📚 Year 13 OCR Engineering: Unit Test Mock Paper Walkthrough | Year 13 OCR 工程：单元测试模拟卷解析
This walkthrough is designed to help Year 13 OCR Engineering students consolidate their knowledge ahead of unit tests. The mock paper covers mechanics of materials, electrical principles, control systems, and manufacturing processes. Each section presents a typical exam-style question, followed by a step-by-step solution in both English and Chinese, linking theory to practical application.
本次解析旨在帮助 Year 13 OCR 工程学生巩固单元测试前的知识。模拟卷涵盖材料力学、电学原理、控制系统和制造工艺。每个部分展示一道典型考试题，然后以中英双语逐步解题，将理论与实践相结合。
1. Stress and Strain in a Tie Bar | 拉杆中的应力与应变
A solid circular tie bar of diameter 16 mm is subjected to an axial tensile force of 24 kN. The material has a Young’s modulus of 210 GPa. Calculate the direct stress in the bar and the resulting strain.
一根直径 16 mm 的实心圆截面拉杆承受 24 kN 的轴向拉力。材料的杨氏模量为 210 GPa。计算杆中的正应力以及由此产生的应变。
First, determine the cross-sectional area: A = πd²/4 = π × (16 × 10⁻³ m)² / 4 ≈ 2.011 × 10⁻⁴ m². Stress σ = F/A = 24 × 10³ N / 2.011 × 10⁻⁴ m² ≈ 1.193 × 10⁸ Pa = 119.3 MPa.
首先，计算横截面积：A = πd²/4 = π × (16 × 10⁻³ m)² / 4 ≈ 2.011 × 10⁻⁴ m²。应力 σ = F/A = 24 × 10³ N / 2.011 × 10⁻⁴ m² ≈ 1.193 × 10⁸ Pa = 119.3 MPa。
Strain ε is found from Hooke’s Law: σ = Eε, so ε = σ/E = 119.3 × 10⁶ Pa / 210 × 10⁹ Pa ≈ 5.68 × 10⁻⁴. This is a dimensionless quantity, meaning the bar elongates by about 0.057% of its original length.
根据胡克定律求应变：σ = Eε，因此 ε = σ/E = 119.3 × 10⁶ Pa / 210 × 10⁹ Pa ≈ 5.68 × 10⁻⁴。这是一个无量纲量，表示杆伸长约为其原长的 0.057%。
2. Factor of Safety for a Stepped Shaft | 阶梯轴的安全系数
A stepped shaft made of mild steel has a yield strength of 250 MPa. The smaller diameter section is 20 mm and carries a torque of 80 Nm and a bending moment of 45 Nm simultaneously. Assuming a maximum shear stress failure criterion, calculate the factor of safety based on the Tresca theory.
一根用低碳钢制成的阶梯轴屈服强度为 250 MPa。较小直径段为 20 mm，同时承受 80 Nm 的扭矩和 45 Nm 的弯矩。假设采用最大剪应力失效准则，根据 Tresca 理论计算安全系数。
First, find the bending stress: σ = M y / I, with y = d/2 = 10 mm, I = πd⁴/64 for a circular section. I = π(0.02)⁴/64 = 7.854 × 10⁻⁹ m⁴. Then σ = 45 × 0.01 / 7.854 × 10⁻⁹ = 57.3 MPa.
首先计算弯曲应力：σ = M y / I，其中 y = d/2 = 10 mm，圆截面 I = πd⁴/64。I = π(0.02)⁴/64 = 7.854 × 10⁻⁹ m⁴。则 σ = 45 × 0.01 / 7.854 × 10⁻⁹ = 57.3 MPa。
Shear stress due to torque: τ_tor = T r / J, with J = πd⁴/32 = 1.571 × 10⁻⁸ m⁴, r = 0.01 m. τ_tor = 80 × 0.01 / 1.571 × 10⁻⁸ = 50.9 MPa.
扭矩产生的剪应力：τ_tor = T r / J，其中 J = πd⁴/32 = 1.571 × 10⁻⁸ m⁴，r = 0.01 m。τ_tor = 80 × 0.01 / 1.571 × 10⁻⁸ = 50.9 MPa。
For a shaft under combined bending and torsion, the principal stresses are σ₁,₂ = (σ/2) ± √[(σ/2)² + τ²]. Here σ = 57.3 MPa, τ = 50.9 MPa. σ₁ = 28.65 + √(28.65² + 50.9²) = 28.65 + 58.4 = 87.05 MPa; σ₂ = 28.65 – 58.4 = -29.75 MPa.
对于承受弯曲与扭转组合的轴，主应力 σ₁,₂ = (σ/2) ± √[(σ/2)² + τ²]。这里 σ = 57.3 MPa，τ = 50.9 MPa。σ₁ = 28.65 + √(28.65² + 50.9²) = 28.65 + 58.4 = 87.05 MPa；σ₂ = 28.65 – 58.4 = -29.75 MPa。
Tresca’s maximum shear stress criterion states that yielding occurs when τ_max = (σ₁ – σ₂)/2 reaches τ_yield = σ_yield/2. τ_max = (87.05 – (-29.75))/2 = 58.4 MPa. The yield shear stress is 250/2 = 125 MPa. Factor of safety = τ_yield / τ_max = 125 / 58.4 ≈ 2.14.
Tresca 最大剪应力准则指出，当 τ_max = (σ₁ – σ₂)/2 达到 τ_yield = σ_yield/2 时发生屈服。τ_max = (87.05 – (-29.75))/2 = 58.4 MPa。屈服剪应力为 250/2 = 125 MPa。安全系数 = τ_yield / τ_max = 125 / 58.4 ≈ 2.14。
3. DC Circuit Analysis with Kirchhoff’s Laws | 利用基尔霍夫定律分析直流电路
For the circuit below, two batteries and three resistors are connected. E₁ = 10 V, R₁ = 2 Ω; E₂ = 6 V, R₂ = 1 Ω; load resistor R₃ = 4 Ω. The batteries are connected with positive terminals facing each other. Determine the current through R₃ using mesh analysis.
对于以下电路，两个电池和三个电阻相连。E₁ = 10 V，R₁ = 2 Ω；E₂ = 6 V，R₂ = 1 Ω；负载电阻 R₃ = 4 Ω。两电池正极相对连接。使用网孔分析法求通过 R₃ 的电流。
Define mesh currents I₁ clockwise in the left loop, I₂ clockwise in the right loop. The left loop equation: 10 – 2 I₁ – 4 (I₁ – I₂) = 0 → 10 = 6 I₁ – 4 I₂. Right loop: -6 – 1 I₂ – 4 (I₂ – I₁) = 0 → -6 = -4 I₁ + 5 I₂.
定义左网孔顺时针电流为 I₁，右网孔顺时针电流为 I₂。左回路方程：10 – 2 I₁ – 4 (I₁ – I₂) = 0 → 10 = 6 I₁ – 4 I₂。右回路：-6 – 1 I₂ – 4 (I₂ – I₁) = 0 → -6 = -4 I₁ + 5 I₂。
Solve the simultaneous equations. From the second: 4 I₁ = 5 I₂ + 6 → I₁ = (5 I₂ + 6)/4. Substitute into first: 10 = 6[(5 I₂ + 6)/4] – 4 I₂ = (30 I₂ + 36)/4 – 4 I₂ = 7.5 I₂ + 9 – 4 I₂ = 3.5 I₂ + 9. Thus 1 = 3.5 I₂ → I₂ = 0.286 A; I₁ = (5×0.286+6)/4 = 1.857 A. Current through R₃ is I₁ – I₂ = 1.571 A.
解联立方程。由第二个方程：4 I₁ = 5 I₂ + 6 → I₁ = (5 I₂ + 6)/4。代入第一个方程：10 = 6[(5 I₂ + 6)/4] – 4 I₂ = (30 I₂ + 36)/4 – 4 I₂ = 7.5 I₂ + 9 – 4 I₂ = 3.5 I₂ + 9。因此 1 = 3.5 I₂ → I₂ = 0.286 A；I₁ = (5×0.286+6)/4 = 1.857 A。通过 R₃ 的电流为 I₁ – I₂ = 1.571 A。
4. Transfer Function of a Simple Mechanical System | 简单机械系统的传递函数
A mass-spring-damper system is described by m d²x/dt² + c dx/dt + k x = F(t). Given m = 2 kg, c = 5 Ns/m, k = 40 N/m, find the transfer function G(s) = X(s)/F(s) and comment on its damping ratio.
一个质量-弹簧-阻尼系统由 m d²x/dt² + c dx/dt + k x = F(t) 描述。已知 m = 2 kg，c = 5 Ns/m，k = 40 N/m，求传递函数 G(s) = X(s)/F(s) 并评论其阻尼比。
Take Laplace transforms with zero initial conditions: (m s² + c s + k) X(s) = F(s). Hence G(s) = 1 / (2 s² + 5 s + 40). This is a second-order system.
在零初始条件下进行拉普拉斯变换：(m s² + c s + k) X(s) = F(s)。因此 G(s) = 1 / (2 s² + 5 s + 40)。这是一个二阶系统。
The standard form is G(s) = (1/k) ω_n² / (s² + 2ζω_n s + ω_n²). By comparing, ω_n² = k/m = 40/2 = 20 → ω_n = √20 ≈ 4.472 rad/s. Then 2ζω_n = c/m = 5/2 = 2.5 → ζ = 2.5 / (2 × 4.472) ≈ 0.28. Since ζ < 1, the system is underdamped.
标准形式为 G(s) = (1/k) ω_n² / (s² + 2ζω_n s + ω_n²)。比较得 ω_n² = k/m = 40/2 = 20 → ω_n = √20 ≈ 4.472 rad/s。然后 2ζω_n = c/m = 5/2 = 2.5 → ζ = 2.5 / (2 × 4.472) ≈ 0.28。由于 ζ < 1，系统处于欠阻尼状态。
5. Operational Amplifier Active Filter Design | 运算放大器有源滤波器设计
Design a low-pass active filter using an op-amp with a non-inverting configuration to have a cut-off frequency of 2 kHz and a pass-band gain of 10. The feedback resistor R_f is to be 10 kΩ. Determine suitable values for the input resistor R_i and the filter capacitor C if a first-order RC network is placed at the input.
设计一个使用运算放大器的低通有源滤波器，采用同相配置，截止频率为 2 kHz，通带增益为 10。反馈电阻 R_f 取 10 kΩ。如果将一阶 RC 网络放置在输入端，确定合适的输入电阻 R_i 和滤波电容 C 的值。
For a non-inverting amplifier, gain A = 1 + R_f / R_i. Desired A = 10, so 10 = 1 + 10 kΩ / R_i → R_i = 10 kΩ / 9 ≈ 1.111 kΩ (use nearest standard value 1.1 kΩ).
对于同相放大器，增益 A = 1 + R_f / R_i。所需 A = 10，因此 10 = 1 + 10 kΩ / R_i → R_i = 10 kΩ / 9 ≈ 1.111 kΩ（使用最接近的标准值 1.1 kΩ）。
The low-pass RC filter at the input sets the cut-off frequency: f_c = 1 / (2π R C). Here we can use the same R as the input resistor for loading considerations, but a separate R_filt can be chosen. Let R = 1.1 kΩ. Then C = 1 / (2π × 2000 Hz × 1100 Ω) ≈ 72.3 nF. A standard value of 68 nF or 82 nF could be selected, adjusting R slightly if needed. The filter’s transfer function will combine the passive RC roll-off with the amplifier’s gain.
输入端的低通 RC 滤波器设定截止频率：f_c = 1 / (2π R C)。考虑到负载，我们可以使用与输入电阻相同的 R，但也可另选 R_filt。取 R = 1.1 kΩ。则 C = 1 / (2π × 2000 Hz × 1100 Ω) ≈ 72.3 nF。可选用标准值 68 nF 或 82 nF，必要时微调 R。滤波器的传递函数将结合无源 RC 衰减和放大器增益。
6. Thermodynamics – First Law Applied to a Piston-Cylinder Assembly | 热力学 – 第一定律应用于活塞-气缸装置
A gas expands in a piston-cylinder assembly from an initial volume of 0.01 m³ to 0.025 m³ at a constant pressure of 150 kPa. During the process, 2.5 kJ of heat is added. Calculate the work done by the gas and the change in internal energy.
某气体在活塞-气缸装置中从初始体积 0.01 m³ 膨胀到 0.025 m³，压力恒定为 150 kPa。过程中加入热量 2.5 kJ。计算气体做的功以及内能的变化。
Work done at constant pressure: W = p ΔV = 150 × 10³ Pa × (0.025 – 0.01) m³ = 150,000 × 0.015 = 2250 J = 2.25 kJ.
恒压下的功：W = p ΔV = 150 × 10³ Pa × (0.025 – 0.01) m³ = 150,000 × 0.015 = 2250 J = 2.25 kJ。
The first law of thermodynamics: ΔU = Q – W. Q = +2.5 kJ (heat added). So ΔU = 2.5 kJ – 2.25 kJ = 0.25 kJ. The internal energy increases by 0.25 kJ.
热力学第一定律：ΔU = Q – W。Q = +2.5 kJ（加入热量）。所以 ΔU = 2.5 kJ – 2.25 kJ = 0.25 kJ。内能增加 0.25 kJ。
7. Manufacturing Process Selection – CNC vs. Injection Moulding | 制造工艺选择 – 数控加工与注塑成型
A company needs to produce 50,000 plastic brackets per year. The bracket has a complex geometry with undercuts. Compare CNC machining and injection moulding in terms of cost, lead time, and material properties for this scenario. Recommend a process.
某公司需要每年生产 50,000 个塑料支架。该支架几何形状复杂，带有侧凹。就此情景，比较数控加工和注塑成型在成本、交付周期和材料性能方面的差异。推荐一种工艺。
For high-volume plastic parts with undercuts, injection moulding is generally more economical despite higher initial tooling costs. The mould can incorporate slides to form undercuts. Cycle times are short (seconds per part), giving a low unit cost at 50,000 units. CNC machining would require multiple setups and fixturing; cycle time per part would be significantly longer, leading to higher labour and machine costs at volume.
对于带有侧凹的高产量塑料件，注塑成型尽管初始模具成本高，但通常更经济。模具可加入滑块成型侧凹。成型周期短（每件数秒），在 50,000 件产量下单位成本低。数控加工需要多次装夹和夹具，单件加工时间会显著增长，导致在大批量下人工和机床成本更高。
Injection moulding also offers better repeatability and can use a range of thermoplastics with good mechanical properties. CNC machining from a solid block may achieve tighter tolerances initially, but tool wear and material waste make it less sustainable. Lead time for mould fabrication can be weeks, but once in production, output is high. The recommended process is injection moulding.
注塑成型还提供更好的重复性，并可使用多种具有良好力学性能的热塑性塑料。尽管数控加工从整块材料切削初始可获得更严格的公差，但刀具磨损和材料浪费使其可持续性较差。模具制造的交货期可能需要数周，但一旦投产，产能很高。推荐工艺为注塑成型。
8. Pneumatic Sequential Control Circuit Design | 气动顺序控制回路设计
Design a purely pneumatic circuit to implement the sequence A+ B+ A- B- (where A+ means extend cylinder A, A- means retract). Use roller-operated limit valves for position sensing and double-piloted 5/2 directional control valves.
设计一个纯气动回路来实现顺序 A+ B+ A- B-（其中 A+ 表示气缸 A 伸出，A- 表示缩回）。使用滚轮操作行程阀进行位置检测和双气控 5/2 换向阀。
The sequence is: extend A, extend B, retract A, retract B. Start by pressing a start valve. When A is fully extended, it trips a limit valve that initiates B+. When B is fully extended, it trips another limit valve to cause A-. When A is fully retracted, its limit valve triggers B-. The circuit uses two 5/2 valves (one per cylinder), each with two pilot signals. Limit valves are placed at the end positions of each cylinder to send pilot signals to the opposing cylinder’s valve.
该顺序为：A 伸出，B 伸出，A 缩回，B 缩回。按下启动阀开始。当 A 完全伸出时，触发一个行程阀启动 B+。当 B 完全伸出时，触发另一个行程阀使 A-。当 A 完全缩回时，其行程阀触发 B-。回路使用两个 5/2 阀（每气缸一个），每个有两个先导信号。行程阀放置在每个气缸的端部位置，将先导信号发送到对方气缸的阀。
Additionally, to avoid signal overlap, limit valves should be configured as idle-return (spring-return) 3/2 valves, and cascade or shifting in-line valves may be used to ensure clean sequential operation if signals persist. However, with the natural sequence A+ B+ A- B-, overlapping signals can be avoided with careful limit-valve placement. The circuit is a fundamental example of sequential logic in pneumatic systems.
此外，为避免信号重叠，行程阀应配置为常断弹簧复位 3/2 阀，如果信号持续存在，可能需要使用级联或串接换向阀以确保清晰顺序操作。但对于 A+ B+ A- B- 这样的自然顺序，通过合理布置行程阀可以避免信号重叠。该回路是气动系统中顺序逻辑的基础实例。
9. Material Selection Using Ashby Charts | 运用 Ashby 图进行材料选择
A lightweight cantilever beam must have high stiffness and minimal mass. Using the material index M = E^{1/2} / ρ, compare aluminium alloy (E = 70 GPa, ρ = 2700 kg/m³), steel (E = 210 GPa, ρ = 7800 kg/m³), and CFRP (E = 150 GPa, ρ = 1550 kg/m³). Which material is best for this performance index?
一根轻质悬臂梁需要高刚度和最小质量。采用材料指数 M = E^{1/2} / ρ，比较铝合金（E = 70 GPa，ρ = 2700 kg/m³）、钢（E = 210 GPa，ρ = 7800 kg/m³）和碳纤维增强塑料 CFRP（E = 150 GPa，ρ = 1550 kg/m³）。哪种材料在该性能指数下最优？
Calculate M for each: Al alloy: √(70×10⁹) / 2700 = (√70 × 10^{4.5}) / 2700 ≈ (8.367 × 31623) / 2700? Let’s compute carefully: E in Pa. √(70×10⁹) = √(70×10⁹) = √70 × 10^{4.5} ≈ 8.367 × 31622.8 = 264,575. Then divide by 2700: 264,575 / 2700 ≈ 98.0 (m³/s units simplified).
分别计算 M：铝合金：√(70×10⁹) / 2700 = (√70 × 10^{4.5}) / 2700 ≈ 264,575 / 2700 ≈ 98.0。
Steel: √(210×10⁹) / 7800 = √210 × 10^{4.5} = 14.491 × 31622.8 ≈ 458,214 / 7800 ≈ 58.7.
钢：√(210×10⁹) / 7800 ≈ 458,214 / 7800 ≈ 58.7。
CFRP: √(150×10⁹) / 1550 = √150 × 10^{4.5} = 12.247 × 31622.8 ≈ 387,300 / 1550 ≈ 249.9. Clearly CFRP has the highest M, making it the best choice for minimum mass at a given stiffness. This aligns with aerospace applications.
CFRP：√(150×10⁹) / 1550 ≈ 387,300 / 1550 ≈ 249.9。显然 CFRP 的 M 值最高，使其成为给定刚度下质量最小的最佳选择。这与航空航天应用相符。
10. Digital Logic – Finite State Machine for Conveyor Control | 数字逻辑 – 传送带控制的有限状态机
A conveyor system has two sensors: S1 (item present at entry) and S2 (item at exit). A motor M runs when an item is between S1 and S2. Design a Moore FSM using D flip-flops to control the motor, ensuring safe behaviour if both sensors are triggered simultaneously.
一个传送带系统有两个传感器：S1（入口有物品）和 S2（物品在出口）。当物品在 S1 和 S2 之间时，电机 M 运行。使用 D 触发器设计一个 Moore 型状态机来控制电机，确保在两个传感器同时触发时的安全行为。
States can be: A (idle, M off), B (item passed S1, M on), C (item reached S2, M off). Transitions: A → B when S1=1; B → C when S2=1; C → A when S1=0 and S2=0. Simultaneously S1 and S2 (error) from state A should go to A (off) or an error state. We’ll assign: from A if S1=1 and S2=1, stay in A (off).
状态可设为：A（空闲，M 关）、B（物品经过 S1，M 开）、C（物品到达 S2，M 关）。转移：当 S1=1 时 A → B；当 S2=1 时 B → C；当 S1=0 且 S2=0 时 C → A。若从状态 A 同时出现 S1 和 S2（错误），则保持 A（关）。
Using two D flip-flops, states are encoded: A=00, B=01, C=10. Next-state logic derived from truth table. Moore output M = 1 only in B. Implementation requires logic gates to compute D₁ and D₂ based on current state Q₁Q₂ and inputs S1, S2. The design ensures the motor never turns on if both sensors are blocked, preventing collisions.
使用两个 D 触发器，状态编码：A=00，B=01，C=10。从真值表推导次态逻辑。Moore 输出 M 仅在状态 B 时为 1。实现需要逻辑门根据当前状态 Q₁Q₂ 和输入 S1、S2 计算 D₁ 和 D₂。该设计确保当两个传感器同时被遮挡时电机不会启动，防止碰撞。
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