Bending Makes Stress Linear in Depth, σ=My/I
A bookshelf sags a little under the weight of books, and a diving board flexes when someone stands on it. This is bending: not pulled, not twisted, but curved sideways. Something interesting happens inside a bent beam. The top fibers shorten and get pressed (compression), while the bottom fibers lengthen and get pulled (tension). Somewhere between them is a surface that neither stretches nor shrinks, called the neutral axis. The stress is zero at the neutral axis and grows linearly with distance from it. Tie that into one line and you get σ = MyI. Where torsion was linear in radius, bending is linear in depth. And this formula explains why a beam is used standing tall rather than lying flat.
First, see what happens inside a bent beam. Drag the curvature to flex it. Think of the beam as a bundle of thin fibers. When it sags downward, the top side becomes the inside of the curve and shortens (compression), while the bottom becomes the outside and lengthens (tension). In the widget you see the split: the top glows blue for compression, the bottom red for tension. A single member holds compression and tension at the same time. And right at the boundary between them runs one line where nothing happens at all.
That line where nothing happens is the neutral axis. Since it neither stretches nor shrinks, its strain is zero, and so is its stress. So what about moving away from it? The farther up you go the more it is pressed, and the farther down the more it is pulled, exactly in proportion to the distance. Drag the probe to change the depth. Twice as far from the neutral axis means twice the stress. That is why the stress distribution forms a bowtie: zero in the middle, fanning out above and below. The key is that the stress is linear in the depth y, the same structure as torsion being linear in radius. It means the outermost fiber is the most at risk.
Now write that distribution as a formula: σ = MyI. M is the bending moment, how hard the beam is being bent. y is the distance from the neutral axis, and the stress being proportional to y lives right here. I is the second moment of area, the geometric quantity that takes the place of torsion's J. Drag M. The whole distribution scales up, and at the farthest fiber y=c the stress is largest: σmax = McI. Compare it with tension's σ = PA and the difference is clear. Tension gives the same stress everywhere in the section, but bending carries the position y, so it grows toward the outside. That is why, for the same section, bending makes the edges yield first.
Here, see properly what I is. I = ∫y²dA, each patch of area multiplied by the square of its distance to the neutral axis, all summed. Because the distance is squared, material far from the neutral axis gets an enormous weight. So the deeper the section, the more explosively I grows. For a rectangle, I = bh³12, the height cubed. In the widget, hold the area fixed and drag only the height. With the same amount of material, standing it taller sends I shooting up. This is why a beam is used standing rather than lying flat, and why an I-beam thins out the middle and piles material at the top and bottom: the material doing the real work is the one far from the neutral axis.
In practice you often only need σmax. So we bundle I and the farthest distance c into the section modulus S = Ic. Then σmax = McI becomes the clean σmax = MS. A single S tells the whole bending strength of that section. The larger S is, the lower the stress for the same moment. Drag the height. Even at the same area, a deeper section has a larger S, so σmax drops. That is why a steel catalogue lists an S value for each member. A designer divides the moment M to be carried by the allowable stress to get the required S, then picks the lightest section that exceeds it. Bending design effectively runs on this one line.