Exercise set

Beam Deflection Exercises

Structural analysis exercises for beam deflection covering simply supported beams, cantilevers, superposition, slope, serviceability limits, and stiffness interpretation.

Branch: Civil and Construction Engineering
Content: Exercise set
Updated: Jun 20, 2026
Revision: v1.1.0 · reviewed

These exercises practise beam deflection calculations and the engineering interpretation of stiffness. The purpose is not only to substitute values into formulas. A beam deflection result should be checked against load path, support assumptions, material stiffness, section properties, span, serviceability limits, and construction reality.

Assume small deflection, linear elastic behaviour, prismatic members, and constant $EI$ unless stated otherwise. In real structural design, these assumptions must be checked. Cracking, composite action, connection flexibility, shear deformation, creep, construction sequence, and support settlement can all change deflection.

How to use these exercises

For each problem, do four checks:

Draw the beam, supports, and load.
Identify whether the formula matches the support condition.
Check units before calculating.
Interpret whether the result is plausible for serviceability.

The most common errors are using the wrong span, confusing point load and distributed load formulas, mixing millimetres and metres, and treating deflection as a strength check. Deflection is usually a serviceability check. A beam can be strong enough and still too flexible.

For each result, state whether the calculation supports member selection, serviceability acceptance, retrofit need, connection review, or a request for a more detailed structural model. A deflection value is meaningful only when the support condition, load case, stiffness assumption, and acceptance limit are explicit.

Exercise 1: Simply supported beam with central point load

A simply supported steel beam has span $L = 6.0\ \text{m}$ and carries a central point load $P = 18\ \text{kN}$ . The beam has:

E = 200\ \text{GPa}

I = 85 \times 10^6\ \text{mm}^4

Find the maximum deflection.

Solution

For a simply supported beam with a central point load:

\displaystyle \delta_{max} = \frac{PL^3}{48EI}

Use consistent units:

P = 18,000\ \text{N}

L = 6,000\ \text{mm}

E = 200,000\ \text{N/mm}^2

Substitute:

\displaystyle \delta_{max} = \frac{18,000(6,000)^3}{48(200,000)(85 \times 10^6)}

\delta_{max} = 4.76\ \text{mm}

The deflection ratio is:

\displaystyle \frac{L}{\delta} = \frac{6,000}{4.76} = 1260

That is a relatively stiff response for many floor-beam serviceability checks, but the acceptance limit depends on the project and code basis.

Exercise 2: Simply supported beam with uniform load

A timber joist spans $L = 4.2\ \text{m}$ and carries a uniformly distributed service load of $w = 3.2\ \text{kN/m}$ . The joist has:

E = 11,000\ \text{N/mm}^2

I = 45 \times 10^6\ \text{mm}^4

Find the maximum midspan deflection.

Solution

For a simply supported beam with uniform load:

\displaystyle \delta_{max} = \frac{5wL^4}{384EI}

Convert:

w = 3.2\ \text{kN/m} = 3.2\ \text{N/mm}

L = 4,200\ \text{mm}

Substitute:

\displaystyle \delta_{max} = \frac{5(3.2)(4,200)^4}{384(11,000)(45 \times 10^6)}

\delta_{max} = 26.1\ \text{mm}

If the serviceability limit were $L/360$ :

\displaystyle \delta_{allow} = \frac{4,200}{360} = 11.7\ \text{mm}

The joist would not satisfy that limit. The engineering response might be a deeper section, reduced span, additional support, stronger material, composite action, or revised load assumption.

Exercise 3: Cantilever with end load

A cantilever bracket projects $L = 900\ \text{mm}$ from a wall and carries a service end load $P = 1.4\ \text{kN}$ . The member has:

E = 200,000\ \text{N/mm}^2

I = 6.5 \times 10^6\ \text{mm}^4

Find the tip deflection and free-end rotation.

Solution

For a cantilever with end load:

\displaystyle \delta_{tip} = \frac{PL^3}{3EI}

\displaystyle \theta_{end} = \frac{PL^2}{2EI}

Convert $P = 1,400\ \text{N}$ .

Tip deflection:

\displaystyle \delta_{tip} = \frac{1,400(900)^3}{3(200,000)(6.5 \times 10^6)}

\delta_{tip} = 0.262\ \text{mm}

Slope:

\displaystyle \theta_{end} = \frac{1,400(900)^2}{2(200,000)(6.5 \times 10^6)}

\theta_{end} = 0.000436\ \text{rad}

The deflection is small, but the connection must still be checked. In cantilevers, connection rotation can dominate member bending deflection if the support is not truly fixed.

Exercise 4: Superposition of two loads

A simply supported beam has span $L = 5.0\ \text{m}$ , stiffness $EI = 1.8 \times 10^{13}\ \text{N mm}^2$ , a central point load $P = 12\ \text{kN}$ , and a uniform load $w = 1.5\ \text{kN/m}$ . Find the total midspan deflection.

Solution

Use superposition because the model is linear elastic.

Point load deflection:

\displaystyle \delta_P = \frac{PL^3}{48EI}

Uniform load deflection:

\displaystyle \delta_w = \frac{5wL^4}{384EI}

Convert:

P = 12,000\ \text{N},\quad L = 5,000\ \text{mm},\quad w = 1.5\ \text{N/mm}

Point load:

\displaystyle \delta_P = \frac{12,000(5,000)^3}{48(1.8 \times 10^{13})} = 1.74\ \text{mm}

Uniform load:

\displaystyle \delta_w = \frac{5(1.5)(5,000)^4}{384(1.8 \times 10^{13})} = 0.68\ \text{mm}

Total:

\delta_{total} = 1.74 + 0.68 = 2.42\ \text{mm}

Superposition is valid here because stiffness is constant, deflections are small, and material response is linear.

Exercise 5: Required moment of inertia

A simply supported beam spans $7.5\ \text{m}$ and carries a uniform service load $w = 4.0\ \text{kN/m}$ . Use $E = 200,000\ \text{N/mm}^2$ . Find the minimum moment of inertia required to satisfy $L/360$ deflection.

Solution

The allowable deflection is:

\displaystyle \delta_{allow} = \frac{7,500}{360} = 20.8\ \text{mm}

Rearrange:

\displaystyle \delta_{max} = \frac{5wL^4}{384EI}

\displaystyle I = \frac{5wL^4}{384E\delta_{allow}}

Substitute:

\displaystyle I = \frac{5(4.0)(7,500)^4}{384(200,000)(20.8)}

I = 39.6 \times 10^6\ \text{mm}^4

This is a serviceability stiffness requirement. The selected beam must also satisfy bending stress, shear, lateral-torsional buckling, bearing, connections, vibration, fire, corrosion, and constructability requirements.

Engineering interpretation

Deflection is a stiffness result, not only a calculation output. It depends strongly on span because many formulas include $L^3$ or $L^4$ . A modest increase in span can create a large increase in deflection. Increasing depth is often efficient because moment of inertia grows rapidly with section depth.

When reviewing beam deflection, ask:

Does the model match the real support condition?
Is the load service load or factored load?
Are units consistent?
Does the member remain uncracked or linear?
Are long-term effects relevant?
Is connection or support flexibility important?
Is vibration more critical than static deflection?
Is the stated acceptance limit taken from the governing code, client criterion, facade tolerance, or equipment requirement?
Are construction-stage deflections, camber, creep, shrinkage, settlement, and composite action relevant to the final service condition?
Can the calculation be traced to the selected section properties, material grade, span definition, load combination, and connection assumption?

These checks turn beam formulas into engineering judgement.

REF

Disciplines

Beam Deflection Exercises

How to use these exercises

Exercise 1: Simply supported beam with central point load

Solution

Exercise 2: Simply supported beam with uniform load

Solution

Exercise 3: Cantilever with end load

Solution

Exercise 4: Superposition of two loads

Solution

Exercise 5: Required moment of inertia

Solution

Engineering interpretation

See also