BE Civil Engineering (IOE, TU) Estimating and Costing (IOE, CE 703) Question Paper 2077 Nepal

Geometry of the cross-section

The embankment cross-section is a symmetrical trapezium:

• Top (formation) width $b = 7.0\ \text{m}$
• Height $h = 1.5\ \text{m}$
• Side slope $s = 2$ (horizontal per unit vertical)

Bottom width $B = b + 2sh = 7.0 + 2(2)(1.5) = 7.0 + 6.0 = 13.0\ \text{m}$

Cross-sectional area of trapezium:

$$A = h\,(b + s h) = 1.5\,(7.0 + 2\times1.5) = 1.5\times(7.0+3.0) = 1.5\times10.0 = 15.0\ \text{m}^2$$

(Check: area of trapezium $=\tfrac{1}{2}(b+B)h = \tfrac{1}{2}(7.0+13.0)\times1.5 = \tfrac12\times20\times1.5 = 15.0\ \text{m}^2$ ✓)

(a) Volume of earthwork

1. Mid-sectional (mean section) method

Since the height is constant, the mid-section area equals the end areas $= 15.0\ \text{m}^2$.

$$V = A \times L = 15.0 \times 300 = \mathbf{4500\ m^3}$$

2. Prismoidal / trapezoidal method

With equal end areas $A_1 = A_2 = 15.0\ \text{m}^2$ and mid-area $A_m = 15.0\ \text{m}^2$, the prismoidal formula gives:

$$V = \frac{L}{6}\,(A_1 + 4A_m + A_2) = \frac{300}{6}(15 + 60 + 15) = 50 \times 90 = \mathbf{4500\ m^3}$$

(b) Why the answers agree

The ground is level and the height (hence the cross-sectional area) is constant over the whole length. Both methods then reduce to (uniform area) × (length), so they must coincide. They differ when the section area varies non-linearly along the length (e.g. when ground undulates or the bank height changes), because the mean-area method assumes a linear variation of area while the prismoidal method accounts for the curvature of the solid and is the more accurate.

(c) Loose (truck) volume required

Let the loose volume be $V_L$. After hauling, the soil compacts to $0.90$ of its loose volume in the embankment:

$$0.90\,V_L = 4500 \Rightarrow V_L = \frac{4500}{0.90} = \mathbf{5000\ m^3}$$

The 12% bulking from bank (in-situ) state describes the bank→loose change at the borrow pit but the controlling requirement here is the loose-to-compacted shrinkage, so the volume of loose soil that must be carried is 5000 m³.

Step 1 — Materials

Bricks (with 2% wastage): $500 \times 1.02 = 510$ nos

$$\text{Cost} = 510 \times 14 = \text{Rs } 7140$$

Mortar. Dry mortar required $= 0.30\ \text{m}^3$. (The 0.30 already represents the dry volume needed for the wet mortar in joints, so we proportion the dry volume directly.)

For 1:6 mortar, total parts $= 1 + 6 = 7$.

Cement volume $= \dfrac{1}{7}\times0.30 = 0.04286\ \text{m}^3$

Cement bags $= \dfrac{0.04286}{0.035} = 1.224$ bags

$$\text{Cement cost} = 1.224 \times 850 = \text{Rs } 1040.7$$

Sand volume $= \dfrac{6}{7}\times0.30 = 0.2571\ \text{m}^3$

$$\text{Sand cost} = 0.2571 \times 2200 = \text{Rs } 565.7$$

Step 2 — Labour

Step 3 — Prime cost

Step 4 — Add-ons

Water, tools & sundries @ 3% $= 0.03 \times 10946.4 = 328.4$ Sub-total $= 10946.4 + 328.4 = 11274.8$

Contractor's OH & Profit @ 10% $= 0.10 \times 11274.8 = 1127.5$

$$\text{Rate per m}^3 = 11274.8 + 1127.5 = \mathbf{Rs\ 12402.3\ per\ m^3}$$

Rounded rate ≈ Rs 12,402 per m³ of brickwork in 1:6 mortar.

Centre-line concept

In the centre-line method we take the length of the centre line of the wall and multiply by the cross-sectional area. For a closed rectangular building the centre line forms a smaller rectangle, and using the centre line automatically accounts for the corners: each external corner adds material while each internal corner removes the same amount, and these cancel exactly when we use the mean (centre) line. No separate corner deduction is needed.

(a) Centre-line length

Internal size $= 5.0 \times 4.0\ \text{m}$, wall thickness $t = 0.30\ \text{m}$.

Centre-line dimensions $=$ internal $+$ one wall thickness (half thickness on each opposite side):

$$L_{cl} = 5.0 + 0.30 = 5.30\ \text{m}, \quad B_{cl} = 4.0 + 0.30 = 4.30\ \text{m}$$

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

$$P_{cl} = 2\,(5.30 + 4.30) = 2 \times 9.60 = \mathbf{19.20\ m}$$

(b) Volume of brickwork (superstructure)

$$V_{bw} = P_{cl} \times t \times h = 19.20 \times 0.30 \times 3.0 = \mathbf{17.28\ m^3}$$

(c) Volume of RCC footing

Footing runs along the same centre line, width $0.60\ \text{m}$, depth $0.30\ \text{m}$:

$$V_{rcc} = P_{cl} \times 0.60 \times 0.30 = 19.20 \times 0.18 = \mathbf{3.456\ m^3}$$

Corner-handling check

If we instead used the long-wall / short-wall method:

• Long walls (out-to-out) length $= 5.0 + 2(0.15) = 5.30\ \text{m}$ each → 2 × 5.30 = 10.60 m
• Short walls (in-to-in) length $= 4.0 - 2(0.15) = 3.70\ \text{m}$ each → 2 × 3.70 = 7.40 m
• Total $= 10.60 + 7.40 = 18.00\ \text{m}$ of wall measured along its own line… but adding back the corner overlaps recovers $19.20\ \text{m}$ of mean line. The centre-line method gives the corner-corrected length 19.20 m in a single step, confirming the result.

(a) Straight-line depreciation

Original cost $C = \text{Rs } 30{,}00{,}000$ Scrap value $S = 10\% \times 30{,}00{,}000 = \text{Rs } 3{,}00{,}000$ Total life $n = 60$ years

Annual depreciation:

$$D = \frac{C - S}{n} = \frac{30{,}00{,}000 - 3{,}00{,}000}{60} = \frac{27{,}00{,}000}{60} = \mathbf{Rs\ 45{,}000/year}$$

Depreciation in 25 years $= 25 \times 45{,}000 = \text{Rs } 11{,}25{,}000$

Present depreciated value of building:

$$V_b = C - 25D = 30{,}00{,}000 - 11{,}25{,}000 = \mathbf{Rs\ 18{,}75{,}000}$$

(b) Present total value of property

$$V_{total} = V_{land} + V_b = 50{,}00{,}000 + 18{,}75{,}000 = \mathbf{Rs\ 68{,}75{,}000}$$

(c) Capitalized value (investor's view)

Year's purchase (YP) at 6% (perpetuity basis):

$$YP = \frac{100}{\text{rate}} = \frac{100}{6} = 16.667$$

Capitalized value:

$$V_{cap} = \text{Net annual rent} \times YP = 4{,}80{,}000 \times 16.667 = \mathbf{Rs\ 80{,}00{,}000}$$

Comparison & comment. The capitalized (income) value of Rs 80,00,000 exceeds the cost-based value of Rs 68,75,000 by about Rs 11.25 lakh (≈16%). This means the property earns a higher return than the cost approach suggests — the location/rental demand is strong, so on a pure investment basis the property is worth more than its depreciated replacement cost. An investor seeking a 6% yield could justify paying up to Rs 80 lakh.

(a) Definition and tendering process

Tender: A tender is a formal written offer submitted by a contractor, in response to an invitation, to execute specified works (or supply goods/services) at quoted rates and within stated conditions. Acceptance of a tender forms a binding contract.

Stages of competitive (open) tendering:

1. Preparation of drawings, specifications, BOQ and cost estimate by the client/engineer.
2. Invitation for Bids (IFB) / Notice published (e.g. in national daily and on PPMO's e-GP portal).
3. Issue/sale of bid documents to interested bidders.
4. Pre-bid meeting and clarifications/addenda (if any).
5. Submission of bids with bid security before the deadline.
6. Opening of bids publicly at the stated time in presence of bidders.
7. Evaluation — preliminary (completeness, security), technical, then financial; arithmetic correction.
8. Award to the lowest substantially-responsive evaluated bidder; issue Letter of Acceptance.
9. Signing of contract agreement after the successful bidder furnishes performance security.

(b) Documents forming the tender / contract document

(c) Earnest money vs. performance security

A related item, retention money, is typically 5% deducted from each running bill (up to a ceiling) and released after the defect-liability period — it supplements the performance security.

Estimate: An estimate is the calculated, probable cost of a project, computed in advance from drawings and specifications using known rates of materials, labour and other charges. It tells the owner whether the scheme is financially feasible and how much money to arrange.

Four types of estimates:

1. Preliminary (approximate) estimate — rough cost based on per-unit (per m², per bed, per km) rates, used to decide whether to proceed and to obtain administrative approval.
2. Detailed estimate — item-wise quantities × rates worked out from full drawings; the most accurate, used for tendering and execution.
3. Revised estimate — prepared when the original is exceeded by more than ~5% or scope changes, to seek fresh sanction.
4. Supplementary estimate — for additional works/items found necessary after the original estimate was sanctioned, prepared and approved separately.

(Others: quantity estimate, annual repair/maintenance estimate.)

Importance of specifications:

• Define the quality of materials and workmanship, fixing the standard the contractor must meet.
• Form a basis of the rate analysis and hence the cost; rates have meaning only against a specification.
• Serve as a legal/contract reference for acceptance, rejection and dispute resolution.
• Ensure the finished work matches the owner's intention even though drawings alone cannot show quality.

Dry volume

For 1 m³ of finished (wet) concrete, dry volume of ingredients:

$$V_{dry} = 1.54\ \text{m}^3$$

Total parts for 1:2:4 $= 1 + 2 + 4 = 7$.

Cement

$$V_{cement} = \frac{1}{7}\times1.54 = 0.22\ \text{m}^3$$ $$\text{Bags} = \frac{0.22}{0.035} = 6.286 \approx \mathbf{6.29\ bags}\ (\approx 314\ \text{kg})$$

Sand (fine aggregate)

$$V_{sand} = \frac{2}{7}\times1.54 = 0.44\ \text{m}^3 = \mathbf{0.44\ m^3}$$

Coarse aggregate

$$V_{agg} = \frac{4}{7}\times1.54 = 0.88\ \text{m}^3 = \mathbf{0.88\ m^3}$$

Summary

(Check: 0.22 + 0.44 + 0.88 = 1.54 m³ ✓)

Gross area of wall (one face)

$$A_{gross} = 6.0 \times 3.0 = 18.0\ \text{m}^2$$

Deductions for openings (one face)

Door: $1.0 \times 2.1 = 2.10\ \text{m}^2$ Window: $1.2 \times 1.5 = 1.80\ \text{m}^2$ Total openings $= 2.10 + 1.80 = 3.90\ \text{m}^2$

For 12 mm plaster on one face, the full opening area is deducted on that face (no separate reveal addition is required at this level of measurement).

Net plaster area

$$A_{net} = 18.0 - 3.90 = \mathbf{14.10\ m^2}$$

Net area of 12 mm cement plaster on one face = 14.10 m².

(Note: as per IS/standard practice, for openings up to 0.5 m² no deduction is made and jambs are not added; both openings here exceed that, so full deduction applies and the result stands at 14.10 m².)

Definition

A sinking fund is an amount set aside annually (and invested at compound interest) so that, by the end of an asset's useful life, the accumulated fund equals the money required to replace the asset. It spreads the future replacement cost over the life as equal yearly instalments.

Annual instalment

The sinking-fund formula:

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

where $S = 20{,}00{,}000$, $i = 0.05$, $n = 40$.

Compute $(1+i)^n = 1.05^{40}$.

$$1.05^{40} = 7.0400$$

So $(1.05^{40} - 1) = 6.0400$.

$$I = 20{,}00{,}000 \times \frac{0.05}{6.0400} = 20{,}00{,}000 \times 0.008278 = \mathbf{Rs\ 16{,}556\ per\ year}$$

Annual sinking-fund instalment ≈ Rs 16,556.

Advantage of centre-line method: It is quick and reduces calculation, since one centre-line length serves footing, masonry, DPC, plinth etc.

Limitation: For plans with cross-walls, junctions or odd angles, a correction (deduction of half the wall thickness at each T-junction/cross-junction) is needed, and for irregular shapes it becomes error-prone — the long-wall short-wall method is then safer.