Finding the area of the circular support –
Π x [(140-120)/2]2 = 314mm2 = 0.314m2
Therefore the stress acting on the surface area of the circular support is (Using s = F/A)
s = 30211.44/0.314 = 96214.78Nm-2
As there are four supporting strut ‘legs’, the total force would be equally distributed along them as such:
30211.44/4 = 7552.86N per ‘leg’
a = 7552.86 x cos30 = 6540.96N
b = 7552.86 x cos60 = 3776.43N
c = 7552.86 x cos60 = 3776.43NThe second boom/box support has the combined forces of the previous boom/box supports weight as well as the total weight of the boom with maximum load and counterweight.
Using previously done calculations of structure weight –
Therefore, the approximate mass of the boom support can be assumed to be ~ 27.6kg = 270.48N
So the total force acting down on the second boom support is 270.48 + 30211.44 = 30481.92N
Overall force acting downwards on the structure is – 52.5 + 27.6 + 27.6 + (13.4x4) = 161.3kg
Which is 1580.74N of force, as well as the force of the weight and counterweights which gives
9800 + 19870.48 + 1580.74N = 31251.22N
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