Year 12 CIE Engineering Unit Test Mock Analysis: Materials, Mechanics & Fluids | 十二年级CIE工程单元测试模拟卷解析：材料、力学与流体

📚 Year 12 CIE Engineering Unit Test Mock Analysis: Materials, Mechanics & Fluids | 十二年级CIE工程单元测试模拟卷解析：材料、力学与流体

This article provides a complete walkthrough of a 50‑mark unit test mock paper designed for Year 12 CIE Engineering. It covers materials behaviour, statics, fluid statics and energy methods. Every question is broken down into clear, step‑by‑step solutions, and each explanation is given in both English and Chinese to help you master the essential techniques.

本文为十二年级CIE工程设计的50分单元测试模拟卷提供完整解析，涵盖材料行为、静力学、流体静力学和能量方法。每道题都被拆解为清晰的逐步解答，且每个解释均提供中英双语对照，帮助你掌握核心解题技巧。

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1. Exam Structure and Topics | 考试结构与主题

The mock paper consists of five compulsory questions, each worth 10 marks. You are expected to complete the test within 60 minutes. The topics are drawn from the Year 12 CIE Engineering syllabus: mechanical properties of materials, shear and factor of safety, bending of beams, hydrostatic pressure, and work–energy with spring compression.

模拟卷包含五道必答题，每题10分，要求在60分钟内完成。主题取自十二年级CIE工程教学大纲：材料力学性能、剪切与安全系数、梁的弯曲、静水压力以及包含弹簧压缩的功–能原理。

A formula sheet is not provided, so you must know key relationships such as stress = force/area, bending stress = M/Z, hydrostatic force F = ρgAh̄ e awarded for the correct final answer, but working steps carry the majority of the marks.

卷中不提供公式表，因此你必须熟记应力 = 力/面积、弯曲应力 = M/Z、静水总压力 F = ρgAh̄ 和能量守恒等关键关系。最终正确答案有分值，但解题步骤占大部分分数。

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2. Q1: Ultimate Tensile Stress, Fracture Stress and Ductility | 题目1：极限拉伸应力、断裂应力与延展性

Question: A steel rod of original diameter 12 mm and gauge length 50 mm is tested in tension. The maximum recorded load is 45 kN and fracture occurs at 32 kN. At fracture the extension is 18 mm and the minimum neck diameter is 8.5 mm. Determine: (a) the ultimate tensile stress, (b) the fracture stress, (c) percentage elongation, and (d) percentage reduction in area.

题目：一根原始直径12 mm、标距50 mm的钢棒进行拉伸试验。最大记录载荷为45 kN，断裂载荷为32 kN。断裂时伸长量为18 mm，缩颈最小直径为8.5 mm。求：(a) 极限拉伸应力，(b) 断裂应力，(c) 延伸率，(d) 断面收缩率。

Step 1: Calculate the original cross‑sectional area. A₀ = πd₀²/4 = π × (12 × 10⁻³ m)² / 4 ≈ 1.131 × 10⁻⁴ m².

步骤1：计算原始横截面积。A₀ = πd₀²/4 = π × (12 × 10⁻³ m)² / 4 ≈ 1.131 × 10⁻⁴ m²。

σ_uts = F_max / A₀

Step 2: Ultimate tensile stress σ_uts = 45 000 N / 1.131 × 10⁻⁴ m² ≈ 398 × 10⁶ Pa = 398 MPa.

步骤2：极限拉伸应力 σ_uts = 45 000 N / 1.131 × 10⁻⁴ m² ≈ 398 × 10⁶ Pa = 398 MPa。

Step 3: Fracture stress uses the original area (unless otherwise stated). σ_frac = 32 000 N / 1.131 × 10⁻⁴ m² ≈ 283  e fracture stress would use the neck area, but CIE often expects nominal fracture stress.)

步骤3：断裂应力使用原始面积（除非指定其他方式）。σ_frac = 32 000 N / 1.131 × 10⁻⁴ m² ≈ 283 MPa。（注：真实断裂应力会用到缩颈面积，但CIE通常要求名义断裂应力。）

% Elongation = (ΔL / L₀) × 100%

Step 4: Percentage elongation = (18 mm / 50 mm) × 100% = 36%.

步骤4：延伸率 = (18 mm / 50 mm) × 100% = 36%。

% Reduction in Area = ( (A₀ − A_min) / A₀ ) × 100%

Step 5: Minimum area A_min = π × (8.5 × 10⁻³ m)² / 4 ≈ 5.674 × 10⁻⁵ m². Reduction = (1.131 × 10⁻⁴ − 5.674 × 10⁻⁵) / 1.131 × 10⁻⁴ × 100% ≈ 49.8%.

步骤5：最小面积 A_min = π × (8.5 × 10⁻³ m)² / 4 ≈ 5.674 × 10⁻⁵ m²。断面收缩率 = (1.131 × 10⁻⁴ − 5.674 × 10⁻⁵) / 1.131 × 10⁻⁴ × 100% ≈ 49.8%。

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3. Q2: Shear Stress and Factor of Safety in a Rivet | 题目2：铆钉中的剪切应力与安全系数

Question: A single rivet of diameter 10 mm is used in a lap joint. The applied load is 20 kN and the rivet is in single shear. The shear strength of the rivet material is 180 MPa. Find the average shear stress induced in the rivet and the resulting factor of safety.

题目：一个直径10 mm的铆钉用于搭接接头，铆钉受单剪切，承受载荷20 kN。铆钉材料的剪切强度为180 MPa。求铆钉内产生的平均剪切应力和相应的安全系数。

Step 1: The shear area for a single‑shear rivet is the cross‑sectional area of the shank. A = πd²/4 = π × (10 × 10⁻³ m)² / 4 ≈ 7.854 × 10⁻⁵ m².

步骤1：单剪切铆钉的剪切面积为杆身横截面积。A = πd²/4 = π × (10 × 10⁻³ m)² / 4 ≈ 7.854 × 10⁻⁵ m²。

τ_avg = F / A

Step 2: Average shear stress τ_avg = 20 000 N / 7.854 × 10⁻⁵ m² ≈ 2.547 × 10⁸ Pa = 254.7 MPa.

步骤2：平均剪切应力 τ_avg = 20 000 N / 7.854 × 10⁻⁵ m² ≈ 2.547 × 10⁸ Pa = 254.7 MPa。

Step 3: Factor of safety = shear strength / τ_avg = 180 MPa / 254.7  d fail under the given load. In design, a larger diameter or multiple rivets are required.

步骤3：安全系数 = 剪切强度 / τ_avg = 180 MPa / 254.7 MPa ≈ 0.71。系数低于1.0意味着该铆钉在给定载荷下会发生破坏。设计时需采用更大直径或多颗铆钉。

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4. Q3: Simply Supported Beam — Reactions, Bending Moment, Bending Stress | 题目3：简支梁——支反力、弯矩与弯曲应力

Question: A horizontal beam of length 4 m rests on simple supports at its ends. A vertical point load of 10 kN acts at the centre. The beam has a rectangular cross‑section 80 mm wide × 120  lculate the support reactions, the maximum bending moment, and the maximum bending stress.

题目：一根长4 m的水平梁两端简支。中点上作用一个10 kN的竖向集中荷载。梁的矩形截面宽80 mm、高120 mm。计算支反力、最大弯矩和最大弯曲应力。

Step 1: By symmetry, each reaction Rₐ = R_b = 10 kN / 2 = 5 kN.

步骤1：由对称性，每个支反力 Rₐ = R_b = 10 kN / 2 = 5 kN。

M_max = PL / 4 (for a central point load)

Step 2: Maximum bending moment occurs at mid‑span. M_max = (10 kN × 4 m) / 4 = 10 kN·m = 10 000 N·m.

步骤2：最大弯矩发生在跨中。M_max = (10 kN × 4 m) / 4 = 10 kN·m = 10 000 N·m。

Step 3: Section modulus Z for a rectangular section: Z = bd²/6 = (80 mm × (120 mm)²) / 6. Convert to metres for consistency: Z = (0.08 m × (0.12 m)²) / 6 = (0.08 × 0.0144) / 6 = 0.001152 / 6 = 1.92 × 10⁻⁴ m³.

步骤3：矩形截面的截面模量 Z：Z = bd²/6 = (80 mm × (120 mm)²) / 6。改用米保持一致：Z = (0.08 m × (0.12 m)²) / 6 = (0.08 × 0.0144) / 6 = 0.001152 / 6 = 1.92 × 10⁻⁴ m³。

σ_max = M_max / Z

Step 4: Maximum bending stress σ_max = 10 000 N·m / 1.92 × 10⁻⁴ m³ ≈ 5.21 × 10⁷ Pa = 52.1 MPa (compression on one face, tension on the other).

步骤4：最大弯曲应力 σ_max = 10 000 N·m / 1.92 × 10⁻⁴ m³ ≈ 5.21 × 10⁷ Pa = 52.1 MPa（一面受压，一面受拉）。

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5. Q4: Hydrostatic Force on a Submerged Rectangular Gate | 题目4：淹没矩形闸门上的静水压力

Question: A vertical rectangular gate of width 2 m and height 3 m is hinged at the top edge. The top edge is 1 m below the free water surface (ρ = 1000 kg  ce on the gate and the vertical distance from the water surface to the centre of pressure.

题目：一个宽2 m、高3 m的竖直矩形闸门在上缘铰接。上缘位于自由水面以下1 m处（ρ = 1000 kg m⁻³）。求作用在闸门上的静水总压力以及水面到压力中心的垂直距离。

Step 1: Depth to the centroid h̄ = 1 m + 3 m/2 = 2.5 m. Gate area A = 2 m × 3 m = 6 m².

步骤1：形心水深 h̄ = 1 m + 3 m/2 = 2.5 m。闸门面积 A = 2 m × 3 m = 6 m²。

F = ρ × g × A × h̄

Step 2: Hydrostatic force F = 1000 kg m⁻³ × 9.81 m s⁻² × 6 m² × 2.5 m = 147 150 N ≈ 147 kN.

步骤2：静水总压力 F = 1000 kg m⁻³ × 9.81 m s⁻² × 6 m² × 2.5 m = 147 150 N ≈ 147 kN。

Step 3: For a rectangle, moment of inertia about centroid I_G = b h³/12 = 2 m × (3 m)³ / 12 = 4.5 m⁴. Centre of pressure depth y_cp = h̄ + (I_G

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