Maintenance manuals of today's civil aviation aircraft (from both Boeing (e.g., B-737) and Airbus (e.g., A320)) have special chapter to describe "standard torque value" for fasteners and connections. These values are determined by material, size, etc.

I wonder where and how these values come from. Could anyone provide me with detail about this?


2 Answers 2


The standardized values are related to the yield strength and elasticity of the fastner. Bolts in normal tolerance holes are effectively spring clamps (bolts in close tolerance holes can be better thought of as spring clamps + shear pins) and when you are tightening a bolt you are using a spiral wedge, the threads, to force the bolt to stretch to clamp the parts together.

You want to be able to stretch the bolt to apply as much clamping force as possible without exceeding its yield, or permanent deformation threshold. It's like stretching a spring too far, so that when unloaded, it is longer than before. You would say that the spring is ruined if that happens. Same with a bolt. If you overtorque it and exceed its yield strength, it will be a bit longer than before when loosened and will have lost some of its ultimate strength after being permanently elongated.

So the standard torque tables are limits on how much the bolt can be stretched, converted to the equivalent rotational force applied to the spiral wedge of the threads. The values are those that keep the axial tension force below the yield threshold, with some safety margin. The standard limits are a function of the material strength and elasticity of the material that a particular bolt is made from.

A lot of the time torque values are given that differ from the standard values because the designer wanted to customize the clamping force for some other purpose, such as to limit clamping force on the base material or gasket or washer, or limit shear loads on parent material threads (like a steel bolt in an aluminum casting). The normal engineering practice is to quote a torque value on the engineering drawing (and hence in the Maintenance Manual procedure) when it differs from the standard tables.

So as a general rule, when a torque value is from the standard torque tables, it's a limit for the bolt itself and is related to its tensile yield limit. When a special torque value is given, it's a limit to control force applied to something other than the bolt itself for various other reasons.

  • $\begingroup$ This may be worth a separate question, but when the issue is the tensile yield limit of the bolt, is it a matter of what the shaft can withstand, or the threads? In standard fasteners, is the thread pitch and depth, and length of the nut, specified such that the shaft would fail before the thread gets stripped? $\endgroup$
    – sdenham
    Feb 17, 2019 at 23:39
  • 2
    $\begingroup$ Depends on how many threads are in the female element to distribute the shear loads on the bolt threads. A nut that is designed for tension loads will have enough threads distributing the shear force to have a total thread shear strength that is higher than the tension yield limit on the bolt. Aircraft nuts designed for tension are fairly deep. Nuts designed for shear applications only are shallow and are not torqued very high, like the nut on a bolt that acts as a shaft for a bellcrank. Yield shear strength of threads are usually limiting when a steel bolt goes in something soft like alum. $\endgroup$
    – John K
    Feb 18, 2019 at 0:54
  • $\begingroup$ +1 I want to add something about thread geometry - but I don't know where to edit it in. For the different geometries there might be some that seize (self weld) at some torque and you want to not exceed that. There are also tapered bolts which will have a torque based on the compressive strength of the material, not the tensile. $\endgroup$
    – Stian
    Feb 18, 2019 at 8:56

Fasteners have been standardized by the ISO. These standards cover e.g. bolt sizes, thread pitch, and material grades.

This ensures that bolts made in factory A fit nuts made in factory B, for all factories worldwide. It also ensures that when you buy a bolt, you can expect it to behave a certain way: a bolt of size X and grade Y will be capable of at least force Z.

ISO 898 defines material grades (called "property classes"):

For metric bolts strength is according ISO 898 Mechanical properties of fasteners made of carbon steel and alloy steel described by "property classes" with designations 4.6, 4.8, 5.8, 8.8, 9.8, 10.9 and 12.9.

This property class is usually embossed on the bolt head. Each property class has requirements for the amount of load a bolt can handle, etc.

The tightening torque tables are derived from these requirements: the torque in the table is typically calculated for bolts that are loaded to 75% of their tensile strength.

Who is responsible for making up these standard torque table?

ISO has set the material properties, but I haven't found a torque table as part of the ISO 898 standard.

There is an SAE standard torque table for aerospace applications: SAE AS 1310B-1996. Unfortunately none of the ~6 torque tables I looked at for this answer provided their origin. There may be more standards.

Standard torque tables make some assumptions. One of these is that the object being bolted has to be strong enough to take the applied force.

And this is where the standard torque table feeds into the design process. Normally you choose a bolt size and grade depending on the materials you want to fasten (their material properties and thickness). It's then up to the manufacturer to either follow the standard torque table (which you want, because it's the easiest on the mechanics that will work on the aircraft), or to specify a different torque for an individual bolt.

  • $\begingroup$ Who is responsible for making up these standard torque table? ISO , aircraft manufacturer, fasterner/connections manufacturer, or other ognizations? $\endgroup$
    – lyl
    Feb 18, 2019 at 1:51
  • $\begingroup$ I've amended my answer. $\endgroup$
    – Hobbes
    Feb 18, 2019 at 18:11
  • $\begingroup$ Many thanks for your info! $\endgroup$
    – lyl
    Feb 19, 2019 at 4:31

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