📚 Year 12 Cambridge Engineering Unit Test Mock Paper Walkthrough | 剑桥工程12年级单元测试模拟卷解析
This walkthrough takes you through a typical unit test for Year 12 Cambridge Engineering, covering core topics such as stress and strain, statics, truss analysis, material selection, and basic electrical principles. Each question is fully solved with clear step-by-step reasoning, helping you to reinforce concepts and exam technique.
本文详细解析一套面向剑桥工程12年级的单元测试模拟卷，涵盖应力与应变、静力学、桁架分析、材料选择和基本电学原理等核心主题。每道题都配有清晰的分步推理，帮助你巩固概念，提升应试技巧。
1. Question 1: Tensile Stress, Strain and Young’s Modulus | 问题1：拉伸应力、应变与杨氏模量
A steel rod of diameter 10 mm and original length 2.0 m is subjected to a tensile force of 15 kN and extends by 1.2 mm. Calculate the tensile stress in the rod, the tensile strain, and Young’s modulus of the material.
一根直径10 mm、原始长度2.0 m的钢杆承受15 kN的拉伸载荷，伸长量为1.2 mm。计算杆的拉伸应力、拉伸应变以及材料的杨氏模量。
Step 1 – Cross‑sectional area: For a circular rod, A = πd²/4. Convert d to metres: 10 mm = 0.01 m. A = π×(0.01)²÷4 = 7.854×10⁻⁵ m².
步骤1 – 横截面积：圆杆面积 A = πd²/4。将直径换算为米：10 mm = 0.01 m。A = π×(0.01)²÷4 = 7.854×10⁻⁵ m²。
Step 2 – Tensile stress: σ = F/A = 15 000 N ÷ 7.854×10⁻⁵ m² = 1.91×10⁸ Pa. In engineering units this is 191 MPa (1 MPa = 10⁶ Pa).
步骤2 – 拉伸应力：σ = F/A = 15 000 N ÷ 7.854×10⁻⁵ m² = 1.91×10⁸ Pa。工程上常写作191 MPa（1 MPa = 10⁶ Pa）。
Step 3 – Tensile strain: ε = ΔL/L₀ = 1.2 mm ÷ 2000 mm = 0.0006 = 6.0×10⁻⁴ (no units).
步骤3 – 拉伸应变：ε = ΔL/L₀ = 1.2 mm ÷ 2000 mm = 0.0006 = 6.0×10⁻⁴（无单位）。
Step 4 – Young’s modulus: E = σ/ε = 191×10⁶ Pa ÷ 6.0×10⁻⁴ = 3.18×10¹¹ Pa ≈ 318 GPa. This value is a little higher than typical structural steel, but the method remains valid for any linear‑elastic material.
步骤4 – 杨氏模量：E = σ/ε = 191×10⁶ Pa ÷ 6.0×10⁻⁴ = 3.18×10¹¹ Pa ≈ 318 GPa。此值略高于典型结构钢，但计算适用于任何线弹性材料。
2. Question 2: Reaction Forces on a Simply Supported Beam | 问题2：简支梁的支座反力
A horizontal beam of length 4.0 m is simply supported at both ends. It carries a single point load of 5.0 kN downwards at its midpoint. Ignoring the self‑weight of the beam, determine the vertical reaction forces at each support.
一根长4.0 m的水平简支梁在两端被支撑，在中点承受5.0 kN的集中力竖直向下作用。忽略梁的自重，求每个支座处的竖向反力。
By symmetry, the reactions must be equal because the load is applied at the centre. Alternatively, take moments about the left support (A): clockwise moment due to the load = 5 kN × 2 m = 10 kN·m. For equilibrium, the anti‑clockwise moment due to the right reaction R_B must be equal: R_B × 4 m = 10 kN·m, giving R_B = 2.5 kN. Vertically, ΣF_y = 0 → R_A + R_B – 5 = 0 → R_A = 2.5 kN.
由对称性可知两反力相等，因为荷载作用于中点。也可取左支座A为矩心：荷载产生的顺时针力矩 = 5 kN × 2 m = 10 kN·m。平衡条件要求右支座反力R_B产生的逆时针力矩相等：R_B × 4 m = 10 kN·m，得R_B = 2.5 kN。再由竖向合力为零：R_A + R_B – 5 = 0 → R_A = 2.5 kN。
3. Question 3: Principle of Moments | 问题3：力矩原理
A uniform metre ruler pivoted at its 50 cm mark is balanced when a 2.0 N weight is hung at the 20 cm mark and an unknown weight W is hung at the 80 cm mark. Find W.
一把均匀的米尺在其50 cm刻度处支起，当在20 cm刻度处悬挂2.0 N的重物并在80 cm刻度处悬挂未知重量W时，尺子达到平衡。求W。
Take moments about the pivot. The 2.0 N force acts 30 cm to the left of the pivot (50 – 20 = 30 cm), producing an anti‑clockwise moment. W acts 30 cm to the right (80 – 50 = 30 cm), producing a clockwise moment. For equilibrium: 2.0 N × 0.30 m = W × 0.30 m, therefore W = 2.0 N. The ruler’s own weight acts at the pivot and produces no moment.
对支点取矩。2.0 N力在支点左侧30 cm处（50 – 20 = 30 cm），产生逆时针力矩；W在支点右侧30 cm处（80 – 50 = 30 cm），产生顺时针力矩。平衡时：2.0 N × 0.30 m = W × 0.30 m，所以 W = 2.0 N。尺子自身重力通过支点，无力矩。
4. Question 4: Method of Joints for a Simple Truss | 问题4：简单桁架的节点法
A symmetrical truss has two pinned supports at A and C, and a ridge joint B. Members AB and BC are each 45° to the horizontal. A vertical downward load of 10 kN is applied at B. Assuming all joints are frictionless pins, determine the force in member AB and state whether it is tension or compression.
一对称桁架在A、C处为铰支座，顶点为B。杆件AB和BC与水平面各成45°角。B点承受10 kN的垂直向下荷载。假设所有节点均为无摩擦铰，求杆件AB的内力并指明其为拉力还是压力。
Isolate joint B. Assume both AB and BC are in tension (pulling away from B). The vertical equilibrium gives: F_AB sin45° + F_BC sin45° = 10 kN (upwards). By symmetry, F_AB = F_BC. Let this force be F. Then 2F sin45° = 10 kN. sin45° = √2/2 ≈ 0.707. F = 10 / (2 × 0.707) ≈ 7.07 kN. Since the calculated force is positive and pulls away from the joint, member AB is in tension at 7.07 kN. (A compression member would push towards the joint.)
隔离B节点。假设AB和BC均受拉力（从节点向外拉）。竖向平衡方程：F_AB sin45° + F_BC sin45° = 10 kN（向上为正）。由于对称，F_AB = F_BC，记作F。于是 2F sin45° = 10 kN。sin45° = √2/2 ≈ 0.707。F = 10 / (2 × 0.707) ≈ 7.07 kN。所得力为正且指向离开节点，故AB受拉力，大小为7.07 kN。（若为压力则会指向节点。）
5. Question 5: Material Selection – Specific Stiffness | 问题5：材料选择 – 比刚度
An engineer needs a light, stiff tie rod. Three candidate materials are available: steel (E = 200 GPa, density ρ = 7800 kg/m³), aluminium alloy (E = 70 GPa, ρ = 2700 kg/m³), and titanium alloy (E = 110 GPa, ρ = 4500 kg/m³). Which material offers the highest specific stiffness (E/ρ) and is therefore the best choice based on this criterion?
工程师需要一根轻质高刚度的拉杆。现有三种候选材料：钢（E = 200 GPa，ρ = 7800 kg/m³）、铝合金（E = 70 GPa，ρ = 2700 kg/m³）和钛合金（E = 110 GPa，ρ = 4500 kg/m³）。哪种材料的比刚度（E/ρ）最高，从而依此标准成为最佳选择？
Calculate E/ρ for each material using consistent SI units (Young’s modulus in Pa and density in kg/m³):
Steel: (200×10⁹) / 7800 ≈ 25.6×10⁶ m²/s².
Aluminium: (70×10⁹) / 2700 ≈ 25.9×10⁶ m²/s².
Titanium: (110×10⁹) / 4500 ≈ 24.4×10⁶ m²/s².
The aluminium alloy has the highest specific stiffness, closely followed by steel. Therefore, on a stiffness‑per‑mass basis, the aluminium alloy is the preferred choice.
用统一的SI单位计算各材料的E/ρ（杨氏模量用Pa，密度用kg/m³）：
钢：(200×10⁹) / 7800 ≈ 25.6×10⁶ m²/s²。
铝合金：(70×10⁹) / 2700 ≈ 25.9×10⁶ m²/s²。
钛合金：(110×10⁹) / 4500 ≈ 24.4×10⁶ m²/s²。
铝合金的比刚度最高，钢紧随其后。因此，在单位质量的刚度上，铝合金是更佳选择。
6. Question 6: Series and Parallel Resistor Circuits | 问题6：电阻的串联与并联
A 12 V DC supply is connected to two resistors: R₁ = 10 Ω and R₂ = 20 Ω. (a) First, the resistors are connected in series. Find the total resistance, circuit current, and the voltage across each resistor. (b) Then, the resistors are connected in parallel. Find the total resistance, total current drawn from the supply, and the current through each resistor.
一个12 V直流电源连接到两个电阻：R₁ = 10 Ω，R₂ = 20 Ω。(a) 首先，电阻串联连接。求总电阻、电路电流和每个电阻两端的电压。(b) 然后，电阻并联连接。求总电阻、电源提供的总电流以及流过每个电阻的电流。
(a) Series: R_total = R₁ + R₂ = 10 + 20 = 30 Ω. Current I = V / R_total = 12 V / 30 Ω = 0.4 A. Voltage across R₁: V₁ = I×R₁ = 0.4 × 10 = 4 V. Voltage across R₂: V₂ = I×R₂ = 0.4 × 20 = 8 V. Note that V₁ + V₂ = 12 V.
(a) 串联：R_total = R₁ + R₂ = 10 + 20 = 30 Ω。电流 I = V / R_total = 12 V / 30 Ω = 0.4 A。R₁两端电压 V₁ = I×R₁ = 0.4 × 10 = 4 V。R₂两端电压 V₂ = I×R₂ = 0.4 × 20 = 8 V。注意 V₁ + V₂ = 12 V。
(b) Parallel: 1/R_total = 1/R₁ + 1/R₂ = 1/10 + 1/20 = 3/20 → R_total = 20/3 ≈ 6.67 Ω. Total current I_total = V / R_total = 12 / 6.67 ≈ 1.8 A. Current through R₁: I₁ = V / R₁ = 12 / 10 = 1.2 A. Through R₂: I₂ = V / R₂ = 12 / 20 = 0.6 A. The sum of branch currents equals the total current.
(b) 并联：1/R_total = 1/R₁ + 1/R₂ = 1/10 + 1/20 = 3/20 → R_total = 20/3 ≈ 6.67 Ω。总电流 I_total = V / R_total = 12 / 6.67 ≈ 1.8 A。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。支路电流之和等于总电流。
7. Question 7: Electrical Power and Energy Consumption | 问题7：电功率与能量消耗
An electric heater is rated at 2.0 kW and designed for a 230 V mains supply. (a) Determine the current drawn by the heater and its resistance. (b) Calculate the electrical energy, in kilowatt‑hours (kWh), consumed if the heater operates for 30 minutes.
一台电热器额定功率为2.0 kW，额定电压为230 V。(a) 求电热器消耗的电流及其电阻。(b) 若电热器工作30分钟，计算消耗的电能（用千瓦时表示）。
(a) Power P = V × I, so I = P / V = 2000 W / 230 V ≈ 8.70 A. Resistance R = V / I = 230 / 8.70 ≈ 26.4 Ω, or using P = V²/R → R = V²/P = 230²/2000 = 52900/2000 = 26.45 Ω.
(a) 功率 P = V × I，所以 I = P / V = 2000 W / 230 V ≈ 8.70 A。电阻 R = V / I = 230 / 8.70 ≈ 26.4 Ω，或用 P = V²/R → R = V²/P = 230²/2000 = 52900/2000 = 26.45 Ω。
(b) Energy E = Power × time. Time must be in hours: 30 min = 0.5 h. E = 2.0 kW × 0.5 h = 1.0 kWh. This is the amount registered by an electricity meter.
(b) 电能 E = 功率 × 时间。时间须换算为小时：30 min = 0.5 h。E = 2.0 kW × 0.5 h = 1.0 kWh。这正是电表计量的数值。
8. Question 8: Interpreting Engineering Drawings – Orthographic Projection | 问题8：解读工程图纸 – 正交投影
An orthographic drawing (third‑angle projection) shows a front view and a top view of a rectangular block. The front view displays a hidden circle near the centre, represented by dashed lines. The top view shows a circle with a cross‑hair centre mark. Identify the feature and describe how it appears in each view.
一张正交投影图（第三角投影）展示了一个矩形块的前视图和俯视图。前视图的中心附近显示了一个用虚线表示的隐藏圆；俯视图显示了一个带有十字中心标记的圆。请识别该特征，并描述其在各视图中的表现形式。
The feature is a circular through‑hole drilled vertically through the block. In the top view, the hole appears as a true circle because we look along its axis. The cross‑hair centre mark indicates the hole’s centre. In the front view, the hole is seen from the side, so its outline is hidden; it is drawn with dashed lines (hidden detail) as a rectangle– the side projection of the cylindrical hole. This is a standard convention in technical drawing.
该特征是一个垂直贯穿整个块体的圆形通孔。在俯视图中，由于视线沿孔轴线方向，孔显示为一个真实的圆，十字中心线标识其圆心。在前视图中，孔从侧面观察，轮廓不可见，因此用虚线（隐藏细节）绘制为矩形——这是圆柱孔的侧向投影。这是工程制图中的标准表达惯例。
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