Shear stresses in beams

Why it matters

Bending stresses size the flanges of a beam, but the shear force has to be carried too, mainly by the web. In aircraft the spar webs, ribs and skin panels are thin, so shear often governs their thickness and their buckling. Shear flow, the shear force per unit length along a line, is also what sizes the rivets and bonds that join stringers, caps and skins.

Key ideas

Where beam shear stress comes from. In a beam with varying bending moment, the bending stresses on two nearby cross-sections are different. A slice of the beam above some level y is therefore not in equilibrium in the longitudinal direction unless a horizontal shear force acts on its cut face. By complementary shear, an equal shear stress acts vertically on the cross-section at that level. This is why shear stress in a beam varies over the depth even though the shear force V is a single number.

The shear formula τ = V·Q/(I·b). Q is the first moment, about the neutral axis, of the area of the section lying beyond the level where τ is wanted (above it or below it, the result is the same), and b is the width of the section at that level. Assumptions: linear-elastic material, the bending formula holds, τ is uniform across the width b, and the section is not too wide or thin-walled in an odd way.

Rectangular section. Q = (b/2)(h²/4 − y²), so τ varies parabolically, zero at the top and bottom fibres and maximum at the neutral axis, where τ_max = 1.5·V/A (50 % above the average V/A).

Solid circular section. τ_max = (4/3)·V/A at the neutral axis.

I-sections. The formula gives a sudden jump in τ at the flange–web junction because b changes from the flange width to the web thickness. Most of the shear (often 90 % or more) is carried by the web, and a common approximation is τ_web ≈ V/(h_w·t_w). The flanges carry most of the bending moment.

Maximum shear stress is not always at the neutral axis. It is at the neutral axis for rectangles, circles and I-sections, but for shapes whose width narrows away from the neutral axis faster than Q drops (for example a triangle, or a diamond) the peak lies elsewhere. For a triangular section it is at mid-height, and equals 1.5·V/A.

Shear flow. q = τ·b = V·Q/I is the longitudinal shear force per unit length (N/m) on a cut. In built-up beams it sizes fasteners: fastener pitch s = F_allow/q for one fastener per row of strength F_allow. In thin-walled sections q flows around the wall like a fluid, which leads to the shear centre in the thin-walled structures subject.

Limits. Very short, deep beams and regions near point loads and supports do not follow the simple theory well.

Formulas

τ = V·Q / (I·b) τ = shear stress at the level considered (Pa), V = shear force (N), Q = first moment of the area beyond the level about the neutral axis (m³), I = second moment of area of the whole section (m⁴), b = width at that level (m).

τ = (V / (2·I))·(h²/4 − y²) Rectangle of depth h (m) at distance y (m) from the neutral axis.

τ_max = 1.5·V / A Rectangle, at the neutral axis; A = b·h (m²).

τ_max = (4/3)·V / A Solid circle, at the neutral axis; A = π·d²/4.

τ_web ≈ V / (h_w·t_w) Approximate average web shear stress in an I-section; h_w = web depth, t_w = web thickness (m).

q = V·Q / I Shear flow (N/m), the longitudinal shear force per unit length.

s = F_allow / q Fastener pitch (m) for a joint carrying shear flow q with allowable load F_allow (N) per fastener.

Worked examples

Example 1 (standard). A rectangular beam 80 mm wide and 200 mm deep carries a shear force of 40 kN. Find the average shear stress, the maximum shear stress, and the shear stress 50 mm above the neutral axis.

1. A = b·h = 80 × 200 = 16 000 mm²; τ_avg = V/A = 40 000/16 000 = 2.5 MPa.
2. τ_max = 1.5·V/A = 1.5 × 2.5 = 3.75 MPa, at the neutral axis.
3. I = b·h³/12 = 80 × 200³/12 = 5.333 × 10⁷ mm⁴.
4. τ = (V/(2I))·(h²/4 − y²) = 40 000/(2 × 5.333 × 10⁷) × (100² − 50²) = 3.75 × 10⁻⁴ × 7500 = 2.81 MPa.

Answer: τ_avg = 2.5 MPa, τ_max = 3.75 MPa, τ(50 mm) ≈ 2.81 MPa.

Example 2 (GATE level). A symmetric I-section has flanges 100 mm × 10 mm and a web 180 mm deep and 10 mm thick (overall depth 200 mm). It carries V = 50 kN. Find τ just below the top flange in the flange, just inside the web at the junction, and at the neutral axis.

1. I = 100 × 200³/12 − 90 × 180³/12 = 6.667 × 10⁷ − 4.374 × 10⁷ = 2.293 × 10⁷ mm⁴.
2. Q of one flange about the neutral axis: 100 × 10 × (100 − 5) = 95 000 mm³.
3. In the flange at the junction (b = 100 mm): τ = 50 000 × 95 000/(2.293 × 10⁷ × 100) = 2.07 MPa.
4. In the web at the junction (b = 10 mm): τ = 50 000 × 95 000/(2.293 × 10⁷ × 10) = 20.7 MPa.
5. At the neutral axis: Q = 95 000 + 10 × 90 × 45 = 135 500 mm³; τ = 50 000 × 135 500/(2.293 × 10⁷ × 10) = 29.6 MPa.
6. Approximate check: V/(h_w·t_w) = 50 000/(180 × 10) = 27.8 MPa, between the junction and peak values.

Answer: 2.07 MPa (flange), 20.7 MPa (web at junction), 29.6 MPa (neutral axis). The tenfold jump at the junction comes from the change in width.

Common mistakes

• Taking Q as the first moment of the whole section (which is zero) instead of the area beyond the level.
• Using the flange width in place of the web thickness at a web point, or vice versa.
• Using Q = b·h²/4 for a rectangle at the neutral axis; the correct value is b·h²/8.
• Assuming τ_max is always at the neutral axis for every shape.
• Treating V/A as the maximum shear stress; it is only the average.

For GATE AE

Expect numericals on τ_max for rectangular and circular sections, the ratio τ_max/τ_avg, shear stress at the flange–web junction of I- and T-sections, and shear flow for fastener spacing in built-up beams. Conceptual questions test the shape of the distribution. Practise computing Q quickly for composite areas, and link this topic to shear flow in thin-walled sections.

Quick check

1. What is τ_max/τ_avg for a rectangular section?
2. What is it for a solid circular section?
3. What is Q at the top fibre of any section?
4. Why does τ jump at the flange–web junction of an I-beam?

Answers: 1. 1.5. 2. 4/3. 3. Zero, so τ = 0 there. 4. Because the width b in τ = V·Q/(I·b) changes suddenly while Q does not.