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Mechanics of Materials

The Normal Stress on the Zero-Shear Face Is the Principal Stress

σ1,2 = (σx+σy)/2 ± R, angle tan2θp, max shear τmax = R, brittle vs ductile failure

In D1 you saw σ and τ swing together as you rotate the element. Among them was a very special angle: the one where the shear stress τ is exactly zero. That face is called a principal plane, and the normal stress on it is a principal stress. Why does it matter? At that angle the normal stress reaches its largest value. The greatest principal stress σ1 and the smallest σ2, these two numbers are the extremes of stress the point must endure. The whole complicated, swinging stress state compresses down to just these two values. And exactly 45 degrees from there the shear is at its maximum, which is τmax. In this lesson we tidy those principal stresses, the principal angle, and the maximum shear into clean formulas, and carry it through to why a material breaks in that direction.

First, pin down where the principal axis is. A principal plane is one where the shear is zero, so set τ′ in the transformation equation to zero and solve for the angle. That gives tan2θp = 2τxy(σx−σy). This angle θp is the principal direction. The important thing is that the principal axis is not fixed; it turns with the stress state. Drag τxy. As the shear grows, the principal axis tilts further. With zero shear, the principal axis is just the x and y axes as they are. So the principal axis is the natural direction the stress state chooses for itself, the coordinate system in which a point's stress looks simplest, where the shear vanishes and only push and pull remain.

Now the values of the principal stresses. Putting θp into the transformation equation and simplifying gives a surprisingly clean form: σ1,2 = (σx+σy)2 ± R. Two values centered on the average (σx+σy)2, spread to either side by a radius R. Here R = √(((σx−σy)/2)² + τxy²). Drag τxy. The average sits still while R grows, and σ1 and σ2 move far apart from the average to either side. It makes intuitive sense: a large shear means the point is in a rougher stress state, and the principal stresses spread wider to match. This average-plus-or-minus-radius structure, does it not look familiar? Two points at a radius from a center. This leads straight into the next lesson's Mohr's circle.

Once you have the principal planes, the maximum-shear plane comes for free: it is always exactly 45 degrees from the principal plane. In the widget, the left element is the principal element. It has no shear arrows; only σ1 and σ2 push and pull on its faces. The right element is that one rotated 45 degrees, the max-shear element. Here there are big shear arrows, and the normal stress on both faces is the same average value (σx+σy)2. The size of that maximum shear is τmax = R = (σ1−σ2)2, the spread of the principal stresses cut in half. Drag τxy and the two elements grow together. To sum up, the stress at a point has two faces: the principal plane where shear vanishes and the normal stress is extreme, and the plane 45 degrees away where the shear is extreme.

Why does this matter so much? Carry it through to failure. Different materials have different weaknesses. A brittle material like chalk or cast iron is weak against pulling, that is, normal tension. So it splits perpendicular to the maximum principal stress σ1. A tensile specimen of it snaps straight across the axis. A ductile material like steel, on the other hand, yields first to sliding, that is, shear. So it tears by slipping along the maximum-shear plane, the face tilted 45 degrees from the principal plane. Toggle between the two. Under the same tension, the orientation of the failure surface is completely different. That is why it matters to find both the principal stress σ1 and the maximum shear τmax: which one is dangerous is decided by the material. For brittle, compare σ1 to the limit; for ductile, compare τmax.

Finally, gather it all on one card. Given σx, σy, τxy, first find the radius R = √(((σx−σy)/2)² + τxy²). Then the principal stresses are σ1,2 = (σx+σy)2 ± R, the principal angle is tan2θp = 2τxy(σx−σy), and the maximum shear is τmax = R. Just these four lines fully solve the stress state at a point. Drag τxy and watch the four values move together. R is the key quantity that ties everything: as R grows, the principal spread and the maximum shear grow with it. But these formulas, used by memory, invite mistakes in signs and angles. So in the next lesson we turn these four lines, as a whole, into a single picture, Mohr's circle, a circle centered on the average with radius R, on which every transformation becomes simply a rotation, an almost magical tool.

In PracticeTo sum up: the principal stresses are the normal stresses on the zero-shear faces. First find the radius R = √(((σx−σy)/2)² + τxy²), and the principal stresses σ1,2 = (σx+σy)2 ± R, the principal angle tan2θp = 2τxy(σx−σy), and the maximum shear τmax = R follow in a row. The average is invariant under rotation, and the max-shear plane sits 45 degrees from the principal plane. Failure is decided by the material: brittle perpendicular to σ1, ductile along the 45-degree plane of τmax. In engineering, this principal-stress analysis is the key tool for pressure vessels, rotating shafts, welds, and judging brittle versus ductile failure. But the formulas are error-prone with signs, so in the next lesson we move to seeing all of this as a single circle, Mohr's circle, on which, with center at the average and radius R, every transformation becomes a rotation.
Mechanics of Materials
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