Structural Analysis Fundamentals / Complex stress and strain

Stress on an inclined plane

Why the answer depends on the plane, and deriving the transformation equations.

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Why this matters

A bar in simple tension is not only carrying direct stress. Cut it at 45° and you find shear stress — which is exactly why a ductile steel test piece fails on a slanted plane rather than straight across. Knowing the stress on every plane is what lets you predict failure.

By the end of this lesson you should be able to

Explain what 'stress at a point' actually means
Derive σ and τ on a plane at angle θ
Show that a bar in pure tension still carries shear

What you should already know

Direct stress and shear stress (Module 7)
Complementary shear stress (Module 10)
Resolving forces and equilibrium (Module 2)

Ask what the stress is at a point in a loaded body and the honest answer is: it depends which plane you ask about. Slice through the point one way and you might find pure tension. Slice through the same point at a different angle and you find a smaller direct stress plus some shear.

So the state of stress at a point is described by the components on two perpendicular planes: σx, σy and the shear τxy. From those three numbers you can work out the stress on any other plane through that point — and that is what the transformation equations do.

From first principles

Stresses on an inclined plane

We want to show: the normal and shear stress on a plane at angle θ, in terms of σx, σy and τxy.

Cut a small triangular wedge out of the material, with two faces along the x and y directions and the third at the angle θ you are interested in. You already know the stresses on the two square faces. The wedge must be in equilibrium, so the forces on the inclined face have to balance whatever the other two faces are doing. Resolve everything perpendicular and parallel to the inclined face, and the answer drops out. The only fiddly part is remembering that stress is force per unit area, so each face contributes a force equal to its stress times its own area — and the three faces have different areas.

Worked example

Stress on a plane at 30°

Given

σx = 80 N/mm² (tension)
σy = −40 N/mm² (compression)
τxy = 30 N/mm²

Find

The direct and shear stress on a plane whose normal is at 30° to the x axis.

Assumptions

Plane stress
Stresses uniform over the element

Predict first

A bar carries pure tension σ, so σx = σ, σy = 0 and τxy = 0. What is the shear stress on a plane at 45°?

Practice

At a point σx = 100 N/mm², σy = 40 N/mm² and there is no shear on those planes. What is the direct stress on a plane inclined at 30° to the plane on which σx acts, in N/mm²?

Practice

For the same stress state, what is the magnitude of the shear stress on that 30° plane, in N/mm²?

Practice

For the same stress state, what is the direct stress on the plane at 45°, in N/mm²?

Summary

Stress at a point depends on the plane; three components describe the whole state
The transformation equations come from equilibrium of a wedge — no material properties involved
Both equations use 2θ, not θ
Even pure tension produces shear stress, greatest at 45°