The basic formula is τ = F / A — shear stress equals the applied force divided by the area resisting that force. This gives average shear stress for the simplest case: a fastener or pin loaded straight across. More complex geometries — beams, shafts in torsion, soil — use different variations of the same underlying idea, covered below.
The Basic Formula
τ = F / A
- τ = shear stress (N/mm² or lbs/in²)
- F = applied force (N or lbs)
- A = area resisting the shear
Single Shear: Bolt or Pin
For a round fastener loaded in single shear (one shear plane), the area is the cross-section of the fastener:
τ_avg = F / (π r²) = 4F / (π d²)
- r = radius, d = diameter
Worked Example
A 10 mm diameter bolt fails in shear under a 20,000 N load.
Shear area: A = π(10)² / 4 = 78.54 mm²
Shear stress: τ = 20,000 / 78.54 ≈ 254.6 MPa
This is a sample calculation to illustrate the method, not a measured material property — real shear strength for a given bolt grade should come from the manufacturer’s spec or a material test, not back-calculated from an assumed failure load.
Bearing Stress: The Check You Shouldn’t Skip
Shear isn’t the only failure mode to check on a bolted or pinned joint — the plate or bracket material around the hole can also fail in bearing (the hole elongating or the material crushing). The bearing area stress equation:
Bt = F / (t × d)
- Bt = bearing area stress (N/mm² or lbs/in²)
- t = plate/bracket thickness
- d = fastener diameter
Checking shear alone and skipping the bearing check is a common oversight — a joint can pass a shear calculation and still fail by the hole elongating in a thinner plate.
Double Shear
A fastener loaded in double shear (two shear planes, such as a clevis pin through a fork joint) resists the same force over twice the area. The formula doesn’t change — only the area does: use the combined area of both shear planes (double the single-shear area), which halves the calculated shear stress for the same applied force.
Formula Variants by Application
The F/A form only applies directly to simple, uniform shear loading. Other common cases need their own formula:
| Application | Formula | Notes |
|---|---|---|
| Bolt/pin, single shear | τ = 4F/(πd²) | Covered above |
| Beam (transverse shear) | τ = VQ/(It) | V = shear force, Q = first moment of area, I = moment of inertia, t = width — not simply V/A |
| Shaft in torsion | τ = Tr/J | T = torque, r = radius, J = polar moment of inertia |
| Soil (Mohr-Coulomb) | τ_f = c + σ’·tan(φ) | c = cohesion, σ’ = effective normal stress, φ = internal friction angle |
These variants are outside the scope of a single fastener calculation — each has its own derivation and assumptions. If your application is a beam, a shaft, or a soil condition rather than a simple pin/bolt joint, use the matching formula rather than forcing τ=F/A onto a geometry it doesn’t fit.
Estimating Ultimate Shear Strength from Tensile Strength
When a material’s shear strength isn’t directly available, a commonly used engineering approximation is:
Ultimate shear strength ≈ 0.6 × Ultimate Tensile Strength (UTS)
A related variant uses 0.57 × tensile yield strength (TYS), derived from a simplified von Mises / distortion-energy approach.
This is a rule-of-thumb estimate from engineering discussion, not a standard or a substitute for tested material data. For anything load-bearing or safety-critical, use the material’s actual shear strength from a datasheet or standard (ASTM specifies several shear test methods, including ASTM B769), not this approximation.
What You Need to Know Before Calculating Anything
The formula alone isn’t enough — you need to know which failure mode you’re actually checking (shear vs. bearing vs. bending), how many shear planes the joint has (single vs. double), and whether your geometry is a simple pin/bolt, a beam, a shaft, or something else entirely. Getting the wrong formula for the geometry is a more common error than getting the arithmetic wrong within the right formula.
Related reading: Peel Strength vs. Shear Strength
