📚 CAIE Engineering Year 13 Unit Test Mock Paper Analysis | CAIE工程13年级单元测试模拟卷解析
This article provides a full breakdown of a mock unit test designed for the CAIE Year 13 Engineering syllabus. Each question is carefully selected to mirror the style and depth of real assessment, covering core topics such as stress and strain, beam analysis, circuit theory, thermodynamics, trusses, torsion, dynamics, fluid mechanics, material selection, and control systems. Detailed step-by-step solutions are given in both English and Chinese to support bilingual learners and reinforce understanding of key engineering principles.
本文为专为CAIE 13年级工程教学大纲设计的单元测试模拟卷提供完整解析。每道题目都精心挑选，模拟真实评估的风格与深度，涵盖应力应变、梁分析、电路理论、热力学、桁架、扭转、动力学、流体力学、材料选择与控制系统等核心主题。文中提供详细的逐步求解过程，并采用中英双语对照，以帮助双语学习者巩固关键工程原理。
1. Question 1: Tensile Stress and Strain | 第1题：拉伸应力与应变计算
Problem: A steel rod of diameter 20 mm carries a tensile load of 50 kN. Young’s modulus E = 200 GPa. Determine (a) the tensile stress, (b) the tensile strain.
题目：一根直径20 mm的钢杆承受50 kN的拉伸载荷。杨氏模量E = 200 GPa。求 (a) 拉伸应力，(b) 拉伸应变。
Cross-sectional area A = πd²/4 = π(20×10⁻³ m)²/4 = 3.142×10⁻⁴ m².
横截面积 A = πd²/4 = π(20×10⁻³ m)²/4 = 3.142×10⁻⁴ m²。
Tensile stress σ = F/A = 50×10³ N / 3.142×10⁻⁴ m² = 1.59×10⁸ Pa = 159 MPa.
拉伸应力 σ = F/A = 50×10³ N / 3.142×10⁻⁴ m² = 1.59×10⁸ Pa = 159 MPa。
Using Hooke’s law, strain ε = σ/E = 159×10⁶ Pa / 200×10⁹ Pa = 7.95×10⁻⁴ (dimensionless).
根据胡克定律，应变 ε = σ/E = 159×10⁶ Pa / 200×10⁹ Pa = 7.95×10⁻⁴（无量纲）。
The rod experiences a stress well below typical steel yield strength, and the strain is small, confirming elastic behaviour.
该钢杆的应力远低于典型钢材屈服强度，且应变很小，证实处于弹性范围内。
2. Question 2: Shear Force and Bending Moment Diagrams | 第2题：剪力与弯矩图
Problem: A simply supported beam of length 4 m carries a central point load of 10 kN. Draw the shear force and bending moment diagrams, and state the maximum bending moment.
题目：一简支梁跨度4 m，跨中承受10 kN集中载荷。绘制剪力图和弯矩图，并求出最大弯矩。
Reactions at supports: by symmetry, Rₐ = R₆ = 5 kN.
支座反力：由对称性，Rₐ = R₆ = 5 kN。
Shear force: +5 kN from left support to midspan, then drops by 10 kN to -5 kN to right support. Diagram is two horizontal lines with a step at centre.
剪力：从左支座至跨中为+5 kN，随后突降10 kN，变为-5 kN直到右支座。图形为两段水平线，跨中处有阶跃。
Bending moment: M(x) = 5x kN·m for 0 ≤ x ≤ 2 m. At midspan (x=2 m), Mₘₐₓ = 5×2 = 10 kN·m. Diagram is a triangle peaking at centre.
弯矩：对于0 ≤ x ≤ 2 m，M(x) = 5x kN·m。在跨中(x=2 m)处，Mₘₐₓ = 5×2 = 10 kN·m。弯矩图为三角形，中央峰值最大。
Maximum bending moment = 10 kN·m, occurring directly under the point load.
最大弯矩为10 kN·m，出现在集中载荷正下方。
3. Question 3: DC Circuit Analysis | 第3题：直流电路分析
Problem: Two resistors R₁ = 10 Ω and R₂ = 20 Ω are connected in parallel across a 12 V battery. Find the total current drawn from the battery and the current through each resistor.
题目：两个电阻R₁=10 Ω和R₂=20 Ω并联后接到12 V电池两端。求电池提供的总电流以及流过各电阻的电流。
Equivalent resistance for parallel: 1/Rₑ = 1/10 + 1/20 = 3/20 ⇒ Rₑ = 20/3 ≈ 6.67 Ω.
并联等效电阻：1/Rₑ = 1/10 + 1/20 = 3/20 ⇒ Rₑ = 20/3 ≈ 6.67 Ω。
Total current Iₜₒₜ = V / Rₑ = 12 V / (20/3 Ω) = 1.8 A.
总电流 Iₜₒₜ = V / Rₑ = 12 V / (20/3 Ω) = 1.8 A。
Current through R₁: I₁ = V / R₁ = 12/10 = 1.2 A. Current through R₂: I₂ = V / R₂ = 12/20 = 0.6 A. Sum = 1.8 A, verifying Kirchhoff’s current law.
流过R₁的电流：I₁ = V / R₁ = 12/10 = 1.2 A。流过R₂的电流：I₂ = V / R₂ = 12/20 = 0.6 A。总和为1.8 A，验证了基尔霍夫电流定律。
4. Question 4: Isothermal Expansion of an Ideal Gas | 第4题：理想气体的等温膨胀
Problem: Two moles of an ideal gas expand isothermally at 300 K from an initial volume of 0.1 m³ to a final volume of 0.3 m³. Calculate the work done by the gas.
题目：2 mol理想气体在300 K下等温膨胀，初始体积0.1 m³，最终体积0.3 m³。计算气体所做的功。
For isothermal process, work done W = nRT ln(V₂/V₁). R = 8.31 J/(mol·K).
对于等温过程，做功 W = nRT ln(V₂/V₁)。R = 8.31 J/(mol·K)。
W = 2 × 8.31 × 300 × ln(0.3/0.1) = 4986 × ln(3) ≈ 4986 × 1.0986 = 5477 J.
W = 2 × 8.31 × 300 × ln(0.3/0.1) = 4986 × ln(3) ≈ 4986 × 1.0986 = 5477 J。
The gas does 5.48 kJ of work on its surroundings during this expansion.
在膨胀过程中，气体对外做功约5.48 kJ。
5. Question 5: Truss Analysis by Method of Joints | 第5题：节点法分析桁架
Problem: A simple triangular truss ABC has pin supports at A and B, with a vertical load of 5 kN applied at apex C. Span AB = 3 m, height from AB to C = 2 m. Find the forces in members AC and BC.
题目：一个简单的三角形桁架ABC，A和B为铰支座，顶点C承受5 kN竖直载荷。跨度AB=3 m，C到AB的垂直高度=2 m。求杆件AC和BC的内力。
By symmetry, vertical reactions at A and B: Rₐᵥ = R₆ᵥ = 2.5 kN. Isolate joint C: vertical equilibrium: FAC sinθ + FBC sinθ = 5 kN. Horizontal: FAC cosθ = FBC cosθ, so forces equal.
由对称性，A和B的竖直反力：Rₐᵥ = R₆ᵥ = 2.5 kN。隔离节点C：竖直平衡：FAC sinθ + FBC sinθ = 5 kN。水平方向：FAC cosθ = FBC cosθ，因此两力相等。
Geometry: AC = √(1.5²+2²)=2.5 m, sinθ = 2/2.5 = 0.8, cosθ = 1.5/2.5 = 0.6.
几何关系：AC = √(1.5²+2²)=2.5 m，sinθ = 2/2.5 = 0.8，cosθ = 1.5/2.5 = 0.6。
2 × FAC × 0.8 = 5 → FAC = 3.125 kN (compression, since it points toward joint). FBC = 3.125 kN compression.
2 × FAC × 0.8 = 5 → FAC = 3.125 kN（受压，因为指向节点）。FBC = 3.125 kN 受压。
6. Question 6: Torsion in a Solid Shaft | 第6题：实心轴的扭转
Problem: A solid circular shaft of diameter 30 mm transmits a torque of 200 N·m. Determine the maximum shear stress. If the shaft is 1 m long and G = 80 GPa, calculate the angle of twist.
题目：一直径30 mm的实心圆轴传递200 N·m的扭矩。求最大剪应力。若轴长1 m，G=80 GPa，计算扭转角。
Polar second moment of area for solid circle: J = (π/32)d⁴ = (π/32)(0.03)⁴ = 7.95×10⁻⁸ m⁴.
实心圆截面的极惯性矩：J = (π/32)d⁴ = (π/32)(0.03)⁴ = 7.95×10⁻⁸ m⁴。
Maximum shear stress τₘₐₓ = T·r / J, with r = d/2 = 0.015 m. τₘₐₓ = 200 × 0.015 / 7.95×10⁻⁸ = 37.7 MPa.
最大剪应力 τₘₐₓ = T·r / J，其中 r = d/2 = 0.015 m。τₘₐₓ = 200 × 0.015 / 7.95×10⁻⁸ = 37.7 MPa。
Angle of twist φ = TL / GJ = (200 × 1) / (80×10⁹ × 7.95×10⁻⁸) = 0.0314 rad ≈ 1.8°.
扭转角 φ = TL / GJ = (200 × 1) / (80×10⁹ × 7.95×10⁻⁸) = 0.0314 rad ≈ 1.8°。
7. Question 7: Dynamics – Connected Masses | 第7题：动力学——连接体问题
Problem: A 5 kg mass hangs vertically, connected by a light inextensible string over a frictionless pulley to a 3 kg mass on a smooth horizontal table. Find the acceleration of the system and the tension in the string. Take g = 9.81 m/s².
题目：5 kg的物体竖直悬挂，通过一根轻质不可伸长的绳子跨过无摩擦滑轮与光滑水平桌上的3 kg物体相连。求系统加速度和绳中张力。取g=9.81 m/s²。
For mass on table (3 kg): T = 3a. For hanging mass (5 kg): 5g – T = 5a.
对桌上物体（3 kg）：T = 3a。对悬挂物体（5 kg）：5g – T = 5a。
Substitute: 5×9.81 – 3a = 5a → 49.05 = 8a → a = 6.13 m/s².
代入：5×9.81 – 3a = 5a → 49.05 = 8a → a = 6.13 m/s²。
Tension T = 3 × 6.13 = 18.4 N.
张力 T = 3 × 6.13 = 18.4 N。
8. Question 8: Bernoulli’s Equation for Fluid Flow | 第8题：流体力学的伯努利方程
Problem: Water flows through a horizontal pipe that narrows from diameter 0.1 m to 0.05 m. At the wider section, pressure is 200 kPa and velocity is 3 m/s. Find the pressure at the narrow section. Density ρ = 1000 kg/m³.
题目：水流经一水平管道，管径从0.1 m收窄至0.05 m。在粗段，压强为200 kPa，流速为3 m/s。求窄段压强。密度ρ=1000 kg/m³。
From continuity, A₁v₁ = A₂v₂. A₁ = π(0.05)², A₂ = π(0.025)². v₂ = v₁×(A₁/A₂) = 3 × (0.05²/0.025²) = 3 × 4 = 12 m/s.
根据连续性方程，A₁v₁ = A₂v₂。A₁ = π(0.05)²，A₂ = π(0.025)²。v₂ = v₁×(A₁/A₂) = 3 × (0.05²/0.025²) = 3 × 4 = 12 m/s。
Bernoulli: p₁ + ½ρv₁² = p₂ + ½ρv₂² (horizontal, same height). 200×10³ + 0.5×1000×3² = p₂ + 0.5×1000×12².
伯努利方程：p₁ + ½ρv₁² = p₂ + ½ρv₂²（水平，同高）。200×10³ + 0.5×1000×3² = p₂ + 0.5×1000×12²。
204500 = p₂ + 72000 → p₂ = 132.5 kPa.
204500 = p₂ + 72000 → p₂ = 132.5 kPa。
The pressure drops as velocity increases, consistent with energy conservation.
随着速度增大，压强下降，符合能量守恒。
9. Question 9: Material Selection Using Ashby Charts | 第9题：使用Ashby图进行材料选择
Problem: Explain how you would use an Ashby chart to select the lightest material for a stiff beam of given length L, which must support a central load F without exceeding a maximum deflection δ. Derive the material index.
题目：解释如何利用Ashby图，为一根给定长度L的刚性梁选择最轻材料，该梁需承受中心载荷F且最大挠度不超过δ。推导材料指标。
The design requirement: stiffness S = F/δ must be achieved. For a beam in bending, stiffness is proportional to EI/L³, where E is Young’s modulus and I is second moment of area. For a given cross-section shape, I ∝ A², so mass m ∝ ρAL, and stiffness S ∝ EA²/L³.
设计要求：必须达到刚度S = F/δ。对于受弯梁，刚度与EI/L³成正比，E为杨氏模量，I为截面惯性矩。给定截面形状，I ∝ A²，因此质量m ∝ ρAL，刚度S ∝ EA²/L³。
Eliminating A gives m ∝ (ρ/E¹ᐟ²) S¹ᐟ² L⁵ᐟ². For minimum mass at fixed stiffness, we need to maximise the material index M = E¹ᐟ²/ρ, or its reciprocal E/ρ if density is in denominator.
消去A得到 m ∝ (ρ/E¹ᐟ²) S¹ᐟ² L⁵ᐟ²。在刚度固定的条件下追求最小质量，需要最大化材料指标 M = E¹ᐟ²/ρ，或考虑ρ在分母时使用E/ρ²。
On an Ashby chart of Young’s modulus vs density, use a selection line with slope 2 (since E ∝ ρ² for constant performance), corresponding to E¹ᐟ²/ρ = constant. Materials above the line are candidates; the lightest can be identified.
在杨氏模量-密度Ashby图上，使用斜率为2的筛选线（因为等性能下E ∝ ρ²），对应E¹ᐟ²/ρ = 常数。线上方的材料为候选，可识别最轻的材料。
10. Question 10: Control Systems – Closed-Loop Transfer Function | 第10题：控制系统——闭环传递函数
Problem: A unity-feedback system has an open-loop transfer function G(s) = 10/[s(s+2)]. Derive the closed-loop transfer function and comment on stability.
题目：一单位反馈系统的开环传递函数为G(s) = 10/[s(s+2)]。推导闭环传递函数，并评论其稳定性。
Closed-loop transfer function T(s) = G(s) / [1 + G(s)] = [10/(s(s+2))] / [1 + 10/(s(s+2))] = 10 / [s(s+2) + 10] = 10 / (s² + 2s + 10).
闭环传递函数 T(s) = G(s) / [1 + G(s)] = [10/(s(s+2))] / [1 + 10/(s(s+2))] = 10 / [s(s+2) + 10] = 10 / (s² + 2s + 10)。
The characteristic equation is s² + 2s + 10 = 0. Roots are s = [-2 ± √(4 – 40)]/2 = -1 ± j3. The poles are complex with negative real part, indicating stable response with damped oscillations.
特征方程为 s² + 2s + 10 = 0。根为 s = [-2 ± √(4 – 40)]/2 = -1 ± j3。极点为具有负实部的复数，表明系统稳定且呈现阻尼振荡响应。
The system is inherently stable and will settle to a step input with an overshoot determined by the damping ratio ζ = 1/√10.
系统固有稳定，对阶跃输入的响应将收敛，超调量由阻尼比 ζ = 1/√10 决定。
Published by TutorHao | Engineering Revision Series | aleveler.com
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