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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long12 marks

Design a concrete mix by the IS 10262 method for the following requirements and report the final mix proportions by mass and the quantities per cubic metre.

  • Characteristic compressive strength at 28 days, fck=30 MPaf_{ck} = 30\ \text{MPa} (M30)
  • Maximum nominal size of aggregate = 20 mm
  • Workability = 75 mm slump
  • Exposure = moderate (max free w/c = 0.50, min cement = 300 kg/m³)
  • Specific gravity: cement = 3.15, coarse aggregate = 2.70, fine aggregate = 2.65
  • Both aggregates in saturated-surface-dry condition
  • Standard deviation s=5.0 MPas = 5.0\ \text{MPa}
  • Adopt the free w/c ratio = 0.45 (governed by strength)
  • Water content for 20 mm aggregate at 50 mm slump (zone II sand) = 186 kg/m³; increase 3% per 25 mm extra slump. No admixture.
  • For w/c = 0.50 and zone II sand, volume of coarse aggregate per unit volume of total aggregate = 0.62. Adjust by +0.01 for every 0.05 decrease in w/c.

Determine target mean strength, cement content, water content, and the final batch quantities of cement, water, fine and coarse aggregate per m³ (assume entrapped air = 2%).

Step 1 — Target mean strength

fck=fck+1.65s=30+1.65×5.0=30+8.25=38.25 MPaf_{ck}' = f_{ck} + 1.65\,s = 30 + 1.65 \times 5.0 = 30 + 8.25 = 38.25\ \text{MPa}

Step 2 — Water/cement ratio

Strength requirement gives w/c = 0.45. Durability limit for moderate exposure is 0.50. Adopt the lower: w/c = 0.45.

Step 3 — Water content

Base = 186 kg/m³ at 50 mm slump. Required slump 75 mm → 25 mm extra → +3%.

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

Step 4 — Cement content

C=W(w/c)=191.60.45=425.8 kg/m3C = \frac{W}{(w/c)} = \frac{191.6}{0.45} = 425.8\ \text{kg/m}^3

Check minimum cement (300 kg/m³): 425.8 > 300 → OK. Adopt C = 425.8 kg/m³.

Step 5 — Coarse aggregate volume fraction

Table value at w/c = 0.50 is 0.62. w/c reduced by 0.05 → increase by 0.01.

Vol. fraction of CA=0.62+0.01=0.63\text{Vol. fraction of CA} = 0.62 + 0.01 = 0.63 Vol. fraction of FA=10.63=0.37\text{Vol. fraction of FA} = 1 - 0.63 = 0.37

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

Total volume = 1 m³. Air = 2% = 0.02 m³.

Volume of cement:

Vc=C3.15×1000=425.83150=0.13518 m3V_c = \frac{C}{3.15 \times 1000} = \frac{425.8}{3150} = 0.13518\ \text{m}^3

Volume of water:

Vw=191.61000=0.19160 m3V_w = \frac{191.6}{1000} = 0.19160\ \text{m}^3

Volume of all aggregate:

Vagg=1(Vc+Vw+Vair)=1(0.13518+0.19160+0.02)=10.34678=0.65322 m3V_{agg} = 1 - (V_c + V_w + V_{air}) = 1 - (0.13518 + 0.19160 + 0.02) = 1 - 0.34678 = 0.65322\ \text{m}^3

Step 7 — Aggregate masses

Coarse aggregate:

MCA=Vagg×0.63×2.70×1000=0.65322×0.63×2700=1111.2 kg/m3M_{CA} = V_{agg} \times 0.63 \times 2.70 \times 1000 = 0.65322 \times 0.63 \times 2700 = 1111.2\ \text{kg/m}^3

Fine aggregate:

MFA=Vagg×0.37×2.65×1000=0.65322×0.37×2650=640.4 kg/m3M_{FA} = V_{agg} \times 0.37 \times 2.65 \times 1000 = 0.65322 \times 0.37 \times 2650 = 640.4\ \text{kg/m}^3

Step 8 — Final mix per m³

MaterialQuantity (kg/m³)
Cement425.8
Water191.6
Fine aggregate640.4
Coarse aggregate1111.2

Mix ratio by mass (cement : FA : CA) = 425.8:640.4:1111.2425.8 : 640.4 : 1111.2 = 1 : 1.50 : 2.61, with w/c = 0.45.

Quick density check: 425.8+191.6+640.4+1111.2=2369.0 kg/m3425.8 + 191.6 + 640.4 + 1111.2 = 2369.0\ \text{kg/m}^3, a reasonable fresh density for normal-weight concrete.

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

Define workability of fresh concrete and explain the factors affecting it. Describe with neat sketches any TWO field/laboratory tests used to measure workability, clearly stating the range of workability each is suitable for. Distinguish between segregation and bleeding and state two preventive measures for each.

Workability — definition

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 loss of homogeneity. It is essentially the amount of internal work required to fully compact the concrete to maximum density.

Factors affecting workability

  1. Water content — most important; more water → more fluidity (but lowers strength/durability).
  2. Aggregate properties — shape (rounded > angular), texture (smooth > rough), maximum size (larger size reduces water demand).
  3. Aggregate/cement ratio — leaner mixes are less workable for the same water.
  4. Grading of aggregate — well-graded aggregate improves workability.
  5. Use of admixtures — plasticizers/superplasticizers greatly increase workability.
  6. Cement properties and fineness — finer cement needs more water.
  7. Ambient conditions and time — high temperature, wind and delay reduce workability.

Tests for workability

(i) Slump test — suitable for medium to high workability (slump 25–175 mm).

     ___                top dia 100 mm
    /   \
   /     \   height 300 mm    --> cone removed, concrete slumps down
  /_______\              bottom dia 200 mm
  Measure 'slump' = drop of top surface from original height (mm).

A frustum mould (200 mm bottom, 100 mm top, 300 mm high) is filled in 4 layers, each tamped 25 times, struck off, lifted vertically, and the subsidence measured.

(ii) Compaction factor test — suitable for low workability (drier mixes) where slump is insensitive.

  [Upper hopper]  -- concrete released --
        |
  [Lower hopper]
        |
  [Cylinder]  -> weigh partially compacted (W1) then fully compacted (W2)
Compaction factor=W1W2=weight of partially compacted concreteweight of fully compacted concrete\text{Compaction factor} = \frac{W_1}{W_2} = \frac{\text{weight of partially compacted concrete}}{\text{weight of fully compacted concrete}}

Typical values 0.78 (very low) to 0.95 (high). Lower value → lower workability.

(Vee-Bee / flow table are also acceptable.)

Segregation vs Bleeding

AspectSegregationBleeding
DefinitionSeparation of coarse aggregate from mortar (or paste from aggregate) so the mix is no longer uniformA form of segregation where water rises to the surface as solids settle
CauseOver-wet/too-dry mix, excessive vibration, dropping from height, poor gradingHigh w/c, over-vibration, lack of fines
EffectHoneycombing, weak/non-uniform concretePorous, weak surface laitance; water channels reduce durability

Preventive measures

Segregation: (1) use well-graded aggregate with adequate fines and correct water content; (2) avoid free fall from height and over-vibration; place close to final position.

Bleeding: (1) reduce w/c and add fine material / air-entraining or pozzolanic admixtures; (2) avoid over-vibration and use finer cement / proper grading.

fresh-concreteworkabilitysegregation-bleeding
3long10 marks

A solid brick masonry column of cross-section 400 mm×500 mm400\ \text{mm} \times 500\ \text{mm} has an effective height of 3.0 m3.0\ \text{m}. The masonry units have a compressive strength of 10 MPa10\ \text{MPa} and the mortar is grade M1 (1:6 cement:sand). Using the basic compressive stress for this combination as fb=0.96 MPaf_b = 0.96\ \text{MPa} (from IS 1905 tables), determine the permissible axial load on the column. Take area reduction factor for a column kak_a as applicable, slenderness based on least lateral dimension, eccentricity assumed nominal (e/t ≤ 1/24), and shape modification factor kp=1.0k_p = 1.0. Use stress reduction factor ksk_s from: SR 6 → 1.0, SR 8 → 0.95, SR 10 → 0.89, SR 12 → 0.84, SR 14 → 0.78 (interpolate linearly).

Step 1 — Slenderness ratio (SR)

For a column, SR is based on the least lateral dimension t=400 mmt = 400\ \text{mm}.

SR=hefft=3000400=7.5SR = \frac{h_{eff}}{t} = \frac{3000}{400} = 7.5

Step 2 — Stress reduction factor ksk_s (slenderness)

Interpolate between SR 6 (1.0) and SR 8 (0.95):

ks=1.0+(7.56)(86)(0.951.0)=1.0+1.52(0.05)=1.00.0375=0.9625k_s = 1.0 + \frac{(7.5 - 6)}{(8 - 6)}(0.95 - 1.0) = 1.0 + \frac{1.5}{2}(-0.05) = 1.0 - 0.0375 = 0.9625

(Eccentricity is nominal, e/t ≤ 1/24, so no further reduction of ksk_s for eccentricity.)

Step 3 — Area reduction factor kak_a

Applicable when sectional area A<0.2 m2A < 0.2\ \text{m}^2:

A=0.400×0.500=0.20 m2A = 0.400 \times 0.500 = 0.20\ \text{m}^2 ka=10.2(A1) ... using IS form ka=0.7+1.5Ak_a = 1 - 0.2\left(\frac{A}{1} \right)\ \text{... using IS form } k_a = 0.7 + 1.5A ka=0.7+1.5×0.20=0.7+0.30=1.00k_a = 0.7 + 1.5 \times 0.20 = 0.7 + 0.30 = 1.00

Since A=0.20 m2A = 0.20\ \text{m}^2 (the threshold), ka=1.00k_a = 1.00 (no reduction).

Step 4 — Shape modification factor

Given kp=1.0k_p = 1.0.

Step 5 — Permissible compressive stress

fc=kskakpfb=0.9625×1.00×1.00×0.96=0.924 MPaf_c = k_s \cdot k_a \cdot k_p \cdot f_b = 0.9625 \times 1.00 \times 1.00 \times 0.96 = 0.924\ \text{MPa}

Step 6 — Permissible axial load

P=fc×A=0.924 N/mm2×(400×500) mm2P = f_c \times A = 0.924\ \text{N/mm}^2 \times (400 \times 500)\ \text{mm}^2 P=0.924×200000=184800 N=184.8 kNP = 0.924 \times 200000 = 184800\ \text{N} = 184.8\ \text{kN}

Permissible axial load on the column ≈ 184.8 kN ( ≈ 185 kN ).

masonry-column-designcompression-memberslenderness
4long8 marks

Explain the concept of durability of concrete and discuss the principal mechanisms of concrete deterioration in aggressive environments: (a) sulphate attack, (b) chloride-induced reinforcement corrosion, and (c) alkali-aggregate reaction. For each, state one practical mitigation measure.

Durability — concept

Durability of concrete is its ability to resist weathering action, chemical attack, abrasion and other deterioration processes while retaining its form, quality and serviceability over the design life in the environment to which it is exposed. A durable concrete is one that performs satisfactorily under anticipated exposure conditions during its service life. Key controlling factors are a low water/cement ratio, dense impermeable microstructure, adequate cement content, proper curing, and sufficient cover to reinforcement.

(a) Sulphate attack

Mechanism: Sulphates (from soil, groundwater, seawater) react with hydrated cement constituents. SO42SO_4^{2-} reacts with calcium hydroxide to form gypsum, and with calcium aluminate hydrate (C-A-H) to form ettringite (calcium sulphoaluminate). These products are expansive:

C3A+3CaSO4+32H2OC3A3CaSO432H2O  (ettringite)C_3A + 3CaSO_4 + 32H_2O \rightarrow C_3A\cdot3CaSO_4\cdot32H_2O \;(\text{ettringite})

The expansion causes cracking, spalling and softening (loss of cohesion).

Mitigation: Use Sulphate Resisting Portland Cement (SRPC) with low C3AC_3A content (and a low w/c).

(b) Chloride-induced corrosion of reinforcement

Mechanism: Chlorides (de-icing salts, marine/coastal exposure) penetrate the cover and break down the passive oxide film on steel once a threshold concentration is reached. Electrochemical corrosion then proceeds:

Anode: FeFe2++2e,Cathode: O2+2H2O+4e4OH\text{Anode: } Fe \rightarrow Fe^{2+} + 2e^-, \quad \text{Cathode: } O_2 + 2H_2O + 4e^- \rightarrow 4OH^-

Rust products occupy 2–6 times the volume of parent steel, generating tensile stresses that crack and spall the cover (and reduce bar cross-section).

Mitigation: Provide adequate cover with low-permeability concrete (low w/c, use of mineral admixtures such as fly ash/silica fume) and limit chloride content of ingredients.

(c) Alkali–Aggregate Reaction (AAR / ASR)

Mechanism: Reactive (amorphous) silica in certain aggregates reacts with the alkalis (Na+,K+,OHNa^+, K^+, OH^-) of the pore solution to form an alkali-silica gel that absorbs water and swells, producing internal expansive pressure → map (pattern) cracking and pop-outs.

Mitigation: Use low-alkali cement (≤ 0.6% Na2ONa_2O equivalent) and/or non-reactive aggregates, or partial replacement with pozzolans (fly ash) that bind alkalis.

Summary table

DeteriorationKey product/processMitigation
Sulphate attackGypsum + ettringite (expansion)SRPC, low w/c
Chloride corrosionPassive film breakdown, expansive rustCover + dense concrete, mineral admixtures
AAR / ASRExpansive silica gelLow-alkali cement, non-reactive aggregate
durabilityconcrete-deteriorationsulphate-chloride-attack
5long10 marks

Cube compressive strength test results (150 mm cubes, 28 days) for a batch of M25 concrete are: 31.2, 28.6, 33.4, 29.8, 30.0 MPa. (a) Compute the mean strength and the standard deviation. (b) Check the acceptance criteria for M25 as per IS 456: mean of any group of 4 consecutive cubes ≥ fck+0.825sf_{ck} + 0.825\,s (with s assumed = 4.0 MPa for M25) AND individual ≥ fck3f_{ck} - 3. (c) A rebound hammer gives an average rebound number of 38 on a 150 mm cube whose calibration curve is f=1.2R12f = 1.2R - 12 (MPa). Estimate its compressive strength and comment on agreement with the destructive value.

(a) Mean and standard deviation

Data (MPa): 31.2, 28.6, 33.4, 29.8, 30.0; n=5n = 5.

Sum =31.2+28.6+33.4+29.8+30.0=153.0= 31.2 + 28.6 + 33.4 + 29.8 + 30.0 = 153.0

xˉ=153.05=30.6 MPa\bar{x} = \frac{153.0}{5} = 30.6\ \text{MPa}

Deviations from mean and their squares:

xix_ixixˉx_i-\bar{x}(xixˉ)2(x_i-\bar{x})^2
31.2+0.60.36
28.6-2.04.00
33.4+2.87.84
29.8-0.80.64
30.0-0.60.36
(xixˉ)2=0.36+4.00+7.84+0.64+0.36=13.20\sum (x_i-\bar{x})^2 = 0.36+4.00+7.84+0.64+0.36 = 13.20

Sample standard deviation (n−1 basis):

s=13.2051=3.30=1.82 MPas = \sqrt{\frac{13.20}{5-1}} = \sqrt{3.30} = 1.82\ \text{MPa}

Mean = 30.6 MPa, standard deviation ≈ 1.82 MPa.

(b) Acceptance criteria (IS 456) for M25, fck=25f_{ck}=25, assume s=4.0s=4.0 MPa

Criterion 1 — mean of group ≥ fck+0.825sf_{ck} + 0.825\,s:

25+0.825×4.0=25+3.30=28.30 MPa25 + 0.825 \times 4.0 = 25 + 3.30 = 28.30\ \text{MPa}

Mean of the (only) group available = 30.6 MPa ≥ 28.30 → satisfied.

Criterion 2 — every individual result ≥ fck3f_{ck} - 3:

253=22 MPa25 - 3 = 22\ \text{MPa}

Lowest individual = 28.6 MPa ≥ 22 → satisfied.

Both criteria met → concrete is acceptable as M25.

(c) Rebound hammer estimate

f=1.2R12=1.2×3812=45.612=33.6 MPaf = 1.2R - 12 = 1.2 \times 38 - 12 = 45.6 - 12 = 33.6\ \text{MPa}

Comment: The NDT (surface) estimate of 33.6 MPa is about 10% higher than the average destructive cube value (30.6 MPa). This is typical — the rebound hammer measures only surface hardness, is sensitive to surface carbonation, moisture, smoothness and aggregate near the surface, and tends to over- or under-estimate the bulk strength. It is a useful comparative/screening tool but should be correlated with cores or cubes for absolute strength; the agreement here (within ~3 MPa) is reasonable.

hardened-concretecompressive-strengthndt-rebound
B

Section B: Short Answer Questions

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6 questions
6short5 marks

List the four principal Bogue compounds of Portland cement and briefly describe the role of each in setting, early strength, ultimate strength and heat of hydration.

The four principal Bogue compounds

CompoundFormulaAbbrev.Typical %
Tricalcium silicate3CaOSiO23CaO\cdot SiO_2C3SC_3S (alite)45–55
Dicalcium silicate2CaOSiO22CaO\cdot SiO_2C2SC_2S (belite)20–30
Tricalcium aluminate3CaOAl2O33CaO\cdot Al_2O_3C3AC_3A8–12
Tetracalcium aluminoferrite4CaOAl2O3Fe2O34CaO\cdot Al_2O_3\cdot Fe_2O_3C4AFC_4AF6–10

Roles

  • C3SC_3S (alite): Hydrates relatively fast; responsible for early strength (up to ~7–28 days) and a major part of ultimate strength. Liberates a moderately high heat of hydration (~500 J/g).
  • C2SC_2S (belite): Hydrates slowly; contributes little early strength but is responsible for later/ultimate strength gain (beyond 28 days). Low heat of hydration (~260 J/g), good durability.
  • C3AC_3A: Reacts very rapidly with water and would cause flash set — gypsum is added to control it. Contributes to initial set, gives high early heat (~870 J/g) but little strength; high C3AC_3A reduces sulphate resistance.
  • C4AFC_4AF: Reacts moderately fast, gives the grey colour, contributes little to strength but improves resistance to sulphate; moderate heat (~420 J/g).

Summary: C3AC_3A governs setting (with gypsum control); C3SC_3S governs early strength and contributes large heat; C2SC_2S governs long-term strength with low heat; C4AFC_4AF acts as a flux/filler with modest contribution.

cementbogue-compoundshydration
7short5 marks

A sample of fine aggregate (1000 g) gives the following sieve analysis. Compute the fineness modulus and comment on whether the sand is coarse, medium or fine.

IS SieveMass retained (g)
4.75 mm20
2.36 mm80
1.18 mm150
600 µm300
300 µm320
150 µm110
Pan20

Step 1 — Cumulative % retained

Total = 1000 g. Compute % retained on each sieve and cumulate (FM uses the standard sieves down to 150 µm; pan is excluded from the sum).

SieveRetained (g)% retainedCumulative % retained
4.75 mm202.02.0
2.36 mm808.010.0
1.18 mm15015.025.0
600 µm30030.055.0
300 µm32032.087.0
150 µm11011.098.0
Pan202.0

Step 2 — Sum of cumulative % retained (standard sieves)

=2.0+10.0+25.0+55.0+87.0+98.0=277.0\sum = 2.0 + 10.0 + 25.0 + 55.0 + 87.0 + 98.0 = 277.0

Step 3 — Fineness modulus

FM=cumulative % retained100=277.0100=2.77FM = \frac{\sum \text{cumulative \% retained}}{100} = \frac{277.0}{100} = 2.77

Step 4 — Comment

General classification of sand by FM:

  • Fine sand: FM 2.2 – 2.6
  • Medium sand: FM 2.6 – 2.9
  • Coarse sand: FM 2.9 – 3.2

With FM = 2.77, the sample is a medium sand, suitable for general concrete work (broadly corresponds to grading Zone II of IS 383).

aggregatesgradingfineness-modulus
8short5 marks

Classify concrete admixtures and explain the working mechanism and one application each of (a) superplasticizers and (b) air-entraining agents. Also state two benefits of using fly ash as a mineral admixture.

Classification of admixtures

Admixtures are materials (other than cement, water, aggregate) added to concrete before/during mixing.

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

(a) Superplasticizers (high-range water reducers)

Mechanism: They are long-chain organic polymers (sulphonated melamine/naphthalene formaldehyde, or polycarboxylic ethers) that adsorb on cement particles and impart a strong negative charge / steric hindrance, causing the flocculated cement grains to deflocculate and disperse. This releases trapped water and reduces inter-particle friction, greatly increasing fluidity.

Effect/Application: Can reduce water by 15–30% at constant workability (→ higher strength/durability), or produce flowing/self-compacting concrete at constant w/c. Application: high-strength concrete, SCC, heavily reinforced/congested sections, pumped concrete.

(b) Air-entraining agents

Mechanism: Surfactants that stabilise billions of tiny (10–300 µm) discrete air bubbles uniformly through the paste. The bubbles act as relief chambers for expanding ice and as ball bearings improving workability.

Effect/Application: Greatly improves freeze–thaw durability and reduces bleeding/segregation (slight strength loss ~5% per 1% air). Application: pavements/structures in cold climates subject to freeze–thaw and de-icing salts.

Two benefits of fly ash

  1. Improved long-term strength and durability — the pozzolanic reaction consumes free Ca(OH)2Ca(OH)_2 to form additional C-S-H, densifying the matrix and lowering permeability (better resistance to sulphate/chloride/AAR).
  2. Lower heat of hydration and improved workability — reduces thermal cracking in mass concrete and, being spherical, acts as a lubricant reducing water demand; also economical/eco-friendly (cement replacement).
admixturessuperplasticizersmineral-admixtures
9short5 marks

Write short notes on any TWO of the following special concretes, covering their key characteristics, constituent requirements and one typical application: (a) High-Performance Concrete (HPC), (b) Self-Compacting Concrete (SCC), (c) Fibre-Reinforced Concrete (FRC).

(Answer any two — all three given for completeness.)

(a) High-Performance Concrete (HPC)

  • Characteristics: Concrete engineered for superior performance — high strength (typically ≥ 60 MPa), very low permeability, high durability and dimensional stability, beyond what conventional concrete achieves.
  • Constituents: Very low w/c (≤ 0.35) achieved with superplasticizers, high cement content plus silica fume / fly ash / GGBS, well-graded strong aggregate, careful quality control and curing.
  • Application: High-rise columns, long-span bridges, marine and aggressive-exposure structures.

(b) Self-Compacting Concrete (SCC)

  • Characteristics: Highly flowable, non-segregating concrete that spreads and fills formwork under its own weight without vibration, passing through congested reinforcement. Key properties: filling ability, passing ability, segregation resistance (tested by slump-flow, V-funnel, L-box).
  • Constituents: High powder/fines content (cement + fly ash/limestone filler), high-range water reducer (polycarboxylate) and often a viscosity-modifying agent (VMA), smaller maximum aggregate size and higher paste volume.
  • Application: Heavily reinforced/complex sections, precast units, tunnel linings, areas where vibration is impractical.

(c) Fibre-Reinforced Concrete (FRC)

  • Characteristics: Concrete with discrete short fibres distributed randomly, improving tensile/flexural strength, toughness, ductility, crack control and impact/fatigue resistance (post-crack load carrying).
  • Constituents: Conventional concrete plus fibres — steel, glass, polypropylene/synthetic, or natural fibres — typically 0.5–2% by volume, with adequate aspect ratio (l/d) and good bond.
  • Application: Industrial floors, pavements, tunnel shotcrete, precast elements, slabs on grade resisting shrinkage cracking.
special-concretehigh-performance-concreteself-compacting-concrete
10short5 marks

Discuss the properties of good building bricks and the functions of mortar in masonry. A brick wall is built with bricks of crushing strength 7.5 MPa in 1:6 cement-sand mortar. State qualitatively how the masonry strength relates to the individual brick and mortar strengths, and why masonry strength is much lower than the brick strength.

Properties of good building bricks

  1. Should be well-burnt, uniform in shape, size and colour with sharp, square edges and a uniform deep-red appearance.
  2. Adequate compressive (crushing) strength — minimum ~3.5 MPa for common bricks (higher for load-bearing).
  3. Low water absorption — not more than 15–20% of dry weight after 24 h immersion.
  4. Hard, durable and free from cracks, flaws, lumps of lime or organic matter.
  5. Should give a clear ringing sound when struck, low efflorescence, and adequate resistance to weathering/abrasion.

Functions of mortar in masonry

  1. Binds the masonry units together into a monolithic mass and transfers/distributes loads uniformly.
  2. Fills/seals the joints, making the masonry weather-tight and preventing penetration of water and air.
  3. Accommodates dimensional irregularities of units, providing a level bedding so units are not point-loaded.
  4. Provides bond for reinforcement (in reinforced masonry) and contributes to appearance (pointing).

Relation of masonry strength to brick and mortar strength

Masonry compressive strength increases with both brick strength and mortar strength, but it is governed predominantly by the brick (unit) strength; mortar has a smaller, secondary influence. Empirically masonry prism strength is only a fraction (roughly 25–50%) of the brick crushing strength.

Why masonry is much weaker than the brick: Under axial compression the (softer, more deformable) mortar in the bed joint tends to expand laterally more than the stiffer brick. The bond restrains the mortar but, by Poisson interaction, induces lateral tension in the bricks (and lateral compression/confinement in the mortar). Bricks are weak in tension, so this lateral tensile stress causes vertical splitting cracks in the units well before the brick's own crushing strength is reached. Additional reductions come from slenderness, eccentricity, joint thickness, and workmanship. Hence for 7.5 MPa bricks in 1:6 mortar the usable basic stress is on the order of ~1 MPa (about an order of magnitude lower than the brick strength).

masonry-materialsbricksmortar
11short5 marks

A load-bearing solid brick wall of thickness 230 mm230\ \text{mm} and actual height 3.45 m3.45\ \text{m} carries a uniformly distributed axial load. The wall is restrained top and bottom (effective height factor = 0.75). The basic permissible compressive stress for the brick-mortar combination is fb=0.50 MPaf_b = 0.50\ \text{MPa}. Taking the stress reduction factor for slenderness ksk_s from SR 8 → 1.0, SR 10 → 0.97, SR 12 → 0.84 (interpolate), and shape & area factors = 1.0, find the safe load per metre run of the wall.

Step 1 — Effective height and slenderness ratio

Effective height:

heff=0.75×3.45=2.5875 m=2587.5 mmh_{eff} = 0.75 \times 3.45 = 2.5875\ \text{m} = 2587.5\ \text{mm}

For a wall, SR is based on effective height / effective thickness (solid wall: effective thickness = actual thickness = 230 mm):

SR=hefft=2587.5230=11.25SR = \frac{h_{eff}}{t} = \frac{2587.5}{230} = 11.25

Step 2 — Stress reduction factor ksk_s

Interpolate between SR 10 (0.97) and SR 12 (0.84):

ks=0.97+(11.2510)(1210)(0.840.97)=0.97+1.252(0.13)k_s = 0.97 + \frac{(11.25 - 10)}{(12 - 10)}(0.84 - 0.97) = 0.97 + \frac{1.25}{2}(-0.13) ks=0.97+0.625×(0.13)=0.970.08125=0.88880.889k_s = 0.97 + 0.625 \times (-0.13) = 0.97 - 0.08125 = 0.8888 \approx 0.889

Step 3 — Permissible compressive stress

With shape factor kp=1.0k_p = 1.0 and area factor ka=1.0k_a = 1.0:

fc=kskakpfb=0.889×1.0×1.0×0.50=0.4444 MPaf_c = k_s \cdot k_a \cdot k_p \cdot f_b = 0.889 \times 1.0 \times 1.0 \times 0.50 = 0.4444\ \text{MPa}

Step 4 — Safe load per metre run

Consider 1 m (1000 mm) length of wall; area per metre run:

A=1000 mm×230 mm=230000 mm2A = 1000\ \text{mm} \times 230\ \text{mm} = 230000\ \text{mm}^2 P=fc×A=0.4444 N/mm2×230000 mm2=102,220 NP = f_c \times A = 0.4444\ \text{N/mm}^2 \times 230000\ \text{mm}^2 = 102{,}220\ \text{N} P102.2 kN per metre runP \approx 102.2\ \text{kN per metre run}

Safe (permissible) axial load ≈ 102 kN per metre run of the wall.

masonry-wall-designslendernessload-bearing-wall

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