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?
To answer some questions.
Orientation is vertical.
Target Wind Speed is 240 km/h
Fixed into aluminium 6061 using a silicone adhesive.
Fibreglass tube is pultruded matting in polyester matrix.
Wind Speed: 240 kmh, 66.67 m/s
Wind Pressure: (0.6x(Vel2)x1.2) = 3200.00 Pa
Fibreglass Projected Area...
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...
I have plugged all these into an excel spreadsheet and I am now playing around with units metric/imperial.
Youngs modulus of E= 30000Mpa converts to 4351132lbs/in2.
I have corrected to rod OD to 16mm ID 14mm
Which has an I = 1328.9 metric = 0.00319 imperial
Density of 1663kg/m^3 = 0.06007969...
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.
