📚 GCSE CCEA Engineering: Quick-Reference Formula Handbook | GCSE CCEA 工程：公式定理速查手册
This handbook covers the key equations and principles required for the CCEA GCSE Engineering specification. Engineers use these formulas to analyse structures, calculate electrical values, determine mechanical advantage, and design safe systems. Mastering these fundamentals is essential for both written examinations and practical design tasks.
本手册涵盖 CCEA GCSE 工程大纲所需的关键公式和原理。工程师运用这些公式分析结构、计算电学量、确定机械利益并设计安全系统。掌握这些基础知识对于笔试和实践设计任务都至关重要。
1. Force, Weight and Mass | 力、重量与质量
In engineering statics, a force is any push or pull acting on a body. Forces are measured in newtons (N). For an object to remain in equilibrium, the resultant force must be zero.
在工程静力学中，力是作用在物体上的推或拉。力的单位是牛顿 (N)。要使物体保持平衡，合力必须为零。
W = m × g
W = m × g （重量 = 质量 × 引力场强度）
Weight is the gravitational force acting on a mass. Here, W is weight (N), m is mass (kg), and g is the gravitational field strength. On Earth, g is approximately 9.8 N/kg, and in most GCSE calculations it is taken as 10 N/kg unless specified otherwise.
重量是作用在质量上的引力。其中 W 为重量 (N)，m 为质量 (kg)，g 为引力场强度。地球上的 g 约为 9.8 N/kg，大多数 GCSE 计算中若无特别说明可取 10 N/kg。
Example: A steel bracket has a mass of 20 kg. The weight acting downwards is W = 20 × 10 = 200 N. This load is used in beam support calculations.
示例：一个钢制支架的质量为 20 kg，向下的重量为 W = 20 × 10 = 200 N。该载荷用于梁支座计算。
2. Moments and Equilibrium | 力矩与平衡
A moment is the turning effect of a force about a pivot. It depends on the size of the force and the perpendicular distance from the pivot to the line of action of the force.
力矩是力绕支点的转动效应，它取决于力的大小和支点到力作用线的垂直距离。
M = F × d
M = F × d （力矩 = 力 × 垂直距离）
M is the moment in newton-metres (N m), F is the force in newtons (N), and d is the perpendicular distance in metres (m). Clockwise moments are usually taken as positive, and anticlockwise as negative, or vice versa, as long as convention is consistent.
M 为力矩，单位牛顿·米 (N m)，F 为力 (N)，d 为垂直距离 (m)。通常规定顺时针力矩为正，逆时针力矩为负，反之亦可，但需保持一致。
For a system in rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments: Σ Mcw = Σ Macw. This principle is used to find unknown forces in levers, beams and other mechanisms.
对于转动平衡的系统，顺时针力矩之和等于逆时针力矩之和：Σ Mcw = Σ Macw。该原理可用于计算杠杆、梁和其他机构中的未知力。
Example: A 0.6 m spanner applies a force of 30 N at its end to loosen a nut. The moment applied is M = 30 × 0.6 = 18 N m.
示例：一把 0.6 m 的扳手在末端施加 30 N 的力来拧松螺母，产生的力矩为 M = 30 × 0.6 = 18 N m。
3. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量
When a material is loaded, it experiences stress and strain. Stress quantifies the internal force per unit area, while strain measures the deformation relative to the original dimension.
当材料受载时，会产生应力和应变。应力衡量单位面积上的内力，应变则衡量相对于原始尺寸的变形。
σ = F / A
σ = F / A （应力 = 力 / 横截面积）
ε = Δl / l₀
ε = Δl / l₀ （应变 = 伸长量 / 原始长度）
σ (sigma) is stress in pascals (Pa) or N/m², F is the applied force (N), and A is the cross-sectional area (m²). ε (epsilon) is strain (dimensionless), Δl is the change in length (m), and l₀ is the original length (m).
σ 为应力，单位帕斯卡 (Pa) 或 N/m²，F 为施加的力 (N)，A 为横截面积 (m²)。ε 为应变 (无量纲)，Δl 为长度变化量 (m)，l₀ 为原始长度 (m)。
Young’s modulus E links stress and strain for a material within its elastic limit:
杨氏模量 E 将材料在弹性范围内的应力和应变联系起来：
E = σ / ε = (F l₀) / (A Δl)
E = σ / ε = (F l₀) / (A Δl)
E has units of pascals (Pa). A stiffer material has a higher Young’s modulus. This relationship is used when selecting materials for structural components.
E 的单位是帕斯卡 (Pa)。刚度大的材料具有较高的杨氏模量。在选择结构零件材料时会用到此关系。
4. Work, Energy and Power | 功、能与功率
Work is done when a force moves an object through a distance in the direction of the force. Energy is the capacity to do work, and power is the rate at which work is performed.
当力使物体沿力的方向移动一段距离时，力就做了功。能量是做功的能力，功率是做功的快慢。
W = F × s
W = F × s （功 = 力 × 沿力方向的距离）
P = W / t = F × v
P = W / t = F × v （功率 = 功 / 时间 = 力 × 恒定速度）
Work done W is in joules (J), force F in newtons (N), distance s
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