# Wings as Beams

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#### wsimpso1

##### Super Moderator
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This discussion topic came up in the thread on beams and how they work. I think that we need to talk about how wings work as beams.

I ran out of time, as the wife and I are leaving on a trip. If the writing is not up to my usual standards, I apologize. Have at, and I will be back on when the trip is finished...

For the moment, let's stick with positive AOA, positive g type of lift, and establish the two frames of reference. One frame of reference is based on direction of airflow, the other the wing’s chordline. The airplane is flying horizontally; lift is in the vertical direction, and the wing AOA can vary from zero to about 18 degrees. Wings carry loads in the vertical direction varying from zero to four, six, or even more times the weight of the airplane. Eighteen degrees is about as much AOA as the wing chord line can make before stall, some a little more, others a little less. In some airplanes, wings can be flown to much higher AOA, but that is post-stall - lift decreases and drag goes up by an order of magnitude when AOA gets higher than about 18 degrees... Getting high CsubL and high speed simultaneously is outside of the envelope we design for in little airplanes.

To frame the problem, lift and drag are described using the same equation:

dynamic pressure * coefficient * projected area

Coefficient of lift on airfoil sections can go to around 1.8, while Cd is not plotted beyond 0.03. Max drag at AOA less than stall is on the order of 1/60 of max lift. That is for sections. Real 2D wings do about 78% that well on lift and drag of real wing is bigger, but there are still about 30 times as many units of lift as there is drag possible in most wings. The forces aft on the wing due to drag are pretty small compared to the lift that can be obtained.

There is also pitching moment, and which is significant and it is the genesis of torque in the wing – its equation is similar to the above:

dynamic pressure * coefficient * chord * projected area

The coefficient is usually between 0 and -0.10, chord in feet is around 3 to 6 feet, so pitxhing moment in ft-lbs is bigger than drag in lbs.

The lift occurring at high AOA is still wing frame of reference UP, but the wing chord line can be tilted up as much as about 18 degrees. So let’s think about positive g AOA near stall. The sine of 18 degrees is about 0.31, while cosine is about 0.95. The lift the wing is making vertically is distributed 31% forward along the chordline and 95% up perpendicular to the chordline. This 31% of wing lift forward accounts for some early wing failures with the wing wracked forward and not up. Let's follow through how these loads are handled in our wings, which are beams in both vertical and horizontal directions.

The lift components forward and up on the wing get distributed along the wing approximately elliptically. Since the lift is spread out chordwise along the wing, and there is pitching moment from the lift, there is also torsion applied to the wing, usually pitching the wing nose down. These loads do get distributed. Then there is some drag which gets subtracted from the forward component of lift. All of these loads are zero at the wingtip and accumulate towards the root end.

Ok, we have lift, pitching moment and drag, and then they load up the wing’s structure. Lift perpendicular to the chord line can get huge, forward component of lift plus drag can get to about 30% of total lift, and moment is smaller...

Wings are Beams

Beams carry shear load, bending and, when needed, torque. So what is a beam? What makes it carry torque? A little on those questions first.

A beam is intended primarily to carry shear and bending loads. Usually you are looking to carry those loads in some kind of efficient manner, whether the criteria are cost or weight or space or resonance avoidance. Sometimes you have functions you want to perform besides just carrying loads. Like lifting and controlling an airplane. You can build up several elements into a beam, even if some of those elements look like beams. And the total beam can look like other things, like airplane wings.

Beams become stressed when carrying shear, bending and torsion. In a wing where the elements are well attached to each other, all of the parts move together in carrying that shear, bending and torsion. Much of how the loads are distributed has to do with what fraction of the total stiffness against bending and torsion each element has.

Imagine two channels, one really beefy, another somewhat thinner. Bolt them together web to web. The bending stiffness of the two are combined to make a larger bending stiffness. And the fraction of bending moment carried by each will be the ratio of each piece’s bending stiffness divided by the whole bending stiffness. Bending load is spread among the pieces according to their relative stiffness.

This combination can be extended to the whole wing. Imagine a big beam, a smaller one, and the wing skin, all nicely tied together so that they deform together under bending and torsion… The bending stiffness in the fore-aft direction, bending stiffness in the up-down direction and torsional stiffness about the centroid all can be calculated. The terms are EIxx, EIyy, and GJ. Here is an interesting one, Ixx + Iyy = J.

How to calculate EI and GJ? There are equations for common shapes, other stuff has to be done numerically. Ixx = Sum (dA*y^2), Iyy = sum(dA*x^2), and J = sum(dA*r^2).

E and G? These are characteristics of the materials. I always have to look up everything except steel.

You can carry bending moments a bunch of ways. The lightest structures for this usually look like the shapes we are accustomed to for beams. And you will find them inside wings…

Now let’s imagine a metal skinned cantilever wing. There is one very identifiable spar near or on the thickest part of the wing with upper and lower caps and a web or two tying them together. There ribs in front of and behind the main spar giving the wings shape and extending back to where a much lighter spar is installed. The drag spar is usually just a light formed sheet metal structure. Then there are skins attached to the spars and ribs. Take a cross section through the wing and you cut the spars and skin – the ribs are few enough that they hardly contribute to the wing, and are usually ignored for bending stiffness. Properly done these parts all move together under bending and torsion. We can thus figure out the cross section’s ability to carry load and deflections while doing so.

We can calculate the area of vertical elements – they carry the vast majority of shear, and usually you only include the vertical elements of the main spar. Timoshenko and others talk about this phenomenon in their books on mechanics of materials.

We can calculate the centroid of the wing and then the I’s and J for the whole cross section. There are simple formulae for the normal beam looking parts, but you have to do some piece wise calcs to do the skin. Once you have I’s and positions for all the pieces you can calculate stresses everywhere and figure out if you made something too heavy or if something needs a little more beef.

Something to remember here, lift perpendicular to the chord line can get big, but the wing is a wide hollow beam only a few inches thick resisting that moment. Since depth of beams is so important, we need a significant beam to do this job.

Next we have somewhat lower force pulling forward on the same wing, but the beam is several feet deep in that direction. Because x^2 is big, dA does not need to be very big to be adequate to carrying these moments.

Usually if the main spar is beefy enough to carry lift perpendicular to the chord line, you have a drag spar, and some kind of bracing tying the spars together (skins and ribs or ribs and drag/anti drag wires), your wing can be made to carry this fore/aft load pretty lightly.

The spar is tailored by doing this operation at the root, and then at intervals out the wing until you get to spar and skin dimensions that you will not go below. Have fun.

Sorry, this is as far as I got...

See you guys later..

Billski

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#### BBerson

##### Light Plane Philosopher
HBA Supporter
If the coefficient of lift (CL) is 1.0 and the coeffient of pitching moment (Cm) is .1, does that mean the pitching moment of the wing in foot pounds is 1/10th that of the wing lift in pounds?

I was using Richard Hiscock's chapter on pitching moment. He said:
"We should note that neither the total lift on the wing or the angle of attack is required to determine the moment."
So using his example formula I get -586 foot pounds of torque total. So I guess that moment is the same at 1g or 4g.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
You gotta include wing chord in the calc.

As to moment changing with g, some are pretty much fixed across the range of AOA, while others move around some.

The big thing is that q changes with V^2. So moment , like drag, and possible lift, can go up quite a bit with speed.

#### Vigilant1

##### Well-Known Member
Thanks for starting the thread, I'm enjoying it. If I may impose/suggest a little:
1) when the time is right, feel free to include an example.
2) similarly, when appropriate (and for extra credit ) maybe include a composite wing structure and that groovy matrix math.

Sorry for the piling on. As the cynic says:. "Every good deed spawns a dozen more requests.'

#### BBerson

##### Light Plane Philosopher
HBA Supporter
Yes, his formula includes Cm, chord, dynamic pressure and wing area.
I just don't know if his example is design moment at 1 g or 4g.

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#### Jimstix

##### Well-Known Member
BB, the prior page says this is an extrapolation from wind tunnel data to simulate Cm at 90 mph. All the wind tunnels I have ever seen operate at 1g. Keep up the good work! Jim

#### BBerson

##### Light Plane Philosopher
HBA Supporter
So I figure 586 foot pounds of torque total for my design. Or 293 foot pounds for each wing at root.
Not sure how to load test yet. Maybe try the full torque at the tip.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
If the coefficient of lift (CL) is 1.0 and the coeffient of pitching moment (Cm) is .1, does that mean the pitching moment of the wing in foot pounds is 1/10th that of the wing lift in pounds?
Not quite. If CL is one CM is -0.1 and the chord is 1, then pitching Moment in ft-lb will be 10% of the lift in lbs. if chord is 2, then 20%, and so on.

#### BBerson

##### Light Plane Philosopher
HBA Supporter
I am still confused about pitching moment (never really looked at it before).
Page four of Theory of Wing sections has the formula. It looks like the only variables is dynamic pressure, wing area, and chord. So I assume at zero g or any g the pitching moment is same. Since lift isn't a factor.
And Hiscock's said: "We should note that neither the total lift on the wing or the angle of attack is required to determine the moment." That doesn't seem right, so I am confused.

And yet the symmetrical NACA 0012 has no pitching moment at any angle of attack or lift.

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#### wsimpso1

##### Super Moderator
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Log Member
Yes, his formula includes Cm, chord, dynamic pressure and wing area.
I just don't know if his example is design moment at 1 g or 4g.
The g's do not matter a lot to Moment. Cm for any one foil mostly stays put, unless you deploy flaps. Once you treat Cm as fixed (for any given foil, chord, and wing area), Wing pitching moment becomes a function of q...

G's don't come into it. You can go from 1g to 4g, provided you are going at least twice stall speed, by simply increasing AOA, and pitching moment of the wing won't change significantly during the event.

Anticipating the next question: Whole airplane pitching moment goes up and requires increasing elevator force with g's because - for a stable airplane - we have CG ahead of the neutral point. The airplane weighs more when we pull g's, and weight times the static margin is a big piece of whole airplane pitching moment.

I think it would help understanding to look up the plots in TOWS for any airfoil. They have plots for each airfoil showing the relationships between AOA and Coefficients of Lift, Drag, and Moment.

Billski

#### wsimpso1

##### Super Moderator
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Log Member
So I figure 586 foot pounds of torque total for my design. Or 293 foot pounds for each wing at root.
Not sure how to load test yet. Maybe try the full torque at the tip.
Wing tests with sandbags; pick a load case, AOA, lift, moment. Figure out the spanwise lift distribution to get the lift. IF all that weight went on the quarter chord line, you would be simulating zero pitching moment. If it is centered a little aft of that, it will generate a pitching moment. You have to figure out just how far aft gives the right moment.

The usual test case is at Vne and max positive g, which is usually a pretty low AOA, the wing is least resistant to bending that way, and pitching moment of the wing is biggest then too. Some folks will also test at Va and max g, AOA is at stall angle, pitching moment is quite a bit smaller then because Va squared is a lot lower than Vne squared.

Billski

#### wsimpso1

##### Super Moderator
Staff member
Log Member
I am still confused about pitching moment (never really looked at it before).
Page four of Theory of Wing sections has the formula. It looks like the only variables is dynamic pressure, wing area, and chord. So I assume at zero g or any g the pitching moment is same. Since lift isn't a factor.
And Hiscock's said: "We should note that neither the total lift on the wing or the angle of attack is required to determine the moment." That doesn't seem right, so I am confused.

And yet the symmetrical NACA 0012 has no pitching moment at any angle of attack or lift.
All that stuff is right. With any foil and wing area and chord, Cm does not usually change much with AOA. Q changes a bunch, and that changes moment from the wing...

#### BBerson

##### Light Plane Philosopher
HBA Supporter
Yes, I looked at the pitching moment of NACA 0012 which is always zero at any angle of attack.
For the whole airplane the usual negative wing pitching moment must increase with speed. But instead of up trim, the pilot usually trims down. (I think, been a few years since I have flown)

Good! I only need to use 293 foot pounds for the basic torsion test. Aileron torque load is a "nother" matter.
I will be testing Ercoupe type diagonal ribs for torsion. Thanks.

#### BBerson

##### Light Plane Philosopher
HBA Supporter
Wing tests with sandbags; pick a load case, AOA, lift, moment. Figure out the spanwise lift distribution to get the lift. IF all that weight went on the quarter chord line, you would be simulating zero pitching moment. If it is centered a little aft of that, it will generate a pitching moment. You have to figure out just how far aft gives the right moment.

The usual test case is at Vne and max positive g, which is usually a pretty low AOA, the wing is least resistant to bending that way, and pitching moment of the wing is biggest then too. Some folks will also test at Va and max g, AOA is at stall angle, pitching moment is quite a bit smaller then because Va squared is a lot lower than Vne squared.

Billski
Yeah, that makes sense. With the wing upside down and loaded a little aft it will test the negative pitching moment. Neat, never thought of that.
Then with the load moved forward and the wing upside down and tilted 14° it will simulate the anti drag load. Got it.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
Yes, I looked at the pitching moment of NACA 0012 which is always zero at any angle of attack.
For the whole airplane the usual negative wing pitching moment must increase with speed. But instead of up trim, the pilot usually trims down. (I think, been a few years since I have flown)

Good! I only need to use 293 foot pounds for the basic torsion test. Aileron torque load is a "nother" matter.
I will be testing Ercoupe type diagonal ribs for torsion. Thanks.
What happens with increasing speed at 1 g is the AOA decreases with speed- q goes up dramatically with speed, so does whatever moment and drag you have, but you are decreasing AOA to keep total lift the same. At just barely above stall speed and 1 g the nose is up, and at twice that speed, the Cl has to be 1/4 what it was, so CL has to go down a bunch. So you roll the elevator down and roll in some trim to keep it there...

Now if you stayed at one g AOA from stall, and doubled your speed, you will either be doing a loop or gotten into a 60 degree banked level turn, or some other maneuver that uses 4 g. The elevator position and AOA and all of the coefficients stay the same, but Forces on everything quadrupled...

Billski

#### Pops

##### Well-Known Member
Log Member
Yeah, that makes sense. With the wing upside down and loaded a little aft it will test the negative pitching moment. Neat, never thought of that.
Then with the load moved forward and the wing upside down and tilted 14° it will simulate the anti drag load. Got it.

#### BBerson

##### Light Plane Philosopher
HBA Supporter
Looking at Hiscock's chapter 5. The max pitching moment is at Vd (velocity dive), substantially more than cruise, of course. I may just load one diagonal rib to destruction and get some idea of the rib design needed for torsion.

The Ercoupe outer panel is about 60 pounds (58 I think).
My design is one third of the Ercoupe gross weight so my outer panel should be 20 pounds.
The Ercoupe ribs seem flimsie for such force. I can't see making such ribs but even thinner for one third gross weight.
Probably go with truss type.

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#### poormansairforce

##### Well-Known Member
Yeah, that makes sense. With the wing upside down and loaded a little aft it will test the negative pitching moment. Neat, never thought of that.
Then with the load moved forward and the wing upside down and tilted 14° it will simulate the anti drag load. Got it.
Question: I always thought that this should be done in a single test with the wing tilted down as this would represent actual load condition? Or did I read that wrong?

#### BBerson

##### Light Plane Philosopher
HBA Supporter
The single test is typical. Since I have no experience with Ercoupe type diagonal ribs for torsion, I want to test torsion in a separate test. I never heard of anyone doing such a test.

#### wsimpso1

##### Super Moderator
Staff member
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The V-n envelope has a bunch of positive and negative g places you could test. Normally, you run a few analytically and then pick the one that gave you smallest FOS for test. I do not know up front if Va with its high AOA and large forward component of lift will be worse or the high speed Vd with low AOA but same g's will be worse on the fabric covered triangulated rib scheme...

Tough problem without FEA with all of the buckling tools turned on... Sounds like a mission for either running it analytically or for testing both cases just to limit load first... In order for your fractional test to do you any good, I would think you would need at minimum two ribs and the spar/ D-tube, maybe four ribs, then give it the wing pitching moment at the root to see if it will go unstable and buckle... Then you are still facing a build/test/ iterate cycle. Ugh.

Billski