### Fastener Bolt Thermal Expansion Contraction Equation and Calculator

**Fastener and Thread Design Menu**

ISO Hardware Engineering Data

ANSI Hardware Engineering Data

ISO Hardware Engineering Data

ANSI Hardware Engineering Data

**Mechanics / Strength of Materials Menu**

Fastener / bolt tension may be achieved by heating a fastener ( bolt ) and nut such that the bolt expands an amount allowing for axial tension to occur when the installed fastener / bolt and nut cools.

The temperature necessary to create the axial stress or force, *F _{t }*, (when the stress is below the elastic limit) can be found with the following equation:

or

**F _{t} = (T - T_{o}) E e **

Where:

*T* = Temperature given in:
degrees Fahrenheit

*F _{t} *= Axial Tension or Force required
given in: psi (pound per square inch)

*e*= Coefficient of linear expansion given in: in./in.-F

E = Youngs Modulus of the fastener given in: psi (pound per square inch)

*T*= Temperature which the fastener is cooled to is given in: Degrees Fahrenheit

_{o }T - T

_{o}= change of temperature realized by fastener.

In finite-element analysis, thermal change of the fasteners may require consideration to determine preload mesh elements in tension or compression. The above equation may be used to determine required temperature changes in such instances.

In practice, the fastener is heated above the required temperature, to compensate for the cooling occurring while the nut is being installed, and the nut is tightened or a pre-torque is applied. The axial tension will develop when the fastener and nut cools.

Another technique to consider, the nut is tightened snugly on the bolt and the bolt is heated in place. When the bolt has elongated sufficiently, as indicated by inserting a feeler gage between the nut and the bearing surface, the nut is tightened or torqued. The bolt develops the required tension as it cools; however, preload may be lost if the joint temperature increases appreciably while the bolt is being heated.

Example:

An axial force
resulting in a tensile force of 32,000 psi is required for
your application at 85F. Known: E = 36.36 x 10^{6},
and e = 6.0 x 10^{-6 }in./in. - F

To determine the required temperature, you substitute the known values into the equation given above:

T = 32,000 / [ (36.36

x 10^{6})(6.0 x 10^{-6 }in./in.)]
+ 85 = **231.681 °****F**