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A

Section A: Long Answer Questions

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5 questions
1long10 marks

Describe the hydration of Portland cement. Name the four major Bogue compounds, write their abbreviations, and explain the role of each in the development of strength and heat of hydration. A sample of Ordinary Portland Cement gave the following oxide composition by mass: CaO=64.0%CaO = 64.0\%, SiO2=21.0%SiO_2 = 21.0\%, Al2O3=5.5%Al_2O_3 = 5.5\%, Fe2O3=3.5%Fe_2O_3 = 3.5\%. Using Bogue's equations, compute the percentages of the four compounds.

Hydration of Portland cement

When water is added to cement, the anhydrous compounds react chemically (an exothermic reaction) to form hydrated products that bind the aggregate together. The principal reactions are:

  • Silicates (C3SC_3S, C2SC_2S) react with water to give C–S–H gel (calcium silicate hydrate, the main strength-giving phase) plus calcium hydroxide Ca(OH)2Ca(OH)_2 (portlandite).
2C3S+6HC3S2H3+3CH2C_3S + 6H \rightarrow C_3S_2H_3 + 3CH 2C2S+4HC3S2H3+CH2C_2S + 4H \rightarrow C_3S_2H_3 + CH
  • C3AC_3A reacts very rapidly; gypsum is added to retard the flash set, forming ettringite.
  • C4AFC_4AF hydrates to give calcium aluminoferrite hydrates; contributes little to strength.

The four Bogue compounds

CompoundFormulaAbbrev.Role
Tricalcium silicate3CaOSiO23CaO\cdot SiO_2C3SC_3S (alite)Early strength (up to 28 d); high heat
Dicalcium silicate2CaOSiO22CaO\cdot SiO_2C2SC_2S (belite)Later/ultimate strength; low heat
Tricalcium aluminate3CaOAl2O33CaO\cdot Al_2O_3C3AC_3AFlash set, very high heat, low durability (sulphate attack)
Tetracalcium aluminoferrite4CaOAl2O3Fe2O34CaO\cdot Al_2O_3\cdot Fe_2O_3C4AFC_4AFLittle strength; gives grey colour, acts as flux

Bogue's equations

C3S=4.071CaO7.600SiO26.718Al2O31.430Fe2O3C_3S = 4.071\,CaO - 7.600\,SiO_2 - 6.718\,Al_2O_3 - 1.430\,Fe_2O_3 C2S=2.867SiO20.7544C3SC_2S = 2.867\,SiO_2 - 0.7544\,C_3S C3A=2.650Al2O31.692Fe2O3C_3A = 2.650\,Al_2O_3 - 1.692\,Fe_2O_3 C4AF=3.043Fe2O3C_4AF = 3.043\,Fe_2O_3

Step 1 — C3SC_3S

C3S=4.071(64.0)7.600(21.0)6.718(5.5)1.430(3.5)C_3S = 4.071(64.0) - 7.600(21.0) - 6.718(5.5) - 1.430(3.5) =260.544159.60036.9495.005=58.99%= 260.544 - 159.600 - 36.949 - 5.005 = 58.99\%

Step 2 — C2SC_2S

C2S=2.867(21.0)0.7544(58.99)=60.20744.502=15.71%C_2S = 2.867(21.0) - 0.7544(58.99) = 60.207 - 44.502 = 15.71\%

Step 3 — C3AC_3A

C3A=2.650(5.5)1.692(3.5)=14.5755.922=8.65%C_3A = 2.650(5.5) - 1.692(3.5) = 14.575 - 5.922 = 8.65\%

Step 4 — C4AFC_4AF

C4AF=3.043(3.5)=10.65%C_4AF = 3.043(3.5) = 10.65\%

Results:

CompoundPercentage
C3SC_3S58.99 %
C2SC_2S15.71 %
C3AC_3A8.65 %
C4AFC_4AF10.65 %
cementhydrationcement-tests
2long10 marks

Explain the factors affecting the workability of fresh concrete and Abrams' water-cement ratio law. A concrete mix is designed with the proportions 1:1.8:3.21 : 1.8 : 3.2 (cement : sand : coarse aggregate) by mass and a water-cement ratio of 0.480.48. For 1m31\,m^3 of compacted concrete with total absolute volume of solids and water filling the volume, take specific gravities: cement 3.153.15, sand 2.652.65, coarse aggregate 2.702.70. Estimate the mass of cement required per m3m^3 (assume no entrained air).

Workability

Workability is the ease with which concrete can be mixed, transported, placed, compacted and finished without segregation or bleeding.

Factors affecting workability:

  1. Water content — most significant; more water increases workability but lowers strength.
  2. Water-cement ratio — higher w/c gives more fluidity.
  3. Aggregate properties — size, shape (rounded > angular), texture, grading.
  4. Aggregate-cement ratio — leaner mixes are less workable.
  5. Admixtures — plasticizers/superplasticizers increase workability.
  6. Fineness of cement — finer cement needs more water.
  7. Time and temperature — higher temperature reduces workability (faster setting).

Abrams' water-cement ratio law

For a fully compacted concrete, the strength is inversely proportional to the water-cement ratio:

fc=AB(w/c)f_c = \frac{A}{B^{(w/c)}}

where AA and BB are empirical constants. Strength decreases as w/c increases (for workable, fully compacted mixes).

Cement content calculation

Let mass of cement =C= C (kg). With ratio 1:1.8:3.21:1.8:3.2 by mass:

  • Sand =1.8C= 1.8C, Coarse aggregate =3.2C= 3.2C, Water =0.48C= 0.48C.

Absolute volumes (volume = mass / (S.G. × 1000) in m3m^3):

Vcement=C3.15×1000V_{cement} = \frac{C}{3.15 \times 1000} Vsand=1.8C2.65×1000V_{sand} = \frac{1.8C}{2.65 \times 1000} VCA=3.2C2.70×1000V_{CA} = \frac{3.2C}{2.70 \times 1000} Vwater=0.48C1.00×1000V_{water} = \frac{0.48C}{1.00 \times 1000}

Sum of absolute volumes =1m3= 1\,m^3 (no air):

C1000(13.15+1.82.65+3.22.70+0.48)=1\frac{C}{1000}\left(\frac{1}{3.15} + \frac{1.8}{2.65} + \frac{3.2}{2.70} + 0.48\right) = 1

Compute each term:

  • 1/3.15=0.317461/3.15 = 0.31746
  • 1.8/2.65=0.679251.8/2.65 = 0.67925
  • 3.2/2.70=1.185193.2/2.70 = 1.18519
  • water =0.48000= 0.48000

Sum =0.31746+0.67925+1.18519+0.48000=2.66190= 0.31746 + 0.67925 + 1.18519 + 0.48000 = 2.66190

Therefore:

C1000(2.66190)=1    C=10002.66190=375.67kg\frac{C}{1000}(2.66190) = 1 \implies C = \frac{1000}{2.66190} = 375.67\,kg

Mass of cement required = ≈ 376 kg per m3m^3.

(For reference: sand =676.2kg= 676.2\,kg, coarse aggregate =1202.1kg= 1202.1\,kg, water =180.3kg= 180.3\,kg per m3m^3.)

concretemix-designwater-cement-ratio
3long8 marks

Describe the structure of timber (cross-section of an exogenous tree) and the defects in timber. Explain why seasoning is necessary and distinguish between natural and kiln seasoning.

Structure of timber (exogenous tree cross-section)

From outside to centre:

  Bark (outer + inner / bast)
  Cambium layer  -> living cells, growth
  Sapwood (alburnum) -> light, carries sap, contains moisture
  Heartwood (duramen) -> dark, dead cells, strongest, durable
  Pith (medulla) -> innermost soft core
  Medullary rays -> radial lines transferring sap
  Annual rings -> one ring per year (age & growth rate)
  • Heartwood is preferred for construction (strong, durable, decay-resistant).
  • Annual rings indicate age; closely spaced rings = strong, slow-grown timber.

Defects in timber

  1. Due to natural forces / growth: knots, shakes (heart, cup, ring, star shakes), twisted fibres, rind galls, burr.
  2. Due to fungi: dry rot, wet rot, blue stain, heart rot.
  3. Due to insects: beetles, marine borers, termites (white ants).
  4. Due to seasoning: bow, cup, twist, warp, split, check, honeycombing.
  5. Due to conversion: chip marks, wane, diagonal grain.

Seasoning of timber

Seasoning = controlled reduction of moisture content of green timber to a level in equilibrium with the surrounding atmosphere (≈ 10–15 %).

Why necessary:

  • Reduces weight; increases strength, stiffness and durability.
  • Reduces shrinkage, warping and splitting after use.
  • Makes timber resistant to fungal/insect attack.
  • Allows it to take paint, polish and preservatives well.

Natural (air) seasoning vs Kiln seasoning

AspectNatural (air) seasoningKiln seasoning
MethodStacking timber in open shed; air circulationDrying in closed chamber with controlled heat & humidity
TimeSlow (months to years)Fast (days)
ControlNo control on temp/humidityFull control
CostCheapCostly (equipment + fuel)
QualityLess uniform; may develop defectsUniform, low moisture, defect-free if done well
Final MCLimited to ~15–18%Can reach very low MC (8–10%)
timberwood-productsseasoning
4long8 marks

Explain the stress-strain behaviour of mild steel under tension with a labelled description of the curve, defining limit of proportionality, yield point, ultimate strength and breaking point. A mild steel bar of diameter 16mm16\,mm and gauge length 80mm80\,mm is tested. The yield load is 52.3kN52.3\,kN, the ultimate load is 78.0kN78.0\,kN, and the final gauge length after fracture is 98.4mm98.4\,mm. Compute the yield stress, ultimate tensile strength and percentage elongation.

Stress-strain behaviour of mild steel (tension)

Stress
  ^                         U (ultimate)
  |                 ____----"""""----___
  |          B,C _/                   \ F (fracture)
  |          __/  (yield plateau)      *
  |       A /
  |      / (P limit of proportionality)
  |    /
  |  /  (linear elastic, Hooke's law)
  | /
  +--------------------------------------> Strain
  • O–P (Limit of proportionality): stress ∝ strain (Hooke's law); slope = Young's modulus EE.
  • P–A (Elastic limit): still elastic but no longer linear.
  • A → upper/lower yield (B, C): material yields; sudden increase in strain at nearly constant stress — yield point.
  • C–U (strain hardening): stress rises again to the ultimate tensile strength at U.
  • U–F: necking; cross-section reduces, load falls, then fracture at F (breaking/cup-and-cone).

Calculations

Cross-sectional area:

A=π4d2=π4(16)2=π4(256)=201.06mm2A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(16)^2 = \frac{\pi}{4}(256) = 201.06\,mm^2

Yield stress:

σy=PyA=52.3×103N201.06mm2=260.1N/mm2\sigma_y = \frac{P_y}{A} = \frac{52.3 \times 10^3\,N}{201.06\,mm^2} = 260.1\,N/mm^2

Ultimate tensile strength:

σu=PuA=78.0×103201.06=387.9N/mm2\sigma_u = \frac{P_u}{A} = \frac{78.0 \times 10^3}{201.06} = 387.9\,N/mm^2

Percentage elongation:

%elong=LfL0L0×100=98.48080×100=18.480×100=23.0%\%\,elong = \frac{L_f - L_0}{L_0}\times 100 = \frac{98.4 - 80}{80}\times 100 = \frac{18.4}{80}\times 100 = 23.0\%

Results:

  • Yield stress σy=260.1N/mm2  (MPa)\sigma_y = \mathbf{260.1\,N/mm^2 \;(MPa)}
  • Ultimate tensile strength σu=387.9N/mm2  (MPa)\sigma_u = \mathbf{387.9\,N/mm^2 \;(MPa)}
  • Percentage elongation =23.0%= \mathbf{23.0\%}
steelmetalsstress-strain
5long8 marks

Compare bricks and building stones as walling materials with respect to strength, durability, cost, workability and availability. Describe the field and laboratory tests for bricks. A first-class brick of size 240×115×70mm240 \times 115 \times 70\,mm fails under a compressive load of 310kN310\,kN in a compression test. Determine its compressive strength and state whether it meets the minimum for a first-class brick (10.5N/mm2\geq 10.5\,N/mm^2 as per common practice / IS).

Bricks vs Building stones

PropertyBricksBuilding stones
StrengthModerate (3.5–35 MPa)High (often 60–200 MPa)
DurabilityGood if well burntExcellent (granite, basalt)
CostCheaper, uniformCostlier (quarrying, dressing, transport)
WorkabilityEasy (uniform size, light)Difficult (heavy, needs dressing)
Weight/handlingLight, easy to layHeavy
AvailabilityMade anywhere clay availableSite/quarry dependent
ArchitectureGood for arches, thin wallsMassive, monumental work

Tests for bricks

Field tests:

  1. Shape & size — uniform, sharp edges, rectangular.
  2. Colour — uniform deep red/copper colour.
  3. Hardness — no impression by finger-nail scratch.
  4. Soundness — two bricks struck together give clear ringing sound, no breakage.
  5. Structure — broken brick shows compact, homogeneous, no flaws/lumps.

Laboratory tests:

  1. Compressive strength test — crushing in compression testing machine.
  2. Water absorption test — 24 h immersion; first-class ≤ 20% by weight.
  3. Efflorescence test — checks soluble salts (rated nil/slight/moderate/heavy).
  4. Dimension tolerance test — measure stack of 20 bricks.
  5. Warpage test — flatness using steel rule and wedge.

Compressive strength calculation

Loaded (bed) face area = length × width:

A=240×115=27600mm2A = 240 \times 115 = 27600\,mm^2 σc=PA=310×103N27600mm2=11.23N/mm2\sigma_c = \frac{P}{A} = \frac{310 \times 10^3\,N}{27600\,mm^2} = 11.23\,N/mm^2

Compressive strength = ≈ 11.23 N/mm² (MPa).

Since 11.23>10.5N/mm211.23 > 10.5\,N/mm^2, the brick satisfies the minimum strength requirement for a first-class brick.

bricksbuilding-stonesmaterial-properties
B

Section B: Short Answer Questions

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6 questions
6short6 marks

Define fineness modulus (FM) of aggregate and state its significance. A sieve analysis of 1000g1000\,g of fine aggregate gives the following retained masses. Compute the fineness modulus.

IS SieveMass retained (g)
4.75 mm20
2.36 mm80
1.18 mm150
600 µm280
300 µm290
150 µm150
Pan30

Fineness modulus

Fineness modulus (FM) is an index of the fineness/coarseness of an aggregate, obtained by summing the cumulative percentages retained on the standard sieves (150 µm and coarser) and dividing by 100. A higher FM means coarser aggregate.

Significance: indicates particle-size distribution and helps in proportioning aggregates for concrete mixes; sand is normally classified by zones using FM (fine sand ~2.2–2.6, coarse sand ~2.9–3.2).

Computation

Total mass = 1000 g, so % retained = mass retained / 10.

SieveRetained (g)% RetainedCumulative % Retained
4.75 mm202.02.0
2.36 mm808.010.0
1.18 mm15015.025.0
600 µm28028.053.0
300 µm29029.082.0
150 µm15015.097.0
Pan303.0(not counted)

Sum of cumulative % retained (150 µm and above):

Σ=2.0+10.0+25.0+53.0+82.0+97.0=269.0\Sigma = 2.0 + 10.0 + 25.0 + 53.0 + 82.0 + 97.0 = 269.0 FM=Σ100=269.0100=2.69FM = \frac{\Sigma}{100} = \frac{269.0}{100} = \mathbf{2.69}

The sand has FM=2.69FM = 2.69, indicating a medium sand suitable for concrete.

aggregatesconcretegrading
7short6 marks

Classify lime used in construction and distinguish between fat lime and hydraulic lime. Explain the slaking of lime. What is meant by mortar and list the desirable properties of a good mortar.

Classification of lime

Based on impurities (mainly clay/silica + alumina) lime is classified as:

  1. Fat lime (pure / high-calcium / quick lime): ≥ 95% CaO; from pure limestone; slakes vigorously; sets only by carbonation in air (non-hydraulic).
  2. Hydraulic lime: contains clay (silica, alumina); sets under water. Sub-divided:
    • Feebly hydraulic (5–10% clay)
    • Moderately hydraulic (11–20% clay)
    • Eminently hydraulic (21–30% clay)
  3. Poor (lean) lime: > 30% impurities; weak, slow setting.

Fat lime vs Hydraulic lime

AspectFat limeHydraulic lime
SourcePure limestoneImpure (clayey) limestone
SettingIn air (carbonation) onlySets under water
SlakingVigorous, large volume increaseSlow, slight expansion
StrengthLowerHigher
UsePlastering, whitewashingMasonry, foundations, wet works

Slaking of lime

Slaking is the chemical reaction of quicklime (CaO) with water to form slaked lime (calcium hydroxide), an exothermic reaction:

CaO+H2OCa(OH)2+heatCaO + H_2O \rightarrow Ca(OH)_2 + heat

The lump quicklime swells, crumbles into a fine powder/putty, and a large amount of heat is liberated.

Mortar

Mortar is a workable paste of a binder (cement/lime), fine aggregate (sand) and water, used to bind masonry units, fill joints, and plaster surfaces.

Desirable properties of good mortar:

  • Good adhesion/bond to masonry units.
  • Adequate strength when set.
  • Workability and ease of placing.
  • Water retentivity (does not dry too fast).
  • Durability and resistance to weathering.
  • Low shrinkage and good water-tightness.
  • Economical and quick setting (as required).
limemortarbinders
8short6 marks

Differentiate between bitumen, tar and asphalt. Name and briefly describe any four laboratory tests performed on bitumen and state what property each measures.

Bitumen vs Tar vs Asphalt

AspectBitumenTarAsphalt
OriginResidue from petroleum (crude oil) refiningDestructive distillation of coal/woodMixture of bitumen + inert mineral matter (natural or manufactured)
CompositionHydrocarbons (paraffins, naphthenes)Hydrocarbons rich in free carbonBitumen + aggregate/filler
ColourBlack, soluble in CS₂Black, more pungentBlack/brown
Temperature susceptibilityLess temperature-susceptibleMore temperature-susceptibleDepends on bitumen content
UseRoad binder, waterproofingLimited (carcinogenic), surface dressingRoad pavements, surfacing

Four laboratory tests on bitumen

  1. Penetration test: measures the depth (in tenths of mm) to which a standard needle penetrates the sample in 5 s at 25 °C under 100 g load. → Measures consistency/hardness (grading, e.g. 60/70).
  2. Softening point test (Ring & Ball): temperature at which bitumen softens enough for a steel ball to fall through a ring. → Measures temperature susceptibility / resistance to flow at high temperature.
  3. Ductility test: the distance (cm) a standard briquette stretches before breaking at 27 °C, 5 cm/min. → Measures ductility / ability to deform without cracking.
  4. Flash and fire point test: lowest temperatures at which vapours flash momentarily / sustain burning. → Measures safe heating temperature / volatility.

(Other valid tests: viscosity test, specific gravity test, loss on heating, solubility test.)

bitumenasphaltroad-materials
9short5 marks

What are the constituents of paint and the functions of each? Distinguish between paint and varnish, and list the characteristics of a good paint.

Constituents of paint and their functions

  1. Base (white lead, zinc oxide, titanium dioxide, etc.): the principal solid pigment; gives body, opacity (hiding power) and durability.
  2. Vehicle / binder (drying oils such as linseed oil, resins): holds pigment particles together, forms the film, gives adhesion and durability.
  3. Pigment / colouring matter: provides desired colour and opacity.
  4. Solvent / thinner (turpentine, mineral spirit): reduces viscosity for easy application; evaporates after application.
  5. Drier: accelerates drying/hardening of the film by oxidation.
  6. Extender / filler (chalk, barytes): cheapens paint and modifies properties (reduces cracking, adds bulk).

Paint vs Varnish

AspectPaintVarnish
AppearanceOpaque, colouredTransparent / translucent
PigmentContains pigmentNo pigment (or little)
PurposeProtect + decorate (hide surface)Protect + give glossy finish, show grain
CompositionBase + vehicle + pigment + solvent + drierResin + solvent (+ drying oil)
UseWalls, metals, woodWood (to display natural grain), over paint

Characteristics of a good paint

  • Spreads easily and freely; good workability.
  • Forms a thin, uniform, durable film with good adhesion.
  • Dries in reasonable time to a hard, glossy surface.
  • Good covering/hiding power (economical coverage).
  • Resistant to weather, moisture and not easily affected by atmosphere.
  • Does not crack, peel, fade or show brush marks.
paintsvarnishesprotective-coatings
10short4 marks

Write short notes on: (a) types of glass used in buildings, and (b) thermoplastics vs thermosetting plastics with one example each.

(a) Types of glass used in buildings

  • Soda-lime (sheet/float) glass: common window/glazing glass; cheap, transparent.
  • Toughened (tempered) glass: heat-treated, 4–5× stronger; shatters into small blunt pieces; doors, façades.
  • Laminated glass: two glass sheets bonded with PVB interlayer; safety/security glazing, holds together when broken.
  • Wired glass: mesh embedded; fire-resistant glazing.
  • Insulating (double-glazed) glass: two panes with air/gas gap; thermal & sound insulation.
  • Float glass: flat, distortion-free; base for most processed glass.
  • Tinted / reflective glass: controls solar heat and glare.

(b) Thermoplastics vs Thermosetting plastics

AspectThermoplasticThermosetting plastic
Effect of heatSoftens on heating, hardens on cooling (reversible)Sets permanently on first heating; cannot be re-softened (irreversible)
StructureLinear/branched chains, weak secondary bondsCross-linked 3-D network
ReusabilityCan be remoulded/recycledCannot be remoulded
Strength/rigiditySofter, flexibleHarder, rigid, heat-resistant
ExamplePVC (also polythene, acrylic)Bakelite (phenol-formaldehyde; also melamine, epoxy)
glassplasticsmaterial-properties
11short9 marks

(a) What are smart materials? Briefly describe any three smart materials used in civil engineering. (b) Explain the bonding types in engineering materials (ionic, covalent, metallic) and how they relate to material properties. (c) A concrete cube of side 150mm150\,mm is tested at 28 days and fails at a load of 787.5kN787.5\,kN. Determine the characteristic compressive strength and identify the likely concrete grade.

(a) Smart materials

Smart (intelligent) materials sense changes in their environment (stress, temperature, moisture, electric/magnetic field) and respond in a controlled, useful and often reversible way.

Three examples in civil engineering:

  1. Piezoelectric materials: generate electric charge under mechanical stress (and deform under voltage); used in structural health monitoring sensors and energy harvesting.
  2. Shape memory alloys (SMA, e.g. Nitinol): recover their original shape on heating; used in seismic dampers, self-centring braces.
  3. Self-healing concrete: contains bacteria/encapsulated agents that precipitate calcite to seal cracks, improving durability. (Other valid: electrochromic glass, thermochromic materials.)

(b) Bonding types and properties

BondMechanismTypical properties
IonicTransfer of electrons between metal & non-metal (e.g. NaCl); electrostatic attractionHard, brittle, high melting point, poor electrical conductivity (good when molten/dissolved)
CovalentSharing of electron pairs (e.g. diamond, silica)Very hard, high strength, high melting point, usually poor conductors
MetallicPositive ions in a 'sea' of delocalized electrons (e.g. steel, aluminium)Ductile, malleable, good electrical & thermal conductivity, opaque/lustrous

Thus bonding directly controls hardness, ductility, conductivity and melting point — e.g. the metallic bond makes steel ductile and weldable, while covalent silica makes glass hard but brittle.

(c) Characteristic compressive strength

Cube face area:

A=150×150=22500mm2A = 150 \times 150 = 22500\,mm^2

Compressive strength:

fc=PA=787.5×103N22500mm2=35.0N/mm2f_c = \frac{P}{A} = \frac{787.5 \times 10^3\,N}{22500\,mm^2} = 35.0\,N/mm^2

Compressive strength = 35.0 N/mm² (MPa).

Since the 28-day characteristic cube strength is 35N/mm235\,N/mm^2, the concrete corresponds to grade M35.

smart-materialsconcrete-propertiesmaterial-structure

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