|
Analysis of Loads on the Pietenpol Air Camper Wing Attach Fittings
Since I have modified the wing attach fittings on my Pietenpol Air Camper I became concerned about the forces acting on those fittings. Specifically, I wanted to know what forces were acting on the fittings at the cabane struts since I have modified these fittings for the folding wing, and at the lift strut fittings since I am using a “V” strut configuration instead of the parallel configuration on the plans. To do this you first need to determine the lift distribution spanwise along the wing. Many aerodynamics texts will tell you that the lift distribution along a rectangular planform wing like that of the Air Camper takes the form of an ellipse. This is because of the loss of lift at the wing tips. What I could not find however was what proportions to use for the ellipse. An ellipse can vary in shape from a near circle to a near line. I reasoned that the proportions of the ellipse should follow the aspect ratio of the wing. This logic agrees with what we know, that high aspect ratio wings, like in sailplanes, are more efficient than lower aspect ratio wings, like in the Air Camper. So, we can now determine the lift distribution of the Air Camper as a 5 by 29 ellipse. If my calculus skills were better I could probably use an equation to determine the actual lift values along the wing, but it’s been too many years, so I had to resort to the spreadsheet at the end of this paper. The formula for a 5 by 29 ellipse is shown below.
With the lift distribution now known, the next thing to figure out is how do these forces affect the fittings? There are probably other ways to get the answer, but for me the easiest is to look at the effects of torque. Let’s look at some examples of how this works. In the first example we have two points, A and B, 10 feet apart, with a weightless (theoretical) board lying across them. We put a 50 lb weight directly over point B.
Now it may seem a bit obvious that point B has 50 lbs and point A has 0, but let’s see how we can show it using torque. Using point A as a pivot point we get a 50 lb weight on a 10 foot arm giving a torque of 500 ft lbs (10 ft x 50 lbs). Working backwards from what we just did we take 500 ft lbs divided by 10 feet to show that there is a force of 50 lbs at point B. Not too useful in this example, but it shows the concept. Conversely, if we take point B as the pivot point we see that there is 0 lbs weight at point A, 10 feet away. Therefore the torque is 0 and there is no force at point A. Let’s look at another configuration. We will use the same setup as in the previous example, but this time we will reposition the 50 lb weight to the center of the board, 5 feet from each end point.
This again may seem obvious that each point, A and B, will feel a 25 lb force, but let’s use the torque calculations once again. Using either point as the pivot point you get 50 lbs at a distance of 5 feet for 250 ft lbs of torque. 250 ft lbs pushing on the opposite point, 10 feet away gives a force of 25 lbs. Very simple so far, but now let’s add another weight and cantilever the board like on a wing spar.
Looking at the torque about point A you get 100 lbs at 0 feet away for 0 torque, plus 50 lbs 20 feet away for 1000 ft lbs and a total of 1000 ft lbs. 1000 ft lbs pushing down on point B, 10 feet away, puts 100 lbs force at point B. Using point B as the pivot point we get 100 lbs at 10 feet for 1000 ft lbs, minus 50 lbs at 10 feet in the other direction for 500 ft lbs resulting in a torque of 500 ft lbs. This torque of 500 ft lbs pushing down on point A 10 feet away gives a force of 50 lbs at point A. Cool!
Now let’s apply this same technique to the Air Camper wing. Using the cabane fitting as the pivot point we can calculate the total torque as the sum of each individual torque value along the wing. The spreadsheet at the end of this paper shows the lift at each inch along the wing as a percentage of the total lift. It also calculates the torque at each inch by multiplying the lift times the distance from the pivot. Summing these numbers all the way to the wing tip gives 51.74 inch lbs. Applying this torque to the lift strut fitting at a distance of 72 inches gives a force of 0.719 lbs. Using the lift strut as the pivot point we calculate the individual torque values in both directions from this point, (as we did fir point B above) and then sum them to get the total torque of 7.69 inch lbs. Applying this to the cabane strut fitting 72 inches away we get 0.107 lbs force at the cabane. To get the total force on the cabane we must add the sum of lift on the center section which is 0.174 giving a total cabane force of 0.281 lbs. But we’re not quite done. The force of torque is perpendicular to a radial line from the pivot point, or in this case perpendicular to the spar. But, the lift strut is not perpendicular to the spar. It angles back to the fuselage and this angle causes more force on the lift strut. To get the force on the lift strut we must determine the distance of a line that is perpendicular to the lift strut and goes through the cabane fitting. In the case of my Air Camper modification that distance is 36 inches, or half of the distance from the pivot point to the wing strut fitting. Therefore the force on the wing strut fitting is doubled from 0.719 to 1.437 lbs. We almost have it now. We just need to determine how much the lift distribution varies from the front to the rear spar cordwise. On the Air Camper wing both spars are located fairly well far forward. The center if lift on an airfoil is also located somewhat forward. Therefore, to simplify the calculations I will assume that the spars share the equally share the lift loads. So, we divide the force on the cabane fittings in two, with half for the front cabane fitting and half for the rear. We do the same for the lift strut fittings that attach the lift struts to the wing. But, because my modification uses a “V” strut we must use the full lift strut force for the lower end of the struts. The table below shows the forces for the three fittings as percentages of total lift, and for an 1100 lb aircraft at 1, 2, and 3 Gs. I also calculated the forces on the fittings of a plans built Air Camper, and for my Kolb MKII.
Loads on Fittings for 1100 lb Aircraft, Folding Wing
Loads on Fittings for 1100 lb Aircraft, Original Wing
Loads on Fittings for 800 lb Kolb MKII
So what have I learned? Well, I see that my four foot center section puts more load on the cabane struts and less on the lift struts than the plans built two foot center section would. No surprise. I see that the highest stress point is where the lift struts attach to the fuselage. I had already known this, and I had doubled the fuselage cross strap at the attach point from 1 x 1/16 to 1 x 1/8. The pivoting fitting that I am using at the fuselage attach point is rated for 3200 lbs, or 4 Gs. Perhaps the thing that surprised me the most was the amount of force on the light weight fittings on my Kolb. If the Kolb fittings can handle the loads, then I am very confident I the fittings on the Air Camper.
Air Camper Wing Lift Distribution and Torque Values
|