Concept of a typical pin joint

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Design of bolted, pinned or riveted members

Once one has found the shear load on the most highly loaded bolt, pin or rivet, one then has to find the diameter d of this and then suitable dimension for the members being connected.

 

1. Bolt or pin diameter: assume that it fails in shear (singleor double) through its diameter. Double shear is shown below. This involvesshear across the two areas A shown

 

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Failure occurs when τ= Sy/2n where Syis the shear strength of the pin or bolt and n is the factor of safety. Solve for d and round up to next preferred size. For bolts one should use the shank area, not the thread area At.For a short pin such as that shown above, the bending stress is not normally the cause of failure. Also, the equation σ= Mc/I is not valid for such a short pin

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2Tensile failure of the member. This may be either simple tension as shown below or bending

 

 

Where Sy is the shear strength of the member

 

 

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When the tensile stress is due to bending use:

σ= F/A + Mc/I being sure to get the correct bending moment M and moment of inertia I about the centroid of the critical cross section

 

3. Bearing stress by the pin on the member. This is calculated using the projected area of the member surface that is loaded by the pin. This is shown below as A = dt. In the example shown there are two such areas so the bearing stress is given by :

σc= -F/Atotal= -F/2dt and is a compressive stress

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Failure occurs when σc= -Syc/nwhere Sycis the compressive yield strength of the member material. This is usually equal to the tensile yield strength for ductile materials

 

 

 

Note that there will be an equal and opposite force on the pin or bolt. In most cases the yield strength of the pin or bolt or pin is greater than that of the member so this is not an issue

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4Tear out failure. An example if this is shown below. The bestway to avoid this is to ensure that the distance from the center of the pin or bolt is at least equal to 1.5 times the bolt diameter

 

 

All of the equations given here are quite approximate and should be used with large factors of safety (at least 2.0). Also, all of these equations are for static loads only and should only be used with ductile materials.

If more accurate results are needed for very critical joints, one has to resort to FEA. This gets quite complex as one has to solve the contact stress problem