Required rivet count against shear and bearing

What does this quantity describe?

From the classical allowable stresses τa,allow ≈ 0.6·Rm/S for shear and σl,allow ≈ 1.5·Rm/S for bearing (Rm tensile strength, S safety factor), two independent minimum rivet counts follow: na = F/(τa,allow·A0) from shear failure and nl = F/(σl,allow·d0·tmin) from bearing failure. Both are rounded up to a whole number; the rivet count actually chosen is the larger of the two.

This is like sizing a component that can fail in two different ways: the required design is calculated separately for each failure mode, then whichever imposes the stricter requirement is chosen — similar to a beam checked against both bending strength and deflection and sized by the stricter criterion.

Formula and variables

na = F·S/(0.6·Rm·A0); nl = F·S/(1.5·Rm·d0·tmin)

• na = F·S/(0.6·Rm·A0)
• nl = F·S/(1.5·Rm·d0·tmin)
• governs: n = max(na, nl)

Choose the inputs correctly

Load F, rivet diameter d0, governing plate thickness tmin, tensile strength Rm and safety factor S jointly determine both rivet counts, since τa,allow and σl,allow both derive from the same Rm and S.

How to use the calculator

Enter force, diameter, plate thickness, tensile strength and safety factor. The calculator shows both required rivet counts na and nl; always choose the larger for the design, rounded up to a whole number.

Worked example

20,000 N, a 10 mm rivet diameter, 8 mm governing plate thickness, Rm = 400 MPa (formed steel rivets) and S = 1.5 give na = 2 (shear) and nl = 1 (bearing); 2 rivets should be chosen since shear failure governs here.

Understand the result and units

na exceeding nl shows that shear failure drives the design in this example — increasing plate thickness alone would not reduce rivet count, while a larger rivet diameter or more rivets directly address the shear issue.

Force is given in N, diameter and plate thickness in mm, tensile strength in MPa, and safety factor and rivet counts are dimensionless.

Why both checks are always needed

Whether shear or bearing first causes a riveted joint to fail depends on the specific geometry: a large rivet diameter with a thin plate favors bearing failure, a small diameter with a thick plate favors shear failure. Since both failure modes can occur independently and no general rule predicts which becomes critical first, a safe design always requires both checks, choosing the larger of the two resulting rivet counts.

Typical applications

The calculation is used for initial sizing of riveted joints, to quickly estimate how many rivets of a given size a load requires before the final arrangement is checked in detail.

Assumptions, limits and common mistakes

The model applies to a single-shear joint with even load sharing. For double-shear joints, na should be halved accordingly. Edge distances, hole spacing and plate net-section stress are not checked here and need separate verification, as does possible eccentric loading of larger rivet groups.

Common mistake: A common mistake is checking only one failure mode and overlooking the other — depending on geometry, either shear or bearing can govern. It is also easy to skip rounding up the computed result, even though only a whole rivet count makes practical sense.