Interactive Lecture
Principal Axes, Neutral Axis Location, and Stress Distribution
by Dr Arnaldo Delli Carri
When a beam has an asymmetric cross-section or when bending moments are applied about axes other than principal axes, the neutral axis does not align with the applied moment, and stress distribution becomes more complex than simple beam bending theory.
In asymmetric bending, the neutral axis is not necessarily parallel to either applied moment — its orientation depends on the cross-section geometry and the moment direction. You need the principal axes to find it, and without it you can't locate the maximum stress.
For symmetric sections with moment applied about a principal axis:
Bending Stress (Flexure Formula):
\[\sigma = \frac{My}{I}\]
Sign convention: Positive M produces tension on positive y side (bottom), compression on negative y side (top).
Where:
Neutral Axis: Coincides with centroidal axis perpendicular to applied moment
Principal axes are orthogonal axes through the centroid about which the product moment of area is zero.
Product Moment of Area:
\[I_{xy} = \int\int xy \, dA\]
For principal axes: \(I_{xy} = 0\)
Principal Moments of Inertia:
\[I_1, I_2 = \frac{I_x + I_y}{2} \pm \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2}\]
Principal Angle:
\[\tan(2\theta_p) = \frac{2I_{xy}}{I_x - I_y}\]
When moments are applied about non-principal axes or for asymmetric sections:
General Bending Stress Formula:
\[\sigma = \frac{(M_x I_y + M_y I_{xy})\,y - (M_y I_x + M_x I_{xy})\,x}{I_x I_y - I_{xy}^2}\]
Simplified for principal axes:
\[\sigma = \frac{M_1 y_1}{I_1} - \frac{M_2 y_2}{I_2}\]
where \(y_1\) and \(y_2\) are the perpendicular distances from principal axes 1 and 2 respectively
The neutral axis is the locus of points where \(\sigma = 0\):
Neutral Axis Equation:
\[\frac{M_1 y_1}{I_1} - \frac{M_2 y_2}{I_2} = 0\]
Neutral Axis Inclination:
\[\tan(\alpha) = \frac{M_2 I_1}{M_1 I_2}\]
where \(\alpha\) is the inclination of the neutral axis measured from principal axis 1
When bending is about a principal axis, the neutral axis coincides with that axis — it lies parallel to the moment vector, not perpendicular to it. The neutral axis only ends up perpendicular to the load when the section is symmetric and loaded in its plane of symmetry. For asymmetric bending the neutral axis rotates away from the principal axis, and locating it is the whole problem.
Similar to stress transformation, moments of inertia can be visualised using Mohr's circle:
Transformation Equations:
\[I_{\xi} = \frac{I_x + I_y}{2} + \frac{I_x - I_y}{2}\cos(2\theta) - I_{xy}\sin(2\theta)\]
\[I_{\eta} = \frac{I_x + I_y}{2} - \frac{I_x - I_y}{2}\cos(2\theta) + I_{xy}\sin(2\theta)\]
\[I_{\xi\eta} = \frac{I_x - I_y}{2}\sin(2\theta) + I_{xy}\cos(2\theta)\]
For asymmetric sections under bending:
Explore how neutral axis location and stress distribution change with section geometry and moment direction.
| Property | Equation | Description |
|---|---|---|
| Bending Stress | \(\sigma = \frac{M_x y}{I_x} - \frac{M_y x}{I_y}\) | General formula (principal axes) |
| Product of Inertia | \(I_{xy} = \int\int xy \, dA\) | Measure of asymmetry |
| Principal Moments | \(I_{1,2} = \frac{I_x + I_y}{2} \pm R\) | \(R = \sqrt{((I_x-I_y)/2)^2 + I_{xy}^2}\) |
| Principal Angle | \(\tan(2\theta_p) = \frac{2I_{xy}}{I_x - I_y}\) | Orientation of principal axes |
| Neutral Axis Angle | \(\tan(\alpha) = \frac{M_2 I_1}{M_1 I_2}\) | Inclination from principal axis 1 |
An L-shaped angle section (100mm × 100mm × 10mm) experiences a bending moment of 5 kN·m about the horizontal axis. The section properties about centroidal axes are: \(I_x = 1.8 \times 10^6\) mm⁴, \(I_y = 1.8 \times 10^6\) mm⁴, \(I_{xy} = 1.2 \times 10^6\) mm⁴. Calculate:
(a) Principal moments of inertia
(b) Principal angle
(c) Maximum bending stress
Show Solution
(a) Principal Moments of Inertia:
\[I_{avg} = \frac{I_x + I_y}{2} = \frac{1.8 + 1.8}{2} = 1.8 \times 10^6 \text{ mm}^4\]
\[R = \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2} = \sqrt{0 + (1.2 \times 10^6)^2} = 1.2 \times 10^6 \text{ mm}^4\]
\[I_1 = 1.8 \times 10^6 + 1.2 \times 10^6 = 3.0 \times 10^6 \text{ mm}^4\]
\[I_2 = 1.8 \times 10^6 - 1.2 \times 10^6 = 0.6 \times 10^6 \text{ mm}^4\]
(b) Principal Angle:
\[\tan(2\theta_p) = \frac{2I_{xy}}{I_x - I_y} = \frac{2(1.2 \times 10^6)}{1.8 - 1.8} = \infty\]
This means \(2\theta_p = 90°\), so \(\theta_p = 45°\)
(c) Maximum Bending Stress:
The moment acts about the x-axis (\(M_x = 5\,\text{kN·m}\), \(M_y = 0\)). Resolve it into the principal directions (\(\theta_p = 45°\)):
\(M_1 = M\cos(45°) = 5 \times 0.707 = 3.54\,\text{kN·m}\)
\(M_2 = -M\sin(45°) = -3.54\,\text{kN·m}\)
For this L100×100×10 angle the centroid lies 28.7 mm from each outer face. The tip of the upright leg is the critical fibre; in the principal frame its coordinates are \(u = 37.2\) mm along axis 1 and \(v = 63.6\) mm along axis 2. The stress there is
\[\sigma = \frac{M_1 v}{I_1} - \frac{M_2 u}{I_2} = \frac{3.54\times10^6 \times 63.6}{3.0\times10^6} - \frac{(-3.54\times10^6) \times 37.2}{0.6\times10^6}\]
\[\sigma = 75 + 220 = 295\,\text{MPa (tension)}\]
The heel of the angle carries the largest compression, \(\sigma = -\dfrac{M_2 u}{I_2} \approx -239\,\text{MPa}\) at \(u = -40.6\) mm.
Answer: (a) \(I_1 = 3.0\times10^6\,\text{mm}^4\), \(I_2 = 0.6\times10^6\,\text{mm}^4\); (b) \(\theta_p = 45°\); (c) \(\sigma_{max} \approx 295\,\text{MPa}\) tension at the leg tip (−239 MPa compression at the heel).
A beam with principal moments \(I_1 = 50 \times 10^6\) mm⁴ and \(I_2 = 20 \times 10^6\) mm⁴ is subjected to moments \(M_1 = 8\) kN·m and \(M_2 = 3\) kN·m about the principal axes. Find the angle of the neutral axis relative to axis 1.
Show Solution
Neutral Axis Angle:
\[\tan(\alpha) = \frac{M_2 I_1}{M_1 I_2} = \frac{3 \times 50 \times 10^6}{8 \times 20 \times 10^6}\]
\[\tan(\alpha) = \frac{150}{160} = 0.9375\]
\[\alpha = \arctan(0.9375) = 43.2°\]
The neutral axis is inclined 43.2° anticlockwise from principal axis 1. Notice it swings toward the axis carrying the larger \(M/I\) contribution.
Answer: \(\alpha \approx 43.2°\) from principal axis 1.
For a T-section with moment applied 30° from horizontal, calculate the stress at a point (x = 30mm, y = 40mm) from the centroid. Given: \(I_x = 2.0 \times 10^6\) mm⁴, \(I_y = 0.8 \times 10^6\) mm⁴, \(I_{xy} = 0\), \(M = 10\) kN·m.
Show Solution
Moment Components:
\(M_x = M\sin(30°) = 10 \times 0.5 = 5\) kN·m
\(M_y = M\cos(30°) = 10 \times 0.866 = 8.66\) kN·m
Stress Formula (Ixy = 0):
\[\sigma = \frac{M_x y}{I_x} - \frac{M_y x}{I_y}\]
\[\sigma = \frac{5 \times 10^6 \times 40}{2.0 \times 10^6} - \frac{8.66 \times 10^6 \times 30}{0.8 \times 10^6}\]
\[\sigma = 100 - 324.75 = -224.75 \text{ MPa (compression)}\]
Answer: \(\sigma \approx -225\,\text{MPa}\) (compression).