Free beam maximum moment problem 6

[Sean99]

Hello All, I'm trying to figure out the maximum moment for a free beam that has a few forces on it. It's meant to model a tailsitter aircraft that has two motors producing thrust on the ends of its main spar. The idea is to understand the maximum moment on the main spar.

Please review the picture provided.

The 10 N forces is mean to symbolize the thrust produce by each motor, and the 2 N is mean to symbolize payload weight, The length of the beam is 0.5 meters. How can I go about calculating the maximum moment on this spar?

Thanks for everyone's help!!

 

[SWComposites]

Did you not have a Statics class in undergrad?
First problem is you do not have a balanced free body.

 

[desertfox]

Hi Sean99

Will apart from the fact your beam is currently floating because it’s not in equilibrium, there are no dimensions on the beam.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[desertfox]

Hi Sean99

You can’t solve the maximum moment because it’s not in equilibrium, so if the plane is on the ground then there are other forces acting to react the out of balance forces.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[hydtools]

Looks like a dynamics problem. Unbalanced forces result in acceleration, f= m×a. Beam weight noted as 5N.

Ted

 

[Sean99]

desertfox, I guess You're right, this isn't so simple. I was thinking too much about a similar problem- solving for the maximum moment on an aircraft wing, which can be simplified into a cantilever beam.

Well thanks for pointing me in the dynamics direction.

hydtools, thanks for your response. Indeed, this will go down the dynamics rabbit hole. I have some reviewing to do.

 

[Snickster]

One way of possibly looking at it is that the engines are oversized to provide the ability to accelerate at approx. up to 1G to higher speeds or levels with full load. The engine maximum thrust of 20N is balanced by the load of the beam and payload which is a total of 9N. So under non accelerating conditions the engines will be thottled to produce 9N thrust. However the engines can be powered up to produce an acceleration to a higher speed of up to 20N thrust. In this case add the resulting G force to all loads evenly with resulting loads downward equal to 20N. The loads of the concentrated loads will act at the applied concentrated load point and the beam load will be a distributed load.

Solve by summing moments about a selected point as in usual beam case using equations for a beam simply supported at ends. I believe that you will necessarily find that the sum of moments is alway zero at any section and maximum beam moment occurs in middle.

 

[Jboggs]

Sean99,
Based on the information you provided, and not knowing anything about you, SWComposite's question was a legitimate one. I saw no reason for your insulting response. We get all kinds of folks on this forum, some with post-doctorate degrees and some with no technical education whatsoever. So it really isn't fair to make any pre-judgments about a poster's education level.

As far as your beam problem, others have already pointed out that it is not stable (unbalanced forces). I will also point out that since you mentioned it is a "tailsitter" aircraft (more commonly called a "taildragger") one could assume that the thrust of the engines is in a relatively horizontal direction rather than vertical (as in drones and helicopters). If that is the case then the beam weight component (acting in a vertical direction) is a non-issue, at least for a 2D analysis.

As you know, the first step in solving a problem is to get all your assumptions correct. That's all we're trying to do.

 

[Sean99]

Hello Snickster, I tried your method, please see the two photos I've attached. I did it in two different ways, the way you explained with the resulting forces being equal to 20 N downwards didnt make much sense to me, or maybe it did with that law, all actions have an equal and opposite...but I thought the resulting forces downwards should be equal to the net forces upwards which would be 11 N , or (20 - (2+2+5)). So I calculated it for both your scenario and mine. Please review them and let me know what you think.

EDIT: oh wait a minute...this doesnt make any sense if the resulting forces equal what's upwards then there would be no acceleration..

 

[Sean99]

the other photo with my assumption-

Jboggs, I think swcomposites question offered nothing and was an insult in and of itself. Maybe you'll forget the integral to sin(x) and youll ask me, and I'll say "didnt you ever take a calc class?? its one of the first integrals you learn in calc 2!"

Also, it's not a taildragger. It is indeed referred to a "tailsitter", look up tailsitter aicraft on google to see the difference.

That's true, I do need to get my assumptions right.

 

[LiftDivergence]

FYI you can analyze a dynamic problem quasi-statically using D'Alembert's Principle.

Keep em' Flying
//Fight Corrosion!

 

[3DDave]

One problem in your analysis is that the distributed load is from some acceleration due to the excess thrust, but the forces you applied initially assume the airplane is at rest. Once the airplane accelerates those loads will change.

 

[Sean99]

LiftDivergence, thanks for your comment, ill look into it.

3D Dave, right you are. What you're saying only applies to my own analysis and not the page titled "Snickster way" correct?

 

[3DDave]

Typically one first establishes the masses, not the weights, before starting a dynamics calculation. If there are weights they are divided by g-sub-c (probably 9.81 m/sec^2 in this case) to get the masses.

I've worked on such calculations and never used explicit fractions. I didn't see that Snickster suggested using fractions.

If the total weight changes then all the fractions have to change. For the most part aircraft are designed to G loading rather than max thrust.

The calculation is much simpler if you ignore the spacing of the attached masses and put them in the middle of the beam. It will be a tiny increase in the moment with a large decrease in the effort to account for it. By superposition you don't even need to create it as a single diagram - do one for the uniform load and another for the concentrated load and add the shear and moments. Again, this will help immensely as the design changes.

 

[rb1957]

we normally analyze static bodies. If the body is accelerating (due to a force imbalance, then add a body force at the CG. you know the mass, so can determine the acceleration. "10N" is thrust from engines ? "2N" is ?? weight is "5N" down, inertial reaction (opposite to the resultant acceleration) is 15N at the CG ... no?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Snickster]

Hello Snickster, I tried your method, please see the two photos I've attached. I did it in two different ways, the way you explained with the resulting forces being equal to 20 N downwards didnt make much sense to me, or maybe it did with that law, all actions have an equal and opposite...but I thought the resulting forces downwards should be equal to the net forces upwards which would be 11 N , or (20 - (2+2+5)). So I calculated it for both your scenario and mine. Please review them and let me know what you think.

EDIT: oh wait a minute...this doesnt make any sense if the resulting forces equal what's upwards then there would be no acceleration.

I agree with the forces and resulting shear diagram you show in the sketch: "Snickters Way"

When there is an acceleration produced by a unbalance of external forces, that acceleration of a body results in an opposing inertial force of every mass of the body equal to F= mass x acceleration, so that the sum of the internal inertial forces developed by acceleration of the body equals the difference in the external forces dF = M (dA) . I believe this is D'lameberts principle as mention by someone else above. In this way you can convert it into a statics problem.

For instance if you are driving in your car and weigh 200 lbs and accelerate at 32.2 ft/sec then you have 1 G of acceleration on you so your body feels a inertial force of 200 lbs (1 G) and is pushed back in your seat and if you have a spring between you and the seat it will compress until it develops 200# in accordance with the spring constant of the spring. This force actually balances the acceeration force pushing in the direction of motion so that all masses in the car and of the car inertial forces are balanced with the acceleration forces assuming no other external forces.

In the case of the aircraft to accelerate it straight upward it would require a force equal to the (Sum F) = M x A

Summation of external forces equals applied thrust forces upward minus weight of aircraft downward = Thrust - 9N.

So you have 11N available for acceleration, the 9N of thrust is just holding the aircraft level.

Now if your acceleration is completely horizontal there is no acting against gravity so 20N full thrust external applied load = M x A internal inertial resisting force (not considering external resistance forces of wind drag etc).

 

[Sean99]

Snickster, thank you for your response. What you've said makes sense to me and I'm glad I did what you suggested right in the "snicksters way" page. However, if we only have 11 N available for acceleration, I don't understand why then the summation of downward forces to calculate the total moment on the beam wouldn't be equal to 11 newtons. Then I would do the same thing as I did in the "snicksters way" page, with the 5 newton weight being replaced by (5/9) * 11, and each 2 N load replace by (2/9) * 11.

I guess what im asking is, why are we equating the total downward force equal to 20 newtons, if we only have 11 newtons available for acceleration?

Thank you!

3Ddave, thanks for your response. I'm not sure where youre going with this, so we would have the weight of 5 newtons acting right the center of the beams, and the two, 2 N weights would also act at the center, and then how do we go about scaling the weights to accommodate the 20 N thrust and thus have our makeshift static, supported at both ends beam?

rb1957, could you elaborate how exactly we would scale the weights?

 

[3DDave]

Step 1) The mass of the beam remains distributed - you evaluate that based on the system acceleration of 1G of a simply supported beam
Step 2) The mass of the concentrated masses is simply relocated to the center - you evaluate that based on the system acceleration of 1G of a simply supported beam.

Step 3) add the contributions and multiply by the expected acceleration.

The forces at the ends of the beams will be solely the amount for each case to produce that acceleration.

Anxiously awaiting seeing the finished drone.

 

[Snickster]

Snickster, thank you for your response. What you've said makes sense to me and I'm glad I did what you suggested right in the "snicksters way" page. However, if we only have 11 N available for acceleration, I don't understand why then the summation of downward forces to calculate the total moment on the beam wouldn't be equal to 11 newtons. Then I would do the same thing as I did in the "snicksters way" page, with the 5 newton weight being replaced by (5/9) * 11, and each 2 N load replace by (2/9) * 11.

I guess what im asking is, why are we equating the total downward force equal to 20 newtons, if we only have 11 newtons available for acceleration?

Because when the aircraft is not accelerating and at constant level then there is still bending moment in the beam based on two end upward forces of 4.5N each = 9N, and downward forces equal to payload 4N plus beam weight 5N = 9N. When the aircraft accelerates upward against gravity then there is an addition 11N output of the engines available which increases the total upward force to 20N and with 20N downward caused by static payload and beam weight = 9N, plus dynamic inertial loads = 11N.

 

[rb1957]

do we have 20N thrust and 9N weight (5N +2*2N) so a net force of 11N, a resulting acceleration on 11N/(9N/g) = 11/9*g ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.