10.4 Flexural Bending Stress and Transverse Shear Stress

10.4 Flexural Bending Stress and Transverse Shear Stress

Bending is the dominant loading mode for beams, and transverse shear accompanies it whenever the bending moment varies along the span. NCEES lists both under stress and strain caused by bending loads and by transverse shear forces, and a complete beam check requires evaluating each.

Flexural Bending Stress and Transverse Shear Stress in Beams

Flexural Bending Stress (Navier's Equation)

Under pure bending or transverse loading, internal bending moment $M$ creates normal stresses $\sigma$ that vary linearly across the beam depth relative to the Neutral Axis (NA), where normal stress is zero.

$\sigma = - \frac{M y}{I}$

where:

• $M$ is the internal bending moment at the cross-section
• $y$ is the perpendicular distance from the neutral axis (positive upwards)
• $I$ is the area moment of inertia of the cross-section about the neutral axis

Maximum tensile and compressive bending stresses occur at the extreme fibers ($y = \pm c$):

$\sigma_{max} = \frac{M c}{I} = \frac{M}{S}$

where $S = \frac{I}{c}$ is the Elastic Section Modulus of the cross-section.

Cross-Sectional Properties for Beam Bending

Parallel Axis Theorem

To find the moment of inertia $I_x$ of a composite shape about its centroidal axis, use the Parallel Axis Theorem:

$I_x = \bar{I}_{xc} + A d_y^2$

where $\bar{I}_{xc}$ is the centroidal moment of inertia of a sub-area $A$, and $d_y$ is the perpendicular distance between the centroid of area $A$ and the composite neutral axis.

Transverse Shear Stress (Jourawski's Formula)

Transverse shear forces $V$ produce horizontal and vertical shear stresses $\tau$ in beams, calculated via Jourawski's Formula:

$\tau = \frac{V Q}{I b}$

where:

• $V$ is the transverse shear force at the section
• $Q = A' \bar{y}'$ is the First Moment of Area above (or below) the level where shear stress is evaluated
• $I$ is the total moment of inertia of the cross-section
• $b$ is the width of the cross-section at the evaluation level

Maximum Shear Stress Formulas for Standard Sections

• Rectangular Beam (width $b$, height $h$): $\tau_{max} = \frac{3 V}{2 A}$ (occurs at the neutral axis $y = 0$)
• Solid Circular Beam (radius $R$): $\tau_{max} = \frac{4 V}{3 A}$ (occurs at the neutral axis)

Which Mode Governs? Bending vs. Transverse Shear

Both stresses arise from the same load, but they scale differently with span, and the comparison decides which one designs the beam.

For a simply supported rectangular beam of span $L$, width $b$, depth $h$, under uniform load $w$:

$\sigma_{\max} = \frac{M_{\max}}{S} = \frac{wL^2/8}{bh^2/6} = \frac{3wL^2}{4bh^2}$

$\tau_{\max} = 1.5\,\frac{V_{\max}}{A} = 1.5\,\frac{wL/2}{bh} = \frac{3wL}{4bh}$

Their ratio depends only on the span-to-depth ratio:

$\frac{\sigma_{\max}}{\tau_{\max}} = \frac{L}{h}$

Design consequence: a typical floor joist at $L/h = 20$ is a bending problem, and its shear stress is negligible. A short, heavily loaded transfer beam or a bearing bracket at $L/h = 3$ must be checked for shear, and timber and reinforced concrete — both weak in shear relative to bending — fail this way in practice.

Note also where each maximum occurs, and that the two locations are complementary:

This complementarity is why the extreme fiber is checked for bending and the neutral axis for shear — and why a point at mid-depth of the web in a wide-flange beam is the critical location for shear while the flange tips are critical for bending.

Maximum Shear Stress Factors

$\tau_{\max} = k\,\frac{V}{A}$

Worked Example: Checking Both Modes

A rectangular timber beam 100 mm wide × 250 mm deep spans 3.0 m simply supported and carries $w = 14$ kN/m. Allowable stresses are $\sigma = 12$ MPa in bending and $\tau = 1.1$ MPa in shear. Is it adequate?

Section properties: $A = 100(250) = 25{,}000\ \text{mm}^2$ $I = \frac{100(250)^3}{12} = \frac{100(1.5625\times10^7)}{12} = 1.302\times10^8\ \text{mm}^4$ $S = \frac{I}{c} = \frac{1.302\times10^8}{125} = 1.042\times10^6\ \text{mm}^3$

Bending check: $M_{\max} = \frac{wL^2}{8} = \frac{14(3.0)^2}{8} = 15.75\ \text{kN}\cdot\text{m} = 1.575\times10^7\ \text{N}\cdot\text{mm}$ $\sigma = \frac{M}{S} = \frac{1.575\times10^7}{1.042\times10^6} = 15.1\ \text{MPa} > 12\ \text{MPa} \quad\textbf{FAILS}$

Shear check: $V_{\max} = \frac{wL}{2} = \frac{14(3.0)}{2} = 21.0\ \text{kN} = 21{,}000\ \text{N}$ $\tau = 1.5\frac{V}{A} = 1.5\frac{21{,}000}{25{,}000} = 1.26\ \text{MPa} > 1.1\ \text{MPa} \quad\textbf{FAILS}$

Both fail, and the utilization ratios reveal which is worse: bending at $15.1/12 = 126%$, shear at $1.26/1.1 = 115%$. With $L/h = 3{,}000/250 = 12$, bending governs as the ratio rule predicts.

Fixing it. Increasing depth to 300 mm: $S = \frac{100(300)^2}{6} = 1.50\times10^6\ \text{mm}^3 \;\Rightarrow\; \sigma = \frac{1.575\times10^7}{1.50\times10^6} = 10.5\ \text{MPa}\ ✓$ $\tau = 1.5\frac{21{,}000}{30{,}000} = 1.05\ \text{MPa}\ ✓$

Both pass. Note the leverage: a 20% depth increase cut bending stress 30% (since $S \propto h^2$) but cut shear stress only 17% (since $A \propto h$). Depth is a far more effective lever against bending than against shear — which is why shear-critical members are widened or web-stiffened rather than deepened.