Skip to main content
ExamExplained

HSC Engineering Studies practice questions for 2026 (all four modules)

HSCEngineering StudiesStudy guide15 min read

A bank of HSC Engineering Studies practice questions across the four Year 12 modules: Civil Structures, Lifting Devices, Personal and Public Transport, and Aeronautical Engineering. Worked calculations and full solutions modelled on NESA exam patterns.

Jump to a section
  1. How to use this question bank
  2. Module 1: Civil Structures (1-5)
  3. Module 2: Lifting Devices (6-10)
  4. Module 3: Personal and Public Transport (11-13)
  5. Module 4: Aeronautical Engineering (14-15)

How to use this question bank

HSC Engineering Studies is a single three-hour paper across four Year 12 modules. These practice questions are grouped by module and modelled on the patterns NESA repeats: calculate a reaction or member force, find a stress or extension, size a gear train or pulley system, and reason about materials selection.

Three rules:

  1. Show your working. Markers award method marks for a correct setup, so a careful solution that lands on a slightly wrong number scores better than a bare correct answer.
  2. Carry units and use sensible significant figures. A correct number in the wrong units loses marks. Convert mm2^2 to m2^2, GPa to Pa and rpm to rad/s before substituting, or work consistently in N, mm and MPa (recall 1 MPa=1 N/mm21 \text{ MPa} = 1 \text{ N/mm}^2).
  3. Draw the free-body diagram. For every statics question, a labelled free-body diagram earns a mark in its own right and prevents sign errors.

Module 1: Civil Structures (1-5)

  1. A simply supported beam spans 6.0 m. It carries a point load of 18 kN located 2.0 m from the left support, plus a uniformly distributed load of 5.0 kN/m over the full span. Calculate the left and right support reactions. (4 marks)

  2. A steel tie rod of cross-sectional area 300 mm2300 \text{ mm}^2 and original length 2.5 m carries a tensile load of 45 kN. Take E=200E = 200 GPa. Calculate the stress, the strain and the extension. (5 marks)

  3. A circular aluminium bar of diameter 25 mm carries an axial tensile load of 40 kN. Take E=70E = 70 GPa. Calculate (a) the cross-sectional area, (b) the stress in MPa, and (c) the strain. (5 marks)

  4. A symmetric triangular truss has a horizontal bottom chord ACAC of length 8.0 m between a pin support at AA and a roller support at CC. The apex BB is 3.0 m above the midpoint of ACAC. A vertical downward load of 24 kN acts at BB. Calculate the force in member ABAB and state whether it is in tension or compression. (5 marks)

  5. Distinguish between a ductile and a brittle material in terms of their stress-strain curves, and give one structural example of each. (3 marks)

Module 2: Lifting Devices (6-10)

  1. A block-and-tackle system has five rope segments supporting the load block and lifts a 3000 N load at 75 percent efficiency. Calculate (a) the velocity ratio, (b) the actual mechanical advantage, and (c) the effort force. (4 marks)

  2. A two-stage gear reduction has stage ratios of 5:1 and 4:1. The motor delivers 30 N m of torque at 1440 rpm at an overall efficiency of 88 percent. Calculate (a) the overall gear ratio, (b) the output torque, and (c) the output speed in rpm. (5 marks)

  3. A hydraulic jack has an input piston of diameter 15 mm and an output piston of diameter 90 mm. An input force of 250 N is applied. Calculate (a) the output force, and (b) the distance the load rises if the input piston moves 72 mm. Assume the jack is ideal. (5 marks)

  4. A crane hoist rope has a minimum breaking load of 240 kN. AS1418 requires a factor of safety of 5. Calculate (a) the safe working load of a single rope, and (b) the maximum hook load using a 4-fall reeving. (4 marks)

  5. Explain the trade-off between force and distance in a pulley system, referring to the conservation of energy. (3 marks)

Module 3: Personal and Public Transport (11-13)

  1. A car of mass 1200 kg accelerates from rest to 27 m/s in 9.0 s on a level road. Calculate (a) the acceleration, and (b) the net driving force required, using Newton's second law. (4 marks)

  2. A vehicle transmission has a gearbox ratio of 3.5:1 in first gear and a final-drive (differential) ratio of 4.1:1. Calculate the overall gear ratio between the engine and the driven wheels. (3 marks)

  3. Explain why composite materials such as carbon-fibre-reinforced polymer are used in high-performance vehicle bodies, referring to one advantage and one disadvantage compared with steel. (4 marks)

Module 4: Aeronautical Engineering (14-15)

  1. State the four forces acting on an aircraft in steady level flight and the equilibrium relationship between them. (3 marks)

  2. Explain, using Bernoulli's principle, how an aerofoil generates lift, and identify one reason why aluminium alloys are widely used in aircraft airframes. (4 marks)

Solutions
Q1
UDL resultant =wL=5.0×6.0=30= w L = 5.0 \times 6.0 = 30 kN at midspan (3.0 m). Moments about the left support (clockwise positive): ∑ML=(18)(2.0)+(30)(3.0)−RR(6.0)=0\sum M_L = (18)(2.0) + (30)(3.0) - R_R(6.0) = 0, so RR=(36+90)/6.0=126/6.0=21R_R = (36 + 90)/6.0 = 126/6.0 = 21 kN. Vertical equilibrium: RL+RR=18+30=48R_L + R_R = 18 + 30 = 48 kN, so RL=48−21=27R_L = 48 - 21 = 27 kN. Left 27 kN, right 21 kN.
Q2
A=300×10−6 m2A = 300 \times 10^{-6} \text{ m}^2. Stress σ=F/A=45×103/(300×10−6)=1.5×108 Pa=150\sigma = F/A = 45 \times 10^3 / (300 \times 10^{-6}) = 1.5 \times 10^8 \text{ Pa} = 150 MPa. Strain ε=σ/E=1.5×108/200×109=7.5×10−4\varepsilon = \sigma/E = 1.5 \times 10^8 / 200 \times 10^9 = 7.5 \times 10^{-4}. Extension ΔL=εL=7.5×10−4×2.5=1.875×10−3 m=1.9\Delta L = \varepsilon L = 7.5 \times 10^{-4} \times 2.5 = 1.875 \times 10^{-3} \text{ m} = 1.9 mm.
Q3
(a) A=πd2/4=π(0.025)2/4=π(6.25×10−4)/4=4.91×10−4 m2A = \pi d^2 / 4 = \pi (0.025)^2 / 4 = \pi (6.25 \times 10^{-4})/4 = 4.91 \times 10^{-4} \text{ m}^2 (491 mm2^2). (b) σ=F/A=40×103/(4.91×10−4)=8.15×107 Pa=81.5\sigma = F/A = 40 \times 10^3 / (4.91 \times 10^{-4}) = 8.15 \times 10^7 \text{ Pa} = 81.5 MPa. (c) ε=σ/E=8.15×107/70×109=1.16×10−3\varepsilon = \sigma/E = 8.15 \times 10^7 / 70 \times 10^9 = 1.16 \times 10^{-3}.
Q4
Reactions: by symmetry RA=RC=24/2=12R_A = R_C = 24/2 = 12 kN upward. Member ABAB runs from AA (0, 0) to BB (4.0, 3.0), length =4.02+3.02=5.0= \sqrt{4.0^2 + 3.0^2} = 5.0 m, with sin⁡θ=3.0/5.0=0.6\sin\theta = 3.0/5.0 = 0.6. At joint AA, assume ABAB in tension; vertical equilibrium ∑Fy=RA+FABsin⁡θ=12+FAB(0.6)=0\sum F_y = R_A + F_{AB}\sin\theta = 12 + F_{AB}(0.6) = 0, so FAB=−12/0.6=−20F_{AB} = -12/0.6 = -20 kN. Negative means ABAB is in compression, carrying 20 kN.
Q5
A ductile material (such as mild structural steel) shows a long linear elastic region, then a yield point and a substantial plastic region with strain hardening before fracture, so it deforms visibly and absorbs much energy before failing. A brittle material (such as cast iron or concrete in tension) shows an almost straight line with little or no plastic region, then sudden fracture at low strain. Ductile example: grade 250 structural steel. Brittle example: cast iron (or concrete in tension).
Q6
(a) VR=n=5VR = n = 5. (b) AMA=η×VR=0.75×5=3.75AMA = \eta \times VR = 0.75 \times 5 = 3.75. (c) Feffort=Fload/AMA=3000/3.75=800F_{\text{effort}} = F_{\text{load}}/AMA = 3000/3.75 = 800 N.
Q7
(a) GR=5×4=20GR = 5 \times 4 = 20. (b) Tout=Tin×GR×η=30×20×0.88=528T_{\text{out}} = T_{\text{in}} \times GR \times \eta = 30 \times 20 \times 0.88 = 528 N m. (c) ωout=ωin/GR=1440/20=72\omega_{\text{out}} = \omega_{\text{in}}/GR = 1440/20 = 72 rpm.
Q8
Area ratio =(d2/d1)2=(90/15)2=62=36= (d_2/d_1)^2 = (90/15)^2 = 6^2 = 36. (a) Fout=Fin×36=250×36=9000F_{\text{out}} = F_{\text{in}} \times 36 = 250 \times 36 = 9000 N. (b) Volume conserved, so d2=d1×A1/A2=72×(1/36)=2.0d_2 = d_1 \times A_1/A_2 = 72 \times (1/36) = 2.0 mm.
Q9
(a) SWL=MBL/FoS=240/5=48SWL = MBL/FoS = 240/5 = 48 kN per rope. (b) With 4-fall reeving the four segments share the hook load, so the maximum hook load =4×48=192= 4 \times 48 = 192 kN (about 19.6 tonnes), provided the hook block and fittings are rated to at least 192 kN.
Q10
In a pulley system the effort force is reduced by the mechanical advantage, but the effort must move a correspondingly greater distance, equal to the load distance times the velocity ratio. Energy is conserved: ignoring losses, the input work (effort times input distance) equals the output work (load times load distance). Friction makes the input work larger than the output work, which is why a real system needs a higher effort than the ideal calculation predicts. You can reduce the force needed, but never the total work required.
Q11
(a) a=Δv/Δt=27/9.0=3.0 m/s2a = \Delta v / \Delta t = 27 / 9.0 = 3.0 \text{ m/s}^2. (b) Net force F=ma=1200×3.0=3600F = ma = 1200 \times 3.0 = 3600 N (this is the net driving force; the actual engine force at the wheels must also overcome rolling resistance and drag).
Q12
Overall gear ratio is the product of the gearbox ratio and the final-drive ratio: GR=3.5×4.1=14.35GR = 3.5 \times 4.1 = 14.35, that is about 14.4:1 between the engine and the driven wheels.
Q13
Carbon-fibre-reinforced polymer has a very high strength-to-weight ratio, so a body panel of the same strength is much lighter than steel, improving acceleration, fuel economy and handling. The main disadvantages are high material and manufacturing cost, and brittle failure behaviour (it can shatter rather than deform plastically, and damage can be hard to detect), unlike steel which deforms progressively and is cheap to repair.
Q14
The four forces are lift (upward), weight (downward), thrust (forward) and drag (rearward). In steady level flight the aircraft is in equilibrium: lift equals weight, and thrust equals drag, so there is no net force and the aircraft neither accelerates nor changes altitude.
Q15
An aerofoil is shaped so that air travels faster over the curved upper surface than along the lower surface. By Bernoulli's principle, the faster-moving air over the top is at lower pressure than the slower air beneath, and this pressure difference acting over the wing area produces a net upward force, lift. Aluminium alloys are widely used in airframes because they offer a high strength-to-weight ratio combined with good corrosion resistance and ease of forming and joining, keeping the aircraft light while strong.

Sources & how we know this

Based on the official NESA syllabus. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

ExamExplained