Strut braced wing structural analysis

[David Lewis]

Example Aeronca Champ kit #301:
span = 23.7 in (= 0.602 m)
weight = 67 g (= 0.66 N)
load factor = 3.5
Lift = 3.5 x 0.66 N = 2.31 N

Graph below depicts the load diagram:
Horizontal axis = distance (metres)
Vertical axis = lift concentration (newtons per metre)
Area under load curve = 2.31 N
Max lift concentration (at centerline)
= 4 * lift / pi * span
= 4 * 2.31 N / pi * .602 m
= 4.89 N/m

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[David Lewis]

We are interested in the load supported by the wing (blue shaded area).
Each pinion generates 1.05 N of lift.
Centroid of the distributed load is 0.138 m from centerline.

First step is to calculate the reaction forces.
To do that, we replace the distributed load by the equivalent resultant force.

[David Lewis]

There are two support reaction forces:
F (supplied by the wing strut), and Fwr at the wing root (supplied by the cabin side).
We are interested in the vertical component of F (denoted Fv).

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Summing moments at the wing root:
Fv * 0.162 = 1.05 * 0.117
Fv = 0.758 N

F = sin 22.6 degrees/0.758 = 1.97 N
(tension load on the strut)

Fwr = lift - Fv
= 1.05 - 0.758 = 0.292 N

[David Lewis]

Normally, shear loads on small model airplane wings are not investigated because, if the wing is strong enough in bending, it will automatically be strong enough in shear. However, it's easier to plot the bending moment diagram if we construct the shear force diagram first. Wherever shear force is zero, the absolute magnitude of bending moment reaches a maximum.

Beginning at the wing root, bending moment is zero. (If the wing were not supported by the strut, we would be able to move the wing tip up and down, and the wing would rotate about the root with no significant resistance. The wing root can be conceptualized as pin connected to the cabin side.)

Shear load at the wing root is the same as the reaction force Fwr (= 0.292 N). Since the reaction force points down, and the load just to the right points up, shear load at the wing root is conventionally assigned a negative polarity.

First point on the shear force diagram at x = 0 is -0.292N.

As we move along the wing, the absolute value of shear force will get smaller because wing load forces point up. Shear will go to zero when the area under the load curve reaches 0.292 N. My computer aided drafting program indicates that occurs 61 mm from the root.

Second point on shear diagram at x = 61 mm = 0 N of shear.
(A bending moment maximum also happens here).

To calculate bending moment at x = 61 mm, replace the distributed load from x = 0 to x = 61 mm with a concentrated 0.292 N force pointing up. The point of application coincides with the centroid of the area under the load curve from x = 0 to x = 61 mm. This centroid is located 30.6 mm from the root.
0.292 N * 30.6 mm = 8.9 mN-m. By convention, moment is negative when it causes a beam to arch upward.

Bending moment at x = 61 mm is -8.9 mN-m.

In the diagram, the magnitude of force F is the area under the load curve from x = 0 to x = b. The point of application for force F is the centroid of this area. The point of contraflexure occurs when the area under the load curve from x = 0 to x = b multiplied by distance a equals 0.292 N times distance b:

F * a = 0.292 * b (in other words, the sum of moments at x = b is 0)

According to my CAD program, when b = 124 mm, F = 0.578 N, and a = 63.7 mm.
0.578 N * 63.7 mm = 36.8 mN-m
0.292 N * 124 mm = 36.2 mN-m (close enough)

Bending moment at x = 124 mm = 0 mN-m.

[davidchoate]

David, this has nothing to do with wing design, but horizontal stabilizer size in relation to wing area. I recently converted a Dumas Aeronca 30" WS. To RC. I built it exactly to scale. Except for adding right and down thrust. I had a expert pilot at my RC Club maiden. It. It flew, but very. Unstablely. He said. He was not going to be able to Land it properly. Also ; what is a good. Decolage to set the tail and wing. I have been told 2deg positive is good.

[David Lewis]

David Choate, Hope it's OK with you. I've moved your question to a new thread (airplane stability). That way we can keep each engineering topic neatly organized and easy to find.

Meanwhile, back at the ranch, the second bending moment maximum occurs at the strut attachment point.
To calculate this moment, replace the distributed load from the strut to the wing tip with a concentrated load.
My CAD program shows 0.32 N at 48 mm from the strut attachment point.
0.32 N * 48 mm = 15.4 mN-m.

00diagram2.JPG

Bending moment at x = 162 mm = 15.4 mN-m.
We now have enough data to plot the bending moment diagram:

00diagram3.JPG

As an exercise, we can finish plotting the shear force diagram.
Shear immediately to the right of the strut attachment equals the area under the load curve from the strut to the wing tip (0.32 N as found above).

At the strut, the shear force is
0.32 N - Fv = 0.32 - 0.758 = -0.44 N

00diagram4.JPG

[David Lewis]

We now have enough info to design the WING SPAR and LIFT STRUT static test jigs.
From the moment diagram, we see maximum moment of 15.4 mN-m occurs at the strut attachment (118 mm from the wing tip).

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How much weight hanging from the wing tip is needed to create 15.4 mN-m of bending moment at the strut attachment point?
x = 15.4 mN-m / 118 mm = 0.131 N = 13.4 g.
You then multiply 13.4 g by 1.5 factor of safety (=20 g).

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We see that stress arising from aerodynamic loads is low. The 1/8" x 1/4" balsa wing spar is not needed. The wing should be designed to withstand crash loads, which is harder to do since analysis is based on experience rather than simple formulae.
Likewise, to test the tension strength of the lift struts and their attachment points, we apply a force of 1.05 N x 1.5 (=161 g) located 117 mm from the wing root.

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[davidchoate]

Recently I have found that it is possible to create a one piece wing for the trainer type Planes. I Know that on a typicle Guillows kit; You just glue the wing panels to the side of the Plane, and wedge and glue in the wing strut to hold it all together. I have since learned about cantalevering, and how it makes the strength.And I must thank You for saying once " although models are scaled much smaller than the fullsize, aerodynamically they work the same. But the Plane is smaller in every way, but the air molecules remain full size". That one sentence, for Me anyway, made Me able to understand so much more easy. But anyway. On the small models I find the struts just for scale looks, and (with a one piece wing) not necessary.But on a full scale Plane It's obviously needeed. Thanx again David.