torque/stress calcs

[Stewart Corman]

I was trying to post this on Otherpower, but somewhere it went to Never-Neverland.

In another thread, someone asked if tubing spars were sufficiently strong to withstand the force of the wind on the blades and admittedly, I hadn't done any calcs on this and just went by gut feel. Please feel free to poke holes in any assumptions/equations/conversions I have used.


Thought that any mechanikers out there might like to chime in.
Mark, this is pertinent if you just fly the three blades I made for you.


Here are some specifics and presumptions using some worst case scenarios:

turbine diameter 18 feet
three blades/TSR=6
furl at 20mph
steel spar tubular dimensions : 1.25OD, 0.065 wall thickness
soft steel because (hot dipped ?) galvanized

I will now work backwards to get the loads:
at 20mph, 100% wind energy = 10KW, so 40% eff = 4KW
for TSR = 6 , rotation @ 20mph = 315rpm

power conversion: 4KW/(746w/hp) = 5.4 hp
torque= hpx5252/rpm = 5.4 x 5252 / 315 = 90ft-lbs
since 3 blades, 30ft-lbs exerted by each blade

modulus of elasticity of mild steel = 29,000 Kpsi
yield strength steel = 35,000 psi

from the tubing geometry:
section modulus Z = .068 in^3

maximum stress for cantelevered beam is :
s = weight x length / Z where W x l = torque???
if so, then
s = 30ft-lbs x 12in/ft / .068 in^3 = 5,300 psi

if above are correct, then this tubing is at 15% of the 35,000 yield
How do you apply safety factors??

I have little concern for the first set of 10 foot blades which obviously have much less power associated with them

note that real tubing is closer to 1.375 diameter (slightly better specs by 23%)
Actual length of stressed member is really 6 inches shorter (outer ubolt hole circle)
If I choose 6 blade and TSR <= 5, lower rotation to 220rpm max is more torque, but the force is distributed over 2x more spars (or not exactly?)

I have not accounted for the bending back of the blades towards the tower (or have I??)

Did I miss anything obvious??

BTW, yes ..this is rocket science!

Stew Corman from sunny Endicott

[Mark Parsons]

Hi Stew,

It looks like your calculations are for tubing stretch along centripedal axis. This force would be greatest under fault conditions of unloaded turbine and maximum RPM. Max RPM would be obtained with unloaded TSR. Don't see that in your calculations.

How about axial deformation from wind force?

I would expect the largest force by a magnitude or more to be the centripedal force wanting to throw the blades off the ends of the spar. How about the shear modulus on the screws that hold the blades to the spar?

An engineer's work is never done... :eek:

Regards,
Mark

[Stewart Corman]

finally got Otherpower reposted and someone did bring up a significant point:

Most of the lift will tend to bend the blades back toward the tower


very good point ... wasn't thinking along those lines

But we do have some numbers to play with:
I need to incorporate the fact that at 20mph, an 18 foot turbine by calculation exerts 250lbs thrust on the tower.

I can assume a point load in the center of the 6 foot paddle on a 30 inch spar, which would then call for a 5 1/2 foot lever arm or 250lbs/3 blades x 5.5 feet = 460 ft-lbs torque backwards from wind


Or I could presume a distributed load along the length of the 6 foot paddle?

Looks like I have a few more calcs to do, but it appears that the tubing will NOT be sufficient for an 18 foot configuration @ 20mph w/o inserting another 12" slip fit tube inside to beef it up

Here is is obvious that 6 blade configuration is of great benefit to distribute the forces over more spars.


Stew