Pre-U Edexcel Engineering: Unit Test Mock Paper Walkthrough | Pre-U Edexcel 工程：单元测试模拟卷解析

📚 Pre-U Edexcel Engineering: Unit Test Mock Paper Walkthrough | Pre-U Edexcel 工程：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for the Pre-U Edexcel Engineering syllabus. Each section presents a typical exam-style question, followed by a step-by-step solution. Key concepts in mechanics, materials, electronics, thermodynamics, and project management are reinforced through worked examples. Use this guide to identify common pitfalls and deepen your understanding of core engineering principles.

本文为 Pre-U Edexcel 工程课程单元测试模拟卷提供详细解析。每个小节呈现一道典型试题，并配有逐步求解过程。通过例题，巩固力学、材料、电子学、热力学及项目管理等关键概念。借助本指南，识别常见错误，加深对核心工程原理的理解。

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1. Forces and Equilibrium | 力与平衡

Question: A simply supported beam of length 4 m carries a point load of 600 N at its midpoint and a uniformly distributed load (UDL) of 200 N/m across the full span. Calculate the vertical reaction forces at the two supports.

题目：一根长 4 m 的简支梁，中点承受 600 N 集中荷载，同时整跨承受 200 N/m 的均布荷载 (UDL)。求两支座的垂直反力。

Step 1: Replace the UDL with an equivalent point load. Total UDL force = 200 N/m x 4 m = 800 N, acting at the midpoint. Together with the point load, the total downward force is 800 + 600 = 1400 N at the centre.

步骤1：将均布荷载等效为一个集中力。总均布力 = 200 N/m x 4 m = 800 N，作用在中点。加上集中荷载，中点总向下力为 800 + 600 = 1400 N。

Step 2: Apply equilibrium conditions. Sum of vertical forces: R_A + R_B – 1400 = 0. Take moments about support A: R_B x 4 – 1400 x 2 = 0. Solve to get R_B = 700 N, then R_A = 700 N.

步骤2：应用平衡条件。竖直力之和：R_A + R_B – 1400 = 0。对 A 点取矩：R_B x 4 – 1400 x 2 = 0。解得 R_B = 700 N，从而 R_A = 700 N。

ΣF_y = 0, ΣM_A = 0 → R_A = R_B = 700 N

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2. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量

Question: A steel rod of diameter 10 mm and original length 2 m elongates by 1.2 mm under a tensile load of 15 kN. Determine the stress, strain, and Young’s modulus of the material.

题目：一根直径 10 mm、原长 2 m 的钢杆，在 15 kN 拉伸荷载下伸长了 1.2 mm。试求材料的应力、应变和杨氏模量。

Step 1: Compute cross-sectional area: A = πd²/4 = π x (0.01 m)² / 4 = 7.854 x 10⁻⁵ m². Stress σ = F / A = 15000 N / 7.854e-5 m² ≈ 1.91 x 10⁸ Pa = 191 MPa.

步骤1：计算截面积：A = πd²/4 = π x (0.01 m)² / 4 = 7.854 x 10⁻⁵ m²。应力 σ = F / A = 15000 N / 7.854e-5 m² ≈ 1.91 x 10⁸ Pa = 191 MPa。

Step 2: Strain ε = ΔL / L₀ = 1.2 mm / 2000 mm = 0.0006 (or 6 x 10⁻⁴).

步骤2：应变 ε = ΔL / L₀ = 1.2 mm / 2000 mm = 0.0006 (即 6 x 10⁻⁴)。

Step 3: Young’s modulus E = σ / ε = 191e6 Pa / 6e-4 ≈ 3.18 x 10¹¹ Pa = 318 GPa. This is close to the typical value for steel.

步骤3：杨氏模量 E = σ / ε = 191e6 Pa / 6e-4 ≈ 3.18 x 10¹¹ Pa = 318 GPa。该值接近钢材的典型数据。

E = (F/A) / (ΔL/L₀) = 318 GPa

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3. Material Selection by Specific Strength | 按比强度选材

Question: Compare aluminium alloy (density 2700 kg/m³, yield strength 280 MPa) with low-carbon steel (density 7850 kg/m³, yield strength 250 MPa). Which material has a higher specific strength, and why is this important for aerospace components?

题目：比较铝合金（密度 2700 kg/m³，屈服强度 280 MPa）与低碳钢（密度 7850 kg/m³，屈服强度 250 MPa）。哪种材料的比强度更高？这对航空部件为何重要？

Step 1: Specific strength = yield strength / density. For Al alloy: 280e6 / 2700 ≈ 103,700 N·m/kg. For steel: 250e6 / 7850 ≈ 31,800 N·m/kg.

步骤1：比强度 = 屈服强度 / 密度。铝合金：280e6 / 2700 ≈ 103,700 N·m/kg。钢：250e6 / 7850 ≈ 31,800 N·m/kg。

Aluminium alloy has a significantly higher specific strength, providing greater strength per unit mass. In aerospace, lightweight structures directly improve fuel efficiency and payload capacity, making aluminium alloys a preferred choice for airframes.

铝合金的比强度高得多，能够以更小的质量提供相同强度。在航空领域，轻质结构直接提升燃油效率与有效载荷，因此铝合金成为机身框架的首选材料。

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4. DC Circuit Analysis | 直流电路分析

Question: Three resistors R₁=10 Ω, R₂=20 Ω, R₃=30 Ω are arranged with R₁ and R₂ in parallel, and that combination in series with R₃. A 12 V battery is connected. Find the total current drawn and the power dissipated in R₃.

题目：三个电阻 R₁=10 Ω, R₂=20 Ω, R₃=30 Ω，其中 R₁ 与 R₂ 并联，再与 R₃ 串联，接至 12 V 电池。求总电流和 R₃ 上的功耗。

Step 1: Equivalent resistance of parallel branch: 1/R_p = 1/10 + 1/20 = 3/20 → R_p = 20/3 ≈ 6.67 Ω. Total R_total = R_p + R₃ = 6.67 + 30 = 36.67 Ω.

步骤1：并联支路等效电阻：1/R_p = 1/10 + 1/20 = 3/20 → R_p = 20/3 ≈ 6.67 Ω。总电阻 R_total = 6.67 + 30 = 36.67 Ω。

Step 2: Total current I_total = V / R_total = 12 V / 36.67 Ω ≈ 0.327 A. Power in R₃: P₃ = I²R = (0.327 A)² x 30 Ω ≈ 3.21 W.

步骤2：总电流 I_total = 12 V / 36.67 Ω ≈ 0.327 A。R₃ 功耗 P₃ = I²R = (0.327 A)² x 30 Ω ≈ 3.21 W。

I = 12 / (30 + (10 x 20)/(10+20)) = 0.327 A

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5. Shear Force and Bending Moment | 剪力与弯矩

Question: A cantilever beam of length 3 m is fixed at one end and free at the other. It carries a concentrated load of 500 N at the free end and a UDL of 200 N/m over its entire length. Draw the shear force and bending moment diagrams, indicating key values.

题目：一根长 3 m 的悬臂梁，一端固定，一端自由。自由端受 500 N 集中力，全梁承受 200 N/m 均布荷载。绘出剪力与弯矩图，并标出关键值。

Step: The shear force at any section x from the free end: SF = 500 + 200x. At fixed end (x=3 m), SF = 500+600=1100 N. Bending moment: M = -[500x + 200x(x/2)] = -(500x + 100x²). At fixed end, M = -(1500+900) = -2400 N·m.

解析：距自由端 x 处的剪力：SF = 500 + 200x。固定端处 (x=3 m) SF = 1100 N。弯矩：M = -[500x + 200x(x/2)] = -(500x + 100x²)。固定端处 M = -2400 N·m。剪力图自 500 N 线性增至 1100 N，弯矩图呈下凹抛物线，最大值在固定端。

SF(x) = 500 + 200x, M(x) = -(500x + 100x²)

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6. Carnot Efficiency and Heat Engine | 卡诺效率与热机

Question: A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. What is the maximum theoretical (Carnot) efficiency? If the actual engine achieves 70% of this ideal efficiency and receives 2 kW of heat, determine the mechanical power output.

题目：一热机在 500 K 热源与 300 K 冷源间工作。最大理论（卡诺）效率是多少？若实际效率为理想值的 70%，且吸热功率为 2 kW，求机械输出功率。

Carnot efficiency: η_carnot = 1 – T_c / T_h = 1 – 300/500 = 0.40 or 40%. Actual efficiency η = 0.70 x 0.40 = 0.28 (28%). Power output = η x Q_in = 0.28 x 2000 W = 560 W.

卡诺效率 η_carnot = 1 – T_c / T_h = 1 – 300/500 = 0.40 (40%)。实际效率 = 0.70 x 0.40 = 0.28 (28%)。输出功率 = 0.28 x 2000 W = 560 W。

η_max = 1 – 300/500 = 0.40, P_out = 560 W

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7. Control System Transfer Function | 控制系统传递函数

Question: An open-loop system has a transfer function G(s) = K / [s(s+2)]. It is placed in a unity feedback loop. Derive the closed-loop transfer function T(s). For K=8, determine the steady-state error to a unit step input.

题目：开环传递函数为 G(s) = K / [s(s+2)]，接入单位负反馈。推导闭环传递函数 T(s)。当 K=8 时，求单位阶跃输入的稳态误差。

Step: T(s) = G(s) / [1 + G(s)] = [K/(s(s+2))] / [1 + K/(s(s+2))] = K / [s² + 2s + K]. For K=8, T(s) = 8/(s²+2s+8). Steady-state error for step input: ess = 1/(1+K_p), where K_p = lim_s→0 G(s) = ∞ (type 1 system). Thus ess = 0.

步骤：T(s) = G/(1+G) = K / [s² + 2s + K]。K=8 时，T(s) = 8/(s²+2s+8)。该系统为 I 型系统，对阶跃输入的稳态误差为 0，因为位置误差常数 K_p 无穷大。

T(s) = K / (s² + 2s + K), e_ss = 0 for step

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8. Dynamics on an Inclined Plane | 斜面动力学

Question: A 10 kg block slides down a 30° incline. The coefficient of kinetic friction is 0.3. Determine the acceleration of the block. (g = 9.81 m/s²)

题目：一个 10 kg 滑块沿 30° 斜面下滑，动摩擦系数为 0.3。求加速度 (g = 9.81 m/s²)。

Downward component of weight: mg sin30° = 10 x 9.81 x 0.5 = 49.05 N. Normal force N = mg cos30° = 10 x 9.81 x 0.866 ≈ 84.96 N. Friction f = μN

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