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Flexure Calculator

Bend a beam and the inside of it has to do some work. The curved-in side gets compressed, the curved-out side stretches, and a thin layer through the middle stays the same length. That thin layer is the neutral axis. Bending stress, or flexure, is what you get when you measure how hard those outer fibers are being pulled or squeezed.

This calculator runs two formulas. The general flexure formula, σ=MyI\sigma = \frac{My}{I}, works for any cross-section once you know the moment of inertia. For plain rectangular beams there's a shortcut, σmax=6Mbh2\sigma_{max} = \frac{6M}{bh^2}, that needs only the width and height. Useful when you're sizing lumber or rectangular bar stock and don't want to compute II separately.

Where the stress lives in the cross-section

Stress in a bent beam doesn't sit at a single value across the section. It varies linearly with distance from the neutral axis: zero in the middle, maximum at the top and bottom faces. That's why orientation matters so much. Lay a 2x6 flat across two sawhorses and it deflects dramatically under your weight. Stand the same board on edge and it barely budges, because the outer fibers are now further from the neutral axis and the moment of inertia is roughly nine times higher.

Using the calculator

Pick a formula set at the top. The General set takes the bending moment, the distance from the neutral axis, and the area moment of inertia. The Rectangular Beam set swaps II and yy for width and height, which most people can measure with a tape.

Fill in every input except the one you want to solve for. Whichever field you leave blank is what comes back, whether that's σ\sigma, MM, yy, II, bb, or hh. Units convert in place, so mixing Pa\text{Pa} with mm4\text{mm}^4 or psi\text{psi} with in4\text{in}^4 is fine.

Working through the math

Start with the general formula.

σ=M⋅yI\sigma = \frac{M \cdot y}{I}

σ\sigma is the bending stress at some point in the section, MM is the internal bending moment, yy is the distance from the neutral axis to that point, and II is the area moment of inertia. Put a 500 N·m moment on a small steel section with I=4×106 mm4I = 4 \times 10^{6} \ \text{mm}^4 and look at the top fiber, y=50 mmy = 50 \ \text{mm} above the neutral axis:

σ=500×0.054×10−6=6,250,000 Pa=6.25 MPa\sigma = \frac{500 \times 0.05}{4 \times 10^{-6}} = 6{,}250{,}000 \text{ Pa} = 6.25 \text{ MPa}

6.25 MPa, well under mild steel's yield strength of around 250 MPa. The beam has plenty of margin.

For a solid rectangle you can skip computing II. Substitute c=h2c = \frac{h}{2} and I=bh312I = \frac{b h^3}{12} into σmax=McI\sigma_{max} = \frac{Mc}{I} and the h3h^3 on top cancels with the hh on the bottom to give:

σmax=6Mbh2\sigma_{max} = \frac{6M}{b h^2}

Because hh is squared in the denominator, doubling the height drops the maximum stress to about a quarter for the same load. That's the geometry argument for tall, narrow joists.

Where flexure shows up

Pretty much anywhere something bends. Floor joists under your feet, the wing spar of a Cessna, the chassis rail of a pickup, the steel I-beams holding up an office floor. Engineers compute the maximum bending stress, compare it to the material's yield strength, and apply a safety factor of around 1.5 to 3 depending on the application.

It also explains a few things you can watch happen. A diving board curves most where the diver stands and breaks at the root, which is exactly where the bending moment is largest. A long bookshelf sags in the middle. A fishing rod bending against a fish is the same formula playing out along a tapered cross-section.

Things to watch out for

Use the maximum bending moment along the beam, not the value at the supports. For a simply-supported beam the moment at the supports is zero. For most loading patterns the peak lives at midspan or directly under a point load. Sketch a free-body diagram or run the loads through a beam calculator first.

For II, double-check you're using the value about the bending axis. The polar moment of inertia is a different number and using it here gives nonsensical answers.

FAQ

Is bending stress the same as tensile stress?

Both are normal stresses, but tensile stress is uniform across a section under an axial pull. Bending stress varies linearly: zero at the neutral axis, maximum at the outer fibers.

What if yy comes out negative?

The sign tells you which side of the neutral axis you're on. Positive yy usually points to the tension side, negative to the compression side. For a failure check you only care about the magnitude.

Why is the rectangular formula so much simpler?

Because cc and II both have clean expressions for a rectangle. Drop them into σmax=McI\sigma_{max} = \frac{Mc}{I}, simplify, and you get 6Mbh2\frac{6M}{b h^2} with no separate II calculation.

Does this account for shear stress?

No. Shear is a separate calculation and is usually small compared to bending for long, slender beams. For short, deep beams it's worth checking shear separately.

Author

hexacalculator design team

Our team blends expertise in mathematics, finance, engineering, physics, and statistics to create advanced, user-friendly calculators. We ensure accuracy, robustness, and simplicity, catering to professionals, students, and enthusiasts. Our diverse skills make complex calculations accessible and reliable for all users.

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