# Self teaching basic average shear stress calculation

Koth
Hello All

I am designing some very basic components for a motorcycle project at home, and because I am interested, thought I'd use it as an opportunity to refresh my memory in order to reduce the risk of the parts failing. To begin, I want to do some basic hand calcs if possible so have identified the very basic formula for calculation of average stress ( = F/A). Furthermore, I need to use a basic safety factor calculation to determine if the specified material will fail so have identified fs = Ys/Ds (Ys = material strength, Ds = shear stress).

I am concerned about a protruding piece of material that is subject to a load that will tend to shear the protrusion from the main body of the component (I will come back with a sketch if necessary). The load is applied perpendicular to the shear plane. Cross section area of the shear plane is 10.6 mm². Load is 400N.

Using the above formula I have ended up with a shear stress of 37.74 MPa.

Naturally, I have a few questions and would be extremely grateful if somebody could give some advice please?

Q1) Is this basic formula acceptable for this scenario?
Q2) When inserting values into the safety factor formula the material datasheet (6061-T6 alloy) offers Ultimate Tensile Strength, Yield Tensile Strength and Shear Strength. I have tried to determine which to use on the basis that I do not want the lug to deform beyond it's Yield point but I'm confused. Is the Shear Strength listed on the DS Ultimate or Yield? Or should I use the rule of thumb that I have found based on a calculated value of 0.577*Yield Tensile Strength? I have also seen a rule of thumb advising 0.6*UTS but assume this is for failure beyond the materials elastic limit?

I understand that a more complex analysis may be required so thank you in advance for your patience with this. As I said, any general advice much appreciated.

Koth