Shear flow in open thin-walled sections

Why it matters

Stringers, spar sections, rib flanges and cut-out edge members are open thin-walled sections, and the transverse shear they carry travels round the wall as shear flow. Knowing its distribution tells you where the web is most highly stressed, how much load the rivets between web and flange must carry, and where a shear load must act to avoid twisting the section (the shear centre, next topic).

Key ideas

Shear flow. In a thin wall of thickness t the shear stress τ is taken as uniform through the thickness and tangential to the mid-line. The shear flow is q = τ·t, the shear force per unit length of mid-line (N/m or N/mm). It is the natural quantity for thin walls because, at a junction, the shear flows in and out must balance, like flow in pipes.

Where it comes from. Take a small element of wall of length δz along the beam and δs round the wall. If the bending moment changes along the beam, the direct stress σ_z changes, and the out-of-balance axial force must be carried by a change in shear flow around the wall: ∂q/∂s + t·∂σ_z/∂z = 0. Substituting the bending stress and using S = dM/dz gives the general open-section formula below.

Open sections. A section is open if it has free edges (channel, I, Z, angle, slit tube). At a free edge the shear flow is zero, because there is no material beyond it to exchange shear with. So q can be found by integrating from a free edge, and no additional unknown appears: an open section is statically determinate in shear.

Assumptions. Linear elastic, homogeneous material; thin walls (τ constant through t); shear loads applied through the shear centre so that there is no twist (otherwise add the torsion solution); direct stresses from the general bending theory; small deformations.

Symmetric sections. If x is an axis of symmetry and S_y acts parallel to y, I_xy = 0 and the formula reduces to q_s = −(S_y/I_xx)·∫t·y ds. Numerically this is the familiar q = S·Q/I, with Q the first moment of the cut-off area about the neutral axis; the sign gives the direction relative to s.

Shape of the distribution. In a flange perpendicular to the load, y is constant so q varies linearly from zero at the free edge. In a web parallel to the load, y varies so q is parabolic, maximum at the neutral axis. The flange flows run in opposite directions in top and bottom flanges and give equal and opposite horizontal forces; the web shear flow integrates to the applied shear.

Thickness. Doubling t everywhere doubles both the first and second moments, so q is unchanged and τ = q/t halves.

Formulas

q = τ·t

• q: shear flow (N/m), τ: shear stress (Pa), t: wall thickness (m).

q_s = −[(S_x·I_xx − S_y·I_xy) / (I_xx·I_yy − I_xy²)]·∫₀ˢ t·x ds − [(S_y·I_yy − S_x·I_xy) / (I_xx·I_yy − I_xy²)]·∫₀ˢ t·y ds

• S_x, S_y: shear forces parallel to x and y (N), applied through the shear centre; I_xx, I_yy, I_xy: centroidal second moments (m⁴); s: distance round the mid-line from a free edge (m); x, y: centroidal coordinates of the wall (m). Positive q acts in the direction of increasing s.

q_s = −(S_y / I_xx)·∫₀ˢ t·y ds

• Same, when I_xy = 0 and only S_y acts.

q = S·Q / I

• Magnitude form for symmetric sections: Q is the first moment about the neutral axis of the wall area between the free edge and the point (m³), I: second moment of the whole section (m⁴).

∫ q ds (over a wall) = force carried by that wall

• Resultants of the shear flow must equal the applied shear forces (check).

Worked examples

Example 1 (standard): channel section. Given: channel with web height h = 200 mm, flange width b = 100 mm, thickness t = 2 mm throughout (mid-line dimensions, thin-walled), S_y = 10 kN through the shear centre, parallel to the web.

1. I_xx = t·h³/12 + 2·b·t·(h/2)² = 2 × 200³/12 + 2 × 100 × 2 × 100² = 1.333 × 10⁶ + 4.0 × 10⁶ = 5.333 × 10⁶ mm⁴.
2. Flange, from the free edge: q = (S_y/I_xx)·t·s·(h/2), linear in s. At the web junction (s = b): q = 10⁴/5.333 × 10⁶ × 2 × 100 × 100 = 37.5 N/mm.
3. Web, starting with 37.5 N/mm at the corner: q = 37.5 + (S_y/I_xx)·t·[(h/2)² − y²]/2. At the neutral axis (y = 0): q_max = 37.5 + 1.875 × 10⁻³ × 2 × 10⁴/2 = 37.5 + 18.75 = 56.25 N/mm.
4. Maximum shear stress: τ = q/t = 56.25/2 = 28.1 N/mm².
5. Check: web resultant = 37.5 × 200 + 1.875 × 10⁻³ × (100² × 200 − 200³/12) = 7500 + 2500 = 10 000 N = S_y. Answer: q = 37.5 N/mm at the corners, 56.25 N/mm at the neutral axis; τ_max = 28.1 MPa.

Example 2 (GATE level): Z-section, I_xy ≠ 0. Given: the Z-section of the previous topic: web h = 200 mm, flanges b = 80 mm, t = 8 mm; top flange from (80, 100) to (0, 100), bottom flange from (0, −100) to (−80, −100). I_xx = 18.13 × 10⁶ mm⁴, I_yy = 2.731 × 10⁶ mm⁴, I_xy = 5.12 × 10⁶ mm⁴, D = I_xx·I_yy − I_xy² = 2.330 × 10¹³ mm⁸. S_y = 20 kN through the shear centre, S_x = 0. Start s at the top flange tip.

1. Coefficients: S_y·I_xy/D = 20 000 × 5.12 × 10⁶/2.330 × 10¹³ = 4.395 × 10⁻³ N/mm⁴ (multiplies ∫t·x ds) and −S_y·I_yy/D = −2.344 × 10⁻³ N/mm⁴ (multiplies ∫t·y ds).
2. Top flange, x = 80 − s, y = 100: ∫t·x ds = 8·(80·s − s²/2), ∫t·y ds = 800·s.
3. At the corner (s = 80): ∫t·x ds = 8 × 3200 = 25 600 mm³, ∫t·y ds = 64 000 mm³, so q = 4.395 × 10⁻³ × 25 600 − 2.344 × 10⁻³ × 64 000 = 112.5 − 150.0 = −37.5 N/mm.
4. Down the web to the neutral axis adds −2.344 × 10⁻³ × 8 × 100²/2 = −93.75 N/mm, so q_NA = −131.25 N/mm.
5. The minus sign means the flow is opposite to s: up the web. Check: integrating the web flow gives 20.0 kN upward, equal to S_y, and the top flange flow (which changes sign at s = 53.3 mm) has zero net horizontal resultant here. Answer: |q| = 37.5 N/mm at the corners, 131.25 N/mm at the neutral axis; τ_max = 131.25/8 = 16.4 MPa.

Common mistakes

• Starting the integration somewhere other than a free edge in an open section, without an unknown starting value.
• Using q = SQ/I for an unsymmetric section where I_xy ≠ 0.
• Forgetting the corner value carried from the flange into the web; q is continuous round a corner.
• Mixing up shear flow (N/mm) and shear stress (N/mm²).
• Thinking a thicker wall reduces shear flow; it reduces shear stress.
• Applying the open-section result when the load does not pass through the shear centre; the section then also twists.

For GATE AE

Expect the shear flow at a flange-web junction or at the neutral axis of a channel, I or T section, maximum shear stress, and the shape of the distribution (linear in flanges, parabolic in webs). Some questions combine this with the shear-centre calculation. Practise thin-walled I_xx quickly and always check that the web flow integrates to the applied shear.

Quick check

1. What is the shear flow at the free edge of an open section?
2. How does q vary along a flange perpendicular to the load?
3. Wall thickness is doubled everywhere. What happens to q and to τ?
4. In Example 1, what force does each flange carry horizontally? Answers: 1. zero; 2. linearly from zero at the free edge; 3. q unchanged, τ halves; 4. ½ × 37.5 N/mm × 100 mm = 1875 N, opposite in the two flanges.