A-Level CAIE Engineering Unit Test Mock Paper Walkthrough | A-Level CAIE 工程：单元测试模拟卷解析

📚 A-Level CAIE Engineering Unit Test Mock Paper Walkthrough | A-Level CAIE 工程：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test paper designed for the CAIE A-Level Engineering syllabus. The paper focuses on core principles from mechanics of materials, thermodynamics, fluid mechanics, electrical engineering, and manufacturing. Each section presents a representative examination-style problem, a step-by-step solution using correct engineering notation, and commentary highlighting common pitfalls and key conceptual links. The aim is to sharpen problem-solving skills and deepen understanding of how theoretical knowledge is applied under timed conditions.

本文提供了一套针对 CAIE A-Level 工程教学大纲设计的单元测试模拟卷的详细解析。模拟卷聚焦于材料力学、热力学、流体力学、电气工程和制造工艺的核心原理。每一节都呈现一道具有代表性的考试题型，给出使用正确工程符号的分步解答，并附上强调常见错误和关键概念联系的评注。目的是提升解题技巧，加深对如何在限时条件下应用理论知识这一问题的理解。

1. Tensile Stress and Strain Calculation | 拉伸应力与应变计算

Problem: A solid cylindrical titanium rod of diameter 8.0 mm and original gauge length 150 mm is subjected to an axial tensile force of 12.0 kN. The rod elongates by 0.36 mm. Determine the engineering stress and engineering strain developed in the rod.

问题：一根直径为 8.0 mm、原始标距长度为 150 mm 的实心圆柱形钛棒承受 12.0 kN 的轴向拉伸力，伸长了 0.36 mm。求棒中产生的工程应力和工程应变。

Solution step 1 — cross-sectional area: The area of a circle is A = πd²/4. Substituting the diameter in metres: A = π × (0.008 m)² / 4 = 3.1416 × 6.4×10⁻⁵ / 4 = 5.0265 × 10⁻⁵ m².

解答步骤 1 — 横截面积：圆的面积公式为 A = πd²/4。代入以米为单位的直径：A = π × (0.008 m)² / 4 = 3.1416 × 6.4×10⁻⁵ / 4 = 5.0265 × 10⁻⁵ m²。

Step 2 — engineering stress: Engineering stress σ is defined as force divided by original cross-sectional area. σ = F / A₀ = 12 000 N / 5.0265×10⁻⁵ m² = 238.7 × 10⁶ Pa ≈ 239 MPa.

步骤 2 — 工程应力：工程应力 σ 定义为力除以原始横截面积。σ = F / A₀ = 12 000 N / 5.0265×10⁻⁵ m² = 238.7 × 10⁶ Pa ≈ 239 MPa。

Step 3 — engineering strain: Engineering strain ε is the change in length divided by the original gauge length. ε = ΔL / L₀ = 0.36 mm / 150 mm = 0.0024 (dimensionless; often expressed as 0.24% strain).

步骤 3 — 工程应变：工程应变 ε 是长度变化量除以原始标距长度。ε = ΔL / L₀ = 0.36 mm / 150 mm = 0.0024（量纲为 1，常表示为 0.24% 的应变）。

Common mistake: Using diameter in mm without converting to metres, or confusing engineering strain with true strain. In this elastic region the difference is negligible, but always show working in SI units to avoid numerical errors.

常见错误：使用毫米作单位且未换算成米，或将工程应变与真实应变混淆。在弹性阶段两者的差别可忽略不计，但务必始终采用国际单位制运算，以避免数值计算错误。

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2. Young’s Modulus and Material Selection | 杨氏模量与材料选择

Problem: Using the data from the previous problem (σ = 239 MPa, ε = 0.0024), calculate the Young’s modulus of the titanium alloy. Explain why a high modulus-to-density ratio is critical for aerospace components such as wing spars.

问题：利用上一题的数据（σ = 239 MPa，ε = 0.0024），计算该钛合金的杨氏模量。说明高的模量-密度比对诸如翼梁等航空航天构件为何至关重要。

Solution: Young’s modulus E is the gradient of the stress-strain curve in the linear elastic region: E = σ / ε. E = 239 × 10⁶ Pa / 0.0024 = 99.6 × 10⁹ Pa ≈ 100 GPa.

解答：杨氏模量 E 是应力-应变曲线在线弹性阶段的斜率：E = σ / ε。E = 239 × 10⁶ Pa / 0.0024 = 99.6 × 10⁹ Pa ≈ 100 GPa。

The calculated E of 100 GPa matches typical values for titanium alloys. In aerospace applications, the specific modulus (E/ρ) is a key performance index. A high value means the material provides stiffness at low weight, reducing structural mass and improving fuel efficiency. Titanium excels here compared with steel, as its density is around 4500 kg/m³ versus 7800 kg/m³ for steel, giving a superior specific modulus.

计算所得的 E = 100 GPa 与钛合金的典型值相符。在航空航天应用中，比模量（E/ρ）是一项关键性能指标。高的比模量意味着材料能以较低重量提供刚度，从而减轻结构质量并提高燃油效率。与钢相比，钛的密度约为 4500 kg/m³，而钢为 7800 kg/m³，因此钛具有更优的比模量。

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3. Factor of Safety in Design | 设计中的安全系数

Problem: The titanium rod in the previous example has a yield strength of 880 MPa. The design specification requires that the maximum operating stress does not exceed 60% of the yield stress. Calculate the factor of safety based on yield strength and comment on the adequacy of the current working stress (239 MPa).

问题：上例中的钛棒屈服强度为 880 MPa。设计规范要求最大工作应力不得超过屈服应力的 60%。计算基于屈服强度的安全系数，并评估当前工作应力（239 MPa）是否合适。

Solution: The allowable stress σ_allowable = 0.60 × σ_yield = 0.60 × 880 MPa = 528 MPa. The factor of safety (FoS) is defined as FoS = σ_yield / σ_working, or alternatively σ_allowable / σ_working if a limiting percentage is given. Here FoS = 528 MPa / 239 MPa ≈ 2.21.

解答：许用应力 σ_allowable = 0.60 × σ_yield = 0.60 × 880 MPa = 528 MPa。安全系数（FoS）可定义为 FoS = σ_yield / σ_working，或当给定了限值百分比时，也可用 σ_allowable / σ_working。此处 FoS = 528 MPa / 239 MPa ≈ 2.21。

An FoS greater than 1.0 confirms the design is safe. The working stress of 239 MPa is well below the 528 MPa limit, leaving a comfortable margin for unexpected loads, material imperfections, and fatigue over the component’s service life. In aerospace engineering, typical factors of safety range from 1.5 to 2.5, so this design is fully compliant.

安全系数大于 1.0 表明设计是安全的。239 MPa 的工作应力远低于 528 MPa 的限值，为意外载荷、材料缺陷以及部件服役期内产生的疲劳预留了充裕的裕量。在航空航天工程中，典型安全系数范围为 1.5 至 2.5，因此本设计完全符合要求。

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4. Shear Force and Bending Moment Diagrams | 剪力图与弯矩图

Problem: A simply supported beam of length 3.0 m carries a central point load of 4.0 kN. Sketch the shear force and bending moment diagrams and calculate the maximum bending moment.

问题：一根长度为 3.0 m 的简支梁在跨中承受 4.0 kN 的集中荷载。试绘制剪力图和弯矩图，并计算最大弯矩。

Solution: For a symmetrically loaded beam, the reactions at each support are R_A = R_B = 4.0 kN / 2 = 2.0 kN upwards. The shear force is constant between the support and the load point: from 0 < x < 1.5 m, V = +2.0 kN; after the load, V jumps to -2.0 kN. The bending moment increases linearly from zero at the support to a maximum at mid-span. M_max = R_A × (L/2) = 2.0 kN × 1.5 m = 3.0 kN·m.

解答：对于对称受荷梁，各支座反力为 R_A = R_B = 4.0 kN / 2 = 2.0 kN 向上。支座至荷载作用点之间的区段剪力恒定：0 < x < 1.5 m 时，V = +2.0 kN；越过荷载后，V 跃变为 -2.0 kN。弯矩从支座处的零沿线性增大至跨中最大值。M_max = R_A × (L/2) = 2.0 kN × 1.5 m = 3.0 kN·m。

Diagram interpretation: The shear force diagram shows a positive rectangle followed by a negative rectangle, with a step change of -4.0 kN at the centre. The bending moment diagram is a triangle peaking at 3.0 kN·m. In an exam, accurately marking the sign conventions and slope of the lines is essential for full marks.

图形解读：剪力图呈现一个正矩形区段后接一个负矩形区段，在跨中处出现 -4.0 kN 的阶跃变化。弯矩图为三角形，峰值为 3.0 kN·m。在考试中，准确标明符号规定和线条斜率是取得满分的关键。

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5. Beam Deflection using Standard Formulae | 运用标准公式计算梁的挠度

Problem: For the beam in Section 4, assume the beam has a second moment of area I = 8.5 × 10⁻⁶ m⁴ and is made of steel with E = 200 GPa. Using the standard deflection formula for a central point load on a simply supported beam, δ_max = (P L³) / (48 E I), calculate the maximum deflection.

问题：对于第 4 节中的梁，假设其截面二次矩为 I = 8.5 × 10⁻⁶ m⁴，材质为钢，E = 200 GPa。运用简支梁跨中受集中荷载的标准挠度公式 δ_max = (P L³) / (48 E I)，计算最大挠度。

Solution: Substitute P = 4.0 × 10³ N, L = 3.0 m, E = 200 × 10⁹ Pa, I = 8.5 × 10⁻⁶ m⁴. Numerator: P L³ = 4000 × 27 = 108,000 N·m³. Denominator: 48 × 200×10⁹ × 8.5×10⁻⁶ = 48 × 200×10⁹ × 8.5×10⁻⁶ = 48 × 1.7×10⁶ = 81.6 × 10⁶ N·m². δ_max = 108,000 / (81.6 × 10⁶) = 1.324 × 10⁻³ m = 1.32 mm.

解答：代入 P = 4.0 × 10³ N，L = 3.0 m，E = 200 × 10⁹ Pa，I = 8.5 × 10⁻⁶ m⁴。分子：P L³ = 4000 × 27 = 108,000 N·m³。分母：48 × 200×10⁹ × 8.5×10⁻⁶ = 48 × 200×10⁹ × 8.5×10⁻⁶ = 48 × 1.7×10⁶ = 81.6 × 10⁶ N·m²。δ_max = 108,000 / (81.6 × 10⁶) = 1.324 × 10⁻³ m = 1.32 mm。

This deflection is small relative to the span (span/deflection ≈ 2270), which is acceptable for most structural steelwork. However, if the beam were aluminium (E ≈ 70 GPa), deflection would roughly triple, illustrating why stiffness must be checked during material substitution.

该挠度相对于跨度而言很小（跨度/挠度 ≈ 2270），对大多数钢结构工程来说是可接受的。然而，如果梁采用铝合金（E ≈ 70 GPa），挠度将大致增加两倍，这说明了为什么在更换材料时必须验算刚度。

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6. Thermal Expansion and Clearance Fit | 热膨胀与间隙配合

Problem: A steel shaft of diameter 50.00 mm at 20 °C is to be inserted into a aluminium housing with a coefficient of linear expansion α_steel = 12 × 10⁻⁶ /°C and α_al = 23 × 10⁻⁶ /°C. If the operating temperature reaches 120 °C, what initial diametral clearance must be provided at 20 °C to avoid seizure, assuming the housing bore expands as a uniform ring?

问题：一根直径为 50.00 mm（20 °C）的钢轴需装入铝制壳体中，钢的线膨胀系数 α_steel = 12 × 10⁻⁶ /°C，铝的 α_al = 23 × 10⁻⁶ /°C。若工作温度可达 120 °C，为避免卡死，在 20 °C 时必须预留多少初始直径间隙？假设壳体孔为均匀圆环膨胀。

Solution: Temperature rise ΔT = 120 – 20 = 100 °C. Increase in shaft diameter Δd_shaft = 50.00 × 12×10⁻⁶ × 100 = 0.060 mm. Increase in housing bore diameter Δd_housing = 50.00 × 23×10⁻⁶ × 100 = 0.115 mm. To avoid seizure, the housing expansion must exceed the shaft expansion by the initial clearance c. At 120 °C, bore diameter = 50.00 + 0.115 = 50.115 mm; shaft diameter = 50.00 + 0.060 = 50.060 mm. The minimum clearance at 120 °C should be zero, so initial clearance c = 50.115 – 50.060 = 0.055 mm. In practice a small positive running clearance is added, so a design clearance of about 0.1 mm would be chosen.

解答：温升 ΔT = 120 – 20 = 100 °C。轴径增量 Δd_shaft = 50.00 × 12×10⁻⁶ × 100 = 0.060 mm。壳孔直径增量 Δd_housing = 50.00 × 23×10⁻⁶ × 100 = 0.115 mm。为避免卡死，壳体膨胀量必须大于轴膨胀量，其差值由初始间隙 c 提供。在 120 °C 时，孔径 = 50.00 + 0.115 = 50.115 mm；轴径 = 50.00 + 0.060 = 50.060 mm。120 °C 时的最小间隙应为零，故初始间隙 c = 50.115 – 50.060 = 0.055 mm。实际中需附加微小的运行间隙，因此通常会选取约 0.1 mm 的设计间隙。

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7. Hydrostatic Pressure on a Vertical Surface | 作用在竖直壁面上的流体静压力

Problem: A rectangular gate of width 2.0 m and height 1.5 m is hinged at the bottom and retains water on one side. The water depth is exactly equal to the gate height. Determine the resultant hydrostatic force acting on the gate and the depth of the centre of pressure from the water surface. Take water density ρ = 1000 kg/m³ and g = 9.81 m/s².

问题：一扇宽 2.0 m、高 1.5 m 的矩形闸门底部装有铰链，一侧挡水，水深恰好等于闸门高度。求作用在闸门上的总流体静压力合力以及压力中心距水面的深度。取水的密度 ρ = 1000 kg/m³，g = 9.81 m/s²。

Solution: Resultant hydrostatic force F = ρ g A x̄, where A is the wetted area and x̄ is the depth to the centroid of the area from the surface. For a vertical rectangle fully submerged to its top edge at surface: A = width × height = 2.0 × 1.5 = 3.0 m². The centroid is at half the height: x̄ = 1.5/2 = 0.75 m. F = 1000 × 9.81 × 3.0 × 0.75 = 22,072.5 N ≈ 22.1 kN.

解答：总流体静压力合力 F = ρ g A x̄，其中 A 为受润面积，x̄ 为面积形心距水面的深度。对于一扇完全浸没且上缘恰好位于水面的竖直矩形平面：A = 宽 × 高 = 2.0 × 1.5 = 3.0 m²。形心位于高度的一半处：x̄ = 1.5 / 2 = 0.75 m。F = 1000 × 9.81 × 3.0 × 0.75 = 22,072.5 N ≈ 22.1 kN。

Centre of pressure depth h_p = x̄ + I_g / (A x̄), where I_g is the second moment of area of the gate about its centroidal horizontal axis. For a rectangle, I_g = (width × height³) / 12 = (2.0 × 1.5³) / 12 = 0.5625 m⁴. Then h_p = 0.75 + 0.5625 / (3.0 × 0.75) = 0.75 + 0.25 = 1.0 m below the surface.

压力中心的深度 h_p = x̄ + I_g / (A x̄)，其中 I_g 为闸门面积对水平形心轴的惯性矩。对于矩形，I_g = (宽 × 高³) / 12 = (2.0 × 1.5³) / 12 = 0.5625 m⁴。因此 h_p = 0.75 + 0.5625 / (3.0 × 0.75) = 0.75 + 0.25 = 1.0 m（水面以下）。

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8. DC Circuit Analysis with Kirchhoff’s Laws | 应用基尔霍夫定律分析直流电路

Problem: A circuit consists of a 20 V battery connected to two branches in parallel. Branch 1 contains a 10 Ω resistor in series with a 5 Ω resistor. Branch 2 contains a 15 Ω resistor. Using Kirchhoff’s voltage and current laws, find the current through each resistor.

问题：一个电路由一组 20 V 电池与两条并联支路相连组成。支路 1 包含一个 10 Ω 和一个 5 Ω 电阻串联，支路 2 包含一个 15 Ω 电阻。运用基尔霍夫电压和电流定律，求流过每个电阻的电流。

Solution: First simplify the parallel network: The total resistance of branch 1, R₁ = 10 + 5 = 15 Ω. Branch 2 resistance R₂ = 15 Ω. The combined parallel resistance R_total = (R₁ × R₂) / (R₁ + R₂) = (15 × 15) / (30) = 7.5 Ω. Total current from the battery I_total = V / R_total = 20 / 7.5 = 2.667 A. This total current splits equally between the two 15 Ω branches, so I₁ = I₂ = 1.333 A each. Therefore, the current through the 15 Ω resistor is 1.333 A; within branch 1, the same 1.333 A flows through both the 10 Ω and 5 Ω resistors because they are in series.

解答：首先简化并联网络：支路 1 的总电阻 R₁ = 10 + 5 = 15 Ω。支路 2 的电阻 R₂ = 15 Ω。并联总电阻 R_total = (R₁ × R₂) / (R₁ + R₂) = (15 × 15) / (30) = 7.5 Ω。电池提供的总电流 I_total = V / R_total = 20 / 7.5 = 2.667 A。该总电流在两条 15 Ω 支路间平分，因此 I₁ = I₂ = 1.333 A。于是流过 15 Ω 电阻的电流为 1.333 A；而在支路 1 中，相同的 1.333 A 流过 10 Ω 和 5 Ω 电阻，因为它们串联。

Verification: Potential difference across parallel combination is 20 V, so across branch 1: V = I₁ × 15 Ω = 20 V, similarly for branch 2. Within branch 1, voltage across the 10 Ω resistor is 13.33 V and across the 5 Ω resistor is 6.67 V, summing to 20 V. This confirms Kirchhoff’s voltage law.

验证：并联组合两端的电位差为 20 V，因此支路 1 两端：V = I₁ × 15 Ω = 20 V，支路 2 同理。在支路 1 内部，10 Ω 电阻上的电压为 13.33 V，5 Ω 电阻上为 6.67 V，合计 20 V，符合基尔霍夫电压定律。

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9. Hardness Testing and Material Properties | 硬度试验与材料性能

Problem: In a Brinell hardness test, a 10 mm diameter hardened steel ball is pressed into a specimen with a load of 3000 kg for 15 seconds. The diameter of the indentation left is measured as 4.20 mm. Calculate the Brinell hardness number (BHN). Also explain why the test is unsuitable for very thin workpieces.

问题：在一次布氏硬度试验中，使用直径为 10 mm 的淬硬钢球，在 3000 kg 载荷下保持 15 秒压入试样。测得残留压痕直径为 4.20 mm。计算布氏硬度值（BHN），并解释该试验为何不适用于极薄的工件。

Solution: BHN is given by BHN = (2P) / [π D (D – √(D² – d²))], where P is load in kgf, D is ball diameter in mm, d is indentation diameter in mm. Here P = 3000 kgf, D = 10 mm, d = 4.20 mm. First compute √(D² – d²) = √(100 – 17.64) = √82.36 = 9.075 mm. Then D – √(D² – d²) = 10 – 9.075 = 0.925 mm. The denominator π D (0.925) = π × 10 × 0.925 = 29.06. BHN = (2 × 3000) / 29.06 = 6000 / 29.06 ≈ 206.

解答：BHN 值公式为 BHN = (2P) / [π D (D – √(D² – d²))]，其中 P 为载荷（kgf），D 为球直径（mm），d 为压痕直径（mm）。此处 P = 3000 kgf，D = 10 mm，d = 4.20 mm。先计算 √(D² – d²) = √(100 – 17.64) = √82.36 = 9.075 mm。然后 D – √(D² – d²) = 10 – 9.075 = 0.925 mm。分母 π D (0.925) = π × 10 × 0.925 = 29.06。BHN = (2 × 3000) / 29.06 = 6000 / 29.06 ≈ 206。

The test is unsuitable for thin workpieces because the indentation depth is a significant fraction of the material thickness, potentially causing deformation of the opposite surface or catastrophic failure of the specimen. Generally, the workpiece thickness should be at least 10 times the indentation depth to obtain a meaningful bulk hardness reading. For thin sections, microhardness tests like Vickers are preferred.

该试验不适用于薄工件，原因是压痕深度占材料厚度的比例很大，可能导致另一侧表面变形或试样发生灾难性破坏。一般而言，工件厚度至少应为压痕深度的 10 倍，才能获得可靠的本体硬度读数。对于薄截面，最好使用维氏等显微硬度试验。

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10. Engineering Tolerances and Fits | 工程公差与配合

Problem: A hole is specified as 25.00 H7 and a shaft is to be matched to provide a transition fit. The H7 hole limits are: lower deviation 0 μm, upper deviation +21 μm. The intended shaft has limits: upper deviation +15 μm, lower deviation -5 μm. Calculate the maximum and minimum clearances or interferences and state the type of fit achieved.

问题：一孔标注为 25.00 H7，需选配一根轴以实现过渡配合。H7 孔的极限偏差为：下偏差 0 μm，上偏差 +21 μm。拟采用的轴极限偏差为：上偏差 +15 μm，下偏差 -5 μm。计算最大和最小间隙或过盈量，并说明获得的配合类型。

Solution: Hole basic size 25.00 mm. Hole limits: maximum = 25.021 mm, minimum = 25.000 mm. Shaft limits: maximum = 25.015 mm, minimum = 24.995 mm. Maximum clearance = largest hole – smallest shaft = 25.021 – 24.995 = 0.026 mm (26 μm). Minimum clearance (or maximum interference) = smallest hole – largest shaft = 25.000 – 25.015 = -0.015 mm, i.e., an interference of 15 μm.

解答：孔的基本尺寸为 25.00 mm。孔极限尺寸：最大 = 25.021 mm，最小 = 25.000 mm。轴极限尺寸：最大 = 25.015 mm，最小 = 24.995 mm。最大间隙 = 最大孔 – 最小轴 = 25.021 – 24.995 = 0.026 mm（26 μm）。最小间隙（或最大过盈）= 最小孔 – 最大轴 = 25.000 – 25.015 = -0.015 mm，即 15 μm 的过盈。

Because the limits produce both a possible clearance and a possible interference, the fit is classified as a transition fit. In precision assembly, this type of fit allows for slight interference or clearance, enabling location while maintaining easy disassembly if needed. The selection of an H7/h6 or H7/k6 shaft would be typical for such requirements.

由于极限偏差既可能产生间隙又可能产生过盈，该配合属于过渡配合。在精密装配中，这种配合允许轻微的过盈或间隙，既能

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