BE Civil Engineering (IOE, TU) Estimating and Costing (IOE, CE 703) Question Paper 2079 Nepal

Step 1 — Height of filling at each station

Height of fill $h = \text{F.L.} - \text{G.L.}$

Step 2 — Cross-sectional area at each station

For a symmetrical embankment of formation width $b=10\ \text{m}$, side slope $s:1$ with $s=2$, on level transverse ground:

$$A = bh + s\,h^2 = 10h + 2h^2$$

Step 3 — Trapezoidal (average-area) method

With common distance $d = 30\ \text{m}$ and $n=6$ sections (5 intervals):

$$V = d\left[\frac{A_0 + A_5}{2} + (A_1 + A_2 + A_3 + A_4)\right]$$ $$= 30\left[\frac{0.000 + 59.500}{2} + (17.920 + 43.680 + 66.880 + 66.880)\right]$$ $$= 30\left[29.750 + 195.360\right] = 30 \times 225.110 = 6753.30\ \text{m}^3$$

Trapezoidal volume = 6753.30 m³

Step 4 — Prismoidal formula

The prismoidal rule requires an odd number of sections (even number of intervals). We have 6 sections / 5 intervals. Apply the prismoidal rule to the first 4 intervals (stations 0–4, 5 sections) and treat the last interval (4–5) by the trapezoidal rule.

Prismoidal rule for stations 0–4 (even number of intervals = 4):

$$V_1 = \frac{d}{3}\left[A_0 + A_4 + 4(A_1 + A_3) + 2(A_2)\right]$$ $$= \frac{30}{3}\left[0.000 + 66.880 + 4(17.920 + 66.880) + 2(43.680)\right]$$ $$= 10\left[66.880 + 4(84.800) + 87.360\right]$$ $$= 10\left[66.880 + 339.200 + 87.360\right] = 10 \times 493.440 = 4934.40\ \text{m}^3$$

Last interval (4–5) by trapezoidal rule:

$$V_2 = d\cdot\frac{A_4 + A_5}{2} = 30 \times \frac{66.880 + 59.500}{2} = 30 \times 63.190 = 1895.70\ \text{m}^3$$ $$V_{\text{prismoidal}} = V_1 + V_2 = 4934.40 + 1895.70 = 6830.10\ \text{m}^3$$

Prismoidal volume = 6830.10 m³

Step 5 — Comment

Difference $= 6830.10 - 6753.30 = 76.80\ \text{m}^3$ (about $1.1\%$). The prismoidal formula accounts for the curvature of the cross-section variation and is more accurate. The trapezoidal method under-estimates here because the area profile is concave (areas rise then level off). For tender/billing purposes the prismoidal result is preferred; the trapezoidal result may be corrected by the prismoidal correction where applicable.

Step 1 — Quantity of dry materials for 1 m³ wet concrete

Total dry volume $= 1.54\ \text{m}^3$. Proportion $1:1.5:3$, sum of ratios $=1+1.5+3 = 5.5$.

• Cement $= \dfrac{1}{5.5}\times 1.54 = 0.2800\ \text{m}^3$
• Sand $= \dfrac{1.5}{5.5}\times 1.54 = 0.4200\ \text{m}^3$
• Aggregate $= \dfrac{3}{5.5}\times 1.54 = 0.8400\ \text{m}^3$

Number of cement bags $= \dfrac{0.2800}{0.0347} = 8.069 \approx 8.07\ \text{bags}$.

Step 2 — Material cost

Cement: $8.07 \times 850 = 6859.50$

Step 3 — Labour cost

Step 4 — Machinery

Mixer + vibrator hire $= 600.00$ per m³.

Step 5 — Build up the rate

O&P $= 0.15 \times 12912.50 = 1936.88$

Total $= 12912.50 + 1936.88 = 14849.38$

Result

Rate of M20 RCC (excluding steel) = NRs. 14,849 per m³ (rounded).

Note on specifications

The rate assumes proper batching by volume with a measured water-cement ratio, machine mixing for at least 2 minutes, mechanical vibration for compaction, and curing for not less than 14 days as per standard specifications. Formwork/shuttering and reinforcement are paid for under separate items.

Step 1 — Centre-line length

Internal dimensions $= 5.00 \times 4.00\ \text{m}$. Wall thickness $t = 0.30\ \text{m}$.

Centre-to-centre dimensions (add one wall thickness to each internal dimension because the centre line lies half a wall thickness outside each face on both sides):

• Length $= 5.00 + 0.30 = 5.30\ \text{m}$
• Width $= 4.00 + 0.30 = 4.30\ \text{m}$

Total centre-line length for the rectangular plan:

$$L_{cl} = 2(5.30 + 4.30) = 2 \times 9.60 = 19.20\ \text{m}$$

(For a continuous rectangular wall run the centre line closes on itself, so no corner correction is needed — the +t/−t corrections at the four corners cancel.)

Step 2 — Gross brickwork volume

$$V_{gross} = L_{cl} \times t \times H = 19.20 \times 0.30 \times 3.00 = 17.280\ \text{m}^3$$

Step 3 — Deductions for openings

Door: $1.00 \times 2.10 \times 0.30 = 0.630\ \text{m}^3$

Windows (2 No.): $2 \times (1.20 \times 1.50 \times 0.30) = 2 \times 0.540 = 1.080\ \text{m}^3$

Total deduction $= 0.630 + 1.080 = 1.710\ \text{m}^3$

Step 4 — Net brickwork

$$V_{net} = 17.280 - 1.710 = 15.570\ \text{m}^3$$

Result

Net superstructure brickwork = 15.57 m³.

Verification by long-wall/short-wall method:

• Long walls (out-to-out) $= 5.00 + 2(0.15) = 5.30\ \text{m}$ each → matches centre-to-centre because thickness folds in symmetrically; both methods give the same $19.20\ \text{m}$ running length, confirming the result.

(a) Depreciated value — straight-line method

Construction cost $C = 24{,}00{,}000$. Scrap value $S = 10\% \times C = 2{,}40{,}000$.

Total depreciable amount $= C - S = 24{,}00{,}000 - 2{,}40{,}000 = 21{,}60{,}000$.

Annual depreciation $D = \dfrac{C - S}{n} = \dfrac{21{,}60{,}000}{60} = 36{,}000$ per year.

Depreciation in 20 years $= 20 \times 36{,}000 = 7{,}20{,}000$.

Depreciated (present) value of building:

$$V_b = C - 20D = 24{,}00{,}000 - 7{,}20{,}000 = 16{,}80{,}000$$

Present value of building = NRs. 16,80,000.

(b) Total present value of property

$$V = V_b + V_{land} = 16{,}80{,}000 + 30{,}00{,}000 = 46{,}80{,}000$$

Total present value = NRs. 46,80,000.

(c) Annual sinking-fund deposit

Sinking-fund (annual deposit) to accumulate amount $A$ in $n$ years at rate $i$:

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

Here $A = 24{,}00{,}000$, $i = 0.08$, $n = 60$.

$(1.08)^{60}$: $\ln(1.08)=0.076961$; $60\times0.076961 = 4.61768$; $e^{4.61768} = 101.257$.

So $(1.08)^{60} - 1 = 101.257 - 1 = 100.257$.

$$I = 24{,}00{,}000 \times \frac{0.08}{100.257} = 24{,}00{,}000 \times 0.0007980 = 1915.1$$

Annual sinking-fund deposit ≈ NRs. 1,915 per year.

(The very small deposit reflects the long 60-year accumulation period and compounding at 8%.)

(a) Components of a complete set of tender documents

1. Notice Inviting Tender (NIT) / Invitation for Bids — public advertisement giving brief scope, eligibility, cost of documents, bid security, submission deadline and opening date.
2. Instructions to Bidders (ITB) — how to prepare, seal, submit and the evaluation criteria.
3. Conditions of Contract — General Conditions (GCC) and Particular/Special Conditions (PCC/SCC) defining rights, obligations, payment terms, disputes, time, penalties.
4. Bill of Quantities (BoQ) with measured items and the priced schedule of rates.
5. Technical Specifications — material, workmanship and testing standards.
6. Drawings — site/layout, structural and detailing drawings.
7. Bid Forms and Schedules — bid letter, bid security form, schedule of unit rates, work programme, list of equipment and key personnel.
8. Form of Contract Agreement and forms of performance security and advance payment guarantee.
9. Appendices/Annexes — site information, soil report, addenda issued.

(b) Earnest money vs. performance security

(c) Three methods of awarding a contract

1. Open competitive (lowest evaluated responsive bid): the contract is awarded to the bidder who submits the lowest substantially responsive and qualified bid; most common and transparent.
2. Negotiated contract: the client negotiates terms and price directly with one or a few contractors, used for specialized, urgent or emergency works.
3. Limited / selective tendering: invitations are issued only to a pre-qualified or shortlisted list of competent contractors, balancing competition with assured capability.

Definition

An estimate is the calculated, anticipated cost (and the quantities of work and materials) of a project, worked out in advance from the drawings and specifications before the work is actually executed. It guides budgeting, sanction of funds and tendering.

Four types of estimates

1. Preliminary / approximate estimate: a rough cost found from approximate data (e.g., plinth-area rate, per-unit rate such as cost per bed for a hospital). Purpose: to decide the financial feasibility of a scheme and obtain administrative approval before detailing.
2. Detailed estimate: the most accurate estimate, prepared item-by-item from complete drawings and specifications with full quantity take-off and rate analysis. Purpose: to obtain technical sanction, invite tenders and control the work.
3. Revised estimate: a fresh detailed estimate prepared when the sanctioned cost is exceeded by more than a set limit (commonly 5%) or the scope changes. Purpose: to obtain revised financial sanction.
4. Supplementary estimate: an additional detailed estimate prepared for extra works that become necessary during execution, in addition to the original. Purpose: to sanction and fund additional/unforeseen works.

(Other valid types: maintenance estimate, annual repair estimate, complete/abstract estimate.)

Step 1 — Main bars (16 mm)

Total length $= 6 \times 6.50 = 39.00\ \text{m}$.

Unit weight $= \dfrac{16^2}{162} = \dfrac{256}{162} = 1.5802\ \text{kg/m}$.

Weight $= 39.00 \times 1.5802 = 61.63\ \text{kg}$.

Step 2 — Stirrups (8 mm)

Number of stirrups $= \dfrac{6000}{150} + 1 = 40 + 1 = 41$ Nos.

Total length $= 41 \times 1.20 = 49.20\ \text{m}$.

Unit weight $= \dfrac{8^2}{162} = \dfrac{64}{162} = 0.3951\ \text{kg/m}$.

Weight $= 49.20 \times 0.3951 = 19.44\ \text{kg}$.

Step 3 — Total weight

$$W = 61.63 + 19.44 = 81.07\ \text{kg}$$

Total weight of steel = 81.07 kg (≈ 81.1 kg).

General vs. detailed specifications

Three purposes of specifications

1. Define quality: they fix the standard/quality of materials and workmanship so the desired durability and performance are achieved.
2. Basis for rate analysis and cost: the rate of each item depends on its specification; they enable correct estimation and fair payment.
3. Legal/contractual control: being part of the contract, they let the client/engineer enforce quality, reject sub-standard work, and settle disputes.

(a) Year's Purchase

For a perpetual income, Year's Purchase:

$$Y.P. = \frac{100}{\text{rate of interest}} = \frac{100}{8} = 12.5$$

Year's Purchase = 12.5.

(b) Capitalized value

$$\text{Capitalized value} = \text{Net annual income} \times Y.P. = 4{,}80{,}000 \times 12.5$$ $$= 60{,}00{,}000$$

Capitalized (present) value of the property = NRs. 60,00,000.

(Interpretation: at an 8% expected return, an investor would pay up to NRs. 60 lakh so that the NRs. 4.80 lakh annual income yields 8% on the outlay: $4{,}80{,}000 / 60{,}00{,}000 = 0.08$.)

Units of measurement and mode of payment

Rule of thumb: work in which thickness varies or is substantial is paid by volume (m³); thin, surface-finish or uniform-thickness work is paid by area (m²); steel is paid by mass.

Step 1 — Gross plaster area (inside faces of 4 walls)

Internal perimeter $= 2(5.00 + 4.00) = 18.00\ \text{m}$.

Gross area $= \text{perimeter} \times \text{height} = 18.00 \times 3.20 = 57.600\ \text{m}^2$.

Step 2 — Deduction for openings

Door $= 1.00 \times 2.10 = 2.100\ \text{m}^2$.

Windows (2 No.) $= 2 \times (1.20 \times 1.50) = 2 \times 1.800 = 3.600\ \text{m}^2$.

Total deduction $= 2.100 + 3.600 = 5.700\ \text{m}^2$.

Step 3 — Net plaster area

$$A_{net} = 57.600 - 5.700 = 51.900\ \text{m}^2$$

Net inside plaster area = 51.90 m².

(Note: in practice the door reveal/jamb plaster and window sill may be added per the method of measurement; here, as specified, full opening areas are deducted once for the single inside face.)