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Section A: Long Answer Questions

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5 questions
1long10 marks

A load-bearing brick wall transmits a uniformly distributed line load of 180 kN/m180\ \text{kN/m} (including self-weight of wall) to the ground. The wall is to rest on a continuous (strip) footing. The supporting soil has a safe bearing capacity of 150 kN/m2150\ \text{kN/m}^2.

(a) Explain the function of a footing and the criteria that govern its proportioning. (b) Determine the required width of the strip footing. (c) If the footing is constructed in plain cement concrete (PCC) and the allowable angle of dispersion of load through the concrete is 4545^\circ, determine the minimum depth (thickness) of the PCC footing measured from the bottom of a 300 mm300\ \text{mm} thick wall. Sketch the load dispersion.

(a) Function and proportioning criteria of a footing

A footing is the enlarged base of a foundation that spreads the concentrated load from a wall or column over a larger area of soil so that the resulting bearing pressure does not exceed the safe bearing capacity (SBC) of the soil, and settlement remains within tolerable limits.

Proportioning criteria:

  • Area of base chosen so that soil pressure \le SBC (prevents shear failure / excessive settlement).
  • Depth/thickness chosen so the footing can transfer load safely (for PCC, by 4545^\circ load dispersion; for RCC, by bending, one-way & two-way shear).
  • The base must be centred under the resultant load so pressure is uniform (no tilting).

(b) Required width of strip footing

For a continuous footing, consider 1 m1\ \text{m} run.

Load per metre run=180 kN/m\text{Load per metre run} = 180\ \text{kN/m} Required area per metre=LoadSBC=180150=1.20 m2 per metre run\text{Required area per metre} = \frac{\text{Load}}{\text{SBC}} = \frac{180}{150} = 1.20\ \text{m}^2 \text{ per metre run}

Since length per run is 1 m1\ \text{m}, the width is:

B=1.20 m21 m=1.20 mB = \frac{1.20\ \text{m}^2}{1\ \text{m}} = 1.20\ \text{m}

Required footing width B=1.20 mB = 1.20\ \text{m} (1200 mm).

(c) Minimum depth of PCC footing (45° dispersion)

The load from the 300 mm300\ \text{mm} wall must spread out to the full footing width of 1200 mm1200\ \text{mm}. The projection (offset) on each side beyond the wall face:

Offset=Btwall2=12003002=9002=450 mm\text{Offset} = \frac{B - t_{wall}}{2} = \frac{1200 - 300}{2} = \frac{900}{2} = 450\ \text{mm}

For a 4545^\circ dispersion line, tan45=1\tan 45^\circ = 1, so depth = offset:

D=Offset×tan45=450×1=450 mmD = \text{Offset} \times \tan 45^\circ = 450 \times 1 = 450\ \text{mm}

Minimum PCC footing depth D=450 mmD = 450\ \text{mm}.

Sketch of load dispersion:

        |<-- 300 -->|   (wall)
   =====+===========+=====
        \           /        ^
         \   45°   /         |  D = 450 mm
          \       /          v
  ========+======+=========+========
  |<-450->|<-300->|<-450->|
  |<------ B = 1200 mm ----------->|

The 4545^\circ lines from each wall face must reach the footing edges, confirming D=450 mmD = 450\ \text{mm}.

foundationsbearing-capacityfooting-design
2long10 marks

A dog-legged staircase is to be provided in a residential building between two floors having a clear floor-to-floor height of 3.30 m3.30\ \text{m}. The stairwell available is 2.50 m×5.00 m2.50\ \text{m} \times 5.00\ \text{m} (width × length). Adopt a riser of 165 mm165\ \text{mm} and a tread (going) of 270 mm270\ \text{mm}.

(a) Define the terms riser, tread, going, flight and landing. (b) Determine the total number of risers and treads required, and distribute them into two equal flights. (c) Determine the width of each flight, the length occupied by the going, and check whether the staircase fits within the stairwell length. Comment on whether the riser–tread combination satisfies the common rule 2R+T=550700 mm2R + T = 550\text{–}700\ \text{mm}.

(a) Definitions

  • Riser: the vertical face of a step; the riser height is the vertical distance between two consecutive treads.
  • Tread: the horizontal upper surface of a step on which the foot is placed.
  • Going: the horizontal distance between the faces of two consecutive risers (effective tread depth, excluding nosing).
  • Flight: an unbroken series of steps between two landings.
  • Landing: the level platform provided at the top, bottom, or between flights to allow rest and change of direction.

(b) Number of risers and treads

Number of risers=Floor-to-floor heightRiser=3300165=20 risers\text{Number of risers} = \frac{\text{Floor-to-floor height}}{\text{Riser}} = \frac{3300}{165} = 20\ \text{risers}

For a dog-legged stair with two equal flights:

Risers per flight=202=10\text{Risers per flight} = \frac{20}{2} = 10

Number of treads is always one less than the risers in each flight (the last riser lands on the landing/floor):

Treads per flight=101=9\text{Treads per flight} = 10 - 1 = 9 Total treads=9×2=18\text{Total treads} = 9 \times 2 = 18

20 risers total; two flights of 10 risers each, 9 treads per flight.

(c) Width of flight, going length, and fit check

Width available = 2.50 m2.50\ \text{m}, shared by two parallel flights with no central gap:

Width of each flight=2.502=1.25 m\text{Width of each flight} = \frac{2.50}{2} = 1.25\ \text{m}

This exceeds the 0.90 m0.90\ \text{m} minimum for residential stairs — acceptable.

Length occupied by the going of one flight (9 treads):

Lgoing=9×270=2430 mm=2.43 mL_{going} = 9 \times 270 = 2430\ \text{mm} = 2.43\ \text{m}

Stairwell length = 5.00 m5.00\ \text{m}. The single flight going (2.43 m) plus a landing of, say, 1.25 m1.25\ \text{m} at the turn:

2.43+1.25=3.68 m<5.00 m2.43 + 1.25 = 3.68\ \text{m} < 5.00\ \text{m}

There is ample room (a generous landing or a small entry space of 5.003.68=1.32 m5.00 - 3.68 = 1.32\ \text{m} remains). The staircase fits comfortably within the stairwell.

Rule check:

2R+T=2(165)+270=330+270=600 mm2R + T = 2(165) + 270 = 330 + 270 = 600\ \text{mm}

Since 550600700550 \le 600 \le 700, the combination is comfortable and satisfies the rule.

stairsgeometric-designbyelaws
3long10 marks

A rectangular hall measures 8 m×12 m8\ \text{m} \times 12\ \text{m} in plan. It is to be covered by a symmetrical gable (couple-close) pitched roof of slope 3030^\circ, with the ridge running along the 12 m12\ \text{m} length.

(a) Differentiate between a flat roof and a pitched roof, giving two situations where each is preferred. (b) Compute the true sloping length of one rafter (rise and slope length), ignoring eaves projection. (c) Compute the total surface area of the two sloping roof planes, and the number of CGI (corrugated galvanised iron) sheets required if each effective sheet covers 1.8 m21.8\ \text{m}^2 and a 10%10\% allowance is made for laps and wastage. (d) Name the principal members of a king-post truss and state the maximum span for which it is economical.

(a) Flat roof vs pitched roof

AspectFlat roofPitched roof
Slope< 10° (nearly horizontal)> 10° (steep)
Drainageneeds careful slope & water-proofingsheds rain/snow quickly
Use of topusable terracenot usable

Flat roofs are preferred in (i) dry/warm climates and (ii) where roof terrace/future vertical extension is needed. Pitched roofs are preferred in (i) heavy rainfall/snowfall regions and (ii) for large-span sheds and traditional sloped-roof architecture.

(b) Rafter geometry

Half span (horizontal run of one rafter):

run=82=4 m\text{run} = \frac{8}{2} = 4\ \text{m}

Rise:

rise=run×tan30=4×0.5774=2.309 m\text{rise} = \text{run} \times \tan 30^\circ = 4 \times 0.5774 = 2.309\ \text{m}

Slope (rafter) length:

L=runcos30=40.8660=4.619 mL = \frac{\text{run}}{\cos 30^\circ} = \frac{4}{0.8660} = 4.619\ \text{m}

Rafter slope length 4.62 m\approx 4.62\ \text{m}, rise 2.31 m\approx 2.31\ \text{m}.

(c) Roof area and number of CGI sheets

Each sloping plane has dimensions L×(ridge length)=4.619 m×12 mL \times (\text{ridge length}) = 4.619\ \text{m} \times 12\ \text{m}.

Area of one plane:

A1=4.619×12=55.43 m2A_1 = 4.619 \times 12 = 55.43\ \text{m}^2

Total for two planes:

A=2×55.43=110.86 m2A = 2 \times 55.43 = 110.86\ \text{m}^2

Add 10%10\% for laps and wastage:

Anet=110.86×1.10=121.94 m2A_{net} = 110.86 \times 1.10 = 121.94\ \text{m}^2

Number of sheets:

N=121.941.8=67.768 sheets (round up)N = \frac{121.94}{1.8} = 67.7 \Rightarrow 68\ \text{sheets (round up)}

Total roof area 110.86 m2\approx 110.86\ \text{m}^2; required CGI sheets = 68.

(d) King-post truss

Principal members: tie beam (bottom horizontal member), two principal rafters (sloping top members), the central king post (vertical tension member from ridge to mid-span of tie beam), and two struts connecting the king post to the principal rafters. The ridge and purlins rest on the rafters.

A king-post truss is economical for spans up to about 58 m5\text{–}8\ \text{m}. For the present 8 m8\ \text{m} span it is suitable; larger spans use a queen-post or steel truss.

roofspitched-roofestimation
4long10 marks

A one-brick-thick (230 mm) external wall is 6.0 m6.0\ \text{m} long and 3.0 m3.0\ \text{m} high. It contains one door opening of 1.0 m×2.1 m1.0\ \text{m} \times 2.1\ \text{m} and one window opening of 1.2 m×1.5 m1.2\ \text{m} \times 1.5\ \text{m}. Modular bricks of nominal size 200×100×100 mm200 \times 100 \times 100\ \text{mm} (i.e. including a 10 mm mortar joint) are used.

(a) State two functions of a wall and explain the difference between a load-bearing wall and a partition wall. (b) Compute the net masonry volume of the wall. (c) Compute the number of modular bricks required, and the dry mortar volume if mortar occupies 25%25\% of the gross masonry volume. (d) If the dry mortar of part (c) is cement–sand 1:61:6, compute the volume of cement and sand required (dry-bulk density of cement =1440 kg/m3= 1440\ \text{kg/m}^3, 11 bag =50 kg= 50\ \text{kg}).

(a) Functions and wall types

Two functions of a wall: (i) to enclose/divide space and provide privacy, security and protection from weather; (ii) to support and transmit floor/roof loads to the foundation (in load-bearing construction) and resist lateral loads.

  • Load-bearing wall: carries its self-weight plus floor/roof loads down to the foundation; thickness governed by load and slenderness. Cannot be removed without structural support.
  • Partition wall: a thin, non-load-bearing wall used only to divide interior space; carries only its self-weight.

(b) Net masonry volume

Gross wall area:

Agross=6.0×3.0=18.0 m2A_{gross} = 6.0 \times 3.0 = 18.0\ \text{m}^2

Openings:

Adoor=1.0×2.1=2.1 m2,Awindow=1.2×1.5=1.8 m2A_{door} = 1.0 \times 2.1 = 2.1\ \text{m}^2, \quad A_{window} = 1.2 \times 1.5 = 1.8\ \text{m}^2 Aopenings=2.1+1.8=3.9 m2A_{openings} = 2.1 + 1.8 = 3.9\ \text{m}^2

Net area:

Anet=18.03.9=14.1 m2A_{net} = 18.0 - 3.9 = 14.1\ \text{m}^2

Net volume (wall thickness 0.23 m0.23\ \text{m}):

V=14.1×0.23=3.243 m3V = 14.1 \times 0.23 = 3.243\ \text{m}^3

Net masonry volume =3.243 m3= 3.243\ \text{m}^3.

(c) Number of bricks and mortar volume

Volume of one modular brick including its mortar joint (nominal size):

v=0.200×0.100×0.100=0.002 m3v = 0.200 \times 0.100 \times 0.100 = 0.002\ \text{m}^3

Number of bricks (nominal volume fills the wall completely):

N=Vv=3.2430.002=1621.51622 bricksN = \frac{V}{v} = \frac{3.243}{0.002} = 1621.5 \Rightarrow 1622\ \text{bricks}

Mortar (dry) volume at 25%25\% of gross masonry volume:

Vmortar=0.25×3.243=0.811 m3V_{mortar} = 0.25 \times 3.243 = 0.811\ \text{m}^3

Bricks required 1622\approx 1622; dry mortar volume 0.81 m3\approx 0.81\ \text{m}^3.

(d) Cement and sand for the 1:6 mortar

Using the dry mortar volume Vmortar=0.811 m3V_{mortar} = 0.811\ \text{m}^3 with ratio 1:61:6 (total 7 parts):

Cement volume:

Vc=17×0.811=0.1159 m3V_c = \frac{1}{7} \times 0.811 = 0.1159\ \text{m}^3 mc=0.1159×1440=166.9 kg=166.950=3.344 bagsm_c = 0.1159 \times 1440 = 166.9\ \text{kg} = \frac{166.9}{50} = 3.34 \Rightarrow 4\ \text{bags}

Sand volume:

Vs=67×0.811=0.695 m3V_s = \frac{6}{7} \times 0.811 = 0.695\ \text{m}^3

Cement 166.9 kg\approx 166.9\ \text{kg} (4 bags); sand 0.695 m3\approx 0.695\ \text{m}^3.

wallsbrick-masonryestimation
5long10 marks

(a) Define dampness in buildings and list four sources of dampness. (b) Explain four common defects/ill-effects caused by dampness in a building. (c) Describe, with a labelled sketch in text/ASCII, how a horizontal damp-proof course (DPC) at plinth level together with treatment of the basement floor prevents rising dampness. Mention two suitable DPC materials and the typical thickness adopted. (d) State four requirements of an ideal damp-proofing material.

(a) Dampness and its sources

Dampness is the presence and movement of unwanted moisture within the components of a building (walls, floors, roofs), usually drawn in from the ground or admitted through defective construction.

Four sources of dampness:

  1. Rising damp — ground moisture rising by capillary action through foundations/walls.
  2. Rain penetration — through roofs, parapets, defective joints, and exposed wall faces.
  3. Leaking/condensation — from defective plumbing, sanitary fittings, or condensation of water vapour on cold surfaces.
  4. Poor drainage / improper slope allowing water to accumulate against walls.

(b) Four defects/ill-effects of dampness

  1. Efflorescence — white salt deposits on plaster/masonry, causing disintegration of the surface.
  2. Deterioration of finishes — blistering, flaking and softening of paint, plaster and distemper.
  3. Decay of timber & corrosion of steel — rot/warping of doors, windows and joists; rusting of reinforcement and fixtures.
  4. Unhealthy living conditions — growth of mould/fungi, musty smell, and breeding of mosquitoes; promotes disease.

(c) Working of horizontal DPC at plinth level

A continuous, impervious horizontal layer (the DPC) is laid in the wall at plinth level (about 150 mm150\ \text{mm} above ground level) and throughout the basement floor, breaking the capillary path so ground moisture cannot rise into the superstructure.

        Wall (superstructure)
        ||||||||||||||||||
  GL    =====DPC=========   <- horizontal DPC at plinth (~150 mm above GL)
  ~~~~~~|              |~~~~~~  ground level
        | plinth/found |
        |   masonry    |
  ------+--------------+------
  ::::::: basement floor DPC :::::::  <- floor membrane turned up to meet wall DPC
  ====== PCC bed / footing ======

The wall DPC and the floor membrane are lapped/turned up so that the moisture barrier is continuous at the wall–floor junction; this is essential, otherwise damp bypasses the DPC at the corner.

Suitable DPC materials: bituminous felt / bitumen mastic, dense cement concrete (1:2:4) with waterproofing compound, or polythene/PVC sheet. Typical thickness: mastic/bitumen layer 2.5 mm\approx 2.5\ \text{mm}; cement-concrete DPC course 2540 mm\approx 25\text{–}40\ \text{mm}.

(d) Requirements of an ideal damp-proofing material

  1. It should be impervious (completely watertight).
  2. It should be durable and last the life of the structure, resisting wear and chemical/biological attack.
  3. It should be strong enough to bear the imposed loads without crushing and be dimensionally stable (free from cracks).
  4. It should be flexible/workable, capable of being laid continuously and lapped at joints, and reasonably economical.
damp-proofingbuilding-defectsconstruction-detailing
B

Section B: Short Answer Questions

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6 questions
6short6 marks

A room of internal size 4.0 m×5.0 m4.0\ \text{m} \times 5.0\ \text{m} and height 3.0 m3.0\ \text{m} is to be cement plastered on all four internal walls (ignore openings for this estimate). A 12 mm12\ \text{mm} thick plaster of cement–sand ratio 1:51:5 is to be used. Estimate the wet mortar volume and the quantity of cement (in bags) required, taking a 20%20\% increase for converting wet to dry mortar and dry-bulk density of cement as 1440 kg/m31440\ \text{kg/m}^3 (1 bag = 50 kg).

Plaster area (perimeter × height):

P=2(4.0+5.0)=18.0 mP = 2(4.0 + 5.0) = 18.0\ \text{m} A=P×h=18.0×3.0=54.0 m2A = P \times h = 18.0 \times 3.0 = 54.0\ \text{m}^2

Wet mortar volume (thickness 12 mm = 0.012 m):

Vwet=54.0×0.012=0.648 m3V_{wet} = 54.0 \times 0.012 = 0.648\ \text{m}^3

Dry mortar volume (add 20%):

Vdry=0.648×1.20=0.7776 m3V_{dry} = 0.648 \times 1.20 = 0.7776\ \text{m}^3

Cement volume (ratio 1:5, so cement = 1/6 of dry volume):

Vcement=11+5×0.7776=0.77766=0.1296 m3V_{cement} = \frac{1}{1+5} \times 0.7776 = \frac{0.7776}{6} = 0.1296\ \text{m}^3

Mass and bags of cement:

m=0.1296×1440=186.6 kgm = 0.1296 \times 1440 = 186.6\ \text{kg} Bags=186.650=3.734 bags\text{Bags} = \frac{186.6}{50} = 3.73 \Rightarrow 4\ \text{bags}

Wet mortar 0.648 m3\approx 0.648\ \text{m}^3; cement required 186.6 kg=4\approx 186.6\ \text{kg} = 4 bags.

plasteringestimationfinishing
7short6 marks

(a) Sketch (in text/ASCII) a fully panelled door and label at least six components. (b) State two functions each of a door and a window. (c) Give the commonly adopted size (width × height) of a single-leaf main entrance door and of an internal door for a residential building.

(a) Labelled panelled door (ASCII)

  +==========================+   <- Lintel above
  | head/top rail            |
  |  +------+   +------+      |
  |  |panel |   |panel |      |  <- Stiles (vertical members) at left & right edges
  |  +------+   +------+      |  <- Mullion (central vertical) between panels
  |  ---- lock/intermediate rail ----
  |  +------+   +------+      |
  |  |panel |   |panel |      |
  |  +------+   +------+      |
  | bottom rail              |
  +==========================+
     ^Frame (jambs) all around; hinges on one stile; handle/lock on the other.

Labelled components: (1) Frame/jamb, (2) Top (head) rail, (3) Bottom rail, (4) Stile, (5) Mullion, (6) Panel, (7) Lock/intermediate rail, (8) Hinges.

(b) Functions

Door: (i) provides access/exit and connection between spaces; (ii) gives privacy, security and control of light/sound/heat when closed.

Window: (i) admits natural light and ventilation (fresh air); (ii) gives outside view and helps regulate light/air while providing weather protection.

(c) Common sizes

  • Single-leaf main entrance door: about 1.0 m×2.1 m1.0\ \text{m} \times 2.1\ \text{m} (1000 mm × 2100 mm).
  • Internal room door: about 0.9 m×2.1 m0.9\ \text{m} \times 2.1\ \text{m} (900 mm × 2100 mm); bathroom/WC door: about 0.75 m×2.0 m0.75\ \text{m} \times 2.0\ \text{m}.
doors-windowscomponentsclassification
8short5 marks

(a) Name the basic components of a ground floor from bottom to top. (b) Compare any three flooring finishes among the following on the basis of durability, cost and typical use: cement-concrete floor, terrazzo (mosaic) floor, and ceramic-tile floor.

(a) Components of a ground floor (bottom to top)

  1. Subgrade / compacted earth (well-rammed natural soil).
  2. Sub-base / soling — sand or compacted hardcore/brick-bat soling.
  3. Base course — lean cement concrete (e.g. 1:4:8) bed.
  4. Floor finish / topping — the wearing surface (cement screed, tiles, terrazzo, etc.). A DPC membrane is provided below the base course to stop rising damp.

(b) Comparison of flooring finishes

FinishDurabilityCostTypical use
Cement-concrete floorGood, hard-wearing but can crack/dustLow (cheapest)Stores, godowns, utility & service areas
Terrazzo (mosaic) floorVery durable, smooth, easy to cleanMedium–high (skilled labour)Residences, offices, hospitals, schools
Ceramic-tile floorDurable, water/stain resistant, hygienicMedium–high (varies with tile)Kitchens, toilets, bathrooms, living areas

Summary: cement-concrete is cheapest and best for rough/utility use; terrazzo gives a decorative, very durable monolithic finish; ceramic tiles give a hygienic, water-resistant finish ideal for wet areas.

floorsflooring-materialscomparison
9short5 marks

A residential plot has an area of 250 m2250\ \text{m}^2. The local building byelaws permit a maximum ground coverage of 60%60\% and a maximum Floor Area Ratio (FAR) of 2.52.5. (a) Define ground coverage and FAR. (b) Determine the maximum permissible ground-floor built-up area and the maximum total built-up (floor) area. (c) If each storey is built to the full permissible ground coverage, find the maximum number of storeys allowed by the FAR.

(a) Definitions

  • Ground coverage: the percentage of the plot area covered by the building's footprint at ground level =built-up area at ground floorplot area×100=\dfrac{\text{built-up area at ground floor}}{\text{plot area}}\times 100.
  • Floor Area Ratio (FAR): the ratio of the total covered (built-up) floor area of all storeys to the plot area =total floor areaplot area=\dfrac{\text{total floor area}}{\text{plot area}}.

(b) Maximum permissible areas

Ground-floor built-up area:

Aground=0.60×250=150 m2A_{ground} = 0.60 \times 250 = 150\ \text{m}^2

Maximum total built-up (floor) area:

Atotal=FAR×plot area=2.5×250=625 m2A_{total} = \text{FAR} \times \text{plot area} = 2.5 \times 250 = 625\ \text{m}^2

Ground coverage =150 m2= 150\ \text{m}^2; total permissible floor area =625 m2= 625\ \text{m}^2.

(c) Maximum number of storeys

If each storey has the full ground-coverage area of 150 m2150\ \text{m}^2:

n=AtotalAground=625150=4.17n = \frac{A_{total}}{A_{ground}} = \frac{625}{150} = 4.17

Since a partial storey is not built, round down:

n=4 storeysn = 4\ \text{storeys}

Maximum number of storeys =4= 4 (using 4×150=600 m2625 m24 \times 150 = 600\ \text{m}^2 \le 625\ \text{m}^2; a 5th full storey would give 750 m2>625750\ \text{m}^2 > 625, not allowed).

byelawsfarground-coverage
10short4 marks

(a) Differentiate between shallow and deep foundations with one example of each. (b) Briefly state two situations in which a pile foundation becomes necessary.

(a) Shallow vs deep foundation

BasisShallow foundationDeep foundation
DepthDepth \le width (generally < 3 m); Terzaghi: DfBD_f \le BDepth \gg width; reaches strong strata far below
Load transfermainly by bearing at baseby end-bearing and/or skin friction along the shaft
Usewhen good soil is near the surfacewhen surface soil is weak/compressible
Exampleisolated/spread footing, raftpile, pier, well/caisson

(b) Situations requiring a pile foundation

  1. When the upper soil layers are weak/compressible (soft clay, loose fill, made-up ground) and firm strata lie deep, piles transfer load to the deeper hard layer.
  2. When heavy concentrated loads must be carried, the water table is high, or uplift/lateral forces (e.g. from waterfront, expansive soils, bridge piers) make spread footings impractical.
foundationsdeep-foundationclassification
11short4 marks

(a) Define a 'bond' in brick masonry and state why it is important. (b) Differentiate between English bond and Flemish bond on any three points.

(a) Bond in brick masonry

A bond is the systematic arrangement of bricks in successive courses so that the vertical (continuous) joints are broken/staggered and adjacent bricks overlap.

Importance: it prevents continuous vertical joints, ties the masonry together so that the load is distributed over a larger area, and gives the wall strength, stability and a good appearance. Without proper bonding the wall would behave as independent columns of brick and could fail.

(b) English bond vs Flemish bond

PointEnglish bondFlemish bond
Arrangement of coursesalternate courses of headers and stretcherseach course has alternate headers and stretchers in the same course
Strengthstronger; best for load-bearing/thick wallscomparatively weaker for one-brick walls
Appearanceless attractivebetter, more pleasing appearance
Use of broken bricks (bats)fewer bats neededmore bats/closers needed; needs more skilled labour

In short, English bond is preferred where strength matters, while Flemish bond is chosen where appearance and economy of facing bricks matter.

wallsbondsbrick-masonry

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The BE Civil Engineering (IOE, TU) Building Technology (IOE, CE 502) 2080 paper carries 80 full marks and is meant to be completed in 180 minutes, across 11 questions.
Is practising this Building Technology (IOE, CE 502) past paper free?
Yes — reading and attempting this Building Technology (IOE, CE 502) past paper on Kekkei is completely free.