Bending and Shear Stress

Pure bending

A beam segment carrying a bending moment with no shear force is in pure (simple) bending — for example, the middle portion of a simply supported beam with two equal loads placed symmetrically. The theory below is exact for pure bending and is used, with negligible error, where shear is also present.

When a beam sags, its top fibres shorten and its bottom fibres lengthen. Somewhere between lies a layer that neither shortens nor lengthens: the neutral layer. Its intersection with a cross-section is the neutral axis (NA).

Assumptions of the theory of simple bending

1. The material is homogeneous, isotropic and obeys Hooke's law; stresses are within the elastic limit.
2. $E$ is the same in tension and compression.
3. Plane sections remain plane after bending (Bernoulli–Euler hypothesis).
4. The beam is initially straight and every layer is free to expand or contract independently.
5. The radius of curvature is large compared with the depth.
6. The loads act in a plane of symmetry of the section (so the beam bends without twisting).

The flexure formula

Consider a layer at distance $y$ from the neutral layer, which bends to radius $R$. Its original length equals the neutral-layer length $Rd\theta$; after bending it is $(R+y)d\theta$. So its strain is

Stress is proportional to the distance from the neutral axis.

• Force equilibrium (no net axial force): $\int \sigma dA=\frac{E}{R}\int ydA=0$ → $\int ydA=0$ → the neutral axis passes through the centroid of the section.
• Moment equilibrium: $M=\int \sigma ydA=\frac{E}{R}\int y^{2}dA=\frac{E}{R}I$.

Section modulus and moment of resistance

The maximum stress occurs at the extreme fibre, $y_{max}$:

$Z$ is the section modulus. For a permissible stress $\sigma$, the moment of resistance is $M_r=\sigma Z$. The beam is safe if $M\le M_r$. A larger $Z$ means a stronger section in bending.

Economical sections

Most of the bending resistance comes from material far from the neutral axis, where stress is high. Material near the NA is lightly stressed. I-sections and box sections put most material in the flanges, giving a large $Z$ for a small area — the most economical shapes for steel beams.

Comparing shapes of equal area

• Square vs circle (same area): $Z_{square}/Z_{circle}=1.18$ — the square is stronger.
• Square with a side horizontal vs diagonal vertical: $Z$ ratio $=\sqrt{2}=1.414$ — the square resting on a side is stronger.
• I-section vs rectangle of the same area and depth: the I-section is several times stronger.

Strongest and stiffest rectangular beam cut from a round log of diameter $D$

With $b^2+d^2=D^2$:

• Strongest (maximum $Z=b{d}^2/6$): $b=D/\sqrt{3}$, $d=D\sqrt{2/3}$, so $d/b=\sqrt{2}$.
• Stiffest (maximum $I=b{d}^3/12$): $b=D/2$, $d=\sqrt{3}D/2$, so $d/b=\sqrt{3}$.

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