Cross-disciplinary Integrated Question Training for Year 8 CIE Engineering | Year 8 CIE 工程：跨学科综合题型训练

📚 Cross-disciplinary Integrated Question Training for Year 8 CIE Engineering | Year 8 CIE 工程：跨学科综合题型训练

In Year 8 CIE Engineering, students are expected to apply knowledge from multiple disciplines—such as physics, mathematics, materials science, and design—to solve real-world problems. This integrated question training provides a series of challenging, cross-curricular tasks to develop analytical thinking and practical problem-solving skills. Each question is followed by a step-by-step solution that mirrors the type of reasoning required in examinations and project work.

在 Year 8 CIE 工程中，学生需要综合运用物理、数学、材料科学和设计等多学科知识来解决实际问题。本综合题型训练提供一系列跨学科的挑战性任务，旨在培养分析思维和实践解决问题的能力。每个问题后附有分步解答，反映了考试和项目工作中所需的推理方式。

1. Mechanics and Mathematics: Cantilever Beam Load Calculation | 力学与数学：悬臂梁负载计算

A cantilever beam of length 2 m supports a point load of 500 N at its free end. The beam has a solid rectangular cross‑section of width 0.1 m and height 0.2 m. By treating the beam as a linear elastic structure, calculate: (a) the maximum bending moment, (b) the maximum bending stress, and (c) the resulting maximum strain. Young's modulus for the beam material is 200 GPa.

一根悬臂梁长 2 m，在其自由端承受 500 N 的集中载荷。梁的截面为实心矩形，宽 0.1 m、高 0.2 m。假设梁为线弹性结构，计算：(a) 最大弯矩；(b) 最大弯曲应力；(c) 相应最大应变。梁材料的杨氏模量为 200 GPa。

Solution:

解答：

nt at the fixed support is M = F × L. Substituting the values: M = 500 N × 2 m = 1000 N·m.

步骤1：计算最大弯矩 M。对于自由端受集中载荷的悬臂梁，固定端弯矩为 M = F × L。代入数值：M = 500 N × 2 m = 1000 N·m。

Step 2: Determine the second moment of area I of the rectangular cross‑ = 0.2 m gives I = (0.1 × (0.2)³) / 12 = (0.1 × 0.008) / 12 = 0.0008 / 12 = 6.67 × 10⁻⁵ m⁴.

步骤2：计算矩形截面的截面惯性矩 I。公式为 I = (b × h³) / 12。代入 b = 0.1 m，h = 0.2 m 得：I = (0.1 × (0.2)³) / 12 = (0.1 × 0.008) / 12 = 0.0008 / 12 = 6.67 × 10⁻⁵ m⁴。

Step 3: Compute the maximum bending stress σ using the flexure formula σ = (M × y_max) / I, where y_max is the distance from the neutral axis to the outermost fibre (h/2). Here y_max 0.1 m. Thus σ = (1000 × 0.1) / 6.67 × 10⁻⁵ = 100 / 6.67 × 10⁻⁵ ≈ 1.50 × 10⁶ Pa = 1.50 MPa.

步骤3：使用弯曲公式 σ = (M × y_max) / I 计算最大弯曲应力，其中 y_max 为中性轴到最外纤维的距离 (h/2)。此处 y_max = 0.1 m。因此 σ = (1000 × 0.1) / 6.67 × 10⁻⁵ = 100 / 6.67 × 10⁻⁵ ≈ 1.50 × 10⁶ Pa = 1.50 MPa。

Step 4: Relate stress to strain via Hooke' ess). This very small strain confirms that the beam remains well within the elastic range.

步骤4：通过胡克定律 ε = σ / E 联系应力与应变。E = 200 GPa = 2 × 10¹¹ Pa，最大应变为 ε = 1.50 × 10⁶ / 2 × 10¹¹ = 7.5 × 10⁻⁶（无量纲）。这样极小的应变证实梁完全处于弹性范围内。

M = F × L
I = b h³ / 12
σ = M y_max / I
ε = σ / E

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2. Electrical Circuits: Series‑Parallel Resistors and LED | 电路：串并联电阻与 LED

parallel network. One branch of the parallel network contains a red LED (forward voltage V_f s a forward current of 20 mA for normal brightness. Determine the value of R2 and the total current drawn from the battery.

一个 9 V 电池与一个 R1 = 100 Ω 的串联电阻连接，然后为并联网络供电。并联网络的一个支路包含一个红色 LED（正向电压 V_f = 2 V）和一个未知电阻 R2 串联；另一个支路包含一个固定电阻 R3 = 150 Ω。LED 需要 20 mA 的正向电流才能正常发光。求 R2 的值以及电池提供的总电流。

Solution:

解答：

Step 1: Define the voltage across the parallel combination as V_p. The LED branch carries I_LED = 20 mA = 0.02 A. Applying Kirchhoff's voltage law to the LED branch gives: V_p = V_f + I_LED × R2 = 2 + 0.02 R2.

步骤1：设并联部分两端电压为 V_p。LED 支路电流为 I_LED = 20 mA = 0.02 A。对该支路应用基尔霍夫电压定律：V_p = V_f + I_LED × R2 = 2 + 0.02 R2。

Step 2: The current through R3 is I3 = V_p / 150. Hence the total current from the battery I_total = I3 + I_LED = V_p/150 + 0.02.

步骤2：通过 R3 的电流为 I3 = V_p / 150。因此电池总电流 I_total = I3 + I_LED = V_p/150 + 0.02。

Step 3: Apply Kirchhoff's voltage law to the entire loop: 9 V = I_total × 100 + V_p. Substitute I_total: 9 = (V_p/150 + 0.02) × 100 + V_p = (100/150)V_p + 2 + V_p = (2/3)V_p + V_p + 2 = (5/3)V_p + 2.

步骤3：对整个回路应用基尔霍夫电压定律：9 V = I_total × 100 + V_p。代入 I_total：9 = (V_p/150 + 0.02) × 100 + V_p = (100/150)V_p + 2 + V_p = (2/3)V_p + V_p + 2 = (5/3)V_p + 2。

Step 4: Solve for V_p: (5/3)V_p = 7 → V_p = 7 × 3/5 = 4.2 V. Then R2 = (V_p - 2) / 0.02 = (4.2 - 2) / 0.02 = 2.2 / 0.02 = 110 Ω. Finally I_total = 4.2/150 + 0.02 = 0.028 + 0.02 = 0.048 A = 48 mA.

步骤4：解出 V_p：(5/3)V_p = 7 → V_p = 7 × 3/5 = 4.2 V。然后 R2 = (V_p - 2) / 0.02 = (4.2 - 2) / 0.02 = 2.2 / 0.02 = 110 Ω。最终 I_total = 4.2/150 + 0.02 = 0.028 + 0.02 = 0.048 A = 48 mA。

V_p = I_LED R2 + V_f
I_total = I_branch1 + I_branch2

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3. Material Properties: Bridge Material Selection | 材料特性：桥梁材料选择

A pedestrian bridge must support a maximum tensile load of 100 kN. Four candidate materials are available, as shown in the table. A safety factor of 2 is required. Determine the minimum cross‑sectional area for each material and calculate the mass of a 1‑metre‑long tension member. Considering both weight and cost, recommend the most suitable material. Justify your choice.

一座人行桥须承受最大 100 kN 的拉伸载荷。现有四种候选材料（见下表），要求安全系数为 2。计算每种材料所需的最小横截面积，并求 1 米长受拉构件的质量。综合重量和成本，推荐最合适的材料并说明理由。

Material	密度 Density (kg/m³)	抗拉强度 Tensile Strength (MPa)	成本 Cost (£/kg)
低碳钢 Mild Steel	7850	400	0.50
铝合金 Aluminium Alloy	2700	300	1.20
松木 Pine Wood	500	40	0.30
碳纤维复合材料 CFRP	1600	1500	15.00

Solution:

解答：

r. Required area A = Force / σ_allow = 100,000 N / (σ_allow). Calculate for each material:

步骤1：许用应力 σ_allow = 抗拉强度 / 安全系数。所需面积 A = 力 / σ_allow = 100,000 N / (σ_allow)。对每种材料计算：

el: σ_allow = 400/2 = 200 MPa = 200 × 10⁶ Pa. A = 1 × 10⁵ / 2 × 10⁸ = 5 × 10⁻⁴ m² = 5 cm².

低碳钢：σ_allow = 400/2 = 200 MPa = 200 × 10⁶ Pa。A = 1 × 10⁵ / 2 × 10⁸ = 5 × 10⁻⁴ m² = 5 cm²。

Aluminium: σ_allow = 300/2 = 150 MPa. A = 1 × 10⁵ / 1.5 × 10⁸ = 6.67 × 10⁻⁴ m² ≈ 6.67 cm².

铝合金：σ_allow = 300/2 = 150 MPa。A = 1 × 10⁵ / 1.5 × 10⁸ = 6.67 × 10⁻⁴ m² ≈ 6.67 cm²。

Pine: σ_allow = 40/2 = 20 MPa. A = 1 × 10⁵ / 2 × 10⁷ = 5 × 10⁻³ m² = 50 cm².

松木：σ_allow = 40/2 = 20 MPa。A = 1 × 10⁵ / 2 × 10⁷ = 5 × 10⁻³ m² = 50 cm²。

CFRP: σ_allow = 1500/2 = 750 MPa. A = 1 × 10⁵ / 7.5 × 10⁸ = 1.33 × 10⁻⁴ m² ≈ 1.33 cm².

碳纤维复合材料：σ_allow = 1500/2 = 750 MPa。A = 1 × 10⁵ / 7.5 × 10⁸ = 1.33 × 10⁻⁴ m² ≈ 1.33 cm²。

e m = A × 1 m × density. Mild steel: m = 5 × 10⁻⁴ × 7850 = 3.925 kg, cost = 3.925 × 0.50 = £1.96. Aluminium: m = 6.67 × 10⁻⁴ × 2700 = 1.80 kg, cost = 1.80 × 1.20 = £2.16. Pine: m = 50 × 10⁻⁴ × 500 = 2.5 kg, cost = 2.5 × 0.30 = £0.75. CFRP: m = 1.33 × 10⁻⁴ × 1600 = 0.213 kg, cost = 0.213 × 15 = £3.20.

步骤2：每米质量 m = A × 1 m × 密度。低碳钢：m = 5 × 10⁻⁴

Published by TutorHao | Year 8 工程 Revision Series | aleveler.com

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