Learning topic
Stresses in Bending
Calculate normal bending stress with σ = My/I and shear stress with the Zhuravsky formula, including section modulus, neutral axis, and critical beam sections.
In transverse bending, a beam cross-section may simultaneously carry normal stresses caused by the bending moment $M$ and shear stresses caused by the shear force $Q$. Their distributions across the section differ, so the critical points for $\sigma$ and $\tau$ do not necessarily coincide.
Normal stresses
For simple bending of a beam within the linear-elastic model, normal stress varies linearly with distance from the neutral axis:
$$\sigma_x(y)=-\frac{My}{I}.$$
The maximum absolute stress at an extreme fiber is:
$$|\sigma_{\max}|=\frac{|M|y_{\max}}{I}=\frac{|M|}{W},\qquad W=\frac{I}{y_{\max}}.$$
- $M$ — bending moment;
- $I$ — second moment of area about the neutral axis;
- $y$ — distance from the neutral axis;
- $y_{\max}$ — distance to the relevant extreme fiber;
- $W$ — section modulus.
The sign of $\sigma_x$ depends on the sign convention for $M$ and $y$; one side of the section is in tension and the other in compression.
In the classical simple-bending model, the neutral axis passes through the centroid and coincides with a principal centroidal axis of the section. Normal stress varies linearly: $\sigma=0$ at the neutral axis and reaches its largest absolute values at the extreme fibers.
Section-modulus form
For a strength check, it is convenient to use $|\sigma_{\max}|=|M|/W$. If the section is unsymmetrical about the neutral axis, the distances to the extreme fibers may differ, so the tensile and compressive sides should be checked separately.
Shear stresses
For transverse bending, the shear stress at a point of the cross-section is determined by the Zhuravsky formula:
$$\tau=\frac{QS}{Ib}.$$
- $Q$ — shear force at the section;
- $S$ — first moment of the cut-off part of the area about the neutral axis;
- $I$ — second moment of area of the entire section about the neutral axis;
- $b$ — local section width at the level where $\tau$ is evaluated.
At a free upper or lower boundary of a solid section, $S$ tends to zero and therefore $\tau$ is zero. For a rectangular section, the maximum shear stress occurs at the neutral axis.
For a rectangular section, the shear-stress distribution over the depth is parabolic and its maximum at the neutral axis is $\tau_{\max}=3Q/(2A)$. In an I-section, a large portion of the shear force is carried by the web because its local width is small while the first moment of the adjacent area can be substantial.
Circular section
For a solid circular section, shear stress is also zero at the external boundary and reaches its maximum near the neutral axis. For the classical solid circle, $\tau_{\max}=4Q/(3A)$.
Strength assessment
Normal stresses are checked at sections with large $|M|$ and at extreme fibers. Shear stresses are checked at sections with large $|Q|$ and at characteristic points of the cross-section. If both $\sigma$ and $\tau$ are significant at the same point, a multiaxial failure criterion may be required.
Limits of the classical model
Navier's formula assumes linear-elastic behavior, small deformation, and the classical assumptions of beam theory. Near concentrated loads, supports, holes, and abrupt changes of section, the local stress distribution can differ substantially from the elementary model.
About this topic
Transverse bending generates both normal flexural stresses and transverse shear stresses within beam cross-sections. This page presents the derivation of Navier's flexure formula for normal stress calculations using section modulus, alongside Jourawski's formula for shear stress distributions across rectangular, I-shaped, and circular profiles.