Year 11 CCEA Engineering: Interdisciplinary Exam Practice | 跨学科综合题型训练

📚 Year 11 CCEA Engineering: Interdisciplinary Exam Practice | 跨学科综合题型训练

Welcome to this interdisciplinary revision resource for CCEA GCSE Engineering. In these exercises, you will apply concepts from mathematics, physics, chemistry, and design to solve realistic engineering problems. Mastering these integrated skills is essential for success in the exam.

欢迎来到这个 CCEA GCSE 工程的跨学科复习资源。在这些练习中，你将运用数学、物理、化学和设计等概念来解决实际的工程问题。掌握这些综合技能对考试成功至关重要。

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1. Stress, Strain and Young Modulus | 应力、应变与杨氏模量

Engineers must predict how materials behave under load. Stress (σ) is internal force per unit area, strain (ε) is extension divided by original length, and Young modulus (E) links them: E = σ / ε. These calculations combine mechanics and algebra.

工程师必须预测材料在载荷下的行为。应力 (σ) 是内部力除以面积，应变 (ε) 是伸长量除以原长，杨氏模量 (E) 将它们联系起来：E = σ / ε。这些计算结合了力学与代数。

ss, strain if it stretches by 2.4 mm, and the Young modulus of the steel. (Use g = 9.8 m/s²)

综合问题：一根直径 12 mm、原长 5 m 的钢缆吊起 800 kg 的质量。若其伸长 2.4 mm，计算拉伸应力、应变以及钢的杨氏模量。（取 g = 9.8 m/s²）

Step 1: Find the applied force.

步骤 1：求作用力。

F = m × g = 800 × 9.8 = 7840 N

Step 2: Cross-sectional area A = π(d/2)² = π(0.012/2)² = π(0.006)² ≈ 1.131 × 10^-4 m². (Convert mm to m)

步骤 2：横截面积 A = π(d/2)² = π(0.012/2)² = π(0.006)² ≈ 1.131 × 10^-4 m²。（将 mm 转换为 m）

σ = F / A = 7840 / (1.131 × 10^-4) ≈ 6.93 × 10⁷ Pa = 69.3 MPa

Step 3: Strain uses ΔL = 2.4 mm and L₀ = 5000 mm.

步骤 3：应变使用 ΔL = 2.4 mm 和 L₀ = 5000 mm。

ε = ΔL / L₀ = 2.4 / 5000 = 4.8 × 10^-4

Step 4: Young modulus E = σ / ε = 6.93 × 10⁷ / (4.8 × 10^-4) ≈ 1.44 × 10¹¹ Pa = 144 GPa. This lies within the typical range for structural steel.

步骤 4：杨氏模量 E = σ / ε = 6.93 × 10⁷ / (4.8 × 10^-4) ≈ 1.44 × 10¹¹ Pa = 144 GPa。该值位于结构钢的典型范围内。

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2. Thermal Expansion and Induced Stress | 热膨胀与诱导应力

Materials expand when heated, following ΔL = α L₀ ΔT. If expansion is restrained, thermal stress develops: σ_thermal = E α ΔT. This links thermal physics and mechanics.

材料受热膨胀，遵循 ΔL = α L₀ ΔT。若膨胀受到约束，会产生热应力：σ_thermal = E α ΔT。这连接了热物理与力学。

Problem: A copper pipe 15 m long is installed at 10°C. Its temperature rises to 40°C. Copper: α = 16.5 × 10^-6 /°C, E = 120 GPa. (a) Find the free expansion ΔL. (b) If the ends are rigidly fixed, what compressive stress develops?

问题：一根 15 m 长的铜管在 10°C 安装。温度升至 40°C。铜：α = 16.5 × 10^-6 /°C，E = 120 GPa。(a) 求自由膨胀量 ΔL。(b) 若两端刚性固定，产生的压应力多大？

Solution (a): ΔT = 40 - 10 = 30°C.

解答 (a)：ΔT = 40 - 10 = 30°C。

ΔL = α L₀ ΔT = 16.5 × 10^-6 × 15 × 30 = 7.425 × 10^-3 m = 7.43 mm

(b) Thermal stress uses the restrained condition:

(b) 热应力使用约束条件：

σ = E α ΔT = 120 × 10⁹ × 16.5 × 10^-6 × 30 = 59.4 × 10⁶ Pa = 59.4 MPa

Such stress could cause buckling; engineers add expansion joints to absorb movement.

这样的应力可能导致屈曲；工程师会加入伸缩缝来吸收位移。

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3. Voltage Dividers with Sensors | 传感器与分压器

A voltage divider produces an output that depends on resistance ratios: V_out = V_in × (R₂ / (R₁ + R₂)). Substituting a sensor (thermistor, LDR) allows the circuit to respond to physical change, integrating electronics with control.

分压器产生的输出电压取决于电阻比：V_out = V_in × (R₂ / (R₁ + R₂))。将传感器（热敏电阻、光敏电阻）接入，可使电路响应物理变化，融合电子与控制。

Problem: A thermistor R_T and a 10 kΩ fixed resistor R₁ form a divider supplied by 5 V. R_T is in the R₂ position (output across R_T). At 20°C, R_T = 12 kΩ. At 30°C, R_T drops to 5 kΩ. Calculate V_out at both temperatures and comment.

问题：热敏电阻 R_T 与 10 kΩ 固定电阻 R₁ 组成分压器，供电 5 V。R_T 在 R₂ 位置（输出跨接在 R_T 两端）。20°C 时 R_T = 12 kΩ，30°C 时降至 5 kΩ。计算两种温度下的 V_out 并加以评论。

At 20°C:

在 20°C 时：

V_out = 5 × (12 / (10 + 12)) = 5 × (12/22) ≈ 2.73 V

At 30°C:

在 30°C 时：

V_out = 5 × (5 / (10 + 5)) = 5 × (5/15) ≈ 1.67 V

As temperature rises, thermistor resistance falls, decreasing V_out. This voltage change could trigger a cooling fan through a comparator.

随着温度上升，热敏电阻阻值下降，V_out 减小。此电压变化可通过比较器触发冷却风扇。

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4. Mechanical Systems and Efficiency | 机械系统与效率

Mechanical advantage (MA) and velocity ratio (VR) describe machine performance. Efficiency η = (MA / VR) × 100% accounts for energy lost to friction. This combines physics and design evaluation.

机械效益 (MA) 和速度比 (VR) 描述机器性能。效率 η = (MA / VR) × 100% 考虑了摩擦损失的能量，融合了物理与设计评估。

Problem: A pulley block with 4 supporting rope sections lifts a load of 2000 N. The effort applied is 600 N. (a) Calculate the actual MA. (b) The VR for this arrangement is 4. Determine the system efficiency. (c) If the efficiency were 100%, what effort would be needed?

问题：一个有

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