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I am affixing a servo motor with a machined enclosure to a base plate with drilled holes. The current design uses eight M-2.5 bolts in a figure U pattern (3-2-3) around the enclosure.

Tightening that many bolts/screws takes time, and the small gauge makes them finickier than larger-gauge bolts (missed nut thread starts, etc.)

If I were to change this design to use fewer, bigger, bolts, what would be the pros/cons of this? Is the amount of position slop significantly different between, say, three M-6 bolts, and eight M-2.5 bolts? Is there a formula for the affixing force that N bolts of size S will provide?

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It's really not as simple as a rule of thumb - there are a lot of factors in each application. I'm going to assume that your bolt application is a fairly traditional situation where you are bolting one piece of material to another (one shear plane) not a more complex sandwich (Isolation pads, transition plates, etc.)

In most bolted connections, the bolts are intended to provide a clamping force normal to the faying surfaces to allow a large friction force to develop between the two materials being bolted. As such, while we pretty much always check that the bolts can hold the load in shear, for design of the connection for performance, the clamping action is a greater consideration. If your faying surfaces are very flat and clean, and your two materials are very stiff, you can imagine that a single, large bolt would suffice for any problem as the clamping force would apply equal friction across the entire faying surface. One problem with using a single bolt is that if the joint does slip, it could slip in a direction that loosens the nut against the bolt, leading to a catastrophic failure.

In reality, usually our two surfaces are somewhat flexible, dirty, and not flat. Because of this, a bolt only successfully applies a clamping force for a small area around itself, so joints that resist a moment (like most motor mounts) will not be very effective with a single bolt. Instead, adding more bolts, farther apart from each other, creates 'moment couples' where because of the distance between each bolt, the actual slip resistance required at each bolt is less. In general, for connections resisting a moment, you want to maximize the overall size of the bolt pattern within reason.

There are, of course, a bunch of other factors. As you suggest, since the absolute tolerance is larger on larger bolts, they do generally require more sloppy holes, meaning they won't inherently provide alignment that is as good as smaller bolts. However, if you align your components independently (by measuring or with a jig,) and tighten the bolts, you can still keep the component in the right place just as well. Conversely, because holes for smaller bolts generally are less oversized, aligning a pattern of many small bolts requires much more precise machining of your parts than aligning a couple of larger bolts. This is because of the smaller oversizing factor primarily, but is compounded by the fact that the more holes you have, the more likely your worst case is to happen (where two holes are just within tolerance in the opposite directions of each other.) Of course, you could drill unusually oversized holes for smaller bolts, but you would find that the head (and/or washer) would not have very much contact with the base material.

As far as cost, for modestly sized parts the costs of machining the parts almost certainly costs more than the cost of the fasteners themselves, so a few larger bolts would be a better option - slightly more expensive bolts, but fewer holes to drill. The size of a hole to drill has much less impact on cost than the time to locate a new hole, especially if it deep enough to require multiple steps (like a spotting drill or center drill) and therefore a tool change. In addition, depending on your scale, materials, and thickness, sometimes smaller holes are actually more expensive as they have to be drilled less aggressively to avoid tool breakage. Two big exceptions to this statement would be if your pieces are being mass produced by casting, injection molding, or a similar volumetric process, or if they are being cut by a profiling process like waterjet or laser cutting, where linear inches are the main driver of cost. AS you point out, the time to assemble the device is mostly governed by the number of bolts rather than their size - for a given length of thread - a large bolt is actually faster to tighten. So this also favors fewer, larger bolts.

As for a formula governing the clamping force, it's nothing too special. Once you establish the pretension on each bolt as installed, you simply multiply that by the static coefficient of friction for your faying surface combination. The hard part is establishing the pretension that you will accomplish in each bolt - there are formulas that will give you tension as a function of torque, lead angle, and materials, but they are known not to be very accurate. The best way to find this value would be by direct measurement after tightening the bolts using the same method you will use in production (torque, feel, turn-of-the-nut, etc.)

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    $\begingroup$ A great answer. Thank you! For me, it turns out that the machining cost difference is zero, because I run the machning myself on a CNC mill. However, the mounting time of getting eight M-2.5 bolts, nuts, and lock washers aligned is a lot more than three M-6 (and the M-2.5 are small enough that they are more annoying per bolt than M-6, too!) $\endgroup$
    – Jon Watte
    Jun 8 '15 at 18:11
  • $\begingroup$ That makes sense. At that point the machining cost is just your time and wear on the machine and cutters. $\endgroup$
    – Ethan48
    Jun 8 '15 at 22:36
  • $\begingroup$ Any idea why female vs male faster heads (or vice versa) are used? Many times I've seen male hex heads, but female torx heads. Is the gender specifically cosmetic or is one "better"? $\endgroup$ Mar 21 '20 at 8:09
  • $\begingroup$ @TheMuffinMan well for the same head size and depth, male hex heads will generally let you transmit more torque. They are also cheaper to manufacture, so they are usually the simplest default if there's not a compelling reason to go with an internal form. Socket head, Torx, and similar are usually used when you need the head to be smaller in diameter, more cosmetic, or easier to clean. $\endgroup$
    – Ethan48
    Mar 21 '20 at 22:47
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There are a few principal advantages to having more bolts.

The first is that the loads are more evenly distributed, especially when the rigidity of the fixture itself is a bit marginal and when it is important to ensure that no separation occurs eg in flange joints of high pressure fluid systems.

Secondly having more bolts for the same nominal load means smaller hole diameter and so small flange diameter which may be useful when you need to package things as tightly as possible (eg in automotive engine applications.)

Thirdly more fasteners can improve redundancy ie if you have 4 bolts and 1 is below spec or improperly assembled then you lose 25% of the design strength if you have 10 bolts and one is wrong then you only lose 10%.

The other side of the coin is that using a lot of small fasteners to support a load much larger than their individual capacity can cause cascade failures if you get unforeseen loading conditions and you can get joints 'unzipping'.

On the other hand there may be instances where a large number of fasteners complicate assembly and maintenance, especially if access is limited and it is likely that the the fasteners might corrode or otherwise jam or seize. Similarly smaller diameter fasteners may have a smaller acceptable torque range than larger ones.

It's also worth bearing in mind that the best practice in designing bolted fixtures is for the bolts to work by clamping two surfaces together so that shear forces are resisted by friction between the mating surfaces rather than being carried directly by the bolts. Similarly bolts usually require moderate clearance in no threaded holes and so aren't suitable as the sole means of providing accurate alignment between two parts. Where this is required it's usual to have something like a stud or notch arrangement to provide a positive means of alignment.

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  • $\begingroup$ Thanks for the additional answer, which more clearly spells out some things only briefly touched on in the accepted answer. $\endgroup$
    – Jon Watte
    Jan 17 '16 at 5:33
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In the machinery industry the holo-krome bolt selectors are used regularly to get tightening torque and resultant bolt tensions: the green one is for inch and blue is for metric: https://www.google.com/#q=holo-krome+screw+selector+cards

If feasible, the miniature sizes such as 2.5mm are avoided. Simplified, bolt strength is proportional to cross section area, or PI*R^2. In comparing a 6mm to a 2.5mm, the correspond radii are 3 and 1.25, The corresponding strength ratios are approximately 3^2 and 1.25^2, or 9 to 1.56, or a ratio of 5.8.

The bigger screws may need blue locktite to keep from becoming loose due to vibration: the length of the screw under tension generally needs to be 4x the diameter to be considered vibration resistant.

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  • $\begingroup$ Well, I don't always have an option on the M2.5, because that's what the part I'm integrating with use :-( One side question for me is whether I in this case design an adapting fixture that screws in (and red-loctites) the 2.5MM pattern, and then adapts to my own parts using M6, or whether I just make my own parts in M2.5 without the adapter. The answer to this question, will inform the answer to that question, too. $\endgroup$
    – Jon Watte
    Jan 18 '16 at 23:32
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Its better always to use more small bolts than few bigger ones because if some of them fail, its better to have more of them.

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    $\begingroup$ I'm not sure that's universally true. For example, one M10 bolt is much stronger than two M1 bolts. The point of failure of the bigger bolt is a lot higher, which has to be multiplied out by the count. There is also the assembly cost -- a solution that solves the problem at a lower cost is better. There's some kind of formula or relation that tells me which is better in any particular combination. That's one kind of formula that I'm after with this question. $\endgroup$
    – Jon Watte
    Jun 7 '15 at 16:22
  • $\begingroup$ This can't be true, since it argues for a infinite number of tiny bolts. I'm not a mechanical engineer, but clearly there are tradeoffs you haven't considered. $\endgroup$ Jun 8 '15 at 16:59

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