📚 A-Level Cambridge Engineering: Case Study Practical Exercises | A-Level Cambridge 工程：案例分析实战演练
In A-Level Cambridge Engineering, case studies are essential for applying theoretical knowledge to real-world problems. This article provides a series of practical exercises covering stress analysis, circuit design, material selection and design processes, helping you master the analytical skills required for the exam.
在A-Level剑桥工程中，案例研究是将理论知识应用于实际问题的关键。本文提供了一系列实战练习，涵盖应力分析、电路设计、材料选择和设计流程，帮助你掌握考试所需的分析技能。
1. Case Study Methodology | 案例分析方法
Every engineering case study follows a structured approach: define the problem, gather relevant data, apply scientific principles, evaluate potential solutions, and present a justified conclusion. This method ensures systematic analysis and clear communication.
每个工程案例研究都遵循结构化方法：界定问题、收集相关数据、应用科学原理、评估可能的解决方案，并给出有依据的结论。这种方法确保分析系统化、表述清晰。
Start by identifying the key requirements, constraints and failure criteria. For a structural case, note the load types, material properties and safety factors. For an electronic case, list the input/output specifications and component limitations.
首先明确关键要求、约束条件和失效准则。对于结构案例，注意载荷类型、材料性能和安全系数。对于电子案例，列出输入/输出规格和元件限制。
Always check units and convert to SI before substitution. Use free-body diagrams, circuit schematics or flowcharts to visualise the system. The final answer must always refer back to the initial problem statement.
代入数据前务必检查单位并转换为国际单位制。使用受力图、电路原理图或流程图来可视化系统。最终答案必须始终回扣最初的问题陈述。
2. Case Study 1: Cantilever Beam Stress Analysis | 案例一：悬臂梁应力分析
A steel cantilever beam of length L = 2.0 m supports a point load F = 500 N at its free end. The cross-section is rectangular with breadth b = 50 mm and depth h = 100 mm. Determine the maximum bending stress and check whether it exceeds the yield strength of 250 MPa.
一根钢制悬臂梁，长度 L = 2.0 m，在自由端承受集中载荷 F = 500 N。截面为矩形，宽度 b = 50 mm，高度 h = 100 mm。计算最大弯曲应力，并检查是否超过 250 MPa 的屈服强度。
Step 1: Bending moment at the fixed support. Mmax = F × L = 500 N × 2.0 m = 1000 Nm.
步骤 1：固定支座处的弯矩。Mmax = F × L = 500 N × 2.0 m = 1000 Nm。
Step 2: Second moment of area for a rectangle about the horizontal neutral axis. I = bh3 / 12 = (0.050 m × (0.100 m)3) / 12 = 4.167 × 10-6 m4.
步骤 2：矩形截面对水平中性轴的惯性矩。I = bh3 / 12 = (0.050 m × (0.100 m)3) / 12 = 4.167 × 10-6 m4。
Step 3: Maximum distance from neutral axis, ymax = h/2 = 0.050 m. Maximum bending stress using flexure formula: σ = M y / I.
σ = 1000 Nm × 0.050 m / 4.167×10-6 m4 = 12.0 × 106 Pa = 12.0 MPa
步骤 3：到中性轴的最大距离 ymax = h/2 = 0.050 m。使用弯曲公式求最大弯曲应力：σ = M y / I。
σ = 1000 Nm × 0.050 m / 4.167×10-6 m4 = 12.0 × 106 Pa = 12.0 MPa
The calculated stress (12.0 MPa) is far below the yield strength (250 MPa). The factor of safety against yielding is 250 / 12.0 ≈ 20.8, so the design is very conservative.
计算得到的应力（12.0 MPa）远低于屈服强度（250 MPa）。抗屈服的安全系数为 250 / 12.0 ≈ 20.8，因此设计非常保守。
3. Case Study 2: Truss Bridge Load Calculation | 案例二：桁架桥负载计算
A simple triangular truss supports a vertical load P = 10 kN at joint C. Joints A and B are pin supports on the same horizontal level, separated by 2.0 m. Joint C is positioned 1.5 m directly above the midpoint of AB. Find the axial forces in members AB, AC and BC.
一个简单三角桁架在节点 C 承受垂直载荷 P = 10 kN。节点 A 和 B 为同一水平线上的铰支座，相距 2.0 m。节点 C 位于 AB 中点正上方 1.5 m 处。求杆件 AB、AC 和 BC 的轴力。
Geometry: length of AB = 2.0 m, height = 1.5 m. Inclined members AC and BC each have length √(1.02 + 1.52) = 1.803 m. Angle θ = tan-1(1.5/1.0) = 56.31°.
几何条件：杆 AB 长 2.0 m，高度 1.5 m。斜杆 AC 和 BC 各长 √(1.02 + 1.52) = 1.803 m。角度 θ = tan-1(1.5/1.0) = 56.31°。
Method of joints – Joint A: unknown forces FAB (horizontal) and FAC (inclined). Equilibrium: ΣFx = 0 → FAB + FAC cos56.31° = 0. ΣFy = 0 → FAC sin56.31° + RA = 0. By symmetry RA = RB = 5 kN upward. Solve: FAC = -5 / sin56.31° = -6.01 kN (compression). Then FAB = -FAC cos56.31° = 3.34 kN (tension).
节点法 – 节点 A：未知力 FAB（水平）和 FAC（倾斜）。平衡方程：ΣFx = 0 → FAB + FAC cos56.31° = 0。ΣFy = 0 → FAC sin56.31° + RA = 0。由对称性 RA = RB = 5 kN 向上。解得：FAC = -5 / sin56.31° = -6.01 kN（受压）。于是 FAB = -FAC cos56.31° = 3.34 kN（受拉）。
Joint C: downward load 10 kN, forces FAC and FBC (equal magnitude, both compression). Verify: 2 × 6.01 sin56.31° = 10.0 kN, equilibrium satisfied. Member BC is also in compression with 6.01 kN.
节点 C：向下载荷 10 kN，力 FAC 和 FBC（大小相等，均受压）。验证：2 × 6.01 sin56.31° = 10.0 kN，平衡条件满足。杆 BC 同样承受 6.01 kN 压力。
4. Case Study 3: Electronic Circuit Troubleshooting | 案例三：电子电路故障排除
A voltage divider consists of R1 = 10 kΩ and R2 = 20 kΩ connected to a 9 V DC supply. The expected output voltage across R2 is Vout = 9 × 20/(10+20) = 6 V. However, a voltmeter reads only 3 V. Diagnose the fault.
一个分压器由 R1 = 10 kΩ 和 R2 = 20 kΩ 组成，接至 9 V 直流电源。预期 R2 两端的输出电压为 Vout = 9 × 20/(10+20) = 6 V。但电压表读数仅为 3 V。诊断故障。
Possible causes: (1) R1 open circuit – no current, Vout would be 0 V, not matching. (2) R2 open circuit – current flows through R1 only, Vout equals supply 9 V, not matching. (3) R1 short circuit – R1 0 Ω, then Vout = 9 × 20/20 = 9 V, not matching. (4) R2 short circuit – then Vout = 0 V. (5) A parallel extra load resistance RL across R2 could reduce effective resistance. Check if RL = 10 kΩ is accidentally connected. Then equivalent R2,eq = (20×10)/(20+10) = 6.67 kΩ. Vout = 9 × 6.67/(10+6.67) = 3.6 V, close to 3 V. Slight meter loading or tolerance gives 3 V.
可能原因：(1) R1 开路——无电流，Vout 为 0 V，不符。(2) R2 开路——电流仅流过 R1，Vout 等于电源 9 V，不符。(3) R1 短路——R1 0 Ω，Vout = 9 × 20/20 = 9 V，不符。(4) R2 短路——Vout = 0 V。(5) R2 上并联了额外负载电阻 RL，会降低等效电阻。检查是否意外连接了 RL = 10 kΩ。则等效 R2,eq = (20×10)/(20+10) = 6.67 kΩ。Vout = 9 × 6.67/(10+6.67) = 3.6 V，接近 3 V。轻微的仪表负载或容差即可得到 3 V。
Conclusion: the fault is likely an unintended load resistor of about 10 kΩ in parallel with R2. Remove the extra path to restore 6 V output.
结论：故障很可能是 R2 上意外并联了约 10 kΩ 的负载电阻。移除额外通路即可恢复 6 V 输出。
5. Case Study 4: Material Selection for a Bicycle Frame | 案例四：自行车车架材料选择
Select the most suitable material for a lightweight, stiff bicycle frame from steel, aluminium alloy, titanium alloy and carbon-fibre composite. Evaluate density, Young’s modulus, yield strength and cost.
从钢、铝合金、钛合金和碳纤维复合材料中选择最适合轻质、高刚度自行车车架的材料。评估密度、杨氏模量、屈服强度和成本。
Typical properties:
典型性能：
• Steel: density ρ = 7800 kg/m3, E = 210 GPa, yield strength σy = 500 MPa, moderate cost. High stiffness and strength but heavy.
• 钢：密度 ρ = 7800 kg/m³，E = 210 GPa，屈服强度 σy = 500 MPa，成本适中。高刚度、高强度，但较重。
• Aluminium alloy: ρ = 2700 kg/m³, E = 70 GPa, σy = 300 MPa, low cost. Light but lower stiffness; tube diameter can be increased to compensate.
• 铝合金：ρ = 2700 kg/m³，E = 70 GPa，σy = 300 MPa，低成本。轻但刚度较低；可通过增大管径补偿。
• Titanium alloy: ρ = 4500 kg/m³, E = 110 GPa, σy = 900 MPa, high cost. Excellent strength-to-weight ratio and corrosion resistance.
• 钛合金：ρ = 4500 kg/m³，E = 110 GPa，σy = 900 MPa，高成本。卓越的比强度及耐腐蚀性。
• Carbon-fibre composite: ρ = 1600 kg/m³, E = 130 GPa (directional), σy ~ 600 MPa, high cost. Very light and tailorable, but brittle and expensive to manufacture.
• 碳纤维复合材料：ρ = 1600 kg/m³，E = 130 GPa（定向），σy ~ 600 MPa，高成本。极轻且可定制，但脆，制造成本高。
For a performance race frame, carbon fibre offers the best specific stiffness (E/ρ ≈ 81 MPa·m3/kg) and is therefore often chosen. Aluminium is a budget-friendly alternative when tube dimensions are optimised.
对于竞赛级车架，碳纤维提供最佳的比刚度（E/ρ ≈ 81 MPa·m³/kg），因此常被选用。铝合金在管材尺寸优化后则是经济实惠的选择。
6. Case Study 5: Thermodynamic System Efficiency | 案例五：热力学系统效率评估
A heat engine operates between a hot reservoir at T1 = 500°C (773 K) and a cold reservoir at T2 = 30°C (303 K). In one cycle it absorbs QH = 150 kJ of heat and produces net work Wnet = 80 kJ. Determine the Carnot efficiency and the actual thermal efficiency, and comment on losses.
一台热机工作在高温热源 T1 = 500°C（773 K）和低温热源 T2 = 30°C（303 K）之间。一个循环中吸收热量 QH = 150 kJ，产生净功 Wnet = 80 kJ。求卡诺效率和实际热效率，并评论损失。
Carnot efficiency: ηCarnot = 1 − T2 / T1 = 1 − 303 / 773 = 0.608, i.e. 60.8%.
卡诺效率：ηCarnot = 1 − T2 / T1 = 1 − 303 / 773 = 0.608，即 60.8％。
Actual efficiency: ηactual = Wnet / QH = 80 kJ / 150 kJ = 0.533, or 53.3%. This is lower than the Carnot limit, as expected due to irreversibilities such as friction, heat transfer across finite temperature differences and internal dissipative processes.
实际效率：ηactual = Wnet / QH = 80 kJ / 150 kJ = 0.533，即 53.3%。这低于卡诺极限，符合不可逆损失（如摩擦、有限温差传热和内部耗散过程）的预期。
The rejected heat QC = QH − Wnet = 70 kJ is discharged to the cold reservoir. Even a perfect reversible engine would reject 150 − (0.608×150) = 58.8 kJ, so the extra 11.2 kJ represents irreversibility.
排出的热量 QC = QH − Wnet = 70 kJ 释放至冷源。即使是完美的可逆热机也会排出 150 − (0.608×150) = 58.8 kJ，因此多出的 11.2 kJ 即代表不可逆性。
7. Case Study 6: Control System Feedback Loop | 案例六：控制系统反馈
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