📚 Year 13 CIE Engineering: Unit Test Mock Paper Walkthrough | Year 13 CIE 工程：单元测试模拟卷解析
Mock unit tests are a powerful way to consolidate Year 13 CIE Engineering concepts, exposing the subtle interplay between mechanics, materials, thermodynamics and electrical principles. This walkthrough dissects a carefully designed paper, showing you how to structure answers, apply key equations and avoid classic pitfalls under timed conditions.
单元模拟测试是巩固 Year 13 CIE 工程知识的有效方式，能揭示力学、材料学、热力学与电学原理之间的微妙联系。本文将逐一拆解一份精心设计的模拟卷，展示如何在限时条件下组织答案、运用关键方程并避开常见陷阱。
1. Static Equilibrium: Simply Supported Beam Reactions | 静力平衡：简支梁支座反力
A beam of length 5.0 m is simply supported at A (left) and B (right). A concentrated vertical load of 10 kN acts downwards at a point 2.0 m from A. Determine the reaction forces at the supports, RA and RB.
一根长 5.0 m 的梁在左端 A 和右端 B 简支。在距 A 点 2.0 m 处作用一个向下的 10 kN 集中力。试求支座反力 RA 和 RB。
Take moments about A to eliminate RA. Clockwise moment from the load is 10 kN × 2.0 m = 20 kN·m. For equilibrium, the sum of moments about any point must be zero, so RB × 5.0 m = 20 kN·m, giving RB = 4.0 kN upwards.
对 A 点取矩可消去 RA。荷载产生的顺时针力矩为 10 kN × 2.0 m = 20 kN·m。平衡要求对任一点的合力矩为零，因此 RB × 5.0 m = 20 kN·m，解得 RB = 4.0 kN 向上。
Resolve vertically: RA + RB = 10 kN, thus RA = 6.0 kN upwards. Always check with moments about B: RA × 5.0 – 10 kN × 3.0 m = 30 – 30 = 0, confirming the result.
竖向力平衡：RA + RB = 10 kN，所以 RA = 6.0 kN 向上。通过对 B 点取矩验证：RA × 5.0 – 10 kN × 3.0 m = 30 – 30 = 0，结果正确。
2. Stress, Strain and Young’s Modulus in a Tension Rod | 拉伸杆中的应力、应变与杨氏模量
A solid circular steel rod of diameter 12 mm and original length 1.8 m carries an axial tensile load of 25 kN. Given Young’s modulus E = 210 GPa, calculate the tensile stress, strain and the total elongation. Assume the material remains within the elastic limit.
一根直径 12 mm、原始长度 1.8 m 的实心圆钢杆承受 25 kN 的轴向拉伸荷载。杨氏模量 E = 210 GPa，试计算拉伸应力、应变及总伸长量。假设材料保持在弹性范围内。
Cross-sectional area A = πd²/4 = π × (12 × 10⁻³ m)² / 4 = 1.131 × 10⁻⁴ m². Direct stress σ = F / A = 25 000 N / 1.131 × 10⁻⁴ m² ≈ 221 MPa (or 2.21 × 10⁸ Pa). The answer is typically expressed in N/mm²: 221 N/mm².
横截面积 A = πd²/4 = π × (12 × 10⁻³ m)² / 4 = 1.131 × 10⁻⁴ m²。正应力 σ = F / A = 25 000 N / 1.131 × 10⁻⁴ m² ≈ 221 MPa（即 2.21 × 10⁸ Pa），常用单位 N/mm² 表示为 221 N/mm²。
Using Hooke’s law, σ = Eε, strain ε = σ / E = 221 × 10⁶ Pa / 210 × 10⁹ Pa = 1.052 × 10⁻³ (dimensionless). Extension ΔL = ε × original length = 1.052 × 10⁻³ × 1.8 m = 1.894 × 10⁻³ m, or about 1.9 mm.
由胡克定律 σ = Eε，应变 ε = σ / E = 221 × 10⁶ Pa / 210 × 10⁹ Pa = 1.052 × 10⁻³（无因次）。伸长量 ΔL = ε × 原长 = 1.052 × 10⁻³ × 1.8 m = 1.894 × 10⁻³ m，约 1.9 mm。
3. Truss Analysis by Method of Joints | 节点法分析桁架
Consider a simple triangular truss with joints labelled P, Q and R. P is a pin support, Q a roller support, and the truss forms a right-angled triangle with PQ horizontal (3 m), PR vertical (4 m) and QR as the hypotenuse. A vertical load of 15 kN acts downwards at joint R. Determine the forces in members PR and QR, stating whether they are in tension or compression.
考虑一个简单三角形桁架，节点标记为 P、Q 和 R。P 为固定铰支座，Q 为滚动支座，桁架呈直角三角形：PQ 水平（3 m），PR 竖直（4 m），QR 为斜边。在节点 R 处作用一向下的 15 kN 竖向荷载。求杆件 PR 和 QR 的内力，并说明是受拉还是受压。
First, find support reactions. By symmetry of geometry and vertical load, taking moments about P gives Q_vertical × 3 = 15 × 0 → actually R is directly above P? Wait – if PR is vertical, R is above P. Q is horizontally 3 m from P. The load is at R. Moment about P: 15 kN × 0 horizontal distance? The line of action is through P, so no moment from load about P. Therefore all vertical reaction must be at P, so P_vertical = 15 kN up, Q_vertical = 0. But Q must provide horizontal restraint? Since no horizontal load, Q_horizontal = 0. We need forces in members.
首先求支座反力。几何对称且荷载位于 P 正上方 R 点，对 P 取矩时荷载力臂为零，因此荷载对 P 的力矩为零。所以 P 处承担全部竖向反力 15 kN 向上，Q 的竖向反力为零；无水平荷载时 Q 的水平反力也为零。
Analyse joint R: three forces meet – load 15 kN down, force in PR (assume tension, pulling away from R) acting downwards? Actually member PR connects P (bottom) to R (top). At joint R, force in PR, if tension, would pull R downwards. Force in QR acts along hypotenuse. Resolve vertically: let F_PR be force in PR (positive if tension). Downward component of any force in QR: if θ is angle at Q, sin θ = 4/5, cos θ = 3/5. For equilibrium at R, vertical: –15 – F_PR + F_QR × (4/5) = 0 (taking up as positive, but PR tension pulls down). Better: sum of vertical forces = 0: –15 kN + F_QR sin θ + component of PR. However PR is vertical, so if tension it pulls R down, i.e. force on R is downwards. So equation: –15 – F_PR + F_QR × (4/5) = 0. Horizontal: –F_QR × (3/5) = 0 → F_QR = 0. Then –15 – F_PR = 0 → F_PR = –15 kN. Negative means opposite to assumed tension, so PR is in compression with 15 kN. QR is zero force member? That seems odd. Let’s reconsider geometry: If P is at origin, R is at (0,4). Q is at (3,0). Load at R is vertical. Joint R: members PR (vertical) and QR (diagonal). No other force. For equilibrium, horizontal component must be zero, so QR’s horizontal component must be zero → only possible if force in QR is zero. Then PR must carry full 15 kN, but PR vertical, so compression 15 kN. This makes sense as a simple cantilever truss? Actually with Q as a roller, Q cannot provide horizontal resistance, but here no horizontal load, so member QR is unstressed. That is a valid answer. So PR: 15 kN compression, QR: 0 kN.
分析节点 R：汇交三力——15 kN 向下，杆 PR 的内力（假设受拉，力由节点向外），斜杆 QR 的内力。设 F_PR 为 PR 内力（受拉为正），向下为正不太方便，改用向上为正：竖向平衡：+ F_QR sin θ + F_PR? PR 是竖直杆，若杆受拉，对节点 R 的作用力向下（杆要拉回节点），所以 F_PR 给节点的力向下；若受压，则向上。设杆内力以受拉为正，则杆对节点的力方向与杆背离节点。PR 连接 P（下）和 R（上），若杆受拉，则对 R 施力向下。竖向平衡： –15 – F_PR + F_QR sin θ = 0。水平方向：杆 QR 对节点 R 的水平力为 –F_QR cos θ（假设杆 QR 受拉，力指向 Q，水平向左），平衡要求 F_QR cos θ = 0，得 F_QR = 0。代入竖向：F_PR = –15 kN，负号表示实际受压，大小为 15 kN。故杆 PR 受 15 kN 压力，杆 QR 为零力杆。
4. Bending Moment and Flexural Stress in a Cantilever | 悬臂梁的弯矩与弯曲应力
A horizontal cantilever beam of length 1.2 m carries a uniformly distributed load (UDL) of 3 kN/m over its entire span. The beam has a rectangular cross-section of width 50 mm and depth 80 mm. Calculate the maximum bending moment and the maximum bending stress.
一根长 1.2 m 的水平悬臂梁全长承受 3 kN/m 的均布荷载。截面为矩形，宽 50 mm、高 80 mm。试求最大弯矩与最大弯曲应力。
For a cantilever with UDL, the maximum bending moment occurs at the fixed support: M_max = wL² / 2 = (3 kN/m) × (1.2 m)² / 2 = 3 × 1.44 / 2 = 2.16 kN·m.
悬臂梁受均布荷载时，最大弯矩发生在固定端：M_max = wL² / 2 = (3 kN/m) × (1.2 m)² / 2 = 3 × 1.44 / 2 = 2.16 kN·m。
Section modulus for a rectangle Z = b d² / 6, where b = 50 mm, d = 80 mm. So Z = 50 × 80² / 6 = 50 × 6400 / 6 = 53 333 mm³ = 53.333 × 10⁻⁶ m³. Maximum bending stress σ_max = M / Z. Use consistent units: M = 2.16 × 10³ N·m = 2160 N·m. Z in m³: 53.333 × 10⁻⁶ m³. σ_max = 2160 / 53.333 × 10⁻⁶ = 40.5 × 10⁶ Pa = 40.5 MPa.
矩形截面系数 Z = b d² / 6，b = 50 mm，d = 80 mm，Z = 50 × 80² / 6 = 50 × 6400 / 6 = 53 333 mm³ = 53.333 × 10⁻⁶ m³。最大弯曲应力 σ_max = M / Z，代入统一单位：M = 2.16 × 10³ N·m = 2160 N·m，得 σ_max = 2160 / (53.333 × 10⁻⁶) = 40.5 × 10⁶ Pa = 40.5 MPa。
Always check that the stress does not exceed the material’s yield strength. This is a typical design check in CIE Engineering unit tests.
始终要验证该应力不超过材料的屈服强度，这是 CIE 工程单元测试中典型的设计校核环节。
5. Kinetic Energy and Thermal Energy Conversion | 动能与热能转换
A block of mass 8 kg slides from rest down a frictionless incline of height 5 m. At the bottom, it strikes a stationary steel container holding 2 kg of water. All the kinetic energy of the block is converted to heat, which is entirely absorbed by the water. Calculate the temperature rise of the water. Specific heat capacity of water = 4200 J/(kg·K).
一个 8 kg 的物块从静止沿光滑斜面下滑，竖直高度差 5 m。在斜面底部，它撞击一个装有 2 kg 水的静止钢容器。物块的全部动能转化为热量，且完全被水吸收。试计算水的温升。水的比热容为 4200 J/(kg·K)。
Gravitational potential energy lost = mgh = 8 kg × 9.81 m/s² × 5 m = 392.4 J. This equals kinetic energy at the bottom, which becomes thermal energy Q = 392.4 J.
物块损失的重力势能 = mgh = 8 kg × 9.81 m/s² × 5 m = 392.4 J。该能量等于斜面底部的动能，并完全转化为热能 Q = 392.4 J。
Q = m_water × c × ΔT. Rearranging: ΔT = Q / (m × c) = 392.4 J / (2 kg × 4200 J/(kg·K)) = 392.4 / 8400 ≈ 0.0467 K. The temperature rise is approximately 0.047 °C.
Q = m_水 × c × ΔT，整理得 ΔT = Q / (m × c) = 392.4 J / (2 kg × 4200 J/(kg·K)) = 392.4 / 8400 ≈ 0.0467 K，温升约为 0.047 °C。
In exam conditions, always state assumptions: no heat loss to surroundings, container’s heat capacity negligible, and water does not evaporate. This question tests energy conservation and real-world limitations.
考试中务必写明假设：无热损至环境、容器热容可忽略、水不蒸发。该题考查能量守恒与现实限制。
6. Kirchhoff’s Laws in a Two-Loop Circuit | 基尔霍夫定律在两回路电路中的应用
A dc circuit contains two loops. The left loop has a 12 V battery with internal resistance 0.5 Ω and a resistor R₁ = 4 Ω. The right loop shares a common resistor R₂ = 3 Ω with the left loop and also contains a 6 V battery (polarity opposing the 12 V) and R₃ = 2 Ω. Determine all branch currents. Use Kirchhoff’s voltage and current laws.
某直流电路包含两个回路。左回路有 12 V 电池（内阻 0.5 Ω）和电阻 R₁ = 4 Ω。右回路与左回路共用电阻 R₂ = 3 Ω，并包含一个与 12 V 极性相反的 6 V 电池和 R₃ = 2 Ω。试用基尔霍夫电压和电流定律求各支路电流。
Label mesh currents: I₁ clockwise in left loop, I₂ clockwise in right loop. Through shared R₂, net current downwards is (I₁ – I₂). KVL left: –12 + 0.5 I₁ + 4 I₁ + 3(I₁ – I₂) = 0 → simplify: 7.5 I₁ – 3 I₂ = 12. KVL right: 3(I₂ – I₁) + 2 I₂ + 6 = 0 → –3 I₁ + 5 I₂ = –6.
设定网孔电流：左回路顺时针 I₁，右回路顺时针 I₂。流经共用电阻 R₂ 的净电流向下为 (I₁ – I₂)。左回路 KVL：–12 + 0.5 I₁ + 4 I₁ + 3(I₁ – I₂) = 0，化简得 7.5 I₁ – 3 I₂ = 12。右回路 KVL：3(I₂ – I₁) + 2 I₂ + 6 = 0，得 –3 I₁ + 5 I₂ = –6。
Solve simultaneously. Multiply first equation by 5: 37.5 I₁ – 15 I₂ = 60. Multiply second by 3: –9 I₁ + 15 I₂ = –18. Add: 28.5 I₁ = 42 → I₁ ≈ 1.474 A. Substitute back: 7.5×1.474 – 3 I₂ = 12 → 11.055 – 3 I₂ = 12 → –3 I₂ = 0.945 → I₂ ≈ –0.315 A. The negative sign for I₂ means actual current flows anticlockwise in the right mesh.
联立求解。第一式乘 5：37.5 I₁ – 15 I₂ = 60。第二式乘 3：–9 I₁ + 15 I₂ = –18。相加得 28.5 I₁ = 42，I₁ ≈ 1.474 A。代回：7.5×1.474 – 3 I₂ = 12，得 11.055 – 3 I₂ = 12，–3 I₂ = 0.945，I₂ ≈ –0.315 A。负号表明 I₂ 实际方向为逆时针。
Branch currents: left battery supplies I₁ = 1.474 A; through R₂ the current is I₁ – I₂ = 1.474 – (–0.315) = 1.789 A downwards; right mesh current magnitude in R₃ is 0.315 A opposite to clockwise, so electrons flow counterclockwise. Always verify power balance as a cross-check.
支路电流：左电池输出 1.474 A；流经 R₂ 的电流为 I₁ – I₂ = 1.474 – (–0.315) = 1.789 A 向下；右回路中 R₃ 的电流大小为 0.315 A，方向与顺时针相反。可用功率平衡进行交叉验证。
7. Digital Logic Gate Simplification | 数字逻辑门化简
Simplify the Boolean expression X = A·B + A·B’ + A’·B using a Karnaugh map or algebraic manipulation, and draw the minimal gate circuit using only two basic gates.
试用卡诺图或代数法化简布尔表达式 X = A·B + A·B’ + A’·B，并画出仅用两个基本门的最简门电路。
Algebraic simplification: X = A·B + A·B’ + A’·B = A·(B + B’) + A’·B = A·1 + A’·B = A + A’·B. Apply the redundancy rule: X = A + B (since A + A’·B = A + B). This is the OR function. Verification with truth table confirms X is true when either A or B is true.
代数化简：X = A·B + A·B’ + A’·B = A·(B + B’) + A’·B = A·1 + A’·B = A + A’·B。利用吸收律 A + A’·B = A + B，得最简形式 X = A + B，即或门功能。真值表验证：当 A 或 B 为真时 X 为真。
The minimal circuit therefore requires a single OR gate. In exam, you may be asked to implement using only NAND or NOR gates; for instance, A OR B can be made from NAND: (A NAND A) NAND (B NAND B), but the question only asked for minimal gate count, so a single OR gate suffices.
故最简电路仅需一个或门。考试中可能要求仅用与非门或或非门实现，例如用与非门实现 A+B 需 ((A NAND A) NAND (B NAND B))，但本题仅要求最少门数，因此单个或门即可。
8. Material Selection Using Performance Indices | 采用性能指标进行材料选择
A design requires a light, stiff cantilever beam of length L, carrying a fixed end load F. The beam must not deflect by more than a specified limit. Derive the material index for minimum mass, given the deflection constraint: δ = FL³ / (3EI) must be ≤ δ_max. The beam has a square cross-section of side b.
某设计需求一根轻质、高刚度的悬臂梁，长度 L，承受固定端荷载 F，挠度不得超过 δ_max。挠度公式为 δ = FL³ / (3EI)。梁截面为正方形，边长为 b。试推导最小质量的材料性能指标。
Mass m = ρ × volume = ρ × b² L. Second moment of area I = b⁴ / 12. Deflection constraint: FL³ / (3E × b⁴/12) ≤ δ_max → 4FL³ / (Eb⁴) ≤ δ_max. Rearranging for free variable b: b⁴ must be at least 4FL³ / (E δ_max). Substitute into mass expression: m = ρ L × b² = ρ L × [4FL³/(E δ_max)]^{1/2}. Group material properties: m ∝ ρ / √E. To minimise mass, maximise the index √E / ρ.
质量 m = ρ × 体积 = ρ × b² L。惯性矩 I = b⁴ / 12。挠度约束：FL³ / (3E × b⁴/12) ≤ δ_max → 4FL³ / (Eb⁴) ≤ δ_max。整理自由变量 b：b⁴ ≥ 4FL³ / (E δ_max)。代入质量表达式：m = ρ L × b² = ρ L × [4FL³/(E δ_max)]^{1/2}。材料参数归并：m ∝ ρ / √E。为最小化质量，应最大化性能指标 √E / ρ。
This index is widely used for stiffness-limited design at minimum weight. Typical candidates: aluminium alloys have low ρ and moderate E; carbon-fibre reinforced polymers perform exceptionally well with high √E/ρ. The process illustrates the Ashby methodology tested in CIE Engineering.
该指标常用于以最小重量为目标的刚度限制设计。典型候选材料：铝合金 ρ 较低、E 适中；碳纤维增强聚合物凭借高 √E/ρ 表现优异。该过程展示了 CIE 工程中考查的 Ashby 材料选择方法。
| Material | Density ρ (kg/m³) | Young’s modulus E (GPa) | Index √E / ρ (m³/kg)^0.5 |
|---|---|---|---|
| Aluminium alloy | 2700 | 70 | √(70×10⁹)/2700 ≈ 3.1 |
| Steel (mild) | 7800 | 210 | √(210×10⁹)/7800 ≈ 1.85 |
| CFRP (unidirectional) | 1600 | 130 | √(130×10⁹)/1600 ≈ 7.1 |
9. Mock Test Reflection and Exam Strategy | 模拟测试反思与备考策略
After working through these questions, students should recognise the common threads: clear free-body diagrams, systematic application of conservation laws, correct unit conversions, and explicit statements of assumptions. Time management is crucial – allocate reading time to identify straightforward numerical problems versus multi-step derivations.
完成以上题目后，学生应梳理出共通脉络：清晰的受力图、守恒定律的系统应用、正确的单位换算，以及明确写出假设。时间管理至关重要——利用阅读时间区分简单计算题与多步推导题。
CIE examiners expect precise units and conversion factors. Always convert mm to m before using in bending stress formulas; check if MPa or GPa is required. Practising under timed conditions helps internalise these steps, turning the unit test into a confidence booster.
CIE 考官期望精确的单位和换算因子。在使用弯曲应力公式前，始终将 mm 转换为 m；确认要求使用 MPa 还是 GPa。限时训练有助于将这些步骤内化，使单元测试成为自信心的助推器。
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