Machine Design: practice problems and exam questions

Designing machine parts that transmit power safely: gear trains, shafts in torsion, combined stresses with Mohr’s circle, and the strength and stability of members.

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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. Gears & gear trainsspeed ratio · torque and power · compound gear traincalculations
  2. Torsion of shaftsshear stress · angle of twist · power transmissioncalculations
  3. Mohr’s circle & stress transformationprincipal stresses · maximum shear and principal direction · stress on an inclined planecalculations
  4. Axial stress & strainnormal stress · strain and elongation · stepped barcalculations
  5. Column bucklingpinned ends · other end conditions · design and critical stresscalculations
  6. Mechanical vibrationsnatural frequency · damping ratio · springs in series and parallelcalculations

Exam tips for Machine Design

  • Power = torque × angular speed: gearing down raises torque by the same ratio.
  • Find the principal stresses before you compare with a material strength.
  • Convert rpm to rad/s with ω = 2π·n/60.

Worked examples

Gears & gear trainsEasy
d1​ = 160 mmd2​ = 400 mm1500 rpmn2​ = ?

A motor turning at 1500 rpm drives a 160 mm pulley. A belt connects it to a 400 mm pulley. Assuming no slip, find the speed n2 of the driven pulley.

n1 = 1500 rpmd1 = 160 mmd2 = 400 mm

Find n2 (rpm)

Show hints
  1. The belt moves at one speed, so both rims have the same speed.
  2. π·d₁·n₁ = π·d₂·n₂.
  3. n₂ = 1500·160/400.
Show the answer and the step-by-step solution

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Equal rim speeds

n₁·d₁ = n₂·d₂

Driven speed

n2 = 1500·160/400 = 600 rpm

Answer

n2 = 600 rpm
Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 1.5 mφ = ?

A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 750 N·md = 40 mmL = 1.5 mG = 39 GPa

Find φ (°)

Show hints
  1. Use φ = TL/(GJ), which gives radians.
  2. Use N and mm throughout: T in N·mm, L in mm, G in MPa (GPa × 1000), J in mm⁴.
  3. J = 251327 mm⁴.
Show the answer and the step-by-step solution

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar second moment of area

J = π·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1500/(39000·251327) = 0.1148 rad
φ = 0.1148·180/π = 6.576°

Answer

φ = 6.576°

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

Shaft torsion calculatorAxial stress and strain calculatorEuler buckling calculatorMohr’s circle calculatorGear ratio calculatorNatural frequency and damping calculatorThermal expansion calculatorBelt length calculator (open and crossed belts)

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