Maximum bending moment in blade

[Eagle9]

Imagine that the electro motor is rotating the blade in air at sea level:
http://img822.imageshack.us/img822/4655/222on.jpg
During this rotation the air impedes the blade’s rotation and if the blade rotates too quickly or if it is made of fragile material rod will be simply broken or bent. Can anybody tell me how can we calculate if the rod (with some certain mass, length, volume and rotational speed) withstands air’s resistance or not? Material’s which property should be taken into consideration when calculating this? Specific strength or tensile strength? :shy: This task was partially solved https://www.physicsforums.com/showthread.php?t=502817" where I learned to calculate the Torque (T) but what should be done now? I need somehow to connect Torque’s value and material’s strength and bending moments; could you please tell me how to do it?

 

[Eagle9]

Does not anybody know the answer?

 

[DickL]

There are 2 forces acting on the blade (ignoring aero lift), the aero drag force, and centrifugal force.

The aero force will bend the blade in the same manner as a non-uniform distributed load will bend a cantilevered beam. The aero force will vary with the square of speed of the blade from the center (0) outward (V=omega * r).

The centrifugal force will vary with the radius. The CF will reduce the bending moment due to the aero drag.

You might look at examples of beam loading with both transverse and axial loading for some insights into how to set up the solution.

 

[Eagle9]

DickL

There are 2 forces acting on the blade (ignoring aero lift), the aero drag force, and centrifugal force.

The aero force will bend the blade in the same manner as a non-uniform distributed load will bend a cantilevered beam. The aero force will vary with the square of speed of the blade from the center (0) outward (V=omega * r).

The centrifugal force will vary with the radius. The CF will reduce the bending moment due to the aero drag.

So, these two kinds of forces will oppose to each other, right?

You might look at examples of beam loading with both transverse and axial loading for some insights into how to set up the solution.

And where can I find such examples?

 

[DickL]

The 2 loadings do not exactly oppose each other. The CF pulls on the blade, putting it into tension. The aero drag bends the blade, creating tension on the leading edge and compression on the trailing edge.

Some beams fail due to failure along the compression side, others due to tension failures. Material properties have a lot to do with which fails first. I suspect, but am not certain, that most hollow beams (tubes, extrusions, etc.) fail due to crippling which is a form of compression failure in the hollow shape. Having a tensile loading reduces the compressive stress and can thereby help the apparent beam strength.

Look in handbooks. I usually turn to my Baumeister & Marks Standard Handbook for Mechanical Engineers, but it is decades old. Another 1 I have is Roark & Young Formulas for Stress and Strain. Although there may not be a ton of such books, there are many to choose from. Or if you are taking, or have taken, a mechanics or structures course, then go back to first principles and put the analysis together.

Analytically, it is an interesting problem with both dynamics elements (CF & aero drag) and statics (combination of tensile loading and bending).

 

[Eagle9]

DickL

The 2 loadings do not exactly oppose each other. The CF pulls on the blade, putting it into tension. The aero drag bends the blade, creating tension on the leading edge and compression on the trailing edge.

I see

Some beams fail due to failure along the compression side, others due to tension failures. Material properties have a lot to do with which fails first. I suspect, but am not certain, that most hollow beams (tubes, extrusions, etc.) fail due to crippling which is a form of compression failure in the hollow shape. Having a tensile loading reduces the compressive stress and can thereby help the apparent beam strength.

In my case there should be monolithic (that is no hollowness) blade made of very good material-Carbon nanotubes, it will have more chances to withstand.

Or if you are taking, or have taken, a mechanics or structures course

Oh, no, I have got nothing to do with engineering or Physics this is just the task that I want to solve…….

Analytically, it is an interesting problem with both dynamics elements (CF & aero drag) and statics (combination of tensile loading and bending).

It seems to me that this problem should have already been solved-recall the helicopters. It has got rotating blades and when designing/manufacturing it (blade) the engineers need to take into consideration the strength of the blade and air’s resistance.

Look in handbooks. I usually turn to my Baumeister & Marks Standard Handbook for Mechanical Engineers

This one?
http://www.knovel.com/web/portal/basic_search/display?_EXT_KNOVEL_DISPLAY_bookid=346

Roark & Young Formulas for Stress and Strain

This one:
https://www.amazon.com/dp/0071210598/?tag=pfamazon01-20

But are there these formulas to see (not in this book but at some web-site)? Or online-calculators would be even better

 

[DickL]

Helicopter rotor loadings and their attachment to the hub are considerably more complex than just the aero drag and CF loadings. They are subjected to aero lift, which is constantly changing in magnitude and direction (at times even downward). Typically the aero lift changes as the blade goes through a revolution, due to control inputs that are cyclic and velocity changes (advancing or retreating relative to the helo's motion). In rotor systems with more than 2 blades, there are also 2 hinges that allow flapping motions (up/down) and lead/lag motions (forward/backward). With such a connection to the hub, the CF is important in keeping the blade from swing too far up or down and to transfer the lift force to the hub.

And then the aero forces for helo's are complicated by swirl effects, flow disturbances from the body and other nearby rotors (tail rotors, etc.), and flight maneuvers.

I don't know if there are any on-line rotor analyses. I haven't look for them, but do try a search. Terms such as helicopter rotor analysis, rotor blade forces, rotor dynamics may help the search.

I think you will want to go to a balanced rotor system, not the single blade shown above. The unbalance loading of a single blade could well be dangerous. If your rotor system is going to be of any significant size and speed (1 ft or 300 mm diameter, more than 50 rpm) you may want to have some help from an engineer that understands rotating systems.

 

[Eagle9]

DickL

you may want to have some help from an engineer that understands rotating systems

Is it possible to find them in web-forums? :shy:

Helicopter rotor loadings and their attachment to the hub are considerably more complex than just the aero drag and CF loadings. They are subjected to aero lift, which is constantly changing in magnitude and direction (at times even downward). Typically the aero lift changes as the blade goes through a revolution, due to control inputs that are cyclic and velocity changes (advancing or retreating relative to the helo's motion). In rotor systems with more than 2 blades, there are also 2 hinges that allow flapping motions (up/down) and lead/lag motions (forward/backward). With such a connection to the hub, the CF is important in keeping the blade from swing too far up or down and to transfer the lift force to the hub.

And then the aero forces for helo's are complicated by swirl effects, flow disturbances from the body and other nearby rotors (tail rotors, etc.), and flight maneuvers.

I don't know if there are any on-line rotor analyses. I haven't look for them, but do try a search. Terms such as helicopter rotor analysis, rotor blade forces, rotor dynamics may help the search.

Well, I mentioned the helicopter’s blade just for the reason to say that such/similar task should have already been solved, that’s it but in my case there is one blade (that is at electro motor’s one side) and it does not move/flight but it is fastened to the Earth….

 

[saqibhameed]

I have got a question related to this discussion please, i am working to calculate maximum bending stress of straight (symmetric) vertical axis wind turbine blade. i have evaluated the normal (perpendicular to chord) and axial (parallel to chord) forces on a single blade when revolving in real life conditions. my question is what other forces do i need to add into these forces which are acting on this blade? centrifugal force?

 

[W R-P]

Find the bending moments in each plane (due to the parallel, then perpendicular forces). Then find the maximum resultant using Pythagoras' theorem. This can be taken as your maximum bending moment.

 

[saqibhameed]

thank you for helping.. but what about Maximum Bending Stress then? because Max. Bend. Stress = Mmax.ymax/I ... what would be the value of I and Ymax ??

 

[W R-P]

You say it's a VAWT right?so you can model the blade as a cantilever with the wind loading at the blade tip.Then use the bending equations
http://www.roymech.co.uk/Useful_Tables/Form/Stress_Strain.html

ANYONE PLEASE CORRECT ME IF I'VE GONE WRONG.

 

[Eagle9]

W R-P

Find the bending moments in each plane (due to the parallel, then perpendicular forces). Then find the maximum resultant using Pythagoras' theorem. This can be taken as your maximum bending moment.

Could you please tell me how this can be done in my case? I have got two forces: There are 2 forces acting on the blade, the aero drag force, and centrifugal force. Actually, in the post 3 it was written that “The aero force will vary with the square of speed of the blade from the center (0) outward (V=omega * r)”, so if radius (that is, the length of the rod) is 1 000 meters and omega is equal to 2 then aero force=2 000, is it correct? :shy:

You say it's a VAWT right?so you can model the blade as a cantilever with the wind loading at the blade tip.Then use the bending equations
http://www.roymech.co.uk/Useful_Tabl...ss_Strain.html

Which one? :shy:

 

[W R-P]

I'm sorry to say that calculation is wrong.
Some airfoils(blade profile) will give you these forces as a function of the radius eg. L=0.902(r^2).But since this isn't given find the lift and drag forces like this:

Lift is given as L= Cl(0.5x air density x WIND SPEED^2), where Cl is the lift coefficient for the blade profile(at the blade setting angle you used).

Drag is similar, D= Cd(0.5x air density x WIND SPEED^2), where Cd is the drag coefficient for the blade profile(at the blade setting angle you used.

When you find these forces, they will be highest at the tip.
So in each plane you will have eg. M1= (max Lift)x blade length, and M2= (max drag)x blade length.
The maximum resultant = sqrt[(M1^2)+(M2^2)]

Any other opinions are welcome

 

[nvn]

The maximum resultant = sqrt[(M1^2)+(M2^2)].

Unfortunately, that will not work. As alluded to earlier by saqibhameed, there is not necessarily a readily-available area moment of inertia (I) value corresponding to an arbitrary resultant moment direction. I do not know what the cross section looks like. If the cross section has at least one axis of bilateral symmetry, then you can compute the bending stress using the orthogonal moments of inertia, using superposition, by transforming your applied moments to the orthogonal axes. If the cross section does not have at least one axis of bilateral symmetry, then if I recall correctly, you must compute the principal moments of inertia, and transform your applied moments to the principal axes (and then use superposition).