Question 2 of 6: Estimating — trench excavation, footing forms and footing concrete

Question 2: Estimating — trench excavation, footing forms and footing concrete (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The take-off data, the trench section and the two RSMeans lines reproduced on the paper.

Find. For each of the three tasks — excavation, formwork and concrete — the quantity in its own unit, the duration in working days at the stated outputs, and the bare cost; then the estimate total.

[Figure not reproduced: Figure 2.1 — foundation plan redrawn to wall centre-lines. The trench follows every heavy line: the external outline, the 80 m divider and the 25 m spine between the two upper rooms. See the official exam paper.]

Approach. Measure one centre-line length for the whole trench, then price the three tasks off that single length — volume for excavation, two-sided contact area for the forms, and footing volume for the concrete — taking durations from the RSMeans daily outputs and the stated concrete production rate.

1. Measure the trench centre-line once. Every quantity in this question is a constant cross-section multiplied by the same length, so the length is measured first and used three times. The external outline is a 80 m × 70 m rectangle with a bay cut into the top edge; a re-entrant bay does not change the perimeter of the top run (\(30 + 20 + 30 = 80\) m) but does add the two 25 m returns:$$L_{\text{ext}} = 2(80 + 70) + 2(25) = 300 + 50 = 350 \text{ m}$$Adding the internal walls — the full-width divider and the spine from the divider up to the soffit of the bay, \(50 - 25 = 25\) m:$$L = 350 + 80 + 25 = \boxed{455 \text{ m of trench}}$$
2. Excavation quantity. The trench is \(0.30 + 0.75 + 0.30 = 1.35\) m wide and \(0.600 + 0.450 = 1.05\) m deep, so$$V_{\text{exc}} = L \times w \times d = 455 \times 1.35 \times 1.05 = \boxed{645 \text{ m}^3} \;(644.96 \text{ m}^3)$$The depth falls inside the 0.3–1.2 m band of the RSMeans line, so that line may be used without adjustment.
3. Excavation duration and bare cost. The 0.29 m³ backhoe crew produces 115 m³ per eight-hour day, and the bare cost is the labour plus equipment columns (there is no material in an excavation line):$$t_{\text{exc}} = \frac{644.96}{115} = 5.61 \text{ days},\qquad C_{\text{exc}} = 644.96 \times \$4.76 = \boxed{\$3{,}070.02}$$Splitting that between the columns gives labour \(644.96 \times \$3.02 = \$1{,}947.79\) and equipment \(644.96 \times \$1.74 = \$1{,}122.23\), which add back to the total — the arithmetic check on a Means line. Allow six working days on the programme.
4. Formwork quantity. The forms are the two vertical faces of the footing over its full 450 mm depth; the wall forms are explicitly excluded and the footing has no end forms of consequence on a continuous run:$$A_{\text{form}} = 2 \times 0.45 \times 455 = \boxed{409.5 \text{ m}^2 \text{ of contact area}}$$
5. Formwork duration and bare cost. Because the material is used twice, the "2 use" line applies — the material column drops from 12.80 to 7.00 per m² since each panel is amortised over two pours, and the crew output rises from 34.84 to 40.88 m² per day because the second use needs no fabrication:$$t_{\text{form}} = \frac{409.5}{40.88} = 10.02 \text{ days},\qquad C_{\text{form}} = 409.5 \times \$25.37 = \boxed{\$10{,}389.02}$$Pricing the same area on the single-use line would cost \(409.5 \times \$34.59 = \$14{,}164.61\), so re-using the material saves \(\$3{,}775.59\) — about 27 percent of the formwork bill, and the reason the question specifies two uses.
6. Concrete quantity, duration and cost. Only the footing is concreted:$$V_{\text{con}} = 455 \times 0.75 \times 0.45 = \boxed{153.6 \text{ m}^3}$$At 5 m³ per hour over an eight-hour day the placing rate is 40 m³ per day, so$$t_{\text{con}} = \frac{153.56}{40} = 3.84 \text{ days}$$The concreting cost is given as a lump sum of \(\$20{,}000\) with material included, which works out at \(\$20{,}000 / 153.56 = \$130.24\) per m³ — a plausible supplied-and-placed rate, which is the sanity check that the lump sum is being read correctly.
7. Assemble the estimate. Adding the three bare costs,$$C_{\text{total}} = \$3{,}070.02 + \$10{,}389.02 + \$20{,}000 = \boxed{\$33{,}459.04}$$and, if the three tasks are run strictly in sequence, \(5.61 + 10.02 + 3.84 = 19.5\) working days, say twenty. In practice the three would be lapped along the trench, so the sequential figure is an upper bound on the foundation programme.