Unsymmetrical bending of beams

Why it matters

Wing boxes, stringers, Z-sections and angle sections used in aircraft are rarely symmetric, and wing loads act both vertically (lift) and horizontally (drag). Under these conditions the simple formula σ = M·y/I gives the wrong stress and the beam deflects out of the plane of loading. Every booms-and-skin wing calculation in this subject starts from the general bending formula developed here.

Key ideas

When bending is "unsymmetrical". Bending is symmetrical only when the applied moment acts about a principal axis of the cross-section. It is unsymmetrical when (a) the section has no axis of symmetry, so the product of inertia I_xy ≠ 0 in the chosen axes, or (b) the moment is inclined to the principal axes, as with combined lift and drag on a wing.

Assumptions. Plane sections remain plane (pure bending, Bernoulli–Navier), linear elastic homogeneous material, small deflections, direct stress σ_z only. Axes x and y pass through the centroid of the section; z is along the beam.

Section properties. I_xx = ∫y² dA, I_yy = ∫x² dA, I_xy = ∫x·y dA about centroidal axes. I_xy is zero if either axis is an axis of symmetry; it can be positive or negative. Principal axes are the centroidal axes for which I_xy = 0; they are found from tan 2θ = −2·I_xy/(I_xx − I_yy) (θ measured from x towards y). Principal second moments are the maximum and minimum second moments of the section.

General bending stress. With the sign convention that a positive M_x produces tension in the region of positive y and a positive M_y produces tension in the region of positive x, the direct stress at (x, y) is the formula in the Formulas section. If I_xy = 0 it reduces to σ_z = M_x·y/I_xx + M_y·x/I_yy, simple superposition of two symmetrical bending cases.

Neutral axis. The neutral axis is the line where σ_z = 0. It always passes through the centroid (for pure bending with no axial load), but in unsymmetrical bending it is not perpendicular to the plane of the applied moment, i.e. not parallel to the moment vector. The maximum stress occurs at the point farthest from the neutral axis, not necessarily farthest from the x-axis.

Deflection. The beam deflects perpendicular to the neutral axis, so a vertical load on a Z-section or angle produces a sideways deflection too. This is why stringers twist and bow and why unsymmetrical sections need lateral support.

Thin-walled approximation. For thin walls, second moments are computed along the wall mid-line, ignoring terms in t³. A thin flange of width b and thickness t at distance d from an axis contributes b·t·d² to the second moment about that axis.

Formulas

σ_z = [(M_y·I_xx − M_x·I_xy) / (I_xx·I_yy − I_xy²)]·x + [(M_x·I_yy − M_y·I_xy) / (I_xx·I_yy − I_xy²)]·y

• σ_z: direct stress (Pa, tension +), M_x, M_y: bending moments about centroidal x and y axes (N·m), sign convention as above; I_xx, I_yy, I_xy: second moments and product of area about centroidal axes (m⁴); x, y: coordinates of the point (m).

σ_z = M_x·y / I_xx + M_y·x / I_yy

• Same symbols. Valid only when x and y are principal axes (I_xy = 0).

y / x = −(M_y·I_xx − M_x·I_xy) / (M_x·I_yy − M_y·I_xy)

• Slope of the neutral axis through the centroid.

tan α = (I_xx / I_yy)·tan θ

• Doubly symmetric section, moment vector inclined at θ to the x principal axis; α is the inclination of the neutral axis to the x-axis.

tan 2θ_p = −2·I_xy / (I_xx − I_yy)

• θ_p: angle of a principal axis from the x-axis.

Worked examples

Example 1 (standard): rectangle under an inclined moment. Given: solid rectangle, width b = 100 mm (along x), depth h = 200 mm (along y). A moment M = 20 kN·m acts with its vector at θ = 30° to the x-axis.

1. Second moments: I_xx = b·h³/12 = 100 × 200³/12 = 6.667 × 10⁷ mm⁴, I_yy = h·b³/12 = 200 × 100³/12 = 1.667 × 10⁷ mm⁴. The axes are axes of symmetry, so I_xy = 0.
2. Components: M_x = M·cos 30° = 17.32 × 10⁶ N·mm, M_y = M·sin 30° = 10.0 × 10⁶ N·mm.
3. Stress at the corner (x = 50 mm, y = 100 mm), where both components give tension: σ = M_x·y/I_xx + M_y·x/I_yy = 17.32 × 10⁶ × 100/6.667 × 10⁷ + 10.0 × 10⁶ × 50/1.667 × 10⁷ = 25.98 + 30.00 = 55.98 N/mm².
4. Neutral axis: tan α = (I_xx/I_yy)·tan θ = 4 × 0.5774 = 2.309, so α = 66.6° from the x-axis, far from the 30° of the moment vector. Answer: σ_max = 56.0 MPa (tension at one corner, compression at the opposite corner); neutral axis at 66.6°. For comparison, the same 20 kN·m about x alone gives only 30 MPa.

Example 2 (GATE level): thin-walled Z-section. Given: Z-section with vertical web of mid-line height h = 200 mm and two flanges of width b = 80 mm, all of thickness t = 8 mm (thin-walled mid-line model). The top flange runs from (0, 100) to (80, 100) and the bottom flange from (0, −100) to (−80, −100). Applied M_x = 10 kN·m, M_y = 0.

1. I_xx = t·h³/12 + 2·b·t·(h/2)² = 8 × 200³/12 + 2 × 80 × 8 × 100² = 5.333 × 10⁶ + 12.8 × 10⁶ = 18.13 × 10⁶ mm⁴.
2. I_yy = 2·t·b³/3 = 2 × 8 × 80³/3 = 2.731 × 10⁶ mm⁴ (web adds nothing in the thin-wall model).
3. I_xy: top flange t·(100)·(b²/2) = 8 × 100 × 3200 = 2.56 × 10⁶, bottom flange the same sign (x and y both negative), so I_xy = 5.12 × 10⁶ mm⁴.
4. Denominator: I_xx·I_yy − I_xy² = 4.952 × 10¹³ − 2.621 × 10¹³ = 2.330 × 10¹³ mm⁸.
5. Coefficients: −M_x·I_xy/D = −10⁷ × 5.12 × 10⁶/2.330 × 10¹³ = −2.197 N/mm³ on x; M_x·I_yy/D = 10⁷ × 2.731 × 10⁶/2.330 × 10¹³ = 1.172 N/mm³ on y.
6. So σ_z = −2.197·x + 1.172·y (N/mm², x and y in mm).
7. Web-flange corner (0, 100): σ = 117.2 N/mm². Top flange tip (80, 100): σ = −175.8 + 117.2 = −58.6 N/mm². By antisymmetry the bottom corner is −117.2 and the bottom tip +58.6.
8. Neutral axis: y/x = 2.197/1.172 = 1.875, i.e. at 61.9° to the x-axis. Answer: σ_max = ±117.2 MPa at the web-flange corners; flange tips −58.6 / +58.6 MPa. Ignoring I_xy would give M_x·y/I_xx = 55.1 MPa everywhere along the top flange, which is wrong in magnitude and, at the tip, in sign.

Common mistakes

• Using σ = M·y/I for a section with I_xy ≠ 0 (Z-sections, angles, unsymmetric booms).
• Getting the sign of I_xy wrong: it is the sign of x·y for each area element about the centroid.
• Mixing sign conventions for M_x and M_y; fix one convention and stick to it in the formula.
• Taking moments about axes that do not pass through the centroid.
• Saying the neutral axis is perpendicular to the load plane, or that it misses the centroid. It passes through the centroid but is inclined.
• Reading the maximum stress at the point farthest from the x-axis instead of farthest from the neutral axis.

For GATE AE

Expect calculation of I_xx, I_yy, I_xy for thin-walled Z, channel or angle sections, stress at a given point from the general formula, neutral-axis inclination, and principal-axis angle. Conceptual items ask when the simple formula applies and how deflection relates to the neutral axis. Practise the thin-walled section property calculation carefully; most errors are there.

Quick check

1. When does σ = M_x·y/I_xx give the right stress?
2. A section has I_xx = 4·I_yy and the moment vector is at 45° to x. At what angle is the neutral axis?
3. Does the neutral axis pass through the centroid in unsymmetrical bending?
4. What is I_xy for a channel section about its centroidal axes when x is the axis of symmetry? Answers: 1. when x and y are principal axes and only M_x acts; 2. tan α = 4 × 1 = 4, α = 76.0°; 3. yes (pure bending, no axial load); 4. zero.