📚 Year 12 WJEC Engineering: Unit Test Mock Exam Walkthrough | WJEC 12年级工程：单元测试模拟卷解析
This walkthrough provides a detailed breakdown of a typical Year 12 WJEC Engineering unit test, covering essential topics such as mechanics, materials, electronics, and thermodynamics. Each section presents a question style commonly encountered in the exam, followed by a step-by-step solution and key commentary. The bilingual format ensures you can grasp the principles in both English and Chinese, reinforcing your understanding for high-stakes assessment.
本解析详细讲解了一份典型的 WJEC 12年级工程单元测试卷，涵盖力学、材料、电子学和热力学等重要知识点。每个小节都展示了一种常见题型，并给出分步解答与关键点评。中英双语的形式有助于你扎实掌握原理，为高利害考试做好充分准备。
1. Resolution of Forces and Vectors | 力的分解与矢量合成
A force of 500 N acts at an angle of 30° to the horizontal. Calculate the horizontal and vertical components. Additionally, two forces act on a point: 200 N along the positive x-axis and 300 N at 60° to the x-axis. Determine the magnitude and direction of the resultant force.
一力大小为 500 N，与水平方向呈 30° 夹角。计算其水平和竖直分量。此外，两个力作用于一点：200 N 沿 x 轴正方向，300 N 与 x 轴呈 60°。求合力的大小与方向。
For the single force: horizontal component Fh = 500 cos 30° = 433 N, vertical component Fv = 500 sin 30° = 250 N. Always resolve into perpendicular components using trigonometric functions and check your calculator is in degree mode.
对于单一力：水平分量 Fh = 500 cos 30° = 433 N，竖直分量 Fv = 500 sin 30° = 250 N。一律用三角函数将力分解为相互垂直的分量，并确认计算器处于角度模式。
For the two-force system: Resolve the 300 N at 60°: x-component = 300 cos 60° = 150 N, y-component = 300 sin 60° ≈ 259.8 N. Sum all x-components: 200 N + 150 N = 350 N. Sum all y-components: 0 + 259.8 N = 259.8 N. The resultant magnitude R = √(350² + 259.8²) ≈ 435.9 N. Direction θ = tan⁻¹(259.8 / 350) ≈ 36.6° above the horizontal. State the answer as 436 N at 36.6° to the horizontal.
对于双力系统：将 300 N 在 60° 处分解：x 分量 = 300 cos 60° = 150 N，y 分量 = 300 sin 60° ≈ 259.8 N。全部 x 分量求和：200 N + 150 N = 350 N。全部 y 分量求和：0 + 259.8 N = 259.8 N。合力大小 R = √(350² + 259.8²) ≈ 435.9 N。方向 θ = tan⁻¹(259.8 / 350) ≈ 36.6°（高于水平）。最终答案需写明 436 N，与水平夹角 36.6°。
2. Stress-Strain and Material Properties | 应力-应变与材料性能
A tensile test is performed on a steel specimen with a diameter of 10 mm and a gauge length of 50 mm. The load-extension curve shows a linear region, a yield point, and ultimate failure. The yield load is 22 kN. Using this data, calculate the yield stress and explain how Young’s modulus can be determined from the linear portion. Also, distinguish between yield stress and ultimate tensile strength (UTS).
对一直径为 10 mm、标距为 50 mm 的钢试样进行拉伸试验。载荷-伸长曲线显示出线性区、屈服点和最终断裂。屈服载荷为 22 kN。利用数据计算屈服应力，并解释如何从线性段求出杨氏模量。同时，区分屈服应力与极限抗拉强度（UTS）。
Cross-sectional area A = π × (5 mm)² = 78.5 mm² = 78.5 × 10⁻⁶ m². Yield stress σy = Fy / A = 22×10³ N / 78.5×10⁻⁶ m² = 280 MPa (approximately). Young’s modulus E is the gradient of the linear stress-strain region: E = Δσ / Δε. In the linear portion, stress is proportional to strain, so E can be found from the slope of the force-extension graph using E = (F/A) / (ΔL/L₀). Remember to convert all units to N, m, and Pa.
横截面积 A = π × (5 mm)² = 78.5 mm² = 78.5 × 10⁻⁶ m²。屈服应力 σy = Fy / A = 22×10³ N / 78.5×10⁻⁶ m² = 280 MPa（约）。杨氏模量 E 是线弹性段应力-应变的斜率：E = Δσ / Δε。在线性段，应力与应变成正比，因此 E 可由力-伸长图斜率结合 E = (F/A) / (ΔL/L₀) 求得。务必统一单位：N、m、Pa。
Yield stress marks the onset of permanent plastic deformation, whereas UTS is the maximum stress the material can withstand before necking and failure. On a stress-strain diagram, yield stress appears just after the linear limit, while UTS is the peak of the curve. In design, yield stress is often used with a safety factor to define allowable stress.
屈服应力代表永久塑性变形的起始点，而极限抗拉强度是材料在颈缩和断裂前能承受的最大应力。在应力-应变图中，屈服应力位于线性极限之后，UTS 则是曲线的峰值。在设计中，屈服应力常与安全系数一起用来确定许用应力。
3. DC Circuits and Kirchhoff’s Laws | 直流电路与基尔霍夫定律
Analyse the circuit shown: a 12 V battery with internal resistance negligible is connected to a network. Resistor R1 = 4 Ω is in series with a node, after which R3 = 6 Ω connects back to the negative terminal. A 6 V battery is connected with R2 = 2 Ω, also linked to the common node. Use Kirchhoff’s laws to solve for the currents in each branch. Assume conventional current flow.
分析所示电路：一个内阻可忽略的 12 V 电池与网络连接。电阻 R1 = 4 Ω 串接至节点，节点后由 R3 = 6 Ω 接回负极。一个 6 V 电池与 R2 = 2 Ω 连接，同样接入该公共节点。应用基尔霍夫定律求解各支路电流。采用常规电流方向。
Define branch currents: let I₁ be the current from the 12 V battery through R1, I₂ the current from the 6 V battery through R2, and I₃ the current through R3. At the node, Kirchhoff’s Current Law gives: I₁ + I₂ = I₃. For loop analysis, assign loop currents i₁ (left loop) and i₂ (right loop). For loop containing 12 V, R1 and R3: 12 − 4i₁ − 6(i₁ − i₂) = 0 → 12 − 10i₁ + 6i₂ = 0. For loop with 6 V, R2 and R3: −6 − 2i₂ − 6(i₂ − i₁) = 0 → −6 − 8i₂ + 6i₁ = 0 (note sign depending on direction). Solve the simultaneous equations: 10i₁ − 6i₂ = 12 and 6i₁ − 8i₂ = 6. Dividing the second by 2: 3i₁ − 4i₂ = 3. Solving yields i₁ = 1.8 A, i₂ = 0.6 A. Then I₁ = i₁ = 1.8 A, I₂ = i₂ = 0.6 A, I₃ = i₁ − i₂ = 1.2 A.
定义支路电流：令 I₁ 为从 12 V 电池流过 R1 的电流，I₂ 为从 6 V 电池流过 R2 的电流，I₃ 为流过 R3 的电流。在节点处，KCL 得：I₁ + I₂ = I₃。回路分析时，设回路电流 i₁（左回路）和 i₂（右回路）。含 12 V、R1 与 R3 的回路：12 − 4i₁ − 6(i₁ − i₂) = 0 → 12 − 10i₁ + 6i₂ = 0。含 6 V、R2 与 R3 的回路：−6 − 2i₂ − 6(i₂ − i₁) = 0 → −6 − 8i₂ + 6i₁ = 0。求解方程组：10i₁ − 6i₂ = 12 及 6i₁ − 8i₂ = 6。第二式除以 2：3i₁ − 4i₂ = 3。解得 i₁ = 1.8 A，i₂ = 0.6 A。因此 I₁ = 1.8 A，I₂ = 0.6 A，I₃ = i₁ − i₂ = 1.2 A。在答案中须说明电流方向。
Confirm the solution by checking power or voltage drops: voltage across R3 is I₃ × 6 = 7.2 V. The potential at the node should be consistent. Always verify your results with a different method, e.g., mesh analysis or node voltage.
用功率或压降验证结果：R3 两端电压为 I₃ × 6 = 7.2 V。节点电位应一致。务必用别的方法（如网孔电流法或节点电压法）进行验证。
4. Shear Force and Bending Moment in Beams | 简支梁的剪力与弯矩
A simply supported beam of length 4 m carries a concentrated load of 10 kN at the mid-span. Neglect the weight of the beam. Draw the shear force diagram (SFD) and bending moment diagram (BMD), and calculate the maximum bending moment.
一长度为 4 m 的简支梁，跨中承受一集中载荷 10 kN。忽略梁的自重。画出剪力图（SFD）和弯矩图（BMD），并计算最大弯矩。
First, determine support reactions. Taking moments about the left support A: RB × 4 m = 10 kN × 2 m ⇒ RB = 5 kN. Vertically, RA + RB = 10 kN ⇒ RA = 5 kN. Sign convention: upward forces and anticlockwise moments positive.
首先计算支座反力。对左支座 A 取矩：RB × 4 m = 10 kN × 2 m ⇒ RB = 5 kN。竖直方向：RA + RB = 10 kN ⇒ RA = 5 kN。符号规定：向上取正，逆时针力矩取正。
Shear force: from the left end to the load point (0 ≤ x < 2 m), SF = +5 kN. At x = 2 m just left of the load SF = +5 kN, just right it drops by 10 kN to −5 kN. Then from 2 m to the right end, SF = −5 kN. Bending moment: for 0 ≤ x ≤ 2 m, M = RA × x = 5x kN·m. At mid-span, Mmax = 5 × 2 = 10 kN·m. For the right half, moment decreases linearly to zero at the right support. The BMD is two straight lines forming a triangle with apex at the load point. Always label axes and key values.
剪力：从左端至载荷作用点（0 ≤ x < 2 m），SF = +5 kN。在 x = 2 m 处，载荷左侧 SF = +5 kN，右侧跃降 10 kN 至 −5 kN。随后从 2 m 到右端 SF = −5 kN。弯矩：对于 0 ≤ x ≤ 2 m，M = RA × x = 5x kN·m。跨中处 Mmax = 5 × 2 = 10 kN·m。右半段弯矩线性递减至右支座为零。BMD 为两条直线构成三角形，顶角在载荷点。务必标注坐标轴和关键数值。
In the exam, you may be asked to sketch the diagrams rather than calculate every point. Clearly indicate the maximum bending moment location and demonstrate the relationship that the slope of the BMD equals the shear force. This reinforces understanding of structural behavior.
在考试中，可能要求画出示意草图而非逐点计算。要清楚标明最大弯矩位置，并体现出 BMD 斜率等于剪力这一关系，以强化对结构行为的理解。
5. Selection of Engineering Materials | 工程材料的选择
A lightweight structural frame is required for a portable bridge. Evaluate the suitability of low carbon steel and 6061 aluminium alloy based on density, yield strength, Young’s modulus, and relative cost. Use the data provided to justify your recommendation.
一台便携式桥梁需要轻质结构框架。根据密度、屈服强度、杨氏模量和相对成本，评价低碳钢与 6061 铝合金的适用性。利用所给数据说明你的选材建议。
| Property | Low Carbon Steel | 6061 Al Alloy |
|---|---|---|
| Density (kg/m³) | 7850 | 2700 |
| Yield Stress (MPa) | 250 | 240 |
| Young’s Modulus (GPa) | 210 | 69 |
| Published by TutorHao | Year 12 工程 Revision Series | aleveler.com
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