Rotor Blade Straps

[automan1223]

I made a set of blade straps to reduce the rotor disc from 30' down to 28'. The skywheels blades are very heavy,


the new hub bar is a fixed pitch type like found in the rotordyne setups.


The aluminum hub bar is 1.250 thick milled down to 1" thk to accomodate coning angle, and has (3) 1/2 bolts securing the 2-2" wide x .250thk steel strap material. There are (4) - 3/8" bolts securing the blade to the straps in their original locations.

The steel material is your typical hot rolled spec but I feel that since I have pitch and track worked out


I will use grade 4130 bar ? to replace the hot rolled bar.


Is this material good enough for continued use ?


Is the old hot rolled stuff ?


how many lbs in tension is the straps with a 13.5' blade section turning 350rpm ?



Hence my earlier question about centrifugal forces acting on the blades. I know dragon wings have only 2 bolts securing straps to hub bar which is much lighter material, but the blades are also quite a bit lighter too.


Suggestions MR Beaty ?

Jonathan
Oriental NC

[C. Beaty]

Jonathan, I don’t know your blade weight and RPM so can only guess at the centrifugal force. Probably in the range of 8,000 to 10,000 lb.

Here’s how you figure it:

Weigh a blade and see where it balances (the CG).

Let’s say it weighs 35 lb. and balances at mid span; 6’ from root end (12’ blade?).

RPM = 325

If you have 2’ of hub from the center of rotation, the CG of the blade is 8’ from the center of rotation (R).

CF = (W x R x rpm x rpm)/2900

= (35 x 8 x 325 x 325)/2900 = 10,198 lb.

Mild steel, off the top of my head has a tensile strength of 50,000 lb./sq. in.

A ¼” by 2.5” strap has a cross section area of .25 x 2.5 = 0.625 sq. in.

But the cross section area is reduced by the material removed for the ½” bolt holes. I suppose no two bolts are in line crosswise so the reduction of area is 0.5 x 0.25 = 0.125 sq. in.

Your net cross section area is 0.625-0.125 = 0.5 sq. in.

At 50,000 psi, your strap tensile strength is 50,000 x 0.5= 25,000lb.

You have 2 straps so the total breaking strength in tension is 50,000lb.

If my guess about weight and RPM was correct, you have a safety factor of 50,000/10,198 = 4.9.

I don’t think you’ll be able to pull the straps apart.

[automan1223]

al hammer was kind enough to zap me this link from the skywheels site itself.

http://www.bladerunneraircraft.com/data.html

it comes in with some figures and such.

you said

"But the cross section area is reduced by the material removed for the ½” bolt holes. I suppose no two bolts are in line crosswise so the reduction of area is 0.5 x 0.25 = 0.125 sq. in"

I have 2 bolts at the end of the hub bar like the dragon wings do. I believe that means "crosswise" the 3rd bolt is back towards the teeter block and is in the middle of the two or in the middle of the bar. So I am guessing I have a bit less than that due to cross sectional losses of 1"

The straps are only the same width as the bar 2" so if I do the math right, I have only 1" of each strap really holding due to the bolt holes. x 2 straps. so in effect :


I have a single 2"x.250 piece of steel holding the blade on to the hub. If I have this correct. I have a total cross section of .500 according to your formula TOTAL.... .500 total cross section is half of 50,000 lbs is 25,000 and that means I have a safety factor of a little more than 2.5 to 1

which does not sound all that great.

Then there is the issue of the hub bar which I believe is 6061t6 aluminum. the cross section of the bar is 1" thick x 2" wide. Subtracting bolt losses. Tens strength listed at about minimum of 40 to 42 ksi in the aircraft spruce catalog but again I am not sure If I am using the right data.

Or am I over simplyfying this because I have 3 bolts holding this together. I am horrible with the math I really appreciate your help.

Thoughts ?

Jonathan

[C. Beaty]

I’m afraid you’re on thin ice if in fact you are using 2” wide straps with ½” bolt holes across from each other.

I was going by hub width of 2.5” and assumed your straps were the same width. Also, the early Rotordyne straps used a staggered hole pattern so no 2 bolts were in line.

To shear 3 ea. ½” AN bolts in 6 places requires something like 88,000 lb., a bit of overkill in the bolt department.

To shear 3 ea. 3/8” AN bolts in 6 places requires about 50,000 lb.

[automan1223]

Chuck, I am going to take a photo, My concerns are 2 fold, the straps and the bar.

Give me 10 mins.

Jonathan

[automan1223]

Ok I just measured eveything again. Hub bar and blade straps are 2.5" wide.

Hub bar is milled down to 1" from 1.250 (effective area)

Straps are 2.5 x .250

bolts are as previously spec'd

see attached photos.

[C. Beaty]

OK, that’s a little better. Pull apart strength should be about 30,000 lb.

In addition, I really don’t know what your blades weigh or how fast they spin. I was merely pulling numbers out of thin air to give you an example of how to do it.

Weigh your blades, tell me where they balance, how long they are and how fast they spin. Then I can come up with real numbers.

[automan1223]

Chuck I really appreciate your offer and I will do that asap. I feel there is a lot of important stuff to be learned here and I dont want to be on the wrong end of this important equation.

The blades spin 330-345rpm.

Each blade measures short of 13' 6" (skwheels have that faired out end cap)

I will get balance data asap.


Thanks again.

Jonathan