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Bolt Hole Clearance 1

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kilbchoi

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Jun 18, 2001
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Hi,

Anybody who let me know hole clearance to allow all
bolts transfer shear load?

For instance regarding M6 (in metric).

Thanks,

K. Choi
 
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Hi K. Choi,

there is no simple answer to your question.

First you have to determine the type of fasterner in the joint and the type of loading on your joint. Is it a bolt, a HILOK, a lokbolt, a rivet or even a srew (only used in non flight critical structure with no threads in bearing !). Is it an ultimate case per FAR 25.561, is it a fatigue case, is it an attachedment for flight critical items, is it just a secondary strucutres joint. All these criteria will require different type of hole size for the same size of fastener. A/C companies might have different requirements for the same application !

Gunther
 
kilbchoi: Here's my summary and interpretation of several documents regarding bolt holes (not Hi-Lok nor rivet holes). Also consult your design code or company design manual, if one exists.

(1) Tight fit. A so-called "tight fit" bolt hole clearance is for high shear applications in primary, major frame member connections. For typical aerospace steel bolts (standard or close tolerance) having bolt ultimate shear strength 620 MPa or greater, in aluminum plates roughly as thick as the bolt nominal diameter or thinner, the following maximum bolt hole diameter is adequate to transmit high shear load in primary, major frame member connections.

D[sub]n[/sub] +tol[sub]a[/sub]/-0.01 mm,

tol[sub]a[/sub] = 0.118 D[sub]n[/sub]/(8.20+D[sub]n[/sub]) mm,

where D[sub]n[/sub] = bolt nominal diameter (mm), and tol[sub]a[/sub] = "tight fit" bolt hole positive diametral tolerance (mm). Round tol[sub]a[/sub] to nearest hundredth mm (0.0005 inch). The resulting "tight fit" bolt hole clearance, sometimes referred to as a tight "slip fit," is about the tightest bolt hole clearance that will still create a removable (noninterference) connection.

An example of the above equation is, 6.00 +0.05/-0.01 mm. Another example is, D[sub]n[/sub] = 0.2500 inch = 6.35 mm, thus tol[sub]a[/sub] = 0.0515 mm (0.00203 inch); round to nearest 0.0005 inch; thus, bolt hole diameter is 0.2500 +0.0020/-0.0005 inch.

Above "tight fit" equation assumes (a) no threads in shear plane, and (b) that you are using a factor of safety in your analysis of 1.50 or greater and a joint fitting factor of 1.15. Domain for above equation is D[sub]n[/sub] = 3 to 36 mm.

(2) Close fit. A so-called "close fit" bolt hole clearance is for medium shear or relatively high redundancy applications, and has the following bolt hole diameter, where D[sub]n[/sub] = bolt nominal diameter (mm).

D[sub]n[/sub] +tol[sub]b[/sub]/-0.00 mm,

tol[sub]b[/sub] = 0.0080 D[sub]n[/sub] + 0.050 mm.

(3) Clearance fit. A so-called "clearance fit" is for low shear, or medium shear and high redundancy, applications and has the following bolt hole diameter.

D +/-tol[sub]c[/sub] mm,

tol[sub]c[/sub] = 0.5(fac1)(1-fac2)(tol[sub]b[/sub]),

where D = D[sub]n[/sub] + [(tol[sub]c[/sub])(1+fac2)/(1-fac2)], fac1 = 2.0 (or some companies might say 2.5), fac2 = 0.5 (or some companies might say 0.6), and D[sub]n[/sub] = bolt nominal diameter (mm).

In conclusion, note that the LMC (least material condition) clearance is LMC bolt hole diameter minus LMC bolt shank diameter. Keep in mind, this LMC condition is something like a 2.58*sigma (2.58 times standard deviation) phenomenon; i.e., if we assume a normal distribution, there's probably about a 99% chance the actual clearance will fall within +/-2.58*sigma or 86% of the full LMC clearance value. Point being, the full LMC clearance value has a relatively low probability of occurrence.
 
This is a good article. Too bad that this type of data has such a hard time surviving at the shop floor/maintenance level. Most inspectors and technicians lack the training to understand this data, and hence do not understand why it is so important that a hole be drilled or reamed correctly.
 
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