Shear Stress Distribution

The shear formula

For a beam in pure transverse shear, the shear stress at a horizontal cut at distance y from the neutral axis is:

      τ(y) = V · Q(y) / (I · b)

where:

• V is the transverse shear force at the section.
• I is the moment of inertia of the entire cross-section about the neutral axis.
• b is the width of the cross-section at the cut.
• Q(y) is the first moment of area of the portion above (or below) the cut, taken about the neutral axis: Q = ∫y dA over that portion.

The geometry of Q is the key insight. Q grows from zero at the top of the beam (no area above the cut) to a maximum at the neutral axis (half the cross-section's area, with its centroid at h/4), then back to zero at the bottom. So τ traces the same shape — zero at the surfaces, peak at the centroid.

Worked example: a rectangular beam

For a rectangle of width b and depth h, with the neutral axis at mid-depth, take a cut at distance y from the neutral axis. The area above the cut is b·(h/2 − y); its centroid is at distance (h/2 + y)/2 from the neutral axis. So:

      Q(y) = b · (h/2 − y) · (h/2 + y)/2 = (b/2) · (h²/4 − y²)

And τ at that cut is:

      τ(y) = V · (b/2)(h²/4 − y²) / (I · b)
           = V (h²/4 − y²) / (2I)
           = (3V / (2bh)) · (1 − 4y²/h²)
           = (3/2) · (V/A) · (1 − 4y²/h²)

At y = 0 (neutral axis): τ_max = 1.5·V/A. At y = ±h/2 (surfaces): τ = 0. The distribution is parabolic.

Numerical example: 100 mm × 200 mm beam carrying V = 50 kN.

      Cross-section area A = 100 × 200 = 20,000 mm² = 0.02 m²
      Average τ = V/A = 50,000 / 0.02 = 2.5 MPa
      Peak τ at neutral axis = 1.5 × 2.5 = 3.75 MPa
      Stress at quarter-depth (y = h/4 = 50 mm):
         τ = 1.5 × 2.5 × (1 − 4(50)²/200²) = 1.5 × 2.5 × 0.75 = 2.81 MPa

     Top of beam ─────────────  τ = 0
                  ╲           ╱
                   ╲         ╱
                    ╲       ╱
                     ╲     ╱
                      ╲   ╱     ← parabolic
       Neutral axis ──╳──── ──  τ = 1.5·V/A (peak)
                      ╱   ╲
                     ╱     ╲
                    ╱       ╲
                   ╱         ╲
                  ╱           ╲
     Bottom ─────────────────  τ = 0

Distribution by cross-section shape

Cross-section	Distribution shape	τ_max / (V/A)	Where τ peaks	Notes
Rectangular	Parabola (smooth, symmetric)	1.5	Neutral axis	Standard derivation; canonical
Solid circular	Parabola (similar to rect)	1.33 (4/3)	Neutral axis (full diameter)	Lower factor; circle distributes shear better
I-beam (wide-flange)	Step jumps at flange-web junction; ≈ uniform in web	≈ 1.05–1.10 (web only)	Neutral axis (in web)	Flanges carry ≈ zero shear
Thin-walled (tube)	Sinusoidal around circumference	2.0	Neutral axis (sides)	Higher factor due to thin walls
T-section	Asymmetric — discontinuity at stem-flange	Varies, ~1.5	Below stem-flange interface	Asymmetric Q profile
Hollow rectangle	Parabolic in webs; near zero in flanges	1.5–1.6 (per web)	Mid-height of webs	Two webs share total shear

Why an I-beam's web carries everything

For a wide-flange I-beam loaded vertically, the bending stress is largest at the flanges (far from the neutral axis) but the shear stress is largest in the web. Why? At a cut just below the top flange, Q includes the entire top flange's area at large y — so Q is big. But the width b at that cut suddenly drops from the flange width (say 200 mm) to the web thickness (say 10 mm). Since τ = VQ/(Ib), reducing b by 20× while Q stays similar gives a 20× jump in shear stress as you cross the flange-web boundary.

     ┌────────────────────────┐  ←  flange: b = 200, τ ≈ 0.05·V/A
     │                        │
     └─────┐            ┌─────┘  ← step jump at flange-web junction
           │   WEB      │
           │            │       ←  web: b = 10, τ ≈ 1.0·V/A
           │            │
           │            │
     ┌─────┘            └─────┐
     │                        │
     └────────────────────────┘  ←  bottom flange

For a W14×90 wide-flange (depth 14 in, web thickness 0.44 in, web height ≈ 12.6 in) carrying V = 100 kips:

      Web area = 12.6 × 0.44 = 5.54 in²
      Approximate τ_web ≈ V / (h_web · t_web) = 100 / 5.54 = 18.0 ksi

The exact shear formula gives τ ≈ 18.5 ksi at the neutral axis — within 3% of the simple web-only approximation. So for I-beam design, engineers reliably use τ ≈ V/(h_w · t_w), ignoring the flanges entirely.

Shear flow in built-up sections

When a cross-section is built up from separate pieces (a wood T-beam from two boards, a steel girder from welded plates, a glulam from layered planks), the connection between pieces must transmit horizontal shear or the section will slip. The shear flow per unit length is:

      q = V · Q / I   (units: N/m, lb/in)

where Q is now the first moment of just the piece being connected, taken about the entire section's neutral axis.

Example: a built-up wood T-beam. Top board 200 mm × 50 mm, web board 50 mm × 200 mm, joined at the upper edge of the web. Find the bolt spacing required if each bolt carries 10 kN in shear and V = 30 kN.

      Centroid of combined section: weighted average gives ȳ ≈ 162 mm from bottom
      Q for top board about neutral axis: A_top · d = (200×50) · (225 − 162) = 6.3 × 10⁵ mm³
      I_combined ≈ 1.7 × 10⁸ mm⁴
      q = V·Q/I = 30,000 × 6.3×10⁵ / 1.7×10⁸ = 111 N/mm = 111 kN/m
      Bolt spacing = 10,000 N / 111 N/mm = 90 mm centre-to-centre

If you space bolts farther than 90 mm apart, the section slips and the effective I drops sharply.

Real-world applications

• Wood beam shear failure. Wood is anisotropic; its shear strength parallel to grain is only 4–10% of its compressive strength along grain. Short, heavily-loaded wood beams develop horizontal cracks at the neutral axis where shear peaks — a failure mode unfamiliar to engineers used to steel.
• Plate girder web buckling. Tall thin webs in I-section bridges carry near-uniform high shear stress, which can buckle the plate diagonally before the steel yields. Vertical and intermediate stiffeners welded to the web prevent this.
• Glulam laminated timber. Engineered timber bonded from many thin laminations achieves high I, but the glue lines must transfer shear flow q = VQ/I. A 90 mm × 360 mm glulam beam with 25 mm laminations needs glue strength of ≈ 1 MPa minimum, easily met by modern PRF adhesives.
• Sandwich panels in aircraft. A composite sandwich with carbon-fiber face sheets and honeycomb core — the face sheets carry bending, the core carries shear. The core must transmit shear at q = V·Q_face/I, typically 0.1–1 MPa — easily achieved by aluminum or aramid honeycomb at densities of 50–100 kg/m³.
• Reinforced concrete beam stirrups. Concrete cracks under tension at 45° to the beam axis (diagonal tension from shear). Vertical steel stirrups cross these cracks and carry the shear that concrete can't. Stirrup spacing is computed from the shear flow demand at each section.

Variants and corrections

Non-Newtonian (viscous) shear correction. For polymers and some aerospace composites, the shear modulus depends on strain rate. Quasi-static shear formulas overpredict deflection by 15–30% in dynamic loading because the material stiffens at high strain rates. Hyperelastic and viscoelastic FEA models correct for this.

Composite (multi-material) sections. When E varies across the cross-section (steel reinforcement in concrete, fibre-reinforced matrix), the shear formula generalizes to τ = VQ_eff / (I_eff · b), where Q_eff and I_eff are computed with each material weighted by its modular ratio E_i/E_ref.

Curved beams. The neutral axis of a strongly curved beam (radius of curvature comparable to the depth) shifts away from the centroidal axis — Winkler's correction. The shear distribution is no longer symmetric and τ_max shifts toward the inner radius. Crane hooks and chain links live in this regime.

Thin-walled open vs closed sections. A closed thin-walled tube (rectangular HSS) has a uniform shear flow around its perimeter. An open thin-walled section (a channel or angle) develops a shear centre offset from the centroid; loading through the centroid causes torsion as well as bending.

Common failure modes and design errors

• Using V/A instead of 1.5·V/A. A common rookie error — picking the simple "average" shear stress instead of the peak. Acceptable for I-beams (where the web carries near-uniform shear) but wrong by 50% for solid sections.
• Ignoring shear flow in built-up sections. Bolting two boards together with screws spaced at 1 m won't transfer the required shear flow; the section slips and acts as two independent beams instead of one composite beam. The effective I drops by a factor of 4 or more.
• Forgetting the shear centre on open thin-walled sections. A C-channel loaded at its centroid twists as well as bends. Loading must pass through the shear centre (offset from the centroid for open sections) to avoid coupled torsion.
• Diagonal cracking in concrete. Designers used to bending-only thinking sometimes under-detail stirrups in deep concrete beams, leading to diagonal tension cracks at 45° to the beam axis — the failure mode the 1971 ACI 318 code was rewritten to prevent.
• Web buckling in plate girders. Tall slender webs in long-span steel bridges buckle diagonally under shear before yielding. Slenderness limits (h/t_w ≤ 260·√(k/F_y)) and intermediate stiffeners prevent this — but old riveted girders without stiffeners must be checked carefully under modern loads.
• Misapplying simple beam theory to deep beams. When span/depth < 5, shear deformations contribute significantly to deflection and the linear-bending-stress assumption breaks down. Use shear-deformable (Timoshenko) beam theory or a 2D elasticity solution.