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Section A: Long Answer Questions

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5 questions
1long10 marks

Describe the chemical composition of ordinary Portland cement and the role of the four major Bogue compounds in hydration and strength development. Using the Bogue equations, compute the percentages of C3SC_3S, C2SC_2S, C3AC_3A and C4AFC_4AF for a cement having 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=3.0%Fe_2O_3 = 3.0\%.

Chemical composition of OPC

OPC is produced by burning a calcareous material (limestone) and an argillaceous material (clay) at about 1450C1450^{\circ}C. The principal oxides are lime (CaOCaO), silica (SiO2SiO_2), alumina (Al2O3Al_2O_3) and iron oxide (Fe2O3Fe_2O_3), with minor amounts of MgOMgO, SO3SO_3 and alkalis. During clinkering these oxides combine into four major compounds (Bogue compounds).

Role of the four Bogue compounds

CompoundSymbolRole
Tricalcium silicateC3SC_3SHydrates fast; main contributor to early strength (up to 28 days) and considerable heat of hydration.
Dicalcium silicateC2SC_2SHydrates slowly; gives later-age strength (beyond 28 days), low heat.
Tricalcium aluminateC3AC_3AVery rapid reaction (flash set); high heat; vulnerable to sulphate attack. Gypsum is added to control its set.
Tetracalcium alumino-ferriteC4AFC_4AFHydrates moderately; contributes little to strength; gives grey colour.

Bogue 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(3.0)C_3S = 4.071(64.0) - 7.600(21.0) - 6.718(5.5) - 1.430(3.0) =260.544159.60036.9494.290=59.71%= 260.544 - 159.600 - 36.949 - 4.290 = \mathbf{59.71\%}

Step 2 — C2SC_2S

C2S=2.867(21.0)0.7544(59.705)=60.20745.041=15.17%C_2S = 2.867(21.0) - 0.7544(59.705) = 60.207 - 45.041 = \mathbf{15.17\%}

Step 3 — C3AC_3A

C3A=2.650(5.5)1.692(3.0)=14.5755.076=9.50%C_3A = 2.650(5.5) - 1.692(3.0) = 14.575 - 5.076 = \mathbf{9.50\%}

Step 4 — C4AFC_4AF

C4AF=3.043(3.0)=9.13%C_4AF = 3.043(3.0) = \mathbf{9.13\%}

Result

CompoundPercentage
C3SC_3S59.71 %
C2SC_2S15.17 %
C3AC_3A9.50 %
C4AFC_4AF9.13 %

The high C3SC_3S content indicates good early-strength characteristics typical of OPC.

cementbogue-compoundshydration
2long12 marks

Design a concrete mix by the absolute-volume method for the following data and determine the mass of each ingredient per cubic metre of concrete.

  • Target mean strength requires water-cement ratio w/c=0.50w/c = 0.50
  • Water content = 186 kg/m3186\ \text{kg/m}^3
  • Specific gravities: cement =3.15= 3.15, fine aggregate =2.65= 2.65, coarse aggregate =2.70= 2.70
  • Entrapped air =2%= 2\%
  • Ratio of fine aggregate to total aggregate (by absolute volume) =0.38= 0.38

Show the absolute-volume balance and state the final mix proportions by mass.

Method

In the absolute-volume method the volumes of all solids, water and air sum to 1 m31\ \text{m}^3.

Step 1 — Cement content

C=waterw/c=1860.50=372 kg/m3C = \frac{\text{water}}{w/c} = \frac{186}{0.50} = 372\ \text{kg/m}^3

Step 2 — Absolute volumes of water, cement, air

  • Water volume =1861000=0.1860 m3= \dfrac{186}{1000} = 0.1860\ \text{m}^3
  • Cement volume =3723.15×1000=0.11810 m3= \dfrac{372}{3.15\times 1000} = 0.11810\ \text{m}^3
  • Air volume =0.02×1=0.0200 m3= 0.02\times 1 = 0.0200\ \text{m}^3

Step 3 — Volume available for total aggregate

Vagg=1(0.1860+0.11810+0.0200)=10.32410=0.67590 m3V_{agg} = 1 - (0.1860 + 0.11810 + 0.0200) = 1 - 0.32410 = 0.67590\ \text{m}^3

Step 4 — Split into fine and coarse (by absolute volume)

  • Fine aggregate volume =0.38×0.67590=0.25684 m3= 0.38\times 0.67590 = 0.25684\ \text{m}^3
  • Coarse aggregate volume =0.62×0.67590=0.41906 m3= 0.62\times 0.67590 = 0.41906\ \text{m}^3

Step 5 — Convert aggregate volumes to mass

  • Fine aggregate =0.25684×2.65×1000=680.6 kg= 0.25684\times 2.65\times 1000 = 680.6\ \text{kg}
  • Coarse aggregate =0.41906×2.70×1000=1131.5 kg= 0.41906\times 2.70\times 1000 = 1131.5\ \text{kg}

Mix per m3\text{m}^3

IngredientMass (kg)
Cement372
Water186
Fine aggregate680.6
Coarse aggregate1131.5

Proportions by mass (cement : FA : CA)

372:680.6:1131.5=1:1.83:3.04(w/c=0.50)372 : 680.6 : 1131.5 = 1 : 1.83 : 3.04 \quad (w/c = 0.50)

Check: total absolute solid + water + air volume =0.11810+0.25684+0.41906+0.1860+0.0200=1.000 m3= 0.11810+0.25684+0.41906+0.1860+0.0200 = 1.000\ \text{m}^3. OK.

mix-designabsolute-volumewater-cement-ratio
3long10 marks

Define workability of fresh concrete and list the factors affecting it. Describe the slump test and the compaction-factor test, stating the workability range each is suited to. Explain how segregation and bleeding affect durability of hardened concrete.

Workability

Workability is the ease with which freshly mixed concrete can be mixed, transported, placed, compacted and finished without segregation. It is governed mainly by the water content and the internal friction of the mix.

Factors affecting workability

  1. Water content (most influential).
  2. Water-cement ratio and aggregate-cement ratio.
  3. Size, shape, texture and grading of aggregate (rounded, well-graded aggregate improves workability).
  4. Use of admixtures (plasticizers/superplasticizers increase it).
  5. Use of supplementary cementitious materials (fly ash improves it; silica fume reduces it).
  6. Ambient temperature and time after mixing.

Slump test

A frustum mould (top 100 mm100\ \text{mm}, bottom 200 mm200\ \text{mm}, height 300 mm300\ \text{mm}) is filled in 3 layers, each tamped 25 times. On lifting the cone, the slump (vertical subsidence in mm) is measured.

   |---100---|
   _________            true slump
  /         \           |
 /           \          v  measured drop
/_____________\  -----  ===
 |---200---|

Suitable for medium to high workability (slump 25-150 mm); poor for very dry or very wet mixes.

Compaction-factor test

Concrete is dropped through two hoppers into a cylinder; the compaction factor = (weight of partially compacted concrete) / (weight of fully compacted concrete). More sensitive at the low-workability (dry-mix) end, where the slump test is insensitive.

Effect of segregation and bleeding on durability

  • Segregation: separation of coarse aggregate from mortar gives a non-uniform, honeycombed structure with voids, lowering strength and increasing permeability, hence reduced durability.
  • Bleeding: upward migration of water creates a weak, porous, laitance-rich top surface and water channels under aggregates/reinforcement, increasing permeability, promoting carbonation, corrosion of steel and scaling. Both reduce the impermeability that protects concrete from aggressive agents.
fresh-concreteworkabilitydurability
4long10 marks

A solid brick masonry column of cross-section 400 mm×500 mm400\ \text{mm} \times 500\ \text{mm} and effective height 3.0 m3.0\ \text{m} carries an axial load. The basic compressive strength of masonry is fb=1.5 N/mm2f_b = 1.5\ \text{N/mm}^2. The load is applied with an eccentricity of 40 mm40\ \text{mm} along the 500 mm500\ \text{mm} direction. Using the reduction-factor approach, determine the safe axial load. Take the stress-reduction factor for combined slenderness and eccentricity from the table below.

SR ratioe/t=0e/t = 0e/t=0.08e/t = 0.08
61.000.90
80.950.84
100.890.78

Step 1 — Slenderness ratio (SR)

Least lateral dimension t=400 mmt = 400\ \text{mm}.

SR=effective heightleast lateral dimension=3000400=7.5SR = \frac{\text{effective height}}{\text{least lateral dimension}} = \frac{3000}{400} = 7.5

Step 2 — Eccentricity ratio

Eccentricity is along the 500 mm500\ \text{mm} direction, so the relevant thickness is t=500 mmt = 500\ \text{mm}.

et=40500=0.08\frac{e}{t} = \frac{40}{500} = 0.08

Step 3 — Stress-reduction factor ksk_s (interpolation)

For e/t=0.08e/t = 0.08, interpolate between SR=6SR = 6 (ks=0.90k_s = 0.90) and SR=8SR = 8 (ks=0.84k_s = 0.84) at SR=7.5SR = 7.5:

ks=0.90(7.56)(86)(0.900.84)=0.900.75(0.06)=0.900.045=0.855k_s = 0.90 - \frac{(7.5-6)}{(8-6)}(0.90-0.84) = 0.90 - 0.75(0.06) = 0.90 - 0.045 = 0.855

Step 4 — Permissible compressive stress

fperm=ks×fb=0.855×1.5=1.2825 N/mm2f_{perm} = k_s \times f_b = 0.855 \times 1.5 = 1.2825\ \text{N/mm}^2

Step 5 — Cross-sectional area

A=400×500=200000 mm2A = 400 \times 500 = 200000\ \text{mm}^2

Step 6 — Safe axial load

P=fperm×A=1.2825×200000=256500 NP = f_{perm}\times A = 1.2825 \times 200000 = 256500\ \text{N} P=256.5 kN\boxed{P = 256.5\ \text{kN}}

The column can safely carry an axial load of about 256.5 kN.

masonry-columnaxial-loadslenderness
5long8 marks

Classify chemical admixtures used in concrete and explain the working mechanism of plasticizers and superplasticizers. Briefly describe TWO special concretes — self-compacting concrete (SCC) and fibre-reinforced concrete (FRC) — stating their key advantages.

Classification of chemical admixtures

  1. Water-reducing admixtures (plasticizers) — reduce water demand for a given workability.
  2. High-range water reducers (superplasticizers) — give very high water reduction / very high workability.
  3. Retarders — delay setting (useful in hot weather, long hauls).
  4. Accelerators — speed up setting/hardening (e.g. cold weather, calcium chloride for plain concrete).
  5. Air-entraining agents — introduce fine air bubbles for freeze-thaw resistance.
  6. Others — waterproofing, grouting, gas-forming, corrosion-inhibiting, colouring admixtures.

Mechanism of plasticizers / superplasticizers

Cement particles tend to flocculate because of surface charges, trapping mixing water.

  • Plasticizers (lignosulphonates) adsorb on cement grains and impart a negative charge, causing electrostatic repulsion that disperses the flocs and releases trapped water, increasing workability (water reduction 5-12%).
  • Superplasticizers (sulphonated melamine/naphthalene formaldehyde, or polycarboxylate ethers) act by stronger electrostatic and, for PCE, steric hindrance from side chains, giving water reduction of 20-40% and enabling high-strength / flowing concrete.

Special concretes

Self-Compacting Concrete (SCC)

Highly flowable concrete that spreads and fills formwork under its own weight without vibration, passing through congested reinforcement without segregation (achieved with superplasticizers + viscosity modifiers + high fines). Advantages: no compaction labour, excellent finish, works in congested sections, less noise.

Fibre-Reinforced Concrete (FRC)

Concrete with dispersed steel, glass, polypropylene or natural fibres. Fibres bridge cracks and resist crack propagation. Advantages: improved tensile/flexural strength, ductility, toughness, impact and fatigue resistance, and control of shrinkage cracking.

admixturesspecial-concrete
B

Section B: Short Answer Questions

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6 questions
6short5 marks

Explain the principle of the rebound (Schmidt) hammer test and the ultrasonic pulse velocity (UPV) test for assessing concrete quality. In a UPV test the pulse travels a path length of 300 mm300\ \text{mm} in 66 μs66\ \mu s (direct transmission). Compute the pulse velocity and comment on the concrete quality using the grading: excellent >4.5> 4.5, good 3.53.5-4.54.5, doubtful 3.03.0-3.53.5, poor <3.0 km/s< 3.0\ \text{km/s}.

Rebound (Schmidt) hammer

A spring-loaded mass strikes a plunger held against the concrete surface; the rebound number depends on the surface hardness, which correlates with compressive strength. It is a quick, non-destructive surface-hardness test, calibrated against a graph.

Ultrasonic pulse velocity (UPV)

An ultrasonic pulse is sent through the concrete between a transmitter and a receiver. The velocity of the pulse is related to the elastic modulus and density, hence to homogeneity, voids/cracks and quality.

Calculation

V=path lengthtransit time=0.300 m66×106 sV = \frac{\text{path length}}{\text{transit time}} = \frac{0.300\ \text{m}}{66\times 10^{-6}\ \text{s}} V=4545.5 m/s=4.55 km/sV = 4545.5\ \text{m/s} = \mathbf{4.55\ \text{km/s}}

Comment

Since V=4.55 km/s>4.5 km/sV = 4.55\ \text{km/s} > 4.5\ \text{km/s}, the concrete quality is rated excellent.

ndtrebound-hammerupv
7short5 marks

Three standard concrete cubes (150 mm150\ \text{mm}) tested at 28 days failed at loads of 720 kN720\ \text{kN}, 765 kN765\ \text{kN} and 750 kN750\ \text{kN}. Compute the individual compressive strengths and the mean strength. State the IS 516 acceptance rule that no individual result should fall below the mean by more than 15%15\%, and check whether all results are acceptable.

Cross-sectional area

A=150×150=22500 mm2A = 150 \times 150 = 22500\ \text{mm}^2

Individual strengths (f=PA)\left(f = \dfrac{P}{A}\right)

  • Cube 1: 720×10322500=32.00 N/mm2\dfrac{720\times 10^3}{22500} = 32.00\ \text{N/mm}^2
  • Cube 2: 765×10322500=34.00 N/mm2\dfrac{765\times 10^3}{22500} = 34.00\ \text{N/mm}^2
  • Cube 3: 750×10322500=33.33 N/mm2\dfrac{750\times 10^3}{22500} = 33.33\ \text{N/mm}^2

Mean strength

fmean=32.00+34.00+33.333=99.333=33.11 N/mm2f_{mean} = \frac{32.00 + 34.00 + 33.33}{3} = \frac{99.33}{3} = 33.11\ \text{N/mm}^2

Acceptance check (no individual result below 0.85fmean0.85\,f_{mean})

0.85×33.11=28.15 N/mm20.85 \times 33.11 = 28.15\ \text{N/mm}^2

Lowest individual result =32.00 N/mm2>28.15 N/mm2= 32.00\ \text{N/mm}^2 > 28.15\ \text{N/mm}^2.

All three results satisfy the rule; the test results are acceptable. Mean strength = 33.11 N/mm².

hardened-concretecompressive-strengthtesting
8short5 marks

The sieve analysis of a 1000 g1000\ \text{g} sample of fine aggregate gave the following retained masses. Compute the fineness modulus and classify the sand zone (FM range: very fine <2.2< 2.2, fine 2.22.2-2.62.6, medium 2.62.6-2.92.9, coarse 2.92.9-3.23.2).

Sieve (mm)4.752.361.180.600.300.15Pan
Retained (g)2011018025023016050

Step 1 — Cumulative % retained

Total sample =1000 g= 1000\ \text{g}.

Sieve (mm)Retained (g)Cum. retained (g)Cum. % retained
4.7520202.0
2.3611013013.0
1.1818031031.0
0.6025056056.0
0.3023079079.0
0.1516095095.0
Pan501000

Step 2 — Fineness modulus

FM = (sum of cumulative % retained on the standard sieves 4.75 to 0.15) / 100. Pan is excluded.

=2.0+13.0+31.0+56.0+79.0+95.0=276.0\sum = 2.0 + 13.0 + 31.0 + 56.0 + 79.0 + 95.0 = 276.0 FM=276.0100=2.76FM = \frac{276.0}{100} = 2.76

Result

FM=2.76\boxed{FM = 2.76}

Since 2.6<2.76<2.92.6 < 2.76 < 2.9, the sand is classified as medium sand.

aggregategradingfineness-modulus
9short5 marks

List the desirable properties of good building bricks and describe the field/laboratory tests used to assess brick quality. Briefly differentiate between English bond and Flemish bond in brick masonry.

Desirable properties of good bricks

  1. Regular shape, sharp edges, uniform size.
  2. Uniform deep-red colour, well-burnt.
  3. Compressive strength not less than about 3.5 N/mm23.5\ \text{N/mm}^2 (common bricks).
  4. Water absorption not more than 20%20\% (by mass) after 24 h immersion.
  5. Low efflorescence (salt deposits).
  6. Should give a clear ringing sound when struck, and not break when dropped from about 1 m.
  7. Good resistance to weathering and fire.

Tests for brick quality

  • Compressive strength test (crushing in a compression machine).
  • Water absorption test (24 h immersion; % increase in mass).
  • Efflorescence test (immerse and observe salt deposits on drying).
  • Soundness test (strike two bricks; clear ring = sound).
  • Hardness test (scratch with a fingernail; no impression for good brick).
  • Dimension/shape test (stacking 20 bricks and measuring).
  • Structure test (broken brick should be compact, homogeneous, no voids).

English bond vs Flemish bond

FeatureEnglish bondFlemish bond
ArrangementAlternate courses of headers and stretchersEach course has alternating headers and stretchers
StrengthStronger; preferred for load-bearing wallsSlightly weaker
AppearanceLess attractiveMore attractive
EconomyUses more bricks/mortarMore economical, fewer broken bats but needs skilled labour
masonry-materialsbricksbonds
10short5 marks

Explain the main causes of deterioration of concrete: sulphate attack, alkali-aggregate reaction, carbonation and corrosion of reinforcement. State TWO practical measures to improve the durability of concrete in an aggressive (coastal) environment.

Causes of deterioration

Sulphate attack

Sulphates (from soil, groundwater, seawater) react with hydrated C3AC_3A and calcium hydroxide to form ettringite and gypsum, which are expansive, causing cracking, spalling and softening. Controlled by using sulphate-resisting cement (low C3AC_3A) and low w/cw/c.

Alkali-aggregate reaction (AAR)

Reactive silica in certain aggregates reacts with alkalis (Na2ONa_2O, K2OK_2O) in cement to form an expansive gel that absorbs water and swells, producing map cracking. Controlled by using low-alkali cement, non-reactive aggregate or pozzolans.

Carbonation

Atmospheric CO2CO_2 reacts with Ca(OH)2Ca(OH)_2 to form CaCO3CaCO_3, lowering pore-water pH from ~13 to below ~9. This destroys the passive layer on steel, making it vulnerable to corrosion.

Corrosion of reinforcement

Loss of passivity (from carbonation or chloride ingress) lets steel corrode; rust occupies more volume, generating internal pressure that cracks and spalls the cover concrete.

Measures for an aggressive coastal environment

  1. Use low water-cement ratio (0.45\le 0.45) and adequate cement content for a dense, impermeable concrete, with sufficient cover to reinforcement.
  2. Use sulphate-resisting / blended cement (with fly ash, slag or silica fume) to resist chloride and sulphate ingress, plus good compaction and curing.
durabilitysulphate-attackcorrosion
11short5 marks

A load-bearing brick wall 230 mm230\ \text{mm} thick and 4.0 m4.0\ \text{m} long carries a uniformly distributed axial load (including self-weight) of 180 kN180\ \text{kN} per metre run at its base. Determine the actual compressive stress at the base of the wall and check its adequacy if the permissible compressive stress is 0.9 N/mm20.9\ \text{N/mm}^2.

Step 1 — Load per metre run on a 1 m strip

Consider a 1 m1\ \text{m} (1000 mm) length of wall. Axial load on this strip P=180 kN=180×103 NP = 180\ \text{kN} = 180\times 10^3\ \text{N}.

Step 2 — Bearing area of the strip

A=thickness×length=230×1000=230000 mm2A = \text{thickness} \times \text{length} = 230 \times 1000 = 230000\ \text{mm}^2

Step 3 — Actual compressive stress

f=PA=180×103230000=0.7826 N/mm2f = \frac{P}{A} = \frac{180\times 10^3}{230000} = 0.7826\ \text{N/mm}^2

Step 4 — Adequacy check

f=0.78 N/mm2<fperm=0.9 N/mm2f = 0.78\ \text{N/mm}^2 < f_{perm} = 0.9\ \text{N/mm}^2

The actual stress (0.78 N/mm²) is less than the permissible stress (0.9 N/mm²); the wall is SAFE in axial compression.

masonry-wallaxial-stressload-bearing

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