On a nuclear submarine the bolts holding the turbine engine in place are to be tightened to a certain tension. The maximum permitted tension on each bolt is, let’s say, X ksi. However, we know that under certain situations (dives, turbulence, bounce) the turbine mass will potentially add additional tension on each bolt holding it in place, lets say, Y ksi. What should the bolts be tightened to (ksi tension) in terms of X and Y?
Nearly able to help you …
We had an ex nuclear sub tech guy working here until last week … he just quit to do research on nuclear fusion laser something … sorry
I’m not sure if this is any different a question really than what tightness a crankshaft main bearing cap should be tightened to, since it will undergo a huge cyclic stress.
The maximum permitted tension is not normally near the yield strength of the bolt, of course. I would expect that it would be much less than half, as that would seriously impact the Factor of Safety of the bolt.
This is sorta a complicated question. You need to factor in what force is needed to properly hold the piece under the static load, and what is the maximum dynamic load the piece will undergo. I don’t know a good way to answer your question in an abstract sense - but then maybe I’m not so good at abstract problems either. 
told me that the bolt/nut torque values are a product of the hardness of the bolt, the size, and the thread count.
later, Tom.
Here we get into the discussion of bolt preload…
The preload torque on a bolt is a product of the bolt diameter, the friction coefficient of the threads, and the axial preload FORCE on the bolt.
To design a bolted connection properly, the bolts must be properly preloaded so that the bolted members handle the static and dynamic external loads of the connection. In most cases the members take about 80% of the load. If the bolts were to take the brunt of the load, the connection would be downrated considerably.
As a rule of thumb the bolts are usually tightened to somewhere around 75% to 90% of their Proof Strength. The Proof Strength of a bolt is lower than the Yield Strength and is dependent on the strength grade of the bolt. There are ANSI and SAE standards that determine mfg. methods and designs for bolts to achieve certain strength grades in bolts.
To address the original question, the submarine bolt tightening specification is designated taking into account the maximum external loads on the bolted connection, the number of bolts, the strength grade of the bolts, and the proper preload to place the majority of the external load on the bolted members and not on the bolts themselves.
I may have left something out, as I always do, so any mechanical engineers that want to rip me a new one, proceed but take out the KY please!
Well, I am a mechanical engineer, but designing bolted joints is not really my bag. To design them right is actually somewhat more difficult than you might think. That being said, I can probably add a few things. First, to do it right, you need a LOT more information than the OP provided. For example:
[ul][li]What size are the bolts?[/li][li]What type are the bolts?[/li][li]How “dynamic” is this Y parameter (i.e., is it high-cycle fatigue?)[/li][li]Clarify what Y is: is it actually additional tension on the bolt, or is it additional tension on the joint?[/li][li]What safety factor is required?[/li][li]Do all bolts on this turbine see the same loading?[/li][li]How stiff is the bolt?[/li][li]How stiff is the joint (meaning the plates or whatnot that the bolt sandwiches)?[/ul][/li]Just for fun, though, I’ll assume that X is the maximum permitted load, factoring in fatigue and safety factor and differences among individual bolts. I’ll also assume that the joint is 10 times as stiff as the bolt (probably not a bad guess). I’ll also assume that Y is tension on the JOINT (not the BOLT), and that X>Y. The formula for load on the bolt is:
Load = Preload + Y/(1 + K[sub]j[/sub]/K[sub]b[/sub])
where
K[sub]j[/sub] = stiffness of the joint
K[sub]b[/sub] = stiffness of the bolt
so
Load = Preload + Y/(1 + 10) = Preload + Y*.0909
So the Preload must be less than (X - .0909Y)
Also, to keep the joint from separating, the Preload must be greater than (1-.0909)Y:
0.909Y < Preload < (X - 0.0909Y)
I imagine anywhere in this range would be adequate; the exact preload would depend on which factors of safety you would like to maximize.
Standard disclaimer, rejecting responsibility should you actually use these calculations.
zut–I’m a mechanical engineer too, but I was not sure that equations were what DeutschFox was looking for.
But I agree with your post.
It’s not ksi, it’s psi or kg/sm2 (kg/sm squared).
peace–What do you mean? Depends on whether you’re talking about force, stress, or torque how you designate preload.
Also, ksi = psi*1000
What unit is kg/m^2 ??? Is that a Navy spec.?
jeel: Your pardon, I misinterpreted your last statement. However, I’d like to add that I’m not sure that those equations were what DeutschFox was looking for, either. I just took a stab at answering something. I suppose if the OP cares to, he can clarify the question.
peace: ksi is short for kilopounds per square inch. You’ll often see stress in units of ksi; e.g., a yield stress of 30 ksi (versus 30,000 psi, which is the same thing). It’s just a bit of shorthand engineering jargon. Somewhat less frequently, you’ll see forces expressed in kips (kilopounds).
I’ll back up jeel’s statement and also state that part of my job for 5 1/2 years was doing stress analysis on bolted connections on aerospace structures. At the company I worked for, the “rule-of-thumb” was to tighten bolts to 90% of yield strength (yeah, for the materials we used we assumed it to be the same as proof strength) for a metal to metal joint. If there was some sort of elastomer in between–for example, a gasket–the recommended maximum torque was 75% of bolt yield strength (it may have been as low as 60%, now that I think about it) to account for creep effects. Another rule of thumb, one we were pretty adamant about, was that the preload should be at least twice the peak vibration load. I will note that different companies and different industries have different rules of thumb.
Of course, you want to make sure that the applied loads (due to pressure, external force, thermal expansion, etc.) combined with the preload do not yield or otherwise fail the bolt. The formulas for determing the total load on a bolt are too ugly to present here, but if you can get a hold of it, check out chapter 8 (especially section 8-5) of Mechanical Engineering Design by Shigley and Mischke, Fifth Edition, for more of the nasty details. When I was analyzing components for the Space Station, I had to use formulas provided by NASA (I can’t remember the document number). They were similar to those found in Shigley, with thermal effects added.
Peace, ksi is a valid unit of pressure. It’s primarily used in stress analysis. 1 ksi = 1000 psi = 1 kip/in[sup]2[/sup]. “Kip” is short for “kilopound.”
And zut just has to go and post while I’m writing. 
jeel: I suspect peace meant Newtons per square meter (Pascals, of course). Kilograms per square meter is an awfully odd unit. Now that I think about it, though, I note that the OP cites “ksi tension”, which is, of course, a misapplication of units, and should be “ksi of stress” or “kips of tension”.
Strainger: 1)Nyah nyah. 2)For my own education, is the 90% YS rule-of-thumb irrespective of the dynamic/static component of the load? The reference book I had implied that bolts which are primarily stressed dynamically (i.e. fatigue loaded) should be preloaded a lot less. This confused me, since I know that engine head bolts, for instance, are often torque-to-yield.
The simple answer I think DeutschFox is looking for is:
The specified torque will provide a clamping load that is calculated by the design engineer (I’m an ME too, btw) to take all of the expected loading conditions into consideration. No added torque is required. In fact, the bolt is likely to be more highly loaded while it’s being torqued than at any operating condition - if it doesn’t break then, it probably never will. Care has to be taken to KEEP the bolt properly tensioned, of course.
Loads in the direction that tend to separate a bolted joint will act mostly to reduce clamping load, not increase tension, until the joint actually separates. If your G loading is that high on your sub, you have a helluva lot of other problems to worry about, too.
Yes, I know this is the “Trust me” argument, and you don’t have to accept it. But I KNOW you don’t want to plow through the math, either - it’s as ugly as Strainger says.
Maybe peace meant kg[sub]f[/sub]/m[sup]2[/sup].
zut, regarding dynamic loading, you want to maximize your preload to minimize the possiblity of jiggling the bolt loose, to put it in layman’s terms (joint separation, if you prefer). So yes, that 90% number* is irrespective of the loading type. For the most part, external loading doesn’t cause any problems regarding yielding or fatigue failure since approximately 80% of the external load is carried by the joint members (only 20% is carried by the bolt) in most cases. Very rarely, however, you run into a case where you have to back off on the torque value to keep the total load from failing the bolt. Sometimes, getting the torque value high enough to where the joint doesn’t separate under external load and low enough to where the bolt doesn’t fail can be quite a balancing act. The stress engineer will often end up recommending a design change, such as larger bolts or more bolts, if he can’t find an adequate torque value.
And yes, jeel is right in that these numbers should be based on proof strength, not yield strength. Boy, is my face red. Unless you have a table that tells you otherwise, a good conservative assumption is that proof strength is 85% of yield strength.
Apologies to the OP since I’m probably making this more complex than he wanted it to be. I enjoy this sort of thing too much, in a sado-masochistic kind of way.
[sup][sub]*90% of proof strength for permanent installation, 75% of proof strength for reused connections, and 60% of proof strength for joints with elastomeric materials.[/sub][/sup]
Me too Strainger…I think that’s part of what made us become engineers in the first place
Funny - you’re the first 2 ME’s I’ve ever encountered who didn’t think bolt load calculations were a pain in the tuchis.
Guess I should have been more specific there, Elvis. I enjoy talking about and researching bolted joint analysis. But as far as performing the calculations, I prefer to just stick the equations in TKSolver or Excel, although doing a bolted joint analysis in ANSYS is kind of fun.
I ain’t an engineer, just noticed something strange… My apologies for ksi: I mistook it for kilograms per square inch. Strictly speaking, pounds (E.,>L. origin) should keep their own company and not mix with Greek kilo (chilo)*. There should be no kilopounds or decigallons. Go completely metric if you like the convenience of preffixes. Kg/m^2= kilograms per square meter, for unit of pressure in metric (it’s different in SI)
- Purists say that roots from different languages are illegitimate. Therefore, no kilobits
Seem to recall that there is no affect on a bolt in tension, until the external loading (maximum of static and dynamic) placed on it exceeds the internal bolt tension loading. This could result in either loosening or deformation, depending on the tension of the bolt. If this is true, and I’m not sure right now, there is no reason not to tighten the bolt such that it is loaded/torqued to approximately its material yield strength (maybe ~90% considering other factors) insuring “tightness”, why not? … I am trying to find an experiment I once saw demonstrating this … maybe I could get some help on this …