Internal Forces at a Section: Normal, Shear, Moment

Cut a beam and expose the three internal resultants that keep it from falling apart.

StaticsMechanical Engineering Year 1Free preview
⏱️ About 16 min

Every beam that isn't snapping in half is fighting back with three hidden forces at every cross-section: a normal force resisting stretch, a shear force resisting sliding, and a bending moment resisting the tendency to fold. Learn to cut the beam, draw those three forces, and compute them — it's the gateway to every beam design calculation that follows.

💡
The big idea: The method of sections for beams is nothing more than equilibrium applied to a piece of the beam. Cut the beam at the section of interest, replace the material you removed with three internal resultants (N, V, M), and write ΣF=0 and ΣM=0 for the remaining piece. The sign convention is a bookkeeping choice, but you must apply it consistently or your diagrams will be upside-down.
🎯 By the end, you'll be able to
  • Apply the method of sections to find N, V, and M at a specified location
  • State the standard sign convention for positive normal force, shear force, and bending moment
  • Recognize that for beams with only transverse loads, the normal force is typically zero
📎 Helpful to know first

The cut: exposing what's hidden

Imagine slicing cleanly through a beam at some distance x from the left end. To keep the left segment in equilibrium, the right segment must have been pushing or pulling on it across that cut. Those invisible interactions are what we call the internal forces, and there are exactly three of them in planar beam theory:

  • Normal force N — acts perpendicular to the cut surface, resisting tension or compression.
  • Shear force V — acts parallel to the cut surface, resisting the sliding of one slice past the next.
  • Bending moment M — a couple that resists rotation of the slice, keeping the beam from folding.

To find their values, simply write the equilibrium equations for whichever side of the cut is easier. Both sides must give the same answer — if they don't, you made an algebra or sign error.

A simply supported beam cut at distance x from the left pin support, exposing internal normal force N, shear force V, and bending moment M N V M P x A B
Positive internal resultants on the left face of a cut: N pulls away from the section (tension), V acts downward, and M acts counterclockwise to produce sagging.
🔑 The standard sign convention
  • Positive N (tension): pulls outward from the section — to the left on the left face, to the right on the right face.
  • Positive V: acts downward on the left face and upward on the right face. This corresponds to clockwise rotation of a differential element.
  • Positive M: acts counterclockwise on the left face and clockwise on the right face. This produces sagging (compression on top fibres, tension on bottom fibres), the shape of a smile.

A common beginner mistake is to flip the shear sign halfway through a problem. Pick the convention above, draw it on your first cut, and stick with it for the entire beam.

Beams with only vertical loads

In most introductory beam problems, every applied load and every reaction is purely vertical. With no horizontal external forces, horizontal equilibrium of the whole beam immediately gives Ax = 0. When you then cut the beam, the normal force N is zero everywhere — there is simply nothing to stretch or compress the beam axially. That lets you focus on shear and moment, but you should still write ΣFx = 0 at the cut to confirm N = 0 explicitly. It's a good habit that catches errors in more complex problems where axial loads do appear.

📝 Worked example: A 10 m simply supported beam has a pin at A and a roller at B. A single downward point load of 20 kN is applied 4 m from A. Find the internal shear force V and bending moment M at a section 2 m from A.
  1. Reactions: ΣM_A = 0 → B_y · 10 − 20 · 4 = 0 → B_y = 8 kN. ΣF_y = 0 → A_y + 8 − 20 = 0 → A_y = 12 kN.
  2. Cut the beam at x = 2 m (this lies to the left of the load, so only A_y appears on the left segment).
  3. Draw the left segment: A_y = 12 kN upward at A; unknown V and M at the cut.
  4. ΣF_y = 0: 12 − V = 0 → V = 12 kN (positive, so it acts downward on the left face as assumed).
  5. ΣM_cut = 0 (counterclockwise positive): M − 12 · 2 = 0 → M = 24 kN·m (positive, so it acts counterclockwise on the left face, producing sagging).
  6. ΣF_x = 0 confirms N = 0.
✓ V = 12 kN, M = 24 kN·m, N = 0.
✏️ Practice: Using the same beam as the example (10 m span, 20 kN load at 4 m from A), what is the internal shear force V, in kN, at a section 6 m from A?
kN
Solution
  1. At x = 6 m, the cut is to the right of the load. The left segment now sees A_y = 12 kN up and the 20 kN load down.
  2. ΣF_y = 0: 12 − 20 − V = 0 → V = −8 kN.
  3. The negative sign means the shear acts opposite to the assumed positive direction — upward on the left face.
✏️ Practice: For the same beam and cut at x = 6 m, what is the internal bending moment M, in kN·m?
kN·m
Solution
  1. For the left segment at x = 6 m: ΣM_cut = 0 → M − 12 · 6 + 20 · (6 − 4) = 0.
  2. M = 72 − 40 = 32 kN·m.
🎮 Interactive: internal forces at a cut LIVE
Predict first: Move the section slider to x = 2 m, then to x = 6 m. How do V and M change when you cross the load?

An interactive beam simulator showing shear and moment values at a movable section along a simply supported beam with a center point load.

The simulator shows V and M at any section along the beam. Notice the shear drops abruptly at the point load, while the moment stays continuous.

Check your understanding

1. For a beam with only vertical loads and vertical reactions, what is the normal force N everywhere?
With no horizontal external forces, ΣFx = 0 at every cut gives N = 0 everywhere.
2. Positive bending moment M produces which deformed shape?
Positive M causes compression in the top fibres and tension in the bottom fibres — the beam sags downward in a smile shape.
3. Why must the left and right segments of a cut give the same values of N, V, and M?
By Newton's third law, the internal force on the left face is equal and opposite to the force on the right face. The magnitude is the same; only the drawn direction flips.
🔎 Related lessons
Distributed Loads — Equivalent Resultants Shear & Moment Diagrams — the Direct Method · soon
✅ Key takeaways
  • The method of sections exposes three internal resultants at any cut: normal force N, shear force V, and bending moment M.
  • Positive N is tension; positive V is downward on the left face; positive M produces sagging.
  • For purely vertical loading, N = 0 everywhere, letting you focus on V and M.
➡️ With N, V, and M defined at a single section, the next step is to find them at every section along the beam — drawing the complete shear and moment diagrams.
Want to test yourself on this? Try the Mechanical Aptitude test →
🎓 Go deeper: external courses & trusted references

External sites are listed for reference only. This course is independent and has no affiliation with, or endorsement from, the institutions named.