Calculate direct, beam, and torsional shear stress. Includes stress distribution, safety factors, and bolt sizing tables.
The **Shear Stress Calculator** covers the three most common cases where a load tries to slide one material layer past another: direct shear (τ = F/A), beam shear (τ = VQ/Ib), and torsional shear (τ = Tr/J). Each mode uses the correct geometry so you can see how force, area, and shape affect the stress.
That matters because bolts, pins, beam webs, adhesive joints, and shafts all fail in different shear patterns. A fastener in direct shear is usually checked by average stress, a beam is checked by the shear distribution across its depth, and a shaft in torsion is checked at the outer surface. The calculator reports the maximum stress and a simple safety factor so you can compare the result against a material limit.
Shear stress is the right check when a part can tear or slide along a plane rather than simply bend. This calculator helps size bolts and pins, estimate beam web shear, and check shaft torsion without switching between separate formulas. It is especially useful when you need the governing shear value and a quick pass/fail comparison to the material strength.
Direct shear: τ = F / A Beam shear (rectangular): τ_max = 3V / (2A) = 1.5 × V/(bh) Torsional shear: τ_max = Tr / J = 16T / (πd³) Polar moment: J = πd⁴/32 (solid circular) First moment: Q_max = bh²/8 (rectangular at neutral axis) Variables: F/V = force (N), A = area (m²), T = torque (N·m), r = radius (m), J = polar moment (m⁴), b = width (m), h = height (m), d = diameter (m)
Result: 99.5 MPa shear stress
A 16 mm diameter bolt (area = π×8² = 201 mm² = 2.01×10⁻⁴ m²) in single shear with 20 kN load: τ = 20,000 / 2.01×10⁻⁴ = 99.5 MPa. For Grade 8.8 bolt (τ_allow ≈ 120 MPa), safety factor = 120/99.5 = 1.2.
Shear capacity checks are required for every structural member and connection. In steel design (AISC specifications), beam web shear capacity, bolt shear strength, and weld shear capacity are explicit limit states. In concrete design (ACI 318), beam shear capacity often governs member sizing, and stirrup reinforcement is designed specifically to resist shear. In wood design, horizontal shear along the grain is frequently the governing failure mode for joists and beams.
The common simplification τ = V/A (average shear stress) is adequate for connections but insufficient for beam design, where the parabolic distribution means the actual maximum stress is 50% higher (for rectangular sections). I-beams have an even more concentrated distribution, with nearly all shear carried by the web.
In practice, structural elements rarely experience pure shear. Most are subjected to combined bending and shear, or combined tension and shear. The Mohr's circle construction determines principal stresses and maximum shear stress from any combination of normal and shear stress. The von Mises equivalent stress (σ_eq = √(σ² + 3τ²)) provides a single comparison value against the yield strength for ductile materials under combined loading.
Bolted connections with multiple fasteners distribute the applied load among the bolts. For eccentrically loaded bolt groups, the instantaneous center method or elastic method determines the load on the most critically loaded bolt. The shear stress in that bolt, combined with any tension from prying action, must be checked against the bolt's allowable combined stress.
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Direct shear assumes uniform stress distribution across the shear plane (F/A) — applicable to bolts, pins, and adhesive joints. Beam shear has a parabolic distribution across the depth (VQ/Ib) — maximum at the neutral axis. The beam formula is more accurate for members bending under transverse loads.
For a bolt in single shear, the shear area is the cross-sectional area at the shear plane: A = πd²/4 using the nominal or stress area diameter. For double shear, multiply by 2. Standard bolt stress areas are tabulated in engineering handbooks.
Also called the Tresca criterion, it predicts yielding when the maximum shear stress reaches the material shear yield strength. This is the shear stress in the most critically oriented plane. For uniaxial tension, τ_max = σ/2, so shear yield = σ_y/2. This is slightly more conservative than the von Mises criterion.
At the neutral axis, the first moment of area Q is maximum (most material above or below contributes to the shearing effect). At the top and bottom surfaces, Q = 0 because there is no area beyond that boundary. This parabolic distribution is derived from equilibrium of a small beam element.
Power = Torque × Angular velocity (P = Tω). For a given power and RPM, the required torque is T = P/(2πn/60). Higher RPM means less torque for the same power, allowing smaller shaft diameters. This is why high-speed transmissions use smaller shafts.
The torsion mode is set up for a solid circular shaft. For a hollow shaft, compute the polar moment yourself with J = π(d_o⁴ − d_i⁴)/32 and enter the equivalent result manually.