BE Civil Engineering (IOE, TU) Estimating and Costing (IOE, CE 703) Question Paper 2080 Nepal

Centre-line method uses one common centre line for the whole wall layout. Because every layer is concentric about the same centre line, the centre-line length is constant for all layers.

Step 1 — Centre-line length. Wall thickness $= 0.30\text{ m}$, so the centre line lies $0.15\text{ m}$ inside each face. Centre-line dimensions $= (5.40 + 0.30) \times (4.20 + 0.30) = 5.70 \times 4.50$.

$$L_c = 2\,(5.70 + 4.50) = 2 \times 10.20 = 20.40\text{ m}$$

(For a single rectangular room there are no internal cross walls, so no junction deductions are needed.)

Step 2 — (a) Earthwork in excavation. Width $= 0.90\text{ m}$, depth $= 0.30\text{ m}$.

$$V_{earth} = L_c \times b \times d = 20.40 \times 0.90 \times 0.30 = 5.508\text{ m}^3$$

Earthwork in excavation $= 5.51\text{ m}^3$.

Step 3 — (b) Foundation concrete (1:4:8). Same plan as excavation bed: width $0.90\text{ m}$, depth $0.30\text{ m}$.

$$V_{conc} = 20.40 \times 0.90 \times 0.30 = 5.508\text{ m}^3$$

Foundation concrete $= 5.51\text{ m}^3$.

Step 4 — (c) Brick masonry. First footing: $20.40 \times 0.60 \times 0.30 = 3.672\text{ m}^3$ Second footing: $20.40 \times 0.40 \times 0.30 = 2.448\text{ m}^3$ Superstructure: $20.40 \times 0.30 \times 3.00 = 18.360\text{ m}^3$

$$V_{brick} = 3.672 + 2.448 + 18.360 = 24.480\text{ m}^3$$

Total brick masonry $= 24.48\text{ m}^3$.

Summary table

Rate analysis for 1 m³ first-class brick masonry in 1:6 cement mortar.

A. Material cost

B. Labour cost

C. Material + Labour

$$7{,}899.00 + 2{,}340.00 = 10{,}239.00$$

D. Water charges & sundries @ 1.5%

$$0.015 \times 10{,}239.00 = 153.585 \approx 153.59$$

E. Sub-total (material + labour + water)

$$10{,}239.00 + 153.59 = 10{,}392.59$$

F. Overhead & profit @ 15%

$$0.15 \times 10{,}392.59 = 1{,}558.89$$

G. Rate per m³

$$10{,}392.59 + 1{,}558.89 = 11{,}951.48$$

Rate of 1 m³ first-class brick masonry in 1:6 ≈ Rs. 11,951.50 (say Rs. 11,952).

Overall beam length $= 4.00 + 2(0.30) = 4.60\text{ m}$ (centre-to-centre of bearings). Reinforcement is taken over the full $4.60\text{ m}$ less end cover; for simplicity bars run the full member length $4.60\text{ m}$ and cover is absorbed in the hooks.

(1) Bottom main bars — 4 × Ø16

• Straight length $= 4.60\text{ m} - 2 \times 0.025 = 4.55\text{ m}$
• Hook/bend allowance $= 2 \times 9d = 2 \times 9 \times 16 = 288\text{ mm} = 0.288\text{ m}$
• Cutting length per bar $= 4.55 + 0.288 = 4.838\text{ m}$
• Total length $= 4 \times 4.838 = 19.352\text{ m}$
• Unit wt $= 16^2/162 = 256/162 = 1.5802\text{ kg/m}$
• Weight $= 19.352 \times 1.5802 = 30.578\text{ kg}$

(2) Top anchor bars — 2 × Ø12 (straight)

• Cutting length per bar $= 4.60 - 2 \times 0.025 = 4.55\text{ m}$
• Total length $= 2 \times 4.55 = 9.10\text{ m}$
• Unit wt $= 12^2/162 = 144/162 = 0.8889\text{ kg/m}$
• Weight $= 9.10 \times 0.8889 = 8.089\text{ kg}$

(3) Stirrups — Ø8 @ 150 c/c Core dimensions: $b = 300 - 2(25) = 250\text{ mm}$; $D = 450 - 2(25) = 400\text{ mm}$.

• Perimeter $= 2(250 + 400) = 1300\text{ mm}$
• Hooks $= 2 \times 10d = 2 \times 10 \times 8 = 160\text{ mm}$
• Cutting length per stirrup $= 1300 + 160 = 1460\text{ mm} = 1.460\text{ m}$
• Number of stirrups $= \dfrac{4600}{150} + 1 = 30.67 + 1 \approx 32$ nos.
• Total length $= 32 \times 1.460 = 46.72\text{ m}$
• Unit wt $= 8^2/162 = 64/162 = 0.3951\text{ kg/m}$
• Weight $= 46.72 \times 0.3951 = 18.460\text{ kg}$

Bar Bending Schedule

Total weight of steel in the beam $\approx 57.13\text{ kg}$ (say 57.13 kg).

Given: Construction cost $C = 60{,}00{,}000$; land $= 25{,}00{,}000$; life $n = 50$ yr; scrap $= 8\%$ of $C$; interest $i = 5\% = 0.05$.

(a) Annual sinking fund instalment. Scrap value $S = 0.08 \times 60{,}00{,}000 = 4{,}80{,}000$. Amount to be accumulated (depreciated cost) $= C - S = 60{,}00{,}000 - 4{,}80{,}000 = 55{,}20{,}000$. Sinking fund coefficient:

$$I = \frac{i}{(1+i)^n - 1} = \frac{0.05}{(1.05)^{50} - 1}$$

$(1.05)^{50} = 11.4674$, so $(1.05)^{50} - 1 = 10.4674$.

$$I = \frac{0.05}{10.4674} = 0.0047767$$

Annual instalment $= 55{,}20{,}000 \times 0.0047767 = 26{,}367.4$ Annual sinking fund instalment $\approx$ Rs. 26,367.

(b) Present value of building after 20 years (straight-line). Annual depreciation $= \dfrac{C - S}{n} = \dfrac{55{,}20{,}000}{50} = 1{,}10{,}400$ per year. Depreciation in 20 years $= 20 \times 1{,}10{,}400 = 22{,}08{,}000$. Present value of building $= C - \text{(accumulated depreciation)} = 60{,}00{,}000 - 22{,}08{,}000 = 37{,}92{,}000$. Depreciated value of building after 20 years = Rs. 37,92,000.

(c) Capitalised value of the property. Year's purchase $YP = \dfrac{1}{\text{rate}} = \dfrac{1}{0.07} = 14.2857$. Capitalised (net) value $= \text{Net annual income} \times YP = 6{,}30{,}000 \times 14.2857 = 90{,}00{,}000$. Capitalised value of the property = Rs. 90,00,000.

(Note: the land value of Rs. 25,00,000 is a non-depreciating asset; only the structure depreciates in part (b). The capitalised value in (c) reflects the whole income-producing property.)

Step 1 — Depth of cutting at each station $h = \text{Ground R.L.} - 100.00$:

Step 2 — Cross-sectional area $A = Bh + sh^2 = 7.5h + 2h^2$:

Interval $L = 30\text{ m}$, number of intervals $= 4$.

Step 3 — (a) Trapezoidal (average end-area) method.

$$V = L\left[\frac{A_0 + A_4}{2} + (A_1 + A_2 + A_3)\right]$$ $$= 30\left[\frac{23.000 + 1.580}{2} + (14.420 + 7.280 + 5.220)\right]$$ $$= 30\left[12.290 + 26.920\right] = 30 \times 39.210 = 1176.30\text{ m}^3$$

Trapezoidal volume $= 1176.30\text{ m}^3$.

Step 4 — (b) Prismoidal method (Simpson's rule; even number of intervals = 4).

$$V = \frac{L}{3}\left[A_0 + A_4 + 4(A_1 + A_3) + 2(A_2)\right]$$ $$= \frac{30}{3}\left[23.000 + 1.580 + 4(14.420 + 5.220) + 2(7.280)\right]$$ $$= 10\left[24.580 + 4(19.640) + 14.560\right]$$ $$= 10\left[24.580 + 78.560 + 14.560\right] = 10 \times 117.700 = 1177.00\text{ m}^3$$

Prismoidal volume $= 1177.00\text{ m}^3$.

Comparison. Difference $= 1177.00 - 1176.30 = 0.70\text{ m}^3$, i.e. about $0.06\%$. The prismoidal method is more accurate; here the surfaces are nearly linear so the two results almost coincide. The trapezoidal method slightly under-estimates the volume in this case.

Detailed estimate vs. approximate estimate

Two methods of approximate estimate:

1. Plinth area method — cost = plinth area × plinth-area rate (Rs./m²). Used for buildings at the planning/sanction stage.
2. Cubical content (volume) method — cost = built-up volume × rate per m³. Used for buildings where height varies; more reliable than plinth-area method for tall structures.

(Other acceptable methods: service-unit method for hospitals/schools/cinemas, per-km rate for roads, bay method for industrial sheds.)

Specification is a written statement describing the nature and class of work, quality and quantity of materials, workmanship, methods of execution and the results to be achieved — information that cannot be fully shown on drawings.

General specification gives a brief, overall description of the class and quality of work for the whole project (e.g. "first-class building with RCC framed structure, brick infill in 1:6, NP-2 finishing"). It is used to convey the general standard and to prepare approximate estimates.

Detailed specification describes each item of work item-by-item in full: materials, proportions, sizes, method of mixing/laying/curing, tolerances and mode of measurement (e.g. exact mix, slump, cover, curing days for RCC 1:1.5:3). It governs actual execution and payment.

Importance in contract documents:

• Defines quality and removes ambiguity about materials and workmanship.
• Forms the legal/technical basis for accepting or rejecting work.
• Enables correct rate analysis and fair payment.
• Supplements drawings, which show dimensions but not quality.

Contents of a complete tender/contract document:

1. Notice inviting tender (NIT) / tender notice.
2. Instructions to bidders and general conditions of contract.
3. Special conditions of contract.
4. Form of tender / bid form (with priced bill of quantities).
5. Bill of quantities (BoQ) and schedule of rates.
6. Drawings (architectural, structural, services).
7. Specifications (general and detailed / technical).
8. Schedule of completion time, milestones and penalty (liquidated damages) clause.
9. Form of agreement and bonds.

Earnest money deposit (EMD): a sum (commonly $1$–$2.5\%$ of the estimated cost) submitted with the bid as a guarantee that the bidder is serious and will sign the contract if awarded. It is refunded to unsuccessful bidders and adjusted/returned for the successful one; it is forfeited if a successful bidder refuses to execute the contract.

Performance (security) deposit: an additional sum (commonly $5$–$10\%$ of the contract amount) furnished by the successful contractor before/after signing the agreement as security for satisfactory performance. It is released after successful completion and the defect-liability period; it is forfeited (wholly or partly) for breach or substandard work.

Step 1 — Gross inside wall area (perimeter × height). Internal perimeter $= 2(4.50 + 3.60) = 2 \times 8.10 = 16.20\text{ m}$. Gross plaster area $= 16.20 \times 3.00 = 48.60\text{ m}^2$.

Step 2 — Deductions for openings. Door $= 1.00 \times 2.10 = 2.10\text{ m}^2$. Two windows $= 2 \times (1.20 \times 1.20) = 2 \times 1.44 = 2.88\text{ m}^2$. Total deduction $= 2.10 + 2.88 = 4.98\text{ m}^2$.

(As per IS/standard practice, for plaster the full opening area is deducted on each face because reveals/soffits roughly compensate; here only the inside face is plastered so full deduction applies.)

Step 3 — Net plaster area.

$$A_{net} = 48.60 - 4.98 = 43.62\text{ m}^2$$

Net area of internal plastering $= 43.62\text{ m}^2$.

Step 1 — Dry volume of materials for 1 m³ wet concrete. Dry volume $= 1.54 \times 1 = 1.54\text{ m}^3$.

Step 2 — Sum of mix proportions. $1 + 2 + 4 = 7$ parts.

Step 3 — Cement. Volume of cement $= \dfrac{1}{7} \times 1.54 = 0.22\text{ m}^3$. Mass $= 0.22 \times 1440 = 316.8\text{ kg}$. Bags $= \dfrac{316.8}{50} = 6.336 \approx 6.34$ bags.

Step 4 — Sand (fine aggregate). $= \dfrac{2}{7} \times 1.54 = 0.44\text{ m}^3$.

Step 5 — Coarse aggregate. $= \dfrac{4}{7} \times 1.54 = 0.88\text{ m}^3$.

Result for 1 m³ M15 (1:2:4):

(Check: $0.22 + 0.44 + 0.88 = 1.54\text{ m}^3$ dry volume — consistent.)

(Answer any two.)

(i) Lump-sum contract: The contractor agrees to complete the whole specified work for a single fixed sum, based on drawings and specifications, irrespective of actual quantities. Suitable when scope is well defined; risk of quantity variation lies with the contractor.

(ii) Item-rate (unit-price) contract: The contractor quotes a rate for each item in the bill of quantities; payment = measured quantity × quoted rate. Most common for civil works because it adjusts automatically to actual quantities executed.

(iii) Scrap value vs. salvage value: Scrap value is the value of a fully dismantled structure's materials (e.g. steel, broken bricks) at the end of life, after deducting demolition cost (typically ≈10% of cost). Salvage value is the value of an item sold/usable as it is at the end of its useful life without demolition.

(iv) Year's purchase (YP): The capital sum required to obtain a net annual income of one rupee at a given rate of interest; $YP = \dfrac{1}{\text{rate of interest}}$ (for perpetual income). Used to convert annual rental income into capitalised value.