COVER STORY

Wing spars handle the shear and bending moment due to the air loads.

mate the resulting loads. ribute these loads over the

ulate the resulting loads on rent pieces. rmine the material, size, and the part. ulate the resulting stresses in

pare these stresses with the m allowable for the material

ze the part as necessary. through 3 are addressed in Rudder attach points must be robust enough to handle the torsion created by rudder deflection.

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Airpla The shear load between two riveted pieces tries to slice each rivet in two.

The structure of the empennage must handle loads from nearly every direction.

ne Structures Make sure the part will handle the load Neal Willford, EAA  We started looking at the process of properly sizing an airplane’s structure to withstand the maximum loads expected in normal operation. This month, we will continue the discussion, and a new spreadsheet is available to download from the EAA Sport Aviation page on the EAA website at www.eaa.org that will help estimate the strength of structural pieces under tension and compression. EAA Sport Aviation

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Recall that the structural design process follows these steps: 1. Determine the overall load factors. 2. Estimate the resulting loads. 3. Distribute these loads over the airplane. 4. Calculate the resulting loads on the different pieces. 5. Determine the material, size, and shape of the part. 6. Calculate the resulting stresses in the part. 7. Compare these stresses with the maximum allowable for the material used. 8. Resize the part as necessary. Steps 1 through 3 are addressed in References 2 through 5, and step 4 was the focus of the last article. The simplest form of loads examined were axial whose magnitude and direction lie along the major axis of a part. Those that pull on a part are tension loads; those that push on it are compression loads. We also saw that a moment is generated on a part when the applied loads act at some angle to the part’s major axis (such as with air loads along a wing). Finally, we walked through the steps in calculating the reactions on a wing spar due to air loads (Figure 1).

External Loads

With the loads determined, it’s time

to consider steps 5 through 8. Figure 1 shows a strut-braced spar and the reactions due to the constant air load of 25 pounds per inch. The loads along the spar can now be determined. Even though the span loading is constant in this case, the load at different locations along the spar is not. The resulting load is perpendicular to the wing axis and is called a transverse shear load, or just shear load for short. We can find the shear load on the spar by starting at the wingtip and summing the load along the span as we move inboard. For example, the shear load at the tip is zero. By moving inboard 1 inch, the shear load increases to 25 pounds, and at 10 inches inboard the shear load has increased to 250 pounds and so forth. Figure 1 shows the resulting shear load diagram along the spar, where I used a sign convention of positive for loads pointing up. There is a linear increase in the shear load until point 2 (the strut attach point), where it suddenly goes from a positive 1,750 pounds to a negative 1,850 pounds. This 3,600pound discontinuity is caused by the downward vertical component of the strut load. This is where the shear load on the spar is at its maximum. Moving inboard of point 2, we can see that the positive 25 pounds/

Figure 1. Shear and moment diagram for the loads shown on the strut-braced wing spar.

inch span loading reduces the magnitude of the shear load on the spar until we reach point 1. The spar shear at this point is a negative 600 pounds, indicating that the direction of the shear load is down. Since the sum of the forces on an object must equal zero if it is to remain stationary, there must be an upward load of 600 pounds at point 1. Looking at the reactions on the spar shown in the top of Figure 1 indicates that the vertical load at point 1 is indeed upward. The bending moment along the spar can be found by using the same method. Starting at the tip and moving inboard, the moment at any point on the spar is equal to the sum of the shear load from the tip to that point on the spar. This is plotted as the bottom diagram in Figure 1. The moment builds slowly at first and then increases rapidly as we move inboard. It reaches a maximum of 61,250 inch-pounds at point 2. Inboard of this point the moment decreases, because the shear that we are now summing is negative and consequently subtracts from the moment. The moment continues to decrease until it is equal to zero at point 1. This is correct, since the spars of a strutbraced wing attach to the fuselage with a single horizontal bolt, which is unable to resist any moment. If we were instead analyzing a cantilever wing, there would not be the negative 3,600-pound load at point 2, so the shear on the spar would continue to increase until it reached a maximum of 3,000 pounds

Figure 2. Examples of tensile, shear, and bearing stress.

-1850 LBS -61250 INCH-LBS MOMENT 100 POUNDS

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100 POUNDS

at point 1. Similarly, the moment diagram would continue to rapidly increase until it reached a maximum of 180,000 inch-pounds, also at point 1. This is nearly three times higher than the maximum moment for the strut-braced version, and consequently would require a much beefier spar. The strut does cause a 4,500pound compression load on the spar that can’t be ignored. Moving the strut further outboard would cause the maximum bending moment to decrease. However, the compression load on the spar due to the strut would increase. After a certain point the outboard bending moment, compression load due to the strut, and air loads on the wing inboard of the strut can cause the moment inboard of the strut to start actually increasing. This is called secondary bending and will be discussed in a future article. There is another kind of moment that an airplane’s structure experiences. Instead of bending the wing or fuselage, it twists it around the part’s axis instead. This is called a torsional moment, or torsion for short. Most wings use airfoils that cause a torsion moment that tries to twist the leading edge down and the trailing edge up. Lowering the flaps or using full aileron deflection at high speeds further aggravates this twisting tendency. In most designs, the wing skins battle this torsion. The fuselage tail cone also experiences torsion when the rudder is deflected or is flown in a sideslip.

Internal Loads

The external loads that we have considered so far are not dependent on the airplane’s construction material nor the size and shape of its parts. As these loads are borne by the structure, though, they do become important. The internal loads in the various parts are called stresses and are the applied loads per unit area. When a part experiences an axial tension or compression load, the stress in the part is equal to the EAA Sport Aviation

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applied load divided by the cross-sectional area. The cross-sectional area is perpendicular to the direction of the applied load. An example of this is shown in Figure 2, where two 0.05by 1-inch straps are joined together with a one-eighth-inch rivet. If we applied a 100-pound tension load on each end, the tension (or tensile) stress in the straps would be equal to 100 pounds divided by the strap’s cross-sectional area of 0.05 square

inches, resulting in a tensile stress of 20,000 pounds per square inch. The maximum tensile stress in the strap is actually at the location with the smallest cross-sectional area, which is where the eighth-inch hole is. In this section the tensile stress is 22,857 pounds per square inch. If the strap assembly experienced a 100-pound compression load instead, the resulting compressive stresses would be calculated the same way. However, as we will see lat-

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er, there are other factors that will influence the maximum compressive load a part can take. In our example, though, the rivet shank does exert a highly localized compressive load on the strap. This load is applied to the area adjacent to the rivet and has the tendency to elongate the hole. This is called bearing stress and is equal to the applied load divided by the area represented by the rivet diameter times the strap thickness. Recall that a force acting perpendicular to a part is called a transverse shear load. This is evident on the rivet in Figure 2, where the shear load on each strap tries to slice the rivet in two. The average shear stress is equal to the shear load divided by the cross-sectional area of the rivet shank. The shear stress in the rivet is not constant, but varies from zero at the edge of the rivet to a maximum shear stress at the widest portion of the rivet. The straps also experience shear stress in the area between the rivet and the edge of the part as the rivet tries to tear out of the strap. This is why in sheet metal design the minimum distance from the rivet centerline to the edge of the part is at least twice the rivet diameter. Another type of shear stress that concerns the airplane designer is torsional shear, and it is the result of the torsional moments discussed earlier. The magnitude of this stress depends on a variety of factors, including the skin thickness and the area bounded by the skin. We will be looking in more detail at torsional stress in future articles.

Material Properties

Designers use a variety of materials to build airplanes, primarily wood, aluminum, steel, and composites. Often the material selection is determined before the detail design work begins and is based on the designer’s construction preferences or manufacturing capabilities. Some people love building with wood, so only wood will do. Others prefer building with metal or composites. The performance goals may also dictate the 36

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construction materials. For ultimate performance, composite construction is hard to beat. That does not mean that good performance cannot be obtained with other materials, though, as the RVs, Falco, Tailwind, and others have exhibited with their metal, wood, and tube and rag construction. Each material has different strength limits the designer needs to know. In the late 1930s, the U.S. government realized the importance of establishing minimum structural design values for the different materials used for airplane construction. The result of the Army-Navy-Commerce committee was the publication ANC-5, Strength of Aircraft Elements. This useful bulletin contained the material properties for wood, steel, and aluminum alloys and the strength capabilities for various sizes of tubing. Eventually wood material properties were covered separately in ANC-18, Design of Wood Aircraft Structures, which is a must-have book for those interested in designing with wood. In time, ANC-5 was replaced by Mil Handbook 5 (available for download at www.hdbk-5.battelle.org/ MIL5HCN1.pdf), which contains the material strength information for most of the metals used in airplanes today. The benefit of using the design information from this handbook is that if the material manufacturer has made the steel or aluminum to the correct specifications, there is a 95 percent probability that 99 percent of the material will be at least as strong as the “A basis” values shown in Mil Handbook 5. Each of these handbooks provides the ultimate tensile, compressive, shear, and bearing stresses for the various materials. These are the stresses at which the material will break. Aluminum and steel are ductile materials that will permanently deform (but not break) at a lower stress value. This lower stress is the yield stress for the material. You may recall from reference 2 that the airplane’s structure may not have permanent deformation (i.e.,

yielded material) with the limit loads applied. Similarly, the ultimate stress of the material may not be exceeded when the ultimate loads are applied. We know that a part is strong enough when we apply the ultimate loads and it doesn’t break. When doing stress analysis, this is shown by determining the margin of safety for each part, which is equal to: Margin of Safety=

Allowable Stress -1 Applied Stress

The allowable stress is the maximum a material can withstand before failing. The margin of safety must be a positive value or the structure will likely fail before the ultimate stress is achieved. Two other material parameters provided in the handbooks that are important to the designer are the modulus of elasticity and the shear modulus. All materials exhibit some measure of springiness, and will stretch when pulled on and compress when pushed. For example, if we pulled on a 1-inch long AN-4 bolt with a 1,000-pound load, it would stretch 0.0007 inches. The relationship between tensile stress and how much the material stretches due to that stress is called the modulus of elasticity. The larger the value, the less the material will stretch. Shear modulus is similar to the modulus of elasticity, except that it is a measure of how much a material will twist for a given amount of torsion. For steel and aluminum, its value is about 38 percent of the modulus of elasticity for each material.

Compressive Strength

An airplane’s structure would be very light if the loads in all the parts were in tension. Unfortunately, this is not the case, and compression complicates the design. For example, if we took an aluminum yardstick and loaded it in tension, it could support more than 2,800 pounds before failing. If instead we loaded it in compression, the yardstick would deflect laterally and collapse at a load less than 40 pounds.

The reason for the 70-fold difference has nothing to do with aluminum’s compressive strength. Long, slender pieces loaded in compression are referred to as columns, and the type of failure the yardstick would experience is called column buckling. Typical examples of columns in an airplane are wing struts, push/ pull control tubes, and the structural members of a truss fuselage. More than 200 years ago, Swiss mathematician Leonhard Euler developed an equation that could predict column-buckling loads. There were several factors Euler found that contributed to a column’s compressive strength. First was that the buckling load is inversely proportional to the column’s length squared, which means if we cut the yardstick in two and put it in compression again, it would now hold four times the load before buckling. Second, he found that the buck-

ling load was directly proportional to the modulus of elasticity of the column’s material. The modulus of elasticity of steel is about three times that of aluminum, so a steel yardstick with the same dimensions would be able to support three times the compressive load before buckling. How the ends of the column are constrained also is a factor. In my yardstick example, each end is fixed but free to rotate, which is referred to as a pin joint. If I had rigidly clamped each end of the yardstick, it would have supported a compressive load four times greater before buckling. Even though the welded joints on a steel tube fuselage would appear to be rigidly clamped, they can still deform under load, so it is recommended that they be considered pin joints when they are analyzed. The final factor is the moment of inertia of the column’s cross section. The moment of inertia depends both

on the cross-section’s area and how the area is distributed. The farther the area is from the center of gravity of the cross section, the larger the moment of inertia becomes. Mathematicians have determined the formulas for calculating the moment of inertia for a variety of common shapes, which can be found in most airplane stress analysis books. For rectangular shapes like my yardstick example, the moment of inertia is proportional to the cube of the minimum dimension. For tubes, it is proportional to the tube diameter raised to the fourth power. This indicates that it is more desirable to increase the tubing diameter instead of the wall thickness when possible. Euler’s equation does a good job estimating the buckling load for long columns, but as they get shorter it starts over-predicting the buckling strength. This is because the columns start experiencing local

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crippling or compressive failure of the material. An example of this is collapsing a beverage can by standing on it. Systematic testing of wood columns and metal tubes allowed researchers to derive empirical equations for design purposes. This month's spreadsheet incorporates these equations and can be used to estimate the strength for wood, steel, and aluminum columns.

Wing Spar Stresses

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The wing spar (or spars) is one of the most critical pieces of an airplane's structure. Its job is to transmit the transverse shear and bending moment due to the air loads to the wing's attach points on the fuselage. Wings with aluminum, plywood, or composite skins are called stressed skin designs and often just have a single main spar to handle the shear and bending moment due to the air loads. They usually also have a rear spar, but it primarily serves as an attach point for the ailerons and flaps, not to transmit bending moment to the fuselage. The wing skins are designed to resist all the wing's torsional loads. Fabric-covered wings do not have skins capable of resisting the torsional loads, so this requires that the rear spar also share in transmitting the shear and bending moment to the fuselage. Reference 2's spreadsheet (available from the EAA's website) will estimate the amount of shear and bending moment each spar will experience, so those designing such a wing may find it helpful. Some fabric-covered wings use

a leading edge skin that is firmly attached (either glued or riveted) to the forward spar. This leading edge skin is designed to be strong enough to resist the wing's torsional loads, allowing the wing aft of the main spar to be fabric covered to save weight. Like stressed skin wings, the forward spar is designed to handle all the shear and bending moment due to the air loads and the rear spar's role is mainly an aileron attach point. Built-up spars consist of a relatively thin web that is sized to resist the shear loads and spar caps that resist the bending moment. The resulting axial load in the spar caps is approximately equal to the bending moment at each position along the wingspan divided by the distance between the "center of gravity" of the areas of the upper and lower spar cap. The resulting tensile or compressive stress in each cap is equal to the axial load divided by the cap area at each station. This is an approximate means of determining the cap stresses, though, and we will provide a more accurate method in determining the spar stresses in future articles, when we will be looking at the particulars of designing using wood, metal, and composites. Armed with this knowledge, you should have a better idea of how to design and size the critical parts of an airplane—and a better understanding of the stresses imposed on your airplane in flight.

more at www.eaa.org

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References: ESSEntial to every mission' Your Aviation Service Partners DUATS Technical Support and Order line: 800.345.3828 e-mail: [email protected] or visit www.duats.com Computer Sciences Corporation Federal Sector 15000 Conference Center Dr., Chantilly, VA 20151-3819 FlightPrep information: www.flightprep.com

1. "Airplane Structures 101, Part I," Willford, Neal, EAA Sport Aviation, January 2005. 2. "Estimating Air Loads, " Willford, Neal, EAA Sport Aviation, June 2003. 3. "Down To Earth," Willford, Neal, EAA Sport Aviation, September 2004. 4. Design Standards for Advanced Ultra-Light Aeroplanes, DS 10141E, available at www.lamac.ca/DS10141%20page.htm. 5. Code of Federal Regulations Airworthiness Standards, Part 23, available at www.faa.gov/regulations.

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