Minimum Fibreglass Tube Engagement 2

[paulmech01]

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

 

[paulmech01]

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 far into the aluminium tube is the connection required to be before it is of no greater benefit?

 

[MintJulep]

How much wind?

Is the contraption vertical, horizontal or somewhere inbetween?

What grade of aluminum?

What type of glass? What is the layup?

Epoxy or polyester?

How is the fiberglass tube joined to the aluminum?

 

[btrueblood]

And what is the outcome if the joint fails - will people get hurt? Will someone's property be damaged?

 

[Tmoose]

My first cut rule of thumb is 4 diameters of engagement. A few minutes later the conversation must get serious.

At some point the stress riser ( in the Fibreglass ?)at the edge of the aluminum tube will be the limiting factor. Generally making the edge flexible by thinning the material is a help. Fancy shapes to avoid a "staright across" edge can help.

http://www.cyclingnews.com/photos/2005/tech/probikes/?id=landis_phonak_bmc/dsc00703

 

[vonlueke]

paulmech01: Based solely on the aluminum tube hoop stress at the mouth of your aluminum tube, and assuming the tube is vertical and is Al 6061 with a lifespan of N = 1e6 cycles, I'm currently getting a required engagement length of b = L/{0.5 + [3.12/(FS*w)^0.5]}, where b = engagement length (mm), L = fibreglass tube total length including engaged portion (mm), FS = factor of safety, and w = uniformly-distributed wind load on tube (N/mm).

As a fictitious example, if L = 5500 mm, FS = 2, and w = 0.0841 N/mm, then b = 678.4 mm.

You'll also need to check the bending stress on the fibreglass tube (sigma = M*c/I) at the aluminum tube mouth, and the bending stress on the aluminum tube at its base or support, to ensure the tubes can support the applied bending moment (with a factor of safety).

You'll also need to check the compressive stress on the fibreglass tube at the aluminum tube mouth (and I don't have an equation for that currently), to see if it governs.

Do you have a wind speed yet?

 

[GregLocock]

Is this a wound glass-fibre spar or mast ? Sorry, I can't remember the proper term.

If so then one of your major issues is local shear strength/crush at the lip of the tube. There are several solutions.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[paulmech01]

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: (76.5mm x 5500mm) =0.421 m2
Total Force Due To Wind: (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 data.

 

[GregLocock]

You have made at least one mistake.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[paulmech01]

OK, Where?

 

[vonlueke]

You divided your total load by area, instead of length. By the way, I get a pressure of 2325 Pa, which gives w = 0.178 N/mm.

Aside from that, I've decided my above equation might be inaccurate, especially if FS > 1, because it somewhat uses an assumption of rigid-body mechanics. If I can, I'll try to investigate further.

 

[paulmech01]

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?

 

[vonlueke]

You're using a higher Cfig (or Cd) than I used, so that's conservative and OK. You get w = 1347/5500 = 0.245 N/mm, whereas I get w = 0.85(0.5*1.23*66.67^2)(0.0765) = 0.178 N/mm.

Without a detailed analysis, my best guess for required engagement length is currently 6*D, though 7*D would of course be better.

 

[Compositepro]

The type of joint you describe will usually fail right where the fiberglass tube enters the aluminum. The contact pressure in flex will press in on the wall and start a buckle that will then cause a compression failure. If it is a pultruded pole it will probably split length wise before compression failure because these tubes have little or no hoop reinforcement. Pultruded poles "ovalize" easily and this causes them to lose a lot of stiffness.

 

[paulmech01]

How do I calculate the, Hoop stress:

1. In the aluminium tube?
2. In the Fibreglass tube?

 

[vonlueke]

paulmech01: The equation in my first post was found to be incorrect.

I investigated further and found that when one tube is eight times more flexible (per unit length) than the other tube (as in your case), the required engagement length appears to be 5*D. Greater than 5*D seems to exhibit little or no change, theoretically. (Note that 6*D might be effective if both tubes are relatively stiff and of comparable stiffness, which isn't the case in your assembly.)

The peak longitudinal shear stress on your fibreglass tube is 1.47 MPa. I think the longitudinal shear strength of the polyester resin, alone, is no less than 8 MPa (maybe higher), so this indicates your fibreglass tube wouldn't split longitudinally. And, your pultruded tube has E-glass matting (fabric), so the weft fibre adds additional longitudinal shear strength, as well as some hoop strength, since there would be significant reinforcement fibre in the hoop direction.

It appears that the contact pressure exerted on your fibreglass tube by the lip of your aluminum tube is concentrated in only a small, 12-mm-wide band, and peaks at roughly 53 MPa on the fibreglass at the top 5 mm of your aluminum tube. (The aluminum tube is below yield.) I'm not sure, but I think slightly over 20 MPa of uniform external pressure exerted over a larger area of the fibreglass tube might cause a local stability problem (?), assuming the fibreglass tube modulus of elasticity is 14 GPa. I don't know if the smallness of the 53 MPa spot will (a) precipitate a stability problem, or (b) be able to sufficiently redistribute causing no problem. If you're concerned, you might consider gluing a 15 mm high, 5 or 6 mm thick piece of aluminum tube inside the fibreglass tube adjacent to the outer aluminum tube top edge, which might prevent the fibreglass tube from being able to collapse inward locally at that contact pressure point.

Regarding your last question, hoop stress is typically calculated as force per unit area on a pipe wall projected to a plane and divided by wall thickness; e.g., p*D/(2*t) for uniform internal or external pressure. But the stress in your tubes is due to bending, not uniform pressure, so the stress changes to mostly shear stress as you go around the tube 90 deg to the neutral axis. In other words, it may not be straightforward (nor meaningful) to calculate a hoop normal stress. The contact stress at the aluminum tube lip is almost a point load (or cosine contact band) exerted on the fibreglass tube, and dissipates as you move away from that point.

 

[vonlueke]

paulmech01: Correction and clarification to my third paragraph: The longitudinal shear stress on the fibreglass tube above (outside) the aluminum tube is 1.47 MPa, but I investigated a little further and found the peak longitudinal shear stress on the fibreglass tube inside the aluminum tube (30 mm below the aluminum tube mouth) increases to 25 MPa. So the longitudinal shear strength of the fibreglass tube would need to exceed this value (times a factor of safety) to avoid the possibility of longitudinal splitting inside the engaged length. The neutral axis longitudinal shear stress increases within the engaged length because, as you would expect, the aluminum tube exerts a large reaction couple (shear force) within the engaged length to resist the applied moment.

Also I noticed, the contact stress on that 53 MPa spot, mentioned earlier, drops to 26.5 MPa by the time it gets to the wall midsurface, just 3 mm below the contact surface. The hoop stress as you go around 90 deg on the fibreglass tube from there is about -22 MPa. The hoop stress on the aluminum tube top edge at the neutral axis is about 55 MPa.