📚 AQA Year 13 Engineering: Case Study Practical Drill | AQA工程：案例分析实战演练
Case study questions in the AQA Year 13 Engineering exam require you to apply principles of statics, mechanics of materials, and design evaluation to a real‑world component. This drill walks you through a complete structural analysis of a cantilever crane jib, from load specification to optimisation. By mastering this workflow, you will be able to tackle any beam‑based case study with confidence.
AQA 13 年级工程考试中的案例分析题，要求你将静力学、材料力学和设计评估知识应用于真实构件。本次演练将带你完整分析一个悬臂起重机吊臂，从载荷规格到优化。掌握这套流程后，你将能自信地应对任何基于梁的案例分析。
1. Case Description and Load Specification | 案例描述与载荷规格
A horizontal cantilever beam of length L = 4.0 m is rigidly fixed at one end and free at the other. The beam carries a concentrated vertical load P = 15 kN at the free tip. The cross‑section is rectangular with breadth b = 0.15 m and depth h = 0.30 m. The material is structural steel with a yield strength σy = 250 MPa and Young’s modulus E = 200 GPa. The beam is used indoors and no dynamic amplification is required. We need to verify the structural adequacy and suggest potential improvements.
一根水平悬臂梁，长度 L = 4.0 m，一端刚性固定，另一端自由。自由端承受集中竖向载荷 P = 15 kN。梁的横截面为矩形，宽度 b = 0.15 m，高度 h = 0.30 m。材料为结构钢，屈服强度 σy = 250 MPa，杨氏模量 E = 200 GPa。梁用于室内，不需要考虑动力放大系数。我们需要验证结构是否安全，并提出可能的改进建议。
2. Free Body Diagram and Reaction Forces | 受力图与支反力
Isolate the beam and apply the equations of equilibrium. At the fixed support we have a vertical reaction force RA and a restraining moment MA. Summing forces vertically: ΣFy = 0 → RA − P = 0 → RA = 15 kN upwards. Taking moments about the fixed end: ΣMA = 0 → MA − P × L = 0 → MA = 15 kN × 4.0 m = 60 kN·m (anticlockwise). These reactions form the boundary conditions for the internal force diagrams.
将梁隔离并应用平衡方程。在固定端存在竖向支反力 RA 和约束弯矩 MA。竖向力平衡：ΣFy = 0 → RA − P = 0 → RA = 15 kN 向上。对固定端取矩：ΣMA = 0 → MA − P × L = 0 → MA = 15 kN × 4.0 m = 60 kN·m（逆时针）。这些反力构成了内力图的边界条件。
3. Shear Force and Bending Moment Diagrams | 剪力图与弯矩图
Cut the beam at a distance x from the free end. The shear force V(x) is constant and equal to +P (or −P depending on sign convention; here we take downward positive for load effects). For the usual sign convention with shear positive when it causes clockwise rotation, we write V = −15 kN from tip to support. The bending moment M(x) varies linearly: M(x) = −P × x, giving zero at the tip and −60 kN·m at the fixed end. The maximum bending moment magnitude is therefore 60 kN·m at the support.
在距自由端 x 处截开。剪力 V(x) 为常数，等于 +P（或依符号规定取 −P；按照使微段顺时针转动时剪力为正的惯例，从自由端到支座 V = −15 kN）。弯矩 M(x) 线性变化：M(x) = −P × x，自由端为 0，固定端为 −60 kN·m。因此最大弯矩绝对值为 60 kN·m，出现在固定端。
- Shear force envelope: constant |V| = 15 kN
- Bending moment envelope: linear, maximum 60 kN·m at support
- 剪力包络：常数 |V| = 15 kN
- 弯矩包络：线性，支座处最大 60 kN·m
4. Bending Stress Calculation | 弯曲应力计算
For a rectangular section, the elastic section modulus Z about the neutral axis is Z = bh²/6. Substituting b = 0.15 m, h = 0.30 m: Z = (0.15 × 0.30²) / 6 = (0.15 × 0.09) / 6 = 0.0135 / 6 = 0.00225 m³. The maximum bending stress σmax = Mmax / Z. Use Mmax = 60 × 10³ N·m. Then σmax = 60 × 10³ / 0.00225 = 26.67 × 10⁶ Pa = 26.67 MPa.
对于矩形截面，绕中性轴的弹性截面模量 Z = bh²/6。代入 b = 0.15 m，h = 0.30 m：Z = (0.15 × 0.30²) / 6 = (0.15 × 0.09) / 6 = 0.0135 / 6 = 0.00225 m³。最大弯曲应力 σmax = Mmax / Z。取 Mmax = 60 × 10³ N·m，则 σmax = 60 × 10³ / 0.00225 = 26.67 × 10⁶ Pa = 26.67 MPa。
σmax = 26.67 MPa ≪ σy = 250 MPa
The calculated stress is far below the yield strength, indicating that the beam is grossly overdesigned for strength alone. We must also check stiffness (deflection) and possible buckling modes.
计算应力远低于屈服强度，表明该梁单纯从强度角度看设计过于保守。我们仍需校核刚度（挠度）和可能的屈曲模式。
5. Deflection Analysis | 挠度分析
The tip deflection of a cantilever with a point load at the free end is given by δ = PL³ / (3EI). First compute the second moment of area I for the rectangular section: I = bh³/12 = 0.15 × (0.30)³ / 12 = 0.15 × 0.027 / 12 = 0.00405 / 12 = 3.375 × 10⁻⁴ m⁴. Then E = 200 × 10⁹ Pa, P = 15 × 10³ N, L = 4.0 m.
自由端受集中力悬臂梁的端部挠度公式为 δ = PL³ / (3EI)。先计算矩形截面的面积二次矩 I：I = bh³/12 = 0.15 × (0.30)³ / 12 = 0.15 × 0.027 / 12 = 0.00405 / 12 = 3.375 × 10⁻⁴ m⁴。E = 200 × 10⁹ Pa，P = 15 × 10³ N，L = 4.0 m。
Calculate PL³ = 15 × 10³ × 64 = 960 × 10³ N·m³. 3EI = 3 × 200 × 10⁹ × 3.375 × 10⁻⁴ = 3 × 200 × 10⁹ × 3.375 × 10⁻⁴ = 600 × 10⁹ × 3.375 × 10⁻⁴ = 600 × 3.375 × 10⁵ = 2025 × 10⁵ = 2.025 × 10⁸ N·m². Therefore δ = 960 × 10³ / (2.025 × 10⁸) = 0.00474 m = 4.74 mm.
计算 PL³ = 15 × 10³ × 64 = 960 × 10³ N·m³。3EI = 3 × 200 × 10⁹ × 3.375 × 10⁻⁴ = 3 × 200 × 10⁹ × 3.375 × 10⁻⁴ = 600 × 10⁹ × 3.375 × 10⁻⁴ = 600 × 3.375 × 10⁵ = 2025 × 10⁵ = 2.025 × 10⁸ N·m²。因此 δ = 960 × 10³ / (2.025 × 10⁸) = 0.00474 m = 4.74 mm。
A typical serviceability limit for crane jibs is L/360 = 4000/360 ≈ 11.1 mm. The actual deflection is 4.74 mm, well within the limit. However, dynamic loads or precision requirements may call for a smaller deflection.
起重机吊臂的常用使用极限为 L/360 = 4000/360 ≈ 11.1 mm。实际挠度 4.74 mm 完全满足。但若存在动载或精度要求，可能需要更小的挠度。
6. Factor of Safety and Failure Criteria | 安全系数与失效准则
Based on yield, the factor of safety FoS = σy / σmax = 250 / 26.67 ≈ 9.37. This is extremely conservative for a static indoor application; typical FoS for structural steel in bending is 1.5–2.5 under normal loads. The beam also experiences shear stress τmax = 3V/(2A), where A = bh = 0.045 m², V = 15 kN → τmax = 3 × 15000 / (2 × 0.045) = 500 kPa = 0.5 MPa, negligible compared with shear yield (≈ 0.6σy = 150 MPa). Overall, static failure is not a concern, but fatigue (if cyclic), buckling of the slender compression flange, or local crippling could govern a more refined design.
基于屈服的安全系数 FoS = σy / σmax = 250 / 26.67 ≈ 9.37。对于室内静载应用这极为保守；结构钢弯曲时正常载荷下的典型安全系数为 1.5–2.5。梁还承受剪应力 τmax = 3V/(2A)，其中 A = bh = 0.045 m²，V = 15 kN → τmax = 3 × 15000 / (2 × 0.045) = 500 kPa = 0.5 MPa，与剪切屈服应力（≈ 0.6σy = 150 MPa）相比可忽略。总体而言，静态破坏不是问题，但疲劳（若循环加载）、受压翼缘的屈曲或局部挤压可能在更精细的设计中起控制作用。
7. Material Selection Alternatives | 材料选择替代方案
Because the current design is over‑strength, we might explore a lighter material such as aluminium alloy 6061‑T6, with σy = 276 MPa and E = 69 GPa. Keeping the same section, the bending stress remains identical because it is geometry‑dependent, so FoS ≈ 10.4. However, the lower E will increase deflection. Recalculating δ with E = 69 GPa: δAl = δsteel × (200/69) = 4.74 × 2.90 ≈ 13.7 mm, which now slightly exceeds L/360 (11.1 mm). The designer would need to increase the section depth by about 15–20% to restore stiffness. Aluminium offers corrosion resistance and a 65% weight reduction, which may justify the change in mobile crane applications.
由于当前设计强度过剩，我们可以探索更轻的材质，例如铝合金 6061‑T6，其 σy = 276 MPa，E = 69 GPa。若保持相同截面，弯曲应力不变（取决于几何形状），故 FoS ≈ 10.4。但较低的 E 会增大挠度。用 E = 69 GPa 重新计算：δAl = δsteel × (200/69) = 4.74 × 2.90 ≈ 13.7 mm，此时略微超出 L/360 (11.1 mm)。设计者需要将截面高度增加约 15–20% 以恢复刚度。铝合金提供耐腐蚀性和 65% 的重量减轻，这在移动式起重机应用中可能具有合理性。
| Property | Steel | Al 6061‑T6 |
|---|---|---|
| Density (kg/m³) | 7850 | 2700 |
| E (GPa) | 200 | 69 |
| σy (MPa) | 250 | 276 |
| Deflection for same section (mm) | 4.74 | 13.7 |
| Mass per unit length (kg/m) | 353 | 121.5 |
Table: Comparison of key properties for the original steel section and an equivalent aluminium alloy section.
表：原始钢截面与等效铝合金截面的关键性能对比。
8. Manufacturing and Cost Considerations | 制造与成本考量
The steel beam can be fabricated from a standard hot‑rolled rectangular hollow section or a solid bar. Solid rectangular bar is easy to source but heavy. Alternatively, welding stiffeners or tapering the beam could reduce weight while maintaining strength. Aluminium extrusions can provide complex shapes with integrated stiffeners, but tooling costs are higher. For a one‑off crane, steel fabrication is often cheaper; for mass‑produced mobile lifts, aluminium’s weight saving reduces fuel and motor costs over the lifetime.
钢梁可采用标准热轧矩形空心型材或实心棒材制造。实心矩形棒容易采购但重量大。也可通过焊接加劲肋或变截面设计减轻重量同时保持强度。铝合金挤压型材可提供带有整体加劲肋的复杂形状，但模具成本较高。对于单件起重机，钢制构件通常更便宜；对于批量生产的移动式升降机，铝合金的减重可在全寿命周期内降低燃油和电机成本。
9. Optimisation Suggestions | 优化建议
Given the excessive FoS, the cross‑section can be reduced. If we aim for a bending stress of 150 MPa (FoS ≈ 1.67), the required section modulus Zreq = Mmax / 150 = 60×10³ / 150×10⁶ = 400 × 10⁻⁶ m³ = 4.0×10⁻⁴ m³. For a rectangular section, Z = bh²/6. Maintaining b = 0.15 m, the required depth h = √(6Z/b) = √(6×4.0×10⁻⁴ / 0.15) = √(0.016) = 0.1265 m. Checking deflection with h = 0.127 m: I = 0.15×(0.127)³/12 = 2.56×10⁻⁵ m⁴, δ = 15×10³×64 / (3×200×10⁹×2.56×10⁻⁵) ≈ 0.0625 m = 62.5 mm, which is far too large. Stiffness therefore governs the design. The beam must be deeper than 0.15 m to limit deflection to, say, 10 mm. Solving δ = PL³/(3E×(bh³/12)) ≤ 0.01 m, we obtain h³ ≥ (4PL³)/(E b × 0.01) = (4×15×10³×64) / (200×10⁹×0.15×0.01) ≈ (3.84×10⁶) / (3.0×10⁷) = 0.128, h ≥ 0.504 m. That would be an uneconomical solid section; a built‑up I‑beam or truss structure would be more efficient.
鉴于安全系数过高，截面可以减小。若希望弯曲应力达到 150 MPa（FoS ≈ 1.67），所需截面模量 Zreq = Mmax / 150 = 60×10³ / 150×10⁶ = 400 × 10⁻⁶ m³ = 4.0×10⁻⁴ m³。对于矩形截面，Z = bh²/6。保持 b = 0.15 m，所需高度 h = √(6Z/b) = √(6×4.0×10⁻⁴ / 0.15) = √(0.016) = 0.1265 m。用 h = 0.127 m 校核挠度：I = 0.15×(0.127)³/12 = 2.56×10⁻⁵ m⁴，δ = 15×10³×64 / (3×200×10⁹×2.56×10⁻⁵) ≈ 0.0625 m = 62.5 mm，过大。因此刚度成为控制因素。为将挠度限制在，例如 10 mm，梁高度必须大于 0.15 m。解 δ = PL³/(3E×(bh³/12)) ≤ 0.01 m，得 h³ ≥ (4PL³)/(E b × 0.01) = (4×15×10³×64) / (200×10⁹×0.15×0.01) ≈ (3.84×10⁶) / (3.0×10⁷) = 0.128，h ≥ 0.504 m。这将是极不经济的实心截面；采用格构式工字梁或桁架结构更具效率。
10. Conclusion and Exam Tips | 结论与考试技巧
This case study illustrates a systematic approach: define loads, calculate reactions, draw shear and moment diagrams, compute maximum bending stress and deflection, check against allowable limits, evaluate material alternatives, and iterate toward an optimised design. In the AQA Engineering exam, always show clear free body diagrams, state your formulas, and comment on the reasonableness of your results. Marks are awarded for method and interpretation, not just the final number.
本案例展示了一套系统方法：定义载荷、计算支反力、绘制剪力和弯矩图、计算最大弯曲应力与挠度、与许用极限对比、评估替代材料并迭代优化设计。在 AQA 工程考试中，应始终绘制清晰的受力图，列出所用公式，并对结果的合理性加以评述。得分点在于方法和解读，而不仅仅是最终数值。
Published by TutorHao | Engineering Revision Series | aleveler.com
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