📚 Year 13 WJEC Engineering: Unit Test Mock Paper Analysis | Year 13 WJEC 工程：单元测试模拟卷解析
This article provides a detailed breakdown of a typical Year 13 WJEC Engineering unit test mock paper. We will go through key questions, common pitfalls, and model solutions to help you strengthen your exam technique and subject knowledge. By understanding the structure and expectations, you can approach the real assessment with confidence.
本文详细解析一份典型的 Year 13 WJEC 工程单元测试模拟卷。我们将梳理关键题目、常见错误和示范解答，帮助同学们强化应试技巧与学科知识。读懂试卷结构和评分期望，你便能更有信心地应对正式考核。
1. Question 1: Material Properties and Surface Treatments | 材料特性与表面处理
The first section tested the ability to select appropriate materials and treatments for a bracket exposed to outdoor conditions. The bracket was initially specified as low carbon steel, and students had to justify a protective coating or an alternative material to prevent red rust.
第一部分考查为户外支架选择合适的材料与处理方案。原设计为低碳钢，考生需论证采用保护涂层或替代材料来防止红锈生成。
A strong answer identified hot-dip galvanising as a sacrificial coating. Zinc corrodes preferentially, protecting the steel even if the coating is scratched. Painting alone was deemed insufficient for long-term outdoor use unless combined with a primer system.
高质量的回答明确指出热浸镀锌能形成牺牲涂层。锌优先腐蚀，即使涂层划伤仍能保护钢材。仅靠油漆涂装不足以满足长期户外要求，除非配合完整的底漆系统。
Alternatively, students could suggest switching to austenitic stainless steel (e.g., 304 grade). This eliminated the need for coating but increased material cost and required specific welding procedures to avoid sensitisation. Marks were awarded for linking choice to the product’s service environment and manufacturing constraints.
替代方案是改用奥氏体不锈钢（如 304 牌号）。这免去了涂层需求，但增加了材料成本，且需要特定的焊接工艺防止敏化。评分关键是将选择与产品使用环境和制造约束联系起来。
2. Question 2: Bending Stress in a Cantilever Beam | 悬臂梁的弯曲应力计算
Candidates were given a cantilever beam of length L = 0.5 m, supporting a point load F = 800 N at the free end, with a rectangular cross-section of width 40 mm and depth 60 mm. The task was to calculate the maximum bending stress.
题目给出长 0.5 m 的悬臂梁，自由端承受 800 N 集中力，截面为宽 40 mm、深 60 mm 的矩形。要求计算最大弯曲应力。
The maximum bending moment occurs at the fixed support: M = F × L. Substituting values gives M = 800 × 0.5 = 400 Nm. Next, the second moment of area I for a rectangle is (b × d³)/12.
最大弯矩发生在固定端：M = F × L。代入得 M = 800 × 0.5 = 400 Nm。接着，矩形截面的惯性矩 I = (b × d³)/12。
I = (0.040 × 0.060³) / 12 = 7.2 × 10⁻⁷ m⁴
Using the bending equation σ = M y / I, with y = d/2 = 0.030 m from the neutral axis, the maximum stress σ_max = (400 × 0.030) / 7.2×10⁻⁷ = 1.67×10⁷ Pa ≈ 16.7 MPa. Many students lost marks by mixing units (mm and m). Always convert to metres before calculating I.
应用弯曲公式 σ = M y / I，y 取中性轴距外表面 0.030 m，得最大应力 σ_max = (400 × 0.030) / 7.2×10⁻⁷ ≈ 16.7 MPa。不少考生因单位混淆（毫米与米）而失分。计算 I 前务必统一为米制。
3. Question 3: Manufacturing Process Selection for a Bracket | 支架的制造工艺选择
This question asked to compare sand casting and CNC milling for producing 500 aluminium brackets per year. Each bracket featured internal pockets and threaded holes.
该题要求对比砂型铸造和 CNC 铣削，用于年产 500 件带内腔和螺纹孔的铝支架。
Sand casting offered low tooling cost and the ability to form complex internal geometries in a single pour. However, surface finish is poor and tolerances are loose (±1.5 mm typical). Threaded holes must be machined afterwards, adding secondary operations.
砂型铸造工装成本低，可一次浇注出复杂内腔。但表面粗糙，公差较松（典型 ±1.5 mm）。螺纹孔需后续机加工，增加二次工序。
CNC milling from billet yields excellent surface finish (Ra 0.8 µm attainable) and tight tolerances (±0.05 mm). Thread milling can be integrated, reducing handling. The main drawback is high material waste and longer machining time per part, making it less economical for medium batches if design simplicity can be achieved.
从坯料 CNC 铣削可获得优良表面（可达 Ra 0.8 µm）和严格公差（±0.05 mm）。螺纹铣削可一并完成，减少搬运。主要缺点是材料浪费多、单件加工时间长；若设计可适当简化，其对中等批量而言经济性较差。
The best answers recommended casting for the main body with subsequent CNC finishing on critical features, justifying the hybrid approach with cost and quality trade-offs.
最佳答案建议采用铸造毛坯加 CNC 精加工关键特征的混合工艺，并从成本和质量权衡角度加以论证。
4. Question 4: Interpreting Tolerances and Dimensioning | 公差与尺寸标注解读
A drawing of a shaft and housing assembly was provided, featuring geometric tolerances such as circular runout and positional tolerance. Students had to explain the meaning of a positional tolerance frame: ⊕ ∅0.1 Ⓜ A B C.
试卷提供了一张轴孔配合图纸，包含圆跳动和位置度等几何公差。考生需解释位置度公差框 ⊕ ∅0.1 Ⓜ A B C 的含义。
The symbol indicates that the axis of the hole must lie within a cylindrical tolerance zone of diameter 0.1 mm, located perfectly relative to datums A, B and C. The Ⓜ modifier means the tolerance applies at the feature’s maximum material condition (MMC) – as the hole size increases from its MMC, bonus tolerance becomes available.
该符号表示孔轴线必须位于直径为 0.1 mm 的圆柱形公差带内，该公差带相对于基准 A、B、C 处于理想位置。Ⓜ 修饰符表示公差在特征的最大实体状态（MMC）下生效——当孔径从 MMC 扩大时，可获得额外公差。
Many responses confused MMC with least material condition (LMC) or failed to link the tolerance to a virtual condition gauge. Clear, step-by-step interpretation is essential for full marks.
不少答案混淆了最大实体状态与最小实体状态（LMC），或未将公差与实效边界量规关联。清晰、逐步的解读是获满分的关键。
5. Question 5: Pneumatic Circuit Design | 气动回路设计
A circuit was required to control a double-acting cylinder using two pushbuttons: one for extension, one for retraction. The cylinder must stay in position when neither button is pressed.
题目要求用两个按钮控制双作用气缸：一个伸出、一个缩回。未按下任何按钮时，活塞应保持原位。
The standard solution uses a 5/2-way valve with spring return to centre (closed centre), or a memory valve with pilot-operated signals. Students who proposed a 5/2 bistable valve with impulse signals scored well, but had to ensure no conflicting signals occurred if both buttons were pressed simultaneously.
标准方案采用 5/2 位弹簧复位（中位封闭）阀，或带先导信号的记忆阀。提出使用 5/2 双稳态阀加脉冲信号的学生得分较高，但必须确保双按钮同时按下时不会出现信号冲突。
Common errors included forgetting flow control valves for speed adjustment, or omitting a start-up condition that would unexpectedly move the cylinder. Always annotate the circuit diagram clearly and explain the valve’s function at each step.
常见错误包括忘记调速用的单向节流阀，或遗漏导致气缸意外动作的初始状态。必须清晰注释回路图，并分步说明阀门功能。
6. Question 6: Quality Control and SPC | 质量控制与统计过程控制
Given a set of measured diameters from a turning process, students were asked to plot an X̄-R chart and determine if the process was in statistical control. Control limits for the range chart and mean chart had to be calculated.
题目给定车削工序的一组直径测量值，要求绘制 X̄-R 控制图并判断过程是否受控。需计算极差图和均值图的控制界限。
The necessary constants for sample size n=5 were provided: A₂ = 0.577, D₃ = 0, D₄ = 2.114. The grand mean X̄̄ was 30.12 mm, mean range R̄ was 0.08 mm. Thus UCL_R = D₄ × R̄ = 2.114 × 0.08 = 0.169 mm. Since all ranges were below UCL, the variation appeared stable.
已知样本容量 n=5 的常数：A₂ = 0.577, D₃ = 0, D₄ = 2.114。总均值 X̄̄ = 30.12 mm，平均极差 R̄ = 0.08 mm。因此 UCL_R = 0.169 mm。所有极差点均低于上控制限，表明变异稳定。
For the X̄ chart, UCL = X̄̄ + A₂×R̄ = 30.12 + 0.577×0.08 = 30.166 mm, LCL = 30.074 mm. One sample mean fell outside these limits, indicating an assignable cause of variation possibly from tool wear. A high-quality answer suggested recalibrating the machine and increasing inspection frequency.
均值图控制限：UCL = 30.166 mm，LCL = 30.074 mm。有一个样本均值出界，提示可能存在刀具磨损等特殊原因变异。优秀答案建议重新校准机床并增加检测频次。
7. Question 7: Design for Manufacture and Assembly (DFMA) | 面向制造与装配的设计
A design problem required reducing the part count of a small hinge assembly from 12 to 5 components. The original design used separate washers, bushings and fasteners.
设计任务要求将一个小铰链组件的零件数从 12 减至 5。原设计包含独立的垫圈、衬套和紧固件。
Applying DFMA principles, students could suggest integrating the bushing into one hinge leaf using a self-lubricating polymer overmould, and using snap-fit pins instead of screws and nuts. This eliminates fasteners and washers, dramatically cutting assembly time.
运用 DFMA 原则，可将衬套通过自润滑聚合物包塑集成到一片铰链叶上，并用卡扣销代替螺栓螺母。如此一来省去紧固件和垫片，大幅缩短装配时间。
The redesign must consider load direction and wear. A well-justified choice of material (e.g., glass-filled nylon for the snap feature) and a statement on how the reduced part count lowers failure modes earned top marks. Students who sketched a clear schematic also gained credit.
重新设计需考虑载荷方向和磨损。合理选择材料（如玻纤尼龙用于卡扣）并说明减少零件数如何降低失效模式可获得高分。清晰示意图亦可加分。
8. Question 8: Power Transmission and Gear Ratios | 动力传输与齿轮比
A compound gear train was given: motor driving gear A (20 teeth), which meshes with gear B (60 teeth). On the same shaft as B, gear C (25 teeth) drives output gear D (100 teeth). Calculate the overall gear ratio and output torque if the motor delivers 1.2 Nm at 2800 rpm.
题给复合齿轮系：电动机驱动齿数 20 的齿轮 A，啮合齿数 60 的齿轮 B；同轴上齿数 25 的齿轮 C 驱动齿数 100 的输出齿轮 D。计算总传动比及输出扭矩，已知电机输出 1.2 Nm、2800 rpm。
The ratio between A and B is 60/20 = 3:1 reduction. The second stage C to D is 100/25 = 4:1 reduction. The overall ratio from motor to load is 3 × 4 = 12:1. Output speed = 2800 / 12 ≈ 233 rpm. Ignoring losses, output torque = motor torque × overall ratio = 1.2 Nm × 12 = 14.4 Nm.
A 与 B 的减速比为 60/20 = 3:1；第二级 C 到 D 为 100/25 = 4:1。总传动比 = 3 × 4 = 12:1。输出转速 = 2800 / 12 ≈ 233 rpm。忽略损耗，输出扭矩 = 1.2 × 12 = 14.4 Nm。
Eff_total = Eff_stage1 × Eff_stage2, assuming 95% per stage: 0.95 × 0.95 = 0.9025; T_out_actual ≈ 14.4 × 0.9025 ≈ 13.0 Nm
To be precise, assume each mating pair efficiency of 95%. The total efficiency is the product: 0.95 × 0.95 ≈ 0.90. Corrected torque ≈ 14.4 × 0.90 ≈ 13.0 Nm. Highlighting the need to consider efficiency in real systems was a key differentiator.
更精确的做法是假设每对啮合效率 95%，总效率 ≈ 0.90，修正扭矩约 13.0 Nm。强调实际系统中必须考虑效率，是得高分的关键区分点。
9. Question 9: Structural Analysis – Stress and Strain Energy | 结构分析——应力与应变能
A rod of original length 2 m and cross-sectional area 250 mm² is subjected to a tensile load of 50 kN, causing an extension of 2.5 mm. Students had to compute direct stress, strain, Young’s modulus, and the strain energy stored.
一根长 2 m、截面积 250 mm² 的圆杆受 50 kN 拉伸，伸长了 2.5 mm。要求计算正应力、应变、杨氏模量及储存的应变能。
Direct stress σ = F / A = (50×10³) / (250×10⁻⁶) = 200×10⁶ Pa = 200 MPa. Strain ε = ΔL / L₀ = 2.5×10⁻³ / 2 = 1.25×10⁻³. Young’s modulus E = σ / ε = 200×10⁶ / 1.25×10⁻³ = 160 GPa.
正应力 σ = F / A = 200 MPa；应变 ε = ΔL / L₀ = 1.25×10⁻³；杨氏模量 E = σ / ε = 160 GPa。
Strain energy U = (1/2) × F × ΔL = 0.5 × 50×10³ × 2.5×10⁻³ = 62.5 J. Alternatively U = (σ² / 2E) × Volume. Volume = A × L = 250×10⁻⁶ × 2 = 5×10⁻⁴ m³, then U = (200×10⁶)² / (2 × 160×10⁹) × 5×10⁻⁴ = 62.5 J. Both methods accepted.
应变能 U = ½ F ΔL = 62.5 J；也可用 U = (σ²/2E)×体积 = 62.5 J，两种方法均可。许多学生忘记将面积转换为 m²，导致误差，务必逐项核对单位。
10. Question 10: Control Systems and Sensors | 控制系统与传感器
A conveyor system must sort packages by weight using a load cell and a pneumatic actuator. The logic controller must receive an analogue signal, compare it to a setpoint, and activate a solenoid valve if weight exceeds 5 kg. Students drew the block diagram and explained sensor selection.
某传送系统需用称重传感器和气动执行器按重量分拣包裹。控制器接收模拟信号，与设定值比较，当重量超过 5 kg 时激活电磁阀。考生需绘制框图并解释传感器选型。
A strain gauge load cell with bridge amplifier was the logical choice for reliability and analogue output. The amplifier signal (0–10 V or 4–20 mA) is fed into a comparator or PLC analogue input. The block diagram should show closed-loop feedback if a check-weighing confirmation was included.
应变片式称重传感器配桥式放大器因其可靠性和模拟输出而成为合理选择。放大信号（0–10 V 或 4–20 mA）送入比较器或 PLC 模拟输入端。若包含复称确认，框图应体现闭环反馈。
Key points: specify hysteresis and response time of the pneumatic valve; ensure the actuator’s force capability matches the package mass; and incorporate an interlock to prevent pushing while the weight platform is settling. Missing these practical details limited answers to lower bands.
要点：指明气动阀的迟滞和响应时间；确保执行器力值匹配包裹质量；加入联锁防止称重平台未稳定时推料。缺少这些实践细节的答案得分受限。
11. Common Pitfalls and Exam Strategies | 常见陷阱与应试策略
Many marks were lost through unit conversion errors, particularly in moment of inertia calculations where mm and m were mixed. Always write the conversion explicitly: 1 mm⁴ = 1 × 10⁻¹² m⁴.
大量失分源于单位换算错误，尤其惯性矩计算中毫米与米的混用。务必明确转换：1 mm⁴ = 1 × 10⁻¹² m⁴。
In design questions, avoiding generic statements like ‘use a stronger material’ and instead specifying the exact grade with justification (e.g., ‘BS 970 080M40 can be heat treated to improve wear resistance’) gained higher marks. The same applied to coating specifications.
设计题中，避免笼统表述如“用更强材料”，应给出具体牌号并论证（如“BS 970 080M40 可热处理以提高耐磨性”），涂层规格亦然，方能获高分。
In extended writing, structure your answer logically: (1) Identify the problem/requirements, (2) Propose two or more feasible solutions, (3) Evaluate each against criteria (cost, manufacturing, environment), (4) Justify the final decision. This mirrors the design process and matches mark scheme expectations.
在长答题中应逻辑清晰：首先识别问题/需求，其次提出至少两种可行方案，然后按成本、制造、环境等标准评估，最后论证最终选择。这符合设计流程，也契合评分标准。
Finally, time management is crucial. Do not spend 30 minutes on a 6-mark question. Practise under timed conditions to calibrate your pace, and leave 5 minutes to review numerical answers for unit consistency and significant figures.
最后，时间管理至关重要。勿在 6 分题上耗费 30 分钟。在限时条件下练习以校准答题节奏，并留出 5 分钟复查数值答案的单位一致性和有效数字。
Published by TutorHao | Engineering Revision Series | aleveler.com
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