FE Other Disciplines — Complete Study Guide (2026) cover

学习指南 · 2026 版

FE Other Disciplines — Complete Study Guide (2026)

All 14 NCEES knowledge areas — statics, dynamics, fluids, thermodynamics, math, materials, electrical, and professional practice.

篇幅
100 页
书内练习题
120
考试事实核对于
2026年9月
文件最后更新
2026年10月
交付
PDF + EPUB · 英文 · 即时下载

关于本书

All 14 NCEES knowledge areas — statics, dynamics, fluids, thermodynamics, math, materials, electrical, and professional practice — taught to the free FE Reference Handbook, worked step by step.

考试费:每次 $225,缴给 NCEES。补考要再交一次。本指南:$19.99,一次买断。

考试事实、适合谁与变化

关于这场考试本身

NCEES Fundamentals of Engineering (FE) Exam — 考试事实
主管机构National Council of Examiners for Engineering and Surveying (NCEES) —— 考试由 Pearson VUE 承办

来源: NCEES — FE exam

题目数量110 道题

来源: NCEES — FE exam

考试时限320 分钟

来源: NCEES — FE exam

及格标准NCEES 未公布

我们查阅但未找到该数据的来源: NCEES — FE exam

费用
  • $225 — 考试费 (NCEES, 每次考试)

来源: NCEES — FE exam

考试语言NCEES 未公布

我们查阅但未找到该数据的来源: NCEES — FE exam

考试事实:每一项均附来源

翻开这本书看看

三页真实内页,直接从你将下载的 PDF 渲染而来 —— 依次是一页速查、一页讲解、一道带解析的例题。没有任何一页是为了好看而重画的。

  • 速查页
    Part III — Cram Sheet · PDF 第 96 页
  • 怎么讲
    Chapter 8 — Safety, Health, and Environment · PDF 第 34 页
  • 一道题的完整解析
    Part II — Practice Questions · PDF 第 58 页

包含什么,不包含什么

包含

  • All 14 official FE knowledge areas at their real exam weights, taught in order
  • Governing equations + worked step-by-step examples, Handbook-referenced

不包含

  • 不寄送纸质书 —— 这是一份你下载后可自行打印的文件
  • 购书不含视频课程、讲师或一对一辅导 —— 站上的免费视频是另一回事
完整的包含项与不包含项

包含

  • Common traps + "what the FE tests here" for every topic
  • 120 practice questions with full explanations, plus a free 423-question online bank on the site
  • A condensed formula & fact cram sheet for exam day
  • PDF (print & tab it) + EPUB (phone / e-reader)

不包含

  • 不含考试报名费与考点费用,这些仍需向官方机构缴纳

目录

查看 18 个部分及各部分起始页
  1. Part I — Fourteen teaching chapters, one for each NCEES第 2 页
  2. Part I — Teaching Chapters第 6 页
  3. Chapter 1 — Fluid Mechanics第 7 页
  4. Chapter 2 — Statics第 12 页
  5. Chapter 3 — Dynamics第 16 页
  6. Chapter 4 — Strength of Materials第 19 页
  7. Chapter 5 — Thermodynamics and Heat Transfer第 23 页
  8. Chapter 6 — Mathematics第 27 页
  9. Chapter 7 — Probability and Statistics第 30 页
  10. Chapter 8 — Safety, Health, and Environment第 33 页
  11. Chapter 9 — Engineering Economics第 36 页
  12. Chapter 10 — Materials第 39 页
  13. Chapter 11 — Basic Electrical Engineering第 43 页
  14. Chapter 12 — Chemistry第 46 页
  15. Chapter 13 — Engineering Ethics and Societal Impacts第 49 页
  16. Chapter 14 — Instrumentation and Controls第 52 页
  17. Part II — Practice Questions第 55 页
  18. Part III — Cram Sheet第 96 页

取自你将下载的那份 PDF,并标明各章起始页 —— 不是在此手工录入的。

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在本页直接阅读第 1 章
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第 1 章 · 占考试 12% · 约 4 分钟读完
Fluid Mechanics
向下滚动 ↓

Fluid mechanics is where fluency pays off most. Almost every FE fluids problem collapses into one of a few tools: fluid statics (pressure with depth), continuity (conservation of mass), the Bernoulli energy equation, or a friction-loss model. The exam skill is recognizing which tool a problem calls for. Once you name the right tool, the arithmetic is usually short.

1.1 Fluid properties and specific weight

A fluid is described first by its density ρ (mass per unit volume, kg/m³) and its specific weight γ = ρ·g (weight per unit volume, N/m³).

  • For water at standard conditions, ρ = 1000 kg/m³, so γ = 1000 × 9.81 = 9,810 N/m³ ≈ 9.81 kN/m³.
  • Specific gravity SG is the ratio of a fluid's density to that of water: SG = ρfluid / ρwater. An oil with SG = 0.8 has ρ = 800 kg/m³.
  • Viscosity μ (dynamic, Pa·s) measures resistance to shear. Kinematic viscosity is ν = μ/ρ (m²/s).

The FE almost always hands you ρ, g, and SG directly. The skill being tested is converting cleanly among density, specific weight, and specific gravity.

1.2 Fluid statics: pressure with depth

In a static fluid, gauge pressure grows linearly with vertical depth:

p = ρ·g·h = γ·h

where h is the vertical distance below the free surface. Pressure at a point acts equally in all directions and does not depend on container shape or total fluid volume.

Absolute pressure adds atmospheric pressure: pabs = pgauge + patm (patm ≈ 101.3 kPa).

Manometers apply the same rule column by column. A mercury column (SG 13.6) of height 0.2 m represents p = 13,600 × 9.81 × 0.2 ≈ 26.7 kPa.

The classic trap: mixing gauge and absolute pressure, or using the slant length of a tube instead of the true vertical height h.

1.3 Buoyancy and submerged forces

Archimedes' principle: the buoyant force on a submerged or floating body equals the weight of the fluid it displaces:

Fb = ρfluid · g · V_displaced

A body floats when its average density is less than the fluid's, sinking until displaced weight equals body weight.

For a flat submerged surface, the resultant hydrostatic force is F = pcentroid · A = γ · hc · A, where h_c is the depth of the centroid. This force acts at the center of pressure, which lies below the centroid because pressure increases with depth. Recognizing that the force uses the pressure at the centroid — not the maximum pressure at the bottom — is a common point-earner.

1.4 Continuity and volumetric flow

Conservation of mass for steady incompressible flow gives the continuity equation:

A₁·V₁ = A₂·V₂

Velocity rises where area shrinks — halving the pipe area doubles the velocity.

  • Volumetric flow rate: Q = A·V (m³/s). For a round pipe, A = π·D²/4.
  • Mass flow rate: ṁ = ρ·Q (kg/s).

Worked example. Water flows at V = 3 m/s in a pipe of diameter D = 0.1 m. Find Q.

A = π/4 × (0.1)² = 7.854 × 10⁻³ m² Q = A·V = 7.854 × 10⁻³ × 3 = 0.0236 m³/s

The frequent mistake: forgetting that area scales with diameter squared, so a small change in diameter produces a large change in velocity.

1.5 The Bernoulli energy equation

Along a streamline with no losses, total head is conserved:

p/γ + V²/(2g) + z = constant

Each term is a "head" in meters — pressure head, velocity head, and elevation head. Energy simply trades among the three.

Worked example (Torricelli). A tank drains through a small opening a depth h = 5 m below the free surface. Find the jet velocity.

At the surface, pressure and elevation head convert entirely to velocity head at the jet: V = √(2·g·h) = √(2 × 9.81 × 5) = √98.1 ≈ 9.9 m/s

Real systems append a head-loss term h_L to the downstream side to account for friction:

p₁/γ + V₁²/(2g) + z₁ = p₂/γ + V₂²/(2g) + z₂ + h_L

1.6 Flow regime, friction, and open channels

The dimensionless Reynolds number sets the flow regime:

Re = ρ·V·D/μ = V·D/ν

  • Re < ~2,100 → laminar (smooth, ordered)
  • Re > ~4,000 → turbulent
  • In between → transitional

Pipe friction loss uses Darcy–Weisbach:

h_L = f · (L/D) · V²/(2g)

where the friction factor f depends on Re and relative roughness (read from a Moody chart). For laminar flow, f = 64/Re directly.

Open-channel flow uses Manning's equation for velocity as a function of slope, roughness, and hydraulic radius. On the FE, the first move is almost always to compute Re, because it tells you which loss model and friction correlation are valid.

Common traps (Fluid Mechanics)

  • Using slant height instead of vertical depth in p = γ·h.
  • Mixing gauge and absolute pressure.
  • Forgetting Q depends on D² (area), so velocity changes dramatically with diameter.
  • Applying Bernoulli where friction is significant (long pipes) without a head-loss term.
  • Using ρ where γ is needed (off by a factor of g).

What the FE tests here

Expect several statics/manometer problems, at least one continuity problem, one or two Bernoulli/Torricelli problems, and a Reynolds-number or Darcy–Weisbach friction problem. The Handbook has all the equations — your job is to identify the regime and pick the tool.

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详细信息

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一次购买,永久下载访问。此电子书是完整的 FE Other Disciplines (NCEES) 学习指南,含 PDF 与 EPUB。仅为教育性摘要,非专业或法律意见 —— 请始终以官方来源核实当前规定。最后更新:October 2026。

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