For pressure I have used:
=(0.5xDensityAir)xVdes2xCfig
=(0.5 x 1.2kg/m3) x 66.66^2 x 1.2
=3199.36Pa
Multiplied by Area Gives a Force(N):
Area 76.5 x 5500mm = 0.421 m2
Area x Pressure = 1346.93 N
Is this not correct?
...(Pressure x Area) 1346.40 N
This gives me a value for W as (1346.4 N / 421000 mm)= 0.00306 n/mm.
When used in "Vonluekes" equation, b = L/{0.5 + [3.12/(FS*w)^0.5]} gives a B = 136.2mm.
Max Bending Moment: (Dist from Centroid to base connection) 3702.60Nm.
Any further sugestions based on this...
The Fibreglass tube is sleeved into the aluminium tube.
The fibreglass is quite long (about 5-6m) out of the aluminium tube in a cantilever situation, and has a load due to wind acting upon it. This generates a bending moment down at the connection into the aluminium tube.
My question is how...
What is the minimum Tube Engagement that is necessary for a connection between a Fibreglass Tube into an aluminium tube that is in cantilever.
Fibreglass tube is 76.5mm OD 6mm Wall
Aluminium Tube is 77.0 ID 6mm wall
Is this just a consumables site.
What product do you mean exactly, the Super Alloy 1?
What is its composition, (ie silver, tin)?
What is the flux used with it made from?
I am trying to make an electrical connection between an electrical cable and aluminium, currently brass pins are soldered to the aluminium and the wire is soldered to the pin.
We need to move away from the solder used to join the brass pin to the aluminium as it contains lead, but no ther lead...
The core of the cable that is connected to the aluminium is tinned copper.
Could this be spot welded or induction or some other resistance weld joined to the aluminium.
The aluminuim is quite thin, Dia 12mm x 1mm wall.
I am trying to make an electrical connection between an electrical cable and aluminium, currently brass pins are soldered to the aluminium and the wire is soldered to the pin.
We need to move away from the solder used to join the brass pin to the aluminium as it contains lead, but no ther lead...
Thanks Steve,
I have just borrowed Thompsons, Theory of Vibration, I think I really do need a text on this topic.
The result looks good, I'll make sure it all seems Ok then I will get back to you guys with result.
Thank Tom, that's handy.
Is this working on the same theory that we have been discussing.
ie.
Per Den Hartog's Mechanical vibrations
wn = a * sqrt(E*I/(mu*l^4)
also is this formula correct for calculating the vortex shedding frequency for a closed end circular hollow section dia 16mm x 720mm...
...2nd Mode 345.43 rads/sec = 54.97Hz
3rd Mode 1899.87 rads/sec = 302.37Hz
Calculating the vortex shedding frequency from the given formula f = 0.22 * V/D with:
D = 16 mm
V = 22.22 (80km/h)
gives:
f = 305.53 Hz (Pretty close to the 3rd mode)
also if velocity = 22 m/s then freq = 302.5 Hz
Now...
The tube is closed at both ends.
The angle of the airflow can change depending on where this is mounted.
Position a = perpendicular flow.
Position b = 45deg up from perpendicular.
I should clarify, the tube is mounted one end, and is generating a vibration when traveling at approx 80km/h, due to wind/airflow which I believe and have been advised due to vortex shedding.
Does above formula output a frequency?
What does mu denote in the above formula?
I need to calculate the natural frequency of a 19mm Dia x 1.6wall x 720mm long Fibreglass Tube. I am trying to calculate the critical speed of of this member as it is developing vibrations at around 80km/h (wind speed). I would to see if the critical speed is close to this 80km/h.
If it is I...
I had a quick look at this thesis, it pretty full on.
Hopefully I can relate the frequency back to Velocity and might nearly be there.
Also relation of phsical properties and there effect.
That was my first thoought, but I don;t think that marketing want this look, and this will increase drag.
How can the frequency and vortex shedding be calculated?
