Elastic Section Modulus & Bending Stress
Relates the applied bending moment to the maximum peak flexural stress on the outermost fibers of a structural cross-section using section modulus S.
\[ S = \frac{I}{y_{max}} \implies \sigma_{max} = \frac{M}{S} = \frac{M \cdot y_{max}}{I} \]
Plain Text Formula:
S = I / y_max ==> sigma_max = M / S = (M * y_max) / I
📐 Variable Definitions & Parameters
| Symbol | Variable Name | Standard Unit | Physical Definition |
|---|---|---|---|
| \[ S \] | Elastic Section Modulus | m³ or cm³ | Geometric resistance to bending deformation |
| \[ I \] | Second Moment of Area | m⁴ | Moment of inertia about the neutral axis |
| \[ y_{max} \] | Distance to Extreme Fiber | m | Perpendicular distance from neutral axis to extreme surface |
| \[ M \] | Bending Moment | N·m or kN·m | Applied internal flexural moment |
| \[ \sigma_{max} \] | Maximum Bending Stress | MPa | Peak tensile or compressive surface stress |
💡 Worked Engineering Example
Problem Statement:
A rectangular timber beam (width b=100 mm, height h=200 mm) experiences a moment of 12 kN·m. Find peak bending stress.
Step-by-Step Solution:
- S = (b * h^2) / 6 = (100 * 200^2) / 6 = 666,667 mm³
- Convert M: 12 kN·m = 12,000,000 N·mm
- Stress = 12,000,000 / 666,667 = 18.0 MPa
🎯 Result: The maximum flexural stress is 18.0 MPa.