Shear centre of thin-walled sections

The shear centre of thin-walled sections: definition, the moment method, symmetry rules, and results for channels, unequal I-sections, angles and slit tubes.

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Why it matters

A shear load produces pure bending only if it passes through one particular point of the section, the shear centre. Anywhere else it also applies a torque, and thin open sections such as channels and Z-stringers are very weak in torsion. The position of the shear centre (and its spanwise line, the flexural axis) controls wing twist under load, aileron effectiveness and flutter, and decides how brackets and stiffeners should be attached.

Key ideas

Definition. The shear centre (flexural centre) is the point in the cross-section through which a transverse shear load must act for the beam to bend without twisting. Equivalently, it is the point about which the resultant moment of the bending shear flows is zero. It is a property of the section geometry (and of the stiffness distribution), not of the load.

How to find it.

  1. Apply a shear force S_y (and separately S_x) assumed to act through the unknown shear centre.
  2. Find the shear flow distribution from the open-section (or closed-section) formula.
  3. Take moments of the shear flows about any convenient point (a corner where several walls meet removes their moments).
  4. Equate to S_y × (distance of the line of action from that point). Solve for the coordinate. Repeat with S_x for the other coordinate.

Symmetry rules.

  • If a section has an axis of symmetry, the shear centre lies on it.
  • Doubly symmetric sections (I-section with equal flanges, rectangular box, tube): shear centre = centroid.
  • Antisymmetric sections (Z-section with equal flanges): shear centre = centroid, at the centre of the web.
  • Sections made of thin walls that all meet at one point (angle, T, cruciform): shear centre at that intersection, because every shear flow passes through it and has no moment about it.

Where it lies for common sections.

  • Channel: on the axis of symmetry, outside the section, on the side of the web away from the flanges, at e = 3b²t_f/(6b·t_f + h·t_w) from the web mid-line. Wider flanges move it further out.
  • Singly symmetric I-section with unequal flanges: on the web, nearer the stiffer (wider) flange; the flanges share a horizontal shear in proportion to their own second moments about the web axis.
  • Thin circular tube with a longitudinal slit: on the diameter through the slit, at 2r from the centre on the side opposite the slit.

Shear centre versus centroid. The centroid is where axial load produces no bending; the shear centre is where shear produces no twist. They coincide only for doubly symmetric or antisymmetric sections.

Load off the shear centre. A shear S applied at distance d from the shear centre is equivalent to S at the shear centre plus a torque S·d. Add the bending shear flow and the torsion solution (Saint-Venant for open sections, Bredt for closed).

Formulas

S_y·ξ_s = Σ (moments of the shear flows about a chosen point)

  • S_y: shear force (N), ξ_s: perpendicular distance from the chosen point to the shear centre (m). General definition; the same with S_x gives η_s.

e = 3·b²·t_f / (6·b·t_f + h·t_w)

  • Channel. b: flange width (m), h: web height (m), t_f, t_w: flange and web thicknesses (m), e: distance of the shear centre from the web mid-line, on the side away from the flanges (m). Thin walls, mid-line dimensions.

e = 3·b² / (6·b + h)

  • Channel with uniform thickness.

e = b²·h²·t_f / (4·I_xx)

  • Channel in terms of I_xx = t_w·h³/12 + b·t_f·h²/2.

h₁ = h·I₂ / (I₁ + I₂)

  • I-section with unequal flanges: h₁ is the distance of the shear centre from flange 1 (m), h: distance between flange mid-lines (m), I₁, I₂: second moments of flanges 1 and 2 about the web axis (m⁴), web contribution neglected.

e = 2·r

  • Slit thin circular tube of mean radius r: distance from the centre, on the side opposite the slit.

Worked examples

Example 1 (standard): channel section. Given: web h = 200 mm, flanges b = 100 mm, t = 2 mm throughout (the section of the previous topic), S_y = 10 kN. From the previous topic, the flange shear flow rises linearly from 0 to 37.5 N/mm at the corners.

  1. Force in each flange: F = ½ × 37.5 × 100 = 1875 N, horizontal, opposite in top and bottom flanges.
  2. Take moments about the web mid-line (the web flow has no moment about it). The two flange forces form a couple: F·h = 1875 × 200 = 375 000 N·mm.
  3. This couple must be balanced by S_y acting at distance e from the web: S_y·e = 375 000, so e = 375 000/10 000 = 37.5 mm.
  4. Check with the formula: e = 3·b²/(6·b + h) = 3 × 100²/(600 + 200) = 37.5 mm. Answer: e = 37.5 mm from the web mid-line, on the side away from the flanges.

Example 2 (GATE level): I-section with unequal flanges. Given: top flange b₁ = 100 mm, bottom flange b₂ = 50 mm, web height between flange mid-lines h = 200 mm, all walls t = 2 mm. The section is symmetric about the vertical web axis, so the shear centre lies on the web; find its height.

  1. Apply a horizontal shear S_x. The web, lying on the y-axis, contributes almost nothing to I_yy, so the flanges carry S_x in proportion to their own second moments about the web axis.
  2. I₁ = t·b₁³/12 = 2 × 100³/12 = 1.667 × 10⁵ mm⁴, I₂ = t·b₂³/12 = 2 × 50³/12 = 2.083 × 10⁴ mm⁴.
  3. Forces: F₁ = S_x·I₁/(I₁ + I₂) = 0.889·S_x, F₂ = 0.111·S_x.
  4. Moments about the top flange mid-line: S_x·h₁ = F₂·h, so h₁ = h·I₂/(I₁ + I₂) = 200 × 2.083 × 10⁴/1.875 × 10⁵ = 22.2 mm.
  5. For comparison, the centroid is at ȳ = (A₂·h + A_w·h/2)/(A₁ + A₂ + A_w) = (100 × 200 + 400 × 100)/(200 + 100 + 400) = 85.7 mm below the top flange. Answer: shear centre 22.2 mm below the top flange mid-line, close to the wider flange and well above the centroid (85.7 mm).

Common mistakes

  • Assuming the shear centre is at the centroid for every section. It is only for doubly symmetric and antisymmetric ones.
  • Placing a channel's shear centre between the flanges. It lies outside, beyond the web.
  • Taking moments about a point and forgetting the moment of one of the walls; choosing the corner where walls meet avoids this.
  • Thinking the shear centre depends on load magnitude. It is a geometric property.
  • Using uniform-thickness formulas when flange and web thicknesses differ.
  • Forgetting the extra torque S·d when the load is off the shear centre.

For GATE AE

Expect the shear-centre location of a channel, unequal I, slit tube or angle, and conceptual questions on symmetry rules and on the effect of applying load away from the shear centre. Practise taking moments of flange forces about the web, and remember the rule that walls meeting at a point put the shear centre at that point.

Quick check

  1. Where is the shear centre of an equal-leg angle section?
  2. A channel has b = 50 mm, h = 100 mm, uniform t. Find e.
  3. Where is the shear centre of a Z-section with equal flanges?
  4. If a shear load acts 10 mm from the shear centre, what extra load must be analysed? Answers: 1. at the intersection of the leg mid-lines (the corner); 2. e = 3 × 2500/(300 + 100) = 18.75 mm from the web; 3. at the centroid, the mid-point of the web; 4. a torque S × 10 mm.

Interview questions

All Aircraft Structures interview questions

Try answering each one aloud before you open it.

  1. 1.What is the shear centre in the context of thin-walled sections?Concept

    The shear centre is a point in the cross-section of a beam or structural member where the application of transverse shear force does not cause the member to twist. It is crucial in the design of thin-walled sections to ensure that loads are applied in a way that minimizes torsional effects.

  2. 2.Explain why the shear centre is important in aircraft structures.Concept

    In aircraft structures, the shear centre is important because it helps in designing components that can withstand loads without twisting. This is critical for maintaining the aerodynamic shape and structural integrity of the aircraft. Proper alignment of loads with the shear centre ensures that the aircraft's structural components perform efficiently and safely under various loading conditions.

  3. 3.How is the shear centre determined for a simple thin-walled open section?Concept

    Assume a shear force S_y acts through the unknown shear centre, find the shear flow distribution from the open-section formula starting at a free edge, and take moments of the shear flows about a convenient point such as a corner. Setting that moment equal to S_y times its unknown lever arm gives one coordinate; repeating with S_x gives the other. Any axis of symmetry contains the shear centre, which often removes one step.

  4. 4.What happens if the load is not applied through the shear centre of a thin-walled section?Application

    If the load is not applied through the shear centre, the section will experience a twisting moment in addition to bending. This can lead to increased stresses and potential structural failure if not properly accounted for in the design. It can also affect the aerodynamic performance of aircraft components.

  5. 5.Why are closed sections often preferred over open sections in aircraft structures?Application

    Closed sections are often preferred because they have a higher torsional stiffness compared to open sections. This means they are less prone to twisting when subjected to loads that do not pass through the shear centre. This property is particularly beneficial in maintaining the structural integrity and aerodynamic shape of aircraft components.

  6. 6.Describe how the shear flow distribution is related to the shear centre in thin-walled sections.Concept

    The shear flow distribution in a thin-walled section is related to the shear centre because it determines how shear forces are transmitted through the section. The shear centre is the point where the resultant of the shear flow does not cause a twisting moment. Understanding this distribution is key to locating the shear centre and ensuring that loads are applied correctly.

  7. 7.Does the material affect the location of the shear centre?Application

    For a homogeneous isotropic section it is purely geometric. If the stiffness varies round the section, as in composite or mixed-material structures, the bending shear flows follow the modulus-weighted section properties, so the shear centre moves towards the stiffer parts; in closed sections the shear-stiffness distribution (G·t) also enters through the compatibility condition. Anisotropic laminates with bending-twisting coupling need a more general analysis.

  8. 8.Where is the shear centre of a thin-walled closed rectangular box 100 mm deep and 50 mm wide with uniform wall thickness?Numerical

    The box is doubly symmetric, and the shear centre lies on every axis of symmetry, so it is at the centroid, the geometric centre of the rectangle. No calculation is needed; the closed-cell analysis with the twist condition would give the same point. Only unsymmetric thicknesses or shapes move it.

  9. 9.For a thin-walled channel section, how does the position of the shear centre change with flange width?Application

    For uniform thickness the shear centre lies outside the section, beyond the web, at e = 3b²/(6b + h) from the web mid-line. As the flange width b increases, the flange shear forces and their couple grow faster than the bending stiffness, so e increases and the shear centre moves further away from the web. With b = 0 (a flat plate) it would lie on the web.

  10. 10.A thin-walled L-section has unequal legs. Where is its shear centre?Numerical

    At the intersection of the mid-lines of the two legs, i.e. at the corner. In each thin leg the shear flow runs along the leg, so its resultant acts along the leg's mid-line and passes through the corner. The moment of all shear flows about the corner is therefore zero for any shear load, which is the definition of the shear centre. This holds for equal or unequal legs and for T and cruciform sections too.

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