bending stress σ=My/I
A bookshelf sags a little under the weight of books, and a diving board bends when a person stands on it. That is bending. It is neither a pull nor a twist. It is a sideways curve. Inside a bent beam something interesting happens. The upper fibers shorten and go into compression, and the lower fibers lengthen and go into tension. Somewhere between them is a face that neither lengthens nor shortens, and that is the neutral axis. The stress is 0 at the neutral axis and grows linearly as you move away. Tied into one line, that is σ = MyI. If torsion was linear in the radius, bending is linear in the depth. And this formula explains why a beam is used standing on edge, not laid flat. Depth, not a flat lay, is what the formula rewards.
First, see what happens inside a bent beam. Move the curvature and bend the beam. It helps to think of the beam as a bundle of thin fibers. When the beam sags downward, the upper fibers become the inside, their path shortens, and they are compressed, while the lower fibers become the outside and lengthen. The top shows as compression and the bottom as tension. Compression and tension sit together inside one member. And on the boundary between them there is one line where nothing happens. The quiet line between them is the next story.
That line where nothing happens is the neutral axis. It neither lengthens nor shortens, so the strain is 0, and therefore the stress is 0 too. What happens as you move away? Going up, the fibers are compressed more, and going down they are stretched more, and exactly in proportion to the distance. Move the probe and change the depth, and at twice the distance from the neutral axis the stress is twice as large. So the stress distribution is 0 in the middle and opens upward and downward like a bow tie. The heart of it is that the stress is linear in the depth y. It is the same structure as torsion being linear in the radius, and it means the outermost fiber is the most dangerous.
Now write that distribution as a formula. σ = MyI. M is the bending moment, how hard the beam is being curved, and y is the distance from the neutral axis, so the stress being proportional to y sits right here. I is the second moment of area, the geometric value that takes the place of J in torsion. Move M and the whole distribution grows in proportion, and the stress is largest at the farthest fiber y=c. σmax = McI. Compare it with tension's σ = PA and the difference is clear. Tension has the same stress everywhere on the section, but bending has the position y in it, so it grows toward the outside. Even on the same section, bending yields from the edge first.
Here, see properly what I is. I = ∫y²dA, each patch of area multiplied by the square of its distance to the neutral axis and added up. Because the square of the distance is in it, material far from the neutral axis receives an enormous weight. So the deeper the section, the more explosively I grows. For a rectangle, I = bh³12, the cube of the height. Leave the area as it is and move only the height, and the same amount of material gives a larger I the more you stand it up. That is why a beam is used on edge and not laid flat. It is also why an I-beam hollows the middle thin and piles the flesh at the top and bottom. The material that actually works is the material far from the neutral axis.
In practice it is often enough to look only at the maximum stress. So the section modulus S = Ic bundles I and the farthest distance c into one number. Then the maximum stress, which was McI, becomes cleanly σmax = MS. One S tells the bending strength of that section as a whole. The larger S is, the lower the stress for the same moment. Move the height and, even at the same area, a deeper section has a larger S and the maximum stress falls. That is why a steel catalog lists an S for each member. A designer divides the moment M to be carried by the allowable stress, finds the S that is needed, and picks the lightest section that exceeds that S. Bending design turns, in effect, on this one line.