Calculate shear area from bolt, plate, or tube dimensions, or find required area from shear force, stress, safety factor, and shear planes.
Shear Area Formula
The calculator uses one of two formula sets depending on the mode you pick.
Mode 1: Area from dimensions. The cross-section perpendicular to the applied shear force is computed from the part geometry, then multiplied by the number of parts and the number of shear planes.
Solid round: A = n * p * pi * d^2 / 4 Rectangular: A = n * w * t Hollow tube: A = n * p * pi * (Do^2 - Di^2) / 4
- A = total shear area
- n = number of identical fasteners or sections
- p = shear planes per fastener (1 for single shear, 2 for double shear)
- d = diameter of the solid round
- w = width along the shear plane
- t = thickness
- Do, Di = outer and inner diameter of the tube
Mode 2: Required area. Given a load, an allowable shear stress, and a safety factor, the calculator solves for the minimum total area that keeps stress at or below the allowable value.
A_required = F * SF / tau_allow A_per_plane = A_required / (n * p) d_equivalent = sqrt(4 * A_per_plane / pi)
- F = total shear force on the joint
- SF = safety factor applied to the load
- tau_allow = allowable shear stress of the material
- n = fasteners sharing the load
- p = shear planes per fastener
- d_equivalent = minimum solid round diameter that delivers the per-plane area
The shape selector swaps in the right geometry formula. The shear-planes input multiplies the section area because a bolt in double shear resists with two cross-sections at once. In Mode 2, the safety factor scales the load up before dividing by stress, so a larger SF returns a larger required area.
Reference Values
Use these as starting points, not as a substitute for a material datasheet or governing code.
| Material | Typical allowable shear stress | Common rule |
|---|---|---|
| Mild steel (A36) | ~145 MPa (21 ksi) | ~0.4 × yield |
| Medium-carbon steel | ~200 MPa (29 ksi) | ~0.577 × yield |
| Aluminum 6061-T6 | ~207 MPa (30 ksi) ultimate | Use ~125 MPa allowable |
| Stainless 304 | ~215 MPa (31 ksi) | ~0.6 × yield |
| Structural timber, parallel to grain | 1 to 2 MPa allowable | Species-dependent |
| Configuration | Shear planes | Effective area for one bolt of diameter d |
|---|---|---|
| Lap joint (single shear) | 1 | π d² / 4 |
| Butt joint with two cover plates | 2 | π d² / 2 |
| Clevis pin in a yoke | 2 | π d² / 2 |
| Punched hole in plate | 1 (perimeter × t) | π d × t |
Worked Examples
Example 1: Single bolt in single shear. A 10 mm bolt connects two plates in a lap joint. Shear area is one circular cross-section.
A = π × (10)² / 4 = 78.54 mm²
Example 2: Required area for a 24 kN load. Allowable shear stress is 200 MPa, safety factor is 1.5, two bolts share the load in double shear.
A_total = 24,000 N × 1.5 / 200 MPa = 180 mm²
A_per_plane = 180 / (2 × 2) = 45 mm²
d_min = sqrt(4 × 45 / π) = 7.57 mm. A 10 mm bolt covers it with margin.
FAQ
Single shear vs double shear: which area do I use? Use one cross-section per shear plane. A bolt in double shear has twice the area resisting the load, so it carries twice the force at the same stress.
Should I use the nominal or tensile-stress area for threaded bolts? For shear through the threads, use the minor-diameter or tensile-stress area. For shear through the unthreaded shank, use the nominal diameter. The calculator works in either case if you enter the matching diameter.
Where does the safety factor go? It multiplies the load. Doubling SF doubles the required area for the same allowable stress. If your code uses a resistance factor instead, divide the allowable stress by that factor and set SF to 1.
Why does the punched-hole case use perimeter times thickness? Punching shears the metal along the cylindrical surface around the hole, not across a circular cross-section. Area equals π d × t, which is what you would enter into the rectangular mode using w = π d and t = thickness.
