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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long12 marks

Design a concrete mix (by the absolute volume method) for an M30 grade reinforced concrete structure using the following data:

  • Characteristic compressive strength: fck=30 N/mm2f_{ck} = 30\ \text{N/mm}^2 at 28 days
  • Specific gravity of cement = 3.15, coarse aggregate = 2.70, fine aggregate = 2.65
  • Maximum nominal size of coarse aggregate = 20 mm
  • Workability required = 75 mm slump
  • Degree of quality control: good (assume standard deviation s=5.0 N/mm2s = 5.0\ \text{N/mm}^2)
  • Exposure: moderate (maximum free w/c ratio = 0.50, minimum cement content = 300 kg/m³)
  • Assume free water content = 186 kg/m³ for 20 mm aggregate at 50 mm slump; add 3% water per additional 25 mm slump
  • Volume of coarse aggregate per unit volume of total aggregate = 0.62 (for w/c = 0.50, zone II sand)
  • Assume entrapped air = 2%

Determine the target mean strength, the w/c ratio, the cement, water, fine and coarse aggregate contents per cubic metre, and state the final mix proportion by mass.

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\times 5.0 = 30 + 8.25 = 38.25\ \text{N/mm}^2

Step 2 — Selection of w/c ratio

From the strength requirement a w/c of about 0.48 corresponds to ~38 N/mm². The exposure (moderate) caps the free w/c at 0.50. The lower value governs, so adopt w/c = 0.48.

Step 3 — Water content

Base water = 186 kg/m³ at 50 mm slump. Required slump = 75 mm, i.e. 25 mm extra → add 3%:

W=186×(1+0.03)=186×1.03=191.6 kg/m3W = 186\times(1+0.03) = 186\times 1.03 = 191.6\ \text{kg/m}^3

Step 4 — Cement content

C=W(w/c)=191.60.48=399.2 kg/m3C = \frac{W}{(w/c)} = \frac{191.6}{0.48} = 399.2\ \text{kg/m}^3

Check minimum cement (300 kg/m³): 399.2 > 300 → OK.

Step 5 — Aggregate volumes (absolute volume method)

Total volume = 1 m³. Entrapped air = 2% = 0.02 m³.

Volume of cement: 399.23.15×1000=0.1267 m3\dfrac{399.2}{3.15\times 1000} = 0.1267\ \text{m}^3

Volume of water: 191.61×1000=0.1916 m3\dfrac{191.6}{1\times 1000} = 0.1916\ \text{m}^3

Volume of all-in aggregate:

Vagg=1(0.1267+0.1916+0.02)=10.3383=0.6617 m3V_{agg} = 1 - (0.1267 + 0.1916 + 0.02) = 1 - 0.3383 = 0.6617\ \text{m}^3

Step 6 — Split into coarse and fine

Volume of CA = 0.62 of total aggregate; volume of FA = 0.38.

Coarse aggregate volume = 0.62×0.6617=0.4103 m30.62\times 0.6617 = 0.4103\ \text{m}^3

Fine aggregate volume = 0.38×0.6617=0.2514 m30.38\times 0.6617 = 0.2514\ \text{m}^3

Mass of CA = 0.4103×2.70×1000=1107.8 kg/m30.4103\times 2.70\times 1000 = 1107.8\ \text{kg/m}^3

Mass of FA = 0.2514×2.65×1000=666.3 kg/m30.2514\times 2.65\times 1000 = 666.3\ \text{kg/m}^3

Step 7 — Final quantities per m³

IngredientMass (kg/m³)
Cement399.2
Water191.6
Fine aggregate666.3
Coarse aggregate1107.8

Mix proportion by mass (Cement : FA : CA)

399.2399.2:666.3399.2:1107.8399.2=1:1.67:2.78,w/c=0.48\frac{399.2}{399.2} : \frac{666.3}{399.2} : \frac{1107.8}{399.2} = \mathbf{1 : 1.67 : 2.78},\quad \text{w/c} = \mathbf{0.48}
mix-designconcrete-proportioningwater-cement-ratio
2long10 marks

A solid brick masonry wall, 230 mm thick and 3.0 m high (storey height), carries an axial load of 180 kN per metre run from a reinforced concrete slab. The wall is effectively held in position and partially restrained in direction at both top and bottom (effective height factor = 0.85). The slab bears over the full thickness of the wall (eccentricity at top ≈ 0). The basic compressive stress of the masonry is 0.96 N/mm20.96\ \text{N/mm}^2.

(a) Compute the slenderness ratio of the wall.

(b) Using the stress reduction factor ksk_s from the table below, the area reduction factor kak_a, and assuming the shape modification factor kp=1.0k_p = 1.0, check whether the wall is safe.

Slenderness ratio (SR)Eccentricity 0
61.00
80.95
100.89
120.84
140.78

Area reduction factor ka=0.7+1.5Ak_a = 0.7 + 1.5A where AA is the loaded area in m² (applies only when A<0.2 m2A < 0.2\ \text{m}^2; otherwise ka=1.0k_a = 1.0).

(a) Slenderness ratio (SR)

Effective height: heff=0.85×H=0.85×3000=2550 mmh_{eff} = 0.85\times H = 0.85\times 3000 = 2550\ \text{mm}

Effective thickness (solid wall): teff=230 mmt_{eff} = 230\ \text{mm}

SR=heffteff=2550230=11.09SR = \frac{h_{eff}}{t_{eff}} = \frac{2550}{230} = \mathbf{11.09}

(b) Permissible stress and check

Stress reduction factor ksk_s — interpolate between SR = 10 (0.89) and SR = 12 (0.84) for eccentricity 0:

ks=0.89(0.890.84)×11.09101210=0.890.05×0.545=0.890.0273=0.863k_s = 0.89 - (0.89-0.84)\times\frac{11.09-10}{12-10} = 0.89 - 0.05\times 0.545 = 0.89 - 0.0273 = 0.863

Area reduction factor — loaded area per metre run = 1.0 m×0.230 m=0.230 m20.2 m21.0\ \text{m}\times 0.230\ \text{m} = 0.230\ \text{m}^2 \ge 0.2\ \text{m}^2, so ka=1.0k_a = 1.0.

Shape factor kp=1.0k_p = 1.0.

Permissible compressive stress:

fperm=kskakp×fb=0.863×1.0×1.0×0.96=0.828 N/mm2f_{perm} = k_s\,k_a\,k_p\times f_b = 0.863\times 1.0\times 1.0\times 0.96 = 0.828\ \text{N/mm}^2

Actual stress (load per metre run over the wall cross-section per metre):

factual=180×103 N1000 mm×230 mm=180000230000=0.783 N/mm2f_{actual} = \frac{180\times 10^3\ \text{N}}{1000\ \text{mm}\times 230\ \text{mm}} = \frac{180000}{230000} = 0.783\ \text{N/mm}^2

Check: factual=0.783 N/mm2<fperm=0.828 N/mm2f_{actual} = 0.783\ \text{N/mm}^2 < f_{perm} = 0.828\ \text{N/mm}^2.

The wall is SAFE (the applied stress is about 95% of the permissible stress, leaving a small margin).

masonry-designwall-designslenderness
3long10 marks

(a) Explain the factors that affect the compressive strength of hardened concrete, with emphasis on the role of the water-cement ratio (Abrams' law) and the gel/space ratio. (5 marks)

(b) A standard concrete cube of 150 mm size failed in a compression testing machine at a total load of 742 kN. Compute the cube compressive strength. If the equivalent cylinder strength is taken as 0.80 times the cube strength, estimate the cylinder strength. (3 marks)

(c) During a rebound hammer (Schmidt hammer) survey, a member gave a mean rebound number that, from the calibration curve, corresponds to an in-situ strength of 28 N/mm². Briefly state two limitations of relying on the rebound number alone, and name one complementary NDT method. (2 marks)

(a) Factors affecting compressive strength of hardened concrete

  1. Water-cement ratio (Abrams' law): For fully compacted concrete, strength is inversely related to the w/c ratio: fc=ABw/cf_c = \dfrac{A}{B^{w/c}}, where AA and BB are empirical constants. A lower w/c gives a denser hydrated paste with fewer capillary pores, hence higher strength. Excess water beyond that needed for hydration leaves capillary voids on evaporation.
  2. Gel/space ratio: Strength depends on the ratio of the volume of hydrated cement gel to the total space available (gel + capillary pores): fc=axnf_c = a\,x^n where xx is the gel/space ratio (typically n3n\approx 3). A higher gel/space ratio (more hydration, less capillary space) yields higher strength independent of age and w/c.
  3. Degree of hydration / age: Strength increases with curing age as hydration proceeds.
  4. Cement type and content; aggregate quality, grading, and bond (transition zone).
  5. Degree of compaction: Voids drastically reduce strength (≈5–6% strength loss per 1% voids).
  6. Curing temperature and moisture, and testing conditions (rate of loading, specimen shape/size, moisture state).

(b) Cube strength and cylinder strength

Cross-sectional area of cube = 150×150=22500 mm2150\times 150 = 22500\ \text{mm}^2.

fcube=PA=742×103 N22500 mm2=32.98 N/mm233.0 N/mm2f_{cube} = \frac{P}{A} = \frac{742\times 10^3\ \text{N}}{22500\ \text{mm}^2} = 32.98\ \text{N/mm}^2 \approx \mathbf{33.0\ \text{N/mm}^2}

Cylinder strength = 0.80×32.98=26.4 N/mm20.80\times 32.98 = \mathbf{26.4\ \text{N/mm}^2}.

(c) Rebound hammer limitations and complementary method

Limitations (any two): (i) it assesses only the surface zone (top ~30 mm), so it does not reflect the interior or core strength; (ii) results are strongly affected by surface moisture, carbonation, smoothness, aggregate type near the surface, and hammer orientation, so site calibration is essential. Complementary NDT method: Ultrasonic Pulse Velocity (UPV) test (the combined SonReb method improves reliability); coring for a direct strength check is also acceptable.

hardened-concretecompressive-strengthndt
4long10 marks

(a) Explain the durability of concrete with respect to sulphate attack and chloride-induced corrosion of reinforcement. State the principal preventive measures for each. (5 marks)

(b) Classify chemical admixtures used in concrete and explain the mechanism of action of (i) a plasticizer/superplasticizer and (ii) an air-entraining agent. (3 marks)

(c) Write short notes on any TWO special concretes: self-compacting concrete (SCC), fibre-reinforced concrete (FRC), and high-performance concrete (HPC). (2 marks)

(a) Durability

Sulphate attack: Sulphates (from soil, groundwater, or seawater) react with hydrated cement compounds. SO₄²⁻ reacts with calcium hydroxide to form gypsum, and with the calcium aluminate hydrate / monosulphate to form ettringite (calcium sulphoaluminate). Both products are expansive, causing internal pressure, cracking, spalling, softening, and loss of strength. Preventive measures: use Sulphate-Resisting Portland Cement (SRPC, low C₃A) or blended cements with PPC/PSC (fly ash/slag); low w/c ratio (≤0.45) for low permeability; adequate cement content and dense, well-compacted concrete; protective coatings in severe exposure.

Chloride-induced corrosion: Chloride ions (from de-icing salts, marine environment, or contaminated aggregates/water) penetrate the cover, depassivate the protective oxide film on the steel once a threshold concentration is reached, and initiate electrochemical corrosion. The rust occupies a larger volume than steel, generating expansive forces that crack and spall the cover. Preventive measures: adequate, dense cover; low w/c and use of mineral admixtures (silica fume, fly ash, slag) to reduce permeability and chloride diffusion; limit chloride content in mix ingredients; corrosion-inhibiting admixtures, coated/galvanised or stainless reinforcement; surface coatings/cathodic protection in aggressive zones.

(b) Chemical admixtures

Classification: plasticizers (water reducers), superplasticizers (high-range water reducers), retarders, accelerators, air-entraining agents, and combinations (e.g., retarding/accelerating water reducers), plus special types (waterproofing, corrosion inhibitors, etc.).

(i) Plasticizer/superplasticizer: Surface-active (anionic) molecules adsorb onto cement particles and impart a negative charge, causing mutual electrostatic repulsion (and, for polycarboxylate ethers, steric hindrance). This deflocculates the cement, releases trapped water, and disperses particles — giving either higher workability at the same w/c or the same workability at a reduced w/c (hence higher strength).

(ii) Air-entraining agent: Surfactants that stabilise billions of tiny (10–300 µm), well-distributed, discrete air bubbles in the paste. These voids relieve hydraulic pressure during freeze–thaw cycles, improve workability and cohesion, and reduce bleeding/segregation (with a small strength penalty).

(c) Special concretes (any two)

  • Self-compacting concrete (SCC): Highly flowable, non-segregating concrete that spreads and fills formwork and encapsulates reinforcement under its own weight without vibration; achieved with superplasticizers, viscosity-modifying agents, high fines, and limited coarse aggregate.
  • Fibre-reinforced concrete (FRC): Concrete with dispersed fibres (steel, glass, polypropylene, etc.) that bridge cracks, improving tensile/flexural strength, toughness, ductility, impact and fatigue resistance, and crack control.
  • High-performance concrete (HPC): Concrete engineered for superior properties (high strength and/or high durability, low permeability) using low w/c, silica fume and other mineral admixtures, and high-range water reducers.
durabilityadmixturesspecial-concrete
5long10 marks

A short brick masonry column of cross-section 460 mm × 460 mm and effective height 3.45 m is subjected to an axial load. The masonry units have a compressive strength giving a basic permissible compressive stress of 1.10 N/mm21.10\ \text{N/mm}^2. Take the stress reduction factor for the relevant slenderness ratio (eccentricity 0) from the table, and compute the area reduction factor where applicable.

Slenderness ratioksk_s (e=0)
61.00
80.95
100.89

Area reduction factor: ka=0.70+1.5Ak_a = 0.70 + 1.5A for loaded area A<0.2 m2A < 0.2\ \text{m}^2; otherwise ka=1.0k_a = 1.0.

Determine (a) the slenderness ratio, (b) the permissible compressive stress, and (c) the safe axial load-carrying capacity of the column.

(a) Slenderness ratio

For a column, the least lateral dimension governs teff=460 mmt_{eff} = 460\ \text{mm}.

SR=heffteff=3450460=7.5SR = \frac{h_{eff}}{t_{eff}} = \frac{3450}{460} = \mathbf{7.5}

(b) Permissible compressive stress

Stress reduction factor — interpolate between SR = 6 (1.00) and SR = 8 (0.95):

ks=1.00(1.000.95)×7.5686=1.000.05×0.75=1.000.0375=0.9625k_s = 1.00 - (1.00-0.95)\times\frac{7.5-6}{8-6} = 1.00 - 0.05\times 0.75 = 1.00 - 0.0375 = 0.9625

Area reduction factor — column cross-sectional area:

A=0.460×0.460=0.2116 m20.2 m2ka=1.0A = 0.460\times 0.460 = 0.2116\ \text{m}^2 \ge 0.2\ \text{m}^2 \Rightarrow k_a = 1.0

(The section is larger than 0.2 m², so no area reduction applies.)

Shape factor taken as kp=1.0k_p = 1.0.

fperm=kskakp×fb=0.9625×1.0×1.0×1.10=1.059 N/mm2f_{perm} = k_s\,k_a\,k_p\times f_b = 0.9625\times 1.0\times 1.0\times 1.10 = 1.059\ \text{N/mm}^2

(c) Safe axial load capacity

Gross cross-sectional area = 460×460=211600 mm2460\times 460 = 211600\ \text{mm}^2.

Psafe=fperm×A=1.059 N/mm2×211600 mm2=224,083 NP_{safe} = f_{perm}\times A = 1.059\ \text{N/mm}^2\times 211600\ \text{mm}^2 = 224{,}083\ \text{N} Psafe224 kN\boxed{P_{safe} \approx \mathbf{224\ \text{kN}}}
masonry-columncolumn-designaxial-load
B

Section B: Short Answer Questions

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6 questions
6short5 marks

List the four major Bogue compounds of ordinary Portland cement and briefly describe the role of each in the strength development and heat of hydration of cement.

The four major Bogue compounds of OPC:

CompoundFormulaAbbrev.Typical %
Tricalcium silicate (alite)3CaO·SiO₂C₃S45–60
Dicalcium silicate (belite)2CaO·SiO₂C₂S15–30
Tricalcium aluminate3CaO·Al₂O₃C₃A6–12
Tetracalcium aluminoferrite4CaO·Al₂O₃·Fe₂O₃C₄AF6–8

Roles:

  • C₃S (alite): Hydrates rapidly; chiefly responsible for early strength (up to ~7–28 days); liberates a high heat of hydration. Major strength contributor.
  • C₂S (belite): Hydrates slowly; responsible for later-age (long-term) strength beyond 28 days; low heat of hydration; contributes to durability.
  • C₃A: Reacts very fast (flash set) and gives high early heat; contributes little to ultimate strength; gypsum is added to retard its flash set. High C₃A reduces sulphate resistance.
  • C₄AF: Hydrates moderately fast but contributes little to strength; gives cement its grey colour; moderate heat; improves resistance to sulphate attack relative to C₃A.
cementconcrete-ingredientsbogue-compounds
7short5 marks

(a) Define workability of fresh concrete and list the factors affecting it. (2 marks)

(b) A slump cone test recorded an initial concrete height of 300 mm (cone height) and a subsided height of 235 mm after lifting the mould. Compute the slump and classify the workability (very low / low / medium / high). State one suitable application for this workability. (3 marks)

(a) Workability

Workability is the property of fresh concrete that determines the ease and homogeneity with which it can be mixed, placed, compacted, and finished without segregation or bleeding. Factors affecting it: water content/water–cement ratio, aggregate grading, shape and texture, maximum aggregate size, aggregate–cement ratio, use of admixtures (plasticizers, air-entrainers), cement properties and fineness, mix proportions, and ambient temperature/time since mixing.

(b) Slump and workability class

Slump=cone heightsubsided height=300235=65 mm\text{Slump} = \text{cone height} - \text{subsided height} = 300 - 235 = \mathbf{65\ \text{mm}}

A slump of 65 mm falls in the medium workability range (typically 50–100 mm).

Suitable application: normal reinforced concrete work — beams, slabs, columns, and footings placed with conventional vibration. (Very low/low slump suits pavements and mass concrete; high slump suits heavily reinforced/congested sections.)

fresh-concreteworkabilityslump-test
8short5 marks

(a) Explain the phenomenon of bulking of sand and its practical significance in volume batching. (2 marks)

(b) The dry rodded unit weight of a fine aggregate is 1600 kg/m31600\ \text{kg/m}^3. When sand is at a certain moisture content its loose volume increases by 28% due to bulking. If 1.0 m31.0\ \text{m}^3 of dry sand is required, what loose (bulked) volume of the moist sand must be batched to deliver the same mass of dry sand? (3 marks)

(a) Bulking of sand

Bulking is the increase in the apparent (loose) volume of fine aggregate when it contains moisture. A thin film of water around each sand particle pushes the particles apart through surface tension, increasing the bulk volume. Bulking increases with moisture content up to about 4–6% (where it can reach 20–40%) and then decreases, becoming zero when the sand is fully saturated/inundated. Significance: In volume batching, moist sand occupies more volume for the same mass, so if not corrected the mix gets less actual sand than intended, making it richer in cement and prone to incorrect proportions. The sand volume must be increased to compensate, or batching by mass should be preferred.

(b) Volume to be batched

Bulking increases the loose volume by 28%. To still place the mass equivalent of 1.0 m³ of dry sand, the moist sand must be batched at the bulked volume:

Vbulked=Vdry×(1+bulking fraction)=1.0×(1+0.28)=1.28 m3V_{bulked} = V_{dry}\times(1+\text{bulking fraction}) = 1.0\times(1+0.28) = \mathbf{1.28\ \text{m}^3}

Check via mass: mass of dry sand required = 1.0×1600=1600 kg1.0\times 1600 = 1600\ \text{kg}. Batching 1.28 m³ of the bulked sand (whose loose unit weight has fallen to 1600/1.28=1250 kg/m31600/1.28 = 1250\ \text{kg/m}^3) delivers 1.28×1250=1600 kg1.28\times 1250 = 1600\ \text{kg} of dry sand. Confirmed.

aggregateconcrete-ingredientsbulking-of-sand
9short5 marks

(a) State the desirable properties of good building bricks. (2 marks)

(b) Explain, with neat sketches described in words, the difference between English bond and Flemish bond in brick masonry, and state which is generally stronger. (3 marks)

(a) Properties of good building bricks

  • Uniform size, shape, and sharp, square edges; uniform deep red/copper colour.
  • Adequate compressive strength (commonly ≥ 3.5 N/mm² for common burnt-clay bricks; higher for structural grades).
  • Low water absorption (preferably ≤ 20% of dry weight by 24-hour immersion).
  • Hard, well-burnt, sound (gives a clear metallic ringing sound when struck); free of cracks, flaws, lumps, and nodules of lime.
  • Low efflorescence (soluble salts), good durability, weather and fire resistance, and good thermal insulation.

(b) English bond vs. Flemish bond

English bond: Consists of alternate courses of headers and stretchers. One course shows only stretchers (long faces), the next shows only headers (short faces). Continuity of vertical joints is broken using queen closers placed next to the quoin header. It provides excellent overlap and breaking of vertical joints through the wall thickness.

Course 1 (stretchers): [____][____][____][____]
Course 2 (headers):    [_][_][_][_][_][_][_][_]
Course 3 (stretchers): [____][____][____][____]

Flemish bond: Each course shows headers and stretchers laid alternately in the same course, with each header centred over the stretcher below. It gives a more pleasing, symmetrical appearance and uses fewer facing bricks, but has more continuous (less broken) internal joints.

Every course: [____][_][____][_][____][_]
              (stretcher-header-stretcher-header...)

Stronger bond: English bond is generally stronger and is preferred for load-bearing/structural walls, because it provides better breaking of vertical joints and superior transverse strength; Flemish bond is chosen mainly for its better appearance and economy.

masonry-materialsbricksbond
10short5 marks

(a) Define curing of concrete and list four common methods of curing. (2 marks)

(b) Distinguish between drying shrinkage and creep of concrete, and state two factors that increase creep. (3 marks)

(a) Curing

Curing is the process of maintaining adequate moisture and a favourable temperature in concrete for a sufficient period after placing, so that hydration of cement continues and the concrete develops its design strength, durability, and impermeability while controlling shrinkage cracking.

Common methods (any four):

  1. Water/ponding (immersion) for slabs and pavements.
  2. Wet covering — gunny bags, hessian, sand, or straw kept continuously moist.
  3. Sprinkling / fog spraying.
  4. Membrane curing using curing compounds or polythene sheeting (sealing moisture in).
  5. Steam curing / accelerated curing (for precast elements).

(b) Drying shrinkage vs. creep

AspectDrying shrinkageCreep
CauseLoss of adsorbed/capillary water to the atmosphereSustained (constant) applied stress over time
Load dependenceIndependent of external loadOccurs only under sustained load
TimeContinues as moisture is lostContinues with time under load (much partly recoverable on unloading)
NatureVolumetric contractionTime-dependent increase in strain at constant stress

Two factors that increase creep: (i) higher sustained stress level and higher w/c ratio / lower-strength concrete; (ii) loading the concrete at an early age (less mature), low ambient humidity, and higher temperature. (Higher cement paste content and finer/poorer aggregate stiffness also increase creep.)

curingcreep-shrinkagehardened-concrete
11short3 marks

In an Ultrasonic Pulse Velocity (UPV) test on a concrete column, the transit time of the ultrasonic pulse over a path length of 300 mm was measured as 68 μs68\ \mu s (direct transmission). Compute the pulse velocity in km/s and comment on the quality of the concrete using the guideline: above 4.5 km/s = excellent, 3.5–4.5 km/s = good, 3.0–3.5 km/s = medium (doubtful), below 3.0 km/s = poor.

Pulse velocity

V=path lengthtransit time=300 mm68 μs=0.300 m68×106 sV = \frac{\text{path length}}{\text{transit time}} = \frac{300\ \text{mm}}{68\ \mu s} = \frac{0.300\ \text{m}}{68\times 10^{-6}\ \text{s}} V=4411.8 m/s=4.41 km/sV = 4411.8\ \text{m/s} = \mathbf{4.41\ \text{km/s}}

Comment: 4.41 km/s lies in the 3.5–4.5 km/s band, so the concrete quality is rated GOOD (sound, well-compacted concrete, just below the 'excellent' threshold).

ndtultrasonic-pulse-velocityquality-control

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