Bending Stress

Calculate bending stress, design rectangular cross-sections, and analyse moment–curvature relationships for beams

Inputs

M

Bending Moment

N·m

Z

Section Modulus

mm³

Formula Interpretation

Bending Stress at Fibre y

\sigma = \varepsilon E = \dfrac{y}{\rho} E

At any fibre a distance y from the neutral axis, the strain is ε = y/ρ (where ρ is the radius of curvature). Multiplied by Young''s modulus E, this gives the bending stress. Stress is zero at the neutral axis and maximum at the extreme fibres.

Moment–Curvature Relationship

\begin{aligned} M &= \dfrac{E}{\rho} I \\ \rho &= \dfrac{EI}{M} \end{aligned}

Integrating the fibre stress distribution over the cross-section gives the bending moment M = (E/ρ)·I. Rearranging, the radius of curvature ρ = EI/M. A larger EI (flexural rigidity) means less curvature (stiffer beam) for the same moment.

Design Formula — Section Modulus

\begin{aligned} M &= \sigma_b Z \\ \sigma_b &= \dfrac{M}{Z} \end{aligned}

Combining the two preceding equations eliminates ρ and E, leaving the practical design formula. For a given bending moment M, the maximum bending stress is σ_b = M/Z. To limit stress below [σ], choose a section with Z ≥ M/[σ].

Rectangle: I and Z

\begin{aligned} I &= \dfrac{bh^3}{12} \\ Z &= \dfrac{bh^2}{6} \end{aligned}

For a rectangular section of width b and height h. With the constraint h = n·b (aspect ratio), the section modulus becomes Z = (n²/6)·b³, giving b = ∛(6Z/n²) once the required Z is known.

Knowledge Points

Neutral Layer and Neutral Axis

When a beam bends, the layer that neither stretches nor compresses is the neutral layer. Its intersection with any cross-section is the neutral axis (N–N). For a homogeneous symmetric section this coincides with the centroidal axis. All bending stress is measured from this axis.

Linear Stress Distribution

Bending stress varies linearly across the depth: zero at the neutral axis, maximum compression at the top fibre ( $\sigma_c = M \cdot y_c/I$ ) and maximum tension at the bottom fibre ( $\sigma_t = M \cdot y_t/I$ ). For symmetric sections $y_c = y_t = h/2$ , so $\sigma_c = \sigma_t = M/Z$ .

Section Modulus Z Governs Strength

The formula $\sigma_b = M/Z$ shows that for a given bending moment, maximising $Z$ minimises stress. Orientating a rectangular beam with the larger dimension vertical (large $h$ ) is the most effective way to increase $Z$ and hence reduce bending stress.

Flexural Rigidity EI Governs Stiffness

The product $EI$ — Young''s modulus times second moment of area — is the flexural rigidity. It controls deflection: $\delta \propto 1/(EI)$ . A stiffer material (high $E$ ) or a deeper section (high $I$ ) both reduce deflection. Strength ( $\sigma$ ) and stiffness ( $\delta$ ) require separate checks.

Worked Example

A simply supported beam of span 1000 mm carries a 4000 N load at 400 mm from A and a 2000 N load at 700 mm from A. The allowable bending stress is 60 MPa. Determine the required rectangular cross-section dimensions with b:h = 1:2.

Knowns

• Span: l = 1000 mm
• Load 1: 4000 N at C = 400 mm from A
• Load 2: 2000 N at D = 700 mm from A
• Allowable stress: [σ] = 60 MPa
• Aspect ratio: h = 2b (n = 2)

Solution

Step 1 — Support Reactions

R_A = R_B = 3000\,\text{N}

Step 2 — Bending Moments at C and D

\begin{aligned} M_C &= -3000 \times 400 \times 10^{-3} = -1200\,\text{N}{\cdot}\text{m} \\ M_D &= -3000 \times 700 \times 10^{-3} + 4000 \times 300 \times 10^{-3} = -900\,\text{N}{\cdot}\text{m} \end{aligned}

Step 3 — Maximum Bending Moment

M_{\max} = -1200\,\text{N}{\cdot}\text{m} \quad (\text{at C})

Step 4 — Required Section Modulus (Formula ③)

Z = \dfrac{M}{\sigma_b} = \dfrac{1200}{60 \times 10^6} = 20{,}000 \times 10^{-9}\,\text{m}^3 = 20{,}000\,\text{mm}^3

Step 5 — Cross-Section Dimensions (h = 2b)

Z = \dfrac{1}{6}b(2b)^2 = \dfrac{2}{3}b^3 = 20{,}000 \implies b^3 = 30{,}000 \implies b = \sqrt[3]{30{,}000} \approx 31.1\,\text{mm}

Step 6 — Height

h = 2b = 2 \times 31.1 \approx 62.2\,\text{mm}

Result: $M_{max}$ = 1200 N·m at C. Required $Z$ = 20 000 mm³. Rectangular section: $b$ ≈ 31.1 mm, $h$ ≈ 62.2 mm ( $h$ = 2 $b$ ).

Extended Knowledge

•Structural steel I-sections concentrate material in the flanges, far from the neutral axis, maximising I and Z for a given cross-sectional area. The flange-dominated stress distribution makes the web lightly stressed — it primarily resists shear, not bending.
•Concrete is strong in compression but weak in tension (ratio ≈ 10:1). Steel reinforcing bars are placed in the tension zone to carry the tensile bending stress. The neutral axis shifts upward relative to the midheight, and the stress distribution is no longer symmetric.
•When a beam also carries an axial force $N$ , the total normal stress at any fibre is $\sigma = N/A \pm M \cdot y/I$ . The neutral axis shifts away from the centroid, and one extreme fibre may reach the yield stress before the other. This ''eccentric loading'' scenario is common in columns.
•At the elastic limit, only the extreme fibres reach yield stress $\sigma_y$ . Beyond this, yielding spreads inward. The fully-plastic moment $M_p = \sigma_y \cdot Z_p$ , where $Z_p$ is the plastic section modulus. The shape factor $f = Z_p / Z$ (e.g. 1.5 for a rectangle) indicates the reserve capacity beyond elastic limit.
•In cyclic loading, even stresses well below the static yield strength can cause fatigue failure. Notches, holes, and abrupt cross-section changes create local stress concentrations ( $K_t > 1$ ) that significantly reduce fatigue life. The nominal bending stress $\sigma_b = M/Z$ must be multiplied by $K_t$ for fatigue assessment.