BE Civil Engineering (IOE, TU) Estimating and Costing (IOE, CE 703) Question Paper 2076 Nepal

Step 1 — Compute cross-sectional areas (A = 7.5 × depth):

Using $A = 7.5 \times \text{depth}$, with cutting positive and filling negative areas tabulated separately.

Step 2 — Cutting (chainage 0 to 100, common distance $d = 20\ \text{m}$).

Areas: $A_0=9.00,\ A_1=12.00,\ A_2=15.00,\ A_3=11.25,\ A_4=6.00,\ A_5=0.00$.

Trapezoidal rule:

$$V = d\left[\frac{A_0+A_n}{2} + (A_1+A_2+A_3+A_4)\right]$$ $$V_{cut}=20\left[\frac{9.00+0.00}{2} + (12.00+15.00+11.25+6.00)\right]$$ $$=20\,[4.50 + 44.25]=20\times48.75=\mathbf{975\ m^3}$$

Prismoidal (Simpson's) rule — needs an even number of intervals; here 0→100 is 5 intervals (odd). Apply Simpson over the first 4 intervals (0→80) and trapezoid for the last (80→100).

Simpson 0→80:

$$V=\frac{d}{3}\big[A_0+A_4+4(A_1+A_3)+2(A_2)\big]$$ $$=\frac{20}{3}\big[9.00+6.00+4(12.00+11.25)+2(15.00)\big]$$ $$=\frac{20}{3}\big[15.00+4(23.25)+30.00\big]=\frac{20}{3}\big[15.00+93.00+30.00\big]=\frac{20}{3}(138.00)=920.00\ m^3$$

Trapezoid 80→100: $V=20\times\frac{6.00+0.00}{2}=60.00\ m^3$.

$$V_{cut,\,prismoidal}=920.00+60.00=\mathbf{980\ m^3}$$

Step 3 — Filling (chainage 100 to 200, $d = 20\ \text{m}$).

Areas: $A_0=0.00,\ A_1=5.25,\ A_2=9.00,\ A_3=12.00,\ A_4=10.50,\ A_5=7.50$.

Trapezoidal rule:

$$V_{fill}=20\left[\frac{0.00+7.50}{2}+(5.25+9.00+12.00+10.50)\right]$$ $$=20\,[3.75+36.75]=20\times40.50=\mathbf{810\ m^3}$$

Prismoidal rule — Simpson over 100→180 (4 intervals) + trapezoid 180→200.

Simpson 100→180:

$$=\frac{20}{3}\big[0.00+10.50+4(5.25+12.00)+2(9.00)\big]$$ $$=\frac{20}{3}\big[10.50+4(17.25)+18.00\big]=\frac{20}{3}\big[10.50+69.00+18.00\big]=\frac{20}{3}(97.50)=650.00\ m^3$$

Trapezoid 180→200: $V=20\times\frac{10.50+7.50}{2}=20\times9.00=180.00\ m^3$.

$$V_{fill,\,prismoidal}=650.00+180.00=\mathbf{830\ m^3}$$

Step 4 — Summary.

Comment: The prismoidal (Simpson's) rule is more accurate because it fits a second-degree (parabolic) curve through three consecutive sections, capturing the curvature of the area variation, whereas the trapezoidal rule assumes a straight-line (linear) variation between sections and thus introduces error wherever the ground/area profile is curved. The trapezoidal rule generally underestimates volume for a convex area profile. For precise earthwork billing, the prismoidal volumes (cut ≈ 980 m³, fill ≈ 830 m³) should be adopted.

Step 1 — Quantities of dry materials for 1 m³ wet concrete (1:1.5:3).

Dry volume $=1.54\ \text{m}^3$. Sum of proportions $=1+1.5+3=5.5$.

• Cement volume $=\dfrac{1}{5.5}\times1.54=0.28\ \text{m}^3$.
• Sand volume $=\dfrac{1.5}{5.5}\times1.54=0.42\ \text{m}^3$.
• Coarse aggregate $=\dfrac{3}{5.5}\times1.54=0.84\ \text{m}^3$.

Step 2 — Convert cement to bags.

Number of bags $=\dfrac{0.28}{0.0347}=8.07\ \text{bags}$. Adopt $8.07$ bags.

Step 3 — Material cost.

Cement: $8.07\times800=6{,}456.00$. Sand: $0.42\times1{,}800=756.00$. Aggregate: $0.84\times2{,}400=2{,}016.00$. Sum $=9{,}228.00$.

Step 4 — Water and sundries (1.5% of material cost).

$0.015\times9{,}228.00=138.42$.

Step 5 — Labour.

• Mason: $1\times1{,}000\times0.30=300.00$.
• Unskilled: $4\times700\times0.30=4\times210=840.00$.
• Operator allowance: $250.00$.
• Labour subtotal $=300.00+840.00+250.00=1{,}390.00$.

Step 6 — Equipment hire. $=600.00$.

Step 7 — Prime cost (before OH&P).

$$\text{PC}=9{,}228.00+138.42+1{,}390.00+600.00=11{,}356.42$$

Step 8 — Overhead and profit (15%).

$0.15\times11{,}356.42=1{,}703.46$.

Step 9 — Total rate per m³.

$$11{,}356.42+1{,}703.46=13{,}059.88\approx\mathbf{Rs.\ 13{,}060\ per\ m^3}$$

Specification note: M20 concrete shall be machine-mixed for not less than 2 minutes, transported without segregation, placed in layers not exceeding 150 mm, compacted with a needle vibrator, and cured for at least 14 days by ponding/wet covering. The above rate is exclusive of reinforcement, formwork and surface finishing, which are paid under separate items.

Step 1 — Centre-line length.

For a single rectangular room, the wall centre line forms a rectangle. Wall thickness $=0.30\ \text{m}$, so add half-thickness on each internal side:

• Centre-line length (long direction) $=5.0+0.30=5.30\ \text{m}$ (internal 5.0 + 2×0.15).
• Centre-line length (short direction) $=4.0+0.30=4.30\ \text{m}$.

Total centre-line length (perimeter of centre-line rectangle):

$$L=2(5.30+4.30)=2\times9.60=\mathbf{19.20\ m}$$

Note on the centre-line method: for a continuous closed run with one type of wall, no deduction/addition for junctions is required because the rectangle's own corners already account for the overlap correctly (each corner is counted once). The single centre-line length $L=19.20\ \text{m}$ is used for all concentric items (footing, plinth, wall).

Step 2 — PCC footing (1:3:6).

Width $=0.70\ \text{m}$, depth $=0.30\ \text{m}$.

$$V_{PCC}=L\times b\times d=19.20\times0.70\times0.30=\mathbf{4.032\ m^3}$$

Step 3 — Brick masonry below plinth (foundation/plinth masonry).

Thickness $=0.30\ \text{m}$, height $=0.60\ \text{m}$.

$$V_{found}=19.20\times0.30\times0.60=\mathbf{3.456\ m^3}$$

Step 4 — Brick masonry in superstructure (above plinth).

Thickness $=0.30\ \text{m}$, height $=2.90\ \text{m}$.

$$V_{super}=19.20\times0.30\times2.90=\mathbf{16.704\ m^3}$$

Summary table.

Final answers: Centre-line length $=\mathbf{19.20\ m}$; PCC $=\mathbf{4.032\ m^3}$; foundation masonry $=\mathbf{3.456\ m^3}$; superstructure masonry $=\mathbf{16.704\ m^3}$ (total brick masonry $=20.16\ m^3$).

Part (a) — Cost (land + depreciated building) method.

Scrap value of building $=10\%$ of Rs. 65,00,000 $=0.10\times6{,}500{,}000=650{,}000$.

Total depreciable amount $=6{,}500{,}000-650{,}000=5{,}850{,}000$.

Annual depreciation (straight-line) $=\dfrac{5{,}850{,}000}{80}=73{,}125$ per year.

Accumulated depreciation in 20 years $=73{,}125\times20=1{,}462{,}500$.

Present (depreciated) value of building:

$$=6{,}500{,}000-1{,}462{,}500=\mathbf{Rs.\ 50{,}37{,}500}$$

Value of land $=250\times90{,}000=22{,}500{,}000=$ Rs. 2,25,00,000.

Total property value (cost method):

$$=5{,}037{,}500+22{,}500{,}000=\mathbf{Rs.\ 2{,}75{,}37{,}500}$$

Part (b) — Income (rental capitalization) method.

For income treated as perpetual, Year's Purchase $=\dfrac{1}{\text{rate of return}}=\dfrac{1}{0.08}=12.5$.

Capitalized value:

$$V=\text{Net annual income}\times\text{Y.P.}=480{,}000\times12.5=\mathbf{Rs.\ 60{,}00{,}000}$$

Comparison and comment.

The two figures differ greatly because they measure different things. The cost method includes the full market value of the land (Rs. 2.25 crore), which dominates the total, plus the bricks-and-mortar value of the structure. The income method capitalizes only the net rent the property currently earns; at an 8% yield, Rs. 4.8 lakh of annual rent supports a capital value of only Rs. 60 lakh. The large gap indicates the property is under-rented relative to its capital (land) value — the rent yields far below 8% on the cost-method value ($480{,}000/27{,}537{,}500\approx1.7\%$). For an owner-occupied or land-rich property the cost method is the more meaningful indicator of worth; for a pure investment decision the income method governs. A prudent valuer would report both and reconcile, noting the latent redevelopment value of the land.

(a) Tender — definition and contract-document contents.

A tender is a formal, written offer submitted by a contractor in response to an invitation (notice inviting tender, NIT) to execute a specified work, supply goods, or render services at quoted rates and within stated conditions. When accepted by the employer, it forms a binding contract.

A complete tender/contract document typically contains:

1. Notice Inviting Tender (NIT) / Invitation for Bids — work description, estimated cost, bid security, time of completion, dates of issue/submission/opening.
2. Instructions to Bidders (ITB) — eligibility, how to fill and submit the bid, validity period.
3. General Conditions of Contract (GCC) and Special/Particular Conditions of Contract (SCC) — rights, duties, payment terms, dispute resolution, penalties.
4. Technical Specifications — quality of materials and workmanship.
5. Drawings — plans, sections, details.
6. Bill of Quantities (BOQ) / Schedule of Rates / Priced Schedule — items, quantities, unit rates, amounts.
7. Form of Bid / Form of Agreement — to be signed by both parties.
8. Forms of securities — bid security, performance security, advance-payment guarantee formats.
9. Schedule of completion / work programme and list of approved makes/brands.
10. Addenda/corrigenda issued before submission, if any.

(b) Earnest money (bid security) vs. performance security.

Bid security is returned to unsuccessful bidders after award and to the successful bidder after it furnishes performance security; performance security is released after successful completion and expiry of the defect-liability (maintenance) period.

(c) Types of contract.

1. Lump-sum contract: The contractor agrees to complete the entire work for a single fixed sum, based on drawings and specifications. Suitable when the scope is well defined and unlikely to change (e.g. a standard, fully designed building).
2. Item-rate (unit-price / BOQ) contract: The contractor quotes rates for each item of work; payment is the measured quantity × quoted rate. Suitable when quantities cannot be precisely fixed in advance (e.g. earthwork, foundations in uncertain ground) — most public civil works use this form.
3. Cost-plus contract: The employer reimburses the actual cost of the work plus a fee (a fixed amount or a percentage of cost). Suitable for urgent or ill-defined work where the scope cannot be estimated beforehand (e.g. emergency/disaster repair, novel works).

Four types of estimates:

1. Preliminary (approximate / rough-cost) estimate: A quick estimate prepared before detailed design using approximate methods (plinth-area rate, per-unit rate such as cost per bed for a hospital). Purpose: to judge financial feasibility and obtain administrative approval and budget provision.
2. Detailed estimate: A complete, item-by-item estimate worked out from final drawings and specifications using measured quantities and analysed rates, supported by detailed measurement (abstract) sheets. Purpose: to obtain technical sanction, invite tenders and control expenditure; it is the most accurate estimate.
3. Revised estimate: A fresh detailed estimate prepared when the sanctioned estimate is exceeded by a defined limit (e.g. more than 5%) due to rate changes or material increase, without change of scope. Purpose: to obtain revised sanction for the increased cost.
4. Supplementary estimate: An additional detailed estimate prepared for extra works or additions found necessary during execution, in addition to the original. Purpose: to sanction and fund additional/changed scope.

(Other valid types include quantity estimate, annual repair/maintenance estimate, and complete estimate.)

Step 1 — Cutting length of one bar.

Hook length per end $=9d=9\times16=144\ \text{mm}=0.144\ \text{m}$. Two hooks $=2\times0.144=0.288\ \text{m}$.

$$\text{Cutting length}=6.0+0.288=\mathbf{6.288\ m}$$

Step 2 — Total length of 5 bars.

$$5\times6.288=\mathbf{31.44\ m}$$

Step 3 — Unit weight of 16 mm bar.

$$w=\frac{d^2}{162}=\frac{16^2}{162}=\frac{256}{162}=1.5802\ \text{kg/m}$$

Step 4 — Total weight.

$$W=31.44\times1.5802=\mathbf{49.68\ kg}$$

Final answers: cutting length per bar $=6.288\ \text{m}$; total length $=31.44\ \text{m}$; total steel weight $\approx\mathbf{49.68\ kg}$.

General vs. detailed specifications.

Two purposes of specifications:

1. Define quality and standard of work: they fix the type and quality of materials, proportions and workmanship, ensuring the work meets the intended standard and enabling fair rate analysis (rates depend on specified quality).
2. Serve as a legal/contractual basis: they form part of the contract documents, so they govern acceptance, measurement and payment, and act as the reference for resolving disputes about quality or method.

Step 1 — Gross internal wall area (four faces, full height).

Internal perimeter $=2(3.6+4.2)=2\times7.8=15.6\ \text{m}$.

Gross plaster area $=\text{perimeter}\times\text{height}=15.6\times3.0=46.80\ \text{m}^2$.

Step 2 — Deductions for openings (one face, full area).

• Door: $1.0\times2.1=2.10\ \text{m}^2$.
• Two windows: $2\times(1.2\times1.5)=2\times1.80=3.60\ \text{m}^2$.
• Total deduction $=2.10+3.60=5.70\ \text{m}^2$.

Step 3 — Net plaster area.

$$A_{net}=46.80-5.70=\mathbf{41.10\ m^2}$$

Answer: Net internal wall plastering area $=\mathbf{41.10\ m^2}$.

Note: Per IS measurement convention, for 12 mm plaster the full opening area is deducted on the side being measured (no jamb/reveal addition for this thin plaster), which is the simplification applied here.

Concept. A sinking fund is a fund created by setting aside a fixed sum (instalment) every year, which accumulates at compound interest so that the total grows to a required amount (e.g. the cost of replacement or the capital to be redeemed) by the end of a fixed period. In valuation it represents the annual provision an owner must make to recoup the depreciating value of a structure by the end of its useful life.

Formula. The annual sinking-fund instalment $I$ to accumulate an amount $S$ in $n$ years at interest rate $i$ is:

$$I=S\cdot\frac{i}{(1+i)^n-1}$$

Given: $S=\text{Rs. }30{,}00{,}000$, $n=25$, $i=0.06$.

Step 1 — Compute $(1+i)^n$.

$$(1.06)^{25}=4.29187$$

Step 2 — Compute the factor.

$$\frac{i}{(1+i)^n-1}=\frac{0.06}{4.29187-1}=\frac{0.06}{3.29187}=0.018226$$

Step 3 — Annual instalment.

$$I=3{,}000{,}000\times0.018226=54{,}678\approx\mathbf{Rs.\ 54{,}678\ per\ year}$$

Answer: An annual sinking-fund deposit of approximately Rs. 54,678 (about Rs. 54,700) will accumulate to Rs. 30,00,000 in 25 years at 6% compound interest.

Method used: the mid-section / mean cross-sectional area method (a single uniform cross-section is multiplied by the length, as the trench has constant depth and width). The cross-section is a symmetrical trapezium because of the side slopes.

Step 1 — Top width of the trapezoidal section.

Side slope $1\text{H}:4\text{V}$ means for every 4 m of depth the trench widens 1 m horizontally on each side; i.e. horizontal spread per side $=\dfrac{1}{4}\times\text{depth}=\dfrac{1}{4}\times1.6=0.40\ \text{m}$.

Top width $=b+2\times0.40=0.9+0.80=1.70\ \text{m}$.

Step 2 — Cross-sectional area (trapezium).

$$A=\frac{1}{2}(b_{bottom}+b_{top})\times d=\frac{1}{2}(0.90+1.70)\times1.6$$ $$=\frac{1}{2}\times2.60\times1.6=1.30\times1.6=2.08\ \text{m}^2$$

(Equivalently $A=(b+s\,d)\,d=(0.9+0.25\times1.6)\times1.6=(0.9+0.40)\times1.6=2.08\ \text{m}^2$, with $s=1/4=0.25$.)

Step 3 — Volume of excavation.

$$V=A\times L=2.08\times30=\mathbf{62.4\ m^3}$$

Answer: Earthwork in excavation $=\mathbf{62.4\ m^3}$, computed by the mean (uniform) cross-section method using a trapezoidal section.