principal stresses explained
In D1, turning the element made the normal stress and the shear stress sway together. Inside that sway there was one very special angle. The angle where the shear stress is exactly 0. That face is a principal plane, and the normal stress on it is a principal stress. At that very angle the normal stress is largest. The largest principal stress σ1 and the smallest σ2, these two numbers are the two ends of the stress a point must bear. A stress state that swayed in a complicated way compresses, in the end, into these two values. Turn exactly 45 degrees from there and the shear is maximum, and that is the maximum shear. This lesson gathers that sway into formulas, and follows through to why a material breaks in that direction. When the face turns, the values the point must bear turn with it.
First, fix where the principal axes are. A principal plane is a face where the shear is 0, so set τ′ of the transformation to 0 and solve for the angle. Then tan2θp = 2τxy(σx−σy) comes out, and this angle is the direction of the principal axes. What matters is that the principal axes are not fixed. They turn together with the stress state. Raise the shear and the principal axes lean more and more, and if the shear is 0 the principal axes are simply the x and y axes as they stand. The principal axes are the natural direction that the stress state sets for itself. The coordinates in which the stress at a point looks simplest, the direction where the shear vanishes and only push and pull remain.
Now the values of the principal stresses. Put the principal angle into the transformation and a surprisingly clean form comes out. σ1,2 = (σx+σy)2 ± R. Two values spread to either side of the average (σx+σy)2 by the radius R. Here R = √(((σx−σy)/2)² + τxy²). Move the shear and the average stays still while R grows, and σ1 and σ2 move away from the average to either side. It makes sense. A large shear means a rough stress state, and the width of the principal stresses widens by that much. This structure of an average ± a radius looks familiar. Two points a radius away from one center, and that leads straight into Mohr's circle in the next lesson. A wider spread means a rougher state at that point.
Once the principal plane is found, the plane of maximum shear comes free. It is always exactly 45 degrees away from a principal plane. The left element is the principal element. There are no shear arrows, and only σ1 and σ2 push and pull the faces. The right element is the maximum-shear element turned 45 degrees from there, and it has large shear arrows while the normal stress on both faces equals the average (σx+σy)2. The size of that maximum shear is τmax = R = (σ1−σ2)2, the width of the principal stresses cut in half. Move the shear and the two elements grow together. The stress at a point has two faces. The principal plane, where the shear vanishes and the normal stress is extreme, and the plane of maximum shear, 45 degrees around, where the shear is extreme.
Why this matters becomes clear when it leads to failure. Each material has a different weakness. A brittle material such as chalk or cast iron is weak in pulling, that is in normal tension. So it splits perpendicular to the largest principal stress σ1, and a tension specimen snaps straight across the axis. A ductile material such as steel yields first to sliding, that is to shear. So it slides and breaks along the plane of maximum shear, a face tilted 45 degrees from a principal plane. Switch the two with the button and, under the same tension, the direction of the broken face changes completely. So both the principal stress σ1 and the maximum shear have to be found. Which one is dangerous is decided by the material, so a brittle one is compared at σ1 and a ductile one at the maximum shear. Look at the weak side of the material before comparing the limit.
Last, gather everything on one page. Given the three stresses, 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. These four lines solve the stress state at a point completely. Move the shear and watch the four values move together, and it shows that R is the quantity that ties everything. When R grows, the width of the principal stresses and the maximum shear grow together. Memorized, these formulas keep tripping on signs and angles. So the next lesson turns these four lines into one picture, Mohr's circle. A circle of radius R about the average, on which every transformation is simply a rotation. Four lines gather the answer for one point.