Shear Stresses in Beams

Shear in beams — the idea

A beam carrying transverse loads has a shear force $V$ at most sections. That force is resisted by shear stresses spread over the cross-section. Unlike bending stress, shear stress is not uniform: it is zero at the top and bottom surfaces and largest near the neutral axis.

Transverse shear always comes with horizontal shear between layers. Stack loose planks and load them — they slide over one another. Glue them and they act as one deep beam; the glue now carries horizontal shear. By complementary shear, the horizontal and vertical shear stresses at any point are equal.

Shear stress formula

Consider a beam element $dx$ long. The bending moment changes from $M$ to $M+dM$ across it. Isolate the part of the element above a level at distance $y$ from the neutral axis. The bending stresses on its two ends differ, and the difference must be balanced by a horizontal shear force on the cut face.

• Net horizontal force on the isolated part $=\frac{dM}{I}\int ydA=\frac{dM}{I}A\bar{y}$
• This equals $\tau bdx$, where $b$ is the width at the cut.

Since $dM/dx=V$:

Assumptions: the shear stress is uniform across the width at a given level, and the bending formula holds. The result is exact for narrow rectangles and a good approximation for most practical sections.

Rectangular section ($b\times d$)

At distance $y$ from the NA: $A=b\left(\frac{d}{2}-y\right)$ and $\bar{y}=\frac{1}{2}\left(\frac{d}{2}+y\right)$, so

• The distribution is parabolic.
• $\tau =0$ at the top and bottom ($y=\pm d/2$).
• Maximum at the NA: ${\tau }_{max}=\frac{V{d}^2}{8I}=\frac{3}{2}\frac{V}{bd}$.

Circular section (diameter $d$)

The distribution is also parabolic, with maximum at the NA:

Triangular section (base $b$, height $h$, base horizontal)

Here the width changes with depth. The maximum shear stress is not at the neutral axis but at mid-height:

The neutral axis lies at $2h/3$ from the apex.

Summary of maximum-to-average ratios