PE Civil — Complete Study Guide (2026)

Vocabulary Is Half the Battle

Construction engineering turns a design into a built project — on time and on budget. The PE tests CPM scheduling, earned-value cost control, earthwork and quantity take-offs, equipment production, temporary works, and jobsite safety. Most questions are short arithmetic once you know the right definition, so precise vocabulary — float, swell, EV, CPI — is genuinely half the battle. Learn the terms exactly and the numbers fall out.

5.1 CPM Scheduling and Float

The Critical Path Method models a project as a network of activities. A forward pass computes early start and early finish; a backward pass computes late finish and late start:

Quantity	Definition
Early finish	EF = ES + duration
Late start	LS = LF − duration
Total float	LF − EF = LS − ES
Free float	(successor's ES) − (activity EF)

Total float is how long an activity can slip without pushing the project end; free float is how long it can slip without delaying any successor's early start. The critical path is the longest path through the network; its activities have zero total float, and it sets the shortest possible project duration. Shortening the project therefore means shortening critical activities. In precedence (activity-on-node) diagrams the default relationship is finish-to-start, and milestones carry zero duration. Crashing buys time by adding resources; the crash cost slope identifies the cheapest activity to accelerate:

cost slope = (crash cost − normal cost) / (normal duration − crash duration)

Worked Example 5.1 — Total float. An activity has ES = 5, duration = 4 (so EF = 9), and the schedule allows LF = 12.

• LS = LF − duration = 12 − 4 = 8. Total float = LF − EF = 12 − 9 = 3 days (= LS − ES = 8 − 5 = 3, consistent).

Worked Example 5.2 — Cheapest to crash. Activity A: normal 10 days/$5,000, crash 7 days/$8,000. Cost slope?

• ($8,000 − $5,000)/(10 − 7) = $3,000/3 = $1,000 per day. Crash the critical activity with the lowest such slope first.

Common traps. Crashing a non-critical activity (it saves nothing). Confusing total float with free float. Forgetting the critical path is the longest path but yields the shortest duration.

What the PE tests here. Forward/backward pass; total and free float; identifying the critical path; crash cost slope and least-cost acceleration.

5.2 Earned Value and Cost Control

Earned-value management compares three dollar figures at a moment in time:

Term	Meaning
Planned value (PV)	Budgeted cost of work scheduled
Earned value (EV)	Budgeted cost of work actually done
Actual cost (AC)	Real money spent

Two indices flag trouble, both dividing EV by a baseline:

CPI = EV/AC (cost) SPI = EV/PV (schedule)

Below 1.0 is unfavorable (over budget, or behind schedule); above 1.0 is favorable. The variances:

CV = EV − AC SV = EV − PV (negative SV = behind schedule)

Forecasting uses the CPI:

EAC = BAC / CPI

Worked Example 5.3 — Reading project health. At a checkpoint: PV = $80,000, EV = $70,000, AC = $90,000.

• CPI = EV/AC = 70,000/90,000 = 0.78 (over budget). SPI = EV/PV = 70,000/80,000 = 0.875 (behind schedule).
• CV = 70,000 − 90,000 = −$20,000; SV = 70,000 − 80,000 = −$10,000.

Worked Example 5.4 — Forecast at completion. Budget at completion BAC = $100,000, CPI = 0.80.

• EAC = BAC/CPI = 100,000/0.80 = $125,000 — the project is trending $25,000 over.

Common traps. Swapping the numerator/denominator in CPI or SPI (EV is always on top). Reading a CPI below 1.0 as good. Confusing cost variance with schedule variance.

What the PE tests here. Defining PV, EV, AC; computing CPI, SPI, CV, SV; interpreting above/below 1.0; forecasting EAC from CPI.

5.3 Earthwork: Swell, Shrinkage, and Volumes

Soil changes volume as it is handled, and the exam keys on three states — bank (in place), loose (excavated/hauled), and compacted (in the fill):

Loose volume = bank × (1 + swell) Load factor = 1/(1 + swell) (loose → bank) Compacted volume = bank × (1 − shrinkage)

Excavation loosens soil, so it swells; placed fill is compacted denser than bank, so it shrinks — meaning you must excavate more bank than the neat fill volume. Roadway volumes between cross-sections use the average-end-area method:

V = [(A₁ + A₂)/2] × L / 27 (to cubic yards, with A in ft², L in ft)

which slightly overestimates versus the more exact prismoidal formula. A mass (haul) diagram plots cumulative cut-minus-fill along the alignment to balance earthwork and set haul direction.

Worked Example 5.5 — How much to excavate for a fill. A fill needs 5,000 bank-yd³ of compacted material; the soil shrinks 10%.

• Wait — read carefully. Compacted = bank × (1 − shrinkage). We need 5,000 yd³ compacted, so bank required = 5,000 / (1 − 0.10) = 5,000/0.90 = 5,556 bank-yd³ must be excavated.

Worked Example 5.6 — Average-end-area volume. Two stations 100 ft apart have cut areas A₁ = 120 ft² and A₂ = 180 ft².

• V = [(120 + 180)/2] × 100 / 27 = 150 × 100 / 27 = 15,000/27 = 556 yd³.

Common traps. Applying swell where shrinkage belongs (excavation swells, placed fill shrinks). Forgetting to divide by 27 for cubic yards. Using the neat fill volume as the bank excavation quantity without the shrinkage correction.

What the PE tests here. Swell/shrinkage/load-factor conversions among bank, loose, and compacted; average-end-area volumes; mass-diagram reasoning.

5.4 Concrete and Material Quantities

Quantity take-offs are unit-conversion discipline. The essential fact:

1 cubic yard = 27 cubic feet

For a slab, volume = length × width × thickness; for a round column, volume = (π/4)·D²·H. Self-weight or tonnage comes from unit weight (concrete ≈ 150 pcf, asphalt ≈ 145 pcf): a layer's weight is area × thickness × unit weight, to tons by dividing by 2,000.

Worked Example 5.7 — Slab concrete order. A slab 40 ft × 30 ft × 6 in thick.

• Volume = 40 × 30 × 0.5 ft = 600 ft³. In yards: 600/27 = 22.2 yd³ — round up when ordering.

Worked Example 5.8 — Asphalt tonnage. A road lift 3,000 ft long, 24 ft wide, 3 in thick; asphalt 145 pcf.

• Volume = 3,000 × 24 × 0.25 = 18,000 ft³. Weight = 18,000 × 145 = 2,610,000 lb. Tons = 2,610,000/2,000 = 1,305 tons.

Common traps. Forgetting to convert thickness from inches to feet. Dividing by the wrong factor (27 for yd³, 2,000 for tons). Losing the unit chain in a multi-step order.

What the PE tests here. Cubic-yard concrete volumes; the 27 ft³/yd³ conversion; tonnage from unit weight; multi-step quantity-to-cost chains.

5.5 Equipment Production and Compaction

Equipment output is cycles per hour times payload per cycle, adjusted for real conditions:

Production = (volume per cycle) × (60 / cycle time in min) × efficiency

A common efficiency is the 50-minute working hour (factor 50/60 ≈ 0.83), accounting for delays and breaks. Balancing a hauling fleet, the number of trucks a loader keeps busy equals truck cycle time / loading time. Field compaction is checked against the lab Proctor:

relative compaction = (field dry density / max Proctor dry density) × 100% (spec often 95%)

Roller choice matches the soil — a sheepsfoot (padfoot) roller kneads cohesive clay; vibratory smooth-drum rollers suit granular soils — and thinner lifts compact more uniformly than thick ones.

Worked Example 5.9 — Excavator production. Bucket 1.5 yd³, cycle time 0.5 min, 50-min working hour.

• Production = 1.5 × (60/0.5) × (50/60) = 1.5 × 120 × 0.833 = 150 yd³/hr.

Worked Example 5.10 — Trucks to match a loader. Truck cycle time 20 min, loading time 4 min.

• Trucks needed = 20/4 = 5 to keep the loader continuously busy.

Common traps. Omitting the efficiency factor. Confusing cycle time (minutes) with cycles per hour. Using field/max backwards in relative compaction. Matching a sheepsfoot roller to sand or a smooth drum to clay.

What the PE tests here. Production from cycle time and payload; efficiency factors; fleet balancing; relative compaction against the Proctor; roller-to-soil matching.

5.6 Temporary Works, Estimating, and Safety

Formwork must resist fresh concrete's lateral pressure, which at the full-liquid-head limit is:

p = γ·h (about 150·h psf)

A faster placement rate raises the pressure, because the concrete stays fluid deeper before it sets; form ties carry that pressure. Rigging follows statics — a two-leg sling's leg tension is:

T = (load/2) / sin θ (θ measured from horizontal)

so as the sling flattens (θ → 0), tension climbs steeply toward infinity. On bidding, know the two markup conventions, which the exam deliberately confuses:

Markup on cost: price = cost × (1 + m) Target margin on selling price: price = cost / (1 − m)

Contingency covers identified-but-unquantified risk; a change order formally modifies scope, price, or time. Safety is largely OSHA: excavations 5 ft or deeper generally require a protective system (sloping, shoring, or shielding); Type C soil is sloped no steeper than 1.5H:1V; a "competent person" can both identify hazards and is authorized to correct them.

Worked Example 5.11 — Sling leg tension. A 4,000 lb load hangs from a two-leg sling at θ = 30° from horizontal.

• T = (load/2)/sin θ = (4,000/2)/sin 30° = 2,000/0.5 = 4,000 lb per leg — each leg carries the full load at 30°, a vivid warning about flat sling angles.

Worked Example 5.12 — Markup versus margin. Cost is $80,000. (a) 20% markup on cost; (b) 20% margin on selling price.

• (a) Price = 80,000 × 1.20 = $96,000.
• (b) Price = 80,000 / (1 − 0.20) = 80,000/0.80 = $100,000. The same "20%" gives different prices — the exam exploits exactly this.

Common traps. Treating markup-on-cost and margin-on-price as the same. Forgetting that a flatter sling angle increases leg tension. Missing that a higher placement rate raises formwork pressure. Misremembering the OSHA 5 ft trigger or the 1.5H:1V Type C slope.

What the PE tests here. Formwork pressure and placement rate; sling statics; markup versus margin; contingency and change orders; OSHA excavation depth, slope, and competent-person rules.