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12.2 Shear stresses in beams

12.2 Shear stresses in beams

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🔗Statics and Strength of Materials
Unit & Topic Study Guides

Shear Stress in Beams

Shear Stress Distribution

When a beam carries a transverse load, internal shear forces develop at every cross-section. Shear stress is the intensity of that internal force per unit area, and it acts parallel to the face of the cross-section rather than perpendicular to it (which would be bending stress).

The distribution of shear stress across a cross-section is not uniform. It varies with position, and the shape of that distribution depends on the geometry of the cross-section. A few key points to remember:

  • Shear stress is zero at the top and bottom free surfaces of the beam, because there's no adjacent material to develop a shear flow against.
  • Shear stress is typically greatest at or near the neutral axis, where bending stress happens to be zero.
  • For a rectangular cross-section, the distribution is parabolic, peaking at the center.

A common misconception from the bending stress topic: students sometimes assume that the most "dangerous" point in a cross-section is always at the extreme fibers (top/bottom). That's true for bending stress, but for shear stress the critical location is usually at the neutral axis. Bending and shear are at their worst at different points in the cross-section.

Factors Influencing Shear Stress

  • Cross-section geometry is the dominant factor. The shape determines how QQ and tt vary across the section, which controls the stress distribution.
  • Material properties like the shear modulus (GG) relate shear stress to shear strain (τ=Gγ\tau = G\gamma), but they don't change the stress distribution itself for elastic analysis. They matter more when you're calculating deflections due to shear.
  • For symmetrical sections (rectangular, circular), maximum shear stress occurs at the centroid (neutral axis).
  • For I-beams and sections with thin webs, the shear stress is carried almost entirely by the web. The maximum shear stress occurs in the web, and it's roughly uniform across the web height. The flanges carry very little shear. This is why the web is the critical element for shear in steel I-beams.

Maximum Shear Stress Location

Where the maximum shear stress falls depends on the cross-section shape:

Cross-SectionLocation of τmax\tau_{max}Why
RectangularNeutral axis (mid-height)QQ is largest and tt is constant
CircularNeutral axis (center)Same reasoning; parabolic distribution
I-beamIn the web, near the neutral axisThe web is thin (small tt), which drives τ=VQ/(It)\tau = VQ/(It) up
T-section or channelWhere the section is thinnest relative to QQRequires calculation; not always at the neutral axis
Shear Stress Distribution, mechanical engineering - Determining the first moment of area (Q) when calculating the shear ...

Other Factors Affecting the Location

  • Loading conditions determine the shear force VV at each cross-section along the beam's length. The cross-section with the largest VV is where you'll find the highest overall shear stress.
  • Non-prismatic beams (beams whose cross-section changes along their length) can shift the critical location because II, QQ, and tt all change with position.

So finding the true maximum shear stress in a beam is a two-part problem: find where VV is largest along the beam's length (from the shear diagram), and find where τ\tau is largest within that cross-section.

Shear Stress Calculation

The Shear Formula

The general shear stress at any point in a beam's cross-section is:

τ=VQIt\tau = \frac{VQ}{It}

where:

  • VV = internal shear force at the cross-section
  • QQ = first moment of area of the portion of the cross-section above (or below) the point where you're calculating stress, taken about the neutral axis
  • II = second moment of area (moment of inertia) of the entire cross-section about the neutral axis
  • tt = width of the cross-section at the point where you're calculating stress
Shear Stress Distribution, Beam Reactions and Diagrams – Strength of Materials Supplement for Power Engineering

Step-by-Step Calculation Process

  1. Draw the shear force diagram for the beam and identify the cross-section where VV is largest (or wherever you need to evaluate).
  2. Locate the neutral axis of the cross-section and compute the moment of inertia II about it.
  3. Choose the point in the cross-section where you want the shear stress (often the neutral axis for τmax\tau_{max}).
  4. Calculate QQ: take the area of the cross-section above (or below) your chosen point, and multiply it by the distance from that area's centroid to the neutral axis. Formally: Q=yˉAQ = \bar{y}' \cdot A', where AA' is the partial area and yˉ\bar{y}' is the distance from its centroid to the neutral axis.
  5. Measure tt: the width of the cross-section at your chosen point.
  6. Substitute into τ=VQ/(It)\tau = VQ/(It).

Example: Rectangular Cross-Section

For a rectangular beam of width bb and height hh, find τmax\tau_{max} at the neutral axis.

The area above the neutral axis is a rectangle of width bb and height h/2h/2. Its centroid sits at h/4h/4 above the neutral axis.

  • Q=Ayˉ=(bh2)(h4)=bh28Q = A' \cdot \bar{y}' = \left(b \cdot \frac{h}{2}\right)\left(\frac{h}{4}\right) = \frac{bh^2}{8}
  • I=bh312I = \frac{bh^3}{12}
  • t=bt = b

Substituting:

τmax=Vbh28bh312b=3V2bh=3V2A\tau_{max} = \frac{V \cdot \frac{bh^2}{8}}{\frac{bh^3}{12} \cdot b} = \frac{3V}{2bh} = \frac{3V}{2A}

This result is worth memorizing: the maximum shear stress in a rectangular beam is 1.5 times the average shear stress (V/AV/A). For a circular cross-section, the corresponding factor is 4/34/3.

Shear Force vs. Shear Stress

These two quantities are related but distinct, and mixing them up is a common mistake.

  • Shear force (VV) is a resultant internal force acting on an entire cross-section. It has units of force (N or lb). You find it from equilibrium using free-body diagrams and shear diagrams.
  • Shear stress (τ\tau) is force per unit area at a specific point within the cross-section. It has units of pressure (Pa or psi). You find it from the shear formula.

The shear formula τ=VQ/(It)\tau = VQ/(It) is the bridge between them. A single value of VV produces a distribution of τ\tau across the cross-section.

Why Constant VV Doesn't Mean Constant τ\tau

Consider a simply supported beam with a single concentrated load PP at midspan. The shear diagram shows V=+P/2V = +P/2 from the left support to midspan, then V=P/2V = -P/2 from midspan to the right support.

Within any cross-section in the left half, VV is the same everywhere: P/2P/2. But the shear stress still varies from zero at the top and bottom surfaces to a maximum at the neutral axis. The variation comes from QQ and tt, which depend on where in the cross-section you look, not on where along the beam you are.

This is the key distinction: the shear diagram tells you how VV changes along the beam's length, while the shear formula tells you how τ\tau changes across the beam's cross-section at any given location.