Beam Bending Stress Calculator

Calculate bending stress in beams from bending moment, moment of inertia, and cross-section geometry.

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Bending Stress Varies Linearly Through Beam Depth

In elastic beam bending, fibers on one side of the neutral axis stretch while fibers on the other side compress. The flexure formula σ=My/I gives normal stress at distance y from the neutral axis. Stress is zero at the neutral axis and reaches its largest magnitude at the extreme fibers where |y| is greatest.

The second moment of area I captures how cross-section geometry resists bending. Placing material farther from the neutral axis increases I strongly, which is why I-beams and deep sections can resist bending efficiently. The simple formula assumes elastic behavior and beam-theory conditions.

σ=My/I,   σmax=Mc/I=M/S
SymbolMeaningWhy it appears / units
MBending momentN·m at the section being checked.
yDistance from neutral axism; signed through the beam depth.
ISecond moment of aream⁴.
S=I/cSection modulusm³; convenient for maximum bending stress.

A moment diagram identifies where bending stress is likely largest along a beam, while the cross-section identifies where it is largest through the depth. Both loading and geometry therefore matter.

The section geometry is the first bending-stress check. Confirm that I is taken about the actual bending axis and that y is the distance to the point where stress is wanted. At the neutral axis y=0, bending stress must be zero; moving toward the extreme fiber must increase its magnitude linearly.

Worked Examples

Example 1: Rectangle 100×150mm, M=5000N·m
I=100×150³/12=28.1e6mm⁴, c=75
Result: σ=5000×1e6×75/28.1e6=13.3 MPa
Low bending stress
Example 2: Steel I-beam M=50kN·m, I=50×10⁶mm⁴, h=300mm
c=150mm
Result: σ=50e6×1e6×150/50e6=150 MPa
Steel yield: ~250 MPa — moderate
Example 3: Rectangular beam section
b=50mm, h=100mm → I=bh³/12
Result: I=4.17×10−6m⁴
Depth contributes with the third power to I for a rectangle.
Example 4: Stress from moment
M=2kN·m, c=0.05m, I=4.17×10−6m⁴
Result: σmax≈24.0MPa
Extreme-fiber stress follows directly once section geometry is known.

Common Mistakes

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Using polar moment J instead of bending moment I

J is associated with circular-shaft torsion. Beam bending uses the area second moment about the relevant neutral axis.

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Checking the wrong bending-moment location

Maximum stress often occurs where |M| is largest, which must be found from statics and the moment diagram.

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Ignoring yielding or local effects

The linear flexure formula is not a complete model after yielding or near stress concentrations, holes, notches, or very short/deep beams.

Frequently Asked Questions

Why maximum stress at outer fibre?
Bending creates strain proportional to distance from neutral axis (centroid). Maximum tension at bottom, maximum compression at top (or vice versa). Neutral axis has zero bending stress.
Section modulus S=I/c significance?
σ = M/S. Higher S → lower stress for same M. I-beams concentrate material far from neutral axis to maximize I/c (section modulus) with minimum material — structurally efficient shape.
Why is bending stress zero at the neutral axis?
Under pure elastic bending, the neutral-axis fibers have zero longitudinal strain, so their corresponding normal stress is zero.
Why are I-beams efficient?
They place much of the material far from the neutral axis, increasing I and section modulus without requiring a solid rectangular block of the same depth.
What determines whether the top surface is in tension or compression?
It depends on the sign of the bending moment and coordinate convention. Sagging and hogging moments reverse which side is stretched.
Is bending stress the same as shear stress?
No. Bending creates longitudinal normal stress, while transverse shear forces create a different shear-stress distribution across the section.
Where is the maximum bending stress in a beam section?
For the elementary flexure formula σ=My/I, stress magnitude increases with distance y from the neutral axis. The maximum bending stress therefore occurs at the extreme fiber, where y=c. Use the correct second moment of area about the bending axis and keep M, c, and I in one consistent unit system.

Formula Explorer connections

Interpretation: This engineering-physics relationship connects load, material property, geometry, deformation or system response. Assumption: Confirm material linearity, geometry, support conditions and safety convention. Small deformation, elastic behavior and ideal loading are common assumptions.

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