Shear

Shear stress acts parallel to a surface. Learn direct shear in pins and bolts (single and double shear), punching, shear strain and modulus, and the shear-stress distribution in beams (VQ/Ib).

Scissors, bolts in a lap joint, a punch making a hole, a key in a shaft — in all of these the load tries to slide one layer of material over the next. That sliding intensity is shear stress τ.

Most materials are weaker in shear than in tension (for ductile steel, shear yield ≈ 0.58 × tensile yield), so shear often governs connections.

An analogy

Pushing the top of a deck of cards sideways: the cards slide over each other. That's shear deformation.

Direct shear

Divide the shear force by the area resisting it. A bolt joining two plates in a lap joint is in single shear (one plane); in a clevis (fork) it is in double shear — two planes share the load, halving the stress.

Punching and blanking

A punch shears a ring of material of area π·d·t. The press force needed is τ_u × πdt, where τ_u (ultimate shear strength) is about 0.8 × S_u for steel.

Shear in beams

Beams in bending also carry shear. The shear stress is zero at the top and bottom surfaces and peaks at the neutral axis. For rectangles τ_max = 1.5V/A; for I-beams nearly all shear is carried by the web (τ ≈ V/A_web).

Complementary shear and shear flow

Shear on one face always implies equal shear on the perpendicular face. In built-up and thin-walled sections, shear flow q = VQ/I (N/mm) sizes welds, rivets and adhesives joining flanges to webs, and locates the shear centre.

Formulas

• Average direct shear: τ = V / A (double shear: A → 2A)
• Punching force: F = τ_u · π d t
• Beam shear stress: τ = V Q / (I b)
• Rectangle maximum: τ_max = 1.5 · V / A

Worked example: Bolt in double shear

Given: Ø16 bolt in a clevis; V = 40 kN

Find: Shear stress.

• A = π × 16²/4 = 201.1 mm²
• Double shear: τ = 40 000/(2 × 201.1) = 99.5 MPa

Answer: ≈ 99.5 MPa

Worked example: Punching a hole

Given: Ø20 hole in 5 mm plate; τ_u = 350 MPa

Find: Punch force.

• Shear area = π × 20 × 5 = 314.2 mm²
• F = 350 × 314.2 = 110 kN

Answer: ≈ 110 kN (11 tonnes)

Common mistakes

• Wrong: Using single-shear area for a double-shear joint. — Right: Count the shear planes; each carries part of the load.
• Wrong: Assuming shear stress is uniform in a beam section. — Right: It's parabolic in rectangles; peak = 1.5 × average.

Applications

• Bolted & pinned joints: Clevis pins, hinge pins and bolts in bearing-type joints are sized for shear.
• Shear pins: Deliberately weak pins protect gearboxes and snow-blower augers by shearing first.
• Sheet-metal presses: Punching and blanking forces select press tonnage.

Key takeaways

• τ = V/A; count shear planes.
• Punching: F = τ_u·πdt.
• Beam shear: τ = VQ/Ib, max 1.5V/A for rectangles.
• Shear yield ≈ 0.58 × tensile yield.

Going further

• Shear deformation in beams: For short, deep beams, Timoshenko theory adds shear deflection to bending deflection.
• Adhesive joints: Lap joints concentrate shear at the overlap ends; tapering and fillets reduce peak stress.
• Shear buckling: Thin webs of deep girders can buckle under shear — stiffeners are welded to prevent it.