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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long12 marks

Design a concrete mix using the IS 10262 procedure for the following requirements and report the final batched quantities per cubic metre.

  • Characteristic compressive strength at 28 days: fck=30 N/mm2f_{ck} = 30\ \text{N/mm}^2 (M30)
  • Maximum nominal size of aggregate: 20 mm
  • Degree of quality control: good (assumed standard deviation s=5.0 N/mm2s = 5.0\ \text{N/mm}^2)
  • Workability: 75 mm slump
  • Exposure: moderate (maximum free w/c = 0.50, minimum cement = 300 kg/m³)
  • Specific gravity: cement = 3.15, coarse aggregate = 2.70, fine aggregate = 2.65
  • Fine aggregate confirms to Zone II
  • Free (chemical) water demand from tables = 186 kg/m³ for 50 mm slump; correction +3% water per 25 mm extra slump
  • Adopted free w/c ratio = 0.45
  • Volume of coarse aggregate per unit volume of total aggregate (Zone II, 20 mm, w/c 0.45) = 0.62; no pumping correction.

Take entrapped air = 2%.

Step 1 — Target mean strength

ftarget=fck+1.65s=30+1.65(5.0)=30+8.25=38.25 N/mm2f_{target} = f_{ck} + 1.65\,s = 30 + 1.65(5.0) = 30 + 8.25 = 38.25\ \text{N/mm}^2

Step 2 — Selection of w/c ratio

The w/c from strength considerations is taken as 0.45 (given). This is lower than the durability limit of 0.50 for moderate exposure, so adopt w/c = 0.45 (governs).

Step 3 — Water content

Base water for 50 mm slump (20 mm aggregate) = 186 kg/m³. Required slump = 75 mm, i.e. 25 mm extra ⇒ +3% water.

W=186×1.03=191.58191.6 kg/m3W = 186 \times 1.03 = 191.58 \approx \mathbf{191.6\ kg/m^3}

Step 4 — Cement content

C=W(w/c)=191.580.45=425.7 kg/m3C = \frac{W}{(w/c)} = \frac{191.58}{0.45} = 425.7\ \text{kg/m}^3

Check minimum cement for moderate exposure = 300 kg/m³. Since 425.7>300425.7 > 300, OK. Adopt C = 425.7 kg/m³.

Step 5 — Proportion of coarse and fine aggregate volumes

Volume of coarse aggregate per unit total aggregate volume = 0.62 ⇒ fine aggregate fraction = 10.62=0.381 - 0.62 = 0.38.

Step 6 — Absolute volume calculation (per 1 m³)

Total volume = 1 m³; entrapped air = 2% = 0.02 m³.

Volume of cement:

Vc=CGc×1000=425.73.15×1000=0.1351 m3V_c = \frac{C}{G_c \times 1000} = \frac{425.7}{3.15 \times 1000} = 0.1351\ \text{m}^3

Volume of water:

Vw=191.581×1000=0.1916 m3V_w = \frac{191.58}{1 \times 1000} = 0.1916\ \text{m}^3

Volume of all aggregate (saturated surface dry):

Vagg=1(Vc+Vw+Vair)=1(0.1351+0.1916+0.02)=10.3467=0.6533 m3V_{agg} = 1 - (V_c + V_w + V_{air}) = 1 - (0.1351 + 0.1916 + 0.02) = 1 - 0.3467 = 0.6533\ \text{m}^3

Step 7 — Mass of coarse aggregate

MCA=Vagg×0.62×GCA×1000=0.6533×0.62×2.70×1000M_{CA} = V_{agg} \times 0.62 \times G_{CA} \times 1000 = 0.6533 \times 0.62 \times 2.70 \times 1000 =0.6533×0.62×2700=1093.6 kg/m3= 0.6533 \times 0.62 \times 2700 = 1093.6\ \text{kg/m}^3

Step 8 — Mass of fine aggregate

MFA=Vagg×0.38×GFA×1000=0.6533×0.38×2.65×1000M_{FA} = V_{agg} \times 0.38 \times G_{FA} \times 1000 = 0.6533 \times 0.38 \times 2.65 \times 1000 =0.6533×0.38×2650=657.9 kg/m3= 0.6533 \times 0.38 \times 2650 = 657.9\ \text{kg/m}^3

Step 9 — Final mix proportions (per m³, SSD basis)

IngredientQuantity (kg/m³)
Cement425.7
Water191.6
Fine aggregate657.9
Coarse aggregate1093.6

By mass the ratio C : FA : CA = 1:1.55:2.571 : 1.55 : 2.57 at w/c = 0.45.

Final answer: Cement 425.7 kg/m³, Water 191.6 kg/m³, Fine aggregate 657.9 kg/m³, Coarse aggregate 1093.6 kg/m³.

concrete-mix-designis-10262water-cement-ratio
2long10 marks

A solid load-bearing brick masonry wall in a two-storey building carries a uniformly distributed axial load. Design data:

  • Actual thickness of wall t=230 mmt = 230\ \text{mm}
  • Effective height H=3.0 mH = 3.0\ \text{m}, effective length is large (wall continuous)
  • Wall is laterally restrained top and bottom
  • Eccentricity of vertical load = 0 (axial)
  • Basic compressive stress of masonry fb=0.50 N/mm2f_b = 0.50\ \text{N/mm}^2 (for mortar grade and brick strength used)
  • The wall is solid (no openings near the loaded zone)

Determine (a) the slenderness ratio, (b) the stress reduction factor ksk_s, and (c) the permissible axial load per metre run of wall. Use the IS 1905 working-stress approach.

For effective height with both ends restrained, take effective height =0.75H= 0.75 H. Stress reduction factor: for slenderness ratio (SR) values 6→1.00, 8→0.95, 10→0.89, 12→0.84 (interpolate linearly).

Step 1 — Effective height and slenderness ratio

With the wall laterally restrained (held in position) at top and bottom, effective height:

heff=0.75H=0.75×3000=2250 mmh_{eff} = 0.75 H = 0.75 \times 3000 = 2250\ \text{mm}

Slenderness ratio (SR) for a wall is based on effective height (height governs here, length large):

SR=hefft=2250230=9.78SR = \frac{h_{eff}}{t} = \frac{2250}{230} = 9.78

(a) Slenderness ratio = 9.78 (well below the limit of 27 for load-bearing walls — OK).

Step 2 — Stress reduction factor ksk_s (interpolation)

Given SR = 8 → 0.95 and SR = 10 → 0.89.

ks=0.95+(0.890.95)×9.788108=0.95+(0.06)×1.782k_s = 0.95 + (0.89 - 0.95)\times\frac{9.78 - 8}{10 - 8} = 0.95 + (-0.06)\times\frac{1.78}{2} ks=0.950.06×0.89=0.950.0534=0.89660.897k_s = 0.95 - 0.06 \times 0.89 = 0.95 - 0.0534 = 0.8966 \approx \mathbf{0.897}

(b) Stress reduction factor ks=0.897k_s = 0.897.

Step 3 — Area eccentricity factor

Since eccentricity = 0, the eccentricity factor kp=1.0k_p = 1.0. (No reduction for eccentric loading.)

Step 4 — Permissible compressive stress

fperm=fb×ks×kp=0.50×0.897×1.0=0.4485 N/mm2f_{perm} = f_b \times k_s \times k_p = 0.50 \times 0.897 \times 1.0 = 0.4485\ \text{N/mm}^2

Step 5 — Permissible load per metre run

Cross-sectional area per metre run:

A=t×L=230 mm×1000 mm=230,000 mm2A = t \times L = 230\ \text{mm} \times 1000\ \text{mm} = 230{,}000\ \text{mm}^2

Permissible axial load:

P=fperm×A=0.4485×230,000=103,155 N=103.2 kNP = f_{perm} \times A = 0.4485 \times 230{,}000 = 103{,}155\ \text{N} = 103.2\ \text{kN}

(c) Permissible axial load per metre run = 103.2 kN/m.

Summary: SR = 9.78, ks=0.897k_s = 0.897, permissible load 103.2 kN/m\approx \mathbf{103.2\ kN/m}.

masonry-designbrick-wallslenderness-ratio
3long10 marks

(a) Explain the hydration of Portland cement, naming the four principal Bogue compounds and describing the role of each in early and later strength development and in heat liberation.

(b) A mass-concrete pour uses a cement whose Bogue composition is: C₃S = 50%, C₂S = 22%, C₃A = 9%, C₄AF = 8%. Using the approximate heats of hydration C₃S = 500 J/g, C₂S = 260 J/g, C₃A = 866 J/g, C₄AF = 420 J/g (at complete hydration), estimate the total heat of hydration per gram of cement and comment on its suitability for mass concreting.

(a) Hydration of Portland cement

When water is added to cement the anhydrous compounds react exothermically to form hydration products, chiefly calcium silicate hydrate (C-S-H gel) and calcium hydroxide (CH). The C-S-H gel is the main binding phase responsible for strength.

The four principal Bogue compounds:

  1. Tricalcium silicate (C₃S, alite) — hydrates rapidly; responsible for early strength (first 7–28 days) and liberates a large amount of heat.
  2. Dicalcium silicate (C₂S, belite) — hydrates slowly; contributes to later-age strength (beyond 28 days) and gives off comparatively little heat.
  3. Tricalcium aluminate (C₃A) — hydrates almost instantly; gives very high early heat and rapid setting (gypsum is added to retard it). Contributes little to strength and is vulnerable to sulphate attack.
  4. Tetracalcium alumino-ferrite (C₄AF) — hydrates moderately fast, gives moderate heat, contributes little to strength but improves sulphate resistance and gives cement its grey colour.

Low-heat cement minimises C₃S and C₃A and raises C₂S.

(b) Total heat of hydration

H=(fraction×specific heat)H = \sum (\text{fraction} \times \text{specific heat})
Compound%FractionHeat (J/g)Contribution (J/g cement)
C₃S500.50500250.0
C₂S220.2226057.2
C₃A90.0986677.94
C₄AF80.0842033.6
H=250.0+57.2+77.94+33.6=418.7 J/gH = 250.0 + 57.2 + 77.94 + 33.6 = 418.7\ \text{J/g}

Total heat of hydration ≈ 418.7 J/g (≈ 419 kJ/kg).

Comment: Ordinary Portland cement typically liberates about 400–500 J/g. The value here (≈419 J/g) is typical of OPC and is on the higher side for mass concrete. The relatively high C₃S (50%) and C₃A (9%) make this cement prone to a large early temperature rise, risking thermal cracking in a large pour. For mass concreting it would be preferable to use low-heat or Portland-pozzolana cement (lower C₃S, lower C₃A, higher C₂S), partial cement replacement by fly ash/GGBS, and pre-cooling of ingredients.

hydrationheat-of-hydrationbogue-compounds
4long10 marks

(a) Compare the rebound (Schmidt) hammer test and the ultrasonic pulse velocity (UPV) test as non-destructive testing methods for concrete, listing the principle, what each measures, two advantages and two limitations of each.

(b) In a UPV survey, an ultrasonic pulse takes 58.5 μs58.5\ \mu s to travel through a 350 mm thick concrete element (direct transmission). Compute the pulse velocity in km/s and classify the concrete quality using the standard grading: > 4.5 km/s excellent, 3.5–4.5 good, 3.0–3.5 medium/doubtful, < 3.0 poor. Also, if a second reading on a suspect zone gives 3.10 km/s over the same 350 mm, find the transit time.

(a) Comparison of rebound hammer and UPV

AspectRebound (Schmidt) HammerUltrasonic Pulse Velocity
PrincipleA spring-driven mass rebounds off a plunger pressed on the surface; rebound number relates to surface hardnessAn ultrasonic pulse (transducer) travels through concrete; velocity depends on density/elastic modulus
MeasuresSurface hardness → approximate compressive strengthPulse velocity → uniformity, density, presence of cracks/voids, EE
Advantage 1Simple, fast, very cheap, portableAssesses the full thickness/interior, not just surface
Advantage 2Large number of readings quickly for uniformity checksDetects internal cracks, voids, honeycombing
Limitation 1Only surface layer (≈30 mm); affected by carbonation, moisture, aggregate typeNeeds access to two faces for best (direct) results; couplant required
Limitation 2Strength correlation is approximate; needs calibrationReinforcement and moisture content affect readings

The two are often combined (SonReb method) to improve strength estimation.

(b) Pulse velocity calculation

Path length L=350 mm=0.350 mL = 350\ \text{mm} = 0.350\ \text{m}, transit time t=58.5 μs=58.5×106 st = 58.5\ \mu s = 58.5 \times 10^{-6}\ \text{s}.

V=Lt=0.35058.5×106=5982.9 m/s=5.98 km/sV = \frac{L}{t} = \frac{0.350}{58.5 \times 10^{-6}} = 5982.9\ \text{m/s} = \mathbf{5.98\ km/s}

Since 5.98>4.5 km/s5.98 > 4.5\ \text{km/s}, the concrete quality is excellent.

Suspect zone — transit time for V = 3.10 km/s:

t=LV=0.3503100=1.129×104 s=112.9 μst = \frac{L}{V} = \frac{0.350}{3100} = 1.129 \times 10^{-4}\ \text{s} = \mathbf{112.9\ \mu s}

The suspect-zone velocity (3.10 km/s) falls in the medium/doubtful band, and the transit time nearly doubles (112.9 µs vs 58.5 µs), indicating likely internal voids or microcracking that warrant further investigation.

ndtrebound-hammerultrasonic-pulse-velocity
5long8 marks

A square brick masonry column has a finished cross-section of 460 mm×460 mm460\ \text{mm} \times 460\ \text{mm} and an effective height of 2.8 m2.8\ \text{m}. The basic permissible compressive stress of the masonry is 0.70 N/mm20.70\ \text{N/mm}^2. The load is axial. For a column the slenderness ratio is based on the least lateral dimension.

Stress reduction factors ksk_s: SR 6→1.00, 8→0.95, 10→0.89. Area reduction factor for small sections: if the horizontal sectional area is less than 0.2 m20.2\ \text{m}^2, multiply permissible stress by (0.70+1.5A)(0.70 + 1.5A) where AA is area in m². Determine the safe axial load the column can carry.

Step 1 — Slenderness ratio (column, least dimension)

SR=heffleast dimension=2800460=6.09SR = \frac{h_{eff}}{\text{least dimension}} = \frac{2800}{460} = 6.09

Step 2 — Stress reduction factor ksk_s

SR = 6.09 is just above 6. Interpolate between SR 6 → 1.00 and SR 8 → 0.95:

ks=1.00+(0.951.00)×6.09686=1.000.05×0.092k_s = 1.00 + (0.95 - 1.00)\times\frac{6.09 - 6}{8 - 6} = 1.00 - 0.05 \times \frac{0.09}{2} ks=1.000.05×0.045=1.000.00225=0.99780.998k_s = 1.00 - 0.05 \times 0.045 = 1.00 - 0.00225 = 0.9978 \approx \mathbf{0.998}

Step 3 — Area reduction factor (small section check)

Sectional area:

A=0.460×0.460=0.2116 m2A = 0.460 \times 0.460 = 0.2116\ \text{m}^2

Since A=0.2116 m2>0.2 m2A = 0.2116\ \text{m}^2 > 0.2\ \text{m}^2, the small-area reduction does not apply. Area factor ka=1.0k_a = 1.0.

Step 4 — Eccentricity factor

Load is axial ⇒ kp=1.0k_p = 1.0.

Step 5 — Permissible compressive stress

fperm=fb×ks×ka×kp=0.70×0.998×1.0×1.0=0.6985 N/mm2f_{perm} = f_b \times k_s \times k_a \times k_p = 0.70 \times 0.998 \times 1.0 \times 1.0 = 0.6985\ \text{N/mm}^2

Step 6 — Safe axial load

Area in mm² = 460×460=211,600 mm2460 \times 460 = 211{,}600\ \text{mm}^2.

P=fperm×A=0.6985×211,600=147,803 N=147.8 kNP = f_{perm} \times A = 0.6985 \times 211{,}600 = 147{,}803\ \text{N} = 147.8\ \text{kN}

Safe axial load ≈ 147.8 kN.

masonry-columnaxial-loadis-1905
B

Section B: Short Answer Questions

Attempt all questions.

6 questions
6short5 marks

Define workability of fresh concrete and list the factors affecting it. Describe the slump test, stating its apparatus, procedure and the types of slump observed, and mention one situation where the slump test is unsuitable.

Workability is the property of fresh concrete that determines the ease with which it can be mixed, transported, placed, compacted and finished without segregation or bleeding, for a given amount of work. It is closely linked to the consistency and the energy required for full compaction.

Factors affecting workability:

  • Water content / water–cement ratio
  • Aggregate properties: size, shape, texture, grading
  • Aggregate–cement ratio (mix proportions)
  • Use of admixtures (plasticizers/superplasticizers)
  • Fineness of cement and use of supplementary cementitious materials
  • Time and ambient temperature

Slump test:

  • Apparatus: Slump cone (frustum: bottom dia 200 mm, top dia 100 mm, height 300 mm), tamping rod (16 mm dia, 600 mm long), base plate.
  • Procedure: Place the cone on a level surface; fill in 4 layers, each tamped 25 times; strike off the top; lift the cone vertically; measure the vertical subsidence (slump) of the concrete.
  • Types of slump: (i) True slump – mass subsides evenly (valid result); (ii) Shear slump – top portion shears off and slips sideways (test should be repeated); (iii) Collapse slump – concrete collapses completely (very wet/harsh mix).

Unsuitable situation: The slump test is unsuitable for very dry/stiff mixes (low workability, e.g. zero-slump concrete) and for very wet/collapse mixes; for such concrete the compaction factor or Vee-Bee consistometer test is preferred.

workabilityslump-testfresh-concrete
7short5 marks

Classify concrete admixtures and explain the working mechanism and uses of (i) superplasticizers (high-range water reducers) and (ii) air-entraining agents. State one possible adverse effect of overdosing a superplasticizer.

Classification of admixtures:

  • Chemical admixtures: plasticizers (water reducers), superplasticizers (high-range water reducers), retarders, accelerators, air-entraining agents, water-proofing/permeability-reducing admixtures.
  • Mineral (supplementary cementitious) admixtures: fly ash, GGBS, silica fume, rice husk ash, metakaolin.

(i) Superplasticizers (high-range water reducers): Mechanism: They are surface-active polymers (sulphonated naphthalene/melamine formaldehyde or polycarboxylate ethers) that adsorb onto cement particles and impart a strong negative charge or steric repulsion, deflocculating the cement grains. The released entrapped water greatly increases fluidity. Uses: (a) produce flowing/self-compacting concrete at the same w/c; (b) achieve very low w/c (high strength) at the same workability; reduce water by 15–30%.

(ii) Air-entraining agents: Mechanism: Surfactants that stabilise millions of tiny, well-distributed air bubbles (≈10–300 µm) in the paste. These bubbles act as pressure-relief voids for water expanding on freezing. Uses: Greatly improve freeze–thaw durability and resistance to de-icing salts; also improve workability and reduce bleeding/segregation.

Adverse effect of superplasticizer overdose: Severe segregation/bleeding and excessive set retardation (delayed setting), and possible loss of cohesion of the mix.

admixturessuperplasticizersconcrete-chemistry
8short5 marks

Explain the durability of concrete with reference to (a) sulphate attack and (b) corrosion of reinforcement. For each, state the mechanism and two preventive measures.

Durability is the ability of concrete to resist weathering action, chemical attack and abrasion while retaining its desired engineering properties over its service life. It is governed largely by permeability, which depends on the w/c ratio, compaction and curing.

(a) Sulphate attack Mechanism: Sulphates (from soil, groundwater, seawater) react with hydrated cement compounds. Sulphate ions react with calcium hydroxide and with hydrated calcium aluminate (C₃A) to form gypsum and ettringite (calcium sulphoaluminate), which are expansive. The expansion causes cracking, spalling and progressive disintegration of concrete. Preventive measures:

  1. Use sulphate-resisting Portland cement (SRC) with low C₃A content, or PPC/PSC.
  2. Use low w/c ratio and dense, well-compacted concrete (low permeability); adequate cover and curing.

(b) Corrosion of reinforcement Mechanism: The high alkalinity (pH ≈ 13) of concrete forms a passive protective film on steel. Carbonation (CO₂ lowering pH) or chloride ingress destroys this film. An electrochemical cell forms (anode/cathode); iron oxidises to rust, which occupies a larger volume, generating internal pressure that cracks and spalls the cover. Preventive measures:

  1. Provide adequate cover and low-permeability concrete (low w/c, good compaction and curing).
  2. Limit chloride content; use corrosion-inhibiting admixtures, coated (epoxy/galvanised) bars, or protective coatings/cathodic protection.
durabilitysulphate-attackcorrosion
9short5 marks

Three concrete cubes of side 150 mm from the same batch were tested in compression at 28 days and failed at loads of 720 kN, 690 kN and 765 kN. (a) Compute the individual cube strengths and the mean strength. (b) The mix is designated M25 (target mean strength 33 N/mm²). Using the IS 516 / IS 456 acceptance rule that the mean of the group must not be less than fck+0.825×f_{ck} + 0.825 \times established s (take established s=4s = 4 N/mm²) AND no individual result less than fck3f_{ck} - 3 N/mm², check acceptance.

Step 1 — Cube cross-sectional area

A=150×150=22,500 mm2A = 150 \times 150 = 22{,}500\ \text{mm}^2

Step 2 — Individual cube strengths (f=PAf = \dfrac{P}{A}, with PP in N):

CubeLoad (kN)Load (N)Strength (N/mm²)
1720720,000720000/22500=32.00720000/22500 = 32.00
2690690,000690000/22500=30.67690000/22500 = 30.67
3765765,000765000/22500=34.00765000/22500 = 34.00

Step 3 — Mean strength

fˉ=32.00+30.67+34.003=96.673=32.22 N/mm2\bar{f} = \frac{32.00 + 30.67 + 34.00}{3} = \frac{96.67}{3} = 32.22\ \text{N/mm}^2

Step 4 — Acceptance checks (M25, fck=25f_{ck}=25, s=4s = 4)

Mean criterion:

fck+0.825s=25+0.825(4)=25+3.30=28.30 N/mm2f_{ck} + 0.825\,s = 25 + 0.825(4) = 25 + 3.30 = 28.30\ \text{N/mm}^2

Mean = 32.22 N/mm² \ge 28.30 N/mm² → satisfied.

Individual criterion:

fck3=253=22 N/mm2f_{ck} - 3 = 25 - 3 = 22\ \text{N/mm}^2

Lowest individual = 30.67 N/mm² \ge 22 N/mm² → satisfied.

Conclusion: Both criteria are met, so the concrete is acceptable as M25. (The mean of 32.22 N/mm² is also close to the target mean strength of 33 N/mm², confirming satisfactory quality control.)

hardened-concretecompressive-strengthtesting
10short5 marks

Write short notes on any TWO of the following special concretes, covering definition, key constituents/principle and one typical application each: (i) Self-Compacting Concrete (SCC), (ii) Fibre-Reinforced Concrete (FRC), (iii) High-Performance Concrete (HPC).

(i) Self-Compacting Concrete (SCC) Definition/principle: A highly flowable, non-segregating concrete that spreads and fills formwork and encapsulates reinforcement under its own weight without any vibration. It balances high deformability (flowability) with adequate segregation resistance (viscosity). Key constituents: High powder/fines content, high-range water reducer (polycarboxylate superplasticizer), often a viscosity-modifying agent (VMA), and a limited coarse-aggregate content. Tested by slump-flow, V-funnel and L-box. Application: Heavily reinforced/congested sections, complex shapes, and pours where vibration is difficult (e.g. deep beams, columns, tunnel linings).

(ii) Fibre-Reinforced Concrete (FRC) Definition/principle: Concrete containing discrete, uniformly distributed fibres (steel, glass, polypropylene, etc.) that bridge cracks, improving tensile strength, ductility, toughness, impact and fatigue resistance, and controlling shrinkage cracking. Key constituents: Normal concrete plus fibres (volume fraction typically 0.5–2%) characterised by aspect ratio (length/diameter). Application: Industrial floors and pavements, tunnel linings (shotcrete), bridge decks, and crack-control in slabs.

(iii) High-Performance Concrete (HPC) Definition/principle: Concrete engineered to give superior performance in strength and/or durability beyond ordinary concrete — high strength, low permeability, long service life. Achieved through a very low w/c ratio with superplasticizers and supplementary cementitious materials (silica fume, fly ash). Key constituents: Low w/c (often < 0.35), high-range water reducer, silica fume/fly ash, well-graded quality aggregates. Application: High-rise columns, long-span bridges, marine and aggressive-exposure structures.

(Any two of the above earn full marks.)

special-concreteself-compacting-concretefibre-reinforced-concrete
11short5 marks

(a) List the desirable properties of good building bricks and define water absorption and efflorescence as quality tests. (b) A brick of dry mass 3120 g is immersed in water for 24 hours and its saturated mass becomes 3556 g. Compute the percentage water absorption and comment on whether it meets the common limit of 20% for first-class bricks.

(a) Desirable properties of good building bricks

  • Uniform size, shape and sharp, well-defined edges
  • Uniform deep red/copper colour, well burnt (not over- or under-burnt)
  • Sufficient compressive strength (first-class ≥ 10.5 N/mm² for IOE/Nepali practice, min ~7 N/mm² as per IS)
  • Low water absorption (≤ 20% for first-class by 24-hour immersion)
  • Free from cracks, flaws, stones and lumps of lime
  • Should give a clear metallic ringing sound when struck
  • Should not show efflorescence; adequate hardness (no scratch by fingernail)

Water absorption = the increase in mass when a dry brick is immersed in water for 24 h, expressed as a percentage of the dry mass; it indicates porosity and durability.

Efflorescence = the deposit of white salts on the brick surface when soluble salts in the brick migrate to the surface and crystallise on drying; tested by repeated wetting/drying and rated nil/slight/moderate/heavy.

(b) Water absorption calculation

Dry mass M1=3120 gM_1 = 3120\ \text{g}, saturated mass M2=3556 gM_2 = 3556\ \text{g}.

Water absorption=M2M1M1×100=355631203120×100\text{Water absorption} = \frac{M_2 - M_1}{M_1}\times 100 = \frac{3556 - 3120}{3120}\times 100 =4363120×100=13.97%14.0%= \frac{436}{3120}\times 100 = 13.97\% \approx \mathbf{14.0\%}

Comment: Since 14.0% < 20%, the water absorption is within the permissible limit, so the brick satisfies the first-class brick requirement for water absorption.

masonry-materialsbricksmortar

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How many marks is the BE Civil Engineering (IOE, TU) Concrete Technology and Masonry Structures (IOE, CE 605) 2076 paper?
The BE Civil Engineering (IOE, TU) Concrete Technology and Masonry Structures (IOE, CE 605) 2076 paper carries 80 full marks and is meant to be completed in 180 minutes, across 11 questions.
Is practising this Concrete Technology and Masonry Structures (IOE, CE 605) past paper free?
Yes — reading and attempting this Concrete Technology and Masonry Structures (IOE, CE 605) past paper on Kekkei is completely free.