Shear stress in beams

A beam under transverse loading does not just bend — it also shears. Cut any cross-section of a loaded beam and the internal shear force $V$ acting there produces a horizontal (and, by pairing, vertical) shear stress that is not spread evenly over the section. It is zero at the top and bottom fibres and largest near the neutral axis — the exact opposite of how bending stress behaves. For everyday slender beams the bending stress usually governs, but for short, deep beams, or timber beams (weak along the grain), shear can be the failure mode that actually controls the design. This article derives the shear-stress formula from equilibrium of a beam element, applies it to rectangular, circular and I-shaped sections, and works through both a solid-section and an I-section example.

The core insight

Because bending moment $M$ varies along a beam wherever shear force $V\ne0$ , horizontal equilibrium of a sub-area of the cross-section forces a shear stress $\tau = VQ/(Ib)$ that is zero at the extreme fibres and maximum at the neutral axis.

The core idea

Why shear stress exists at all. Take a short beam element of length $dx$ where the bending moment changes from $M$ to $M+dM$ across the two faces (this happens exactly where $V = dM/dx \ne 0$ ). The bending stress on the left and right faces of any sub-area above a chosen level therefore differs slightly. For that sub-area to be in horizontal equilibrium, a shear stress must act along the horizontal cut at its base, balancing the unequal normal-stress resultants pushing on its two vertical faces.

The shear-stress formula. Carrying that equilibrium argument through (full derivation below) gives:

\tau = \frac{VQ}{Ib}

where: $V$ is the internal shear force at the section, $I$ is the second moment of area of the entire cross-section about the neutral axis, $b$ is the width of the section at the level where $\tau$ is evaluated, and $Q$ is the first moment of area of the portion of the section above (or below) that level, taken about the neutral axis, $Q = \bar y' A'$ .

Where $\tau$ is zero and where it peaks. $Q$ vanishes at the extreme fibre (there is no area left beyond it) and grows toward the neutral axis, so for a typical solid section $\tau = 0$ at top and bottom and $\tau$ is maximum at the neutral axis — the reverse of the bending-stress picture, where the extreme fibres carry the most stress and the neutral axis carries none.

Shortcut formulas for standard shapes. For a solid rectangular section, the distribution across the depth is parabolic, and

\tau_{max} = \frac{3V}{2A}

i.e. $1.5$ times the average shear stress $V/A$ . For a solid circular section, $\tau_{max} = \dfrac{4V}{3A}$ , i.e. $4/3$ times the average. Both formulas apply only to their named shapes.

I-sections: the web carries almost everything. In an I-beam, the web is the narrow strip straddling the neutral axis, so it is exactly where $\tau = VQ/(Ib)$ is largest — the thin $b$ there keeps the stress high. The wide flanges, in contrast, have a much larger $b$ , so $\tau$ drops sharply the instant the cut crosses from web into flange. Because the web alone typically carries over 90% of $V$ , designers commonly use the approximation $\tau_{avg} \approx V/A_{web}$ instead of computing the full distribution.

Figure

Shear stress distribution across the depth of a rectangular beam section shown as a parabola that is zero at the top and bottom fibres and maximum at the neutral axis

Derivation ADeriving the shear-stress formula from equilibrium

The goal is to show, using only horizontal-force equilibrium of a sub-element of a beam and the flexure formula for bending stress, that the shear stress at any level of a cross-section equals $\tau = VQ/(Ib)$ , where $Q$ is the first moment of the area beyond that level and $b$ is the width there.

Step 1. Isolate a beam slice of length $dx$ where the bending moment is $M$ on the left face and $M+dM$ on the right face, since $dM/dx = V \ne 0$ over this slice. From the flexure formula, the bending stress at height $y$ from the neutral axis is $\sigma = My/I$ on the left face and $\sigma = (M+dM)y/I$ on the right face — slightly larger on the right because $dM$ is positive there.

Step 2. From this slice, isolate only the sub-area $A'$ lying above a chosen level $y_1$ (a horizontal cut through the section). The normal force pushing on its left face is $F_1 = \int_{A'} \sigma\,dA = \dfrac{M}{I}\int_{A'} y\,dA = \dfrac{M Q}{I}$ , where $Q = \int_{A'} y\,dA = \bar y' A'$ is the first moment of $A'$ about the neutral axis. On the right face, by the same logic, $F_2 = \dfrac{(M+dM)Q}{I}$ .

Step 3. Since $F_2$ exceeds $F_1$ by $dF = \dfrac{dM\,Q}{I}$ , the sub-area is not in horizontal equilibrium from the normal forces alone. The missing horizontal force must come from a shear stress $\tau$ acting on the horizontal cut at the base of $A'$ , over an area $b\,dx$ (where $b$ is the section width at level $y_1$ ). Equilibrium requires $\tau (b\,dx) = dF = \dfrac{dM\,Q}{I}$ .

Step 4. Divide through by $b\,dx$ and use $dM/dx = V$ : $\tau = \dfrac{1}{b}\cdot\dfrac{dM}{dx}\cdot\dfrac{Q}{I} = \dfrac{VQ}{Ib}$ . This is the shear-stress formula, valid at any level of any cross-section provided $Q$ is the first moment of the area beyond that level and $b$ is the width measured exactly at that level.

Worked example

A rectangular beam cross-section, $b = 80\ \text{mm}$ wide and $h = 240\ \text{mm}$ deep, carries a shear force $V = 30\ \text{kN}$ . Find the maximum shear stress, and the shear stress at $y = 60\ \text{mm}$ above the neutral axis.

Step 1. Section properties. Moment of inertia $I = \dfrac{bh^3}{12} = \dfrac{(0.08)(0.24)^3}{12} = 9.216\times10^{-5}\ \text{m}^4$ , and area $A = 0.08\times0.24 = 0.0192\ \text{m}^2$ .

Step 2. Maximum shear stress, at the neutral axis, using the rectangular shortcut: $\tau_{max} = \dfrac{3V}{2A} = \dfrac{3(30{,}000)}{2(0.0192)} = 2.344\times10^{6}\ \text{Pa} = 2.34\ \text{MPa}$ .

Step 3. Shear stress at $y = 0.06\ \text{m}$ : the area above this level is a strip from $y=0.06$ to $y=0.12\ \text{m}$ (the top fibre), of height $0.06\ \text{m}$ and width $0.08\ \text{m}$ . $A' = 0.08\times0.06 = 0.0048\ \text{m}^2$ , and its centroid is at $\bar y' = (0.06+0.12)/2 = 0.09\ \text{m}$ , so $Q = A'\bar y' = 0.0048\times0.09 = 4.32\times10^{-4}\ \text{m}^3$ .

Step 4. Apply the formula: $\tau = \dfrac{VQ}{Ib} = \dfrac{(30{,}000)(4.32\times10^{-4})}{(9.216\times10^{-5})(0.08)} = 1.758\times10^{6}\ \text{Pa} \approx 1.76\ \text{MPa}$ . This is less than $\tau_{max} = 2.34\ \text{MPa}$ , as expected since $y=0.06\ \text{m}$ is off the neutral axis, not on it. Answer: $\tau_{max} = 2.34\ \text{MPa}$ at the neutral axis, and $\tau = 1.76\ \text{MPa}$ at $60\ \text{mm}$ above it.

Competitive stretch

An I-beam has a web $10\ \text{mm}$ thick and $200\ \text{mm}$ deep (between flanges), carrying $V = 60\ \text{kN}$ . Using the web-area approximation, $A_{web} = (0.01)(0.20) = 2.0\times10^{-3}\ \text{m}^2$ , so $\tau_{avg,web} = V/A_{web} = 60{,}000/(2.0\times10^{-3}) = 30\times10^{6}\ \text{Pa} = 30\ \text{MPa}$ . Designers accept this shortcut because the web alone carries over $90\%$ of $V$ , so computing the full $\tau = VQ/(Ib)$ distribution across the web and flanges buys little extra accuracy for the effort.

Common traps

Using the full width of the cross-section for $b$ when the point of interest is on a section with varying width (like an I-section) — $b$ must be the width at that exact level, which is why $\tau$ jumps at the flange–web junction.
Confusing $Q$ , the first moment of only the partial area above or below the point of interest, with the first moment of the entire section, which is always zero about the centroidal axis.
Assuming shear stress peaks at the extreme fibres like bending stress does — it is actually zero there and maximum at the neutral axis for typical solid sections.
Applying $\tau_{max} = 3V/2A$ or $\tau_{max} = 4V/3A$ to a shape other than a rectangle or circle respectively — these are shortcut results, not the general formula.

Ask the tutor: If computing $Q$ for an unfamiliar composite cross-section, or seeing why shear stress peaks at the neutral axis instead of the edges, isn't clicking yet, ask the AI tutor to walk through the derivation and a worked $Q$ calculation with you.

Why is shear stress maximum at the neutral axis instead of at the top and bottom like bending stress?

Because shear stress at a level depends on $Q$ , the first moment of the area beyond that level. At the extreme fibre there is no area left beyond it, so $Q=0$ and $\tau=0$ . As you move toward the neutral axis, more area lies beyond the cut and its centroid also sits farther from the axis for parts of the section, so $Q$ grows and peaks near the neutral axis (exactly at it, for standard solid shapes). Bending stress does the opposite because it depends directly on distance $y$ from the axis, which is largest at the extreme fibres and zero at the axis.

Do I always need to compute Q and I separately, or can I just use the shortcut formulas?

The shortcuts $\tau_{max}=3V/2A$ (rectangle) and $\tau_{max}=4V/3A$ (circle) only give the maximum shear stress at the neutral axis for those two specific shapes. If you need the shear stress at any other level, or the section is not rectangular or circular (an I-section, a T-section, a composite shape), you must go back to $\tau=VQ/(Ib)$ and compute $Q$ , $I$ and $b$ for that particular level yourself.

Why does the shear stress jump suddenly at the flange-web junction of an I-section?

Because $\tau = VQ/(Ib)$ has $b$ in the denominator, and $b$ changes abruptly there — the web is narrow, but the flange spans the full width of the I-section. Just below the junction (in the web) $b$ is small, so $\tau$ is large; just above it (in the flange) $b$ is suddenly much larger, so $\tau$ drops sharply even though $Q$ and $I$ barely change across that boundary. This jump is why the web, not the flange, is treated as carrying essentially all the shear.

Shear stress in beams

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads