Just as an addition to the previous post. Results if the pressure is seen over the full height of the cylinder.
(Zero radial displacement at both ends, zero rotational restraint at top and full rotational restraint at base)
Distributed shear at base = 68.42 lbf/in
Distributed BM at base =...
Did some cylinder calc's using assumptions (method not checked). I think earlier in the post you said you'd like to make your part out of Aluminum.
Used inputs.
E = 10007.6 ksi (used 69 GPa)
Poisson's ratio = 0.33
Vertical cylinder length = 12.008in (assumed, used 305mm)
Unpressurized length at...
Having a quick look at the assy, and knowing the assy is to be pressurized, the initial thoughts are to determine the interfacing actions of the parts at the bolted joint. You can view your part as a thin walled cylinder connected to a circular plate at the top. The pressure will tend to make...
Could you include a picture of the component when assembled (define boundary conditions, + bolt preload) to its interfacing structure. Also, can you show in a diagram how the component is loaded.
Could you show a picture of the part and how the tube is going to be loaded. How many bolts, are they evenly pitched? I can only mention that thread radial loading will occur (under preload and axial loading conditions). If the remaining wall thickness deflects radially under these loading...
As I'm in a different industry, I wouldn't want to give you anything that isn't recognized. I do suggest asking your work colleagues about the use of the Bolt Group Analysis (BGA) method, and if there's a standard approach to the analysis of a bolted flange attachment (possible inclusion of...
As a hand calc, you could beam your two Ref. point loads / torques to the frame (use of BGA). Your frame looks symmetric, so look at one side (neglect cross members). It looks like your torques will load the center of the frame the most. If shown to be the highest applied frame loads, assume the...
You could view the FBD as something like this. Pb is the bolt tension load, keeping the two parts together, and Pa1 or Pa2 will be the abutment load, causing the prying load. Obviously there's internal pressure. The shear is due to the differential pressure load. There's bolt bending and there's...
At a first glance, the bolts will be unsupported over the shim thickness. Internal pressure will put the bolts into shear, and, due to the unsupported length, I'd suggest a bending calc'n. Due to internal pressure, the horizontal sections of the orange and blue parts will tend to bow radially...
I think this is part of the way to calculating the total deflection. This is just for bending, the horizontal beam will also have shear deflection and the vertical will have a compression displacement (section areas, E and G needed). 'Delta L' is 1/2 of the vertical column width.
I believe there's a similar 'extension' for LibreOffice, I think it's called iMath. I haven't used it as I use Mathcad, but I may have a look at it some time.
Just a pointer, you may think it's irrelevant. The joint is a preloaded bolt with material being clamped in the middle. If the clamped material was removed and a centrifugal force was applied, the nut would move vertically (assume zero rotation, fixed - fixed deflected shape, as though the mat'l...
If the design isn’t yet finalised, you may want to consider a clamp hinged on one side (pivot pin). Attached is a go at determining the band clamp bending and contact distribution for such a design. In the case of trying to reduce the bending over the bolted flange length, I can only suggest...
If friction is not overcome, then you can effectively assume the mating surfaces of the nut and clamped material are bonded (or fused). Then the force traversing the interface will enter the clamped material, causing a diminishing shear displacement through the thickness and (increasing shear...
