📚 IGCSE CAIE Engineering: Case Study Practical Exercises | IGCSE CAIE 工程：案例分析实战演练
In IGCSE CAIE Engineering, case studies are a powerful way to connect theoretical knowledge with real-world applications. This article provides a series of practical exercises designed to sharpen your analytical skills, covering material selection, structural analysis, circuits, mechanisms, manufacturing, sustainability, and safety. Each section breaks down a typical exam-style scenario and demonstrates how to approach it step by step.
在 IGCSE CAIE 工程科目中，案例分析是将理论知识与实际应用联系起来的有力方式。本文提供了一系列实战演练，旨在锻炼你的分析能力，涵盖材料选择、结构分析、电路、机构、制造工艺、可持续性和安全等领域。每个小节拆解一个典型的考试风格情境，并逐步演示如何应对。
1. Understanding Case Study Requirements | 理解案例分析要求
Before diving into calculations, carefully read the given scenario to identify what the question is asking. Look for keywords such as 'select a suitable material', 'calculate the load', 'suggest an improvement', or 'evaluate the environmental impact'. Highlight all numerical data, constraints, and performance criteria.
在深入计算之前，请仔细阅读所给情境，明确题目在问什么。寻找关键词，例如 "选择合适的材料"、"计算负载"、"提出改进建议" 或 "评估环境影响"。标出所有数值数据、限制条件和性能标准。
A typical IGCSE Engineering case study will combine several topics. For instance, a bridge design question may involve material properties, force analysis, manufacturing processes, and cost. Your answer must show a logical flow from data gathering to final recommendation, supported by clear reasoning and relevant formulas.
一个典型的 IGCSE 工程案例会将多个主题结合起来。例如，一座桥梁设计问题可能涉及材料特性、力的分析、制造工艺和成本。你的答案必须展示从数据收集到最终建议的逻辑流程，并用清晰的推理和相关公式加以支撑。
Always define your variables and state any assumptions. Marks are awarded for correct units, significant figures, and the use of standard engineering notation. Practice by annotating past paper case studies to become fluent in extracting critical information.
始终定义变量并说明所有假设。正确的单位、有效数字和标准工程表示法都会得分。通过批注历年真题中的案例来练习，以便熟练提取关键信息。
2. Material Selection Case | 材料选择案例
A bicycle frame designer must choose between low-carbon steel, aluminium alloy, and carbon fibre composite. The requirements are: high strength-to-weight ratio, corrosion resistance, ease of manufacturing, and cost below 80 per unit material. Given data: steel density 7.8 g/cm³, tensile strength 400 MPa; aluminium density 2.7 g/cm³, strength 300 MPa; carbon fibre density 1.6 g/cm³, strength 600 MPa; relative cost indices: steel 1, aluminium 3, carbon fibre 12.
一位自行车车架设计师必须在低碳钢、铝合金和碳纤维复合材料之间做出选择。要求是：高比强度、耐腐蚀、易于制造，且单位材料成本低于 80。给定数据：钢密度 7.8 g/cm³，抗拉强度 400 MPa；铝密度 2.7 g/cm³，强度 300 MPa；碳纤维密度 1.6 g/cm³，强度 600 MPa；相对成本指数：钢 1，铝 3，碳纤维 12。
/kg, carbon fibre 600/1.6 = 375 kN·m/kg. Carbon fibre has the best strength-to-weight ratio, but its cost index is 12, which may exceed the budget when taking manufacturing complexity into account.
首先，计算每种材料的比强度（强度/密度）：钢约为 400/7.8 ≈ 51.3 kN·m/kg，铝 300/2.7 ≈ 111.1 kN·m/kg，碳纤维 600/1.6 = 375 kN·m/kg。碳纤维的比强度最优，但其成本指数为 12，考虑到制造复杂性时可能会超出预算。
Aluminium offers a good balance: specific strength is more than double that of steel, and it naturally resists corrosion, reducing the need for coating. With a cost index of 3, it is likely to meet the 80-unit target. Therefore, aluminium alloy is the recommended choice based on the given criteria.
铝合金提供了一个良好的平衡：比强度是钢的两倍多，且天然耐腐蚀，减少了涂层需求。成本指数为 3，很可能满足 80 单位的目标。因此，根据给定标准，推荐选择铝合金。
3. Structural Analysis and Load Calculations | 结构分析与负载计算
A simply supported beam of length 4 m carries a central point load of 500 N. Determine the reaction forces at each support and the maximum bending moment. Draw a conclusion about the required section modulus if the allowable bending stress is 150 MPa.
一根简支梁长 4 m，跨中承受 500 N 的集中荷载。求各支座的反力以及最大弯矩。如果许用弯曲应力为 150 MPa，得出所需截面模量的结论。
central, by symmetry, each reaction force R = 500/2 = 250 N. The maximum bending moment Mmax occurs at the centre: Mmax e rearrange to Z = M/σ = 500 N·m / 150 × 10⁶ Pa = 500 × 10³ N·mm / 150 N/mm² = 3333.3 mm³.
由于荷载居中，由对称性可知各支座反力 R = 500/2 = 250 N。最大弯矩 Mmax 出现在跨中：Mmax = (F × L) / 4 = (500 N × 4 m) / 4 = 500 N·m。使用弯曲公式 σ = M/Z，其中 Z 为截面模量，变形得 Z = M/σ = 500 N·m / 150 × 10⁶ Pa = 500 × 10³ N·mm / 150 N/mm² = 3333.3 mm³。
bd²/6. You could select dimensions accordingly, also checking deflection limits. Always include a factor of safety in real design exercises.
因此该梁的截面模量必须至少为 3.33 cm³。如果梁是矩形截面，宽度为 b、高度为 d，则 Z = bd²/6。你可以相应选择尺寸，同时检查挠度限值。在实际设计练习中，始终要计入安全系数。
4. Circuit and Electronic System Analysis | 电路与电子系统分析
A sensing circuit uses a thermistor in a potential divider with a fixed 10 kΩ resistor. The supply voltage is 5 V. At 25 °C the thermistor resistance is 12 kΩ. Calculate the output voltage Vout across the fixed resistor. What happens to Vout as temperature rises to 50 °C if the thermistor resistance drops to 5 kΩ?
一感测电路在分压器中使用热敏电阻和一个固定的 10 kΩ 电阻。电源电压为 5 V。25 °C 时热敏电阻阻值为 12 kΩ。计算固定电阻两端的输出电压 Vout。如果温度升高到 50 °C 时热敏电阻阻值降至 5 kΩ，Vout 会如何变化？
The potential divider formula is Vout = Vin × (R2 / (R1 + R2)). Here let R1 be the thermistor, R2 the fixed 10 kΩ. At 25 °C: Vout = 5 × (10k / (12k + 10k)) = 5 × (10/22) ≈ 2.27 V. At 50 °C: Vout = 5 × (10k / (5k + 10k)) = 5 × (10/15) ≈ 3.33 V. Thus Vout increases with rising temperature because the thermistor resistance decreases, giving a larger share of the voltage to the fixed resistor.
分压器公式为 Vout = Vin × (R2 / (R1 + R2))。这里设 R1 为热敏电阻，R2 为固定 10 kΩ。25 °C 时：Vout = 5 × (10k / (12k + 10k)) = 5 × (10/22) ≈ 2.27 V。50 °C 时：Vout = 5 × (10k / (5k + 10k)) = 5 × (10/15) ≈ 3.33 V。因此，随着温度升高，Vout 上升，因为热敏电阻阻值下降，使固定电阻获得更大的电压份额。
This behaviour can be used to trigger a cooling fan. You might be asked to sketch the output voltage against temperature or to suggest a comparator circuit with a reference voltage for automatic control. Always label your graphs and show threshold levels clearly.
此特性可用于触发冷却风扇。你可能会被要求绘制输出电压随温度变化的草图，或建议一个带有参考电压的比较器电路以实现自动控制。始终标记图表并清晰显示阈值水平。
5. Mechanical Systems and Mechanisms | 机械系统与机构
A gear train consists of a driver gear with 20 teeth, an idler gear of 30 teeth, and a driven gear of 15 teeth. If the driver rotates at 120 rpm clockwise, calculate the speed and direction of the driven gear. Determine the overall velocity ratio.
一齿轮系由 20 齿的主动轮、30 齿的惰轮和 15 齿的从动轮组成。如果主动轮以 120 rpm 顺时针旋转，计算从动轮的转速和方向。确定总速度比。
driver) = 15/20 = 3/4. Since this is less than 1, it is a step-up in speed (overdrive). Driven speed = driver speed × (driver teeth/driven teeth) = 120 × (20/15) = 160 rpm. The direction: driver clockwise, idler will be anti-clockwise, then driven will be clockwise again. So the driven gear rotates at 160 rpm clockwise.
惰轮不影响速度比；如果是外啮合，它只改变方向。速度比 = 从动轮齿数 / 主动轮齿数 = 15/20 = 3/4。由于此值小于 1，这是增速传动（超速）。从动轮转速 = 主动轮转速 × (主动轮齿数/从动轮齿数) = 120 × (20/15) = 160 rpm。方向：主动轮顺时针，惰轮将逆时针，然后从动轮再次顺时针。因此从动轮以 160 rpm 顺时针旋转。
r / ω = 50 / 12.57 ≈ 3.98 N·m. Similar analysis can be extended to belt drives and chain systems.
如果传递的功率为 50 W，计算主动轴上的扭矩。角速度 ω = 120 rpm × (2π/60) = 4π rad/s ≈ 12.57 rad/s。扭矩 = 功率 / ω = 50 / 12.57 ≈ 3.98 N·m。类似的分析可扩展到带传动和链传动系统。
6. Manufacturing Processes and Quality Control | 制造工艺与质量管理
Die casting has high initial tooling cost but low per-part cost; CNC machining has no tooling cost but higher unit cost. Given fixed cost of casting tooling = £8000, variable cost per casting = £0.45; CNC variable cost = £1.20 per part. Compare the total costs for one week's production and recommend the economical process.
一家公司决定每周生产 5000 个铝支架。考虑了两种工艺：压铸和用实心坯料进行数控加工。压铸的初始模具成本高但单件成本低；数控加工无模具成本但单件成本较高。已知压铸模具固定成本 = 8000 英镑，每件压铸可变成本 = 0.45 英镑；数控加工可变成本 = 每件 1.20 英镑。比较一周生产的总成本，并推荐经济的工艺。
Total cost for die casting = 8000 + 0.45 × 5000 = 8000 + 2250 = £10,250. Total cost for CNC = 0 + 1.20 × 5000 = £ 0 = 0.75 Q → Q ≈ 10,667 units. Since 5000 is below the break-even, CNC is more economical.
压铸总成本 = 8000 + 0.45 × 5000 = 8000 + 2250 = 10250 英镑。数控加工总成本 = 0 + 1.20 × 5000 = 6000 英镑。因此对于此批量，数控加工更便宜。盈亏平衡点为 8000 + 0.45 Q = 1.20 Q → 8000 = 0.75 Q → Q ≈ 10667 件。由于 5000 小于平衡点，数控加工更为经济。
In an exam, you should also discuss quality control methods such as statistical process control (SPC) and go/no-go gauges. CNC machining gives better dimensional accuracy and surface finish for low volumes, whereas die casting offers consistency for high volumes.
在考试中，你还应讨论质量控制方法，如统计过程控制 (SPC) 和通/止规。数控加工在低产量时可提供更好的尺寸精度和表面光洁度，而压铸则为高产量提供一致性。
7. Sustainability and Environmental Impact | 可持续性与环境影响
Evaluate the life cycle of a smartphone casing made from virgin ABS plastic versus a casing made from 30% post-consumer recycled ABS. Consider raw material extraction, energy in manufacturing, transportation, and end-of-life disposal. Use approximate carbon footprint data: virgin ABS 3.4 kg CO₂ per kg, recycled ABS 2.1 kg CO₂ per kg; energy for moulding is identical at 5 MJ/kg.
评估由原生 ABS 塑料制成的智能手机外壳与由 30% 消费后再生 ABS 制成的外壳的生命周期。考虑原材料开采、制造能耗、运输和报废处置。使用近似的碳足迹数据：原生 ABS 3.4 kg CO₂/kg，再生 ABS 2.1 kg CO₂/kg；注塑能耗相同，均为 5 MJ/kg。
O₂/MJ) = 5 × 0.015 × 0.5 = 0.0375 kg CO₂; total = 0.0885 kg CO₂. Recycled blend casing: 70% virgin, 30% 0.0375; total ≈ 0.0827 kg CO₂. The recycled version reduces carbon footprint by about 6.5%.
假设外壳质量为 0.015 kg。原生外壳：材料 CO₂ = 0.015 × 3.4 = 0.051 kg CO₂；能耗 CO₂（假设电网排放因子 0.5 kg CO₂/MJ）= 5 × 0.015 × 0.5 = 0.0375 kg CO₂；总计 = 0.0885 kg CO₂。含再生料的外壳：70% 原生，30% 再生。加权材料 CO₂ = (0.7×3.4 + 0.3×2.1) × 0.015 = (3.01) × 0.015 = 0.04515 kg CO₂；能耗相同 0.0375；总计 ≈ 0.0827 kg CO₂。回收版本可减少约 6.5% 的碳足迹。
Further improvements could include design for disassembly, using bioplastics, or reducing material thickness. In case studies, always link your discussion to the 6 R's: Reduce, Reuse, Recycle, Rethink, Refuse, Repair.
进一步的改进可包括面向拆卸的设计、使用生物塑料或减小材料厚度。在案例分析中，始终将你的讨论与 6R 原则联系起来：减少、重用、回收、再思考、拒绝、修复。
8. Safety and Risk Assessment | 安全与风险评估
A workshop task involves using a pillar drill to make 10 mm holes in a steel bracket clamped to the table. Identify at least three hazards and outline corresponding control measures using the hierarchy of control. Relate your answer to the IGCSE Engineering syllabus which expects awareness of PPE, guarding, and safe operating procedures.
一项车间任务涉及使用台钻在夹紧在工作台上的钢支架上钻 10 mm 的孔。识别至少三种危险，并运用控制层级概述相应的控制措施。将你的答案联系到 IGCSE 工程教学大纲，该大纲要求了解个人防护装备、防护装置和安全操作规程。
Hazard 1: Rotating drill bit can catch loose clothing or long hair. Control: enforce a dress code (tie back hair, remove dangling items) – this is an administrative control. Hazard 2: Swarf and flying chips can cause eye injury. Control: wear safety goggles (PPE) and use a machine guard (engineering control). Hazard 3: Unsecured workpiece may spin dangerously. Control: securely clamp the bracket to the machine table (engineering control) and provide training on correct clamping methods (administrative).
危险 1：旋转的钻头可能卷入宽松衣物或长发。控制：强制着装规定（束发、摘除悬垂物品）——属于管理控制。危险 2：切屑和飞溅的铁屑可导致眼部受伤。控制：佩戴安全护目镜（个人防护装备）并使用机器防护罩（工程控制）。危险 3：未固定的工件可能危险地旋转。控制：将支架牢固地夹紧在工作台上（工程控制）并提供正确夹紧方法的培训（管理控制）。
A risk matrix can be used to assess severity and likelihood. For instance, the risk of chuck key left in the chuck is a high-severity, medium-likelihood event. The key control is to use a self-ejecting chuck key and a pre-start check. All control measures must be documented in a safe system of work.
可以使用风险矩阵来评估严重性和可能性。例如，将卡盘钥匙遗留在卡盘中的风险是严重性高、可能性中等的事件。关键控制是使用自弹出式卡盘钥匙和启动前检查。所有控制措施必须记录在安全工作体系中。
9. Comprehensive Analysis: Bridge Design Case | 综合分析：桥梁设计案例
A pedestrian bridge of span 8 m is to be built across a stream. It must support a uniform live load of 2 kN/m² and its own weight. Two truss designs are proposed: a Warren truss made from tubular steel and a Howe truss made from timber. Steel costs £2/kg, timber £0.8/kg, but steel has a longer life and requires less maintenance. Use a simplified analysis to compare bending moments and select the more sustainable option considering a 30-year life cycle.
一座跨度为 8 m 的人行天桥将横跨一条小溪。它必须支撑 2 kN/m² 的均布活荷载和自身重量。提出了两种桁架设计方案：用钢管制成的沃伦桁架和用木材制成的豪氏桁架。钢材成本 2 英镑/kg，木材 0.8 英镑/kg，但钢材寿命更长且维护更少。通过简化分析比较弯矩，并考虑 30 年生命周期选择更可持续的方案。
t resist this moment. The steel truss weighs 500 kg total, timber truss 800 kg. Additional dead load moment from self-weight must be considered: self-weight UDL for steel = (500 × 9.81 / 8) N/m ≈ 613 N/m = 0.613 kN/m; for timber = (800 × 9.81 / 8) ≈ 981 N/m = 0.981 kN/m. Adjusted total UDL: steel 4.613 kN/m, timber 4.981 kN/m. New Msteel = 4.613 × 64 / 8 = 36.9 kN·m; Mtimber = 4.981 × 64 / 8 = 39.85 kN·m.
假设桥面宽度 2 m，因此均布荷载 UDL = 2 kN/m² × 2 m = 4 kN/m。作为简支梁，最大弯矩 M = wL²/8 = 4 × 8²/8 = 32 kN·m。两种桁架都必须抵抗此弯矩。钢桁架总重 500 kg，木桁架 800 kg。必须考虑自重产生的附加恒载力矩：钢的自重 UDL = (500 × 9.81 / 8) N/m ≈ 613 N/m = 0.613 kN/m；木材 = (800 × 9.81 / 8) ≈ 981 N/m = 0.981 kN/m。调整后的总 UDL：钢 4.613 kN/m，木材 4.981 kN/m。新的钢弯矩 M = 4.613 × 64 / 8 = 36.9 kN·m；木弯矩 M = 4.981 × 64 / 8 = 39.85 kN·m。
Now calculate material cost and embodied carbon. Steel: 500 kg × £2 = £1000; timber: 800 kg × £0.8 = £640. However, steel may need painting every 5 years, adding £200 each time (total £ ears), while timber may need replacement after 15 years, doubling the material cost (total £ cyclable, whereas timber is renewable but the replacement cycle increases resource use. The best choice must balance cost, safety, and environmental impact.
现在计算材料成本和隐含碳。钢：500 kg × £2 = 1000 英镑；木材：800 kg × £0.8 = 640 英镑。然而，钢材可能每 5 年需要涂装一次，每次增加 200 英镑（30 年总计 1000 英镑），而木材可能在 15 年后需要更换，使材料成本翻倍（总计 1280 英镑）。钢材方案尽管初始成本较高，但全寿命成本较低且更耐用。在可持续性方面，钢材高度可回收，而木材可再生但更换周期增加了资源消耗。最佳选择必须平衡成本、安全性和环境影响。
10. Practical Exercise Strategies | 实战演练与解题技巧
When tackling a case study in the exam, begin by allocating time proportionally to the marks. Use bullet points to structure your response, but write in full sentences where explanation is required. Always refer to the data provided and show your working clearly, even for simple arithmetic, because marks are given for method.
在考试中解答案例分析时，首先根据分值按比例分配时间。用项目符号组织回答，但在需要解释的地方用完整的句子书写。始终引用提供的数据，并清晰展示计算过程，即使是简单的算术，因为方法步骤也会给分。
Practice quantitative analysis next to qualitative justification. For example, if you recommend a material, give the numerical comparison (e.g., specific strength) and then state why this matters for the function of the product. Use diagrams liberally: a simple free-body diagram, a circuit schematic, or a chart can clarify your reasoning and earn communication marks.
在定性论证旁边练习定量分析。例如，如果你推荐一种材料，给出数值对比（如比强度），然后说明这对产品功能的重要性。随意使用图表：一个简单的受力分析图、电路原理图或图表可以阐明你的推理并获得表达能力分。
Finally, review common pitfall: forgetting unit conversions (mm to m, kW to W). Create a checklist: identify knowns, unknowns, formulas, appropriate assumptions, and a final sense check. Doing regular mock case studies will build speed and confidence for the IGCSE CAIE Engineering examination.
最后，回顾常见陷阱：忘记单位换算（毫米转米，千瓦转瓦）。制作一份检查清单：识别已知量、未知量、公式、适当的假设以及最终的合理性检查。定期进行模拟案例分析练习将为 IGCSE CAIE 工程考试积累速度和信心。
Published by TutorHao | Engineering Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图