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Queensferry

Module 14 · Lesson 14.4

Stress trajectories

Following the principal stresses through a beam, and why cracks go where they go.

Why this matters

Mohr's circle gives the principal stresses at one point. Do it at every point and join up the directions, and you get a map of how force actually travels through a member. It explains the diagonal cracks near the ends of a concrete beam, the direction reinforcement wants to run, and why a hole in the wrong place is so much worse than a hole in the right one.

By the end of this lesson you should be able to

  • Define a stress trajectory
  • Combine bending and shear stresses at a point in a beam
  • Explain why the trajectories are horizontal at the extreme fibres and at 45° on the neutral axis
  • Relate the trajectories to observed crack patterns

What you should already know

  • Principal stresses and Mohr's circle (this module)
  • Bending stress σ = My/I (Module 9)
  • Shear stress τ = VQ/It (Module 10)

A stress trajectory is a curve drawn so that at every point along it the tangent is the direction of a principal stress. Because the two principal directions are always at right angles, the trajectories form two families crossing each other everywhere at 90°.

By definition there is no shear stress on a plane normal to a trajectory. So the trajectories show the paths along which the material is being purely stretched or purely compressed — the routes the load is actually taking through the member.

For a simply supported beam under transverse load, take any point and note the two stresses acting there:

  • A bending stress σ = My/I, acting along the beam, which is greatest at the extreme fibres and zero on the neutral axis.
  • A shear stress τ = VQ/It, which is zero at the extreme fibres and greatest on the neutral axis.

The two vary in opposite senses, and that is what makes the picture interesting.

Work through the three characteristic locations.

At the top and bottom surfaces. τ = 0, because there is no surface to supply the complementary shear. With no shear on those planes, they are the principal planes. So the trajectories run horizontally, parallel to the beam surface.

On the neutral axis. σ = 0 and only shear acts — a state of pure shear. The principal stresses in pure shear are ±τ at 45°. So the trajectories cross the neutral axis at 45°, one family in tension and the other in compression.

In between. Both stresses act, and the principal directions rotate smoothly from horizontal at the surfaces to 45° at the neutral axis.

The result is two families of curves: one running from the bottom face near mid-span, where the tension is greatest, arcing upwards towards the supports at increasing angles; the other its mirror image in compression. Together they look remarkably like an arch and a tie — which is exactly what a beam is, once you look at it this way.

Predict first

At a point exactly on the neutral axis of a beam under transverse load, at what angle to the beam axis do the principal stresses act?

Worked example

Principal stresses through the depth of a beam

Given

  • Rectangular beam 100 mm wide × 300 mm deep, so A = 30 000 mm² and I = 225 × 10⁶ mm⁴
  • At the section considered: bending moment M = 30.0 kNm and shear force V = 60.0 kN

Find

The principal stresses and their directions at the top fibre and on the neutral axis.

    Practice

    A rectangular beam 100 mm wide × 300 mm deep carries a shear force of 60.0 kN at a section. What is the shear stress on the neutral axis, in N/mm²?

    Practice

    At that same point on the neutral axis the bending stress is zero. What is the magnitude of the major principal stress, in N/mm²?

    Practice

    At the extreme top fibre of the same beam, the bending stress is 20.0 N/mm² compression and the shear stress is zero. At what angle to the beam axis does the major principal stress act? Give the angle in degrees.

    Summary

    • A stress trajectory follows a principal stress direction; the two families always cross at 90°
    • There is no shear stress across a trajectory, so it shows the path the load takes
    • At the extreme fibres τ = 0, so the trajectories run horizontally
    • On the neutral axis σ = 0, giving pure shear with principal stresses at 45°
    • Directions rotate smoothly between the two
    • Concrete cracks across the tensile trajectories: vertical at mid-span, diagonal near supports
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    This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint