Well I suppose you go ahead an analyze the resulting structure to get internal forces and deflections?
I think you need to be a bit more specific with your question, I am sure that your not looking for such a trivial answer.

"2) If you apply Load 1 on Structure 1, then add a structural element to it and finally apply a new Load 2, then the final state of forces and displacements is obtained by adding L1 on Structure 1 to L2 on structure 2." ... no, I don't see the displacements of Structure1 being relevant to Structure2.
Structure1 under Load1 gives Displacement1 (or 11)
Structure2 under Load1 gives Displacement1' (or 12)
Strcuture2 under Load2 gives Displacement2 (or 22)
Structure2 under load1 and Load2 is Displacement1'+Displacement2

"But what if after the first load you add a new part and then remove an old one?" remove the old what ?

And this assumes small displacements. Superposition can fail under large displacements.

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

rb1957, remove any part of the original structure after you added a new part. Could be a beam, a plate, a shell, whatever.

driftLimiter, but how exactly do you go about that?
If you apply the load to the new structure, it will not lead to the results you are loooking for.

To be honest I have no idea what your asking. How is the structural model supposed to know or care about what some other version of it was? Are you trying to envelope the results from multiple configurations? If thats what your after you need a structural analysis toolkit that allows for phased construction.

I have no idea what you are talking about.
Seems like what we normally do in the process of optimizing structures.

I think you must use good engineering judgment about how the modified structure will behave. When in doubt, be conservative.

BA

"rb1957, remove any part of the original structure after you added a new part. Could be a beam, a plate, a shell, whatever."

so you're asking "But what if after the first load you add a new part and then remove an old one?" to paraphrase ... create structure 2 by adding to and removing pieces of Structure 1 ... a new structure needs a new FEA. Structure1 displacements are irrelevant to Structure2.

Now you can remove the effect of a structural component by, knowing the load in it in Structure1, applying an opposite load to the Strcuture1' ...
Structure1 under Load1 gives Displacement1, and the load in ElementA is PA
Structure1' is Strcuture1 with ElementA removed
Strcuture1' under Load-PA gives Displacement2
summing Displacement2 and Displacement1 gives you the displacements for Structure1' under Load1 ... although why you'd bother is beyond me.

Something similar we do is "Fail Safe Structure Analysis" where we'll fail discrete elements and see how the load redistributes through the structure.
Running several models under the same loads.

As before, superposition assumes small displacements.

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

driftLimiter

The model "knows" what the previous version is because the new part whas added while the load was applied! Stresses, strains, displacements... the whole shabang. Then you have a new structure with old parts already under stress and a new one that is not. Finally, you remove another part of the structure. Like I said, the order matters.
1) Apply load
2) Add new part to structure
3) Remove old one.

rb1957, you not only create a new structure... you create a new structure in which some elements are stressed and others aren't. You keep missing the critical part, which is the fact the you do not deload at any point in time.
Simple removal of items in a structure while under a given load is really the same as if you deload and reload in between. It really doesn't change final results.
But if you ADD new parts to a structure that is already loaded and the load is never removed, the results are quite different from deloading, then adding the new part and loading again. Any part that you add to a structure while loaded will not be under any stress or displacement.

Maybe some of you might be confused by the fact that in the particular case of structures that both begin and end as isostatic, the order of events in time doesn't really make any difference whatsoever to the final results. But in the case of hyperstatic (indeterminate) structures, the order matters.

BTW, rb1957... your proposal does not include the adding of a new part. That is crucial. If you only remove stuff, then the problem is trivial.

Quote (Danielsp)

You keep missing the critical part, which is the fact the you do not deload at any point in time.

When did you state that? It's not reasonable to assume it. Normal practice is to remove as much load as possible while making these sorts of modifications. Two reasons off the top of my head: 1) safety of the workers (welding on a transfer girder carrying a few hundred kips doesn't sound like a good idea to me) and 2) simplification of analysis.

But to your question...which I think is most applicable to shoring applications...you just have to maintain awareness of the state of stress in each member in the system as you modify it.

1) Analyze the system and determine stresses in the members at critical points (or just come up with the bending and shear diagrams so you can derive stresses at critical points later).
2) Add a support. If the load doesn't change and you didn't jack it up at all, your stresses will remain the same. This is where an understanding of the situation is required. Is there full live load on it when the post is installed? Full dead load? If full live load, then it gets interesting because when live load is removed you'll have uplift at the new support. So you have to figure out the most plausible scenario(s) (and/or code required scenario(s)) and determine the state of stresses for those.
3) Remove the other support. If you have multiple scenarios, you'll be getting into multiple permutations to figure out the final stress. As long as you're in the linear elastic range, you can just use superposition here.

Probably best to use judgement to figure out what the 'worst case' permutations will be and envelope it.

phamENG, I edited my opening post to make it even clearer. There are no multiple scenarios, just the one load.

I see no way to use superposition here, what exactly would you superpose to what?

In that case, it's really simple.

For a case where you start with a simply supported beam, add a support somewhere between the first two, and then remove one of the first two (you didn't define the structure or the location/direction of the load)

1) Analyze the original beam, find the displacement of the beam at the point of the new support.
2) Add the support. Since the beam is already carrying the full load and you said we're not unloading the original structure, the load on the new support is zero and the deflected shape remains unchanged.
3) When the support at one end is removed, the beam will find a new equilibrium and, since it's a determinate structure, you won't get any "locked in" stresses (unless you want to go deep in the weeds an look at steel residual stresses, creep in wood/concrete, etc.). What you will have is a slightly sloped beam - because you do have a locked in displacement where the support was added.

Yes, phamENG. Youe are absolutely right, that is the case for isostatic structures - only for those, though.
But as I stated in my 3 Aug 22 12:45 post, hiperstatic structures will keep some locked in stresses. That is the whole point.

redundancy (and it's hyperstatic) doesn't matter. If you change the structure (subtractive or additive) you change the internal loadpaths.

Of course, if you have a determinate structure and remove a support (or the loadpath to a support) then you have a mechanism (and not a structure).

If you have a redundant structure, then you can remove internal components and still (maybe) have a structure. As I said above this is "Fail Safe" analysis.

But you need to run each of your structures.

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

Hiperstatic? I'm guessing you mean hyperstatic? Or maybe it's spelled different where you are. Don't know. It's a bit esoteric...I've never met anyone who actually uses that term. Everyone I've ever met or worked with calls it 'indeterminate.'

You'd still find the stresses the same way, but you have to use indeterminate analysis techniques. You'll account for any 'locked in' stresses with the displacement at the new support location.

Next time, post a sketch.

Yes, phamENG... I'm sorry for my misspelling.

It takes 5min in bluebeam or some other software to draw a proper sketch.

No worries...I'm not from Brazil...different terminology and spelling is used in different places. Gets in the way a lot on an international forum.

Quote (If you load a structure then remove some part of it while keeping the same load,)

A couple of projects come to mind... one about 30 years back, the client had a large PEMB structure. He wanted to remove one of the outside supports for a new door. There was a lot of reinforcing required of the exterior bay and the third one in.

Just recently I looked at a PEMB added roof structure. It was a PEMB where the exterior supports were on a substantial masonry parapet. The parapet was about 5' high and the horizontal thrust was accommodated by using diagonal bracing angles to the floor. The client wanted to remove the diagonals. It was a 4 span structure and removing the diagonals would require reinforcing the entire PEMB structure.

So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

Brad805 and phamENG , I'm sorry... I just didnt see how a sketch would be relevant. It is a general problem, could be applied to any indeterminate structure you like.

phamENG, I have the proper software to solve indeterminate structures (https://atirsoft.com/strap/). But the solution always assumes that there are no locked in stresses.
I am just trying to figure out the general solution to problems like this and found nothing online.

Starting simple, consider a structure where you remove a support. This happens fairly frequently during construction. You can model this by applying loads to the 'new' structure (ie without the support in question), at the old support location, that are equal but opposite to the support reaction that previously existed. This gives the effect of removing the support. Add the results to the previous results from when the support was in place.

You can extend this principle to whatever you remove from the structure.

Quote:

1) If you load a structure then remove some part of it while keeping the same load, then the new moments, shear forces and strains can be easily obtained by simply ignoring the original structure. You load the new structure and that's it.

Not the case when there are time-dependent effects like shrinkage and creep. What I described above is more general.

Yes, dik. That appears to be one instance of the problem at hand: a structure that you wish to reinforce before removing some part, and the load is present at all times.

steveh49, just removing is trivial. The thing is when you

1) Load the structure
2) ADD a new part without removing the load
3) Only then remove a support or whatever other structure part that you wish.

As I said...terminology varies, but a sketch of the situation would have put us all on the right track from the get-go.

Does that software allow you to force a displacement of a support? If so, that's all you need. If it doesn't, then you'll have to do it by hand.

There seems to be alot of confusion in this thread in terms of what part of the structure is there, what part of the structure isn't there, what load is there, what load isn't there, etc.

At any point in time, there is a structure in place and a load in place. If there are phases of construction and/or demolition when any of the structure or the loading changes, you can simply create multiple structural models to reflect such conditions. Add a beam, create a new model. Remove a beam, create a new model. This approach will result in a snap shot of the structure at any given point in time. As you have probably realized from the confusion above, a picture is worth 1000 words, we cannot see what you are seeing.

rb1957, your take on determinate structures is right. But not on the indeterminate ones.
Fails-safe analysis refers to structures in which you only remove parts. How do you calculate stresses and displacements in the situation I described?

Yes, phamENG. I can force displacements. How would you proceed exactly?

Quote:

steveh49, just removing is trivial. The thing is when you

1) Load the structure
2) ADD a new part without removing the load
3) Only then remove a support or whatever other structure part that you wish.

Apply the negative reaction load to whatever your new structure is, including your added part. That's the general principle.

Edit: I'll expand on how this works. Remove the support but apply the old reactions as external loads. You get the same results as when the support was in place. Now apply opposite loads. There is now zero net force at the old support location: it has been removed. But the negative loads were applied to the new structure with the added part, so the stresses and deflections due to the negative load are calculated from analysis of the new structure.

I'm starting to feel like a broken record.

1)Analyze the original structure.
2)Determine the displacement at the point you plan to install the new support.
3)Run the model again with the support removed and a forced displacement at the new support equal to the deflection calculated in step 2.

It's appreciated pham... the learning process requires redundancy.

So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

Quote (the load is present at all times)

Another problem often encountered by engineers is changing load conditions... it's just part of the game we play.

So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

steveh49, by negative reaction load you mean minus all the loads exchanged between the removed part and the remaining structures?

phamENG, I'm not adding new supports, but new parts to the same structure. Would it work the same?

"rb1957, your take on determinate structures is right. But not on the indeterminate ones." ... no, I understand indeterminate structures very well. The point to my "maybe" was in an indeterminate structure you can remove a structural element that can fail more than one loadpath, and the resulting structure could become a mechanism. I tried to cover al the "what ifs" (if I hadn't I'm sure someone would've pointed this out to me).

yes, fail safe is subtractive changes ... I used it as an example.

any changes to the structure mean rerunning the model. Ok, maybe not if you are just reducing the size of a member and reason that the change in loadpaths is negligible.

what is the big deal in rerunning the model ?

Something you can do is divide the model into "super-elements". These condense the structure to stiffnesses at the boundary. No change inside the super-element then no need to rerun.

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

I'm done until there's a detailed problem statement and images/sketches to illustrate.

Quote:

I'm not adding new supports, but new parts to the same structure. Would it work the same?

The principle is the same. If you removed a section of plate, you would apply loads opposite to the stresses on the boundary of the removed section of plate.

like pham, I have no idea what the issue is.

I don't know why we're talking about removing elements are then applying their load.

Why would someone ...
1) have a structure and a set of internal loads, then
2) want to simulate removing an element by (with the unloaded structure) removing an element then applying the the reverse internal load of that element to get the effect of removing this element, then
3) adding the results of 2) to 1); instead of rerunning the external loads with on the structure without the element.

but OP says this is simple, and wants to know the impact of adding an element. Add your element and rerun your model.
You could maybe get somewhere by noting how the nodes (at the end of your new element) displace under your external loads and then apply this as enforced displacements on the model with the added element ??

But this is philosophy ...

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

steveh49, I believe you're right and that the principle works also for indeterminate structures. At least for a test situation I calculated by hand, which is this:

Suppose you have n springs of constant K in paralell supporting a load P, then each spreing would take P/n under displacement x0 = P/(Kn).
If you add a new spring while the structure is loaded, no force or displacement will change and the new spring will be under no load.
But after you remove one of the original springs under stress, what will be the total displacement x of the springs?

In the final situation we have n-1 old springs with displacement x and a new one under displacement x-x0. Equilibrium requires that the total spring force is still P, so

(n-1)Kx +K(x-x0) = P
But Kx0 = P/n, so
(n-1)Kx +Kx -P/n = P
nKx = P +P/n
Kx = P/n +P/n² is the new force in each of the old springs, and the new spring has force K(x-x0) = P/n²

Those are exactly the values you'd get if you added
a) original structure (n springs) under load P, that's P/n for old springs and 0 for new (yet to come) spring
b) new structure (also n springs) under load P/n (replacing a removed spring), that's P/n² for old springs and new alike.

Quote (Danielsp)

If you add a new spring while the structure is loaded, no force or displacement will change and the new spring will be under no load.

This seems like an academic stretch to me. If you add an exactly similar spring to the arrangement it would take load and reduce the load on the other springs. If however you added a spring with a different neutral depth then perhaps no elongation of that spring. But how does this all relate to a real world structure I think is everyone's question. It really depends on the element/s your considering, the loading, and the support conditions.

driftLimiter, general principles might seem like academic stretches, but they are all we have in order to make models.

The fact that you add a spring after the original structure is deformed is indeed equivalent to using a spring with a different neutral depth.

How does this relate to a real world structure?
Well, if you only build stuff then load it... it doesn't. I know that's the vast majority of cases in professional practice and also in theoretical structural analysis, both static and dynamic.
But if you happen to need to demolish a specific part of a structure and wanna make sure the whole thing doesnt come down... you better build a reinforcement somewhere first. Then, it would be nice if you could put some numbers on final displacements, moments, shears and axial forces to make sure the structure can take them.

Also, dik gently provided us with a picture of a real world application of those principles.

Quote (Danielsp)

The fact that you add a spring after the original structure is deformed is indeed equivalent to using a spring with a different neutral depth.

I don't think that this is true in general. It might be true in a specific loading or support condition. Lets say your hanging your load P from n springs. You add another identical spring to the arrangement, you must stretch it by the amount of the displacement of the original structure. Then after you stretch it and attach it to the load, the displacement would reduce and the force in each spring would reduce proportionally.

If you want a good answer you will have to be more specific.

rb1957, I can rerun the model as much as I want, but simply applying the original load to either the original or the new structures (they can even be identical, apart from locked in stresses!) will not provide the results of the structure in the situation I described. Look at the spring example I used, or the one by steveh49 .

Yes, driftLimiter. If I pre-stretched the new spring by just the right amount, then I could just rerun the model.
But that is not the situation I described, which is: you insert a new, unloaded part to an existing, loaded structure and only then you remove something from the original structure.

If all the elements remain elastic in all the different loading scenarios, then static equilibrium and displacement compatibility must always be satisfied. If this is satisfied in your modeling then your results will be consistent with reality. Adding certain elements in a displaced condition might change the stiffness distribution in your structure and thus change the load path. You can used a phased construction analysis technique to model the effect of changing elements/Loads in a displaced structure. This is often done for composite concrete construction.

steveh49, apparently you nailed it, although I'd still like to create a test case that does not involve adding or removing supports (unfortunately they are less intuitive and harder to judge). Anyways, it seemed great structural thinking of your part right there!

BTW, going a step further... the implementation of that solution is not so easy on the software I use. I'd have to collect each of the forces and moments in the hundreds of common nodes on the interface of the parts to be removed (potentially a few DOF per node) then insert those as loads. It's not an automated process, so I was wondering whether the final result could be a linear combination of the partial structures under full external load, as opposed to finding and inserting the internal load.

For example, in the springs example I described, if you multiply the original structure solution by 1, and the new structure solution by 1/n, and add those numbers, you get the right results. See:

a) spring forces of original structure under load P: P/n; P/n...; P/n; and 0 on the future spring

b) spring forces of new srtucture under load P if originally unstressed: P/n; P/n;... and P/n also on the new one

1*a + (1/n)*b =
P/n +P/n²; P/n +P/n²... and P/n² on the new one

getting the results of originalstructure *1 +newstructure *1/n would be quite easy. But of course the coefficients 1 and 1/n would be different if you removed 2 springs, or if they had different stiffnesses, etc. So, I was wondering if there is a simple way of finding out those coefficients.

if you mean, take steve's 2nd diagram (a SS beam with a load) and now add a support, mid-span.
Then yes there's a method for doing this, called unit force method.

If you want to add a support mid-span, then apply a force to the beam that reverses the SS displacement.
The unit force method says apply a unit force (1lbf) at the mid-span location, and determine the displacement at the mid-span position, du.
Then knowing the deflection at the mid-span from the SS beam problem, D, now you know the force to apply to reverse the displacement, D/du*1lbf.

This is doable for very simple structures, but very quickly becomes too much work.

You get the same result by analyzing the 3 support beam directly.

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

driftLimiter, the software I use does alow for different stages of construction. But you have to apply Load 1 to initially unloaded structure 1, load 2 to initially unloaded structure 2, etc and then add them all together. I don't see how to apply that to locked in stresses apart from the use of internal forces such as described by steveh49.

get some better s/ware ! what are you using ?

why do you have to "apply Load 1 to initially unloaded structure 1, load 2 to initially unloaded structure 2, etc and then add them all together" ?

If you mean to "fudge" supports and elements and such (ie Load 1 is the load that would be element 1 (being added to the structure), load 2 is load in element 2, etc), then you are missing a term. Multiple redundancies interact. Say you have a beam on 4 supports, so you can solve aa a determinate beam on two supports, extract the displacements at the two supports you removed. Then separately apply unit loads at these two locations, and get the displacements due to a load at the support (like load1 gives unit displacement at 1, call it d11, and load2 at 2 gives d22; yes?) But there is also a displacement at 2 due to the load at 1, call it d21, and vis a versa a displacement at 1 due to the load2, d12. Now you have two simultaneous equations P1*d11+P2*d12 = D1 and P1*d21+P2*d22 = D2 (or something like) to get P1 and P2.

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

rb1957, I'm using STRAP. I don't make the buying decisions around here.

I don't *have* to "apply Load 1 to initially unloaded structure 1, load 2 to initially unloaded structure 2, etc and then add them all together", but that sequence of events is automated on the software. I just indicate that I want a combination of different loads from different structures with the coefficients I determine, and it does the job. On the other hand, if I want to use the internal forces and moments of 5 DOF for 100 nodes, I have to first extract all that data from a huge table of all nodes and DOFs, then put it into some text format to input it as a load before processing. I'd be doing the bulk of the work myself, instead of just clicking some buttons. As you said, creating the equations using unit loads from the ground up would be unfeasible for any substantial number of nodes.

you poor bastard (that's not derogatory, that's the Australian adjective/noun/...) ... i feel your pain, brother. Have you asked the "Strap" helpdesk ?

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

rb1957, they are useless. They can only help with specific funcionalities of the software and how they work. This case involves some theoretical framework (which is what I wanted to discuss here), so it would be completely out of their scope of work.

does another code to what you want ? (I still can't really figure it out ??)

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

I may have confused things by posting a statically determinate example. Here's a trivial indeterminate example. Danielsp, I know you want to do something more complex and you've understood the principle - I'm posting this for others reading along.

Not even real code, RB. A simple spreadsheet will do.

Steveh, if I get it right it's missing a final stage of removing something. In my case it's always adding and removing beams or plates, but I suppose supports work the same way.

Yeah, maybe not a good example either. Too trivial.

A closer example to what I am doing is this. Suppose I wanna transform the first structure into the second while loaded




1) These would be the bending moments of the structures if built unloaded




2) So after loading the original structure, I build the new span and it looks like this (hinge not real, just added to create the real moment diagram)




Removing the right span would amount (to the remaining structure) to adding this load to this structure




3) So these are the moments introduced to the new structure by demolishing the right side span




Therefore, we arrive at final moments by adding situation 3 to the original moments of the original structure

And the moments at the nodes would be 0; -1,04; 6,32; 0; 0

Hope I got it right, Seteveh

Another way to do this would be to replace the right support by its force in the original structure, 0,93kN downwards, so it would look like this (spans are 3m wide each).




But demolishing the right span amounts to the same as cutting the right support lose, so I add the necessary load to cancel its effect, which is




When adding both, I arrive at the same results as the other method

Is the sequence below correct? At the point in time you cut the member in stage 1 the end of the remaining section undergo a rotation. When you install the new segments you will build whatever end rotation you have at that time into your joint. I do not see the stress changing as you show unless you deform the new sections somehow during install. Further changes to the boundary conditions or loading will change how it behaves, but the 10kN force always seems present.

No, Brad. As I pointed out, you demolish after building. If it was the other way around, it would be trivial. And there is no rebuilding.

ok, now some pictures.

but the original 2nd structure has a moment of 3.33 at the LH support,

but your method seems to say 1.04 ?

ie the last 2 pix don't equal the first, no?

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

or are you trying to derive the reactions for the beam on 4 supports ??

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

why would the last pic ...


be different to the first ??



the RH unloaded overhang does nothing to the problem ?

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

Ok, I get it now. If you add the x1/x2/x3 values I would not be surprised if someone would not check this. It is a very easy problem to run if you have a better staged analysis package.

Word of advice. Start with a simple sketch in the future. While you are heavily interested in the problem others come here for different reasons. Not many want to wade thru endless textual descriptions when a decent drawing will convey the question very quickly.

rb, the 3,33 and 5,56 pic shows what WOULD be the bending moments IF the 10kN load had just added to the two left spans in the usual manner: no history, no previous deformations, no existing displacements or rotations, no initial 3rd span, only a plain 2-span problem. Just for the sake of comparison with the actual values at the end.

Let me see if I understand your example please correct me where I go astray.

1) Load structure in original (2) span configuration.
2) Build in new span to the left.
3) Remove the span on the right.

You are assuming that when you install the span on the left, that it has no forces in because it is joined to the already rotated end of the original span.

So theoretically, what would normally be a straight and continuous joint becomes a 'kinked' joint that is still fully fixed. And if you say unload the structure after making that joint you would have stresses and deformations 'locked in'. If you twist the end of the new span to match the rotation of the deflected beam then it would bounce back to neutral after you unload it no locking in.

In any case no matter what the structure has locked in it, external and internal equilibrium has to be satisfied.

Brad, I didnt use a drawing because I was afraid it would tie the discussion to the specific case.
The real situation I am analyzing, btw, looks very little like the one I showed. It is a slab, a hole of which will be demolished after other regions are reinforced. Actually, after the initial reinforcement there will be 3 demolishing+rebuilding stages. So, I hope all we have discussed really is a general principle.

The building order and your account of my assumption are correct, drift.

lets label the four supports ... A, B, C, D (for the LH).

so 1) has B, C, D effective. It could have an unloaded overhang to A without any change.

and you're trying to get to 2) (A, B, C loaded) by saying ...
1) starting with 1) ... with PB1, PC1, PD1 and MC1, then
2) apply -PD to 2) (with the RH overhang) and get PA2, PB2, PC2, and MB2 and MC2, then
3) add together ... PA = PA2, PB = PB1+PB2, PC = PC1+PC2, PD = PD-PD = 0, MB = MB2, MC = MC1+MC2 = 0

I see the logic, but I don't think so. It is intriguing that MC2 = -MC1.

I think if you want to go from 1) (BCD) to 2) (ABC) you need to go through 4) (ABCD) and superimpose 2) with -PD.

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

If we could model the joint with a predefined rotation then we could solve your problem quite nicely, not sure if this feature is available....

The new span will have the same rotational stiffness so the final structure will displace proportionally and elastically as it should, but we have already locked a rotation into the joint which would tend to increase deflections in the final condition. [img https://res.cloudinary.com/engineering-com/image/upload/v1659631029/tips/STAGED-1_ew5fh2_bolxx4.png]

Drift, that would be easy to model. Just use a fixed instead of pinned support. But it would just be unfeasible to build.

ok, good point ... but what are you building that you need to jump through these hoops ?

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

If that is hard to build and easy to model, then I'm lost as to what your thinking is easy to build and what is hard to model about this problem.

You agreed when I said

Quote (Driftlimiter)

You are assuming that when you install the span on the left, that it has no forces in because it is joined to the already rotated end of the original span.

If you model that joint as a fixed then this entire problem is moot. I feel as if we are talking around and around here not getting any closer to a solution because the goalposts keep moving.

I wonder if we should be using the 3 moment equation, modelling the beam as interacting spans ?

1) 2 spans, BC and CD; with load P
2) 2 spans, AB and BC; with load P
3) 3 spans, AB, BC, and CD; with load P
4) 2 spans, AB and BC; with load -D3 (the negative load from support D of the three span beam)
3) + 4) = 2) ... maybe !?

I think you're taking the negative of load D from 1) and applying it to ABC, then adding this to 1) to get 2). And I don't think that's right.

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

rb, It's a 35m diameter 2m thick slab at the bottom of a subway access well 33m deep.
There is already a 40cm slab in place but it's flat and part of final slab will be built below the existing slab and part over it (yeah, the 2m slab has a lower level some 2,5m below and it is connected to the upper level).
We can't simply remove the existing slab because it's actually working as a 2D strut for the concrete walls around it.
So the solution is, build the upper part of the 2m slab on top of the 40cm slab, then open a hole on the thin slab, build the 2m slab region below that hole. Rinse and repeat 3x because it would be too big of a hole to break on our 40cm strut slab.

drift, I guess steveh49 already solved the problem.
I created an example based on his ideas at 4 Aug 22 11:50.

ok, so now we have the problem ... how to extend a slab. I'll leave that to you Civils, but why not assume two independent slabs (a pin joint) ?

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

rb, the slabs are not in different spans. On a vertical projection, they occupy the same area, a full circle. The final shape is partly above and partly below the existing one.

Drift, at least for me, solving the physical problem was not easy. So much so that I came here looking for answers, and even after some discussion, someone else solved it. Apparently, it wasn't easy for most people.

To prevent the rotation of a structure node in real life and then build something around that node does not seem easy for me. Especially when compared to letting the node free to rotate.

It sounds to me like you're reinforcing an existing slab.

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

Rb yes, that's a reasonable description.
In that spirit, parts of the reinforcement would be on top of the existing slab and parts of the reinforcement will be under it.
So, this is how it goes:

1) Existing slab is already loaded
2) Upper reinforcement is built
3) Part of the existing slab is demolished in order to build the lower reinforcement
4) Lower reinforcement is built

That is my present version of the "load, remove and then add" problem which motivated my query.

What is the structure you are concerned with regarding the load sequence? The old slab? The new slab? If it’s reinforced concrete it should be ductile and this level of fuss over exactly which member is doing exactly what may not be warranted…

You’re chopping out a third of the circle at a time and replacing with a thicker slab? Is that what you mean by a hole?

How does the 400mm slab work? How is it supported? Where are these walls that it is bracing?

Tomfh, I'm worried about both. I'm chopping a third of the hole at a time. The hole is a rectangle of some 5,5x25,5m that starts at the perimeter and runs through the center.
The existing slab is fully supported by the soil below. The walls are at the perimeter of the circle.

If the slab is fully supported on soil then what load is it seeing?

Presumably the new slab will also be supported on the same soil?

And the hole is only a small portion of the original slab? How thick is the new slab outside of the hole area?

Maybe draw a sketch of what going on. It’s a bit hard to understand.

The slab is actually a strut for the walls. The upper level of the new slab will be on top of the existing one. The lower, under it.
The hole has around 240m² in a total of some 1000m². Outside the whole, the new slab is 2m thick.

please stop and send a drwg. we were talking about slabs which I think we thought of as round, ok, it's rectangular ... oh and now has a hole in it.

for the life of me I don't understand why you can't put all the reinforcement on top, or why you can't disassemble the existing slab piecemeal and replace it (piecemeal) with the new one ?

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

Yes I’m really struggling to understand it myself.

I feel we’re missing some important information. A slab increasing in thickness from 400mm to 2000mm? What is really going on here?

Folks, unfortunartely I'm not at liberty to release more detailed data. I was really just trying to discuss general principles, not one specific structure.

then I guess we've been as helpful as we can be.

We don't need the project name (nor particularly location, unless that is relevant).

If you can't show us the specific design, we can't talk (sensibily) in terms of generalisations. How relevant was all that discussion on beams (to reinforcing a slab) ?
Sure we can think it quite the change (from 400mm thick to 2000mm) but you can say, "yeah, that's what we want to do". Of course this can raise questions in our minds and we could say "there are things here we don't understand and don't want to mislead people (into doing dangerous things), so pass."

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

Rb, the problem had already been fully understood and solved by steveh49 a couple of days ago. I didn't quite understand what was everybody else's goal on the topic after this. He answered my question, but I'd be glad to continue discussing the issue if that's what you'd like to do.

As I stated on my messages, I was just addressing the absolutely general situation of how to solve any structure undergoing the situation "load, add part and then remove part". Yes, it was originally motivated by the slab situation but at no point I asked about the slab itself. Steveh49 showed how to approach this using a simple beam structure, but the principles he stated are universal and therefore can be applied to slabs, shells or whatever else. I just made a quick passing mention about the slabs to Brad because he insisted on some sketches and I pointed out that they were irrelevant since I was after general solutions, not solving my specific structure.

At the time, I said, "Brad, I didnt use a drawing because I was afraid it would tie the discussion to the specific case." And apparently I was just exactly right. After that comment, everybody started to talk just about slabs while my query was about structures in general.

Really ? The last thing SteveH contributed was "Yeah, maybe not a good example either. Too trivial."

Since then we discovered you were talking about a slab, which I thought completely changed the subject.

But if you're happy, I'm not dissatisfied.

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

Rb, yes... that's his last message. After he laid out the general concept and created a simple example.
As I said, I was never talking specifically about slabs or a slab. The whole time I was asking about a general problem, but suddenly everybody else changed the subject. I guess that's because most people dont want to deal with really universal principles since they are evidently harder.

you said it was a slab ... "rb, It's a 35m diameter 2m thick slab at the bottom of a subway access well 33m deep." (17:29 on the 4th Aug)

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

Yes, rb... I did say it was a slab. In passing. But I wasn't discussing the slabs, even in that message.
I mentioned it for the sole reason that... you asked me to! ("what are you building that you need to jump through these hoops ?" 4 Aug 22 16:58)