Internal forces
A beam that is perfectly in equilibrium can still snap in the middle. Equilibrium of the whole says nothing about what any part of it is carrying — and the only way to find out is to cut the thing open and look.
Everything so far has treated a body as a single object that either balances or does not. That was a deliberate simplification, and it has a price: a structure can satisfy every equilibrium equation you can write for it and still fail, because failure is local. The load that matters is not the one on the beam, it is the one at the worst point along the beam. Finding it means asking a question the whole-body diagram cannot answer, and the way in was hiding in the free-body diagram all along.
Cutting the member
Drawing the boundary for a free-body diagram sorted forces into internal and external, and the sorting depended entirely on where the boundary went. Slide it so that it passes through the middle of a member rather than around it, and forces that were internal a moment ago are crossing the boundary: external, drawn, and solvable. That is the whole idea. There is nothing more to the method of sections than choosing to cut somewhere interesting.
Nothing physical happens at the cut. The beam is not damaged and the material is not removed; you have simply decided to account for one part of it and not the other. What the discarded part was doing is then owed to you as unknowns on the exposed face, and equilibrium of the part you kept is enough to find them.
What is actually distributed over that face is a traction — a push or pull per unit area, varying from point to point across the depth of the section, and there are infinitely many of them. You do not want infinitely many unknowns. You want the answer from force system resultants, which said that any collection of forces whatever, however many and however placed, is equivalent to one force and one couple at a chosen point:
Take the point to be the centroid of the cut face and the entire distribution collapses to exactly two vectors. This is not a modelling assumption or an approximation. It is a theorem, and it is the reason the subject is tractable: no matter how complicated the stress on that face turns out to be, statics only ever needs to know its resultant force and its resultant couple.
Which means the number of unknowns is fixed before you know anything about the beam. In a plane, a force has two components and a couple has one, so a cut costs you three unknowns and hands you the three planar equilibrium equations to pay for them. Every section, anywhere, on any member, is a determinate problem.
A beam, held up, carrying a load.
Normal force, shear and moment
The force and the couple on the face are more useful split up, because the three pieces do three physically different things to the member. Resolve the force along the axis of the member and across it:
The component along the axis is the normal force \(N\). It stretches the member or squashes it, and it is the only one of the three that a rope could carry. The component across the axis is the shear \(V\), which tries to slide the two faces past one another, like the two blades of a pair of scissors closing on the section. The couple is the bending moment \(M\), which curves the member: it shortens the fibres on one side of the section and lengthens them on the other.
Three quantities in a plane. In three dimensions there are six, for the same reason there were six equilibrium equations — a force with three components and a couple with three. The extra couple about the member’s own axis is the torque, and it twists rather than bends, which is why shafts are analysed for it separately.
Keeping the three separate is worth the trouble because they are resisted by different features of the cross-section and they fail in different ways. A section can be perfectly adequate in tension and hopeless in bending. Reporting a single “internal force” would throw away exactly the information a designer needs, and the quantity that usually governs is the one that varies fastest along the member.
A frequent early confusion is between the shear \(V\) at a section and the reactions at the supports. They have the same units and they are often numerically equal near a support, but they are different things: a reaction is an external force applied by the world to the beam, and \(V\) is what one part of the beam is doing to the next part along. The clue is that a reaction sits at a fixed place, while \(V\) has a value at every point.
One force on the face, at whatever angle it likes.
The two faces of a cut
A cut makes two faces, one on each piece, and there is no way to choose between them. Whatever the right-hand piece was applying to the left-hand piece, the left-hand piece was applying back in equal measure and opposite sense. Newton’s third law is not a separate rule to remember here; it is the statement that the material at the section was never doing two things at once.
So the two sets are related by nothing more than a sign:
and that has a practical consequence worth more than the algebra. You may analyse whichever piece you prefer and you will get the same section forces, because the two answers are the same answer written from opposite sides. Pick the side with fewer forces on it. If a cantilever is cut near its free end, the free end has almost nothing on it and the built-in end has three reactions you would have had to find first; cutting well can make finding the reactions unnecessary altogether.
It also explains why internal forces never appeared in any earlier diagram. Draw a boundary around the whole beam and every internal pair lies inside it, so every one of them cancels in \(\sum \mathbf{F}\) and in \(\sum \mathbf{M}\). Equilibrium of the whole is completely blind to them. That is not a flaw in the method — it is what makes rigid-body statics work at all — but it is why a separate question has to be asked to see them.
The mistake this invites is drawing the two faces with the same set of arrows, both pointing the same way, and then being surprised when the two sides of the beam disagree. If you have drawn \(V\) downward on the left piece, it must be upward on the right piece. Draw the pair together once, properly, and the habit sticks.
One cut, and it has two faces.
The sign convention
Since the two faces carry opposite arrows, “the shear is upward” is not a statement about the beam. It is a statement about which piece you happened to keep, and it changes if you keep the other one. A convention that assigned signs by coordinate direction would therefore give a different sign at the same point depending on which way you worked, which is useless.
The convention in use avoids this by naming the deformation instead. A quantity is positive when the pair of arrows, taken together on a short segment, deforms it in the agreed way:
Positive \(N\) pulls outward on both faces, so the segment is stretched. Positive \(V\) acts upward on the left face and downward on the right, so the segment is turned clockwise. Positive \(M\) is the pair that bends the segment concave upward — sagging, with the top fibres shortening and the bottom fibres stretching. Each of those three descriptions survives being read from either side, which is the entire point.
Sagging positive is worth taking seriously rather than memorising, because it is the sign that engineering conclusions are drawn from. In a simply supported beam the moment sags everywhere, so the underside is in tension, which is where reinforcement goes in a concrete beam. Over an internal support in a continuous beam the moment hogs, the tension moves to the top, and the steel moves with it. The sign of \(M\) tells you which face of the member is being pulled apart, and getting it backwards puts the reinforcement in the compression zone.
In practice, always draw the unknowns in their positive sense, whatever you suspect the true answer to be, and let the algebra decide. A negative result then means the arrow points the other way and the beam hogs rather than sags — a piece of information, not an error. Assuming a sense to make the arithmetic come out positive is how sign errors get hidden rather than found.
Positive N pulls outward on both faces.
Working a section
The procedure is short and the order matters. Find the support reactions first, from the free-body diagram of the entire member, using the moment centres that isolate them one at a time as in equilibrium of a rigid body. Then cut at the section of interest, keep the simpler side, and draw \(N\), \(V\) and \(M\) on the exposed face in their positive senses. Then apply the three equations.
Taking moments about the cut itself is the move that makes it quick, because \(N\) and \(V\) both pass through that point and drop out, leaving \(M\) alone:
Three equations, each yielding one unknown, and never a simultaneous pair. Take a 6 m simply supported beam with 12 kN hung 4 m from the left support. The reactions are 4 kN and 8 kN. Cut at 5 m and keep the right-hand metre, which carries nothing but the 8 kN reaction: vertical balance gives \(V = -8\) kN and moments about the cut give \(M = 8\) kN·m. Keeping the left-hand side instead means carrying two forces through the arithmetic to reach identical answers.
Two cautions about where the cut goes. Never cut exactly at a point load or at a support, because \(V\) is genuinely different on the two sides of such a point and the question “what is \(V\) there” has no single answer; cut a little to one side and say which side you mean. And a distributed load must be replaced by its resultant only after the cut is made, using the part of it that lies on the piece you kept — replacing the whole load by its resultant first and then cutting through the resultant is the most common way to get a confidently wrong answer.
One section gives one number. Real design needs the worst section, and finding it by cutting at a hundred places is not a method. Let \(x\) stay a symbol rather than a number and the same three equations return \(V(x)\) and \(M(x)\) as functions along the whole member — which is exactly what shear and moment diagrams are, and where the maximum finds itself.