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I have been trying to evaluate the stress on aluminium propeller:

Diameter: 78", 2600RPM

Have fed load numbers from here, applied distributed mass model onto the blades.

Static Thrust (2600 rpm):
MT 2-blade, 83" composite, 920 pounds
MT 3-blade, 80" composite, 900 pounds
McCauley 2-blade, 82" aluminium, 860 pounds
McCauley 3-blade, 78" aluminium, 825 pounds

The results are rather scary - 30 MPa out of 55 max, for 6061 class Aluminium (the visual deformation is increased 20x for better representation), resulting in +-1cm vertical bending.

Is this normal?

Is there any info on industry's practice for safety margin (numbers show its less that 50%)? Or any kind of test procedures applicable to manufacturers?

enter image description here

The shear stresses are negligible - 150 Pa out of 150 MPa enter image description here

The two propeller case is really scary looking Below is a distributed mass (375Kg) - 80% of Max. yield stress enter image description here

And centrifugal force - over-stressed by 200%: enter image description here

Finally, a combined non-Linear analysis (incl. torque) - over-stressed by 250%: (notice that everything not in blue will bend and likely crack, with only the blade tips surviving the torture)

enter image description here

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    $\begingroup$ Apart from the thrust load, did you take into account the centrifugal force...? $\endgroup$
    – xxavier
    Apr 20, 2017 at 10:35
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    $\begingroup$ Are you sure you are using the correct blade profile in your calculations? GIGO $\endgroup$
    – Romeo_4808N
    Apr 20, 2017 at 15:25
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    $\begingroup$ The static case may not even be limiting. Vibration and fatigue are probably more critical and are covered by 14 CFR §25.907. There are also acceleration loads. $\endgroup$
    – fooot
    Apr 20, 2017 at 16:10
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    $\begingroup$ I know this sounds silly, but perhaps some of the numbers in the calculation are off by a magnitude or the unit is wrong? $\endgroup$
    – kevin
    Apr 20, 2017 at 16:42
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    $\begingroup$ If you are looking for some useful feedback, it would be handy for you to post your full set of applied boundary conditions and mesh (inc screenshots). There are indications here to me that you have incorrectly restrained your model; and that you have possibly copied your geometry from a wind turbine instead of an aeroplane propellor. The two function in different ways and cannot be transposed from one function to the other. Also, your scales need some work. Standardise the scales to a fixed range instead of whatever is the max experienced. $\endgroup$
    – KuangGradeMark11
    Apr 20, 2017 at 19:10

2 Answers 2

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The plots of your propellers seem to use a regular airfoil all the way to the root. Real propellers become almost cylindrical there, however. Also, aluminium propellers are hollow, so the outer part is only a thin shell, saving mass and reducing centrifugal loads significantly.

After all, real aluminium propellers work, so there must be a way to keep the stress in operation below the yield stress. FAR 23.907(c) even demands that a non-wooden propeller must be life-safe:

The applicant must perform an evaluation of the propeller to show that failure due to fatigue will be avoided throughout the operational life of the propeller using the fatigue and structural data obtained in accordance with part 35 of this chapter and the vibration data obtained from compliance with paragraph (a) of this section.

Below is a drawing of a wooden propeller, but the outer shape of a metal propeller is not much different.

Drawing of a wooden propeller for the Fokker D VII

Drawing of a wooden propeller for the Fokker D VII (picture source)

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  • $\begingroup$ Section 35 (AIRWORTHINESS STANDARDS: PROPELLERS) is exactly what I have been looking for. I will try to run the similar tests on some WW2 propellers. $\endgroup$
    – FlegmatoidZoid
    Apr 21, 2017 at 6:09
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    $\begingroup$ @FlegmatoidZoid: Cool, then please let us know what results you get. $\endgroup$
    – Peter Kämpf
    Apr 21, 2017 at 6:56
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For the design of propeller consider 1.5 times of rpm. This assumption gives double the centrifugal force, thrust and drag acting in the propeller.

Consider the endurance limit based on following data. This helps to avoid fatigue problems

enter image description here

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    $\begingroup$ If you could indicate the source of your data, that would increase the credibility of your answer significantly! $\endgroup$
    – bogl
    Oct 30, 2018 at 9:02
  • $\begingroup$ the exact value is 1.41. Apply the value in equation to understand the changes. $\endgroup$
    – Shinu_John
    Oct 30, 2018 at 10:13

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