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6.2 · Choose section cuts and sign conventions

Learn to choose section cuts and sign conventions through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Internal Forces and Beam Diagrams

A practical method for finding internal forces and moments in planar members

A section cut is an imaginary cut through a member. It lets you examine the equilibrium of one part of a structure and determine the forces and moment that the removed part would exert on it. The method is useful only when the cut is chosen and labelled carefully: the free-body diagram must show the loads and supports on the isolated part, as well as the internal resultants at the cut. This lesson focuses on choosing those cuts and keeping signs consistent. All examples use planar, statically determinate beams.

What you will learn

  • Choose a section cut that isolates the part of a member needed for analysis.
  • Draw a free-body diagram of one side of a cut and show the internal resultants.
  • Use a consistent sign convention for axial force, shear force, and bending moment.
  • Solve the equilibrium equations and check force and moment balance.

1. What a section cut represents

Imagine a beam as one rigid body in equilibrium. If you cut it at a chosen location, you can isolate either the left piece or the right piece. The remaining piece would have exerted forces and a moment on the isolated piece at the cut. Replace that removed interaction with three unknowns: axial force NN, shear force VV, and bending moment MM. These are internal resultants; they are not extra applied loads on the uncut beam.
For a useful cut, first identify the location where you want the internal resultants. Then choose the side of the cut with fewer unknown support reactions or fewer applied loads. A cut should normally lie between load or support locations, not directly through a point force or support, because the free-body diagram at such a location can be ambiguous. If internal resultants are needed in more than one interval, use a separate cut and equilibrium diagram for each interval.
A section cut does not create a new physical support. It exposes the interaction across an imagined boundary. On the free-body diagram of one isolated piece, draw all external loads and support reactions acting on that piece, then draw the three internal resultants at the cut.
  • Choose the cut at the location where the internal resultants are required.
  • Isolate the side with the simpler equilibrium equations.
  • Include only forces and reactions that act on the chosen piece.

2. Axes and a consistent sign convention

For the beam examples, take +x+x to the right and +y+y upward. Take counterclockwise moments as positive. State these choices before writing equations; then use them for every force and moment in the free-body diagram.
Use this section-resultant convention: positive NN pulls away from the cut, so it represents tension. On an isolated left piece, positive VV acts upward at the cut and positive MM acts counterclockwise. On an isolated right piece, the corresponding positive directions reverse: positive VV acts downward and positive MM acts clockwise. These opposite directions represent the same internal interaction viewed on opposite sides of the cut.
You may instead assume directions that are convenient, as long as you label them and use them consistently. If an answer is negative, the actual direction is opposite the assumed direction. A negative result does not by itself mean the equilibrium method failed.
For a planar rigid-body free-body diagram, use force balance in the two coordinate directions and moment balance about a convenient point. Choosing the cut itself as the moment point often removes the internal force terms from the moment equation, leaving the internal moment and moments of external forces.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Positive axial force is tension under the stated convention.
  • Positive shear and moment directions reverse between left-piece and right-piece diagrams.
  • Use the same axes and moment sign throughout each calculation.

3. A reliable procedure for each cut

Begin with the whole structure if support reactions are needed. Find those reactions from equilibrium before making a cut. Next, mark the cut and select one side. Redraw that side as a free body rather than relying on the original whole-structure sketch.
At the cut, draw NN, VV, and MM in the directions required by the convention for the chosen side. Write equilibrium equations with signed components. For moments, use a clear lever arm and the chosen counterclockwise-positive rule. Solve the equations, retain units, and interpret any negative result as a reversed direction.
Finally, check the isolated piece independently. The signed horizontal and vertical force sums should each be zero, and the moment sum about a stated point should also be zero. A force check alone is not enough: an incorrect lever arm or moment sign can leave force balance correct while moment balance fails.
N, V, MN,\ V,\ M
  • Find whole-body reactions first when they are needed.
  • Redraw the isolated segment and show all three internal resultants.
  • Verify both force balance and moment balance.

Worked example

Example 1: Cut through a simply supported beam

A 6 m beam is supported by a pin at A and a roller at B. A 12 kN downward point load acts 4 m from A. Find the internal resultants at a cut 2 m from A, using the left piece.
  1. Find the support reactions
    Use the whole beam first. There is no horizontal applied load, so the pin's horizontal reaction is zero. Taking moments about A gives the vertical reaction at B.
    By(6 m)−(12 kN)(4 m)=0B_y(6\,\mathrm{m})-(12\,\mathrm{kN})(4\,\mathrm{m})=0
  2. Use the left-piece free body
    The moment equation gives By=8 kNB_y=8\,\mathrm{kN}. Vertical equilibrium of the whole beam then gives Ay=4 kNA_y=4\,\mathrm{kN}. On the left piece, the only external force is AyA_y; the cut is 2 m from A. Apply the left-piece convention: positive shear is upward and positive moment is counterclockwise.
    Ay=4 kN,By=8 kNA_y=4\,\mathrm{kN},\quad B_y=8\,\mathrm{kN}
  3. Solve and check
    Horizontal balance gives N=0N=0. Vertical balance gives V=−4 kNV=-4\,\mathrm{kN}, so the actual shear is 4 kN downward on the left piece. Taking moments about the cut, the reaction at A has a 2 m lever arm and gives a clockwise moment; the internal moment must balance it. The result is positive under the stated counterclockwise convention.
    N=0,V=−4 kN,M=8 kN⋅mN=0,\quad V=-4\,\mathrm{kN},\quad M=8\,\mathrm{kN\cdot m}
Answer: At the cut, N=0N=0, V=−4 kNV=-4\,\mathrm{kN}, and M=8 kN⋅mM=8\,\mathrm{kN\cdot m}. Thus the shear acts downward on the left piece and the internal moment acts counterclockwise.
Check: Vertical forces balance: 4−4=0 kN4-4=0\,\mathrm{kN}. Moments about the cut balance: the 4 kN reaction 2 m to the left gives −8 kN⋅m-8\,\mathrm{kN\cdot m}, which is balanced by the +8 kN⋅m+8\,\mathrm{kN\cdot m} internal moment.

Worked example

Example 2: Use the right piece and its reversed signs

For the beam in Example 1, find the internal resultants at a cut 5 m from A by isolating the right piece. Use the right-piece sign convention.
  1. Select the simpler side
    The cut is at 5 m, so the right piece is only 1 m long and contains the known reaction By=8 kNB_y=8\,\mathrm{kN}. The 12 kN load and support A are on the other side and do not appear on this free-body diagram. There is no horizontal force, so axial force is zero.
    N=0N=0
  2. Apply vertical balance
    On a right-piece diagram, positive shear is downward. The upward roller reaction and the signed shear must balance. A negative shear value therefore means the actual force acts upward on this right piece.
    8 kN+V=08\,\mathrm{kN}+V=0
  3. Apply moment balance at the cut
    The upward reaction at B is 1 m to the right of the cut, so it creates a counterclockwise moment. Positive internal moment on the right piece is clockwise. Their signed moments balance as shown.
    (8 kN)(1 m)−M=0(8\,\mathrm{kN})(1\,\mathrm{m})-M=0
Answer: N=0N=0, V=−8 kNV=-8\,\mathrm{kN}, and M=8 kN⋅mM=8\,\mathrm{kN\cdot m}. The shear is actually upward on the right piece, and the internal moment is clockwise.
Check: Vertical balance is 8−8=0 kN8-8=0\,\mathrm{kN}. Moment balance about the cut is +8−8=0 kN⋅m+8-8=0\,\mathrm{kN\cdot m}, with counterclockwise taken as positive.

Worked example

Example 3: Place the cut between two point loads

A 5 m beam has a pin at A and a roller at B. Downward loads of 10 kN and 5 kN act at 2 m and 4 m from A, respectively. Find the resultants at a cut 3 m from A using the left piece.
  1. Find the reactions
    Take moments about A for the whole beam. The two downward loads have moment arms of 2 m and 4 m. Then use vertical force balance to find the reaction at A.
    (5 m)By−(10 kN)(2 m)−(5 kN)(4 m)=0(5\,\mathrm{m})B_y-(10\,\mathrm{kN})(2\,\mathrm{m})-(5\,\mathrm{kN})(4\,\mathrm{m})=0
  2. Isolate the left segment
    The reaction is By=8 kNB_y=8\,\mathrm{kN}, so vertical balance gives Ay=7 kNA_y=7\,\mathrm{kN}. The cut is between the loads: the left piece contains the 10 kN load but not the 5 kN load or the roller reaction. There are no horizontal loads.
    Ay=7 kN,By=8 kN,N=0A_y=7\,\mathrm{kN},\quad B_y=8\,\mathrm{kN},\quad N=0
  3. Solve shear and moment
    Vertical balance on the left piece includes AyA_y, the 10 kN downward load, and positive shear upward at the cut. Taking moments about the cut, the reaction is 3 m away and the point load is 1 m away. Their signed moments determine the cut moment.
    7−10+V=0,M−(7)(3)+(10)(1)=07-10+V=0,\quad M-(7)(3)+(10)(1)=0
Answer: N=0N=0, V=3 kNV=3\,\mathrm{kN}, and M=11 kN⋅mM=11\,\mathrm{kN\cdot m}. Both shear and moment are positive for the chosen left-piece convention.
Check: Vertical forces sum to 7−10+3=0 kN7-10+3=0\,\mathrm{kN}. Moments about the cut sum to −21+10+11=0 kN⋅m-21+10+11=0\,\mathrm{kN\cdot m}.

Common mistakes and how to avoid them

Drawing internal resultants on the whole-beam diagram without isolating either side of the cut.
Correction: Redraw one segment as its own free body and show only loads and reactions acting on that segment, together with the cut resultants.
Using the same positive shear and moment arrows on left-piece and right-piece diagrams.
Correction: Under the convention in this lesson, positive shear and moment arrows reverse on opposite faces. State which piece is isolated before assigning directions.
Treating a negative answer as an invalid solution.
Correction: A negative signed result indicates that the actual direction is opposite the assumed positive direction.
Checking force balance but not moment balance.
Correction: Check both coordinate force sums and a moment sum about a clear reference point. Include signed lever arms and moment directions.
Including a load that lies on the other side of the cut.
Correction: Inspect the chosen segment carefully. Only loads and reactions physically acting on the isolated piece belong on its free-body diagram.

Lesson summary

  • Choose a cut where internal resultants are needed, then isolate the side with the simpler free-body diagram.
  • Show axial force, shear, and moment at the cut, and state the sign convention for the selected side.
  • Use planar equilibrium with consistent axes, signs, and units.
  • Interpret negative results as reversed directions and verify force and moment balance.

Check your understanding

Question 1

A left beam segment is isolated. You define positive shear as upward at the cut. The calculation gives V=−6 kNV=-6\,\mathrm{kN}. What does this mean?
  1. The actual shear is 6 kN downward on the left segment.
  2. The actual shear is 6 kN upward on the left segment.
  3. The beam is not in equilibrium.
  4. The shear must be reported as zero.
Show answer and explanation
The actual shear is 6 kN downward on the left segment.
A negative signed value means the actual direction is opposite the assumed positive direction. Here, positive shear was upward, so the actual shear is downward.

Question 2

Why might you isolate the right side instead of the left side of a section cut?
  1. The right side always has zero internal moment.
  2. It may contain fewer unknown reactions or applied loads.
  3. The right side changes the beam's support type.
  4. It removes the need to use moment equilibrium.
Show answer and explanation
It may contain fewer unknown reactions or applied loads.
Either side can be used, but choosing the side with fewer unknowns often makes the equilibrium equations simpler.

Question 3

For a cut 3 m from A, a 7 kN upward reaction at A and a 10 kN downward load 2 m from A act on the left piece. Positive shear is upward. What is the signed shear at the cut?
  1. −3 kN-3\,\mathrm{kN}
  2. 3 kN3\,\mathrm{kN}
  3. 17 kN17\,\mathrm{kN}
  4. 0 kN0\,\mathrm{kN}
Show answer and explanation
3 kN3\,\mathrm{kN}
Vertical equilibrium gives 7−10+V=07-10+V=0, so V=3 kNV=3\,\mathrm{kN}, positive in the upward direction.

Key terms

Section cut
An imagined cut used to isolate part of a member and expose the internal interaction at that location.
Internal resultant
A force or moment representing the combined interaction across a cut; for a planar beam, the resultants are axial force, shear force, and bending moment.
Sign convention
A stated choice of positive directions for forces and moments used consistently in equations.
Free-body diagram
A drawing of an isolated body or segment showing the external forces, reactions, and internal resultants acting on it.

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