Engineering Fundamentals and Applied Science
The Engineering Fundamentals and Applied Science section tests whether you can reduce a practical problem to first principles - statics, mechanics of materials, thermodynamics, fluid flow, basic circuits - and carry units correctly through to a defensible number.
Core rules
- Dimensional homogeneity: both sides of any equation must reduce to the same base units. N = kg·m/s², Pa = N/m², J = N·m, W = J/s. Convert MPa to Pa (×10⁶) before substituting.
- Equilibrium first: draw the free-body diagram, then apply ΣF = 0 and ΣM = 0. Reactions come from equilibrium, not intuition.
- Stress and strain: σ = F/A, ε = ΔL/L, and σ = Eε only below the proportional limit. Design to allowable stress σ_allow = S_y/n.
- Steady-flow energy: Q − W = ṁ[(h₂−h₁) + (V₂²−V₁²)/2 + g(z₂−z₁)]; pump shaft power = ṁgH/η.
- Electrical: P = VI for DC; balanced three-phase P = √3·V_L·I_L·cos φ.
Worked example
A 10 mm diameter steel tie rod (E = 200 GPa, S_y = 250 MPa) carries 20 kN in tension with a required design factor of 1.5.
- A = π(0.010)²/4 = 7.854 × 10⁻⁵ m².
- σ = 20 000 / 7.854 × 10⁻⁵ = 254.6 MPa - above yield, so the rod would permanently stretch.
- Allowable stress = 250/1.5 = 166.7 MPa, so A_req = 20 000 / (166.7 × 10⁶) = 1.20 × 10⁻⁴ m².
- d = √(4A/π) = 12.4 mm, so select the next standard size, 14 mm.
Traps
Gauge versus absolute pressure; using radius where diameter belongs; multiplying MPa directly by newtons; applying Hooke's law past yield; ignoring restrained thermal strain σ = EαΔT.