I'm not very proficient at aero- or fluidynamics or stress calculations, but I want to know how to calculate the stress on parts of a given rocket, given the height(or air density), orientation and velocity vector of said rocket.

For example one could ask: 'Is the stage connector which can sustain this much stress capable of flying through max Q?' where one has the data on how much stress it can endure but not on how much it will experience.

Edit 2: Firstly, what forces do act on a rocket and parts of that rocket? Aerodynamic drag, gravity, acceleration from the thrusters, obviously, but which are important in this calculation? And secondly, while I may be able to google the formulas for all of these forces and more, how do I 'convert' from force to mechanical stress? This is the real problem, I think.

I do not need the 100% perfect solution, an approximation suffices.

Thanks in advance, JAG

Edit: maybe a simple usecase would be helpful: given a very oversimplified rocket consisting of four parts (engine, tank, separator and payload with fairing) of known stress endurance (although I do not know how to express this - oopsy) the mechanical stress would be different on every part when taking a trajectory, where the rocket is not pointing in the direction of travel at this point than flying straight up. How would i go about calculating this stress, both drag-induced and otherwise for every of the four parts and compare it to the allowed values?

Meta: Firstly, to give some context: This is research for building a game (actually a mod). So I need this to calculate whether a rocket would explode or tear apart at every point in time. Secondly, should I delete obsolete information which got updated by an edit or keep everything for reference? It kinda gets too long and messy...

  • $\begingroup$ Space agencies and rocket designers use finite element analysis software such as NASTRAN to calculate the loads and stresses on structural parts. en.wikipedia.org/wiki/Nastran For shuttle, the loads (for ascent at least) were pre-calculated and plotted out on graphs of angle-of-attack and sideslip with boundaries drawn on them to see when if "the rocket is not pointing in the direction of travel" a limit was exceeded. These graphs had the unusual name of "squatcheloids". $\endgroup$ Commented Sep 8, 2018 at 0:29
  • $\begingroup$ Well, as I do not have the capabilities nor the need to use such a software, this doesn't really help me, sorry. With 'How does NASA do it' I meant how does the software calculate this. You seem informed, however. So could you give some insight about this? $\endgroup$
    – JustAGuy
    Commented Sep 8, 2018 at 17:06
  • $\begingroup$ Sorry, you asked how NASA did it, before you edited that out of your question. $\endgroup$ Commented Sep 8, 2018 at 19:52

2 Answers 2


Here's the 10,000 ft level explanation. Stress is force per unit area. To calculate stress in a part (a structural member) you have to know the dimensions of that part. Then you calculate all the forces acting on that part and divide by the area. Then, to know if it bends or breaks, you have to know the limits of the material you are using. Do you have that kind of details about this rocket? I doubt it.

So for your 4 component game level analysis here's a suggestion.

Maybe you should try to calculate the forces acting on each of your 4 components and then see what forces are transmitted between the 4 components. Set some sensible limits (i.e. trial and error) at which the components tear apart from each other. Start by drawing a free body diagram of each component. Remember the force sums between each component must be zero unless the components are moving relative to each other!


the quickest approximation can be done using drag

D = C * rho* V^2 * A /2

Where C is an experimental coefficient, rho is air density, V is velocity, and A is reference area.

This gives the sum of drag forces on a body.

Now, doing some hand waving, and ignoring boundary layer theory, we will assume (incorrectly) that drag acts equally on all points of our rocket. So, D/da = drag/surface_area

You can (accurate to maybe an order of magnitude) say that a part of a rocket's body experiences D*total_surface_area/surface_area_of_part units of shear force.

Edit: to answer everyone's questions/comments about force/acceleration at the "other end" of the rocket:

It doesn't matter. You see how there's no acceleration or force component on the RHS of the drag equation? The drag equation doesn't care about the hot gas being expelled out back, it cares about air molecules slamming into the nose, and pulling against the outer shell of the rocket.

Edit: I'm focused on the forces on the surface of a rocket. (since @justaguy mentioned "aero- or fluidynamics [sic]" and fairings- which are external aero/fluid components). If we care about the shear/normal stress on internal components (which is a very real concern), this isn't a aero/fluid problem anymore and we can consider the g-forces experienced during acceleration. Then, don't worry about drag so much as net acceleration (thrust + drag)

  • 1
    $\begingroup$ Is acceleration a factor? $\endgroup$ Commented Sep 7, 2018 at 14:01
  • $\begingroup$ @MohammadAthar remember that there's some force at the other end of the rocket as well! Take a look at these questions and their answers for example: 1, 2, 3 $\endgroup$
    – uhoh
    Commented Sep 7, 2018 at 15:09
  • $\begingroup$ Reference area is the surface area perpendicular to the velocity vector, right? Or which part is meant by that? Also, can C be derived (approximately) from anything or is it purely experimental? $\endgroup$
    – JustAGuy
    Commented Sep 7, 2018 at 16:24
  • $\begingroup$ the reference area is indeed the "face" area of the body. C is derived experimentally, but for simply geometries, tables exist: en.wikipedia.org/wiki/Drag_coefficient $\endgroup$ Commented Sep 7, 2018 at 17:11
  • $\begingroup$ Well thanks for your answer, edits and comments, but I am indeed looking for the general/internal stress. I just thought it would be more related to aerodynamics. I am gonna edit the question, but your answer was really helpful already! $\endgroup$
    – JustAGuy
    Commented Sep 7, 2018 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.