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Queensferry

Module 3 · Lesson 3.1

Cutting the member open

How internal forces arise, and how to find them at any section.

Why this matters

Loads are applied on the outside. Materials fail on the inside. The cut-section method is the bridge between the two, and it is the single most reusable idea in the whole subject.

By the end of this lesson you should be able to

  • Cut a member and draw the free body of one side
  • Find the normal force, shear force and bending moment at the cut
  • Explain why internal actions change along the length

Imagine sawing a loaded beam in half. It would immediately fall apart, because the two halves were holding each other. Whatever forces the material was providing across that plane are the internal forces.

To find them, cut the beam at the section you care about, throw one side away, and ask what forces you would have to apply at the cut face to keep the remaining piece exactly where it is. Those forces are what the material was carrying.

At a cut in a plane member there are three of them. Normal force N acts along the member and either stretches or squashes it. Shear force V acts across the member and tries to slide one face past the other. Bending moment M is the turning effect that bends the member. In three dimensions there is also torsion, which twists it.

The same eight metre beam with a dashed cut line at three metres, showing that the left-hand piece must be held in equilibrium by internal forces at the cut.5 kN/m40 kN45.00 kN35.00 kNcut
Cut at 3.0 m. Everything to the left of the dashed line must be held still by the forces acting across that face.

Try it

Cut the beam and look inside

Move the cut along the beam. The internal shear force and bending moment are whatever it takes to keep the left-hand piece in equilibrium.

m from the left support

Left reaction
25.00 kN
Shear V(x)
17.00 kN
Moment M(x)
42.00 kN·m
Beam cut at 2.0 metres from the left support4 kN/m24 kN25.00 kN31.00 kNcutShear force V (kN)-31.00Bending moment M (kN·m)75.00

Equilibrium of the left-hand piece

Vertical: 25.00 kN up, minus 8.00 kN of load applied so far, leaves V = 17.00 kN carried across the cut.

Moments about the cut give M = 42.00 kN·m (sagging).

Predict first

Why do the internal forces change as you move the cut along the beam?

Worked example

Internal forces at a section

Internal forces at a section5 kN/m40 kN45.00 kN35.00 kNcut

Given

  • The 8.0 m beam from Module 3: RA = 45 kN, RB = 35 kN
  • UDL of 5 kN/m over the whole span
  • Point load of 40 kN at 3.0 m
  • Cut taken at x = 3.0 m, just to the left of the point load

Find

The shear force and bending moment at the cut.

Assumptions

  • No horizontal loads, so the normal force is zero

    Practice

    A simply supported beam spans 8.0 m and carries a UDL of 6 kN/m over its whole length. What is the shear force 2.0 m from the left support?

    Practice

    For that same beam, what is the bending moment 2.0 m from the left support?

    Practice

    A member carries a purely axial tensile force of 85 kN and no transverse load. What is the shear force at any section along it?

    The cut-section method gives the normal force just as readily as the shear and moment, and it is worth drawing its own diagram when a member carries load along its length.

    Cut the member, take one side, and resolve along the member instead of across it. The normal force is the algebraic sum of the axial components to that side, taken as positive in tension.

    A column picking up load at each floor is the standard case. The normal force is constant between floors and steps by the load applied at each one, so the normal force diagram is a series of steps — exactly like a torque diagram, and for the same reason: point actions produce jumps, distributed ones produce slopes. A column carrying only its own self-weight, for instance, has a linearly varying normal force.

    Worked example

    Normal force diagram for a column, and superposition

    Given

    • A three-storey column carrying 200 kN at roof level, 300 kN at second floor and 300 kN at first floor
    • Separately: the same column also carries a bending moment from a wind case

    Find

    The normal force in each storey, and how the two load cases combine.

      Practice

      A column carries 200 kN at roof level, 300 kN at second floor and 300 kN at first floor. What is the magnitude of the normal force in the bottom storey, in kN?

      Practice

      A member carries an axial compression producing 5.00 N/mm² and a bending moment producing ±8.00 N/mm². By superposition, what is the magnitude of the largest compressive stress, in N/mm²?

      Summary

      • Internal forces are whatever the material must supply to hold the cut piece in equilibrium
      • N acts along, V acts across, M bends, and torsion twists
      • Either side of the cut gives the same answer — use it as a check
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      This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint