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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long10 marks

Design a concrete mix for an M30 grade reinforced concrete column using the IS 10262 (concrete mix proportioning) method for the following data:

  • Grade designation: M30, OPC 43 grade, specific gravity of cement = 3.15
  • Maximum size of aggregate (angular coarse aggregate) = 20 mm
  • Workability: slump = 75 mm; exposure = moderate
  • Specific gravity of coarse aggregate = 2.70; of fine aggregate = 2.65
  • Fine aggregate conforms to Zone II
  • Standard deviation s=5.0 N/mm2s = 5.0\ \text{N/mm}^2
  • Assume the mix is to be made without any chemical admixture.

Determine the target mean strength, the free water–cement ratio, and the final mix proportions by mass (cement : water : fine aggregate : coarse aggregate) per cubic metre of concrete. Use a maximum free water content of 186 kg/m3186\ \text{kg/m}^3 for 20 mm aggregate at 50 mm slump and the standard correction of +3% water per 25 mm increase in slump. Take the volume of coarse aggregate per unit volume of total aggregate (Zone II, 20 mm) as 0.62.

Step 1 — Target mean strength fckf_{ck}'

fck=fck+1.65s=30+1.65×5.0=30+8.25=38.25 N/mm2f_{ck}' = f_{ck} + 1.65\,s = 30 + 1.65 \times 5.0 = 30 + 8.25 = \mathbf{38.25\ \text{N/mm}^2}

Step 2 — Free water–cement ratio (w/c)

For M30 and moderate exposure, from the strength–(w/c) relationship for 43 grade OPC, a free w/c of about 0.45 gives the required target mean strength of 38.25 N/mm². The maximum permissible w/c for moderate exposure (IS 456) is 0.50, so adopt w/c = 0.45 (0.45 < 0.50, OK).

Step 3 — Free water content

Base water for 20 mm aggregate at 50 mm slump = 186 kg/m³.

Slump correction: required slump = 75 mm, i.e. (75 − 50)/25 = 1 step of 25 mm +3%\Rightarrow +3\%.

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

Step 4 — Cement content

C=Ww/c=191.60.45=425.7 kg/m3C = \frac{W}{w/c} = \frac{191.6}{0.45} = \mathbf{425.7\ \text{kg/m}^3}

Minimum cement for moderate exposure = 300 kg/m³ < 425.7 kg/m³ → OK.

Step 5 — Proportion of volume of coarse and fine aggregate

Volume of coarse aggregate per unit volume of total aggregate (Zone II, 20 mm) = 0.62.

For a pumpable/RCC column the table value may be retained as 0.62 (no adjustment for w/c here since w/c ≈ 0.45 is close to the reference 0.50; a small +0.01 correction is within tolerance and is neglected).

VCA=0.62,VFA=10.62=0.38V_{CA} = 0.62, \qquad V_{FA} = 1 - 0.62 = 0.38

Step 6 — Mix calculation by absolute volume (per 1 m³)

Total volume = 1 m³; assume entrapped air = 2% = 0.02 m³ for 20 mm aggregate.

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=191.61000=0.1916 m3V_w = \frac{191.6}{1000} = 0.1916\ \text{m}^3

Volume of all aggregate (solid):

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

Step 7 — Mass of coarse and fine aggregate

Coarse aggregate:

CA=Vagg×VCA×GCA×1000=0.6533×0.62×2.70×1000=1093.6 kg/m3CA = V_{agg} \times V_{CA} \times G_{CA} \times 1000 = 0.6533 \times 0.62 \times 2.70 \times 1000 = \mathbf{1093.6\ \text{kg/m}^3}

Fine aggregate:

FA=Vagg×VFA×GFA×1000=0.6533×0.38×2.65×1000=657.8 kg/m3FA = V_{agg} \times V_{FA} \times G_{FA} \times 1000 = 0.6533 \times 0.38 \times 2.65 \times 1000 = \mathbf{657.8\ \text{kg/m}^3}

Step 8 — Final mix proportions (per m³, by mass)

IngredientMass (kg/m³)Ratio (by mass)
Cement425.71.00
Water191.60.45
Fine aggregate657.81.55
Coarse aggregate1093.62.57

Final mix proportion (C : FA : CA) = 1 : 1.55 : 2.57, w/c = 0.45.

Trial batches must follow to confirm workability and 28-day strength.

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

A solid load-bearing brick masonry wall in a single-storey building is 230 mm thick and 3.0 m clear in height between lateral supports (slab at top and floor at bottom, both providing effective lateral restraint). The wall carries an axial vertical load (including self-weight) of 90 kN per metre run. The masonry is built with bricks of crushing strength 10 N/mm² in cement mortar of grade M1 (1:6). Using IS 1905 (code of practice for structural use of unreinforced masonry):

(a) Determine the slenderness ratio of the wall and the corresponding stress reduction factor ksk_s. (b) Taking the basic permissible compressive stress fbf_b from the relation for brick strength 10 N/mm² with M1 mortar as 0.96 N/mm20.96\ \text{N/mm}^2, and an area reduction factor ka=1.0k_a = 1.0 (wall area > 0.2 m²), check whether the wall is safe in axial compression. Use an eccentricity factor kp=1.0k_p = 1.0 (axial load).

Given: thickness t=230t = 230 mm, height H=3.0H = 3.0 m, axial load P=90P = 90 kN/m, fb=0.96f_b = 0.96 N/mm².

Step 1 — Effective height and effective thickness

The wall is laterally supported top and bottom by floor/roof: effective height heff=0.75Hh_{eff} = 0.75H for full restraint (slab continuous), but for a slab simply resting, IS 1905 gives heff=1.0Hh_{eff} = 1.0H. Taking the common design value of full restraint:

heff=0.75×3.0=2.25 m=2250 mmh_{eff} = 0.75 \times 3.0 = 2.25\ \text{m} = 2250\ \text{mm} teff=t=230 mm (solid wall)t_{eff} = t = 230\ \text{mm (solid wall)}

Step 2 — Slenderness ratio (SR)

SR=heffteff=2250230=9.78SR = \frac{h_{eff}}{t_{eff}} = \frac{2250}{230} = \mathbf{9.78}

This is well below the maximum of 27 for load-bearing walls → OK.

Step 3 — Stress reduction factor ksk_s

From IS 1905 Table (SR vs ksk_s, eccentricity e/t = 0):

  • SR = 8 → ks=1.00k_s = 1.00
  • SR = 10 → ks=0.97k_s = 0.97

Interpolating at SR = 9.78:

ks=1.00(9.788)(108)(1.000.97)=1.001.782(0.03)=1.000.0267=0.973k_s = 1.00 - \frac{(9.78-8)}{(10-8)}(1.00-0.97) = 1.00 - \frac{1.78}{2}(0.03) = 1.00 - 0.0267 = \mathbf{0.973}

Step 4 — Permissible compressive stress

fperm=fb×ks×ka×kp=0.96×0.973×1.0×1.0=0.934 N/mm2f_{perm} = f_b \times k_s \times k_a \times k_p = 0.96 \times 0.973 \times 1.0 \times 1.0 = \mathbf{0.934\ \text{N/mm}^2}

Step 5 — Actual stress in the wall

Per metre run, area of cross-section A=1000×230=230000 mm2A = 1000 \times 230 = 230000\ \text{mm}^2.

factual=PA=90×103 N230000 mm2=0.391 N/mm2f_{actual} = \frac{P}{A} = \frac{90 \times 10^3\ \text{N}}{230000\ \text{mm}^2} = \mathbf{0.391\ \text{N/mm}^2}

Step 6 — Safety check

factual=0.391 N/mm2<fperm=0.934 N/mm2f_{actual} = 0.391\ \text{N/mm}^2 < f_{perm} = 0.934\ \text{N/mm}^2

The wall is SAFE in axial compression. Utilisation = 0.391/0.934 ≈ 42 %, so the section has a large reserve; thickness could be optimised but is governed by minimum-thickness/practical requirements.

masonry-wallload-bearingslenderness-ratio
3long10 marks

Explain the factors affecting the compressive strength of hardened concrete. In a laboratory, three standard 150 mm cubes of the same M25 mix were tested in compression at 28 days and failed at loads of 565 kN, 590 kN and 548 kN respectively.

(a) Compute the individual cube strengths, the mean strength, and assess acceptance per the IS 456 single-sample criteria (individual cube not less than fck4f_{ck} - 4 and mean not less than fck+0.825sf_{ck} + 0.825 s, taking assumed s=4 N/mm2s = 4\ \text{N/mm}^2). (b) A 150 mm cube gave 30 N/mm². Estimate the equivalent 150 mm × 300 mm cylinder strength using the typical cylinder/cube ratio of 0.80.

Part 0 — Factors affecting compressive strength of hardened concrete

  1. Water–cement ratio — the single most important factor; strength falls as w/c rises (Abrams' law).
  2. Degree of compaction — voids drastically reduce strength (≈5–6 % strength loss per 1 % voids).
  3. Cement type and content; quality of aggregate (shape, texture, grading, strength of parent rock).
  4. Curing (moisture and temperature) and age — strength increases with continued hydration.
  5. Admixtures, transition-zone (ITZ) quality, and testing variables (specimen shape/size, rate of loading, capping, moisture state).

Part (a) — Cube strengths and acceptance

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

f1=565×10322500=25.11 N/mm2f_1 = \frac{565\times10^3}{22500} = 25.11\ \text{N/mm}^2 f2=590×10322500=26.22 N/mm2f_2 = \frac{590\times10^3}{22500} = 26.22\ \text{N/mm}^2 f3=548×10322500=24.36 N/mm2f_3 = \frac{548\times10^3}{22500} = 24.36\ \text{N/mm}^2

Mean strength:

fˉ=25.11+26.22+24.363=75.693=25.23 N/mm2\bar f = \frac{25.11 + 26.22 + 24.36}{3} = \frac{75.69}{3} = \mathbf{25.23\ \text{N/mm}^2}

Acceptance check (M25, fck=25f_{ck}=25):

  • Individual minimum: fck4=254=21 N/mm2f_{ck} - 4 = 25 - 4 = 21\ \text{N/mm}^2. Lowest cube = 24.36 N/mm² > 21 → OK.
  • Mean minimum: fck+0.825s=25+0.825×4=25+3.30=28.30 N/mm2f_{ck} + 0.825 s = 25 + 0.825\times4 = 25 + 3.30 = 28.30\ \text{N/mm}^2.

Mean = 25.23 N/mm² < 28.30 N/mm² → the mean criterion is NOT satisfied.

Conclusion: Although every individual result exceeds the lower limit, the sample mean falls short of fck+0.825sf_{ck}+0.825s. The batch does not meet the statistical acceptance criterion and the concrete in question is liable to rejection / further investigation (core tests, load test).

Part (b) — Cylinder strength estimate

Using cylinder/cube ratio ≈ 0.80:

fcyl=0.80×30=24 N/mm2f_{cyl} = 0.80 \times 30 = \mathbf{24\ \text{N/mm}^2}

The cylinder strength is lower because of its greater height-to-diameter ratio (less platen confinement / end-restraint effect than a cube).

hardened-concretecompressive-strengthtesting
4long8 marks

(a) Classify chemical admixtures used in concrete and describe the action and a typical use of any three classes. (b) What is a Superplasticizer (high-range water reducer)? Explain how it can be used either to increase workability at constant w/c or to increase strength at constant workability, and name the two main chemical families. (c) Define 'durability' of concrete and list four mechanisms that attack concrete durability with one preventive measure each.

(a) Classification of chemical admixtures (IS 9103 / ASTM C494 types):

  1. Water-reducing (plasticizers, Type A): disperse cement flocs (electrostatic/steric repulsion), giving the same workability at lower water → higher strength and durability. Use: general structural concrete.
  2. Retarders (Type B): delay setting (e.g., sugar, lignosulphonates) by slowing C₃A/C₃S hydration. Use: hot-weather concreting, large pours, ready-mix transport.
  3. Accelerators (Type C): speed up setting/early strength (e.g., CaCl₂ for plain concrete, non-chloride for RCC). Use: cold-weather, repair, early formwork removal.
  4. Air-entraining agents: create stable micro air bubbles for freeze–thaw resistance and workability.
  5. Superplasticizers (Type F/G): high-range water reducers (15–30 % water reduction).

(Any three classes described earns full marks for this part.)

(b) Superplasticizer (high-range water reducer)

A superplasticizer is an admixture that produces a large reduction in mixing water (15–30 %) without loss of workability, by strong dispersion of cement particles.

  • Constant w/c → more workability: keep water and cement fixed, add SP → flowing/self-levelling concrete (high slump) for congested reinforcement or pumping.
  • Constant workability → more strength: keep slump fixed but cut water (and keep cement) → lower w/c → denser paste → higher strength and durability.

Two main chemical families: (i) Sulphonated naphthalene formaldehyde (SNF) and sulphonated melamine formaldehyde (SMF) condensates (older, electrostatic action); (ii) Polycarboxylic ether (PCE) polymers (modern, steric + electrostatic, longer slump retention).

(c) Durability

Durability is the ability of concrete to resist weathering action, chemical attack and abrasion while retaining its desired engineering properties (serviceability) over its design life with minimum maintenance.

Attack mechanismPreventive measure
Sulphate attack (ettringite/gypsum expansion)Use sulphate-resisting cement (SRC) / low-C₃A cement, low w/c
Chloride-induced reinforcement corrosionLow w/c, adequate cover, corrosion-inhibiting admixture, dense impermeable concrete
Carbonation (loss of passivity)Low permeability, adequate cover, good curing
Alkali–silica reaction (ASR)Use low-alkali cement, non-reactive aggregate, fly ash/GGBS
Freeze–thawAir entrainment, low w/c

(Any four mechanisms with measures suffice.)

admixturesspecial-concretedurability
5long8 marks

A short, square brick masonry column carries a purely axial working load of 320 kN. The column is built with bricks of crushing strength 12.5 N/mm² in cement-lime mortar of grade M2, for which the basic permissible compressive stress is fb=1.10 N/mm2f_b = 1.10\ \text{N/mm}^2. The column effective height is 2.4 m. Design the plan size of the column (square, in multiples of half-brick i.e. 115 mm modules giving sizes like 230, 345, 460 mm). Apply the area reduction factor ka=0.7+1.5Ak_a = 0.7 + 1.5A (with AA in m²) when the horizontal cross-sectional area is less than 0.2 m², and check slenderness using stress reduction factor ksk_s.

Given: P=320P = 320 kN, fb=1.10f_b = 1.10 N/mm², heff=2.4h_{eff} = 2.4 m.

Step 1 — First trial size (ignore kak_a, ksk_s)

Required area ≈ P/fb=320×1031.10=290909 mm2=0.291 m2P/f_b = \dfrac{320\times10^3}{1.10} = 290909\ \text{mm}^2 = 0.291\ \text{m}^2.

Try a square of side 460 mm (4 half-bricks): A=460×460=211600 mm2=0.2116 m2A = 460\times460 = 211600\ \text{mm}^2 = 0.2116\ \text{m}^2. Side 345 mm gives 0.119 m² (too small once factors applied). Start the check with 460 mm × 460 mm.

Step 2 — Slenderness ratio and ksk_s

SR=hefft=2400460=5.22  (<6)SR = \frac{h_{eff}}{t} = \frac{2400}{460} = 5.22 \;(<6)

For SR ≤ 6, ks=1.0k_s = 1.0 (no slenderness reduction).

Step 3 — Area reduction factor kak_a

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

Step 4 — Permissible stress and capacity

fperm=fbkska=1.10×1.0×1.0=1.10 N/mm2f_{perm} = f_b \, k_s \, k_a = 1.10 \times 1.0 \times 1.0 = 1.10\ \text{N/mm}^2 Pcap=fperm×A=1.10×211600=232760 N=232.8 kNP_{cap} = f_{perm}\times A = 1.10 \times 211600 = 232760\ \text{N} = 232.8\ \text{kN}

Pcap=232.8P_{cap} = 232.8 kN < 320 kN → unsafe. Increase the section.

Step 5 — Next trial: 575 mm × 575 mm (5 half-bricks)

A=575×575=330625 mm2=0.3306 m2  (>0.2 m2ka=1.0)A = 575\times575 = 330625\ \text{mm}^2 = 0.3306\ \text{m}^2 \;(>0.2\ \text{m}^2 \Rightarrow k_a=1.0) SR=2400575=4.17<6ks=1.0SR = \frac{2400}{575} = 4.17 < 6 \Rightarrow k_s = 1.0 Pcap=1.10×330625=363688 N=363.7 kNP_{cap} = 1.10 \times 330625 = 363688\ \text{N} = \mathbf{363.7\ \text{kN}}

Pcap=363.7P_{cap} = 363.7 kN > 320 kN → SAFE.

Step 6 — Result

Adopt a 575 mm × 575 mm square masonry column. Utilisation = 320/363.7 ≈ 88 %, an efficient design.

(Note: the ka=0.7+1.5Ak_a = 0.7 + 1.5A rule is shown for completeness; here every adequate trial has A>0.2 m2A>0.2\ \text{m}^2, so kak_a stays 1.0. A 460 mm column would only reach kak_a-territory if A<0.2 m2A<0.2\ \text{m}^2, which it is not.)

masonry-columnaxial-loaddesign
B

Section B: Short Answer Questions

Attempt all questions.

6 questions
6short5 marks

(a) Name the four major Bogue compounds of Portland cement and state which one is mainly responsible for early strength and which for ultimate strength. (b) Define fineness of cement and explain briefly how increased fineness affects hydration, strength gain and heat of hydration.

(a) Bogue compounds (with abbreviations):

CompoundFormulaAbbrev.Role
Tricalcium silicate3CaO·SiO₂C₃S (alite)Early strength (up to ~28 days), main strength contributor; high heat
Dicalcium silicate2CaO·SiO₂C₂S (belite)Later/ultimate strength (beyond 28 days); low heat
Tricalcium aluminate3CaO·Al₂O₃C₃AFlash set (controlled by gypsum); very high heat; poor sulphate resistance
Tetracalcium aluminoferrite4CaO·Al₂O₃·Fe₂O₃C₄AFLittle strength; gives grey colour
  • Early strength → C₃S.
  • Ultimate (long-term) strength → C₂S.

(b) Fineness of cement

Fineness is the measure of the particle size / specific surface area of cement (expressed as surface area in m²/kg by the Blaine air-permeability test, or as residue on a 90-micron sieve).

Effect of increased fineness:

  • Faster hydration — more surface area exposed to water → quicker reaction.
  • Higher early strength and better workability/cohesion of paste.
  • Greater (and faster) heat of hydration, increasing risk of thermal/shrinkage cracking in mass concrete.
  • Increased water demand and somewhat greater drying shrinkage; also more prone to deterioration if stored (rapid aeration).
cementfinenesshydration
7short5 marks

(a) Define workability of fresh concrete and name any three tests used to measure it, stating the workability range each is suitable for. (b) In a compacting factor test, the partially compacted concrete weighed 9.8 kg and the fully compacted concrete (same mould) weighed 11.2 kg. Compute the compacting factor and comment on the workability.

(a) Workability

Workability is the property of freshly mixed concrete that determines the ease and homogeneity with which it can be mixed, placed, compacted and finished without segregation or bleeding.

TestSuitable workability range
Slump testMedium to high (slump 25–150 mm); not for very dry mixes
Compacting factor testLow to medium (dry/lean mixes where slump is near zero)
Vee-Bee consistometerVery low / very dry, stiff mixes
(Flow table)High to very high (flowing concrete)

(Any three with ranges earn full marks.)

(b) Compacting factor calculation

Compacting factor=weight of partially compacted concreteweight of fully compacted concrete=9.811.2=0.875\text{Compacting factor} = \frac{\text{weight of partially compacted concrete}}{\text{weight of fully compacted concrete}} = \frac{9.8}{11.2} = \mathbf{0.875}

Comment: A compacting factor of about 0.85–0.92 corresponds to medium workability. The value 0.875 indicates a mix of medium workability suitable for normally reinforced sections compacted by vibration. (Low workability ≈ 0.78; high ≈ 0.95.)

fresh-concreteworkabilityslump-test
8short4 marks

Briefly describe two non-destructive testing (NDT) methods for in-situ assessment of concrete: the rebound (Schmidt) hammer and the ultrasonic pulse velocity (UPV) test. For UPV, a pulse travels a path length of 300 mm through a concrete element in 75 microseconds; compute the pulse velocity in km/s and comment on the concrete quality using the standard grading.

Rebound (Schmidt) hammer test

A spring-loaded mass is released against a plunger held on the concrete surface; the rebound number (R) depends on the surface hardness, which correlates with compressive strength. It is quick and cheap but assesses only the surface layer and needs calibration; results are affected by surface finish, moisture, carbonation and aggregate type. Used for uniformity assessment and comparative strength estimation.

Ultrasonic pulse velocity (UPV) test

An ultrasonic pulse (≈ 50–60 kHz) is sent through the concrete between a transmitter and receiver; the transit time is measured and velocity = path length / time. Higher velocity → denser, sounder, more homogeneous concrete; it detects voids, cracks and assesses uniformity through the full depth of the member.

Calculation

V=Lt=0.300 m75×106 s=4000 m/s=4.0 km/sV = \frac{L}{t} = \frac{0.300\ \text{m}}{75 \times 10^{-6}\ \text{s}} = 4000\ \text{m/s} = \mathbf{4.0\ \text{km/s}}

Quality grading (IS 13311 Part 1):

Pulse velocity (km/s)Concrete quality
> 4.5Excellent
3.5 – 4.5Good
3.0 – 3.5Medium / doubtful
< 3.0Poor

Since V=4.0V = 4.0 km/s falls in the 3.5–4.5 range, the concrete quality is GOOD.

ndtrebound-hammerupv
9short5 marks

(a) Explain the phenomenon of 'bulking of sand' and its practical importance in volume batching. (b) Dry sand has a bulk density of 1600 kg/m31600\ \text{kg/m}^3. At a certain moisture content the sand bulks so that its loose volume increases by 28%. A mix calls for 0.40 m³ of sand measured loose (in the bulked condition). Determine the equivalent mass of dry sand actually supplied, and the percentage error if the bulking were ignored.

(a) Bulking of sand

Bulking of sand is the increase in the apparent (loose) volume of sand caused by surface moisture. Films of water around the sand particles keep them apart by surface tension, pushing the grains apart and increasing voids; the volume can increase by 20–40 % at about 4–6 % moisture. Beyond saturation (sand fully wet/inundated), the films merge and the volume returns to nearly the dry value.

Practical importance: In volume batching, a fixed measuring box of damp sand actually contains less solid sand than dry sand of the same volume. Ignoring bulking leads to under-batching of sand (a richer, sandy-deficient mix), poorer grading and inconsistent strength. Hence either an allowance for bulking is made (extra volume added) or, better, sand is batched by mass.

(b) Calculation

Dry-sand bulk density = 1600 kg/m³; bulking increases loose volume by 28 %.

The 0.40 m³ measured is the bulked (loose, damp) volume. The equivalent dry loose volume is:

Vdry=Vbulked1+0.28=0.401.28=0.3125 m3V_{dry} = \frac{V_{bulked}}{1 + 0.28} = \frac{0.40}{1.28} = 0.3125\ \text{m}^3

Mass of dry sand actually supplied:

m=Vdry×ρdry=0.3125×1600=500 kgm = V_{dry} \times \rho_{dry} = 0.3125 \times 1600 = \mathbf{500\ \text{kg}}

If bulking were ignored, one would assume 0.40 m³ of dry sand:

massumed=0.40×1600=640 kgm_{assumed} = 0.40 \times 1600 = 640\ \text{kg}

But only 500 kg is really present. Shortfall = 640 − 500 = 140 kg.

Error=640500640×100=140640×100=21.9%\text{Error} = \frac{640 - 500}{640}\times 100 = \frac{140}{640}\times100 = \mathbf{21.9\%}

Ignoring bulking would over-estimate the sand by about 22 %, i.e. the mix would actually receive ~22 % less sand than intended.

aggregatebulking-of-sandmoisture
10short4 marks

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

(i) Fibre-Reinforced Concrete (FRC)

  • Composition: ordinary concrete with discrete short fibres (steel, polypropylene, glass, or natural) dispersed uniformly, typically 0.5–2 % by volume.
  • Key property: greatly improved post-cracking ductility / toughness and resistance to cracking; fibres bridge cracks and arrest their propagation, improving tensile/flexural behaviour and impact/fatigue resistance.
  • Application: industrial floors and pavements, tunnel linings (shotcrete), bridge decks, precast elements.

(ii) Self-Compacting Concrete (SCC)

  • Composition: highly flowable mix with high-range water reducer (PCE superplasticizer), high fines/powder content and a viscosity-modifying agent; balanced to resist segregation.
  • Key property: flows and consolidates under its own weight without vibration, completely filling formwork and encapsulating reinforcement (high filling and passing ability). Tested by slump-flow / V-funnel / L-box.
  • Application: heavily congested reinforcement, complex/inaccessible formwork, architectural concrete, precast.

(iii) High-Performance Concrete (HPC)

  • Composition: low w/c (often <0.4), superplasticizer and supplementary cementitious materials (silica fume, fly ash, GGBS) for a dense microstructure.
  • Key property: high strength and superior durability / low permeability (and good workability) — performance criteria beyond ordinary concrete.
  • Application: high-rise columns, long-span bridges, marine and aggressive-environment structures.

(Any two of the above answered fully earn the marks.)

special-concretefibre-reinforcedself-compacting
11short6 marks

(a) State the functions of mortar in masonry and list the common grades of cement mortar (M1, M2, M3) with their nominal cement:sand proportions. (b) Explain how the strength of a masonry assemblage (prism) depends on the relative strength of the brick and the mortar, and why a mortar that is much stronger than the brick is not desirable. (c) A brick masonry prism is built with bricks of crushing strength 15 N/mm². Using the empirical relation that prism (masonry) compressive strength is approximately fm0.25fbfmof_m \approx 0.25\,\sqrt{f_b \cdot f_{mo}} where fbf_b is brick strength and fmof_{mo} the mortar cube strength (both in N/mm²), estimate the masonry strength if a 1:6 mortar of cube strength 3 N/mm² is used.

(a) Functions of mortar and grades

Functions of mortar:

  • Binds the masonry units into a monolithic whole and transmits/distributes loads uniformly between units.
  • Fills the joints, making the wall weather-tight and preventing air/water penetration.
  • Accommodates dimensional irregularities of units and provides a uniform bedding so stress concentrations are avoided.
  • Provides bond for any reinforcement and contributes to appearance (pointing).

Common cement-mortar grades (nominal cement:sand by volume):

GradeCement : SandTypical use
M11 : 6General load-bearing masonry, moderate load
M21 : 5 (or 1:4)Higher load-bearing masonry
M31 : 3 (or 1:4)Heavily loaded / important masonry

(Proportions vary slightly between codes; the trend richer M1→M3 is the key point.)

(b) Brick–mortar interaction in prism strength

Under vertical load the mortar joint tends to expand laterally more than the stiffer brick (Poisson effect). Because the brick restrains the mortar, the mortar is put into a triaxial (confined) compression while the brick is put into lateral tension. Failure of masonry is therefore usually by vertical splitting (tension) cracking of the brick, not crushing of the mortar.

Consequently:

  • Masonry strength is always less than the brick crushing strength and increases only slowly with mortar strength.
  • A mortar much stronger (and stiffer) than the brick is not desirable: it gives little extra prism strength (brick still governs), is uneconomical, and a rigid mortar causes brittle cracking and poor accommodation of movement/settlement. A slightly weaker, more workable mortar yields better bond, ductility and crack distribution. Hence "the mortar should be no stronger than necessary."

(c) Estimate of masonry strength

fm0.25fbfmo=0.2515×3=0.2545f_m \approx 0.25\sqrt{f_b \cdot f_{mo}} = 0.25\sqrt{15 \times 3} = 0.25\sqrt{45} 45=6.708\sqrt{45} = 6.708 fm0.25×6.708=1.68 N/mm2f_m \approx 0.25 \times 6.708 = \mathbf{1.68\ \text{N/mm}^2}

The estimated masonry (prism) compressive strength is about 1.68 N/mm², i.e. only ~11 % of the brick crushing strength — confirming that masonry strength is governed largely by the brick and the joint interaction, not the mortar alone.

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