📚 GCSE WJEC Engineering: Case Study Practical Exercises | GCSE WJEC 工程：案例分析实战演练
Mastering case study questions is essential for success in GCSE WJEC Engineering. These extended-response tasks require students to apply theoretical knowledge to real-world scenarios, demonstrating analytical, evaluative, and design skills. This article provides a series of practical exercises, each targeting key topics from the specification, with step-by-step guidance on how to structure high-quality answers.
掌握案例分析题对于在 GCSE WJEC 工程考试中取得成功至关重要。这些长篇论述题要求学生将理论知识应用于现实场景，展示分析、评估和设计能力。本文提供一系列实战演练，每个练习都针对考纲中的关键主题，并逐步指导如何组织高质量答案。
1. Understanding the Engineering Case Study Approach | 理解工程案例分析的方法
WJEC Engineering papers often feature a detailed context—such as designing a product for a specific user or selecting a manufacturing method—followed by sub-questions that explore materials, processes, electronics, structures, or sustainability. Students must read the scenario carefully and link each answer to the given context.
WJEC 工程试卷通常提供一个详细的背景——例如为特定用户设计产品或选择制造方法——然后给出考察材料、工艺、电子、结构或可持续性等方面的小问。学生必须仔细阅读情境，将每个答案与给定背景联系起来。
A structured approach begins with identifying command words (e.g., ‘explain’, ‘evaluate’, ‘calculate’, ‘justify’), underlining key data, and planning responses that include technical vocabulary and relevant equations.
结构化的方法首先要识别指令词（如“解释”“评估”“计算”“论证”），标记关键数据，并规划包含技术术语和相关公式的答案。
For design and make tasks, always justify choices against design criteria, such as cost, performance, durability, and environmental impact. Wasting time on generic answers that ignore the specific context will lose marks.
对于设计与制造任务，始终依据设计标准（如成本、性能、耐用性和环境影响）来论证选择。忽略具体背景而写下泛泛答案会白白丢分。
2. Case Study 1: Material Selection for a Lightweight Bicycle Frame | 案例分析1：轻量化自行车车架的材料选择
Scenario: A manufacturer wants to produce a high-performance racing bicycle. The frame must be as light as possible, yet stiff enough to transfer pedalling force efficiently. Cost is not the primary concern, but the frame must be manufacturable using current technology.
情景：一家制造商想生产高性能竞赛自行车。车架必须尽可能轻，同时要足够坚固以高效传递踩踏力。成本不是主要考虑，但车架必须能用现有技术制造。
Typical exam questions could be: ‘Explain why carbon fibre would be a suitable material for the frame’ and ‘Compare the properties of aluminium alloy and titanium alloy for this application.’
典型的考题可能是：“解释为什么碳纤维适合作为车架材料”以及“比较铝合金和钛合金在此应用中的性能”。
Carbon fibre-reinforced polymer (CFRP) offers an exceptional strength-to-density ratio (specific strength). Its density of around 1.6 g/cm³ is much lower than aluminium (2.7 g/cm³) or titanium (4.5 g/cm³). This allows a lighter frame without sacrificing stiffness.
碳纤维增强聚合物 (CFRP) 具有出色的强度-密度比（比强度）。其密度约 1.6 g/cm³，远低于铝 (2.7 g/cm³) 或钛 (4.5 g/cm³)。这使得车架更轻而不牺牲刚度。
However, CFRP is anisotropic, meaning its strength is directional. Engineers must orient fibres to handle stresses from pedalling and impacts. Aluminium alloy is isotropic, easier to recycle, and has lower material cost. A comparison using specific strength is useful.
然而，CFRP 呈各向异性，意味着强度有方向性。工程师必须定向排列纤维以应对踩踏和冲击应力。铝合金是各向同性的，更易回收，且材料成本较低。使用比强度进行比较很有帮助。
Specific Strength = Tensile Strength ÷ Density
比强度 = 抗拉强度 ÷ 密度
Aluminium alloy 6061 has a tensile strength of about 310 MPa and density 2.7 g/cm³, giving a specific strength of roughly 115 kN·m/kg. Titanium alloy Ti-6Al-4V reaches ~900 MPa with density 4.43 g/cm³, specific strength ~203 kN·m/kg. CFRP can achieve over 1500 MPa along fibres, with a specific strength exceeding 900 kN·m/kg.
铝合金 6061 抗拉强度约 310 MPa，密度 2.7 g/cm³，比强度约 115 kN·m/kg。钛合金 Ti-6Al-4V 强度约 900 MPa，密度 4.43 g/cm³，比强度约 203 kN·m/kg。CFRP 沿纤维方向强度可超过 1500 MPa，比强度超过 900 kN·m/kg。
3. Case Study 2: Manufacturing Process for a Plastic Phone Case | 案例分析2：塑料手机壳的制造工艺
Scenario: A company plans to mass-produce a sleek phone case from ABS plastic. The part must have a smooth, glossy surface, precise dimensions, and a low unit cost. The mould cost can be high, but the production run is in the hundreds of thousands.
情景：一家公司计划用 ABS 塑料批量生产一款时尚手机壳。零件必须具有光滑的高光表面、精确的尺寸，并且单件成本要低。模具费可以较高，但生产批量有数十万件。
Injection moulding is the obvious choice. ABS granules are heated, melted, and injected under high pressure into a multi-cavity steel mould. The mould includes cooling channels to solidify the plastic quickly, achieving cycle times around 25–35 seconds.
注塑成型是显而易见的选择。ABS 颗粒被加热熔化，在高压下注射进多型腔钢制模具。模具内设有冷却通道，使塑料快速固化，循环时间约 25–35 秒。
To ensure a smooth surface, the mould cavity is polished, and the gate is placed on a hidden edge. Uniform wall thickness of 2–3 mm is maintained throughout the design to prevent sink marks and warping. Ejector pins must be located on internal surfaces.
为确保表面光滑，型腔经过抛光处理，浇口放置在隐蔽边缘上。整个设计保持 2–3 mm 的均匀壁厚，以防止缩痕和翘曲。顶针必须布置在内表面上。
Production rate = (Number of cavities × 3600) ÷ Cycle time (seconds)
生产率 = （型腔数 × 3600）÷ 周期时间（秒）
For a 4-cavity mould and a 30 s cycle, the hourly output is (4 × 3600) ÷ 30 = 480 parts. When justifying the process, mention the reduction in labour costs and the excellent repeatability of injection moulding.
对于 4 腔模具和 30 秒周期，每小时产量为 (4 × 3600) ÷ 30 = 480 件。在论证该工艺时，要提到人工成本的降低以及注塑成型出色的重复精度。
4. Case Study 3: Electronic Circuit Design for a Night Light | 案例分析3：夜灯的电子电路设计
Scenario: Design a circuit for an automatic night light that turns on an LED when ambient light falls below a certain level. The system uses a 5 V power supply, an LDR, a fixed resistor, an NPN transistor, and a current-limiting resistor for the LED.
情景：为一个自动夜灯设计电路，在环境光线低于一定水平时点亮 LED。系统使用 5 V 电源、一个 LDR、一个固定电阻、一个 NPN 晶体管和一个 LED 限流电阻。
The LDR and fixed resistor R1 form a voltage divider. In the dark, the LDR resistance rises to around 100 kΩ. If R1 is 10 kΩ, the voltage at the junction is given by:
LDR 和固定电阻 R1 组成一个分压器。黑暗中 LDR 电阻升至约 100 kΩ。若 R1 为 10 kΩ，则连接点的电压由下式给出：
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