I'm struggling to calculate with any confidence the allowable shear force for a metric bolt holding two aluminium pieces together.

I only need an approximate figure, since I will allow a very large margin of safety.

Assume that the pieces being joined are sufficiently strong. Assume the bolt is threaded throughout its length. It is acceptable for the thread to be damaged, but no permanent deformation to the bolt core. Assume a single bolt holds the two pieces such that the pieces can freely rotate, but that they won't.

The two pieces being joined are are 5mm and 3mm thick. The bolt is M8 with 1mm thread. Assume the bolt is made of a weaker steel.

What is the value of the shear force for the above?

So far, the most progress I have made is by using this online calculator, but I think it's for tension strength. Nevertheless, the "proof load" comes out at 8.82kN for for "property grade" 4.6. I'm guessing the area strength is of similar magnitude.


2 Answers 2


The Guide to Design Criteria for Bolted and Riveted Joints 1974 by Fisher and Struik, you can use the approximation:

$$S_{Ult-Shear}\sim 0.62\cdot S_{Ult}$$

and from the Distortion Energy Theory:

$$S_{YieldShear}\sim 0.58\cdot S_{Yield}$$

The shear force to yield the bolt is:

$$F = S_{YieldShear}\cdot A_{bolt}$$

Where $$A_{bolt}$$

is the cross sectional area of the bolt in the shear plane

Since the shear plane is across the threaded portion of the bolt, use the minor diameter for an external M8x1 thread, 6.596mm, to calculate the area.

If you're interested in the force required to break the bolt, simply substitute the Ultimate Shear Strength for Yield Shear Strength.

In all likelihood though, the aluminum will tear long before you get to the point of shearing the bolt.


For metric bolts the tensile strength can be calculated from the grade of the bolt expressed as A.B (eg 8.8).

The first number is the ultimate tensile stress in 100s of Mega Pascals and the number after the point is the yield stress in tenths of the UTS.

For example an 8.8 bolt has a UTS of 800 MPa and the yield stress is 8/10 of that (80%). Multiply by the cross sectional area to get the corresponding tensile loads. If you stick to those units you get an answer in Newtons.

Typically for ductile metals shear strength will be about 50% of tensile strength.

If the joint is able to slip you also need to consider the stresses on the hole, which in practice may be the limiting factor, especially if yo are using a steel bolt to join aluminium.


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