Lessons · Engineering · shear stress in a pin
Shear in a pin: the force trying to slice it
A pin loaded across its axis carries shear stress τ = V / A over its cross-section. In double shear two sections share the load, so the stress halves.
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
What it is for
A clevis pin on a hydraulic cylinder sheared clean through and the boom came down. The pin was sized for tension, which it never sees; it is cut across, and the area that resists cutting is the circle of the pin, once or twice.
How to think about it
Count the faces being cut. A pin through two plates is cut once; through a clevis with the load in the middle, twice. The area is the pin's circle times the number of faces, and the shear stress is the force over that.
Worked example
τ = V / A, V the force across the pin, A the area being cutShear stress is the same shape as normal stress; the difference is which way the force goes.
A 10 mm pin in single shear holding 6 kN: A = π × 10² / 4 = 78.5 mm²One circle of pin resists.
τ = 6000 / 78.5 = 76.4 MPaThe shear stress on that one face.
In double shear two faces share it: τ = 6000 / (2 × 78.5) = 38.2 MPaSame pin, same load, half the stress, because the pin has to be cut twice to let go.
Your turn
An 8 mm pin, single shear, 4 kN. Write the shear area.
A = π × ² / 4 = 50.3 mm²
Solve one, graded on the server
The trap
Counting one shear face when the pin sits in a clevis and has two, or two when it has one. Get the faces wrong and the pin is either twice as strong as you think or half.