Water Resources HubFE section 14 of 16 · free theory
Construction Engineering
Schedules, quantities, equipment, and earthwork — the Construction slice of the FE Civil exam, built around calculations you can do by hand in a few minutes.
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CPM scheduling: forward pass, backward pass, float
The critical path method turns a list of activities and dependencies into a project duration. Run two passes over the network:
Forward: ES = max(EF of predecessors), EF = ES + duration Backward: LF = min(LS of successors), LS = LF − duration
Total float = LS − ES = LF − EF Free float = min(ES of successors) − EF
| ES, EF | early start / early finish |
| LS, LF | late start / late finish |
| Total float | how long an activity can slip without delaying the project |
| Free float | how long it can slip without delaying any successor |
The critical path is the longest-duration path through the network — its activities have zero total float, and its length is the project duration. Gantt (bar) charts show the same schedule as horizontal bars over time; they display timing and overlaps but not dependencies, which is why CPM networks sit underneath them.
Worked example Forward/backward pass and critical path
Given:
- A (3 days) precedes B (4 days) and C (5 days); B and C both precede D (2 days).
Solution:
- Forward pass: A: ES 0, EF 3. B: ES 3, EF 7. C: ES 3, EF 8. D: ES = max(7, 8) = 8, EF 10. Project duration = 10 days.
- Backward pass: D: LF 10, LS 8. B: LF 8, LS 4. C: LF 8, LS 3. A: LF = min(4, 3) = 3, LS 0.
- Floats: A: 0; B: 4 − 3 = 1 day; C: 0; D: 0. The zero-float path A→C→D is critical (3 + 5 + 2 = 10 days). B can slip 1 day without affecting the finish.
Answer: Project duration 10 days; critical path A→C→D; activity B has 1 day of total float.
Estimating and quantity takeoff
Estimating starts with takeoff: measuring quantities from the drawings, then pricing them. The exam-level mechanics are unit conversions and careful arithmetic rather than pricing strategy:
Volume (CY) = Volume (ft³) / 27 1 CY = 27 ft³
| Bank CY (BCY) | material in its natural, in-place state |
| Loose CY (LCY) | material after excavation (swelled) |
| Compacted CY (CCY) | material after placement and compaction (shrunk) |
Equipment is rated in loose cubic yards per hour; earthwork pay quantities are usually bank (in place). Convert with the swell factor: LCY = BCY × (1 + swell). A typical soil swell of 25% means 100 BCY becomes 125 LCY.
Equipment productivity
Equipment output is bucket (or blade) capacity divided by cycle time, scaled to a working hour. The 50-minute hour (job efficiency 50/60 ≈ 0.833) is the standard allowance for non-productive time:
Production (LCY/h) = (60 × capacity (LCY) × efficiency) / cycle time (min)
Trucks required = truck cycle time / loader cycle time (round up)
| Cycle time | load + haul + dump + return (and spot/wait) for the controlling unit, min |
| Efficiency | 50-min hour → 0.833 unless stated otherwise |
Trucks are sized so the loader never waits: enough trucks to cover one full truck cycle per loader cycle, rounded up — 7.2 trucks means 8 trucks. Production is always governed by the slowest link: quote the loader or the hauler rate, whichever is lower.
Worked example Dozer production in loose and bank yards
Given:
- Dozer blade capacity 5 LCY, cycle time 1.2 min, 50-minute hour, soil swell 25%.
Solution:
- Production = 60 × 5 × (50/60) / 1.2 = 250 / 1.2 = 208.3 LCY/h.
- Convert to bank yards: 208.3 / 1.25 = 166.7 BCY/h.
Answer: ≈ 208 LCY/h (≈ 167 BCY/h after 25% swell).
Earthwork volumes
Volumes between two cross-sections come from the average end area method; the prismoidal formula adds the mid-section for a better answer when sections vary non-linearly:
V = L (A1 + A2) / 2 V = L (A1 + 4Am + A2) / 6
| V | volume, ft³ (divide by 27 for CY) |
| L | distance between the end sections, ft |
| A1, A2 | end cross-section areas, ft² |
| Am | mid-section area, ft² (prismoidal only) |
Prismoidal is exact for prismoids and a better approximation when the mid-section differs from the average of the ends. If Am happens to equal (A1 + A2)/2 — sections varying linearly — the two methods agree exactly.
Worked example Average end area vs prismoidal
Given:
- Two sections 100 ft apart: A1 = 120 ft², A2 = 180 ft², mid-section Am = 145 ft².
Solution:
- Average end area: V = 100 × (120 + 180)/2 = 15,000 ft³ = 15,000/27 = 555.6 CY.
- Prismoidal: V = 100 × (120 + 4 × 145 + 180)/6 = 100 × 880/6 = 14,666.7 ft³ = 543.2 CY.
- The mid-section is leaner than the average of the ends, so prismoidal correctly trims the volume by about 12 CY.
Answer: 555.6 CY by average end area; 543.2 CY by prismoidal.
Cost control concepts
The exam tests the vocabulary of tracking, not full earned-value arithmetic. Planned value is what the schedule said you would spend; earned value is what the completed work was worth; actual cost is what you really spent. Cost variance (earned − actual) and schedule variance (earned − planned) tell you, at a glance, whether the job is over budget, behind schedule, or both — and the exam usually asks only which of the two is true and in which direction.