Year 13 Cambridge Engineering: Unit Test Mock Paper Analysis | 剑桥工程 Year 13 单元测试模拟卷解析

📚 Year 13 Cambridge Engineering: Unit Test Mock Paper Analysis | 剑桥工程 Year 13 单元测试模拟卷解析

This article provides a detailed analysis of a mock unit test designed for Year 13 Cambridge A-Level Engineering. The paper covers core topics including mechanics of materials, thermodynamics, fluid mechanics, control systems, and manufacturing processes. Each question is broken down with step-by-step solutions, common pitfalls, and key revision points. The aim is to help students consolidate their understanding and improve exam technique.

本文详细解析了一份为剑桥 A-Level 工程学 Year 13 设计的单元测试模拟卷。试卷涵盖材料力学、热力学、流体力学、控制系统与制造工艺等核心主题。每道题均配有逐步解析、常见易错点以及关键复习要点，旨在帮助学生巩固理解并提升应试技巧。

1. Stress and Strain in Composite Bars | 复合材料杆的应力与应变

Question 1: A mild steel rod of diameter 16 mm and length 1.2 m is surrounded by a copper tube of external diameter 24 mm and the same length. The compound bar is subjected to an axial tensile load of 40 kN. Given E_steel = 210 GPa and E_copper = 120 GPa, calculate the stress in each material.

题目1：一根直径为 16 mm、长度为 1.2 m 的低碳钢棒，其外部套有一根外径为 24 mm、长度相同的铜管。复合材料杆承受 40 kN 的轴向拉伸载荷。已知钢的弹性模量 E_steel = 210 GPa，铜的 E_copper = 120 GPa，试计算每种材料中的应力。

For a compound bar under an axial load, the strain in both materials is equal because they are perfectly bonded and have the same length. Let ε be the common strain. The total load is shared: F_total = (A_s × E_s + A_c × E_c) × ε. First, compute areas: A_s = π(16/2)² = π × 8² = 201.06 mm². A_c = π(12² – 8²) = π(144 – 64) = π × 80 = 251.33 mm². Then solving for ε: ε = F / (A_s E_s + A_c E_c ⁹)] = 40 000 / (42 222.6 + 30 159.6) = 40 000 / 72 382.2 = 5.526×10⁻⁴. Stress in steel = E_s ε = 210×10⁹ × 5.526×10⁻⁴ = 116.0 MPa. Stress in copper = 120×10⁹ × 5.526×10⁻⁴ = 66.3 MPa.

复合材料杆承受轴向载荷时，由于两种材料完全粘合并具有相同长度，其应变相等。设 ε 为公共应变。总载荷由两部分分担：F_total = (A_s × E_s + A_c × E_c) × ε。首先计算截面积：A_s = π(16/2)² = π × 8² = 201.06 mm²；A_c = π(12² – 8²) = π(144 – 64) = π × 80 = 251.33 mm²。然后求解 ε：ε = F / (A_s E_s + A_c E_c [(201.06×10⁻⁶ m² × 210×10⁹ Pa) + (251.33×10⁻⁶ × 120×10⁹)] = 40 000 / 72 382.2 = 5.526×10⁻⁴。钢的应力 = 116.0 MPa；铜的应力 = 66.3 MPa。

A common mistake is to assume the loads share equally or to forget converting diameters to radii. Always check compatibility of deformation for parallel bonded elements.

常见错误是假设载荷平均分配，或忘记将直径转换为半径。对于并联粘合元件，务必校核变形协调条件。

---

2. Beam Deflection by Macaulay’s Method | 用麦考利法计算梁的挠度

d-span. Using Macaulay’s method, derive the expression for deflection and find the maximum deflection. Take EI = 4×10⁶ N·m².

题目2：一根跨度为 3 m 的简支梁，全长承受 5 kN/m 的均布载荷，并在跨中作用有 10 kN 的集中力。使用麦考利法推导挠度表达式，并求最大挠度。取 EI = 4×10⁶ N·m²。

ach (5×3 + 10)/2 = 12.5 kN. The bending moment equation with Macaulay’s notation: M(x) = R_A x – w x²/2 – W⟨x – a⟩, where a = 1.5 m. Here R_A = 12.5 kN, w = 5 kN/m, W = 10 kN. Thus, EI d²y/dx² = 12.5x – 5x²/2 – 10⟨x – 1.5⟩. Integrate once for slope: EI dy/dx = 6.25x² – (5/6)x³ – 5⟨x – 1.5⟩² + C₁. Integrate again for deflection: EI y = (6.25/3)x³ – (5/24)x⁴ – (5/3)⟨x – 1.5⟩³ + C₁x + C₂. Apply boundary conditions: at x = 0, y = 0 ⇒ C₂ = 0. At x = 3, y = 0 ⇒ plug x=3, note ⟨3 – 1.5⟩ = 1.5 so ⟨1.5⟩ on at x = 1.5 m (by symmetry). Calculate numerical value. The result: maximum deflection ≈ –5.48 mm (downward).

由对称性得支座反力均为 (5×3 + 10)/2 = 12.5 kN。用麦考利符号写弯矩方程：M(x) = R_A x – w x²/2 – W⟨x – a⟩，其中 a = 1.5 m。于是 EI d²y/dx² = 12.5x – 5x²/2 – 10⟨x – 1.5⟩。积分得转角方程：EI dy/dx = 6.25x² – (5/6)x³ – 5⟨x – 1.5⟩² + C₁。再积分得挠度方程：EI y = (6.25/3)x³ – (5/24)x⁴ – (5/3)⟨x – 1.5⟩³ + C₁x + C₂。利用边界条件：x=0 时 y=0 ⇒ C₂=0；x=3 时 y=0，代入计算可得 C₁，再由对称性知最大挠度位于 x=1.5 m 处，解得最大挠度约为 –5.48 mm（向下）。

The Macaulay bracket ensures that terms are ignored when the argument inside the bracket is negative. This method elegantly handles discontinuities in loading.

麦考利括号确保当括号内为负时忽略该项，这种方法能优雅地处理载荷不连续的情况。

---

3. Thermodynamic Efficiency of a Gas Turbine Cycle | 燃气轮机循环的热力学效率

Question 3: An ideal Brayton cycle operates with a pressure ratio of 12:1. The inlet temperature to the compressor is 300 K, and the turbine inlet temperature is 1400 K. Assuming constant specific heats (c_p kJ/kg·K, γ = 1.4), calculate the thermal efficiency and the net work output per kg of air.

题目3：一理想布雷顿循环的压力比为 12:1，压气机进口温度为 300 K，涡轮进口温度为 1400 K。设比热为定值（c_p = 1.005 kJ/kg·K，γ = 1.4），计算热效率和每千克空气的净功输出。

For ideal Brayton cycle: T₂ = T₁ (r_p)^((γ-1)/γ) = 300 × (12)⁰.2857 = 300 × 2.052 = 615.6 K. T₄ = T₃ / (r_p)^((γ-1)/γ) = 1400 / 2.052 = 682.3 K. Compressor work w_c = c_p (T₂ – T₁) = 1.005 × (615.6 – 300) = 317.2 kJ/kg. Turbine work w_t = c_p (T₃ – T₄) = 1.005 × (1400 – 682.3) = 721.3 kJ/kg. Net work w_net = w_t – w_c = 404.1 kJ/kg. Heat added q_in = c_p (T₃ – T₂) = 1.005 × (1400 – 615.6) = 788.3 kJ/kg. Efficiency η = w_net / q_in = 404.1 / 788.3 = 0.5126 or 51.26%.

理想布雷顿循环：T₂ = T₁ (r_p)^((γ-1)/γ) = 300 × 12⁰.2857 = 615.6 K；T₄ = T₃ / (r_p)^((γ-1)/γ) = 1400 / 2.052 = 682.3 K。压气机耗功 w_c = c_p (T₂ – T₁) = 1.005 × (615.6 – 300) = 317.2 kJ/kg；涡轮做功 w_t = c_p (T₃ – T₄) = 1.005 × (1400 – 682.3) = 721.3 kJ/kg。净功 w_net = w_t – w_c = 404.1 kJ/kg；加热量 q_in = c_p (T₃ – T₂) = 788.3 kJ/kg。效率 η = 404.1/788.3 ≈ 51.26%。

The back-work ratio (w_c / w_t) is about 44%, which is typical for gas turbine cycles. Note that the efficiency depends only on pressure ratio for an ideal cycle.

压气机耗功与涡轮做功之比约为 44%，这是燃气轮机循环的典型特征。注意，理想循环的效率仅取决于压力比。

---

4. Pipe Flow and Darcy–Weisbach Friction Loss | 管流与达西-魏斯巴赫沿程损失

diameter 150 mm and length 500 m. Determine the head loss due to friction using the Darcy–Weisbach equation. Assume the friction factor f = 0.025 based on the Moody chart.

题目4：水以 0.02 m³/s 的流量流经一直径 150 mm、长 500 m 的水平铸铁管。利用达西-魏斯巴赫公式求沿程摩擦水头损失。取穆迪图摩擦系数 f = 0.025。

Velocity V = Q / A = 0.02 / (π × 0.075²) = 0.02 / 0.01767 = 1.132 m/s. Darcy–Weisbach: h_f = f (L/D) (V² / (2g)). So h_f .62) = 0.025 × 3333.3 × 0.0653 = 5.44 m. Thus the friction head loss is about 5.44 metres.

流速 V = Q / A = 0.02 / (π × 0.075²) = 1.132 m/s。达西-魏斯巴赫公式：h_f = f (L/D) (V²/(2g))。代入得 h_f = 0.025 × (500/0.15) × (1.132²/(2×9.81)) = 0.025 × 3333.3 × 0.0653 ≈ 5.44 m。因此沿程水头损失约为 5.44 米。

In a real pipeline system, minor losses from bends and valves would also need to be added. The friction factor can be verified by calculating the Reynolds number and relative roughness.

在实际管道系统中，还需考虑弯头和阀门引起的局部损失。摩擦系数可通过计算雷诺数和相对粗糙度来验证。

---

5. PID Controller Tuning Using Ziegler–Nichols | 用 Ziegler–Nichols 法整定 PID 控制器

L = 2 s and a time constant T = 10 s, with a steady-state gain K = 1.5. Using the Ziegler–Nichols open-loop method, determine the parameters for a PI controller.

题目5：某热力系统的开环阶跃响应测试得出反应曲线：死区时间 L = 2 s，时间常数 T = 10 s，稳态增益 K = 1.5。用 Ziegler–Nichols 开环法求 PI 控制器的参数。

Ziegler–Nichols open-loop tuning for PI: K_c (T / (K L)) = 0.9 × (10 / (1.5 × 2)) = 0.9 × (10 / 3) = 0.9 × 3.333 = 3.0. Integral time T_i = 3.33 L = 3.33 × 2 = 6.66 s. So the PI settings are: proportional gain K_c = 3.0, integral time T_i = 6.66 s (or integral gain K_i = K_c / T_i = 0.45 s⁻¹).

Ziegler–Nichols 开环整定 PI 控制器：K_c = 0.9 (T/(K L)) = 0.9 × (10/(1.5×2)) = 0.9 × 3.333 = 3.0；积分时间 T_i = 3.33 L = 3.33 × 2 = 6.66 s。因此 PI 参数为：比例增益 K_c = 3.0，积分时间 T_i = 6.66 s（或积分增益 K_i = 0.45 s⁻¹）。

These values serve as a starting point; fine-tuning may be required to achieve desired stability margins. Note that open-loop methods assume the process can be approximated by a first-order plus dead time model.

这些参数可作为初始设定，可能需要进一步细调以满足期望的稳定裕度。开环法假设过程可用一阶加纯滞后模型近似。

---

广告 / Sponsored

6. Critical Load for a Steel Column (Euler Buckling) | 钢柱的临界载荷（欧拉屈曲）

length of 2.5 m. Determine the critical buckling load using Euler’s formula. E = 210 GPa.

题目6：两端铰支的实心圆截面钢柱，直径 40 mm，有效长度 2.5 m。用欧拉公式计算临界屈曲载荷。E = 210 GPa。

For pin-ended column, effective length factor K = 1, so L_e ond moment of area I = π d⁴ / 64 = π × 0.04⁴ / 64 = 1.257×10⁻⁷ m⁴. Slenderness ratio λ = L_e / r = 2.5 / 0.01 = 250. Critical load P_cr = π² E I / L_e 210×10⁹ × 1.257×10⁻⁷) / 2.5² = (9.8696 × 210×10⁹ × 1.257×10⁻⁷) / 6.25 = (260.7×10²) / 6.25? Let' pute: E I = 210e9 × 1.257e-7 = 26 397 N·m². π² E I ≈ 9.8696 × 26397 = 260 540 N·m². Then P_cr = 260 540 / 6.25 = 41 686 N ≈ 41.7 kN.

两端铰支柱的有效长度系数 K=1，故 L_e = 2.5 m。实心圆截面的最小回转半径 r = d/4 = 10 mm = 0.01 m。面积 A = π×0.02² = 1.257×10⁻³ m²。截面惯矩 I = π d⁴/64 = 1.257×10⁻⁷ m⁴。长细比 λ = L_e/r = 250。临界载荷 P_cr = π² E I / L_e² = (π²×210×10⁹×1.257×10⁻⁷) / 2.5² = 260 540 / 6.25 = 41 686 N ≈ 41.7 kN。

Because the slenderness ratio is high (250), Euler buckling governs; the stress at buckling is well below the yield stress. A larger diameter would increase the load capacity dramatically.

由于长细比较大（250），欧拉屈曲起控制作用；屈曲时的应力远低于屈服应力。增大直径可显著提高承载能力。

---

7. AC Circuit Power and Power Factor Correction | 交流电路功率与功率因数校正

Question 7: An industrial load draws 50 kW at a lagging power factor of 0.65 from a 415 V, 50 Hz three-phase supply. Calculate the capacitor rating required to improve the power factor to 0.95 lagging.

题目7：某工业负载从 415 V、50 Hz 的三相电源吸收 50 kW 的有功功率，滞后功率因数为 0.65。计算将功率因数提高到 0.95（滞后）所需的电容器容量。

The required reactive power compensation Q_c = P (tan φ₁ – tan φ₂). cos φ₁ = 0.65 ⇒ φ₁ = arccos 0.65 = 49.46°, tan φ₁ = 1.169. cos φ₂ = 0.95 ⇒ φ₂ = arccos 0.95 = 18.19°, tan φ₂ = 0.3287. Thus Q_c = 50 (1.169 – 0.3287) = 50 × 0.8403 = 42.0 kVAR. Since it' AR. The capacitance per phase depends on delta or star connection. For delta connection, Q_c/phase = 14 kVAR, and C = Q_c / (3 × V_line² × ω). With V_line (3 × 172 225 × 314.16) = 42 000 / 162 230 000 = 2.59×10⁻⁴ F = 259 μF per phase (if delta).

需要补偿的无功功率 Q_c = P (tan φ₁ – tan φ₂)。功率因数 0.65 对应 φ₁ = 49.46°，tan φ₁ = 1.169；功率因数 0.95 对应 φ₂ = 18.19°，tan φ₂ = 0.3287。因此 Q_c = 50 × (1.169 – 0.3287) = 42.0 kVAR。三相系统中电容器组总容量为 42 kVAR。若采用三角形接法，每相电容 C = 42 000 / (3 × 415² × 314.16) ≈ 259 μF。

Improving power factor reduces line current, I²R losses, and utility penalties. Capacitor banks are typically installed in parallel close to inductive loads.

提高功率因数可降低线路电流、减少 I²R 损耗并避免电费罚款。电容器组通常并联安装在感性负载附近。

---

8. Cam Design: Displacement, Velocity and Acceleration | 凸轮设计：位移、速度与加速度

Question 8: A disc cam rotating at 300 rpm operates a rocker follower with a rise of 20 mm over 90° of cam rotation, followed by a dwell of 60°. The rise profile is parabolic. Determine the maximum velocity and acceleration of the follower during the rise.

题目8：一盘形凸轮以 300 rpm 旋转，驱动一摇杆从动件，凸轮转过 90° 时从动件上升 20 mm，随后驻留 60°。上升段采用抛物线运动规律。求上升过程中从动件的最大速度和加速度。

ent for 0 ≤ θ ≤ β/2: s = 2h (θ/β)², where h = 20 mm. Maximum velocity occurs at θ = β/2: v_max = 2hω/β = 2 × 0.02 m × 31.42 / 1.5708 = 0.8 m/s? Let' 568/1.5708 = 0.8 m/s. Maximum acceleration occurs at boundaries (start and end of rise): a_max ²/β² = 4 × 0.02 × (31.42)² / (1.5708)² = 0.08 × 987.2 / 2.467 = 78.98 / 2.467 = 32.0 m/s².

凸轮角速度 ω = 300×2π/60 = 31.42 rad/s。升程角 β = 90° = π/2 = 1.5708 rad。抛物线运动规律，在 0 ≤ θ ≤ β/2 区间位移 s = 2h (θ/β)²，其中 h = 20 mm。最大速度出现在 θ = β/2 处：v_max = 2hω/β = 2×0.02×31.42/1.5708 = 0.8 m/s。最大加速度出现在升程始末：a_max = 4h ω²/β² = 4×0.02×(31.42)²/(1.5708)² = 32.0 m/s²。

Parabolic profiles are used when smooth acceleration characteristics are required. Note that the acceleration is finite but changes abruptly, which may cause mild vibration at high speeds.

抛物线运动规律适用于要求加速度平滑的场合。注意，加速度有限但存在突变，高速时可能引起轻微振动。

---

9. Spur Gear Strength: Lewis Formula | 直齿圆柱齿轮强度：路易斯公式

is 120 MPa. Using the Lewis equation, check the design and suggest a suitable module if necessary. Take the Lewis form factor Y = 0.32 for 20 teeth.

题目9：一个 20 齿、模数 4 mm 的直齿圆柱齿轮，在 1200 rpm 下传递 5 kW 功率。齿宽为 30 mm，材料许用弯曲应力为 120 MPa。用路易斯公式校核设计，必要时推荐合适的模数。取齿形系数 Y = 0.32（20 齿）。

Tangential force F_t mm = 0.08 m. Pitch line velocity v = π d N / 60 = π × 0.08 × 1200 / 60 = 5.027 m/s. Then F_t = 5000 W / 5.027 m/s = 994.6 N. Lewis bending stress σ_b = F_t / (b × m × Y) with b in mm, m in mm, but careful: σ_b = F_t Pa < 120 MPa, the design is safe (or overly conservative). However, a smaller module could be used; but we need to check dynamic effects. For a more compact gear, module could be reduced but must satisfy strength.

圆周力 F_t = 功率/速度。分度圆直径 d = m × Z = 4×20 = 80 mm = 0.08 m。分度圆线速度 v = π d N / 60 = π×0.08×1200/60 = 5.027 m/s。于是 F_t = 5000 W / 5.027 m/s = 994.6 N。路易斯弯曲应力 σ_b = F_t / (b × m × Y)（单位 N/mm²），b = 30 mm，m = 4 mm，Y = 0.32，则 σ_b = 994.6 / (30×4×0.32) = 994.6 / 38.4 = 25.9 MPa。由于 25.9 MPa < 120 MPa，设计安全（甚至过于保守），可考虑减小模数，但需校核动载效应。

The Lewis equation assumes the full load is carried by a single tooth, which is conservative. Modern standards also incorporate velocity and dynamic factors for a more accurate assessment.

路易斯公式假设全部载荷由单个轮齿承担，较为保守。现代标准还会引入速度和动载系数以进行更精确的评估。

---

10. Manufacturing Process Selection for a Gearbox Shaft | 变速箱轴的制造工艺选择

Question 10: Select and justify a suitable manufacturing process for producing a medium-carbon steel gearbox shaft of 500 mm length, with varying diameters and keyways. Consider both primary shaping and finishing processes.

题目10：选择并论证一种制造工艺，用于生产一根长度为 500 mm、具有多段直径和键槽的中碳钢变速箱轴。同时考虑初步成形和精加工工艺。

he raw billet is heated and formed close to final shape, reducing material waste. Then rough turning on a CNC lathe rapidly removes excess stock. After heat treatment (e.g., quenching and tempering to achieve desired hardness), finish turning and grinding are applied to achieve tight tolerances (e.g., h6 tolerance) and surface finish (Ra 0.8 μm). Keyways are machined using a broaching machine after the cylindrical surfaces are finished. For mass production, CNC grinding and broaching ensure consistency. This hybrid process (forge → rough machine → heat treat → finish machine → broach) balances cost, strength, and precision.

初步成形选用锻造（如模锻），因其能使金属流线与轴的外形一致，提高疲劳强度。将坯料加热后锻压至接近最终形状，可减少材料浪费。然后采用 CNC 车床粗加工，快速去除余量。热处理（如淬火+回火）达到所需硬度后，进行精车和磨削以实现紧公差（如 h6 级）和表面粗糙度（Ra 0.8 μm）。键槽在圆柱面精加工后用拉床加工。批量生产中，CNC 磨削和拉削可保证一致性。这种复合工艺（锻造→粗加工→热处理→精加工→拉削）平衡了成本、强度与精度。

Alternative processes like casting are unsuitable because shafts require high tensile and fatigue strength. Additive manufacturing could be considered for prototyping, but current production environments favour conventional machining for cost-effectiveness.

铸造等替代工艺不适合，因轴类零件需要高抗拉和疲劳强度。增材制造可用于原型制作，但当前生产环境下，传统机加工更具成本效益。

---

Published by TutorHao | Engineering Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）