How Do Airframe Stresses Change During Rocket Motor Ignition and Flight?

[denverdoc]

I'm trying to put together a proposal for a small amatuer rocket flight, and could use some assistance. This is not a terrorist strike aimed at your backyard, but flown under the auspices of the Triopli Rocket Association that provides safety guidelines and insurance to members in good standing.
See: http://www.tripoli.org/

My question has to do with the stresses between an external airframe and one or more enclosed motors that are attached to the skin of the airframe. Generally the configuration consists of an inner cylinder which is glovefit sized for the motor that is then glued into centering rings. In turn these annuli are glued to the inside of the airframe at varius stations.

This particular configuration is a bit different. But I believe the same principles apply.

The stresses would seem to vary. At full ignition, there would be a strong axial shear component equal to the thrust of the motor trying to tear loose and leave the airframe behind. Now my question is assuming that the airframe gets dragged along for the ride, is there any change in the net forces.

For one the drag of the vehicle is not felt by the motor(s) directly. This would then add to the shear forces exerted on the adhesive or mechanical attachments to the motor?

OTOH the airframe is developing velocity and so its inertia is helping to relieve the stress--I think but am not sure whether "jerk" is operative here--in other words--hit the motor with a sledge hammer, the shear forces are peak, but once it all starts to move, the same relative force is not as apt to tear things loose.

When the motor burns out, then another development: the drag forces become paramount on the af, while the inertia of the motor is unabated.

So another jerk?

Clearly my thinking on the subject is muddled, and any help is appreciated.

 

[AlephZero]

You have hit on a real issue that you need to address to do a proper stress analysis. The key to getting it right is to realize the rocket is accelerating (or not, when it's still on the launch pad, or course).

The effect of the acceleration is to create a force (the so called d'Alembert force) on each part of the structure equal to -mass*acceleration.

When you include those forces, you will get the correct loads and stresses between the parts of the structure.

As a simple example, consider the rocket as a uniform tube of mass M with the motor attached to the base. In flight (ignoring gravity) the axial force in the tube will vary linearly, zero at the top and equal to the thrust at the bottom.

Hope that gets you thinking along the right lines.

If by "jerk" you mean the rate of change of acceleration, that's not important compared with the accleration itself.

 

[denverdoc]

That helps--am I right in thinking that d'Alembert's principle is equivalent to Newton's sixth corrolary which broadens inertial frames to include those subject to parallel constant acceleration(s)?

For instance, if you have a uniformly accelerating pickup truck with a box in the bed, how long does it take to hit the tailgate given a particular Mu and distance x from the tailgate?

One of course can do this conventionally, or introduce a virtual acceleration such that the truck is at rest and any free bodies are now subject to this acceleration--which leads to the solution immediately. Is this akin to what D'Alembert was doing?

 

[AlephZero]

Yes d'Alembert's principle is equivalent to Newton's laws of motion, it's just a convenient way of doing the accounting to include all the inertia effects correctly.

First you treat the complete rocket as a rigid body to find its accleration from the resultant force on the whole structure - e.g. accel = (thrust - drag - weight)/mass if it's going vertically upwards.

Then you apply fictitious forces of -mass*accel to each part individually (e.g. to the motor and the body).

You now have a balanced set of forces and you can forget about accelerations for the rest of the analysis.

For a general motion where the rocket has a linear acceleration, is rotating, and the rotation is accelerating, you also include the "centrifugal/centripetal forces" etc in the same way.

Ignoring weight and drag for a simple example: if the rocket body has mass M, the motor has mass m, and the thrust is T, the acceleration of the rocket is T/(m+M).

The d'Alembert force on the rocket body is -MT/(m+M), and that is the force the rocket body exerts on the motor.

The d'Alembert force on the motor is -mT/(m+M). The resultant force on the motor is T - mT/(m+M) = +MT/(m+M) which is the force the motor exerts on the rocket body.

Of course you can interpret this in terms of the acceleration in a non-accelerating reference frame: the fraction m/(m+M) of the thrust is accelerating the motor itself, and M/(m+M) is accelerating the rocket.

 

[denverdoc]

AZ,

Very helpful of you, I'm going to give this some thought when I have some time off tomorrow...see if I can't finish my problem.