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2.3 Engineering Sciences

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FE Civil Fundamentals of Engineering — Civil

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1 Introduction & FE Civil Exam Strategy

1.1 FE Civil Exam Facts (2026) 1.2 Mastering the NCEES FE Reference Handbook 1.3 Computational Tools & Numerical Methods 1.4 Study Plan, Pacing & Exam-Day Logistics

2 Mathematics, Probability, Statistics & General Topics

2.1 Mathematics (Algebra, Trigonometry, Geometry & Calculus) 2.2 Probability & Statistics 2.3 Engineering Sciences 2.4 Ethics & Professional Practice 2.5 Engineering Economics

3 Statics, Dynamics & Mechanics of Materials

3.1 Statics I: Resultants, Equilibrium & Trusses 3.2 Statics II: Friction, Centroids & Moment of Inertia 3.3 Dynamics (Kinematics & Kinetics) 3.4 Mechanics of Materials I: Stress, Strain & Axial Loading 3.5 Mechanics of Materials II: Beams, Torsion, Bending & Columns 3.6 Civil Engineering Materials

4 Fluid Mechanics, Hydraulics & Water Resources

4.1 Fluid Properties & Fluid Statics 4.2 Fluid Dynamics: Continuity, Bernoulli & Momentum 4.3 Hydraulics: Pipe Flow, Open Channel & Pumps 4.4 Hydrologic Systems 4.5 Environmental Engineering (Water & Wastewater)

5 Geotechnical & Structural Engineering

5.1 Soil Classification & Index Properties 5.2 Soil Mechanics: Effective Stress, Consolidation & Shear Strength 5.3 Foundations, Retaining Walls & Slope Stability 5.4 Structural Analysis 5.5 Structural Design (Steel & Reinforced Concrete) 5.6 Loads, Load Combinations & Design Codes

6 Transportation, Construction & Surveying

6.1 Transportation: Geometric Design 6.2 Transportation: Traffic Engineering & Pavement 6.3 Construction: Scheduling, Earthwork & Estimating 6.4 Construction: Means, Methods & Safety 6.5 Surveying

2.3 Engineering Sciences

Key Takeaways Explain This Topic AI

Work W = F·d (force times distance along the motion); energy and work share units of joules (N·m) or ft·lb; power P = W/t = work per unit time, in watts (J/s).

Ohm's law V = I·R relates voltage, current, and resistance; electrical power P = VI = I²R = V²/R.

Resistors in series add directly R = R₁+R₂+…; in parallel the reciprocals add 1/R = 1/R₁+1/R₂+…, so parallel resistance is always less than the smallest resistor.

Sensible heat Q = m·c·ΔT relates heat to mass, specific heat, and temperature change; energy is conserved across mechanical, electrical, and thermal forms.

Last updated: June 2026

Scope and the Conservation Theme

Engineering Sciences is 4–6 questions on the FE Civil exam and gathers the cross-disciplinary physics that other areas assume: work, energy, and power , basic electricity , and basic thermodynamics/heat . Civil engineers meet these in pumps (power), site power and instrumentation (circuits), and concrete curing or pavement temperature (heat). The single most powerful idea is conservation of energy — energy is neither created nor destroyed, only converted among mechanical, electrical, and thermal forms.

Because the exam is open to the NCEES FE Reference Handbook , success means selecting the correct relation and keeping a consistent unit system; SI (joules, watts, kelvin) is cleanest, but be ready for US Customary units (ft·lb, Btu).

Work, Energy & Power

Work is force acting through a distance in the direction of motion: W = F·d (more generally W = ∫F·dx, and W = F·d·cosθ when force and motion are not aligned). Units: joule (J = N·m) in SI, ft·lb in USCS. Energy is the capacity to do work and shares those units:

  • Quantity Formula Notes
  • Kinetic energy KE = ½mv² mass m (kg), velocity v (m/s)
  • Potential energy PE = mgh g = 9.81 m/s², height h
  • Work W = F·d along direction of force
  • Power P = W/t = F·v rate of doing work

Power is the time rate of work, P = W/t , measured in watts (W = J/s); 1 horsepower = 746 W. For a steady force moving at velocity v, P = F·v.

The work-energy principle ties these together: the net work done on a body equals its change in kinetic energy, W_net = ΔKE = ½mv₂² − ½mv₁². For conservative systems, mechanical energy is conserved: KE₁ + PE₁ = KE₂ + PE₂, so a mass dropped from height h arrives with v = √(2gh). These relations let you solve impact, braking-distance, and falling-object problems without integrating forces. The recurring trap is mixing weight and mass: in SI, mass is in kilograms and weight W = mg in newtons; in US Customary, a 'pound' can be force (lbf) or mass (lbm), and you must insert g_c = 32.2 lbm·ft/(lbf·s²) to keep units consistent.

Basic Electricity — Ohm's Law and Resistance

Direct-current circuits obey Ohm's law: V = I·R , where V is voltage (volts), I is current (amperes), and R is resistance (ohms, Ω). Electrical power has three equivalent forms: P = V·I = I²R = V²/R (watts). Combine resistors with these rules:

Series: R_eq = R₁ + R₂ + R₃ + … (same current through each; voltages add). Series resistance is always larger than any one resistor.

Parallel: 1/R_eq = 1/R₁ + 1/R₂ + … (same voltage across each; currents add). Parallel resistance is always smaller than the smallest resistor.

For just two parallel resistors a handy form is R_eq = (R₁·R₂)/(R₁ + R₂). The classic trap is adding parallel resistors directly instead of adding reciprocals.

Two conservation laws govern circuits and are listed in the Handbook. Kirchhoff's current law (KCL): the sum of currents into a node equals the sum out (charge conservation). Kirchhoff's voltage law (KVL): the sum of voltage rises and drops around any closed loop is zero (energy conservation). In a series circuit the same current flows everywhere and the source voltage divides across the resistors (voltage divider: V_R = V_source·R/R_total); in a parallel circuit the same voltage appears across every branch and the current divides. Energy consumed over time is E = P·t, billed in kilowatt-hours (1 kWh = 3.6 MJ).

Basic Thermodynamics & Heat

The first law of thermodynamics is conservation of energy: ΔU = Q − W (internal energy change equals heat added minus work done by the system). On the FE Civil exam the heat concept most likely to appear is sensible heat : Q = m·c·ΔT , where m is mass (kg), c is specific heat (J/kg·°C; water ≈ 4186 J/kg·°C), and ΔT is the temperature change. 8·T(°C) + 32. Heat transfers by conduction (Fourier's law, q = −kA·dT/dx), convection , and radiation . A ΔT is identical in °C and K (a 10 °C rise equals a 10 K rise), but absolute temperatures are not — a frequent mix-up.

Worked Example — Pump Power

A pump lifts water to height h = 20 m at a mass flow rate of ṁ = 50 kg/s. Find the ideal power required.

Step 1 — energy per second (power) added equals the rate of potential-energy gain: P = ṁ·g·h.

Step 2 — substitute: P = (50 kg/s)(9.81 m/s²)(20 m) = 9,810 W ≈ 9.81 kW .

If the pump is 80% efficient, the input power is P_in = P/η = 9.81/0.80 ≈ 12.3 kW .

Cross-check (electricity): if that 12.3 kW pump runs on a 480-V circuit, the current is I = P/V = 12,300/480 ≈ 25.6 A — a direct use of P = VI. Keeping units consistent (watts, not kilowatts, when dividing by volts) prevents a factor-of-1000 error.

Worked Example — Steady Conduction

For one-dimensional steady conduction through a flat wall, Fourier's law integrates to Q = k·A·ΔT/L , where k is thermal conductivity (W/m·K), A is the cross-sectional area, ΔT is the temperature difference across the wall, and L is the wall thickness. The combination L/(k·A) is the thermal resistance R_th, so Q = ΔT/R_th — a direct analogy to Ohm's law (Q ↔ I, ΔT ↔ V, R_th ↔ R).

A concrete wall has k = 1.7 W/m·K, area A = 10 m², thickness L = 0.20 m, with inside surface at 25 °C and outside at 5 °C (ΔT = 20 °C = 20 K).

Step 1 — apply Fourier's law: Q = k·A·ΔT/L = (1.7)(10)(20)/0.20.

Step 2 — evaluate: Q = (1.7 × 10 × 20)/0.20 = 340/0.20 = 1,700 W of heat flow through the wall.

Because a ΔT of 20 °C equals 20 K, the result is identical whether you work in Celsius or kelvin — only temperature differences enter the equation.

Worked Example — Series & Parallel Resistance

Resistors of R₁ = 4 Ω and R₂ = 12 Ω are placed in parallel , and that combination is then put in series with R₃ = 5 Ω across a 24-V source. Find the total resistance and the source current.

  • Step Operation Result
  • Parallel pair R₁₂ = (R₁·R₂)/(R₁+R₂) = (4·12)/(4+12) 48/16 = 3 Ω
  • Add series resistor R_eq = R₁₂ + R₃ = 3 + 5 8 Ω
  • Source current I = V/R_eq = 24/8 3 A
  • Power dissipated P = V·I = 24 × 3 72 W

The parallel result (3 Ω) is correctly smaller than the smaller branch (4 Ω), confirming the reciprocal rule. The full 3 A flows through R₃, but it divides between the parallel branches in inverse proportion to their resistances: the 4-Ω branch carries three times the current of the 12-Ω branch.