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Section A: Long Answer Questions

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5 questions
1long10 marks

A road in cutting is to be constructed in level ground. The depths of cutting at five consecutive cross-sections, each spaced 30 m30\text{ m} apart along the centre line, are 1.2 m, 1.8 m, 2.4 m, 1.6 m1.2\text{ m},\ 1.8\text{ m},\ 2.4\text{ m},\ 1.6\text{ m} and 0.9 m0.9\text{ m} respectively. The formation width of the road is 10 m10\text{ m} and the side slopes of the cutting are 2:12:1 (horizontal : vertical). The ground is level transversely at every section.

(a) Compute the quantity of earthwork (cutting) by the trapezoidal (average end area) rule.

(b) Compute the same quantity by the prismoidal (Simpson's) rule and comment on which value is generally adopted for payment and why. (7+3)(7+3)

Cross-sectional area at each station (level section, two-sided slope):

For a level section, A=d(b+sd)A = d\,(b + s\,d) where b=10 mb=10\text{ m} (formation width), s=2s=2 (side slope), and dd is the depth of cutting.

StationDepth dd (m)A=d(10+2d)A=d(10+2d) (m²)
01.21.2(10+2.4)=14.881.2(10+2.4)=14.88
11.81.8(10+3.6)=24.481.8(10+3.6)=24.48
22.42.4(10+4.8)=35.522.4(10+4.8)=35.52
31.61.6(10+3.2)=21.121.6(10+3.2)=21.12
40.90.9(10+1.8)=10.620.9(10+1.8)=10.62

(a) Trapezoidal (average end area) rule

V=L[A0+A42+A1+A2+A3]V=L\left[\frac{A_0+A_4}{2}+A_1+A_2+A_3\right] V=30[14.88+10.622+24.48+35.52+21.12]V=30\left[\frac{14.88+10.62}{2}+24.48+35.52+21.12\right] V=30[12.75+81.12]=30×93.87V=30\,[\,12.75+81.12\,]=30\times 93.87 Vtrap=2816.10 m3\boxed{V_{trap}=2816.10\ \text{m}^3}

(b) Prismoidal (Simpson's 1/3) rule — there are 5 ordinates (4 intervals, even number), so the rule applies directly:

V=L3[A0+A4+4(A1+A3)+2A2]V=\frac{L}{3}\left[A_0+A_4+4(A_1+A_3)+2A_2\right] V=303[14.88+10.62+4(24.48+21.12)+2(35.52)]V=\frac{30}{3}\left[14.88+10.62+4(24.48+21.12)+2(35.52)\right] V=10[25.50+4(45.60)+71.04]=10[25.50+182.40+71.04]V=10\,[\,25.50+4(45.60)+71.04\,]=10\,[\,25.50+182.40+71.04\,] Vprism=2789.40 m3\boxed{V_{prism}=2789.40\ \text{m}^3}

Comment: The prismoidal rule treats the solid between sections as a true prismoid and is therefore more accurate. The trapezoidal value is higher by 2816.102789.40=26.70 m32816.10-2789.40=26.70\text{ m}^3 (about 1%). For final payment in earthwork the prismoidal volume is generally adopted; the difference (trapezoidal − prismoidal) is the prismoidal correction and is deducted from the trapezoidal estimate when greater precision is required.

earthwork-estimatetrapezoidal-prismoidalmethods-of-measurement
2long10 marks

Prepare the rate analysis for 1 m³ of cement concrete of nominal mix 1:2:4 used in plain footing, using the following data:

  • Dry-to-wet volume factor (bulkage/voids) =1.54=1.54
  • 1 bag of cement =0.035 m3=0.035\text{ m}^3; rate of cement =NRs. 800=\text{NRs. }800 per bag
  • Sand =NRs. 1,800=\text{NRs. }1{,}800 per m³; coarse aggregate =NRs. 2,200=\text{NRs. }2{,}200 per m³
  • Labour for mixing, placing and compacting per m³: 1 mason @ NRs. 1,200/day and 4 labourers @ NRs. 900/day
  • Add 15% of (material + labour) cost for contractor's overhead and profit

Determine the rate per cubic metre of concrete. (4+4+2)(4+4+2)

Step 1 — Dry quantities of materials per 1 m³ of wet concrete

Sum of proportions =1+2+4=7=1+2+4=7. Dry volume =1.54 m3=1.54\text{ m}^3.

  • Cement =17×1.54=0.22 m3=\dfrac{1}{7}\times1.54=0.22\text{ m}^3
  • Sand =27×1.54=0.44 m3=\dfrac{2}{7}\times1.54=0.44\text{ m}^3
  • Coarse aggregate =47×1.54=0.88 m3=\dfrac{4}{7}\times1.54=0.88\text{ m}^3

Cement in bags =0.220.035=6.286 bags=\dfrac{0.22}{0.035}=6.286\text{ bags}

Step 2 — Material cost

MaterialQuantityRateAmount (NRs.)
Cement6.286 bags800/bag6.286×800=5028.576.286\times800=5028.57
Sand0.44 m³1800/m³0.44×1800=792.000.44\times1800=792.00
Aggregate0.88 m³2200/m³0.88×2200=1936.000.88\times2200=1936.00
Material subtotal7,756.57

Step 3 — Labour cost

LabourNo.Rate/dayAmount (NRs.)
Mason112001200.00
Labourer49003600.00
Labour subtotal4,800.00

Step 4 — Build-up of rate

  • Material + Labour =7756.57+4800.00=12,556.57=7756.57+4800.00=12{,}556.57
  • Overhead & profit @15% =0.15×12556.57=1,883.49=0.15\times12556.57=1{,}883.49
  • Total rate =12556.57+1883.49=14,440.06=12556.57+1883.49=14{,}440.06
Rate per m3 of 1:2:4 concreteNRs. 14,440\boxed{\text{Rate per m}^3\ \text{of }1{:}2{:}4\ \text{concrete}\approx \text{NRs. }14{,}440}
rate-analysiscement-concretematerial-labour
3long10 marks

A single rectangular room has internal dimensions 4.0 m×3.0 m4.0\text{ m}\times3.0\text{ m}. The superstructure walls are of brick masonry, 0.30 m0.30\text{ m} thick (one-and-a-half brick), built to a clear height of 3.0 m3.0\text{ m} above plinth (ignore openings for this part).

(a) Using the centre-line method, compute the volume of brick masonry in the walls.

(b) Estimate the number of modern bricks required, taking 500 bricks per m³ of finished masonry, and the volume of cement mortar at 30% of brick-masonry volume.

(c) State one situation where the centre-line method needs a correction and how it is applied. (5+3+2)(5+3+2)

(a) Centre-line method

Wall thickness t=0.30 mt=0.30\text{ m}, so the centre line lies t/2=0.15 mt/2=0.15\text{ m} outside each internal face.

Centre-line length of a single closed rectangular wall:

Lcl=2[(Li+t)+(Bi+t)]=2[(4.0+0.30)+(3.0+0.30)]L_{cl}=2\big[(L_i+t)+(B_i+t)\big]=2\big[(4.0+0.30)+(3.0+0.30)\big] Lcl=2[4.30+3.30]=2×7.60=15.20 mL_{cl}=2\,[4.30+3.30]=2\times7.60=15.20\text{ m}

(For a single rectangular building no junction correction is needed because the perimeter is one continuous wall.)

Volume of brick masonry:

V=Lcl×t×h=15.20×0.30×3.0V=L_{cl}\times t\times h=15.20\times0.30\times3.0 V=13.68 m3\boxed{V=13.68\ \text{m}^3}

(b) Number of bricks and mortar

  • Bricks =13.68×500=6,840 bricks=13.68\times500=\mathbf{6{,}840\ bricks}
  • Cement mortar =0.30×13.68=4.10 m3=0.30\times13.68=\mathbf{4.10\ m^3}

(c) When correction is needed: In the centre-line method, wherever walls of the same thickness meet at a junction (T-junction or cross-junction), the masonry at the meeting length is counted twice. The correction is to deduct half the wall thickness for each junction from the centre-line length, i.e. subtract t2\tfrac{t}{2} per side of every junction. For a simple single rectangular room there is no such junction, so no deduction applies here.

brick-masonrycentre-line-methodquantity-estimate
4long10 marks

A simply supported RC roof slab is 4.0 m4.0\text{ m} (span) × 5.0 m\times\ 5.0\text{ m} (width) in plan. The reinforcement is:

  • Main bars: 12 mm dia @ 150 mm c/c, running along the 4.0 m span
  • Distribution bars: 10 mm dia @ 200 mm c/c, running along the 5.0 m width
  • Clear cover =25 mm=25\text{ mm} on all edges; provide a standard hook/bend allowance of 9d9d at each end of every bar.

Prepare the bar-bending computation and find the total weight of steel. Use unit weights w=d2162 kg/mw=\dfrac{d^2}{162}\text{ kg/m} (12 mm =0.888=0.888 kg/m, 10 mm =0.617=0.617 kg/m). (5+5)(5+5)

Step 1 — Number of bars

Main bars are distributed across the 5.0 m width:

nmain=5.02(0.025)0.15+1=4.950.15+1=33+1=34 barsn_{main}=\frac{5.0-2(0.025)}{0.15}+1=\frac{4.95}{0.15}+1=33+1=34\ \text{bars}

Distribution bars are distributed across the 4.0 m span:

ndist=4.02(0.025)0.20+1=3.950.20+1=19.75+120 barsn_{dist}=\frac{4.0-2(0.025)}{0.20}+1=\frac{3.95}{0.20}+1=19.75+1\approx 20\ \text{bars}

Step 2 — Cutting length of each bar (length == clear span between covers ++ two end bends of 9d9d)

Main (12 mm), runs along 4.0 m span:

Lmain=(4.02×0.025)+2×9×0.012=3.95+0.216=4.166 mL_{main}=(4.0-2\times0.025)+2\times9\times0.012=3.95+0.216=4.166\text{ m}

Distribution (10 mm), runs along 5.0 m width:

Ldist=(5.02×0.025)+2×9×0.010=4.95+0.180=5.130 mL_{dist}=(5.0-2\times0.025)+2\times9\times0.010=4.95+0.180=5.130\text{ m}

Step 3 — Total length and weight

BarDiaNo.Length each (m)Total length (m)Unit wt (kg/m)Weight (kg)
Main12344.166141.644141.6440.888125.78125.78
Distribution10205.130102.600102.6000.61763.3063.30
Total steel=125.78+63.30\text{Total steel}=125.78+63.30 189.08 kg (say 189 kg)\boxed{\approx 189.08\ \text{kg}\ (\text{say }189\ \text{kg})}

Adding a customary 3% wastage/laps allowance gives 194.8 kg\approx194.8\text{ kg} for ordering.

bar-bending-schedulercc-slabsteel-measurement
5long10 marks

A reinforced-concrete office building was constructed at a total cost of NRs. 25,00,000 (excluding land). Its estimated useful life is 50 years and the salvage (scrap) value at the end of life is NRs. 2,50,000.

(a) Find the annual depreciation and the book value after 20 years by the straight-line method.

(b) Determine the annual sinking-fund instalment that must be set aside, at 8% compound interest, to recover the depreciable amount over the building's life.

(c) Briefly distinguish between scrap value and salvage value. (4+4+2)(4+4+2)

(a) Straight-line depreciation

Depreciable amount =CS=25,00,0002,50,000=22,50,000=C-S=25{,}00{,}000-2{,}50{,}000=22{,}50{,}000.

Annual depreciation:

D=CSn=22,50,00050=NRs. 45,000 per yearD=\frac{C-S}{n}=\frac{22{,}50{,}000}{50}=\mathbf{NRs.\ 45{,}000\ per\ year}

Book value after 20 years:

BV20=C20D=25,00,00020×45,000=25,00,0009,00,000BV_{20}=C-20D=25{,}00{,}000-20\times45{,}000=25{,}00{,}000-9{,}00{,}000 BV20=NRs. 16,00,000\boxed{BV_{20}=\text{NRs. }16{,}00{,}000}

(b) Sinking-fund instalment (annual deposit to accumulate CSC-S in nn years at rate ii):

I=(CS)i(1+i)n1I=(C-S)\cdot\frac{i}{(1+i)^n-1}

With i=0.08, n=50i=0.08,\ n=50:

(1.08)50=46.9016  (1.08)501=45.9016(1.08)^{50}=46.9016\ \Rightarrow\ (1.08)^{50}-1=45.9016 0.0845.9016=0.0017429\frac{0.08}{45.9016}=0.0017429 I=22,50,000×0.0017429I=22{,}50{,}000\times0.0017429 INRs. 3,921 per year\boxed{I\approx \text{NRs. }3{,}921\ \text{per year}}

(c) Scrap vs. salvage value:

  • Scrap value is the value of the dismantled material (e.g. broken concrete, old steel) when the structure is demolished at the end of its life, after deducting the cost of demolition — usually about 10% of the cost.
  • Salvage value is the value a utility/asset still fetches when sold without being dismantled, i.e. it can still be useful to another buyer at the end of its useful life to the present owner.
valuationdepreciationsinking-fund
B

Section B: Short Answer Questions

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6 questions
6short5 marks

Define estimate and explain, with one practical use of each, any four types of estimates prepared during the life of a civil engineering project. (1+4)(1+4)

Estimate: An estimate is the anticipated/approximate computation of the quantities of various items of work and the probable cost of a project, prepared before execution from drawings, specifications and prevailing rates. It also gives a forecast of materials, labour, plant and time required.

Four types of estimates:

  1. Preliminary (approximate / rough cost) estimate — made from the plinth-area or per-unit (per-bed, per-km) rate to decide the financial feasibility of a scheme and to obtain administrative approval.
  2. Detailed estimate — the complete item-wise estimate prepared from finalized drawings and detailed measurements; it forms the basis of tendering and execution.
  3. Revised estimate — a fresh detailed estimate prepared when the original estimate is exceeded by more than 5% (or when the scope changes), giving reasons for the increase.
  4. Supplementary estimate — prepared for additional works that become necessary while the original work is in progress; submitted along with the original for sanction.

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

types-of-estimatespreliminary-detailed
7short6 marks

Differentiate between general specifications and detailed specifications. Write a brief detailed specification for first-class brickwork in cement mortar (1:6) in superstructure, covering materials, proportion, soaking, laying and curing. (2+4)(2+4)

General vs. Detailed specification

General specificationDetailed specification
Gives a general/brief idea of the class and nature of work for the whole projectDescribes each item of work fully and exactly
Used for preliminary estimating and to convey overall qualityUsed for execution and as a basis of payment
No item-wise quantities or workmanship detailStates materials, proportion, workmanship, method and tolerances item by item

Detailed specification — First-class brickwork in cement mortar (1:6), superstructure:

  • Bricks: First-class, well-burnt, of uniform deep-red colour, regular size, sharp edges, free from cracks and flaws; minimum crushing strength as specified; water absorption not more than 20% of dry weight.
  • Mortar: Cement and sand in the ratio 1:6 by volume; sand to be clean, sharp and well graded; cement of standard make; mortar to be mixed in a mechanical mixer (or on a clean platform) and used within 30 minutes of adding water.
  • Soaking: Bricks shall be fully soaked in clean water for at least 6–12 hours (until air bubbling stops) before use.
  • Laying: Bricks laid in English bond in horizontal courses with frogs upward; every course truly level and plumb; vertical joints broken (staggered); joint thickness not exceeding 10 mm and completely filled with mortar; courses checked with plumb-bob and spirit level.
  • Curing: The completed brickwork shall be kept wet by curing for at least 7 days. Height of brickwork raised in a day shall not exceed 1.0–1.5 m. Joints raked to 10–12 mm depth where plastering/pointing is to follow.
  • Measurement: Measured in m³ of finished work, deducting openings as per the code of measurement.
specificationsgeneral-detailed-specification
8short6 marks

List the essential documents that make up a tender/contract for a building work. Distinguish between an item-rate (unit-price) contract and a lump-sum contract, giving one situation where each is preferred. (3+3)(3+3)

Essential tender / contract documents:

  1. Title page and index
  2. Notice inviting tender (NIT) / invitation for bids
  3. Instructions to bidders and general & special conditions of contract
  4. Drawings (architectural, structural, services)
  5. General and detailed technical specifications
  6. Bill of Quantities (BoQ) / schedule of rates
  7. Form of agreement, form of bid, and bid security/EMD details
  8. Schedule of issue of materials and tools, time of completion and penalty (liquidated damages) clauses

Item-rate vs. Lump-sum contract

Item-rate (unit-price)Lump-sum
Contractor quotes a rate for each item; payment = actual measured quantity × quoted rateContractor quotes one total price for the whole work as per drawings/specifications
Final cost varies with measured quantities; detailed measurement neededFinal cost is fixed; little site measurement of quantities
Risk of quantity variation borne by the ownerRisk of quantity variation borne by the contractor
Preferred when quantities cannot be assessed precisely in advance (e.g. earthwork, foundations in uncertain strata)Preferred when drawings and quantities are complete and unlikely to change (e.g. a standard, well-defined building)
tender-contract-documentstypes-of-contract
9short5 marks

A room has internal dimensions 4.0 m×3.0 m4.0\text{ m}\times3.0\text{ m} and clear height 3.0 m3.0\text{ m}. The four internal wall faces and the ceiling are to be plastered with 12 mm cement plaster. One door of 1.0 m×2.1 m1.0\text{ m}\times2.1\text{ m} and two windows each 1.2 m×1.5 m1.2\text{ m}\times1.5\text{ m} are provided in the walls. Compute the net area of plastering. (5)(5)

Wall (vertical) plaster area — internal faces

Inner perimeter =2(4.0+3.0)=14.0 m=2(4.0+3.0)=14.0\text{ m}.

Gross wall area == perimeter ×\times height =14.0×3.0=42.0 m2=14.0\times3.0=42.0\text{ m}^2.

Deductions for openings (for 12 mm plaster, full opening area is deducted):

  • Door =1.0×2.1=2.10 m2=1.0\times2.1=2.10\text{ m}^2
  • Windows =2×(1.2×1.5)=2×1.80=3.60 m2=2\times(1.2\times1.5)=2\times1.80=3.60\text{ m}^2
  • Total deduction =2.10+3.60=5.70 m2=2.10+3.60=5.70\text{ m}^2

Net wall plaster =42.05.70=36.30 m2=42.0-5.70=36.30\text{ m}^2.

Ceiling plaster =4.0×3.0=12.0 m2=4.0\times3.0=12.0\text{ m}^2.

Total net plaster area

=36.30+12.0=36.30+12.0 =48.30 m2\boxed{=48.30\ \text{m}^2}
plasteringmeasurement-of-civil-works
10short5 marks

Explain the long-wall and short-wall method and the centre-line method of taking out quantities. State the relationship between the long-wall/short-wall lengths and the centre-line length for a single rectangular building, and mention one advantage of each method. (3+2)(3+2)

Long-wall and short-wall (separate / individual wall) method

  • Walls running in one direction are taken as long walls and those in the perpendicular direction as short walls.
  • Long-wall length is measured out-to-out (centre-line length ++ one wall thickness, i.e. +t2+\tfrac{t}{2} at each end).
  • Short-wall length is measured in-to-in (centre-line length - one wall thickness, i.e. t2-\tfrac{t}{2} at each end).
  • Quantity =(length×breadth×height)=\sum(\text{length}\times \text{breadth}\times \text{height}) for long and short walls separately.

Centre-line method

  • A single centre-line length is taken for the whole building: Lcl=2(Li+Bi+2t)L_{cl}=2(L_i+B_i+2t) type expression, and quantity =Lcl×t×h=L_{cl}\times t\times h.
  • Junction corrections (t2-\tfrac{t}{2} per junction side) are applied where same-thickness walls meet.

Relationship (single rectangular building):

Total long-wall length+Total short-wall length=2Lcl\text{Total long-wall length} + \text{Total short-wall length} = 2\,L_{cl}

because each long wall is +t2+\tfrac{t}{2} longer and each short wall is t2-\tfrac{t}{2} shorter than the centre line, so the additions and subtractions cancel and the combined length equals twice the centre-line length.

Advantages:

  • Long-wall/short-wall: gives wall-wise quantities directly, convenient when walls differ in thickness or height.
  • Centre-line: much quicker for buildings with walls of uniform thickness, as one length serves all items (excavation, footing, masonry).
methods-of-measurementlong-wall-short-wallcentre-line
11short3 marks

Define valuation and briefly state any three methods of valuation of a property. (1+2)(1+2)

Valuation: Valuation is the technique of estimating the fair, present worth (market value) of a property — land, building or other engineering asset — for purposes such as buying/selling, taxation, rent fixation, mortgage, insurance or acquisition.

Three methods of valuation:

  1. Rental (capitalized-value) method — the net annual rental income is capitalized using a suitable year's purchase to give the capital value; suited to rented/income-yielding property.
  2. Cost (land-and-building / replacement) method — value == cost of land ++ present cost of constructing the building - depreciation; used when reliable rental data are unavailable.
  3. Comparative (market / sales-comparison) method — value is derived by comparing recent sale prices of similar properties in the same locality, with adjustments for differences.

(Other valid methods: development method, profit-based method, depreciation method.)

valuationmethods-of-valuation

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