Bending stress is calculated with the flexure formula: σ = My/I. This equation gives you the stress at any point in a beam’s cross-section when it’s being bent by an applied load. The three variables you need are the bending moment (M) at the section you’re analyzing, the distance (y) from the neutral axis to the point of interest, and the moment of inertia (I) of the cross-section. Once you understand each variable and how to find it, the calculation itself is straightforward.
The Flexure Formula
The core equation for bending stress is:
σ = My / I
- σ = bending stress at the point you’re evaluating (in psi or MPa)
- M = the bending moment at the cross-section (in lb·in or N·m)
- y = the distance from the neutral axis to the point where you want the stress (in inches or mm)
- I = the second moment of area, commonly called the moment of inertia, of the cross-section (in in⁴ or mm⁴)
When you want the maximum bending stress, you substitute the largest bending moment in the beam for M and the farthest distance from the neutral axis (often called “c”) for y. That gives you:
σ_max = M_max × c / I
This formula works under a key assumption: the material behaves elastically (it springs back when the load is removed), and the cross-section is symmetric enough that strain varies linearly from the neutral axis outward. For most standard structural beams under normal loads, this holds true.
Step-by-Step Process
A University of Washington engineering reference lays out the sequence cleanly. For a beam with a constant cross-section and known supports:
- Step 1: Find the support reactions and draw the shear and moment diagrams for your beam.
- Step 2: Read the maximum bending moment (positive or negative) from the moment diagram.
- Step 3: Locate the centroid of the cross-section. This is where the neutral axis sits.
- Step 4: Calculate the moment of inertia (I) of the section about that centroid.
- Step 5: Measure the distance (c) from the centroid to the outermost fiber of the section.
- Step 6: Plug into σ_max = M_max × c / I.
The rest of this article walks through the trickiest parts: finding M, finding I, and understanding the neutral axis.
Finding the Bending Moment (M)
The bending moment depends on how the beam is supported and how it’s loaded. For standard cases, you can use well-known formulas rather than drawing a full moment diagram from scratch.
Simply Supported Beams
A simply supported beam rests on a pin at one end and a roller at the other, free to rotate at both supports. Two of the most common loading cases:
- Uniform distributed load (w) across the entire span (L): M_max = wL² / 8, occurring at midspan.
- Single concentrated load (P) at the center: M_max = PL / 4, also at midspan.
Cantilever Beams
A cantilever is fixed at one end and free at the other. The maximum moment always occurs at the fixed support:
- Uniform distributed load (w) across the full length (L): M_max = wL² / 2.
- Single concentrated load (P) at the free end: M_max = PL.
Notice that cantilevers produce much larger moments than simply supported beams for the same load and span. A cantilever with a point load at the tip has a maximum moment four times larger than the same load centered on a simply supported beam of the same length.
If your loading doesn’t match a standard case, you’ll need to solve for the support reactions using equilibrium equations and then construct a shear and moment diagram to locate and quantify the maximum moment.
Understanding the Neutral Axis and “y”
When a beam bends, one side compresses and the other side stretches in tension. Somewhere in between, there’s a line that neither compresses nor stretches. That’s the neutral axis, and it passes through the centroid of the cross-section.
For a symmetric shape like a rectangle or a circle, the centroid is simply at the geometric center. A 6-inch-tall rectangular beam has its neutral axis 3 inches from either edge. For an asymmetric section like a T-beam, you need to calculate the centroid by taking the area-weighted average of each component’s position.
The variable “y” in the flexure formula is measured from this neutral axis. To find the maximum stress, you use the largest value of y, which is the distance from the neutral axis to whichever edge of the cross-section is farthest away. In a symmetric section, both edges are equally far, so the maximum tensile and compressive stresses are equal in magnitude. In an asymmetric section, the stress is higher on whichever side is farther from the centroid.
Calculating the Moment of Inertia (I)
The moment of inertia (technically the second moment of area) quantifies how the material in a cross-section is distributed relative to the bending axis. More material farther from the neutral axis means a larger I, which means lower stress for the same moment. This is why I-beams are so efficient: they concentrate material in the flanges, far from the center.
Formulas for common shapes, with b = width, h = height, and r = radius:
- Rectangle: I = bh³ / 12
- Circle: I = πr⁴ / 4
For a rectangular beam that’s 2 inches wide and 6 inches tall: I = (2)(6³) / 12 = 36 in⁴. For a circular rod with a 2-inch radius: I = π(2⁴) / 4 ≈ 12.57 in⁴.
For composite shapes like I-beams or channels, you break the section into simple rectangles, calculate I for each piece about its own centroid, then use the parallel axis theorem (I = I_own + A × d²) to shift each piece’s inertia to the overall centroid. Add them all up for the total I.
A Worked Example
Say you have a simply supported beam spanning 120 inches, carrying a uniform distributed load of 50 lb/in, with a rectangular cross-section that’s 3 inches wide and 8 inches tall.
Bending moment: M_max = wL² / 8 = (50)(120²) / 8 = 90,000 lb·in.
Moment of inertia: I = bh³ / 12 = (3)(8³) / 12 = 128 in⁴.
Distance to extreme fiber: c = 8 / 2 = 4 inches (symmetric section).
Maximum bending stress: σ_max = (90,000)(4) / 128 = 2,812.5 psi.
That’s the peak stress at the top and bottom faces of the beam, at midspan where the moment is greatest. The top face is in compression, the bottom in tension (for a beam bending downward under load).
The Section Modulus Shortcut
If you’re always solving for maximum stress, you can simplify the formula by combining I and c into a single value called the elastic section modulus: S = I / c. Then the formula becomes:
σ_max = M / S
This is especially convenient for standard structural shapes because steel and aluminum design manuals list the section modulus for every catalog profile. You can look up S for a W10×22 steel beam, for example, and skip the I and c calculations entirely. In the worked example above, S = 128 / 4 = 32 in³, and σ_max = 90,000 / 32 = 2,812.5 psi, same result with one less step.
Units to Keep Consistent
The most common source of errors in bending stress calculations is mismatched units. In the US Customary system, stress comes out in psi (pounds per square inch) when M is in lb·in, y is in inches, and I is in in⁴. In the metric system, if M is in N·mm, y in mm, and I in mm⁴, stress comes out in MPa (since 1 N/mm² = 1 MPa). If your moment is in N·m, convert to N·mm by multiplying by 1,000 before plugging in.
Comparing Results to Material Strength
Once you’ve calculated the bending stress, you need to know whether the beam can handle it. The key comparison is your calculated stress versus the material’s yield strength, which is the stress level where permanent deformation begins. Typical yield strength ranges: structural steel falls between 250 and 550 MPa (roughly 36,000 to 80,000 psi), while aluminum alloys range from about 100 to 400 MPa (14,500 to 58,000 psi).
In practice, you don’t design right up to the yield point. Engineers apply a safety factor that accounts for uncertainties in loading, material quality, and real-world conditions. The allowable stress equals the yield strength divided by the safety factor. Design codes specify which safety factors to use for different materials and applications. If your calculated bending stress exceeds the allowable stress, you need a stronger material, a larger cross-section, or a shorter span.