When a beam bends under load, normal stresses develop across its cross-section. These stresses aren't uniform. They vary linearly from zero at the neutral axis to their maximum values at the outermost fibers.

The pattern works like this: on one side of the neutral axis, fibers are being compressed (shortened), while on the other side, fibers are being stretched (elongated). For a simply supported beam with a downward load, compressive stresses develop on the top and tensile stresses develop on the bottom.

The stress at any point is directly proportional to its distance from the neutral axis. A point twice as far from the neutral axis experiences twice the stress.
The distribution flips direction across the neutral axis, creating a linear gradient from maximum compression to maximum tension.
Both the magnitude and direction of the applied bending moment control this distribution.

Maximum Stress Location

Maximum normal stresses always occur at the extreme fibers, the points farthest from the neutral axis. Where exactly those points are depends on the cross-section shape:

Rectangular beam: maximum stresses at the top and bottom surfaces
I-beam: maximum stresses at the top and bottom flanges
Circular beam: maximum stresses at the very top and bottom of the circle

For a doubly symmetric cross-section, the maximum tensile and compressive stresses have equal magnitudes but opposite signs. Identifying these locations is the first step in checking whether a beam can safely carry its load.

Neutral Axis Location

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Symmetric Cross-Sections

The neutral axis is the line across the cross-section where normal stress equals zero. No tension, no compression. Everything pivots around this line.

For symmetric cross-sections (rectangular, circular, I-shaped), the neutral axis passes through the centroid of the cross-section. This makes life easier because you can find it by geometry alone:

Rectangular beam: the neutral axis sits at mid-height, $h/2$ from either surface
Circular beam: the neutral axis passes through the center of the circle
I-beam: the neutral axis coincides with the horizontal centerline of the web

Unsymmetric Cross-Sections

For unsymmetric shapes like T-beams or L-beams, the neutral axis still passes through the centroid, but the centroid is no longer at the obvious geometric center. You need to calculate its location using the first moment of area.

Here's the process:

Divide the cross-section into simple shapes (rectangles, etc.).
Choose a reference axis (typically the bottom edge of the section).
For each sub-shape, calculate its area $A_i$ and the distance $\bar{y}_i$ from the reference axis to its own centroid.
Compute the centroid location: $\bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i}$

The neutral axis passes through this centroid. For a T-beam, the neutral axis typically sits closer to the flange (the wider part) because the flange contributes more area. For an L-beam, it shifts toward the corner with the larger leg.

Maximum Normal Stress Calculation

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Flexure Formula

The flexure formula is the central equation for this topic:

$\sigma = \frac{My}{I}$

$\sigma$ = normal stress at the point of interest
$M$ = internal bending moment at the cross-section
$y$ = distance from the neutral axis to the point of interest
$I$ = moment of inertia of the cross-section about the neutral axis

To find the maximum normal stress, plug in $y_{max}$ (the distance from the neutral axis to the extreme fiber) along with the bending moment $M$ at that section:

$\sigma_{max} = \frac{M \cdot y_{max}}{I}$

You'll sometimes see this written using the section modulus $S = I / y_{max}$ , which simplifies the expression to $\sigma_{max} = M / S$ . This form is especially handy for design problems where you're selecting a beam size.

Moment of Inertia

The moment of inertia $I$ quantifies how the cross-sectional area is distributed relative to the neutral axis. A larger $I$ means lower stress for the same bending moment, which is why I-beams (with material concentrated far from the neutral axis) are so efficient.

Standard formulas for common shapes:

Rectangle (base $b$ , height $h$ ): $I = \frac{bh^3}{12}$
Circle (radius $r$ ): $I = \frac{\pi r^4}{4}$
Hollow rectangle / I-shape: subtract the inner rectangle's $I$ from the outer, e.g., $I = \frac{bh^3}{12} - \frac{b_1 h_1^3}{12}$

For composite or irregular sections, use the parallel axis theorem:

$I = \bar{I} + A d^2$

where $\bar{I}$ is the moment of inertia of a sub-shape about its own centroid, $A$ is its area, and $d$ is the distance between the sub-shape's centroid and the overall neutral axis. Sum this for each sub-shape to get the total $I$ .

Bending Moment and Normal Stress

Proportional Relationship

The flexure formula $\sigma = My/I$ shows that stress and bending moment are directly proportional (with $y$ and $I$ held constant at a given point). Double the bending moment, and you double the stress at every point in the cross-section. This linear relationship is what makes superposition work for beam problems: you can analyze load cases separately and add the results.

Stress Distribution along the Beam

The bending moment generally varies along the length of a beam, so the normal stress varies along the length too. The critical design section is wherever the bending moment reaches its maximum magnitude.

Sign convention matters. A positive bending moment (beam sagging) produces tensile stress at the bottom fibers and compressive stress at the top. A negative bending moment (beam hogging) reverses this: compression on the bottom, tension on the top.

Common cases to know:

Simply supported beam, uniform distributed load: maximum bending moment (and maximum normal stress) occurs at midspan, with $M_{max} = \frac{wL^2}{8}$ .
Cantilever beam, point load at the free end: maximum bending moment occurs at the fixed support, with $M_{max} = PL$ .
Simply supported beam, point load at midspan: maximum bending moment at midspan, with $M_{max} = \frac{PL}{4}$ .

At any cross-section, once you know $M$ from your shear and moment diagrams, you can plug it into the flexure formula to find the stress distribution at that location. The sections with the largest $|M|$ are the ones most likely to fail, so those are where you focus your design checks.