📚 Year 11 CIE Engineering: Case Study Practical Drill | Year 11 CIE 工程：案例分析实战演练
Case study questions are a vital part of the CIE Engineering examination, testing your ability to apply theoretical knowledge to real-world design scenarios. This practical drill will guide you through a structured approach, from reading a design brief to justifying material choices, performing mechanical calculations, and evaluating safety. By working through two complete case studies, you will develop the analytical habits needed to score top marks.
案例分析题是 CIE 工程考试的核心组成部分，考查你将理论知识应用于真实设计场景的能力。本次实战演练将通过系统化的方法，带你完成从解读设计任务书到论证材料选择、进行力学计算和评估安全性的全过程。通过两个完整的案例演练，你将养成高分必备的分析习惯。
1. Understanding the Case Study Context | 理解案例背景
Begin by carefully reading the brief. Identify the product’s purpose, its operating environment, and any specific performance targets mentioned, such as load capacity, speed, or lifespan. Underline keywords like ‘lightweight’, ‘corrosion-resistant’, or ‘high stiffness’ — these directly hint at material and design priorities.
仔细阅读任务书。明确产品的用途、工作环境以及任何明确的性能指标，例如承载能力、速度或使用寿命。用下划线标出”轻量化””耐腐蚀””高刚度”等关键词——它们直接暗示了材料与设计的优先方向。
Next, list the stakeholders: who will use this product, who will manufacture it, and who will maintain it? A bicycle brake must be safe for the rider and easy for a mechanic to adjust; a crane hook must protect both the operator and the load. This broad view ensures you don’t miss human factors.
接着列出利益相关者：谁将使用该产品、谁将制造它、谁将维护它？自行车刹车必须对骑行者安全且便于技工调整；起重机吊钩必须同时保护操作员和负载。这种全局视角能确保你不会忽略人因要素。
2. Identifying Key Requirements and Constraints | 识别关键需求与约束
Distinguish between functional requirements (what the product must do) and constraints (limits on size, cost, weight, or manufacturing method). For example, a brake caliper must deliver a clamping force of at least 300 N, but its mass must stay below 200 g and its production cost under £5 per unit. These numbers form the boundary of your design space.
区分功能性需求（产品必须做什么）和约束条件（尺寸、成本、重量或制造方法的限制）。例如，刹车卡钳必须提供至少 300 N 的夹紧力，但其质量必须低于 200 g，单件生产成本需控制在 5 英镑以下。这些数字构成了你的设计空间边界。
Also capture standards and regulations. Engineering products often need to meet ISO, EN, or national safety codes. A lifting hook, for instance, is governed by standards like ISO 8539:2009, which imposes a minimum safety factor of 4 against yield. Referencing such standards demonstrates professional awareness.
同时记录标准与法规。工程产品常需符合 ISO、EN 或国家安全规范。例如，吊钩受 ISO 8539:2009 等标准管控，要求对屈服强度的最小安全系数为 4。引用这类标准可展现专业意识。
3. Material Selection Fundamentals | 材料选择基础
Use a systematic method: first screen materials by the ‘must-have’ properties, then rank them by ‘wishes’. For a brake caliper, must-have properties include yield strength above 200 MPa and good fatigue resistance; a wish might be low thermal expansion. Create a shortlist of three candidates and justify your final choice with data.
使用系统化方法：先用”必备”属性筛选材料，再按”期望”属性排序。对于刹车卡钳，必备属性包括屈服强度高于 200 MPa 和良好的抗疲劳性；期望属性可能是低热膨胀。列出三种候选材料，并用数据论证最终选择。
| Material | Yield Strength (MPa) | Density (g/cm³) | Corrosion Resistance | Relative Cost |
|---|---|---|---|---|
| Aluminium 6061-T6 | 276 | 2.70 | Good | Medium |
| Low Carbon Steel | 250 | 7.85 | Poor (needs coating) | Low |
| Glass-filled Nylon | 80 | 1.35 | Excellent | Low |
In a weight-sensitive application like a bicycle, the excellent strength-to-weight ratio of Al 6061-T6 makes it a strong candidate, provided the cost target is met. If corrosion is the main concern, glass-filled nylon might serve for non-critical brackets.
在自行车这种重量敏感的应用中，6061-T6 铝合金优异的比强度使其成为强有力的候选，前提是成本达标。如果腐蚀是主要关注点，玻璃纤维填充尼龙可用于非关键支架。
4. Mechanical Stress and Strain Analysis | 应力与应变分析
When a component is loaded, internal stress σ develops. For a simple axial load, stress is force divided by cross-sectional area. The resulting strain ε is the extension per unit length. These are linked by Young’s modulus E for materials within the elastic limit.
当构件受载时，内部会产生应力 σ。对于简单的轴向载荷，应力等于力除以横截面积。由此产生的应变 ε 是单位长度的伸长量。在弹性极限内，二者通过杨氏模量 E 联系。
σ = F / A ε = ΔL / L₀ E = σ / ε
For example, a cylindrical rod of diameter 10 mm (area ≈ 78.5 mm²) carrying 2000 N experiences a stress of about 25.5 MPa. If the rod is aluminium (E ≈ 69 GPa), the strain is roughly 0.00037, meaning a 100 mm length extends by only 0.037 mm — a small but critical value.
例如，一根直径 10 mm（面积约 78.5 mm²）的圆柱杆承受 2000 N 时，应力约为 25.5 MPa。如果材料是铝（E ≈ 69 GPa），应变约为 0.00037，意味着 100 mm 长度仅伸长 0.037 mm——数值虽小，却十分关键。
5. Force Analysis and Free Body Diagrams | 受力分析与自由体图
A clear free body diagram (FBD) is the engineer’s most powerful tool. Isolate the part of interest, draw all forces acting on it (applied loads, reactions, friction), and apply equilibrium: ΣF = 0 and ΣM = 0. This reveals internal forces you’ll later use in stress calculations.
清晰的自由体图是工程师最有力的工具。隔离感兴趣的部分，画出所有作用力（外加载荷、支反力、摩擦力），并应用平衡条件：ΣF = 0 和 ΣM = 0。这将揭示后续应力计算所需的内部力。
Consider a simple lever: an input force Fᵢ at distance dᵢ from the pivot and an output force Fₒ at distance dₒ. Balancing moments gives Fᵢ × dᵢ = Fₒ × dₒ. Thus, a mechanical advantage (MA) = dᵢ / dₒ amplifies force when dᵢ > dₒ.
以一个简单杠杆为例：输入力 Fᵢ 到支点距离 dᵢ，输出力 Fₒ 距支点 dₒ。力矩平衡得出 Fᵢ × dᵢ = Fₒ × dₒ。因此，机械效益 MA = dᵢ / dₒ，当 dᵢ > dₒ 时可放大作用力。
6. Manufacturing Processes and Tolerances | 制造工艺与公差
The selected manufacturing process affects both cost and mechanical properties. Forging aligns grain flow and improves fatigue life but requires expensive tooling, making it suitable for high-volume safety parts like crane hooks. Die casting offers complex shapes at lower cost but may introduce porosity, weakening the part.
所选的制造工艺会影响成本和力学性能。锻造可理顺金属流线并提高疲劳寿命，但需要昂贵模具，适合吊钩等大批量安全零件。压铸能以较低成本获得复杂形状，但可能引入气孔，削弱零件强度。
Tolerances must be specified realistically. A brake caliper piston requires a close sliding fit (e.g., H7/g6) to prevent fluid leakage, whereas a bracket for a cable guide can accept a wider tolerance (±0.5 mm). Always relate tolerance choice to function and cost.
公差必须现实地规定。刹车卡钳活塞需采用紧密的滑动配合（如 H7/g6）以防液体泄漏，而线管导向支架可接受较宽的公差（±0.5 mm）。始终将公差选择与功能和成本关联起来。
7. Safety Factors and Risk Assessment | 安全系数与风险评估
The factor of safety (FoS) is the ratio of the material’s yield strength (or ultimate strength) to the maximum working stress. FoS accounts for uncertainties in load, material properties, and consequences of failure. A crane hook carrying a suspended load over people requires an FoS of 5–6, whereas a non-critical bracket might use 2.
安全系数（FoS）是材料屈服强度（或极限强度）与最大工作应力之比。它用于应对载荷、材料特性的不确定性和失效后果。在人员上方悬吊重物的起重机吊钩需要 5–6 的安全系数，而非关键支架可能仅用 2。
Safety Factor (n) = σyield / σworking
A working stress of 45 MPa in a steel with yield 250 MPa gives n ≈ 5.6. This would be acceptable for a lifting device according to many codes. However, if stress concentrations at a sharp corner raise local stress to 120 MPa, the local safety factor drops to 2.1, which may be unsafe.
工作应力为 45 MPa，钢材屈服强度 250 MPa，得出 n ≈ 5.6。按多数规范这对起重设备是合格的。但若尖角处的应力集中使局部应力升至 120 MPa，局部安全系数将降至 2.1，可能不安全。
8. Practical Case 1: Bicycle Brake Caliper | 实战案例一：自行车刹车卡钳
A side-pull caliper brake uses a cable to pull two arms together. The rider’s hand force on the lever is about 80 N; after the lever’s mechanical advantage of 4:1, the cable tension becomes 320 N. The caliper arms have an effective lever ratio of 1:1.5, converting cable pull into pad clamping force.
侧拉式卡钳刹车用钢线拉动两臂合拢。骑行者施加于刹把的手力约为 80 N；经过刹把 4:1 的机械效益，钢线张力变为 320 N。卡钳臂具有 1:1.5 的有效杠杆比，将钢线拉力转化为刹车块夹紧力。
Fpad = 320 N × 1.5 = 480 N
Each pad presses against the rim with 480 N. With a friction coefficient μ = 0.4, the tangential friction force per pad is Ff = μ × 480 ≈ 192 N. Both pads together produce a total braking force of 384 N at the rim, providing strong deceleration for a 100 kg rider+bike system.
每个刹车块以 480 N 压向轮圈。摩擦系数 μ = 0.4，每侧切向摩擦力约 Ff = 192 N。两侧刹车块在轮圈处产生总共 384 N 的制动力，足以使 100 kg 的骑行者+自行车系统迅速减速。
The caliper arm is a cantilever beam. Its critical section, near the pivot, is 8 mm wide and 12 mm deep. The bending moment here is M = 480 N × 35 mm = 16800 N·mm. The section modulus Z = (b × h²)/6 = (8×12²)/6 = 192 mm³. Bending stress σb = M / Z ≈ 87.5 MPa, well below Al 6061-T6 yield (276 MPa), giving a local safety factor above 3.
卡钳臂可视为悬臂梁。其临界截面靠近转轴，宽 8 mm、厚 12 mm。该处弯矩 M = 480 N × 35 mm = 16800 N·mm。截面模量 Z = (b × h²)/6 = (8×12²)/6 = 192 mm³。弯曲应力 σb = M / Z ≈ 87.5 MPa，远低于 6061-T6 的屈服强度（276 MPa），局部安全系数超过 3。
9. Practical Case 2: Crane Hook Design | 实战案例二：起重机吊钩设计
A forged steel crane hook must lift 5000 N (approx. 510 kg). The hook’s critical cross-section, at the inner radius, is trapezoidal with an area of 160 mm². The direct tensile stress is σtensile = 5000 N / 160 mm² = 31.25 MPa. However, the load is not purely axial; bending adds significant stress.
一个锻钢吊钩需起吊 5000 N（约 510 kg）。吊钩内半径处的临界截面为梯形，面积 160 mm²。直接拉伸应力 σtensile = 5000/160 = 31.25 MPa。但载荷并非纯轴向；弯曲会叠加显著应力。
Using a simplified curved beam approach, the maximum stress at the inner fibre can be estimated as σmax ≈ k × σtensile, where k is a stress concentration factor. For a typical hook with r/D ≈ 0.2, k may be around 3.5. Thus σmax ≈ 109 MPa. The hook is made from alloy steel with σyield = 620 MPa, yielding an overall safety factor of 620/109 ≈ 5.7, which meets the required FoS ≥ 5.
采用简化的曲梁方法，内侧纤维的最大应力可估算为 σmax ≈ k × σtensile，其中 k 为应力集中系数。对于 r/D ≈ 0.2 的典型吊钩，k 约为 3.5。因此 σmax ≈ 109 MPa。吊钩材料为合金钢，屈服强度 620 MPa，总体安全系数 = 620/109 ≈ 5.7，满足 FoS ≥ 5 的要求。
To improve fatigue life, the hook is shot-peened after forging, introducing compressive residual stress on the surface. Regular non-destructive testing (magnetic particle inspection) is prescribed to detect any cracks before they propagate.
为提高疲劳寿命，吊钩锻后进行喷丸处理，在表面引入残余压应力。规程要求定期进行无损检测（磁粉探伤），以便在裂纹扩展前及时发现。
10. Exam Tips and Common Pitfalls | 考试技巧与常见误区
Always show your working step by step. Even if the final numerical answer is wrong, clear methodology can earn most of the marks. Begin with a labelled FBD, write the governing equation, substitute values with units, and then state the result. Never skip units — they are a powerful error check.
务必逐步展示计算过程。即使最终数值答案错误，清晰的方法也能获得大部分分数。从带标注的自由体图开始，写出控制方程，代入带单位的数值，然后给出结果。绝不遗漏单位——它们是强大的查错工具。
When justifying a material choice, do not just name it. Compare at least two alternatives with data, then link properties to the design requirement. For instance, ‘I chose Al 6061-T6 over low carbon steel because its density is 66% lower, reducing the brake’s unsprung mass and improving suspension response, while still providing adequate strength.’
当论证材料选择时，不要只说出名称。用数据比较至少两种替代材料，然后将性能与设计要求关联。例如：”我选择 6061-T6 铝合金而非低碳钢，因为其密度低 66%，可降低刹车的非悬挂质量并改善悬架响应，同时仍提供足够强度。”
Finally, manage your time. A typical case study question may be worth 20 marks; allocate about 20 minutes. Read, highlight constraints (2 min), plan your approach (2 min), execute calculations and written explanation (12 min), review and check units (4 min). Practising under timed conditions is the best preparation.
最后，管理好时间。一道典型的案例分析题可能值 20 分；分配约 20 分钟。阅读并标出约束条件（2 分钟），规划方法（2 分钟），执行计算和书面解释（12 分钟），复查并检查单位（4 分钟）。计时实战是备考的最佳方式。
Published by TutorHao | Engineering Revision Series | aleveler.com
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