To determine the performance of a bolted joint it is necessary to calculate the joint stiffness.  That is the deflection of the joint under a bolt loading condition When geometry of the bolted joint is an annulus with an OD less than 2,5 x the bolt diameter the joint stiffness can be conveniently calculated using k = EA/l .  When the joint includes a non-metallic gasket or material with a low modulus of elasticity then this item may have such a low stiffness value that the stiffness of the metal parts of the connection will have very low impact on the overall stiffness.


It is very difficult to obtain theoretical values of joint stiffness because a number of factors affect the result including.

Any of these factors can have a significant effect on the resulting joint stiffness

The notes below identify methods which approximate the loading regime which has been found to occur by various studies using different methods.

A bolted joint can include a number of separate parts.  The relationship between the individual stiffness values of the parts and the total stiffness of the clamped joint is as shown below..

The sketch below shows a bolted surface.  It has been shown using using ultrasonics and FEA that the pressure in a bolted surface is greatest under the bolt head and reduces as the distance from the bolt interface increases.  A method of be approximating the force distribution is based on use of the cone geometry as shown.  Various cone half-apex angles are used but for systems using a washer-face annulus and with hardened steel, cast iron or aluminium an angle of 25^o to 35^o considered reasonable.   These notes relate to half apex angle of 30^o

Referring to the contraction of a small element dx of cone subject to a load P as shown.

The area is determined as follows

Substituting this into the equation for contraction of an element..

To obtain the total deflection the equation above is integrated between 0 and t.

Using integration manipulation techniques this results in

The spring rate of the truncated cone is therefore ..

The bolt hole diameter is d, the diameter of the applied pressure e.g.the washer diameter, is and the plate thickness is t.

For a half apex angle of 30^o which provides a reasonable value for many engineering metals, the equation simplifies to..

Important Note: Widely different stiffness values result from different studies.  It has also been proved that different loading conditions and surface conditions also affect the resulting stiffness.

The variation in half apex angle α can be notice from the different studies completed.  Differences result from various reason including D/d ratios differences and surface finish difference

Showing stiffness values for steel plate E = 207 kN/mm² and a half apex angle α of 30 ^o calculated using equation X above..

d= bolt dia and D = 1,5 d = approximately as would result from a typical bolt or cap screw .

Showing stiffness values for steel plate E = 207 kN/mm² and a half apex angle α of 60 ^o calculated using Equation X above.  This table is provided to identify the variation is stiffness using two different, but still relevant, values of α

d= bolt dia and D = 1,5 d = approx as would result from a typical bolt or cap screw .

The force distribution on a typical bolted joint comprising two plates is shown below:

It is necessary to obtain a value for the stiffness for each part of the joint.  In the case above for each of the two plates.

If the plates are of the same material and similar thickness then the overall stiffness of the joint is..

FEA work has been completed (Wileman ,Choudury & Green)for similar 2 part bolted joint with typical washers (D_w = 1,5d) and joint members of the same material and an approximate curve fit curve for the resulting relationship has been derived..

Values for A and B are provided in the following table...

Consider two plates 20mm thick clamped with a 10m bolt using the table with a half apex angle α of 30 ^o

k for each plate from Table A above = 3 503 kN/mm .... 1/ k _t = 1/k + 1/k .
Therefore k _t = k/2 = 1 7513 kN/mm.

Using the formula above with A = 0,78715,B = 0,62873,d = 10, l=40 and E = 207 OOO
k _t = E.d.A exp (Bd/l) = 1 906 kN/mm