Bending Stress Calculator
Enter the bending moment and beam cross-section dimensions to find the maximum bending stress, moment of inertia, and section modulus. Supports seven cross-section shapes, six dimension units, and six stress output units. Results update instantly and a "show your work" panel walks through the flexure formula step by step.
Formula
Worked example
A rectangular beam 100 mm wide and 150 mm deep carries a bending moment of 5 kN m. I = (100 x 150^3) / 12 = 28,125,000 mm^4. c = 150/2 = 75 mm. sigma = (5,000,000 N mm x 75 mm) / 28,125,000 mm^4 = 13.33 MPa.
What is bending stress?
When a beam carries a load, the cross-section at any point experiences an internal bending moment. This moment stretches the fibres on one side of the neutral axis (tension) and compresses the fibres on the other side (compression). The resulting normal stress varies linearly across the depth of the section, reaching its maximum value at the outermost fibre, the point farthest from the neutral axis. That maximum value is what engineers call the bending stress, or flexural stress, and it must stay below the material yield strength (divided by an appropriate safety factor) to prevent permanent deformation or fracture.
The flexure formula
The standard engineering beam theory (Euler-Bernoulli) gives the bending stress at any fibre y from the neutral axis as: sigma = M y / I. At the extreme fibre (y = c) the stress is maximum: sigma_max = M c / I. Because I / c appears repeatedly, engineers define the section modulus S = I / c, reducing the formula to sigma_max = M / S. A larger S means the section can carry more moment for the same stress, which is why structural shapes like I-beams are designed to push as much material as possible away from the neutral axis.
Choosing an efficient cross-section
For pure bending in one plane, the most efficient shapes concentrate material at the top and bottom flanges, far from the neutral axis. I-beams achieve this while keeping the overall weight low. Rectangular hollow sections (RHS) perform similarly and resist bending in two planes. Solid circles are compact but relatively inefficient per kilogram because material near the centroid adds little to the moment of inertia. T-sections have an asymmetric neutral axis, so the extreme fibre distance differs on the tension and compression sides, and the controlling fibre stress must be checked separately for each. When selecting a section, always compare S values alongside weight per unit length to find the most material-efficient option for the design moment.
Safety factors and design checks
Bending stress alone is rarely the only check in a structural design. Shear stress at the neutral axis, deflection under service loads, local flange buckling, and lateral-torsional buckling are all equally important limit states that most design codes require you to verify. Typical safety factors against yielding range from 1.5 (LRFD at ultimate) to 2.5 or more for brittle materials. Fatigue loading, dynamic effects, and stress concentrations at holes or welds require additional reduction factors. This calculator gives the elastic bending stress on a prismatic cross-section with no holes or notches; apply the relevant code factors before drawing design conclusions.
Moment of inertia formulas by cross-section
| Shape | I formula | c (extreme fibre) |
|---|---|---|
| Rectangle (b x h) | b h³ / 12 | h / 2 |
| Square (a x a) | a⁴ / 12 | a / 2 |
| RHS (b x h, wall t) | (b h³ - bᵢ hᵢ³) / 12 | h / 2 |
| Solid circle (d) | π d⁴ / 64 | d / 2 |
| Hollow circle (d, di) | π (d⁴ - di⁴) / 64 | d / 2 |
| I-beam (sym.) | (b h³ - (b-tw) hw³) / 12 | h / 2 |
| T-section | Parallel-axis theorem (asymmetric) | Larger of ybar, H-ybar |
Standard closed-form expressions for the second moment of area about the horizontal centroidal axis.
Frequently asked questions
What is the flexure formula and where does it come from?
The flexure formula sigma = M c / I comes from Euler-Bernoulli beam theory, which assumes that plane cross-sections remain plane after bending and that the material behaves linearly elastic. Under these assumptions, strains (and therefore stresses) vary linearly across the depth of the section, reaching zero at the neutral axis and their peak at the extreme fibre. Integrating the resulting stress distribution over the cross-sectional area gives the moment equilibrium condition, and solving for stress at c yields the formula. It is accurate for slender beams where the depth-to-span ratio is small and is the standard starting point in structural steel, timber, and concrete design codes.
What is the section modulus and why does it matter?
The section modulus S = I / c is a single number that captures a cross-section's bending efficiency. Because sigma_max = M / S, a higher S means a lower peak stress for the same applied moment. When comparing different beam profiles of similar size, the one with the highest S carries the most moment before reaching yield. Standard section tables in design codes (AISC, EN 1993, etc.) list S directly so engineers can select beams by required section modulus without recalculating I and c each time.
How does the neutral axis location affect bending stress?
For symmetric sections such as rectangles, squares, I-beams, and circles, the neutral axis passes through the centroid and is equidistant from both extreme fibres, so the tensile and compressive peak stresses are equal. For asymmetric sections such as T-sections, the neutral axis is closer to the larger element, and the two extreme fibre distances (c_top and c_bot) differ. The critical stress is at the larger c, but if the material has different allowable stresses in tension and compression (as concrete does), you must check both fibres against their respective limits.
What units should I use for bending moment and section dimensions?
You can use any consistent set of units as long as you apply them throughout. Common engineering combinations are: N and mm (gives stress in MPa, which equals N/mm^2); kN and m (gives stress in kPa, then divide by 1000 for MPa); lbf and in (gives stress in psi). This calculator converts all inputs to SI internally before computing, then converts back to your chosen stress unit for output, so you can freely mix moment units (kN m, ft lbf, etc.) with dimension units (mm, in, etc.).
Does this calculator account for shear stress or deflection?
No. This tool computes only the normal bending stress from the flexure formula. In most beams, shear stress peaks at the neutral axis and is zero at the extreme fibres, so it does not add to the bending stress directly. However, combined bending plus shear can create a critical stress state at intermediate fibres, which requires a principal-stress or Von Mises check. Deflection depends on the full span, loading pattern, and support conditions, and requires a separate beam-deflection calculation. Always perform a complete set of structural checks before finalising a design.