# How Much Should a Bolted Joint be Tightened?

April 3, 2023
Engineers often ignore external forces acting on the joint when specifying the amount of torque for installing the bolts. Here’s a way to avoid that.

This article was updated April 3, 2023. It was originally published June 7, 2016.

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Workers often tighten bolts to 75% to 80% of the bolt proof load. This works for many joints, but in some cases, external tensile loads reduce bolt clamping forces to zero. Something other than the 75% or 80% rule of thumb is needed.

The external tensile force needed to reduce clamping to zero (within the bolt’s and cramped parts’ elastic limits) is determined by:

(Fe/FPR)0 = (Fb/FPR)0

= ((1+r)/r)(Fi/FPR)

where r = Kp/Kb, Kp is the clamped part’s spring constant (lb/in), and Kb is the bolt’s spring constant (lb/in); Fe is the external load on the joint (lb), FPR is the bolt proof load (lb), Fb is bolt load (lb); and Fi is bolt preload (lb). The subscript “0” identifies a value at zero clamping load. This equation can be used to calculate the maximum force that can be applied to a specific joint without the parts separating.

For common values of r, the force can determined from the graph below, which is based on the equation.

The bolt preload that produces a given non-zero clamping force can be determined from the above graph in combination with the equation:

Fe/FPR = (1-b)(Fe/FPR)0

Where b =|Fp/Fi|

The corresponding bolt load can be computed from:

Fb/FPR = (Fe/FPR)0Fi/FPR)(b/r)

Here’s an example:

For a bolted joint where r is 10 and Fi/FPR is 0.75, the graph gives:

(Fe/FPR)0 = (Fb/FPR)0

For a clamping force of 25% of the preload (b = 0.25):

(Fe/FPR) = (1-0.25)(0.825)

=0.619

And then:

(Fb/FPR) = 0.825 - (0.75)(0.5)/10

= 0.806

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