Bending Stress from Section Modulus (σ = M/S)

σ=MS\sigma = \frac{M}{S}

Worked example: 10 kN·m through S = 8.333e-5 m^3 → 120 MPa — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Stress in bending →

UniversityMechanics of Materials

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning. Find 1 more lesson on this formula.

See your Report Card
Compete with your friends
share your results
Learning zone

Bending Stress from Section Modulus (σ = M/S) explained

σMS

This is how beams are actually selected in the field. Work out the maximum moment, divide by the allowable stress, and you have the section modulus you must buy: S ≥ M/σ_allow. A 20 000 ft·lbf moment at an allowable 24 ksi needs S = 240 000 in·lbf ÷ 24 000 psi = 10 in³ — flip to the table, find the lightest shape with S above 10 in³, done. In SI the same check on a 10 kN·m moment through a section with S = 8.333 × 10⁻⁵ m³ gives σ = 10 000 ÷ 8.333 × 10⁻⁵ = 120 MPa.

Enter S as a plain number in m³ (1 in³ = 1.6387 × 10⁻⁵ m³) and M in newton-metres; the answer comes back in pascals. The trap is that passing the stress check is not the same as passing the design. A beam sized purely on S may still deflect visibly, may buckle sideways if its compression flange is unbraced, and may crush its web where it lands on a bearing plate. Stress, deflection, lateral-torsional buckling and bearing are four separate checks, and in long shallow spans deflection usually governs first.

Bending Stress from Section Modulus (σ = M/S) formula

σ=MS\sigma = \frac{M}{S}
Where
  • σ\sigma= Bending stress (kPa)
  • MM= Bending moment (N·m)
  • SS= Section modulus (mm³)

Missing one of these? Work it out first, then come back