📚 IGCSE Cambridge Engineering: Case Study Practical Exercises | IGCSE 剑桥工程：案例分析实战演练
Engineering case studies are a powerful tool for bridging the gap between theory and real‑world practice. In the IGCSE Cambridge Engineering course, you will encounter scenarios that test your ability to analyse, calculate, and propose practical solutions. This article takes you through a series of case study exercises, each designed to reinforce key concepts and sharpen your problem‑solving skills.
工程案例研究是连接理论与现实世界的桥梁。在IGCSE剑桥工程课程中，你会遇到需要分析、计算并提出可行方案的场景。本文通过一组案例演练，帮你巩固核心概念并提升解决问题的能力。
1. Engineering Case Study Methodology | 工程案例分析方法
Every case study follows a logical sequence: define the problem, collect relevant data, apply engineering principles to analyse the situation, develop possible solutions, and evaluate them against constraints such as cost, safety, and materials. This systematic method ensures that no critical aspect is overlooked.
每个案例都遵循逻辑顺序：界定问题，收集数据，用工程原理分析，提出方案，并根据成本、安全和材料等约束条件进行评价。这种系统方法确保不遗漏关键点。
In an exam or project, you should begin by listing known quantities, sketching a diagram if applicable, and writing down the governing equations. Always state your assumptions clearly — for example, ‘neglect friction’ or ‘assume uniform cross‑section’.
考试或项目中，应先列出已知量，必要时画草图，写下控制方程。清晰说明假设，例如“忽略摩擦”或“假设截面均匀”。
2. Case 1: Structural Analysis of a Truss Bridge | 案例一：桁架桥梁结构分析
A simply supported truss bridge of span 8 m carries a central point load of 10 kN. The truss consists of two sloping top members at 30° to the horizontal, a horizontal bottom tie, and a vertical king post. Your task is to determine the forces in each member and identify which are in tension and compression.
一座简支桁架桥跨度8 m，中央承受10 kN集中荷载。桁架由两根与水平呈30°的斜向顶杆、一根水平底梁和一根竖杆组成。任务是求各杆内力，并判断受拉还是受压。
Using method of joints at the left support: vertical equilibrium gives the reaction RA = 5 kN. At the joint, the vertical component of sloping member F₁ balances the reaction, so F₁ × sin30° = 5 kN, hence F₁ = 10 kN (tension).
用节点法分析左支座：竖向平衡得反力 RA = 5 kN。节点处，斜杆 F₁ 的竖向分力与反力平衡，F₁ × sin30° = 5 kN，因此 F₁ = 10 kN（受拉）。
Horizontal equilibrium at the same joint gives the force in the bottom tie F₂ = F₁ × cos30° = 10 × 0.866 = 8.66 kN (compression). The vertical king post carries the difference in vertical forces and is found to be 0 kN under symmetrical loading — it acts mainly as a stabiliser.
同节点的水平平衡给出底梁内力 F₂ = F₁ × cos30° = 10 × 0.866 = 8.66 kN（受压）。竖杆在对称荷载下竖向力差为零，内力为 0 kN，主要起稳定作用。
The computed forces can be summarised as:
计算结果总结如下：
F₁ = 10 kN (tension), F₂ = 8.66 kN (compression)
3. Case 2: Fault Diagnosis in a DC Circuit | 案例二：直流电路故障诊断
A simple DC circuit consists of a 9 V battery, a switch, and two resistors R₁ = 100 Ω and R₂ = 200 Ω connected in series. When the switch is closed, the lamp (rated 6 V, 0.3 A) placed across R₂ does not light up. Diagnose the possible faults using a multimeter.
简单直流电路：9 V电池、开关、串联的 R₁ = 100 Ω 和 R₂ = 200 Ω。闭合开关后，并联在 R₂ 两端的灯泡（额定6 V, 0.3 A）不亮。用万用表诊断故障。
The expected total resistance is 300 Ω, giving a current I = 9 V / 300 Ω = 0.03 A. The voltage across R₂ would be V₂ = I × 200 Ω = 6 V, which matches the lamp rating. The fact that the lamp does not light suggests either an open‑circuit or a short‑circuit condition.
预期总电阻300 Ω，电流 I = 9 V / 300 Ω = 0.03 A。R₂ 两端电压 V₂ = 0.03 A × 200 Ω = 6 V，恰好符合灯泡额定值。不亮说明存在断路或短路故障。
First, measure voltage across the battery terminals — if zero, the battery is dead. Next, measure voltage across R₂: if it reads 9 V, R₂ is open‑circuit (no current flows, full voltage appears across the break). If it reads very low voltage (near 0 V), R₂ might be shorted. Testing continuity and resistance with the power off confirms the fault.
先测电池端电压——若为零，电池失效。再测 R₂ 两端电压：若显示9 V，说明 R₂ 断路（无电流，开路点承受全部电压）；若电压极低（接近0 V），R₂ 可能短路。断电后测通断和电阻值可确认故障。
4. Case 3: Gear Train Speed and Torque Calculation | 案例三：齿轮传动速度与扭矩计算
An electric motor runs at 3000 rpm and drives a machine through a simple gear train. The driver gear has 20 teeth, and the driven gear has 60 teeth. The motor delivers a torque of 2 Nm. Calculate the output speed and the ideal output torque, assuming 100% efficiency.
电动机转速3000 rpm，通过简单齿轮系驱动机器。主动轮齿数20，从动轮齿数60。电机输出扭矩2 Nm。计算输出转速和理想输出扭矩（假设效率100%）。
The speed ratio (velocity ratio) is given by the tooth ratio: VR = N₂ / N₁ = 60 / 20 = 3. Therefore, the output speed ω₂ = ω₁ / VR = 3000 rpm / 3 = 1000 rpm.
齿数比即速度比：VR = N₂ / N₁ = 60 / 20 = 3。所以输出转速 ω₂ = ω₁ / 3 = 3000 rpm / 3 = 1000 rpm。
In an ideal gear train (no energy loss), power is conserved: T₁ × ω₁ = T₂ × ω₂. Rearranging, T₂ = T₁ × (ω₁ / ω₂) = 2 Nm × (3000 / 1000) = 6 Nm. Hence, torque is multiplied by the same ratio.
理想齿轮系（无能量损失）功率守恒：T₁ × ω₁ = T₂ × ω₂。移项得 T₂ = T₁ × (ω₁ / ω₂) = 2 Nm × 3 = 6 Nm。扭矩被放大相同的倍数。
In reality, friction reduces efficiency to about 90–95%, so actual output torque would be around 5.4–5.7 Nm. This case illustrates how gear ratios are selected to match speed and torque requirements.
实际中摩擦使效率降至90-95%，实际输出扭矩约5.4–5.7 Nm。该案例展示了如何选择传动比以匹配转速和扭矩需求。
5. Case 4: Hydraulic Lift Force Calculation | 案例四：液压升降力计算
A hydraulic car lift uses Pascal’s principle. The small piston has a diameter of 20 mm, and the large lifting piston has a diameter of 150 mm. A force of 200 N is applied to the small piston. Determine the maximum load the lift can raise, assuming the pistons are at the same height.
液压汽车升降机利用帕斯卡原理。小活塞直径20 mm，大活塞直径150 mm。在小活塞上施加200 N的力。假定活塞同高，计算可举升的最大负载。
First, compute piston areas: A₁ = π × (10 mm)² = 314.2 mm²; A₂ = π × (75 mm)² = 17 671.5 mm². The pressure p is the same throughout the fluid: p = F₁ / A₁ = 200 N / 314.2 mm² = 0.6366 N/mm² (MPa).
先算面积：A₁ = π × (10 mm)² = 314.2 mm²；A₂ = π × (75 mm)² = 17 671.5 mm²。流体各处压力相等：p = F₁ / A₁ = 200 N / 314.2 mm² ≈ 0.6366 N/mm²。
The force on the large piston is F₂ = p × A₂ = 0.6366 N/mm² × 17 671.5 mm² ≈ 11 250 N. In kilograms, the lift can support approximately 11 250 N ÷ 9.81 m/s² ≈ 1147 kg (including the platform weight).
大活塞受力 F₂ = p × A₂ = 0.6366 N/mm² × 17 671.5 mm² ≈ 11 250 N。换算成质量，约能举起11 250 N ÷ 9.81 m/s² ≈ 1147 kg（含平台重量）。
This dramatic force multiplication explains why hydraulic systems are used in heavy machinery. The trade‑off is that the small piston must move a much greater distance to displace enough fluid to lift the load.
这种显著的力量放大效果解释了为何重机械普遍使用液压。代价是小活塞必须移动更长距离以排出足够流体来举升负载。
6. Case 5: Material Selection for a Bicycle Frame | 案例五：自行车车架的材料选择
You are tasked with choosing a material for a lightweight bicycle frame. Three candidates are considered: low‑alloy steel (density 7800 kg/m³, yield strength 350 MPa, modulus 210 GPa, relative cost 1), 6061 aluminium alloy (density 2700 kg/m³, yield strength 275 MPa, modulus 69 GPa, cost 3), and carbon fibre composite (density 1600 kg/m³, strength 600 MPa, modulus 130 GPa, cost 15).
要求选择轻量化自行车车架材料。候选三种：低合金钢（密度7800 kg/m³，屈服强度350 MPa，弹性模量210 GPa，成本系数1）、6061铝合金（2700 kg/m³，275 MPa，69 GPa，成本3）、碳纤维复合材料（1600 kg/m³，600 MPa，130 GPa，成本15）。
To minimise mass while maintaining strength, compare the strength‑to‑weight ratio (specific strength). For steel: 350 / 7800 ≈ 0.045 MN·m/kg; aluminium: 275 / 2700 ≈ 0.102 MN·m/kg; carbon fibre: 600 / 1600 = 0.375 MN·m/kg. Carbon fibre clearly outperforms the others.
为在保持强度的同时减轻质量，比较比强度（强度/密度）。钢：350 / 7800 ≈ 0.045 MN·m/kg；铝：275 / 2700 ≈ 0.102 MN·m/kg；碳纤维：600 / 1600 = 0.375 MN·m/kg。碳纤维明显胜出。
Stiffness is also critical; compare specific modulus (E/ρ). Steel: 210 / 7.8 ≈ 26.9 GPa/(g/cm³); aluminium: 69 / 2.7 ≈ 25.6; carbon fibre: 130 / 1.6 ≈ 81.3. Again, carbon fibre is superior. However, cost is significant: carbon fibre is 15 times more expensive than steel, limiting its use to high‑performance bikes.
刚度同样关键，比较比模量（E/ρ）。钢：210/7.8≈26.9；铝：69/2.7≈25.6；碳纤维：130/1.6≈81.3。碳纤维再次领先。但成本显著，碳纤维比钢贵15倍，因此仅限于高性能自行车。
A sensible recommendation for a mid‑range bicycle is aluminium alloy, combining moderate cost, good strength, and low density. The case highlights that material selection is always a compromise between performance and budget.
中档自行车的合理建议是铝合金，兼顾适中成本、良好强度和低密度。此案例突显选材总是性能与成本的折中。
7. Case 6: PID Control Tuning for a Drone | 案例六：无人机PID控制调节
A quadcopter drone uses a PID controller to maintain stable hovering. The response is oscillatory, with the drone bobbing up and down. The current gains are Kp = 2.0, Ki = 0.5, Kd = 0.1. Your task is to adjust the gains to reduce oscillation and achieve a steady altitude hold.
四轴无人机用PID控制器维持稳定悬停。响应振荡，无人机上下起伏。当前增益：Kp = 2.0，Ki = 0.5，Kd = 0.1。要求调节增益以减小振荡并实现稳定定高。
First, identify the problem: persistent oscillations indicate too much proportional gain. Reduce Kp to 1.2. The integral term Ki is helping to eliminate steady‑state error but may contribute to overshoot; leave it unchanged initially.
先识别问题：持续振荡说明比例增益过大。将 Kp 降至1.2。积分项 Ki 有助于消除稳态误差，但可能增加超调，先保持不变。
Increase the derivative gain Kd to 0.4 to add damping, which predicts future error and counteracts sudden changes. After tuning, the drone stabilises within ±5 cm of the set altitude, compared to the previous ±30 cm oscillation.
提高微分增益 Kd 至0.4以增加阻尼，微分项可预测未来误差并抑制突变。调节后，无人机稳定在设定高度±5 cm内，而之前振荡达±30 cm。
The final PID parameters are Kp = 1.2, Ki = 0.5, Kd = 0.4. This case demonstrates a practical tuning technique: start with a lower Kp, gradually increase Ki for zero steady error, and add Kd to dampen overshoot.
最终参数为 Kp = 1.2, Ki = 0.5, Kd = 0.4。此案例提供了实用整定方法：先降低 Kp，逐步增加 Ki 消除静差，再加入 Kd 抑制超调。
8. Case 7: Manufacturing Process Selection for a Bracket | 案例七：支架制造工艺选择
A small aluminium bracket is needed for a consumer product. Quantities range from 500 to 10 000 units per year. The options are sand casting (low tooling cost, poor surface finish, low production rate) and CNC machining (high tooling cost, excellent accuracy, fast per‑part cycle). Select the most economical process for the given volumes.
某消费品需要小铝支架，年产量500至10 000件。工艺可选砂型铸造（模具成本低，表面粗糙，生产率低）和CNC机加工（装夹成本高，精度极好，单件加工快）。为给定产量选择最经济的工艺。
For 500 units, the casting tooling cost is £500 and unit cost is £4, giving a total of £500 + 500 × £4 = £2500. CNC programming and fixture cost is £2000, with a unit cost of £2, total £2000 + 500 × £2 = £3000. Casting is cheaper by £500.
500件时，铸造模具费£500，单件成本£4，总成本 £500 + 500×£4 = £2500。CNC编程与夹具费£2000，单件£2，总成本 £2000 + 500×£2 = £3000。铸造便宜£500。
For 10 000 units, casting total = £500 + 10 000 × £4 = £40 500; CNC total = £2000 + 10 000 × £2 = £22 000. CNC machining becomes significantly cheaper at high volume due to the lower variable cost.
10 000件时，铸造总成本 = £500 + 10 000×£4 = £40 500；CNC总成本 = £2000 + 10 000×£2 = £22 000。大批量时CNC因变动成本低而显著便宜。
The break‑even quantity is found by equating total costs: 500 + 4Q = 2000 + 2Q, giving Q = 750 units. Below 750, choose casting; above, choose CNC machining. This break‑even analysis is a vital tool in manufacturing decision‑making.
盈亏平衡点：500 + 4Q = 2000 + 2Q，解得 Q = 750件。低于750件选铸造，高于则选CNC。这种盈亏平衡分析是制造决策的重要工具。
Published by TutorHao | Engineering Revision Series | aleveler.com
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