IGCSE CAIE Engineering: Interdisciplinary Integrated Question Practice | IGCSE CAIE 工程：跨学科综合题型训练

📚 IGCSE CAIE Engineering: Interdisciplinary Integrated Question Practice | IGCSE CAIE 工程：跨学科综合题型训练

Engineering on the IGCSE CAIE syllabus is inherently interdisciplinary, bridging mechanical principles, electrical theory, material science, manufacturing processes, and economic decision‑making. The most challenging examination questions require students to connect these fields – for example, calculating motor torque while considering electrical input power, or selecting materials based on both strength and cost constraints. This revision resource presents ten targeted practice scenarios, each blending multiple disciplines. By working through them, you will strengthen your ability to analyse complex engineering problems and produce well‑reasoned, integrated solutions.

IGCSE CAIE 工程学科本质上是跨学科的，融合了机械原理、电学理论、材料科学、制造工艺和经济决策。最具挑战性的考题要求学生将这些领域联系起来——例如，在计算电机扭矩时同时考虑电输入功率，或基于强度和成本双重约束选择材料。这份复习资料提供了十个有针对性的综合作业情景，每一个都融合了多学科知识。通过演练，你将增强分析复杂工程问题并给出有据可依的综合解答的能力。

1. Static Equilibrium in a Crane Arm | 起重机臂的静力平衡

A crane arm is modelled as a uniform beam 4.0 m long, pinned at its base A and supported by a steel cable attached 2.5 m from A. The cable makes an angle of 30° with the beam. A load of 2400 N hangs from the free end B. Treat the beam as massless and determine the tension T in the cable and the horizontal and vertical reaction components at the pin A.

一台起重机臂可简化为长4.0 m的均匀梁，底部在A点铰接，并由一根钢索在距A点2.5 m处牵拉，钢索与梁成30°夹角。自由端B悬挂2400 N载荷。忽略梁自重，求钢索张力T及铰点A的水平与垂直反力分量。

Taking moments about A eliminates the unknown pin reactions. The perpendicular distance from A to the cable’s line of action is 2.5 sin30° = 1.25 m. The load exerts a clockwise moment of 2400 × 4.0 = 9600 Nm. For rotational equilibrium, the anti‑clockwise moment from T must equal this:

对A点取矩可消去未知铰接力。钢索作用线到A的垂直距离为2.5 sin30° = 1.25 m。载荷产生顺时针力矩2400 × 4.0 = 9600 Nm。由转动平衡，T产生的逆时针力矩必须与之相等：

T × 1.25 = 9600 ⇒ T = 7680 N

Now resolve forces horizontally and vertically. The cable’s horizontal component is T cos30° = 7680 × 0.866 ≈ 6650 N to the left, so the pin must provide an equal rightward reaction Rₕ = 6650 N. Vertically, T sin30° = 3840 N upwards plus the upward pin reaction Rᵥ must balance the 2400 N downwards load: Rᵥ + 3840 = 2400 → Rᵥ = −1440 N, meaning it acts downwards.

现在分解水平与垂直力。钢索的水平分量为T cos30° = 7680 × 0.866 ≈ 6650 N向左，因此铰点须提供向右的水平反力Rₕ = 6650 N。垂直方向上，T sin30° = 3840 N向上，加上向上的Rᵥ须平衡2400 N向下的载荷：Rᵥ + 3840 = 2400 → Rᵥ = −1440 N，负号表示方向向下。

2. Motor‑Driven Winch: Electrical–Mechanical Integration | 电机驱动绞车：电气—机械综合

A 12 V DC motor draws 6.5 A while lifting a 500 N weight using a drum of radius 0.12 m. The motor’s efficiency under load is 68%. Calculate (a) electrical input power, (b) mechanical output power, (c) the torque delivered to the drum, and (d) the lifting speed.

一台12 V直流电机在通过半径0.12 m的卷筒提升500 N重物时输入电流为6.5 A，电机在负载下的效率为68%。计算：(a) 电输入功率，(b) 机械输出功率，(c) 作用在卷筒上的扭矩，(d) 提升速度。

(a) Input power P_in = V × I = 12 × 6.5 = 78 W. (b) Output power P_out = 0.68 × 78 ≈ 53.0 W. (c) The torque required to raise the load equals the tension (weight) times drum radius: T = 500 N × 0.12 m = 60 Nm. (d) The output power also equals torque × angular velocity ω, so ω = P_out / T = 53.0 / 60 ≈ 0.883 rad s⁻¹. The linear speed v = ωr = 0.883 × 0.12 ≈ 0.106 m s⁻¹.

(a) 输入功率P_in = V × I = 12 × 6.5 = 78 W。(b) 输出功率P_out = 0.68 × 78 ≈ 53.0 W。(c) 提升载荷所需扭矩等于张力（重物重量）乘以卷筒半径：T = 500 N × 0.12 m = 60 Nm。(d) 输出功率也等于扭矩×角速度ω，因此ω = P_out / T = 53.0 / 60 ≈ 0.883 rad s⁻¹，线速度v = ωr = 0.883 × 0.12 ≈ 0.106 m s⁻¹。

3. Material Selection for a Lightweight Cantilever Bracket | 轻型悬臂支架的材料选择

An engineer must choose a material for a bracket that carries a static end‑load of 2000 N at a distance of 0.8 m from the support. The allowable maximum bending stress is 90 MPa for all candidates. A preliminary design gives the required section modulus Z = M/σ = (2000×0.8)/(90×10⁶) = 1.78×10⁻⁵ m³. The table below summarises candidate materials. Determine the lightest option that meets the strength requirement and discuss additional constraints.

工程师需要为支架选材，支架在距支撑0.8 m处承受2000 N的静态端部载荷。所有候选材料的许用弯曲应力均为90 MPa。初步设计给出的截面模量Z = M/σ = (2000×0.8)/(90×10⁶) = 1.78×10⁻⁵ m³。下表概括了候选材料。确定满足强度要求的最轻选项，并讨论其他约束条件。

The required volume V = Z × (shape factor, assumed constant), so mass m = ρV is proportional to ρ for a fixed Z. Ranking by density (mass) gives GFRP (1.85 relative), aluminium (2.70), steel (7.85). The lightest is GFRP, but cost per kg is highest. A full engineering decision must also consider corrosion resistance, manufacturability, and total lifecycle cost – a cross‑disciplinary trade‑off.

所需体积V = Z × (形状系数，假设相同)，所以质量m = ρV在固定Z下正比于密度。按密度（质量）排序为：GFRP (1.85)、铝合金 (2.70)、钢 (7.85)。最轻的是GFRP，但其每千克成本最高。完整的工程决策还需考虑耐腐蚀性、可制造性以及全生命周期成本——这是一个跨学科的权衡过程。

4. Energy‑Chain Analysis: Solar‑Powered Water Pump | 能量链分析：太阳能水泵

Photovoltaic panels provide an output of 55 W under operating conditions. This drives a pump that lifts water through a height of 12 m. The required flow rate is 0.8 L s⁻¹. Take the density of water as 1000 kg m⁻³ and gravitational field strength as 10 N kg⁻¹. (a) Calculate the hydraulic power needed. (b) Determine the overall system efficiency from solar input to useful hydraulic output. (c) Suggest why the actual efficiency can never reach 100%.

光伏板在工作条件下提供55 W输出，驱动水泵将水提升12 m。要求流量为0.8 L s⁻¹。取水的密度为1000 kg m⁻³，重力场强度为10 N kg⁻¹。(a) 计算所需的水力功率。(b) 求从太阳能输入到有用水力输出的系统总效率。(c) 说明实际效率为何不可能达到100%。

(a) Convert flow rate to m³ s⁻¹: 0.8 L s⁻¹ = 0.0008 m³ s⁻¹. Hydraulic power P_hyd = ρgQH = 1000 × 10 × 0.0008 × 12 = 96 W. (b) Since the available solar electric power is only 55 W, the pump cannot meet the demand under these conditions – the system is undersized. If we hypothetically use a larger panel delivering the required 96 W input, efficiency would be 96/96 = 100%, but with real panels the output varies. In reality the pump would deliver a lower flow rate; the efficiency based on 55 W input would be (P_hyd / 55) × 100, but P_hyd is the actual hydraulic power produced, not the demanded 96 W. If the pump actually lifts 0.45 L s⁻¹, P_hyd = 1000×10×0.00045×12 = 54 W, giving an efficiency of 54/55 ≈ 98%. (c) Losses occur in the panel (heat), motor (friction, I²R), pump (fluid friction), and piping (turbulence). Every energy conversion step introduces irreversibilities.

(a) 流量换算为0.8 L s⁻¹ = 0.0008 m³ s⁻¹。水力功率P_hyd = ρgQH = 1000 × 10 × 0.0008 × 12 = 96 W。(b) 可用的太阳能电力仅为55 W，因此该泵无法满足需求——系统容量不足。若假想使用更大面板提供96 W输入，效率为96/96 = 100%，但实际面板输出多变。实际中泵将输出更低的流量；基于55 W输入计算的效率需采用实际水力功率。若泵实际提升0.45 L s⁻¹，P_hyd = 1000×10×0.00045×12 = 54 W，效率约为54/55 ≈ 98%。(c) 能量损失发生在电池板（发热）、电机（摩擦、I²R）、泵（流体摩擦）及管道（湍流）中。每一次能量转换都会引入不可逆损失。

5. Design Problem: Access Ramp for Wheelchair Users | 设计问题：轮椅使用者坡道

You are asked to specify a permanent access ramp to a community centre. List four functional requirements and four constraints that must guide the design. Provide a short justification for each constraint, linking to materials, ergonomics, or cost.

你需要为一所社区中心设计一条永久性无障碍坡道。列出四项功能要求和四项约束条件，并为每项约束简要说明理由，关联到材料、人机工程或成本。

Functional requirements: (1) Allow safe wheelchair passage up a vertical rise of 0.9 m; (2) Provide a slip‑resistant surface

Published by TutorHao | IGCSE 工程 Revision Series | aleveler.com

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