Structural Analysis: practice problems and exam questions

Analysis of statically determinate beams and trusses, shear force and bending moment diagrams, deflections and the stability of columns.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Beam support reactions1 point load · 2 point loads · point load + UDLcalculations
  2. Shear force & bending momentpoint load · full-span UDL · point load + UDLcalculations
  3. Truss: method of jointssymmetric · asymmetric · vertical + horizontal loadcalculations
  4. Beam deflectioncantilever, end load · uniform load · superposition and designcalculations
  5. Column bucklingpinned ends · other end conditions · design and critical stresscalculations

Exam tips for Structural Analysis

  • Sketch the shear force and bending moment diagrams: the largest moment is where the shear force crosses zero.
  • Use the method of sections when you need the force in only one or two truss members.
  • Combine standard deflection cases by superposition instead of integrating from scratch.

Worked examples

Beam support reactionsEasy
P = 55 kNABRA​RB​ = ?2 m2 m

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.

L = 4 mP = 55 kN at 2 m

Find RB (kN)

Show hints
  1. Use static equilibrium: ΣFy = 0 and ΣM = 0. Two unknowns, two equations.
  2. Take moments about A. RA passes through A, so it drops out and you get RB directly.
  3. Total downward load is 55 kN, so RA + RB = 55 kN.
Show the answer and the step-by-step solution

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.

P = 55 kNABRA​RB​2 m2 m

Free-body diagram

Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·4 = 55·2
RB = 110/4 = 27.5 kN
ΣFy = 0: RA = 55 − 27.5 = 27.5 kN

Answer

RB = 27.5 kN
Shear force & bending momentMedium
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

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • 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.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Beam reaction calculatorBeam deflection calculatorShear force and bending moment diagram calculatorEuler buckling calculatorBending stress calculator (σ = My/I)Truss calculator (method of joints)

Related courses

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