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Mechanics of Materials

Stress Is Internal Force per Unit Area

At an imaginary cut, the internal force each patch of area carries, σ = P / A

A bar is being pulled from both ends. From the outside, all you see are the two end forces. But what about inside the bar? Every cross-section is carrying that same force. To actually see it, you make one imaginary cut through the bar. The cut face reveals the internal force, the pull that was holding the two sides together. Stress is exactly that: at the cut, how much internal force each patch of area carries. Not the force itself, but the density of force. This single view is where all of mechanics of materials begins, and it is why the same load breaks a thin spot first.

First, just pull the bar. Drag P and the bar stretches; the harder you pull, the more it stretches. This is tension: the two ends being pulled apart in opposite directions. Since the wall holds one end, the wall pulls back with an equal force, called the reaction. For now, just get the feel: a force is passing through the bar, and the bar is resisting it. That resisting interior is what we will open up in the next scene.

Now cut the bar in your mind, just once. You are not really cutting it; you are imagining "what if I cut here." Take only the right piece and look: P pulls at its end, yet the piece does not fly off. That is because, at the cut face, the left piece was holding it back with an equal force. This holding force is the internal force. Drag the cut to move it. Wherever you cut, the internal force always equals P. The external load passes straight through the bar's interior. This imaginary cut is the only way to see what is inside.

Even for the same internal force, a thick bar and a thin bar have very different insides. The internal force P is shared across the whole cross-section, so when the area is large, each patch carries a smaller share. That share per patch is stress. σ = PA. Drag the area A. P stays the same, but as A grows the color turns blue for low stress, and as A shrinks it turns red for high stress. That is why, under the same pull, the thin part becomes dangerous first: stress spikes only there. What decides whether something breaks is not the force, but the stress.

How the force acts on the cut face matters too. There are two cases. When the force acts perpendicular to the face, it pulls the face straight apart or presses it together. This is normal stress σ: tension if pulling, compression if pushing. When instead the force acts parallel along the face, it slides the face. This is shear stress τ. Toggle between the two with the buttons. Normal pulls the blocks apart; shear pushes them out of line. Both are force divided by area, but the direction differs, so the way the material resists, and the way it fails, differ too. When a bolt or a pin snaps, shear is usually the culprit.

Now for a feel for the unit. Stress is force divided by area, so its unit is N/m², which we call the pascal, Pa. But a pascal is tiny, so in engineering we use the megapascal, MPa, a million of them. The handy fact to remember: 1 MPa = 1 N/mm². One newton on one square millimeter. Drag P to pull the rod and watch where σ lands on the scale. Air pressure is about 0.1 MPa, concrete is tens of MPa, and steel yields around 250 MPa. Once you attach this sense of position to the numbers, you can tell at a glance whether a result makes sense.

In PracticeTo sum up: stress is the internal force carried per unit area at an imaginary cut. σ = PA. The external load passes straight through the bar, so the internal force equals P everywhere, and dividing that internal force by the cross-sectional area gives the stress. Force perpendicular to a face is normal stress σ; force along a face is shear stress τ. The unit is MPa, and 1 MPa = 1 N/mm². In engineering, to judge whether a part will hold, you always look at stress, not force, because under the same load the stress shoots up where the section is narrow, around holes, at stepped corners, and that is where failure starts. In the next lesson, we follow how much this pull actually stretches the bar, through strain.
Mechanics of Materials
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