the surface condition factor will be:

$f = 4.51 \times 400^{-0.265} = 0.9218$

Size factor:

$s = 1.24 (1000 d)^{-0.107} = 0.5921 d^{-0.107}$

Loading factor:

$l = 0.85 \times axial$

Temperature factor:

$t = 1$

Reliability factor:

$k = 0.814 (99\% \ reliability \ assumed\ )$

Fatigue strength for infinite cycles:

$S_f = 0.9218 \times (0.5921 d ^{-0.107}) \times 0.85 \times 0.814 \times 200\\S_f = 75.53 d^-0.107 \ MPa$

Using modified Goodman relation with assumed factor of safety:

$n = 4\\\\\frac{136d^{-2}}{75.53d^{-0.107}\times 10^6} + \frac{136d^{-2}}{400 \times 10^6}=\frac{1}{4}\\\\d = 2.5\ mm$

Shear Failure:

$\xi = 2.2 \times \frac{49.05}{\pi \times 0.25 \times 0.0025^2}\\\\ \xi = 22\ MPa$

Axial load of the bolt is 49.05 N.

$\sigma = 2.2 \times \frac{49.05}{\pi \times 0.25 \times 0.0025^2}\\\\ \sigma = 22\ MPa$

Principal stress are:

$\sigma _1 = \frac{\sigma}{2} + \sqrt{\frac{\sigma^2}{4} + \xi^2} = 35.6 MPa\\\\\sigma _2 = \frac{\sigma}{2} - \sqrt{\frac{\sigma^2}{4} + \xi^2} = 13.6 MPa$

$\sigma _v = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} = 44MPa$

We know, $\sigma_{max} = 44\ MPa$

$\sigma_m = 22 \ MPa\\\sigma_a = \sigma_{max} - \sigma_{m} = 22 \ MPa$

Using endurance strength values:

$\frac{22}{75.53 (0.0025)^{-0.107}} + \frac{22}{400} = \frac{1}{n}\\\\n=4.8$

The bolt is safe in shear.

We will need to check:

1. Natural frequency of supporting system

2. Range of excitation frequencies