### Prop Loads on a Corvair

Posted:

**Mon Sep 20, 2004 3:05 pm**The following is from the excellent web site of John Brannen, unfortunately

the images attached would not copy so please visit the web site, there is more to see: http://www.freewebs.com/brannen/index.html

On the Corvaircraft e-mail list that I belong to there was some concern about running 72Ã¢â‚¬? props on a direct drive Corvair engine due to the asymmetric prop loads and the gyroscopic loads. I decided to perform an analysis of the loads that are experienced. The individual loads in question are the asymmetric and gyroscopic loads mentioned above and also the torsion load due to engine torque. Each load load is calculated individually and then the combined loading is calculated. The stresses due to the combined loading are then compared to the material specifications for the crank to determine the fatigue life of the crank under these loading conditions. The link below is to a Microsoft Excel Spreadsheet that I set up for this problem. The text below gives a description as to the logic behind the spreadsheet and calculations.

ASYMMETRIC LOADS

I spent some time looking through all of my textbooks from college, the aircraft design and engineering books that I have accumulated since college and also searching the internet for information on calculating asymmetric propeller thrust or P-factor. I found nothing. I then decided to see if I could logically derive an approximation for it. Here is my approach.

First, imagine that the prop is actually a disk that produces a certain total thrust. If the total thrust that the prop produces at a given flight condition is known, the average thrust produced by a pie shaped segment of the prop disk can be calculated. Now that we know the average thrust each segment produces, we can calculate the moment that the segment creates about the vertical and horizontal axes of the prop disk if we know where the "thrust center" of that segment is. For conservatism and a first look, I decided to assume that the thrust of the segment is centered on the wedge at the prop radius. The moment arm of each segment can be calculated easily for each segment using simple trigonometry. The figure below shows what I am talking about.

Now that I had the geometry part finished, I had to decide how to quantify how much thrust each segment produces when the airplane is in a high angle-of-attack. First, the whole premise of P-factor is that when the airplane is in high angle-of-attack flight, the descending blade (3 o'clock) is operating at a higher induced angle of attack than the ascending blade thus creating more lift (thrust) on the descending side as opposed to the ascending side. How much is the question. Assume that the angle of attack of the airplane is 10 degrees, then the difference in the angle of attack of the descending blade when compared to the ascending blade is 10*2=20 degrees. One thing you have to understand that in subsonic flight, the flow-field ahead of the prop is affected before the prop actually gets to that point. In other words, the airflow is already turning to try to go through the prop disk at a perpendicular angle before it reaches the prop disk. This will have the effect of reducing the angle of attack on the descending blade and increasing the angle of attack on the ascending blade thereby reducing the difference between the two. I decided to examine the lift coefficient of the Clark Y airfoil at itÃ¢â‚¬â„¢s maximum and at the AOA 20 degrees less than that. By doing this, the represented AOA of the airplane would be something greater than 10 degrees. The max lift coefficient occurs about 16 degrees AOA and is 1.5. The lift coefficient at Ã¢â‚¬â€œ4 degrees AOA is about 0.2. So, the ascending blade operates at a lift coefficient that is about 13.33 percent of the descending blade if we neglect upfield flow effects. This only occurs at the 3 and 9 oÃ¢â‚¬â„¢clock positions since the angle of attack at the 12 and 6 oÃ¢â‚¬â„¢clock positions is equal. If you were to follow one blade around in itÃ¢â‚¬â„¢s path, you will see that the angle of attack varies at every position and can be represented by a sinusoidal variation with the highest value occurring at 3 oÃ¢â‚¬â„¢clock (descending) position and the lowest value occurring at the 9 oÃ¢â‚¬â„¢clock position (ascending). The 12 and 6 oÃ¢â‚¬â„¢clock positions will be equal and at the median value between the 3 and 9 oÃ¢â‚¬â„¢clock positions. Getting the median value depends on how much total thrust the prop produces and this can be calculated using the equation Thrust = n*P/V where n is the prop efficiency, P is the power applied to the prop, and V is the forward velocity. As can be seen from this equation, the thrust of the prop decreases as the airspeed, or forward velocity, of the airplane increases. To maximize this, I chose 40 mph (58.6 ft/sec) as the airspeed for my calculations. I assumed that the engine could make 100 hp (55000 ft-lb/sec). The prop efficiency in cruise is about 0.85 for a fixed pitch prop, but at minimum speed, the prop efficiency drops significantly. I prowled though some aero textbooks I looked at data for several different airplanes with fixed pitch props flying at minimum speed (just above stall), the average prop efficiency in this region was about 0.4. So, the thrust produced by a 100 hp airplane flying at 40 mph with 0.4 prop efficiency is about 375 pounds. Putting all of this together, I got the following graph of the thrust produced by 10 degree prop segments. If you add up the thrust of all of the individual segments you get 375 which is the thrust of the whole prop. Note here that 0 degree's relates to the 3 o'clock position and 180 degrees relates to the 9 o'clock position.

Now that I have the geometry and the thrust that each segment produces, I can calculate the moment that each segment creates on the crank and then sum these moments to get the total moment. The moment is resolved into 2 components, the moment about the vertical axis of the prop disk and second the moment about the horizontal axis of the prop disk. By inspection, the moment about the horizontal axis should be zero and was verified by calculating it in a spreadsheet. The moment about the vertical axis came out to be 5162 in-lbs.

GYROSCOPIC LOADS

To calculate the bending load imposed on the crank due to gyroscopic effects the following equation is used.:

GyroMoment=I*w*b

Where: I=moment of inertia of the prop, w=the rotation speed of the prop, and b=turn/pitch rate in radians/sec.

I initially used a conservative approach and calculated the moment of inertia assuming that the mass distribution from prop centerline to tip was constant. This was conservative because the actual mass distribution is biased toward the centerline. I then decided that I could do this better. I decided that the moment of inertia of the prop can be calculated relatively closely by assuming that the mass distribution is greatest at the centerline and tapers linearly to the tip. Imagine a constant width bar whose thickness tapers from some thickness at one end to a smaller thickness at the other end. This is what I used to mode the mass distribution of the prop. I used a tip thickness that is 1/10th of the thickness at the root. This is conservative since the cross sections of this bar are not shaped for the airfoil and are not tapered in plan form. I took this mass distribution and calculated the moment of inertia contribution of discretized sections and then summed them up. In other words I mathematically chopped up the bar into 20 equal length pieces, calculated the mass of each piece, and knowing what distance each piece was from center, I calculated the moment of inertia contribution of each and then added this up. I came up with a moment of inertia of one blade of 0.55. I also checked the moment of inertia calculation by saying that the thickness at the root and tip were the same, and as expected, the result matched the moment of inertia equation of a slender rod at 0.931. Using I=1.1 (remember there are 2 blades), w=314, and b=1, the bending moment from gyro loads is 362 ft-lbs, or 4350 in-lbs.

COMBINED STRESS CALCULATION

The bending moment loads were added to get a total bending moment of 4350+5162= 9512 in-lbs. The engine also introduces a torque load into the crank that must be considered. I used a torque of 170 ft-lbs = 2040 in-lbs..

To see if the crankshaft can take the loads, we should calculate the peak stresses in the crank. The stress from the bending will be highest at the outside diameter of the crank. To calculate this stress for the circular crank cross section, I used the equation sigma = M*c/I, where sigma is the tensile stress, M is the applied moment, c is the distance from the centroid of the crank, and I is the area moment of inertia of the crank. For a hollow shaft, the area moment of inertia is I=pi/4*(R1^4-R0^4) where R1=outside radius of crank and R0=inside radius of crank.

I=3.1416/4*(0.75^4-0.5^4) = 0.199 so, the maximum outer fiber tensile stress due to bending is

sigma=9512*0.75/0.199 = 35773 psi

We also want to see what that torque means to the stress, so calculate the shear stress using tau = T*r/Ip, where tau is the shear stress, T is the applied torque, r is the radius of the crank, and Ip is the polar moment of inertia of the crank. For a circular section, Ip=2I. So:

tau=(170*12)*0.75/(2*.199)=3836 psi

To calculate the combined stresses due to both of these, the combined principal stress is

Principalstress=0.5*(sigma)Ã‚Â±[(sigma/2)^2+tau^2]^0.5

Which gives 0.5*35773Ã‚Â±[(35773/2)^2+3836^2]^0.5 = 36179 and -407 psi for the principal stresses. The maximum shear stress is tau(max) = [(sigma/2)^2+tau^2]^0.5. which gives a max shear of 18293. None of these numbers even come close to the tensile ultimate, or tensile yield values for the crank material. The material properties of interest for 1045 steel are: ultimate tensile = 89900 psi, Tensile yield = 70300 psi, and fatigue limit = 45000 psi. I have been informed that the early cranks were 1045 steel where the later model cranks were 5140 steel. The material properties for 5140 are: ultimate tensile = 113000 psi, tensile yield = 66000 psi. The data I have does not list fatigue limit for 5140, but a general rule of thumb for steels is that the fatigue limit is about 50% of the ultimate tensile.

FATIGUE LIFE

The last consideration is to see what the fatigue life of the crank is. Fatigue life depends on the material, number of cycles and the cyclic stress. If we use the above maximum principal stress as the cyclic stress, and look at the S-N curve for 1045 steel, we see that the stress level is below the "knee" of the S-N curve otherwise known as the fatigue limit. This is the stress level at which the fatigue life is considered infinite. Therefore, the fatigue life of the crank for these loading conditions should be infinite.

I personally believe that the assumptions IÃ¢â‚¬â„¢ve made in these calculations are very conservative and, I would have no reservations running a lightweight 72Ã¢â‚¬? prop on a Corvair motor. Am I planning on doing it? No. Does that mean that my calculations are worthless? That is for you to decide for yourself. I also said before that each person needs to weigh each choice for their airplane themselves. If you are uncomfortable with any of these calculations, you should go with what has been tested in flight.

the images attached would not copy so please visit the web site, there is more to see: http://www.freewebs.com/brannen/index.html

On the Corvaircraft e-mail list that I belong to there was some concern about running 72Ã¢â‚¬? props on a direct drive Corvair engine due to the asymmetric prop loads and the gyroscopic loads. I decided to perform an analysis of the loads that are experienced. The individual loads in question are the asymmetric and gyroscopic loads mentioned above and also the torsion load due to engine torque. Each load load is calculated individually and then the combined loading is calculated. The stresses due to the combined loading are then compared to the material specifications for the crank to determine the fatigue life of the crank under these loading conditions. The link below is to a Microsoft Excel Spreadsheet that I set up for this problem. The text below gives a description as to the logic behind the spreadsheet and calculations.

ASYMMETRIC LOADS

I spent some time looking through all of my textbooks from college, the aircraft design and engineering books that I have accumulated since college and also searching the internet for information on calculating asymmetric propeller thrust or P-factor. I found nothing. I then decided to see if I could logically derive an approximation for it. Here is my approach.

First, imagine that the prop is actually a disk that produces a certain total thrust. If the total thrust that the prop produces at a given flight condition is known, the average thrust produced by a pie shaped segment of the prop disk can be calculated. Now that we know the average thrust each segment produces, we can calculate the moment that the segment creates about the vertical and horizontal axes of the prop disk if we know where the "thrust center" of that segment is. For conservatism and a first look, I decided to assume that the thrust of the segment is centered on the wedge at the prop radius. The moment arm of each segment can be calculated easily for each segment using simple trigonometry. The figure below shows what I am talking about.

Now that I had the geometry part finished, I had to decide how to quantify how much thrust each segment produces when the airplane is in a high angle-of-attack. First, the whole premise of P-factor is that when the airplane is in high angle-of-attack flight, the descending blade (3 o'clock) is operating at a higher induced angle of attack than the ascending blade thus creating more lift (thrust) on the descending side as opposed to the ascending side. How much is the question. Assume that the angle of attack of the airplane is 10 degrees, then the difference in the angle of attack of the descending blade when compared to the ascending blade is 10*2=20 degrees. One thing you have to understand that in subsonic flight, the flow-field ahead of the prop is affected before the prop actually gets to that point. In other words, the airflow is already turning to try to go through the prop disk at a perpendicular angle before it reaches the prop disk. This will have the effect of reducing the angle of attack on the descending blade and increasing the angle of attack on the ascending blade thereby reducing the difference between the two. I decided to examine the lift coefficient of the Clark Y airfoil at itÃ¢â‚¬â„¢s maximum and at the AOA 20 degrees less than that. By doing this, the represented AOA of the airplane would be something greater than 10 degrees. The max lift coefficient occurs about 16 degrees AOA and is 1.5. The lift coefficient at Ã¢â‚¬â€œ4 degrees AOA is about 0.2. So, the ascending blade operates at a lift coefficient that is about 13.33 percent of the descending blade if we neglect upfield flow effects. This only occurs at the 3 and 9 oÃ¢â‚¬â„¢clock positions since the angle of attack at the 12 and 6 oÃ¢â‚¬â„¢clock positions is equal. If you were to follow one blade around in itÃ¢â‚¬â„¢s path, you will see that the angle of attack varies at every position and can be represented by a sinusoidal variation with the highest value occurring at 3 oÃ¢â‚¬â„¢clock (descending) position and the lowest value occurring at the 9 oÃ¢â‚¬â„¢clock position (ascending). The 12 and 6 oÃ¢â‚¬â„¢clock positions will be equal and at the median value between the 3 and 9 oÃ¢â‚¬â„¢clock positions. Getting the median value depends on how much total thrust the prop produces and this can be calculated using the equation Thrust = n*P/V where n is the prop efficiency, P is the power applied to the prop, and V is the forward velocity. As can be seen from this equation, the thrust of the prop decreases as the airspeed, or forward velocity, of the airplane increases. To maximize this, I chose 40 mph (58.6 ft/sec) as the airspeed for my calculations. I assumed that the engine could make 100 hp (55000 ft-lb/sec). The prop efficiency in cruise is about 0.85 for a fixed pitch prop, but at minimum speed, the prop efficiency drops significantly. I prowled though some aero textbooks I looked at data for several different airplanes with fixed pitch props flying at minimum speed (just above stall), the average prop efficiency in this region was about 0.4. So, the thrust produced by a 100 hp airplane flying at 40 mph with 0.4 prop efficiency is about 375 pounds. Putting all of this together, I got the following graph of the thrust produced by 10 degree prop segments. If you add up the thrust of all of the individual segments you get 375 which is the thrust of the whole prop. Note here that 0 degree's relates to the 3 o'clock position and 180 degrees relates to the 9 o'clock position.

Now that I have the geometry and the thrust that each segment produces, I can calculate the moment that each segment creates on the crank and then sum these moments to get the total moment. The moment is resolved into 2 components, the moment about the vertical axis of the prop disk and second the moment about the horizontal axis of the prop disk. By inspection, the moment about the horizontal axis should be zero and was verified by calculating it in a spreadsheet. The moment about the vertical axis came out to be 5162 in-lbs.

GYROSCOPIC LOADS

To calculate the bending load imposed on the crank due to gyroscopic effects the following equation is used.:

GyroMoment=I*w*b

Where: I=moment of inertia of the prop, w=the rotation speed of the prop, and b=turn/pitch rate in radians/sec.

I initially used a conservative approach and calculated the moment of inertia assuming that the mass distribution from prop centerline to tip was constant. This was conservative because the actual mass distribution is biased toward the centerline. I then decided that I could do this better. I decided that the moment of inertia of the prop can be calculated relatively closely by assuming that the mass distribution is greatest at the centerline and tapers linearly to the tip. Imagine a constant width bar whose thickness tapers from some thickness at one end to a smaller thickness at the other end. This is what I used to mode the mass distribution of the prop. I used a tip thickness that is 1/10th of the thickness at the root. This is conservative since the cross sections of this bar are not shaped for the airfoil and are not tapered in plan form. I took this mass distribution and calculated the moment of inertia contribution of discretized sections and then summed them up. In other words I mathematically chopped up the bar into 20 equal length pieces, calculated the mass of each piece, and knowing what distance each piece was from center, I calculated the moment of inertia contribution of each and then added this up. I came up with a moment of inertia of one blade of 0.55. I also checked the moment of inertia calculation by saying that the thickness at the root and tip were the same, and as expected, the result matched the moment of inertia equation of a slender rod at 0.931. Using I=1.1 (remember there are 2 blades), w=314, and b=1, the bending moment from gyro loads is 362 ft-lbs, or 4350 in-lbs.

COMBINED STRESS CALCULATION

The bending moment loads were added to get a total bending moment of 4350+5162= 9512 in-lbs. The engine also introduces a torque load into the crank that must be considered. I used a torque of 170 ft-lbs = 2040 in-lbs..

To see if the crankshaft can take the loads, we should calculate the peak stresses in the crank. The stress from the bending will be highest at the outside diameter of the crank. To calculate this stress for the circular crank cross section, I used the equation sigma = M*c/I, where sigma is the tensile stress, M is the applied moment, c is the distance from the centroid of the crank, and I is the area moment of inertia of the crank. For a hollow shaft, the area moment of inertia is I=pi/4*(R1^4-R0^4) where R1=outside radius of crank and R0=inside radius of crank.

I=3.1416/4*(0.75^4-0.5^4) = 0.199 so, the maximum outer fiber tensile stress due to bending is

sigma=9512*0.75/0.199 = 35773 psi

We also want to see what that torque means to the stress, so calculate the shear stress using tau = T*r/Ip, where tau is the shear stress, T is the applied torque, r is the radius of the crank, and Ip is the polar moment of inertia of the crank. For a circular section, Ip=2I. So:

tau=(170*12)*0.75/(2*.199)=3836 psi

To calculate the combined stresses due to both of these, the combined principal stress is

Principalstress=0.5*(sigma)Ã‚Â±[(sigma/2)^2+tau^2]^0.5

Which gives 0.5*35773Ã‚Â±[(35773/2)^2+3836^2]^0.5 = 36179 and -407 psi for the principal stresses. The maximum shear stress is tau(max) = [(sigma/2)^2+tau^2]^0.5. which gives a max shear of 18293. None of these numbers even come close to the tensile ultimate, or tensile yield values for the crank material. The material properties of interest for 1045 steel are: ultimate tensile = 89900 psi, Tensile yield = 70300 psi, and fatigue limit = 45000 psi. I have been informed that the early cranks were 1045 steel where the later model cranks were 5140 steel. The material properties for 5140 are: ultimate tensile = 113000 psi, tensile yield = 66000 psi. The data I have does not list fatigue limit for 5140, but a general rule of thumb for steels is that the fatigue limit is about 50% of the ultimate tensile.

FATIGUE LIFE

The last consideration is to see what the fatigue life of the crank is. Fatigue life depends on the material, number of cycles and the cyclic stress. If we use the above maximum principal stress as the cyclic stress, and look at the S-N curve for 1045 steel, we see that the stress level is below the "knee" of the S-N curve otherwise known as the fatigue limit. This is the stress level at which the fatigue life is considered infinite. Therefore, the fatigue life of the crank for these loading conditions should be infinite.

I personally believe that the assumptions IÃ¢â‚¬â„¢ve made in these calculations are very conservative and, I would have no reservations running a lightweight 72Ã¢â‚¬? prop on a Corvair motor. Am I planning on doing it? No. Does that mean that my calculations are worthless? That is for you to decide for yourself. I also said before that each person needs to weigh each choice for their airplane themselves. If you are uncomfortable with any of these calculations, you should go with what has been tested in flight.