Pre-U OCR Engineering: Mock Unit Test Analysis & Solutions | Pre-U OCR 工程：单元测试模拟卷解析

📚 Pre-U OCR Engineering: Mock Unit Test Analysis & Solutions | Pre-U OCR 工程：单元测试模拟卷解析

Mastering mock unit tests is key to success in Pre-U OCR Engineering. This article walks through a complete mock paper, providing detailed solutions, highlighting common pitfalls, and reinforcing core principles across statics, circuits, materials, thermodynamics, and systems.

掌握模拟单元测试是在 Pre-U OCR 工程中取得成功的关键。本文详解一套完整的模拟卷，提供详细解答，突出常见错误，并巩固静力学、电路、材料、热力学和系统等核心原理。

---

1. Force Resolution & Support Reactions | 力的分解与支座反力

A typical question presents a simply supported beam carrying a point load. You must resolve forces at an angle and calculate vertical and horizontal support reactions.

一道典型题目给出简支梁承受集中载荷，要求分解斜向力并计算竖直与水平支座反力。

Example: a 100 N force acts at 30° to the horizontal on a beam. The vertical component is 100 sin 30° = 50 N, and the horizontal component is 100 cos 30° = 86.6 N. Summing forces horizontally gives the horizontal reaction at the pin support.

示例：一个 100 N 的力以与水平方向成 30° 作用在梁上。竖直分量为 100 sin 30° = 50 N，水平分量为 100 cos 30° = 86.6 N。水平方向力求和即得到销支座的水平反力。

Common mistake: forgetting that a pin provides both vertical and horizontal reactions, while a roller provides only a reaction perpendicular to the surface. Always check your sum of moments about a point to find unknown reactions.

常见错误：忘记销支座同时提供竖直和水平反力，而滚动支座只提供垂直于支承面的反力。始终应该对某点取矩求和，以求出未知反力。

ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0

ΣFₓ = 0，ΣFᵧ = 0，ΣM = 0

---

2. Shear Force & Bending Moment Diagrams | 剪力图与弯矩图

Constructing shear force (SF) and bending moment (BM) diagrams reveals critical points in a beam. The mock test asks for these diagrams for a cantilever with a uniformly distributed load (UDL).

绘制剪力（SF）和弯矩（BM）图能揭示梁上的关键点。模拟卷要求为承受均匀分布载荷（UDL）的悬臂梁绘制这些图。

For a cantilever of length L with UDL w per unit length, the shear force at a distance x from the free end is V(x) = w x, starting at zero and increasing linearly. The bending moment is M(x) = −w x² / 2, with the maximum negative moment at the fixed support: Mₘₐₓ = −w L² / 2.

对于长为 L、承受单位长度均布载荷 w 的悬臂梁，距自由端 x 处的剪力为 V(x) = w x，从零开始线性增大。弯矩为 M(x) = −w x² / 2，最大负弯矩出现在固定端：Mₘₐₓ = −w L² / 2。

Always label axes, indicate sign conventions (sagging positive), and highlight areas where shear crosses zero—these correspond to local BM maxima. Many candidates lose marks for missing values at supports.

绘图时务必标注坐标轴、标明符号约定（下凹为正），并标出剪力为零的位置——这些位置对应局部弯矩极值。许多考生因遗漏支座处的数值而失分。

---

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

Tensile test data is a staple in OCR Engineering. The mock question provides a force-extension graph for a steel wire of original length L₀ and cross‑sectional area A₀, asking for Young’s modulus.

拉伸试验数据是 OCR 工程考试的基本内容。模拟题给出了一根原长为 L₀、截面积为 A₀ 的钢丝的力-伸长量曲线，要求计算杨氏模量。

Engineering stress σ = F / A₀ and engineering strain ε = ΔL / L₀. Young’s modulus E = σ / ε, typically taken from the linear elastic region. If at a load of 2 kN the extension is 0.25 mm, with L₀ = 100 mm and A₀ = 5 × 10⁻⁶ m², then E = (2000 / 5×10⁻⁶) / (0.25 / 100) = 1.6 × 10¹¹ Pa.

工程应力 σ = F / A₀，工程应变 ε = ΔL / L₀。杨氏模量 E = σ / ε，通常取自线弹性区域。若载荷为 2 kN 时伸长量为 0.25 mm，且 L₀ = 100 mm、A₀ = 5 × 10⁻⁶ m²，则 E = (2000 / 5×10⁻⁶) / (0.25 / 100) = 1.6 × 10¹¹ Pa。

Key pitfalls: using instantaneous area instead of original, confusing strain units, or selecting a point beyond the proportional limit. Clearly show conversions to base SI units.

关键陷阱：使用瞬时面积而非原始面积、混淆应变单位，或选取超出比例极限的数据点。必须清晰展示基本 SI 单位的换算。

σ = F / A₀, ε = ΔL / L₀, E = σ / ε

σ = F / A₀，ε = ΔL / L₀，E = σ / ε

---

4. Ohm’s Law & DC Circuit Simplification | 欧姆定律与直流电路简化

Simple resistive networks appear in every examination. You must calculate equivalent resistance, branch currents, and power dissipation correctly.

简单的电阻网络每一次考试都会出现。你必须正确计算等效电阻、支路电流和功耗。

For resistors in series: Rₛ = R₁ + R₂ + … ; in parallel: 1/Rₚ = 1/R₁ + 1/R₂ + … Always reduce the network stepwise, redrawing the simplified circuit at each stage. The mock item asks for the total resistance of a 10 Ω and 15 Ω resistor in parallel followed by a 6 Ω series resistor.

串联电阻：Rₛ = R₁ + R₂ + …；并联电阻：1/Rₚ = 1/R₁ + 1/R₂ + … 应始终逐步化简网络，每一步重新绘制简化电路。模拟题要求计算 10 Ω 与 15 Ω 并联后再串联一个 6 Ω 电阻的总阻值。

Parallel pair: Rₚ = (10×15)/(10+15) = 6 Ω. Total R = 6 + 6 = 12 Ω. Then, applying Ohm’s law V = I R, if connected to 24 V, the supply current is 24 / 12 = 2 A. Power P = I² R = 2² × 12 = 48 W.

并联等效：Rₚ = (10×15)/(10+15) = 6 Ω。总电阻 R = 6 + 6 = 12 Ω。然后，应用欧姆定律 V = I R，若接至 24 V 电源，则供电电流为 24 / 12 = 2 A。功率 P = I² R = 2² × 12 = 48 W。

A frequent error is misapplying the current divider formula. Instead, find voltage across the parallel branch first: Vₚ = I × Rₚ = 2 × 6 = 12 V. Then each branch current is Vₚ / Rᵢ.

常见错误是误用分流公式。正确做法是先求并联支路两端电压：Vₚ = I × Rₚ = 2 × 6 = 12 V，然后各支路电流为 Vₚ / Rᵢ。

---

5. Kirchhoff’s Laws & Mesh Analysis | 基尔霍夫定律与网孔分析

When circuits contain multiple voltage sources or complex loops, Kirchhoff’s voltage law (KVL) and current law (KCL) are essential. The mock exam includes a two-loop circuit requiring mesh currents.

当电路包含多个电压源或复杂回路时，基尔霍夫电压定律（KVL）和电流定律（KCL）至关重要。模拟卷包含一个需要网孔电流的两回路电路。

Assign mesh currents I₁ and I₂ in clockwise directions. For each mesh, sum voltage rises: in mesh 1, 10 = 4I₁ + 2(I₁ − I₂); in mesh 2, −5 = 6I₂ + 2(I₂ − I₁). Solve the simultaneous equations: 10 = 6I₁ − 2I₂ and −5 = 8I₂ − 2I₁.

设定顺时针方向的网孔电流 I₁ 和 I₂。对每一网孔，列出电压升之和：网孔 1，10 = 4I₁ + 2(I₁ − I₂)；网孔 2，−5 = 6I₂ + 2(I₂ − I₁)。求解联立方程：10 = 6I₁ − 2I₂ 和 −5 = 8I₂ − 2I₁。

Solution gives I₁ = 1.5 A, I₂ = 0.25 A. The current through the shared 2 Ω resistor is I₁ − I₂ = 1.25 A. Always validate by checking power balance.

解得 I₁ = 1.5 A，I₂ = 0.25 A。流过共用 2 Ω 电阻的电流为 I₁ − I₂ = 1.25 A。始终通过校核功率平衡来验证结果。

ΣVₗₒₒₚ = 0, ΣIᵢₙ = ΣIₒᵤₜ

ΣVₗₒₒₚ = 0，ΣIᵢₙ = ΣIₒᵤₜ

---

6. AC Circuit Analysis & Power Factor | 交流电路分析与功率因数

AC analysis introduces reactance, impedance, and phase angle. The mock question supplies a series R‑L circuit connected to a 50 Hz, 230 V supply, with R = 30 Ω and L = 0.1 H. Tasks: calculate inductive reactance, impedance, current, and power factor.

交流分析涉及电抗、阻抗和相位角。模拟题给出一个串联 R‑L 电路，接至 50 Hz、230 V 电源，R = 30 Ω，L = 0.1 H。要求计算感抗、阻抗、电流和功率因数。

Inductive reactance Xₗ = 2πfL = 2π × 50 × 0.1 = 31.42 Ω. Impedance Z = √(R² + Xₗ²) = √(30² + 31.42²) = 43.45 Ω. Current I = V / Z = 230 / 43.45 = 5.29 A. Power factor cos φ = R / Z = 30 / 43.45 = 0.69 lagging.

感抗 Xₗ = 2πfL = 2π × 50 × 0.1 = 31.42 Ω。阻抗 Z = √(R² + Xₗ²) = √(30² + 31.42²) = 43.45 Ω。电流 I = V / Z = 230 / 43.45 = 5.29 A。功率因数 cos φ = R / Z = 30 / 43.45 = 0.69（滞后）。

Always specify ‘leading’ or ‘lagging’. Purely inductive circuit has PF=0 lagging; capacitive circuits cause leading PF. Multimeter readings in AC are RMS values, so V and I are RMS unless stated otherwise.

务必注明“超前”或“滞后”。纯电感电路功率因数为 0（滞后）；电容电路引起超前功率因数。交流万用表读数为有效值，因此除非另有说明，电压和电流均为有效值。

Xₗ = 2πfL, Z = √(R² + X²), cos φ = R / Z

Xₗ = 2πfL，Z = √(R² + X²)，cos φ = R / Z

---

7. Electromagnetic Systems & Torque | 电磁系统与转矩

Motor and generator principles feature regularly. A typical mock problem describes a single rectangular coil in a magnetic field, calculating generated EMF and torque.

电动机和发电机原理经常出现。一道典型的模拟题描述单矩形线圈在磁场中的情形，并要求计算产生的电动势和转矩。

For a coil of N turns, area A m², rotating at angular velocity ω in a uniform flux density B, the instantaneous EMF is e = N B A ω sin ωt. Peak EMF is Eₘₐₓ = N B A ω. If N=100, B=0.5 T, A=0.01 m², ω=300 rad/s, then Eₘₐₓ = 100 × 0.5 × 0.01 × 300 = 150 V.

对于匝数 N、面积 A m² 的线圈，在均匀磁通密度 B 中以角速度 ω 旋转，瞬时电动势为 e = N B A ω sin ωt。峰值电动势 Eₘₐₓ = N B A ω。若 N=100，B=0.5 T，A=0.01 m²，ω=300 rad/s，则 Eₘₐₓ = 100 × 0.5 × 0.01 × 300 = 150 V。

Torque on the coil: τ = N B I A sin α. At the instant when the plane of the coil is parallel to the field (α=90°), torque is maximum: τₘₐₓ = N B I A. Many candidates confuse the angle for torque with that for flux linkage.

线圈上的转矩：τ = N B I A sin α。当线圈平面与磁场平行时（α=90°），转矩最大：τₘₐₓ = N B I A。许多考生将转矩的角度与磁链的角度混淆。

---

8. Thermodynamics & Ideal Gas Law | 热力学与理想气体定律

Thermal physics problems in the mock test require using pV = nRT, specific heat capacity, and latent heat. A closed gas cylinder undergoes heating; calculate final pressure and energy transferred.

模拟卷中的热学问题要求运用 pV = nRT、比热容和潜热。一个密闭气罐受热；要求计算最终压强和传递的能量。

Fixed mass of air (n = 0.2 mol), initial volume V₁ = 5 × 10⁻³ m³, temperature rises from 300 K to 450 K, volume constant. Using p₁ / T₁ = p₂ / T₂, initial pressure p₁ = nRT₁ / V₁ = (0.2 × 8.31 × 300) / 5×10⁻³ = 99.7 kPa. Then p₂ = p₁ × (450/300) = 149.6 kPa.

固定质量空气（n = 0.2 mol），初始体积 V₁ = 5 × 10⁻³ m³，温度从 300 K 升至 450 K，体积不变。由 p₁ / T₁ = p₂ / T₂，初始压力 p₁ = nRT₁ / V₁ = (0.2 × 8.31 × 300) / 5×10⁻³ = 99.7 kPa。则 p₂ = p₁ × (450/300) = 149.6 kPa。

Heat supplied at constant volume: Q = n Cₓ ΔT. For a diatomic gas, Cₓ = (5/2)R = 20.8 J mol⁻¹ K⁻¹. So Q = 0.2 × 20.8 × 150 = 624 J. Always use Kelvin for gas law calculations.

定容加热量：Q = n Cₓ ΔT。对于双原子气体，Cₓ = (5/2)R = 20.8 J mol⁻¹ K⁻¹。因此 Q = 0.2 × 20.8 × 150 = 624 J。气体定律计算务必使用开尔文温标。

pV = nRT, Q = m c ΔT, Q = m L

pV = nRT，Q = m c ΔT，Q = m L

---

9. Material Properties & Ashby Charts | 材料性能与阿什比图

Selecting a material for an engineering component often involves Ashby charts. The mock question presents a log‑log plot of Young’s modulus versus density and asks which material class is best for a light, stiff beam.

为工程部件选材时常使用阿什比图。模拟题

Published by TutorHao | Pre-U 工程 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）