Beam deflection: practice problems with step-by-step solutions

Free Beam deflection practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: cantilever, end load, uniform load, superposition and design.

Free practice problemEasy
P = 3 kNδL = 1.5 m

A cantilever carries a point load P at its free end. Find the deflection δ at the free end.

P = 3 kNL = 1.5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴
mm
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What you practise

  1. Easycantilever, end load
  2. Mediumuniform load
  3. Hardsuperposition and design

Key formulas

Point load at midspan
δ = PL³/(48EI)
Uniform load, simply supported
δ = 5wL⁴/(384EI)
Cantilever, end load
δ = PL³/(3EI)
Cantilever, uniform load
δ = wL⁴/(8EI)
Superposition
δ = δ1 + δ2

Learn Beam deflection from scratch

Before this: Shear force & bending moment, Centroid & second moment of area

1What deflection is

Every beam bends a little under load. The distance a point of the beam moves down is its deflection δ, usually in mm.

PδL

A beam can be strong enough and still bend too much. Floors then feel bouncy, plaster cracks and doors stick. A common limit for floors is span/360: a 3.6 m floor may deflect at most 10 mm.

How much a beam deflects depends on four things:

  • the load: twice the load gives twice the deflection;
  • the span: deflection grows very fast with length (L³ or L⁴);
  • the material’s Young’s modulus E: stiffer material bends less;
  • the section’s second moment of area I: a deeper section bends less.

E and I always appear together as the product EI, the flexural stiffness of the beam.

Self-check You replace a beam by one with twice the second moment of area I. Everything else stays the same. What happens to the deflection?
  1. It becomes a quarter
  2. It doubles
  3. It halves
  4. It stays the same

Answer: It halves. Deflection is inversely proportional to EI. Doubling I halves δ.

2The standard formulas

Deflections come from integrating the bending moment twice, but for the common cases the result is tabulated. Four cases cover most questions:

  • Cantilever, point load P at the free end: δ = P·L³ / (3·E·I)
  • Cantilever, UDL w over the whole length: δ = w·L⁴ / (8·E·I)
  • Simply supported, point load P at midspan: δmax = P·L³ / (48·E·I)
  • Simply supported, UDL w over the whole span: δmax = 5·w·L⁴ / (384·E·I)
PδL

For a cantilever the largest deflection is at the free end; for a simply supported beam with symmetric loads it is at midspan.

When a beam carries several loads, the deflections from each load on its own add up. This is superposition.

Self-check A cantilever with an end load has a tip deflection of 2 mm. The cantilever is made twice as long. What is the new deflection?
  1. 16 mm
  2. 32 mm
  3. 4 mm
  4. 8 mm

Answer: 16 mm. δ is proportional to L³, so doubling L multiplies δ by 2³ = 8: 2·8 = 16 mm. 4 mm treats it as linear; 32 mm uses L⁴, which is the UDL case.

3Units: work in N and mm

The formulas only work with consistent units. The safe choice is newtons and millimetres:

  • P in N: kN × 1000.
  • w in N/mm: 1 kN/m is exactly 1 N/mm.
  • L in mm: m × 1000.
  • E in N/mm²: GPa × 1000. Steel: 200 GPa = 200 000 N/mm².
  • I in mm⁴: “36 × 10⁶ mm⁴” means 36 000 000 mm⁴.

The deflection then comes out in mm. Because L is raised to the third or fourth power, the numbers get large. A calculator with powers of ten keeps them under control.

Self-check A beam is made of steel, E = 200 GPa. Which value of E do you use with N and mm?
  1. 0.2 N/mm²
  2. 200 000 000 N/mm²
  3. 200 000 N/mm²
  4. 200 N/mm²

Answer: 200 000 N/mm². 1 GPa = 1000 MPa = 1000 N/mm², so 200 GPa = 200 000 N/mm². Using 200 makes the deflection 1000 times too large.

4The method step by step

  1. Identify the case: cantilever or simply supported, point load or UDL.
  2. Convert P, w, L, E and I to N and mm.
  3. Pick the matching formula.
  4. Calculate δ in mm.
  5. For several loads: calculate each one on its own and add them.
  6. For a design question: set δ equal to the allowed value (for example L/360) and solve for I.

5A worked example

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

Worked example

A steel cantilever (E = 200 GPa, I = 36 × 10⁶ mm⁴) is 3 m long and carries a point load of 8 kN at its free end. Find the deflection at the free end.

1

Identify the case

P = 8 kNδL = 3 m

A cantilever with a point load at the free end: δ = PL³/(3EI).

2

Convert to N and mm

P = 8000 N, L = 3000 mm, E = 200 000 N/mm², I = 36 000 000 mm⁴.

3

Flexural stiffness

E·I = 200 000 · 36 × 10⁶ = 7.2 × 10¹² N·mm²
4

Tip deflection

L³ = 3000³ = 2.7 × 10¹⁰ mm³
δ = 8000 · 2.7 × 10¹⁰ / (3 · 7.2 × 10¹²) = 10 mm
5

Check the answer

10 mm on a 3 m cantilever is L/300. That is a realistic size for a loaded steel beam: neither a fraction of a millimetre nor tens of centimetres.

Answer

The free end deflects δ = 10 mm.

6Common mistakes

  • Mixing units: L in metres with E in N/mm² gives nonsense. Convert everything to N and mm first.
  • Leaving E in GPa, which makes δ 1000 times too large.
  • Using the wrong case, for example PL³/(48EI) for a cantilever. Check the supports and the load type.
  • Forgetting the 5 in 5wL⁴/(384EI).
  • In superposition, using a different formula for each load but forgetting one of the loads.
Self-check A simply supported beam carries a midspan point load that alone gives 4 mm, and a UDL that alone gives 6 mm. What is the midspan deflection?
  1. 10 mm
  2. 2 mm
  3. 24 mm
  4. 6 mm

Answer: 10 mm. By superposition the deflections from each load add: 4 + 6 = 10 mm. Taking only the larger one, the difference or the product are all wrong.

7How exam questions look

  • Easy: cantilever with an end load. Find δ = PL³/(3EI). Variants: a simply supported beam with a central load, the end slope, or the load that gives a certain deflection.
  • Medium: uniform load. A cantilever (wL⁴/8EI) or a simply supported beam (5wL⁴/384EI). Variants: off-centre loads and end slopes.
  • Hard: superposition and design. Add the deflections of a point load and a UDL, or find the smallest I that keeps δ below L/360.

For the design case: Ireq = 5wL⁴ / (384·E·δallow), with δallow = L/360.

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

Worked examples

Example 1Easy
P = 19 kNABδ2 m2 m

A simply supported beam carries a point load P at midspan. Find the midspan deflection δmax.

P = 19 kNL = 4 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

Find δmax (mm)

Show hints
  1. For a simply supported beam with a central point load the midspan deflection is δ = PL³/(48EI).
  2. Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
  3. EI = 2.4 × 10¹² N·mm².
Show the answer and the step-by-step solution

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 19000 N, L = 4000 mm
EI = 2.4 × 10¹² N·mm²

Midspan deflection

δmax = PL³/(48EI) = 19000·4000³/(48·2.4 × 10¹²) = 10.56 mm

Answer

δmax = 10.56 mm
Example 2Medium
P = 26 kNABδ2 m4 m

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2 m from A. Find the deflection δB at the free end B.

P = 26 kNa = 2 mL = 6 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

Find δB (mm)

Show hints
  1. The part between A and the load behaves like a cantilever of length a with an end load: δ = Pa³/(3EI), θ = Pa²/(2EI).
  2. Beyond the load there is no moment, so that part stays straight: δB = Pa³/(3EI) + Pa²/(2EI)·(L − a) = Pa²(3L − a)/(6EI).
  3. a = 2000 mm, EI = 1.68 × 10¹³ N·mm².
Show the answer and the step-by-step solution

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 26000 N, a = 2000 mm, L = 6000 mm
EI = 1.68 × 10¹³ N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δP = Pa³/(3EI) = 4.127 mm
θP = Pa²/(2EI) = 0.003095 rad
δB = δP + θP·(L − a) = Pa²(3L − a)/(6EI) = 16.51 mm

Answer

δB = 16.51 mm
Example 3Hard
w = 4 kN/mABδ3 m

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth hreq of the section.

w = 4 kN/mL = 3 mE = 11 GPa (timber)b = 100 mm

Find hreq (mm)

Show hints
  1. Set the midspan deflection 5wL⁴/(384EI) equal to the limit and solve for the required I.
  2. Then use I = bh³/12 for the rectangle and solve for h.
  3. Ireq = 4.602 × 10⁷ mm⁴.
Show the answer and the step-by-step solution

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 4 N/mm, L = 3000 mm, E = 11000 N/mm²

Deflection limit

δallow = L/360 = 3000/360 = 8.333 mm

Required second moment of area

Ireq = 5wL⁴/(384·E·δallow) = 5·4·3000⁴/(384·11000·8.333) = 4.602 × 10⁷ mm⁴

Required depth

bh³/12 = Ireq → h = (12·4.602 × 10⁷/100)1/3 = 176.8 mm

Answer

hreq = 176.8 mm

In practice, choose the next standard depth above this value.

Questions and answers

What is Beam deflection?

Every beam bends a little under load. The distance a point of the beam moves down is its deflection δ, usually in mm.

A beam can be strong enough and still bend too much. Floors then feel bouncy, plaster cracks and doors stick. A common limit for floors is span/360: a 3.6 m floor may deflect at most 10 mm.

How do I solve Beam deflection problems step by step?

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

What are common mistakes in Beam deflection?
  • PL³/(3EI) is the cantilever with an end load. A simply supported beam with a central load deflects PL³/(48EI).
  • The coefficient 5/384 belongs to a uniform load (with wL⁴). For a central point load use 1/48.
  • That is the deflection under the load. Beyond the load the beam is straight but tilted, so add θ·(L − a).
What do the Beam deflection exercises cover?
  • Easy: cantilever, end load
  • Medium: uniform load
  • Hard: superposition and design

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

Shear force & bending momentCentroid & second moment of area

Calculators

Beam deflection calculator

Courses with this topic

Mechanics of MaterialsStructural AnalysisFE Mechanical exam practiceFE Civil exam practice

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Printable worksheet

Beam deflection worksheet with answers
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