Shear force & bending moment: practice problems with step-by-step solutions

Also called: shear and moment diagrams · SFD and BMD

Free Shear force & bending moment practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: point load, full-span UDL, point load + UDL.

Free practice problemEasy
P = 60 kNAB2.5 m2.5 m

The simply supported beam carries a single point load P. Find the shear force Vx just to the right of the load. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 60 kN at 2.5 m
kN
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What you practise

  1. Easypoint load
  2. Mediumfull-span UDL
  3. Hardpoint load + UDL

How to solve Shear force & bending moment problems

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.

Common mistakes

  • Including loads from both sides of the cut.
  • Using the wrong sign convention for sagging and hogging moments.
  • Answering with the shear force when the bending moment is asked.

Key formulas

Shear and moment
V = ΣF on one side, M = ΣF·d on one side
Bending stress
σ = M·y/I
Rectangle
I = bh³/12
Section modulus
W = I/ymax
Simply supported, uniform load
Mmax = wL²/8

Learn Shear force & bending moment from scratch

Before this: Beam support reactions

1What happens inside a beam

In the reactions lesson you found the forces on a beam from outside. Now look inside: what does the material at one point of the beam have to carry?

Imagine sawing the beam in two at some point. Each half would fall. To keep it in place, the other half must hold it with two internal actions:

  • the shear force V, which stops one side from sliding up or down past the other (in kN);
  • the bending moment M, which stops one side from rotating relative to the other (in kN·m).
PAB2 m2 m2 m

A beam that bends into a smile, with the bottom stretched, is sagging. A beam that bends into a frown, like a balcony, is hogging.

Engineers need M because the bending stress in the beam grows with it, and V because the beam must not shear through near the supports.

Self-check A simply supported beam carries a load downwards. Which side of the beam is stretched?
  1. Both sides equally
  2. Neither side, the beam only shortens
  3. The bottom: the beam is sagging
  4. The top: the beam is hogging

Answer: The bottom: the beam is sagging. A beam on two supports bends into a smile under a downward load. The bottom fibres get longer (tension), the top fibres shorter (compression).

2The method of sections

To find V and M at a distance x from A, cut the beam there and keep only the part to the left of the cut. That part is in equilibrium on its own.

RA​VMAx

The usual sign rules: an upward force on the left part counts as positive shear, and a sagging moment is positive. This gives two short recipes:

  • Shear force: V = RA minus all downward loads left of the cut.
  • Bending moment: M = RA·x minus, for each load left of the cut, the load times its distance to the cut.

A UDL portion left of the cut acts as one force w·ℓ at its own middle, just as in the reactions lesson.

You may keep the right part instead. You get the same V and M, which is a useful check.

Self-check RA = 10 kN. The first load is 5 m from A. What is the bending moment at x = 3 m?
  1. 10 kN·m
  2. 13 kN·m
  3. 3.3 kN·m
  4. 30 kN·m

Answer: 30 kN·m. No load is left of the cut, so M = RA·x = 10·3 = 30 kN·m. 10 kN is the shear force there, not the moment.

3Diagrams and the maximum moment

If you calculate V and M at every x, you get the shear force diagram and the bending moment diagram. Their shapes follow simple rules:

  • Between point loads, V is constant and M is a straight line.
  • Under a UDL, V slopes down in a straight line and M is a curve (a parabola).
  • A point load makes V jump by the size of the load.
  • M is largest where V changes sign, passing through zero.

Two results come up so often that they are worth knowing:

  • Point load P at midspan: Mmax = P·L / 4
  • UDL w over the whole span: Mmax = w·L² / 8
Self-check A simply supported beam of 6 m carries a UDL of 4 kN/m over its whole span. What is the largest bending moment?
  1. 12 kN·m
  2. 144 kN·m
  3. 18 kN·m
  4. 36 kN·m

Answer: 18 kN·m. Mmax = wL²/8 = 4·36/8 = 18 kN·m at midspan. 36 uses /4 (the point-load rule); 12 kN is the reaction wL/2.

4The method step by step

  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section x and keep the left part.
  3. Shear force: V = RA minus all downward loads on the left part.
  4. Bending moment: M = RA·x minus each load on the left part times its distance to the cut.
  5. For the maximum moment: find where V = 0 and calculate M there.

Draw the left part as a small separate sketch. It makes it obvious which loads to include.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A simply supported beam of span 6 m carries a UDL of 4 kN/m over its whole length. Find the shear force V and the bending moment M at x = 2 m from A.

1

Look at the beam

w = 4 kN/mABRA​RB​2 m4 m
2

Support reactions

The total load is 4 · 6 = 24 kN, in the middle. By symmetry each support carries half:

RA = RB = 12 kN
3

Keep the left part

w = 4 kN/m12 kNVMA2 m

The UDL on the left part is 4 · 2 = 8 kN. It acts at its own middle, 1 m from the cut.

4

Shear force

V = 12 − 8 = 4 kN
5

Bending moment

M = 12 · 2 − 8 · 1 = 16 kN·m

It is positive, so the beam sags here.

6

Check the answer

Use the right part instead: it is 4 m long, with RB = 12 kN and a UDL of 16 kN at 2 m from the cut.

M = 12 · 4 − 16 · 2 = 16 kN·m

The full diagram is a parabola with its peak wL²/8 = 18 kN·m at midspan. 16 kN·m at x = 2 m fits.

1618 kN·mMx = 2x = 3
Answer

V = 4 kN and M = 16 kN·m (sagging).

6Common mistakes

  • Including loads from both sides of the cut. Use only the part you kept.
  • Measuring the lever arm from A instead of from the cut. In M, each distance is from the load to the cut.
  • Answering with V when M is asked, or the other way round. Check the unit: kN or kN·m.
  • Taking the whole UDL instead of only the part left of the cut.
  • Mixing sign rules. With the usual rules, a simply supported beam under downward loads has positive (sagging) moments everywhere.
Self-check A 4 m simply supported beam carries 20 kN at midspan, so RA = 10 kN. What is M at x = 1 m?
  1. 0 kN·m
  2. 10 kN
  3. 10 kN·m
  4. 20 kN·m

Answer: 10 kN·m. Only RA is left of the cut: M = 10·1 = 10 kN·m. 0 wrongly subtracts the load, which is right of the cut. 20 kN·m is PL/4, the maximum at midspan. 10 kN is the shear force.

7How exam questions look

  • Easy: one point load. Find M at the load, or V just beside it. Variants use a cantilever, where M is largest at the wall.
  • Medium: a UDL over the full span, as in the example. Find V or M at a given x, or the maximum moment.
  • Hard: a point load and a UDL together, or a partial UDL. Find M at a section or the maximum moment where V = 0.

Some versions use a beam that overhangs a support. The overhang hogs, so M is negative there.

Always start with the reactions and write them on your sketch.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
40 kNABCDEFG2 m2 m2 m2 m2 m2 m

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

40 kN @ 2 m
  1. 1A (x = 0 m)
  2. 2B (x = 2 m)
  3. 3C (x = 4 m)
  4. 4D (x = 6 m)
  5. 5E (x = 8 m)
  6. 6F (x = 10 m)
  7. 7G (x = 12 m)
Show hints
  1. Peaks of the bending moment sit where the shear force changes sign, at a fixed wall, or over a support next to an overhang.
  2. Find the support reactions first: R1 = 33.33 kN.
  3. Compare the moments at the candidate marks; the largest size is 66.67 kN·m.
Show the answer and explanation

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R1 = 33.33 kN, R2 = 6.667 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

MA = 0, MB = 66.67 kN·m, MC = 53.33, MD = 40, ME = 26.67, MF = 13.33, MG = 0

Answer

Mark B: MB = 66.67 kN·m
Example 2Medium
w = 5 kN/mABsection0.5 m5.5 m

Find the bending moment Mx at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 5 kN/m from 0 to 6 mx = 0.5 m

Find Mx (kN·m)

Show hints
  1. First find the support reactions from equilibrium of the whole beam.
  2. Then cut the beam at x = 0.5 m and look only at the part to the left of the cut. Sum the moments of those forces about the cut.
  3. RA = 15 kN.
Show the answer and the step-by-step solution

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·6 = 30 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 5·6·3
RB = 90/6 = 15 kN
ΣFy = 0: RA = 30 − 15 = 15 kN
w = 5 kN/mABRA​ = 15RB​ = 15section0.5 m5.5 m

Cut at x = 0.5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 15·0.5 − 5·0.5·0.25 = 6.875 kN·m

Answer

Mx = 6.875 kN·m
Example 3Hard
w = 6 kN/mP = 45 kNABsection1.5 m3.5 m1 m

Find the bending moment Mx at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 45 kN at 1.5 mw = 6 kN/m from 0 to 6 mx = 5 m

Find Mx (kN·m)

Show hints
  1. First find the support reactions from equilibrium of the whole beam.
  2. Then cut the beam at x = 5 m and look only at the part to the left of the cut. Sum the moments of those forces about the cut.
  3. RA = 51.75 kN.
Show the answer and the step-by-step solution

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·6 = 36 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 45·1.5 + 6·6·3
RB = 175.5/6 = 29.25 kN
ΣFy = 0: RA = 81 − 29.25 = 51.75 kN
w = 6 kN/mP = 45 kNABRA​ = 51.75RB​ = 29.25section1.5 m3.5 m1 m

Cut at x = 5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = RA·x − (each load left of the cut × its distance to the cut)
Mx = 51.75·5 − 45·(5 − 1.5) − 6·5·2.5 = 26.25 kN·m

Answer

Mx = 26.25 kN·m

Questions and answers

What is Shear force & bending moment?

In the reactions lesson you found the forces on a beam from outside. Now look inside: what does the material at one point of the beam have to carry?

Imagine sawing the beam in two at some point. Each half would fall. To keep it in place, the other half must hold it with two internal actions:

How do I solve Shear force & bending moment problems step by step?
  1. Find the support reactions from equilibrium of the whole beam.
  2. Cut the beam at the section you need and keep one side.
  3. Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
  4. For the maximum moment, find where the shear force is zero.
What are common mistakes in Shear force & bending moment?
  • Including loads from both sides of the cut.
  • Using the wrong sign convention for sagging and hogging moments.
  • Answering with the shear force when the bending moment is asked.
What do the Shear force & bending moment exercises cover?
  • Easy: point load
  • Medium: full-span UDL
  • Hard: point load + UDL

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Beam support reactions

Continue with

Beam deflection

Calculators

Shear force and bending moment diagram calculatorBending stress calculator (σ = My/I)

Courses with this topic

StaticsMechanics of MaterialsStructural AnalysisFE Mechanical exam practiceFE Civil exam practice

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Printable worksheet

Shear force & bending moment worksheet with answers
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