About Flight 587 tail loads

timor · Nov 5, 2002

Hi all,

I was reading a document from NTSB web site about the Flight AA587 Airbus A300-600R. The tailfin was ripped off due to using rudder when the plane was yawed. The presentation of vertical tail loads can be seen from page 3 of the document:

http://www.ntsb.gov/Events/2001/AA587/exhibits/242282.pdf

Here the diagrams show the torsion, bending and shear as a function of time when the rudder is fully deflected and kept there for about 17 seconds. What is not presented is the effect of full opposite rudder after about 3 secs when in full sideslip. So I'm trying to figure it out from the diagram this way:

Torsion: Effect of rudder about 0.5*Limit Load
Effect of air loads about 0.5*LL
Bending: Effect of rudder 0.8*LL, air loads 1.2*LL
Shear: Effect of rudder 0.6*LL, air loads 1.2*LL

I think the effect of rudder deflection alone can be seen taking the values after about 0.5 sec. after deflection (straight lines to about 0.5 sec.)
In case of opposite rudder the rudder and the air loads act from opposite directions in torsion, and in the same direction in bending and shear. So the total loads are:

Torsion: 0.5-0.5=0
Bending: 0.8+1.2= 2.0 times the limit load
Shear: 0.6+1.2=1.8 times the limit load.

Is this kind of figuring about right ?

Timo

KSshopper · Nov 7, 2002

The loads presented in this diagram are dynamic, not steady state. The steady state bending and shear are about 1% - 2% limit load. Although steady state loads can be combined algebraically, the dynamic loads are governed by a set of coupled differential equations that require more information than presented in these charts.

timor · Nov 7, 2002

Yes, my point was, that in this case you can just combine the dynamic values algebraically with quite good accuracy. After my first post I found article from Aviation Week's web site. Quote:

"The point at which the tail broke was 29% beyond that required by certification regulations. Airbus engineers estimate the tail broke at 1.93 times the limit load, whereas it is only required to withstand an ultimate load of 1.5 times limit load. In fact, during ground static tests the tail broke at 1.93 times limit load...
The loads were so high despite the rudder travel limiter because the back-and-forth pulsing of the rudder built up large sideslip angles that greatly increased fin loads. Forces exceeded ultimate load on the penultimate swing. Several methods of calculating the loads indicate the fin broke at 1.17-1.35 million lb.-ft. of root bending, 79,000-92,300 lb. of side force, and 39,800-69,500 lb.-ft. of torsion"

http://www.aviationnow.com/avnow/news/channel_awst.jsp?view=story&id=news/aw1101key.xml

So my calculation is in the right ballpark, except for the torsion. The article suggest that the sideslip angles were even higher (and rudder deflection smaller) than in the Airbus paper, that would explain the extra torsion.

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